= , and coeff. of fric. = tan 0.
Ng
151. A train starts from a station at^f and runs on a level to a station
at B, I ft. away. If the speed is not to exceed v ft. per sec., show that
the time between the two stations is
l_ Wv P + B
"r
v g 2 (P-R)(B + R)'
W being the gross weight of the train, P the mean uniform pull exerted
by the engine, R the road resistance, and B the retarding effect of the
brakes.
Also, if the speed is not limited, show that the least time in which
the train can run between the specified points is
^ W_ P + B
' ~i (P R)(B+R)
and that the maximum speed attained is
w ?, s --*-v~*
EXAMPLES. 233
152. A locomotive capable of exerting a uniform pull of 2 tons, with
a 24-in. stroke, 2o-in. cylinder, and 6o-in. driving-wheels, hauls a train
between two stations 3 miles apart. The gross weight of the train and
locomotive = 200 tons; the road resistance = 12 Ibs. per ton (of 2000
Ibs.) ; the brakes, when applied, press with two thirds of the weight on
the wheels of the engine and brake-van, viz., 90 tons, the coefficient of
friction being .18. Find (a) the least time between the stations; (b) the
distance in which the train is brought to rest; (c) the maximum speed
attained ; (d) the pressure of the steam ; (e) the weight upon the driving-
wheels.
Ans. 0) 513.8 sec. ; (6) 990 ft.; (c) 42 miles per hour; (d) 25
Ibs. per sq. in. ; (, and 2/> 2 ) lying in the planes of principal stress.
This ellipse is called the ellipse of stress, and the stress on any
plane AB at O is the semi-diameter of the ellipse drawn in a di-
rection making an angle ip with the axis OC, fy being given by
tan i/j = cot y. (Eq. (8).) . . . . (n)
6. Constant Components of p r . Take the planes of
principal stress as planes of reference (Fig. 214).
F
A
FIG.
242 THEORY OF STRUCTURES.
Draw ON perpendicular to AB, and take ON -^
Let the obliquity of OR = = RON = 90 - - y.
Join NR. Then
NR = OR" + ON 2 - 2OR . ON cos
= Pr + (^~} - A(A + A) sin (0 + 7
But A 8 = A 2 sin2 r +A 2 cos ' x> and
sin (^ + y) = sin ^'cos y + cos ^ sin y = ^ cos a y + -- sin" y.
(See eqs. (8).)
.-. NR* = A a sin 8 y + A' cos 2 r + ( 2
-(A+A)(Asin 8 y+Acos 8
(A+AV . . fA-AV
= V T ) -AA- I 2 |-
(12)
Hence, /^^ intensity of stress OR at any point O of the plane
A OB is the resultant of two constant intensities
and
the former being perpendicular to the plane.
THE ANGLE ONR. 243
7. The Angle ONK = 2y.
sin ONR OR p r
sin ~ NR ~~ p l A'
2
But
sin = cos (& + y) = cos ^ cos y sin ^ sin y
= ^~ sin r cos = sih 2 *
sn
2A 2
.-. sin OA 7 ^? = sin 2y, or (7^7? 2y. . . (13)
Let NR (Fig. 214) produced in both directions meet OA in
F and O in (7.
The angle OFN = 180 - ONR - NOF
= 180 - 2y - (90 - Y) = 90 - y = FON.
.'. NF = NO ; so, NG = NO =
.'. A 7 " is the middle point of F.
Also
RF=FN- NR = ON- NR =
and
RG = RN+NG = RN+ ON =
N. B. The shear at O
A -A
cos (2x - 90) = (A -A) sin Y cos X -
244 THEORY OF STRUCTURES.
8. Maximum Shear. ON has no component along AB.
Hence, the shear on AB is NR cos (angle between NR and
AB), and is evidently a maximum when the angle is nil. Its
value is then NR, or Pl ~ P \
9. Application to Shafting. At any point in a plane sec-
tion of a strained solid, let r be the intensity of stress, and 6
its obliquity.
At the same point in a second plane let s be the intensity
of stress, and 0' its obliquity.
By Art. 6, r and s are the resultants of two constant
stresses
and
+ /-/, + A) cos*'.
Subtracting one equation from the other,
First. Consider the case of combined torsion and bending,
as when a length of shafting bears a heavy pulley at some point
between the bearings.
Let p be the intensity of stress (compression or tension)
due to the bending moment M b .
APPLICATION TO SHAFTING.
245
Let q be the intensity of shear due to the twisting mo-
ment M t .
p and q act in planes at right angles to each other.
.-. r cos 6 = p y r sin 6 = q s, and 6' = 90.
/. r a =/ a + * and s = q.
Hence, by eq. (16),
and by eq. (15),
and
The max. shear =
__ A -
-V?
+2*;.
also
(17)
(19)
(20)
(21)
= (Chap. VI.) and f = r (Chap. IX.)
for a shaft of radius r.
t '\; . . . (22)
246
and
THEORY OF STRUCTURES.
(23)
&==(-
Perhaps the most important example of the application of
the above principle is the case of a
shaft acted upon by a crank (Fig. 215).
A force P applied to the centre C of
the crank-pin is resisted by an equal
and opposite force at the bearing JB,
forming a couple of moment P. CB = M.
This couple may be resolved into a
FIG. 215. bending couple of moment M b = P. AB
= P. BC cos d = J/cos tf, and a twisting couple of moment
M t = P . AC = P . BC sin d = M sin d ; tf being the angle
ABC.
', . . (24)
2M
and the max. shear = ,
nr
(25)
If the working tensile or compressive stress (/>,) and the
working shear stress ( ? ) are given, the corresponding
values of r may be obtained from eqs. (22) and (23) or eqs.
(24) and (25) ; the greater value being adopted for the radius
of the shaft.
Second. Consider the case of combined torsion and tension
or compression.
Let the tensile or compressive force be P.
P
p, the intensity of the tension or compression, = - t ;
<7, " " " shear = -5-4 .
PRINCIPAL AND CONJUGATE STRESSES. 247
and
10. Conjugate Stresses. Consider the equilibrium of an
indefinitely small parallelepiped
abed (Fig. 216) of a strained body,
the faces ab, cd being parallel to
the plane XOX, and the faces ad,
be to the plane YOY.
Let the stresses on ab, cd act
parallel to the plane YOY. The
total stresses on ab and cd are
equal in amount, act at the centres of the faces, are parallel to
YOY, and therefore neutralize one another.
Hence the total stresses on ad and be must also neutralize
one another. But they are equal in amount, and act at the
middle points of ad, be; they must therefore be parallel to
XOX.
Hence, if two planes traverse a point in a strained body,
and if the stress on one of the planes is parallel to the other
plane, then the stress on the latter is parallel to the first
plane.
Such planes are called planes of conjugate stress, and the
stresses themselves are called conjugate stresses.
Principal stresses are of course conjugate stresses as well.
Conjugate stresses have equal obliquities, each obliquity
being the complement of the same angle.
n. Relations between Principal and Conjugate
Stresses (Fig. 217). Take any line ON =
248 THEORY OF STRUCTURES.
P P
With N as centre and a radius = , describe a semi-
circle.
Let be the common obliquity of a pair of conjugate
stresses.
FIG. 217.
Draw ORS, making an angle with ON, and cutting the
semicircle in the points R and 5.
Join NR, NS.
OR and OS are evidently a pair of conjugate stresses.
Draw NV perpendicular to RS and bisecting it in V.
Draw the tangent OT\ join NT.
Let OR = r, OS = s. Then
rs = OR. OS = OT* =
and
=(A+A)c os 0. (29)
The maximum value of the obliquity, i.e., of 0, is the angle
TON.
Call this angle 0. Then
NT A -A / x
' ' ' ' ' (30)
PRINCIPAL AND CONJUGATE STRESSES.
249
Let OR, OR' be a pair of con-
jugate stresses (Fig. 218).
Let OG, OH be the axes of R ;
greatest and least principal stress,
respectively.
Draw ON normal to OR.
Let the angle GOR = #, RON
= 6, HON = GOR' = 7, as before. Then
FIG. 218.
and by eqs. (8),
~ cot y = tan f/> =cot
cot
,*_ /i ~A _ cot x cot (y + ^) sin 6
~ ~ ~ : sin 2
or
Hence,
angle GON 90 y = ^
and
angle HOR = y + & = {0 +
(32)
. . . (33)
250 THEORY OF STRUCTURES.
12. Ratio of r to s.
r_OR_ OV-RV _ OV-)/NR* -NV*
s ~" OS * OV+RV~ OV+VNR*-NV n -
ONcos 6 - VNR* ON* sin 3
ON cos e + VNR* - ON* sin 2 e
But
NR A - A NT
= sin TON = sin 0.
r cos 8 1/sin 3 i - sin
A =i+sin0'*
as in eq. (38).
Pressure against a Vertical Plane. Let ACB (Fig. 222),
the ground-surface of a mass of earthwork, be inclined to the
horizon at an angle 6.
Consider a particle at a vertical depth CD = x below C.
Let s be the vertical intensity of pressure on the particle
at D.
Let r be the conjugate intensity of pressure on the particle
at D.
This conjugate pressure acts in the direction ED parallel ta
the ground-surface, and its obliquity is B.
Take DE so that
_r_ cos 1/cos' 8 cos*
DC '~ s " cos e _|_ |/ cos * e cos 2 0*
Then ze/ . ."/? represents in direction and
magnitude the intensity of pressure on the
vertical plane DC at the point D, w being the weight of a unit
of volume of the earthwork.
Join CE.
The intensity of pressure at any other point m is evi-
dently w . mn, mn being drawn parallel to DE.
Hence, the total pressure on the plane DC = weight of
prism DCE
w.DC.DE w.DC*r
cos 6 = cos
2 2 S
_ wx* cos 6 I/cos 2 6 cos 3
~2~ cos 6 + Vcos 2 - cos 2 ' ' * (43 '
Again, s is the pressure due to the weight of the vertical
column CD.
.-. s = wx cos 0, p ...... (44)
THEORY OF STRUCTURES.
and
cos 6 t/cos' 6 cos 2
r =. wx cos u . (45)
cos 6 + Vcos* V - cos 2
"By means of this last equation the total pressure on CD
imay be easily deduced as follows :
The pressure on an element dx at a depth x
^cos I/cos' cos' 7
= 77MT = wx cos - dx.
cos + l/cos' J cos 2
.'. total pressure = jrdx = etc.
The total resultant pressure is parallel in direction to the
ground-surface, and its point of application is evidently at
two thirds of the total depth CD.
15. Earth Foundations. CASE I. Let the weight of the
superstructure be uniformly distributed over the base, and let
p Q be the intensity of the pressure produced by it.
If p h is the maximum horizontal intensity of pressure cor-
responding to/ ,
/_ < I +sin
p h i sin '
In the natural ground, let p v be the maximum vertical in-
tensity of pressure corresponding to the horizontal intensity
ph Then
p^ < i + sin
p v -i sin '
Hence
+sin0\ 2
pi = VI sin
If ^ is the depth of the foundation, and w the weight of a
cubic foot of the earth,
p v = wx ;
A
(46)
wx = \i sin
EA R TH FO UNDA TIONS. 259
Let h + x be the height of the superstructure, and let a
cubic foot of it weigh w r . Then
Hence, a minimum value of x is given by
x) _ /i_+j|
" -~~ V7-si
= p, suppose;
wx ' \i sin 0/ ~ /&*'
';!#
*=^_^7F 2 (47)
CASE II. Let the superstructure produce on the base a
uniformly varying pressure of maximum intensity / : and mini-
mum intensity/.,.
By Case I,
W x =i- sn
( .
In the natural ground the minimum horizontal intensity of
pressure is
I sin
p h wx ; - - .
I + sin
When the foundation-trench is excavated, this pressure
tends to raise the bottom and push in the sides. The weight
of the superstructure should therefore be at least equal to the
weight of the material excavated in order to develop a hori-
zontal pressure of an intensity equal
A < * sin
' / 2 = I + sin '
Combining this with the last equation,
Combining (48) and (49),
(Rankine's Civil Engineering, Arts. 237, 239.)
200
THEORY OF STRUCTURES.
16. Retaining-walls. Consider a portion ABMN of a
wall (Fig. 223).
Let Wbe its weight, and let the di-
rection of W cut MN in C.
Let Pbe the resultant of the forces
externally applied to ABNM and tend-
ing to overthrow it. Let D be its
point of application, and let its di-
rection meet that of W in E.
Let F be the centre of pressure (or
resistance) at the bed MN.
FIG. 223. Let O be the middle point of MN.
Let MN t, OF qt, OC rt, q and r being each less
than unity.
Let x' and y', respectively, be the horizontal and vertical
co-ordinates of D with respect to F.
Let the inclination to the horizon of MN = a, of P's
direction = ft.
Conditions of Equilibrium. (a) The moment of P with
respect to F ^ the moment of W with respect to F, or
P(y' cos ft x' sin ft) < W(qt =F rt) cos a ;
(51)
the upper or lower sign being taken according as C falls on the
left or right of O.
In ordinary practice q varies from J to f .
EXAMPLE. A masonry wall (Fig. 224)
of rectangular section, x ft. high, 4 ft. wide,
weighing 125 Ibs. per cubic foot, is built
upon a horizontal base and retains water
(weighing 62^ Ibs. per cubic foot) on one
side level with the top of the wall.
FIG. 224.
= 125 X
= 4 ft.
RE TA IN ING- WALLS. 261
or
*? < 192? (52)
If 0=J, * 2 <48 and x < 6.928 ft.
If = f, *' < 72 and * < 8.485 ft.
() The maximum intensity of pressure at the bed MN must
not exceed the safe working resistance of the material to
crushing. The load upon the bed is rarely if ever uniformly
distributed. It is practically sufficient to assume that the in-
tensity of the pressure diminishes at a uniform rate from the
most compressed edge inwards.
Let / be the maximum intensity of pressure, and R the
total pressure on the bed.
Three cases^may be considered.
CASE I. Let the intensity of the pressure diminish uniformly
from / at M to o at N (Fig. 225).
Take MG perpendicular to MN and =f', join GN.
The pressure upon the bed Gp^
is represented by the triangle ^^
MGN. "\^
^-
MFC
The brdinate through the FlG - 22 s-
centre of gravity of the triangle, parallel to GM, cuts MN in
the centre of pressure F.
*- - - = g*.
CASE II. Let the maximum intensity /> MG in Case I.
Take MH =. f, and the triangle
MHK = R (Fig - 226 )-
The pressure on the bed is
^ now represented by the triangle
MHK.
R = MH. MK = . MK.
M F O K N
FlG - 22fi - The ordinate through the
262 THEORY OF STRUCTURES.
centre of gravity of the triangle MHK parallel to HM cuts MN
in the centre of pressure F.
.-. q t = OF = OM- MF=-- .
But
t 2 R
/. qt = --- - ; and hence
T O C' T
9 = ~ ~ -ft and is evidently > g. . Y (53)
CASE III. Let the maximum intensity/ < MG in Case I.
Take ML =f, and the trapezoid MLSN = R (Fig. 227).
The pressure on the bed is now represented by the trape-
zoid MLSN.
G .-. R = XML + NS)MN
r^.
and
MFC N 2R
FIG. 7 . ^V5 = - /.
The ordinate through the centre of gravity of the trapezoid
parallel to LM cuts MN in the centre of pressure F.
Draw ST parallel to NM.
The moment of MLSN with respect to O
= moment of MTSN with respect to O
+ moment of ZSiTwith respect to (9,
or
TS*
\(ML + NS)MN. OF =
Hence, q = ( t), and is evidently <%.... (54)
RE TAINING- WALLS. 263
Now W must be a function of x, the vertical depth of N
below B ; P also may be a function of x.
Hence if /is given, and the corresponding value of q from
(53) or (54) substituted in (52), x may be found.
When (53) is employed, the value of x found must make
?>>
When (54) is employed, the value of x found must make
r<>
EXAMPLE. The rectangular wall in (), the safe crushing
strength of the material being 10,000 Ibs. per square foot (=/).
R= W= 500*.
By (53),
~ 2 I2O
Substituting in (52),
Hence,
x < 9-03 ft.
Again, q > -- 2j > .4248, and is a fortiori > g.
If (54) is employed,
1/80 \
* = 5b V-
Hence, by (52),
By trial ;tr is found to lie between 12 and 13 ; each of these
values makes q > -J-, which is contrary to (54).
The first is therefore the correct substitution.
(c\ The angle between the directions of the resultant pres-
sure and a normal to the bed must be less than the angle of
friction.
264
THEORY OF STRUCTURES.
Let be the angle of friction, R the mutual normal pressure.
Resolving along the bed and perpendicular to it,
and
Pcos a 4- ft Wsm a < R tan
Psin a + J3+ Wcosa = R',
Pcos a-\- ft IV sin a
Psin
which reduces to
Wcosa
< tan 0,
P(cos ft-\-a cos sin /?-)- sin 0)> W^sin cos a-j-cos sin <*),
or
P cos /? -f or + < JFsin a + 0,
or
or
P(cos ft cos a -(- - sin ft sin or + 0) <
a -f- 0,
tan
Pcos
Psin/* +
. (55)
17. Rankine's Theory of Earthwork applied to Retain-
c ing-walls. Fig. 228 represents a
vertical section of a wall retaining
earthwork. AB is a vertical plane
cutting the ground-surface AC in the
point A.
Consider the equilibrium of the
whole mass of masonry and earth-
work in front of AB.
Let the depth AB x.
The total pressure on AB is, by
(43),
_ wx* cos 6 1/cos 2 6 cos 2
P = cos - .
2 cos + I/cos 3 cos 2
LINE OF RUPTURE. 265
Its point of application is D, and BD = .
Let W be the weight of the whole mass under considera-
tion, and let its direction cut the base of the wall in the point G.
Let F be the centre of pressure in the wall-base.
Taking moments of jPand W 7 about F,
p - cos 6 j[jf , , . . (4)
and
(5)
Corresponding points on the profiles, e.g., P and <2, have a
common subtangent of the constant value , for
= ..... (6)
dy) w
Area PNOA = ydx = f . - = ( F, - ,,), (7)
where /W= F,.
Area QNOB = f'yd = -( Y, - t,), (8)
where QN = Y,.
.: Area QPAB = -( K,+ F, - , + ^) = (T' - 7"), . (9)
where PQ=Y,+ Y,= T'.
RESERVOIR WALLS.
Thus the area of the portion under consideration is equal
to the product of the subtangent and the difference of thick-
ness at top and bottom.
Lines of resistance with reservoir empty. Let g l be the
point in which the vertical through the C. of G. of the portion
OAPN intersects PN. Then
Ng l X area OAPN = f* X ydx y - ;
I/O
So if gi be the point in which the vertical through the
C. of G. of the portion OBQN intersects QN,
Let G be the point in which the vertical through the
C. of G. of the whole mass ABQP intersects PQ. Then
NG X area ABQP = Ng, X area A ONP- Ng, X area BONQ,
or
IV A. W l A. W *
274
THEORY OF STRUCTURES.
The horizontal distance between G and a vertical through
the middle point of AB
=NG-\(f ^jy-'x-w-wjy-^-w*-**
=. one half of the horizontal distance between the verticals
through the middle points of AB and CD.
The locus of G can therefore be easily plotted.
Lines of Resistance with Reservoir Full. Let R be the centre
of resistance in PQ (Fig. 232).
Draw the vertical QS, and consider
the equilibrium of the mass QSAPQ.
Let w' = weight of a cubic foot of
water.
w'x* x
= moment of water-pressure
against QS about R
= moment of weight of QBS
about R -f- moment of
weight of QPAB about R t
or
W X
moment of QBS about R+ ~(T' - T)w.GR.
The first term on the right-hand side of this equation is
generally very small and may be 'disregarded, the error being
on the safe side.
In such case
i w' x*
GR -6jT'-T'
Also the mean intensity of the vertical pressure
w X area APQB
A = or>
RESERVOIR WALLS.
and the maximum intensity of the vertical pressure
275
2R
= */
I -2q
or
= -Jv(l + 6?) = /(! + 6?)(l - y,).
General Case. Let the profile be of any form, and consider
any portion ABQP, Fig. 233.
Take the vertical through Q as T
the axis of x, and the horizontal
line coincident with top of wall as
the axis of y.
The horizontal distance (y) be-
tween the axis of x and the vertical
through the C. of G. of the portion
under consideration is given by the
equation
/ being the width, dx the thickness,
and y the horizontal distance from OQ of the C. of G. of any
layer MN at a depth x from the top.
When the reservoir is empty, the deviation of the centre of
resistance from the centre of base
When the reservoir is full, let q'T be the deviation of the
centre of resistance from the centre of the base, and disregard
the moment of the weight of the water between OQ and the
profile BQ. Then
276
THEORY OF STRUCTURES.
moment of water pr. moment of wt. of ABQP
weight of ABQP
w I tdx
t/o
Hence
w'x*
w I tdx
t/o
21. General Equations of Stress. Let x, y, z be the
co-ordinates with respect to three rectangular axes of any point
O in a strained body.
Consider the equilibrium of an
element of the body in the form
of an indefinitely small parallele-
piped with its edges OA (= dx\
OB( dy), OC( dz] parallel to
the axes of x, y, z. It is assumed
that the faces of the element are
sufficiently small to allow of the
distribution of stress over them
being regarded as uniform. The
FlG - 234 ' resultant force on each face will
therefore be a single force acting at its middle point.
Let X l , Y l , Z^ be the components parallel to the axes x, y, z
of the resultant force per unit area, on the
face EC.
" X^ , Y 9 , Z 9 be the corresponding components for the
taceAC.
" -Yg , Y 3 , Z 3 be the corresponding components for the
face AB.
These components are functions of x, y, z, and therefore
become
for the adjacent face AD ;
GENERAL EQUATIONS OF STRESS.
for the adjacent face BD ;
for the adjacent face DC.
Hence, the total stress parallel to the axis of x
= X.dydz - (x, + d -^dx]dydz + X,dzdx - (x, + ^
IdX, . dX,
= ~(-&+W
Similarly, the total stress parallel to the axis of y
and the total stress parallel to the axis of z
t dZ t dZ
Let p be the density of the mass at O, and let P x ,P y , P g be
the components parallel to the axes of x, y, z of the external
force, per unit mass, at O.
pdxdydzP x is the component parallel to the axis of x of
the external force on the element ;
pdxdydzPy is the component parallel to the axis of y of
the external force on the element ;
pdxdydzP z is the component parallel to the axis of z of
the external force on the element.
278 THEORY OF STRUCTURES.
The element is in equilibrium.
dx l dy dz
dY, , dY, , dY,
dZ \ ,dZ^ >dZ*
dx" dy" dz~
These are the general equations of stress.
Again, take moments about axes through the centre of the
element parallel to the axes of co-ordinates, and neglect terms
involving (dxjdydz, dx(dyfdz, dxdy(dz}*.
and X=
(2)
Adopting Lamp's notation, i.e., taking
N l , N t , N 3 as the normal intensities of stress at O on planes
perpendicular to the axes of x, y, z ;
7", as the tangential intensity of stress at O on a plane
perpendicular to the axis of x if due to a stress
parallel to the axis of j/, or on a plane perpen-
dicular to the axis of y if due to a stress parallel
to the axis of x ; and 7^ , T a similarly, equa-
tions (i) become
^ + ~^'~ Jr ~^ 1
rf'+tL+W
dx dy dz
(3)
GENERAL EQUATIONS OF STRESS.
279
Next consider the equilibrium of a tetrahedral element
having three of its faces parallel to
the co-ordinate planes. Let /, m, n
be the direction-cosines of the normal
to the fourth face.
Also, let X, V, Z be the compo-
nents parallel to the axes of x, y, z
of the intensity of stress R on the
fourth face.
X = IN, + mT, + n T t + \pP x ldx.
But the last term disappears in FlG - 2 35-
the limit when the tetrahedron is indefinitely small, and hence
(4)
These three equations define R in direction and magnitude
when the stresses on the three rectangular planes are known.
Let it be required to determine the planes upon which the
stress is wholly normal. We have
Y=mR, ZnR..
(5)
Substituting these values of X, Y, Z in eqs. (4) and eliminat
ing /, m, n, we obtain
- T?- T?- T?
1 T,T S ) = o ; (6)
R* -
a cubic equation giving three real values for R, and therefore
three sets of values for /, m, and n, showing that there are three
planes at O on each of which the intensity of stress is wholly
normal. These planes are at right angles to each other and
are called principal planes, the corresponding stresses being prin-
cipal stresses. They are the principal planes of the quadric,
= c.
(7)
280
THEORY OF STRUCTURES.
For, the equation to the tangent plane at the extremity of a
radius r whose direction-cosines are /, m, n is
Xrx + Yry + Zrz = c, ....'. (8)
and the equation of the parallel diametral plane is
Xx + Yy + Zz = o (9)
The direction-cosines of the perpendicular to this plane are
X
-
Y
-
Z
-
so that the resultant stress R must act in the direction of this
perpendicular.
Hence the intensities of stress on the planes perpendicular
to the axes of the quadric (7) are wholly normal.
Refer the quadfic to its principal planes as planes of refer-*
ence. All the 7"s vanish and its equation becomes
(10)
Also, the general equations (3) become
%-*
Again,
RELATION BETWEEN STRESS AND STRAIN. 28 1
Consider X 9 Y, Z as the co-ordinates of the extremity of
the straight line representing R in direction and magnitude.
Equation (12) is then the equation to an ellipsoid whose semi-
axes are N lt N 91 N t . As a plane at O turns round O as a fixed
centre, the extremity of a line representing the intensity of
stress R on the plane will trace out an ellipsoid. This ellipsoid
is called the ellipsoid of stress.
Note I. The coefficients in the cubic equation (6) are in-
variants. Thus, NI -f N 9 + N, is constant, or the sum of three
normal intensities of stress on three planes placed at right
angles at any point of a strained body is the same for all
positions of the three planes.
Note 2. The perpendicular/ from O on the tangent plane,
equation (8),
Note 3. Let the stress be the same for all positions of the
plane at 0. Then N, = N^ = N z , and the ellipsoid (12) be-
comes a sphere. The stress is therefore everywhere normal,
and the body must be a perfect fluid. Conversely, if the
stress is everywhere normal, the body must be a perfect fluid,
the ellipsoid becomes a sphere, and therefore N l = N 9 = N a .
22. Relation between Stress and Strain. In Art. 13 it
was shown that when the size and figure of a body are altered
in two dimensions, there is an ellipse of strain analogous to the
ellipse of stress. If the alteration takes place in three dimen-
sions, it may be similarly shown that every state of strain may
be represented by an ellipsoid of strain analogous to the ellip-
soid of stress. The axes of the ellipsoid are the principal axes
of strain, and every strain may be resolved into three simple
strains parallel to these axes.
282
THEORY OF STRUCTURES.
It is assumed that the strains remain very small, that the
stresses developed are proportional
to the corresponding strains, and
that their effects may be superposed.
Consider an element of the un-
strained body in the form of a rect-
x angular parallelepiped, having its
edges PQ (= A), PR (= k}, PS (= I)
parallel to the axes of co-ordinates.
When the body is strained, the
element becomes distorted, the new
edges being P'Q', P'R', P'S'.
Let x, y, z be the co-ordinates of P.
Let x -f- u, y -f- v, z -\- w be the co-ordinates of P'.
By Taylor's Theorem the co-ordinates with respect to P' of
FlG> 236 '
du
dx
dv ,dw
*-, h
dx dx
r> , i.du J , dv \ ,dw
R are k, k( i +-=-) k ;
a \ a i a
du .dv . . dw\
r , , .
S'are/-, /- ,
dz dz
ay i ay
dw\
.
dz j
Hence, strain parallel to axis of x =
dv\
I 7Z~ i "717 / *
.
dy
ISOTROPIC BODIES.
28 5
where A, =
mE
, is the coefficient of dilatation, and
A = mm ~ l
~ (m+i)(m-2J
Again, the straining changes the angle RPS by an amount
-| - , producing two tangential stresses, each equal to
dy dz
G\ -| - I, parallel to the axes of y and z.
\dv dz>
dw dv
Similarly,
T -G(
2 " \dz
du
s-
(21)
G is called the coefficient of rigidity or transverse elasticity,
and is designated n in Thomson and Tait's notation, and >w in
Lamp's notation.
Relation between A, A, and G. Equations (20) and (21) pre-
serve the same forms whatever rectangular axes may be chosen.
Keep the axis of z fixed and turn the axes of x and y
through an angle a.
Let Nf be the normal stress parallel to the new axis of x.
.*. N{ N^ cos 2 a -|- N^ sin 2 a + 2 7" 3 sin a cos a. (22)
Let x 1 ', y' and z/, v' be the new co-ordinates and displace-
ments.
, T . A du' . ^Idv' . dw'\ ,. ..du' .
' N > = A ^' + A (^ + w) = (A ~ ^ + K6 -
_ du . dv . dw du' . dv' . dw' .
For -- + -- + -:-, = = y-, + + - - , is an invariant.
dx dy dz dx dy dz
.286 THEORY OF STRUCTURES.
The values of N{ given by eqs. (22) and (23) must be
identical. Now,
x = x' cos a y' sin a, y = x' sin a -)- y' cos a ; }
z/' = & cos a -f- ^ sin a', t;' = & sin a -\- v cos a . )
du du . dv
.'. _ = cos a + sin a
^,r dx dx
du 2 . dv . a /^ . ^/z/\ .
= -j- cos 2 a + - sin a + !_--[-__ jsm a cos a
^r dy \dy dxl
du . dv T.
- cos 2 a -f- - sin 3 a + ? sin a cos a ;
tfJtr y 6-
and by eq. (23),
AV = (A -*)(^ cos 8 a+^ sin 3 a+5 sin a cos a] + A0. (25)
>^r ay G- '
Also by eqs. (20) and (22),
= (A\)- cos 2 +- sin +r sin a cos a+A0. (26)
\d>x dy A ~~ A /
Eqs. (25) and (26) must be identical.
^ _ A w.fi'
IT = 5(^+1) = " = n - |iK * (27)
Adding together equations (20),
/ , flfe , d
^"dy^ ~dz'
It may be easily shown that the normal stresses can each
be separated into a fluid pressure p and a distorting stress.
AP PLICA TIONS. 287
Hence, putting
AT AT AT__^_ m & (d 11 I dV . 6
9 - 3 P -- ^ m __ ^ [^ -r j- -r -
t> mE
the cubic elasticity -. =- = . r = K. (28)
au dv aw ^(m 2)
dx dy dz
24. Applications. i. Traction. One end of a cylindrical
bar of isotropic material is fixed and the bar is stretched in the
direction of its length. The axis of the bar is the only line
not moved laterally by contraction.
Take this line as the axis of x.
The displacements u, v, w of any point x, y, z may be ex-
pressed in the form
u = ax, v = Py t w = PZ. . . , (29)
By eqs. (20) and (29),
.- 1 (30)
By eqs. (21) and (29), all the tangential stresses vanish.
Hence, since N lt N 9 , N 3 are constant, and since the equa-
tions of internal equilibrium contain only differential coeffi-
cients of the stresses, the hypothesis, eq. (29), satisfies these
equations.
First. Let N^ = o = N 3 ; i.e., let no external force act upon
the curved surface.
.-. - ft A + \(ft + ) = o,
or
P _ A _ I
a ~ T+Ji ~ m
Thus, the coefficient of contraction is less than the coefficient
of expansion.
288 THEORY OF STRUCTURES.
Again, by eqs. (30) and (32),
Z = A-2* = A-* = . (33)
a. am
Second. If the bar, instead of being free to move laterally,
has its surface acted upon by a uniform pressure P, then
N, = N 3 = P.
By eqs. (3 1) and (32),
ft APXN,
a ~ \(N, + 2/>) - AN, '
- * . (35)
For example, let P be sufficient to prevent lateral contrac-
tion. Then ft = o and, by eqs. (31) and (35),
aA = N, = -y = (m - i)P.
2. Torsion. (a) Let a circular cylinder (hollow or solid) of
length / undergo torsion around its axis (the axis of x), and let
t be the angle through which one end is twisted relatively to
the other. A point in a transverse section distant x from the
latter will be twisted through an angle x .
The displacements u, v, w of the point x,y, z in this section
may be expressed in the form
u = o, v = zx-y w = -\-yx- .
By eqs. (20) and (21),
and
APPLICA TIONS. 289
The algebraic sum of the moments of 7!, , J", with respect
to the axis
r being the distance of the point (x, y, z) from the axis.
Hence, the moment M, = Pp (Chap. IX), of the couple
producing torsion
_ ^ * / i J c* __ /~* T /~*f}T
~lJ ' 1
dS being an element of the area at (x,y, z\ I the polar moment
of inertia, and the torsion per unit of length of the cylinder,
or the rate of twist.
The torsional rigidity of a solid cylinder
= = GI = -nR*
6 2
R being the radius of the cylinder.
(b] Torsion of a bar of elliptic section.
The displacements u, v, w may now be expressed in the
form
u = F(y, z), v = 6x2, w = Qxy.
du _ dv _ dw^
' dx dy dz'
7-0, T.= G+
Hence, by the general eqs. (3),
* ....... (36)
2QO THEORY OF STRUCTURES.
.Also, the surface stresses are zero ;
-and hence, by eqs. '(3$),
~dy = Q(zdz -\-ydy) (38)
This equation must hold true at the surface.
Let the equation to the elliptic section be
dz fy
= - .^v,,, (40)
and by eq. (38),
2 dll . dlt _ _ / 72 _ 2X
u dyz satisfies this last equation and also eq. (36), if
Again, the algebraic sum of the moments 71, , T 3 with re-
spect to the axis of x,
2G6
T(*4*q^ < 5I >
For the further treatment of this subject, the student is re-
ferred to St. Venant's edition of Clebsch, and to Thomson and
Tait's Natural Philosophy.
3. Work done in the small strain of a body (Clapeyron's
Theorem) M ultiply eqs. (3) \yyudxdy dz, v dx dy dz, wdxdydz,
and find the triple integral of their sum throughout the whole
of the solid.
The terms involving the components P x t P y , P 2 may be dis-
regarded, as the deformations due to their action are generally
inappreciable.
Also,
///s-*** ;
=ff(N x 'u x - N x "u x "}dydz -fffNSj-dxdydz\
N x ', N x " being the values of N l at the two points in which the
line parallel to the axis of x cuts the surface of the body, and
u x i u x" the corresponding values of u.
Let dS, dS' be the elementary areas of the surface at these
points, and /', I" the cosines of the angles between the normals
to these elements and the axis of x.
The double integral on the right-hand side of the last equa-
tion then becomes
ff(N x 'l'u x 'dS - NJ'l"H x "dS) =
APPLfCA TfOlfS 293
Treating the other terms similarly,
O = 2{(N l l+ T,m + 7a +(T,l+ N,m
Hence, the work done = $2(Xu + Yv + Zw)dS
- ^W, + N,N, + NJt t - T? - 77 - T.*) }
I
M + Njf t + ffjr t - T: - T; - T,'
E being the ordinary modulus of elasticity.
294 THEORY OF STRUCTURES.
EXAMPLES.
1. At a point within a strained solid there are two conjugate stresses,
viz., a tension of 200 Ibs. and a thrust of 150 Ibs. per square inch, the
common obliquity being 30. Find (a) the principal stresses ; (b) the
maximum shear and the direction and magnitude of the corresponding
resultant stress; (c) the resultant stress upon a plane inclined at 30 to
the axis of greatest principal stress.
Ans. (a) A tension of 204.65 Ibs. and a thrust of 146.95 Ibs. per sq. in.
(b) 175.8 Ibs. per sq. in.; 173.2 Ibs. in a direction making an
angle of 40 13' with the axis of greatest principal stress.
( < >
wet
22 " .26
Metal on metal
dry
.15 " .2
wet . .
.3
Metal and wood c
.15
on each other 1
or each on itself '
occasionally lubricated as usual. . . .
constantly lubricated
.07 to .08
.OS
The apparatus employed in carrying out these experiments
consisted of a box which could be loaded at pleasure, and
which was made to slide along a horizontal bed by means of a
cord passing over a pulley and carrying a weight at the end.
The contact-surfaces of the bed and box were formed of
the materials to be experimented upon. The pull was meas-
ured and recorded by a spring dynamometer.
More recent experiments, however, have shown that
Coulomb's laws cannot be regarded as universally applicable,
but that / depends upon the velocity, the pressure, and the
temperature. At very low velocities Morin's results have
been verified (Fleeming Jenkin). At high velocities f rap-
idly diminishes as the velocity increases. Franke, having
carefully examined the results of various series of experi-
ments, especially those of Poire"e, Bochet, and Galton, has
suggested the formula
v being the velocity and/ , a, coefficients depending upon the
nature and condition of the rubbing surfaces.
For example,
f n = .29 and a = .04 for cast-iron on steel with dry sur-
faces.
/ .29 and a = .02 for wrought-iron on wrought-iron with
dry surfaces.
/ = .24 and a .0285 for wrought-iron on wrought-iron
with slightly damp surfaces.
Ball has shown that at very low pressures f increases as
3C2 THEORY OF STRUCTURES.
the pressure diminishes, while Rennie's experiments indicate
that at very high pressures/ rapidly increases with the press-
ure, and this is perhaps partly due to a depression, or to an
abrasion of the rubbing surfaces.
2. Inclined Plane. Let a body of weight P slide uni-
formly up an inclined plane under a force Q inclined at an
angle ft to the plane.
Let F be the friction resisting the mo-
tion, R the pressure on the plane, and a the
plane's inclination.
The two equations of equilibrium are
F = Q cos j3 P sin a
and
R = Q sin ft + P cos ex.
F Qcos/3 Psma
/. -=r = 7=5 : 77-: n = coefficient of friction = /.
R Q sin /) -f- P cos a J
Let the resultant of F and R make an angle with the
normal to the plane. Then
_F_ Q cos ft P sin a Q _ sin (a + 0)
~R~ - Qsin P + Pcosa' P ~ cos (ft - $)'
is called the angle of friction. It has also been called the
angle of repose, since a body will remain at rest on an inclined
plane so long as its inclination does not exceed the angle of
friction.
If a = o = ft, then ^ = tan = /.
The work done in traversing a distance x = Q cos ft . x. If
Q is variable, the work done / Q cos ft . dx
3. Wedge. The wedge, or key, is often employed to con-
nect members of a structure, and is generally driven into posi-
WEDGE.
303
tion by the blow of a hammer. It is also employed to force
out moisture from materials by induc-
ing a pressure thereon.
The figure represents a wedge de-
scending vertically under a continuous
pressure P, thus producing a lateral
motion in the horizontal member C,
\yhich must therefore exert a pressure
Q upon the vertical face AB.
The member H is fixed, and it is
assumed that the motion of the machine
is uniform, so that the wedge and 7 are
in a state of relative equilibrium.
Let R l , R^ be the reactions at the faces DE, DF, respec-
tively, their directions making an angle 0, equal to the angle
of friction, with the normals to the corresponding faces.
Let a be the angle between DE and the vertical, a' the
angle between DF and the vertical.
Consider the wedge, and neglect its weight, which is usually
inappreciable as compared with P.
Resolving vertically,
R l cos (90 a + 0) + RS cos (90 a'
= P=R 1 sin (a + 0) + R, sin (a f -\- 0).
Resolving horizontally,
or
R, sin (90 a + 0) R^ sin (90 a' + 0) = o,
R l cos (or + 0) = ^ 2 cos (a' + 0) (2)
Consider the member C, and neglect its weight.
Resolving horizontally,
R, cos (a + 0) = Q = R, cos (a f + 0). .
(3)
Assuming the wedge isosceles, as is usually the case, a = a',
and hence,
by eq. (2), R l = R tt and by eq. (i), 2R, sin (a + 0) = P. (4)
3O4 THEORY OF STRUCTURES.
Hence, by eqs. (3) and (4),
Q _ cot (OL + 0) __ external resistance overcome
P~ 2 effort exerted " ' 5 '
(N.B. This ratio of resistance to effort is termed the mechan-
ical advantage, or put chase, of a machine.)
Suppose the motion of the machine reversed, so that Q be-
comes the effort and P the resistance.
The reactions R, , R^ now fall below the normals, and the
equations of relative equilibrium are the same as the above,
with substituted for 0.
Thus, = cot (a 0) (6)
The two cases may be included in the expression
-p =*cot(0) (7)
For a given value of P, Q increases with a.
If there were no friction, would be zero, and eq. (7) would
become
Q cot a
Thus, the effect of friction may be allowed for, by assuming
the wedge frictionless, but with an angle increased by 20 in the
first case, and diminished by 20 in the second case.
Again, when P is the effort and Q the resistance, eq. (5)
shows that if a -\- > 90, the ratio -y is negative, which is
impossible, while if a -\- = 90, -~ is zero, and in order to
WEDGE. 305
overcome Q, however small it might be, P would require to be
infinitely great. Hence,
a + must be < 90,
Q
and below this limit -p diminishes as increases.
Similarly, it may be shown from eq. (7) that when Q is the
effort and P the resistance,
must be < a,
and that below this limit 5- increases with 0.
Efficiency. During the uniform motion of the machine, let
any point a descend vertically to the point b. The correspond-
ing horizontal displacement is evidently 2bc.
The motive work = P . ab ;
" useful work =Q.2bc.
Hence, the efficiency = ~ 7- = -7,- . 2 tan a
= tan a cot (a -f- 0), by eq. (5).
This is a maximum for a given value of when
and the max. efficiency = tan ^45 j cot ^45 + -J
For the reverse motion, the efficiency
P.ab
3 o6
THEORY OF STRUCTUKES.
This is a maximum when a =. 45 -{- . Thus the
imax. efficiency = cot ^45 -f- -j tan 145 -) = -
sin
-{- sin 0'
4. Screws. A screw is usually designed to produce a
linear motion or to overcome a resistance in the direction of its
length. It is set in motion by means of a couple acting in a
plane perpendicular to its axis. A reaction is produced be-
tween the screw and nut which must necessarily be equivalent
to the couple and resistance, the motion being steady.
Take the case of a square * -threaded screw. It may be
assumed that the reaction is concentrated along a helical line,
whose diameter, d, is a mean between the external and internal
diameters of the thread, and that its distribution along this
line is uniform. It will also be supposed that the axes of the
couple and screw are coincident, so that there will be no lateral
pressure on the nut.
Let M be the driving couple.
" Q " " axial resistance to be over-
come.
" r " " reaction at any point a of the
helical line, and let be
angle between its direction
and the normal at a ; is
the angle of friction.
" a " " angle between the tangent
at a and the horizontal ; a
is called the pitch-angle.
Since the reaction between the screw and nut must be
equivalent to M and Q, then
* Square-threaded screws work more accurately than those with a V-thread,
but the efficiency of the latter has been shown to be very little less than that of
the former (Poncelet). On the other hand, the V-thread is the stronger, much
less metal being removed in cutting it than is the case with a square thread.
Again, with a V-thread there is a tendency to burst the nut, which does not
obtain in a screw with a square thread.
FIG. 240.
SCRE W 'S. SO/
Q algebraic sum of vertical components of the reac-
tions at all points of the line of contact,
= 2\r cos (a + 0)] = cos (a + )S(r), .... (i)
and M = algebraic sum of the moments with respect to the
axis of the horizontal components of the reactions at all points
of the line of contact,
. . . (2)
Let the couple consist of two equal and opposite forces, P,
acting at the ends of a lever of length/, so that M = Pp.
Hence, by eqs. (i) and (2),
and the mechanical advantage
Q ip
If = o, -p = cot of, and the effect of friction may
be allowed for, by assuming the screw frictionless, but with a
pitch-angle equal to a -\- 0.
Again, let the figure represent one complete turn of the
thread developed in the plane of the
paper. CD is the corresponding length
of the thread ; DE the circumference
it d; CE, parallel to the axis, the pitch 5"
h ; and CDE the pitch-angle a. FG. 241.
The motive work in one revolution = M . 2?r = Pp . 2n.
The useful work done in one revolution = Qh.
Hence, the efficiency = ~^ = 22- co t (a + 0) j^
= cot ( a + 0) = tan a cot ( a + 0)- (4)
3O8 THEORY OF STRUCTURES.
This is a maximum when a = 45 -, its value then being
In practice, however, a is generally much smaller, efficiency
being sacrificed to secure a large mechanical advantage, which,
according to eq. (3), increases as a diminishes.
If a -\- = 90, = o, so that to overcome Q, however
small it may be, would require an infinite effort P.
Suppose the pitch-angle sufficiently coarse to allow of the
screw being reversed. Q now becomes the effort and P the
resistance. The direction of r falls on the other side of the
normal, and the relation between P and Q is the same as
above, being substituted for 0.
Thus,
and therefore the mechanical advantage
P
If a = 0, = o, and to overcome P, however small it
may be, Q would require to be infinite.
.-. a > 0.
If a < 0, reversal of motion is impossible, and the screw
then possesses the property, so important in practice, of serv-
ing to fasten securely together different structural parts, or of
locking machines.
ENDLESS SCREWS.
309
Again, it may be necessary to take into account the friction
between the nut and its seat, as well as the friction at the end
of the screw. The corresponding moments of friction with
respect to the axis are (Art. 8)
f
7
3 4 -
and / (
/ being the coefficient of friction, d, , d^ the external and inter-
nal diameters of the seat, and d' the diameter of the end of
the screw.
5. Endless Screws (Fig. 242). A screw is often made
to work with a toothed wheel, as, for ex-
ample, in raising sluice-gates, when the screw
is also made sufficiently fine to prevent, by
friction alone, the gates from falling back
under their own weight. The theory is very
similar to the preceding. Let the screw drive. (
A tooth rises on the thread, and the wheel
turns against a tangential resistance Q, which
is approximately parallel to the axis of the
screw.
FIG. 242.
Let Fig. 243 represent one complete turn of the thread
developed in the plane of the paper, a
being the pitch-angle as before.
Consider a tooth. It is acted upon
by Q in a direction parallel to the
axis, and by the reaction R between
the thread and tooth, making an angle
(the angle of friction) with the normal
to the thread CD.
FIG. 243.
.-. Q = R cos (a + 0).
Again, the horizontal component of R, viz., R sin (a -|- 0),
has a moment R sin (a -f- 0) with respect to the axis of the
3IO THEORY OF STRUCTURES.
screw, and this must be equivalent to the moment of the driv-
ing-couple, viz., Pp (Art. 4).
= * sin (a
Thus the relation between P and Q is the same as in the pre
ceding article.
Similarly if the wheel acts as the driver,
6. Rolling Friction. The friction between a rolling body
and the surface over which it rolls is called rolling friction.
Prof. Osborne Reynolds has given the true explanation of the
resistance to rolling in the case of elastic bodies. The roller
produces a deformation of the surfaces in contact, so that the
distance rolled over is greater than the actual distance between
the terminal points. This he verified by experiment, and con-
cluded that the resistance to rolling was due to the sliding of
one surface over the other, and that it would naturally increase
or diminish with the deformation. In proof of this he found,
for example, that the resistance to an iron roller on india-
rubber is ten times as great as the resistance when the roller is
on an iron surface. Hence the harder and smoother the sur-
faces, the less is the rolling friction. The resistance is not
sensibly affected by the use of lubricants, as the advantage of
a smaller coefficient of friction is largely counteracted by the
increased tendency to slip. Other experiments are yet re-
quired to show how far the resistance is modified by the
speed.
Generally, as in the case of ordinary roadways, the resist-
ance is chiefly governed by the amount of the deformation of
the surface and by the extent to which its material is crushed.
Let a roller of weight W (Y\g. 244) be on the point of motion
under the action of a horizontal pull R.
ROLLING FRICTION. 31 1
The resultant reaction between the surfaces in contact
must pass through the point of intersection of R and IV.
Let it also cut the surface in the point B.
Let d be the horizontal distance between B and W.
11 p " vertical " " B " R.
Taking moments about B y
Rp = Wd,
or R<
R = the resistance = W-.
P
Coulomb and Morin inferred, j w
as the results of a series of ex- FIG. 2 44 .
periments, that d is independent of the load upon the roller as
well as of its diameter,* but is dependent upon the nature of
the surfaces in contact.
* Dupuit's experiments led him to the conclusion that d is proportional to
the square root of the diameter, but this requires further verification.
Let n be the coefficient of sliding friction.
The resistance of the roller to sliding is /j, W, and " rolling " will be insured
d
if R < fj. W, i.e., if - < tan 0, which is generally the case so long as the direc-
/
tion of R does not fall below the centre of the roller.
Assume that R is applied at the centre. The radius r may be substituted
for/, since (/is very small, and hence
R = W-.
r
An equation of the same form applies to a wheel rolling on a hard roadway
over obstacles of small height, and also when rolling on soft ground. In the
latter case, the resistance is proportional to the product of the weight upon the
wheel into the depth of the rut, and the depth for a small arc is inversely pro-
portional to the radius.
Experiments on the tractional resistance to vehicles on ordinary roads are
few in number and incomplete, so that it is impossible to draw therefrom any
general conclusion.
From the experiments carried out by Easton and Anderson, it would appear
that the value of d in inches varies from 1.6 to 2.6 for wagons on soft ground,
and that the resistance is not sensibly affected by the use of springs. Upon
a hard road, in fair condition, the resistance was found to be irom to \ of
that on the soft ground, the average value of d being \ inch, and was very
sensibly diminished by the use of springs.
312
THEORY OF STRUCTURES.
7. Journal-friction. Experiments indicate that f is not
the same for curved as for plane surfaces, and in the ordinary
cases of journals turning in well-
lubricated bearings the value of /is
probably governed by a combina-
tion of the laws of fluid friction and
of the sliding friction of solids.
The bearing part of the journal
is generally truly cylindrical and is
terminated by shoulders resting
against the ends of the step in
which the journal turns.
Consider a journal in a semicircular bearing with the cap
removed. When the cap is screwed on, the load upon the
journal will be increased by an amount approximately equal
to the tension of the bolts. Let -Pbe the load.
Assume that the line of action of the load is vertical and
that it intersects the axis of the shaft. This load is balanced
by the reaction at the surface of contact, but much uncertainty
exists as to the manner in which this reaction is distributed.
There are two extremes, the one corresponding to a normal
pressure of constant intensity at every point of contact, the
other to a normal pressure of an intensity varying from a
maximum at the lowest point A to a minimum at the edge of
the bearing B.
Let / be the length of the bearing, and consider a small
element AS at any point C, the radius OC (= r) making an
angle 6 with the vertical OA.
First. Let / be the constant normal intensity of pressure.
p =
e. i)
2pir.
Frictional resistance =
= fpl^(AS}=fplnrfP- .
The frictional resistance probably approximates to this
limit when the journal is new.
JO URN A L- FRICTION. 3 1 3
Second. Let / = p Q cos 0,
so that the intensity is now proportional to the depth CD and
varies from a maximum / at A to nil at B. This, perhaps,
represents more accurately the pressure at different points
when the journal is worn.
2/ /r, A cos a . dO = / /r-
t/o 2
and the frictional resistance = 2(fj>4Sl) 2fpjr = fP-.
Hence, the frictional resistance lies between fP- and fP .
2 7t
It may be represented by j*P t IJL being a coefficient of friction
to be determined in each case by experiment.
The total moment of frictional resistance must necessarily
be equal and opposite to the moment M of the couple twisting
the shaft ; i.e.,
M =
Thus, the total reaction at the surface of contact is equiva-
lent to a single force P tangential to a circle of radius /*r having
its centre at O and called the friction-circle.
The work absorbed by axle-friction per revolution
= M.27t =. 2
The work absorbed by axle-friction per minute
N being the number of revolutions and v the velocity per
minute.
314 THEORY OF STRUCTURES.
The work absorbed by frictional resistance produces an
equivalent amount of heat, which should be dissipated at once
in order to prevent the journal from becoming too hot. This
may be done by giving the journal sufficient bearing surface
(an area equal to the product of the diameter and the length
of the bearing), and by the employment of a suitable unguent.
Suppose that h units of heat per square inch of bearing
surface (Id) are dissipated per minute.
Let / inches be the length and d inches the diameter of
the journal.
hdl = heat-units dissipated = heat-units equivalent to
frictional resistance
J being Joule's equivalent, or 778 ft.-lbs.
12/ft PN \2jJl Pv
.*. - = r- and -- = .
fJLTt I ^ Id
p
Let 7-7 =p = pressure per square inch of bearing surface.
Id
\2jh
= a constant.
In Morin's experiments af varied from 2 to 4 in., P from
330 Ibs. to 2 tons, and v did not exceed 30 ft. per minute; so
that/z/ was < 5000, and the coefficient of friction for the given
limits was found to be the same as for sliding friction.
Much greater values of pv occur in modern practice.
Rankine gives p(v + 20) = 44800 as applicable to locomo-
tives.
Thurston gives pv = 60000 as applicable to marine engines
and to stationary steam-engines.
Frictional wear prevents the diminution of /below a certain
JOURNAL FRICTION. 31$
limit at which the pressure per unit of bearing surf ace exceeds
a value/ given by the formula.
where
-
In practice k = % for slow-moving journals (e.g., joint-pins),
and varies from I-J to 3 for journals in continuous motion. The
best practice makes the length of the journal equal to four
diameters (i.e., k = 4) for mill-shafting.
Again, 'if the journal is considered a beam supported at
the ends,
q being the maximum permissible stress per square inch, and
C a coefficient depending upon the method of support and
upon the manner of the loading.
k
/. # oc .
i
For a given value of P, d diminishes as q increases. Also,
it has been shown that the work absorbed by friction is
directly proportional to d.
Hence, for both reasons, d should be a minimum and the
shaft should be made of the strongest and most durable
material. In practice the pressure per square inch of bearing
surface may be taken at about 2 tons per square inch for cast-
iron, 3! tons per square inch for wrought-iron, and 6J tons per
square inch for cast-steel.
It would appear, however, from the recent experiments of
Tower and others, that the nature of the material might become
of minor importance, while that of a suitable lubricant would be
of paramount importance. They show that the friction of
properly lubricated journals follows the laws of fluid friction
much more closely than those of solid friction, and that the
3l6 THEORY OF STRUCTURES.
lubrication might be made so perfect as to prevent any ab-
solute contact between the journal and its bearing. The
journal would therefore float in the lubricant, so that there
would be no metallic friction. The loss of power due to fric-
tional resistance, as well as the consequent wear and tear, would
be very considerably diminished, while the load upon the
journal might be increased to almost any extent.
Tower's experiments also indicate that the friction dimin-
ishes as the temperature rises, a result which had already been
experimentally determined by Him. It was also inferred by
Hirn that, if the temperature were kept uniform, the friction
would be approximately proportional to Vv, and Thurston
has enunciated the law that, with a cool bearing, the friction is
approximately proportional to Vv for all speeds exceeding
100 ft. per minute.
With a speed of 150 ft. per minute and with pressures vary-
ing from 100 to 750 Ibs. per square inch, Thurston found ex-
perimentally that /"varied inversely as the square root of the
intensity of the pressure. The same law, but without any
limitations as to speed or pressure, had been previously stated
by Hirn.
8. Pivots. Pivots are usually cylindrical, with the circular
edge of the base removed and sometimes with the whole of
the base rounded. Conical pivots are employed in special
machines in which, e.g., it is important to keep the axis of the
shaft in an invariable position. Spherical pivots are often
used for shafts subject to sudden shocks or to a lateral move-
ment.
(a) Cylindrical Pivots. If the shafts are to run slowly, the
intensity of pressure (/) on the step should not be so great as
to squeeze out the lubricant. Reuleaux gives the following
rules :
The maximum value of / in Ibs. per square inch should be
700 for wrought-iron on gun-metal, 470 for cast-iron on gun-
metal, and 1400 for wrought-iron on lignum-vitae.
For rapidly-moving shafts,
d=c
PIVOTS.
317
n being the number of revolutions per minute, c a coefficient
to be determined by experiment (=.0045),
and Pthe load upon the pivot.
Suppose the surface of the step to be
divided into rings, and let one of these
rings be bounded by the radii x, x -J- dx.
In one revolution the work absorbed
by the friction of this ring
= ji . 2nx . dx . 2nx.
Hence the total work absorbed in one revolution
where
FIG. 246.
1? - d*
and d l , d^ are the external and internal diameters of the sur-
face in contact.
If the whole of the surface is in contact, d^ = o, and the
work absorbed = \^,nPd r
Again, the moment of friction for the ring
.dx.x =
and the total moment
12
If d^ o, the moment = d l .
Thus, in both cases, the work absorbed by friction = 27t
times the moment of friction.
THEORY OF STRUCTURES.
Let D be the mean diameter of the surface in contact
Let 2y be the width of the surface in contact = d^ d^ .
Then
work absorbed = fj.nP \D -\- J.
Sometimes shafts have to run at high speeds and to bear
heavy pressures, as, e.g., in screw-propellers and turbines. In
order that there may be as little vibration as possible, p must
be as small as practicable, and this is to some extent insured
by using a collar-journal.
Let N be the number of collars, and let d^ , d^ be the exter-
nal and internal diameters of a collar.
Then work absorbed by friction per revolution per collar
L 2n X moment of friction.
o
According to Reuleaux, the mean diameter of a collar
n
~-
n being the number of revolutions per minute.
Also, the width of surface in contact = d l d^ = .48
and the maximum allowable pressure per square inch
(b) Wear. The wear at any point of the elementary ring
must necessarily be proportional to the friction ///>, and also to
the amount of rubbing surface which passes over the point in
a unit of time, i.e., the velocity Ax ; A being the angular ve-
locity of the shaft.
PIVOTS. 319
Hence, the wear at any point is proportional to
(c) Conical Pivots. As before, suppose
the surface of the step to be divided into a
number of elementary rings. Two cases will
be discussed :
First. Assume that the normal intensity
of pressure p at the surface of contact is
constant.
Let x,x-\-dx be the distances of D and
E, respectively, from the axis. IGZ
The total moment of friction
sin a
3 sin a
x^ , x^ being the radii of the top and bottom sections of the
step.
Also, P, the total load on the pivot,
= / pDE sin a . 2nx = 2np I xdx
U JC
r* 1
= I
t/.r a
/*,
I dx
IX * f
Also, P = I pDE sin a . inx
t/.r a
= 2npx
Hence total moment of friction = : - (x. + #,).
2 sm a v *
Schieles Pivots. The object aimed at in these pivots
is to give the step such a form that the wear
and the pressure are the same at all points.
Let be the angle made by the tangent at
-V--/D any point of the step with the axis.
Let y be the distance of the point from
the axis. Then
py a sin 0;
and hence if/ is constant,
y a sin 6 or y cosec = a const.
is the equation of the generating line of the step. This line is
known as the tractrix and also as the anti-friction curve. If
the tangent at D intersects the axis in T,
DT = y cosec 6 = a const.
The curve may be traced by passing from one point to an-
other and keeping the tangent DT of constant length.
The above equation may be written
ds
yr- = a const. = a,
' dy
BELTS AND ROPES.
321
or
ds a dy I fdy \
~d~x = ~~ ~y ~dx = V l + w '
which may be easily integrated, the result being the analytical
equation to the curve, viz.,
-f- Va* y + a const.
Schiele or anti-friction pivots are suitable for high speeds, but
have not been very generally adopted.
9. Belts and Ropes. Let the figure represent a pulley
movable about a journal at O, and let a belt (or rope), acted
upon by forces 7, , 7 2 at the ends, embrace a portion ABC
of the circumference subtending an angle a at the centre.
In order that there may be motion in the direction of the
arrow, 7, must exceed 7 2 by an amount sufficient to overcome
the frictional resistance along the arc of contact and the resist-
ance to bending due to the stiffness of the belt.
Consider first the frictional resist-
ance, and suppose the belt to be on the
point of slipping.
Any small element BB' ( ds) of
the belt js acted upon by a pull T tan-
gential to the pulley at B, a pull T dT
tangential to the pulley at B' , and by a
reaction equivalent to a normal torceRds
at the middle point of BB' , and a tan-
gential force, or frictional resistance,
j&ds.
Let the angle COB = 0, and the an-
gle BOB' = dB. ,
Resolving normally,
FIG. 249.
7/n
(T+ T-dT)sm Rds - o.
.322 THEORY OF STRUCTURES.
Resolving tangentially,
2 ^ '
-jj. being the coefficient of friction.
Now dti being very small, sin is approximately ,
dB
cos is approximately unity, and small quantities of the
second order may be disregarded.
Hence, eqs. (i) and (2) may be written
TdO Rds o, ...... (3)
= (4)
dT
, or
Integrating,
C being a constant of integration.
When 6 = 0, T = T 9 , and hence log.T; = C.
or = ^ ........ (6)
7 2
When = a, T = T l , and hence
^ being the number 2.71828, i.e., the base of the Naperian
system of logarithms.
BELTS AND ROPES. 323
If a is increased by /?, the new ratio of tensions will be
&& times the old ratio ; so that if OL increases in arithmetical
progression, the ratio of tensions will increase in geometrical
progression. This rapid increase in the ratio of the tensions,
corresponding to a comparatively small increase in the arc of
contact, is utilized in " brakes"
for the purpose of absorbing
surplus energy. For example :
A flexible brake consisting
of an iron or steel strap, or,
again, of a chain, or of a series
of iron bars faced with wood
and jointed together, embraces
about three-fourths of the cir-
cumference of an iron or wooden
drum. One end of the brake FlG - 25 '
is secured to a fixed point and the other to the end B of a
lever AOB turning about a fulcrum at O. A force applied at
A will cause the brake to clasp the drum and so produce fric-
tion which will gradually bring the drum to rest.
Let &9 be the angular velocity of the drum before the brake
is applied.
Let / be the moment of inertia of the drum with respect to
its axis.
The kinetic energy of the drum = .
When the brake is applied, the motion being in the direc-
tion of the arrow, let the greater and less tensions at its ends
be T lt T^ t respectively.
Let n be the number of revolutions in which the drum is
brought to rest. Then
i/fi? a = (7;- T^ndn, (8)
d being the diameter of the drum.
Also, if Pis the force applied at A, and if / and q are the
perpendicular distances of O from the directions of P and T 9 ,
respectively,
Pp=T# (9)
3 24 THEOR Y OF S 7^R UCTURES.
Again,
T* = TS*, ......... (10)
u
for dry belts on iron pulleys is .28, and for wire ropes .24; if
the belts are wet, >u is about .38.
Formulae (6) and (7) are also true for non-circular pulleys.
10. Effective Tension. The pull.available for the trans-
mission of power = T l 7!, = 5. Let HP be the horse-
power transmitted, v the speed of transmission in feet per sec-
ond, a the sectional area of the rope or belt, and s the stress
per square inch in the advancing portion of the belt.
Then, if T l and T t are in pounds,
= ., and
'
,
550 550'
The working tensile stress per square inch usually adopted
for leather belts varies from 285 Ibs. (Morin) to 35 5 Ibs. (Claudel),
EFFECT OF HIGH SPEED. 325
an average value being 300 Ibs. In wire ropes, 8500 Ibs. per
square inch may be considered an average working tension.
Hempen ropes for the transmission of power generally vary
from 4^ to 6J- in. in circumference.
II. Effect of High Speed. When the speed of trans-
mission is great, the effect of centrifugal force must be taken
into account.
wads if
The centrifugal force or the element ds = ---- , w being
the specific weight of the belt or rope, and r the radius of the
pulley.
Eq. (3) above now becomes
wads v*
Tdd Rds ---- = o,
g r
or
and hence, by eq. (4),
Integrating,
-r Wa
T
since T = T^ when 6 = o.
Also, T = T, when 6 = a, and therefore
g
or
_ V- i).
o
326 THEORY OF STRUCTURES.
The work transmitted per second
= (T, - 7> = T,v - *>(<- - i),
which is a maximum and equal to |7' 8 (^ a i) when
, and the two tensions are then in the ratio of
2 ^ _(_ i to 3.
The speed for which no work is transmitted, i.e., the limit-
ing speed, is given by
wa , i /*
\v v = o, or v = \f
12. Slip of Belts. A length / of the belt (or rope) becomes
/( i -j-^jon the advancing side and l[\ + -jj on the slack side,
T T
where p l = - and / = - , E being the coefficient of elasticity.
Thus, the advancing pulley draws on a greater length than is
given off to the driven pulley, and its speed must therefore
exceed that of the latter by an amount given by the equation
ifi] _
reduction of speed, or slip _ ~ El \ r E' _ p l
speed of driving pulley / p
The slip or creep of the belt measures the loss of work.
In ordinary practice the loss with leather belting does not ex-
ceed 2 per cent, while with wire ropes it is so small that it may
be disregarded.
PA N Y ' S D YNA MO ME TER.
327
13. Prony's Dynamometer. This dynamometer is one of
the commonest forms of friction-brake. The motor whose
power is to be measured turns a wheel E which revolves be-
tween the wood block B and a band of wood blocks A. To
/INP
FIG. 251.
the lower block is attached a lever of radius / carrying a
weight P at the free end. By means of the screws C, D the
blocks may be tightened around the circumference until the
unknown moment of frictional resistance FR is equal to the
known moment Pp.
The weight P, which rests upon the ground when the
screws are slack, is now just balanced.
The work absorbed by friction per minute = 2nRFn = 2nPpn v
n being the number of revolutions per minute.
14. Stiffness of Belts and Ropes. The belt on reaching
the pulley is bent to the curvature of the periphery, and is
straightened again when it leaves the pulley. Thus, an amount
of work, increasing with the stiffness of the belt, must be ex-
pended to overcome the resistance to bending. As the result
of experiment, this resistance has been expressed in the form
T-B, T being the tension of the belt, a its sectional area, R the
radius of the pulley, and b a coefficient to be determined.
According to Redtenbacher, b = 2.36 in. for hempen ropes.
" " " b = 1.67 " " " "
" " Reuleaux, b = 3.4 " " leather belts.
328
THEORY OF STRUCTURES.
Let the figure represent a sheave in a pulley-block turning
in the direction of the arrow about a
journal of radius r.
Let T t be the effort, T 9 the re-
sistance.
The resistance due to the stiff-
ness of the belt may be allowed for
*T*
qTs by adding -j- to the force 7", . The
ftR OK
FIG. 252. frictional resistance at the journal-
surface is P sin orfP, P being the resultant of T lt 7 1 , .
The motion being steady, taking moments about the centre,
or
If T; and T 9 are parallel, P = T> + T>, and the last equa
tion becomes
Let the pulley turn through a small angle 6.
The counter-efficiency of the sheave
motive work Tfl _ T^
useful work z= =
2fr
, a
~^~
R - fr~~ b R - fr'
In the case of an endless belt connecting a pair of pulleys
of radius R 19 R^ the resistance due to stiffness may be taken
equal to ^-\jf "f"j>"/* ^ Dem g tne mean tension [= ~~ L ~~^
The resistance due to journal-friction = frP\- +^~J-
The useful resistance = T, T^ = S.
WHEEL AND AXLE.
Hence, the counter-efficiency
329
In wire ropes the stress due to bending may be calculated
as follows:
Let x be the radius of a wire. The radius of its axis is
sensibly the same as the radius R of the pulley.
The outer layers of the wire will be stretched, and the inner
shortened, while the axis will remain unchanged in length.
Hence,
x change of length of outer or inner strands unit stress
R ~ length of axis E
x
and the unit stress due to bending = ~ .
15. Wheel and Axle. Let the figure represent a wheel
of radius / turning on an axle of radius r, under the action of
the two tangential forces P and Q, in-
clined to each other at an angle 6.
The resultant R of P and Q must
equilibrate the resultant reaction be-
tween the wheel and axle at the sur-
face of contact.
Let the directions of P and Q
meet in T.
If there were no friction, the re-
sultant reaction and the resultant R
would necessarily pass through O
and T.
Taking friction into account, the
direction of R will be inclined to TO.
Let its direction intersect the circumference of the axle in the
point A. The angle between TA and the normal AO at A,
the motion being steady, is equal to the angle of friction ; call
it 0.
FIG. 253.
330 THEORY OF STRUCTURES.
Taking moments about O,
Pp Qp Rr sin = o
Also,
. . (I)
R* = P* + Q + 2 PQ cos e (2)
Let/= sin =
=, fjL being the coefficient of friction.
Eq. (i) may now be written
Pp- Qp-fRr = 0. . ,
' If P and Q are parallel in direction,
8 = and R = P -4- Q.
Let the figure represent a wheel and axle.
(3)
254.
Let P be the effort and Q the weight lifted, the directions
of Pand Q being parallel.
Let IV be the weight of the " wheel and axle."
Let R 1 and R^ be the vertical reactions at the bearings.
Let/ be the radius of the wheel.
Let q " " " axle.
Let r " " " bearings.
Take moments about the axis. Then
Pp Qq RS sl ' n ~ RS sin = 0. . . (4)
TOOTHED GEARING. 331
But
...... (5)
Hence,
or
P(p-fr) = Q(q+fr)+fWr. ........ (6)
Efficiency. In turning through an angle 0,
motive work = PpQ,
useful work == QqQ,
effirier -
" e -~ = '
and the ratio -p is given by eq. (6).
16. Toothed Gearing. In toothed gearing the friction is
partly rolling and partly sliding, but the former will be disre-
garded, as it is small as compared with the latter.
Let the pitch-circles of a pair of teeth in contact at the
point B touch at the point A ; and consider the action before
reaching the line of centres a a , i.e., along the arc of
approach.
332 THEORY OF STRUCTURES.
The line AB is normal to the surfaces in contact at the
point B.
Let R be the resultant reaction at B. Its direction, the
motion being steady, makes an angle 0, equal to the angle of
friction, with AB.
Let B be the angle between O,O t and AB.
Let the motive force and force of resistance be respectively
equivalent to a force P tangential to the pitch-circle O l , arid
to a force Q tangential to the pitch-circle O, .
Let r l , r a be the radii of the two wheels.
The work absorbed by friction in turning through the small
arc ds
. (i)
Consider the wheel O v , and take moments about the centre,
Pr 1 = ^|r 1 sin(6/-0) + ^sm0f, ... (2)
where AB = x.
Similarly, from the wheel <9 3
0r a = tf K sin (0 - 0) - * sin 0}. ... (3)
Hence,
jtr
n sin (0 - 0) - sin 4>
= - -- - ...... (4)
sin (0 0) -f- - sin
f i
and therefore
'i i \
(r + rl* sin
'i
+ * sn
(5)
sin (0 0) -- sin
Hence, the work absorbed by friction in the arc ds
= Q-
sin (6 0) sin
7*0
TOOTHED GEARING. 333
In precisely the same manner it can be shown that, after
leaving the line of centres, i.e., in the arc of recess,
ft sin (0 -|- 0) sin
p - x
sin (0+0) + - sin
and the work absorbed by friction in the arc ds
= Q -- . .... (8)
sin (0+0) -- sin
^2
The ratio -p and the loss of work given by eqs. (4) and (6)
are respectively greater than the ratio ^~ and the loss of work
given by eqs. (7) and (8), and therefore it is advisable to make
the arc of approach as small as possible.
Again, by eq. (4), motion will be impossible if
sin (6 0) + - sin = o ;
**i
i.e., if cot = cot
r 1 sin
and this can only be true if the direction of R passes through (9 3 .
Simple approximate expressions for the lost work and
efficiency may be obtained as follows:
6 differs very little from 90, and x is small as compared
with r 2 and differs little from the corresponding arc s meas-
ured from A.
Hence the work absorbed by friction in the arc ds
= Q tan - *& =
334 .THEORY OF STRUCTURES.
and the work lost in arc of approach s 1
The useful work done in the same interval = Qs l .
The counter-efficiency (reciprocal of efficiency)
Similarly for the arc of recess s^ ,
the lost work = g/J- + - , - (i i)
*7*j T'j/ 2
and the counter-efficiency = i -}-/*( -J-- ). . (12)
\r, /y 2
If ^ = ,y a = pitch = / = - - 1 = - ? , n^ , 2 being the num-
^1 ?^ 3
her of teeth in the driver and the follower, respectively, the ex-
pressions for the lost work given by eqs. (9) and (11) are iden-
tical, and those for the counter-efficiency given by eqs. (10)
and (12) are also identical.
