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## Q. 10.12

A simply supported beam with uniformly distributed loading and of rectangular cross-section has a maximum bending stress, $\sigma_{\max }$ . Calculate the strain energy stored in the beam. Assume the beam length = L, section width = b, section depth = h and modulus of elasticity of material of beam to be E.

## Verified Solution

Let the uniformly load intensity be $w_0$. Therefore, bending moment at any position on the beam is

$M_x=\frac{w_{ o } L}{2} x-\frac{w_{ o } x^2}{2}=\frac{w_{ o }}{2}\left(L x-x^2\right)$                 (1)

Evidently,

$\left(M_x\right)_{\max }=\left.M_x\right|_{x=L / 2}=\frac{w_{ o } L^2}{8}$

Now,

$\sigma_{\max }=\frac{6\left(M_x\right)_{\max }}{b h^2} \Rightarrow\left(M_x\right)_{\max }=\left\lgroup \frac{b h^2}{6} \right\rgroup \sigma_{\max }$

Therefore,

$\frac{w_{ o } L^2}{8}=\left\lgroup \frac{b h^2}{6} \right\rgroup \sigma_{\max } \Rightarrow w_{ o }=\left\lgroup \frac{4}{3} \right\rgroup\left\lgroup \frac{b h^2}{L^2} \right\rgroup \sigma_{\max }$              (2)

Again,

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

From Eq. (1),

$U_{\text {bending }}=\left\lgroup \frac{1}{2 E \bar{I}} \right\rgroup \int_0^L \frac{w_0^2}{4}\left(L x-x^2\right) d x$

$=\left\lgroup \frac{w_0^2}{8 E \bar{I}} \right\rgroup\left[\frac{L^2 x^3}{3}+\frac{x^5}{5}-\frac{L x^4}{2}\right]_0^L$

$=\left\lgroup \frac{w_0^2 L^2}{8 E \bar{I}} \right\rgroup\left\lgroup \frac{1}{3}+\frac{1}{5}-\frac{1}{2} \right\rgroup =\frac{w_0^2 L^2}{240 E \bar{I}}$

Now putting the value of $w_o$ from Eq. (2), we get

$U_{\text {bending }}=\left\lgroup \frac{L^5}{240 E \bar{I}} \right\rgroup\left\lgroup \frac{16}{9} \right\rgroup \frac{b^2 h^4}{L^4} \sigma_{\max }^2$

$=\frac{L b^2 h^2 \sigma_{\max }^2}{(15)(9) E} \times \frac{12}{b h^3}$

$=\left\lgroup\frac{4}{45} \right\rgroup\left\lgroup\frac{L b h}{E} \right\rgroup \sigma_{\max }^2$

The required strain energy due to bending of the beam is $4 L b h \sigma_{\max }^2 / 45 E$