Question 4.4: Calculate the support reactions in the simply supported beam...

Only a vertical load is applied to the beam, so that only vertical reactions, R_{ A } and R_{ C }, are produced. Suppose that the beam at C is given a small imaginary, that is, a virtual, displacement, \Delta_{v, C }, in the direction of R_{ C } as shown in Fig. 4.7(b). Since we are concerned here solely with the external forces acting on the beam, we may regard the beam as a rigid body. The beam therefore rotates about A so that C moves to C ^{\prime} and B moves to B ^{\prime}. From similar triangles. we see that

\Delta_{v, B }=\frac{a}{a+b} \Delta_{v, C }=\frac{a}{L} \Delta_{v, C }  (i)

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

W_{t}=R_{ C } \Delta_{v, C }-W \Delta_{v, B }  (ii)

Note that the work done by the load, W, is negative, since \Delta_{v, B } is in the opposite direction to its line of action. Note also that the support reaction, R_{ A }, does no work, since the beam rotates only about A. Now substituting for \Delta_{v, B } in Eq. (ii) from Eq. (i), we have

W_{t}=R_{ C } \Delta_{v, C }-W \frac{a}{L} \Delta_{v, C }  (iii)

Since the beam is in equilibrium, W_{t} is zero from the principle of virtual work. Hence, from Eq. (iii),

which gives

which is the result which would have been obtained from a consideration of the moment equilibrium of the beam about A. The determination of R_{ A } follows in a similar manner. Suppose now that, instead of the single displacement \Delta_{v, C }, the complete beam is given a vertical virtual displacement, \Delta_{v}, together with a virtual rotation, \theta_{v}, about A, as shown in Fig. 4.7(c). The total virtual work, W_{t}, done by the forces acting on the beam is now given by

W_{t}=R_{ A } \Delta_{v}-W\left(\Delta_{v}+a \theta_{v}\right)+R_{ C }\left(\Delta_{v}+L \theta_{v}\right)=0  (iv)

since the beam is in equilibrium. Rearranging Eq. (iv),

\left(R_{ A }+R_{ C }-W\right) \Delta_{v}+\left(R_{ C } L-W a\right) \theta_{v}=0  (v)

Equation (v) is valid for all values of \Delta_{V} and \theta_{v} so that

which are the equations of equilibrium we would have obtained by resolving forces vertically and taking moments about A.