MecMovies 4.0 : M6.2: Torsion Theory for Circular Sections

scene 17 of 21

To derive the polar moment of inertia for a solid circular shaft, consider a ring-shaped area $dA$ having a circumference of $2 πρ$ and a thickness of $dρ$ . Evaluate this integral to find the polar moment of inertia $J$ for a solid circular cross section:

$J = \int_{A} ρ^{2} dA = \int_{0}^{R} ρ^{2} (2 πρ) dρ = 2 π \int_{0}^{R} ρ^{3} dρ = \frac{2 π}{4} {[ρ^{4}]}_{0}^{R} ∴ J = \frac{{πR}^{4}}{2} = \frac{{πD}^{4}}{32}$