Thus, the whole work lost during the action of a pair of
teeth
....... (13)
and the counter-efficiency
<
This last equation shows that the efficiency increases with
the number of teeth.
EFFICIENCY OF MECHANISMS.
335
If the follower is an annular wheel, - - must be substi-
tuted for -+- in the above equations. Thus, with an an-
nular wheel the counter-efficiency is diminished and the
efficiency, therefore, increased.
It has been assumed that R and Q are constant, as their
variation from a constant value is probably small. It has also
been assumed that only one pair of teeth are in contact. The
theory, however, holds good when more than one pair are in
contact, an effort and resistance, corresponding to P and Q,
being supposed to act for each pair.
17. Bevel-wheels. Let I A, IB represent the develop-
ments of the axes of the pitch-
circles 77, , 77 3 of a pair of bevel-
wheels when the pitch-cones are
spread out fiat, O l , O z being the
corresponding centres.
The preceding formulae will ap-
ply to bevel-wheels, the radii being
6^7, <9 2 7, and the pitch being meas- ,!,''''
ured on the circumferences I A, IB.
18. Efficiency of Mechanisms.
Generally speaking, the ratio of
the effort P to the resistance Q in a.
mechanism may be expressed as a
function of the coefficient of fric-
tion yu. Thus,
t
If, now, the mechanism is moved so that the points of
application of P and Q traverse small distances Ax, Ay in the
directions of the forces,
.1. ^ .
the efficiency = j~- =
PAx
I Ay
336
THEORY OF STRUCTURES.
Ay
But the ratio -r- depends only upon the geometrical rela-
tions between the different parts of the mechanism, and will
therefore remain the same if it is assumed that /* is zero. In
such a case the efficiency would be perfect, or the motive work
(PAx) would be equal to the useful work (QAy\ and therefore
I =
I Ay
Hence, the efficiency
TABLE OF COEFFICIENTS OF AXLE-FRICTION.
>,
Q
Greasy and
Wet.
Ordinary
Lubrication.
Continuous
Lubrication.
Pure Carriage-
grease.
Lard and
Plumbago.
O
Bell-metal on bell-metal
Brass on brass .
Brass on cast-iron
Cast-iron on bell-metal
161
o6<;
.16
Cast-iron on brass ,
Cast-iron on cast-iron
14
Cast-iron on lignum-vitae.
xSe
Lignum-vitae on cast-iron
Lignum-vitae on lignum-vitae
Wrought-iron on bell-metal
Wrought-iron on cast-iron
.251
.189
.n6
075
J 7
.07
054
.09
.11
15
Wrought-iron on lignum-vitae. . .
.187
125
EXAMPLES. 337
EXAMPLES.
i. In a pair of four-sheaved blocks, it is found that it requires a force
P to raise a weight $P', and a force $P to raise a weight i $P'. Show
that the general relation between the force P and the weight W to be
raised is given by
Find the efficiency when raising the weights 5/" and
2. Find the mechanical advantage when an inch bolt is screwed up
by a i5-in. spanner, the effective diameter of the nut being if in., the
diameter at the base of the thread .84 in., and .15 being the coefficient of
friction.
3. A belt, embracing one-half the circumference of a pulley, transmits
10 H. P. ; the pulley makes 30 revolutions per minute and is 7 ft. in
diameter. Neglecting slip, find TV and Ti ; /a being .125.
4. A -in. rope passes over a 6-in. pulley, the diameter of the axis being;
% in. ; the load upon the axis == 2 x the rope tension. Find the efficiency
of the pulley, the coefficient of axle-friction being .08 and the coefficient
for stiffness .47.
Hence also deduce the efficiency of a pair of three-sheaved blocks.
5. If the pulleys are 50 ft. c. to c. and if the tight is three times the
slack tension, find the length of the belt, the coefficient of friction being
and the diameter of one of the pulleys 12 in.
6. Show that the work transmitted by a belt passing over a pulley
/~T
will be a maximum when it travels at the rate of A/ __ 2 ft. per sec., T*
yn
being the slack tension and m the mass of a unit of length of the belt.
The tight tension on a 2o-in. belt, embracing one-half the circum-
ference of the pulley, is 1200 Ibs. Find the maximum work the belt will
transmit, the thickness of the belt being .2 in. and its weight .0325 Ib.
per cubic inch. (Coefficient of friction = .28.)
7. In an endless belt passing over two pulleys, the least tension is
150 Ibs., the coefficient of friction .28, and the angle subtended by the
arc of contact 148. Find the greatest tension. The diameter of the
larger wheel is 78 in., of the smaller 10 in., of the bearings 3 in. Find
the efficiency. A tightening-pulley is made to press on the slack side
of the belt. Assuming that the working tension is to the coefficient of
elasticity in the ratio of i to 80, find the increment of the arc of contact
THEORY OF STRUCTURES.
on the belt-pulley, the tension of the slack side, and the force of the
tightening-pulley.
8. A belt weighing Ib. per lineal foot, connects two 42-in. pulleys, one
making 240 revolutions per minute. Find the limiting tension for which
work will be transmitted. Also find the tight and slack tensions and
the efficiency when the belt transmits 5 horse-power. Diameter of axle
2 in.; coefficient of friction = .28.
9. A circular saw makes 1000 revolutions per minute and is driven
"by a belt 3 in. wide and \ in. thick, itsweight per cubic inch being .0325
Ib. The belt passes over a lo-in. pulley, embracing one-half the cir-
cumference, and transmits 6 H. P. Find the light and slack tensions,
the coefficient of friction being .28.
10. A flexible band, embracing three-fourths of the circumference of
a brake-pulley keyed on a revolving shaft, has one extremity attached to
the end A of the lever AOB, and the other to \\\t fixed point O (between
A and J3) about which the lever oscillates. The pressure between the
band and pulley is effected by a force applied at right angles to the lever
at the end B. Show that the time in which the axle is brought to rest
is about i\ times as great when revolving in one direction as in the
opposite (/ = .2).
11. In a Prony-brake test of a Westinghouse engine, the blocks were
fixed to a 24-in. fly-wheel with a 6-in. face, and the balance-reading was
48 Ibs.; the distance from centre of shaft to centre of balance, measured
horizontally, was 30 in., and the number of revolutions per minute was
624. Find the H. P. Ans. 14.3.
12. An engine makes 150 revolutions per minute. If the diameter of
the brake-pulley is 45 in. and the pull on the brake is 50 Ibs., find the
B. H. P. Ans. 2.67.
13. A small water-motor is tested by a tail dynamometer. The pul-
ley is 18 in. in diameter; the weight is 60 Ibs.; the spring registers a
pull of 50 Ibs.; the number of revolutions per minute = 500. Find the
B. H. P. Ans. f
14. The power of an engine making n revolutions per minute is
tested by a Prony brake having its arm of length r connected with a
spring-balance which registers a force P. The arm is vertical and the
weight W of the brake is supported by a stiff spring fixed vertically
below the centre of the wheel. What error in B. H. P. would be intro-
duced by placing the spring x ft. away from the central position ?
Ans. Bl ^* , B being the B. H. P.
15. Find work absorbed by friction per revolution by a pivot 3 in.
long and carrying 6 tons, its upper face being 6 in. in diameter, coeffi-
cient of friction .04, and 2 a being 90.
EXAMPLES. 339
1 6. The diameter of a solid cylindrical cast-steel pivot is 2% in. Find
the diameter of an equally efficient conical pivot.
17. The pressure upon a 4-in. journal making 50 revolutions per min-
ute is 6 tons, the coefficient of friction being .05. Find the number of
units of heat generated per second; Joule's mechanical equivalent of
heat being 778 ft.-lbs.
1 8. A water-wheel of 20 ft. diameter and weighing 20,000 Ibs. makes
10 revolutions per minute; the gudgeons are 6 in. in diameter and the
coefficient of friction is .1. Find the loss of mechanical effect due to
friction. If the motive power is suddenly cut off, how many revolutions
will the wheel make before coming to rest ? Ans. f H. P. ; 10.9.
19. A fly-wheel weighing 8000 Ibs. and having a radius of gyration of
10 ft. is disconnected from the engine at the moment it is making 27
revolutions per minute ; it stops after making 17 revolutions. Find the
coefficient of friction, the axle being 12 in. in diameter. Ans. .2325.
20. A railway truck weighing 12 tons is carried on wheels 3 ft. in
diameter ; the journals are 4 in. in diameter, the coefficient of friction
jjV Find the resistance of the truck so far as it arises from the friction
of the journals. Ans. 37^ Ibs.
21. A tramcar wheel is 30 in. in diameter, the axle 2$- in.; the coeffi-
cient of axle-friction .08, of rolling friction .09. Find the resistance pet
ton. Ans. 28.37 Ibs.
22. A bearing 16 in. in diameter is acted upon by a horizontal force
of 50 tons and a vertical force of 10 tons ; the coefficient of friction is -fa.
Find the H. P. absorbed by friction per revolution. Ans. .906 H. P.
23. A steel pivot 3 in. in diameter and under a pressure of 5 tons
makes 60 revolutions per minute in a cast-iron step well lubricated with
oil. How much work is absorbed by friction, the coefficient of friction
being .08 ?
24. A pair of spur-wheels are 4 in. and 2 in. in diameter ; the flanks
of the teeth are radial ; the larger wheel has 16 teeth ; the arc of ap-
proach = arc of recess = | of the pitch. Show how to form the teeth,
and find their efficiency. (Coefficient of friction = .n.)
25. Find the work lost by the friction of a pair of teeth, the number
of teeth in the wheels being 32 and 16, and the diameter of the larger
wheel, which transmits 3 horse-power at 50 revolutions per minute, 3 ft.
26. The driver of a pair of wheels has 120 teeth, and each wheel has
an addendum equal to .28 times the pitch ; the arcs of approach and
recess are each equal to the pitch ; the tooth-flanks are radial. (Coeffi-
cient of friction = .106.) Find the efficiency.
CHAPTER VI.
ON THE TRANSVERSE STRENGTH OF BEAMS.
FIG. 257.
I. To determine the Elastic Moment. Let the plane of
-- |h the paper be a plane of symmetry
_ | c with respect to the beam PQRS.
If the beam is subjected to the
action of external forces in this
plane, PQRS is bent and as-
sumes a curved form P'Q'R'S'.
FIG. 258. The upper layer of fibres, Q'R', is
extended, the lower layer, P'S', is compressed, while of the
layers within the beam, those nearer P'S' are compressed and
those nearer Q'R' are extended. Hence, there must be a layer
M'N' between P'S' and Q'R' which is neither compressed nor
extended. It is called the neutral surface (or cylinder), and
its axis is perpendicular to the plane of flexure. In the present
treatise it is proposed to deal with flexure in one plane only,
and, in general, it will be found more convenient to refer to
M'N' as the neutral line (or axis), a term only used in refer-
ence to a transverse section.
If a force act upon the beam in the direction of its length,
the lower layer P'S', instead of being compressed, may be
stretched. In such a case there is no neutral surface within
the beam, but theoretically it still exists some-
where without the beam.
Let ABCD be an indefinitely small rect-
angular element of the unstrained beam, and
let its length be s. Let A'B'C'D', Fig. 260,
be the element after deformation by the external forces.
340
FIG. 259.
THE ELASTIC MOMENT. 34*
P'Q', the neutral line, being neither com- ,
pressed nor extended, is unchanged in length
and equal to PQ s.
Let the normals at P r and Q' to the neutral
line meet in the point O ; O is the centre of
curvature of P'Q'*
Also, as the flexure of the element is very
small, the normal planes through OP' and OQ'
may be assumed to be perpendicular to all the
layers which traverse the corresponding sec-
tions of the beam, so that they must coincide
with the planes A ' D' and B ' C' ', respectively.
The assumptions made in the above are :
(a) That the beam is symmetrical with
respect to a certain plane.
(b) That the material of the beam is homo-
geneous. FIG. 260.
(c) That sections which are plane before bending remain
plane after bending.
(d) That the ratio of longitudinal stress to the correspond-
ing strain is the ordinary (i.e., Young's) modulus of elasticity
notwithstanding the lateral connection of the elementary
layers.
(e) That these elementary layers expand and contract
freely under tensile and compressive forces.
Consider an elementary layer p'q', of length s', sectional
area a l , and distant y l from the neutral surface.
Let OP' = R = OQ'.
From the similar figures OP'Q and Op'q',
Op' p'q' R+y, s' y v s'-s
Also, if t l is the stress along the layer p'q' ,
34 2 THEORY OF STRUCTURES.
E being the coefficient of elasticity of the material of^ the
beam.
So, if / 2 , # 2 , 7 2 , / 3 , a 3 , jj/ 3 , . . . are respectively the stress,
sectional area, and distance from the neutral surface, of the
several layers of the element,
The total stress along the beam is the algebraic sum of all
these elementary stresses,
Again, the moment of /, about P' = t^ = ^7^ ;
and so on.
Thus, the Elastic Moment for the section A ' D' = the alge-
braic sum of the moments of all the elementary stresses in the
different layers about P' 9
THE ELASTIC MOMENT. 343
Now, 2 (ay*) is the moment of inertia of the section of the
beam through A 'D ', with respect to a straight line passing
through the neutral line and perpendicular to the plane of
flexure, i.e., the plane of the paper. It is usually denoted by
/ or AJ?, A being the sectional area, and k the radius of
gyration. Thus,
E E
the elastic moment = -7,- / = -^Ak*.
K. K.
But the elastic moment is equal and opposite to the bending
moment (M) due to the external forces, at the same section.
Hence
Note.\\. is necessary in the above to use the term alge-
braic, as the elementary stresses change in character, and
therefore in sign, on passing from one side of the neutral sur-
face to the other.
Cor. I. Bearing in mind assumption
(e), the figure represents on an exaggerated
scale the transverse section of the beam
at A'D ', the upper and lower breadths
of the beam, A' A" and H D" , being re-
spectively contracted and stretched, and
being also arcs of circles having a common
centre at O'.
Let R' be the radius of the arc P' P" ,
whose length remains unchanged.
Let mE be the lateral coefficient of
FlG - 2fii - elasticity, m being a numerical coefficient.
As before, for any layer at a distance y from P'P",
/. R' = mR.
344 THEORY OF STRUCTURES.
Thus, within the limits of elasticity, the curvature of the breadth
is that of the length, and does not sensibly affect the re-
sistance of the beam to bending. The influence, however,
upon the bending may become sensible if the breadth is very
large as compared with the depth, as, e.g., in the case of iron
or steel plates.
Cor. 2. If the resolved part of the external forces in the
direction of the length of the beam is nil,
E
the total longitudinal stress = -^(ay) = o, or ^(ay) = o,
showing that P f must be the centre of gravity of the section
through A' D' . Hence, when the external forces produce no
longitudinal stress in the beam, the neutral line is the locus of
the centres of gravity of all the sections perpendicular to the
length of the beam.
Cor. 3. If /, a, y be, respectively, the stress, sectional area,
and distance of a fibre from the neutral line, then
F E t
--ay = /, or -=,-y = - = intensity of stress =f y , suppose,
J\. rL a
EXAMPLE i. A timber beam, 6 in. square and 20 ft. long,
rests upon two supports, and is uniformly loaded with a weight
of 1000 Ibs. per lineal foot. Determine the stress at the centre
at a point distant 2 in. from the neutral line.
Also find the central curvature, E being 1,200,000 Ibs.
/T /^
/=- = 108, M = 1000 X io 1000 X 5 = 5000 ft.-lbs.
= 60,000 inch -Ibs., and y = 2 in.
INTERNAL STRESSES. 345
Hence from the above equations,
1200000 f y
x 108 = 60000 =
Thus R = 2160 in. = 180 ft., and f y nii-fr Ibs. per sq. in.
Ex. 2. A standpipe section, 33 ft. in length and weighing
5720 Ibs., is placed upon two supports in the same horizontal
plane, 30 ft. apart. The internal diameter of the pipe is 30
in., and its thickness j- inch. Determine the additional
uniformly distributed load which the pipe can carry between
the bearings, so that the stress in the metal may nowhere ex-
ceed 2 tons per square inch.
Let J^be the required load in pounds.
3
The weight of the pipe between the bearings = . 5720
= 5200 Ibs.
Thus, the total distributed weight between the bearings
= (^+5200 Ibs.)
Now M = f
and the stress in the metal is necessarily greatest at the central
section.
W+ 5200
M, at the centre, = .30. 12 inch-lbs. ;
f e = 2 X 2240 Ibs., and - = nr*t = . 1 5' .. -.
W-\- 52OO 22 , I
^ . 30. 12 = 2.2240.--. 15 - = 72000 X 22,
8 .72
and hence W 30,000 Ibs.
34-6 THEORY OF STRUCTURES.
Cor. 4. The beam is strained to the limit of safety when
either of the extreme layers A'B', D ' C is strained to the limit
of elasticity. In such a case, the least of the values of for
y
the extreme layers A'B', D' C is the greatest consistent with
the strength of the beam ; and if f c and c are the corresponding
intensity of stress, and distance from the neutral axis,
EXAMPLE. Compare the strengths of two similarly loaded
beams of the same material, of equal lengths and equal sectional
areas, the one being round and the other square.
Let r be the radius of the round beam ; f r9 the intensity of
the skin stress.
Let a be a side of the square beam ; f a9 the intensity of
the skin stress. Then
<2 2 ; /, for round bar, = , and for square bar = .
Also, since the beams are similarly loaded, the bending
moments at corresponding points are equal.
r 4 #12
2
so that
Thus, under the same load, the round beam is strained to
a greater extent than the square beam, and the latter is the
stronger in the ratio of \^SS to
BREAKING WEIGHTS. 347
Cor. 5. The neutral surface is neither stretched nor com-
pressed, so that it is not subjected to any longitudinal stress.
But it by no means follows that this surface is wholly free from
stress, and it will be subsequently seen that the effect of a
shearing force, when it exists, is to stretch and compress the
different particles in diagonal directions making angles of 45
with the surface.
bd* d
Cor. 6. For a rectangular beam// = , and c = .
,.*=// =/**=/..
c d 12 6
If the beam is fixed at one end and loaded at the other
with a weight W, the maximum bending moment = Wl.
If the beam is fixed at one end and loaded uniformly with
a weight wl = W, the maximum bending moment
wP Wl
If the beam rests upon two supports and carries a weight W
Wl
at the centre, the maximum bending moment ,
If the beam rests upon two supports and carries a uniformly
distributed load of wl = W, the maximum bending moment
Wl
Hence, in the first case, IV = =-\
" " second- W=2
" third W=^
" " fourth " W=^-
THEORY OF STRUCTURES.
In general, - W = g- q-j- ;
t/ being some coefficient depending upon the manner of the
loading.
Now, if the laws of elasticity held true up to the point of
rupture, these equations w^ould give the breaking weights (W 7 ),
corresponding to different ultimate unit stresses (/), but the
values thus derived differ widely from the results of experi-
ment. It is usual to determine the breaking weight (W) of a
rectangular beam from the formula W = C -T-, where C is a
constant which depends both upon the manner of the loading
and the nature of the material, and is called the coefficient of
rupture.
The modulus of rupture is the value of/ in the ordinary
bending-moment formula (M = /) when the load on the
beam is its breaking load.
The preceding equations, however, may be evidently em-
ployed to determine the breaking weights in the several cases
by making J -^-q = C. In this case /"is no longer the real stress,
but may be called the coefficient of bending strength.
The values of C for iron, steel, and timber beams, supported
at the two ends and loaded in the centre, are given. in the
Tables at the end of Chapter III.
The corresponding value of f is obtained from the equation
or
EXAMPLE. Determine the central breaking weight of a
EQUALIZATION OF STRESS. 349
red-pine beam, 10 in. deep, 6 in. wide, and resting upon two
supports 20 ft. apart.
The value of C for red pine is about 5700. Hence,
the breaking weight = W 5700 ll?_ = 14,250 Ibs.
2O X 12
2. Equalization of Stress. The stress at any point of a
beam under a transverse load is proportional to its distance
from the neutral plane so long as the elastic limit is not ex-
ceeded. At this limit materials which have no ductility give
way. In materials possessing ductility, the stress may go on
increasing for some distance beyond the elastic limit without
producing rupture, but the stress is no longer proportional to
the distance from the neutral plane, its variation being much
slower. This is due to the fact that the portion in compres-
sion acquires increased rigidity and so exerts a continually
increasing resistance (Chap. Ill) almost if not quite up to
the point of rupture, while in the stretched portion a flow of
metal occurs and an approximately constant resistance to the
stress is developed. Thus, there will be a more or less perfect
equalization of stress throughout the section, accompanied by
an increase of the elastic limit and of the apparent strength,
the increase depending both upon the form of section and the
ductility.
For example, if the tensile elastic limit is the same as the
compressive. the shaded portion of Fig. 262 gives a graphical
__
^'- J==
FIG. 262. FIG. 263. FIG. 264.
representation of the total stress in a beam of rectangular
section when the straining is within the elastic limit. Beyond
this limit, it may be represented as in Fig. 263, and will be
350
THEORY OF STRUCTURES.
intermediate between Fig. 262 and the shaded rectangle of
Fig. 264 which corresponds to a state of perfect equaliza-
tion.
3. Surface Loading.* It may be well to draw attention to
another important assumption upon which is based the mathe-
matical treatment of the problem of Beam Flexure.
It has been assumed that the external forces acting on a
beam can be so applied that they may be considered as dis-
tributed uniformly over the whole section. Thus when a beam
encastre" is loaded at the free end, Fig. 265, the load P is as-
FIG. 265.
sumed to be uniformly distributed over the section ab, i.e.,
each element in the section is supposed to experience the same
amount of strain due to the load, and the reaction of the wall
is also supposed to be uniformly distributed over each element
in the section cd.
It is clear that such suppositions must be far from the
truth.
In practice, the load P must be hung by some means from
the beam, say by a stirrup passing over the top. The whole
load is then concentrated at the line of contact of the stirrup
with the beam, and it is obviously untrue to say that every
* This article was kindly written by Professor Carus-Wilson and is an
abstract of a Paper presented by him to the Physical Society.
SURFACE LOADING. 35 1
element in the section ab is equally strained. But more
than this. It has been assumed that, taking the effect of
the load as distributed uniformly over the section ab, and
a certain deflection thereby produced, the effect of P on
each element of the section ab may be disregarded in com-
parison with the strains involved in the deflection which P
produces.
It will probably be difficult at first to grasp the fact that
certain measurable effects have been actually neglected, but
that this is so may be seen by supposing the beam in question
to be a pine beam, and the stirrup of iron. Experience proves
that with a very moderate load the beam will be indented at a.
But the theory shows that the longitudinal tension at a is
zero and increases to a maximum at d.
Thus, so far from the squeezing effect of the load being
distributed uniformly over the section ab, it is concentrated at
a, and hence it is impossible to neglect it.
Engineers have always recognized the existence of this
" surface-loading" effect in practice, and where possible, have
provided a good " bearing" in order to avoid such local
strains ; but this cannot always be done as, for instance,
in the case of rollers under bridge ends. The theory of flex-
ure is therefore manifestly incomplete if it cannot take into
account the actual manner in which the loads are and must be
applied.
FIG. 266.
It can be shown that the effect of placing a pressure of p
tons per inch run, say in the form of a loaded roller, on a beam
resting upon a flat surface, as in Fig. 266, to prevent it from
352
THEORY OF STRUCTURES.
bending, is to compress every element say along ab with an
intensity given approximately by the equation
where f is the pressure at a distance x from a, the point of con-
tact, and h = ab. This is the equation to a curve be which is
approximately an hyperbola.
When a beam is bent by the application of external forces, a
very close approximation to the true condition may be obtained
by superposing this surface-loading effect on that found for
bending.
Take the case of a beam supported at the ends and loaded
at the centre, and let it be required to find the condition along
ab, Fig. 267.
FIG. 267.
The effect of the bending is to produce compression above
and tension below the point c, and these effects may be repre-
sented by a right line de passing through c.
The surface-loading effect may be represented by an hyper-
bola giving the compression at any point along ab due to the
load. The hyperbola and straight line will intersect in two
SURFACE LOADING. 353
points h and/, which shows that at two points k ',/' along ab the
vertical squeeze produced by the load is of equal intensity to
the horizontal squeeze produced by the bending ; hence an
element at each of these points is subject to cubical compres-
sion only. From a to/' the beam is squeezed vertically, from
/' to h' it is squeezed horizontally, and from^' to b it is stretched
horizontally. The intensities are given at every point by the
difference between the ordinates of the line of bending de and
the curve of loading. It will appear that one effect of surface-
loading is to make the neutral axis rise up under the load and
pass through the point h f , for there is neither compression nor
tension at that point.
This can be verified by examining the condition of a bent
glass beam by polarized light. The neutral axis is pushed up
under the load and there is a black ring passing through the point
/'. If the span is diminished and the load kept constant, it is
clear that ae will become less, while the curve of loading remains
the same, until the line dee ceases to cut the curve ; every
element along ab will then be subjected to horizontal stretch,
and the stretch is greatest at a ; the result obtained by neglect-
ing the surface loading is that only elements from c to b are
stretched, the greatest stretch being at b. The position of the
" neutral points " is given by the equation
T6~m
where y is the distance from the top edge, h equals the depth
ab, m -^-y 4, and a = one-half of the span.
For all elements in ab to be stretched, the ratio of span to
depth, viz., -= , must be equal to or less than 4.25. In other
words, for any beam, and any load, if the span is less than 4^
times the depth, every element in the normal under the load
is stretched horizontally.
354
THEORY OF STRUCTURES.
4. Beam acted upon by a Bending Moment in a Plane
which is not a Principal Plane.
Let*XOX, YOYbe the principal axes of the plane section
^of the beam.
\y
FIG. 268.
Let the axis MOM of the bending moment M make an
angle ot with OX.
M may be resolved into two components, viz.,
M cos a X and M sin a =. Y.
These components may be dealt with separately and the
results superposed.
Thus, the total stress, /, at any point (x y y)
stress due to X + stress due to Y = ~- + =/,
1* -fy
I* , I y being the moments of inertia with respect to the axes
XOX, YOY, respectively.
If the point (xy) is on the neutral axis, then
or
being the angle between the neutral axis and XOX.
SPRINGS.
355
Also see Art. 6, Chap. VIII. In this article 6 is the angle
between the neutral axis and the axis of the couple, i.e.,
6 = ft - a.
5. Springs. (a) Flat Springs. If two forces, each equal
to P but acting in opposite directions in the same straight line,
are applied to the ends of a straight uniform strip of flat steel
spring, the spring will assume one of the forms shown below,
known as the elastic curve. This curve is also the form of the
linear arch best suited to withstand a fluid pressure, Chap.
XIII.
FIG. 274.
FIG. 275.
Consider a point B of the spring distant y from the line of
action of P. Then
Py bending moment at B = -=- ,
THEORY OF STRUCTURES.
R being the radius of curvature at B, and / the moment of
inertia of the section.
If E and /are both constant,
Ry = a constant
is the equation to the elastic curve.
(b) Spiral Springs (as, e.g., in a watch). Let the figure rep-
resent a spiral spring fixed at C and to an arbor at A, and
subjected at every point of its length
to a bending action only.
Consider the equilibrium of any
portion AB of the spring.
The forces at A are equivalent to a
couple of moment M, and to a force P
acting in some direction AD.
This couple and force must balance the elastic moment
at B.
.'. M-\- Py = El X change of curvature at B,
y being the distance of B from the line of action of P, or
R being the radius of curvature at B before winding, and R
that after winding.
Let ds be an elementary length of the spring at B.
Then, for the whole spring,
2(M + Py]ds = 12 - - - 12(40 -
or M2ds -f- P^yds El X total change of curvature between
A and C ;
.-. Ms -\- Psy = EI(B - ),
SP KINGS. 357
s being the length of the spring, y the distance of its C. of G.
from AD, 6 the angle through which the spring is wound up,
and 6 the " unwinding" due to the fixture at C. With a large
number of coils the distance between the C. of G. and A may
be assumed to be nil and then y = o.
Also, if the spring is so secured that there is no change of
direction relatively to the barrel,
= o, and Ms = Eld.
Let the winding-up be effected by a couple of moment
Qq = M, Q being a tangential force at the circumference of a
circle of radius q.
The distance through which Q moves (or deflection of Q)
= g# = s, since M = f,
/being the skin stress, and c the distance of the neutral axis
of the spring from the skin.
Thus, if b is the width of a spring of circular or rectangular
b
section, c = , and hence
the deflection = -j~s.
The work done = -Q x deflection.= -~q6 =
2 2 q 2
/' si y
2 E^ '2 EC* ' 2E a !
k* being the square of the radius of gyration, A the sectional
area of the spring, and Fits volume.
In case of spring of rectangular section - r = - .
c i rcu i ar 4 = -
* 4
358
THEORY OF STRUCTURES.
Again, the spiral spring in Fig. 277 is wholly subjected to
a bending action by means of a twisting couple of moment
M = Qq in a plane perpendicular to the axis of the spring.
Any torsion in the spring itself is now due to the coils not
being perfectly flat.
FIG. 277.
Let R = radius of a coil before the couple is applied.
" R = " " " " after " " u "
6 being the angle of twist ; or
El ~ El ~ R~~ ^ ~ v
N being the number of coils before the couple is applied, and
^y " after " " " u
The distance through which g acts, i.e., the " deflection,"
and the work done
fV V
_
8 E
for spring of rectangular section,
" u " circular
6. Beams of Uniform Strength. A beam having the
same maximum unit stress (/") at every section is said to be a
beam of uniform strength.
BEAMS OF UNIFORM STRENGTH.
359
At any section of a beam AB ( /) denote the bending
moment by M, the depth of the beam by y, and its breadth
by b. Then
being the distance of the skin from the neutral axis, and A
the area of the section.
Evidently c and k are each proportional to y, and A to by.
or
.-. fbf cc J/,
nfbf = M t
n being a coefficient whose value depends upon the form of
section.
Four cases will be considered.
CASE a. Assume that the breadth b is constant, and let
nfb = j. Then
or
y =
Thus AB may be either the lower edge of the beam, the
ordinates of the upper edge being the different values of y, or
it may be a line of symmetry with respect to the profile, in
which case the ordinates are the different values of -.
EXAMPLE I. A cantilever AB loaded at the free end with a
weight W^.
At a distance x from A t
y = pM p w,x.
Theoretically, therefore, the
beam, in elevation, is the area
ACD, the curve CAD being a
B
FIG. 278.
THEORY OF STRUCTURES.
parabola with its vertex at A and having a parameter
The max. depth = 2CB CD = Vp WJ.
The form of this beam is very similar to that adopted for
cranks and for the cast-iron beams of engines. In the latter,
the material is usually concentrated in the flanges, a rib being
reserved along the neutral axis for purposes of connection.
Again, geometrical conditions of transmission require the
teeth of wheels to be of approximately uniform strength.
A cantilever of approximately uniform strength may be ob-
tained by taking the tangents C, DF as the upper and lower
edges of the beam instead of the curves CA, DA. The depth
of the beam at A is then EF \CD = j- VpW~L Although,
theoretically, the depth at A is nil, practically the beam must
have sufficient sectional area at A to bear the shear due to W l ,
and the depth VpWJ will be found ample for this purpose.
Note. The dotted lines show the beams of uniform
strength, when the lower edge is the horizontal line AB.
Ex. 2. A cantilever AB carrying a uniformly distributed
load W,.
At a distance x from A,
or
FIG. 279.
The beam, in elevation, is there-
fore the area A CD, AC, AD being two straight lines, and the
maximum depth being
The sectional area at A is nil, as both the bending moment
and shear at that point are zero.
BEAMS OF UNIFORM STRENGTH.
361
Note. The dotted lines show the cantilever of uniform
strength when AB is the lower edge.
Ex. 3. A cantilever AB carrying a weight W l at the free
end A and also a uniformly distributed load W^,
FIG. 280.
At the distance x from A,
This equation may be written in the form
( ^/V
r + w l ) y
= i.
Theoretically, therefore, the beam, in elevation, is the area
ACD, the curve CAD being an hyperbola having its centre at
I W \
H ^where AH = -yn-l), and semi-axes equal to
U and
v IML
V w
The maximum depth CD = A. p\WJ-\- W~\ = 2BC.
362 THEORY OF STRUCTURES.
A cantilever of approximately uniform strength may be ob-
tained by taking the tangents CE, DF as the upper and lower
edges of the beam instead of the curves CA, DA. It may be
W
easily shown that the depth of this beam at A is
and this will give sufficient sectional area at A to bear the
shear due to W r
Note. The dotted lines show the cantilever of uniform
strength, when the lower edge is the line AB.
Ex. 4. A beam AB supported at A and B, and carrying a
load W^ at the middle point O.
\ At a distance x from 0,
G /^^ 1 JX E
Theoretically, therefore, the
beam, in elevation, is the area
ACBD, the curves CAD, CBD being two equal parabolas,
having their vertices at A and B, respectively, and having
parameters equal to \p W^ .
The maximum depth = CD ~ 2CO = % \ f ~pWJ .
A beam of approximately uniform strength may be ob-
tained by taking the tangents CE, CG as the upper edges
instead of the curves CA, CB, and the tangents DF t DH as
the lower edges instead of the curves DA, DB.
The depth of the beam at A and B is now EF = GH
2 '
and this depth will give a sectional area at the ends of the beam
sufficient to bear the shears at these point, viz., - .
Note. The dotted lines show the beam of uniform strength
when the line AB is the lower edge.
Ex. 5. A beam AB supported at A and B, and carrying a
uniformly distributed load W^
BEAMS OF UNIFORM STRENGTH.
363
At a distance x from the
middle point O,
This equation may be writ- D
^, c FIG. 282.
ten in the form
*4 _/_
1 pwj
4 8
Theoretically, therefore, the beam, in elevation, is an ellipse
ACBD, having its centre at O and axes
AB = / and CD
IpWJ
=v~
The maximum depth is of course the axis CD = 2 CO.
Practically, the beam must have a certain depth at A and
B in order to bear the shears due to the reactions at these
W
points, viz., -. If the horizontal tangents at C and at D are
substituted for the curves, the volume of the new beam is to
the volume of the elliptic beam in the ratio of 4 to n.
Note. The dotted line shows the beam of uniform strength
when its lower edge is the line AB.
Ex. 6. A beam AB supported at A and B, and carrying a
load W } at the middle point O and also a uniformly distributed
load W,.
At a distance x from (9,
This equation may be written in the form
. i WJ^
-;#*.
pi
= I
w-
THEORY OF STRUCTURES.
Theoretically, therefore, the
beam, in elevation, is the area
ACBD, the curves CAD and
CBD being the arcs of ellipses
having the centres at the points
K and L, respectively, where
( w/ W l\ *
The maximum depth CD = 20C = J>* ] '- + -~ \ .
(2 o )
A beam of approximately uniform strength may be obtained
by taking as the upper edge the tangents to the curves at C,
and as the lower edge the tangents to the curves at D.
It may be easily shown that the depth at the ends^4 and B
W A- W
is now CD ' ' ' and this depth will make allowance for
2 W \ -\- KK a
W _L w
the shear - a at these points.
Note. The dotted lines show the beam of uniform strength
when the lower edge is the line AB.
CASE b. Assume that the ratio of the breadth (b) to the
depth (y) is constant, i.e., that transverse sections are similar.
y oc b a ty M,
or the ordinates of the profile of the beam both in plan and
elevation are proportional to the cube roots of the ordinates
of the curve of bending moments.
For concentrated loads the bounding curves are evidently
cubical parabolas.
CASE c. Assume that the depth y is constant. Then
b a M t
so that the ordinates of the beam in plan are directly propor-
tional to the ordinates of the curve of bending moments.
CASE d. Assume that the sectional area yb is constant.
Then
y a 'M,
FLANGED GIRDERS, ETC, 365
and the ordinates in elevation are directly proportional to the
ordinates of the curve of bending moments.
In this beam, the distribution of the material is very de-
fective, as the breadth b ( = - J -) must be infinite when^ = o,
i.e., at the points at which the bending moment is nil.
Timber beams of uniform strength are uncommon, as there
is no economy in their use, the portions removed to bring the
beam to the necessary form being of no practical value.
6. Flanged Girders, etc. Beams subjected to forces, of
which the lines of action are at right angles to the direction of
their length, are usually termed Girders; a Semi-girder, or
Cantilever, is a girder with one end fixed and the other free.
It has been shown that the stress in the different layers of
a beam increases with the distance from the neutral surface, so
that the most effective distribution of the material is made by
withdrawing it from the neighborhood of the neutral surface
and concentrating it in those parts which are liable to be more
severely strained. This consideration has led to the introduction
of Flanged Girders, i.e., girders consisting of one or two flanges
(or tables), united to one or two webs, and designated Single-
webbed or Double-webbed ( Tubular] accordingly.
7
(
FIG. 284. FIG. 285.
T
j
IT T
i JL
FIG. 286. FIG. 287. FIG. 288. FIG. 289. FIG. 290.
The web may be open like lattice-work (Fig. 284), or closed
and continuous (Fig. 285).
The principal sections adopted for flanged girders are :
The Tee (Figs. 286 and 287), the I or Double-tee (Figs. 288
and 289), the Tubular or Box (Fig. 290).
Classification of Flanged Girders. Generally speaking,
flanged girders may be divided into two classes, viz.:
366 THEORY OF STRUCTURES.
I. Girders witJi Horizontal Flanges. In these the flanges
can only convey horizontal stresses, and the shearing force,
which is vertical, must be wholly transmitted to the flanges
through the medium of the web.
If the web is open, or lattice-work, the flange stresses are
transmitted through the lattices.
If the web is continuous, the distribution of stress, arising
from the transmission of the shearing force, is indeterminate,
and may lie in certain curves ; but the stress at every point is
resolvable into vertical and horizontal components. Thus, the
portion of the web adjoining the flanges bears a part of the
horizontal stresses, and aids the flanges to an extent depend-
ent upon its thickness.
With a thin web this aid is so trifling in amount that it
may be disregarded without serious error.
II. Girders with one or both Flanges Curved. In these the
shearing stress is borne in part by the flanges, so that the web
has less duty to perform and requires a proportionately less
sectional area.
Equilibrium of Flanged Girders. AB is a girder in equi-
o librium under the action of external
forces, and has its upper flange com-
pressed and its lower flange ex-
tended. Suppose the girder to be
FlG - 8 9 X - divided into two segments by an
imaginary vertical plane MN. Consider the segment AMN.
It is kept in equilibrium by the external forces on the left of
MN, by the compressive flange stress at N ( = C), by the
tensile flange stress at M ( T), and by the vertical and
horizontal web stresses along MN. The horizontal web
stresses may be neglected if the web is thin, while the vertical
web stresses pass through M and N, and consequently have no
moments about these points.
Let d be the effective depth of the girder, i.e., the distance
between the points of application of the resultant flange stresses
in the plane MN.
Take moments about Jfand //successively. Then
Cd = the algebraic sum of the moments about M of
FLANGED GIRDERS, ETC. 367
the external forces upon AMN =. the bending moment at
MN=M.
So, Td = M\ .'. Cd=M= Td, and C = T.
Hence, the flange stresses at any vertical section of a girder
are equal in magnitude but opposite in kind. The flange
stress, whether compressive or tensile, will be denoted by F.
EXAMPLE. A flanged girder, of which the effective depth
is 10 ft., rests upon two supports 80 ft. apart, and carries a uni-
formly distributed load of 2500 Ibs. per lineal foot. Determine
the flange stress at 10 ft. from the end, and find the area of
the flange at this point, so that the unit stress in the metal
may not exceed 10,000 Ibs. per square inch.
The vertical reaction at each support
80 X 2500
= -- - = 100,000 Ibs.
/. F. 10 = M looooo X 10 2500 X 10 X 5 = 875,000 ft.-lbs.
,:F = 87,500 Ibs.
87500
The required area = --- = 8.75 sq. in.
i oooo
Cor. i. Fd=M=^I = ^L
R y
Cor. 2. At any vertical section of a girder,
let #,,#, be the sectional areas of the lower and upper flanges,
respectively ;
fuft, be the unit stresses in the lower and upper flanges,
respectively. Then
and the sectional areas are inversely proportional to the unit
stresses.
This assumes that F is uniformly distributed over the
areas a lt a 9 , so that the effective depth is the vertical distance
between centres of gravity of these areas. Thus, the flange
stresses at the centres of gravity are taken to be equal to the
3^8 THEORY OF STRUCTURES.
maximum stresses, and the resistance offered by the web to
bending is disregarded. The error due to the former may
become of importance, and it may be found advisable to make
the effective depth a geometric mean between the depths from
outside to outside and from inside to inside of the flanges.
Thus, if these latter depths are h^ , /^ , the effective depth
= Vk& (Art. 7).
EXAMPLE I. At a given vertical section of a flanged girder
the sectional area of the top flange is 10 sq. in., and the cor-
responding unit stress is 8000 Ibs. per square inch. Find the
sectional area of the lower flange, so that the unit stress in it
may not exceed 10,000 Ibs. per square inch.
a t . 10000 ='F= 10 . 8000 ; /. a l = 8 sq. in. and F = 80,000 Ibs.
Ex. 2. A wrought-iron girder weighing w Ibs. per lineal
ft., of / ft. span and d ft. depth, has horizontal flanges and
a uniform cross-section. The weight of the web is equal to the
weight of the flanges. Show that if the coefficient of strength
is 9000 Ibs. per square inch, the limiting value of / is 5400^ ft.,
k being the ratio of depth to span.
wr
Maximum flange stress --,- ;
Area of each flange = -- 5-1 in. ;
9000 . Sa
and
4wT
Total sectional area = ^ in.,
total volume of girder in feet = ' 5--;'
9000. 8^. 144
Hence,
4wr .480
wl = total W e I ght = - 95 ^- w: ,
and
d
I = 5400^- = 5400^.
FLANGED GIRDERS, ETC. 369
Note. The compressive strength of cast-iron is almost six
times as great as the tensile strength, and therefore the area
of the tension flange of a girder of this material should be
about six times that of the compression flange. Considering,
however, the difficulty there is in obtaining sound castings,
and also the necessity to provide sufficient lateral strength, it
by no means follows, nor is it even probable, that the ratio of
ultimate strengths is the best for the working strengths. Some
authorities are of the opinion that girders should be designed
with a view to their elastic strength, and that therefore the
working unit stresses in the case of wrought-iron and steel
should be equal, if this will insure sufficient lateral stability,
and in the ratio of 2 to I or 3 to I for cast-iron, which will give
sufficient lateral stability and make allowance for defective
castings.
The formula W = Cr is often employed to determine the
strength of a cast- or wrought-iron girder which rests upon two x
supports / inches apart, d being its depth in inches, and a the
net sectional area of the bottom flange in square inches. C is
a constant to be determined by experiment. Its average value
for cast-iron is 24 or 26, according as the girder is cast on its
side or with its bottom flange upwards. An average value of
C for wr -ought-iron is 80.
Cor. 3. A girder with horizontal flanges, of length / and
depth d, rests upon two supports, and is uniformly loaded with
a weight w per unit of length.
The bending moment at a vertical plane distant x from the
centre is
. will \ I \i// \ wl
Also, M Fd afd, a being the sectional area of either
flange at the plane under consideration, and f the correspond-
ing unit stress.
wrt AX*\
= (i--).
3/O THEORY OF STRUCTURES,
Let A be the flange sectional area at the centre. Then
Hence
an expression from which the flange sectional area at any point
of the girder may be obtained when the area at the centre is
known.
Cor. 4. F represents indifferently the sum of the horizontal
elastic forces either above or below the neutral axis, and is
therefore proportional to A, the sectional area of the girder ;
d is the distance between the centres of resultant stress and is
proportional to D, the depth of the girder.
/. Mat AD = CAD,
a form frequently adopted for solid rectangular or round gird-
ers, but also applicable to other forms.
Remark. The effective length of a girder may be taken to
be the distance from centre to centre of bearings.
The effective depth depends in part upon the character of
the web, but in the calculation of flange stresses the following
approximate rules are sufficiently accurate for practical pur-
poses :
If the web is continuous and very thin, the effective depth
is the full depth of the girder.
If the web is continuous and too thick to be neglected, the
effective depth is the distance between the inner surfaces of
the flanges.
If the web is open or lattice-work, the effective depth is the
vertical distance between the points of attachment of the
lattices.
If the flanges are cellular, the effective depth is the distance
between the centres of the upper and lower cells.
EXAMPLES OF MOMENTS OF INERTIA. $Jl
7. Examples of Moments of Inertia. (a) Double-tee Sec-
tion. First, suppose the web to be so thin that
it may be disregarded without sensible error. ""T"^
K2'
Let the neutral axis pass through G, the cen- -k
tre of gravity of the section. J 1
Let a^ , a^ be the sectional areas of the lower A
and upper flanges, respectively, and assume that FlG ' 2Q2 '
each flange is concentrated at its centre line.
Let h l , h z be the distances of these centre lines from G.
Let //, + h, = d.
Approximately, / = aji* -\- aji*.
Also, (a l + a f
h i = and h^ --- , defining the position of G.
Again,
192 i 96 144 i 72
a = - . = sq. in. and a. = -- . - = sq. in.
727 7 2 7 4
Also,
M = 250 ft.-tons = 3000 inch-tons.
.-. /, = 2-g- 9 ^- tons per sq. in. and f t = i||-f tons per sq. in.
Third. It is often convenient to calculate the moment of
inertia of a built beam symmetrical with respect to the neutral
axis, as follows :
Let Fig. 293 represent the section of such a beam, com-
posed of equal flanges connected with the web by four equal
angle-irons.
Let the width AF oi the flange = a.
u the side BC(= DE) of an angle-iron = b.
" thickness GH(= KL) of an angle-iron =f.
EF _
, be the outside depth of the section.
, " " depth between flanges.
FIG. 293. h^ "
Let ^ 3 be the depth between the faces MN, M'N'.
" h, " " " " " " ATZ, A' 7 /: 7 .
EXAMPLES OF MOMENTS OF INERTIA.
375
In this value of /, the weakening effect due to the rivet-
holes in the tension flange has been disregarded.* If it is to be
taken into account, let/ be the diameter of the rivets.
The centre of gravity of the section is now moved towards
the compression flange from its original position through a
distance
and the moment of inertia of the net section with respect to
the axis through the new C. of G. is
A' being the net area of the section, and / having the value
given above.
Fourth. The value of 7 for a double-tee section may be
more accurately determined as follows :
Let the area of the top flange be A 1 , and
its depth h r
Let the area of the bottom flange be A^ ,
and its depth // 8 .
Let the area of the web flange be A^ , and
its depth /z 2 .
Let A,-\- Af{- A 3 = A, and /z,+ //+ h z =k. FIG. 294 .
Let G be the centre of gravity of the section.
" G l " " " " top flange. m
" web.
" bottom of flange.
Let y v be the distance of G from the upper edge of the
section.
Let y t be the distance of G from the lower edge of the
section.
Take moments about G s . Then
THEORY OF STRUCTURES.
or
GG = *,(*, + 2*. + *J + *.(*, + *.
2,A
So,
rr -A^+
^ ~
and
Hence,
h k h
2A
2A
+ kj - Afa + ^ 3 ) - A^ - h
2
So,
= GG, + ~= etc.
Again, /, with respect to G,
G,ff+A,. G,ff + A S . G,G',
A&* + A,&,* + A,A*
/, being equal to ' ' - ' ' ~ .
Hence,
7 = 7 ' + ^' +
\A^ + 2A, + A.) + A,(A, + /,,)'.
EXAMPLES OF MOMENTS OF INERTIA. ^77
A 1 A^ 1 +^+2A l A t A,(/l 1 +/l^ 1 +2^+^
+ A,A,'(A t + ky - 2A l A^i,(A, + A,)(A, + A,)
+ A,A,'(A, + A,)' + A,A'(A, + 2A, + A,)'
+ 2A 1 A,A,(A 1 + 2*. + h,)(h, + A,)
1+ 2A 1 A,A 3 (/i 1 + 2/1, + h^(h, + h, + h, + h,)
A.A^ + k$A + A,A,(/i, + k$A 1
+ (4& + AfAJfa + 2h, + h$
W* + WA
- A.A^/^ + 2// a + h$A
Hence, finally,
12
Cor. I. If 7/ t and h t are small compared with ^,, put
Then
y =
2A
h ,
= -- , nearly,
THEORY OF STRUCTURES.
and
A A' + A. (h' - k -^^] + AM
12
,,^A, , A,A,+A,A,
tttT" ~4A
4A
Note. If At is also very small, as in the case of an open
web, then
A A A
i/ 2 = h' j- and / = h' 2 j- 2 , approximately.
si. si.
Cor. 2. Let y a , y b be the distances of G from the upper and
lower edges, respectively ; let f a , f b be the corresponding
maximum working unit stresses.
From the preceding corollary, y b = y^ = - - - -,
2 2/i
or
A. + A. + A, __ y a +y b
2A. + A, 2y b
3 l yb * 2% 1 f b
Hence,
EXAMPLES OF MOMENTS OF INERTIA.
379
and
,,, 1 ,
~ lt 12 +
4A
12A
A + A: + 4 (A, +
+ A,
6 /. +/.
r&,
Fifth. T-section.
Let the area of the flange be A, , and its
depth h,.
Let the area of the web be A 9 , and its
depth h^.
LetAi + A, = A, and h, + h^ = h.
Let G be the centre of gravity of the sec-
tion, G^ of the flange, and G 9 of the web.
FIG.
295-
Let y l be the distance of G from foot of the web.
Then
. l -
2 2
3^0 THEORY OF STRUCTURES.
and
_ ^i + * I A A - A A = *i A A - AA
2 2(A l +A,) 2 ' 2A
Again, ,.
^ ^ ^ii/ AJi . ~ ~ h. AJi
G lG = ^ + k,- yt =^ and G.G =*;- = ..
Hence /, with respect to a horizontal line through G,
= A^+A,. G>G'+A/ + A..G.G; ''
which reduces to
.
12 4A
Cor. I. If h^ is very small as compared with ^ a , put
= AJf + A.(?- -) = (A, + ) k>, nearly,
then
or
and
1 =
or
12 ' 4A r
Cor. 2. Let y a be the distance of the compressed, or upper,
side from the neutral axis.
12 4A
TO DESIGN A GIRDER OF UNIFORM STRENGTH. 381
Let y b be the distance of the stretched, or lower, side from
the neutral axis.
Let/ a be the crushing unit stress, f b the tensile unit stress.
h' 2.A I A.
From the preceding, y a - '- ^- - ; but h' = y a -\-y b ;
m >, ^ = > =
2 A,+A t 2y b 2f b
Hence, /becomes
2 ~7 i 7"' 7
fa+fb fa
Note. Although the preceding approximate methods are
often useful, they can only be regarded as tentative and should
always be checked by an accurate determination of the moment
of inertia and of the position of the neutral axis.
8. To design a Girder of Uniform Strength, of an
I-section with equal Flange Areas, to carry a Given
Load.
Let y be the depth of the girder at a distance x from its
middle point.
Let A be the sectional area of each flange at a distance x
from its middle point.
Let A' be the sectional area of the web at a distance x from
its middle point.
Let M be the bending moment at a distance x from its mid-
dle point.
Let 5 be the shearing force at a distance x from its middle
point. Then *
/being the safe unit stress in tension or compression.
382 THEORY OF STRUCTURES.
Web. Assume that the web transmits the whole of the
shearing force. This is not strictly correct if the flange is
curved, as the flange then bears a portion of the shearing force.
The error, however, is on the safe side.
Theoretically, the web should contain no more material than
is absolutely necessary.
Let f s be the safe unit stress in shear. Then
A'- S
~
and the sectional area is, therefore, independent of the depth.
A.' S
The thickness of the web = = -7,
but this is often too small to be of any practical use.
Experience indicates that the minimum thickness of a plate
which has to stand ordinary wear and tear is about J or T 5 ^- in.,
while if subjected to saline influence its thickness should be
f or |- in. Thus, the weight of the web rapidly increases with
the depth, and the greatest economy will be realized for a cer-
tain definite ratio of the depth to the span.
The thickness of the web in a cast-iron girder usually
varies from I to 2 in.
In the case of riveted girders with plate webs of medium
size, all practical requirements are effectively met by specifying
that the shearing stress is not to exceed one-half of the flange
tensile stress, and that stiffeners are to be introduced at inter-
vals not exceeding twice the depth of the girder when the
thickness of the web is less than one-eightieth of the depth.
Again, it is a common practical rule to stiffen the web of a
plate girder at intervals approximately equal to the depth of
the girder, whenever the shearing stress in pounds per square
Crra \
I -\ ' j, H being the ratio of the
depth of the web to its thickness.
Flanges. First. Assume that the flanges have the same
sectional area from end to end of girder.
TO DESIGN A GIRDER OF UNIFORM STRENGTH. 383
If the effect of the web is neglected,
M
and the depth of the beam at any point is proportional to the
ordinate of the bending-moment curve at the same point.
For example, let the load be uniformly distributed and of
intensity w ; and let / be the span. Then
FIG. 296.
and the beam in elevation is the parabola ACJ3, having its
vertex at C and a central depth CO = TTT"> The depths thus
determined are a little greater than the depths more correctly
given by the equation
M
y =
Second. Assume that the depth y of the girder is constant.
Then
A 1 M
and, neglecting the effect of the web, the area of the flange at
any point is proportional to the ordinate of the curve of bend-
ing moments at the same point.
Let the load be uniformly distributed and of intensity w ;
also, let the flange be of the same uniform width b throughout.
A 1 2 3 5 6 7 B
FIG. 297,
The flange, in elevation, is then the parabola ACB, having
its vertex at C and its central thickness CO ^7-7. Such
Zfyb
THEORY OF STRUCTURES.
beams are usually of wrought-iron or steel, and are built up by
means of plates. It is impracticable to cut these plates in such
a manner as to make the curved boundary of the flange a true
parabola (or any other curve). Hence, the flange is generally
constructed as follows :
Draw the curve of bending moments to any given scale.
By altering the scale, the ordinates of the same curve will
represent the flange thicknesses. Divide the span into seg-
ments of suitable lengths.
From A to I and B to 7 the thickness of the flange is
\a 7/; from I to 2 and 7 to 6 the thickness is 2b = 6e ; from
2 to 3 and 6 to 5 the thickness is $c = ^d\ and from 3 to 5 the
thickness is CO.
The more correct value of A( = -p-1 is somewhat
V fy 67
less than that now determined, but the error is on the safe
side.
Again, at any section,
E 2f
- , and hence R oc y, the depth.
Thus the curvature diminishes as the depth increases, so
that a girder with horizontal flanges is superior in point of
stiffness to one of the parabolic form. The amount of metal
in the web of the latter is much less than in that of the
former. If great flexibility is required, as in certain dyna-
mometers, the parabolic form is of course the best.
9. Deflection of Girders. The principles of economy and
strength require a girder to be designed in such a manner that
every part of it is proportioned to the greatest stress to which
it may be subjected. When such a girder is acted upon by
external forces, it is uniformly strained throughout, and in
bending, the neutral axis must necessarily assume the form of
an arc of a circle, provided the limit of elasticity is not ex-
ceeded. It might be supposed that the curve of deflection is
dependent upon the character of the web, and this is doubtless
the case, but experiments indicate that so long as the flange
DEFLECTION OF GIRDERS. 385
unit stresses are unaltered in amount, the influence of the web
may be disregarded without sensible error.
Let f be the unit stress in the beam at a distance y from
the neutral axis ; let d be the depth of the beam. Then
/ M E
- = - T = -=. = a constant,
y 1 K
assuming that the neutral axis is an arc of a circle of radius R.
But y oc d, and
7 = Ak* a Ad\
Hence f a y a d\ and if the depth is constant, /is also con-
stant and the beam is of uniform strength.
If the area A is constant,
EXAMPLE I. A cantilever bent under the action of exter-
nal forces, so that its neutral axis AB assumes
the form of an arc of a circle having its centre
* O.
Draw the verticals OA, BF, and the horizon-
The vertical deviation of B from the hori-
zontal, viz., BF y is the maximum deflection. ,'/'
Denote it by D. 6
Let radius of circle = R.
Since the deflection is very small, BE is approximately
equal to AB ( = /), the length of the cantilever.
/. r = BE" = AE(2R - AE) = 2RD - D* = 2 RD,
as D 1 may be disregarded without much error.
Also, the deflection at any point distant x from A is evi-
dently ^. If /is the stress in the material at a distance y
from the neutral axis,
f__E L _^DE 2DEy
~y ~ R ~ ' S ~' f ~~ ~lf~ '
386
THEORY OF STRUCTURES.
Ex. 2. A girder resting upon two supports at A and B
is bent under the action of external
forces so that its neutral axis ACB
assumes the form of an arc of a circle
having its centre at O.
Draw the vertical OC, meeting the
horizontal AB in F.
CF is the maximum deflection ;
denote it by D.
Since D is very small, its square
may be disregarded and the horizontal AB may be supposed
equal to the length ACB ( = /) of the girder, without much
error. Then
FIG. 299.
- = AF* = FC(zR - FC) = 2RD - D 1 = 2RD.
Hence,
/ &
Also, since = -5,
y R
f ~
WEy
x
The deflection at a distance x from F = D .
Ex. 3. A timber beam of 20 ft. span, is 12 in. deep and 6
in. wide : what uniformly distributed load ( W) will deflect the
beam I in., E being 1,200,000 Ibs. ?
By Ex. 2,
(2 4 0) 2
.'. R 7200 in.
_E T _ 1200000
8 ' 12 "J^~' 7200
12
I20OOOO 6 . I2 9
720O 12
.-. W= 4800 Ibs.
DEFLECTION OF GIRDERS. 387
Ex. 4. Let s l , f l , d lt and j a , / 2 , 4 , respectively, be the
length, unit stress, and distance from the neutral axis of the
stretched and compressed outside fibres in Examples (i) and
(2).
Let d l -f- d^ = d the total depth of the girder.
Hence, from similar figures,
d, s, R-d,
- and 7- -
/ 2P "" Z?
* A'- ' A
Also,
. /i+/ IL '""* _ ^ + ^
Ex. 5. A truss of span 120 ft. and 15 ft. deep is strained
so that the flange tensile and compressive unit stresses are
10,000 and 8000 Ibs., respectively. Find the deflection, and
difference of length between the extreme fibres.
30000000 i 20
.'. s 1 J 3 = .864 in., and R = 25,000 ft.
Hence also D = = -864 in.
10. Camber. Owing to the play at the joints, a bridge-
truss, when first erected, will deflect to a much greater extent
388 THEORY OF STRUCTURES.
than is indicated by theory, and the material of the truss will
receive a permanent set, which, however, will not prove detri-
mental to the stability of the structure, unless it is increased
by subsequent loads.
If the chords were made straight, they would curve down-
wards, and, although it does not necessarily follow that the
strength of the truss would be sensibly impaired, the appear-
ance would not be pleasing.
In practice it is usual to specify that the truss is to have
such a camber, or upward convexity, that under ordinary loads
the grade line will be true and straight.
The camber may be given to the truss by lengthening the
upper or shortening the lower chord, and the difference of
length should be equally divided amongst all the panels.
The lengths of the web members in a cambered truss are
not the same as if the chords were horizontal, and must be
carefully calculated, otherwise the several parts will not fit
accurately together.
To find an approximate value for the camber, etc. :
Let d be the depth of the truss.
Let s l , ^ 2 be the lengths of the upper and lower chords,
respectively.
Let /! , / 2 be the unit stresses in the upper and lower
chords, respectively.
Let d l , d^ be the distances of the neutral axis from the
upper and lower chords, respectively.
Let R be the radius of curvature of the neutral axis.
Let / be the span of the truss.
d, *. //i d i l ~ ** / 3
- = j- = - and - = = - , approximately,
the chords being assumed to be circular arcs.
Hence, the excess in length, of the upper over the lower
chord,
STIFFNESS. 389
Let x l , x^ be the cambers of the upper and lower chords,
respectively. R-\- d l and R d^ are the radii of the upper
and lower chords, respectively.
By similar figures, the horizontal distance between the ends
r> 1
of the upper chord = -= V, and the horizontal distance be-
tween the ends of the lower chord = - 5~~V.
Hence,
R
and
( j
= x, . 2(R + O, approximately,
^) approximately.
i i * \ A I i
8^* R ' oR\ Ri
r iy
Hence, approximately, the camber = .
Note. The deflection of a well-designed and well-built
truss is often much less than, and should never exceed, I inch
per 100 ft. of span under the maximum load.
II. Stiffness. If D is the maximum deflection of a girder
W D
of span / under a load W, then -, or more usually , is a
measure of the stiffness of the girder.
In practice, the deflection of an iron or a steel girder, under
the working load, should lie between and ^ , i.e., it is
limited to i or 2 in. per 100 ft. of span, and rarely exceeds
, or 1.2 in. per 100 ft. of span.
1000
A timber beam should not deflect more than -^-, or i in.
360
per 30 ft. of span.
390 THEORY OF STRUCTURES.
Let M l be the bending moment at the most deflected
point. Then
Also,
f being a numerical coefficient (in Art. 9, Ex. \,p = i ; in Ex.
2,/ = i).
Thus
I
gives the bending moment M^ to which the girder of a speci-
fied stiffness y may be subjected.
Again, if the material is to bear a certain specified unit
stress y, the maximum bending moment M^ to which the
girder may be subjected is given by the equation
, = = -,
c qd
*
q being a numerical coefficient less than unity, depending upon
the form of the section.
Cczteris paribus, the ratio of depth to span may be fixed
by making the stiffness and strength of equal importance. Then
M M and therefore
*-* I ~ ] J r
~pl\l)~~ qd '
or
d~\l
DISTRIBUTION OF SHEARING STRESS.
391
In practice the proper stiffness of a girder is sometimes
secured by requiring the central depth to lie between and
, its value depending upon the material of which the girder
is composed, its sectional form, and the work to be done.
EXAMPLE. A cast-iron beam of rectangular section and of
20 ft. span carries a uniformly distributed load of 20 tons ; the
coefficient of working strength is 2 tons per sq. in. ; the stiff-
ness is .001 ; E is 8000 tons. Find the dimensions of the beam
viz., b the breadth and d the depth.
20.20 _f r _ bd* _bd*
jyj. !^ .12 j. ^^ 2 ., ^^ !
8 c 63
Also,
.-. bd* = 1800.
20 . 20 EIiD\ 8 . 8000 . bd*
--.12 = M= 7 (-.-) = .(.ooi);
Pl\l I 12 . 20. 12 V
.-. bd* = 27000.
8
Hence,
2700O
-f=
1800
= 15 in. and b = 8 in.
12. Distribution of Shearing Stress. Let Figs. 300 and
301 represent a slice of a beam bounded by two consecutive sec-
A
PX~ j > Q
-^ 4-2 *
0/~
*d i
B- B '
FIG. 300.
tions AB, A'B' , transverse to the horizontal neutral axis O O f .
Let the abscissae of these sections with respect to an origin
39 2 THEORY OF STRUCTURES.
in the neutral axis be x and x -f- dx, so that the thickness of
the slice is dx.
In the limit, since dx is indefinitely small, corresponding
linear dimensions in the two sections are the same.
Let / be the moment of inertia of the section AB (or A ' B'
in the limit) with respect to the neutral axis.
Let c be the distance of A (or A' in the limit) from the
neutral axis.
Let/j , f t be the unit stresses at A and A , respectively.
Consider the portion A CCA' of the slice, CC ' being
parallel to and at a distance Y from the neutral axis. Since
it is in equilibrium, the algebraic sum of the horizontal forces
acting upon it must be nil. These forces are:
The total horizontal force upon ACC,
11 A'C'C', and
shear along the surface CC'.
The horizontal force upon an element PQ of thickness dy
and at a olistance y from the neutral axis
2 being the width PQ. Thus the total horizontal force upon
"ACC
= (* yzdy = ^Ay,
c j Y c
A being the area of ACC, and y the distance of the centre of
gravity of this area from O0 r .
Similarly, "the total horizontal force upon A'C'C'
/ J J ~ ' ., J'
DISTRIBUTION OF SHEARING STRESS.
393
// f\ -
Hence, f -\Ay = difference of the horizontal forces
upon A CC and A'C'C',
=. horizontal shear along CC' ,
= qwdx ;
q being the intensity of this shear, and w the width of the
section at-CC.
Let M and M dM be the bending moments at the two
consecutive sections AB, A ' B '. Then
M=--I and M-dM=--I,
c c
and therefore
Hence,
dM
dM -
= (---)'
\C c I
J
= qwdx,
or
dM Ay S A -
qw = ~dx T = 7 A *>
since = shearing force at the section AB = 5.
dx
EXAMPLE I. Solid rectangular section of width b.
12
or
FIG. 302.
and the intensity of the shear at any point of AB may be rep-
resented by the horizontal distance of the point from the
parabola A VB, having its vertex at F, where OV ' - .
be
394 THEORY OF STRUCTURES.
The maximum intensity of shear is at O and its value is
sin a),
is the unit stress in the material of the beam at a distance y
from the neutral axis due to the bending action at MN of the
external forces on the segment AMN.
Hence, also, the total um\. stress in the material in the plane
MN at a distance y from the neutral axis is
2(P cos a) 2(P cos a)
-~- 2 f y =~ -- 2
the signs depending upon the kind of stress.
It will be observed that this formula is composed of two
intensities, the one due to a direct pull or thrust, the other due
to a bending action. The latter is proportional to the distance
of the unit area under consideration from the neutral axis. It
is sometimes assumed that the same law of variation of stress
holds true over the real or imaginary joints of masonry and
brickwork structures, e.g., in piers, chimney-stacks, walls,
arches, etc. In such cases the loci of the centres of pressure
correspond to the neutral axis of a beam, and the maximum
THEORY OF STRUCTURES.
and minimum values of the intensity occur at the edges of the
joint.
EXAMPLE I. A horizontal beam of length /, depth d, and
sectional area A is supported at the ends, and carries a weight
W at its middle point. It is also subjected to the action of a
force H acting in the direction of its length.
First. Let the line of action of // coincide with the axis of
the beam.
The intensity of the stress in the skin at the centre
= 7".
But c oc d, and 7 = Ak* oc Ad\
I . , Ad
.-. - a Ad ,
c n
n being a coefficient depending upon the form of the section.
If the section is a circle, n = 8 ; if a rectangle, n = 6.
Hence,
HI , n W A
the skin stress = db ~2 I * 4~ ~ ~ZF 3)
si \ 4 " d/
Wl
since M = .
4
Wl
If the load W 7 is uniformly distributed, M = -5-.
o
Thus, a very small load on the beam may considerably in-
crease the intensity of stress, and this intensity will be still
further increased by the deflection of the beam under its load,
so that, in order to prevent excessive straining, it is often
necessary to introduce more supports than are actually required
to make the beam sufficiently stiff.
Second. If the line of action of H is at a distance h from
the neutral axis, an additional bending moment Hh will be in-
troduced.
BEAM ACTED UPON BY OBLIQUE FORCES.
399
Ex. 2. The inclined beam OA, carrying a uniformly dis-
tributed load of w per unit of length, is supported at A and rests
against a smooth vertical surface at O.
The resultant weight wl is vertical
and acts through the centre C of OA ;
the reaction R^ at O is horizontal.
Let the directions of wl and R^ meet
in B. For equilibrium, the reaction ^ 3
at A must also pass through B.
Let the vertical through C meet the
horizontal through A in D.
The triangle ABD is a triangle of forces for the three forces
which meet at B.
FlG< 3 5 '
R
- /
wl
AD AD
n T\ T*
BD 2. DC
Of,
being the angle OAD. Hence
wl
R, cot a.
Consider a section MN, perpendicular to the beam, at a dis-
tance x from O.
The only forces on the left of MN are R 1 and the weight
upon OM. This last is wx, and its resultant acts at the centre
x
of OM, i.e., at a distance - from MN.
The component of R t along the beam
wl cos 2 a
= R^ cos a = -- : - .
2 sin a
The component of R^ perpendicular to the beam
D wl
= R l sm a = cos a.
40O THEORY OF STRUCTURES.
The component of wx along the beam = wx sin a.
The component of wx perpendicular to the beam = wxcos a.
Hence,
wl cos a a
the total compression at NM = : -4- wx sin a = C- ;
2 sin or
the shearing force at MN = cos a wx cos a = S x
the bending moment at MN= x cos a wx cos a = M x \
and
These expressions may be interpreted graphically as already
described, C x , S x being represented by the ordinates of straight
lines, and M x , f y by the ordinates of parabolas.
f y , for example, consists of two parts which may be treated
independently. Draw OE and AF
perpendicular to OA,a.nd respectively
equal or proportional to
wl cos 3 a wl cos 2 a wl
-7 . and - : + r sin a.
2A sin a 2A sin a ' A
Join EF. The unit stress at any
FlG - 36- point of the beam due to direct com-
pression is represented by the ordinate (drawn parallel to OE
or AF) from that point to EF.
Upon the line GG' drawn through the middle point B per-
pendicular to OA, take .BG = BG ', equal or proportional to
y wl*
y-r- cos a. According as the stress due to the bending action
1 o
at any point of the beam is compressive or tensile, it is repre-
sented by the ordinate (drawn parallel to OE and AF) from
that point to the parabola OGA or OG'A ; G and ', respec-
tively, being the vertices, and GG' a common axis.
SIMILAR GIRDERS PRINCIPAL PROPERTIES. 4OI
By superposing these results, the parabolas EHF, EH'F are
obtained, the ordinates of which are respectively proportional
to the values of f y for the compressed and stretched parts of
the beam, i.e., for the parts above and below the neutral!
surface.
14. Similar Girders. Two girders are said to be similar
when the linear dimensions of the one bear the same constant
proportion to the corresponding linear dimensions of the other.
Thus, if ft, ft', d, $', A, X' are corresponding breadths, depths,
and lengths of two similar girders,
ft 6 X
~8' = e of Aj /, Q, . . . by means of certain constant multipliers.
404 THEORY OF STRUCTURES.
Cor. I. If the two girders are similar and of the same
material,
p = g = r = ju, E = J5', and p=i.
Hence,
from (y), Q = <2/A and the weights vary directly as the
cubes of the linear dimensions ;
" (e), M' = MJJ?, and the bending moments vary directly
as the fourth powers of the linear di-
mensions;
/' s/
" (C) and (;?), -j P = , and the flange unit stresses vary
/ s
directly as the web unit stresses;
" ( z )> "A = ^ 8 > an d t ^ ie deflections vary directly as the
squares of the linear dimensions ;
W r
" W ^M7 == ^ ' an< ^ ^ e breaking weights vary directly as
the squares of the linear dimensions.
Cor. 2. Let the girders be of the same material, of equal
length, of equal rectangular sectional areas, and equally loaded.
Let b, b lt and d, d,, respectively, be the breadths and
depths of the girders. Then
b l = qb and d l = rd.
Hence,
b^d^ = qrbd.
But b,d } = bd\ / qr i. Also, p i.
ALLOWANCE FOR THE WEIGHT OF A BEAM. 405
f i
Thus from C, y = - = q ;
R' A
0andi, -- = r' = ;
,- --- r ~~
W ~ ~ f '
b
If d l = b, then b l ^/, and r = -,.
_
Hence, -- - ?> and ^ == = =
17. To make Allowance for the Weight of a Beam. A
beam is sometimes of such length that its weight becomes of
importance as compared with the load it has to carry, and
must be taken into account in determining the dimensions of
the beam.
The necessary provision may be made by increasing the
width of the beam designed to carry the external load alone,
the width being a dimension of the first order in the expression
for the elastic moment.
Assume that the weight of the beam and the external load
are reduced to equivalent uniformly distributed loads.
Let W t be the external load.
" b e " " breadth of a beam designed to support this
load only. }
" B e " " weight of the beam.
" W " " total load, the weight of the beam being
taken into account.
