In the case of biaxial stress, the maximum value of shear stress is given by:
Try it first: pick an option, then read the explanation.
Correct answer: D. none of these
Asked in SSC JE 24/01/2018 (Morning)
Detailed Explanation
With principal stresses σ₁ and σ₂ (and no applied shear), the maximum in-plane shear stress acts on planes at 45° and equals the radius of Mohr's circle.
\[ \tau_{max} = \frac{\sigma_1 - \sigma_2}{2} \]
Why “none of these” Is Correct
The correct expression is half the difference of the normal stresses, which is not among (a)–(c). So the answer is none of these.
| Quantity | Formula |
|---|---|
| Principal stresses | σ₁,₂ = (σx + σy)/2 ± √(((σx − σy)/2)² + τxy²) |
| Principal planes | tan 2θp = 2τxy/(σx − σy); shear stress on them is zero |
| Maximum in-plane shear | τmax = (σ₁ − σ₂)/2 = radius of Mohr's circle |
| Planes of maximum shear | At 45° to the principal planes; normal stress on them = (σ₁ + σ₂)/2 |
Why the Other Options Are Wrong
- A. the difference of the normal stresses: The full difference σ₁ − σ₂ is the Mohr's circle diameter, twice τmax.
- B. the sum of the normal stresses: The sum has nothing to do with maximum shear.
- C. half the sum of the normal stresses: Half the sum is the centre of Mohr's circle — the normal stress on the max-shear plane.
Key Point
Biaxial: τmax = (σ₁ − σ₂)/2 (in-plane).
Key Concepts for Students
- Absolute max shear in 3-D = (σmax − σmin)/2, including σ₃ = 0.
- If σ₁ and σ₂ are both tensile, the absolute maximum is σ₁/2.
- Equal σ₁ = σ₂ ⇒ zero in-plane shear.
Practice more: this is question 24 in Principal Stress and Principal Planes MCQs – Part 2. Try the whole set and check every answer instantly.