A conical bar tapers uniformly from a diameter of 15 mm to a diameter of 40 mm in a length of 400 mm. What is the elongation of the bar under an axial tensile force of 100 kN ? (Take E = 2 × 10⁵ N/mm²)
- A0·242 mm
- B0·121 mm
- C0·424 mmCorrect
- D0·212 mm
Solution
Concept
Elongation of a conical frustum
Formula
δ = 4 P L / (π E d₁ d₂)
Step-by-step solution
1. δ = 4 × 100 000 × 400 / (π × 2×10⁵ × 15 × 40). 2. = 1.6×10⁸ / (3.77×10⁸) = 0.424 mm.
Explanation
For a circular bar tapering uniformly, the elongation is 4PL/(πEd₁d₂).
Why the other options are wrong
0.212 and 0.121 are one-half and one-quarter of the answer, 0.242 comes from using the mean diameter.
Shortcut
δ = 4PL/(πE d₁ d₂).
Common mistake
Using the mean diameter (gives 0.2).
Correct answer: C) 0·424 mm