" b " " corresponding breadth of the beam.
g weight " " "
Then W-B=W e ,
b B W W- B W e
and
b e ~ B~ w, " W.-B. ~ w e - B;
4O6 THEORY OF STRUCTURES.
wj>. wjs. w;
Hence, 6,-.-^-, B -- -^^ and W -- W^J;
EXAMPLE. Apply the preceding results to a cast-iron
girder of rectangular section resting upon two supports 30 ft.
apart. The girder is 12 in. deep and carries a uniformly dis-
tributed load of 30,000 Ibs.
Take 4 as a factor of safety ; b e is given by
I2OOOO bd*
where C 30,000 Ibs., d = 12 in., and / = 360 in. ;
/. b e 5 in.
Hence,
B e = -^7 -^ X 30 X 450 = 5625 Ibs. ;
144
W e B e 30000 5625 = 24,375 Ibs. ;
30000,5
24375
lbs> ;
W= ^ ^
EXAMPLES.
1. An iron bar is bent into the arc of a circle of 500 ft. diameter;
the coefficient of elasticity is 30,000,000 Ibs. Find the moment of resist-
ance of a section of the bar and the maximum intensity of stress in the
metal, (a) when the bar is round and i in. in diameter, (<) when the bar
is square having a side of r inch.
If the metal is not to be strained above 10,000 Ibs. per sq. in., find (c}
the diameter of the smallest circle into which the bar can be bent.
Ans.(a) ^Tt in. -Ibs.; 5000 Ibs. (b) 833^ in. -Ibs.; 5000 Ibs. (c) 250 ft.
2. A piece of timber 10 ft. long, 12 in. deep, 8 in. wide, and having a
working strength of 1000 Ibs. per sq. in., carries a load, including its
own weight, of w Ibs. per lineal foot. Find the value of w, (a) when the
timber acts as a cantilever ; (b) when it acts as a beam supported at the
ends. Find (c) stress in material 3 in. from neutral axis at fixed end of
cantilever and at middle of beam.
Am, (a) 320 Ibs.; (b) 1280 Ibs.; (c] 500 Ibs. per sq. in.
3. Is it safe for a man weighing 160 Ibs. to stand at the centre of a
spruce plank loft, long, 2 in. wide, and 2 in. thick, supported by vertical
ropes at the ends ? The safe working strength of the timber is 1200 Ibs.
per sq. in.
Ans. No ; the maximum safe weight at the centre is 53^ Ibs.
4. Compare the uniformly distributed loads which can be borne by
two beams of rectangular section, the several linear dimensions of the
one being n times the corresponding dimensions of the other. Also
compare the moments of resistance of corresponding sections.
Ans. n* ; n*.
5. A cast-iron beam of rectangular section, 12 in. deep, 6 in. wide, and
16 ft. long, carries, in addition to its own weight, a single load P\ the
coefficient of working strength is 2000 Ibs. per sq. in. Find the value of
P when it is placed (a) at the middle point; (b) at 2$- ft. from one end.
Ans. (a) 8475 Ibs. ; (b) 11,300 Ibs.
6. A round and a square beam of equal length and equally loaded are
to be of equal strength. Find the ratio of the diameter to the side of the
square. Ans. \/~$6 :
407
408 THEORY OF STRUCTURES.
7. Compare the relative strengths of two beams of the same length and
material (a) when the sections are similar and have areas in the ratio
of i to 4; (b} when one section is a circle and the other a square, aside of
the latter being equal to the diameter of the former.
Ans. (a) i to 8 ; (b) 56 to 33.
8. Compare the strength of a cylindrical beam with the strength of
the strongest (a) rectangular and (b) square beam that can be cut from it.
Ans. (a) 112 : 99 |/^~; (b) 33 : 14 fa
9. A boiler-plate tube 36 ft. long, 30 in. inside diameter, weighs 4200
IDS. and rests upon supports 33 ft. apart. Find the maximum intensity
of stress in the metal. What additional weight may be suspended from
the centre, assuming that the stress is nowhere to exceed 8000 Ibs. per
sq. in. ?
Ans. 741 i- Ibs. per sq. in.; 1 8,854 -|f| Ibs.
10. Compare the relative strengths of two rectangular beams of equal
length, the breadth (ft) and depth (d) of one being the depth (b) and
breadth (d) of the other. Ans. d* : P.
11. A yellow-pine beam 14 in. wide, 15 in. deep, and resting upon sup-
ports 129 in. apart, was just able to bear a weight of 34 tons at the cen-
tre. What weight will a beam of the same material, of 45 in. span and
5 in. square, bear ? Ans. \\\ tons.
12. A cast-iron rectangular girder rests upon supports 12 ft. apart
and carries a weight of 2000 Ibs. at the centre. If the breadth is one-
half the depth, find the sectional area of the girder so that the intensity
of stress may nowhere exceed 4000 Ibs. per sq. in.
Ans. 1 8 sq. in.; if weight of girder is to be taken into account,
the depth d is given by d s i.oi2$d* 216 = o.
13. Find the depth of a wrought-iron girder 6 in. wide which might be
substituted for the cast-iron girder in the preceding question, the coeffi-
cient of strength for the wrought-iron being 8000 Ibs. per sq. in.
Ans. 4.762 in.; if weight of girder is to be included, the depth d
is given by d* .54 the specific weight of the material.
Ans. ^ - > n being a coefficient depending upon the form of
the section.
56. If the beam in the preceding question is to be supported at its two
ends, what will its limiting length be ? Ans. , .
57. Find the limiting length,of a cedar cantilever of rectangular sec-
tion, k being 40, w = 36 Ibs. per cu. ft., and/ = 1800 Ibs. per sq. in.
Ans. 60 ft.
58. A steel cantilever 2 in. square has an elastic strength of 15 tons
per sq. in. What must its limiting length be so that there may be no
set f Ans. 23.4 ft.
59. Find the limiting length of a wrought-iron beam of circular sec-
tion, k being 64 and the elastic strength 8 tons per sq. in. What will this
length be if a beam of I-section, having equal flange areas and a web
area equal to the joint area of the flanges, is substituted for the circular
section ? Ans. 84 ft. ; 224 ft.
60. A rectangular cast-iron beam having its length, depth, and
breadth in the ratio of 60 to 4 to i, rests upon supports at the two ends.
Find the dimensions of the beam so that the intensity of stress under its
own weight may nowhere exceed 4500 Ibs. per sq. in.
Ans. I = 128 ft. ; d = 8-jV ft. ; b = 2 T ^ ft.
61. A beam supported at the ends can just bear its own weight W\.o-
W
gether with a single weight at the centre. What load may be placed
at the centre of a beam whose transverse section is similar but m* as
great, its length being n times as great ? If the beam could support only
its own weight, what would be the relation between m and n ?
(m* m^ n
Ans. W ; m = -.
\n 2 ) 2
62. The flanges of a rolled joist are each 4 in. wide by \ in. thick ;
the web is 8 in. deep by inch thick. Find the position of the neutral
axis, the maximum intensities of stress per square inch being 10,000
Ibs. in tension and 8000 Ibs. in compression. Ans. k\. = 3$ ; hi ~ 4$.
EXAMPLES. 415
63. A continuous lattice-girder is supported at four points, each of the
side spans being 140 ft. n in. in length, 22 ft. 3 in. in depth, and weigh-
ing .68 ton per lineal foot. On one occasion an excessive load lifted the
end of one of the side spans off the abutment. Find the consequent in-
tensity of stress in the bottom flange at the pier, where its sectional
area is 127 sq. in. Ans. 2.3893 tons per sq. in.
64. A railway girder is 101.2 ft. long, 22.25 ft- deep, and weighs 3764
Ibs. per lineal foot. Find the maximum shearing force and flange stresses
at 25 ft. from one end when a live load of 2500 Ibs. per lineal foot crosses
the girder.
65. A floor with superimposed load weighs 160 Ibs. per sq. ft. and is
carried by tubular girders 17 ft. c. to c. and 42 ft. between bearings.
Find the depth of the girders (neglecting effect of web), the safe inch-
stress in the metal being 9000 Ibs., and the sectional area of the tension
flange at the centre 32 sq. in. Ans. 24.99 in -
66. Design a timber cantilever of approximately uniform strength from
the following data: length = 12 ft.; square section; load at free end
= i ton ; coefficient of working strength = | ton per sq. in. What
must be the dimensions at the fixed and free ends so that the cantilever
might carry an additional uniformly distributed load of 2 tons ?
Ans. Side = 15.1 in. at fixed end and = 10 in. at free end;
side = 19.1 in. at fixed end and = (19.1 in.) at free end.
67. Show that the curved profile of a cantilever of uniform strength
designed to carry a load W at the free end, is theoretically a cubical
parabola. Also show that by taking the tangents to the profile at the
fixed end as the boundaries of the cantilever, a cantilever of approxi-
mately uniform strength is obtained having a depth at the free end
equal to two-thirds of the depth at the fixed 6nd.
68. Design a wheel-spoke 33 in. in length to be of approximately uni-
form strength, the intensity of stress being 4000 Ibs. per sq. in. ; the load
at the end of the spoke is a force of 1000 Ibs. applied tangentially to the
wheel's periphery, and the section of the spoke is to be (a) circular, (b)
elliptical, the ratio of the depth to the breadth being 2|.
Ans. (a) Depth at hub = 6.982 in., at periphery = 4.634 in.
(b) " = 9.435 " " = 9-35 in.
Breadth " " = 3.7/4 " " = 3-74 in.
69. A beam of 17 ft. span is loaded with 7, 7, 11, and n tons at
points 1,6, ii, and 15 ft. from one end. Determine the depths at these
points, the beam being of uniform breadth and of approximately uni-
form strength ; the coefficient of working strength = 2 tons per sq. in.;
the depth of the section of maximum resistance to bending = 16 in.
16065 277 x i6 s 1087 x i6 a
Ans ' * =
i 7 8 5
670 x i6 2
and d + (S r - S,)r
Also, when x = r -{- s, M = S s s -\- S r r.
Between CandA, S is constant = S t suppose, =
and hence
c" being the constant of integration.
But M = SsS -\- S r r when x = r + s.
.-. c" = Sj + S r r - Sjr + s),
and
M= Ss + 5> + S r r - S t (r -f s).
Hence, at A t o = S t t + 5> + 5>.
6^r. 4. The equation - - = 5 indicates that the shearing
force at a vertical section of a girder is the increment of the
bending moment at that section per unit of length, and is an
important relation in calculating the number of rivets required
for flange and web connections.
2. On the Interpretation of the General Equations.
The bending moment M at any transverse section of a girder
El
may be obtained from the equation M= ,R being the
.A.
DEFLECTION OF BEAMS. 433
radius of curvature of the neutral axis at the section under
consideration. \
Let OA, in Figs. 313 and 314, represent a portion of the
neutral axis of a bent girder.
--^ Jy
FIG. 313. FIG. 314.
Take O as the origin, the horizontal line OX as the axis of
x, and the line O Y drawn vertically downwards as the axis of y.
Let x, y be the co-ordinates of any point P in the neutral
axis.
If R is the radius of curvature at P, then
dx* a d&
"
the sign being -f- or according as the girder is bent as in
Fig. 313 or as in Fig. 314, and being the angle between the
tangent at P and OX.
dy
Now, jk is the tangent of the angle which the tangent line
at P to the neutral axis makes with OX, and the angle is always
very small. Thus, -j- is also very small, and squares and
dy
higher powers of -7 may be disregarded without serious error.
Hence,
I ~ ,
= -jTT approximately,
and the bending-moment equation becomes
M = EI-jj.
434 THEORY OF STRUCTURES.
The integration of this equation introduces two arbitrary
constants, of which the values are to be determined from given
conditions. At the point or points of support, for example,
the neutral axis may be horizontal or may slope at a given
angle.
Let 6 be the slope at P. Since 6 is generally very small,
6 tan 6, approximately,
and hence
or
dB M
From this last equation
A" 0=
and the change of slope between any given limits is represented
by the corresponding area of the bending-moment curve.
dy r
Also, since = 0, y = J 6dx,
d>&
and the deflection is measured by the area of a curve repre-
senting the slope at each point.
Again, by Art. i,
Comparing eqs. (A) and (B), it will be observed that
M
y, Oj and -^-,, i.e., the deflection, slope, and bending moment,
are connected with one another in precisely the same manner
NEUTRAL AXIS OF A LOADED BEAM. 435
as M, S, and p, i.e., the bending moment, shearing force, and
load. Thus, the mutual relations between curves drawn to
represent the deflection, slope, and bending moment must be the
same, mutatis mutandis, as those between the curves of bending
moment, shearing force, and load.
For example, divide the effective bending-moment area into
a number of elementary areas by drawing vertical lines at con-
venient distances apart, and suppose these elementary areas to
represent weights. Two reciprocal figures connecting y, 0, and
M may now be drawn exactly as described in Chap. I, and it
at once follows that
(a) Any two sides of the funicular polygon, or, in the limit
(when the widths of the elementary areas are indefinitely
diminished), any two tangents to the funicular or deflection
curve, meet in a point which is vertically below the centre of
gravity of the corresponding effective moment area.
(b) The segments iH, nH into which the line of weights is
divided by drawing OH parallel to the closing line CD, give the
slopes (= ~2Mdx) at the supports.
N.B. In the case of a semi-girder, the last side of a
polygon is the closing line, and in gives the total change of
slope.
(c) If the polar distance is made equal to El, the intercept
between the closing line and the funicular or deflection curve
measures the deflection.
3. Examples of the Form assumed by the Neutral
Axis of a Loaded Beam.
EXAMPLE I. A semi-girder fixed at one end O so that the
neutral axis at that point is horizontal carries a weight W at
the other end A. At any point (x, y) of the neutral axis
x). . . (A)
Integrating,
l being a constant of integration. But FlG - 3*s-
the girder is fixed at O, so that the inclination of the neutral
THEORY OF STRUCTURES.
axis to the horizon at this point is zero, and thus, when x = o,
dy
-j- is o, and therefore c t = o.
Hence,
% : ^4-D ..... . . (B
Integrating,
2 being a constant of integration. But y = o when x = o,
and therefore a = o.
Hence,
dy
Equation (B) gives the value of , i.e., the slope, at any
point of which the abscissa is x.
Equation (C) defines the curve assumed by the neutral axis,
and gives the value of y, i.e., the deflection, corresponding to
any abscissa x.
Let <*! be the slope, and d l the deflection at A.
From (B),
w
and from (C),
i w r
d , ^~ T-
NEUTRAL AXIS OF A LOADED BEAM.
437
Ex. 2. A semi-girder fixed at one end O carries a uniformly
distributed load of intensity w.
At any point P (x, y) of the
neutral axis,
. . (A) Y
FIG. 316.
Integrating,
='* -<*+$+>..
l being a constant of integration.
dy
But -j- = o when x = o, and therefore ^ = O. Hence,
dx
(B)
Integrating,
a being a constant of integration.
But j = o when x = o, and therefore 2 = O. Hence,
2 2
(Q
Let a be the slope and d^ the deflection at A. Hence, from
(B),
and from (C),
i wr
43$ THEORY OF STRUCTURES.
Ex. 3. A semi-girder fixed at one end carries a uniformly
distributed load of intensity w, and also a single weight Wat the
free end. This is merely a combination of Examples I and 2,
and the resulting equations are :
x) + (l-xf (A)
Also, if A is the slope and D the deflection at the free end,
from (B),
i iwr wi\
tan A = -g^ 1- -^- ] = tan or, -j- tan <* a ;
and from (C),
. The slope (a) and deflection ( 3I7 '
P (x, y) between O and C,
EI -=rjr ; and from (C), ^ = -5 -py .
ID &i 48 /j/
a. B A EX * 5 ' The irder OA rests
-^Jj, | ^' x upon supports at <9, ^4, and carries
P~~~C" a uniformly distributed load of in-
tensity w.
At any point P (x, y) of the
FlG - 318. neutral axis,
Integrating,
P T dy __wl a wx*
~fcc''^~4~ X ~~6~ T~ C I>
\ being a constant of integration.
But -j- = o when x = -, and therefore c. .
dx 24
Hence,
dx " 4 6 24 '
Integrating,
wl , wx* wl*
Ely = x" x 4- c mt
12 24 24
:, being a constant of integration.
But 7 = when x = o, and therefore c^ = o.
Hence
7? = ^r 3 - ^- x. . (C)
12 24 24 V '
NEUTRAL AXIS OF A LOADED BEAM. 441
Let or a be the slope at O, and d^ the deflection at the centre.
Then,
from (B), tan a, = ^ ; and from (C), < = ^ ~.
Ex. 6. A girder rests upon two supports and carries a uni-
formly distributed load of intensity w, together with a single
weight W at the centre. This is merely a combination of
Examples 4 and 5, and the resulting equations are :
dx ' ~ 4 16 4 6
and
(B)
12 ID 12 24 24
Also, if A is the slope at the origin, and D the central
deflection, we have, from (B),
i fwr wr\
tan A = j(-rf- + J = tan ,+ tan a ;
and from (C),
*= +- -<+<
The slope and deflection of an arbitrarily loaded
girder resting upon two supports may be determined in the
manner described in Art. 2.
Let C be the lowest point of the deflection curve. The
tangents at C and O will intersect in a point T which is ver-
tically below the centre of gravity of the bending-moment
area corresponding to OC.
Denote this area by Fand the horizontal distance of centre
44 2 THEORY OF STRUCTURES.
of gravity from OY by x. Let OL be the angle between OT
and CT produced. Then
F d
d being the maximum deflection.
In Ex. 6, e.g., the girder being symmetrically loaded,
i wi i 2 wr i wr 2 i wr 5 /
F= sr- and Fx = ^ 4- ~- EId
242^382 1632 24 82
Ex. 7. Suppose that the end O of the girder in Ex. 5
R R ^ is fixed. The fixture introduces
\j(^\\ a left-handed couple at ; let its
K---.,_ a l ~~~^k * moment be M,.
P Let the reactions at O and A
y be R t , R^ , respectively.
At any point P (x, y] of the
neutral axis,
But M, i.e., 7-r4, is zero when x = /.
-^ ...... (2)
Integrating eq. (i),
There is no constant of integration, as -,- = o when x = o.
dx
Integrating eq. (3),
- Ely = R?*^ - M?- . (4)
1 6 24 J 2
There is no constant of integration, as x and y vanish to-
gether.
NEUTRAL AXIS OF A LOADED BEAM.
But y also vanishes when x = /, so that
443
Hence, by eqs. (2) and (5),
and so tf, = //. . (6)
Thus, the b ending-moment, slope, and deflection equations
are, respectively,
(*>
=<"* --
. I. The bending moment is nil at points given by
5 w , wl*
^f= o = -golx x* -- g-,
F
i.e., when x = or /. Take OF= .
4 4
Since -^-7 = o, .F is a point of inflexion.
If the girder is cut through at this point,
and a hinge introduced sufficiently
strong to transmit the shear (= %wl), FIG. 320.
the stability of the girder will not be impaired.
Hence, the girder may be considered as made up of two
independent portions, viz. :
(a) A cantilever OF-oi length , carrying a uniformly dis-
tributed load of intensity w, together with a weight wl at F.
444 THEORY OF STRUCTURES.
The maximum bending moment on OF is at O, and is
3 f / , wl I wr
= 8 W/ 4 + 78 =: l~'
(b) A girder FA of length , carrying a uniformly distrib-
uted load of intensity w.
The maximum, bending moment on FA is at the middle
This result may also be obtained from eq. (7) by putting
-- o. Whence
dx
O = %wl wx, or x = f/,
.and therefore
The shearing force and bending moment at different points
of the girder may be represented graphically as follows :
The shearing force at any point of which the abscissa is x is
S = f wl wx.
Take OB and A C, respectively equal or proportional to wl
and fw/; join BC. The line BC cuts OA in /?, where OP = f/.
The shearing force at any point is represented by the ordinate
between that point and the line BC.
The bending moment at the point (x, y) is
Take OG, DE, and OF, respectively equal or proportional
wr g I
to -5, --^w/ 2 , and -. The bending moment at any point is
o I2o 4
represented by the ordinate between that point and the parab-
ola passing through G, F, and A, having its vertex at E and
its axis vertical.
NEUTRAL AXIS OF A LOADED BEAM. 445
dy
Cor. 2. The deflection is a maximum when = o, i.e., when
or at the point given by x = -g05 1/33).
Substituting this value of x in eq. (9), the corresponding
value of jj/ may be obtained.
Ex. 8. If both ends of the girder in eq. (7) are fixed, the
wl
reaction at each support is evidently , and the equation of
moments becomes
T^jd^y _ wl wx*
Integrating,
dy wl w
El-f- = x* ** M.x (2)
dx \ 6
dy
No constant of integration is required, as -r- = o when x = o.
-j- is also zero when x = I (also when x = ).
dx V 21
and hence
^' = 1? (3)
Integrating eq. (2),
12 24 24
There is no constant of integration, as x and y vanish together.
The central deflection I i.e., when # = )=* ~^~~cj'
44-6 THEORY OF STRUCTURES.
If the load, instead of being uniformly distributed, is a
weight ^concentrated at the centre, then, for one half oi the
girder,
Integrating,
dv W
/-/- = x* - M,x ...... (6)
dx 4
dy
There is no constant of integration, as -j- = o when x = o.
dy I
-j- is also evidently zero when x = , and hence
W I Wl
= _/>_^-, or M$~-g-. V . (7)
Integrating eq. (6),
W x* W Wl
-Efy=x*-M l - = x*-~x\ ... (8)
12 '2 12 l6
There is no constant of integration, as x and y vanish together.
I Wl
The central deflection = --- p-=-.
192 El
4. Supports not in same Horizontal Plane. In the
preceding examples it has been assumed that the ends of the
girder are in the same horizontal plane. Suppose that one end,
e.g., A, falls below O by an amount jj>, , y l being small as com-
pared with /.
The abscissae of points in the neutral axis are not sensibly
changed, but the conditions of integration are altered. Con-
sider Ex. 4.
Between O and C,
W
NEUTRAL AXIS OF A LOADED BEAM. 447
Integrating,
c l being a constant of integration.
Integrating again,
W
- Ely = *> + c>* ...... (3)
There is no constant due to the last integration, as x and y
vanish together.
Between C and A,
W f A W
"*-
Integrating twice,
-EI d j- x = -~(l-xf + c,, ...... (5)
and
W
-Efy=(l-*r + cs + c.. ..... (6)
c^ , <:, being constants of integration.
The tangent at C is no longer horizontal, but makes a defi-
dy
nite angle 6 with the horizon, so that -j- is now tan 6 when
x . Also, the values of -=- and y at C, viz., tan and d, as
given by eqs. (2) and (3), must be identical with those given by
eqs. (5) and (6), while the value y, at A, as given by eq. (6)
when x = /, is equal to y l . Therefore
W W
? + ^ = - EI tan = - ^t' + c,,
r? + Ci = _ EId = p + C + c
90 '2 96 3 2
and
44^ THEORY OF STRUCTURES.
Hence,
v W v W W
'
fully defining both halves of the neutral axis.
dy
Again, in Ex. 6 it is no longer true that -^- = o when
x = , but the conditions of integration are y = o when x = o,
dy
and y = y l when x = /. These, together with -y- = o when
dx
x = o, are also the conditions in Ex. 7. Other cases may be
similarly treated.
5. To Discuss the Form assumed by the Neutral Axis
of a Girder OA which rests upon Supports at O and A,
and carries a Weight JP at a Point B, distant r from O.
Let OBA be the neutral axis of the deflected girder.
The reactions at O and A are
c . , ,
x / - r
P -j and P-j, respectively.
1 / /
Let BC, the deflection at C t
Let a be the slope of the neutral axis at B.
The portions OB, BA must be treated separately, as the
weight at B causes discontinuity in the equation of moments.
First, at any point (x, y) of OB,
Integrating,
c l being a constant of integration.
NEUTRAL AXIS OF A LOADED BEAM. 449
dy
But -r- = tan a when x = r, and therefore
/ rr*
-El tan a = P j + *, .
Hence,
Integrating,
Ti-T-'v- (3)
There is no constant of integration, as x and y vanish to-
gether.
Also, y = d when x = r.
- r tan a) = - P--. ... (4)
In the same manner, if A is taken as the origin, and AB
treated as above, equations similar to (i), (2), (3), and (4)
will be obtained, and may be at once written down by sub-
stituting in these equations n a for a, P T - for P^^-, lr
I I
for r, and r for / r.
Thus, the equation corresponding to (4) is
- El { d - (I - r) tan (* - a) } = - P '- iLU^ u ( 5)
o
Subtracting (5) from (4),
p
na==-r-r-2r); .... (6)
and from (4),
450 THEORY OF STRUCTURES.
Thus, eqs. (2) and (3) become
and
-Ely = ?'-*>-?(l-r)(2l-r)*,. . (9)
the latter being the equation to the portion OB of the neutral
axis, and the former giving its slope at any point.
Next, at any point (x,y) of BA,
z P j-x - P(x - r) (10)
Integrating,
c a being a constant of integration.
dy
But -j- = tan a when x = r.
dx
pl r ? + ' = ~ EI tan " = ~ f 7 (/ ~ r)(/ ~ 2r
and
Pl-r , , N
c 9 =---r(2t-r).
Hence,
- / = T-r 1 *' - ? ( ^ - r)i - S (/ - r)(2/ - r) -
Integrating,
- Ely = g 1 -^^' ~ g(* - ^ - /(' -
4 being a constant of integration.
But y d when x = r.
NEUTRAL AXIS OF A LOADED BEAM. 45 1
and 4 = o. Hence,
- Ely = > - (* - r)' - Pr (l-r)(2/-r)x, (12)
which is the equation to the portion BA of the neutral axis,
eq. (11) giving its slope at any point.
In the figure r < , and the maximum deflection of the
girder will evidently lie between B and A, at a point given by
dy
putting =o in eq. (12), which easily reduces to
and therefore
X = I
is the abscissa of the most deflected point. The corresponding
deflection is found by substituting this value of x in eq. (12).
If r > , the maximum deflection lies between O and B, at
dy
a point determined by putting -^- = o in eq. (8), which then
easily reduces to
from which
fr(2l - r)
"V" .<
452 THEORY OF STRUCTURES.
Substituting this value of x in eq. (9),
the maximum deflection = gj j I ) .
EXAMPLE. P = 15,000 Ibs., / = 100 ft., r = go ft.
The distance of most deflected point from O
/go X 1 10 ,
= y - = 57-44 ft.,
and the maximum deflection
_ 15000 10 too X iioy 500000
~ W X I5o V 3 / = ~~^T~ (33) *
6. To Discuss the Form of the Neutral Axis of a Girder
OA which rests upon Supports at O and A and carries
several Weights JP, , J* a , 1% , . . . , at points i, 2, 3, . . . , of
which the Distances from O are r r a , r, , . . . , respectively.
A-X
FIG. 322.
It may be assumed that the total effect of all the weights
is the sum of the effects of the separate weights, and thus each
may be treated independently, as in the preceding article.
Let a l , ar a , or 3 , . . . be the slopes at the points i, 2, 3, . . .
of the neutral axis.
Considering P t , the equation to Oi is
- Ely = '-V - '(/ -
NEUTRAL AXIS OF A GIRDER OA. 453
and to lA,
- Ely = ^V - (* - r,)' - T -j(l- r^(2l- r>.
Considering P, , the equation to O2 is
- Ely = ~V - ^/ -
and to 2A,
l
-6 T ~ - - / -
and so on for P z , P 4 , etc.
The total deflection F at any point (x, F) is the sum of the
deflections due to the several loads.
Take, e.g., a point between 3 and 4, and let d l , d^ , d^ , . . .
be the deflections of this point, due to P lt P 9 , P a , . . . , respec-
tively. Then
-Eld, =
-EU =
and so on. Hence,
-
(A)
454 THEORY OF STRUCTURES.
Again, the position of the most deflected point is found by
making -7- = o in the equation to that portion of the neutral
axis between two of the weights in which the said point lies.
The result is a quadratic equation, and the value of x derived
therefrom may be substituted in eq. (A), which then gives the
maximum deflection.
EXAMPLE. A girder of 100 ft. span supports two weights
of 20,000 Ibs. and 30,000 Ibs. at points distant respectively 20 ft.
and 60 ft. from one end.
The most deflected point must evidently lie between the
two weights, and the equation to the corresponding portion of
the neutral axis is
x* 2OOOO
EIY = ^(20000 X 80 + 30000 X 40) g (* 2 ) 3
- -^-(20000 X 20 X 80 X 1 80 + 30000 X 60 X 40 X 140)
14000 , i oooo.
= - x - (x 20) 26400000^.
Kis a maximum when
dY
- o 14000^ ioooo(^ 20) 26400000,
or
X* + lOO^r -7600 = O,
or
x = 50.497 ft.
Remark. Instead of assuming -jr = ^cEI-j-^ , it would be
77 T JA
more accurate to take -jz- = El cos 6 (Art. 2), and the
first integration would make the left-hand side of the slope
equation El sin instead of El tan 6.
MOMENT OF INERTIA VARIABLE.
455
7. Moment of Inertia variable. In the preceding exam-
ples the moment of inertia / has been assumed to be constant.
From the general equations,
-->. j 2 - ,
dx c
c being proportional to the depth of the girder at a transverse
section distant x from the origin.
Hence, for beams of uniform strength, the value of c in
terms of x may be substituted in the last equation, which may
then be integrated.
Again, let Fig. 323 represent a cantilever of length /, spe-
cific weight w, circular section,
and with a parabolic profile, the
vertex of the parabola being at A.
Let 2b be the depth of the
cantilever at the fixed end.
Let the cantilever also carry a
uniformly distributed load of in-
tensity/.
Consider a transverse section of radius z at a distance x
from the fixed end.
Let x, y be the co-ordinates of the neutral axis at the same
section. Then
But *' = -/_
FIG. 323.
or
Integrating,
4~ r
wnV
~6~7~
-7 + *T- - (0
45 6
THEORY OF STRUCTURES.
dy
There is no constant of integration, as - = o when x = o.
Integrating again,
px*
There is no constant of integration, as x and jj/ vanish to-
gether. Thus, equation (i) gives the slope at any point, and
equation (2) defines the neutral axis.
The slope at the free end (* = /) = -m +
The deflection
~&
8. Springs Fixed at One End and Loaded at the other
with a Weight W.
Data. Length = /; breadth = b, and depth = d at fixed
support ; V= volume of spring; f= maximum coefficient of
strength ; A = maximum deflection.
CASE a. Simple rectangular spring.
By Ex. i, Art. 39,
_
-"'
since
Wl _M _2f _ 1207
'
Also,
y 2//
61 '
Hence,
The work done = =
(3)
SPXJNGS.
457
CASE b. Spring of constant depth but triangular in plan.
Let b x be the breadth at a distance x from the fixed end.
Then
b I - x
and / at the same point
l-X
12
12
bd\
FIG. 325.
1207
EbcT'
Integrating twice,
and
Also,
dy _
dx~
12WI
(A\
6WI ,
V
y
A
E$3r* '
6wr /r
T/T7 A
Ebd* "Ed
bd*ffP f*bdl f*V
(5)
Hence,
/. V
The work done =
61 Ed 6E ' $'
r
WA f*V
(6)
N.B. The results I to 6 are the same if the springs are
compound ; i.e., if the rectangular spring is composed of n
simple rectangular springs laid one above the other, and if the
triangular spring is composed of n triangular springs laid one
above the other.
458 THEORY OF STRUCTURES.
CASE c. Spring of constant width but parabolic in elevation.
Let d x be the depth at a distance x from the fixed end.
Then
dl - /
and /at the same point =
FIG. 326.
bd: wti-*\i
12 ' 12 I / /'
d *y W (J \ l2Wl * f , x
,.^ r = _(/_*)=__.(/_,)
Integrating twice,
^ = T ^ ^ * (/ ~ * }l ~ 2/t(/
and hence
*KJL. 4// a
E bd*~ ->K<1 v)
Also,
The work done = =-2
2 ~6 '
9. Girder Encastre at the Ends. The girder BCDEFG
rests upon supports at the ends, is held in position by blocks
forced between the ends and the abutments, and carries a uni-
formly distributed load of intensity w.
GIRDER ENCASTRE AT THE ENDS.
459
It is required to determine the pressure that must be devel-
oped between the blocks and the girder so that the straight
portion between vertical sections at points O and A of the
W.I
FIG. 327.
neutral axis may be in the same condition as if the girder were
fixed at these sections.
Let / be the length of OA.
Let R be the reaction at the surface BC> and r its distance
from O.
Let /fbe the reaction between the block and the end CD,
and h its distance from O.
Let P be the weight of the segment on the left of the ver-
tical section O, and / its distance from O.
Then for the equilibrium of the segment on the left of the
section at O,
2 * 12
/? P
.\K-:*>*~ 2 ,
and
I wl\ wl*
(P \r-Pp--
\ 21 ' 12
the required pressure.
Again, take O as the origin, OA as the axis of x, and a
vertical through O as the axis of _y.
At any point (x, ^/) of the neutral axis,
_ E = x _ _ . ( s ee Ex. 8.)
dx* 2 2 12 ^
460 THEORY OF STRUCTURES.
10. On the Work done in bending a Beam. Let
A'B'C'D' be an originally rectangular ele-
\q' ment of a beam strained under the action of
jfg. external forces.
Let the surfaces AD', B'C' meet in G>;
O is the centre of curvature of the arc P Q of
the neutral axis.
Let OP = R = OQ f .
Let the length of the arc P Q' = dx.
Consider any elementary fibre/'/, of length
dx'y of sectional area a, and distant y from the
neutral axis.
Let t be the stress in p'q' .
The work done in stretching/'/
But --- and
-jr--4
dx 9 -dx y
The work done in stretching/'/ = -
and the work done in deforming the prism A'B'C'D'
iE
Hence, the total work between two sections of abscissae
-^a>
C'-'iEI , El r'"dx
= .T.*- I?"
I J/
But - \ therefore the work between the given limits
TRANSVERSE VIBRATIONS. 461
This expression is necessarily equal to the work of the ex-
ternal forces between the same limits, and is also the semi vis-
viva acquired by the beam in changing from its natural state
of equilibrium.
Cor. If the proof load P is concentrated at one point of a
p
beam, and if d is the proof deflection, the resilience = d.
If a proof load of intensity w is uniformly distributed over
the beam, and if y is the deflection at any point, the resili-
ence = / wydx, the integration extending throughout the
whole length of the beam.
The case of the single weight, however, is the most useful
in practice.
ii. On the Transverse Vibrations of a Beam resting
upon Two Supports in the same Horizontal Plane.
It is assumed
(a) That the beam is homogeneous and of uniform sectional
area.
(b) That the axis (neutral) remains unaltered in length.
(c) That the vibrations are small.
(d) That the particles of the beam vibrate in the vertical
planes in which they are primarily situated. In reality, these
particles have a slight angular motion about the horizontal axis
through the centre of gravity of the section, but for the sake
of simplicity the effect of this motion is disregarded.
Jif M44M
/I A
dx
A-X
Y
Ci Y C'
i S+dS
w dx
FIG. 329.
Let OA be the beam.
Take O as the origin, the neutral line OA as the axis of x,
and the vertical O Fas the axis of y.
Consider an element of the beam, bounded by the vertical
4^2 THEORY OF STRUCTURES.
planes BC, B ' C ', of which the abscissae are x and x -f- dx,
respectively.
Let w be the intensity of the load per unit of length ; hence
wdx is the load upon the given element, and acts vertically
through its centre.
Let 5 be the shearing force at B ; 5 -f- dS the shearing
force at B'.
Let M be the bending moment at B] M -\- dM the bend-
ing moment at B'.
iv d* v
Also, the resistance of the element to acceleration = -.
g at
Hence, at any time t,
w d y
dx -j-f + 5 (S -f- dS) wdx = o,
or
_
d? w dx
-r% + Jfr.S + C, (5)
c being a constant of integration.
dy
When x l ry -j- = tan a.
.-. - El tan a = R r _, |- - v|r + ^.^ + c, . . (6)
Integrating (5),
- Ely = Rr-^ - v>r ~ + M r _t ~ + ex. . (7)
There is no constant of integration, as x and y vanish together.
Also, y = o when x l r .
or
c = R r 1 4- -\- v>r M r , -. (8)
r izr r 4 ^ r 1 / .. \/
THE THEOREM OF THREE MOMENTS. 4 6 5
Substituting this value of c in eq. (6),
- EI tan a = R r ., ^ - w r + Jf r . t j ..... ( 9 )
Similarly, the segment XV gives
Adding eqs. (9) and (10), transposing, and simplifying,
R r -Jr + Rr + fr + i
= WS + fV + ,/',+ . - \M r _J r - PV r+I . (ii)
Finally, combining eqs. (4) and (u),
+ w r+I r r+l ). (12)
If the girder is supported at points, there are w 2 equa-
tions connecting the corresponding bending moments, and two
additional equations result from the conditions of support at
the ends. For example, if the ends merely rest upon the sup-
dy
ports M l o and M n = o ; if an end is fixed, -=- = o at that
point.
The point of maximum bending moment, the points of
inflexion, and the point of maximum deflection in any span are
dM dv
found by making =- = o, M o, and -j- o, respectively.
Thus, for the span OX,
dM
o = R r ^ WyX ;
/. x = f , and maximum B.M. = -- -f- M r . t \
"Wr 2 W r
a quadratic giving x ;
466 THEORY OF STRUCTURES.
c,
a cubic from which x may be found by trial. The maximum
deflection is obtained by substituting the value of x in eq. (7) ;
c being given by eq. (8).
CASE B. Let the loads upon OX, XV, respectively, consist oi
a number of weights P l , P 9 , P 3 , . . . , distant p l , / 2 ,/ 3 , . . . from
O, and Q l , <2 3 , <2, , . . . , distant #,,,,&,... from F. Refer
the neutral axis OA X to the rectangular axes Ox, Oy.
It may be assumed that the total effect of all the weights
is the algebraic sum of the effects of the weights taken sepa-
rately.
Consider the effect of P t at A.
The equation of moments at any point (x, y) of the neutral
axis between O and A is
Integrating,
c l being a constant of integration.
Integrating again,
-Efy^Rr-^+Mr-^ + cs. . . (3)
There is no constant of integration, as* and y vanish together.
The equation of moments at any point (x, y) between A and
(4)
THE THEOREM OF THREE MOMENTS.
Integrating,
-El4jL = R r .-^(x-p l )'+M r -* + c t . ... (5)
Integrating again,
- Ely = R r .~ - (* - A) 3 + M r _ + v + ,, . (6)
dy
Now, at the point A, the values of -7- and jj/, as given by
m&
eqs. (5) and (6), are identical with those given by eqs. (2) and
(3) ; also, in equation (6), y = o when x = l r .
Hence,
and
o = R^ - (/ r - A) 3 + ^r-~ + cj r + c 9 ;
so that
3 = 0,
and
, = *.= -^-, + J(/,-/ 1 ) t -^-,j. (7)
Let a be the slope at X\ then, by eqs. (5) and (7),
- EI tan a = R_ - (/, - py(2l r +/,) + Jf r _. (8)
Similarly, the segment XV gives
-EI^(n- a } = Rj + M r+ ...... (9)
468 THEORY OF STRUCTURES.
Adding eqs. (8) and (9), and transposing,
*,_,/,' + JWV* = & - A)'(2/ r +A)
-f^.^-fJf^/^. (10)
Again, taking moments about Jf,
Rr j r _ />('r-A) + -^-i = ^ = ^WH-X + ^H-:> 00
whence
R r _,/; + ^H-x'V+x
and finally, by eqs. (10) and (12),
M r j r + 2M r (l r + /, +1 ) + M r+1 t r+I = - Pfa - A 2 )- (13)
The effect of each weight may be discussed in the same
manner, and hence the relation between M r _ iy M r , and M r+l
may be expressed in the form
r _ j r + 2 M r (l r + / r+1 ) + M r+l t r+l = - 2fa - p ,
Cor. I. The relation between M r _ T , M r , M r + l for a uni-
formly distributed load maybe easily deduced from eq. (14).
For example, let a uniformly distributed load of intensity w r
cover a length 2a( ~
*~ "-
2 "/- 8/^
i ^-
r -
2
For an arbitrarily distributed load,
K ^(A~/) , M _ P(l t -p) I
^ = ~
R = 2^_Zl} +~= ^ Q(/ * ~ q) - - A+B
If w l = O, or if P and hence ^4 o, then R l is negative.
So if w a = o, or if Q and hence B o, then ^? 3 is nega-
tive.
Hence, if either of the spans is unloaded, the reaction at
the abutment end of the unloaded span is negative and that
end is subjected to a hammering action. This evil may be
obviated :
(a) By loading the spans sufficiently to make R^ and R^
zero or positive.
This result is attained for R,
if 3Wl /,' + 4W/.Y, > a/4 1 , or
and for ^ 3
if 4 vW + 3 ^/, 3 > ^y, 8 , or if
47 2 THEORY OF STRUCTURES.
(b) By using a latching apparatus to keep the ends from
rising.
(c) By employing suitable machinery to exert an upward
pressure, at least equal to the corresponding negative reaction
upon each end, which is thus wholly prevented from leaving its
seat.
Cor. i. When the load is uniformly distributed, the dis-
tance x of the point of inflection in OX from O is given by
M = o = R,x -- , and therefore x - .
2 W 1
Similarly, the distance of the point of inflection in XV from
K= 2 A
U\
If / t = /, = /, then
M=-^ l+ w, R t = 7 -^^ * = =
And if w l = 2/ a = w, then
In the latter case - = / = - 3 , and thus a hinge may
w 1 w 2
be introduced in each span at a distance from the centre pier
equal to one fourth of the span, without impairing the stability
of the girder. Hence, also, the continuous girder of two equal
spans may be considered as consisting of two independent
girders, each of length |/, resting upon end supports, and of
two cantilevers each of length .
Ex. 2. Swing-bridges with two points of support at the
' t' f
h {> I*
FIG. 333.
pivot pier, as, e.g., when they are carried upon rollers running
in a circular path.
A P PLICA TIONS. 47 3
This is a case of a continuous girder of three spans.
Let I, , / 2 , / be the lengths of the spaces, w, , /, , ze/ 3 the
corresponding intensities of the loads, which are assumed to
be uniformly distributed.
Let R lt R^j R 3 , R t be the reactions at the supports ;
M l , M t , M a , M t the corresponding bending moments. Then
-$(w 1 t: + wj;)', . (i)
-}(>X + wA). . (2)
Let the ends of the girders rest upon the supports, and
assume, as is usually the case in practice, that the centre span
is unloaded, i.e., that w^ = o. Then
M l = o and M^ = o.
From (i) and (2),
A = -*/,', ". (3)
and
M,l, + 2 !&,(!, + /,)=-&,!: ..... (4)
Hence,
/./.V,
'
34"
and
_ ^/. 3 4 - W,V, + 4)
- 4(444 + 34' + 444 + 444)'
Taking moments about the second support,
6/,V > + 6/ 1 V J ' + 8/ 1 '4/ 1 ) + Wj 4V,_ _ (7)
4(444 + 34' + 444 + 444)
Taking moments about the third support,
6/.V, + 6/.V.' + 8/,V,Q + w,n (K .
()
4(444 + 34' + 444 + 444)
474 THEORY OF STRUCTURES.
Thus R t and jR t are both positive for all uniform distribu-
tions of load over the side spans, and no hammering action
can take place at the ends.
Again, if the span on the left is unloaded, i.e., if w 1 = o,
M 9 is positive and M 3 negative ; and if the span on the right is
unloaded, i.e., if w 3 = o, M, is negative and M 3 positive.
Thus, at the piers, the flanges of the girder will be sub-
jected to stresses which are alternately tensile and compressive,
and must be designed accordingly. The same result is also
true for arbitrarily distributed loads.
Ex. 3. The weights on the wheels of a locomotive passing
over a continuous girder of two spans, each of 50 ft., taken in
order, are as follows : 15,000 Ibs., 24,000 Ibs., 24,000 Ibs., 24,000
Ibs., 24,000 Ibs. The distances of the wheels, centre to centre,
taken in the same order, are 90 in., 56 in., 52 in., 56 in. Let
it be required to place the wheels in such a position as to give
the maximum bending moment at the centre pier.
The pier must evidently lie between the third and fourth
wheels.
Let x be the distance, in inches, of the weight of 15,000 Ibs.
from the nearest abutment. The remaining two weights on
the span are respectively x -f- 90 and x -(- 146 in. from the
same abutment.
The two weights on the other span are 142^ and 198 ;r
in 1 ., respectively, from the nearest abutment.
Hence, by Case B, Art. 13, if M is the bending moment at
the centre pier,
' - (x + 90)'
APPL1CA TIONS.
475
Making o for maximum value of M t and simplifying,
dx
and therefore
+ 27648* = 2518848,
x = 87.39 in. = 7.28 + ft.
Thus, the B. M. at the centre pier is a maximum when the
first wheel is 7.28 ft. from the nearest abutment.
The maximum B. M. in inch-pounds is obtained by substi-
tuting x = 87.39 in. in the above equation.
15. Maximum Bending Moments at the Points of Sup-
port of Continuous Girders of n equal Spans.
Let the figure represent a continuous girder of n spans, I,
2, 3, ... I being the n I intermediate supports.
o 123 r I r r-\-i I n
"A A A A A A A A A A A~
CASE I. Assume all the spans to be of the same length /,
and let w l , w t , . . . w n _ l , w n be the intensities of loads uni-
formly distributed over the 1st, 2d, ...( i)th and nth spans,
respectively.
By the Theorem of Three Moments,
-(
4 l
_ _/_'
4 2
_/_'
4
/ 2
? 8 + 4^ 4 + ^ 6 = -(w. + w 6 ) ;
(O
(2)
(3)
(4)
(5)
4/6 THEORY OF STRUCTURES.
r
m n _ 3 + 4m n _ 2 + /_, = - -(w n _ 2 + w n . r ) ; (n 2)
r
m n . 2 + 4w_ T = -<>-i + Wn)- (n i)
4
?# and ;# are both zero, as the girder is supposed to be rest-
ing upon the abutments at o and n.
From these (n i) equations, the bending moments m l ,
m^ , . . . m n _i may be found in terms of the distributed loads.
Eliminating m^ from 2 and 3,
r
mi _ x 5 ^ 3 _ 4 ^ 4 = _ - j ( Wf + ^3) - 4K + ^4)}- (*0
Eliminating w g from 4 and x^ ,
Eliminating MI from 5 and
m
.
Finally, by successively eliminating w 6 , m 6
-! -X
/ a
--U w + w 3)-4K +
the upper or lower sign being taken for the terms within the
brackets according as n is odd or even, and the coefficients
^-i , #*- 2 , tf- 3 , bein'g given by the law,
APPLICATIONS. 477
a, = 4a t a, = 209 ;
# a = 4*1 = 4 ;
, = I.
Commencing with equations n 3 and n 2, and proceed-
ing as before,
0_x*i *-i
/i
= - \ a n - 9 (u>i + /,) rf_ 3 (w a + ze; 3 ) + d_ 4 (w s + ;) . . .
1 5O_ 4 + w_ 3 ) =F 4(w-3 + >_,) (^- 2 + V_ f ) } , (^)
the upper or lower sign being taken for the terms within the
brackets according as n is odd or even.
Solving the two equations y and 2,
r
,(* f -xi)= - -|^-i^- 2
(<* n -i + a-3a n -*Wn-i =F
and
^-i(tfVi - i) =
/ a (
- 1 a n _ 2 w,
4
Hence, since w 19 w tt ...w n are positive integers, the value of
m H will be greatest when ;, , / a , w 4 , w e , o; 8 , . . . are greatest
and w a , w 5 , w n , . . . are least ; and the value of m n _ v will be
greatest when w w , w n ^ , w_ 3 , w M _ 4 , . . . are greatest, and w n _ 2 ,
w M - 4 , W M _, . . . are least. In other words, the bending moments
at the 1st and (n i)th intermediate supports have their maxi-
47 8 THEORY OF ^STRUCTURES.
mum values when the two spans adjacent to the support in
question, and then every alternate span, are loaded, and the re-
maining spans unloaded.
#?,,#?,,... m n -i ma y now be easily determined.
Thus, by eq. (i),
r
m * = 7^ + w ^ ~ 4w i
4
r ( 4
= - -7 J K + wj - -i - -a
4 l "
a + . . . I-
But a n _, = 4a n _ 2 a n _ y
and is greatest when w 9 , w 3 , w t , w 1f . . . are greatest and
w 4 , w, , w s , . . . are least.
Similarly, by eqs. (i) and (2),
CONTINUOUS GIRDERS. 479
and is greatest when w l , w a , w^ , w t , w 6 , . . . are greatest and
w a , w 6 , w 7 , w 9 , . . . are least.
Thus, the general principle may be enunciated, that " in a
horizontal continuous girder of n equal spans, with its ends
resting upon two abutments, the bending moment at an inter-
mediate support is greatest when the two spans adjacent to
such support, and the alternate spans counting in both direc-
tions, carry uniforrnly distributed loads, the remainder of the
spans being unloaded."
CASE II. The principle deduced in Case I also holds true
when the loads are distributed in any arbitrary manner.
Consider the effect of a weight w in the rth span concen-
trated at a point distant/ from the (r i)th support.
By the Theorem of Three Moments,
4^+^ = 0; (i)
m 1 + 4m, + m 3 = 0' t (2)
w* + 4^3 + ^4 = o ; . , , , . . (3)
m r . a + 4m r ., + m r = w -* /) = A, suppose ; (r i)
r + m r+l = -w
= w (I p)(2l /)=: B, suppose ; (r)
m r + 4m r+1 + m r+2 = o ; . . . . (r + i)
_i = O ; ....( 2)
= o (n i)
4^0 THEORY OF STRUCTURES.
By equations (i), (2), (3), ... (r - 2),
ill I
-
the upper or lower sign being taken according as r is even or
odd.
By equations (n i), (n 2), (n 3), . . . (r + i),
. m r+2 m r+l m r
= -f - - = -
a n-r-i
The coefficients a are given by the same law as for the co-
efficients a in Case I. Thus,
a r ... a n _ r
m r _ 2 = -- m r _ t and m r+l = - m r .
Substituting these values of m r _ 2 and m r+l in the (r i)th
and rth equations,
and
rtr-i + M r \A - ^A = B = m r .,
a n-r+2 '
where
<*r- 2
and c = ~
Hence, solving the last two equations,
Ac - B Bb-A
* r _ I ^-. _ and ^=_ 7 - .
The ratios - - and ^- are each less than unity, and
&r-i &n-r+i
hence b and c are each < 4 and > 3.
CONTINUOUS GIRDERS. 481
It may now easily be shown that Ac B and Bb A are
each positive. Hence, m r _^ and m r are both of the same sign*
The bending moment m q at any intermediate support on
the left of r i is given by
m q = -| -- m r -i if 9 an d r are the one even and the other odd,
or
= -- --m r -i if 9 an< 3 r are both even or both odd.
Thus the bending moment at the q\.\\ support is increased
in the former case and diminished in the latter.
If q is on the right of r,
m q = -\ *~ q+l m r if q and r are both even or both odd,
a n-r+i
or
m q -=. -- ^^~m r if q and r are the one even and the other odd,
and the bending moment on the ^th support is increased in the
former case and diminished in the latter.
Thus, the general principle may be enunciated, that, "in a
horizontal continuous girder of n equal spans, with its ends
resting upon two abutments, the bending moment at an inter-
mediate support is greatest when the two spans adjacent to
such support, and the alternate spans counting in both direc-
tions, are loaded, the remainder of the spans being unloaded."
CASE III. The same general principle still holds true when
the two end spans are of different lengths.
E.g., let the length of the first span be kl, k being a
numerical coefficient, and let 2(1 -\- k) = x.
Eq. (i) now becomes
m^x -^m^ Q.
482 THEORY OF STRUCTURES.
Proceeding as before,
.m^ _ m^ _ _m* _ _ m, _
}>'* ~ V' " ^3 : " b, ~~
the coefficients b lt b^, b^, . . . being given by the same law as
before, viz.,
=..*;
, = 4 3 - ^ = 4* i ;
4 = 4A ^= 1 5* -4J
5 = 4^ 3 b % = 56^ 15;
The two sets of coefficients (a) and (b) are identical when
x = 4; and when x > 4, all the coefficients b except the first
(b l i) are numerically increased.
Hence, the same general results will follow.
N.B. The equations giving m q are simple and easily ap-
plicable in practice. They may be written
a q B Ac .
m a = - -j if q is on the left of r,
a r _i DC I
and
m q = "Ir + i ~L - if ^ is on the right of r.
If there are several weights on the rth span,
') and B
EXAMPLE. The viaduct over the Osse consists of two end
spans, each of 94 ft., and five intermediate spans, each of 126 ft.
The platform is carried by two main girders which are con-
MAXIMUM BENDING MOMENTS. 483
tinuous from end to end. The total dead load upon the girders
may be taken at one ton (of 2000 Ibs.) per lineal foot.
Denote the supports, taken in order, by the letters a, b, c, d,
e,f,g, h, and let it be required to find the maximum bending
moment at d when the bridge is subjected to an additional
proof load of ij tons per lineal foot.
The spans ab, cd, de, fgol each girder carry ij tons per
lineal foot.
The spans be, ef, gh of each girder carry \ ton per lineal foot.
Denoting the bending moments at a, b, c, d, e, f t g, h, re-
spectively, by m l ,;//,... m s , the intermediate spans by /, the
end spans by /, and remembering that m l = o = m 6 , we have
a = -~
4
r
, + 4m 3 + m< = - -
r
, + 4m t + m,= - -(i J- + i J) ;
/ a
5 + 4^6 + ^ = - -(4 +
But k = T 9 ^- = f , very nearly.
/a 2^
.-. 7m, + 2m, = - - . ^ ; ..... (i)
4 5
-.- ..... (2)
THEORY OF STRUCTURES.
2m t + 7m,
From eqs. (i), (2), (3),
From eqs. (4), (5), (6),
l l L.
(V)
4 2 '
/^ *7
CA>
4 2'
7.
CO
4 2'
(6}
Hence, ; 4 , the maximum required,
16. General Theorem of Three Moments. The most
general form of theorem of three moments may be deduced as
follows :
Oi 2 3
FIG. 334.
Let O, X, V, the (r i)th, rth, and (r -\- i)th supports of a
continuous girder of several spans, be depressed the vertical
distances d, (= O,O\ d, (=O,X), and d, (= 6> 3 F), respectively,
below the proper level Ofl^O^ of the girder.
GENERAL THEOREM OF THREE MOMENTS. 48$
d^ , d^, d z are necessarily very small quantities.
Let OCXDVbe the deflection curve, and let the tangent at
X meet the verticals through O and V in E and F, and the
tangents at O and V in T t and 7", .
Let #, be the change of curvature from O to X (= OTf).
" ^ " " " " " " Fto X(= FT^V).
Let A lf A i be the effective moment areas for the spans OX,
XV, respectively.
Let .TJ be the distance (measured horizontally) of the centre
of gravity of A l from O.
Let x^ be the distance (measured horizontally) of the centre
of gravity of A^ from F.
Let 0,E =?>, Of = y % .
By Art. 2,
, _
r +I h /, + J~^/ 1 4 "4 +I /
But
7i - < _ < - J 8 Z. I _L A-^ . ^
7 7 > r 7* "t" 7 ~~ / IT
/ r / r+I / r / r+I / r /r+x
d^ d^d^ di i (A,x A^\
-JT / r+I -^A / r -/ ;
Again, by Art. 13, Cor. 2,
^
and
^4 r , y4 r+I being the areas of the bending-moment curves for
the spans OX, XV, respectively, on the assumption that they
are independent girders, or cut at O, X, and V, and z rj z r ^
486 THEORY OF STRUCTURES.
being the horizontal distances of the centres of gravity of these
areas from O and V. Hence,
M r _J r + 2M r (l r + / r+I ) + M r+l l r+l
KA* r 6 A * r+l -U
. = bA r j -- vSl r+l -j -- 1-
*r tr+i
Note. If O, X, or V is above O,O^ then d, , af a , or d^ is
negative.
Cor. The forms of the Theorem of Three Moments given in
Cases A and B, Art. 13, may be, immediately deduced from the
last equation.
CASE A.
A ---
^r-
' r+I
CASE B.
17. Advantages and Disadvantages of Continuous
Girders. The advantages claimed for continuous girders are
facility of erection, a saving in the flange material, and the re-
moval of a portion of the weight from the centre of a span to-
PROPERTIES OF CONTINUOUS GIRDERS. 487
wards the piers. Circumstances, however, may modify these
advantages, and even render them completely valueless. The
flange stresses are governed by the position of the points of in-
flexion, which, under a moving load, will fluctuate through a
distance dependent upon the number of intermediate supports
and upon the nature of the loading. In bridges in which the
ratio of the dead load to the live load is small the fluctuation is
considerable, so that for a sensible length of the main girders,
a passing train will subject local members to stresses which are
alternately positive and negative. This necessitates a local
increase of material, as. each member must be designed to bear
a much higher stress than if it were strained in one way only.
Again, the web of a continuous girder, even under a uni-
formly distributed dead load, is theoretically heavier than if
each span were independent, and its weight is still further in-
creased when it has to resist the complex stresses induced by
a moving load.
Hence, in such bridges the slight saving, if there be any,
cannot be said to counterbalance the extra labor of calculation
and workmanship.
In girders subjected to a dead load only, and in bridges in
which the ratio of the dead load to the live load is large, the
saving becomes more marked, and increases with the number
of intermediate supports, being theoretically a maximum when
the number is infinite. This maximum economy may be ap-
proximated to in practice by making the end spans about four-
fifths the intermediate spans.
In the calculations relating to the Theorem of Three Mo-
ments, it has been assumed that the quantity El is constant,
while in reality E, even for mild steel, may vary 10 or 15 per
cent from a mean value, and / may vary still more. It does
not appear, however, that this variation has any appreciable
effect if the depth of the girder or truss changes gradually, but
the effect may become very marked with a rapid change of
depth, as, e.g., in the case of swing-bridges of the triangular
The graphical method of treatment may still be employed
by substituting, for the actual curve of moments, a reduced
488 THEORY OF STRUCTURES.
curve, formed by changing the lengths of the ordinates in the
ratio of the value of El at a datum section to EL
It is often found economical to increase the depth of the
girder over the piers, which introduces a local stiffness and
moves the points of inflexion farther from the supports. A
point of inflexion may be made to travel a short distance by
raising or depressing one of the supports.
In order to insure the full advantage of continuity the ut-
most care and skill are required both in design and workman-
ship. Allowance has to be made for the excessive expansion
and contraction due to changes of temperature, and the piers
and abutments must be of the strongest and best description
so that there shall be no settlement. Indeed, the difficulties
and uncertainties to be dealt with in the construction of con-
tinuous girders are of such a serious if not insurmountable
character that American engineers have almost entirely dis-
carded their use except for draw-spans.
Much, in fact, is mere guesswork, and it is usual in prac-
tice to be guided by experience, which confines the points of
inflexion within certain safe limits.
Under these circumstances it may prove desirable to fix
the points of inflexion absolutely, and the advantages of doing
so are (a) that the calculation of the web stresses becomes
easy and definite^ instead of being complicated and even in-
determinate ; (b) that reversed stresses (for which pin-trusses
are less adapted than riveted trusses) are almost entirely
avoided ; (c) that the stresses are not sensibly affected by
slight inequalities in the levels of the supports ; (d) that the
straining due to a change of temperature takes place under
more favorable conditions.
The fixing may be thus effected :
(a) A hinge may be introduced at the selected point.
The benefit of doing so is very obvious when circumstances
require a wide centre span and two short side spans.
(b) If the web is open, i.e., lattice-work, the point of inflex-
ion in the upper flange may be fixed by cutting the flange at
the selected point and lowering one of the supports so as to
produce a slight opening between the severed parts. The
ADVANTAGES OF CONTINUOUS GIRDERS. 489
position of the point of inflexion in the lower flange is then
defined by the condition that the algebraic sum of the hori-
zontal components of the stresses in the diagonals intersected
by a line joining the two points of inflexion is zero.
It must be remembered, however, that this fixing of the
points of inflexion, or the cutting of the chords, destroys the
property of continuity, and, indeed, is the essential distinction
between a continuous girder and a cantilever.
Four methods may be followed in the erection of a contin-
uous girder, viz.:
1. It may be built on the ground and lifted into place.
2. It may be built on the ground and rolled endwise over
the piers. As the bridge is pushed forward, the forward end
acts as a cantilever for the whole length of a span, until the
next pier is reached. This method of erection is common in
France.
3. It may be built in position on a scaffold.
4. Each span may be erected separately, and continuity pro-
duced by securely jointing consecutive ends, having drawn to-
gether the upper flanges. A more effective distribution of the
material is often made by leaving a little space between the
flanges and forming a wedge-shaped joint.
49 THEORY OF STRUCTURES.
EXAMPLES.
1. Two angle-irons, each 2 in. x 2 in. x in., were placed upon sup-
ports 12 ft. 9 in. apart, the transverse outside distance between the bars
being 9^ in., and were prevented from turning inwards by a thin plate
upon the upper faces. The bars were tested under uniformly distributed
loads, and each was found to have deflected 2 T 5 ^ in. when the load over
the two was 1008 Ibs. Find E and the position of the neutral axis.
Ans. /= J/\ ; E= 17,226,139 Ibs.; neutral axis $ in. from*
upper face.
2. Both bars in the preceding question failed together when the
total load consisted of ioi cwts. (cwt. = 112 Ibs.) uniformly distributed,
and 3 cwts. at the centre. Find the maximum stress in the metal.
Ans. Compressive unit stress = 20,323 Ibs. ;
Tensile unit stress = 39,577 Ibs.
3. Show that the moments of resistance of an elliptic section and of
the strongest rectangular section that can be cut out of the same are in
the ratio of 99 ^3 to 112, and that the areas of the sections are in the
ratio of 33 to 14 |/2.
4. Show that the moments of resistance of an isosceles triangular
section and of the strongest rectangular section that can be cut out of
the same are in the ratio of 27 to 16, and that the areas of the two
sections are in the ratio of 9 to 4.
5. An angle-iron, 3 in. x 3 in. x ^ in., was placed upon supports
12 ft. 9 in. apart, and deflected i- in. under a load of 8 cwts. uniformly
distributed and 2 cwts. at the centre. Find E and the position of the
neutral axis.
Ans. E = 16,079,611 Ibs.; neutral axis iff in. from upper face.
6. The effective length and central depth of a cast-iron girder resting
upon two supports were respectively n ft. 7 in. and 10 in. ; the bottom
flange was 10 in. wide and i J in. thick ; the top flange was 2^ in. wide
and in. thick; the thickness of the web was f in. The girder was
tested by being loaded at points 3f ft. from each end, and failed when
the load at each point was 17^ tons. What were the total central
flange stresses at the moment of rupture?
What was the central deflection when the load at each point was 7|
tons? (E = 18,000,000 Ibs., and the weight of the girder = 3368 Ibs.)
Ans. 164,747.4 Ibs.; .368 in.
EXAMPLES. 491
7. A tubular girder rests upon supports 36 ft. apart. At 6 ft. from
one end the flanges are each 27 in. wide and 2f in. thick, the net area of
the tension flange being 60 in., while the web consists of two fa-in.
plates, 36 in. deep and 18 in. apart. Neglecting the effect of the angle-
irons uniting the web plates to the flanges, determine the moment of
resistance.
The girder has to carry a uniformly distributed dead load of 56 tons,
a uniformly distributed live load of 54 tons, and a local load at the
given section of 100 tons. What are the corresponding flange stresses
per square inch?
How many |--in. rivets are required at the given section to unite the
angle-irons to the flanges?
Ans. 238.13 x coeff. of strength ; 3.3186 tons ; 3.896 tons.
8. A yellow-pine beam, 14 in. wide and 15 in. deep, was placed upon-
supports 10 ft. 9 in. apart, and deflected f in. under a load of 20 tons at
the centre. Find E, neglecting the weight of the beam.
Ans. E = 1,272,112 Ibs.
9. What were the intensities of the normal and tangential stresses at
2 ft. from a support and 2\ in. from neutral plane, upon a plane inclined
at 30 to the axis of the beam in the preceding question?
Ans. 132.83 and 218.91 Ibs.
10. A beam is supported at the ends and bends under its own weight.
Show that the upward force at the centre which will exactly neutralize
the bending action is equal to f or \ of the weight of the beam (w) r
according as the ends are free or fixed.
Find the neutralizing forces at the quarter spans.
Ans. Ends free ^-gw at each or faw at one of the points of
division. *
Ends fixed ^w at each or %w at one of the points of
division.
11. A beam 8 in. wide and weighing 50 Ibs. per cubic foot rests upon
supports 30 ft. apart. Find its depth so that it may deflect f in. under
its own weight. (E i ,200,000 Ibs.) Ans. 9.185 in.
12. A rectangular girder of given length (/) and breadth (&) rests
upon two supports and carries a weight P at the centre. Find its depth
so that the elongation of the lowest fibres may be l Vff of the original
length.
Ans
13. A yellow-pine beam, 14 in. wide, 15 in. deep, and weighing 32 Ibs.
per cubic foot, was placed upon supports 10 ft. 6 in. apart. Under
uniformly distributed loads of 59,734 Ibs. and of 127,606 Ibs. the central
49 2 THEORY OF STRUCTURES.
deflections were respectively .18 in. and .29 in. Find the mean value
of E.
Also determine the additional weight at the centre which will increase
the first deflection by ^ of an inch. Ans. 2,552,980 Ibs.; 24,121 Ibs.
14. In the preceding question find for the load of 59,734 Ibs. the
maximum intensities of thrust, tension, and shear at a point half-way
between the neutral axis and the outside skin in a transverse section at
one of the points of trisection of the beam. Also find the inclinations
of the planes of principal stress at the point.
Ans. 1609.255, 169.562, 119.364 Ibs. ; = 3 48!'.
15. A pitch pine beam, 14 in. wide, 15 in. deep, and weighing 45 Ibs.
per cubic foot, is placed upon supports 10 ft. 9 in. apart, and carries a
load of 20 tons at the centre. Find the deflection and curvature, E
being 1,270,000 Ibs. What stiffness does this give ?
What amount of uniformly distributed load will produce the same
deflection? Ans. ^J T ; 32 tons.
16. In the preceding question find the maximum intensities of thrust,
tension, and shear at points (a) half-way between the neutral axis and
the outside skin, (b) at one third of the depth of the beam, in a trans-
verse section at one of the quarter spans. Also find the inclinations of
the planes of principal stress at these points.
Ans. (a) 95 I - 8 53> 292.969, 329.442 Ibs.; 6=9 34$'.
(b) 658.774, 171.108, 243.833 Ibs.; 6 = 15 5of.
17. A piece of greenheart, 142 in. between supports, 9 in. deep, and
5 in. wide, was tested by being loaded at two points, distant 23 in. from
the centre, with equal weights. Under weights at each point of 4480
Ibs., 11,200 Ibs., and 17,920 Ibs. the central deflections were .13 in., .37
in., .67 in., respectively. Find the mean coefficient of elasticity. The
beam broke under a load of 32,368 Ibs. at each point. Find the
coefficient of bending strength.
1 8. A sample cast-iron girder for the Waterloo Corn Warehouses,
Liverpool, 20 ft. 7-J- in. in length and 21 in. in depth (total) at the centre,
was placed upon supports 18 ft. i in. apart, and tested under a
uniformly distributed load. The top flange was 5 in. x i in., the
bottom flange was 18 in. x 2 in., and the web was ij in. thick. The
girder deflected .15 in., .2 in., .25 in., and .28 in. under loads (including
weight of girder) of 63,763 Ibs., 88,571 Ibs., 107,468 Ibs., and 119,746 Ibs.,
respectively, and broke during a sharp frost under a load of 390,282 Ibs.
Find the mean coefficient of elasticity and the central flange stresses at
the moment of rupture.
Ans. 7 = 3309.122; E= 17,427,327 Ibs.; 20,121 Ibs., 47,168 Ibs.
19. A steel rectangular girder, 2 in. wide, 4 in. deep, is placed upon
EXAMPLES. 493
supports 20 ft. apart. If E is 35,000,000 Ibs., find the weight which, if
placed at the centre, will cause the beam to deflect i in.
Ans. 1296^7 Ibs.
20. A timber joist weighing 48 Ibs per cubic foot, 2 in. wide x 12 in.
deep x 14 ft. long, deflected .825 in. under a load of 887 Ibs. at the
centre. Find E. Ans. 397,880 Ibs.
21. A beam of span / is uniformly loaded. Compare its strength and
stiffness (a) when merely resting upon supports at the ends ; (b) when
fixed at one end and resting upon a support at the other ; (c) when fixed
at both ends. In case (c) two hinges are introduced at points distant y
from the centre ; show that the strength of the beam is economized to
the best effect when y = , and that the stiffness is a maximum when
y = very nearly.
4
Ans. Cases (a) and (). mi :m*m:i
Cases (a) and (c). mi : m a :: 3 : 2
Case (c). Max. economy, mi : m*
Max. stiffness, mi : m*
: D* :: i : .416.
: Da :: 10 : 3.
2 : i
5:24/2.
4:3;
15 : 4 (approx.).
22. A beam AB of span /, carrying a uniformly distributed load of
intensity TV, rests upon a support at B and is imperfectly fixed at A, so
i iif/ a
that the neutral axis at A has a slope of g -j=j . The end B is lower
than A by an amount jzj . Find the reactions. How much must B
be lowered so that the whole of the weight may be borne at A ? Find
the work done in bending the beam.
21 ii 7 wl*
Ans. wl, ivl ; -~ -==.
32 32 48 1
23. A round wrought-iron bar / ft. long and d in. in diameter can
just carry its own weight. Find / in terms of d, (a) the allowable de-
flection being i in. per 100 ft. of span, E being 30,000,000 Ibs. ; (ff) the
allowable stress being 8960 Ibs. per square inch; (V) the stiffness given by
(a) and the strength given by (b) being of equal importance.
Ans. (d) /= |/2 50^? 2 ; (b) I = 4/224^; (c) /= \\d.
24. A square steel bar i ft. long and having a side of length d in. can
just carry its own weight; its stiffness is y^r and the maximum allow-
able working stress is 7 tons per square inch. Find / in terms of d, E
being 13,000 tons. /(in ft.) 13
Ans. -77: : \ ==
/ N
centre span,^> = -f -2 f 100 2o;r 2 -f jr 8 j.
(b) First. Loaded span between support and weight,
Loaded span between weight and hinge,
l2 5 ar ~ 703125.
Unloaded side span horizontal ; centre span
straight between hinges.
Second. Side span.jK = - -f %x ' ) ;
El \ 6 /
x2\ 5000000
centre span,, = -5- -j +
EXAMPLES. 495
30. A uniformly loaded beam, with both ends absolutely fixed, is
hinged at a point dividing the span into segments a and b. Draw curves
of shearing force and bending moment, and compare the strength and
stiffness of the beam when the hinge is (a) at the middle point ; (b) at
a point of trisection ; (c) at a quarter-span. Also, determine the slope of
the segments of these points.
w $a* + 8at>* + 3^ 4 w 3# 4 4- %a*b + 5^ 4
Ans ' * 1= ~ ~ : * 2= - ^
_ 3^ */ _ wa $a* + ^a*b + 4
" 8 " a 3 + P ' 8 s + 8
J/' : J/" : M'" :: 14 : 14 : ii ; >' : >'' : >'":: 6.25 : 3.29 : 2.66.
Slopes in (#) = - ; in (b) = - for segment a,
c 6 E c 81
and = - for segment ;
in (c) = f for segment a, and
E c 176
/ / 9 2 r
= for segment b.
EC 891
31. A horizontal beam rests upon two supports and is loaded with a
weight W at a point dividing the span into segments a and b. Find
the deflection at this point and the work done in bending the beam.
