An element in plane stress is subjected to normal stresses p₁ = 120 N/mm² and p₂ = 45 N/mm² (both are tensile stresses) and shearing stress of 30 N/mm² (simple shear). What is the normal stress (pn) acting as an element rotated through an angle 45° ?
- APn = – 37·5 N/mm²
- BPn = 52·5 N/mm²
- CPn = – 73·5 N/mm²
- DPn = 112·5 N/mm²Correct
Solution
Concept
Normal stress on an oblique plane
Formula
pn = (p₁ + p₂)/2 + (p₁ − p₂)/2 cos 2θ + q sin 2θ
Step-by-step solution
1. (p₁ + p₂)/2 = 82.5 N/mm². 2. cos 90° = 0, so the second term is zero. 3. q sin 90° = 30 × 1 = 30. 4. pn = 82.5 + 30 = 112.5 N/mm².
Explanation
At 45° the cosine term vanishes and the normal stress is the mean stress plus the shear stress (shear sign taken in the sense that increases the normal stress on that plane).
Why the other options are wrong
52.5 N/mm² results if the shear term is subtracted (opposite sign of shear). The negative values are not possible since both stresses are tensile and the mean stress is 82.5.
Shortcut
At 45°: pn = mean ± q.
Common mistake
Taking the opposite sign of the shear stress.
Correct answer: D) Pn = 112·5 N/mm²