A compound bar has a steel bar (3500 mm²) and a brass bar (3000 mm²) of equal length carrying a compressive load of 100,000 N. If Es = 0.2 MN/mm² and Eb = 0.1 MN/mm², the stresses developed are:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 23/01/2018 (Evening)
Detailed Explanation
In a compound bar, both materials shorten by the same amount (same length), so they have the same strain. Stress = E × strain, so the stiffer material carries more stress.
Equal strain: \( \dfrac{\sigma_s}{E_s} = \dfrac{\sigma_b}{E_b} \Rightarrow \sigma_s = 2\sigma_b \)
Equilibrium: \( \sigma_s A_s + \sigma_b A_b = 100000 \Rightarrow 2\sigma_b(3500) + \sigma_b(3000) = 100000 \)
\[ 10000\,\sigma_b = 100000 \Rightarrow \sigma_b = 10,\ \sigma_s = 20\ \text{N/mm}^2 \]
Why “σb = 10 N/mm², σs = 20 N/mm²” Is Correct
Steel carries 70 kN and brass 30 kN, giving σb = 10 N/mm² and σs = 20 N/mm².
Why the Other Options Are Wrong
- B. σb = 8 N/mm², σs = 16 N/mm²: These stresses add up to only 80 kN of load.
- C. σb = 6 N/mm², σs = 12 N/mm²: These add up to only 60 kN.
- D. σb = 5 N/mm², σs = 10 N/mm²: These add up to only 50 kN.
Key Point
Compound bar: equal strain (σ ∝ E) + equilibrium (Σσ·A = P).
Key Concepts for Students
- Modular ratio m = Es/Eb = 2 here.
- Transformed area method: A_eq = As + Ab/m (in steel units).
- RCC columns use the same idea with steel and concrete.
Practice more: this is question 36 in Stress and Strain MCQs – Part 3. Try the whole set and check every answer instantly.