The tensile stresses at a point across two mutually perpendicular planes are 150 N/mm² and 80 N/mm². What are the tangential and normal stresses, respectively, on a plane inclined at 30° to the axis of the minor stress ?
- A112·5 N/mm² ; 10·31 N/mm²
- B122·5 N/mm² ; 20·31 N/mm²
- C132·5 N/mm² ; 30·31 N/mm²Correct
- D142·5 N/mm² ; 40·31 N/mm²
Solution
Concept
Normal and tangential stress on an oblique plane
Formula
σn = σ₁cos²θ + σ₂sin²θ; τ = (σ₁ − σ₂)/2 · sin 2θ
Step-by-step solution
1. σ₁ = 150, σ₂ = 80, θ = 30°. 2. σn = 150 cos²30° + 80 sin²30° = 112.5 + 20 = 132.5 N/mm². 3. τ = (150 − 80)/2 × sin 60° = 35 × 0.866 = 30.31 N/mm². 4. The pair (132.5, 30.31) is option (c).
Explanation
The two values 132.5 and 30.31 appear together only in option (c). Note that the options list the normal stress first and the tangential stress second, in the reverse of the question's wording.
Why the other options are wrong
The other options pair different values (112.5, 122.5, 142.5 with 10.31, 20.31, 40.31) that do not come from the transformation equations.
Shortcut
Normal: 150 cos²θ + 80 sin²θ. Shear: 35 sin 2θ.
Common mistake
Measuring the angle from the other principal axis (would give 97.5 and 30.31).
Correct answer: C) 132·5 N/mm² ; 30·31 N/mm²