W
Ans -
a*t>* IV* a*P I W \
; = x deflection .
EI(a + b) 6EI a + * + 8a*6 - 2a* ) '
; rfv=*f \ T^a iog 'b + " ~6F~ : \
wl* i wl* 2wl*
'
43. Deduce the slope and deflection at the free end
(d) When the depth 2a in (a) of the preceding question is nil, i.e.,
when the profile is an isosceles triangle.
(e) Due to a uniformly distributed load of intensity p over the
cantilever (a). Hence, also, deduce the slope and deflection when the
depth 2a is nil.
(/) Due to a weight W at the free end of (a).
(g) Due to a uniformly distributed load of intensity p upon the
cantilever (<:).
Wl* Wl*
a (2? + $ a b - a*)(b - a)
)
I 90 '
3 pl*
~4~EtP'
^' g 3*-
b-a
2 EPt* 10
44. A cantilever of fength /, specific weight -w, and square in section,
a side of the section being 2b at the fixed and 2a at the free end, bends
under its own weight. Find the slope and deflection of the neutral axis
at the free end. Hence, also, deduce corresponding results when the
cantilever is a regular pyramid.
(b
45. If the section of the cantilever in the preceding question, instead
of being square, is a regular figure with any number of equal sides, show
that the neutral axis is a parabola with its vertex at the point of fixture.
46. The section of a cantilever of length / is an ellipse, the major axis
(vertical} being twice the minor axis. Find the deflection at the end
49^ THEORY OF STRUCTURES.
under a single weight W,f being the coefficient of working strength
and E the coefficient of elasticity. / 297 /V 5 \ J
Ans. --- m?7T
\7ooo h 3 W I
47. A cast-iron beam of an inverted T-section rests upon supports
22 ft. apart; the web is I in. thick and 20 in. deep; the flange is 1.2 in.
thick and 12 in. wide; the beam carries a uniformly distributed load of
99,000 Ibs. Find the maximum deflection, E being 17,920,000 Ibs.
Ans. .822. in. (/ = 1608.65).
48. Find the maximum deflection of a cast-iron cantilever 2 in. wide
x 3 in. deep x 120 in. long under its own weight, E being 17,920,000 Ibs.
Ans. H in.
49. A girder of uniform, strength, of length /, breadth b, and depth d,
rests upon two supports and carries a uniformly distributed load of w Ibs.
per unit of length, which produces an inch-stress of / Ibs. at every point
n 2 f\t b \\
of the material. Show that the central deflection is -- (~ / t
2 E \ 3y
when b is constant and ^variable. Find the deflection when d is con-
stant and b variable. fr
Ans. +.
4,Ed
50. A semi-girder of uniform strength, of length /, breadth b, and
depth d, carries a weight W at the free end which produces an inch-
stress of/ Ibs. at every point of the material. Prove that the maximum
deflection is - [ -) when b is constant and d variable, and that
3 E \6 W)
it is twice as great as it would be if the section were uniform throughout
and equal to that at the support.
What would be the maximum deflection if the semi-girder were
subjected to a uniformly distributed load of w Ibs. per unit of length ?
Ans. .
E J $w
51. The neutral axis of a symmetrically loaded girder, whose moment
of inertia is constant, assumes the form of an elliptic or circular arc.
Show that the bending moment at any point of the deflected girder is
inversely proportional to the cube of the vertical distance between the
point and the centre of the ellipse or circle.
52. A vertical row of water-tight sheet piling, 12 ft. high, is
supported by a series of uprights placed 6 ft. centre to centre and
securely fixed at the base. Find the greatest deviation of an upright
from the vertical when the water rises to the top of the piling. What
will the maximum deviation be when the water is 6 ft. from the top ?
wblf 3110400 wb wbc 218720
Ans. - = ~ : =r. (h - cY + =-,(* - cY = -~- .
3o7 El y>EI 24 El hi
EXAMPLES. 499
53. A vertical row of water-tight sheet piling, 30 ft. high, is supported
by a series of uprights placed 8 ft. centre to centre and securely fixed at
the base, while the upper ends are kept in the vertical by struts sloping
at 45. If the water rises to the top of the piling, find (a) the thrust on a
strut ; (d) the maximum intensity of stress in an upright ; (c) the amount
and position of the maximum deviation of an upright from the vertical.
iuh*
Ans. 45000 |/2 Ibs. ; max. B. M. = -- , and max. intensity of
stress = . 1 - - -7- -- - ; deflection is a max. when
A ( 10 / I5| / 5 j
h 30 7vh* 32
x = = =, and its amount = ^j ---- - .
V5 VS ^ 7SV5
54. The piling in the preceding example is strengthened by a second
series of struts sloping at 45 from the points of maximum deviation.
Find the normal reactions upon an upright and the bending moment at
its foot.
What will be the reactions and bending moment if the second row of
struts starts from the middle of the uprights?
Ans. .00754W/* 2 ; .lyjwk* ; .^2O27 i w/t 3 ; -^wk* ; If^rf* ; fH^// 9 .
55. A continuous girder of three spans, the outside spans being
equal, is uniformly loaded. What must be the ratio of the lengths of
the centre and a side span so that the neutral axis may be horizontal
over the intermediate supports? Ans. 4/T : y'T.
56. What should the ratio be if the centre span is hinged (a) at the
centre; (ff) at the points of trisection ? Ans. (a) 4/2" : i ; () 3 : 2 \/~ 2 .
57. Four weights, each of 6 tons, follow each other at fixed distances
of 5 ft. over a continuous girder of two spans, each equal to 50 ft. If the
second and third supports are i in. and i in., respectively, vertically
below the first support, find the maximum B. M. at the intermediate
support. / El \ ,
Ans. .9855 -- 1 ft. -tons.
\ 40000,1
58. A continuous girder of two equal 5o-ft. spans is fixed at one of
the end supports. The girder carries a uniformly distributed load of
loco Ibs. per lineal foot. Find the reactions and bending moments at
the points of support. How much must the intermediate support be
lowered so that it may bear none of the load ? How much should the
free end be then lowered to bring upon the supports the same loads as at
the first ?
Ans. Reactions = 23,214^, 57,142^, 19,642! Ibs. ;
Bending moments = 178,571^, 267,8574 ft. -Ibs;
500 THEORY OF STRUCTURES.
59. Four loads, each of 12 tons and spaced 5,4, and 5 ft. apart, travel
in order over a continuous girder of two spans, the one of 30 and the other
of 20 ft. Place the wheels so as to throw a maximum B. M. upon the
centre support, and find the corresponding reactions.
Draw a diagram of B. M., and find the maximum deflection of each
span.
60. The loads upon the wheels of a truck, locomotive, and tender,
counting in order from the front, are 7, 7, 10, 10, 10, 10, 8, 8, 8, 8 tons,
the intervals being 5, 5, 5, 5, 5, 9, 5, 4, 5 ft. The loads travel over a
continuous girder of two 5o-ft. spans AB, BC. Place the locomotive,
etc., (a) on the span AB so as to give a maximum B. M. at B\ (8) so as
to give an absolute maximum B. M. at B.
61. A continuous girder of two spans AB, BC has its two ends A
and C fixed to the abutments. The load upon AB is a weight P distant
/ from A, and that upon BC a weight Q distant g from C. The length
of AB = l\ , of BC /a . The bending moments at A, B, C are Mi , M* ,
Ms, respectively. The areas of the bending-moment curves for the
spans AB, BC assumed to be independent girders are^i, At, respect-
ively. Show that
Mill + M t (li + / a ) + M*h = - 2(Ai + A*),
and M*(li + /,) =
If /, = / 3 = /, show that Mi is a maximum if
2l(Pp - Qg) =
62. A continuous girder of two spans AB, BC rests upon supports at
A, B. A uniformly distributed load EF travels over the girder. G\ is
the centre of gravity of the portion BE upon AB, and G* that of the
portion./?/'' upon BC. If the bending moment at B is a maximum, show
that
AE.EB _ Ad
CF . FB ~ CG*'
63. An eight-wheel locomotive travels over a continuous girder of
two loo-ft. spans ; the truck-wheels are 6 ft. centre to centre, the load
upon each pair being 8000 Ibs. ; the driving-wheels are 8 ft. centre to
centre, the load upon each pair being 16,000 Ibs. ; the distance centre to
centre between the front drivers and the nearest truck-wheels is also
8 ft. Place the locomotive so as to throw a maximum B. M. upon the
centre support, and find the corresponding reactions.
64. If an end of a continuous girder of any number of spans is fixed,
show that the relation between the moment of fixture (Mi) and the
W/ 2
bending moment (M*) at the consecutive support, is 2Mi + M? = ,
4
or 2J/i + M* = ~^S[Pp(l p)(2l /)], according as the load upon
EXAMPLES. 5OI
the span (/) between the fixed end and the consecutive support is of
uniform intensity or consists of a number of weights Pi, P*, Pa, . . .
concentrated at points distant A, /, /s, . . . from the fixed end.
65. A continuous girder of two spans AB, BC, carrying a load of
uniform intensity, has one end A fixed, and the other end rests upon the
support at C. If the bending moments at A and B are equal, show that
the spans are in the ratio of 1/3 to f/2, and find the reactions at the
supports, Wi being the load upon AB, and W* that upon BC.
Ans. At A reaction =
B " =
C " =tJF..
66. A viaduct over the Garonne at Bordeaux consists of seven spans,
viz., two end spans, each of 57.375 m., and five intermediate spans, each
of 77.06 m. ; the mam girders are continuous from end to end, and are
each subjected to a dead load of 3050 k. per lineal metre. Determine the
absolute maximum bending moment at the third support from one end.
Also find the corresponding reactions, the points of inflexion, and the
maximum deflection in the first and second spans.
67. A continuous girder consists of two spans, each 50 ft. in length;
the effective depth of the girder is 8 ft. If one of the end bearings
settles to the extent of i in., find the maximum increase in the flange
and shearing stress caused thereby, and show by a diagram the change
in the distribution of the stresses throughout the girder. (Assume the
section of the girder to be uniform, and take E = 25,000,000 Ibs.)
Ans. Increase of maximum B. M. = s -Af[ ~ f i I,
\2i6rf /
" shearing force = ff/,
w being weight per unit of length, and /the moment of inertia.
68. A girder carrying a uniformly distributed load is continuous over
four supports, and consists of a centre span (/ 2 ) and two equal side
spans (/i). Find the ratio of l\ to / 2 , so that the neutral axis at the
intermediate supports may be horizontal. Also find the value of the
ratio when a hinge is introduced (a) at the middle point of the centre
span ; (fr) at the points of trisection of the centre span ; (c) at the middle
points of the half lengths of the centre span.
A I_i ^-! !_! A '-3
' /,- 3 ' / 2 2 ~2'/ 2 '- 9 ; ^~4'
69. In a certain Howe truss bridge of eight panels, the timber cross-
ties are directly supported by the lower chords, and are placed suffi-
ciently close to distribute the load in an approximately uniform manner
over the whole length of these chords, thus producing an additional
stress due to flexure. Assuming that the chords may be regarded as
girders supported at the ends and continuous over seven intermediate
5O2
THEORY OF STRUCTURES.
supports coincident with the panel points, and that these panel points
are in a truly horizontal line, determine (a) the bending moments and
reactions at the panel points ; (b) the maximum intermediate bending
moments; and (c) the points of inflection, corresponding to a load of w
per unit of length, / being the length of a panel.
Ans. (a) At i st support ; 2d support; 3d support;
B. M. = o ; - VW 2 ;
Reaction = R, = a// ; R* = ivl ; R z =
At 4th support ; 5th support.
B. M. = -^W 2 : -H^l\
Reaction = Ri= fffw/; Rs> f ffw/.
Maximum intermediate B. M. = - '-
in istspan ;
6208. 5
= W" / 4
(<:) Points of inflexion in the four spans are given by
* = = ^/; *i(/ + *)
W 388
x)
x)
-*0 2 = o ;
xf = o.
70. A. continuous girder of two equal spans \&fixefl $& one of the end
supports. The girder carries a uniformly distributed load of intensity
w. If the length of each span is /. find the reactions and moment of
fixture. How much must the intermediate support be lowered so that
it may bear none of the load ? How much should the free-end support
then be lowered to bring upon the supports the same loads as before ?
ii 16 13 w/ 2 c wl* 5 //*
wl tvl -
Ans. wl, wl, tvl\
28 14 28
14
24
71. Each of the main girders of the Torksey Bridge is continuous
and consists of two equal spans, each 130 ft. long. The girders are
double-webbed ; the thickness of each web plate is i in. at the centre
and f in. at the abutments and centre pier ; the total depth of the gir-
ders is 10 ft., and the depth from centre to centre of the flanges is 9 ft.
4 in.---FTtidX/?) the reactions at the supports, and also (b) the points of
inflexion, when 2oc\tpns of live load cover one span, the total dead load
upon each span being. 180 tons uniformly distributed. The top flange is
EXAMPLES.
503
cellular ; its gross sectional area at the centre of each span is 51 sq in ,
and the corresponding net sectional area of the bottom flange is 55 sq.
in. Determine (6-) the flange stresses and (d} the position of the neutral
axis. (7=372,500.) Also (e) determine the reactions when, first, B and,
second, Care lowered i in. (E '= 16,900 tons.)
Ans. (a) 155, 350, 55 tons.
() io6 r ^ and 79! ft. from end support.
(c) 6 7 and 7.3 tons per sq. in. in loaded span; 1.13 and
1.22 tons per sq. in. in unloaded span.
(d) 58.3 in. from centre line of top flange.
(e) First. R, =
i EI
Second. Ri = 1 55 ~ ^TT
lEI
= 350- sTTi
I 7
- S ? + ;T-
lEI
= 350+--^-;
4 *
i EI
'=55-o
Where
}./
18625
11232
72. Two tracks, 6 ft. apart, cross the Torksey Bridge, and are sup-
ported by single-webbed plate cross-girders 25 ft. long and 14 in. deep.
If the whole of the weight upon a pair of drivers, viz., 10 tons, be directly
transmitted to one of these cross-girders, draw the corresponding shear-
ing-force and bending-moment diagrams (i) if the ends of the cross-
girder are fixed to the bottom flanges of the main girders ; (2) if they
merely rest on the said flanges. Find the maximum deflection of the
cross-girder and the work done in bending it, in each case.
Ans. (i)
- at 13.208 ft. from one end.
EI
Total work of flexure = .-- ft -tons.
73. A swing-bridge consists of the tail end AB, and of a span ffC, of
length i ft., the pivot being at B. The ballast-box of weight ?Fextends
over a length AD (= 2c ft.), and the weight of the bridge from D to B
is w tons per lineal foot. If DB = x, if p is the cost per ton of the
bridge, and if q is the cost per ton of the ballast, show that the total cost
is a minimum when x + c = (- ), and that the corresponding
weight of the ballast is wx( - i J
504 THEORY OF STRUCTURES.
74 Compare, graphically, the shearing forces and bending moments
along the span BC of the bridge in the preceding question when the
bridge is closed, with their values when the bridge is open. What pro-
vision should be made to meet the change in the kind of stress ?
75. Each of the main girders of a railway bridge resting upon two
end supports and five intermediate supports is fixed at the centre sup-
port, is 3 ft. deep throughout, and is designed to carry a uniformly dis-
tributed dead load of t ton and a live load of | ton per lineal foot. The
end spans are each 51 ft. 8 in. and the intermediate spans each 50 ft. in
the clear. Find the reactions at the supports. The girders are single-
webbed and double-flanged ; the flanges are 12 in. wide and equal in
sectional area, the areas for the intermediate spans being 13 sq. in. and
17 sq. in. at the centre and piers respectively. Find the corresponding
moments of resistance and flange stresses, the web being in. thick.
Ans. Reaction at istand 7th supports = I5HMM at 2C ^ an ^ 5th
supports = 43fff til; at 3d and 5th supports = 35|ffl ;
at 4th support = 3%'?/4\ 8 s tons.
At piers = 693 and flange stresses are 3.59 tons per sq. in.
at 2d support, 2.45 at 3d, and 2.83 at 4th.
At centre == 549 and flange stresses in istspan = 3.2 tons
per sq. in., in 2d = 1.3, and in 3d = 1.78.
76. A continuous beam of four equal spans carries a uniformly dis-
tributed load of w intensity per unit of length. The second support is
depressed a certain distance d below the horizontal, and the reaction at
the 2d support is twice that at the ist. Show that the reactions at the
ist, 2d, 3d, 4th, and 5th supports are in the ratio of the numbers 15, 30,
36, 34, and 1 3 ; find d. With this same value of d find the reactions when
one end infixed.
i wP
Ri =
77. A continuous girder of two equal spans (/) is uniformly loaded.
Show that the ends will just touch their supports if the centre support
w/ 4
is raised
78. If ^i , di , d 3 , di are respectively the deflections of the ist, 2d, 3d,
and 4th panel points in question 69, show that the bending moment at
the middle panel point (J/ 4 ) is given by
EXAMPLES. 505
79. A girder supported at the ends is 30 ft. in the clear and carries
two stationary loads, viz., 7 tons concentrated at 6 ft. and 12 tons at 18
ft. from the left support. Find the position and amount of the maxi-
mum deflection, and also the work of flexure. The girder is built up of
plates and angle-irons and is 24 in. deep. If the moment of resistance
due to the web is neglected, and if the intensity of the longitudinal stress
is not to exceed 5 tons per sq. in., _ what should be the flange sectional
area corresponding to the maximum bending moment.
Ans. Max. deflection = f fr 8 %(x 6) 3 ^g&x, where
x = 1 5.34 ft.
67161.6 ,
Work = ----;,_ ft.-tons.
J^LJ.
Sect, area = 10.32 sq. in.
80. Determine the work of flexure and the necessary flange sectional
area at the centre if the girder in the preceding question is subjected
to a uniformly distributed load of 40 tons, instead of the isolated loads.
1540000,
Ans. Work = - - ft. -tons; sect, area = 15 sq. in,
81. (a) The bridge over the Garonne at Langon carries a double
track, is about 695 ft. in length, and consists of three spans, AB, BC, CD.
The two main girders are continuous and rest upon the abutments at
A and D and upon piers at B and C. The effective length of each of
the spans AB, CD is 208 ft. 6 in., and of the centre span J3C 24.3 ft. The
permanent load upon a main girder is 1277 Ibs. per lineal foot, and the
proof load is 2688 Ibs. per lineal foot. Find the reactions at the sup-
ports (i) when the proof load covers the span AB ; (2) when the proof
load covers the span BC '; (3) when the proof load cover the spans AB
and BC '; (4) when the proof load covers the whole girder.
Draw shearing-force and bending-moment diagrams for each case.
(b) At the piers the web is in. thick and 18 ft. in depth, and each
flange is made up of four plates in. thick and 3 ft. wide. Determine
the flange stresses for cases (i) and (3).
(c) The angle-irons connecting the flanges with the web at the pier
are riveted to the former with i^-in. rivets and to the latter with i-in.
rivets. How many of each kind are required in one line per lineal foot
on both sides of the pier at B, 8000 Ibs. per square inch being safe
shearing stress ?
(d) The effective height of the pier at B is 41 ft., its mean thickness
is 14 ft. 9 in., its width is 42 ft. 9 in., and it weighs 125 Ibs. per cubic
foot. If there is no surcharge on the bridge, and if the coefficient of
friction between the sliding surfaces at the top of the pier is taken at
.15, show that the overturning moment due to the dilatation of the
girders is about -fa of the amount of stability of the pier.
506 THEORY OF STRUCTURES.
(e) Find the points of inflexion and also the maximum deflections in
Case 3.
What practical advantage is derived from the calculation of the
deflection ?
Ans.(a) Case i. R\ = 353469.95; R? = 656955.7;
Rs = 280612.55 ; Ri = 109608.77 Ibs. ;
Mi= 12247115.3; M 3 = 4823424.5 ft.-lbs.
Case 2. R, = 68185 2 = R, Ibs. ;
A> 2 = 6791783 = ^? 3 Ibs.;
M*= 13439537.7 ft.-lbs. = Aft.
Case 3. RI = 312982.65 ; 7? 2 = 1024035 ;
Rs = 647691 ; R t = 69121.47 Ibs. ;
Mi= 1565031.2 ; Aft = 8226621.2 ft.-lbs.
Case 4. .#1=422591.42 =y? 4 lbs.;
7? 2 = 1304647. 55 = R 3 Ibs. ;
J/ 2 = 15455566 = J/a ft.-lbs.
(0) /= 2130816; in case i,/ 2 = 7448.9 Ibs. per sq. in.
/ = 2 933-6 " " " "
in case 3,/ 2 = 9400.3 " " " "
73 = 5003.5 " " " "
(Weakening effect of rivet-holes in tension flange is
neglected.)
(?) 9.1 per lineal foot; 11.5 pe^r lineal foot.
(d) Moment of stability = 23833291^11 ft.-lbs. ;
overturning moment = 1919408.8 ft.-lbs.;
ratio = 12.4.
(e) Points of inflexion: in AB, 157.8 ft. from A', in BC,
at a distance x from B given by x* 2581^+10426^
= o; in CD, at 54.1 ft. from D.
Max. deflections:
In AB, (165/5 jr 4 5 2l6 3-7* 8 + 227693091.6*),
where* is given by 66o|* 2 156491. 3* + 227693091. 6=0 ;
In BC, 7(1 65.2** 853829*" + 9977485. 9* 2 10327286968),
xS/
where * is given by * 2 3876* + 30196^ = o.
82. A beam AB of span / carrying a uniformly distributed load of
intensity w is fixed at A and merely supported at B. The end B is
w/ 4
lowered by an amount -r-=- r . Find the reactions. How much must B
ioA/
be lowered so that the whole of the weight may be borne at A?
i w/ 4
Ans. f|w/ at A, -&wl at. ^ ; - .
8 L1
EXAMPLES.
83. Solve the preceding question supposing the fixture at A to be
imperfect, the neutral axis making with the horizontal an angle whose
tangent is -7^7. Ans. %wl, \wl\ -~ -777.
40 L1 4 '*
84. A wrought-iron girder of I-section, 2 ft. deep, with flanges of equal
area and having their joint area equal to that of the web, viz., 48 sq. in.,
carries ^ ton per lineal foot, is 100 ft. long, consists of five equal spans,
and is continuous over six supports. Find the reactions when the third
support is lowered |- in. How much must this support be lowered so
that the reaction may be nil at (a) the ist support ; (b) the 3d ; (c) the
5th ? How much must the support be raised so that the reaction may
be nil at (d} the 2d ; (e) the 4th ; and (/) the 6th support ? E = 16,500
tons.
Ans. R^ = 2 If ; 7? 2 = 1 5f > ^ 8 = STF
j? 4 = Hff ; Rs = 9 T *V ; -#6 = 4 T 5 tons.
() if in.; mff in.; (c) 2ft in. ;
(d) ijft in. ; 00 2^ in. ; (/) 6* in.
85. If the three supports of any two equal consecutive spans of a
continuous girder of any number of spans are depressed below the
horizontal, show that the relation between the three bending moments
at the supports will be unaffected if the depression of the centre support
is a mean between the depressions of the other two supports.
86. A girder consists of two spans AB, BC, each of length /, and is
continuous over a centre pier B. A uniform load of length 2a (< /) and
of intensity W travels over AB. Find the reactions at the supports for
any given position of the load, and show that the bending moment at
aivl f a"* \ i
the centre pier is a maximum and equal to --.[-I -s] when the
3 1/3 V rj
centre of the load is at a distance I j from A.
87. A continuous girder rests upon three supports and consists of
two unequal spans AB (= /,), BC (= / a ). A uniform load of intensity iv
travels over AB, and at a given instant covers a length AD (= r) of the
span. If R lt R 3 are the reactions at A and C, respectively, show that
2 _ / 2 3 T r *
Draw a diagram showing the shearing force in front of the moving
load as it crosses the girder.
88. If the live load in the preceding question may cover both spans,
show that the shearing force at any point D is a maximum when AD
and BC SLVQ loaded and BD unloaded.
Illustrate this force graphically, taking into account the dead load
upon the girder.
508 THEORY OF STRUCTURES.
89. A continuous-girder bridge has a centre span of 300 ft. and two
side spans, each of 200 ft. The dead load upon each of the main girders
is 1250 Ibs. per lineal foot. In one of the side spans there is also an
additional load of 2500 Ibs. per lineal foot upon each girder. Find the
reactions and points of inflexion. How much must the third support
from the loaded end be lowered so that the pressure upon it may be just
zero ?
Ans. Let W = weight on loaded span = 750,000 Ibs.
R, = -ftfc W Ibs. ; Ri = |ff I W Ibs.;
R, = -/WV ^ Ibs. ; Rt = tffr W Ibs.
M, = \V/- W ft.-lbs.; M*= ViV- ^ ft. -Ibs.
Distance of point of inflexion in loaded span from nearest
end support = i62ff ft.
Distance of point of inflexion in unloaded end span from
nearest end support = I45ff ft.
Distance of point of inflexion in intermediate span from
end support in unloaded span is the value of x in the
equation x* f p,r + AJU^QJUL _ o
56350000 W
3d support must be lowered a distance = - ^ -- .
87 El
90. A continuous girder AC consists of two equal spans AB, BC, each
of length /, and carries a uniformly distributed load of intensity w\ upon
AB, and of intensity / 2 upon BC. Determine the bending moments at
the supports, the maximum intermediate bending moments, and the re-
actions (a) when both ends of the girder are fixed ; (b) when one end A
is fixed and the other free.
Ans. Denoting the reactions and bending moments at A, B, C
by Ri , Mi , y? 2 , Mi , R s , M 3 , respectively :
r / 2
(a) Mi = ( 50/1 + w a ) ; M 9 = -- (wi + /a);
48 24
r RS
M 3 = (wi 5^2); M max . in AB -- 1- Mi , in BC
48 2Wi
* r>
= 4-
2W 2 ID
R* = (o/i + 90/2).
ID
/ 2 / a
(b) Mi = -- 3(3^1 o/O; Mi = ^-(o/i + 20/2);
28 28
/? a /? 2
M 3 = o ; M max . mAB = + M lt in BC =
22/i 20/
I20/a).
EXAMPLES. 509
91. In the preceding question, if Wi = iv* = w, find the points of in-
flexion and the maximum deflection in each case and for each span.
Ans. (a) Points of inflexion for AB or BC are given by
6.r>-6;r/ + / 2 = o.
Max. deflection for AB or BC is given by
Ely- (2tx-x*-r),
in which the value of x is found from
(b) Points of inflexion in AB are given by
I4.r 2 i3.r/ + 2/ 2 = o, and in BC by x =
Max. deflection for AB is given by
- Ely = -==-
I DO
and
28.T 2 39/JT + I2/ 2 =0.
Max. deflection for BC is given by
and
28* 3 33.^ + 4/ 3 = o.
92. A continuous girder ^4C consists of two equal spans AB, BC of
15 m. each. Determine the bending moments at the supports, the maxi-
mum intermediate bending moments, and the reactions (a) when the
load upon each span is 3000 k. per metre ; (&) when the load per metre is
3000 k. upon AB and 1000 k. upon BC. Call M\ , M*, M* the bending
moments and R\ , Ri , R 3 the reactions at A, B, C, respectively, and con-
sider three cases, viz., when both ends of the girder are free, when both
ends are fixed, and when one end is free and the other fixed.
Ans. Case I :
(a) M! = o = M 9 ; M* = 84375 km. ; M max . in AB or BC
47460.9375 km.
fii = R 3 = I6875/& ; 7? 2 = 56250^.
(&) Mi = o = M 3 ; M* = 56250 km. ; M ma x. in AB
= 5 8 593-75 km - in BC 7031.25 km.
R l = 18750 k. ; y? 2 = 37500 k.; 7? 8 = 3750 k.
Case II:
(a) Mi = M-! = M = 56250 km.; M max . in AB or BC
= 28125 km.
r>
7?i = - = 7? 3 = 22500 k.
510 THEORY OF STRUCTURES.
(b) Mi 65625 km. ; M* = 37500 km. ; M 3 = 9375
km. M max . in AB = 33398.4375 km., in BC
= 6445.3125 km.;
/?! = 24375 k.; R* = 30000 k.; R 3 = 5625 k.
Case III :
(a) Mi = 48214!- km.; M 9 = 72321$ km.; M s = o.
M ma . v . in AB= 24537111 km., in BC = 52o88-j-f| km.;
R, = 20892^ k.; 7? 2 = 51428^ k.; R 3 = 482 if k.
(b} Mi = 64285! km.; Mi = 401 78$ km.; M, = o.
Mmax. in AB = 32573-^5- km., in BC = 11623-^
km. ;
A'i = 24107! k.; Ri = 3 107 1 } k., R 3 = 4821! k.
93. Show that a uniformly loaded and continuous girder of two equal
spans, with both ends fixed, is 2.08 times as stiff as if the ends were free
and merely rested on the supports.
94. A single weight travels over the span AB of a girder of two equal
spans, AB, BC, continuous over a centre pier B. Show that the reaction
at C is a maximum when the distance of the weight from A is -~ if the
ends A and C rest upon their supports, and when the distance is \AB if
the two ends are fixed. Find the corresponding bending moments at
the central pier.
PJ ">
Ans. --- -PL
95. A girder with both ends fixed carries two equal loads W at points
dividing the girder into segments a, b, c. Determine the reactions and
bending moments at the supports.
+ babe + 3#V + 2 (= o) the
bending moments.
(a) XT. = 20.72 m. ;
x* is given by I7oojr 2 2 424017^2 + 395089! = o;
x z by I7oo.r 3 2 47767^3 + 238839! =o ; jr 4 = 19.38 m.
l?i = 38348 T V k.; R* = 81 i6of k. ; y? 3 = 341071 k.;
R, = 499 I0 f k. ; ^ = 16473^ k.
M-i 197544^ km. ; M 3 = 5357if km. ;
km.
$12 THEORY OF STRUCTURES.
(b) x l = 19.556 m. ; 37ocxr 2 2 103000^3 + 50357^ = o;
I700.T3 2 41071^3 + 205357^ = o ; XL = 20. 168 m.
7?! = 361787 k. ; ^ 2 = 107821^ k.; fi s = 62964! k.;
/? 4 = 45892^ k. ; ^ 5 = i7H2f k.
M* = 258928^ km. ; Ms = 120535!- km.;
M t = 102678^ km.
(e) xi = 19.64 m. ; -r 2 and jr a are given by
i4^ 2 -375.r-f 1875 = 0.
Mt Mi = 247767% km. ; M a = 165178^ km.
Abs. max. B. M. at 2d support (= max. B. M. at 4th sup-
port) occurs when ist, 2d, and 4th spans are loaded,
and = 264508^1 km.
Abs. max. B. M. at 3d support occurs when 2d and 3d
spans are loaded and = 209821^ km.
100. In the preceding question find the absolute maximum flange
unit stress at the piers, / being .093929232444. Ans. 4.5 k. per sq. mm.
101. The Osse iron viaduct consists of seven spans, viz., two end
spans of 28.8 m. and five intermediate spans of 38 m. ; each main girder
is continuous and carries a dead load of 1450 k. per lineal metre. Find
the bending moments at the supports when a proof load of 2250 k. per
lineal metre for each girder covers all the spans; and also find the
absolute maximum bending moment at the fourth support. Is the
following section of sufficient strength ? two equal flanges, each com-
posed of a 6oo-mm. x 8-mm. plate riveted by means of two loo-mm.
x ico-mm. x i2-mm. angles to a 6oo-mm. x lo-mm. vertical web plate
and two 8o-mm. x 8o-mm. x ii-mm. angles riveted to each horizontal
plate with the ends of the horizontal arms 15 mm. from the edges of
the plates; the whole depth of the section being 4.016 m., and the dis-
tance between the web plates, which is open, being 2.8 m. If insuf-
ficient, how would you strengthen it?
Ans. My = 416,518 km.; Ms = 452,790 km.; M^ = 443,722 km.
Max. B. M. = 542,199 km. /= .14074440467.
.-. = .07009183,
cM*
and max. flange stress = = 7.73 k. per sq. mm.
This is much too large. The section may be strengthened
by adding two 6oo-mm. x 8-mm. plates to each flange.
/ is thus increased by .0783425536, and the flange unit
stress becomes 5 k. per sq. mm.
CHAPTER VIII.
PILLARS.
1. Classification. The manner in which a material fails
under pressure depends not merely upon its nature but also
upon its dimensions and form. A short pillar, e.g., a cubical
block, will bear a weight that will almost crush it into powder,
while a thin plank or a metal coin subjected to enormous com-
pression will be only condensed thereby. In designing struts
or posts for bridges and other structures, it must be borne in
mind that such members have to resist buckling and bending in
addition to a direct pressure, and that the tendency to buckle
or bend increases with the ratio of the length of a pillar to its
least transverse dimension.
Hodgkinson, guided by the results of his experiments,
divided all pillars with truly flat and firmly bedded ends into
three classes, viz. :
(A) Short Pillars, of which the ratio of the length to the
diameter is less than 4 or 5 ; these fail under a direct pressure.
(B) Medium Pillars, of which the ratio of the length to the
diameter exceeds 5, and is less than 30 if of cast-iron or tim-
ber, and less than 60 if of wrought-iron ; these fail partly by
crushing and partly by flexure.
(C) Long Pillars, of which the ratio of the length to the
diameter exceeds 30 if of cast-iron or timber, and 60 if of
wrought-iron ; these fail wholly by flexure.
2. Further Deductions from Hodgkinson's Experi-
ments. A pillar with both ends rough from the foundry so
that a load can be applied only at a few isolated points, and a
pillar with a rounded end so that the load can be applied only
513
514 THEORY OF STRUCTURES.
along the axis, are each one-tJiird of the strength of a pillar of
class B, and from one-third to two-thirds of the strength of a
pillar of class C, the pillars being of the same dimensions.
The strength of a pillar with one end flat and the other
round is an arithmetical mean between the strengths of two
pillars of the same dimensions, the one having both ends flat
and the other both ends round.
Disks at the ends of pillars only slightly increase their
strength, but facilitate the formation of connections.
An enlargement of the middle section of a pillar sometimes
increases its strength in a small degree, as in the case of solid
cast-iron pillars with rounded ends which are made stronger
by about one-seventh ; hollow cast-iron pillars are not affected.
The strength of a disk-ended pillar is increased by about one-
eighth or one-ninth when the middle diameter is lengthened by
50 per cent., but for slight enlargements the increase is imper-
ceptible.
The strength of hollow cast-iron pillars is not affected by a
slight variation in the thickness of the metal, as a thin shell is
much harder than a thick one. The excess above or deficiency
below the average thickness should not exceed 25 per cent.
3. Form. According to Hodgkinson, the relative strengths
of long cast-iron pillars of equal weight and length may be
tabulated as follows :
(a) Pillars with flat ends.
The strength of a solid round pillar being 100,
" " square " is 93 ;
" " triangular " is 1 10.
(#) Pillars with round ends, i.e., ends for hinging or pin
connections.
The strength of a hollow cylindrical pillar being 100,
" " an H-shaped . " is 74.6;
" " a +-shaped " is 44.2.
The strengths of a long solid round pillar with flat ends,
and a long hollow cylindrical pillar with round ends, are ap-
proximately in the ratio of 2.3 to I.
The stiff cst kind of wrought-iron strut is a built tube, the
THE FAILURE OF PILLARS. 515
section consisting of a cell or of cells, which may be circular,
rectangular, triangular, or of any convenient form.
In experimenting upon hollow tubes, Hodgkinson found
that, other conditions remaining the same, the circular was the
strongest, and was followed in order of strength by the square
in four compartments j+j ; the rectangle in two compartments,
FT! ; the rectangle, a ; and the square.
The addition of a diaphragm across the middle of the rect-
angle doubled its resistance to crippling.
4. Modes of Failure. The manner in which the crush-
ing of short pillars takes place depends upon the material, and
the failure may be due to splitting, bulging, or buckling.
(a) Splitting into fragments is characteristic of such crys-
talline, fibrous, or granular substances as glass, timber,
stone, brick, and cast-iron.
The 'compressive strength of these substances is
much greater than their tensile strength, and when they
fail they do so suddenly.
A hard vitreous material, e.g., glass or vitrified
brick, splits into a number of prisms (Fig. 335).
A fibrous material, e.g., timber, and granular materials, e.g.,
cast-iron and many kinds of stone and
brick, shear or slide along planes oblique
to the direction of the thrust, and form
one or more wedges or pyramids (Figs.
336, 337, 238)-
Sometimes a granular or a crystalline substance will sud-
denly give way and be reduced to powder.
(b) Bulging, i.e., a lateral spreading out, is characteristic of
blocks of fibrous materials, e.g., wrought-iron, copper, lead, and
timber, and fracture occurs in the form of longitudinal cracks.
All substances, however, even the most crystalline, will
bulge slightly before they fail, if they possess some degree of
toughness.
(c) Buckling is characteristic of fibrous materials, and the
resistance of a pillar to buckling is always less than its resist-
ance to direct crushing, and is independent of length.
Thin malleable plates usually fail by the bending, pucker-
THEORY OF STRUCTURES.
ing, wrinkling, or crumpling up of the fibres, and the same
phenomena may be observed in the case of timber and of long
bars.
Long plate tubes, when compressed longitudinally, first
bend and eventually fail by the buckling of a short length on
the concave side.
The ultimate resistance to buckling of a well-made and
well-shaped tube is about 27,000 Ibs. per square inch section of
metal, which may be increased to 33,000 or 36,000 Ibs. per
square inch by dividing the tube into two or more compart-
ments.
A rectangular wrought-iron or steel tube offers the greatest
resistance to buckling when the mass of the material is con-
centrated at the angles, while the sides consist of thin plates
or lattice-work sufficiently strong to prevent the bending of
the angles.
Timber offers about twice the resistance to crushing when
dry that it does when wet, as the presence of moisture dimin-
ishes the lateral adhesion of the fibres.
5. Uniform Stress. Let a short pillar be subjected to a
w | pressure of W Ibs. uniformly distributed over its
end and acting in the direction of its axis.
Let 5 be the transverse sectional area of the pil-
lar.
W
Let / = be the intensity of stress per unit
O
of area of any transverse section AB.
TIG. 339. J
Let A'B' be any other section of area S', in-
clined to the axis at an angle 0. The intensity of stress per
W W
unit of area of A'B' = = sin 6 = p sin 0, which may be
o o
resolved into a component/ sin 2 normal to A'B', and a com-
ponent/ sin 6 cos 6, i.e.,/ - , parallel to A'B'. The last
intensity is evidently a maximum when 6 = 45, so that the
plane along which the resistance to shearing is least, and there-
fore along which the fracture of a homogeneous material would
tend to take place, makes an angle of 45 with the axis.
UNIFORM L Y'VAR YING STRESS.
517
None of the materials of construction are truly homo-
geneous, and in the case of cast-iron the irregularity of the
texture and the hardness of the skin cause the angle between
the plane of shear and the direction of the thrust to vary
from 32 to 42. Brick chimneys sometimes fail by the shear-
ing of the mortar, the upper portion sliding over an oblique
plane.
Hodgkinson's experiments upon blocks of different mate-
rials led him to infer that the true crushing strength of a ma-
terial is obtained when the ratio of length to diameter is at
least i-J-; for a less ratio the resistance to compression is un-
duly increased by the friction at the surfaces between which
the block is crushed.
6. Uniformly Varying Stress. The load upon a pillar is
rarely, if ever, uniformly distrib-
uted, but it is practically sufficient
to assume that the pressure in
any transverse section varies uni-
formly.
Any variable external force ap-
plied normally to a plane surface
AA of area 5 may be graphically
represented by a cylinder AABB,
the end BB being the locus of the
extremities of ordinates erected
upon A A, each ordinate being pro-
portional to the intensity of press-
ure at the point on which it is
erected.
Let P be the total force upon *AA, and let the line of its
resultant intersect A A in C\ C is the centre of pressure of A A,
and the ordinate CC necessarily passes through the centre of
gravity of the cylinder.
Again, the resultant internal stress developed in AA is P,
and may of course be graphically represented by the same
cylinder AABB.
Assume that the pressure upon AA varies uniformly; the
surface BB is then a plane inclined at a certain ano-le to AA.
FIG. 340.
5 i8
THEORY OF STRUCTURES.
Take 0, the centre of figure of AA, as the origin, and AA
as the plane of x, y.
Let O y, the axis of y, be parallel to
that line EE of the plane BB which is
parallel to the plane AA.
Through EE draw a plane DD par-
allel to AA, and form the cylinder
A ADD.
The two cylinders A ABB and A ADD
are evidently equal in volume, and OF,
the average ordinate, represents the mean
pressure over AA ; let it be denoted
by A-
At any point R of the plane AA,
erect the ordinate RQP, intersecting the
planes DD, BB, in Q and P, respect-
Y/ ively.
FlG - 341- Let x, y be the co-ordinates of R.
The pressure at R
a being a constant depending upon the variation.
Note. The sign of x is negative for points on the left of O,.
and the pressure at a point corresponding to R\sp Q ax.
Let x n , y be the co-ordinates of the centre of pressure C.
Let AS be an elementary area at any point R.
Then pAS is the pressure upon AS, and 2(J>4S) is the
total pressure upon the surface AA, 2 being the symbol of
summation.
Hence,
and
But / = / + ax.
and
axy)AS\
UNIFORMLY VARYING STRESS.
519
Now is the centre of figure of AA, and therefore 2
and ~2(ydS} are each zero.
Also, 2(4 S) = S, 2(x*4S) is the moment of inertia (/) of
A A with respect to OY, and 2Z(xydS) is the product of inertia
(K) about the axis OZ.
.'. x p S = al x P
and
(l)
(2)
Cor. I. Ip any symmetrical
section y Q is zero, and X Q is the
deviation of the centre of pres-
sure C from the centre of fig-
ure a
Let x l be the distance from
O of the extreme points A of
the section.
The greatest stress in A A is/ + ax l = /\, suppose.
FlG " 342 '
A-!
or
^>
A
It is generally advisable, especially in masonry structures,
to limit x by the condition that the stress shall be nowhere
negative, i.e., a tension. Now the minimum stress is/ ax^ ,
so that to fulfil this condition,
p f > or = ax l , But p l = ax,
< or = 2/..
520 THEORY OF STRUCTURES.
Hence, by eq. (3),
A, i
and therefore
I
< or = I ; i.e., x, < or = -.
Cor. 2. The uniformly varying stress is equivalent to a
single force P along the axis, and a couple of moment
= a Vr + K\
Cor. 3. The line CO is said to be conjugate to OY.
If the angle COX = 0, then cotV = = ~.
7. Hodgkinson's Formulae for the Ultimate Strength of
Long and Medium Pillars. When a Jong pillar is subjected to
a crushing force it first yields sideways, and eventually breaks
in a manner apparently similar to the fracture of a beam under
a transverse load. This similarity, however, is modified by the
fact that an initial longitudinal compression is induced in the
pillar by the superimposed load.
Hodgkinson deduced, experimentally, that the strength of
long solid round iron and square timber pillars, with flat and
firmly bedded ends, is given by an expression of the form
W being the breaking weight in tons of 2240 Ibs. ;
d " " diameter or side of the pillar in inches ;
I " " length of the pillar in feet;
n and m being numerical indices ;
FORMULAE FOR ULTIMATE STRENGTH OF PILLARS. $21
A being a constant varying with the material and with the
sectional form of the pillar.
. For iron pillars ............... n = 3.6 and m = 1.7
" timber pillars ............ n = 4 and m = 2
" cast-iron ............................ A = 44.16
" wrought-iron ........................ A = 133.75
" dry Dantzic oak ..................... A 10.95
" dry red deal ......................... A 7.81
" dry French oak ...................... A 6*9
The strength of long hollow round cast-iron pillars was
found to be given by
- d
w -44.34 -r-
d being the external and d l the internal diameter, both in
inches.
Thus, the strength of a hollow cast-iron pillar is approxi-
mately equal to the difference between the strengths of two
solid cast-iron pillars whose diameters are equal to the external
and internal diameters of the hollow pillar.
The strength of medium pillars may be obtained by the
formula
w Wfs
-
W being the breaking weight in tons of 2240 Ibs. ;
W " " " " " " " " " as derived
from the formula for long pillars ;
/being the ultimate crushing strength in tons per square inch ;
S being the sectional area of the pillar in square inches.
Again, if the ends of a cast-iron pillar are rounded, the
above formulae may be still employed to determine its strength,
A being 14.9 for a solid and 13 for a hollow pillar.
522 THEORY OF STRUCTURES.
8. Gordon's Formula for the Ultimate Strength of a
Pillar. The method discussed in the preceding articles, being
practically very inconvenient, is not generally used,
and the present article will treat of Professor Gordon's
formula, which has a better theoretical basis and is
easier of application.
The effect of a weight W upon a pillar of length /
;2/ n and sectional area 5 may be divided into two parts :
(a) A direct thrust, which produces a uniform com-
W
pression of intensity -= = p l .
(b) A bending moment, which causes the pillar to
FIG. 343. yield in the direction of its least dimension (k).
Letjy be the greatest deviation of the pillar from
the vertical.
The bending moment M at the point of maximum stress
may be represented by Wy.
Let / 2 be the stress in the extreme layers due to this bend-
ing moment.
Now
c being the distance of the layer under consideration from
the neutral axis, ju. a constant depending upon the sectional
form, and b the dimension perpendicular to the plane of flexure.
Wv
and
Butj/cc. (Art. 9, Chap. VI.)!
wr wr / a
' and
VALUES OF a AND / GORDON'S FORMULA. $2$
a being some constant to be determined by experiment.
Hence, the total stress in the most strained fibre is
or
which is Gordon's formula.
Cor. If the weight- upon the pillar causes the stress in any
transverse section to vary uniformly, the direct thrust in the
W\ *2' S ] W
extreme layers is -~r\i -| * / instead of -~, (Cor. i, Art.
J 1 O
6,) ;F O being the greatest deviation of the line of resultant
thrust from the axis of the pillar.
Let k be the radius of gyration of the cross-section. Then
and the expression for the direct thrust may be written
W
Hence, Gordon's formula becomes
/
-
5 A
9. Values of a and /. The following table, giving the
values of the constants a and/* in Gordon's formula, has been
524
THEORY OF STRUCTURES.
prepared by taking an average of the best known results, and
is applicable to round and square pillars with square ends.
f in Ibs.
per sq. in.
a
For cast-iron solid rectangular pillars
80 ooo
-r\r*
" " " round "
80 ooo
ff
T JU
" " hollow rectangular "
80 ooo
TOU
*4
" " " round "
80 ooo
,ri*
For wrought-iron solid rectangular pillars . .
TTO
-JU.
" " " round "
36 ooo
^(TTJT
JL*
" " thick hollow round "
36 ooo
*^M
6? 2OO
^FffT
_JL-
" round " ',
67 2OO
^OtTTT
*JU.
67 2OO
TTTTTT
, 1,^.
For strong-steel solid rectangular pillars
1 14 ooo
^?oiy
d&
" " " round "
1 14 ooo
1400
JL-
" " hollow round "
114 ooo
^DTT
i .
5 ooo
T^ffTJ
_1I
" " round "
5 ooo
^^^
_1_
7 2OO
2s ff
_i
^0
If Gordon's formula is applied to pillars with pin ends, 4*2
takes the place of a ; and if to pillars with one pin end and one
square end, |^z takes the place of a.
10. Graphical Comparison of the Crushing Unit Strength
of Solid Round Cast-iron, Wrought-iron, and Mild-steel
Pillars.
The crushing unit stress is given by/ =
Take the different values of j as abscissae, and the corre-
sponding values of / as ordinates ; the resulting curves are
shown in Fig. 344.
Hence, the strength of a mild-steel pillar always exceeds
that of a wrought-iron pillar but is less than that of a cast-iron
pillar when j < 10.7 ; a wrought-iron pillar is stronger or weaker
than a cast-iron pillar according as r > or < 28.5.
APPLICATIONS OF GORDON'S FORMULA.
80,000 Ibs.
28,486.4 Ibs.
in
40 50 60
FIG. 344.
ii. Application of Gordon's Formula to Pillars of other
Sectional Forms.
In any section whatever, the least transverse dimension for
calculation (i.e., ti) is to be measured in the plane of greatest
flexure.
Thus, it may be taken as the least diameter of the rectangle
circumscribing tee (Fig. 345), channel (Fig. 346), and cruciform
(Fig. 347) sections, and as the perpendicular from the angle to the
opposite side of a triangle circumscribing angle(F\g. 348) sections.
FIG. 345 .
FIG. 34 6.
FIG. 347.
FIG. 348.
From a series of experiments upon wrought-iron pillars of
these sections, f was found to be 42,500 Ibs., and a, .
QOO
In cast-iron struts of a cruciform section /= 80,000 Ibs.
and a -.
400
526 THEORY OF STRUCTURES.
These results are only approximately true, and apply to
pillars fixed at both ends.
12. Rankine's Modification of Gordon's Formula. The
factor a in Gordon's formula is by no means constant, and not
only varies with the nature of the material, with the length of
the pillar, with the condition of its ends, etc., but also with the
sectional form of the pillar. The variation due to this latter
cause may be eliminated, and the formula rendered somewhat
more exact, by introducing the least radius of gyration instead
of the least transverse dimension.
If k is the least radius of gyration,
/ mbh* __ m
r* *
- / T
mass non n
m and n being constants which depend upon the sectional
form. Thus, Gordon's formula for pillars with square ends
may be written
.+:, j
in which , is independent of the sectional form, all variations
of the latter being included in k*. This modified form of
Gordon's formula was first suggested by Rankine.
4a l is substituted for a l if the pillar has two pin ends, and
Q
-a t or 2#, is substituted for a l if the pillar has one pin end and
one square end.
Rankine gives
for wrought iron, f =. 36000 Ibs., - = 36000;
for cast-iron, /= 80000 Ibs., - = 6400 ;
for dry timber, /= 7200 Ibs., - == 3000;
a,
KANKINE'S MODIFICATION OF GORDON'S FORMULA. $2?
In good American practice the safe working unit stress in
bridge compression members is determined by the formula
Safe working unit stress =
f being 8000 Ibs. for wrought-iron and 10,000 Ibs. for steel,
and being 40,000 for two square ends, 30,000 for one square
and one pin end, and 20,000 for two pin ends.
Another formula often employed is,
H\ . /'
/ Tf\
Working stress in Ibs. persq. in. X (4 + ) ==
H being the ratio of length to least breadth, where, in the case
of wrought-iron,
f = 38,500 Ibs. and = 5820 for two square ends;
f 38,500 " " - = 3000 " one square and one pin end.
/' = 37,800 " " - = 1900 " two pin ends.
r_r
^\\e factor of safety, viz., 4 -| , increases with H, and par-
tially provides for the corresponding decrease in the strength
to resist side blows.
EXAMPLES. According to Rankine the ultimate compres-
sive strength of wrought-iron struts, in pounds per square
inch, is
W 36000
36000 *
528 THEORY OF STRUCTURES.
If the section is a solid rectangle, k? = , and hence
36000
r
1 3000 h*
If the section is a solid circle, / 2 = -^, and hence
36000
A=
7-.-
1 2250 tf
If the section is a thin annulus, k* = -^ nearly, and hence
36000
1 +4500^
_If - is small, W=fS.
If is large, ^. ,.
Comparing the last result with eq. (5), Case 4, Art. 16,
which gives a theoretical value of a t , the actual value being
somewhat different.
13. Values of fc a for Different Sections.
(a) Solid rectangle : k* = -~ = , h being the least dimen-
O 1 2
sion.
tx\ if n i v 7 I ( bh * - b ' h '*\
(b) Hollow rectangle: k = -^ = 7^ bh b'h'r ' g
the greatest and least outside dimensions, and b', h' the great-
est and least inside dimensions, respectively.
VALUES OF &* FOR DIFFERENT SECTIONS. $2$
Let / be the thickness of the metal. Then
b' = b 2t and h' = h 2t,
and hence
I bh*-(b-2i)(h-2t}* V
~ 12 bh (b 2t)(h 2t)' 12 b + h>
approximately, when t is small compared with h, i.e., for a thin
hollow rectangle.
For a square cell, J? = -g-.
(c) Solid triangle : ft* = -~- = =, h being the height.
(d] Hollow triangle : 1? = -^- = - , . ^ , 7 . , , ^, ^ being the
base and height of the outside triangle, and ', h' the base and
height of the inside triangle, respectively. Also, T> = 77.
_ ^ P - V* i__tf IP + b f *\
- k ^i$F -&'*?- is\ y r
Hence, for a thin triangular cell, &* = .
r ;r2
(e) Solid cylinder : * = = ;, k being the diameter.
(/) Hollow cylinder: ff = ^ = ~^( A ' + h '^ h and h ' bein g
the external and internal diameters, respectively.
Hence, for a thin cylindrical cell, & = , approximately.
EXAMPLE. Gordon's formula for hollow cylindrical cast-
iron pillars is
w__ _L _ _ _ /_
~ " a>
.
h 3
500 / 3 4000
530 THEORY OF STRUCTURES.
The relation /, = -- - may be assumed to hold for
hollow square struts and also for struts of a cruciform section.
Ex. i. For a hollow square having its diagonal equal to
the internal diameter of the hollow cylinder, i.e., k ',
=' and ' =
Ex. 2. If the side of the square is equal to the external
diameter, i.e., h, then
k* = -z- , and /, =
6 ' ^ ~ 3 /' '
(g) Cruciform section, the arms being equal:
_
/ = h ; S zbk ^ .
12 ' 12 12
~ = ' nearly -
FIG. 349.
Hence, the formula for a cast-iron pillar of cruciform section
may be written
(/z) Angle-iron of unequal ribs, the greater being and the
less A :
VALUES OF &* FOR DIFFERENT SECTIONS.
53 1
ra
Hence, if b = h, i.e., if the ribs are equal, ft? .
(i) Channel-iron, the dimensions being as in Fig. 350 :
2bht\h + tY
T
12 ~ 4(2At + 6t)
2ht . 2bhf
( 2t
= h \ 71 + 4
Also, 5 = bt + 2ht.
=] IF
' nearly '
_ _ ___
1 2(2 At + bt) 4(2 At + bt}* ) '
Let the area of the two flanges A = 2ht, and let the
area of the web =. B bt. Then
_ -
12(A
(fc) \\-iron, breadth of flanges being b, length of web h, and
thickness of metal / :
73j 1 .3 73.
/ = 2~ + = 2--, nearly ; 5=
At.
*' /P ~ I22bt
2bt
I2A+B'
A being the area of the flanges, and .5 the area of the web.
(/) Circular segment, of radius r and length rO :
Hence, for a semicircle, since = ?r,
532 THEORY OF STRUCTURES.
(m) Barlow rail: tf = -- , nearly.
(n) Two Barlow rails, riveted base to base: a = .393;-*,
nearly.
14. American Iron Columns. In 1880 Mr. G. Bouscaren
read before the American Society of Civil Engineers a paper
containing the results of a series of experiments made for the
Cincinnati Southern Railroad upon Keystone, square, Phoenix,
and American Bridge Co.'s columns.
EYSTONE SQUARE PHOENIX ' AM. BRIDGE CO.
FIG. 351. FIG. 352. FIG. 353. FIG. 354.
These experiments show, as those of Hodgkinson and
others have also shown, that the strength of iron and steel
columns is not only dependent on the ratio of length to diam-
eter, and on the forfh of the cross-section, but also on the
proportions of parts, details of design and workmanship, and
on the quality of the material of which the columns are con-
structed.
Further, they seem to lead to the conclusions that Gordon's
formula is more correct as modified by Rankine, and that, in
the case of columns hinged at both ends, Rankine's formula,
with tfj assumed at double the value it has when the formula is
applied to columns with flat ends, is practically correct.
The subjoined table gives the values of the constants
#, and / as deduced from Bouscaren's experiments by Prof.
W. H. Burr.
In 1 88 1 Messrs. Clarke, Reeves & Co. presented to the
American Society of Civil Engineers a paper containing the
results of experiments upon twenty Phcenix columns, which
appeared to show that neither Gordon's nor Rankine's formula
expressed the true strength of a column of the Phcenix type.
In the discussion that followed the reading of this paper, how-
ever, it was demonstrated that, within the range of the experi-
ments, the strength of intermediate lengths and sections of
AMERICAN IRON COLUMNS.
533
/ in Ibs.
For keystone columns with flat ends swelled
" " " " " " straight (open or
closed)
" " " " " " open (swelled straight)
" pin ends swelled
For square columns with flat ends
" " " " pin ends
For Phoenix columns with flat ends
" " '* " round ends
" " " " pin ends
For American Bridge Co.'s columns with flat ends
" " " " " " round ends
" " 4< " " " pin ends
36,000
39>500
38,300
38,300
39,000
39.000
42,000
42,000
42,000
36,000
36,000
36,000
18300
1
18300
1
12000
1
35000
1
17000
1
50000
1
12500
1
22700
1
46000
1
11500
1
21500
Phoenix columns can be obtained either from Rankine's for-
mula by slightly changing the constants, or from very simple
new formulae.
Mr. W. G. Bouscaren showed that by making a. =
i ooooo
W
and/= 38000, the calculated values of -^- agree very nearly
o
with the actual experimental results.
Mr. D. J. Whittemore gave the following (only applicable
for lengths varying from 5 to 45 diameters) as expressing the
probable ultimate strength of these columns :
Wlbs. = (1200 /T)3
H being the ratio of length to diameter.
Mr. C. E. Emery stated that the ultimate strength in each
case is approximately represented by the formula
M/1K 355Q63 + 3095Q#
H+6.i 7 $ -'
H being the ratio of length to diameter.
534 THEORY OF STRUCTURES.
Taking the different values of H as abscissae, and of W as
ordinates, this is the equation of an hyperbola. It agrees very
accurately with the experimental results from 20 diameters
upwards; at 15 diameters the calculated values of W arc
greater than those given by the experiments ; for a less num-
ber of diameters the experimental results are the higher, but
the variations are slight, and are provided for in the factor of
safety.
The following very simple formulae, due to Prof. W. H.
Burr, give results agreeing closely with those obtained in the
experiments :
For values of j < 30, the ultimate strength in pounds per
square inch
= 64700 4600. IJL
V k '
For values of T between 30 and 140, the ultimate strength
in pounds per square inch
= 39640 46 j,
k being the radius of gyration.
15. Long Thin Pillar. Let ACB be the bent axis of a
thin pillar of length /, having two pin ends and carry-
ing a load W at B.
Let d be the greatest deviation of the axis from the
vertical. Then
E
Wd bending moment = ^,-7, . . (i)
being the curvature of the pillar and 7 the moment
K
v of inertia of the most strained transverse section.
FJG. 355 . This equation is only true on the assumptions that
(1) initially, the pillar is perfectly straight ;
(2) initially, the line of action of the load coincides with the
axis of the pillar ;
LONG THIN PILLAR. 535
(3) the material of the pillar is homogeneous.
These assumptions cannot be fulfilled in practice, and varia-
tions from theoretical accuracy may, perhaps, be provided for
by supposing that the line of action of the load is at a small
distance x from the axis of the pillar. The bending-moment
equation then becomes
f^ being the skin stress due to bending at a distance c from the
neutral axis.
Again, assuming that the bent axis is in the form of an arc
of a circle,
f ='/!, ... . . (4)
and consequently
Wx
d =-pI^W> ' (5)
where
' "'^ ' " ''' ''^" P =^f- (Q
If the line of action of the load W coincided with the axis
of the pillar, then x would be nil.
Hence, by eq. (5), so long as the load is less than P, d o,
and the failure of the pillar would be due to direct crush-
ing. If the load is equal to P, d would become indeterminate
I ) and the pillar would remain in a state of neutral equi-
librium at any inclination to the vertical.
It is impossible that W should exceed P, as d would then
be negative ; and therefore a load greater than P would cause
the pillar to bend over laterally until it broke.
536 THEORY OF STRUCTURES.
Thus, P ~ JT- must be the theoretical maximum compres-
sive strength of the pillar.
Again, let A be the area of the section under consideration ;
" p be the total intensity of the skin stress at the
section ;
" f be the intensity of the direct stress due to W
W_
~~ A ;
" /", be the intensity of the stress due to P
P
Then
the sign of f^ being positive for the compressed side of the
pillar and negative for the side in tension.
(8)
k being the radius of gyration.
Let h be the least transverse dimension of the section in
the plane of flexure. Then
c oc h and k also a h.
c _n
*' fi = ~h '
n being a coefficient depending upon the form of the section.
For a rectangle, n = 6 ; for a circle, n = 8 ; also,
Px
LONG THIN PILLAR. 537
Thus, however small x may be,/ continually increases as
the difference between f l and f diminishes. The pillar will
therefore fail for some value of / less than the theoretical
maximum. This is in accordance with experience, as it is
found that a small load causes a moderate flexure in a long
pillar, and that this flexure gradually increases with the load
until fracture takes place.
In no case should / exceed the elastic limit, as in such
case a set would be produced and the deviation x would be
increased.
If the tensile strength of the material of the pillar is small,
as in the case of cast-iron, failure may arise from the tearing of
the stretched layers.
Cor. i. The above also applies to the case of a pillar with
one end fixed and the other free, but the value of P is
2EI
then -^- .
Cor. 2. According to Euler (see following article), the
7T 2
more correct value of P is pEIjt, ^ being I, 2, J, or 4, accord-
ing as the pillar has two pin ends, one fixed end and one end
guided in the direction of the thrust, one fixed and one free
end, or two fixed ends.
El & fh^
P evidently a ^ a EAj$ a -A\j
Hence, (a) the strength of a long pillar is proportional to
the coefficient of elasticity ; (b) the strengths of similar pillars
are as the sectional areas.
Again, f t oc -^ oc d.
But Wd = --/ a / 2 a d.
Hence W is approximately constant, and the weight which
produces moderate flexure is approximately equal to the break-
ing weight.
EXAMPLE. Find the crushing load of a solid mild-steel
pillar 3 in. in diameter and 10 ft. long, with two pin ends.
THEORY OF STRUCTURES.
Also find the deviation (x) of the line of action of a load of
20,000 Ibs. from the axis of the pillar, so that the maximum
intensity of stress may not exceed 10,000 Ibs. per square inch.
By Gordon's formula and the table, page 524,
the crushing load = --r- Viiw = 85292.3 Ibs.
~T TTTFTV -3 y
Again, the theoretical maximum compressive strength P
8 X 28000000 n(tf
~5T 6l875lbs -
/, 61875 99
* P- w /,-/ 41875 67*
Hence
20000 /
10000 = -3-3-
or x = .65 in.
16. Long Columns of Uniform Section. (Ruler's Theory.)
CASE I. Columns with both ends hinged.
f The column OA of length / is bent
[k , under a thrust P and takes the curved
form OMA.
Take O as the origin, the vertical
through O as the axis of x, and the hori-
zontal through O as the axis of y.
Consider a section at any point M
(x,y). If there is equilibrium and if the
line of action of P coincides with the
TT~" axis of the column, the equation of mo-
FIG. 35 6. ments at M is
; :; : ; . : - EI & = M =**> ::.::.,'_,
or
d'y P
LONG COLUMNS OF UNIFORM SECTION. 539
dy
Multiplying each side of the equation by - and integrating,
(2)
b being a constant of integration.
dy adx
Integrating,
sm-'.l-J =
or
y = b sin (ax + c), . . . . (3)
c being a constant of integration.
When x = o, y is also o, and hence b = o or c = o.
If b = o, y is always o, and lateral flexure is impossible.
Take c = o. Then
jj/ = sin ax (4)
Also, when x OA = OMA, nearly, = /, y = o.
.*. o = b sin al,
or
and hence
P=n*El"t (5)
Now the /ft&rf value of P evidently corresponds to n = i r
and hence the minimum thrust which will bend the column
laterally is
540 THEORY OF STRUCTURES.
Cor. I. If the column is made to pass through N points
dividing the vertical OA into N + I equal divisions, then
y = o when x =
and therefore, by eq. (4),
al
or
al
and hence
i) 3 .
FIG. 357.
As before, the least value of P corresponds to n = I, and
is the least force which will bend the column laterally.
Hence, the strength of the column is increased in the ratio
of 4, 9, 16, etc., by causing it to pass through points which divide
its length into 2, 3, 4, etc., equal parts, respectively.
Cor. 2. The value of b may be approximately determined
as follows : \
Let ds = length of element at M.
Let = inclination to vertical of tangent at M.
Then
pressure upon ds P cos 6 = P-r-
and the
. ., S , ,
compression of ds = ds = -=r-;dx,
A being the sectional area of the column.
Hence, the total diminution of the length of the column
!'
. 7 />
: / rf*__ /.
LONG COLUMNS OF UNIFORM SECTION. 541
Again, the length of the column
= f'( l + ^^) dx = f (l + a ' 6 ' cos ' ax} ' dx '
= / ( l H -- cos2 ax\dx, approximately,
Hence, if L is the initial length of the column, i.e., the
length before compression,
and consequently
EIL-l\ I
b = 2 ~
CASE 2. Columns with one end fixed and the other constrained
to lie in the same vertical.
Assume that the lateral deviation is prevented ^
by means of a horizontal force H at the top of a
column. Then
TO ! \
M
-*).. . . (I)
MW
A particular solution of this is
I
,, - FIG. 358.
^ y d u
54 2 THEORY OF STRUCTURES.
and eq. (i) becomes
or
d*u
-&=-* *, *VV (2)
The solution of the last equation is
y / = u = b sin (ax + c), . . : , ' " v" . (3)
b and c being constants of integration.
TT
(4)
dy
But - = o when x = o,
and j = o when ^r = o and when x /.
rr
.* 0= , -\-abcosc\
TT
O = sn
Hence
<2/-|- ^ = o and al = tan c = tan at,
and therefore
= 4-493
= / V /'
which may be written in the form
L
/
(5)
LONG COLUMNS OF UNIFORM SECTION.
It is sufficiently approximate to write
El
543
P= 27T 5
CASE 3. Columns with one end fixed and the other free.
A rigid arm AB is connected with the
free end A of a column, and a vertical B
force P applied at B bends the column
laterally, until its axis assumes the curved
form OMA.
Let AB q, A C = p, and let / be the
length of the column, OC, nearly.
The inclination of AB to the horizon
is so small that the difference in length
between AB and its horizontal projection
may be disregarded. The moment equa-
tion at any point M (x, y) is
or
(i)
dy
Multiplying each side by 2 and integrating,
b being a constant of integration.
dy
But -T- = o when y = o, and hence b = o.
(2)
or
544 THEORY OF STRUCTURES.
Integrating,
c being a constant of integration, or
4) 4- a y
But y = o when x = o, and hence c = o.
.*. p = cos^. .<,;..... . . (4)
Also, y ~ p when ^- = /.
.*. -: cos a/. (5)
If q is very small or nil, the term ~rq~ may be disregarded,
and then
o cos al.
being a whole odd number.
The least value of P corresponds to n = i, and the minimum
pressure which will cause the column to bend laterally is
Cor. I. By eq. (5) the deviation of the top of the column
from the vertical is
i cos al
LONG COLUMNS OF UNIFORM SECTION. 54$
Cor. 2. Let the force applied at B be oblique and let its.
vertical and horizontal components be P and //, respectively.
The moment equation now becomes
A particular solution of this is
*) ..... (10)
Let y = y 1 -f- u.
Substituting in eq. (9),
or
The solution of this equation is
u = b sin (ax -f- c) y y' t
b and c being constants of integration.
TT
(12)
When x = o, / and ^ are each = o ; and when x = /, y = q+
TT
Hence, o =/ + ?+ /-(- ^ sin ^;
rr
O = ^ -\- al> cos c ;
546 THEORY OF STRUCTURES.
three equations giving b, c, and p, and therefore fully deter-
mining y.
CASE 4. Column with both ends fixed.
Let IJL be the end moment of fixture. Then
or
-*> + * = (*- JO, . . (l)
,
where b = -p.
FW. 360. Multiplying each side of the equation by 2^ and
integrating,
d being a constant of integration.
But -- = o when ^ o, and hence d
y*) ...... (2)
or
I Integrating,
or
-^ = cos (** + <:), (3)
c being a constant of integration.
LONG COLUMNS OF UNIFORM SECTION. 547
But jj/ = o when x o and when x I. Hence
i = cos c and I = cos (al-\- c).
and therefore c = o and al =
n being a whole number. Hence,
or
The least value of P corresponds to n = I, and the mini-
mum thrust which will cause the column to bend laterally is
17. Remarks. From the preceding it appears that the
maximum theoretical compressive strength of a column per
unit of area may be expressed in the form
k being the radius of gyration, and A a coefficient whose value
is i, 2, J, or 4, according as the column has two hinged ends,
one end fixed and the other guided in the direction of thrust, one
end fixed and the other free, or two fixed ends.
This formula is easy of application, but Hodgkinson's
experiments show that the value of P as derived therefrom is
too large. This may be partly due to the assumption that the
elasticity of the material is perfect.
The factors of safety to be used with this formula vary
from 4 to 8 for iron and steel and from 4 to 15 for timber.
The objection to the use of flat bars as compression mem-
bers has sometimes been overestimated.
Consider, e.g., the case of a flat bar hinged at both ends.
THEORY OF STRUCTURES.
Let the coefficient of elasticity of the material be 25,000,-
ooo Ibs.
Let the working stress per square inch be 8000 Ibs.
The bar will not bend laterally under pressure so long as
7T
the unit stress < Etft, and
jrV /
8000 < 25000000- 7 f , or -7 < 50.7.
Hence, the length of a flat bar in compression seems to be
comparatively limited. If, however, both ends are securely
fixed, the strength is quadrupled and the admissible length of
bar is doubled, while it may be still further increased by fixing
the bar at intermediate points as indicated in Corollary i, page
540. This shows the marked advantage to be gained by rivet-
ing together the diagonals of lattice-girders at the points where
they cross each other.
p
The value of f . . r (Art. 15) must not exceed the elastic
./i
limit. It is difficult to define with any degree of accuracy the
elastic limit of cast-iron and timber. It is claimed, indeed, that
the latter has no elastic limit, properly so called, but that a
permanent set is produced by every elastic change of form. It
may be assumed, however, that the elasticity of these materials
is practically unaffected so long as they are not loaded to more
than one half of the ultimate crushing load.
Hence, taking
E = 29,000,000 Ibs. and f = 20,000 Ibs. for wrought-iron r
" = 29,000,000 " " /= 33,600" " soft steel,
E 29,000,000 " " / = 56,000 " " hard steel,
=17,000,000 " " / = 40,000 " " cast-iron,
= 1,500,000 " " /= 3,600 " " dry timber,
the pillars will not bend laterally unless the ratio of -> or
a 2r
LONG COLUMNS OF UNIFORM SECTION. 549
(d being the shortest side of a rectangular section and r the
radius of a circular section) exceeds the values given in the
following table :
Material. Value of -4- Value of . Formula.
a 27"
Wrought-iron 34.5 29.9
Soft steel 26.6 24.3
Hard steel 20.3 17.9 } f = ' =
Cast-iron 18.7 16.2
Dry timber 18.5 16
Wrought-iron 48.8 42.3
Soft steel 37-7 34-4
Hard steel 28.8 25.3 V ^ _ _
Cast-iron 26.4 22.9
Dry timber 26. i 22. 7
Wrought-iron 17.2 14.9
Soft steel 13.3 12. i
Hard steel 10.1 8.9
Cast-iron 9.3 8.1
Dry timber 9.2 8
Wrought-iron 69 59.9
Soft steel 53.3 48.7
Hard steel 40.7 35.7 \ f ^.
Cast-iron 374 32.4
Dry timber 37 32
Baker has deduced by experiment the following formulae
for the strength of wrought-iron and steel pillars of from 10 to
30 diameters in length and with fixed ends, the tensile strength
of the metals ranging from 20 to 60 tons (2240 Ibs.) per square
inch :
Let t be the tensile strength of the iron or steel, and H the
ratio of length to diameter.
