Question 4.1: Calculate the support reactions in the cantilever beam AB sh...

The concentrated load, W, induces a vertical reaction, R_{ A } , and also one of moment, M_{ A }, at A. Suppose that the beam is given a small imaginary, that is virtual, rotation, \theta_{v, A}, as shown in Fig. 4.4(b). Since we are concerned here with only external forces, we may regard the beam as a rigid body, so that the beam remains straight and B is displaced to B ^{\prime}. The vertical displacement of B, \Delta_{v, B }, is then given by

or

\theta_{v, A }=\Delta_{v, B } / L (i)

The total virtual work, W_{t}, done by all the forces acting on the beam is given by

W_{t}=W \Delta_{v, B }-M_{ A } \theta_{v, A }  (ii)

Note that the contribution of M_{ A } to the total virtual work done is negative, since the assumed direction of M_{ A } the opposite sense to the virtual displacement, \theta_{v, A }. Note also that there is no linear movement of the beam at A so that R_{ A } does no work. Substituting in Eq. (ii) for \theta_{v, A } from Eq. (i), we have

W_{t}=W \Delta_{v, B }-M_{ A } \Delta_{v, B } / L  (iii)

Since the beam is in equilibrium, W_{t}=0, from the principle of virtual work. Therefore,

so that

which is the result which would have been obtained from considering the moment equilibrium of the beam about A. Suppose now that the complete beam is given a virtual displacement, \Delta_{v}, as shown in Fig. 4.4(c). There is no rotation of the beam, so that M_{ A } does no work. The total virtual work done is then given by

W_{t}=W \Delta_{v}-R_{ A } \Delta_{V}  (iv)

The contribution of R_{ A } is negative, since its line of action is in the direction opposite to \Delta_{v}. The beam is in equilibrium, so that W_{t}=0. Therefore, from Eq. (iv),

which is the result we would have obtained by resolving forces vertically.