The angle between a principal plane and the plane of maximum shear is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2011
Detailed Explanation
Shear on a plane inclined at θ to a principal plane varies as sin 2θ; its maximum is at 2θ = 90°.
\[ \tau_\theta = \frac{\sigma_1 - \sigma_2}{2}\sin 2\theta,\qquad \sin 2\theta = 1 \Rightarrow \theta = 45^\circ \]
Why “45°” Is Correct
The planes of maximum shear are at 45° to the principal planes. (135° describes the same pair of planes, but when 45° is listed it is the expected answer.)
| 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
- B. 90°: 90° is the angle between the two principal planes.
- C. 135°: 135° is equivalent, but 45° is the standard (acute) answer when both are listed.
- D. 60°: 60° does not maximise sin 2θ.
Key Point
Max-shear planes are at 45° to the principal planes.
Key Concepts for Students
- On Mohr's circle they are 90° apart.
- Normal stress on the max-shear plane = (σ₁ + σ₂)/2.
- Compare with the related question where only 135° is offered.
Practice more: this is question 19 in Principal Stress and Principal Planes MCQs – Part 2. Try the whole set and check every answer instantly.