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

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The figure shows a ring-shaped area d A plotted on the radial surface of a shaft with radius R. The radius of the ring is rho and the circumference is 2 pi rho. The thickness of the ring is d ρ. d A = 2 pi rho times d P.

To derive the polar moment of inertia for a hollow circular cross section, the value of $ρ$ ranges between the inside radius $r$ and the outside radius $R$ .

$J = \int_{A} ρ^{2} dA = \int_{r}^{R} ρ^{2} (2 πρ) dρ = 2 π \int_{r}^{R} ρ^{3} dρ = \frac{2 π}{4} {[ρ^{4}]}_{r}^{R} ∴ J = \frac{π}{2} (R^{4} - r^{4}) = \frac{π}{32} (D^{4} - d^{4})$