The section modulus of a beam section may be obtained by dividing its moment of inertia by:
Try it first: pick an option, then read the explanation.
Correct answer: B. the distance of the extreme fibre from the neutral axis
Asked in SSC JE 02/03/2017 (Evening)
Detailed Explanation
The section modulus packages the two geometric quantities in σ = My/I into one number so that σ_max = M/Z.
\[ \sigma_{max} = \frac{M\,y_{max}}{I} = \frac{M}{Z},\qquad Z = \frac{I}{y_{max}} \]
Why “the distance of the extreme fibre from the neutral axis” Is Correct
Z is I divided by the distance of the extreme (most distant) fibre from the neutral axis. For unsymmetric sections there are two values, one for the top and one for the bottom fibre.
| Section | I about centroidal axis | y max | Section modulus Z |
|---|---|---|---|
| Rectangle b × d (axis ∥ b) | bd³/12 | d/2 | bd²/6 |
| Square side a (sides horizontal) | a⁴/12 | a/2 | a³/6 |
| Square side a (diagonal horizontal) | a⁴/12 | a/√2 | a³/(6√2) |
| Solid circle, diameter d | πd⁴/64 | d/2 | πd³/32 |
| Triangle b × h (axis ∥ base) | bh³/36 | 2h/3 | bh²/24 |
| Hollow circle D, d | π(D⁴ − d⁴)/64 | D/2 | π(D⁴ − d⁴)/(32D) |
Why the Other Options Are Wrong
- A. the depth of the section: Dividing by the full depth gives half the correct Z for a symmetric section.
- C. the maximum tensile stress at the section: Stress is not a geometric property; Z depends only on the shape.
- D. the maximum compressive stress at the section: Same as (c): dividing by a stress does not give a section modulus.
Key Point
Z = I / y_max (mm³).
Key Concepts for Students
- Rectangle: Z = bd²/6.
- Unsymmetric section: Z_top ≠ Z_bottom.
- Plastic modulus Z_p is different from elastic Z.
Practice more: this is question 24 in Bending and Shear Stress MCQs – Part 2. Try the whole set and check every answer instantly.