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MOVING LOADS ON KAIL WAY
1 UNDEEBEIDGES
INCLUDING :
DIAGRAMS OF BENDING MOMENTS AND SHEARING FORCES
AND
TABLES OF EQUIVALENT UNIFORM LIVE LOADS.
BY
HARRY BAMFORD, M.Sc., A.M.lNST.C.E.,
\ i
Lecturer on Engineering Drawing and Design, Glasgow University,
Formerly Associate Professor of Hydraulics, McGill University, Montreal.
WHITTAKEK & CO,
2, WHITE HART STREET, PATERNOSTER SQUARE, LONDON, E.G. ;
AND 64-66, FIFTH AVENUE, NEW YORK.
1907.
V
GENERAL
LONDON :
PRINTED BY WILLIAM CLOWES AND SONS, LIMITED,
DUKE STREET, STAMFORD STREET, S.E., AND GKEA.T WINDMILL STREET, W.
PREFACE.
THE attention of the author having been directed, some five years ago, to the
tedious and unscientific methods commonly used in practice in the preparation
of tables of " equivalent uniform live loads " for railway under-line bridges,
it occurred to him that much of the labour which the application of these
methods involved might be saved, and the work simplified, by a direct appli-
cation of the funicular polygon. Working on these lines he has succeeded in
devising a graphical method, whereby, on a single diagram, the maximum shears
and the maximum bending moments and the points along the spans at which
they occur can be determined with facility for a wide range of spans and for
any given type-train. A full description of this method originally appeared in
Engineering on September 7th, 1906, and was followed in quick succession
by three other articles dealing more or less closely with the same subject.
These four articles are now brought together in Chapters IV-VII of the
following work, and three other chapters on Diagrams of Bending Moments
and Shearing Forces in Beams added, in the hope that it may prove useful
to engineering students in general and to designers of railway underbridges in
particular.
The author's thanks are due to Engineering for permission to reproduce the
articles above referred to and for kindly supplying electros of the diagrams.
H. BAMFORD.
Glasgow, November 1907.
173461
CONTENTS.
CHAPTER PAGE
I. BENDING MOMENTS AND SHEARING FORCES IN BEAMS . . * 1
II. DIAGRAMS OF BENDING MOMENTS AND SHEARING FORCES DUE TO
FIXED LOADS . . . * 4
III. DIAGRAMS OF BENDING MOMENTS AND SHEARING FORCES DUE TO
MOVING LOADS . . * ; . . . . . . 21
IV. GRAPHICAL DETERMINATION OF THE MAXIMUM BENDING MOMENT
AND MAXIMUM SHEAR DUE TO A TRAIN-LOAD . . . . 42
Y. ANALYTICAL METHODS OF DETERMINING THE MAXIMUM BENDING
MOMENT AND MAXIMUM SHEAR DUE TO A TRAIN-LOAD . . 54
VI. DIAGRAMS OF MAXIMUM BENDING MOMENTS . . . . ', G5
VII. DIAGRAMS OF MAXIMUM SHEAR '] ..... 71
MOVING LOADS ON RAILWAY UNDERBBIDGES.
CHAPTER I.
BENDING MOMENTS AND SHEAEING FORCES IN BEAMS.
THE bending moment and shearing force at any cross section, or point, as it is
more conveniently termed, of a beam, are generally determined on the assump-
tion that the external forces acting upon the beam are all in one plane
the plane of bending and at right angles to the direction of the beam's
length. Now, for any such system of co-planer forces, the conditions of
equilibrium are :
(1) The algebraical sum of the components of the forces taken in any
direction is nil ; and
(2) The algebraical sum of the moments of the forces about any point in
their plane is nil. If the forces under consideration, however, be not in
equilibrium, then :
(3) The algebraical sum of the components of the forces taken in any
direction is equal to the component of their resultant taken in the same
direction; and
(4) The algebraical sum of the moments of the forces about any point
in their plane is equal to the moment of their resultant about the same
point.
Let AB, therefore, Fig. 1, represent a beam of span L supported at both
ends and loaded with a single load W concentrated at a point E, distant d from
B
MOVING LOADS ON RAILWAY UNDEKBRIDGES.
B, and let P A and P B be the supporting forces at A and B respectively. Then,
from the above conditions of equilibrium, we have :
PA + PB- W=0; or P A + P B = W ;
and, taking moments about B and A respectively,
W.d
P.L-
= 0or P =
L J
W.d
and W.(L - d) - P B .L = ; or P B = W - ^-^ = W- P A .
