Polar Moment Calculator

Background.

A drive shaft, a socket extension and a bicycle's crank axle share one job: carrying a twisting load without twisting too far or yielding. How well a round bar does that depends on its material and on a single geometric property of its cross-section, the polar second moment of area, J. The material decides how much shear stress the metal can take; J decides how much stress and twist a given torque produces in it.

The defining feature of J is that it grows with the fourth power of the diameter. A shaft twice as thick has sixteen times the polar moment, so under the same torque it twists a sixteenth as much and its peak shear stress falls to an eighth.

The same property explains why hollow shafts work so well. Material contributes to J in proportion to the square of its distance from the axis, so metal near the centre does very little. Bore out the middle half of the diameter of an 80 mm shaft and you remove a quarter of the steel but only 6.25% of the polar moment, which is why vehicle drive shafts are usually made as tubes.

The page computes J for a solid circle or a concentric ring from outer and inner diameters in metres, and returns m⁴. It stops at geometry. Stress, twist and strength also need the torque, the length and the material's shear modulus, and none of them is calculated here. Nor is this the mass moment of inertia used for flywheels, which is measured in kg·m². For any shape other than a circle or circular tube, J is the wrong quantity for twisting altogether; such sections use a separate torsion constant that is smaller, often far smaller, than the polar moment.

What is polar moment calculator?

The polar second moment of area, J, describes how a cross-section's area is spread around the axis perpendicular to it: each small piece of area is multiplied by the square of its distance from the centre, and the products are summed. For circles and circular tubes it equals the sum of the two ordinary second moments about perpendicular diameters, J = Ix + Iy, and it is the stiffness term in the torsion equations for round shafts. Its unit is length to the fourth power, here m⁴.

How to use this calculator.

1. 01Enter Outer diameter in metres. Divide a millimetre measurement by 1,000, so an 80 mm bar is 0.08.
2. 02Enter Inner diameter as the bore of a tube in metres, or 0 for a solid bar. It must be smaller than the outer diameter; if you know the wall thickness t instead, the bore is the outer diameter minus 2t.
3. 03Read J in m⁴. Shaft values look tiny in these units, so multiply by 10¹² for mm⁴ or by 10⁸ for cm⁴, the units most section tables use.
4. 04Take J into the torsion formulas: maximum shear stress τ = T·r ÷ J at the outer surface, and angle of twist θ = T·L ÷ (G·J), with T the torque, r the outer radius, L the length and G the shear modulus.

The formula.

J = π(Do⁴ − Di⁴) ÷ 32, where Do and Di are the outer and inner diameters. For a solid circle, summing r² over the whole area gives πD⁴ ÷ 32, the same as πR⁴ ÷ 2 written with the radius. A tube is the large circle with the small one taken away, so its J is simply the difference of the two. The fourth power dominates everything: a 1% error in the measured outer diameter becomes about a 4% error in J. The bore matters far less, because its fourth power is subtracted from a much larger number; when the bore is half the outer diameter, Di⁴ is only a sixteenth of Do⁴. The page needs a positive outer diameter and an inner diameter of zero or more that is strictly smaller than the outer one.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Beer et al., Mechanics of Materials, 8th ed. (PRINT).
2. [2]MIT OpenCourseWare, Mechanics of Materials, torsion of circular shafts.
3. [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.