The ratio of the section modulus of a circular section of diameter D to that of a square section of side B (with B = D) is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 27/01/2018 (Evening)
Detailed Explanation
The book's wording asks for square-to-circle, but its key (3π/16) is the circle-to-square ratio; the square-to-circle ratio, 16/(3π), is not among the options. The stem here is written the way the key works. The original "i only … iv only" options are shown directly.
\[ Z_c = \frac{\pi D^3}{32},\qquad Z_s = \frac{B^3}{6} = \frac{D^3}{6} \]
\[ \frac{Z_c}{Z_s} = \frac{\pi D^3/32}{D^3/6} = \frac{6\pi}{32} = \frac{3\pi}{16} \approx 0.589 \]
Why “3π/16” Is Correct
Circle to square = 3π/16 ≈ 0.59. The square of side D has a much larger section modulus than the circle inscribed in it, because it has more material far from the neutral axis.
| 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. 2π/15: 2π/15 does not come from either section modulus.
- C. 3π/8: 3π/8 is twice the correct ratio.
- D. π/16: π/16 is the polar section modulus factor of a shaft (Zp = πD³/16), not this ratio.
Key Point
Z_circle/Z_square (B = D) = 3π/16 ≈ 0.59; square-to-circle = 16/(3π) ≈ 1.70.
Key Concepts for Students
- Z circle = πd³/32; Z rectangle = bd²/6.
- For equal areas, a square is about 18 % stronger than a circle.
- Read carefully which ratio is asked — inverting it is a common trap.
Practice more: this is question 16 in Bending and Shear Stress MCQs – Part 2. Try the whole set and check every answer instantly.