alleryThe elastic potential energy stored per unit volume in a stretched wire (within the elastic limit) is
A$\tfrac{1}{2}\,\text{stress} \times \text{strain}$
B$\text{stress} \times \text{strain}$
C$\tfrac{1}{2}\,\text{stress}^{2}\times Y$
D$\tfrac{1}{2}\,\text{strain}^{2}/Y$
Answer & Solution
Correct answer: A. $\tfrac{1}{2}\,\text{stress} \times \text{strain}$
Energy density $u = \tfrac{1}{2}\sigma\varepsilon = \tfrac{1}{2}\dfrac{\sigma^2}{Y} = \tfrac{1}{2}Y\varepsilon^2$.
It's literally the area under the linear stress–strain curve.
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