A bar of length L and cross-sectional area A carries a gradually applied tensile load W. The strain energy stored in the bar is:
Try it first: pick an option, then read the explanation.
Correct answer: D. W²L/2AE
Asked in SSC JE 04/03/2017 (Morning); SSC JE 2012
Detailed Explanation
Under a gradually applied load, the load rises from zero to W while the bar stretches by δ. The work done (area under the load–extension line) is stored as strain energy.
\[ U = \tfrac12 W\delta = \tfrac12 W\cdot\frac{WL}{AE} = \frac{W^2L}{2AE} \]
Equivalently \( U = \dfrac{\sigma^2}{2E}AL \) with \( \sigma = W/A \).
Why “W²L/2AE” Is Correct
The strain energy is W²L/2AE — option (d).
Why the Other Options Are Wrong
- A. WL/2AE: WL/2AE is half the elongation, not energy.
- B. WL/AE: WL/AE is the elongation δ.
- C. W²L/AE: W²L/AE misses the factor ½ for gradual loading.
Key Point
Gradual load: U = W²L/2AE = σ²V/2E.
Key Concepts for Students
- A suddenly applied load W doubles the stress (2W/A) and stores four times the strain energy.
- Modulus of resilience = σ_e²/2E.
- Strain energy is always positive.
Practice more: this is question 45 in Stress and Strain MCQs – Part 3. Try the whole set and check every answer instantly.