A material has strength in tension, compression and shear as 30 N/mm², 90 N/mm² and 25 N/mm², respectively. If a specimen of diameter 25 mm is tested in compression, then identify the failure load and failure plane.
- AFailure load is 14726 N and failure plane is 30° to the plane of axial compression
- BFailure load is 24543 N and failure plane is 45° to the plane of axial compressionCorrect
- CFailure load is 36574 N and failure plane is 60° to the plane of axial compression
- DFailure load is 18745 N and failure plane is 90° to the plane of axial compression
Solution
Concept
Failure by shear in a simple compression test
Formula
τmax = σ/2; P = σ A
Step-by-step solution
1. Shear strength is 25 N/mm², so yielding by shear starts when σ/2 = 25, i.e. σ = 50 N/mm² (less than 90). 2. A = π × 25² / 4 = 490.87 mm². 3. P = 50 × 490.87 = 24543 N. 4. Maximum shear acts on planes at 45° to the axis.
Explanation
Under uniaxial compression the maximum shear stress is σ/2 on 45° planes. Since the material is weaker in shear (25) than half its compressive strength (45), it fails in shear at σ = 50 N/mm², before reaching 90.
Why the other options are wrong
14726 N corresponds to 30 N/mm² (tensile strength). 36574 and 18745 do not correspond to a stress of 50 N/mm².
Shortcut
Shear fails first if σ/2 reaches the shear strength: σ = 2 × 25 = 50.
Common mistake
Using the compressive strength 90 N/mm² to find the load.
Correct answer: B) Failure load is 24543 N and failure plane is 45° to the plane of axial compression