Or generally, if M_1 M_2 M_3 are the moments of the external forces to the
left of O, A, and B respectively, and s, t and c the perpendiculars from O,
A and B on the directions of the forces cut by the section, then
Ss = M_1; Tt = M_2 and Cc = M_3.
Still more generally if H is the stress on any bar, h the perpendicular
distance from the join of the other two bars cut by the section, and M is
the moment of the forces on one side of that join,
Hh = M.
[Illustration: Fig. 39.]
[Illustration: Fig. 40.]
22. _Distribution of Bending Moment and Shearing Force._--Let a girder of
span l, fig. 39, supported at the ends, carry a fixed load W at m from the
right abutment. The reactions at the abutments are R_1 = Wm/l and R_2 =
W(l-m)/l. The shears on vertical sections to the left and right of the load
are R_1 and -R_2, and the distribution of shearing force is given by two
rectangles. Bending moment increases uniformly from either abutment to the
load, at which the bending moment is M = R_2m = R_1(l-m). The distribution
of bending moment is given by the ordinates of a triangle. Next let the
girder carry a uniform load w per ft. run (fig. 40). The total load [v.04
p.0551] is wl; the reactions at abutments, R_1 = R_2 = ½wl. The
distribution of shear on vertical sections is given by the ordinates of a
sloping line. The greatest bending moment is at the centre and = M_c =
1/8wl^2. At any point x from the abutment, the bending moment is M =
½wx(l-x), an equation to a parabola.
[Illustration: Fig. 41.]
[Illustration: Fig. 42.]
23. _Shear due to Travelling Loads._--Let a uniform train weighing w per
ft. run advance over a girder of span 2c, from the left abutment. When it
covers the girder to a distance x from the centre (fig. 41) the total load
is w(c+x); the reaction at B is
R_2 = w(c+x)×(c+x)/4c = w/4c(c+x)²,
[Illustration: FIG. 41.]
[Illustration: FIG. 42.]
which is also the shearing force at C for that position of the load. As the
load travels, the shear at the head of the train will be given by the
ordinates of a parabola having its vertex at A, and a maximum F_{max.} =
-½wl at B. If the load travels the reverse way, the shearing force at the
head of the train is given by the ordinates of the dotted parabola. The
greatest shear at C for any position of the load occurs when the head of
the train is at C. For any load p between C and B will increase the
reaction at B and therefore the shear at C by part of p, but at the same
time will diminish the shear at C by the whole of p. The web of a girder
must resist the maximum shear, and, with a travelling load like a railway
train, this is greater for partial than for complete loading. Generally a
girder supports both a dead and a live load. The distribution of total
shear, due to a dead load w_l per ft. run and a travelling load w_l per ft.
run, is shown in fig. 42, arranged so that the dead load shear is added to
the maximum travelling load shear of the same sign.
[Illustration: FIG. 43.]
24. _Counterbracing._--In the case of girders with braced webs, the tension
bars of which are not adapted to resist a thrust, another circumstance due
to the position of the live load must be considered. For a train advancing
from the left, the travelling load shear in the left half of the span is of
a different sign from that due to the dead load. Fig. 43 shows the maximum
shear at vertical sections due to a dead and travelling load, the latter
advancing (fig. 43, a) from the left and (fig. 43, b) from the right
abutment. Comparing the figures it will be seen that over a distance x near
the middle of the girder the shear changes sign, according as the load
advances from the left or the right. The bracing bars, therefore, for this
part of the girder must be adapted to resist either tension or thrust.
Further, the range of stress to which they are subjected is the sum of the
stresses due to the load advancing from the left or the right.
[Illustration: FIG. 44.]
[Illustration: FIG. 45.]
[Illustration: FIG. 46.]
25. _Greatest Shear when concentrated Loads travel over the Bridge._--To
find the greatest shear with a set of concentrated loads at fixed
distances, let the loads advance from the left abutment, and let C be the
section at which the shear is required (fig. 44). The greatest shear at C
may occur with W_1 at C. If W_1 passes beyond C, the shear at C will
probably be greatest when W_2 is at C. Let R be the resultant of the loads
on the bridge when W_1 is at C. Then the reaction at B and shear at C is
Rn/l. Next let the loads advance a distance a so that W_2 comes to C. Then
the shear at C is R(n+a)/l-W_1, plus any reaction d at B, due to any
additional load which has come on the girder during the movement. The shear
will therefore be increased by bringing W_2 to C, if Ra/l+d > W_1 and d is
generally small and negligible. This result is modified if the action of
the load near the section is distributed to the bracing intersections by
rail and cross girders. In fig. 45 the action of W is distributed to A and
B by the flooring. Then the loads at A and B are W(p-x)/p and Wx/p. Now let
C (fig. 46) be the section at which the greatest shear is required, and let
the loads advance from the left till W_1 is at C. If R is the resultant of
the loads then on the girder, the reaction at B and shear at C is Rn/l. But
the shear may be greater when W_2 is at C. In that case the shear at C
becomes R(n+a)/l+d-W_1, if a > p, and R(n+a)/l+d-W_1a/p, if a < p. If we
neglect d, then the shear increases by moving W_2 to C, if Ra/l > W_1 in
the first case, and if Ra/l > W_1a/p in the second case.
[Illustration: FIG. 47.]
[Illustration: FIG. 48.]
26. _Greatest Bending Moment due to travelling concentrated Loads._--For
the greatest bending moment due to a travelling live load, let a load of w
per ft. run advance from the left abutment (fig. 47), and let its centre be
at x from the left abutment. The reaction at B is 2wx²/l and the bending
moment at any section C, at m from the left abutment, is 2wx²/(l-m)/l,
which increases as x increases till the span is covered. Hence, for uniform
travelling loads, the bending moments are greatest when the loading is
complete. In that case the loads on either side of C are proportional to m
and l-m. In the case of a series of travelling loads at fixed distances
apart passing over the girder from the left, let W_1, W_2 (fig. 48), at
distances x and x+a from the left abutment, be their resultants on either
side of C. Then the reaction at B is W_1x/l+W_2(x+a)/l. The bending moment
at C is