![](http://www.classle.net/userfiles/image/strain_energy_4a(1).gif)
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If a beam is subjected to Bending Moment 'M' as shown in Fig. Consider an element dx, the Strain Energy in the element is dU is,
![](http://www.classle.net/userfiles/image/strain_energy_4.gif)
Problem no 1
A simply supported beam of length l carries a concentrated load W at distances of 'a' and 'b' from the two ends. Find expressions for the total strain energy of the beam and the deflection under load.
Solution:
The integration for strain energy can only be applied over a length of beam for which a continuous expression for M can be obtained. This usually implies a separate integration for each section between two concentrated loads or reactions.
For the section AB, The bending moment M at distance 'x' from Point A is,
![M\;=\;\left(\frac{Wb}{l} \right)x](http://www.codecogs.com/images/eqns/0598f201609391b4d422e39e481e4847.gif)
The Strain Energy in the section AB is,
![U_a\;=\;\int_{0}^{a}{\frac{W^2\;b^2\;x^2}{2\;l^2\;EI}\;dx}](http://www.codecogs.com/images/eqns/031ec36a47e9e8ecb5656d97cca7c1dd.gif)
![\therefore\;\;\;\;\;U_a\;=\;\frac{W^2\;b^2}{2\;l^2\;EI}\left[\frac{x^3}{3} \right]_0^a](http://www.codecogs.com/images/eqns/9dc4fbf2c82b4e19de5dd6c3527bf11a.gif)
![\therefore\;\;\;\;\;U_a\;=\;\frac{W^2\;a^3\;b^2}{6\,E\,I\,l^2}](http://www.codecogs.com/images/eqns/f7f3eca73373eeb448143b51e1ca6bbe.gif)
Similarly by taking a variable X measured from C,The Strain Energy stored In the Section BC is
![\therefore\;\;\;\;\;U_b\;=\;\int_{0}^{b}{\frac{W^2\;a^2\;X^2}{2\;l^2\,E\,I}dX}\;=\;\frac{W^2\;a^2\;b^3}{6\,E\,I\,l^2}](http://www.codecogs.com/images/eqns/e16686e6ca19d11ec855072ec037acd8.gif)
Total Strain Energy is,
![U=U_a+U_b=\left( \frac{W^2\;a^2\;b^2}{6\,E\,I\,l^2} \right)(a\;+\;b)](http://www.codecogs.com/images/eqns/52f3afd5973c2226e2d056b09c731f02.gif)
For a Central Load,
![\therefore\;\;\;\;\;U_b\;=\;\int_{0}^{b}{\frac{W^2\;a^2\;X^2}{2\;l^2\,E\,I}dX}\;=\;\frac{W^2\;a^2\;b^3}{6\,E\,I\,l^2}](http://www.codecogs.com/images/eqns/e16686e6ca19d11ec855072ec037acd8.gif)
![U=U_a+U_b=\left( \frac{W^2\;a^2\;b^2}{6\,E\,I\,l^2} \right)(a\;+\;b)](http://www.codecogs.com/images/eqns/52f3afd5973c2226e2d056b09c731f02.gif)
![\therefore\;\;\;\;\;\;U\;=\;\frac{W^2\;a^2\;b^2}{6\;E\;I\;l}](http://www.codecogs.com/images/eqns/516d8daa43bd30eeaff93ddbbcc4e5c9.gif)
But if
is the deflection under the load, the strain energy must be equal to the work done by the load if it is gradually applied.
![\displaystyle \delta](http://www.codecogs.com/images/eqns/b8e4e8a9a6f30943d8743cb8cb575a14.gif)
![\frac{1}{2}W\,\delta =\frac{W^2\;a^2\;b^2}{6\,E\,I\,l}](http://www.codecogs.com/images/eqns/85a6433df4ddda0fbdd1faacba644c8a.gif)
![\therefore\;\;\;\;\;\delta =\frac{W\;a^2\;b^2}{3\,E\,I\,l}](http://www.codecogs.com/images/eqns/042caf35b710e2f2148a925db608cca0.gif)
For a Central Load,
![\displaystyle a\;=\;b\;=\;\frac{l}{2}](http://www.codecogs.com/images/eqns/8278755309aab0a4ad77d0371bc2e786.gif)
![\therefore\;\;\;\;\;\delta \;=\;\left(\frac{W}{3\,E\,I\,l} \right)\left(\frac{l^2}{4} \right)\left(\frac{l^2}{4} \right)](http://www.codecogs.com/images/eqns/fdcef6a045b3c5833d2ef3a15123a261.gif)
Hence, the central deflection due to a point load applied at mid point of the beam is,
![\mathbf{\delta =\frac{W\;l^3}{48\;E\;I}}](http://www.codecogs.com/images/eqns/bebf64be87281e3ab3006c989246738b.gif)
Note:
It should be noted that this method of finding deflection is limited to cases where only one concentrated load is applied ( i.e. doing work)and then only gives the deflection under the load A. For a more general application of strain energy to deflection we can use Castigliano's Theorems.
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