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We assume that the deformation is wholly due to external load, neglecting the effect of the weight of the rod (see next problem). Then a well known formula says, elastic energy per unit volume $$ =\frac{1}{2} \text { stress } \times \text { strain }=\frac{1}{2} \sigma \varepsilon $$ This gives $\frac{1}{2} \frac{m}{\rho} E \varepsilon^{2} \approx 0.04 \mathrm{~kJ}$ for the total deformation energy.
Physical Fundamentals Of Mdchanics
Elastic Deformations of a Solid Body
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