Question 9.15: A simple beam AB of length L supports a uniform load of inte...

(a) Strain energy from the bending moment. The reaction of the beam at support A is qL/2, and therefore the expression for the bending moment in the beam is

M=\frac{qLx}{2}-\frac{qx^{2}}{2}=\frac{q}{2}(Lx-x^{2})                      (d)

The strain energy of the beam (from Eq. 9-80a) is

U=\int{\frac{M^{2}dx}{2EI}}                (9-80a)

U=\int_{0}^{L}{\frac{M^{2}dx}{2EI}}=\frac{1}{2EI}\int_{0}^{L}{\left[\frac{q}{2}(Lx-x^{2} )\right]^{2}dx}=\frac{q^{2}}{8EI}\int_{0}^{L}{\left(L^{2}x^{2}-2Lx^{3}+x^{4}\right)dx}        (e)

from which we get

U=\frac{q^{2}L^{5}}{240EI}            (9-83)

Note that the load q appears to the second power, which is consistent with the fact that strain energy is always positive. Furthermore, Eq. (9-83) shows that strain energy is not a linear function of the loads, even though the beam itself behaves in a linearly elastic manner.
(b) Strain energy from the deflection curve. The equation of the deflection curve for a simple beam with a uniform load is given in Case 1 of Table G-2, Appendix G, as follows:

\nu=-\frac{qx}{24EI}\left(L^{3}-2Lx^{2}+x^{3}\right)   \nu^{\prime}=-\frac{q}{24EI}\left(L^{3}-6Lx^{2}-4x^{3}\right)   \delta_{C}=\delta_{\max}=\frac{5qL^{4}}{384EI}            \theta_{A}=\theta_{B}=\frac{qL^{3}}{24EI}

\nu=-\frac{qx}{24EI}\left(L^{3}-2Lx^{2}+x^{3}\right)                (f)

Taking two derivatives of this equation, we get

\frac{d\nu}{dx}=-\frac{q}{24EI}\left(L^{3}-6Lx^{2}+4x^{3}\right)        \frac{d^{2}\nu}{dx^{2}}=\frac{q}{2EI}\left(Lx-x^{2}\right)

Substituting the latter expression into the equation for strain energy (Eq. 9-80b), we obtain

U=\int{\frac{EI}{2}\left(\frac{d^{2}\nu}{dx^{2}}\right)^{2}}dx                  (9-80b)

=\frac{q^{2}}{8EI}\int_{0}^{L}{\left(L^{2}x^{2}-2Lx^{3}+x^{4}\right)dx}                                            (g)

Since the final integral in this equation is the same as the final integral in Eq. (e), we get the same result as before (Eq. 9-83).