The intensity of longitudinal (bending) stress in a cross-section at any point a distance r from the neutral axis is proportional to:
Try it first: pick an option, then read the explanation.
Correct answer: A. r
Asked in SSC JE 2016
Detailed Explanation
The book places this bending-stress question in the columns chapter. In pure bending, strain varies linearly with distance from the neutral axis, and so does stress (Hooke's law).
\[ \frac{M}{I} = \frac{\sigma}{r} = \frac{E}{R} \;\Rightarrow\; \sigma = \frac{M}{I}\,r \]
Why “r” Is Correct
Stress is directly proportional to r — zero at the neutral axis and maximum at the extreme fibre.
Why the Other Options Are Wrong
- B. 1/r: Stress increases with r, not decreases.
- C. r²: The variation is linear, not quadratic.
- D. 1/r²: This is an inverse-square law, unrelated to bending stress.
Key Point
Bending stress σ = My/I ∝ distance from the neutral axis.
Key Concepts for Students
- Plane sections remain plane → linear strain.
- Max stress at the extreme fibre: σ = M/Z.
- Same linearity holds for torsional shear stress (τ ∝ r).
Practice more: this is question 34 in Columns and Struts MCQs – Part 3. Try the whole set and check every answer instantly.