Of several prismatic beams of equal length and the same material, the one that can carry the maximum load in flexure is the one with the maximum:
Try it first: pick an option, then read the explanation.
Correct answer: C. section modulus
Asked in SSC JE 2013
Detailed Explanation
A beam fails in flexure when the extreme-fibre stress reaches the allowable value. The moment it can then resist is M = σ_allow × Z.
\[ M_r = \sigma_{allow}\,Z,\qquad Z = \frac{I}{y_{max}} \]
Why “section modulus” Is Correct
With the same material and length, the beam with the largest section modulus resists the largest moment, and so carries the largest load. Moment of inertia alone is not enough: a large I with a large y_max may still give a small Z.
| 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. depth of section: Depth helps, but a deep, thin section can still have a small Z.
- B. area of cross-section: Area alone does not decide bending strength; distribution of area does.
- D. moment of inertia: I matters through Z = I/y_max; I alone ignores how far the extreme fibre is.
Key Point
Flexural strength ∝ section modulus Z (M = σZ).
Key Concepts for Students
- Stiffness (deflection) depends on I; strength depends on Z.
- Moving material away from the N.A. raises Z efficiently.
- Z is in mm³.
Practice more: this is question 23 in Bending and Shear Stress MCQs – Part 2. Try the whole set and check every answer instantly.