Calculate the support reactions in the cantilever beam shown in Fig. 4.5(a).

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Accepted Answer

In this case, we obtain a solution by simultaneously giving the beam a virtual displacement, [latex]\Delta_{V, A }[/latex], at A and a virtual rotation, [latex] heta_{v, A }[/latex], at A. The total deflection at B is then [latex]\Delta_{V, A }+ heta_{v, A } L[/latex] and at a distance x from A is [latex]\Delta_{v, A }+ heta_{v, A } x .[/latex] Since the beam carries a uniformly distributed load, we find the virtual work done by the load by first considering an elemental length, [latex]\delta x[/latex], of the load a distance x from A. The load on the element is [latex]w \delta x[/latex] and the virtual work done by this elemental load is given by

[latex]\delta W=w \delta x\left(\Delta_{v, A }+ heta_{v, A } x ight)[/latex]

The total virtual work done on the beam is given by

[latex]W_{t}=\int_{0}^{L} w\left(\Delta_{v, A }+ heta_{v, A } x ight) d x-M_{ A } heta_{v, A }-R_{ A } \Delta_{v, A }[/latex]

which simplifies to

[latex]W_{t}=\left(w L-R_{ A } ight) \Delta_{v, A }+\left[\left(w L^{2} / 2 ight)-M_{ A } ight] heta_{v, A }=0[/latex] (i)

since the beam is in equilibrium. Equation (i) is valid for all values of [latex]\Delta_{V, A }[/latex] and [latex] heta_{v, A }[/latex], so that

[latex]w L-R_{ A }=0 ext { and }\left(w L^{2} / 2 ight)-M_{ A }=0[/latex]

Therefore,

[latex]R_{ A }=w L ext { and } M_{ A }=w L^{2} / 2[/latex]

the results which would have been obtained by resolving forces vertically and taking moments about A.

Question 4.2: Calculate the support reactions in the cantilever beam shown...

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