A simple beam AB of length L supports a uniform load of intensity q (Fig. 9-33). (a) Evaluate the strain energy of the beam from the bending moment in the beam. (b) Evaluate the strain energy of the beam from the equation of the deflection curve. (Note: The beam has constant flexural rigidity EI.)

Question

A simple beam AB of length L supports a uniform load of intensity q (Fig. 9-33). (a) Evaluate the strain energy of the beam from the bending moment in the beam. (b) Evaluate the strain energy of the beam from the equation of the deflection curve. (Note: The beam has constant flexural rigidity EI.)

Accepted Answer

(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

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

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

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

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

from which we get

[latex]U=\frac{q^{2}L^{5}}{240EI}[/latex] (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:

[latex] u=-\frac{qx}{24EI}\left(L^{3}-2Lx^{2}+x^{3} ight)[/latex]

[latex] u^{\prime}=-\frac{q}{24EI}\left(L^{3}-6Lx^{2}-4x^{3} ight)[/latex]

[latex]\delta_{C}=\delta_{\max}=\frac{5qL^{4}}{384EI}[/latex] [latex] heta_{A}= heta_{B}=\frac{qL^{3}}{24EI}[/latex]

[latex] u=-\frac{qx}{24EI}\left(L^{3}-2Lx^{2}+x^{3} ight)[/latex] (f)

Taking two derivatives of this equation, we get

[latex]\frac{d u}{dx}=-\frac{q}{24EI}\left(L^{3}-6Lx^{2}+4x^{3} ight)[/latex] [latex]\frac{d^{2} u}{dx^{2}}=\frac{q}{2EI}\left(Lx-x^{2} ight)[/latex]

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

[latex]U=\int{\frac{EI}{2}\left(\frac{d^{2} u}{dx^{2}} ight)^{2}}dx[/latex] (9-80b)

[latex]U=\int_{0}^{L}{\frac{EI}{2}\left(\frac{d^{2} u}{dx^{2}} ight)^{2}}dx=\frac{EI}{2}\int_{0}^{L}{\left[\frac{q}{2EI}\left(Lx-x^{2} ight) ight]^{2}}dx[/latex]

[latex]=\frac{q^{2}}{8EI}\int_{0}^{L}{\left(L^{2}x^{2}-2Lx^{3}+x^{4} ight)dx}[/latex] (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).

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