If the normal stresses due to longitudinal and transverse loads on a bar are σ₁ and σ₂ respectively, the normal component of stress on a plane inclined at θ to the longitudinal load is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 22/01/2018 (Evening)
Detailed Explanation
Each normal stress contributes to the normal stress on the inclined plane in proportion to the square of the cosine of the angle between the stress and the plane's normal (one cosine to resolve the force, one to account for the larger area).
The plane makes θ with σ₁, so its normal makes (90° − θ) with σ₁ and θ with σ₂:
\[ \sigma_n = \sigma_1\cos^2(90^\circ - \theta) + \sigma_2\cos^2\theta = \sigma_1\sin^2\theta + \sigma_2\cos^2\theta \]
Why “σ₁ sin²θ + σ₂ cos²θ” Is Correct
The normal component is σ₁ sin²θ + σ₂ cos²θ; the shear component on the same plane is ((σ₁ − σ₂)/2) sin 2θ.
| Loading | Normal stress σn on the plane | Shear stress τ on the plane | Maximum shear |
|---|---|---|---|
| Uniaxial σ (θ = angle of the plane's normal to σ) | σ cos²θ | (σ/2) sin 2θ | σ/2 at θ = 45° |
| Biaxial σ₁, σ₂ (θ = angle of the normal to σ₁) | (σ₁ + σ₂)/2 + ((σ₁ − σ₂)/2) cos 2θ | ((σ₁ − σ₂)/2) sin 2θ | (σ₁ − σ₂)/2 at θ = 45° |
| General σx, σy, τxy | (σx + σy)/2 + ((σx − σy)/2) cos 2θ + τxy sin 2θ | ((σx − σy)/2) sin 2θ − τxy cos 2θ | √(((σx − σy)/2)² + τxy²) |
If θ is measured from the plane itself (i.e. from the direction of the stress) instead of from its normal, sin and cos swap: σn = σ sin²θ, τ = σ sinθ cosθ.
Why the Other Options Are Wrong
- A. σ₁ sinθ + σ₂ cosθ: Stresses transform with squared trigonometric terms, not first powers.
- C. (σ₁ − σ₂) sin 2θ / 2: This is the tangential (shear) component.
- D. (σ₁ + σ₂) sin 2θ / 2: The sum of stresses does not multiply sin 2θ in any standard formula.
Key Point
Plane at θ to σ₁: σn = σ₁ sin²θ + σ₂ cos²θ, τ = ((σ₁ − σ₂)/2) sin 2θ.
Key Concepts for Students
- If θ is measured to the normal instead, σn = σ₁ cos²θ + σ₂ sin²θ.
- σn on two perpendicular planes always adds up to σ₁ + σ₂.
- Mohr's circle shows the same relations graphically.
Practice more: this is question 7 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.