A circular section tapering bar is rigidly fixed at both the ends. The diameter changes from 75 mm at one end to 150 mm at the other end, in a length of 1·2 m. What is the maximum stress in the bar if the temperature is raised by 32°C ? (Take E = 2 × 10⁵ N/mm² and α = 12 × 10⁻⁶ per 1°C)
- A123.6 N/mm²
- B143.6 N/mm²
- C163.6 N/mm²
- D153.6 N/mm²Correct
Solution
Concept
Temperature stress in a bar of varying section
Formula
α ΔT L = P ∫ dx / (A E); σmax at the smaller end
Step-by-step solution
1. ∫ dx / A = 4 L / (π d₁ d₂). 2. Free expansion α ΔT L = P × 4L / (π E d₁ d₂), so P = α ΔT E π d₁ d₂ / 4. 3. Stress at the small end σ = P / (π d₁²/4) = α ΔT E (d₂/d₁) = 12×10⁻⁶ × 32 × 2×10⁵ × 2 = 153.6 N/mm².
Explanation
The force is the same everywhere, so the stress is greatest at the smallest section.
Why the other options are wrong
76.8 N/mm² (the uniform-bar value) is not an option; 123.6, 143.6 and 163.6 do not equal 76.8 × 2.
Shortcut
σmax = α ΔT E (d₂/d₁).
Common mistake
Using the mean section.
Correct answer: D) 153.6 N/mm²