Then the ultimate compressive resistance, in pounds per
square inch,
for solid round pillars = (.4 .oo6Ii)(t + 18) ;
for thin tubes = (.44 .OO4//X/ + 18) ;
for tubes with stiffening ribs = (.44 .OO2//)(V 4- 18) ;
for girder sections = (.4 - .oO4//)'/-f- 18).
55O THEORY OF STRUCTURES.
18. Weyrauch's Theory of the Resistance to Buckling.
In order to make allowance for buckling, Weyrauch pro-
poses the two following methods :
METHOD I. Let F l be the necessary sectional area, and .
the admissible unit stress for a strut subjected to loads vary-
ing from a maximum compression B l to a minimum com-
pression By
Let/ 7 ' be the necessary sectional area, and b' the admissible
unit stress for a strut subjected to loads which vary between a
given maximum tension and a given maximum compression,
B' being the numerically absolute maximum load, and B" the
maximum load of the opposite kind.
According to Art. 7, Chap. Ill, if there is no tendency to
buckling,
* = !=T-^IH ..... c>
1 4+-.f) - ,. .
and
"
If there is a tendency to buckling, let / be the length of
the strut, F its required sectional area, and T the mean unit
.stress at the moment of buckling.
Then, according to the theory of long struts,
(3)
d being a coefficient depending upon the method adopted for
securing the ends, E the coefficient of elasticity, and / the
least moment of inertia of the section.
Also, let t be the statical compressive strength of the ma-
terial of the strut, and take t = j*T. Then
RESISTANCE TO BUCKLING WE YRA UCH'S THEORY. 55 1
where
552 THEORY OF STRUCTURES.
and
n/ D/
F ,, = - ~~\> if B " is a compression. (9)
If /* < I, equations (i) and (2) give larger sectional areas
than equations (7), (8), and (9), so that the latter are to be ap-
plied only when /* > I.
METHOD II. General formulae applicable to all values of /*
may be obtained by following the same line of reasoning as
that adopted in the proof of Gordon's formula. It is there
assumed that the total unit stress in the most strained fibre is
72\
A ( l +*!/ A Dem g tne stress due to direct compression, and
r
p^a-j^ that due to the bending action.
So, instead of employing equations (i) and (2) when ju < i,
and equations (7), (8), and (9) when ^ > i, formulae including
all cases may be obtained by substituting for the compressive
forces in equations (i) and (2) their values multiplied by I
Thus, equation (i) becomes
and equation (2) becomes
F= 7 g?7 {, if B' is a compression, (n)
V '(,_ m '
or
B'
com P ress i n -
Equations (7), (8), (9), respectively, give larger values of F
than the corresponding equations (10), (u), and (12).
RESISTANCE 7V BUCKLING IVEYRAUCH' S THEORY. 553
Note. For wrought-iron bars it may be assumed, as in Arts.
5, 6, Chap. Ill, tha't z/, = z/ == 700 k. per sq. cm., and m^ = m
=1.
The value of i, and by eq. (7) the required sectional area is
/^ X 1.188 = AGO. x IsI 88 _ 5 7<9 sq> cm<
Second. F, = Af& = ^(r t f r 2 2 ),
r, being the external and r 2 the internal radius of the section.
554 THEORY OF STRUCTURES.
Let r l 9 cm. and r, = 7.92 cm. Then
*(r? ~ O - 5743 sq. cm.
Also,
' I r' + rf" 143-7264
Hence, by eq. (4),
_ 360X360 4 .
24000 143,7264
Thus, in the latter case, since ft < I, there is no tendency
to buckling.
If the area is determined by equation (10), its value becomes
1.15 X $JL = 65 sq. cm.
19. Flexure of Columns. In Art. 16 the moment equa-
tion has been expressed in the form
and this is sufficiently accurate if the deviation of the axis of the
strut from the vertical is so small that [-T-] may be neglected
without sensible error.
The more correct equation is
p being the radius of curvature.
Consider, e.g., the strut in Art. 16, Case I. Then
P i dO dO
-j^- T y = - = = -j- sin 6,
Er p ds dy
FLEXURE OF COLUMNS. 555
being the inclination of the tangent at M to the axis of *,
and ds an element of the bent strut at M.
... _ tfydy sin Odd.
Integrating,
, ..... (i)
being the value of at a strut end.
a
Let sin = p and sin - = p sin 0. Then
or
y = cos ( 2 )
^
Let Y be the maximum deviation of the axis of the strut
from the vertical, i.e., the value of y when = o or = o.
Then
2 sin ~
-- 2 - ......... (3)
a
Again,
i dB
a Vi // sin 2
Hence, if /be the length of the strut,
d 2 ,
(4)
.F M (0) being an elliptic integral of the first kind.
Let P be the least thrust which will make the strut bend.
As shown in Art. 16,
556 THEORY OF STRUCTURES.
and, by eq. (4), the corresponding value of the modulus yu is
given by
F^}=- 2 (5)
*
Let the actual thrust on the strut be
P=n'P', .-..- . (6)
H* being a coefficient > unity.
The corresponding value of the modulus is given by
By reference to Legendre's Tables it is found that a large
n
increase in the value of //, i.e., of sin or # , is necessary in
order to produce even a small increase in the value of
and therefore of n{ A / f I. Hence, as soon as the thrust
(-V?)
P exceeds the least thrust which will bend the column, viz.,
P', rapidly increases.
The total maximum intensity of stress in the skin of the
strut at the most deflected point
P MB P PYz
z being the distance of the skin from the neutral axis, and /
p
being equal to r.
A
The last term of this equation includes the product fE,
n
which is very large, and also the factor sin , which increases
with # so that the ultimate strength of the material is rapidly
approached, and, in fact, rupture usually takes place before the
column has assumed the position of equilibrium defined by the
slope # at the ends.
FLEXURE OF COLUMNS. 557
If there were no limit to the flexure, the column would
take its position of equilibrium only after a number of oscilla-
tions about this position, and the maximum stress in the
material would be necessarily greater than that given by
eq. (8).
Again,
I (i 2;*' sin 2 0X0
dx = ds cos 6 = .
a V I p sin a
Let X be the vertical distance between the strut ends. Then
0) being an elliptic integral of the second kind.
Hence, the diminution in the length of the strut
20. Flexure of Columns (Findlay). In a paper on the
flexure of columns read before the Canadian Society of Civil
Engineers (Vol. IV, Part I), Findlay expresses the moment
equation in the form
p and being the values of p and when M = o.
558 THEORY OF STRUCTURES.
Hinged Ends. It is assumed that the line of action of the
thrust P is at a distance d from the axis of the strut. Then
'. - w
or
where a* = -~j, p = total stress at the distance z from the
neutral axis, and /= stress due to direct thrust f -jj, so that
.the stress due to bending = p f.
It is also assumed that the form of the axis of the column
before it is acted upon by the thrust P, is a curve of sines
defined by the equation
nx
.y. = Jcos T ,........ (4)
the'origin being half-way between the ends of the strut, and A
being the maximum initial deviation of the axis from the ver-
tical, i.e., the value of y^ when x = o.
d*y An* nx
.*. r-^ = j$- cos -y,
and hence, by eq. (3),
^2.- V 4-d\- A-
A solution of this equation is
nx
, cos ax . f COS T~
cos-
FLEXURE OF COLUMNS. $59
Now - - is always small for such values of f as would con-
stitute a safe working load, and therefore
al aT
cos - i --, approximately,
so that eq. (6) becomes
COS
or
-f
5
i G8x*dA z , . . . Hence, the total moment of resist-
ance of the section
= G6I,
I being the moment of inertia with respect to the axis.
But this moment of resistance (M) is equal and opposite to
the moment of the couple (P,P). Hence,
M=GOI=Pp.
The twisting moment will of course vary with a variable
resistance, and the last equation gives its mean value.
The shaft, however, must be designed (see Cor. 4) for the
maximum couple to whi<*h it may be subjected, and the moment
of this couple (= J/,) may be expressed in terms of the mean
by the equation
ju. being a coefficient to be determined in each case. In a series
of experiments with different engines, Milton found that //
varied from 1.3 to 2.1, but doubtless the variation is often be-
tween still wider limits.
Cor. i. Let /be the stress at the point farthest from the
axis. For a solid round shaft, of diameter D,
and = G9 .
3 2
Let T be the total torsion in degrees. Then
i nT ,
z
J^OR SIGNAL STRENGTH OF SHAFTS.
and hence
D
" L 1 80 2'
or
_
D~ fi.60
Taking the following mean values of G
Material. G f
Cast-iron .............. 6,300,000 5,6oo
Wrought-iron .......... 10,500,000 7,200
Steel .................. 12,000,000 1 1,200
2j = 9.8T for cast iron, = 12.7^ for wrought-iron, = 9.3^
for steel.
Thus, the twist is i each 9.8 diameters in length for cast-
iron, each 12.7 diameters in length for wrought-iron, and each
9.3 diameters in length for steel. This is often much too small,
and in practice the twist is usually limited to T ^- per lineal foot
of length. For a hollow round shaft, D being the external and
D l the internal diameter,
If the thickness (7") of the hollow shaft is small compared
with D,
& - Z> 3 4 = D' - (D - 2Ty = SD*T, approximately,
\
and
5/2 THEORY OF STRUCTURES.
The use of compressed steel admits of shafts being made
hollow. For a solid square shaft, H being the side of the square,
IT
and/, the stress at the end of a diagonal, = G6-=-
H 6
and
GI GH*
In these results it is assumed that G0[ or = =r~ I
\ D H /
is constant at different points of the cross-section, which, how-
ever, is only true for circular sections.
In non-circular sections the stress is more generally greatest
at points in the bounding surface which are nearest to the axis
and least at those points which are farthest from the axis.
St. Venant, who first called attention to this fact, gave the fol-
lowing, amongst others, as the results of his investigations.
Designating by unity the torsional rigidity \^= j-j of a shaft
with circular section, the torsional rigidity of a shaft of equal
/ 271 /~fr
sectional area is .8863, .8863 X \ / n * . l , .7255, or \ / -, ac-
cording as the section is a square, a rectangle with sides in
the ratio n to I, an equilateral triangle, or an ellipse whose
major and minor axes are 2a and 2b, respectively.
Cor. 2. The torsional stress per unit of area- at a distance x
from the axis is GOx.
Hence, if 9 i and x = i, G is the force that will twist a
TORSIONAL STRENGTH OF SHAFTS. 573
unit of area at a unit of distance from the axis through an angle
unity.
Cauchy found analytically that in an isotropic body G is
two-fifths of the coefficient of direct elasticity.
Experiments indicate that G is about three-eighths or one-
third of the coefficient of direct elasticity.
f -f (1 r>4
Cor. 3. For a solid cylinder, Pp = - , R being the radius,
and therefore R* oc -^T> If the shaft is to have a certain speci-
Gv
fied stiffness, i.e., if is fixed, R* oc -?r, and for a given twisting
moment R* oc . Now G is nearly the same for wrought-iron
G
and steel, so that there is little if any advantage to be gained
by the use of the latter.
After passing the elastic limit, the stress varies much more
slowly than as the distance from the axis, and there will be a
partial equalization of stress, the apparent torsional strength
being increased.
Cor. 4. In any transverse section of a solid cylindrical shaft,
the maximum unit stress
M l being the moment of the maximum twisting couple.
This relation is true so long as the stress does not exceed
the elastic limit, and agrees with the practical rule that the
diameter of a cylindrical shaft subjected to torsional forces is
proportional to the cube root of the twisting couple.
The rule is usually expressed in the form
M, = KD\ so that K = ^.
Wohler's experiments show that the value of f depends,
to some extent, upon its fluctuation under the variable twist-
574 THEORY OF STRUCTURES.
ing moment. Ordinarily it should not exceed 7200 Ibs. per square
inch for wrought-iron, in which case K i\ J^ X --f- = I4H-
(Note. If P l is the torsional breaking weight,
_
" D* ~~ I} 3
is the coefficient of torsional rupture?)
Cor. 5. Let* Wbe the work transmitted to a shaft of D in.
diameter, in foot-pounds per minute, N being the correspond
in- number of revolutions. Then
M
12 W = inch-pounds transmitted = 2nMN 2it TV
KD
M l
since M = mean twisting moment = -. Hence,
w
Let HP be the horse-power transmitted per minute. Then
W = 33000 HP. Also for wrought-iron K i| x - ? f.
Hence p -- -^- = D\ and if p = 1.43,
HP
N
-i formula agreeing with the best practice in the case of
wrought-iron shafts subjected to torsional forces only. Such
shafts should, therefore, carry no pulleys.
Cor. 6. The resilience of a cylindrical axle is the product of
one half of the greatest moment of torsion into the correspond-
ing angle of torsion.
Cor. 7. It often happens in practice that a shaft (or beam)
is subjected to a bending as well as to a torsional action.
DISTANCE BETWEEN THE BEARINGS OF SHAFTING. $?$
The combined bending and twisting moments are equiv-
alent (Art. 8, Chap. IV) to the moment
where M b = nM t , M b being the bending and M t the twisting
moment at the given section.
Hence, remembering that the maximum twisting moment
J/, is equal to pM tJ we have for a wrought-iron shaft,
If n = .36 -f- , this becomes
HP
a formula agreeing with the best practice in the case of trans-
mission with bending, as, e.g., in the crank-shafts of marine
engines.
It often happens that n has a still larger value, as, e.g., in
the case of head shafts properly supported against springing.
The usual formula is then
D
3 I HP
= 5V W'
corresponding to n = .72 -{- .
4. Distance between Bearings. The distance between
the bearings of a line of shafting is limited by the considera-
tion that the stiffness of the shaft must be such as will enable
it to resist excessive bending under its own weight and under
any other loads (e.g., pulleys, wheels, etc.) applied to it.
For this reason, the ratio of the maximum deviation of the
axis of the shaft from the straight to the corresponding dis-
tance between bearings should not exceed a certain fraction
whose value has been variously estimated by different writers.
Let / be the distance in feet between bearings, a the
diameter of the shaft in inches, w the weight of the ma-
576 THEORY OF STRUCTURES.
terial of the shaft per cubic foot, and let the applied load be
equivalent to a load per lineal unit of length m times that
of the shaft. Assume a stiffness of T ^Vir an ^ tna t the axis of
the shaft is truly in line at the bearings. The maximum de-
flection of the shaft is given by the formula (Art. 3, Ex. 8.
Chap. VII)
_ I (m + i)(weight of shaft)/ 3 . 1728
- ~
I nd* i 64 / 3 . 1728
-s-UH + l) -- tt>/-~r - ~ - .
384 v 4 144 ar*/ 4 ^
/ " IOO 2E
or
_ s / Ed*
~ V 5 ow ( m +
EXAMPLE. For wrought-iron, E = 3o,ocxD,ooo Ibs. and
w 480 Ibs.
If the applied load, instead of being uniformly distributed
is concentrated at the centre, the maximum deflection
i (m + j)( weight of shaft)/ 3 . 1728
~ 192 El ~~'
and hence
~, Ed'
[QOw(m -f- i
EXAMPLE. For wrought-iron /= 8.5
CYLINDRICAL SPIRAL SPRINGS.
577
5. Efficiency of Shafting. Let it require the whole of
the driving moment to overcome the friction in the case of a
shaft of diameter d and length L. The efficiency of a shaft of
the same diameter and length / = I -= .
But
,. = (Pp) = moment of friction = /* L
w being the specific weight of the material of the shaft, and
the coefficient of friction. Hence,
w
wl
and the efficiency = I 2yw--.
Hence,
S= 2nyn, approx.
Wy =
fnr* GQnr*
2 2
r being the radius of the spring.
6. Cylindrical Spiral Spring. Let the figure represent
a cylindrical spiral spring of length s, supporting
a weight W. Consider a section of the spring at
any point B.
At this point there is a shear W^and a torque
Wy, y being the distance of B from the axis of
the spring, i.e., the radius of the coil.
The effect of W may generally be neglected
as compared with the effect of the moment Wy,
and it may be therefore assumed that the spring
is under torsion at every point. Let there be n
coils. Then
FIG. 363.
578 THEORY OF STRUCTURES.
The elongation of the spring
Syf 2nfny*
__
: ~~
Wy a Wy'S fnr'S
The work done OS = -
A weight hung at the lower end tends to turn as well as
lengthen the spring, and this is due to a slight bending action.
According to Hartnell, f =. 60,000 Ibs. per square inch for
f-in. steel, f = 50,000 Ibs. per square inch for -J-in. steel, and
G varies from 13,000,000 Ibs. for J-in. steel to 11,000,000 Ibs.
for f-in. steel.
Also for wire less than f in. in diameter,
Wny*
and the deflection
EXAMPLE. A wrought-iron shaft in a rolling-mHl makes 50
revolutions per minute and transmits 120 H. P., which is sup-
plied from a waterfall by means of a turbine. Determine the
diameter of the shaft (i) if the maximum stress in the metal
is not to exceed 9000 Ibs. per square inch ; (2) if the angle of
torsion is not to exceed ^ per lineal foot.
As a matter of fact, the diameter of the shaft is 3 j in. at
the bearings and 4 in. in the intermediate lengths. What are
the corresponding maximum inch-stresses in the metal?
Let the twisting couple be represented by force P at the
end of an arm /. Then
P X 27tp x 95 = 120 X 33,ooo ft.-lbs.
120 X 33000 126000 , 126000
.% Pp = ll4-_ _ _ _ ft.-lbs. = X 12 in.-lbs.
27r X 95 19 J 9
126000 X 12
First, - -
and hence
EXAMPLE. 579
, 126000
Second, -- X 12 = Pp =
But 6 -- ~ X y- X ~ ; take G = io,5OO,cxx). Then
126000 X 12 10500000 22 I I I 22
~ T 9 -- = 2 x 7 X 785 x n X 77 x 7^-
Hence,
Z^ 4 =689.45, and D= 5.12 in.
Third, the maximum stresses in the real shaft at the bear-
ings and in the intermediate lengths are respectively given by
126000 stress 22
_ XI2 = __ X _
and
126000 stress 22
^- X 12, -^- Xy X(4).
From the former, the maximum stress = 7682 Ibs. persq. inch.
" latter, " " =6330 " " " "
580 THEORY OF STRUCTURES.
EXAMPLES.
1. A steel shaft 4 in. in diameter is subjected to a twisting couple
which produces a circumferential stress of 15,000 Ibs. What is the stress
(shear) at a point I in. from the centre of the shaft ?
Determine the twisting couple. Ans. 7500 Ibs.; 23,57 if Ibs.
2. A weight of 2| tons at the end of a i-ft. lever twists asunder a
steel shaft if in. in diameter. Find the breaking weight at the end of a
2-ft. lever, and also the modulus of rupture.
Ans. 1 1 tons ; 23, 510 Ibs.
3. A couple of A^ft.-tons twists asunder a shaft of diameter d. Find
the couple which will twist asunder a shaft of the same material and
diameter 2.d. Ans. &N.
4. Compare the couples required to twist two shafts of the same
material through the same angle, the one shaft being / ft. long and
d in. in diameter, the other 2/ ft. long and id in. in diameter.
Compare the couples, the diameter of the latter shaft being .
Ans. i to 8 ; 32 to I.
5. A shaft 1 5 ft. long and 4^ in. in diameter is twisted through an angle
of 2 under a couple of 2000 ft.-lbs. Find the couple which will twist
a shaft of the same material 20 ft. long and i\ in. in diameter through
an angle of 2^. Ans. 12,288 ft. Ibs.
6. A round cast-iron shaft 15 ft. in length is acted upon by a weight
of 2000 Ibs. applied at the circumference of a wheel on the shaft ; the
diameter of the wheel is 2 ft. Find the diameter of the shaft so that the
total angle of torsion may not exceed 2. Ans. 3.53 in.
7. A wrought-iron shaft is subjected to a twisting couple of 12,000 ft.-
lbs. ; the length of the shaft between the sections at which the power is
received and given off is 30 ft.; the total admissible twist is 4. Find
the diameter of the shaft, ju (page 570) being f , and m 10,000,000 Ibs.
Ans. 7.74. in.
8. A wrought-iron shaft 20 ft. long and 5 in. in diameter is twisted
through an angle of 2. Find ihe maximum stress in the material, m
being 10,500,000 ft.-lbs. Ans. 3819.2 ibs. per sq. in.
EXAMPLES. 58l
9. A crane chain exerts a pull of 6000 Ibs. tangentially to the drum
upon which it is wrapped. Find the diameter of a wrought-iron axle
which will transmit the resulting couple, the effective radius of the drum
being 7^ in. ; the safe working stress per square inch being 7200 Ibs.
Ans. 3.17 in.
10. Find the diameter and the total angle of torsion of a 12-ft.
wrought-iron shaft driven by a water-wheel of 20 H. P., making 25
revolutions per minute, m being 10,000,000 Ibs., and the working stress
7200 Ibs. per square inch. Ans. 5.6 in.; 2. 2.
n. A turbine makes 1 14 revolutions per minute, and transmits 92
H. P. through the medium of a shaft 8 ft. 6 in. in length. What must be
the diameter of the shaft so that the total angle of torsion may not ex-
2
ceed , m being 10,500,000 Ibs. ? Ans. 4.7 in.
Determine the side of a square .pine shaft that might be substituted
for the iron shaft.
12. A steel shaft 20 ft. in length and 3 in. in diameter makes 200
revolutions per minute and transmits 50 H. P. Through what angle is
the -shaft twisted ?
A wrought-iron shaft of the same length is to do the same work at
the same speed. Find its diameter so that the stress at the circumference
may not exceed f of that at the circumference of the steel shaft.
Ans. 2. 6; 3.556 in.
13. A vertical cast-iron axle in the Saltaire works makes 92 revolu-
tions per minute and transmits 300 H. P.; its diameter is 10 in. Find
the angle of torsion. Ans. .0144 per lineal foot.
14. In a spinning-mill a cast-iron shaft 8J in. in diameter makes 27
revolutions per minute; the angle of torsion is not to exceed per
lineal foot. Find the work transmitted. Ans. 62.19 H. P.
15. A square wooden shaft 8 ft. in length is acted upon by a force of
200 Ibs., applied at the circumference of an 8 ft. -wheel on the shaft.
Find the length of the side of the shaft, so that the total torsion may
not exceed 2 (in = 400000). What should be the diameter of a round
shaft of equal strength and of the same material ?
Ans. 4.96 in.; 5.09 in.
16. A shaft transmits a given H. P. at ^revolutions per minute with-
out bending. Find the weight of the shaft in pounds'per lineal foot.
/H.P.M
Ans. 2.- .
582 THEORY OF STRUCTURES.
17. The working stress in a steel shaft subjected to a twisting couple
of looo in.-tons is limited to 11,200 Ibs. per square inch. Find its diam-
eter; also find the diameter of the steel shaft which will transmit 5000
H. P. at 66 revolutions per minute,// being f. Ans. 10 in. ; 6.88 in.
1 8. A wrought-iron shaft is twisted by a couple of 10 ft. -tons. Find
its diameter (a) if the torsion is not to exceed i per lineal foot, (ft) if the
safe working stress is 7200 Ibs. per square inch, m = 10,000,000 Ibs.
Ans. (a) 3.7 in.; (&) 5.7 in.
19. A steel shaft 2 in. in diameter makes 100 revolutions per minute
and transmits 25 H. P. Find the maximum working stress and the tor-
sion per lineal foot, m being 10,000,000 Ibs. Also find the diameter of a
shaft of the same material which will transmit 100 H. P. with the same
maximum working stress. Ans. io,o22 T \ Ibs. ; .0478 ; 3.17 in.
20. The crank of a horizontal engine is 3 ft. 6 in. and the connecting-
rod 9 ft. long. At half-stroke the pressure in the connecting-rod is 500
Ibs. What is the corresponding twisting moment on the crank-shaft ?
Ans. 1716^ ft.-lbs.
21. If the horizontal pressure upon the piston end of the connecting
rod in the previous question is constant, find the maximum twisting mo-
ment on the crank-shaft.
/ . sin 0cos0 \
Ans. Pi sin + _ -- j , being given by
n* cos 2 + 4 (sin 2 cos 2 i) + 2 sin 4 0(i + sin 2 0) sin < = o
where n = */- = \ 8 -.
N.B. If sin 2 is neglected as compared with 2 ,
the maximum moment = P sin 0(1+
V n
being very nearly 72.
22. Show that a hollow shaft is both stiffer and stronger than a solid
shaft of the same weight and length.
23. Find the percentage of weight saved by using a hollow instead of
a solid shaft.
Ans. If of equal stiffness = .
m* + i
If of equal strength = 100 \ i /J m \ m * ~ !
1 (" + O a )
m being the ratio of the external to the internal diameter
of hollow shaft.
24. A hollow cast-iron shaft of 12 in. external diameter is twisted by
a couple of 27,000 ft.-lbs. Find the proper thickness of the metal so that
the stress may not exceed 5000 Ibs. per square inch. Ans. .619 in.
EXAMPLES.
25. The external diameter of a hollow shaft is/ times the internal.
Compare its torsional strength with that of a solid shaft of the same ma-
terial and weight.
Ans. -
/ J + i
26. If the solid shaft is 10 in. in diameter, and the internal diameter
of the hollow shaft is 5 inches, find the external diameter and compare
the torsional strengths. AnSt 5 j/g i n< . ^ to 3.
27. A hollow steel shaft has an external diameter d and an internal
d
diameter . Compare its torsional strength with that of (a) a solid
steel shaft of diameter d\ (b) a solid wrought-iron shaft of diameter d\
the safe working stresses of steel and iron being 5 tons and 3! tons re-
spectively. Ans (a) -fa ; () |-|.
28. What twisting moment can be transmitted by a hollow steel shaft
of 8 in. internal and 10 in. external diameter, the working stress being
$ tons per square inch ? Ans. 184^ in.-tons.
29. If/i is the safe torsional working stress of a shaft, and / a is the
safe working stress when the shaft acts as a beam, show that the tor-
sional resistance of the shaft is to its bending resistance in the ratio of
2/ a tO/,.
30. The wrought-iron screw shaft of a steamship is driven by a pair
of cranks set at right angles and 21.7 in. in length; the horizontal pull
upon each crank-pin is 176,400 IDS., and the effective length of the shaft
is 866 in. Find the diameter of the shaft so that (i) the circumferential
stress may not exceed 9000 Ibs. per square inch ; (2) the angle of torsion
r d
may not exceed per lineal foot ; m being 10,000,000 Ibs. The actual
diameter of the shaft is 14.9 in. What is the actual torsion ?
Ans. (i) 14.53 m 'J ( 2 ) H-89 in.; (3) total torsion = 5. 545.
31. The ultimate tensile strength of the iron being 60,000 Ibs. per
square inch, find the actual ultimate strength under unlimited repetitions
of stress. Ans. 54,899 Ibs. (Unwin's formula).
32. What is the torsion in the preceding question when one of the
cranks passes a dead point ?
33. A steel shaft 300 feet in length makes 200 revolutions per minute
and transmits 10 H. P. Determine its diameter so that the greatest
stress in the material may be the same as the stress at the circumference
of an iron shaft i in. in diameter and transmitting 500 ft. -Ibs.
Ans. .807 in. (= in.)
34. Determine the coefficient of torsional rupture for the shaft in
Question 33, 10 being the factor of safety.
35. A wrought-iron shaft in a rolling-mill is 220 feet in length, makes
95 revolutions per minute, and transmits 1 20 H. P. to the rolls; the main
body of the shaft is 4 in. in diameter, and it revolves in gudgeons 3f in.
584 THEORY OF STRUCTURES.
in diameter. Find the greatest shear stress in the shaft proper and in
the portion of the shaft at the gudgeons. Ans. 6330.2 Ibs.; 7508 Ibs.
36. Power is taken from a shaft by means of a pulley 24 inches in
diameter which is keyed on to the shaft at a point dividing the distance
between two consecutive supports into segments of 20 and 80 in. ; the
tangential force at the circumference of the pulley is 5500 Ibs. If the
shaft is of cast-iron, determine its diameter, taking into account the
bending action to which it is subjected. Ans. 4.7 in.
37. Show that the resilience of a twisted shaft is proportional to its
weight. / 2 Volume
Ans. Resilience = .
in 4
38. If a round bar of any material is subjected to a twisting couple,
show that its maximum resilience is two-thirds the maximum resilience
of the material.
39. Determine the diameter of a wrought-iron shaft for a screw
steamer, and the torsion per lineal foot; the indicated H. P. = 1000, the
number of revolutions per minute = 150, the length of the shaft from
thrust bearing to screw = 75 ft., and the safe working stress = 7200 Ibs.
per square inch. Ans. 6.67 in. ; io.5.
40. In a spinning-mill a cast-iron shaft 84 ft. long makes 50 revolu-
tions per minute and transmits 270 H. P. Find its diameter (i) if the
stress in the metal is not to exceed 5000 Ibs. per square inch ; (2) if the
, o
angle of torsion per lineal foot is not to exceed .
Also (3) in the first case find the total torsion.
Ans. (i) 7.02 in. ; (2) 10.23 in. ; (3) 28. 8.
41. A circular shaft is twisted beyond the limit of elasticity. If the
equalization of stress is perfect, show that for a given maximum stress the
twisting couple is greater than it would be if the elasticity were perfect,
in the ratio of 4 to 3.
42. Determine (a) the profile of a shaft of length / which at every
point is so proportioned as to be just able to bear the power it has to
transmit plus the power required to overcome the friction beyond the
point under consideration. Find (ft) the efficiency of such a shaft, and
(c) the efficiency of a shaft made up of a series of n divisions, each of
uniform diameter.
Ans. (a) The radius/ of any section distant x from the driving end
X
is y = re ?> L , r being the radius of the driving end and
L the length of a shaft of uniform diameter, such that
the whole driving moment is required to overcome its
own friction.
EXAMPLES. 585
43. A steel shaft carries a 5~ft. pulley midway between the supports
and makes 6 revolutions per minute, the tangential force on the pulley
being 500 Ibs. Taking the coefficient of working strength at 11,200 Ibs.
per square inch, find the diameter of the shaft and the proper distance
between the bearings.
44. A steel shaft 4 inches in diameter and weighing 490 Ibs. per cubic
foot makes 100 revolutions per minute. If the working stress in the
metal is 11,200 Ibs. per square inch, find the twisting couple and the dis-
tance to which the work can be transmitted ; the coefficient of friction
being .05, and the efficiency of the shaft f.
Ans. 140,800 in. -Ibs. ; 8228^ ft.
45. If the shaft is of steel, and if the loss due to friction is 20 per cent,
find the distance to which work may be transmitted, ju being .05.
Ans. 6582! ft.
46. A wrought-iron shaft 220 ft. between bearings and 4 in. in diam-
eter can safely transmit 120 H. P. at the rate of 95 revolutions per
minute. What is the efficiency of the shaft? (jit = fa.) Ans. .976.
47. The efficiency of a wrought-iron shaft is ; the working stress in
the metal is 7200 Ibs. per square inch ; the coefficient of friction is .125.
How far can the work be transmitted ? Ans. 4320 ft.
48. A spring is formed of steel wire ; the mean diameter of the coils
is i inch ; the working stress of the wire is 50,000 Ibs. per square inch ;
the elongation under a weight of 19/3- Ibs. is 2 inches; the coefficient of
transverse elasticity is 12,000,000 Ibs. Find the diameter of the wire and
the number of coils.
49. Find the weight of a helical spring which is to bear a safe load of 6
tons with a deflection of I inch, G being 12,000,000 Ibs., and f 60.000 Ibs.
50. Find the time of oscillation of a spring, the normal displacement
under a given load being A. AJ
Ans. TI\ ~.
5 1 . Find the deflection under the weight W of a conical helical spring
(a) of circular section ; (<) of rectangular section, the radii of the extreme
coils being ji and j a , and the radial distance from the axis to a point of
the spring at an angular distance from the commencement of the spiral
r 2 R
8000 to oooo
Double-riveted
7
H
Treble-riveted
8 to 85
,,
Steel
Single-riveted
. ^^
I2OOO to I3OOO
14
Double-riveted
7
(,
Treble-riveted
8 to 85
tt
For cast-iron cylinders the working value of /may be taken
at about 2000 Ibs. per square inch.
The total pressure upon each of the flat ends of the cylinder
The longitudinal tension in a thin hollow cylinder
K^P P^ " ,
" 2nrt ~ 2t' ^ }
and is one half of the circumferential stress/.
Cor. i. Let the cylinder be subjected to an external pressure
p' as well as to an internal pressure p. Then
fl=pr-p'r>,
(3)
r' being the radius of the outside surface of the cylinder, /is
a tension or a pressure according as pr
588 THEORY OF STRUCTURES.
Generally, r r' is very small, and the relation (3) may be
written
ft = r(p-p').
2. Thick Hollow Cylinder. If / is large, the stress is no
longer uniformly distributed over the thickness. Suppose that
the assumptions (i) and (3) of Art. I still hold, also that the
cylinder ends are free, and that the annulus forming the section
of the cylinder is composed of an infinite number of concentric
rings. Under these conditions the straining of the cylinder
cannot affect its cylindrical form. Hence, right sections of the
cylinder in the unstrained state remain planes after the strain-
ing, so that the longitudinal strain at every point must be the
same. Two methods will be discussed.
FIRST METHOD. Let dx be the thickness of one of the
rings of radius x, and let dq be the intensity of the circum-
ferential stress.
pr p'r' = difference between the total pressures from
within and without = total circumferential stress = / *" dq.
If it be assumed that the thickness (= r' r) remains un-
changed under the pressure, then the circumferential extension
of each of the concentric rings must be equal to the same con-
stant quantity A, and therefore
dq = Edx - ,
*
E being the coefficient of elasticity. Hence,
. , E\ C r 'dx 1 r'
prp'r'=- I = log,-.
27t J r X 27t *' r
en /= E
2
elastic limit is not exceeded, and therefore
Let / be the tensile unit stress. Then /= E - if the
2nr
THICK HOLLOW CYLINDER. 589
or
r' pr-p'r' i(pr
,. -= x + _- + -( - f - , approx., (4)
if p is small as compared with/; and hence,
t r ' pr-p'r' ilr
= ~ ~~ h
r
In most cases which occur in practice/' is so small as com-
pared with/ that it may be disregarded.
Hence, making/' zero in equation (5),
Formulae (5) and (6) may be employed even if the elastic
limit is exceeded, if f is considered a coefficient of strength
to be determined by experience.
Cor. Rankine, in his Applied Mechanics, obtains by
another method,
//_P
'-~\ f-P'
if/' be neglected. Hence,
, *3
== i +y + 2 7^ ' a PP roximatel y'
if p is small as compared with /, and therefore
59O THEORY OF STRUCTURES.
an equation identical with (6).
SECOND METHOD. Consider a ring bounded by the radii
x, x + dx, at any point.
Let q be the normal (i.e., radial) intensity of stress.
Let /be the intensity of stress tangential to the ring.
" s " " " " " perpendicular to the plane
of the ring.
Let a-, /?, y be the corresponding strains.
Let E and mE be respectively the coefficients of direct and
lateral elasticity.
Then, since E, f, s are principal stresses (Chap. IV),
_ _
~
f %*
r' r' ' x' r r'
3. Spherical Shells. Let the data be the same as before.
The section made by any diametral plane must develop a total
resistance of 2nrtf. Then
2nrtf 7tr*p,
or
*f = pr - (0
Hence, a spherical shell is twice as strong as a cylindrical
shell of the same diameter and thickness of metal, so that the
strongest parts of egg-ended boilers are the ends.
Cor. I. Let the shell be subjected to an external pressure
p' as well as to an internal pressure p. Then
2n -tf= nrp* nr' *#'.
/"is a tension or a pressure according as r*p ^ r'*p'.
Generally, r' r is very small, and the relation (2) may be
written
ft = r -(.P-p'} (3)
592 THEORY OF STRUCTURES.
Cor. 2. For a thick hollow sphere, Rankine obtains
P == 2 Sr" + 2r" a PP roximatelv - (4)
4. Practical Remarks. A common rule requires that the
working pressure in fresh-water boilers should not exceed one-
sixth of the bursting pressure, and in the case of marine boilers
that it should not exceed one-seventh.
An English Board of Trade rule is that the tensile working
stress in the boiler-plate is not to exceed 6000 Ibs. per square
inch of gross section, and French law fixes this limit at 4250
Ibs. per square inch.
The thickness to be given to the wrought-iron plates of a
cylindrical boiler is, According to French law,
/ = .0036/2;- + .1 in.;
according to Prussian law,
/ (^3 _ i) r + i in. .oo$nr + .1 in., approximately,
r being the radius in inches, and n the excess of the internal
above the external pressure in atmospheres.
The thickness given to cast-iron cylindrical boiler-tubes is,
according to French law, five times the thickness of equivalent
wrought-iron tubes ; according to Prussian law,
t _ ^.om _ i)r + -J in. = .01 nr + i in., approximately.
Steam-boilers before being used should be subjected to a
hydrostatic test varying from \\ to 3 times the pressure at
which they are to be worked.
Fairbairn conducted an extensive series of experiments
upon the collapsing strength of riveted plate-iron flues, by
enclosing the flues in larger cylinders and subjecting them to
hydraulic pressure. From these experiments he deduced the
following formula for a ivrought-iron cylindrical flue or tube :
Collapsing pressure I = ^ =403 1 50 ^
in pounds per square inch of surface i "
PRACTICAL RULES FOR BOILERS AND FLUES. 593
t "being the thickness and r the radius in inches, and / the length
in feet.
This formula cannot be relied upon in extreme cases and
when the thickness of the tube is less than f in.
Note. In practice, t 2 may be generally used instead of t 2 - 19 .
The experiments also showed that the strength of an elliptical
tube is almost the same as that of a circular tube of which the
radius is the radius of curvature at the ends of the minor axis.
Hence, if a and b are the major and minor axes of the ellipse^
the above formula becomes
b / 2 ' 19
p = 403150 - a -j-.
By riveting angle- or T-irons around a tube, its length is
virtually diminished and its strength is therefore increased, as
it varies inversely as the length.
The thickness of tubes subjected to external pressure is,
according to French law, twice the thickness of tubes subjected
to interior pressure, but under otherwise similar conditions ;
according to Prussian law the thickness of heating pipes is
/ = .0067^ Vn -f- .05 in., if of sheet-iron,
and
t .Old Vn + .07 in., if of brass.
According to Reuleaux, the thickness (f) of a round flat
plate of radius r, subjected to a normal pressure, uniformly dis-
tributed and of intensity/, is given by the formula
7 t _ A7
according as the plate is merely supported around the rim or
is rigidly fixed around the rim, as, e.g., the end plates of a
cylindrical boiler ; /", as before, is the coefficient of strength.
The corresponding deflections of the plate are
-(-V
6\t>
and -, - M=.
594 THEORY OF STRUCTURES.
EXAMPLES.
1. What should be the thickness of the plates of a cylindrical boHer
6 ft. in diameter and worked to a pressure of 50 Ibs. per square inch, in
order that the working tensile stress may not exceed 1.67 tons per square
inch of gross section ? Ans. .42 in.
2. A cylindrical boiler with hemispherical ends is 4 ft. in diameter
and 22 ft. in length. Determine the thickness of the plates for a steam-
pressure of 4 atmospheres.
3. What is the collapsing pressure of a flue 10 ft. long, 36 in. in
diameter, and composed of i-in. plates? Also of a flue 30 ft. long, 48 in.
in diameter, and T \ in. thick? Ans. 490.84 Ibs.; 91.59 Ibs.
4. Determine the thickness of a 2-in. locomotive fire-tube to support
an external pressure of 5 atmospheres.
5. A copper steam-pipe is 4 in. in diameter and \ in. thick. Find the
working pressure, the safe coefficient of strength for copper being 1000
Ibs. per square inch. Ans. 125 Ibs. per square inch.
6. A /-ft. boiler of ^-in. plates was burst at a longitudinal double-
riveted joint by a pressure of 310 Ibs. per square inch. Find the coef-
ficient of ultimate strength. Ans. 29,760 Ibs.
7. A 50- in. cylindrical boiler of -^ in. plates is made of wrought-
iron whose safe coefficient of strength is 4000 Ibs. per square inch. Find
the working pressure. Ans. 50 Ibs. per square inch.
8. A lo-in. cast-iron water-pipe is subjected to a pressure of 250 Ibs.
per square inch. Find its thickness, the coefficient of working strength
being 2000 Ibs. per square inch. Ans. i in.
9. A steel spherical shell 36 in. in diameter and f in. thick is sub-
jected to an internal fluid pressure of 300 Ibs. per square inch. Find its
coefficient of strength. Ans. 7200 Ibs.
10. A thin, hollow, spherical, elastic envelope, whose internal
radius is R, was subjected to a fluid pressure which caused it to expand
gradually until its radius became R\ . Determine the work done.
11. The plates of a cylindrical boiler 5 ft. in diameter are \ in. thick.
Find to what pressure the boiler may be worked so that the tensile stress
in the plates may not exceed i| tons per square inch of gross section.
EXAMPLES, 595
12. Show that the assumption of a uniform distribution of stress in
the thickness of a cylindrical or spherical boiler is only admissible when
the thickness is very small.
13. A metal cylinder of internal radius r and external radius nr is
subjected to an internal pressure of p tons per square inch. Show that
the total work done in stretching the cylinder circumferentially is
-~~ -j - ft. -tons per square foot of surface, E being the metal's co-
efficient of elasticity.
14. The cast-iron cylinder of an hydraulic press has an external
diameter twice the internal, and is subjected to an internal pressure of
/tons per square inch. Find the pHncipal stresses at the outer and inner
circumferences. Also, if the pressure is 3 tons per square inch, and if the
internal diameter is 10 in., find the work done in stretching the cylinder
circumferentially, E being 8000 Ibs.
Ans. At inner circumference, q =p, a thrust, and/= ^p, a tension.
At outer circumference, q = o, and/ = f /, a tension.
Woik = 126 ft.-lbs. per square foot of surface.
15. The chamber of a 2/-ton breech-loader has an external diameter
of 40 in. and an internal diameter of 14 in. Under a powder pressure of
1 8 tons per square inch, find the principal stresses at the outer and inner
circumferences, and also the work done ; E being 13,000 Ibs.
Ans. At inner, q = 18 tons, compression ; at outer, q = o.
At inner,/ = 23^ tons, tension ; at outer,/ = 5^ tons,
tension.
Work = \\ ft.-tons per sq. ft. of surface.
16. What should be the thickness of a o-in. cylinder (a) which has
to withstand a pressure of 800 Ibs. per square inch, the maximum allow-
able tensile stress being 24,000 Ibs. per square inch ; (ff) which has to
withstand a pressure of 6000 Ibs. per square inch ; the maximum allow-
able tensile stress being 10,000 Ibs. per square inch?
Ans. (a) 1.86 in. ; (b) 4^ in.
17. Show that the radial (or) and hoop (fl) strains in thick hollow
cylinders and spheres are connected by the relation a - .
1 8. Prove that the relation in Ex. 1.7 is satisfied by the values ob-
tained for/ and q in the Second Method of Art. 2, Chap. X.
19. A thick hollow sphere of internal radius r and external radius
nr is subjected to an internal pressure p and an external pressure p .
Determine the principal stresses at a distance x from the centre.
_P'n*-p p_-Jf V . p*?_-_p _ p_-_ n*S_
~ - " 3 ' J ~ " -
THEORY OF STRUCTURES.
20. Assuming that the annulus forming the section of a cylindrical
boiler is composed of a number of infinitely thin rings, show that the
JL
pressure at the circumference of a ring of radius r is per unit of
surface, and that the circumferential stress is h , A and B de-
r mr m + I
noting arbitrary constants, and m being the coefficient of lateral con-
traction. Find the values of A and B, p* and pi being respectively the
internal and external pressures.
21. Show that in the case of a spherical boiler the pressure and cir-
cumferential stress are respectively and + . Find
' r i+m r i (0 l)f** ~ 3
A and B.
22. Solve Questions i, 2, 6, 7, 8, 9, and n on the supposition that / is
not small as compared with r.
23. Taking/ = 4000 Ibs. per square inch and E = 30,000,000 Ibs.,
Find the thickness and deflection of the end plates of the boiler in Ques-
tion 7.
CHAPTER XL
BRIDGES.
1. Classification. Bridges may be divided into four gen-
eral classes, viz. : (A) Bridges with horizontal girders ; (B) Can-
tilever bridges (Art. 15) ; (C) Suspension bridges (Chap. XII);
(D) Arched bridges (Chap. XIII). The present chapter treats
of bridges in Classes A and B only.
2. Comparative Advantages of Curved and Horizontal
Flanges in Girders for Bridges of Class A. The depth is
sometimes varied for the sake of appearance, and it is also
claimed that, an economy of material is effected by giving the
chord a slope, as, e.g., in the case of the Sault Bridge (Art. 19).
Such a truss is intermediate between a truss with horizontal
flanges and one of the parabolic form. The curved or para-
bolic form is not well suited to plate construction, and a dimi-
nution in depth lessens the resistance of the girder to distor-
tion. Again, if the bottom flange is curved, the bracing for
the lower part of the girder is restricted within narrow limits,
and the girder itself must be independent, so that in a bridge
of several spans any advantage which might be derivable from
continuity is necessarily lost. Generally speaking, the best and
most economical form of girder is that in which the depth is
uniform throughout, and in which the necessary thickness of
flange at any point is obtained by increasing the number of
plates.
3. Depth of Girder or Truss (Class A). The depth usu-
ally varies from one-fifteenth to one-seventh (and even more) of
the span. It is generally found advisable to give large girders
597
598
THEORY OF STRUCTURES.
an increased depth, and they should, therefore, be designed to
have a specified strength. If the span is more than twelve times
the depth, the deflection becomes a serious consideration, and
the girder should be designed to have a specified stiffness. The
depth should not be more than about i^ times the width of
the bridge, and is therefore limited to 24 ft. for a single and to
40 ft. for a double-track bridge.
4. Position of Platform. The platform may be supported
either at .ike-'top^r bottom flanges, or in some intermediate
positidfu In favor/ oiVthe last it is claimed that the main girders
may be braced together below the platform (Fig. 365), while
the upper portions s'tfrve as parapets or guards, and also that
,-the vibration communicated by a passing train is diminished.
J.The position, however, is not conducive to rigidity, and a large
amount of metal is required to form the connections.
pi A
FIG. 365.
FIG. 366.
The method of supporting the platform on the top flanges
(Fig. 366) renders the whole depth of the girder available for
bracing, and is best adapted to girders of shallow depth.
Heavy cross-girders may be entirely dispensed with in the case
of a single-track bridge, and the load most effectively distrib-
uted, by laying the rails directly upon the flanges and vertically
above the neutral line. Provision may be made for side spaces
by employing sufficiently long cross-girders, or by means of
short cantilevers fixed to the flanges, the advantage of the
POSITION OF PLATFORM.
599
former arrangement being that it increases the resistance to
lateral flexure, and gives the platform more elasticity.
Figs. 367, 368, 369 show the cross-girders attached to the
bottom flanges, and the desirability of this mode of support
increases with the depth of the main girders, of which the cen-
tres of gravity should be as low as possible. If the cross-girders
are suspended by hangers or bolts below the flanges (Fig. 369),
the depth, and therefore the resistance to flexure, is increased.
FIG. 368.
FIG. 369.
In order to stiffen the main girders, braces and verticals,
consisting of angle- or tee iron, are introduced and connected
with the cross-girders by gusset pieces, etc. ; also, for the same
purpose, the cross-girders may be prolonged on each side, and
the end joined to the top flanges by suitable bars.
When the depth of the main girders is more than about
5 ft., the top flanges should be braced together. But the
minimum clear headway over the rails is 16 ft., so that some
other method should be adopted for the support of the plat-
form when the depth of the main girders is more than 5 ft.
and less than 16 ft.
Assume that the depth of the platform below the flanges is
2 ft., and that the depth of the transverse bracing at the top is
i ft. ; the total limiting depths are 7 ft. and 19 ft., and if I to 8
is taken as a mean ratio of the depth to the span, the corre-
sponding limiting spans are 56 ft. and 152 ft.
600 THEORY OF STRUCTURES.
5. Comparative Advantages of Two, Three, and Four
Main Girders. A bridge is generally constructed with two
main girders, but if it is crossed by a double track a third is
occasionally added, and sometimes each track is carried by two
independent girders.
The employment of four independent girders possesses the
one great advantage of facilitating the maintenance of the
bridge, as one-half may be closed for repairs without interrupt-
ing the traffic. On the other hand, the rails at the approaches
must deviate from the main lines in order to enter the bridge,
so that the width of the bridge is much increased, and far
more material is required in its construction.
Few, if any, reasons can be urged in favor of the introduc-
tion of a third intermediate girder, since it presents all the
objectionable features of the last system without any corre-
sponding recommendation.
The two-girder system is to be preferred, as the rails, by
such an arrangement, may be continued over the bridge with-
out deviation at the approaches, and a large amount of ma-
terial is economized, even taking into consideration the in-
creased weight of long cross-girders.
6. Bridge Loads. In order to determine the stresses in
the different members of a bridge truss, or main girder, it is
necessary to ascertain the amount and character of the load to
which the bridge may be subjected. The load is partly dead,
partly live, and depends upon the type of truss, the span, the
number of tracks, and a variety of other conditions.
The dead load increases with the span, and embraces the
weight of the main girders (or trusses), cross-girders, platform,
rails, ballast, and accumulations of snow.
As to the live load see Art. 19.
7. Trellis or Lattice Girders. The ordinary trellis or
lattice girder consists of a pair of horizontal chords and two
series of diagonals inclined in opposite directions (Fig. 370).
The system of trellis is said to be single, double, or treble, ac-
cording to the number of diagonals met by the same vertical
section.
TRELLIS OR LATTICE GIRDERS.
60 1
Vertical stiffeners, united to the chords and diagonals, may
be introduced at regular intervals.
FIG. 370.
371, 372, 373, 374 show appropriate sections for the
top chord ; the bottom chord may be formed of fished and
riveted plates, or of links and pins.
-ii ir
FIG. 371. FIG. 372. FIG. 373. FIG. 374.
The verticals and diagonals may be of an L, T, I, H, LJ , or
other suitable section, but the diagonals, except in the case of
a single system of trellis, are usually flat bars, riveted together
at the points of intersection.
An objection to this class of girder is the number of the
joints.
The stresses in the diagonals are determined on the assump-
tion that the shearing force at any vertical section is equally
distributed between the diagonals met by that section, which
is equivalent to the substitution of a mean stress for the differ-
ent stresses in the several bars.
E.g., let w be the permanent load concentrated at each apex
in Fig. 370.
Let 9 be the inclination of the diagonals to the vertical.
The reaction at A fyw, and the shearing force at the
section MN = y^w ^.w = 2>2 W -
This shearing force must be transmitted through the diag-
onals.
Hence, the stress in ab due to the permanent load
sec
-w sec
602 THEORY OF STRUCTURES.
Again, let w' be the live load concentrated at an apex.
The greatest shear at mn due to the live load occurs when
every apex between a and 7 is loaded.
This shear = corresponding reaction at I = -f-J-w', and the
stress in ab due to the live load
= i X -f- f ' sec f f w' sec 8.
Hence, the total maximum stress in ab = (|w -f- J J ze/) sec 0.
The greatest stress of a kind opposite to that due to the
dead load is produced in ab when the live load w' is concen-
trated at every apex between I and b.
The shear to be transmitted is then 2\w due to the dead load,
and -j-fw' due to the live load, and the resultant stress in ab
sec 6 = (f w l%w') sec 6.
This stress may be negative, and must be provided for by
introducing a counter-brace or by proportioning the bar to
bear both the greatest tensile and the greatest compressive
stress to which it may be subjected.
The stress in any other bar may be obtained as above.
The chord stresses are greatest when the live load covers
the whole of the girder, and may be obtained by the method
of moments, or in the manner described in the succeeding
articles.
In the above it is assumed that the members of the girder
are riveted together. If they are connected by pins, each of
the diagonal systems may be treated as being independent.
Thus, the system I 2^^34567 transmits to the supports
the stresses due to loads at a, 3, and 5.
The shear due to the dead load, transmitted through ab,
3 w
= reaction at I load at a = w w .
2 2
W
Hence, the stress in ab due to the dead load = - sec 6.
The stress in ab due to the live load is greatest when w' is
concentrated at each of the points 3 and 5,
WARREN GIRDER.
603
The maximum shear due to live load transmitted through ab
= ^w f = fee/,
and the corresponding stress in ab = / sec 9.
Hence, the total maximum stress in ab
as compared with (fcw + |- f /) sec obtained on the first as-
sumption.
8. Warren Girder. The Warren girder consists of two
horizontal chords and a series of diagonal braces forming a
single triangulation, or zigzag. Fig. 375.
N-1
024 n N-2
FIG. 375.
The principles which regulate the construction of trellis
girders are equally applicable to those of the Warren type.
The cross-girders (floor-beams) are spaced so as to occur at
the apex of each triangle.
When the platform is supported at the top chords, the re-
sistance of the structure to lateral flexure may be increased by
horizontal bracing between the cross-girders and by diagonal
bracing between the main girders.
When the platform is supported on the bottom chords,
additional cross-girders may be suspended from the apices in
the upper chords, which also have the effect of adding to the
rigidity of the main girders.
Let w be the dead load concentrated at an apex or joint.
.' u live tt u " " *' " **
" / " " span of the girder.
" k " " depth ". "
" s " " length of each diagonal brace.
" 6 " " inclination of each diagonal brace to the ver-
tical.
" N -\- i be the number of joints.
604 THEORY OF STRUCTURES.
Then
Two cases will be considered.
CASE I. All t he joints loaded.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
Let S n be the shearing force at a vertical section between
the joints n and n -f- i.
Let H n be the horizontal chord stress between the joints
n I and n -f- I-
The total load due to both dead and live loads
= (w-\-w')(N - i).
The reaction at each abutment due to this total load
The shearing forces in the different bays are
UU _!_ it)'
w A- w'
-~-
Tfl -J 7/
(N i), between o and i ;
3), ! .< 2 .
2 x - 5), 2 3 ;
and
The corresponding diagonal stresses are
5 sec 0, .V, sec 0, S M sec 0.
WARREN GIRDER. 605
The last stresses multiplied by sin 6 give the increments of
the chord stresses at each joint. Thus,
H l = tension in o 2 = 5 tan
+ 1 j i / / r
IV / IV f W l> 1\ I
ff t = compression in i 3 = ,S tan 0-|- .S, tan
w + w' IN i w -\- w' I N 3
_ W + tt/ / 2(7V ~ 2)
-fiT 3 = tension in 2 4 = //, + 5", tan 0-\- S 9 tan
-3)
_ _
2 k N
compression in 3 5 = //!, + ^a tan ^ + $* tan
4)
and H n = horizontal stress in chord, between the joints
w -\- w' / n(Nn)
n i and n + i ! -- T - X7 ' being a tension for a
2t K J.\
bay in the bottom chord, and a compression for a bay in the
top chord.
Note. The same results may be obtained by the method
of moments ; e.g., find the chord stress between the joints
n r and n -|- i.
Let a vertical plane divide the girder a little on the tight
of n.
The portion of the girder on the left of the secant plane is
kept in equilibrium by the reaction at the left abutment, the
horizontal stresses in the chords, and the stress in the diagonal
from ;/ to n -\- i.
Take moments about the joint n. Then
606 THEORY OF STRUCTURES.
n _ - _ _
w-\- w' N n
~ 2 N
/. H n = etc.
Diagonal Stresses due to Dead Load.
Let d n be the stress in the diagonal n, n -\- I, due to the
dead load.
The shearing forces in the different bays due to the dead
load are
w w
-(N i), between o and I ; -(N 3), between I and 2 ;
2 2
W W
-(^-5), " 2 " 4; -(N-j), 3 " 4;
and (N 2n i), between n and n -j- I.
The corresponding diagonal stresses are :
w w s
a compression -(N i) sec 6 = -(N i)-, = d^ inoi ;
2 A?
a tension (TV 3) sec 6 = (N $)-r = d l in 1 2
2 2 A?
a compression (A 7 " 5) sec -(N 5)7- ^ 2 in 23 ;
2 2 rt
and the stress in the nth diagonal between n and n + i is
WARREN GIRDER. 607
being a tension or a compression according as the brace slopes
down or up towards the centre. = compression in o i when all the joints are loaded
sw'N(N- i)
~~k^~ N~
D l = tension in I 2 when all the joints except one are loaded
SW '(N-
~ 7~ N ~
^ = compression in 2 3 when all the joints except I and 2 are
D^ = tension in 34 when all the joints except I, 2, and 3 are
In^H ^'(^-3)(^-4)
loaded = J -- - Jr -.
D n = stress in n, n -f- i when all the joints except i, 2, 3, . . .
sw f (N-n)(N-n-i)
and n are loaded = - - ~v~~
DI stress in o I before the load comes upon the girder = 9.
608 THEORY OF STRUCTURES.
s w'
J}/= compression in I 2 when the joint I is loaded = -7 i.
K IN
/Y = tension in 2 3 when the joints I and 2 are loaded = - 3.
>,'= compression in 34 when the joints I, 2, and 3 are loaded
sw'
= kN 6 '
JD n '= stress inn,n-{-i, when the joints I, 2, . . . and ware loaded
s w' n(n-\- i)
= 'k~N~~~2~*
The total maximum stress in the th diagonal of the same
kind as that due to the dead load = d n + D n .
The resultant stress in the nth diagonal when the load
covers the shorter segment = d n D n '.
This resultant stress is of the same kind as that due to the
dead load so long as d n > D n ', and need not be considered since
d n + D n is the maximum stress of that kind.
If D n ' > d M1 it is necessary to provide for a stress in the
given diagonal of a kind opposite to that due to d n -\- D H , and
equal in amount to D n ' d n .
This is effected by counterbracing or by proportioning the
bar to bear both the stresses d H -)- D H and D n ' d n .
CASE II. Only joints denoted by even numbers loaded.
2 ' 4 N~ 2 N
vvvvvv
1 3 N-1
FlG - 376-
2 ^ N-2 N
3 N-T
FIG. 377-
Chord Stresses. The stresses are greatest when the live load
covers the whole of the girder.
WARREN GIRDER.
The total load due to both dead and live loads
609
The reaction at each abutment due to this total load
To find //i, take moments about I. Then
w I w' /
To find H % , take moments about 2.
To find //g, take moments about 3.
u k = _?L( N _ 2 ) 3 __ _ (w
To find HI, take moments about 4.
w 4- w' /
-(7^- 2)4^ -
To find H n , take moments about n, an
Then
let ^^ be even.
and
_
~
(A 7 ' 2)^ ( 2}
4~ 4
|- w' / (#" ~ *)
6 10 THEORY OF STRUCTURES.
Next, let n be odd. Then
- (w + w')x\(n - 2) + ( - 4) + + 5 + 3 +
' L \ ^ N ~ 2 ) n _ (* ~ I )'
')~N\ 4 4
and
w J rW ' I ny 2)n (n i) s
N
N
Note. If is even,
ryy _!_ / I
HN_, the stress in the middle bay, = - g -rtf.
N .
If is odd,
w + w' I N* 4
HN_, the stress in the middle bay, = g T ^7
Diagonal Stresses due to the Dead Load. The shearing forces
w
in the different bays due to the dead load are (N ' 2) between
4
w "W
o and 2, -(N 6) between 2 and 4, -(TV 10) between 4 and
4 4
^, etc.
The corresponding diagonal stresses are
s w
d in o i = T -W 2) = d l in I 2 ;
4 v
i,m 23 =( N ~ 4) =4 in 34;
/ 4 in 4 5 = j ^- 10) n= d b in 56;
* 4
etc., etc., etc.
HOWE TRUSS.
611
Thus the stresses in the diagonals which meet at an unloaded
joint are equal in magnitude but opposite in kind.
Diagonal Stresses due to the Live Load. These are found
as in Case I, and
-A,
N
If is odd, there is a single stress at the foot of each of
these columns.
The maximum resultant stress due to both dead and live
loads is obtained as before.
E.g., the maximum resultant stress in 3 4 when the longer
segment is loaded
and the maximum resultant stress in 3 4 when the shorter seg-
ment is loaded
= 4- D; = A - A'-
Note. 6 is generally 60, in which case s 2 ^..
9. Howe Truss. Fig. 378 is a skeleton diagram of a
Howe truss.
FIG. 378.
The truss may be of timber, of iron, or of timber and iron
combined.
The chords of a timber truss usually consist of three or
more parallel members, placed a little distance apart so as to
allow iron suspenders with screwed ends to pass between them
(Figs. 379 and 380).
6l2
THEORY OF STRUCTURES.
Each member is made up of a number of lengths scarfed
or fished together (Figs. 381 and 382).
The main braces, shown by the full diagonal lines in Fig.
378, are composed of two or more members.
The counter-braces, which are introduced to withstand the
effect of a live load, and are shown by the dotted diagonal
lines in Fig. 378, are either single or are composed of two or
more members. They are set between the main braces, and
are bolted to the latter at the points of intersection.
The main braces and counters abut against solid hard-wood
or hollow cast-iron angle-blocks (Fig. 380). They are designed
to withstand compressive forces only, and are kept in place by
tightening up the nuts at the heads of the suspenders.
At
FIG. 380.
FIG. 381.
FIG. 382.
FIG. 383.
FIG. 384.
The angle-blocks extend over the whole width of the chords ;
if they are made of iron, they may be strengthened by ribs.
If the bottom chord is of iron, it may be constructed on the
same principles as those employed for other iron girders. It
often consists of a number of links, set on edge, and connected
by pins (Figs. 383 and 384). In such a case the lower angle-
blocks should have grooves to receive the bars, so as to prevent
lateral flexure.
If the truss is made entirely of iron, the top chord may be
formed of lengths of cast-iron provided with suitable flanges
by which they can be bolted together. Angle-blocks may also
be cast in the same piece with the chord.
To determine the stresses in the different members, the
HOWE TRUSS. 613
same data are assumed as for the Warren girder, except that
N is now the number of panels.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
Let H n be the chord stress in the nth panel.
The total load due to both dead and live loads
= (w + w'}(N i).
The reaction at each abutment due to this total load
Let a plane MM' divide the truss as in Fig. 378. The por-
tion of the truss on the left of the secant plane is kept in
equilibrium by the load upon that portion, the reaction at the
left abutment, the chord stresses in the nth panels, and the
tension in the nth suspender.
First, let the load be on the top chord and take moments
about the foot of the nth suspender. Then
or
w -f- w' n(N n)
_ w + w' I n(N ri)
ff - ~~~~'
Next, let the load be on the bottom chord and take
moments about the head of the nth suspender. Then
w + w' n(N n)
H n k - L / -- -, as before.
Thus, H n is the same for corresponding panels, whether
the load is on the top or bottom chord.
Diagonal Stresses due to the Dead Load. Let V M f be the
6l4 THEORY OF STRUCTURES.
shearing force in the nth panel, or the tension on the nth sus-
pender due to the dead load.
First, let the load be on the top chord. Then
w (N i \
V H ' = (N i) nw = w\ - n\.
Next, let the load be on the bottom chord. Then
lit //V-U I \
V m ' = --(N _,)_(_ i) w = w(t- - n].
The corresponding diagonal stresses are
and
N+i
Diagonal Stresses due to the Live Load. Let V n " be the
shearing force in the nth panel, or tension on the nth suspen-
der, when the live load covers the longer segment.
First, let the load be on the top chord.
The greatest stress in the nth brace, of the same kind as that
produced by the dead load, occurs when all the panel points
on the right of MM' are loaded. With such load, V n ", the
shearing force on the left of MM ', = the reaction at o
/' N-n
--(7VT _ n i) TT ,
2^ N
and the corresponding diagonal stress, D n ,
___
- k 2 (1V I} N
Hence, the resultant tension on the nth suspender due to
both dead and live loads = V n = V n r + V n " .
N- \ \ w '
- - n + ~(N- n -
HOWE TRUSS. 615
and the resultant maximum compression on the nth brace due
to both dead and live loads
V N-n
The live load tends to produce the greatest stress in the
nth counter when it covers the shorter segment up to and in-
cluding the nth panel point. Even then there will be no stress
in the counter unless the effect of the live load exceeds that
of the dead load in the (n + i)th brace.
The shearing force on the right of MM' = the reaction at N
_ w' n(n -\- i)
Hence,
ZV tne corresponding diagonal stress, = - - 7T ,
w 2 M V
and the resultant stress in the counter = D n f d n+t
s ( w' n(n + i) JV-
Next, let the load be on the bottom chord. Then
and
D = SW -( N ri N - n + l
k 2 ^ N
Hence,
and
Also, the stress in the ^th counter is
N i
6i6
THEORY OF STRUCTURES.
.K common value of is 45, when sec 6 = = 1.414,
and tan 6 = -rr/, l -
NK
The end panels and posts, shown by the dotted lines in
Fig. 378, may be omitted when the platform is suspended from
the lower chords.
10. Single and Double Intersection Trusses. Fig.
385 represents the simplest form of single-intersection (or
FIG. 385.
Pratt) truss ; i.e., a truss in which a diagonal crosses one panel
only. It may be constructed entirely of iron or steel, or may
have the chords and verticals of wood. The verticals are in
compression and the diagonals in tension. The angle-blocks
are therefore placed above the top and below the bottom
chord. Counter-braces, shown by the dotted diagonals, are in-
troduced to withstand the effect of a live load.
If the truss is inverted it becomes one of the Howe type,
and the stresses in the several members of both trusses may
be found in precisely the same manner.
Fig. 386 represents a double-intersection (or Whipple)
XXXXXXX
/
7
FIG. 386.
truss, i.e., a truss in which a diagonal crosses two panels. It
may be constructed entirely of iron or steel. It is of the pin-
connected type, and the two diagonal systems may be treated
independently.
Let 0' be the inclination of AB to the vertical.
" * " " " Ai, CD, . .. to the vertical.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
The reaction at A from the system A BCD . . . = 4(w j r w') ;
" " A " - 4*23... = (H-*'l
SINGLE AND DOUBLE INTERSECTION TRUSSES. 6l/
Wj w' being the dead and live loads concentrated at the panel
points C, 2, E, 4, . . .
The shearing forces in the different bays are :
4(w + w') in AC, from the system ABCD . . .
(w + w') in AC, " " " ^123...
3w w n <2,
|( w _|_ w ') in 2E, " " " ^4 I 2 3
2 ( w + /') in 4, " " "
K W + w ') in 46, " " "
i(w + z/ r ) in G6, " " c<
j(, _j. ,') in 67, " " " ^123...
The corresponding diagonal stresses are :
4(w + w'} sec 6^ in ^.5 ; $%(w + w') sec 6 in ^4 I ;
3(0; -f w/ ) sec ^ in ^^ J 2 i( w + w ') sec ^ in 2 3 J etc -
Hence, the / -f- J /ze/)
sec ft
The live load tends to produce the greatest stress in any
counter 5 8 when it is concentrated at all the panel points of
the system A I 2 3 ... on the left of 8.
The reaction at the right abutment is then /, and the
corresponding stress in the counter = ze/ sec 0. Thus, the
resultant stress in the counter = (f w' %w) sec 0, \w sec 6 being
the stress in 6 7 due to the dead load.
Similarly, the stresses in any other diagonal and counter
may be found.
The Pratt truss composed entirely of iron and with some
of the details of the Whipple truss is sometimes called a
Murphy-Whipple truss. The Linville truss is a Whipple truss
made of wrought-iron, the verticals being tubular columns.
II. Post and Quadrangular Trusses. The peculiarity of
the Post truss (Fig. 387) is that the
struts are inclined at an angle of
about 22 30' to the vertical, with a
FlG 387- view to an economy of material.
The ties cross two panels at an angle of 45 with the vertical.
In the quadrangular truss
(Fig. 388) the bottom chord has
additional points of support half- FIG. 3 88.
way between the panel points.
The Bollman, Fink, and other bridge-trusses have been
referred to in a previous chapter.
12. Bowstring Girder or Truss. The bowstring girder
in its simplest form is represented by Fig. 389, and is an excel-
lent structure in point of strength and economy.
The top chord is curved, and either springs from shoes
(sockets) which are held together by a horizontal tie, or has its
ends riveted to those of the tie.
The strongest bow is one composed of iron or steel cylin-
BOWSTRING TRUSS.
619
drical tubes, but any suitable section may be adopted, and the
inverted trough offers special facilities for the attachment of
verticals and diagonals.
The tie is constructed on the same principles as those em-
ployed for other iron girders, but in its best form it consists of
flat bars set on edge and connected with the shoes by gibs and
cotters.
The platform is suspended from the bow by means of ver-
tical bars which are usually of an I section, and are set with
the greatest breadth transverse, so as to increase the resistance
to lateral flexure. In large bridges the webs of verticals and
diagonals may be lattice-work.
If the load upon the girder is uniformly distributed and
stationary, verticals only are required for its suspension, and
the neutral axis of the bow should be a parabola. An irregu-
larly distributed load, such as that due to a passing train, tends
to change the shape of the bow, and diagonals are introduced
to resist this tendency.
A circular arc is often used instead of a parabola.
To determine the stresses in the different members, assum-
ing that the axis ABC of the top chord is a parabola :
Let w be the dead load per lineal foot.
" w' " " live "
" / " " span of the girder.
" k " " greatest depth BD of the girder.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
The total load due to both dead and live loads (w -j- w')l.
The reaction at each abutment due to this total load
/.
620
THEORY OF STRUCTURES.
Let H be the horizontal thrust at the crown.
T " " " tension in the tie.
Imagine the girder to be cut by a vertical plane a little on
the right of BD. The portion ABD is kept in equilibrium by
the reaction at A, the weight upon AD, and the forces H
and T.
Take moments about B and D. Then
Tk =
and
= Hk,
Let H' be the thrust along the chord at any point P.
Let x be the horizontal distance of P from B.
The portion PB is kept in equilibrium by the thrust H at
B, the thrust H' at P, and the weight (w + w')x between P
and B. Hence,
H* sec' i = H'* = H* + (w + ^v}*x\
i being the inclination of the tangent at P to the horizontal,
and
tw + w'\t r
the thrust at A = +
Diagonal Stresses due to Live Load. Assume that the load is
concentrated at the panel points, and let it move from A
towards C.
If the diagonals slope as in Fig. 390, they are all ties, and
the live load produces the greatest stress in any one of them,
as QS, when all the panel points from A up to and including
Q are loaded.
BOWSTRING TRUSS. 621
Let ;r, y be the horizontal and vertical co-ordinates, respec-
tively, of any point on the parabola with respect to B as
origin.
The equation of the parabola is
Let the tangent at the apex P meet DB produced in L,
and DC produced in E.
Draw the horizontal line PM.
From the properties of the parabola, LM = 2BM.
Let PM x and BM = y.
From the similar triangles LMP and LDE,
.-
~'~~
x+QE'
A,,,
. &L _ l ~ 2x
'' QE~ 1+2X'
Draw EF perpendicular to QS produced, and imagine the
girder to be cut by a vertical plane a little on the right of PQ.
The portion of the girder between PQ and C is kept in
equilibrium by the reaction R at C> the thrust in the bow at P,
the tension in the tie at Q, and the stress in the diagonal QS.
Denote the stress by D n , and let the panel OQ be the nth.
Let be the inclination of QS to the horizontal.
Take moments about E. Then
RX CE,
or
D n = R* cosec B ....... (3)
622 THEORY OF STRUCTURES.
Let N be the total nuntber of panels. Then
is a panel length, and w'-r= is a panel weight.
Also, x = njj -, and hence
l-2x _CE___ Nn
7+ 2x ~ QE ~~ ~~^~*
R, the reaction at C when the n panel points preceding T
are loaded,
n(n
Thus, equation (3) becomes
A-y/(+i)^^cosecft .... (4)
Again, by equation (i),
_
" 2 ^ 2
N*
and
ST' ST
Hence, finally,
/ H -\TI \ (* # I )(^ ~T~ I ) I
M AT 2 < v /\
- (5)
This formula evidently applies to all the diagonals between D
and C.
