A state of pure shear stress is produced by:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 29/01/2018 (Morning)
Detailed Explanation
On a plane at 45° to the principal directions, the normal stress is (σ₁ + σ₂)/2 and the shear stress is (σ₁ − σ₂)/2. For pure shear the normal stress must vanish.
\[ \sigma_1 = +\sigma,\ \sigma_2 = -\sigma:\qquad \sigma_{n,45^\circ} = \frac{\sigma + (-\sigma)}{2} = 0,\qquad \tau_{45^\circ} = \frac{\sigma - (-\sigma)}{2} = \sigma \]
Why “tension in one direction and equal compression in the perpendicular direction” Is Correct
Tension in one direction with equal compression at right angles makes the 45° planes carry shear only. Mohr's circle for this case is centred at the origin with radius σ.
Why the Other Options Are Wrong
- B. equal tension in two directions at right angles: Equal biaxial tension gives a point-circle: no shear on any plane.
- C. equal compression in two directions at right angles: Equal biaxial compression also gives zero shear on every plane.
- D. none of these: Option (a) is correct.
Key Point
Pure shear ⇔ principal stresses +σ and −σ; shear planes at 45°.
Key Concepts for Students
- Torsion of a shaft produces pure shear on the surface.
- Principal stresses in pure shear τ are ±τ at 45°.
- Equal all-round stress gives zero shear on every plane.
Practice more: this is question 5 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.