Question 15.4: Determine the vertical deflection of the free end of the can...

Let us suppose that the actual deflection of the cantilever at B produced by the uniformly distributed load is v_{\mathrm{B}} and that a vertically downward virtual unit load was applied at \mathrm{B} before the actual deflection took place. The external virtual work done by the unit load is, from Fig. 15.10(b), 1 v_{\mathrm{B}}. The deflection, v_{\mathrm{B}}, is assumed to be caused by bending only, i.e. we are ignoring any deflections due to shear. The internal virtual work is given by Eq. (15.21)

W_{\mathrm{i},M}=\sum\int_{L}{\frac{M_{\mathrm{A}}M_{\mathrm{v}}}{E I}}\,\mathrm{d}x       (15.21)

which, since only one member is involved, becomes

W_{\mathrm{i}, M}=\int_{0}^{L} \frac{M_{\mathrm{A}} M_{\mathrm{v}}}{E I} \mathrm{~d} x     (i)

The virtual moments, M_{\mathrm{v}}, are produced by a unit load so that we shall replace M_{\mathrm{v}} by M_{1}. Then

W_{\mathrm{i}, M}=\int_{0}^{L} \frac{M_{\mathrm{A}} M_{1}}{E I} \mathrm{~d} x     (ii)

At any section of the beam a distance x from the built-in end

M_{\mathrm{A}}=-\frac{w}{2}(L-x)^{2} \quad M_{1}=-1(L-x)

Substituting for M_{\mathrm{A}} and M_{1} in Eq. (ii) and equating the external virtual work done by the unit load to the internal virtual work we have

1 v_{\mathrm{B}}=\int_{0}^{L} \frac{w}{2 E I}(L-x)^{3} \mathrm{~d} x

which gives

v_{\mathrm{B}}=-\frac{w}{2 E I}\left[\frac{1}{4}(L-x)^{4}\right]_{0}^{L}

so that

v_{\mathrm{B}}=\frac{w L^{4}}{8 E I}       (as in Ex. 13.2)

Note that v_{\mathrm{B}} is in fact negative but the positive sign here indicates that it is in the same direction as the unit load.