Fig. 1.
J3
e (L-at) - * - -*
Now take any point D between A and E, and consider the equilibrium of
the portion A D of the beam. If, at D, we imagine two opposite forces, P! and
P 2 , each equal and parallel to P A , to be applied, these forces, balancing one
another, will not disturb the equilibrium in any way ; P A and P 1} however, will
constitute a couple of moment P A .AD = P A .#, tending to rotate AD in a
Fig. 2.
clockwise direction, while the unbalanced force P 2 will tend to translate A D
upwards. Considering, in the same way, the equilibrium of the portion B D,
we see that P B at B, and W at E, are equivalent to a force = W P B = PA
acting downwards at D, together with a couple of moment =. P B .B D W.E1)
= P B (L - x) - W (L - x - d) = - (W - P B ) (L - x) + P A -L = I\.x,
tending to rotate B D in a counter-clockwise direction.
MOVING LOADS ON RAILWAY UNDERBRIDGES.
The two equal couples acting in opposite directions upon AD and BD
respectively tend to bend the beam at D, and the two equal and opposite forces
tend to shear the beam at the point. The moment of either couple is called the
bending moment at D, and either force is called the shearing force at the point.
Again, let the beam be loaded in any manner, and assume K! and E 2 to be
the total loads upon the portions A D and B D respectively, and a and b the
distances of their centres of gravity from D (Fig. 4). Then, from the above
condition of equilibrium, it follows that :
Bending moment at D = P A . x RI . a = P B (L x) E 2 . b,
and shearing force at D = P A - E l = R 2 - P B .
Fig. 3.
"-tt
Or the bending moment at any point of a beam is the algebraical sum of the
moments about the point of all the external forces acting upon the portion on
either side of it, and the shearing force at the point is the algebraical sum of the
external forces acting either upon the portion to the left, or upon the portion to
the right of the point.
For convenience we shall regard the bending moment at any point of the
Fig. 4.
4
A
to
s
i>
k Jt X-
i
- (-*) - f;
"beam as positive or negative according as the beam, under the action of the
external forces, tends to bend concavely upwards (as in Fig. 2), or concavely
-downwards at the point, and the shearing force at any point as positive or
negative, according as the portion on the left of the point tends to slide upwards
or downwards relatively to the portion on the right (Fig. 3).
B 2
MOVING LOADS ON HAIL WAY UNDERB1UDGES.
CHAPTEK II.
DIAGRAMS OF BENDING MOMENTS AND SHEARING FORCES DUE TO FIXED LOADS.
Case I. Cantilever fixed at one end and loaded at the other.
Let A B, Fig. 5, represent a cantilever fixed at B and loaded with a single
concentrated load W at A. The bending moment at any point D, distant
x from A, will be M^ = W.x, and at B it will be M L = - W.L, and the
shearing force will be constant throughout and = - W. The diagram of
Fig. 5.
1
-WL
bending moments, therefore, will be a triangle, ale, Fig. 6, and the diagram of
shearing forces a rectangle, as shown in Fig. 7.
Case II. Cantilever loaded with several concentrated loads.
Let the cantilever represented in Fig. 8 be loaded with the concentrated
loads W lf W 2 and W 3 .
MOVING LOADS ON RAILWAY UNDERBRIDGES. 5
Proceeding as in Case I, the diagram of bending moments due to W l
acting alone will be the triangle abb l} Fig. 9, and similarly the diagrams
of bending moments due to W 2 and W 3 respectively will be the triangles
c b-J) 2 and d b 2 b 3) where b^ 2 and b 2 b 3 represent W 2 (a 2 + a 3 ) and W 3 . a 3
respectively.
The diagrams of shears due to W x , W 2 and W 3 respectively will be
rectangles, as shown by a'b'c'd', e'c'fg' and h'fi'k', Fig. 10.
Fig. 8.
Hence, due to Wi, W 2 and W 3 combined, the diagram of bending moments
will be the polygon allude a, Fig. 9, and the diagram of shears the stepped figure
shown in Fig. 10.
Case III. Cantilever loaded throughout its length with a uniformly-
distributed load.
Let w be the load per unit of length, and D a point distant x from A,
Fig. 11 ; then, since the algebraical sum of the forces acting on the portion AD
MOVING LOADS ON It AIL WAY UNDERBEIDGES.
is - w x, and the distance from D of their centre of gravity 6 , the shear at D is-
A
z?
S >c = - w.x and the bending moment, M^, = - w. -.
Fig. 11.
