Question 13.18: The cantilever AB shown in Fig. 13.22(a) carries a uniformly...

Suppose that the reaction at the support B is R_B . Using the principle of superposition we can represent the combined effect of the distributed load and the reaction R_B as the sum of the two loads acting separately as shown in Fig. 13.22(b) and (c). Also, since the vertical deflection of B in Fig. 13.22(a) is zero, it follows that the vertical downward deflection of B in Fig. 13.22(b) must be numerically equal to the vertically upward deflection of B in Fig. 13.22(c). Therefore using the results of Exs (13.1) and (13.2) we have

\left|\frac{R_BL^3}{3EI} \right| = \left|\frac{wL^4}{8EI} \right|

whence

R_B = \frac{3}{8} wL

It is now possible to determine the reactions R_A and M_A at the built-in end using the equations of simple statics. Taking moments about A for the beam in Fig. 13.22(a) we have

M_A = \frac{wL^2}{2} – R_BL = \frac{wL^2}{2} – \frac{3}{8} wL^2 = \frac{1}{8} wL^2

Resolving vertically

R_A = wL − R_B = wL − \frac{3}{8} wL = \frac{5}{8} wL

In the solution of Ex. 13.18 we selected R_B as the redundancy; in fact, any one of the three support reactions, M_A,R_A or R_B , could have been chosen. Let us suppose that M_A is taken to be the redundant reaction. We now represent the combined loading of Fig. 13.22(a) as the sum of the separate loading systems shown in Fig. 13.23(a) and (b) and work in terms of the rotations of the beam at A due to the distributed load and the applied moment, M_A . Clearly, since there is no rotation at the built-in end of a cantilever, the rotations produced separately in Fig. 13.23(a) and (b) must be numerically equal but opposite in direction. Using the method of Section 13.1 it may be shown that

\theta _A   (due to w) = \frac{wL^3}{24EI}     (clockwise)

and

\theta _A   (due to M_A ) = \frac{M_AL}{3EI}     (anticlockwise)

Since

|θ_A(M_A)|=|θ_A(w)|

we have

M_A = \frac{wL^2}{8}

as before.