A rectangular block is subjected to two perpendicular stresses of 10 MPa tension and 10 MPa compression. What are the stresses on planes inclined at 45° with the horizontal plane ?
- ANormal Stress = 10 MPa and Shear Stress = 20 MPa
- BNormal Stress = 0 MPa and Shear Stress = – 10 MPa (counter clockwise direction)Correct
- CNormal Stress = 5 MPa and Shear Stress = – 20 MPa (counter clockwise direction)
- DNormal Stress = 10 MPa and Shear Stress = 5 MPa
Solution
Concept
Stress transformation (Mohr circle)
Formula
σn = (σx + σy)/2 + (σx − σy)/2 · cos 2θ; τ = (σx − σy)/2 · sin 2θ
Step-by-step solution
1. σx = +10, σy = −10, θ = 45°. 2. σn = 0 + 10 cos 90° = 0. 3. |τ| = 10 sin 90° = 10 MPa, with sign by convention (counter-clockwise).
Explanation
This is the state of pure shear in disguise. The normal stress on the 45° planes is zero and the shear stress equals the magnitude of the applied direct stresses.
Why the other options are wrong
The other options give non-zero normal stress or a shear stress larger than the radius of the Mohr circle (10 MPa).
Shortcut
Equal and opposite direct stresses → 45° planes: σ = 0, τ = ±σ.
Common mistake
Taking τ = 20 (the difference of the stresses) instead of half of it.
Correct answer: B) Normal Stress = 0 MPa and Shear Stress = – 10 MPa (counter clockwise direction)