If an element of a stressed body is in a state of pure shear with a magnitude of 40 N/mm², then the magnitude of maximum principal stress at that location will be
- A80 N/mm²
- B40 N/mm²Correct
- C60 N/mm²
- D20 N/mm²
Solution
Concept
Principal stresses in pure shear
Formula
σ1,2 = ± τ
Step-by-step solution
1. In pure shear σx = σy = 0 and τxy = 40 N/mm². 2. σ = ± √(τ²) = ± 40 N/mm². 3. Maximum principal stress = +40 N/mm² (tensile).
Explanation
A state of pure shear is equivalent to equal tension and compression of magnitude τ on planes at 45°. So the principal stresses are +τ and −τ.
Shortcut
Pure shear τ → principal stresses +τ and −τ.
Common mistake
Dividing by 2.
Correct answer: B) 40 N/mm²