T
The diagram of shears, therefore, is a triangle, a'b'c, Fig. 13, and the diagram
of bending moments the figure ale, Fig. 12, where arc is a parabola having its-
vertex at a.
Fig. 12.
.*_
c
Case IV. Cantilever loaded with a uniformly-distributedjoad over a portion
of its length only.
Let the loaded portion A E be of length /, and w the load per unit of length ;
Fig. 13
rf C
-W.
then, for this portion, the curves of bending moment and shear will be the same
as in Case III, viz., a parabola (af, Fig. 15) and straight line (a'f, Fig. 16)
respectively.
Between E and B the shear will be constant and = w.l, and the bending
MOVING LOADS ON RAILWAY UNDERBRIDGES. 7
moment will increase (numerically) at a uniform rate, as shown by the straight
I 2 i l\
line fc, from w ^ at E to w I (L <> ) at B; and the bending moment at
any point D, distant x from A, will be M x = w .1 ( x ^ j.
Case V. Beam supported at both ends and loaded with a single concentrated
load.
Fig. 14.
Let A B, Fig. 17, be the beam and W the load acting at a point E, distant
d from B ; then :
W.d 4t> W(L-d)
PA = - and P B = - ^j
At any point D, distant x from A, the bending moment will be M^, = P A . x
= ~ . x, if D lies between A and E, or M^ = P A x W (x (L d))
if D lies between E and B. The bending moment diagram, therefore, will be a
s
MOVING LOADS ON RAILWAY UNDERBR1DGES.
triangle, arl, Fig. 18, where m r [represents P A .(L d) = P B .^, the bending
moment at E.
Again, the shear for all points between A and E will be = P A , and for all
Figs. 17 and 1ft.
points between E and B it will be = P A - W = - P B , and the diagram of
shears will consist of two rectangles as shown in Fig. 19.
Case Va. Let the beam in the last case extend over the supports by amounts
represented by AE and BF respectively (Fig. 17ft), and assume the supporting-
forces P A and P B uniformly distributed over A E and B F. The intensity (or
Fig. 19.
P P
load per unit of length) of these loads will be w^ = -r - ^ and iv 2 = ^ and the
resultants of the loads will pass through the centres, c and c', of A E and B F.
The diagram of bending moments, therefore, will be as shown in Fig. 18a,
where ek and fm are parabolas (the same as in Case II, but inverted), and
MOVING LOADS ON RAILWAY UNDERBR1DGES.
9
r Jc and r m straight lines passing, when produced, through c and c' respectively.
The bending moment represented by- m r, therefore, will be = P A . c m =
Fig. 17 a.
:C
tt
f
P B .c'm, and the bending moment at any point between the supports will be
the same as for a beam having a span equal to c c', with the supporting forces
concentrated at the points c and c'. The distance c c', therefore, may, so far as
Fig. ISa.
bending moments are concerned, be called the " effective span," and A B the
clear span.
Again, from the diagram of shearing forces represented in Fig. 19&, it is
Fig. 19a.
evident that so far as shearing forces between A and B are concerned, the
supporting forces P A and P B may be regarded as acting at c and c' respectively
or the distance cd will be the effective span in this case also.
10 MOVING LOADS ON RAILWAY UNDERBRIDGES.
The span referred to in the following examples is the effective span.
Case VI. Beam supported at both ends, and loaded with two or more
concentrated loads. Let the beam AB, Fig. 20, be acted upon by loads Wi,
W 2 and W 3 at points E, F and G, distant d lt d 2 and d 3 respectively from B,
Then :
W
r =
s (L - *,)
~L~
Now draw, under A B, the horizontal line a b, Fig. 21, to represent the span,
and from b erect the perpendicular be to represent, on any scale, P A .L, and
mark the points / and g so that ef and fg represent, on the same scale, Wi . d l
and W 2 . d<i respectively. The lines a e, 1 / and 2 g will then cut the verticals-
through E, F and G respectively in points 1, 2 and 3, such that the polygon
a 1 2 3 b will be a diagram of bending moments. For if, through any point I>
in the beam, a vertical line be drawn cutting the lines a e, 1 /, 2 g and a b, in
r, s, n and ra respectively,
m r be ^
- = - T - = P A , or m r = P A . a m = P A . A D :
am L
and similarly, r s = W l .ED, s n = W 2 . F D, and m n = m r r s s n
= P.AD - Wi.ED - W 2 .FD = the bending moment at D. The ordinate,
m n, therefore, of the polygon a 1 2 3 b } at any point m along the span, represents
the bending moment at that point ; or, in other words, the polygon is a diagram
of bending moments.
