Question 10.19: Assuming that the prismatic beam AB as shown in Figure 10.35...

From the given loading, we can compute bending moment, M_x at a position as shown in Figure 10.36, as follows:

From Figure 10.36(b) above

M_x=-\frac{w_{ o } x^2}{2}

Therefore,            U=\left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^L M_x^2 d x=\left\lgroup \frac{6}{E b h^3} \right\rgroup \int_0^L \frac{w_{ o }^2}{4} x^4 d x

=\frac{3}{10} \frac{w_0^2 L^5}{E b h^3}=\frac{3}{10} \frac{w_0^2 L^4}{E b^2 h^4}(L b h)

But, V is the volume of beam = Lbh

\frac{U}{V}=\frac{3}{10} \frac{w_{ o }^2 L^4}{E b^2 h^4}            (1)

Now, \tilde{u}=\sigma^2 / 2 E is strain energy density for rectangular section and is given by

\tilde{u}_{\max }=\frac{\sigma_{\max }^2}{2 E} \quad \text { where } \sigma_{\max }=\frac{6 M_{\max }}{b h^2}

But,              \left|M_{\max }\right|=\frac{w_0 L^2}{2}=\left|M_{x=L}\right|

Thus,          \tilde{u}_{\max }=\frac{18 M_{\max }^2}{E b^2 h^4}=\frac{9 w_{ o } L^4}{2 E b^2 h^4}               (2)

From Eqs. (1) and (2),

\frac{\tilde{u}_{\max }}{(U / V)}=\frac{9}{2} \times \frac{10}{3}=15

or          \tilde{u}_{\max }=15\left\lgroup \frac{U}{V} \right\rgroup