A column of rectangular section b × d is subjected to an eccentric point load P with eccentricity e in the direction of the dimension d. The bending stress developed is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 04/12/2023 (9 am–11 am)
Detailed Explanation
The eccentric load produces a bending moment M = P·e about the axis perpendicular to the direction of eccentricity. The bending stress at the extreme fibre is M divided by the section modulus about that axis.
\[ I = \frac{bd^3}{12},\quad y_{max} = \frac{d}{2},\quad Z = \frac{bd^2}{6} \]
\[ \sigma_b = \frac{M}{Z} = \frac{Pe}{bd^2/6} = \frac{6Pe}{bd^2} \]
Why “6Pe/bd²” Is Correct
Bending takes place across the dimension d (the direction of eccentricity), so d is the "depth" in Z = bd²/6. The extreme-fibre bending stress is 6Pe/bd², added to and subtracted from the direct stress P/bd.
Why the Other Options Are Wrong
- A. 3Pe/bd²: 3Pe/bd² is half the correct value (uses Z = bd²/3).
- C. 24Pe/bd²: 24Pe/bd² is four times too large.
- D. 12Pe/bd²: 12Pe/bd² is M·y/I with y = d instead of d/2.
Key Point
Eccentric load: σ = P/A ± 6Pe/(bd²) for a rectangle, with d measured along the eccentricity.
Key Concepts for Students
- Total stress = P/bd (1 ± 6e/d).
- No tension when e ≤ d/6.
- For biaxial eccentricity add both bending terms: P/A ± M_x/Z_x ± M_y/Z_y.
Practice more: this is question 3 in Bending and Shear Stress MCQs – Part 1. Try the whole set and check every answer instantly.