To find the amount of the respective pressures on the rollers and the pivot, let P (fig. 398) be the position of the pivot and C that of the rollers.

Case III-An Arch (fig. 399). In the preceding investigation each wing of the closed bridge has been treated separately. If it be desired to take advantage of the support afforded by the mutual abutment of the two meeting faces of the bridge, it is evident that the tail end must be lifted in order to develop the full thrust due to the dead weight. Assuming (as would essentially be the case) that the underside of the cantilevers constitute a real or virtual arc of riser and span 2 a, we have by the conditions of equilibrium for three forces and by similar triangles,

The

which gives us an expression for the amount of mutual thrust. upward force, Fo, at C, required to develop this thrust is found by taking moments about B

so that,

b being the distance, BC.

This force, the value of which is identical with

that of the counterpoise, is additional to the reaction at C, due to the load on BC.

from which it is apparent that it may be considerable and that carefully adjusted and solid bearings are essential. It is a matter of some difficulty to secure these in the case of swing bridges, and accordingly it is not usual for the central reaction to be much, if at all, relied upon. In bascule bridges, on the other hand, it is comparatively easy to provide accurate bearing surfaces.

Case IV-A continuous beam supported at three points (fig. 400).—Let

A B C be a girder continuous over three points of support-A, B, and C all on the same level. Take the intermediate support, B, as the origin of co-ordinates, and let y represent the deflection of the beam at the point X due to a uniform load, w, per unit length. Let S be the shearing stress and M the bending moment at the same point.

By a well-known formula establishing the connection between the bending moment (M) the modulus of elasticity (E) the moment of inertia (I) and the radius of curvature (R), we have at any point X—

Now, when the curvature is very small, as is assumed to be the case in the foregoing relationship, we may find a very close approximation for the

where x and y are the co-ordinates of the deflection curve. write

Hence we may

If

Again, let us consider the conditions of equilibrium at the point X. P be a point indefinitely near to B, where the shearing stress is S1 and the bending moment M1, it is clear that for equilibrium of the portion PX, we have

To find what value to attach to the constant (C1) in this expression, we have the following consideration:-Let 3 be the slope of the beam at the origin, B—or, in other words, the inclination of the tangent of the curve to

the horizontal. Then tan 3

=

dy
dx

and, in the limit, tan 3

= B. When this is the case x is so small as to become negligible, and so we can write

The constant is omitted in this case because y = 0 when x=0. Again, since y=0 when x = b, we have—

Now, at a point Q, equally indefinitely near to but on the opposite side of B, we shall find the bending moment identical in value with that at the point P. We can therefore write a similar equation in this case, noting that a has a negative value. Thus—

B

Subtract, and for both M, and M, write M, or the bending moment at the point B, to which they both approximate so closely as to be practically identical with it and each other.

Accordingly,

Again, taking moments about A for the portion A B—

Divide by 6, and substitute in equation (96) above

This equation is known as the Theorem of Three Moments, and its first enunciation is attributed to Clapeyron. By means of the relationship thus established, if the bending moments at two of the points of support of a uniformly loaded beam are known, the third can be deduced. The bending moments at the end supports are sufficiently obvious. If the beam project a w c2 distance, c, beyond the outer support, C, the moment at C is 2

If the beam

do not project, the moment at the point of support is zero. The shearing stresses can then be obtained from the formula already given, viz. :—

From these, the reactions at the points of support are readily forthcoming, Sc, if there be no overhang. If there be an over

for RA

=

SA and Rc

=

Assuming that there is no overhang this equation simplifies into

Equation (98) may be confirmed by an independent investigation which

is worthy of notice, for it gives an expression for the current moment in terms of the moments at the points of support.

If AB (fig. 401) be a portion of a weightless beam between any two supports, PQ, with bending moments, y, and y, at A and B respectively,

Ꭹ

A X B
Fig. 401.

due to some external system of loading, it is clear that the line of moments between A and B will be right, and by a simple application of geometrical principles

If, however, the beam be not weightless, or be loaded with a weight of w lbs. per foot

run, the curve of moments is parabolic and the equation becomes

The foregoing relationship is, of course, conditional upon there being no point of support between A and B. When such is not the case, and there is an upward reaction, R, at the point, X, we must expand the expression still further into

=

w

2

Y (x1 + x2) Y 1 X 2 + Y 2 X 1 + 7 X1 X2 (X1 + X2) Rx (101)
X1 X2 (X1 + x2) − Rx1 x2.

an equation which is identical with the value of R, given above, when the notation has been adapted thereto.

The second equation (100) in the preceding group yields us an expression for the current bending moment at any point, X, intermediate between the points of support.

w b x
2

If we revert to the case in which there are no moments at the end supports, we may derive the amounts of reaction at these points very readily, as follows:—

Also, from a consideration of the conditions of equilibrium to the left of B,