BOWSTRING TRUSS. 623
Similarly, it may be easily shown that the stress in any
diagonal between D and A is given by an expression of pre-
cisely the same form.
Hence, the value of D n in equation (5) is general for the
whole girder.
A load moving from C towards A requires diagonals in-
clined in an opposite direction to those shown in Fig. 390.
Stresses in the Verticals due to the Live Load. Let V H be the
stress in the ?zth vertical PQ due to the live load. This stress
is evidently a compression, and is a maximum when all the
panel points from A up to and including O are loaded.
Imagine the girder to be cut by a plane S' S" very near PO,
Fig. 390. The portion of the girder between S'S" and C is kept
in equilibrium by the reaction R' at C, the thrust in the bow
at P, the tension in the tie at O, and the compression V n in
the vertical.
Take moments about E. Then
V n QE = X'X CE, or V n = R'^^-- ,
and R', the reaction at C when the (n i) panel points from A
to and i
Hence,
up to and including O are loaded, = / -
Vn = l-
2
a general formula for all the verticals.
Let v n be the tension in the nth vertical due to the dead
load. The resultant stress in it when the live load covers AO
is v^r- V H , and if negative, this is the maximum compression
to which PQ is subjected.
\iv n V n is positive, the vertical PQ is never in compression.
The maximum tension in a vertical occurs when the live
load covers the whole of the girder and = w'--*j + the tension
due to the dead load.
Note. The same results are obtained when N is odd.
624 THEORY OF STRUCTURES.
13. Bowstring Girder with Isosceles Bracing.
Diagonal Stresses due to the Dead Load. Under a dead load
the bow is equilibrated and the tie is subjected to a uniform
tensile stress equal in amount to the horizontal thrust at the
crown. The braces merely serve to transmit the load to the
bow and are all ties.
Let T l , 7", be the tensile stresses in the two diagonals
meeting at any panel point Q. Let 0, , # 2 be the inclinations
of the diagonals to the horizontal.
Let W be the panel weight suspended from Q.
The stress in the tie on each side of Q is the same, and
therefore T l , 7^, and Ware necessarily in equilibrium.
Hence,
T* 1X7 I J T* 117
T t W - -TT. ; 77-7, and T z = W-
Sin \y j -J- c/ 2 ) Sin ^i/j p t/ 2 y
Diagonal Stresses due to the Live Load. Let TV be the num-
ber of half panels.
2l
The length of a panel = -ry ; the weight at a panel point
Let the load move from A towards C. All the braces in-
clined like OP are ties, and all those inclined like QP are struts.
The live load produces the greatest stress in OP when it
covers the girder between A and O.
Denote this stress by D n ; OG is the th half-panel.
As before,
RX CE (i)
BOWSTRING GIRDER WITH ISOSCELES BRACING. 62$
I w'nl(n-\-2]
The load upon AO = nw -j-=, and hence R = -j~r' - TT .
The ratio of CE to EF is denoted by the same expression
as in the preceding article. Thus,
_i-
_w'ln-\-2N n[. N
" == ~8~;&+i AT N-n- I ~'.' '
The live load produces the greatest stress in OM when it
covers the girder up to and including D.
Denote the stress by D n f ; DG is now the nth half panel.
Let R f be the reaction at C.
As before,
CE
cosec 6, ...... (3)
being tjae angle MOD.
The weight upon AD = (n i)ze/-r~
and hence
2 A"' '
It may be easily shown, as in the preceding article, that
CE N-n-, A .
TvS - i - y an d cosec = N - TT^TT - -
OE n+i 4nk(Nn)
7"") ^ "^ f \
n ~~~~~%k n N N-n ~
Hence, when the load moves from A towards C, eq. (2)
gives the diagonal stress when n is even, and eq. (4) gives the
stress when n is odd.
If the load moves from C towards A, the stresses are re-
versed in kind, so that the braces have to be designed to act
both as struts and ties.
626
THEORY OF STRUCTURES.
. By inverting Fig. 391, a bowstring girder is obtained
with the horizontal chord in compression and the bow in
tension.
14. Bowstring Suspension Bridge (Lenticular Truss}.
This bridge is a combination of the ordinary and inverted
bowstrings. The most important example is that erected at
Saltash, Cornwall, which has a clear span of 445 feet. The
bow is a wrought-iron tube of an elliptical section stiffened
at intervals by diaphragms, and the tie is a pair of chains.
A girder of this class may be made to resist the action of a
passing load either by the stiffness of the bow or by diagonal
bracing.
B Q
In Fig. 392, let BD = k, B'D = k'.
Let //be the horizontal thrust at B, and T the horizontal pull
at B' ', when the live load covers the whole of the girder. Then
fjei _] not' 7 2
rr 1
L i z./ *
8
First, iet k = k' . Then
TT y
f
1 6
which is one half of the corresponding stress in a bowstring
girder of span / and depth k.
One half of the total load is supported by the bow and one
half is transmitted through the verticals to the tie. Hence,
the stress in each vertical -(w' -\- w"\
w" being the portion of the dead weight per lineal foot borne
by the verticals, and A^the number of panels.
CANTILEVER TRUSSES. 627
The diagonals are strained only under a passing load.
Let PP' be a vertical through , the point of intersection
of any two diagonals in the same panel, and let the load move
from A towards O.
By drawing the tangent at P and proceeding as in. Art. 13,
the expression for the diagonal stress in QS becomes, as before,
Similarly, the stress in the vertical QQ is
/ w'n(n i)l -
Next, let k and k' be unequal.
Let Wbe the weight of the bow, W the weight of the tie.
Then, under these loads,
~%"k = H ^ H ' : ~8~' r W'~~k'* ' ' ^
The verticals are not strained unless the platform is attached
to them along the common chord ADO. In such a case, the
weight of the platform is to be included in W'
The tangents at P and P' evidently meet AO produced in
the same point O ', for EO' is independent of k or k'. Hence,
the stresses in the verticals and diagonals due to the passing
load may be obtained as before.
15. Cantilever Trusses. A cantilever is a structure sup-
ported at one end only, and a bridge of which such a structure
forms part may be called a cantilever bridge. Two cantilevers
BEUD.GE.OVER ST. LAWRENCE AT NIAGARA.
FIG. 393.
may project from the supports so as to meet, or a gap may be
left between them which may be bridged by an independent
628
THEORY OF STRUCTURES.
girder resting upon or hinged to the ends of the cantilevers.
The form of the cantilever is subject to considerable variation.
SUKKUR BRIDGE
FIG. 394.
\ / \ /^T*"^^*^_
FORTH BRIDGE.
FIG. 395-
Figs. 396 to 401 represent the simplest forms of a cantilever
frame. If the member AB has to support a uniformly dis-
FIG. 396.
A J, i 4, -IB
FIG. 398.
FIG. 397.
FIG. 400.
FIG. 401.
FIG. 403.
tributed load as well as a concentrated load at B, intermediate
stays may be introduced as shown by the full or by the dotted
CANTILEVER TRUSSES. 629
lines in Figs. 398 and 399. Should a live load travel over
AB, each stay must be designed to bear with safety the
maximum stress to which it may be subjected.
Figs. 400 and 401 show cantilever trusses with parallel
chords. If the truss is of the double-intersection type, Fig.
401, the stresses in the members terminating in B become in-
determinate. They may be made determinate by introducing
a short link BD, Fig. 402. Thus, if, in DB produced, BG be
taken to represent the resultant stress along the link, and if
the parallelogram HK be completed, BK will represent the
stress along BE, and BH that along BF.
This link device has been employed to equalize the pressure
on the turn-table TT of a swing-bridge (Fig. 403). An " equal-
izer" or " rocker-link" BD, Fig. 404, conveys the stresses trans-
mitted through the members of the truss terminating in D to
the centre posts BT.
Theoretically, therefore, the pressure over TT will be evenly
distributed, whatever the loading may be, if the direction of
BD bisects the angle TB T and if friction is neglected.
The joint between the central span and the cantilever re-
quires the most careful consideration and should fulfil the
following conditions:
(a) The two cantilevers should be free to expand and con-
tract under changes of temperature.
(b) The central span should have a longitudinal support
which will enable it to withstand the effect of the braking of a
train or the pressure of a wind blowing longitudinally.
(c) The wind-pressure on the central span should bear
equally on the two cantilevers.
(d) The connections at both ends should have sufficient
lateral rigidity to check undue lateral vibration. Conditions
(a) and (c) would be fulfilled by supporting the central span
like an ordinary bridge-truss upon a rocker bolted down at one
end and upon a rocker resting on expansion rollers at the
other. This, however, would not satisfy condition (b). It is
preferable to support the span by means of rollers or links at
both ends, and to secure it to one cantilever only on the
central line of the bridge with a large vertical pin, adapted to
630 THEORY OF STRUCTURES.
transmit all the lateral shearing force. A similar pin at the
other end, free to move in an elongated hole, or some equiva-
lent arrangement, as, e.g., a sleeve-joint bearing laterally and
with rollers in the seat, is a satisfactory method of transmitting
the shearing force at that end also. (If there is an end post, it
may be made to act like a hinge so as to allow for expansion,
etc.) The points of contrary flexure of the whole bridge under
wind-pressure are thus fixed, and all uncertainty as to wind-
stresses removed.
Where other spans have to be built adjacent to a large can-
tilever span, it should not be hastily assumed that it is neces-
sarily best to counterbalance the cantilever by a contiguous
cantilever in the opposite direction. If it is possible to obtain
good foundations arid if piers are not expensive, it might be
cheaper to build a number of short independent side spans and
to secure the cantilever to an independent anchorage. If this
is done, care must be taken to give the abutment sufficient sta-
bility to take up the unbalanced thrust along the lower boom
of the cantilever.
Suppose that the cantilever is anchored back by means of
a single back-stay.
Let W = weight necessary to resist the pull of the back-
stay ;
h = depth of end post of cantilever ;
z = horizontal distance between foot of post and
anchorage ;
M bending moment at abutment = Wz.
If it is now assumed that the sectional areas of the post
and back-stay are proportioned to the stresses they have to bear
(which is never the case in practice), the quantity of material in
these members must be proportional to
!?* Wh = W** =
which is a minimum when z
If a horizontal member is introduced between the feet of
CANTILEVER TRUSSES. 631
the back-stay and the post, the quantity of material becomes
proportional to
h h
which is a minimum when z h, i.e., when the back-stay slopes
at an angle of 45. By making the angle between the back-
stay and the horizontal a little less than 45, a certain amount
of material may be saved in the joints of the back-stays and
also in the anchors, which more than compensates for the in-
creased weight of the anchors themselves.
(Note. In these calculations it is also assumed that the top
chord is horizontal, and that the feet of the post and back stay
are in the same horizontal plane. This is rarely the case in
practice.)
According to the above the weight of material necessary
for the back-stay is directly proportional to the bending moment
a the abutment and inversely proportional to the depth of the
cantilever, other things being equal. A double cantilever has,
in general, no anchorage of any great importance.
If the span is very great, a cantilever bridge usually re-
quires less material than any other rigid structure of equal
strength, even though anchorage may have to be provided.
If two large spans are to be built, a. double cantilever, requir-
ing no anchorage, may effect a very considerable saving in
material, although a double pier, of sufficient width for stability
under all conditions of loading, will be necessary.
Again, where false-works are costly or impossible, the
property of the cantilever, that it can be made to support
itself during erection, gives it an immense advantage. If the
design of the cantilever is such that it can be built out rapidly
and cheaply, it will often be the most economical frame in the
end, even if the total quantity of material is not so small as
that required for some other type of bridge. In all engineering'
work, quantity of material is only one of the elements of cost,
and this should be carefully borne in mind when designing a
cantilever bridge, because a want of regard to the method of
632 THEORY OF STRUCTURES.
erection may easily add to its cost an amount much greater
than can be saved by economizing material.
In ordinary bridge-trusses the amount of the web metal is
greatest at the ends and least at the centre, while the amount
of the chord metal is least at the ends and greatest at the
centre. Thus, the assumption of a uniformly distributed dead
load for such bridges is, generally speaking, sufficiently ac-
curate for practical purposes. In the case of cantilever
bridges, however, the circumstances are entirely different. In
these the amount of the metal both in the web and in the
chords is greatest at the support and least* at the end. For
example, the weight of the cantilevers (exclusive of the weight
of platform, viz., -J ton per lineal foot) for the Indus Bridge,
per lineal foot, varies from 6J tons at the supports to I ton at
the outer ends. Hence, the hypothesis of a uniformly dis-
tributed dead load for such structures cannot hold good.
The weight of a cantilever for a given span may be approxi-
mately calculated in the following manner :
Determine the stresses in the several members, panel by
panel
(A) For a load consisting of
(1) a given unit weight, say 100 tons, at the outer end ;
(2) the corresponding dead weight.
(B) For a load consisting of
(1) the specified live load ;
(2) the corresponding panel dead weight.
Thus, the whole weight of a panel will be the sum of the
weights deduced in (A) and (B), and the total weight of the
cantilever will be the sum of the several panel weights.
This process evidently gives at the same time the weights
of cantilevers of one, two, three, etc., panel lengths, the loads
remaining the same.
The panel dead weights referred to in (A) and (B) must, in
the first place, be assumed. This can be done with a large de-
gree of accuracy, as the dead weight must necessarily gradiially
increase towards the support, and any error in a particular
panel may be easily rectified by subsequent calculations.
CAN TILE VER TR USSES.
633
Again, the preceding remarks indicate a method of finding
the most economical cantilever length in any given case.
Take, e.g., an opening spanned by two equal cantilevers and
an intermediate girder. Having selected the type of bridge to
be employed for the intermediate span, estimate, either from
existing bridges or otherwise, the weights of independent
bridges of the same type and of different spans. Sketch a
skeleton diagram of the cantilever, extending over one-half of
the whole span, and apply to it the processes referred to in (A)
and (B).
If L is the length of the cantilever and P that of a panel,
the following table, in which the intermediate span increases
by two panel lengths at a time, may be prepared :
c
1 1 1)
S-, j
P-a
||6
c
rt
u
rt2
li:ll
Jf2.|4
.11
0.23
|||
'"Id
llll
*o C
ss
||
ill?
JltJl
Is
o c
3"'
aT" 5
^ ,
a
&"
&""
f*
H C
L
2P
L-2P
6P
L - 6P
SP
L - SP
etc.
etc.
Weight in col. 3 = one-half vi the weight of the intermediate
girder
+ one-half of the live load it carries if uni-
formly distributed. (The proportion will
be greater than one-half for arbitrarily
distributed loads, and may be easily de-
termined in the usual manner.)
Col. 5 gives the weights obtained as in A.
weight on end of cantilever
Col. 6 = col. 5 X -.
Col. 7 gives the weights obtained as in B.
Col. 8 = col. 2 -f- col. 6 -f col. 7.
It is important to bear in mind that an increase in the weight
of the central span necessitates a corresponding increase in the
634
THEORY OF STRUCTURES.
weights of the cantilevers. Hence, in order that the weight of
the structure may be a minimum, the best material with the
highest practicable working unit stress should be employed for
the centre span.
The table must of course be modified to meet the require-
ments of different sites. Thus, if anchorage is needed, a column
may be added for the weights of the back-stays, etc.
16. Curve of Cantilever Boom. Consider a cantilever
with one horizontal boom OA, and let x, y be the co-ordinates
of any point P in the other boom, O being the origin of co-or-
-a J
\y
FIG. 405.
FIG. 406.
dinates and A the abutment end of the cantilever.
Let Wbe the portion of the weight of an independent span
supported at O.
Let w be the intensity of the load at the vertical section
through P.
Assume (i) that there are no diagonal strains, and, hence,
that the web consists of vertical members only ;
(2) that the stress H in the horizontal boom is
constant, and therefore the bending moment
at P = Hy ;
(3) that the whole load is transmitted through the
vertical members of the web.
Let k be such a factor that kTl is the weight of a member
of length /, subjected to a stress T.
(Note. If / is in feet and T in tons, then k for steel is about
.0003, allowance being made for loss of section or increase of
weight at connections.)
w consists of two parts, viz., a constant part p, due to the
weight of the platform, wind-bracing, etc., which is assumed to
CURVE OF CANTILEVER BOOM. 635
be uniformly distributed ; and a variable part, due to the
weight of the cantilever, which may be obtained as follows:
Weight of element dx of horizontal boom = kHdx.
" " web corresponding to dx = kwydx.
" " element of curved boom corresponding to dx
-).
Hence the variable intensity of weight
and
w =
Again, if M is the bending moment and 5 the shearing
force at the vertical section through P, then
d'M ttS d*y
kH
Integrating twice,
A and B being constants of integration.
dv
When x = o, y = o, and ffg- = W.
Thus, A=o and B = W.
636 THEORY OF STRUCTURES.
Hence,
is the equation to the curve of the boom, and represents an
ellipse with its major axis vertical, and with the lengths of the
(p + 2kH\*
two axes in a ratio equal to ^ -777 -- J .
The depth of the longest cantilever is determined by the
vertical tangent at the end of the minor axis, and corresponds
to the value of y given by making = o in the preceding
equation, which gives y = .
For a given value of H the curve of the boom is independ-
ent of the span. Again, for a given length of cantilever with
a boom of this elliptic form, a value of H may be found which
will make the total weight a minimum, and which will there-
fore give the most economical depth. Such an investigation,
however, can only be of interest to mathematicians, as the
hypotheses are far from being even approximately true in
practice, and the resulting depth would be obviously too great.
Assumption (i) on page 634 no longer holds when a live
load has to be considered. Diagonal bracings must then be
introduced, which become heavier as the depth increases, in
consequence of their increased length. The diagonal bracings
are also largely affected by the length of the panels. If the
panels are short, and if a great depth of cantilever, diminishing
rapidly away from the abutment, is used, the angles of the
diagonal bracing, near the abutment, will be unfavorable to
economy. This difficulty may be avoided by adopting a
double system of triangulation over the deeper part of the
cantilever only, or even a treble system for some distance in
a large span. The objections justly urged against multiple
systems of triangulation in trusses lose most of their force in
large cantilevers. In the first place, the method of erection
by building out insures that each diagonal shall take its proper
share of the dead load ; and in the second place, it should be
CURVE OF CANTILEVER BOOM. 637
remembered that only in large spans could a double system
have anything to recommend it, and then only near the abut-
ment where the stresses are greatest : in such cases the moving
load only produces a small portion of the entire stress in the
web. In practice, a compromise has to be made between dif-
ferent requirements, and the depth must be kept within such
limits as will admit of reasonable proportions in other respects,
while the diagonal ties or struts may be allowed to vary in in-
clination, to some extent, from one panel to another.
Again, in fixing the panel length, care must be taken that
there is no undue excess of platform weight, as this will pro-
duce ?, corresponding increase in the weight of the cantilever.
An excessive depth of cantilever generally causes an in-
crease in the cost of erection.
Both theory and practice, however, indicate that it will be
more advantageous to choose a greater depth for a cantilever
than for an ordinary girder bridge.
An ordinary proportion for a large girder bridge would be
one-ninth to one seventh of the span, and if for the girder were
substituted two cantilevers meeting in the middle of the span,
the depth might with advantage be considerably increased
beyond this proportion at the abutment, if it be reduced to nil
where the cantilevers meet. When a central span is introduced,
resting upon the ends of the two cantilevers, the concentrated
load on the end gives an additional reason for still further in-
creasing the depth at the abutment proportionally to the hngth
of the cantilever. The greatest economical depth has probably
been reached in the Indus bridge, in which the depth at the
abutment = .54 X length of cantilever. Probably the propor-
tion of one-third of the length of the cantilever would be
ample, except where the anchorage causes a considerable part
of the whole weight, but each case must be considered on its
own merits. The reduction of deflection obtained by increas-
ing the depth is also an appreciable consideration.
If a depth be chosen not widely different from that which
makes the quantity of material a minimum, the weight will be
only slightly increased, while it is possible that great structural
advantages may be gained in other directions. In recommend-
638 THEORY OF STRUCTURES.
ing a great depth for a cantilever at its abutment, it is assumed
that the depth will be continuously reduced from the abutment
outwards. If the load were continuously distributed, it is by
no means certain that a cantilever of uniform depth would re-
quire more material than one of varying depth, but it has
already been pointed out to what extent the weight of the
structure itself necessarily varies, and if the concentrated load
at the end were separately considered, the economical truss
would be a simple triangular frame of very great depth. From
economic considerations, it would be well to reduce the depth
of the cantilever at the outer end to nil, but in many cases it
is thought advisable to maintain a depth at this point equal to
that at the end of the central span, so that the latter may be
built out without false-works, under the same system of erection
as is pursued in the case of the cantilever. The post at the
ends of the central span and cantilever is sometimes hinged to
allow for expansion.
17. Deflection. A serious objection urged against can-
tilever bridges is the excessive and irregular deflection to which
they are sometimes subject. They usually deflect more than
ordinary truss-bridges, and the deflection is proportionately
increased under suddenly applied loads. In the endeavor to
recover its normal position, the cantilever springs back with
increased force and, owing to the small resistance offered by
the weight and stiffness at the outer end, there may result,
especially in light bridges, a kicking movement. It must, how-
ever, be borne in mind that the deflection, of which the impor-
tance in connection with iron bridges has always been recog-
nized, is not in itself necessarily an evil, except in so far as it is
an indication or a cause of over-strain.
18. The Statical Deflection, due to a quiescent load,
must be distinguished from what might be called the dynamical
deflection, i.e., the additional deflection due to a load in motion.
The former should not exceed the deflection corresponding to
the statical stresses for which the bridge is designed. The
amount of the dynamical deflection depends both upon the
nature of the lo^ds and upon the manner in which they are
applied, nor are there sufficient data to determine its value
LIVE LOAD. 639
even approximately. It certainly largely increases the statical
stresses and produces other ill effects of which little is known.
Hitherto, the question as to the deflection of framed struc-
tures has received but meagre attention, and formulae deduced
for solid girders have been employed with misleading results.
It would seem to be more scientific and correct to treat each
member separately and to consider its individual deformation.
19. Rollers. One end of a bridge usually rests upon nests
of turned wrought-iron or steel friction rollers running between
planed surfaces. The diameter of a roller should not be less
than 2 inches, and the pressure upon it in pounds per lineal
inch should not exceed 500 Vd if made of wrought-iron, or
600 tfa if made of steel, d being the diameter in inches.
20. Live Load. It is a common practice with many en-
gineers to specify the live load for a bridge as consisting 01 a
number of arbitrary concentrated weights which are more or
less equivalent to the loads thrown upon the locomotive and
car axles.
Figs. 407, 408, and 409 are examples of such practice.
90'
FIG. 407.
FIG. 408.
f
FIG. 409.
With such a live load, the determination of the position of
the locomotive and cars which will give a maximum shear and
a maximum bending moment at any section is much facilitated
by the principles enunciated in Art. 8, Chap. II.
640 THEORY OF STRUCTURES.
If the chords* are parallel, and if 5 is the maximum shear
transmitted through a diagonal inclined at an angle 6 to the
vertical, the maximum stress in that diagonal = vSsec 0, and
the corresponding stress transmitted to a chord through the
diagonal
= Ssec#sin# = 5 tan 6.
A modification is necessary when the chords are not paral-
lel. Consider, e.g., a truss with a horizontal bottom chord
and a top chord composed of inclined members. Retain the
same notation as in the article referred to, and let />, , D 9 be
the stresses corresponding to \hefirst and second distributions,
respectively, in a diagonal met by a vertical section between
the rth and (r -\- i)th weights. Also, let the member of the
upper chord cut by the same section be produced to meet the
horizontal chord produced in the point C.
FIG. 410.
Let AC = h, and let / be the perpendicular from C upon
the diagonal in question.
Taking moments about C,
D,p RJi w,
and
w r (h + / a r x) w r+l (h + / a r ^ x) . . .
- w r+q (A + I a r+q x).
It. is assumed, for simplicity, that no weights leave or ad-
vance upon the bridge.
/. A A ,
LIVE LOAD.
641
according as
a,) . . . w r (k + la r ) =
RJt w,(h + I a, x) w^(h + / a, x) . . .
- w r (h -\-l-a r x) tv r+l (h + I a r+I x) . . .
- w r+q (h + I - a r+q -f x\
or
=
where R q '(l -\- h) = algebraic sum of the moments, with respect
to C f of the weights transferred,
and
Hence,
=
according as
Take, e.g., the truss represented by the accompanying dia-
gram (Sault Ste. Marie Bridge), the live load being that shown
by Fig. 411, i.e., the loading from a Standard Consolidation
engine with four drivers and one leading wheel.
FIG. 411.
Span 239 ft.
Length of centre verticals 40 ft.; of end verticals 27 ft,
THEORY OF STRUCTURES.
Applying the principles referred to in the preceding it is
found that the distributions of live load, concentrated at the
panel points, which will give the maximum stresses in the
several members, may be tabulated as below :
End
Distribu-
Reac-
Load
Load
Load
Load
Load
Load
Load
Load
Load
tions.
tion
at A-
at/ 2 .
at/ 8 .
at/> 4 .
at A-
at/ 6 .
at/ 7 .
at/e-
at/,.
at A.
Case i
2
187990
162920
495<
38700
495oo
459 2 5
38700
43750
45925
36225
43750
36000
36225
36000
36000
36000
36000
36000
36000
3
124230
6400
47209
40200
43400
45800
37100
36000
36000
6400
47200
40200
43400
45800
37100
69410
6400
47200
40200
4^400
45800
37100
6
47400
6400
47200
40200
43400
45800
7
29100
6400
47200
40200
43400
" 8
15380
6400
47200
40200
Dead weight
121500
27000
27000
27000
27000
27000
27000
27000
27000
,27000
N.B. These numbers are convenient whole numbers within
about one-half of one per cent of the calculated results. The
panel length is also assumed to be 24 ft.
In Cases I arid 2 the third driver is at a panel point ; in the
remaining cases the second driver is at a panel point.
The dead weight includes the weight of the ironwork and
flooring. The panel loads may be easily calculated, either
analytically or graphically. For example, let A, B, C, D be
four consecutive panel points, and let the third driver be at B.
Panel load at A
,
= 7500 g + 12000
Panel load at B
/io8 2\
J = 11823, say 11,900 Ibs.
8o+236+288+232
> sa y 49*500 Ibs.
Panel load at C
87+21
288
= 38445, say 38,700 Ibs.
LIVE LOAD.
643
Or, graphically, upon the vertical through B (Fig. 412) take
BM to represent 7500 Ibs., and join AM. Let the vertical
through #, meet AM in , , and the horizontal through AM
in c t . Then a^ represents the portion of 7500 Ibs. borne at
B, and b l c l the portion borne at A.
Also, take BN to represent 12,000 Ibs. ; join ^4 N, CN. Let
the verticals through a^, a 9 , #A
FIG. 412.
represent the portions of each 12,000 Ibs. borne at B, while
AA> ^3 represent the portions borne at A, and 4 4 the portion
borne at C.
Finally, take BO to represent 10,625 Ibs., and join CO. Let
the verticals through a 6 , a 6 meet CO in < 6 , ^ 6 , and the horizon-
tal through O in 6 , c % . Then # B ^ 5 , ajb^ are the portions of each
10,625 Ibs. borne at B, while b^c^, b 6 c 6 are the portions borne at
C. Thus the total weight at B
= <* A + "A + "A
A
A
It is open to grave question whether the extremely nice
calculations required by the assumption of arbitrary weight
calculations are not unnecessary except for floor systems. The
constantly increasing locomotive and car weights and the
variety in type of locomotive would seem to render such cal-
culations, based as they are upon one particular distribution of
load, of no effect.
On the other hand, if it is assumed that the standard live
load consists of a uniform load of, say, 3000 Ibs. to 3600 Ibs.
per lineal foot, with a single weight of, say, 25,000 Ibs. to
35,000 Ibs. for each truss, at the head or at any other specified
644
THEORY OF STRUCTURES.
point, i.e., rolling on the uniform load, the calculations would
be much simplified and the resulting stresses would be at least
as approximately accurate.
Let E be the single concentrated load, T the panel train
load, and D the panel dead load.
Consider a truss of N panels with a single diagonal system,
Fig. 413, and let E be at the rth panel point.
N-t N
FIG. 413.
The shear immediately in front of due to
the shear at same point due to T
~ N 2
the shear at same point due to D
D N(N 2r + I)
-W~ ~^ ~*
Diagonal Stresses. The maximum diagonal stresses may
now be easily tabulated as follows :
TABLE T.
2
O
S-o
o
" ' J5
%
^
%
V
3
Q ^
o
o
6
o
T
1
>
flj
%
Z
5
2 . |
Q
o
Diagonal.
Multiplier (N -
Max. Vertical S
E transmitted
(1)
Multiplier
(N-r-i
M
!p
Total Maximut
Shear due to
transmitted.
Secant 6.
Max. Diagonal S
Live Load.
Multiplier
N(N - 2r
Vertical Shear d
Load transmit
D
N
Diag. Stress due t
j
fj t
A
N.N-T
N-I.N--L T
JV i^
N.Ni
N.N-i D
a 1
*w
2
2 N
1 JV
2
2 N
N-a.N-i T
~~l N
LIVE LOAD. 645
Col. I designates the several diagonals.
Col. 2 gives the multiplier N r for different values of r.
Gol. 3 gives the maximum vertical shears due to E trans-
mitted through the several diagonals. This shear for any
given diagonal is the product of the corresponding multiplier
in col. 2 and -r^r.
Cols. 4 and 5, 9 and 10 give similar quantities for the live
and dead loads.
Col. 6 gives the sums of the shears in cols. 3 and 5, i.e., it
gives the total maximum vertical shears due to live load.
Col. 8 gives the maximum diagonal stresses due to live
load. For any specified diagonal it is the product of the cor-
responding shear in col. 7 and the secant of the angle between
the vertical and the diagonal in question.
Col. 1 1 in like manner gives the maximum diagonal stress
due to dead load.
Col. 12 gives the total maximum diagonal stresses due to
both live and dead loads.
Another column might be added giving the sectional areas
of the diagonals.
In the above table the diagonal stresses due to the live and
dead loads are separately determined, as different coefficients
of strength are sometimes specified for the two kinds of load.
With a suitable compound coefficient of strength, cols. 6, 8,
and ii may be replaced by a column giving the sums of the
corresponding shears in cols. 3, 5, and 10. These sums, multi-
plied by secant #, give the maximum diagonal stresses.
Stresses in the Verticals. The maximum stress in any ver-
tical, say at the rth panel point, is evidently the vertical com-
ponent of the maximum diagonal stress in the rth panel, i.e., it
is the maximum vertical shear in the rth panel.
To be more accurate, this amount should be diminished by
the portion of the weight of the lower chord borne at the foot
of the vertical in question.
Chord Stresses. Take the load at each panel point
= ~
646
THEORY OF STRUCTURES.
TABLE II. (COMPRESSION CHORD.)
-
T3X!
p
/
N
C
1
I
+
o
sl+
ll* 1
rtC H
^
-
11 -
5*21
^1
PH
I-
a
*
fl
gs^s
a
rt
^6^
SsS
-1
2
>
H
C/5
H
Col. i designates the chord panel length.
Col. 3 gives the several vertical shears transmitted to the
chords through the diagonals. They are the product of
W\W ~^~ ^ ~^~ i
Col. 5 gives the chord stresses due to these shears, i.e., the
product of the shears in col. 3 and the corresponding values of
tan in col. 4.
Col. 6 gives the total chord stresses in the several panels.
In any given panel the total chord stress is equal to the chord
stress due to the shear in that panel plus the total chord stress
in the preceding panel.
Another column for the sectional areas of the several
lengths of chord may be added if required, each length being
designed as a strut, hinged or fixed at the ends, according to
the method of construction.
A precisely similar
table may be prepared
for the tension chord.
EXAMPLE i. An
7V^/-panelled deck-truss
of 108 ft. span and 18 ft. deep, with a single diagonal system.
Concentrated load E for each truss = 25,000 Ibs.
Train load T for each truss = 1600 Ibs. per lineal ft.
= 21,600 Ibs. per panel.
Bridge (dead) load D for each truss = 800 Ibs. per lineal ft.
= 10,800 Ibs. per panel.
sec 8 = , tan 6 = f .
c 4 a
LIVE LOAD. 647
TABLE OF MAXIMUM DIAGONAL STRESSES. (See TABLE I.)
uS
T
1
8
fx
2
.2 g-o
i
8?
II
i
(I
co
II
u
"t3
i
00 o
i
k
Q
k
8!
3?*
c
C
x^>
. * J 'd
2 *
* rt
1
&
00
1
v8 oo
OWJiJ
5^3
0>
00
00
OQ
Q
00
N
Y
1
r 1
C/)
S^
*
M
Q"
H
4
7
21875
21
56700
78575
li
982,8}
28
37800
4725
145468}
da
10
.
8
^ OJ-O
nf^-d
1
L
il
v
m
_
N
S 3 g
.
S 3
fc
1
^
c^Q
si
a
II
71
II
u 1 -^
_
*^i-3
'a
I
\
o .
C/5
a
V
H
O 1
8 00
"5
O 1
Jloo
"5J.I
i
K^>
rt's'l
2.2 1
i
X
&l
X
51
H^^
in
|c^HJ
i
C" ^
H 5
!
7
6
5
4
21875
18750
15625
12500
21
6*
^}
56700
17550
9450
6750
78575
36300
25075
19250
I
982I8J
45375
45135
34650
28
"i
4i
37800
16875
"475
6075
47250
21093!-
20655
10935
i 454 68f
66465*
65790
45585
*4
3
9375
J-
1350
10725
1 r
J 935
1.
)j C
1215
20520
?
2
i
6250
1562*
1562*
J 3S
7600
1562*
15624
ii
13680
28124
1953*
- 34
~ll*
- 4725
10125
-'5525
- 8505
-18225
2 7945
5'75
The counterbrace cf is required to take up the resultant
shear of 6250 4725 = 1525 Ibs., which is opposite in kind to
that due to the dead load.
The first line in the table gives the maximum thrust along
the end post (/). It is made up of the stresses transmitted
LIVE LOAD.
649
through the two systems of diagonals when the 25,000 Ibs. is
at the first panel point.
TABLE OF MAXIMUM STRESSES IN VERTICALS.
The maximum stress in an end vertical evidently occurs
when the 25,000 Ibs. is concentrated at its foot.
z/j = 25000 + 10800 = 35800 Ibs. (tension);
^19250+ 6075 = 25325 " (compression);
v 3 = 10725 + 675 = 11400 "
v 4 = 6250 4725= 1525 " "
Chord Stresses. Load at each panel point
= ~+T+D = 35525.
TABLE OF MAXIMUM STRESSES IN COMPRESSION CHORD.
Total
Member
Multiplier.
25|5 = 444 of .
Tan 0.
Chord Stress
due to Shear.
Maximum
Chord
Stress.
( 28
I24337&
93253*
j
Cl
j +12*
55507^1
37745*
I
4163011
566171*
| T 9i50i$i
191501$!
c?
4*
19982^
.*
29974-^
221476^5
Cl
*
j
3330*|
224806^
Note. c l is made up of the thrusts transmitted through
TABLE OF MAXIMUM STRESSES IN TENSION CHORD.
Member.
Multi-
plier.
35525
Tang.
Chord Stress
due to Shear.
Total Maximum
Chord Stress.
8
/I = /S
28
124337*
i
93253*
93253^
/ 3
19}
i
''
37745 T 5 ir
i
56617!*
191501!*
Ex. 3. A through-bridge of the Warren type (Fig. 415)
having the same span and loading as in Exs. i and 2.
FIG. 415.
650 THEORY Of STRUCTURES.
TABLE OF MAXIMUM STRESSES IN DIAGONALS.
$
8
.
1
iii
1
0.
II
if.
t^
M
II
.a
13.
II
-"1
c.
11.
g eg
i
u
1
8 OO
"3
as
00
"3
00 OO
III
1
|II
d\ ~=-di
7
21875
21
56700
28
37800
116375
1.155
I344I4
dsdt
6
18750
15
40500
20
27000
86250
99619
df, = d*
5
15625
10
27OOO
12
16200
58825
67943
dT=d*
4
12500
6
16200
4
5400
34100
39386
dg=div
3
9375
3
8100
4
- 5400
12075
13947
dn=d^
2
6250
i
27OO
12
16200
d\$~=-d\i
I
3125
20
27000
The resultant stresses, d 9 = d l9 , are of an opposite kind to
the corresponding stresses due to the dead load. Thus, the
diagonals upon which they act must be designed so as to bear
both tensile and compressive stresses. The stresses d lt d s ,
d t , . . . are compressions, and d 9 , d^ , \S and {5, where 5 is the total area within the out-
line of the girder. The assumed factor of safety
is to be 4.
American Specifications. (a) The lateral bracing in the
plane of the roadway is to be designed so as to bear a pressure
of 30 Ibs. per square foot upon the vertical surface of one
truss and upon the surface of a train averaging 12 sq. ft. per
lineal foot, i.e., 360 Ibs. per lineal foot ; this latter is to be re-
garded as a live load. The lateral bracing in the plane of the
other chord is to be designed so as to bear a pressure of 50 Ibs.
per square foot upon twice the vertical surface of one truss.
(&) The portal, vertical, and horizontal bracing is to be
654
THEORY OF STRUCTURES.
proportioned for a pressure of 30 Ibs. per square foot upon
twice the vertical surface of one truss and upon the surface of
a train averaging 10 sq. ft. per lineal foot, i.e., 300 Ibs. per
lineal foot, the latter being treated as a live load.
(c) Live load in plane of roadway due to wind-pressure
= 300 Ibs. per lineal foot.
Fixed load in plane of roadway due to wind-pressure
= 150 Ibs. per lineal foot.
Fixed load in plane of other chord due to wind-pressure
= 150 Ibs. per lineal foot.
Lateral Bracing. Consider a truss-bridge with parallel
chords and panels of length/. Let A be the area of the ver-
tical surface of one truss.
According to (a), the lateral bracing in the plane of the
roadway is subjected to (i) a panel live load of 360^ Ibs. and
(2) a panel fixed load of $oA Ibs., while in the plane of the
other chord it is subjected to a panel fixed load of
50 X 2A 100^4 Ibs.
Thus, if the figure represent the bracing in the plane of the
roadway of a ten-panel truss, and if the wind blow upon the
30|A
30 A
80JA 36CJjP
3o|A
S6 P
FlG. 416.
36g P 360 P 360 P 36/ + 30^) tan 8 = 8(360^ + 30^) tan V Ibs. ;
C 4 C 3 + 2|(36o/ + $oA) tan B = \o%($6op + 30^) tan Ibs. ;
tan = \26o 0^ tan fl Ibs.
90 8 being the angle between a diagonal and a chord.
Again, the wind-pressure tends to capsize a train and throws
y
an additional pressure of P-=. Ibs. per lineal foot upon the lee-
ward rail, P being the pressure in pounds per lineal foot on the
train surface, y the vertical distance between the line of action
of P and the top of the rails, and G the gauge of the rails.
THEORY OF STRUCTURES.
Thus, the total pressure on leeward rail
(
w v\
2 +P G) lbS> Pei " Hneal f 0t >
and the total pressure on windward rail
(w y\
- P ~j Ibs. per lineal foot,
w being the weight of the train in pounds per lineal foot.
Hence, the total vertical pressure at a panel point of the
leeward truss
S being the distance between the trusses.
24. Stringers. Each length of stringer between consecu-
tive floor-beams may be regarded as an independent girder
resting upon supports at the ends, and should be designed to
bear with safety the absolute maximum bending moment to
which it may be subjected by the live load. If the beams are
not too far apart, the absolute maximum bending moment will
be at the centre when a driver is at that point. Again, in the
case of the Sault Ste. Marie Bridge, it may be easily shown
that the maximum bending moment is produced when the four
pairs of drivers are between the floor-beams.
Let/ distance of first driver from nearest point of support.
The reaction at this support
= 4tr(824 - 47) = 4^(206 -y).
The bending moment is evidently a maximum at the second
or third driver, and at the second driver
o.o.( 2 o6 y)($6 +7) 12000 X 56 ;
MAXIMUM ALLOWABLE STRESS.
at the third driver
= 104(206 7X108 +7) 12000(52 + 108).
In the first case it is an absolute maximum when y = 75" ;
" " second " " " " " " y 49" ;
its value in each case being 2,i88,i66| in.-lbs.
Hence, the bending moment is an absolute maximum and
equal to 2,i88,i66f in.-lbs., at two points distant 75 in. from
each point of support.
Also, if /! is the moment of inertia of the section of the
stringer at these points, , the distance of the neutral axis from
the outside skin, and/ t the coefficient of strength, then
-(2i88i66f) /i for the inner stringer,
and
-(2i88i66f) = /,- for the outer stringer.
3 c \
The continuity of the stringers adds considerably to their
strength.
25. Maximum Allowable Stress. Denoting by A and B,
respectively, the numerically greatest and least stresses to
which a member is to be subjected, the following rules will give
results which are in accordance with the best practice :
I. Members subjected to Tensile Stresses only.
For ivr ought-iron, maximum stress per square inch
= 10000 Ibs. 8000^1 + -jj Ibs. = ^3.81 + 1.9 -jj tons.
For steel, maximum stress per square inch
= 12000 Ibs. = 10000(^1 -f- -rj Ibs. = [,5.08 -f- 2.54 J tons.
658 THEORY OF STRUCTURES.
II. Members subjected both to Tensile and Comprcssivc Stresses.
For wr ought-iron, maximum stress per square inch
= 8000(1 - -jj Ibs. = (3 8l -- 1.9^) tons.
For steel, maximum stress per square inch
= 10000(1 - j) Ibs. = (5.08 - 2.54-) tons.
III. Members subjected to Compressive Stresses only.
Denote the ratio of the length (/) to the least radius of
gyration (k) by r.
The maximum stress per square inch = 2 Ibs.,
/being 8000 Ibs. for wrought-iron and 10,000 Ibs. for steel, and
being 40,000, 30,000, or 20,000, according as the member has
a
two square (fixed) ends, one square and one pin end, or two
pin ends.
Again, the maximum stress per square inch for steel struts
(/? \
1 + A! ft>s.;
A I
" " square ends (10000 4Or) I + -j Ibs.;
\ A'
" pin ends = (5 - ^)(i + ~) tons ;
" " square ends = ( 5 ^Jl 1 4~ ~/r] tons.
In the last two expressions r < 40. These expressions may
be also employed in the case of alternating stresses, but the
(T)
I -|
CAMBER. 659
26. Camber. Owing to the play at the joints, a girder or
truss will deflect to a much greater extent than is indicated by
theory, and the material will receive a permanent set, which,
however, will not prove detrimental to the stability of the
structure unless it is increased by subsequent loads. If the
chords were initially made straight, they would curve down-
wards ; and although it does not necessarily follow that the
strength of the truss would be sensibly impaired, the appear-
ance would not be pleasing.
In practice it is often specified that the girder or truss is to
have such a camber or upward convexity that under ordinary
loads the grade line will be true and straight ; or, again, that a
camber shall be given to the span by making the panel lengths
of the top chord greater than those of the bottom chord by
.125 in. for every 10 ft.
The lengths of the web members in a cambered truss are
not the same as if the chords were horizontal, and must be care-
fully calculated so as to insure that the several parts will fit
together.
To find an Approximate Value for the Camber, etc.
Let d be the depth of the truss.
Let s 1 , s y be the lengths of the upper and lower chords, re-
spectively.
Let /"j , yj be the unit stresses in upper and lower chords,
respectively.
Let d l , d^ be the distances of the neutral axis from the
upper and lower chords, respectively.
Let R be the radius of curvature of the neutral axis.
Let / be the span of the truss,
Then
4 -/ /I 4 /-*. /,
= - = and = - =
the chords being assumed to be circular arcs.
Hence, the excess in length of the upper over the lower
chord
660 THEORY OF STRUCTURES.
Let ;r, , x^ be the cambers of the upper and lower chords,
respectively ; R -f- d^ and R d^ are the radii of the upper
and lower chords, respectively.
By similar triangles,
the horizontal distance between ) R-\- d
the ends of the upper chord j " R
I ._ K-a
\-^ L
the horizontal distance between i R
the ends of the lower chord
Hence,
\1 "
and
j = x, . 2(R + d,\ approximately,
(- ^ -/) = x^ . 2(R d^), approximately.
,
and ** = SR I --R]-
27. Rivet-connection between Flanges and Web.
The web is generally riveted to angle-irons forming part of
the flanges.
The increment of the flange stress transmitted through
the web from point to point tends to make the angle-irons slide
over the flange surfaces.
Denote the increment by F, and let h be the effective depth
of the girder or truss.
Then, if 5 be the shearing force at any point,
Fh = the increment of the bending moment per unit
of length
ldM\
= l = = o in the case ot a close web,
\dx I
and Fh the increment of the bending moment
= (4M) = Sa in the case of an open web ;
a being the distance between the two consecutive apices or
panel points within which ^ lies.
EYE-BARS AND PINS.
66 1
Hence, if TV be the number of rivets per unit of length for
the close web, or the number between the two consecutive
apices for the open web,
N f s = F = Y for the close web,
and
= T- for the open web,
d being the diameter, of a rivet, and f s the safe coefficient of
shearing strength.
28. Eye-bars and Pins. Eye-bars connected with pins
have been commonly employed in the construction of suspen-
sion cables, the tension chords of ordinary trusses and canti-
levers, and the diagonals of web systems. The requisite sec-
tional area is obtained by placing a number of bars side by
side on the same pin, and, if necessary, by setting two or more
tiers of bars one above another.
FIG. 417.
rFh
rf=h
FIG. 418.
FIG. 419.
The figures represent groups of eye-bars as they often
occur in practice.
If two sets of 2n bars pull upon the pin in opposite direc-
tions, as in Figs, 418 and 419, the bending moment on the pin
will be nPp, P being the pull upon each bar, and / the distance
between the centre lines of two consecutive bars.
THEORY OF STRUCTURES.
Hence,
f being the stress in the material of the pin at a distance c from
the neutral axis, and / the moment of inertia.
In general, the bending action upon a pin connecting a
number of vertical, horizontal, and inclined bars may be de-
termined as follows :
Consider one-half of the pin only.
Let F, Fig. 420^ be the resultant stress in the vertical bars.
It is necessarily equal in magnitude but opposite in direc-
tion to the vertical component of the resultant of the stresses
h-*-
FIG. 420.
in the inclined bars. Let v be the distance between the lines
of action of these two resultants. The corresponding bending
action upon the pin is that due to a couple of which the mo-
ment is Vv.
Let h be the distance between the lines of action of the
equal resultants H of the horizontal stresses upon each side of
the pin. The corresponding bending action upon the pin is
that due to a couple of which the moment is Hh.
Hence, the maximum bending action is that due to a couple
of which the moment is the resultant of the two momenjts Vv
and Hh. viz.,
Eye-bars. In England it has been the practice to roll bars
having enlarged ends, and to forge the eyes under hydraulic
EYE-BARS AND PINS. 663
pressure with suitably shaped dies. In America both hammer-
forged and hydraulic-forged eye-bars are made, the latter being
called weldless eye-bars. Careful mathematical and experimen-
tal investigations have been carried out to determine the proper
dimensions of the link-head and pin, but owing to the very
complex character of the stresses developed in the metal around
the eye, an accurate mathematical solution is impossible.
Let d be the width and /
the thickness of the shank of
the eye-bar represented in Fig.
421. Let 5 be the width of
the metal at the sides of the
eye, and H the width at the
end. Let D be the diameter
of the pin.
The proportions of the head
are governed by the general condition that each and every part
should be at least as strong as the shank.
When the bar is subjected to a tensile stress the pin is
tightly embraced, and failure may arise from any one -of the
following causes :
(a) The pin may be shorn through.
Hence, if the pin is in double shear, its sectional area should
be at least one-half that of the shank.
It may happen that the pin is bent, but that fracture is pre-
vented by the closing up of the pieces between the pin-head
and nut ; the efficiency, however,of the connection is destroyed,
as the bars are no longer free to turn on the pin.
In practice, D for flat bars varies from f^to %d, but usually
lies between %d and ^d.
The diameter of the pin for the end of a round bar is gen-
erally made equal to \\ times the diameter of the bar.
The pin should be turned so as to fit the eye accurately,
but the best practice allows a difference of from -fa to ^4^ f
an inch in the diameters of the pin and eye.
(&) The link may tear across MN.
On account of the perforation of the head, the direct pull
on the shank is bent out of the straight and distributed aver
664 THEORY OF STRUCTURES.
the sections S. There is no reason for the assumption that
the distribution is uniform, and it is obviously probable that
the intensity of stress is greatest in the metal next the hole.
Hence, the sectional area of the metal across MN must be at
least equal to that of the shank, and in practice is always
greater.
5 usually varies from .55^ to .625^.
The sectional area through the sides of the eye in the head of
a round bar varies from i-J times to twice that of the bar.
(c) The pin may be torn through the head.
Theoretically, the sectional area of the metal across PQ
should be one-half that of the shank. The metal in front of
the pin, however, may be likened to a uniformly loaded girder
with both ends fixed, and is subjected to a bending as well as
to a shearing action. Hence, the minimum value of H has been
fixed at \d, and if H is made equal to d, both kinds of action
will be amply provided for.
(d) The bearing surface may be insufficient.
If such be the case, the intensity of the pressure upon the
bearing surface is excessive, the eye becomes oval, the metal is
upset, and a fracture takes place. Or again, as the hole elon-
gates, the metal in the sections 5 next the hole will be drawn
out, and a crack will commence, extending outwards until frac-
ture is produced.
In practice, adequate bearing surface may be obtained by
thickening the head so as to confine the maximum intensity of
the pressure within a given limit.
(e) The head may be torn through the shoulder at XY.
Hence, ^TFis made equal to d.
The radius of curvature R of the shoulder varies from I \d
to J.6d.
Note. The thickness of the shank should be , or d at
4 7
least.
The following table gives the eye-bar proportions common
in American practice :
STEEL EYE-BARS.
66 5
Value of 6".
of d.
of D.
Weldless
Hammered
Bars.
Bars.
.00
67
1-5
33
.00
75
1-5
33
.00
1. 00
1-5
50
.00
1.25
1.6
50
.00
1-33
1-7
.00
1.50
1.8 5
.67
.00
i-75
2.OO
67
.00
2.OO
2.25
75
Also, in weldless bars, //= S\ in hammered bars, H ' = d.
29. Steel Eye-bars. Hydraulic-forged steel eye-bars are
now being largely made. The steel has an ultimate tenacity of
from 60,000 to 68,000 Ibs. per square inch, an elastic limit of not
less than 50 per cent, and an elongation of from 17 to 20 per cent
in a length equal to ten times the least transverse dimension.
The Phcenix Bridge Co. and the Edge Moor Iron Co. give
the following tables of steel eye-bar proportions :
Phoenix Bridge Co.
Edge Moor Iron Co.
Mini- i Excess of
Width of
bar*/.
Diameter of
Pin-hole.
Diam-
eter of
Head.
Width
of
baroT.
Diameter
of
Pin-hole.
Diameter
of
Head.
mum
Thick-
ness
Sectional Area
of Head along
PP over Sec-
of Bar. 1 tion of Bar.
3
2&. 2 lf
7
2
Ii
*4
I
33#
3
3iV 3lf
8
2
2|
si
I
33*
4
3 T V
9
2i
2i
5+
I
33#
4
311, 4rV 4tt
10
2i
3i
6i i
33#
5
5
5
3 tf ' 4
4H 5 T V
sl-5, V*
ii
12
13
3
3
4
2i
4
4f
6*
8 f
9* f
33#
33#
33
6
4H- 4lt
13
4
3
ioi f
33^
6
5f, 5iV 5lf
14
5
41
i f
37^
6
6ft, 6tt. 6 Tf
15
5
5|
I2i
f
37^
7
5 A
15
6
5i
I3i
|
37
7
5lf, 6ft. 6H
16
6
6i
J4i
37^
8
6'if, 7ft, 7U
6f*
17
17
7
7
5*
7i
1 51
17
i
40^
40^
8
6ft, 6tf, 7 A
18
8
5f
17
i
40^
8
7li 8f
19
8
6f
18
i
4<$
8
ai,9t
20
9
9
7 T \- 7H
8|, 8|
2O
21
10
8|
22
10
8J,9f
23
10
10, 10^
24
In both the Phoenix and Edge Moor bars the thickness of the head is the
same as that of the body of the bar, or does not exceed it by more than -fa in.
O66 THEORY OF STRUCTURES.
30. Rivets. A rivet is an iron or steel shank, slightly
tapered at one end (the tail), and surmounted at the other by
a cup or pan-shaped head (Fig. 422). It is used to join steel or
iron plates, bars, etc. For this purpose the rivet is generally
heated to a cherry-red, the shank or spindle is passed through
D
FIG. 422. FIG. 423. FIG. 424. FIG. 425. FIG. 426.
the hole prepared for it, and the tail is made into a button, or
point. The hollow cup-tool gives to the point a nearly hemi-
spherical shape, and forms what is called a snap-rivet (Fig.
423). Snap-rivets, partly for the sake of appearance, are com-
monly used in girder-work, but they are not so tight as conical-
pointed rivets (staff-rivets), which are hammered into shape
until almost cold (Fig. 424).
When a smooth surface is required, the rivets are counter-sunk
(Fig. 425). The counter-sinking is drilled and may extend
through the plate, or a shoulder may be left at the inner edge.
Cold-riveting is adopted for the small rivets in boiler work
and also wherever heating is impracticable, but tightly-driven
turned bolts are sometimes substituted for the rivets. In all
such cases the material of the rivets or bolts should be of su-
perior quality.
Loose rivets are easily discovered by tapping, and, if very
loose, should be at once replaced. It must be borne in mind,
however, that expansions and contractions of a complicated
character invariably accompany hot-riveting, and it cannot be
supposed that the rivets will be perfectly tight. Indeed, it is
doubtful whether a rivet has any hold in a straight drilled hole,
except at the ends.
Riveting is accomplished either by hand or machine, the latter
being far the more effective. A machine will squeeze a rivet,
at almost any temperature, into a most irregular hole, but the
exigencies of practical conditions often prevent its use, except
for ordinary work, and its advantages can rarely be obtained
STRENGTH OF PUNCHED AND DRILLED PLATES. 667
where they would be most appreciated, as, e.g., in the riveting
up of connections.
31. Dimensions of Rivets. The diameter (d) of a rivet
in ordinary girder-work varies from f in. to I inch, and rarely
exceeds i in.
The thickness (/) of a plate in ordinary girder-work should
never be less than J in., and a thickness of f in., or even -^ in.,
is preferable.
Let T be the total thickness through which a rivet passes.
According to Fairbairn,
When / < % in., d should be about 2t.
When / > -J in., d should be about \\t.
According to Unwin,
When / varies from J in. to I in. and passes through
two thicknesses of plate, d lies between f t -f- jV and
When the rivets join several plates, d -- (- .
8 8
According to French practice,
Diameter of head if d.
Length of rivet from head = T -f- \\d.
According to Rankine,
Length of rivet from head = T -f- 2\d.
The rise of the head = \d.
The- diameter of the rivet-hole is made larger than that of
the shank by from ^ to -J in., so as to allow for the expansion
of the latter when hot.
There seems to be no objection to the use of long rivets,
provided they are properly heated and secured.
32. Strength of Punched and Drilled Plates. Experi-
ment shows that the tenacity of iron and steel plates is con-
siderably diminished by punching. This deterioration in
tenacity seems to be due to a molecular change in a narrow
annulus of the metal around the hole. The removal of the
annulus largely neutralizes the effect of the punching, and,
hence, the holes are sometimes punched -J in. less in diameter
than the rivets and are subsequently rimered or drilled out to
the full size. The original strength may also be almost
668 THEORY OF STRUCTURES.
entirely restored by annealing, and, generally, in steel work,
either this process is adopted or the annulus referred to above
is removed.
Punching does not sensibly affect the strength of Landore-
Siemens unannealed plates, and only slightly diminishes the
strength of thin steel plates, but causes a considerable loss of
tenacity in thick steel plates ; the loss, however, is less than
for iron plates.
The harder the material the greater is the loss of tenacity.
Iron seems to suffer more from punching when the holes
are near the edge than when removed to some distance from
it, while mild steel suffers less when the hole is one diameter
from the edge than when it is so far that there is no bulging at
the edge.
The injury caused by punching may be avoided by drilling
the holes. In important girder-work and whenever great
accuracy of workmanship is required, a uniform pitch may be
insured and the full strength of the metal retained by the use
of multiple drills. Drilling is a necessity for first-class work
when the diameter of the holes is less than the thickness of
the plate, and also when several plates are piled. It is impos-
sible to punch plates, bars, angles, etc., in spite of all ex-
pedients, in such a manner that the holes in any two exactly
correspond, and the irregularity becomes intensified in a pile,
the passage of the rivet often being completely blocked. A
drift, or rimer, is then driven through the hole by main force,
cracking and bending the plates in its passage, and separating
them one from another.
The holes may be punched for ordinary work, and in plates
sf which the thickness is less than the diameter of the rivets.
\Vhenever the metal is of an inferior quality, the holes should
be drilled.
33. Riveted Joints. In lap joints (Figs. 427 and 430) the
plates overlap and are riveted together by one or more rows
of rivets which are said to be in single shear, as each rivet has
to be sheared through one section only.
In fish (or btitf) joints (Figs. 428 and 429) the rivets are
in double shear, i.e., must be each sheared through two sections.
RIVETED JOINTS.
669
Thus they are not subjected to the one-sided pull to which
rivets in single shear are liable.
FIG. 427
In fish joints the ends of the plates meet, and the plates
are riveted to a single cover (Fig. 428), or to two covers (Fig.
429), by means of one or more rows of rivets on each side of the
joint.
A fish joint is properly termed a butt joint when the plates
are in compression. The plates should butt evenly against one
another, although they seldom do so in practice. Indeed, the
mere process of riveting draws the plates slightly apart, leav-
ing a gap which is often concealed by caulking. A much
better method is to fill up the space with some such hard sub-
stance as cast-zinc, but the best method, if the work will allow
of the increased cost, is to form a jump joint, i.e., to plane the
eyes of the plates carefully, and then bring them into close
contact, when a short cover with one or two rows of rivets will
suffice to hold them in position.
The riveting is said to be single, double, triple, etc., according
as the joint is secured by one, two, three, or more rows of rivets.
o o o
o o o
O O
o o o
o o o
o o o
^o
00
00
00
oo
ZIGZAG
FIG. 432.
CHAIN
FIG. 431.
Double, triple, etc., riveting may be chain (Fig. 431) or zig-
zag (Fig. 432). In the former case the rivets form straight
lines longitudinally and transversely, while in the latter the
rivets in each row divide the space between the rivets in adja-
cent rows. Experiments indicate that chain is somewhat
stronger than zigzag riveting.
670
THEORY OF STRUCTURES.
Figs. 433 to 435 show forms of joint usually adopted for
bridge-work. In boiler-work the rivets are necessarily very
close together, and if the strength of the solid plate be assumed
to be IOO, the strength of a single-riveted joint hardly exceeds
50, while double-riveting will only increase it to 60 or 70. Fair-
o o
o o o
o o o
o o
FIG. 433.
FIG. 434.
FIG.
435-
bairn proposed to make the joint and unpunched plate equally
strong by increasing the thickness of the punched portion of
the plate, but this is somewhat difficult in practice.
The stresses developed in a riveted joint are of a most com-
plex character and can hardly be subjected to exact mathe-
matical analysis. For example, the distribution of stress will
be necessarily irregular (a) if the pull upon the joint is one-
sided ; (b) when local action exists, or the plates stretch, or in-
ternal strains are in the metal before punching ; (c) if there is a
lack of symmetry in the arrangement of the rivets, so that one
rivet is more severely strained than another; (d) when the
workmanship is defective.
The joint may fail in any one of the following ways :
(1) The rivets may shear.
(2) The rivets may be forced into and crush the plate.
(3) The rivets may be torn out of the plate.
(4) The plate may tear in a direction transverse to that of
the stress.
The resistance to rupture should be the same in each of the
four cases, and always as great as possible.
The shearing and tensile strengths of plate-iron are very
nearly equal. Thus, iron with a tenacity of 20 tons per square
inch has a shearing strength of 18 to 20 tons per square inch.
Rivet-iron is usually somewhat stronger than plate-iron.
Again, the shearing strength of steel per square inch varies
THE ORE TIC A L DED UC TIONS. 67 1
from about 24 tons for steel, with a tenacity of about 30 tons,
to about 33 tons for steel, with a tenacity of about 50 tons ; an
average value for rivet-steel with a tenacity of 30 tons being
24 tons.
Hence, if 4 be a factor of safety, the working coefficient?
become
For wrought-iron \ * tons P er SC * uare inch in shear ' and
( 5 " " " ' ; tension.
_ . (6 tons per square inch in shear, and
F rSteel \n " " "tension
Allowance, however, must be made for irregularity in the dis-
tribution of stress and for defective workmanship, and in
riveting wrought-iron plates together it is a common practice
to make the aggregate section of the rivets at least equal to
and sometimes 20 per cent greater than the net section f the
plate through the rivet-holes.
Hence, the working coefficients are reduced to
4 or 4^ tons per square inch for wrought-iron,
and
5 or 5J " " " " " steel,
according to the character of the joint.
There is very little reliable information respecting the in-
dentation of plates by rivets and bolts, and it is most uncertain
to what extent the tenacity of the plates is affected by such
indentation. Further experiments are required to show the
effect of the crushing pressure upon the bearing area (i.e., the
diameter of the rivet multiplied by the thickness of the plate],
although a few indicate that the shearing strength of the rivet
diminishes after the intensity of the bearing pressure exceeds
a certain maximum limit.
34. Theoretical Deductions.
Let 5 be the total stress at a riveted joint ;
/i>/2> / a >/4> De the safe tensile, shearing, compressive,
and bearing unit stresses, respectively ;
/ be the thickness of a plate, and w its width ,
?2 THEORY OF STRUCTURES,
N\*t the total number of rivets on one side of a joint;
it be the total number of rivets in one row ;
p be the pitch of the rivets, i.e., the distance centre to
centre ;
d be the diameter of the rivets ;
x be the distance between the centre line of the nearest
row of rivets and the edge of the plate.
Value of x. It has been found that the minimum safe
value of x is d, and this in most cases gives a sufficient overlap
( 2x\ while x = \d is a maximum limit which amply pro-
vides for the bending and shearing to which the joint may be
subjected. Thus the overlap will vary from 2d to ^d.
x may be supposed to consist of a length x^ to resist the
shearing action, and a length x^ to resist the bending action.
It is impossible to determine theoretically the exact value of
JF,, as the straining at the joint is very complex, but the metal
in front of each rivet (the rivets at the ends of the joint ex-
cepted) may be likened to a uniformly loaded beam of length
d, depth x^ -- , and breadth /, with both ends fixed. Its
moment of resistance is therefore -?t Lr, j , / being the
maximum unit stress due to the bending. Also, if P is the
load upon the rivet, the mean of the bending moments at the
P
end and centre is ^d.
o
Hence, approximately,
* '--
It will be assumed that the shearing strength of the rivet
is equal to the strength of a beam to resist cross-breaking.
Single-riveted lap and single-cover joints (Figs. 427 and 428).
~~A = (p-d)tf^dtf^ ..... (I)
THEORETICAL DEDUCTIONS. 6/3
*xjj[i = -f % .
d , i / d'f, . .
-* = 2+ 4\/**-J7 ' ' (3)
As already pointed out, these joints are weakened by the
bending action developed, and possibly also by the concentra-
tion of the stress towards the inner faces of the plates.
Single-riveted double-cover joints (Fig. 429).
(4)
tf t = 2 /..
4
(5)
3 rfr 21 - 2 4
These joints are much stronger than joints with single
covers. Also, equation (4) shows that the bearing unit stress
in a double-cover joint is twice as great (theoretically] as in a
single-cover joint (eq. i), so that rivets of a larger diameter
may be employed in the latter than is possible in the former,
d
for corresponding values of -.
THEORY OF STRUCTURES.
Chain-riveted joints (Fig. 431).
5 = N -- / 9 when there is one cover only ; . . (8)
5 = N - / 2 when there are two covers. . . . (9)
This class of joint is employed for the flanges of bridge
girders, the plates being piled as in Figs. 436, 437, 438, and n
being usually 3, 4, or 5.
In Fig. 437 the plates are grouped so as to break joint, and
opinions differ as to whether this arrangement is superior to
the full butt shown in Fig. 438. The advantages of the latter
FIG. 436.
FIG. 437. FIG. 438.
are that the plates may be cut in uniform lengths, and the
flanges built up with a degree of accuracy which cannot be
otherwise attained, while the short and awkward pieces accom-
panying broken joints are dispensed with.
A good practical rule, and one saving much labor and ex-
pense, is to make the lengths of the plates, bars, etc., multiples
of the pitch, and to design the covers, connections, etc., so as
to interfere with the pitch as little as possible.
The distance between two consecutive joints of a group
(Fig. 437) is generally made equal to twice the pitch.
An excellent plan for lap and single-cover joints is to
arrange the rivets as shown in Figs. 431 to 435.
The strength of the plate at the joint is only weakened by
one rivet-hole, for the plate cannot tear a^t its weakest section,
COVERS. 675
i.e., along the central row of rivets (##), until the rivets be-
tween it and the edge are shorn in two.
Let there be m rows of rivets, i 1,22,
3 3, ... (Fig. 439)-
The total number of rivets is evi- (_
dently m\ O^O^O^O
Let /i i & > ? 3 , , , ... be the unit ten- O,XO I !
sile stresses in the plate along the lines I I,
2 2 > 3 3 respectively. Then
i
+
U
FIG. 439.
5 = (w d}tf^ = nfft , for the line i i ;
n ~J- I la.
m rj \ 4 mft^t
From this is finally deduced
eNnkf
The plate thickness may now be found by equation (9) ;
the diameter of rivet from d = kt, and the pitch from
md
p = - . In the above investigations no account has been
taken of the effect of the bearing pressure on the rivets or
plate.
If f c be the allowable bearing pressure per projected square
inch of rivet surface, the following relation must obtain :
. . . . . (14)
This may be written
fc ~ Nd
Then if f e be estimated by this equation, and if it should 1
greater than 43 tons per square inch in a lap joint, or 45 fr
tons in a butt joint, such joint will fail by the rivets shec
before the full strength of the plate is exerted, as Kent
experiments show that with these values of f c the rive
not attain their natural ultimate shearing strength (viz.,/ 5 ,,
fail at shearing stresses much below this.
EFFICIENCY OF RIVETED JOINTS. 68 1
Ar;ain, the maximum allowable ratio of (i.e., k] as the
preliminary datum for the design of a joint, may be fixed by
using the expression
<->
deduced from the obvious relation similar to (14)
eN-d*f s = Ndtf c .
4
win suggests the relation
o o 6 o
UUCJUU
^ c*
min
J'8i^
^j
O O Q O Q
* Q O O
ro O MD - ro
J -O
< -a
t-< Q,
U &
o i
W 5
J S P.V"
... <3 t-N
IAVO
-o oo -
TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 683
il II
.rts
CO
Single intersec
Single intersec
K
cH
6*6
sO
Wen
be ta_bfl
S S w
t* ~
OQ
U
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13 - s s
MMMrO,-O
Svg'S
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s
So v oV
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^5
So
Soj
U
*
^^O %o"b 'Vo'V'w "M V> V* * "N ^ ^o "V
^ ^- in 10 io\o t^t^t^. oooovocS o^
8^85? ^2
684
THEORY OF STRUCTURES.
1 J
5-
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BRIDGES
YEATMAN, <
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PEN-TRUSS
,D, PETERSON,
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c 1 2 c 1 ^2 N. S 2 2
c^ H c/5 c/oH HH Q
O O *J j U | O
& ,.
o ' c c c ^ffi ' c"^
o O O
:
^ ^
W
Q s |s fi
fc ^ -a 1
x
0^ PM"
W W
1
llljf
c/3
H
r^-sl
K ^ ^ ^K oo oooo ^
* ~ ~
T W
^ ^ .^ o
' * *
l-M p^
O
" O ""o
^"^H
s ^
1 0u
VrtVsSi fe SSaV^
miffi&S I
"
O O
ft c
^ c a
W M M M N Wf^fO
wj
TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 68$
X
X
X
X
X
I
o o
. o
.
j'
O.
5 s
6 w C 1-1' U HH U
U N4 **8*^J a g- |~g ~
5 s ^1? s
W
1
gs?
I
.g,!
S !
686
THEORY OF STRUCTURES.
fan S
o ^
H w
c . .
111 If *- g&l
1 8 n^ls|
10 ='Ou 00 'Stt
1 fiklU
xj fftr^sS . .
laa
a
2 a sjss"
Iff*
fffiS
-
.S
5 05
O O O Q Q O "TO O
2> c^oo^ o O O O t^. M
II
sis
L.
* i
< 0,
* -^-00 00 O
O\ t^ N O CO
oo N mop
f" rooo
4) W t) i)
O OJO
11
V
Id
c 5
r-82
t^l
s 2 .e:
JCJo
rj : '.-a rt
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l|t't-
e-'c-c-cs
aisdaajjqSnoj
TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 68?
Ss
So
o.t:
c a
J
unity. Show that the amount of ma-
terial in the struts and ties is a minimum when
tan a = k tan ft.
7. A lattice-girder of 40 ft. span, 5 ft. depth, and with horizontal
chords has a web composed of two systems of right-angled triangles and
is designed to support a dead and a live load, each of -J- ton per lineal
foot. Determine the maximum stresses in the members of the third
bay from one end met by a vertical plane.
Ans. \iriveted: Diagonal stress = V// 2 tons J
Chord stress = 27 tons.
If pin-connected : Diagonal stress = ff 1/2 and f ^2 tons ;
Chord stress = 26 tons.
8. A lattice-truss of 100 ft. span and 10 ft. depth has a web composed
of four systems of right-angled triangles. The maximum stress in the
EXAMPLES. 691
diagonal joining the sixth apex in the upper chord to the fourth apex in
the lower is 16 tons. Find the dead load, the live load being i ton per
lineal foot, assuming the truss to be (a) riveted, (b) pin-connected.
Ans. (a) .554 ton ; (b) 1.062 tons.
9. A lattice-girder of 40 ft. span has a web composed of two systems
of triangles (base = 10 ft.) and is designed to carry a live load of 1600
Ibs. per lineal foot and a dead load of 1200 Ibs. per lineal foot. Defin-
ing the stress-length of a member to be the product of its length into
the stress to which it is subjected find the depth of the truss so that its
total stress-length may be a minimum. Ans. 10.19 ft-
10. Determine the maximum stresses in the members of a lattice-
truss of 40 ft. span and 4 ft. depth, with two systems of triangles (base
= 8 ft.), (a) when riveted together; (b) when pin-connected. Dead load
= J ton per lineal foot, live load = $ ton per lineal foot.
Ans. Bays ist; 2d ; 3d; 4th; 5th.
(a) Diags. 6f 4/2~; 5.534/2"; 4.054/2"; 2.854/2"; if 4/2~tons ;
Tens.chd. 6| ; i8f; 2;f; 33!; 36$.
Comp. chd. Same.
( 7.5^2 p.rVS 4.7^2 j 2.34/1 v\ 2.34/2^
Tens.chd. 6; 18; 27; 33; 36 "
Comp. chd. Same.
ii. The platform of a single-track bridge is supported upon the top
chords of two Warren girders ; each girder is 100 ft. long, and its brac-
ing is formed of ten equilateral tfiangles (base 10 ft.); the dead weight
of the bridge is 900 Ibs. per lineal foot. ; the greatest total stress in the
seventh sloping member from one end when a train crosses the bridge
is 41,394.8 Ibs. Determine the weight of the live load per lineal foot.
Prepare a table showing the greatest stress in each bar and bay when a
single load of 1 5,000 Ibs. crosses the girder.
