Question

Strain energy of a solid body due to a gradually applied load is equal to:
(Consider the cross-sectional area of the body to be uniform throughout)

• A. $0.5\times \text{load}\times \text{extension}$
• B. $0.5\times \frac{YA}{L}\times (\Delta L{)}^2$
• C. Work done by load
• D. $0.5\times \text{stress}\times \text{strain}\times \text{volume}$

Answer 1

Answer: (A), (B), (C), (D)

$0.5\times \text{stress}\times \text{strain}\times \text{volume}$

Solids are analogical to spring for storing energy. When solids are deformed by applying load, they store energy (similar to the case of extension/compression of spring)

The amount of energy stored is equal to work done by the external agent i.e load.
For spring,
Stretching force $F=Kx...\left(1\right)$
where $x$ is the extension in spring.
For elastic solids,
$\text{Load}=\text{Tension Force}$
$F=\frac{YA}{L}\Delta L...\left(2\right)$
where $\Delta L=\text{extension in solid}$
From analogy, comparing Eq. $\left(1\right)$ and $\left(2\right)$,
$K=\frac{YA}{L}...\left(3\right)$
$\&x=\Delta L...\left(4\right)$
Elastic potential energy stored in spring is given as,
$U=\frac{1}{2}K{x}^2=0.5K{x}^2$
So, elastic potential energy for solids from Eq.$\left(3\right)$ and $\left(4\right)$,
$\text{Strain energy}\left(U\right)=0.5\times \frac{YA}{L}\times (\Delta L{)}^2$
$\Rightarrow U=0.5\times \frac{YA\Delta L}{L}\times \Delta L$
$\Rightarrow U=0.5\times \text{Load}\times \text{extension}$

Or $U=0.5\times \frac{\text{Load}}{\text{Area}}\times \frac{\text{extension}}{\text{Length}}\times \text{Area}\times \text{Length}$
Substituting for stress and volume,
$\therefore U=0.5\times \text{(stress)}\times \text{(strain)}\times \text{(volume)}$
Therefore, all four options are correct.