If the normal stresses due to longitudinal and transverse loads on a bar are σ₁ and σ₂ respectively, the tangential component of stress on a plane inclined at θ to the longitudinal load is:
Try it first: pick an option, then read the explanation.
Correct answer: C. (σ₁ − σ₂) sin 2θ / 2
Asked in SSC JE 24/01/2018 (Evening)
Detailed Explanation
Each normal stress produces a shear component on the inclined plane, but in opposite senses, so the shear depends on the difference σ₁ − σ₂.
\[ \tau = \sigma_1\sin\theta\cos\theta - \sigma_2\sin\theta\cos\theta = \frac{\sigma_1 - \sigma_2}{2}\sin 2\theta \]
Why “(σ₁ − σ₂) sin 2θ / 2” Is Correct
The tangential component is (σ₁ − σ₂) sin 2θ / 2. It is maximum at θ = 45°, giving (σ₁ − σ₂)/2.
Why the Other Options Are Wrong
- A. σ₁ sinθ + σ₂ cosθ: First powers of sin and cos do not appear in stress transformation.
- B. σ₁ sin²θ + σ₂ cos²θ: This is the normal component σn.
- D. (σ₁ + σ₂) sin 2θ / 2: The shear depends on the difference, not the sum, of the stresses.
Key Point
Biaxial: τθ = ((σ₁ − σ₂)/2) sin 2θ; zero when σ₁ = σ₂.
Key Concepts for Students
- Equal biaxial stress ⇒ no shear on any plane.
- σ₂ = −σ₁ ⇒ τmax = σ₁ (pure shear).
- τmax = radius of Mohr's circle.
Practice more: this is question 11 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.