A rod which tapers uniformly from 30 mm diameter to 15 mm diameter in a length of 300 mm is subjected to an axial load of 6 kN. If Young’s modulus is 2 × 10⁵ N/mm², what is the elongation of the rod ?
- A0·06 / π mm
- B0·08 / π mmCorrect
- C0·6 / π mm
- D0·8 / π mm
Solution
Concept
Elongation of a uniformly tapering circular bar
Formula
δ = 4 P L / (π E d₁ d₂)
Step-by-step solution
1. δ = 4 × 6000 × 300 / (π × 2×10⁵ × 30 × 15). 2. = 7.2×10⁶ / (π × 9×10⁷) = 0.08 / π mm ≈ 0.0255 mm.
Explanation
For a conical frustum the elongation is 4PL/(πE d₁d₂). The expression simplifies to 0.08/π mm.
Why the other options are wrong
0.06/π, 0.6/π and 0.8/π come from slips of factors 3/4, 7.5 and 10 in the numerator.
Shortcut
δ = 4PL/(π E d₁ d₂).
Common mistake
Using the mean diameter (22.5 mm) which gives a different value.
Correct answer: B) 0·08 / π mm