Question

The moment of inertia of a hollow circular section whose external diameter is 8 cm and internal diameter is 6 cm, about centroidal axis is.
1. 437.5 cm⁴
2. 337.45 cm⁴
3. 237.5 cm⁴
4. 137.45 cm⁴

Answer 1

Correct Answer - Option 4 : 137.45 cm⁴

Concept:

Moment of inertia is the sum of the product of mass of each particle with the square of its distance from the axis of the rotation.

M.O.I for hollow circular cross-section \( = \frac{\pi }{{64}}\left( {d_0^4 - d_i^4} \right)\) 

Where d₀ = outer diameter of cross section

d_i = inner diameter of cross section

Calculation:

d₀ = 8 cm

d_i = 6 cm

\(I = \frac{\pi }{{64}}\left( {{8^4} - {6^4}} \right)\)

= 137.45 cm⁴

Other Important Points:

Formula of moment of inertia for various other figures is given below.

S.No.	Shape of cross-section	INA	Ymax	Z
1	Rectangle	\(I = \frac{{b{d^3}}}{{12}}\)	\({Y_{max}} = \frac{d}{2}\)	\(Z = \frac{{b{d^2}}}{6}\)
2	Circular	\(I = \frac{\pi }{{64}}{D^4}\)	\({Y_{max}} = \frac{d}{2}\)	\(Z = \frac{\pi }{{32}}{D^3}\)
3	Triangular	\(I = \frac{{B{h^3}}}{{36}}\)	\({Y_{max}} = \frac{{2H}}{3}\)	\(Z = \frac{{B{H^3}}}{{24}}\)