The maximum allowable shear stress in a material is 100 kg/cm². A square bar of this material carries a tensile force of 5000 kg. The side of the square should be:
Try it first: pick an option, then read the explanation.
Correct answer: B. 5 cm
Asked in SSC JE 02/03/2017 (Morning)
Detailed Explanation
A bar in pure tension has no shear stress on its cross-section, but it does on inclined planes. On a plane at θ to the cross-section, τ = (σ/2) sin 2θ, which is largest at 45°.
\[ \tau_{max} = \frac{\sigma}{2} = \frac{P}{2a^2} \Rightarrow 100 = \frac{5000}{2a^2} \Rightarrow a^2 = 25 \Rightarrow a = 5\ \text{cm} \]
Why “5 cm” Is Correct
The side needed is 5 cm. The normal stress is then 200 kg/cm², and the maximum shear stress on 45° planes is exactly the allowable 100 kg/cm².
| Plane | Normal stress | Shear stress |
|---|---|---|
| Cross-section (θ = 0°) | σ = P/A | 0 |
| 45° plane | σ/2 | σ/2 (maximum) |
| Longitudinal plane (θ = 90°) | 0 | 0 |
Why the Other Options Are Wrong
- A. 10 cm: 10 cm gives τmax = 25 kg/cm² — very uneconomical.
- C. 12 cm: 12 cm is oversized.
- D. √12 cm: √12 ≈ 3.46 cm gives τmax ≈ 208 kg/cm², above the limit.
Key Point
Uniaxial stress σ: τmax = σ/2 on planes at 45°.
Key Concepts for Students
- Ductile bars in tension fail by shear along 45° planes (cup-and-cone).
- Mohr's circle for uniaxial stress has radius σ/2.
- Using τ = P/A here would be wrong.
Practice more: this is question 65 in Bending and Shear Stress MCQs – Part 5. Try the whole set and check every answer instantly.