The strain energy per unit volume of a solid circular shaft under axial tension is:
Try it first: pick an option, then read the explanation.
Correct answer: A. σ²/(2E)
Asked in SSC JE 2012
Detailed Explanation
The book places this axial strain-energy question in the torsion chapter. Note the loading is axial tension, not torsion.
\[ \frac{U}{V} = \frac12\,\sigma\,\varepsilon = \frac12\,\sigma\cdot\frac{\sigma}{E} = \frac{\sigma^2}{2E} \]
Why “σ²/(2E)” Is Correct
With uniform stress over the whole section, the energy per unit volume is σ²/(2E), whatever the shape of the cross-section.
| Quantity | Solid shaft (diameter D) | Hollow shaft (D outer, d inner) |
|---|---|---|
| Polar moment of inertia J | πD⁴/32 | π(D⁴ − d⁴)/32 |
| Polar modulus Zp = J/R | πD³/16 | π(D⁴ − d⁴)/(16D) |
| Maximum shear stress τ = T/Zp | 16T/(πD³) | 16TD/(π(D⁴ − d⁴)) |
| Angle of twist θ | TL/(GJ) | TL/(GJ) |
| Strain energy per unit volume | τ²/(4G) | (τ²/4G)·(D² + d²)/D² |
Torsion equation: T/J = τ/R = Gθ/L. GJ is the torsional rigidity; GJ/L is the torsional stiffness.
Why the Other Options Are Wrong
- B. σ²/(4E): σ²/4E has the form of the torsion result (τ²/4G), which averages a linearly varying stress — not the axial case.
- C. σ²/(8E): σ²/8E has no basis here.
- D. σ²/(16E): σ²/16E has no basis here.
Key Point
Axial: σ²/2E. Solid shaft in torsion: τmax²/4G. Hollow shaft: (τ²/4G)(D² + d²)/D².
Key Concepts for Students
- Energy = area under the load–deflection line.
- Uniform stress uses material fully.
- Torsion stores only half per unit volume because τ varies.
Practice more: this is question 31 in Torsion MCQs – Part 3. Try the whole set and check every answer instantly.