The moment of inertia of a circular section of diameter d about its diameter is:
Try it first: pick an option, then read the explanation.
Correct answer: D. πd⁴/64
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
The polar moment of inertia of a circle is easy to integrate using thin rings. Two perpendicular diameters share it equally.
\[ I_p = \int_0^{R} r^2\,(2\pi r\,dr) = \frac{\pi R^4}{2} = \frac{\pi d^4}{32} \]
\[ I_{xx} = I_{yy} = \frac{I_p}{2} = \frac{\pi d^4}{64} \]
Why “πd⁴/64” Is Correct
About a diameter, I = πd⁴/64.
| 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. πd³/16: πd³/16 is the polar section modulus (Zp) of a solid shaft.
- B. πd⁴/32: πd⁴/32 is the polar moment of inertia J.
- C. πd³/32: πd³/32 is the section modulus Z, not I.
Key Point
Circle: I = πd⁴/64, J = πd⁴/32, Z = πd³/32, Zp = πd³/16.
Key Concepts for Students
- Perpendicular axis theorem: I_z = I_x + I_y (plane areas).
- Units: I in mm⁴, Z in mm³.
- Hollow circle: I = π(D⁴ − d⁴)/64.
Practice more: this is question 32 in Bending and Shear Stress MCQs – Part 3. Try the whole set and check every answer instantly.