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Strain Energy due to Bending Moment

 



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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,
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 
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}



\therefore\;\;\;\;\;U_a\;=\;\frac{W^2\;b^2}{2\;l^2\;EI}\left[\frac{x^3}{3} \right]_0^a

\therefore\;\;\;\;\;U_a\;=\;\frac{W^2\;a^3\;b^2}{6\,E\,I\,l^2}

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}

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)

\therefore\;\;\;\;\;\;U\;=\;\frac{W^2\;a^2\;b^2}{6\;E\;I\;l}

But if \displaystyle \delta is the deflection under the load, the strain energy must be equal to the work done by the load if it is gradually applied.
\frac{1}{2}W\,\delta =\frac{W^2\;a^2\;b^2}{6\,E\,I\,l}
\therefore\;\;\;\;\;\delta =\frac{W\;a^2\;b^2}{3\,E\,I\,l}  
For a Central Load, \displaystyle a\;=\;b\;=\;\frac{l}{2}
\therefore\;\;\;\;\;\delta \;=\;\left(\frac{W}{3\,E\,I\,l} \right)\left(\frac{l^2}{4} \right)\left(\frac{l^2}{4} \right)



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}}
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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