A simply supported beam of circular cross-section, diameter D and length l, carries a concentrated load W at the centre. The strength (load capacity) of the beam is proportional to:
Try it first: pick an option, then read the explanation.
Correct answer: C. D³/l
Asked in SSC JE 01/03/2017 (Morning)
Detailed Explanation
The load the beam can carry is the load that makes the maximum bending stress equal to the allowable stress σ.
\[ \frac{Wl}{4} = \sigma\,\frac{\pi D^3}{32} \Rightarrow W = \frac{\pi\sigma D^3}{8\,l} \propto \frac{D^3}{l} \]
Why “D³/l” Is Correct
The safe load grows with the cube of the diameter and falls with the span: strength ∝ D³/l.
| 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. 1/D³: Strength increases with D, it does not fall as 1/D³.
- B. 1/D²: 1/D² also wrongly makes a thicker beam weaker.
- D. D²/l: D² is the area dependence (axial or shear), not bending.
Key Point
Circular beam: W ∝ σD³/l (because Z ∝ D³).
Key Concepts for Students
- Doubling D increases bending strength 8 times.
- Doubling the span halves the safe central load.
- Stiffness ∝ D⁴/l³.
Practice more: this is question 33 in Bending and Shear Stress MCQs – Part 3. Try the whole set and check every answer instantly.