Question 15.5: Determine the rotation, i.e. the slope, of the beam ABC show...
Determine the rotation, i.e. the slope, of the beam ABC shown in Fig. 15.11(a) at A.
The actual rotation of the beam at A produced by the actual concentrated load, W, is \theta_{\mathrm{A}}. Let us suppose that a virtual unit moment is applied at A before the actual rotation takes place, as shown in Fig. 15.11(b). The virtual unit moment induces virtual support reactions of R_{\mathrm{V}, \mathrm{A}}(=1 / L) acting downwards and R_{\mathrm{V}, \mathrm{C}}(=1 / L) acting upwards. The actual internal bending moments are
\begin{aligned}& M_{\mathrm{A}}=+\frac{W}{2} x \quad 0 \leq x \leq L / 2 \\& M_{\mathrm{A}}=+\frac{W}{2}(L-x) \quad L / 2 \leq x \leq L\end{aligned}
The internal virtual bending moment is
M_{\mathrm{v}}=1-\frac{1}{L} x \quad 0 \leq x \leq L
The external virtual work done is 1 \theta_{\mathrm{A}} (the virtual support reactions do no work as there is no vertical displacement of the beam at the supports) and the internal virtual work done 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)
Hence
1 \theta_{\mathrm{A}}=\frac{1}{E I}\left[\int_{0}^{L / 2} \frac{W}{2} x\left(1-\frac{x}{L}\right) \mathrm{d} x+\int_{L / 2}^{L} \frac{W}{2}(L-x)\left(1-\frac{x}{L}\right) \mathrm{d} x\right] (i)
Simplifying Eq. (i) we have
\theta_{\mathrm{A}}=\frac{W}{2 E I L}\left[\int_{0}^{L / 2}\left(L x-x^{2}\right) \mathrm{d} x+\int_{L / 2}^{L}(L-x)^{2} \mathrm{~d} x\right] (ii)
Hence
\theta_{\mathrm{A}}=\frac{W}{2 E I L}\left\{\left[L \frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{L / 2}-\frac{1}{3}\left[(L-x)^{3}\right]_{L / 2}^{L}\right\}
from which
\theta_{\mathrm{A}}=\frac{W L^{2}}{16 E I}
which is the result that may be obtained from Eq. (iii)
E I{\frac{\mathrm{d}v}{\mathrm{d}x}}={\frac{W}{16}}(4x^{2}-L^{2}) (iii)
of Ex. 13.5.