When a member is subjected to an axial tensile load, the greatest normal stress is equal to:
Try it first: pick an option, then read the explanation.
Correct answer: C. twice the maximum shear stress
Asked in SSC JE 03/03/2017 (Evening)
Detailed Explanation
Under simple tension, the normal stress is greatest on the cross-section (σ = P/A) and the shear stress is greatest on planes at 45°, where it equals σ/2.
\[ \tau_\theta = \frac{\sigma}{2}\sin 2\theta \Rightarrow \tau_{max} = \frac{\sigma}{2}\ (\theta = 45^\circ) \Rightarrow \sigma_{max} = 2\tau_{max} \]
Why “twice the maximum shear stress” Is Correct
The greatest normal stress equals twice the maximum shear stress. This is why ductile bars in tension (weak in shear) fail on cone-shaped 45° surfaces.
Why the Other Options Are Wrong
- A. half the maximum shear stress: Inverted: it is the shear stress that is half the normal stress.
- B. the maximum shear stress: τmax is only half the normal stress.
- D. none of these: Option (c) is correct.
Key Point
Uniaxial: τmax = σ/2 at 45°; σ = 2τmax.
Key Concepts for Students
- Ductile cup-and-cone fracture follows the max shear planes.
- Brittle materials fail on the cross-section (max normal stress).
- In torsion, σmax = τmax (pure shear).
Practice more: this is question 12 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.