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

from which we get

U=\frac{q^2L^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:

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

Taking two derivatives of this equation, we get

\frac{dv}{dx}=-\frac{q}{24EI}(L^3\ -\ 6Lx^2+4x^3)\quad \quad \frac{d^2v}{dx^2}=\frac{q}{2EI}(Lx\ -\ x^2)

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

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