The shearing force between A and E is constant and = P A ; between E and F
it is = P A Wi, between F and G it is = P A - W l W 2 , and between G and
B it is = P A W l - W 2 - W 3 = - P B . The diagram of shearing forces,
therefore, will be as shown in Fig. 24.
Another method. A more purely graphical method than the above, and
one which is of somewhat greater importance in consequence of its greater
applicability and usefulness, is what may be called "the funicular polygon
method," which may be described as follows.
Along a vertical line of loads a 1-3 b, Fig. 23, lay off the distances
12 MOVING LOADS ON RAILWAY TJNDERBR1DGES.
a. 1-1. 2, 1.2-2.3 and 2.3-36, to represent, on any scale, W l5 W 2 and W 3
respectively ; and taking any pole 0, draw the radius vectors 0-a . 1, 0-1 . 2,
O-2.3 and 0-3. b. Then, starting from any point a in the vertical through
A, Fig. 22, draw a 1, 12, 23 and 3 I parallel respectively to the above radius
vectors, to cut the verticals through W x , W 2 , W 3 and B (Fig. 20) in the points
1, 2, 3 and b } and then draw the radius vector 0-b a parallel to the closing
line, b a, of the funicular polygon a!23b. P B and P A will then be represented
by 3.b-b.a and b.a-a.l (Fig. 23) respectively, and the polygon a!23ba
will be a diagram of bending moments.
For suppose a 1 2 3 b to be a frame of bars supported vertically at a and b
and acted upon by the loads W x , W 2 and W 3 suspended from the joints 1, 2
and 3 respectively, the triangle of forces for joint 1 would then be a. 1-1. 2-0
(Fig. 23), and that for joint a, a.l-0-b.a. Hence, if a. 1-1. 2 represent
Wj, 0-a.l would represent the pull in 1 a (Fig. 22) and b.a-a.l the
supporting force at a ; and as the supporting forces are obviously the same for
both the frame and beam, b . a-a . 1 will represent P A . Similarly, 3 b-b a will
represent P B .
Now draw through the pole a horizontal line to cut the line of loads in
K, and through any point D in the beam a vertical line cutting a 1, 1 2, 2 3
(produced if necessary) and a b in r, s, n, and m respectively. Then, since the
triangles r m a, r s 1 and s n 2, Fig. 22, are respectively similar to the triangles
b. a-a. 1-0, a. 1-1. 2-0 and 1.2-2.3-0, Fig. 23,
b
rm
i
* w
rs
5 n
r a
A
D
PA-
AD
.a-a.
rs
1
a
.1-0 ~
9.1
E
K'
D
or r s =
K
a
.1-1.
sn
2
1
.2-0"
F
D
OK
W 2 .FD
1
.2-2.
3
w,
2
.3-0
K'
K
and
I.Z-Z.3 W 2 'Z.'6-(
PA.AD Wi.ED - Wo. FD Ma;
Iherefore mn = r m rs sn = - * - = T^T^J
OK. UK.
and M ro the bending moment at D, is = m n X 0~K, where m n is measured on
the linear scale on which A B represents the span, and K on the scale
of loads.
MOVING LOADS ON RAILWAY UNDEEBEIDGES.
13
The diagram of shearing forces can be readily constructed by drawing the
base line b'b' through the point b a, Fig. 23, and projecting across from the points
of division in the line of loads as shown in Fig. 24.
Note. If the first and last sides (a 1 and & 3) of the funicular polygon
a 1 2 3 & be produced to meet in T, and a T b be regarded as a frame supported
at a and b, it will be clear, from the triangle of forces a . 1-3 b-0, that a
load = Wi 4- W 2 4- W 3 at T would give the same supporting forces at A and B
as before. Hence W l + W 2 + W 3 at T is equivalent to W 1 at E + W 2 at F 4-
Fig. 25.
A
jj
a
1
* - *~t P- X J '
r 5
Fig. 26.
W 3 at G ; or T is a point in the line of action of the resultant of W 1? W 2
and W 3 .
Case VII. Beam supported at both ends and loaded throughout its length
with a uniformly-distributed load.
Let w be the load per unit of length and P A and P B the supporting forces.
