The strain energy stored in the beam with flexural rigidity EI and loaded as shown in the figure is

Concept:

Strain energy due to bending moment is given by:

\(U=\smallint \frac{{{M^2}dx}}{{2EI}}\)

The reaction at both ends will be:

R_A = R_B = P (due to symmetricity)

Calculation:

Given:

The bending moment diagram of the beam is:

For section AB and CD bending will be variable and equal i.e. M = px

\({U_{AB}} = {U_{CD}} = \smallint \frac{{{M^2}dx}}{{2EI}} = \mathop \smallint \limits_0^L \frac{{{{\left( {Px} \right)}^2}dx}}{{2EI}}\)

And for section BC bending moment is constant i.e. M = PL

\({U_{BC}} = \smallint \frac{{{M^2}dx}}{{2EI}} = \mathop \smallint \limits_0^{2L} \frac{{{{\left( {PL} \right)}^2}dx}}{{2EI}}\)

Total strain energy stored is:

U = U_AB + U_BC + U_CD

\(U = 2\mathop \smallint \limits_0^L \frac{{{{\left( {Px} \right)}^2}dx}}{{2EI}} + \mathop \smallint \limits_0^{2L} \frac{{{{\left( {PL} \right)}^2}dx}}{{2EI}}\)

\(U = \frac{{{P^2}}}{{2EI}}\left[ {\left[ {\frac{{2{x^3}}}{3}} \right]_0^L + [{L^2}x]_0^{2L}} \right]\)

\(U = \frac{{{P^2}}}{{2EI}}\left[ {\frac{{2{L^3}}}{3} + 2{L^3}} \right]\)

\(U = \frac{{{P^2}}}{{2EI}} \times \frac{{8{L^3}}}{3} = \frac{{4{P^2}{L^3}}}{{3EI}}\)