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## Q. 7.10

Determine the deflection at point $B$ of the beam shown in Fig. $7.15(\mathrm{a})$ by the virtual work method. Use the graphical procedure (Table 7.6) to evaluate the virtual work integral.

 Table 7.6 Integrals $\int_{0}^{L} M_{v} M d x$ for Moment Diagrams of Simple Geometric Shapes $M_{v}$ M $M_{v1}M_{1}L$ $\frac{1}{2} M_{v1}M_{1}L$ $\frac{1}{2}\left(M_{v1}+M_{v2} \right) M_{1}L$ $\frac{1}{2}M_{v1} M_{1}L$ $\frac{1}{2} M_{v1}M_{1}L$ $\frac{1}{3} M_{v1}M_{1}L$ $\frac{1}{6}\left(M_{v1}+2M_{v2} \right) M_{1}L$ $\frac{1}{6}M_{v1} M_{1}\left(L+l_{1} \right)$ $\frac{1}{2} M_{v1}M_{1}L$ $\frac{1}{6} M_{v1}M_{1}L$ $\frac{1}{6}\left(2M_{v1}+M_{v2} \right) M_{1}L$ $\frac{1}{6}M_{v1} M_{1}\left(L+l_{2} \right)$ $\frac{1}{2}M_{v1}\left(M_{1}+M_{2} \right) L$ $\frac{1}{6}M_{v1}\left(M_{1}+2M_{2} \right) L$ $\frac{1}{6}\left[M_{v1}\left(2M_{1}+M_{2} \right)+M_{v2}\left(M_{1}+2M_{2} \right) \right]L$ $\frac{1}{6}M_{v1}\left[M_{1}\left(L+l_{2} \right)+M_{2}\left(L+l_{1} \right) \right]$ $\frac{2}{3} M_{v1}M_{1}L$ $\frac{1}{3} M_{v1}M_{1}L$ $\frac{1}{3}\left(M_{v1}+M_{v2} \right) M_{1} L$ $\frac{1}{3}M_{v1} M_{1} \left\lgroup L+\frac{l_{1}l_{2} }{L} \right\rgroup$ $\frac{2}{3} M_{v1}M_{1}L$ $\frac{5}{12} M_{v1}M_{1}L$ $\frac{1}{12}\left(3M_{v1}+5M_{v2} \right) M_{1} L$ $\frac{1}{12}M_{v1} M_{1} \left\lgroup3L+3l_{1}-\frac{l^{2}_{1} }{L} \right\rgroup$ $\frac{1}{3} M_{v1}M_{1}L$ $\frac{1}{4} M_{v1}M_{1}L$ $\frac{1}{12}\left(M_{v1}+3M_{v2} \right) M_{1} L$ $\frac{1}{12}M_{v1} M_{1} \left\lgroup L+l_{1}+\frac{l^{2}_{1} }{L} \right\rgroup$

## Verified Solution

The real and virtual systems, along with their bending moment diagrams ( $M$ and $M_{v}$ ), are shown in Figs. $7.15$ (b) and (c), respectively. As the flexural rigidity $E I$ is constant along the length of the beam, there is no need to subdivide the beam into segments, and the virtual work equation (Eq. $(7.30))$ for the deflection at $B$ can be expressed as

$\begin{array}{l} 1(\Delta)=\int_{0}^{L} \frac{M_{v} M}{E I} d x\\ \end{array}$     (7.30)

$1\left(\Delta_{B}\right)=\frac{1}{E I} \int_{0}^{L} M_{v} M d x$     (1)

To evaluate the integral $\int_{0}^{L} M_{v} M d x$ graphically, we first compare the shape of the $M$ diagram in Fig. $7.15$ (b) with the shapes listed in the left column of Table 7.6. Note that the shape of the $M$ diagram matches the shape located in the sixth row of the table. Next, we compare the shape of the $M_{v}$ diagram (Fig. 7.15(c)) with those given in the top row of the table, and notice that it is similar to the shape in the fifth column. This indicates that the expression for evaluating the integral $\int_{0}^{L} M_{v} M d x$, in this case, is located at the intersection of the sixth row and the fifth column of Table $7.6$, that is,

$\int_{0}^{L} M_{v} M d x=\frac{1}{3} M_{v 1} M_{1}\left(L+\frac{l_{1} l_{2}}{L}\right)$

By substituting the numerical values of $M_{v 1}=2.25 \mathrm{kN} \cdot \mathrm{m}, M_{1}=630 \mathrm{kN} \cdot \mathrm{m}, L=12 \mathrm{~m}, l_{1}=3 \mathrm{~m}$ and $l_{2}=9 \mathrm{~m}$, into the foregoing equation, we compute the integral to be

$\int_{0}^{12} M_{v} M d x=\frac{1}{3}(2.25)(630)\left(12+\frac{3(9)}{12}\right)=6,733.13 \mathrm{kN}^{2} \cdot \mathrm{m}^{3}$

The desired deflection at $B$ can now be conveniently determined by applying the virtual work equation (Eq. 1) as

$(1 \mathrm{kN}) \Delta_{B}=\frac{1}{E I} \int_{0}^{12} M_{v} M d x=\frac{6,733.13 \mathrm{kN}^{2} \cdot \mathrm{m}^{3}}{E I}$

Therefore,

$\begin{array}{l} \Delta_{B}=\frac{6,733.13 \mathrm{kN} \cdot \mathrm{m}^{3}}{E I}=\frac{6,733.13}{70(1,800)}=0.0534 \mathrm{~m}=53.4 \mathrm{~mm} \\ \Delta_{B}=53.4 \mathrm{~mm} \downarrow \end{array}$