The moment of inertia of a triangular section about an axis through its centre of gravity (parallel to the base) is:
Try it first: pick an option, then read the explanation.
Correct answer: B. bh³/36
Asked in SSC JE 02/03/2017 (Morning)
Detailed Explanation
The standard result is about the base, bh³/12. Shift it to the centroid (h/3 above the base) using the parallel axis theorem.
\[ I_G = I_{base} - A\left(\frac{h}{3}\right)^2 = \frac{bh^3}{12} - \frac{bh}{2}\cdot\frac{h^2}{9} = \frac{bh^3}{12} - \frac{bh^3}{18} = \frac{bh^3}{36} \]
Why “bh³/36” Is Correct
About the centroidal axis parallel to the base, I = bh³/36.
| 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. bh³/12: bh³/12 is about the base of the triangle (or the centroidal axis of a rectangle).
- C. bh³/32: bh³/32 does not correspond to the triangle.
- D. none of these: bh³/36 is listed.
Key Point
Triangle: I_G = bh³/36, I_base = bh³/12, I_apex = bh³/4.
Key Concepts for Students
- Parallel axis theorem: I = I_G + Ad².
- Centroid of a triangle is h/3 from the base.
- Z of a triangle (to the apex) = bh²/24.
Practice more: this is question 28 in Bending and Shear Stress MCQs – Part 2. Try the whole set and check every answer instantly.