Elasticity & Stress-Strain: Diagram-Based Questions (5)

Question

The stress-strain diagram of a metal wire shows a linear region from origin to the proportional limit at strain $\varepsilon = 0.002$ and stress $\sigma = 4 \times 10^8\,\text{N/m}^2$. Find Young’s modulus and the elastic potential energy stored per unit volume at the proportional limit.

Solution — Step by Step

Young’s modulus is $2\times 10^{11}\,\text{N/m}^2$ and elastic energy density is $4\times 10^5\,\text{J/m}^3$.

Why This Works

In the linear (Hookean) region, stress is proportional to strain, so the slope $\sigma/\varepsilon$ is a constant — Young’s modulus. Beyond this region the slope flattens and the material behaves plastically.

The energy stored per unit volume is the work done per unit volume during loading, which is the area under the stress-strain curve. For a triangle, that’s $\tfrac{1}{2}\sigma\varepsilon$. Beyond the elastic limit, the area still represents work but most of it isn’t recoverable — it goes into heat and permanent deformation.

Alternative Method

Use the strain-energy formula $U = \tfrac{1}{2}\,Y\,\varepsilon^2$ per unit volume:

$u = \tfrac{1}{2}\times 2\times 10^{11} \times (0.002)^2 = 10^{11}\times 4\times 10^{-6} = 4\times 10^5\,\text{J/m}^3$

Same answer.

Common Mistake

Computing $u = \sigma\varepsilon$ without the $\tfrac{1}{2}$ factor. Students forget the area under the linear region is a triangle, not a rectangle. The answer doubles, and the option that matches is invariably wrong by design.