Area Moment of Inertia: Solid Round Bar

A circle is the same in every direction, so a solid round bar has the same I about every axis through its centre: πd⁴/64. A 100 mm bar gives I = π × 0.1⁴ ÷ 64 = 4.909 × 10⁻⁶ m⁴. The fourth power is brutal in both directions: a 2-inch shaft is sixteen times stiffer in bending than a 1-inch shaft, and a 10% undersize bar has lost a third of its stiffness.

That fourth power is also the argument for hollow sections. Subtracting the inner circle gives I = π(d⁴ − dᵢ⁴)/64, and because the removed core sits close to the neutral axis it was contributing almost nothing: bore a 100 mm bar out to 50 mm and you lose 6% of the stiffness while shedding 25% of the weight. It is why scaffold tube, bicycle frames and drill pipe are all round and hollow. Remember the value here is in m⁴, convert with 1 in⁴ = 4.162314 × 10⁻⁷ m⁴, and do not confuse it with the polar moment J = πd⁴/32, which is exactly twice as large and belongs to torsion, not bending.