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