Then :
14 MOVING LOADS ON RAILWAY UNDERBRIDOES.
the bending moment at any point D, distant x from A, is :
and the shear at D is :
Q . _ P ( v\ (9\
& x r A - w& w[ -<; - x \ . . . (4)
The bending moment, M,., varies with the ordinates to a parabola, a v b,
Fig. 26, of which (1) is the equation. At the centre the bending moment is a
maximum, is = -7, f L . -~ ( < T ) ) = 5, and is represented by the ordinate v o
2 V A \& / ) o
in Fig. 26. The shear, S m varies with the ordinates to a straight line C C',
Fig. 28.
A
3
Z
r\
\
\
*<
-
3
2
i
3' Z' / V * *
Fig. 27, from H ~- at A to ~- at B, and is nil at the centre.
The parabola avb, Fig. 26, may be constructed as follows. Having set off
ov at right angles to a b to represent, on any scale, - ~- , complete the
rectangle, aide, Fig. 28, and divide v d, v e, dl and e a into the same (any)
number of equal parts. Now draw through the points of division 1', 2', 3' ...
in v d and v e lines parallel to v o and join v to the points of division 1, 2, 3 . . .
in e a and d b. Then the intersections of v 1, v 2, v 3 . . . with the lines drawn
through 1', 2', 3' ... respectively, will give points on the parabola required.
Case VIII. Beam supported at the ends and loaded with a uniformly-
distributed load over a portion of its length only.
MOVING LOADS ON RAILWAY UNDERBR1DQES. 15
Let L be the length of the span ; I the length of E F, the portion loaded
with a uniformly-distributed load of w per unit of length, and d, the distance of
F from B. Then :
wl(^ + d)
PA = ~
and P B = -
For any point between A and E, distant x from A, M x = P A x which varies
with x according to the ordinates to a straight line, a e } where e e represents
P A .A E.
For any point, D, between E and F,
M x = P A . A D - W.D E. and at F, M, = P A . A F - W (^ F ) a .
For points between F and B,
M^ = P B (L - x).
To construct the bending moment diagram, therefore, lay off ~b I', Fig. 30, to
represent P A . L, and let a b' cut the vertical drawn through E in e ; e e will then
represent P A . A E, the bending moment at E. Now let the vertical drawn from
any point D, in E F, cut a b in m and a V in q ; then q m will represent P A . A D,
W.(DE) 2
and if we lay off along q m, a length qp to represent , mp will represent
a
P A . A D W f - jp-, the bending moment at D. Other points may be obtained
on the parabolic curve, epf, in the same way, and the diagram completed by
-drawing the straight line/&.
Another method. Consider EF as if it were a uniformly -loaded beam
supported at E and F and draw the parabolic curve e'sf for the uniform
load. The bending moment at any point D of this beam will be
M D = w -4-.E D , , and will be represented by the ordinate m s. Now
join e with/, Fig. 30, and let ef cut qm in n ; then, from similar triangles
on en ED , ED
*= j TTTS, or an = rf.^r
rf ef E F J E F
16
MOVING LOADS ON RAILWAY UNDERBRIDGES.
And since rf and qp represent
and respectively,
p n ( = q n - q p) represents w '% -=-= - ^\ - = M D| or ^ n = m s.
Hence, erecting the ordinates c'c and /'/ to represent P A .AE and P B .BF
(the bending moments at E and F respectively), and drawing the straight lines a e,
ef and fb, and then erecting ordinates from ef equal to the corresponding
ordinates of the parabola e' sf, the diagram of bending moments aepfl will
be obtained.
MOVING LOADS ON RAILWAY UNDERBRIDOES. 17
The diagram of shears will be as shown in Fig. 31.
3. Eelation between the^ bending moment and shearing force at any point of
a loaded beam.
(a) Concentrated loads.
Let the beam be supported at the ends and loaded as represented in Fig. 20
then:
between A and E the shearing force Si = P A ;
E F S a = P A -W l5
F G S 8 =P A -W 1 -W a ;
n G B S 4 = P A - W 1 -W 2 -W 3 = -P B ;
and at E the bending moment M E = P A .AE = Si . A E ;
F M^PA.AF-Wi.EF;
= P A .AE + (PA - WOEF = M E + s 2 .EF ;
G M G =P A .AG-W 1 EG- W 2 .FG;
= PA.AF-W^EFH- (PA-W^W^FG;
= M F + S 3 .FG.
Hence *% = S 1? MF " ME = S 2 and MG " MF = S 3 ; and, in general, if S be
2
Ji _r Jc br
the shearing force in the segment between any two consecutive loads, B L the
length of the segment, and 8 M the increment of the bending moment, then :
8 = - -, or the shear at any point of the beam is equal to the rate of increase
of the bending moment at the point.
(b) Continuous loads.