Ans. t\ t z >t 3 h ft
WWWXAAA/ 277H Ibs. per lin. ft.
FIG. 440.
Stresses in diagonals : 4/3 ; 2d = 10 4/3 ; 3d 5 / -^3
tons.
Cornp. chord : stress in ist bay = 2 F 4/3 ; 2d = 4 ? 4/3 ; 3d = 20 ^3
tons.
Weight at each joint = 5{f tons.
694 THEORY OF STRUCTURES.
1 8. The accompanying truss of 240 ft. span and 30 ft. deep is to be de-
signed for a panel engine load of 24,000 Ibs., a panel train load of 18,000
i Ibs., and a panel bridge load of 12,000
Ibs, Determine graphically the maximum
// stresses in the members met by the ver-
tical MN. Also, draw a stress diagram
for the whole truss when it is covered
FIG. 443. with a uniformly distributed live load of
180,000 Ibs.
19. Loads of 3f , 6, 6, 6, and 6 tons follow each other in order over a ten-
panel truss at distances of 8, 5f, 4^, and 4! ft. apart. Determine the posi-
tion of the loads which will give the maximum diagonal and chord
stresses in the third and fourth panels. Span = 120 ft.
20. Determine the moment of resistance of a floor-beam for the Sault
Ste. Marie Bridge from the following data : Floor-beams, 16' 6" long
and 23' lof" apart ; the dead weight of the flooring, stringers, etc.
= 800 Ibs. per lineal foot of floor-beam; the live load as given in Fig.-
407, Art. 20, page 639 ; the load is transmitted to the floor-beam by four
lines of stringers so spaced as to throw two-thirds of the load upon the
inner pair, which are 3 ft. C. to C.
21. In a truss-bridge the panels are 17 ft. and the floor-beams 13 ft.
in length. Loads of 8, 12, 12, 12, 12, 10, 10, 10, and 10 tons follow each
other in order over the bridge at the distances of 7^, 4^, 4^, 4!, 7^, 5^, 6,
and 5^ ft. apart. Determine the moment of resistance of the beam,
taking the load due to the platform, etc., to be 500 Ibs. per lineal foot.
22. If the bridge in the preceding question is of the riveted type with
a single diagonal system, and with verticals at the panel points, the num-
ber of the panels being ten, find how many i-in. rivets are required in
the third panel from one end to connect the web with the chords, assum-
ing the panel live load to be 30,000 Ibs. and the panel dead load to be
10,000 Ibs.
23. With the loading given by Fig. 409, Art. 20, page 639, design a
floor-beam for a single-track bridge with panels 22 ft. long, the weight of
the platform being 450 Ibs. per square yard, and of each longitudinal 200
Ibs. per lineal yard.
24. Prepare a table giving the stresses of the several members of a
double-intersection through-truss of 154 ft. span, 20 ft. depth, and with
eleven panels. The panel engine, live, and bridge loads are 56,000,
35,000, and 16,800 Ibs., respectively.
25. Prepare a table giving the stresses in the several members of a
double-intersection deck-truss of 342 ft. span, 33 ft. depth, and with
eighteen panels. (Double-track bridge.) The panel engine, train
(or live), and dead loads are 96,000, 54,000, and 36,200 Ibs., respectively.
26. Prepare a table giving the stresses in the several members of a
.EXAMPLES.
.695
through-truss for a double-track bridge of 342 ft. span, 40 ft. depth, and
with nineteen panels. The panel engine, live, and dead loads are 96,000,
53,000, and 43,200 Ibs., respectively (double-intersection).
Cl C2 C3 C4 C5 C6 C7 C8 C9
Ans.
FIG. 444.
Chord
Panel.
Mult.
7326
Mult.
5063
Total Shear
transmitted
to Panel.
Tan-
gent.
Panel Stress
from Shear.
Total
Panel
Stress.
t = t.
18
131868
i53
774639
906507
45
407929
407929
t^
- 7326
8oi
407571*
400245^
45
180110.475
588039
t^
72*
9
* 6
6ij-
9
53*
9
2
42*
9
t 6
- 7326
34*
1 74673*
167347*
9
150612$
tg
23*
9
'10 j -.
9
(
jeo
45
1
c \
i
1
<
80*
45
V
921696
1
72*
9
|
Cn
_
61*
9
c a
53*
9
C 4
42*
9
f|
34*
9
C 6
23*
9
C-r
_
15*
9
c
4*
9
'*
- 3*
9
1
Shear
Diag.
Mult.
553
Mult.
2790
due to
Live
Mult.
2274
Total
Shear.
Secant
Max.
Stress.
Load.
p
18
153
171
.0965
4t
J 7
63*
.0965
^2
16
56*
72*
345
\
i5
75795
70742
48*
42*
118575
2III IO
189317
53*
139851
121659
350961
310976
345
345
12
65689
60636
35*
30*
99045
85095
164734
42*
34*
96645
78453
261379
224184
345
345
301528
di
II
24*
23*
345
dg
10
20^
15*
345
dg
9
4*
345
^10
8
40424
I 2 *
34875
75299
- 3*
- 7959
345
90572
a'll
7
35371
8*
-14*
- 32973
345
35122
/1 = 139200 Ibs.
>2 = 350961 "
/3 = 310976 "
' 4 = 261379 Ibs.
Z' 5 = 224184 "
"6 = 177377 "
= 142972 Ibs.
= 98955 "
= 67340 "
27. Prepare a table showing the stresses in the several members'
(including counters) of a ten-panel double-track through railway bridge
of 184^ ft. span and 34 ft. depth, the live and dead loads being respect-
ively 2250 Ibs. and 1100 Ibs. per lineal foot. (Thamesville Bridge.)
28. Determine the minimum stress-length (stress in a member multi-
plied by its length) for a double-intersection Pratt truss of 154 ft. span and
with eleven panels. The panel loads for engine = 44,000 Ibs., for train =
696 THEORY OF STRUCTURES.
27,500 Ibs., for bridge = 13,200 Ibs. ; coefficient of working strength = 8000
Ibs. per square inch for both compression and tension.
29. A six-panel single-intersection Pratt truss is uniformly loaded.
Assuming the same coefficient of strength both for compression and
tension, show that the economy of material will be greatest when the
diagonals are inclined at 32 25' to the vertical.
30. A double-intersection truss for a single-track through-bridge of
204 ft. span is 29 ft. deep, 20 ft. wide, and has twelve panels. Find the
stresses produced in the members of the leeward truss by a panel wind-
pressure of 5000 Ibs. acting 8 ft. above base of rails (5-ft. gauge).
Ans. Sloping members : ist = 27500 sec oc ; 2d = 12708$ sec a ;
3d = 10208$ sec ft ; 4th = 7708$ sec ft\
5th = 5208$ sec a ; 6th = 2708$ sec ;
7th = 208$ sec ft.
Tension chord : ist panel = 27500 tan a = 2d ; 3d = 40208$ tan a;
4th = 40208$ tan a + 10208$ tan ft;
5th = 40208$ tan a + 17916! tan ft ;
6th = 40208$ tan a + 23125 tan ft.
Compression chord : ist = 40208$ tan a + 10208$ tan ft ;
2d = 40208$ tan a + 17916$ tan ft ;
3d = 40208$ tan n + 23125 tan ft ;
4th = 40208$ tan a + 25833$ tan ft ;
5th = 40208$ tan a + 26041! tan ft.
Verticals : ist = 5000 ; 2d = 7708$ ; 3d = 5208$;
4th = 2708$ ; 5th = 208$ Ibs.
tan a = $| ; tan ft = ff .
31. In the preceding question find the maximum stresses in the
members of the fourth panel met by a vertical plane ; engine panel load
= 85,000 Ibs., train panel load = 40,800 Ibs., bridge panel load 22,500
Ibs.
Ans. Stresses in tension chord = 456^430.45 Ibs. ; in compression
chord 645,311.77 Ibs. ; in sloping members = 206,242.5 Ibs.;
and 139,705.62 Ibs.
32. Each of the two Pratt single-intersection five-panel trusses for a
single-track bridge is 55 ft. centre to centre of end pins and 1 1 ft. 6 in.
deep. Timber floor-beams are laid upon the upper chords 2f ft. centre
to centre ; the width between the chords = 10 ft. Find the proper scant-
ling of the floor-beams for the loading given in Fig. 407, page 639.
Also determine the maximum chord and diagonal stresses in the centre
panel due to the same live load.
33. Prepare a table giving the stresses in the several members of a
double-intersection deck-truss of 342 ft. span, 40 ft. depth, and with
EXAMPLES. 697
nineteen panels. (Double-track bridge.) The panel engine, train (or
live), and dead loads are 96,000, 53,000, and 43,200 Ibs., respectively.
34. Prepare a table giving the stresses in the several members of a
deck-truss for a double-track bridge of 342 ft. span, 33 ft. depth, and
with eighteen panels. The panel engine, live, arid dead loads are 96,000,
54,000, and 36,000 Ibs., respectively.
35. The two trusses for a 16 ft. roadway are each 100 ft. in the clear,
17 ft. 3 in. deep, and of the, type repre-
sented in the figure ; under a live load of
1 1 20 Ibs. per lineal foot the greatest total
stress in AB is 35,400 Ibs. Determine the
permanent load. FIG. 445.
The diagonals and verticals are riveted to angle-irons forming
part of the flanges. How many f-in. rivets are required for the con-
nection of AB and BC at B ? Also, how many are required between
A and C to resist the tendency of the angle-irons to slip longitudinally ?
Working-shear stress = 10,000 Ibs. per square inch.
Ans. 708.9 Ibs. ; 8; 4; n.
36. The compression chord of a bowstring truss is a circular arc of
80 ft. span and 10 ft. rise ; the bracing is of the isosceles type, the bases
of the isosceles triangles dividing the tension chord into eight equal
lengths. Determine the maximum stresses in the members met by a
vertical plane 28 ft. from one end. The live and dead loads are each
i ton per lineal foot.
37. Design a parabolic bowstring truss of 80 ft. span and 10 ft. rise
for a dead load of ton and a live load of i ton per lineal foot. The
joints between the web and the tension chord are to divide the latter
into eight equal divisions.
38. The compression chord of a bowstring truss is a circular arc.
The depth of the truss is 14 ft. at the centre and 5 ft. at each end ; the
span = 100 ft. ; the load upon the truss = 840 Ibs. per lineal foot. Find
the stresses in all the members. Determine also the maximum stresses
in the members met by a vertical 25 ft. from one end when a live load
of looo Ibs. per lineal foot crosses the girder. What counter-braces are
required ?
39. A Pratt truss with sloping end posts has a length of 150 ft. centre
to centre, and a height of 30 ft. centre to centre, with panels 15 ft. long ;
the dead load is 3000 Ibs. per lineal foot, and the live load 1200 Ibs.
Determine the maximum stresses in the end posts, in the third post from
one end, in the middle of the bottom chord, and in the members of the
third panel met by a vertical plane.
40. Design a cross-tie for a double-track open-web bridge, the ties
698 THEORY OF STRUCTURES.
being 18 ft. 5 in. centre to centre, and the live load for the floor system
being 8000 Ibs. per lineal foot.
41. A bowstring roof-truss of 50 ft. span, 15 ft. rise, and five panels is
to be designed to resist a wind blowing horizontally with a pressure of
40 Ibs. per square foot* The depth of the truss at the centre is 10 ft.
Determine, graphically, the stresses in the several members of the truss,
assuming that the roof rests on rollers at the windward support.
42. A bowstring truss of 120 ft. span and 15 ft. rise is of the isosceles
braced type, the bases of the isosceles triangles dividing the tension
chord into twelve equal divisions ; the dead and live loads are ton and
i ton per lineal foot, respectively. Find the maximum stresses in the
members met by vertical planes immediately on the right of the second
and fourth joints in the tension chord.
43. The figure is a skeleton diagram of the Sault Ste. Marie Bridge
(C P. R.). Span = 239 ft. ; there are ten panels, each of 23.9 ft.,, say 24
493000499,000
275,000 275,000 416,000 478,000 484,000
FIG. 446.
ft. ; the length of the end verticals = 27 ft., of the centre verticals = 40
ft.; width on truss centres = 17^ ft. The bridge is designed to bear the
loading given by Fig. 407, page 639. Show that
(a) The stresses in every panel length of each chord are greatest when
third driver is at a panel point ; and find the value of the several
stresses.
: (b} The stresses in the verticals a and the diagonals b are greatest
when the third driver is at a panel point ; and find their values.
/ (f) The stresses in the remaining members of the truss are greatest
when the second driver is at a panel point ; and find their values.
(d) The maximum stresses in the verticals a! vary from a tension of
64,000 Ibs. to a compression of 11,000 Ibs.
(e) The stress in the counter-brace c is nil.
Am. The values of thestresses in the several members are marked
on the diagram. They are deduced from the distributions
given in the table on page 642, and are correct within a very
small percentage.
44. The figure represents a counterbalanced swing-bridge, 16 ft. deep
and wholly supported upon the turn-table at A
and B '; the dead weight is 650 Ibs. per lineal foot
of bridge ; the counterpoise is hung from C and D.
FIG. 447? Find its weight, assuming (a) that the whole of it
is transmitted to B ; (b) that a portion of it is transmitted to A through
EXAMPLES.
699
a member BE, sufficient to make the reactions at A and B equal.
determine the stresses in the several members of the truss.
Also,
Ans. Counterpoise in case (a) = 26,162$ Ibs. ;
in case (b) = 22,186$% Ibs.
Stress transmitted through BE in case ()= 24,012 Ibs.
45. The figure represents a counterbalanced swing-bridge ; the dead
load upon the bridge is 650 Ibs. per lineal
foot ; the counterpoise is suspended from
CD. Find its value, the joint at E being
so designed that the whole of the load
upon the bridge is always transmitted FIG. 44 b.
through the main posts EA, EB, and is evenly distributed between the
points of support at A and B.
Ans. 20,694.3 Ibs.
46. Find the stresses in the several members of the truss in the pre-
ceding question (a) when the bridge is open ; (b) when the bridge is
closed and is subjected to a live load of 3000 Ibs. per lineal foot.
Height of truss at E = 16 ft., at F = 8 ft.
47. Prepare a table giving the stresses in the several members of a
a cs c 3 C4 C6 . single-intersection through-truss of
y\ \ 2 N 3 J\ 4 JX 5 X 6 \ \ \ \ J 54 ft - span> 20 ft - de P th ' and with
/ fi\ff2\f3\ffAfo\j \J \j \J \J \ eleven panels. The panel engine,
IQ U& 3 14 tfl 1C
FIG. 449. live, and dead (or bridge) loads
27,500, 17,600, and 8470 Ibs., respectively.
Ans.
Diag.
Mult.
2500
Mult.
1600
Sum.
Mult.
770
Sum.
sec.
Total M%
Stress. Y
p
IO
25000
45
72000
97000
55
42350
139350
22
170007
d l
9
22500
36
57600
80100
44
3380
113980
22
139056
dn
g
200 oo
28
44800
64800
33
25410
90210
22
110057
<^3
7
17500
21
33600
51100
22
16940
68040
22
83009
6
15000
15
24000
39000
11
8470
47470
22
579 J 4
4 = 47,470 ; ^ 5 = 28,500 Ibs.
48. Compare the relative amounts of iron required in the webs of a
single- and a double-intersection Pratt deck-truss of zoo ft. span and
having eight panels. Panel live load = L, panel dead load = D.
49. The figure represents a pier, square in plan, supporting the ends
of two deck -trusses, each 200 ft. long and 30 ft. deep. The height of the
pier is 50 ft. and is made up of three panels, the
upper and lower being each 17 ft. deep. Ten
square feet of bridge surface and ten square feet
of train surface per lineal foot are subjected to
a wind-pressure of 40 Ibs. per square foot. The
centre of pressure for the bridge is 68 ft., and
for the train 86 ft., above the pier's base. The
wind also produces a horizontal pressure of 4000
Fia. 450. Ibs. at each of the intermediate panel points on
the windward side of the pier. Width of pier = 17 ft. at top and 33! ft.
at bottom. The bridge load = 1600 Ibs. per lineal foot, live load = 3000
Ibs. per lineal foot. Determine
(a) The overturning moment (3180 ft.-tons).
(b) The horizontal force due to the wind at the top of the pier.
(61.6 tons.)
(c) The tension in the vertical anchorage ties at 5 and T. (Nil.}
(d) The vertical and horizontal reactions at T. (275 and 65.6 tons.)
Draw a diagram giving the wind-stresses in all the members, and in-
dicate which are in tension and which in compression.
Ascertain whether the wind-pressure of 40 Ibs. per square foot upon
a train of empty cars weighing 900 Ibs. per lineal foot will produce a
tension anywhere in the inclined posts. What will be the tension in the
anchorage ties ? (20.75 tons.)
Find the stresses in the traction bracing (i) when a loaded train trav-
elling at 30 miles an hour is braked just as the engine is over the pier
and brought to rest in a length of 300 ft. ; (2) when a loaded train with
the engine over the pier is started by a sudden admission into the cylin-
ders of steam at 100 Ibs. per square inch. Stroke of cylinder = 16 in.,
diameter of drivers = 5 ft.
EXAMPLES.
701
50. The figure represents one half of one of the piers of the Bouble
Viaduct. The spans are crossed by two lattice-gir-
ders, 14' 9" deep and having a deck platform. The
height of the pier is 183' 9" and is made up of eleven
panels of equal depth. Width of pier at top = 13'
ii", at bottom = 67' 7". With wind-pressure at
55.3 Ibs. per square foot, the total pressure on the
girder, train, and pier have been calculated to be 20,
16.2, and 20 tons, acting at points 196.2, 210.3, and
92.85 ft., respectively, above the base. The dead
weight upon each half pier is 222^ tons, of which 60
tons is weight of half span, 120 tons the weight of
the half pier, and 42^ tons the weight of the train.
Assuming that the wind-pressure on the pier is a
iiorizontal force of 2 tons at each panel point on the
windward side, and that the weight of the pier may
be considered as a weight of 6 tons at each panel
point, determine
(a) The overturning moment. FIG. 451.
(b) The total horizontal force at the top of the pier due to the wind.
(c) The tension in each of the vertical anchorage ties at 5 and T due
to the wind-pressure.
(d) The vertical and horizontal reactions at T.
Show that the greatest compressive stress occurs in the member RT,
and that it amounts to 422 tons.
Draw a stress diagram giving the stresses in all the members, indi-
cating which are in tension and which in compression. Width of pier
at A = 20 ft., at = 23$ ft., at C = 36^ ft.
What will be the effect of braking the train when running at 30 miles
an hour, so as to bring it to rest within a distance of 220 ft. ? Width
of pier in direction of bridge = 9! ft. at top and = 20 ft. at bottom.
Ans. (a) 9188 ft.-tons; (ti) 39.9 tons; (c} 24! tons, (d) Hori-
zontal reaction = 59.9 tons ; vertical reaction = 247 tons.
The accompanying figure represents a portion of a cantilever truss,
the horizontal distances of the points A, B t C
from the free end being A , / a , I* , respectively.
The boom ABC is inclined at an angle a, and the
boom XYZ at an angle ft, to the horizon. Find
the deflections at the end of the cantilever due to
X Y z (a) an increase kAB in the length of AB\ (2) an
FIG. 452. increase kiB Y in the length of B Y ; (3) a decrease
k*XY in the length of XY\ (4) a decrease kBX in the length of BX.
w
Ans. (i)
BX sin ABX
702
THEORY OF STRUCTURES.
(3)
BX
(4)*
BX cos a
sin
- / 9 (cot BXY - cot
In the preceding question, if k l = k* = k* = ki = k, and if A W is
parallel to ^Jf, and AX to ^ F, show that the angle between WX and
xY F after deformation
= 2>&(cot ABX + cot B YX).
Hence also, if the truss is of uniform depth d, show that the " deviation "
2k
of the boom per unit of length is constant and equal to
d
52. Six bars have to be arranged upon a steel pin ; each bar is I in.
wide and is subjected to a stress of 64,000 Ibs. Should the bars be ar-
64,000 Ibs.
FIG. 454. Method 2.
ranged according to method i or method 2 ? Why ? Determine the di-
ameter of the pin.
53. The accompanying sketch represents one of the pin connections
in a certain bridge which was recently overthrown. The two innermost
bars are web members inclined to the horizon at an angle whose cosine
"^^^
-[42,100 Ib8.
FIG. 455.
is .815. The thickness of the bars and the maximum stresses to which
they are severally subjected are shown on the diagram. Is the 3-in.
wrought-iron pin sufficiently strong?
CHAPTER XII.
SUSPENSION-BRIDGES.
I. Cables. The modern suspension-bridge consists of two
or more cables from which the platform is suspended by iron
or steel rods. The cables pass over lofty supports (piers), and
are secured to anchorages upon which they exert a direct pull.
Chain or link cables are the most common in England and
Europe, and consist of iron or steel links set on edge and
pinned together. Formerly the links were made by welding
the heads to a flat bar, but they are now invariably rolled in
one piece, and the proportional dimensions of the head, which
in the old bridges are very imperfect, have been much im-
proved.
Hoop-iron cables have been used in a few cases, but the
practice is now abandoned, on account of the difficulty attend-
ing the manufacture of endless hoop-iron.
Wire-rope cables are the most common in America, and
form the strongest ties in proportion to their weight. They
consist of a number of parallel wire ropes or strands, compactly
bound together in a cylindrical bundle by a wire wound round
the outside. There are usually seven strands, one forming a
core round which are placed the remaining six. It was found
impossible to employ a seven-strand cable in the construction
of the East River Bridge, New York, as the individual strands
would have been far too bulky to manipulate. The same ob-
jection held against a thirteen-strand cable (thirteen is the next
number giving an approximately cylindrical shape), and it was
finally decided to make the cable with nineteen strands. Seven
of these are pressed together so as to form a centre core, around
which are placed the remaining twelve, the whole being con-
tinuously wrapped with wire.
703
704
THEORY OF STRUCTURES.
In laying up a cable great care is required to distribute the
tension uniformly amongst the wires. This may be effected
either by giving each wire the same deflection or by using
straight wire, i.e., wire which when unrolled upon the floor
from a coil remains straight and shows no tendency to spring
back. The distribution of stress is practically uniform in un-
twisted wire ropes. Such ropes are spun from the wires and
strands without giving any twist to individual wires.
The back-stay is the portion of the cable extending from an
anchorage to the nearest pier.
The elevation of the cables should be sufficient to allow for
settling, which chiefly arises from the deflection due to the load
and from changes of temperature.
The cables may be protected from atmospheric influence
by giving them a thorough coating of paint, oil, or varnish, but
wherever they are subject to saline influence, zinc seems to be
the only certain safeguard.
2. Anchorage, Anchorage Chains, Saddles. The an-
chorage, or abutment, is a heavy mass of masonry or natural
rock to which the end of a cable is made fast, and which re-
sists by its dead weight the pull upon the cable.
FIG. 456. FIG. 457. FIG. 458.
The cable traverses the anchorage as in Figs. 456 to 458,
and passes through a strong, heavy cast-iron anchor-plate, and,
if made of wire rope, has its end effectively secured by turning
it round a dead-eye and splicing it to itself. Much care, how-
ever, is required to prevent a wire-rope cable from rusting on
account of the great extent of its surface, and it is considered
advisable that the wire portion of the cable should always ter-
minate at the entrance to the anchorage and there be attached
to a massive chain of bars, which is continued to the anchor-
plate or plates and secured by bolts, wedges, or keys.
ANCHORAGE, ANCHORAGE CHAINS, SADDLES.
In order to reduce as much as possible the depth to which
it is necessary to sink the anchor-plates, the anchor-chains are
frequently curved as in Fig. 458. This gives rise to an oblique
force, and the masonry in the part of the abutment subjected
to such force should be laid with its beds perpendicular to the
line of thrust.
The anchor-chains are made of compound links consisting
alternately of an odd and an even number of bars. The friction
of the link-heads on the knuckle-plates considerably lessens
the stress in a chain, and it is therefore usual to diminish its
sectional area gradually from the entrance E to the anchor.
This is effected in the Niagara Suspension Bridge by varying
the section of the bars, and in the East River Bridge by vary-
ing both the section and the number of the bars.
The necessity of preserving the anchor-chains from rust is
of such importance that many engineers consider it most
essential that the passages and channels containing the chains
and fastenings should be accessible for periodical examination,
painting, and repairs. This is unnecessary if the chains are
first chemically cleaned and then embedded in good hydraulic
cement, as they will thus be perfectly protected from all at-
mospheric influence.
The direction of an anchor-chain is changed by means of a
saddle or knuckle-plate, which should be capable of sliding to
an extent sufficient to allow for the expansion and contraction
of the chain. This may be accomplished without the aid of
rollers by bedding the saddle upon a four- or five-inch thickness
of asphalted felt.
The chain, where it passes over the piers, rests on saddles,
the object of which is to furnish
bearings with easy vertical curves.
Either the saddle may be constructed
as in Fig. 459, so as to allow the
cable to slip over it with compara-
tively little friction, or the chain may be secured to the saddle,
and the saddle supported upon rollers which work over a per-
fectly true and horizontal bed formed by a saddle-plate fixed
to the pier.
7 o6
THEORY OF STRUCTURES.
3. Suspenders. The suspenders are the vertical or in-
clined rods which carry the platform.
FIG. 460.
FIG. 461.
FIG. 462.
FIG. 463.
In Fig. 460 the suspender rests in the groove of a cast-
iron yoke which straddles the cable. Fig. 461 shows the
suspender bolted to a wrought-iron or steel ring which em-
braces the cable. When there are more than two cables in
the same vertical plane, various methods are adopted to insure
the uniform distribution of the load amongst the set. In Fig.
462, for example, the suspender is fastened to the centre of a
small wrougnt-iron lever PQ, and the ends of the lever are
connected with the cables by the equally strained rods PR
and QS. In the Chelsea bridge the distribution is made by
means of an irregularly shaped plate (Fig. 463), one angle of
which is supported by a joint-pin, while a pin also passes
through another angle and rests upon one of the chains.
The suspenders carry the ends of the cross-girders (floor-
beams), and are spaced from 5 to 20 ft. apart. They should
be provided with wrought-iron screw-boxes for purposes of
adjustment.
4. Curve of Cable. CASE A. An arbitrarily loaded flexible
cable takes the shape of one of the catenaries, but the true
catenary is the curve in which a cable of uniform section and
material hangs under its own weight only.
Let A be the lowest point of the cable, and take the ver-
tical through A as the axis of y.
Take the horizontal through O as the axis of x, the origin
O being chosen so that
Hmp, ...... (i)
/ being the weight of a unit of length of the cable, and H the
horizontal pull at A.
CURVE OF CABLE.
707
m or AO is the parameter, or modulus, of the catenary, and
OG is the directrix.
Let x, y be the co-ordinates of any point P, the length of
the arc AP being s.
Draw the tangent PT and the ordinate PN.
The triangle PNT is evidently a triangle of forces for the
portion AP, PN representing the weight of AP (viz., ps), PT
\
O N
FIG. 464.
\
the tangential pull T at P, and NT the horizontal pull H at
A.
dy n ~ AT PN PS s
* -7-= tan PTN= -=jr =-=-, ... (2)
dx TN H m
which gives the differential equation to the catenary.
It may be easily integrated as follows :
ds
or
dy^ l~ ~~? i
v I + * = *
f. (s)
ds
dx
. ? tog(*
in
c being a constant of integration.
When x = o, s = o, and therefore log m = c.
Hence,
log M^+^ =
708 THEORY OF STRUCTURES.
or
or
m
- \ / \
-e -). ...... (4)
Again, ^ ji
dy s i* *
-y- = - = -(e e );
V
dx m 2
and hence,
tn -
(5)
The constant of integration is zero, since y = m when
x o.
The last equation is the equation to the catenary, while eq.
(4) gives the length of the arc A P.
By equations (4) and (5),
(6)
Draw NM perpendicular to PT, and let the angle PTN =
PNM=e. Then
PM = PN sin = y = s , . . . (7)
/* x m* *
=y 7, = 2 A- and tan 3 = 2 Ai,
Note.li )&, = >&, = ,
and hence
tan ft-as= = tan ^,.
6. Length of Arc of Cable. Let OP = s, Fig. 465,
wx
Since tan = -==,
2
or
sec a OdB = -^^r = ~ approximately.
Integrating between O and P,
.
* /
7. Weight of Cable. The ultimate tenacity of iron wire
is 90,000 Ibs. per square inch, while that of steel rises to
200,000 Ibs., and even more. The strength and gauge of cable
wire may be insured by specifying that the wire is to have a
certain ultimate tenacity and elastic limit, and that a given
number of lineal feet of wire is to weigh one pound. Each of
the wires for the cables of the East River Bridge was to have
an ultimate tenacity of 3400 Ibs., an elastic limit of 1600 Ibs.,
and 14 lineal feet of the wire were to weigh one pound. A very
uniform wire, having a coefficient of elasticity of 29,000,000 Ibs.,
has been the result, and the process of straightening has raised
the ultimate tenacity and elastic limit nearly 8 per cent.
Let W, be the weight of a length a, (= OD) of a cable of
sufficient sectional area to bear safely the horizontal tension H.
Let W^ be the weight of the length s,( = OA) of the cable
of a sectional area sufficient to bear safely the tension 7\ at A.
Let /be the safe inch-stress.
Let q be the specific weight of the cable material.
Then
and
7 1 4 THEORY OF STRUCTURES.
'. --.| -<-*+!<- +.4
or
+ V), nearly.
A saving may be effected by proportioning any given section
to the pull across that section.
At any point (x, y) the pull = H sec 0, and the correspond-
ing sectional area j . The weight per unit of length
= -7. g, and the total weight of the length s l (= OA) is
r sec ds
Hq
f
Hence,
and also
The weight of a cubic inch of steel averages .283 Ib.
The weight of a cubic inch of wrought-iron averages .278 Ib.
IT
The volume in inches of the cable of weight W 1 = \2a l p .
DEFLECTION OF CABLE. 715
W,
-- = .283 lb. or .278 lb.,
according as the cable is made of steel or iron.
Let the safe inch-stress of steel wire be taken at 33,960 Ibs.,
of the best cable-iron at 14,958 Ibs., and of the best chain-links
at 9972 Ibs. Then
W, = Ha, X .283 X = for steel cables ;
W t = Ha, X .278 X -~- g = for iron cables ;
t = Ha, X .278 X ~ = for link cables.
Note. About one-eighth may be added to the net weight of
a chain-cable for eyes and fastenings.
8. Deflection of a Cable due to an Elementary Change
in its Length.
By the corollary of Art. 6 the total length (S) of the cable
A OB is
Now a, and #, are constant ; h^ h 9 is also constant, and
therefore dh^ = dh v Hence,
If the alteration in length is due to a change of / in the
temperature,
dS = etS,
e being the coefficient of linear expansion and = ^ ^~
per degree Fahr. for wrought-iron.
THEORY OF STRUCTURES.
In England the effective range of temperature is about 60
Fahr., while in other countries it is usual to provide for a range
of from 100 to 150 F.
If the alteration is due. to a pull of intensity /per unit of
area,
dS = gS,
E being the coefficient of elasticity of the cable material.
If h, = h, = h,
a 16 h
# , = a t = , and dS = dh.
9. Curve of Cable from which the load is suspended by a
series of sloping rods.
r o E N
FIG. 466.
Let O be the lowest point of such a cable. Let the tangent
at O, and a line through O parallel to the suspenders, be the
axes of x and y, respectively.
Let w' be the intensity of the oblique load. Consider a
portion OP of the cable, and let the co-ordinates of P with
respect to OX, OY be x and y.
Draw the ordinate PN, and let the tangent at P meet ON
in E.
As before, PNE is a triangle of forces, and E is the middle
point of ON. Then
PN 2
_ _ ^ 2 _
~~' ~~ ~ y '
the equation to a parabola with its axis parallel to OY and its
focus at a point S, where ^SO ,- .
CURVE OF CABLE WITH OBLIQUE SUSPENDERS.
Cor. i. Let the axis meet the tangent at O in T f , and let
its inclination to OX be i.
Let A be the vertex, and ON' a perpendicular to the axis.
Then
SO = ST' = SA + A T f = SA + AN'.
But AS .AN' = ON" = N f T'*tan*i
.-. AS = AN' tan 2 i, and SO = AS(i + cot 2 *) =' -..
sm a z
Hence, the parameter 4^45 = ^SO sin 2 /.
Cor. 2. Let P be the oblique load upon the cable between
O and P.
Let Q be the total thrust upon the platform at E.
11 w " " load per horizontal unit of length.
" q " " rate of increase of thrust along platform.
" / " " length of PE.
Then
w
w . , and q = w cot z ;
sm i sin i
p = ff ^
x y
x*
t* y + + *y cos i.
4
Cor. 3. Let s be the length of OP, and let 6 be the inclina-
tion of PE to OY. Then
s = AP- AO
tan (9 - ^ sec(9 - ^
+ log,jtan (90 &) + sec (90 - 0)} - tan (90- *) sec (go *)
- log, { tan (90 - 1 ) + sec (90 - i) } j
//"sin* ( ;-. cot # + cosec <5 )
--- : -J cot cosec 6 cot / cosec i 4- log, - r^
2W ( cot z -f~ cosec z )
7i8
THEORY OF STRUCTURES.
It may be easily shown, as in the Note to Art. 6, that ap-
proximately
. .
s = x 4- y cos t -4-
sin 2 *
t . .
X -j-J/COSZ
10. Pressure upon Piers, etc.
Let 7 1 , be the tension in the main cable at A.
" T; " " " " " back-stay at A
" ?, /? be the inclinations to the horizontal of the tangents
at A to the main cable and back-stay, respectively.
The total vertical pressure upon the pier at A
7 1 , sin a -j- 7", sin y# /?.
The total resultant horizontal force at ^4
= 7i cos f ~ T z cos fi = Q.
If the cable is secured to a saddle which is free to move
horizontally on the top of the pier (Fig. 467),
Q the frictional resistance to the tendency to motion,
or Q = ^R t
fa being the corresponding coefficient of friction.
FIG. 467.
Let A Fig. 468, be the total height of the pier, and let W
its weight.
Let FG be the base o
of the centre of pressure.
4
be its weight.
Let FG be the base of the pier, and K the limiting position
, * f
AUXILIARY OR STIFFENING TRUSS.
Let /, q be the distance of P and W, respectively, from
Then
pp _L Wq
for stability of position Q = -- />""
and for stability of friction, when the pier is of masonry,
. the coefficient of friction of the masonry.
If /^ is sufficiently small to be disregarded, Q is approxi-
mately nil, and 7, cos a = 7", cos fi = H. The pressure upon
the pier is now wholly vertical and is //"(tan a -j- tan /?).
When the cable slides over smooth rounded saddles (Fig.
459), the tensions 7, and 7, are approximately the same.
Thus,
R = T, (sin a + sin ft) and g = T t (cos a cos ft).
It a = fi, Q = o, and the pressure upon the pier is wholly
vertical, its amount being 27, sin a.
The piers are made of timber, iron, steel, or masonry, and
allow of great scope in architectural design.
The cable should in no case be rigidly attached to the pier,
unless the lower end of the latter is free to revolve through a
small angle about a horizontal axis.
II. Auxiliary or Stiffening Truss. The object of a stiff-
ening truss (Fig. 469) is to distribute a passing load over the
cable in such a manner that it cannot be distorted. The pull 1
upon each suspender must therefore be the same, and this vir-
tually assumes that the effect of the extensibility of the cable
and suspenders upon the figure of the stiffening truss may be
disregarded.
7 20 THEORY OF STRUCTURES.
The ends O and A must be anchored, or held down by
pins, but should be free to move horizontally.
Let there be n suspenders dividing the span into (n -\- i)
equal segments of length a.
Let P be the total weight transmitted to the cable, and z
the distance of its centre of gravity from the vertical through O.
Let T be the pull upon each suspender.
Taking moments about O,
/ being the length of OA.
Also, if t is the intensity of pull per unit of span,
r
tl = nT, and hence Pz t .
Let there be a central suspender of length s. There will,
therefore, be - suspenders on each side of the centre.
r
The parameter of the parabola = =- .
Hence, the total length of all the suspenders
If there is no central suspender, i.e., if n is even,
the total length = (n i)(s + -" -jJ.
Denote the total length of suspenders by L. Then
the stress-length = TL = ~PL.
AUXILIARY OR STIFFENING TRUSS. ? '21
Let w be the uniform intensity of the dead load.
CASE I. The bridge partially loaded.
Let w' be the maximum uniform intensity of the live load,
and let this load advance from A and cover a length AB.
Let OB x, and let R l , R^ be the pressures at O and A,
respectively.
For equilibrium,
R^R^ + tl-wl- /(/ - *) = o ; (i)
J_z/-^(/-*) =0. ... (2)
Also, since the whole of the weight is to be transmitted
through the suspenders,
>(l x) ....... (3)
From eqs. (i), (2), and (3),
(4)
which shows that the reactions at O and A are equal in mag-
nitude but opposite in kind. They are evidently greatest when
x = -, i.e., when the live load covers half the bridge, and the
w'l
common value is then r- .
O
The shearing force at any point between O and B distant x'
from O
which becomes j(l x) R l = R^ when x' equal x.
Thus the shear at the head of the live load is equal in magni-
tude to the reaction at each end, and is an absolute maximum
722 THEORY OF STRUCTURES.
when the live load covers half the bridge. The web of the
rf
8
truss must therefore be designed to bear a shear of ^- at the
centre and ends.
Again, the bending moment at any point between O and B
distant x' from O
= R l x' + t x" = ~ l -=(x-xx'), . . (6)
which is greatest when *' =-, i.e., at the centre of OB, its
VOJ / _ JK
value then being f~ x *- Thus, the bending moment is
O /
d * 2
an absolute maximum when ~j~(tx* **) = o, i.e., when x =. -/,.
w'
and its value is then --- /*.
54
The bending moment at any point between B and A dis-
tant x' from O
" - - x? = j(x> - x)(l- x'\ (7)
d
which is greatest when ~r~ t {(^' ~~ *Yt ~ x '}\ = i- e -> when
14-x
x' = , or at the centre of AB, its value then being
w' x
(I x}*. Thus, the bending moment is an absolute niaxi-
/
3
mum when -j-\x(l x)*\ = o, i.e., when x = , and its value is
then + /'.
54
Hence, the maximum bending moments of the unloaded and
loaded divisions of the truss are equal in magnitude but opposite
in direction, and occur at the points of triscction (D, C) of OA
AUXILIARY OR STIFFENING TRUSS. 723
when the live load covers one-third (AC) and two-thirds (AD) of
the bridge, respectively.
Each chord must evidently be designed to resist both
tension and compression, and in order to avoid unnecessary
nicety of calculation, the section of the truss may be kept uni-
form throughout the middle half of its length.
CASE II. A single concentrated load W at any point B of
the truss. W now takes the place of the live load of intensity
. The bending moment is then negative va^
is a maximum when x' = x -, its value then being
Wi
The bending moment at any point between B and A dis
tant x' from O
= R,x' + (t- w )-- W( X ' -x) = -j(x' - /)- -
which is a maximum when
I W I l\
i.e., when x' = x -4- -, and its value is then -Ax } .
1 2 2/\ 21
Note. The stiffening truss is most effective in its action, but
adds considerably to the weight and cost of the whole struc-
ture. Provision has to be made both for the extra truss and
for the extra material required in the cable to carry this extra
load.
AUXILIARY OR STIFFENING TRUSS. 725
Stiffening Truss hinged at the Centre. Provision may be
made for counteracting the straining due to changes of tem-
perature by hinging the truss at the centre E.
Let a live load of intensity w advance from A.
First, let the live load cover a length AB = x f > ).
Let R^ , R^ be the pressures at O, A, respectively.
The equations of equilibrium are
l + R t + (t-iv)/-iv'x = ; .... (I)
--*-'=" W
(3)
Eqs. (2) and (3) being obtained by taking moments about E.
Hence,
w'
t-w=- -(/ 3 - 4** + 2*') ; ... (4)
; (5)
Next, iet the live load cover the length BO ( < -)
Let AB = x as before, and let JR/ t RJ, t' be the new values
of R lt R^ t, respectively.
The equations of equilibrium are now
'X-'V-~*)=o; . . (7)
726 THEORY OF STRUCTURES.
*,'- + (;'-Hg=o; (9)
and hence,
/'- W = 2-(/- *)'[=-(/-, -a-')]' (10)
I W
\i -*)>(= -R) (12)
Diagram of Maximum Shearing Force. The shear at any
point distant z from A in the unloaded portion BO when the
live load covers AB
^R^(t-w)(l-z) ......... (13)
>-, _/)(/_*)}
' - /)(/- z) - w\l- z)
= minus the shear at the same point when AB is
unloaded and the live load covers BO.
For a given value of z the maximum shear, positive or
negative, at any point of OB, is found by making (see eq. (13) )
or
3*) - (/-*K- 4/+ 4*) = O,
or
7 4# 2/
jr=z/ T- (H)
AUXILIARY OR STIFFENING TRUSS.
Hence, by eqs. (4), (5), (13), (14),
727
I 2X
the maximum shear = %w'x , . . (15)
/-*
and may be represented by the ordinate m
{positive or negative) of the curve mnpq.
For example, at the points defined by
*= /, v, */,
the shears are greatest when
*= K |/, i/,
and their values are, respectively,
o.
o
VI
n I
FIG. 470.
Again, the shear at any point distant z from A in the
loaded portion BE when the live load covers AB
= ^ + (/ _ ,)(/- ^) - W '(^ - ^) .... (16)
- w -
w\l x)
= minus the shear at the same point when AB is
unloaded and the live load covers BO.
Hence, by eqs. (4), (5), (16),
\w'
the shear = q= - -^(l - ^z](l - x)\ . (17)
increasing for a given value of z with / x, and, therefore, a
maximum when x = z. Thus,
the maximum shear = =f ~r(^ "~ 4 Jtr )(^~- ^t ( J 8)
and occurs immediately / front of the load when it covers
AB, and immediately behind the load when it covers BO. It
7 28 THEORY OF STRUCTURES.
may be represented by the ordinate (positive or negative] of
the curve orsq.
For example, at the points defined by
the maximum shears given by eq. (18) are, respectively,
o, yfrw7, T Ve>'/, AV*//, \w'l.
Diagram of Maximum Bending Moment. The bending
moment at any point in BO distant z from A when the live
load covers AB
(19)
*}
= minus the bending moment at the same point when
the live load covers BO.
Hence, by eqs. (4), (5), (19), the bending moment
I w' I w'
= - 2 -j(f - At* + 3**X/ -*) =F - -f(l* - Al* + z**)(l -*) 1 .
For a given value of z this is a maximum and equal to
W ! zl zl 2.Z 2lz
T (1-2,) when * =
Thus, the maximum bending moment may be represented
by the ordinate (positive or negative] of a curve.
For example, at the points defined by
* = /, v, */, t/, '-,
AUXILIARY OR STIFFENING TRUSS.
the bending moments are greatest when ^
v, T y, v, v, V-,
their values being, respectively,
The absolute maximum bending moment may be found as
follows :
For a given value of x the bending moment (see eq. (19)) is
a maximum when
or
/ ze;
Hence, the maximum bending moment
h " -
2 * w " - 8 /' 4/*r -f 2* 2 '
It will be an absolute maximum for a value of x found by put-
ting its differential with respect to x equal to nil.
This differential easily reduces to
x = f/ is an approximate solution of this equation, and the cor-
responding maximum bending moment = -^-^w'T.
The preceding calculations show that at every point in its
length the truss may be subjected to equal maximum shears and
equal maximum bending moments of opposite signs.
Again, it may be easily shown, in a similar manner, that
73 THEORY OF STRUCTURES.
when a single weight W travels over the truss,
the maximum positive shear at a distance z from A
W
= 7 r(2/' -5/2
the maximum negative shear
W
either = -(/ 2 - 5 /*+ 42*)
I W
or = - -j
and the maximum bending moment
W
= T*(/ -*)(/- 2,).
12. Suspension-bridge Loads. The heaviest distributed
load to which a highway bridge may be subjected is that due
to a dense crowd of people, and is fixed by modern French
practice at 82 Ibs. per square foot. Probably, however, it is
unsafe to estimate. the load at less than from 100 to 140 Ibs. per
square foot, while allowance has also to be made for the con-
centration upon a single wheel of as much as 36,000 Ibs., and
perhaps more.
A moderate force repeatedly applied will, if the interval
between the blows corresponds to the vibration interval of the
chain, rapidly produce an excessive oscillation (Chap. Ill,
Cor. 2, Art. 24). Thus, a procession marching in step across
a suspension-bridge may strain it far more intensely than a
dead load, and will set up a synchronous vibration which may
prove absolutely dangerous. For a like reason the wind
usually sets up a wave-motion from end to end of a bridge.
The factor of safety for the dead load of a suspension-bridge
should not be less than 2^ or 3, and for the live load it is
advisable to make it 6. With respect to this point it may be
remarked that the efficiency of a cable does not depend so
much upon its ultimate strength as upon its limit of elasticity,
MODIFICATIONS OF THE SIMPLE SUSPENSION-BRIDGE. 731
and so long as the latter is not exceeded the cable remains un-
injured. For example, the breaking weight of one of the 1 5-inch
cables of the East River Bridge is estimated to be 12,000 tons,
its limit of elasticity being 81 18 tons ; so that with i only as a
factor of safety, the stress would still fall below the elastic
limit and have no injurious effect. The continual application
of such a load would doubtless ultimately lead to the destruc-
tion of the bridge.
The dip of the cable of a suspension-bridge usually varies
from ^ to Y 1 ^ of the span, and is rarely as much as y 1 ^, except
for small spans. Although a greater ratio of dip to span would
give increased economy and an increased limiting span, the
passage of a live load would be accompanied by a greater dis-
tortion of the chains and a larger oscillatory movement.
Steadiness is therefore secured at the cost of economy by
adopting a comparatively flat curve for the chains.
13. Modifications of the Simple Suspension-bridge.
The disadvantages connected with suspension-bridges are very
great. The position of the platform is restricted, massive
anchorages and piers are generally required, and any change in
the distribution of the load produces a sensible deformation in
the structure. Owing to the want of rigidity, a considerable
vertical and horizontal oscillatory motion may be caused, and
many efforts have been made to modify the bridge in such a
manner as to neutralize the tendency to oscillation.
(a) The simplest improvement is that shown in Fig. 472,
where the point of the cable most liable to deformation is
attached to the piers by short straight chains AB.
FIG. 472.
(b) A series of inclined stays, or iron ropes, radiating from
the pier-saddles, may be made to support the platform at a
number of equidistant points (Fig. 473). Such ropes were used
in the Niagara Bridge, and still more recently in the East River
73 2 THEORY OF Sl^RUCTURES.
Bridge. The lower ends of the ropes are generally made fast
to the top or bottom chord of the bridge-truss, so that the cor-
responding chord stress is increased and the neutral axis pro-
portionately displaced. To remedy this, it has been proposed
to connect the ropes with a horizontal tie coincident in position
with the neutral axis. Again, the cables of the Niagara and
FIG. 473 .
East River bridges do not hang in vertical planes, but are in-
clined inwards, the distance between them being greatest at
the piers and least at the centre of the span. This drawing in
adds greatly to the lateral stability, which may be still further
increased by a series of horizontal ties.
(c) In Fig. 474 two cables in the same vertical plane are
diagonally braced together. In principle this method is similar
FIG. 474.
to that adopted in the stiff ening truss (discussed in Art. 1 1), but
is probably less efficient on account of the flexible character of
the cables, although a slight economy of material might doubt-
less be realized. The braces act both as struts and ties, and
the stresses to which they are subjected may be easily calcu-
lated.
(d) In Fig. 475 a single chain is diagonally braced to the
platform. The weight of the bridge must be sufficient to insure
FIG. 475.
that no suspender will be subjected to a thrust, or the efficiency
of the arrangement is destroyed. An objection to this as well
MODIFICATIONS OF THE SIMPLE SUSPENSION-BRIDGE. 733
as to the preceding method is that the variation in the curva-
ture of the chain under changes of temperature tends to loosen
and strain the joints.
The principle has been adopted (Fig. 476) with greater per-
fection in the construction of a foot-bridge at Frankfort. The
FIG. 476.
girder is cut at the centre, the chain is hinged, and the rigidity
is obtained by means of vertical and inclined braces which act
both as struts and ties.
(e) In Fig. 477 the girder is supported at several points by
FIG. 477.
straight chains running directly to the pier-saddles, and the
chains are kept in place by being hung from a curved chain by
vertical rods.
(/) It has been proposed to employ a stiff inverted arched
rib of wrought-iron instead of the flexible cable. All straining
action may be eliminated by hinging the rib at the centre and
piers, and the theory of the stresses developed in this tension
rib is precisely similar to that of the arched rib, except that
the stresses are reversed in kind.
(g) The platform of every suspension-bridge should be
braced horizontally. The floor-beams are sometimes laid on
the skew in order that the two ends of a beam may be sus-
pended from points which do not oscillate concordantly, and
also to distribute the load over a greater length of cable.
734 THEORY OF STRUCTURES.
EXAMPLES.
1. The span of a suspension-bridge is 200 ft., the dip of the chains is
80 ft., and the weight of the roadway is i ton per foot run. Find the ten-
sions at the middle and ends of each chain. Ans. 31 \ tons ; 58.94 tons.
2. Assuming that a steel rope (or a single wire) will bear a tension of
15 tons per square inch, show that it will safely bear its own weight over
a span of about one mile, the dip being one-fourteenth of the span.
Ans. Max. tension = 33,074 Ibs.
3. Show that a steel rope of the best quality, with a dip of one-seventh
of the span, will not break until the span exceeds 7 miles, the ultimate
strength of the rope being 60 tons per square inch.
Ans. Max. tension = 59.545 tons per square inch.
4. The river span of a suspension-bridge is 930 ft. and weighs 5976
tons, of which 1439 tons are borne by stays radiating from the summit
of each pier, while the remaining weight is distributed between four
I5~in. steel-wire cables, producing in each at the piers a tension of 2064
tons. Find the dip of the cables. Ans. 66.44 ft.
The estimated maximum traffic upon the river span is 1311 tons
uniformly distributed. Determine the increased stress in the cables.
Ans. 596.4 tons.
To what extent might the traffic be safely increased, the limit of
elasticity of a cable being 8116 tons, and its breaking stress 12,300 tons ?
Ans. To 13,303 tons uniformly distributed.
5. If the span = /, the total uniform load = W, and the dip = ,
show that the maximum tension = 1.58 W, the minimum tension
= 1.5 IV, the length of the chain = i.oiS/, and find the increase of dir>
corresponding to an elongation of i in. in the chain.
6. A cable weighing p Ibs. per lineal foot of lengthjs stretched be-
tween supports in the same horizontal line and 20 ft. apart. If the max-
imum deflection is ft., determine the greatest and least tensions.
Ans. Parameter m = 100 ft.; max. tension = ioo/; min. ten-
sion = ioo^.
7. A light suspension-bridge carries a foot-path 8 ft. wide over a
river 90 ft. wide by means of eight equidistant suspending rods, the dip
being 10 ft. Each cable consists of nine straight links. Find their several
lengths. If the load upon the platform is 120 Ibs. per square foot, and
EXAMPLES. 735
if one-fourth of the load is borne by the piers, find the sectional areas
of the several links, allowing 10,000 Ibs. per square inch.
Ans. Lengths in ft., 10 ; 10.049; 10.198; 10.44; 10.77.
Tensions in Ibs., 45000; 45004/101 ; 45004/104; 4500^109;
4500 4/1 1 6.
Areas in sq. in., 4.5 ; 4.522 ; 4.59 ; 4.698 ; 4.847.
8. A suspension-bridge of 200 ft. span and 20 ft. dip has 4^ sus-
penders on each side ; the dead weight = 3000 Ibs. per lineal foot ; the
live load = 2000 Ibs. per lineal foot. Find the maximum pull on a sus-
pender, the maximum bending moment and the maximum shear on the
stiffening truss. Also, findthe elongation in the chain due to the live load.
Ans. Max. pull = 12,500 Ibs. ; max. shear = 30,000 Ibs.; max.
B.M. = 1,066,666$ ft. -Ibs. ; elongation = 89,600,000 -f- EA, A
being sectional area of a cable, and E the coefficient of elas-
ticity.
9. A foot-path 8 ft. wide is to be carried over a river 100 ft. wide by
two cables of uniform sectional area and having a dip of 10 ft. Assum-
ing the load on the platform to be 112 Ibs. per square foot, find the
greatest pull on the cables, their sectional area, length, and weight.
(Safe stress = 8960 Ibs. per square inch ; specific weight of cable = 480
Ibs. per cubic foot.)
Ans. H = -=T= 56,000 Ibs.; area =6.73 sq. in.;
4/29
length = io2f ft.; weight = 2302.65 Ibs.
10. Find the depression in the cables in the last question due to an
increment of length under a change of 60 F. from the mean temperature.
(Coefficient of expansion =i -f- 144000.) Ans. .0802 ft.
n. Each side of the platform of a suspension-bridge for a span of 100
ft. is carried by nine equidistant suspenders. Design a stiffening truss for
a live load of 1000 Ibs. per lineal foot, and determine the pull upon the
suspenders due to the live load when the load produces (i) an absolute
maximum shear ; (2) an absolute maximum bending moment.
Ans. Max. shear = 6250 Ibs.; max. B.M. = 92,592^ ft.-lbs.; pull
on suspender = (i) 2777! Ibs., (2) = 185 iff Ibs. or 3703-1 rf Ibs.
12. In a suspension-bridge (recently blown down) each cable was de-
signed to carry a total load of 84 tons (including its own weight). The
distance between the piers = 1270 ft.; the deflection of the cable = 91 ft.
Find (a) the length of the cable ; (b) the pull on the cable at the piers
and at the lowest point ; (c) the amounts by which these pulls are changed
by a variation of 40 F. from the mean temperature ; ( anc * 9843! Ibs. at ends of the half truss and at
the points dividing the half span into four equal seg-
ments.
Max. B. M. due to 78,750 Ibs. is at centre of half truss and
= 184,570^- ft. -Ibs. Max. B. M. due to 1050 Ibs. per lineal
foot =176, 1 80 ff|, 221,484!, and 153,808-^1 ft.-lbs. at points
dividing the half truss into four equal segments.
1 8. Show that the total extension of a cable of uniform sectional area
A under a uniformly distributed load of intensity iu is
wl
3 / 16 since tne triangle AGO is evidently a
triangle of forces for the forces acting upon the mass under
consideration.
Also,
W = / wzdx . sin 0.
.'. Cwzxdx . sin = WX - W~Y= X ffvz sin Odx - PY.
t/o W t/o
This is the equation to the line of resistance.
Taking the differential of this equation,
wz'X sin BdX = Xwz' sin BdX + WdX - PdY 9
z' being the depth corresponding to the abscissa X.
dY W AO
Thus the tangents to the curve of pressures and to the
curve of centres of pressure at any given point coincide, and
the curves must therefore also coincide.
CONDITIONS OF EQUILIBRIUM. 745
3. Conditions of Equilibrium. Let the figure represent
a portion of an arch of thickness unity, between any two bed-
joints (real or imaginary] MN, PQ.
Let W be its weight together with that of the superincum-
bent load. Let the direction of
the reaction R' at the joint MN
intersect MN in m and the direc-
tion of J^in n. For equilibrium,
the reaction R" at the joint PQQ.
must also pass through n. Let its
direction intersect PQ in O. In O +W
order that the equilibrium may be
stable, three conditions must be HIG ' 48z<
fulfilled, viz. :
First. The point O must lie between P and Q, so that there
may be no tendency to turn about the edges Pand Q.
Second. There must be no sliding along PQ, and therefore
the angle between the direction of R" and the normal to PQ
must not exceed the angle of friction of the material of which
the arch is composed.
N.B. The angle of friction for stone upon stone is about
30.
Third. The maximum intensity of stress at any point in PQ
must not exceed the safe resistance of the material.
Further, the stress should not change in character, in the
case of masonry and brick arches, but should be a compression
at every point, as these materials are not suited to withstand
tensile forces.
The best position for O would be the middle point of PQ,
as the pressure would then be uniformly distributed over the
area PQ. It is, however, impracticable to insure such a dis-
tribution, and it has been sometimes assumed that the stress
varies uniformly,
With this assumption, let ^Vbe the normal component of R".
Let /be the maximum compressive stress, i.e., the stress at
the most compressed edge, e.g., P.
Let OS q . PQ, S being the middle point of PQ, and q a
coefficient whose value is to be determined.
746 THEORY OF STRUCTURES.
PO
Then if PO < ,
N- f ' PQ
and in the limit when PO = , i.e., when the intensity of
stress varies uniformly from /at Pto nil at Q,
?=* and &**
(See Art. 16, Chap. IV.)
Similarly, if Q is the most compressed edge, the limiting
position of O, the centre of resistance or pressure, is at a point
PQ
O' denned by QO' - ~.
Hence, as there should be no tendency on the part of the
joints to open at either edge, it is inferred that PO or QO'
PQ
should be > , i.e., that the point O should lie within the
middle third of the joint.
Experience, however, shows that the " middle-third "
theory cannot be accepted as a solution of the problem of
arch stability, and that its chief use is to indicate the proper
dimensions of the abutments. Joint cracks are to be found in
more than 90$ of the arches actually constructed, and cases
may be instanced in which the joints have opened so widely
that the whole of the thrust is transmitted through the edges.
In Telford's masonry arch over the Severn, of 150 ft. span,
Baker discovered that there had been a settlement (15 in.)
sufficient to induce a slight reverse curvature at the crown of
the soffit. Again, the position of the centre of pressure at a
joint is indeterminate, and it is therefore impossible as well as
useless to make any calculations as to the maximum intensity
of stress due to the pressure at the joint. What seems to
JOINT OF RUPTURE. 747
happen in practice is, that the straining at the joints generally
exceeds the limit of elasticity, and that the pressure is uni-
formly distributed for a certain distance on each side of the
curve of pressures. Thus, the proper dimensions of a stable
arch are usually determined by empirical rules which have
been deduced as the results of experience. For example,
Baker makes the following statement :
Let T be the thrust in tons or pounds per lineal foot of
width of arch.
Let f be the safe working stress in tons or pounds per
square foot.
An arch will be stable if an ideal arch, with its bounding
i T
surfaces at a minimum distance of from the curve of pres-
sures, can be traced so as to lie within the actual arch. An
advance would be made towards a more correct theory if it
were possible to introduce into the question, the elasticity and
compressibility of the materials of construction. These ele-
ments, however, vary between such wide limits that no
reliance can be placed upon the stresses derivable from their
values.
4. Joint of Rupture. Let I 2, 3 4 be the bounding surfaces
between which the curve of pressures must lie, and let 4 be
2
FIG. 483.
the centre of pressure at the crown. A series of curves of
pressure may be drawn for the same given load, but with
different values of the horizontal thrust h.
Let AfXy be that particular curve which for a value //of the
horizontal thrust is tangent to the surface I 2 at x ; the joint at
x is called the joint of rupture.
The angle which the joint of rupture makes with the
748
THEORY OF STRUCTURES.
horizontal is about 30 in semicircular and 45 in elliptic
arches.
The position of the joint in any given arch may be tenta-
tively found as follows :
Lety be any joint in the surface I 2.
Let Wbe the weight upon the arch between /"and I.
Let X be the horizontal distance between J and the centre
of gravity of W.
Let Y be the vertical distance between J and 4.
It will also be assumed that the thrust at 4 is horizontal.
If the curve of pressure be now supposed to pass through
J, the corresponding value of the horizontal thrust h is given
by
kY= WX.
By means of this equation, values of h may be calculated
for a number of joints in the neighborhood of the haunch, and
the greatest of these values will be the horizontal thrust H for
the joint x. This is evident, as the curve of pressure for a
smaller value of h must necessarily fall below ^xy.
When this happens, the joints will tend to open at the
lower edge of the joint I 4 and at the upper edges of the joints
at x and at 2 3, so that the arch may sink at the crown and
spread, unless the abutments and the lower portions of the
arch are massive enough to counteract this tendency.
If the curve of pressure fall above A^xy, an amount of back-
ing sufficient to transmit the thrust to the abutments must be
provided. The same result may be attained by a uniform in-
crease in the thickness of the arch ring, or by a gradual increase
from the crown to the abutments.
For example, the upper sur-
face (extrados) of the ring for an
arch with a semicircular soffit
A KB, having its centre at O, may
be delineated in the following
manner:
^ Let x define the joint of rup-
ture in the soffit ; then AOx 30.
\
MINIMUM THICKNESS OF ABUTMENT.
749
be
Y
the
In Ox produced take xx r = 2 X KD, KD being the thick-
ness at the crown.
The arc Dx' of a circle struck from a centre in DO pro-
duced may be taken as a part of the upper boundary of the
ring, and the remainder may be completed by the tangent at
x l to the arc Dx' .
5. Minimum Thickness of Abutment. Let
resultant thrust at the horizontal joint BC of a
rectangular abutment ABCD.
Let y be the distance of its point of applica- C
tion from B.
Let // and V be the horizontal and vertical
components of T.
Let w be the specific weight of the material
in the abutment.
Let h be the height AB of the abutment.
Let t be the width AD of the abutment. FIG. 485.
In order that there may be no tendency to turn about the
toe D, the moment of the weight of the abutment with respect
to D plus the moment of V with respect to D must be greater
than the moment of H with respect to D. Or,
or
- + V(t-y)> Hh,
H \ I 2 4- 2V 4- V *
h > V W~* Wk- ufk*
This relation must hold good whatever the height of the
abutment may be ; and if h is made equal to oo ,
which defines a minimum limit for the thickness of the abut-
ment.
750 THEORY OF STRUCTURES.
6. Empirical Formulae. In practice the thickness /at
the crown is often found in terms of s, the span, or in terms of
p, the radius of curvature at the crown, from the formulae
t = c Vs, or t = Vcp,
t, s, and p being all in feet, and c being a constant.
According to Dupuit, / = .36 Vs for a full arch ;
/ = .27 Vs for a segmental arch.
According to Rankine, = l/.i 2p for a single arch ;
/ = V.ijp for an arch of a series.
7. Examples of Linear Arches, or Curves of Pressure.
\(a) Linear Arch in the Form of a Parabola. Suppose that
the cable in Art. 4, Chap. XII, Case B, is exactly inverted,
and that it is stiffened in such a manner as to resist distortion.
Suppose also that the load still remains a uniformly distributed
weight of intensity w per horizontal unit of length. A thrust
will now be developed at every point of the inverted cable
equal to the tension at the corresponding point of the original
cable. Thus the inverted parabola is a linear arch suitable for
a real arch which has to support a load of intensity w per
horizontal unit of length.
The horizontal thrust at the crown = H = wp,
p being the radius of curvature at the crown.
(fr) Linear Arch in the Form of a Catenary. Transformed
Catenary. If the cable in Art. 4, Chap. XIT, Case A, is in-
N T o N verted and stiffened as before, a linear
arch is obtained suitable for" a real
arch which has to support a load dis-
tributed in such a manner that the
weight upon any portion AP is pro-
p , portional to the length of AP, and is
in fact =ps. The area OAPN ' ms.
Thus, a lamina of thickness unity
and specific weight w, bounded by the curve AP, the directrix
ON, and the verticals AO, PN, weighs wms, and may be taken
EXAMPLES OF LINEAR ARCHES. 75 r
to represent the load upon the arch if wms ps, i.e., if wm = J>,
i.e., if the weight of m units of the lamina is w.
The horizontal thrust at the crown H wm = wp,
the radius of curvature (p) at the crown being equal to m.
A disadvantage attached to a linear arch in the form of a
catenary lies in the fact that only one catenary can pass
through two given points, while, in practice, it is often neces-
sary that an arch shall pass through three ^points in order to
meet the requirements of a given rise and span. This difficulty
may be obviated by the use of the transformed catenary.
Upon the lamina PAPNN as base, erect a solid, with its
horizontal sections all the same, and, for simplicity, with its
generating line perpendicular to the base.
Cut this solid by a plane through NN inclined at any re-
quired angle to the base. The intersection of the plane and
solid will define a transformed catenary P'A'P', or a new linear
arch, and the shape of a new lamina P'A'P'NN, under which
the arch will be balanced. This is evident, as the new arch and
lamina are merely parallel projections of the original.
The projections of horizontal lines will remain the same in
length.
The projections of vertical lines will be c times the lengths
of the lines from which they are projected, c being the secant
of the angle made by the cutting plane with the base.
Let .*, Y be the co-ordinates of any point P' of the trans-
formed catenary.
Let x, y be the co-ordinates of the corresponding point P
in the catenary proper.
Then
YPW A'Q M
_ _
= ~
y ~ PN ~ AO
The equation to the catenary proper is
(2)
75 ^ THEORY OF STRUCTURES.
Substituting in the last equation the value of y given by eq. (i),
which is the equation to the transformed catenary.
With this form of Linear arch the depths M over the crown
and Y over the springings, for a span 2x, may be assumed, and
the corresponding value of m determined from eq. (3).
It is convenient, in calculating m, to write eq. (3) in the form
-'- ... (4)
The slope i' at P' is given by
dY Ml*- -c\ Ms
tarn = y = I* e 1 = ,
dx 2m\ I m
s being the length AP of the catenary proper, corresponding
to the length A'P' of the transformed catenary.
'P'N= C
.
754
THEORY OF STRUCTURES.
Thus, at any point of the arch,
the horizontal intensity of pressure
= vertical intensity = normal intensity =/.
Again, the total horizontal pressure on one-half of the arch
= 2(p. CE) = p2(CE) = pr = H,
and the total vertical pressure on one-half of the arch
= 2(p . DE) =
Hence, at any point of the arch the tangential thrust = pr.
Next, upon the semicircle as base, erect a semi-cylinder.
Cut the latter by an inclined plane drawn through a line in the
plane of the base parallel to OA. The intersection of the cut-
ting plane and the semi-cylinder is the semi-ellipse B'AB', in
which the vertical lines are unchanged in length, while the
lengths of the horizontal lines are c times the lengths of the
corresponding lines in the semicircle, c being the secant of the
angle made by the cutting plane with the base. A semi-
elliptic arch is thus obtained, and the forces to which it is sub-
jected are parallel projections of the forces acting upon the
semicircular arch.
These new forces are in equilibrium (see Corollary).
Let P' = the total vertical pressure upon one-half of the
arch ;
H' the total horizontal pressure upon one-half of the
arch ;
EXAMPLES OF LINEAR ARCHES, 755
P
py = vertical intensity of pressure = ^-57 ;
tft
p x ' = horizontal intensity of pressure = ~7=rr
Then
P f = P=H = pr; ........ (i)
P P
y '- OB'~ c.OB~~ cr ""
H f = cH = cP = cP'-, ....... (3)
H'
Hence, by eq. (3),
/_ ___
px ~- OA' ~ OA ~~ : r ~
*L OB '
~P~~ ~OA'
or, the total horizontal and vertical thrusts are in the ratio of
the axes to which they are respectively parallel, and, by eqs.
(2) and (4),
A.' -I OA "
p x '~S~ OB'* ;
or, the vertical and horizontal intensities of pressure are in the
ratio of the squares of the axes to which they are respectively
parallel.
Any two rectangular axes OG, OK in the circle will project
into a pair of conjugate radii OG' , OK' in the ellipse.
Let OG' = r lt OK' = r 2 ;
Q = total thrust along elliptic arch at K\
E> a /"
Then
H r H r
THEORY OF STRUCTURES.
or, the total thrusts along an elliptic arch at the extremities of
a pair of conjugate radii are in the ratio of the radii to which
they are respectively parallel.
The preceding results show that an elliptic linear arch is
suitable for a load distributed in such a manner that the vertical
and horizontal intensities (eqs. (2) and (4) ) at any point of the
arch are unequal, but are uniform in direction and magnitude.
Corollary. It can be easily shown that the projected forces
acting upon the elliptic arch are in equilibrium.
The equations of equilibrium for the forces acting upon the
circular arch may be written
T being the thrust along the arch at the point xy, and X, Y
the forces acting upon the arch parallel to the axes of x and
y, respectively.
If T', X', V be the corresponding projected forces,
~ = ~, Xds = cX'ds', Yds = Y'ds*.
Hence, the above equations may be written
d j^p cdx^ + cX'ds' = o,
and
or
and
d\
( Tid ] + *'***
Hence, the forces T 1 ', X', and Y' are also in equilibrium.
EXAMPLES OF LINEAR ARCHES.
(d) Hydrostatic Arch. Let the figure represent a portion
of a linear arch suited to support a load
which will induce in it a normal pressure at
every point. The pressure being normal
has no tangential component, and the
thrust (7") along the arch must therefore be
everywhere the same.
Consider any indefinitely small element
CD.
It is kept in equilibrium by the equal -FIG ' 489 '
thrusts (7") at the extremities C and D, and by the pressure
/ . CD. The intensity of pressure / being assumed uniform
for the element CD, the line of action of the pressure/. CD
bisects CD at right angles.
Let the normals at C and D meet in O l , the centre of
curvature.
Take Of O,D = p, and the angle CO,D = 2 AS.
Resolving along the bisector of the angle
6 =p. CD pp
or
2TA& = pp. 2 AS \
and hence,
T = pp = a constant. . . . (i)
Thus, a series of curves may be obtained in which p varies
inversely as/, and the hydrostatic arch is that curve for which
\\\e pressure p at any point is directly proportional to the depth
of the point below a given horizontal plane.
Denote the depth by y, and let w be the specific weight of
the substance to which the pressure/ is due. Then
P = wy, .......... . . (2)
and
T pp wyp = a constant. . . . (3)
The curve may be delineated by means of the equation
yp = const ........... (4)
75 8 THEORY OF STRUCTURES.
It may be shown, precisely as in Case (c), that the horizontal
intensity of pressure (p^)
=: the vertical intensity (p y ) =fl (5)
Take as the origin of co-ordinates the point O vertically
above the crown of the arch, in the given horizontal plane.
Let the horizontal line through O be the axis of x.
" " vertical " " " " " " " y.
Any portion AM of the arch is kept in equilibrium by the
O equal thrusts (T) at A and M,
j and by the resultant load P upon
AM, which must necessarily act
in a direction bisecting the angle
ANM.
FIG. 490. Complete the parallelogram
AM, and take SN NM to represent T.
The diagonal NL will therefore represent P.
Let be the inclination of the tangent at M to the hori-
zontal.
The vertical load upon AM vertical component of P
= LK T sin 6 = pp sin = wyp sin = wy p sin 0, . (6)
y ot p being the values of y, p, respectively, at A.
The horizontal load upon AM= horizontal component of P
= NK=SN-KS = T-
= 2pp (sin -) = 2wyp sin = 2wy p (sin -) . . (7)
\ 2i' '
Again, the vertical load upon AM
/ "* pdx w / ydx = wy p sin ; (8)
e/o vo
the horizontal load upon AM
f*y f*y iv / \
= J pdy = wj^ ydy = -(/ - y*) = 2wy p (s'm -j . (9)
EXAMPLES OF LINEAR ARCHES. 759
Equation (8) also shows that the area bounded by the curve
AM, the verticals through M and A, and the horizontal
through is equal to y p a sin #, and is therefore proportional
to sin 0. At the points defined by d = 90 the tangents to
the arch are vertical, and the portion of the arch between these
tangents is alone available for supporting a load. The vertical
and horizontal loads upon one-half the arch are each equal to
WJW
Corollary. The relation given in eq. (i) holds true in any
arch for elements upon which the pressure is wholly normal.
This has been already proved for the parabola and catenary,
in cases (a) and (b).
At the point A' of the elliptic arch,
_ OB'* _ c*r* _
= '-~ r ~~ ~
Hence, the horizontal thrust at A'
= PyP = ~P = P Cr = H -
(e) Geostatic Arch. The geostatic is a parallel projection of
the hydrostatic arch.
