A block of brittle material of cross-section 4 m² carries an axial tensile load of 20 tonnes. The maximum shear stress in the block is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 01/03/2017 (Evening)
Detailed Explanation
Under uniaxial tension, the maximum shear stress is half the normal stress and acts on planes at 45°. Convert units carefully first.
\[ A = 4\ \text{m}^2 = 40\,000\ \text{cm}^2,\quad W = 20\ \text{t} = 20\,000\ \text{kg} \]
\[ \sigma = \frac{20\,000}{40\,000} = 0.5\ \text{kg/cm}^2,\qquad \tau_{max} = \frac{\sigma}{2} = 0.25\ \text{kg/cm}^2 \]
Why “none of these” Is Correct
The maximum shear stress is only 0.25 kg/cm², which matches none of the options — so the answer is none of these.
Why the Other Options Are Wrong
- A. 1250 kg/cm²: 1250 kg/cm² comes from an area of 8 cm² — a unit error.
- B. 1000 kg/cm²: 1000 kg/cm² has no basis with these data.
- C. 500 kg/cm²: 500 kg/cm² has no basis with these data.
Key Point
Uniaxial: τmax = P/(2A). Convert m² to cm² (× 10⁴) before dividing.
Key Concepts for Students
- 1 m² = 10 000 cm²; 1 tonne = 1000 kg.
- Brittle materials usually fail by normal stress, not shear.
- Even if the area were 4 cm², τmax = 2500 kg/cm² — still none of the options.
Practice more: this is question 26 in Principal Stress and Principal Planes MCQs – Part 2. Try the whole set and check every answer instantly.