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## Q. 10.14

For the beam shown in Figure 10.30, find strain energy.

## Verified Solution

The reactions of the supports are shown itself in Figure 10.30 as $R_1 \text { and } R_2$. Clearly,

$R_1=\left\lgroup \frac{P}{2}+\frac{M_{ o }}{L} \right\rgroup \quad \text { and } \quad R_2=\left\lgroup \frac{P}{2}-\frac{M_{ o }}{L} \right\rgroup$

Now, bending moment M at any section at a distance x from end A is

$M_x=\left\{\begin{array}{ll} R_1 x ; & 0 \leq x \leq(L / 2) \\ R_1 x-P\left\lgroup x-\frac{L}{2} \right\rgroup ; & (L / 2) \leq x \leq L \end{array}\right\}$

Putting the values of $R_1 \text { and } R_2$, we get

$M_x=\left\{\begin{array}{ll} \left\lgroup \frac{P}{2}+\frac{M_{ o }}{L} \right\rgroup x ; & 0 \leq x \leq(L / 2) \\ \left\lgroup \frac{M_{ o }}{L}-\frac{P}{2} \right\rgroup x+\frac{P L}{2} ; & (L / 2) \leq x \leq L \end{array}\right\}$                 (1)

The strain energy due to bending is

$U_{\text {bending }}=\left\lgroup \frac{1}{2 E \bar{I}} \right\rgroup \int_0^L M_x^2 d x$

$=\left\lgroup \frac{1}{2 E \bar{I}} \right\rgroup\left[\int_0^{L / 2} M_x^2 d x+\int_{L / 2}^L M_x^2 d x\right]$

$=\left\lgroup \frac{1}{2 E \bar{I}} \right\rgroup \left[\left\lgroup\frac{P}{2}+\frac{M_{ o }}{L} \right\rgroup^2 \int_0^{L / 2} x^2 d x+\int_{L / 2}^L\left\lgroup\frac{M_{ o }}{L}-\frac{P}{2} \right\rgroup^2 x^2 d x\right.$

$\left.+\left\lgroup \frac{P L}{2} \right\rgroup^2 \int_{L / 2}^L d x+\left\lgroup \frac{P L}{2} \right\rgroup\left\lgroup \frac{M_{ o }}{L}-\frac{P}{2} \right\rgroup \int_{L / 2}^L 2 x d x\right]$

$=\left\lgroup\frac{1}{2 E \bar{I}} \right\rgroup \left[\left\lgroup\frac{P}{2}+\frac{M_{ o }}{L} \right\rgroup^2 \frac{L^3}{24}+\frac{1}{3}\left\lgroup\frac{M_{ o }}{L}-\frac{P}{2} \right\rgroup^2\left\lgroup L^3-\frac{L^3}{8} \right\rgroup \right.$

$\left.+\frac{P^2 L^2}{4}\left\lgroup \frac{L}{2} \right\rgroup+\left\lgroup \frac{P L}{2} \right\rgroup \left\lgroup \frac{M_{ o }}{L}-\frac{P}{2} \right\rgroup \left\lgroup L^2-\frac{L^2}{4} \right\rgroup \right]$

Therefore,

$U_{\text {bending }}=\left\lgroup \frac{1}{2 E \bar{I}}\right\rgroup\left[\left\lgroup \frac{P^2 L^3}{96}+\frac{M_{ o }^2 L}{24}+\frac{P M_{ o } L^2}{24}\right\rgroup +\left\lgroup \frac{7 M_{ o }^2 L}{24}+\frac{7 P^2 L^3}{96}-\frac{7 P M_{ o } L^2}{24}\right\rgroup \right.$

$\left.+\frac{P^2 L^3}{8}+\left\lgroup \frac{3 M_0 P L^2}{8}-\frac{3 P^2 L^3}{16} \right\rgroup \right]$

$=\frac{1}{2 E \bar{I}}\left[\left\lgroup \frac{8}{96}+\frac{1}{8}-\frac{3}{16} \right\rgroup P^2 L^3+\frac{M_{ o }^2 L}{3}+\left\lgroup \frac{1}{24}-\frac{7}{24}+\frac{3}{8} \right\rgroup P M_{ o } L^2\right]$

$=\frac{1}{2 E \bar{I}}\left[\frac{P^2 L^3}{48}+\frac{M_{ o }^2 L}{3}+\frac{P M_{ o }^2 L^2}{8}\right]=\frac{P^2 L^3}{96 E \bar{I}}+\frac{M_{ o }^2 L}{6 E \bar{I}}+\frac{P M_{ o }^2 L^2}{16 E \bar{I}}$

Thus, the required strain energy due to bending is

$U_{\text {bending }}=\left\lgroup \frac{P^2 L^3}{96 E \bar{I}}+\frac{M_{ o }^2 L}{6 E \bar{I}}+\frac{P M_{ o }^2 L^2}{16 E \bar{I}} \right\rgroup$