A square element carries a tensile stress σ in the horizontal direction and a compressive stress σ in the vertical direction (normal stresses only). The radius of its Mohr's circle is:
Try it first: pick an option, then read the explanation.
Correct answer: A. σ
Asked in SSC JE 2011
Detailed Explanation
With normal stresses only, the faces of the element are principal planes, and the radius of Mohr's circle is half the difference of the principal stresses.
\[ R = \frac{\sigma_1 - \sigma_2}{2} = \frac{\sigma - (-\sigma)}{2} = \sigma \]
Why “σ” Is Correct
The radius is σ, and the circle is centred at the origin — the state of pure shear, with τmax = σ on the 45° planes.
Why the Other Options Are Wrong
- B. σ/2: σ/2 is the radius for uniaxial stress σ only.
- C. 2σ: 2σ is the diameter, not the radius.
- D. 0.6σ: 0.6σ has no basis.
Key Point
Mohr radius = (σ₁ − σ₂)/2; for +σ and −σ it is σ.
Key Concepts for Students
- Centre at (σ₁ + σ₂)/2 = 0 here.
- This is the pure shear state.
- Mind the sign of compressive stress.
Practice more: this is question 20 in Principal Stress and Principal Planes MCQs – Part 2. Try the whole set and check every answer instantly.