The vertical forces and the lengths of vertical lines are
unchanged.
The horizontal forces and lengths of hori-
zontal lines are changed in a given ratio
c to I.
Let B' A be the half-geostatic curve de- FIG. 491.
rived from the half-hydrostatic curve BA.
The vertical load on AB'
P f = P= thrust along arch at B'. ... (i)
The horizontal load on AB'
= H' = cH thrust along arch at A. . . . (2)
The new vertical intensity
-^'- -*-A2 ,
f > OB'
THEORY OF STRUCTURES.
The new horizontal intensity
H 1 cH
~ =C ^ = ^ .... (4)
Thus, the geostatic arch is suited to support a load so dis-
tributed as to produce at any point a pair of conjugate press-
ures ; pressures, in fact, similar to those developed according
to the theory of earthwork.
Let R l , R^ be the radii of curvature of the geostatic arch
at the points A, B' , respectively, and let r lt r y be the radii of
curvature at the corresponding points A, B of the hydrostatic
arch.
The load is wholly normal at A and B ' . Thus,
H' =p y f R l =^R l =:cH=cpr l . ... (5)
' R* = ^ ....... ..... (6)
Also,
cR, = r,. . ..... '. . . . . . (8)
(/) General Case. Let the figure represent any linear
p arch suited to support a load which is sym-
metrically distributed with respect to the
crown A, and which produces at every point
of the arch a pair of conjugate pressures,
the one horizontal and the other vertical.
Take as the axis of y the vertical through
the crown, and as the axis of x the hori-
FIG. 492. zontal through an origin O at a given dis-
tance from A.
Any portion A M of the arch is kept in equilibrium by the
horizontal thrust H at A, the tangential thrust T at M, and
the resultant load upon AM, which must necessarily act through
the point of intersection N of the lines of action of //and T.
Since the load at A is wholly vertical, H is given by
X.=P,P., - ...... (i)
EXAMPLES OF LINEAR ARCHES. jl
p and p being, respectively, the vertical intensity of pressure
and the radius of curvature at A.
Let MN = T, and take NS = H .
Complete the parallelogram SM\ the diagonal NL is the
resultant load upon AM "in direction and magnitude.
The vertical (KL) and the horizontal (KN) projections of
NL are, therefore, respectively, the vertical and horizontal
loads upon AM.
Denote the vertical load by V, the horizontal by//. Then
(2),
and
H=KN=SN-SK=H ~ Fcot 0, . . (3)
being the angle between MN and the horizon.
dV
p yy the vertical intensity of pressure, -j . ..... (4)
p x , the horizontal intensity of pressure
"> ..... (?)
EXAMPLE. A semicircular arch of radius r, with a hori-
zontal extrados at a vertical distance R from the centre.
The angle between the radius to J/and the vertical = 6.
.'. x r sin #, y = R r cos 0. . . . (i)
dx=r cos Ode, dy = r sin OdO ..... (2)
p y wy = w(R r cos 0), ...... (3)
w being the specific weight of the load. Hence,
V = wf\R - r cos 0)r cos BdB
I rtt r sin 2#\
= wr(R sine-- - ]. ... (4)
7^2 THEORY OF STRUCTURES.
Equations (3) and (4) give H '; for
p. = w(R-r), (5)
ind hence
HQ == wr(R -r) (6)
p x , the horizontal intensity of pressure,
d . ( n r B - sin cos 6 _\ , ,
= -7- (Fcotff) = w\R . -a rcosfl). (7)
dy^ \ 2 sin 9 /
Rankine gives the following method of determining whether
a linear arch may be adopted as the intrados of a real arch.
At the crown a of a linear arch ab measure on the normal a
length aCj so that c may fall within the limits required for
stability (e.g., within the middle third).
At c two equal and opposite forces, of the same magnitude
as the horizontal thrust H at a, and acting at right angles to
ac, may be introduced without altering the equilibrium.
Thus the thrust at a is replaced by an equal thrust at c, and
a right-handed couple of moment H . ac.
Similarly, the tangential thrust T at any point d of ab
may be replaced by an equal and parallel thrust at e, and a
couple of moment T . de.
The arch will be stable if the length of de, which is normal
to ab at dj is fixed by the condition T . de = H . ac, and if the
line which is the locus of e falls within a certain area (e.g.,
within the middle third of the arch ring.
8. Arched Ribs in Iron, Steel, or Timber. In the fol-
lowing articles, the term arched rib is applied to arches con-
structed of iron, steel, or timber. The coefficients of elasticity
are known quantities which are severally found to lie between
certain not very wide limits, and their values maybe introduced
into the calculations with the result of giving to them greater
accuracy. There are other considerations, however, involved
in the problem of the stability of arched ribs which still render
its solution more or less indeterminate.
It has been shown that the curve of pressure, or linear arch,
ARCHED RIB UNDER A VERTICAL LOAD. 763
is a funicular polygon of the extraneous forces which act upon
the real arch. It is, therefore, also the b ending-moment curve,
drawn to a definite scale, for a similarly loaded horizontal
girder of the same span, whose axis is the springing line.
When the arched rib carries a given symmetrically dis-
tributed load, it will be assumed that the linear arch coincides
with the axis of the rib, and that the thrust at any normal
cross-section is axial and uniformly distributed.
The total stress at any point is made up of a number of
subsidiary stresses, of which the most important are : (i) a
direct thrust ; (2) a stress due to flexure ; (3) a stress due to a
change of temperature. Each of these may be investigated
separately, and the results superposed.
9. Bending Moment (M) and Thrust (T) at any Point
of an Arched Rib under a Vertical Load. Let ABC be the
axis of the rib.
Let D and E be points on the same vertical line, E being
D-D
FIG. 493.
on the axis of the rib and D on the linear arch for any given
distribution of load.
Resolve the reaction at A into its vertical and horizontal
components, and denote the latter by H.
Since all the forces, excepting H, are vertical, the difference
between the moments at D and E = H . DE.
But moment at D o. Hence,
moment at E = M= H . DE.
Let the normal at E meet the linear arch in D'. Then, if
T is the thrust along the axis at E,
n r E
Tcos DED' = ff= 7 > approximately,
or
H .DE = T.D'E = M.
764 THEORY OF STRUCTURES.
10. Rib with Hinged Ends ; Invariability of Span.
Let ABC be the axis of a rib supported at the ends on pins or
FIG. 494.
on cylindrical bearings. The resultant thrusts at A and C
must necessarily pass through the centres of rotation. The
vertical components of the thrusts are equal to the corre-
sponding reactions at the ends of a girder of the same span
and similarly loaded, and H is given by the last equation in
the preceding article when DE has been found.
Let ADC be the linear arch for any arbitrary distribution
of the load, and let it intersect the axis of the rib at S. The
curvature of the more heavily loaded portion AES will be
flattened, while that of the remainder will be sharpened.
The bending moment at any point E of the axis tends to
change the inclination of the rib at that point.
Let the vertical through E intersect the linear arch in D
and the horizontal through A in F.
Let 8 be the inclination of the tangent at E to the hori-
zontal.
Let /be the moment of inertia of the section of the rib
at .
Let ds be an element of the axis at E.
_, Mds H.DE.ds
Change of inclination at E dv = ^ = - -=j -- .
If this change of curvature were effected by causing the
whole curve on the left of E to turn about E through an angle
dO, the horizontal displacement of A would be
ARCHED RIB WITH HINGED ENDS.
This is evidently equal to the horizontal displacement of
, and the algebraic sum of the horizontal displacements of all
points along the axis is
H.DE. EF. ds rH. DE . EF. ds
2- m - -=J- - - = Q, . . (i)
since the length AC is assumed to be invariable.
Thus, the actual linear arch must fulfil the condition ex-
pressed by eq. (i), which may be written
rDE.EF.ds
J- - -=0, (2)
since H and E are constant.
If the rib is of uniform section, /is also constant, and eq. (2)
becomes
CDE.EF.ds=o (3)
Also, since DE is the difference between DF and EF,
f(DF ~ EF)EF. ds=o =f^F. EF. ds-J*EF*ds (4)
Remark. Eq. i expresses the fact that the span remains
invariable when a series of bending moments, H . DE, act at
points along the rib. These, however, are accompanied by a
thrust along the arch, and the axis of the rib varies in length
with the variation of thrust.
Let H be the horizontal thrust for that symmetrical loading
which makes the linear arch coincide with the axis of the rib.
Let T be the corresponding thrust along the rib at E.
The shortening of the element ds at E of unit section
T ~ T
E
EXAMPLE I. Let the axis of a rib of uniform section and
hinged at both ends be a semicircle of radius r.
Let a single weight W be placed at a point upon the rib
whose horizontal distance from (9, the centre of the span, is a.
7 66
THEORY OF STRUCTURES.
The " linear arch " (or bending-moment curve) consists of
two straight lines DA, DC.
FIG. 495.
Draw any vertical line intersecting the axis, the linear
arch, and the springing line AC in E' , D', F', respectively.
Let OF' = x, and let dx be the horizontal projection upon
AC of the element ds at E'.
Then
-^ = cosec E'OF' = -== ,
dx E'F '
or
(i)
Applying condition (4),
f D'F'rdx + f D'F 'rdx = f E'F'rdx,
or
f D'F'dx + f D'F'dx = f E'F'dx,
or area of triangle ADC area of semicircle.
And if z be the vertical distance of D from AC,
zr =
ARCHED RIB WITH HINGED ENDS.
767
or
nr
z = = one-half of length of rib.
(2)
nr
(3)
Hence, if h be the horizontal thrust on the arch due to W,
= M = W
r* a'
2r
(4)
Similarly, if there are a number of weights W^ W^ PF 3 , . . ,
upon the rib, and if h^ h^, h^, . . . are the corresponding hori-
zontal thrusts, the total horizontal thrust //will be the sum of
these separate thrusts, i.e.,
(5)
It will be observed that the apices (D lt D^, D s , . . .) of the
several linear arches (triangles) lie in a horizontal line at the
nr
vertical distance from the springing line.
Ex. 2. An arched rib hinged at the ends and loaded with
weights W,, W^ W^ . . .
-. L__ _J i._.J__
\
FIG. 497.
Let i 2 3 4 ... be the line of loads, W^ being represented
by i 2, JF 2 by 2 3, W 3 by 3 4, etc., and let the segments \x,
THEORY. OF STRUCTURES.
iix, respectively, represent the vertical reactions at A and C.
Take the horizontal length xP to represent H, and draw the
radial lines Pi, P2, P$ t . . .
The equilibrium polygon Ag^g^ . . . must be the funicu-
lar polygon of the forces with respect to the pole P, and there-
fore the directions of the resultant thrusts from A to lt E l to
v , 9 to E 3 , . . . are respectively parallel to Pi, P2, ^3, ...
The tangential (axial) thrust and shear at any point p of
the rib, e.g., between E t and E s , may be easily found by draw-
ing Pt parallel to the tangent at/, and 3^ perpendicular to PL
The direct tangential thrust is evidently represented by Pt,
and the normal shear at the same point by 3/. The latter is
home by the web.
If/ is a point at which a weight is concentrated, e.g., t ,
draw Pt't" parallel to the tangent at , and 5/', 6t" perpen-
dicular to Pt't".
Pt' represents the axial thrust immediately on the left of
E+ , and 5/' the corresponding normal shear, while Pt" repre-
sents the axial thrust immediately on the right of E t , and 6t"
the corresponding normal shear.
A vertical line through P can only meet the line of loads
at infinity.
Thus, it would require the loads at A and C to be infinitely
great in order that the thrusts at these points might be vertical.
Practically, no linear arch will even approximately coincide
with the axis of a rib rising vertically at the springings, and
lience neither a semicircular nor a semi-elliptical axis is to be
recommended.
Ex. 3. Let the axis of the rib be a circular arc of span 21
and radius r, subtending an angle 2ot at the centre N.
Let the angles between the radii NE, NE' and the vertical
be ft and 0, respectively.
The element ds at E' = rd6.
Also, E'F' = r(cos 6 cos a) ; AF = / r sin 6 ;
D'F' = (l ~ r sin
ARCHED RIB WITH HINGED ENDS. 769
Applying condition (5),
/ r (cos 8 - cos 'a)*rdO
I rr~(t r sin 6>)r(cos 6 cos a)rdB
+ f j^ (lr sin 6>)r(cos 6 cos a)rdO,
Jt l ~- a
which easily reduces to
r\a(cos 20. -f- 2) f sin 2#J
T -j / a ( s i n ^ ^ cos ) -| (cos 2 or z as in Ex. I.
Ex. 4. Let the axis be a parabola of span 2/ and rise
(Fig. 498, Ex. 3). From the properties of the parabola,
la '
and
ds* = d
or, approximately,
ds dx\\ 2X '
Applying condition (5),
which easily reduces to
an equation giving z or DF.
Note. If the arch is very flat, so that ds may be considered
ARCHED RIB WITH ENDS ABSOLUTELY FIXED. 771
as approximately equal to dx, the term 2j-,^ 2 in the above
equation may be disregarded, and it may be easily shown that
16
or
2 =
32 k
II. Rib with Ends absolutely Fixed. Let ABC be the
axis of the rib. The fixture of the ends introduces two un-
'K L'
FIG. 499.
known moments at these points, and since H is also unknown,
three conditions must be satisfied before the strength of the
rib can be calculated.
Represent the linear arch by the dotted lines KL ; the
points K, L may fall above or below the points A, C.
Let a vertical line DEF intersect the linear arch in D, the
axis pf the rib in E, and the horizontal through A in F.
As in Art. 10, change of inclination at E, or dO, = ^- r .
\ El
But the total change of inclination of the rib between A and
C must be nil, as the ends are fixed.
*Mds
Mas
-7 = =
H.DE.ds
which may be written
(I)
(2)
since H and E are constant.
77 2 THE OR Y OF STRUCTURES.
If the section of the rib is uniform, / is constant and eq.
(2) becomes
Again, the total horizontal displacement between A and C
will be nil if the abutments are immovable. If they yield, the
amount of the yielding must be determined in each case, and
may be denoted by an expression of the form yw//, yu being
some coefficient.
As in Art. 10, the total horizontal displacement
p
~ J
H.DE.EF.ds
H.DE.EF.ds
~ET
But H and E are constant.
*DE . EF. ds
(5)
If the section of the rib is uniform, / is also constant, and
hence
fDE.EF.ds = o or =;... (6)
and since DE is the difference between DF and EF, this last
may be written
ds = o or = . ... (7)
Again, the total vertical displacement between A and C
must be nil.
The vertical displacement of E (see Art. 10)
ARCHED RIB WITH ENDS ABSOLUTELY FIXED.
Hence, the total vertical displacement
*H.DE.AF ,
= r
which may be written
.AF
/JJtL . Sif , , .
-f" *= fe>
since H and are constant. If the section of the rib is also
constant,
.AF.ds. (10)
Eqs. (2), (5), and (9) are the three equations of condition.
In eq. (9) AF must be measured from same abutment
throughout the summation.
The integration extends from A to C.
EXAMPLE I. Let the axis of the rib be a circular arc of
span 2/, subtending an angle 2a at the centre N.
Let a weight W be concentrated on the rib at a point E
whose horizontal distance from the middle point of the span
is a.
Let the radius NE make an angle /3 with the vertical.
The " linear arch " consists of two straight lines DA', DC' .
Let A A = r,, DF = z,CC' = y, .
774 THEORY OF STRUCTURES.
Draw any ordinate E'F' intersecting the linear arch in D f .
Let the radius NE' make an angle with the vertical.
Then
E'F' r(cos 6 cos ).
AF' = l rsintf, and D'F'=(lrs\n
if F f is on the left of F\
and D'F' = (l-rsm
if /*"' is on the right of F.
Also, ^y = rdO.
Applying condition (i),
os 6> cos *)<# ...... (i)
Applying condition (3), and assuming ^ = o,
/ B (cos ^ - cos ) | (/ - r sin ^q^' +^ 1 ^
3
+ Acos - cos a) I (/ r sin ^)y^ +7, I
os 6> - cos r)V0 ..... (2)
ARCHED RIB WITH ENDS ABSOLUTELY FIXED.
Applying condition (5),
+ A/+r sin 0) j (/- rsin tff^+f
= rC a (cos # cos )(/ r sin 0) (i + 2
+ r j^ + (/ - ^-yf~ }(/+*)(>+ 2
y
r *(i - ^) (/ - *) (i + 2^)^
/-a
// / r s
4* - r
.
EFFECT OF A CHANGE OF' TEMPERA TURE. 777
These equations may be at once integrated, and the result-
ing equations will give the values of y lt jj> 2 , z.
If the arch is very flat, so that ds may be taken to be ap-
proximately the same as dx, it may be easily shown that
2 /+ 50 2 I 5# 6
12. Effect of a Change of Temperature. The variation
in the span 2/ of an arch for a change of t from the mean
temperature is approximately = 2etl, e being the coefficient of
expansion.
Hence, if H t is the horizontal force induced by a change of
temperature, the condition that the length AC is invariable is
expressed by the equation
DE.EF.ds
__ 2et i _ a
If the rib is of uniform section, 7 is constant; and since
E is also constant, the equation may be written
. EF. ds 2etl = o.
EXAMPLE I. Let the axis AEC of a rib of uniform section
FIG.
be the arc of a circle of radius r subtending an angle 2a at the
centre.
First, let the rib be hinged at both ends.
7/8 THEORY OF STRUCTURES.
It is evident that the straight line AC is the "linear arch.'*
Then,
J*DE . EF. ds ^j^EF^ds = r* J*\cos B - cos afdB
r*{a(2 -f- cos 201) | sin 2a\.
Also, / r sin a.
H / 3
Note. If the axis is a semicircle, a = 90, and
!L?L
EI 2
2etl = o.
Second, let the rib \>z fixed zk both ends.
The " linear arch" is now a straight line A'C' at a distance
= Z?/*') from ^4 C given by the equation
CDE.ds^O.
.-. CDF. ds =fF. ds y
or
z Cds = r* /"(cos ^ cos a)dO,
or
a.3- = r(sin a a cos a).
Also,
. EF. ds =f(DF . EF^EF^ds = zj*EFds ~ C EF *ds
= 2.3T 2 (sin a a cos <*)~ r*{a(2-\- cos 2) f sin 2a\.
TT f -^
.'. -JT\ 2Jsr\sin a a cos a) r*{a(2 + cos 2a) f sin 2a\ (
2etl = o,
and / = r sin a.
Ex. 2. Let the axis ^7 of a rib of uniform section be a
parabola of span 2/ and rise k. (See Fig. 501 in Ex. I.)
EFFEC7" OF A CHANGE OF l^EMPERATURE. 779
First, let the rib be hinged at both ends.
The straight line AC is the linear arch. Then
CDE .EF.ds = f l F 2
and hence,
Second, let the rib be fixed at both ends.
The linear arch is the line A'C' at a distance z (= DF)
from A C given by the equation
or
E .ds = o = C(DF ~ EF)ds,
DFfds = fEF. ds.
*T(i + 7^)^ = A(i - f!)(i + 2^>)^.
or
2 2
Also,
.EF.ds =&F .EF.ds-^ (*
7^0 THEORY OF STRUCTURES.
Hence,
= o.
Remark. The coefficient of expansion per degree of Fah-
renheit is .0000062 and .0000067 for cast- and wrought-iron
beams, respectively. Hence, the corresponding total expansion
or contraction in a length of 100 ft., for a range of 60 F. from
the mean temperature, is .0372 ft. (= ^/') and .0402 ft. (= ").
In practice the actual variation of length rarely exceeds one-
half Q( these amounts, which is chiefly owing to structural con-
straint.
13. Deflection of an Arched Rib.
FIG. 502.
Let the abutments be immovable.
Let ABC be the axis of the rib in its normal position.
Let ADC represent the position of the axis when the rib is
loaded.
Let BDF be the ordinate at the centre of the span ; join
AB, AD.
Then
* = AD* - AF* = AB- AF\
arc AB
But
arc AB arc AD /
~~ ~ ~E '
/being the intensity of stress due to the change in the length
of the axis.
/. DF* = AB*i - - AF* =
ELEMENTARY DEFORMATION OF AN ARCHED RIB. /8l
AB* 2 -
f\*
= BF*- DF* = (BF- DF)(BF + DF)
= 2BF(BD\ approximately.
^rj is also sufficiently small to be disregarded. Hence,
h I
AB* f V + I* f
BD, the deflection, = -jr= -^ 7 -g , approximately.
14. Elementary Deformation of an Arched Rib.
FIG. 503.
The arched rib represented by Fig. 503 springs from two
abutments and is under a vertical load. The neutral axis PQ
is the locus of the centres of gravity of all the cross-sections, of
the rib, and may be regarded as a linear arch, to which the
conditions governing the equilibrium of the rib are equally ap-
plicable.
Let A A ' be any cross-section of the rib. The segment
AA'P is kept in equilibrium by the external forces which act
upon it, and by the molecular action at A A'.
The external forces are reducible to a single force at C and
to a couple of which the moment M is the algebraic sum of
the moments with respect to C of all the. forces on the right
of C.
The single force at C may be resolved into a component T
along the neutral axis, and a component Sin the plane A A'.
THEORY OF STRUCTURES.
The latter has very little effect upon the curvature of the neu-
tral axis, and may be disregarded as compared with M.
Before deformation let the consecutive cross-sections BE 1
and AA meet in R ; R is the centre of curvature of the arc
CC' of the neutral axis.
After deformation it may be assumed that the plane A A'
remains unchanged, but that the plane BB' takes the position
B"B'". Let AA' and B"B"' meet in R' ; R is the centre of
curvature of the arc CC' after deformation.
Let abc be any layer at a distance z from C.
Let CC = As, CR = R, CR' = R' y and let Aa be the sec-
tional area of the layer abc.
By similar figures,
ac __ R' + z ab _ R+z
~~A r>/ and ~ ~
As R As R
i i i\
The tensile stress in abc
- A be As.z
= E . Aa; = E . Aa -r
ab ab
z? A I I l \ i
= jfi . Aa . x\~pT jrjji very nearly.
The moment of this stress with respect to C
Hence, the moment of resistance at A A '
the integral extending over the whole of the section.
ELEMENTARY DEFORMATION OF AN ARCHED RIB. 783
Again, the effect of the force T is to lengthen or shorten
the element CC', so that the plane BB' will receive a motion
of translation, but the position of -R' is practically unaltered.
Corollary i. Let A be the area of the section AA r .
The total unit stress in the layer abc
T Ms
the sign being plus or minus according as M acts towards or
from the edge of the rib under consideration.
From this expression may be deduced (i) the position of
the point at which the intensity of the stress is a maximum for
any given distribution of the load; (2) the distribution of the
load that makes the intensity an absolute maximum ; (3) the
value of the intensity.
Cor. 2. Let w be the total intensity of the vertical load per
horizontal unit of length.
Let w, be the portion of w which produces only a. direct
compression.
Let //be the horizontal thrust of the arch.
Let P be the total load between the crown and AA' which
produces compression.
Refer the rib to the horizontal OX and the vertical OPY
as the axes of x and y, respectively.
Let x, y be the co-ordinates of C.
Then "
P=ff~-' t but dP = w,dx.
dx
....... (3)
MW
also, :
784 THEORY OF STRUCTURES.
15. General Equations.
Let / be the span of the arch.
Let x, y be the co-ordinates of the point C before deforma-
tion.
Let x\ y' be the co-ordinates of the point C after deforma-
tion.
Let be the angle between tangent at C and OX before
deformation.
Let 0' be the angle between tangent at C and OX after
deformation.
Let ds be the length of the element CC' before deforma-
tion.
Let ds be the length of the element CC' after deformation.
d(T i dO i
Effect of flexure. ~jp ~g> and ~fa ~ R'
Mi i dO' de dO' - dO
Let i be the change of slope at C. Then
Mds Mds
dt = dO dB' =
C
-0' = f.+ /
El ~ Eldx
**M ds
t being the change of slope at P, and a quantity whose value
has yet to be determined.
Again, the general equations of equilibrium at the plane
A A' are
d
for the portion w l , Cor. 2, Art. 14, produces compression only
and no shear.
GENERAL EQUATIONS. 785
S 9 being the still undetermined vertical component of the shear
dv
at P, and ~ the slope at P. Also,
y, ; - *, (8)
J/ being the still undetermined bending moment at P.
Equations (5), (6), (7), and (8) contain the four undeter-
mined constants //, 5 , M , / .
Let M l , 5 t , and z, be the values of M, 5, and z, respectively,
at Q.
Equations of Condition. In practice the ends of the rib are
either j&ritf? or free.
If they are fixed, z' = o ; if they are free, M = o. In either
case the number of undetermined constants reduces to three.
If the abutments are immovable, x l / = o. If the abut-
ments yield, x^ / must be found by experiment. Let ;r, /
= /w being some coefficient. T\\t first equation of condi-
tion is
x l l=o 1 or x,~l=^H. .... (9)
Again, Q is immovable in a vertical direction, and the
second equation of condition is
(10)
Again, if the end Q is fixed, i l = o ; and if free, M l = o ; and
the third equation of condition is
* t = o, or M l =o ...... (il)
Substituting in equations (7) and (8) the values of the three
constants as determined by these conditions, the shearing force
and bending moment may be found at any section of the rib.
Again,
cos Q' = cos (0 i) cos + i s ^ n ^ J
sin 0' = sin (0 i) sin 6 i cos 6.
786 THEORY OF STRUCTURES.
dx' dx ,dy dy' dy dx
" ~j~r ~r + * j and jr = j l ~r- ( I2
ds' ds ds ds ds ds v
Hence, approximately,
d . , .dy d .dx
jrt* x) t -j- and -j( y' y\ i .
ds^ ds ds^ ds
Thus, if JTand Fare respectively the horizontal and verti-
cal displacements,
dX .dy dY .dx
-j- = tr- and r =. tr,
as as ds ds
or
dX . dY
16. Effect of T and of a Change of t in the Temperature.
Also, if there is a change from the mean of t in the tem-
perature, the length ds\i T^J must be multiplied by
(i e/), e being the coefficient of linear expansion.
-p-j et), approximately. (14)
By equations (i 2),
<& = (dx + i . dyy^ = (dx + i . dy)(i -jj &)
and
dy' = (dy - i.dx)^ = (dy - i.dx](i - ~ et).
GENERAL EQUATIONS. 787
.-. dX = d(x' - x )^idy--
and
= d(y f y) = idx - ^ etdy, approximately,
Hence,
and
Note. A nearer approximation than is given by the pre-
ceding results may be obtained as follows:
Let x + dx, y -\- dy be the co-ordinates of a point very
near C before deformation.
Let x' + dx' , y' -f- dy' be the co-ordinates of a point very
near C after deformation.
Then
ds* = dx* + dy* and ds'* = dx'* + dy'\
.-. ds'* - ds* = dx'* - dx* + dy'* - dy'*,
or
(ds'-ds)(ds f + ds) = (dx'-dx)(dx f + dx) + (dy'-dy)(dy : + dy).
. (ds' ds)ds = (dx' dx)dx + (dy' dy)dy y approximately.
. , , .ds dy
.'. dx dx = (ds ds)-j- (dy dy)-~
ax 'ax
and
ds dx
dy' -dy = (ds'
Hence, by equations (12) and (14),
.dy J T(ds
dx' dx t-^j-dx -=r-.( I dx eti-- dx
dx EA\dx' ^dx
788 THEORY OF STRUCTURES.
and
T (ds\dx J lds\dx ,
dy > -dy=~ ^^-dz -
/. ay / 1 i ds\ r x tds\
i -j-dx I -rr-Ti j 1 ax et I I -j- 1 dx
dx / hA\dx I J \dxl
'
and
r IdsVdx , r*tds\*dx
These equations are to be used instead of equations (15)
and (16), the remainder of the calculations being computed
precisely as before.
The following problems are, in the main, the same as those
given in Art. 180 of Rankine's Civil Engineering, I3th edition-
17. Rib of Uniform Stiffness. Let \hzdepth and sectional
form of the rib be uniform, and let its breadth at each point
vary as the secant of the inclination of the tangent at the point
to the horizontal.
Let A), /j be the sectional area and moment of inertia at
the crown.
Let A, I be the sectional area and moment of inertia at any
point C, Fig. 503.
Then
(17)
Also, since the moments of inertia of similar figures vary
as the breadth and as the cube of the depth, and since the
depth in the present case is constant,
(18)
T HsecO H
Again, -j- = - - = -j-, and the intensity of the thrust
is constant throughout.
ARCHED RIB OF UNIFORM DEPTH. 789
Hence, equations (5), (15), and (16), respectively, become
/*
-*=J*
dy
H
(21)
Equation (19) shows that the deflection at each point of the
rib is the same as that at corresponding points of a straight
horizontal beam of a uniform section equal to that of the rib
at the crown, and acted upon by the same bending moments.
Ribs of uniform stiffness are not usual in practice, but the
formulae deduced in the present article may be applied without
sensible error to flat segmental ribs of uniform section.
18. Parabolic Rib of Uniform Depth and Stiffness, with
Rolling Load; the Ends fixed in Direction; the Abut-
ments immovable.
D E O
FIG.
504-
Let the axis of x be a tangent to the neutral curve at its
summit.
Let k be the rise of the curve.
Let x, y be the co-ordinates at any point C with respect
to O.
Then
and
d~x-~ -T\~2-*>'
79 THEORY OF STRUCTURES.
Let -w be the .dead load per horizontal unit of length.
2/ " " live " " " " " "
Let the live load cover a, length DE, = r/, of the span.
'Denote by (A) formulae relating to the unloaded division
OEj and by (B) formulae relating to the loaded division DE.
.Equations (7) and (8), respectively, become
(o Z, JLT \
-)*; ......... (24)
(B) S=S,+ ^f - w} X - w'\ x - (i - r)l\. . . (25)
;. ...... (26)
(B) M = M, + S.x+--w-\x-(i-r)l\>. (27)
Since the ends are fixed,
*'. =0 = *; ...... . . (28)
Hence, by equations (19) and (26),
I ' (UH
I ( x }
(A) ,= --^ ^Ms + S. +(-jr -w)^ J; . (29)
and;hy equations (19) and (27),
i . IZkH
. (30)
i= -
When x = /, i = *', = o, and therefore, by the last equation,
\- - /
) 9 -H-\/ 2 W )A & r l *- (30
ARCHED RIB OF UNIFORM DEPTH. 791
dv
Again, let i = -- . Then
/"' .dy J C l dv dy , .dy C l d*y ,
I ?*"= / Txix dx = l ^x- / v ^ dx '
*/ o y o */ o
But i, = o, and - = --.
r i dy , s& r i u r i r* mj
' / V 1 *^ ~ ~T / vdx =-~r \ 7 ^
ty o t/o t/o */o
By the conditions of the problem, x' x and j/ 7 are
each zero at Q. Hence, equations (20) and (21), respectively,
become
(33)
=-J MX. .......... (34)
Substitute in eqs. (33) and (34) the value of i given by eq.
(30), and integrate between the limits o and /. Then
i* r tun \ r r
o -- h
" '
c o o -- --- ---
EI l ( 6 ' "24 ' \ I* J 120 120
and
ss.tuff
i IMS
-17.1
which may be written
79 2 THEORY OF STRUCTURES.
and
Hence, by eqs. (31), (35), (36),
4 ' ' 3
When x = l,M=M lt and 5=5,.
Hence, by eqs. (25) and (27),
and
\/ a wW a
w \ .
/2 2
-AjH; (37)
: ('-Z')+f^ (38)
15 ,
H=- --- _-^_ *J. . . . (39)
Substituting in these equations the values of S , M , given
above, we have
, . . . (40)
A I l>
and
r = - - M//VI- - -r + -) + -kH. . (41)
12 \2 3 4/3
To find the greatest intensity of stress, etc. The intensity of
T f-f
the stress due to direct compression *=-*-
ARCHED RIB OF UNIFORM DEPTH.
793
The intensity of the stress in the outside layers of the rib
due to bending is the same as that in the outside layers of a
horizontal beam of uniform section A 1 acted upon by the same
moments as act on the rib, for the deflections of the beam and
rib are equal at every point (eq. (19) ). Also, since the rib is
fixed at both ends, the bending moment due to that portion of
the load which produces flexure is a maximum at the loaded
end, i.e., at Q. Hence, the maximum intensity of stress (/,)
occurs at Q, and/, = r- -A/, 7, z l being the distance of the
AI /,
layers from the neutral axis.
H and M l are both functions of r, and therefore /j is an ab-
solute maximum when
But
and
dp,
o
i
dH
8.
\- '
dM,
dr
dH
150
~A
// a
dr -
L A
dr
r}*
dr
4
k
i-t-
15.
7, '
dM.
V44)
Hence, /, is an absolute maximum when
The roots of this equation are
r = i
and
2 4
r = 7
53 A
1
I
(45)
794
THEORY OF STRUCTURES.
11 .
r I makes -j-r- zero, so that the maximum value of p.
dr*
corresponds to one of the remaining roots.
Thus,
thewwwr. thrust '
and
(4 6 )
the max. tension = - H+ -jMj = //', (47)
the values of H and M l being found by substituting in eqs.
(39) and (40
i+^^-
2 4 Aft
or
. , 45
_2_ ' 4
(48)
according as the stress is a thrust or a tension.
If eq. (47) gives a negative result, there is no tension at any
point of the rib.
Note. The moment of inertia may be expressed in the form
q being a coefficient depending upon the form of the section.
Hence,
the maximum intensity of stress = ( H '-|- l ) . . ( : tg)
Corollary I. If the depth of the rib is small as compared
with k, the fraction j will be a small quantity, and the maxi-
mum intensity of stress will approximately correspond to r = -J-.
ARCHED RIB OF UNIFORM STIFFNESS. 795
The denominator in eq. (39) may be taken to be k, and it may
be easily shown that the values of //,//' are
: - (50)
,_ i (,// i i S *,\ 5 rf/ 54 v/r\
' - T, I TV + 7^*4^1 + 3125^: 5' (5I)
. 2. If the numerator in eqs. (48) is greater than the
denominator, then r must be unity. Hence, by eq. (39) and
and by eqs. (38) and (41),
Thus, //,/," can be found by substituting these values of
H and M l in eqs. (46) and (47).
19. Parabolic Rib of Uniform Stiffness, hinged at the
Ends.
Let the rib be similar to that of the preceding article.
Since the ends are hinged, M 9 = o = M l , while i is an un-
determined constant.
The following equations apply :
(A) S=S.+f=_- W J:r; (54)
(B) S=S.+ K- w )*- v /\x-(i-rW; (55)
79^ THEORY OF STRUCTURES.
(A) M =Ss + ? -;. ....... (56)
(B) M =S,x + -w-{x-( l -r- } l\\ (57)
%kH
(58)
!-* - + -- - i*-(-^ (59)
Assume that the horizontal and vertical displacements of
the loaded end are nil.
Substitute in eqs. (20) and (21) the value of i given by eq.
(59). Integrate and reduce, neglecting the term involving the
temperature. Then
_//.J.l _^ JL. : (6o )
S6ff V s ,.
-
From (57), since M, = o,
\/ , 7 r 3
-^- -wj--ze;7-. . . . (62)
Equations (60), (61), and (62) are the equations of condi-
tion.
Subtract (61) from (60). Then
UH \r lr* r*\ H
"/ r -
which may be written
ARCHED RIB OF UNIFORM STIFFNESS. 797
Subtract (63) from (62). Then
Hence,
/'U + V-5r ; +2o}
L " (6S)
Eliminating 5 between (61) and (62),
UH \/ 3
Also, by (55),
r - / - ^V/ = - /> suppose. (67)
Eliminating 5 between (62) and (67),
_ / , =5i= (^_^_<4_. . . . (68)
Eqs. (62), (65), (66), and (68) give the values of H y S ,S lt
and i .
Again, the maximum bending moment M' occurs at a
dM
point given by j- = o in (57), i.e.,
-w'\x-(i-r)l\. .(69)
Subtract (69) from (67). Then
_/>, = 5, = (^?- ;)(/- *) - /(/-*).
79$ THEORY OF STRUCTURES.
Hence, the distance from the loaded end of the point at
which the bending moment is greatest is
w --
Substitute this value of x in (57), and, for convenience, put
w --w pr m>
Then
p
M , = / -
m
'-m)l- W 'rl\
P'/w' - m a
m"\ 2~ '
But by (62), o = S.
Hence, M' ', the maximum bending moment,
\
(71)
As before, the greatest stress (a thrust)
= -H+M= P ;, .... ( 72 )
and the value of r which makes // an absolute maximum is
given by ~- = o. But by (71), M' involves r 10 in the numera-
ARCHED RIB OF UNIFORM STIFFNESS. 799
dpi
tor and r 5 in the denominator, so that -~ = o will be an
ar
equation involving r 14 .
One of its roots is r = i, which generally gives a minimum
value of //. Dividing by r i, the equation reduces to one
of the thirteenth order, but is still far too complex for use. It
is found, however, that r =. % gives a close approximation to the
absolute maximum thrust.
With this value of r, and, for convenience, putting
15 I, i
" "'
By (65),
By (62),
By (68),
By (66),
By (70),
n i w'
(74)
- I W'
(75)
(77)
r^TJ-^-+y
By (70.
. If the rib is merely supported at the ends but not
fixed, the. horizontal displacement of the loaded end may be
800 THEORY OF STRUCTURES.
represented by ^H (Art. 1 1). Thus the term ^H must be
added to the right-hand side of eq. (15).
20. Parabolic Rib of Uniform Stiffness, hinged at the
Crown and also at the Ends. In this case M=o at the
crown, which introduces a fourth equation of condition.
By (57).
L ( UH _ ^_^Tf_l V
~~ s 2 ~r" \~r~ ' w ls ~ 2 \ 2 "" r l '
which may be written
>r-r \ J f i\
'-r + -). (79)
Eliminating S between (79) and (62),
UH
w = w ( 2r -\- ^r i).
Hence,
H = -7}^ w'(2r* 4r+ i)}. . . . (80)
8 (
By (79)
S =*^-(3r* -4/-+I) (81)
By (68),
By (66),
By (70) and (82),
l-z=J =L. . (84)
By (71),
PARABOLIC RIB OF UNIFORM STIFFNESS. 8oi
When r = j,
w'r w f r \ ' (86)
t = - -77-7-, and M' = -?.
384 hi, 64 J
These results agree with those of (73) to (78), if n = i.
In general, when n I,
/
w-\- (5r a 5r 4 -f- 2r 6 ) =. w w\2r* ^r -f- i),
by (65) and (80). Hence,
and the roots are r j-, r = i, r = 4/2.
Hence, w = I only renders the expressions in (86) identical
with the corresponding expressions of the preceding article
when n = J or i.
Again, the intensity of thrust is greatest at the outer flange
of the loaded and the inner flange of the unloaded half of the
rib, and is
r uw a . w 1
The intensity of tension is greatest at the inner flange of the
loaded and the outer flange of the unloaded half of the rib,
and is
w' if w'
The greatest total horizontal thrust occurs when r = I, and
its value is
8O2 THEORY OF STRUCTURES.
21. Maximum Deflection of an Arched Rib. The deflec-
tion must necessarily be a maximum at a point given by i = o.
Solve for x and substitute in (16) to find the deflection y' y;
the deflection is an absolute maximum when -j\y' y) = o.
The resulting equation involves r to a high power, and is too
intricate to be of use. It has been found by trial, however,
that in all ordinary cases the absolute maximum deflection
occurs at the middle of the rib, when the live load covers its
v/hole length, i.e., when x -, and r = I.
CASE I. Rib of Art. 18. For convenience, put I + ~^ = s.
Then, by (39),
By (38) and (41),
-M.= ^ + < V 'f-^^^^ = -M,.. (88)
By (36) and (38),
S.= -6 ............. (89)
By (30), (38), (89),
i=-jTJ \ M X ~ 3 M J + 2M *J\ (9>
Hence, the maximum deflection
C" ., M, f*l x* , x \ MJ'
~ J ( ldx = ~ Tit (* ~ i-T + 2 r) dx = ~ F/32
r w + w' * i 5 e t r
=: - n: 7* T d. , suppose. . . (01 )
384 hi, s ~ 128 s k
MAXIMUM DEFLECTION OF AN ARCHED RIB. 803
The central deflection d^ of a uniform straight horizontal
beam of the same span, of the same section as the rib at the
crown, and with its ends fixed, is
Hence, neglecting the term involving the temperature,
4 = ^4 ......... (93)
CASE II. Rib of Art. 19.
By (65),
By (66) and (62),
By (30), (94), and (95),
Hence, the maximum deflection
n
^. (97)
If the ends of the beam in Case I are free, its central de-
flection
5 *l
El
_
(98)
Thus, the deflection of the arched rib in both cases is less
than that of the beam.
804
THEORY OF STRUCTURES.
22. Arched Rib of Uniform Stiffness fixed at the Ends
and connected at the Crown with a Horizontal Distribut-
ing Girder. The load is transmitted to the rib by vertical
struts so that the vertical displacements of corresponding
points of the rib and girder are the same. The horizontal
thrust in the loaded is not necessarily equal to that in the un-
loaded division of the rib, but the excess of the thrust in the
loaded division will be borne by the distributing girder, if the
rib and girder are connected in such a manner that the hori-
zontal displacement of each at the crown is the same.
The formulae of Art. 18 are applicable in the present case
with the modification that /, is to include the moment of
inertia of the girder.
The maximum thrust and tension in the rib are given by
equations (64) and (65).
Let z' be the depth of the girder, A' its sectional area.
The greatest thrust in the girder = ; -] --- ~. (99)
MJ_
~2EL
The greatest tension in the girder =
2EL
h~- ( I0 )
H and M l being given by equations (66) and (67), respectively.
The girder must have its ends so supported as to be capable
of transmitting a thrust.
23. Stresses in Spandril Posts and Diagonals. Fig. 505
represents an arch in which the spandril consists of a series of
vertical posts and diagonal braces.
1
n n+l
FIG. 505.
Let the axis of the curved rib be a parabola. The arch is
then equilibrated under a uniformly distributed load, and the
diagonals will be only called into play under a passing load.
STRESSES IN SPANDRIL POSTS AND DIAGONALS. 805
Let x, y be the co-ordinates of any point F of the parabola
with respect to the vertex C. Then
4&
y = TT*
Let the tangent at Fmeet CB in Z, and the horizontal BE
in G.
Let BC = k'. Then
BL^BC- CL = BC- CN=k' -y.
Let Nbe the total number of panels.
Consider any diagonal ED between the nth and (n -f- i)th
posts.
Let w be the greatest panel live load.
The greatest compression in ED occurs when the passing
load is concentrated at the first n I panel points.
Imagine a vertical section a little on the left of EF.
The portion of the frame on the right of this section is
kept in equilibrium by the reaction R at P, and by the stresses
in the three members met by the secant plane.
Taking moments about G,
D.GE cos B = R.AG,
D being the stress in DE, and the angle DEP.
Now,
w'n(n i)
~2 7v^'
Also,
" k' - y ' 2y
and hence,
^ z? r* D i k'x -\- xy A r A I > k'x x
GE GB -4- x = - --, and GA = ---
2J 22^
Hence,
2 A^ IT j: + xy
The stresses in the counter-braces (shown by dotted lines in
the figure) may be obtained in the same manner.
806 THEORY OF STRUCTURES.
The greatest thrust in EF ~ w' -|- w.
The greatest tension in EF = ZJcos w, w being the
dead load upon EF.
If the last expression is negative, EF is never in tension.
24. Clerk Maxwell's Method of determining the Re-
sultant Thrusts at the Supports of a Framed Arch. Let
As be the change in the length s of any member of the frame
under the action of a force P, and let a be the sectional area of
the member. Then
.
Ea
the sign depending upon the character of the stress.
Assume that all the members except the one under con-
sideration are perfectly rigid, and let Al be the alteration in
the span / corresponding to As. The ratio is equal to a
constant m, which depends only upon the geometrical form of
the frame.
.. Al = m . As mP-^- .
Ea
Again, P may be supposed to consist of two parts, viz.,/,
due to a horizontal force H between the springings, and / 2 due
to a vertical force V applied at one springing, while the other
is firmly secured to keep the frame from turning.
By the principle of virtual velocities,
/, M
Similarly, - is equal to some constant n, which depends
only upon the form of the frame.
=- (m*H+- mnV\~.
CLERK MAXWELL'S METHOD.
Hence, the total change in / for all the members is
If the abutments yield, let ^Al = }*H, ^ being some co-
efficient to be determined by experiment. Then
If the abutments are immovable, 241 is zero, and
(D)
2 (m*~
the same as the corresponding reaction at the end of a
girder of the same span and similarly loaded. The required
thrust is the resultant of H and F, and the stress in each
member may be computed graphically or by the method of
moments. In any particular case proceed as follows :
(1) Prepare tables of the values of m and n for each member.
(2) Assume a cross-section for each member, based on a
probable assumed value for the resultant of V and H.
(3) Prepare a table of the value of w?-- for each member,
and form the sum '2{m' i ^~
\ Ea
(4) Determine, separately, the horizontal thrust between
the springings due to the loads at the different joints. Thus,
let v l , v^ be the vertical reactions at the right and left supports
due to any one of these loads. Form the sum
using v l for all the members on the right of the load and v^ for
all those on its left. The corresponding thrust may then be
808 THEORY OF STRUCTURES.
found by eq. (C) or eq. (D), and the total thrust H is the sum
of the thrusts due to all the weights taken separately.
(5) Repeat the process for each combination of live and
dead load so as to find the maximum stresses to which any
member may be subjected.
(6) If the assumed cross-sections are not suited to thes^
maximum stresses, make fresh assumptions and repeat the
whole calculation.
The same method may be applied to determine the result-
ant tensions at the supports of a framed suspension-bridge.
Note. The formulae for a parabolic rib may be applied
without material error to a rib in the form of a segment of a
circle. More exact formulae may be obtained for the latter in
a manner precisely similar to that described in Arts. 18-22,
but the integrations will be much simplified by using polar co-
ordinates, the centre of the circle being the pole.
EXAMPLES. 809
EXAMPLES.
1. Assuming that an arch may be divided into elementary portions
by imaginary joint planes parallel to the direction of the load upon the
arch, find the limiting span of an arch with a horizontal upper surface
and a parabolic soffit (latus rectum = 40 ft.), the depth over the crown
being 6 ft. and the specific weight of the load 120 Ibs. per cubic foot;
the thrust at the crown is horizontal (= P) and 4 ft. above the soffit.
2. A masonry arch of 90 ft. span and 30 ft. rise, with a parabolic in-
trados and a horizontal extrados, springs from abutments with vertical
faces and 10 ft. thick, the outside faces being carried up to meet the
extrados. The depth of the keystone is 3 ft. The centre of resistance
at the springing is the middle of the joint, and at the crown 12 in. below
the extrados. The specific weight of the masonry may be taken at 150
Ibs. per cubic foot. Determine (a) the resultant pressure in the vertical
joint at the crown ; (&} the resultant pressure in the horizontal joint at
the springing ; (c) the maximum stress in the vertical joint aligning with
the inside of an abutment.
3. The intrados of an arch of 100 ft, span and 20 ft. rise is the segment
of a circle. The arch ring has a uniform thickness of 3 ft. and weighs
140 Ibs. per cubic foot ; the superincumbent load may be taken at 480
Ibs. per lineal foot of the ring. Determine the mutual pressures at the
key and springing, their points of application being 2 ft. and i ft., re-
spectively, from the intrados. Also find the curve of the centres of pres-
sure.
4. The soffit of an arch of 30 ft. span and 10 ft. rise is a transformed
catenary. The masonry rises 10 ft. over the crown, and the specific
weight of the load upon the arch may be taken at 120 Ibs. per cubic foot.
Determine the direction and amount of the thrust at the springing.
5. A concrete arch has a clear spring of 75 ft. and a rise of 7^ ft. ; the
height of masonry over crown = 5 ft. ; the weight of the concrete = 144
Ibs. per cubic foot. Determine the transformed catenary, the amount
and direction of the thrust at the springing, and the curvatures at the
crown and springing.
Ans. m 23.9 ; thrust = 91,354 Ibs. ; slope at springing = 25! ;
radius of curvature = 114.2 ft. at crown and =248.7 ft. at
springing.
6. Determine the transformed catenary for an arch of 60 ft. span and
15 ft. rise, the masonry rising 6 ft. over the crown and weighing 120 Ibs.
per cubic foot. Also find the amount and direction of the thrust at the
abutments.
8 10 THEORY OF STRUCTURES.
7. Determine the transformed catenary for an arch of 30 ft. span and
7^ ft. rise, the height of masonry over the crown being 4^ ft. ; weight of
the masonry = 125 Ibs. per cubic foot. Also find the thrust at the spring-
ing and the curvature at the crown and the springing.
8. In a parabolic arch of 50 ft. span and 10 ft. rise, hinged at both
ends, a weight of i ton is concentrated at a point whose horizontal dis-
tance from the crown is 10 ft. Find the total thrust along the axis oi
tUe rib on each side of the given point, allowing for a change of 60 from
the mean temperature (e = .0000694).
9. A parabolic arched rib of 100 ft. span and 20 ft. rise is fixed at the
springings. The uniformly distributed load upon one-half of the arch
is loo tons, and upon the other 200 tons. Find the bending moment
and shearing force at 25 ft. from each end.
10. An arched rib with parabolic axis, of 100 ft. span and 12^ ft. rise,
is loaded with I ton at the centre and i ton at 20 ft. from the centre,
measured horizontally. Determine the thrusts and shears along the rib
at the latter point, and show how they will be affected by a change of
100 F. from the mean ; the coefficient of linear expansion being .00125
for 1 80 F.
u. A parabolic arched rib hinged at the ends, of 64 ft. span and 16
ft. rise, is loaded with i ton at each of the points of division of eight
equal horizontal divisions. Find the horizontal thrust on the rib, allowing
for a change of 60 F. from the mean temperature. Also find the maxi-
mum flange stresses, the rib being of double-tee section and 12 in. deep
throughout. (Coefficient of linear expansion per i F. / -f- 144000.)
12. The axis of an arched rib of 50 ft. span, 10 ft. rise, and ninged at
both ends is a parabola. Draw the linear arch when the rib is loaded
with two weights each equal to 2 tons concentrated at two points 10 ft.
from the centre of the span. If the rib is of double-tee section and 24
in. deep, find the maximum flange stresses.
If the arch is loaded so as to produce a stress of 10,000 Ibs. per square
inch in the metal, show that the rib will deflect .029 ft., E being 25,000,000
Ibs.
13. A steel parabolic arched rib of 50 ft. span and 10 ft. rise is hinged
at both ends and loaded at the centre with a weight of 12 tons. Find
the horizontal thrust on the rib when the temperature varies 60 F. from
the mean, and also find the maximum flange stresses, the rib being of
double-tee section and 12 in. deep.
14. A semicircular rib, pivoted at the crown and springings, is loaded
uniformly per horizontal unit of length. Determine the position and
magnitude of the maximum bending moments, and show that the hori-
zontal thrust on the rib is one-fourth of the total load.
15. Draw the linear arch for a semicircular rib of uniform section
EXAMPLES. 8ll
under a load uniformly distributed per horizontal unit of length (a) when
hinged at both ends ; (b) when hinged at both ends and at the centre; (c)
when fixed at both ends.
16. A semi-elliptic rib (axes ia and 2b) is pivoted at the springing.
Find the position and magnitude of the maximum bending moment, the
load being uniformly distributed per horizontal unit of length.
How will the result be affected if the rib is also pivoted at the crown ?
17. Draw the equilibrium polygon for a parabolic arch of 100 ft.
span and 20 ft. rise when loaded with weights of 3, 2, 4, and 2 tons, re-
spectively, at the end of the third, sixth, eighth, and ninth division from
the left support, of ten equal horizontal divisions. (Neglect the weight
of the rib.)
If the rib consist of a web and of two flanges 2^ ft. from centre to
centre, determine the maximum flange stress.
18. Find the flange stresses at the ends of the rib, in the preceding
question, and also at the points at which the weights are concentrated,
when both ends are absolutely fixed.
19. A semicircular rib of 28 ft. span carries a weight of ton at 4 ft.
(measured horizontally) from the centre. Find the thrust and shear at
the centre of the rib and at the point at which the weight is concen-
trated.
20. The axis of an arched rib hinged at both ends, for a span of 50 ft.
and a rise of 10 ft., is a parabola. Draw the equilibrium polygon when
the arch is loaded with two equal weights of 2 tons concentrated at two
points 10 ft. from the centre of the span. Also determine the maximum
flange stress in the rib, which is a double-tee section 2 ft. deep.
21. The load upon a parabolic rib of 50 ft. span and 15 ft. rise, hinged
at both ends, consists of weights of I, 2, and 3 tons at points 15, 25, and
40 ft., respectively, from one end. Find the axial thrusts and the shears
at these points.
Ans. Horizontal thrust = 9.6 tons.
Axial thrusts : above i ton = 9.3 tons ;
below i " =97 "
above 3 tons = 8.3 "
below 3 " = io.r "
Shears : above i ton = 3. i tons ;
below i " = 2.2 "
above 3 tons = 5 "
below 3 " = 2.6 "
22. Draw the linear arch and determine the maximum flange stresses
for an arched rib of 80 ft. span, 16 ft. rise, and loaded with five weights
each of 2 tons at the end of the first, second, third, fourth, and fifth
division, of eight equal horizontal divisions. The rib is of double-tee
8l2 THEORY OF STRUCTURES.
section and 30 in. deep. Also find the shears and the axial thrusts at
the fifth point of division.
23. A wrought-iron parabolic lib of 96 ft. span and 16 ft. rise is
hinged at the two abutments; it is of a double-tee section uniform
throughout, and 24 in. deep from centre to centre of the flanges. Deter-
mine the compression at the centre, and also the position and amount
of the maximum bending moment (a) when a load of 48 tons is concen-
trated at the centre ; (b) when a load of 96 tons is uniformly distributed
per horizontal unit of length.
Determine the deflection of the rib in each case.
24. Design a parabolic arched rib of 100 ft. span and 20 It. rise, hinged
at both ends and at the middle joint ; dead load =40 tons uniformly
distributed per horizontal unit of length, and live load = i ton per hori-
zontal foot.
25. Show how the calculations in the preceding question are affected
when both ends are absolutely fixed.
26. In the framed arch represented by the figure, the span is 120 ft.,
t * ie r ' Se I2 * L> l ^ e c ^ e P t ^ 1 ^ tne truss at the
crown 5 ft., the fixed load at each top joint
FIG. 5 o6. 10 tons, and the moving load 10 tons. De-
termine the maximum stress in each member with any distribution of
load. Show that, approximately, the amount of metal required for the
arch : the amount required for a bowstring lattice-girder of the same
span and 17 ft. deep at the centre : the amount required for a girder of
the same span and 12 ft. deep :: 100 : 155 : 175.
27. The steel parabolic ribs for one of the Harlem River bridges has a
clear opening of 510 ft., a rise of 90 ft., a depth of 13 ft., and are spaced
14 ft. centre to centre. The dead weight per lineal foot is estimated at
33,000 Ibs. and the live load at 8000 Ibs. ; a variation in temperature of
75 F. from the mean is also to be allowed for. Determine the maxi-
mum bending moment (assuming /constant), and the maximum deflec-
tion. R = 26,000,000 Ibs. Show how to deduce the play at the hinges.
28. A cast-iron arch (see figure) whose cross-sections are rectangular
O'g'o' and uniformly 3 in. wide, has a straight horizon-
tal extrados, and is hinged at the centre and at
the abutments. Calculate the normal intensity of
stress at the top 'and bottom edges D, of the
FIG. 507. vertical section, distant 5 ft. from the centre of
the span, due to a vertical load of 20 tons concentrated at a point dis-
tant 5 ft. 4 in. horizontally from B. Also find the maximum intensity
of the shearing stress on the same section, and state the point at which
it occurs. (AB 21 ft. 4 in.).
INDEX.
Allowance for the weight of a beam, 403.
Alternating stresses, 152.
American iron columns, 532.
Anchorage, 704.
Angle of repose, 237.
" torsion, 568.
Angular momentum, 177.
Anti-friction curve, 320.
" pivots, 320.
Arch, 470.
Arch abutment, maximum thickness of,
649.
Arch, conditions of equilibrium of, 745.
" formulae for thickness of, 750.
11 linear, 743, 750, 760.
Arched ribs, 740, 762.
" " deflection of, 780, 802.
Arched ribs, effect of change of tem-
perature on, 770, 786.
Arched ribs, elementary deformation of,
781.
Arched ribs, general equations of equi-
librium of, 784.
Arched ribs, graphical determination of
stresses in, 677.
Arched rib of uniform stiffness, 788,
789, 795, 800, 804.
Arched ribs with fixed ends, 771.
'* " with hinged ends, 764.
Arched ribs with axis in form of circu-
lar arc, 769, 773.
Arched ribs with parabolic axis, 760,
775-
Arched ribs with semicircular axis, 765,
775-
Arches, middle-third theory of, 746.
Auxiliary truss, 719.
Back-stays, 16, 704.
Baker's formulae for strength of pillars,
549-
Balancing, 198.
Beam acted upon by oblique forces, 396.
Beam, transverse strength of, 340, 429.
Beam, transverse vibration of loaded,
461.
Beams, equilibrium of, 93.
" of uniform strength, 358-365.
Bearing surface, 314, 315.
Belts, 321.
" effect of high speed in, 325.
" effective tension of, 324.
" slip of, 326.
" stiffness of, 327.
Bending moment, 96, 118, 434.
Bending moment in plane which is not
a principal plane, 354.
Bending moment, relation between, and
shearing stress, 108.
Bevel-wheels, 335.
Boilers, 586.
Bollman truss, 56, 618.
Bowstring truss, 61, 618.
Brace, i, 25.
Brakes, 323.
Breaking-down point, 149.
Breaking stress, 147.
weights, 348, 399.
Breaking weights of iron girders, 369,
370..
Breaking weights, tables of, 212.
Brickwork, 1.49.
Bridge, bowsinng suspension, 626.
" loads, 600.
" trusses, 17, 52.
chords of, 625.
" depth of, 597.
Bridge trusses, maximum allowable
stress in, 657.
Bridge trusses, stiffness of, 598.
" stringers of, 656.
Bridges, 597.
position of platform of, 598.
Buckling of pillars, 513, 515.
813
8i4
INDEX.
Cable with sloping suspenders, 717.
Cables, 703.
' curves of, 706.
' deflection of, 714.
' length of arc of, 712.
' parameter of, 711.
weight of, 713.
Camber, 388, 659.
Cantilever, 365.
" curve of boom of, 634.
deflection of, 638.
" depth of, 637.
" weight of, 632.
Cast-iron, 147.
Catenary, 34, 706, 750.
Cement, 150.
Centres of gravity, n.
Centre of resistance, I, 743.
Centrifugal force, 181.
Centripetal force, 182.
Clapeyron's theorem, 292.
Coefficient of cubic elasticity, 255.
elasticity, 141, 143.
fluidity, 162.
hardness, 164.
lateral elasticity, 144.
rigidity, 254, 285.
rupture, 248.
torsional rupture, 574.
transverse elasticity, 285.
Collar-beams, 25.
Columns, see Pillars, 513, 538.
flexure of, 554, 5*57.
Compound strain, 236.
Compression, 141.
Conjugate stresses, 247.
Continuous girders, 463.
Continuous girders, advantages and dis-
advantages of, 486.
Continuous girders, maximum bending
moment in, 465.
Coulomb's laws, 568.
Counterbrace, 60.
Counter-efficiency, 328.
Counterforts, 270.
Covers of riveted joints, 665.
Cranes, 13.
bent, 31.
" derrick, 16.
jib, 13-
pit, IS-
Crank effort, 207.
Cubic elasticity, 255.
" strain, 283.
D<.ad load, 143, 600.
Deflection, curve of, 434.
of girders, 384-386, 638.
Deformation, 140, 251, 254.
Dock walls, 270.
Dynamometer, Prony's, 327.
Earth foundations, 258.
Earthwork, 255.
pressure of, 257.
Earthwork, Rankine's theory applied to
retaining walls, 264.
Efficiency of mechanisms, 335.
" of riveted joints, 666.
Elastic curve, 355.
" moment, 96, 340.
Elasticity, 140.
coefficient of, 141, 143.
" cubic, 255.
" lateral, 144.
" transverse, 285,
limit of, 145.
Ellipse of stress, 241.
Ellipsoid of stress, 281.
Empirical rules for wind-pressure, 663.
Encastr6 girders, 458.
Energy, 207.
" curves of, 207.
" fluctuation of, 207.
" kinetic, 167, 169, 170.
" potential, 167.
Envelope of moments, 121.
Equalization of stress, 349.
Equalizer, 629.
Equilibrated polygon, 740.
Equilibrium of beams, 428.
Equilibrium of beams, general equations
of, 428.
Equilibrium of flanged girders, 366.
Euler's theory of the strength of pillars,
537-
Examples, 69-92, 132-139, 216-234,
294-298, 337-339* 407-427, 490-512,
563-567, 580-585, 594~59 6 > 689-702,
734-739, 809-812.
Expansion of solids, 215.
Extension of prismatic bar, 289.
Extrados, 740.
Eyebars, 661.
steel, 665.
Factor of safety, 150.
Fatigue, 152.
Fink truss, 54.
Flanged girders, 365.
" " equilibrium of, 366.
" stiffness of, 384.
Flanges, 365, 597.
" curved, 366.
" horizontal. 366.
Flexure of columns, see Pillars, 554,
557-
Flow of solids, 162.
INDEX.
815
Fluctuation of stress, 151.
Fluid pressure, 162.
Force polygon, 3, 7, 119.
Foundations, earth, 258.
limiting depth of, 258.
of walls, 270.
Fracture, 141.
Framed arch, stresses in, 804.
Framed arch, Clerk Maxwell's method
of determining stresses in a, 806.
Frames, i, 2.
incomplete, 27, 61.
Friction, 300.
angle of, 237.
coefficient of, 300, 313.
journal, 310.
rolling, 310.
Funicular curve, 10.
polygon, 3, 7, 117, II9 .
Gins, 17.
Girder of uniform strength, 381.
Gordon's formulae for pillars, 522.
Hinged girders, 127-131.
Hodgkinson's formulae for the strength
of pillars, 513. 517-521.
Hooke's law, 142.
Howe truss, 58, 611.
Impact, 184.
Impulse, 176.
Incomplete frames, 27.
Inertia, 198.
moment of, 12, 342.
pressure due to, 200.
Inflection, point of, 453-463.
Internal stress, 235.
Isotropic bodies, 283.
Joint of rupture, 747.
Keystone, 741.
Lateral bracing, 654.
Lattice girder, 600.
Launhardt's formula, 153.
Lenticular truss, 626.
Limit of elasticity, 145.
Line of loads. 5.
resistance, 273-276, 741, 750.
" rupture, 265.
Linear arch, 743, 753-760.
Loads, live, in, 115, IIQ> 600,639,641,
Loads, stationary (dead), 118, 600.
Long pillars, 535.
Mansard roof, 6.
Masonry, 149.
Mechanical advantage, 294.
Middle third theory of arches, 746,
Modulus of elasticity, 141.
" rupture, 348.
" transverse elasticity, 254
Moment of forces, 116.
" inertia, 12, 342.
" examples of, 371-81.
Moment of inertia, variable section of
455-
Moment of inflexibility, 96.
" resistance, 96.
Momentum, 176.
Mortar, 150.
Neutral axis, 340.
of a loaded beam, 435-454.
surface, 340.
Oblique resistance, 169.
OsciLatory motion, 190, 195.
Panel points, 52.
r^anels, 54.
D iers, 65.
513.
Euler's formulas for, 527.
failure of, 515.
flexure of, 515.
formulae for American, 527-532.
Gordon's formulae for, 522
Pillars, Hodgkinson's formulae for '517-
521.
Pillars, Rankine's formula for, 526.
Pillars with stress uniformly distributed
516.
Pillars with uniformly varying stress,
Pins, 661.
Piston velocity, curves of, 205
Pivots, 316.
conical, 319
cylindrica., 316.
Schiele's (anti-friction), 320
Plasticity, 141.
Poisson's ratio, 142
Pole, 7.
Polygon of forces, 3, 7.
" pressure, 743.
Pratt truss, 60.
Primitive strength, 153.
Principals, 33, 34.
Prony's dynamometer, 327.
Proof strain, 171.
stress. 171.
Purchase 304.
Purlins, 33, 34.
Radius of gyration, 174, 528-531
p "' of twist, 289.
8i6
INDEX.
Redundant bars, 48.
Reservoir walls, 271.
Resilience, 171.
Retaining walls, 260.
Retaining walls, conditions of equiib-
rium of. 260.
Retaining walls, Rankine's theory ap-
plied to, 264.
Rivet connection between flange and
web, 660.
Riveted joints, 668.
covers of, 675.
design of, 678.
efficiency of, 679.
failure of, 670.
stresses in, 670.
theory of, 671-675.
Riveting, 666.
Rivets, 666.
" dimensions of, 667.
Rocker-link, 629.
Rollers, 35, 639.
Roof trusses, 17.
" distribution of loads on, 39.
types of, 33.
weights of, 37.
Ropes, 321.
Saddles, 704.
Schiele's pivots, 320.
Screws, 306.
" endless, 309.
Sections, method of, 62
Set, 145. -,. ' *~~**v
Shafting, distance. EetvySfen bearings of,
575- ' /'* * 1
Shafting, efficiency of, 577.
intprrfal-streiB&iii, 237.
stiffness of, 573.
" ' torsional strength of, 288-291.
Shear, 141.
" maximum, 121, 237.
Shearing force, 95.
" '. ^ et examples of, 97, 108.
Shearing force and bending moment,
relation between, 108.
Shearing stress, 198.
. " distribution of, 391.
Shear-legs, 17.
Similar girders, 401-404.
Skew-backs, 34, 74-
Soffit, 740.
Spandril, 740.
Specific weight, 143.
Spherical shells, 591.
Spritigings, 740.
Springs, 355, 456-458.
" simple rectangular, 456.
Springs, spiral, 477.
Springs of constant depth, but triangu-
lar in plan, 4.57.
Springs of constant width, but parabolic
in elevation, 457.
Statical breaking strength, 153.
Steel, 148.
Stiffening truss, 719.
" hinged at centre, 725.
Stiffness, 190, 387, 389.
Strain, 140, 251.
Straining cill, 18.
Stress, 141, 251
" and strain, relation between, 281.
" general equations of, 277.
principal, 241.
planes of, 237.
Stresses, conjugate, 248, 250.
Stress-strain curves, 147-149.
line, 144.
Strut, i.
St. Venant's torsion results, 572.
Surface loading, 350.
Suspenders, 706.
Suspension-bridges, 703.
" loads on, 730.
Suspension-bridges, modifications of,
73i.
Suspension-bridges, pressure upon piers
of, 718.
Swing-bridges, 470-472.
Tables of breaking weights and coeffi-
cients of bending strength of timber,
212, 213.
Table of coefficients of axle friction,
336.
Table of coefficients in Gordon's for-
mula, 524.
Table of coefficients in Rankine's mod-
ification of Gordon's formula, 526.
Tables of diagonal and chord stresses,
644-650.
Tables of efficiencies, 587.
" " elliptic integrals, 562.
" " expansion of solids, 215.
" " eyebar dimensions, 665.
" " factors of safety, 214.
Tables of loads for highway bridges,
687.
Tables of strengths, elasticities, and
weights of iron and steel, 210.
Table of strengths, elasticities, and
weights of various alloys, 211.
Tables of weights of modern bridges,
682-687.
Table of weights and crushing strength
of rocks, 214
Table of weights of roof coverings, 67.
INDEX.
8I 7
Table of weights of roof frames, 67.
Tension, 141.
Theorem of three moments, 463-470.
Thick hollow cylinder, 588.
Timber, 149.
Torsion, 141, 568.
" St. Venant's results, 572.
Torsional coefficient of elasticity, 145.
resilience, 574.
Torsional strength of shafts, 288, 289,
571, 572.
Transverse strength, 141.
Transverse vibration of a loaded beam,
461.
Trellis girder, 600.
Tresca's theory of flow of solids, 162.
Tripods, 17.
Truss, 2.
' composite, 31.
king-post, 21.
' queen-post, 25.
' roof, 32.
' triangular, 19.
Trussed beams, 53.
Twist, 141.
Unwin's formula, 159.
Values of k*, 174, 528-531.
Vibration strength, 153.
Voussoir, 741.
Warren truss, 57.
Web thickness, 382.
Wedge, 303
Weights of roof coverings, 67.
" frames, 67.
Weyrauch's formula, 153.
Weyrauch's theory of buckling of pillars,
550.
Wheel and axle, 329.
Whipple truss, 618.
Wind pressure, 38, 67, 629, 651, 653.
Wind-pressure, American specifications
of, 652.
Wind-Pressure Commission rules, 653.
Wind-pressure, empirical regulations,
653.
Wohler's law, 150.
Work, 167.
effective, 178.
external, 168.
internal, 168.
useful, 178.
waste, 178.
Work done in bending a beam, 460.
Work done in small deformation of a
body, 292.
Work of journal friction, 314.
Working load, 150.
" strength, 150.
" stress, 150.
Wrought-iron, 148.
Yield-point, 149.
ERRATA.
Page 12, 1 5th line from top, for A\ , A*, As, read a\ , a*, a*.
23, Fig. 34, for BO read AO, and for AO read BO.
16562 16562
47, 2d line, for , read -=-.
52, 3d line from bottom, for W* read W*.
55, 8th line from top, for 2 7i read 2 7\.
63, 7th line from bottom, for "revolving" read " resolving."
3 V 13 3 V's
64, 3d line from top, for - read
2V 13
127, Fig. 181, line x* should lie between W* and B instead of
between E and W*.
215, i st line, for ACTORS read FACTORS.
219, 4th line from top, for 17,828 Ibs. read 17,328 Ibs.
223, 6th line from top, for 8625 ft.-lbs. read 15,750 ft.-lbs.
239, nth line from top, in 2d equation, for = I read = i.
274, 5th line from top, for CD read QP.
276, ist line (numerator), for ABQP read OBQ.
294, 2d line from bottom, for "thrust" read "tension," and for
"tension" read "thrust."
295, I2th line from bottom, for 11 38' read 3 29'; nth line from
bottom, for 3 26' read 63 26'.
297, loth line from top, for 9.8 ft. read 12.2 ft.; I3th line from
bottom, for 13.19 ft. read y - ft.
298, 7th line from bottom, for 20,720 Ibs. read i4,o49 T Jt Ibs.
326, 3d line from top, for $T*(.
530, ist line under (-), for S = 26/1 I? read S = 2.bh &* ; and
in denominator of next equation, for // 2 read ^.
555, 7th line from bottom, equation, for dQ read d(f>.
565, 3d line of Example 27, for "maximum" read "minimum."
571, bottom line, for i.57/) 2 Tread 1.57 D*Tf.
578, ist line of Example, for 50 read 95.
. " 580, Examples 7 and 8, for m read G.
581, Examples 10, u, and 15, for m read G.
582, Examples 18 and 19, for m read G.
583, Example 30, for m read G.
647, Table of Maximum Stresses in the Verticals, change 3d and
4th lines to read
7/3 = 42625 + 16200 = 58,825 Ibs.
7/4 = 28700 + 5400 = 34,100 "
649, 9th line from top, for 6250 read 7600, and for 1525 read 2875.
650, ist line of Compression Chord Table, for di read d\.
707, Equation (2), for PS read ps.
711, 6th line from bottom, after Art. 4 add " Case B."
714, 9th line from top, for H\ sec 6 read H sec 6.
715, 9th line from bottom, equation, for ^ read .
724, 4th line from top, for iv read x.
758, Equation 7, for 2iuyp sin read 2wyp\s\t\ ) .
u I* I
772, Equation (5), for read juE; Equations (6) and (7), for ^
read nEI, //
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