For a simply supported circular beam with a central load W, the maximum bending stress is equal to the direct tensile stress produced by the same load. The ratio of length to diameter is:
Try it first: pick an option, then read the explanation.
Correct answer: C. 1/2
Asked in SSC JE 23/01/2018 (Evening)
Detailed Explanation
Same idea as for a rectangular beam: compare the mid-span bending stress M/Z with the direct stress W/A, now using circle properties.
\[ \sigma_b = \frac{WL/4}{\pi d^3/32} = \frac{8WL}{\pi d^3},\qquad \sigma_t = \frac{W}{\pi d^2/4} = \frac{4W}{\pi d^2} \]
\[ \frac{8WL}{\pi d^3} = \frac{4W}{\pi d^2} \Rightarrow \frac{L}{d} = \frac{1}{2} \]
Why “1/2” Is Correct
The stresses are equal when L/d = 1/2.
| 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. 1/8: 1/8 makes the bending stress only a quarter of the direct stress.
- B. 1/4: 1/4 makes the bending stress half the direct stress.
- D. 1/3: 1/3 gives σb/σt = 2/3, not 1.
Key Point
Circle, central load: σb/σt = 2L/d; equal when L/d = 1/2 (rectangle: 2/3).
Key Concepts for Students
- Z circle = πd³/32.
- Area circle = πd²/4.
- In real beams L/d is large, so bending stress dominates.
Practice more: this is question 20 in Bending and Shear Stress MCQs – Part 2. Try the whole set and check every answer instantly.