At a point in a strained material, two mutually perpendicular tensile stresses of 2000 kg/cm² and 1000 kg/cm² act. The intensity of shear (tangential) stress on a plane inclined at 15° to the axis of the minor stress is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 01/03/2017 (Morning)
Detailed Explanation
Note on this question: the exam asked for the tensile (normal) stress on the plane, which works out to about 1933 kg/cm² and is not among the options. The commission accepted (b) 250 kg/cm², which is the shear stress on that plane, so the stem here asks for the shear stress.
A plane inclined at 15° to the minor-stress axis has its normal at 15° to the major stress σ₁ = 2000 kg/cm².
\[ \tau = \frac{\sigma_1 - \sigma_2}{2}\sin 2\theta = \frac{2000 - 1000}{2}\sin 30^\circ = 500\times0.5 = 250\ \text{kg/cm}^2 \]
\[ \sigma_n = \frac{\sigma_1 + \sigma_2}{2} + \frac{\sigma_1 - \sigma_2}{2}\cos 2\theta = 1500 + 500\cos 30^\circ \approx 1933\ \text{kg/cm}^2 \]
Why “250 kg/cm²” Is Correct
The shear stress on the plane is 250 kg/cm² — the commission's accepted answer. On Mohr's circle (centre 1500, radius 500), the plane is the point 30° round from σ₁.
Why the Other Options Are Wrong
- A. 125 kg/cm²: 125 kg/cm² is half the shear stress.
- C. 500 kg/cm²: 500 kg/cm² is the maximum shear stress (at 45°), not the value at 15°.
- D. 1000 kg/cm²: 1000 kg/cm² is the minor principal stress (or σ₁ − σ₂).
Key Point
Biaxial: τ = ((σ₁ − σ₂)/2) sin 2θ; at 15°: 250 kg/cm². Normal stress there ≈ 1933 kg/cm².
Key Concepts for Students
- Check which angle convention the question uses (plane vs normal).
- Mohr's circle doubles the angle.
- The commission accepted (b); the stem is reworded to match it.
Practice more: this is question 13 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.