Polar Moment of Inertia: Solid Shaft

Torsion is bending's rotational twin, and J plays the role I plays in bending: it measures how far the section's area lies from the shaft's centre rather than from a line through it. By the perpendicular-axis theorem $J = I_x + I_y$, and for a circle those two are equal, so J = 2I = πd⁴/32 exactly. A 50 mm shaft has J = π × 0.05⁴ ÷ 32 = 6.136 × 10⁻⁷ m⁴.

The same fourth-power leverage applies: go from a 50 mm to a 60 mm shaft and torsional stiffness rises by (60/50)⁴ = 2.07, more than double for a 20% size step. Hollow shafts win even harder here than in bending, since torsional shear is zero at the centre and maximum at the surface: the core is dead weight, which is why propeller shafts and torque tubes are bored out. Two warnings: J is returned in m⁴ as a plain number (1 in⁴ = 4.162314 × 10⁻⁷ m⁴), and this simple formula holds only for circular sections. Saint-Venant showed in 1855 that non-circular bars warp out of plane when twisted, and a square bar's torsional constant is about 0.844 (a/2)⁴ × 2.25, not its polar moment at all.