A bar of uniform cross-section A and length L is suspended from top. The extension of the bar due to self-weight only in terms of Young’s modulus E and unit weight γ is
- AγL²/2ECorrect
- BγL²/AE
- CγL/2E
- DγL/2E
Solution
Concept
Elongation due to self weight
Formula
δ = γ L² / (2E)
Step-by-step solution
1. Weight of the portion below a section at distance x from the top = γ A (L − x). 2. Elongation of element dx = γ (L − x) dx / E. 3. δ = ∫₀ᴸ γ (L − x) dx / E = γ L² / 2E.
Explanation
The load carried by the bar grows towards the top. Integrating gives half of the elongation produced by a load equal to the whole weight acting at the free end.
Why the other options are wrong
γL²/AE contains A, which cancels out. γL/2E has dimensions of strain, not length (options C and D are printed identically).
Shortcut
Self-weight elongation = ½ × (W L / A E) = γL²/2E.
Common mistake
Keeping A in the answer.
Correct answer: A) γL²/2E