Polar Moment of Inertia Formula: J = πd⁴/32 for Shafts

The polar moment of inertia formula for a solid circular shaft is $J = \frac{\pi d^4}{32}$ , which is the same as $\frac{\pi r^4}{2}$ using the radius. For a hollow shaft with outer diameter D and inner diameter d it becomes $J = \frac{\pi (D^4 - d^4)}{32}$ . The unit is metre to the fourth power (m⁴).

This is an area property of a cross section, used in strength of materials to find how a shaft resists twisting. In general $J = \int r^2\, dA$ , the sum of each small area times the square of its distance from the axis through the centroid. It looks like the mass moment of inertia with area in place of mass, which is where the name comes from.

Quick facts

Solid shaft
$J = \frac{\pi d^4}{32} = \frac{\pi r^4}{2}$
Hollow shaft
$J = \frac{\pi(D^4 - d^4)}{32}$
General
$J = \int r^2\, dA = I_x + I_y$
SI unit
m⁴ (often mm⁴ in design tables)
Dimensional formula
[M⁰ L⁴ T⁰]
Torsion equation
$\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}$

Polar moment of inertia formula from the perpendicular axis rule

For a flat cross section, take x and y axes in its plane through the centroid. The perpendicular axis theorem gives $J = I_x + I_y$ . For a circle, $I_x = I_y = \frac{\pi d^4}{64}$ , so $J = 2 \times \frac{\pi d^4}{64} = \frac{\pi d^4}{32}$ .

You can also integrate directly. A thin ring of radius ρ and width dρ has area $2\pi\rho\,d\rho$ . Then $J = \int_0^R \rho^2 \cdot 2\pi\rho\,d\rho = \frac{\pi R^4}{2}$ . With R = d/2 this becomes $\frac{\pi d^4}{32}$ , the same result.

Two kinds of polar moment

Area polar moment versus mass polar moment
Feature	Area polar moment J	Mass moment about the shaft axis
Defined as	$\int r^2\,dA$	$\int r^2\,dm$
Unit	m⁴	kg m²
Solid round section	$\frac{\pi r^4}{2}$	$\frac{1}{2}MR^2$ for a solid cylinder
Used for	Twisting stress and angle of twist	Angular acceleration under torque

Engineering texts often say polar moment of inertia when they mean the area quantity J. Physics texts usually mean the mass quantity. Check the unit in the question: m⁴ means area, kg m² means mass.

Worked examples

Example 1: a solid steel shaft

Find J for a solid shaft of diameter 50 mm.

d = 0.050 m, so $d^4 = 6.25 \times 10^{-6}\ \text{m}^4$ .

$J = \frac{\pi \times 6.25 \times 10^{-6}}{32} \approx 6.14 \times 10^{-7}\ \text{m}^4$

Example 2: a hollow tube

A tube has outer diameter 60 mm and inner diameter 40 mm. Find J.

$D^4 = (0.06)^4 = 1.296 \times 10^{-5}$ , $d^4 = (0.04)^4 = 2.56 \times 10^{-6}$ . Difference: $1.04 \times 10^{-5}\ \text{m}^4$ .

$J = \frac{\pi \times 1.04 \times 10^{-5}}{32} \approx 1.02 \times 10^{-6}\ \text{m}^4$

The tube uses less metal than a solid 60 mm bar yet keeps most of its stiffness, because the material near the centre contributes little to J.

Example 3: shear stress under a torque

The shaft of Example 1 carries a torque of 500 N m. Find the largest shear stress, at the surface.

From the torsion equation, $\tau_{max} = \frac{Tr}{J} = \frac{500 \times 0.025}{6.14 \times 10^{-7}} \approx 2.04 \times 10^7\ \text{Pa}$ , about 20.4 MPa.

Where J is used

The torsion equation $\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}$ links the applied torque T, the shear stress τ at radius r, the modulus of rigidity G, and the angle of twist θ over a length L. A larger J means lower stress and less twist for the same torque. Drive shafts, axles and the twisting wire of a torsion pendulum are all designed this way.

Because J grows with $d^4$ , doubling the diameter makes the shaft sixteen times stiffer in twisting. That fourth power is the main lesson of the formula.

Use the formula only for circular sections. Square or I shaped sections need a different torsion constant.

Mistakes to watch for

Using the radius in the diameter form, which makes the answer 16 times too big.
Leaving millimetres in. 50 mm is 0.05 m, and the fourth power magnifies any slip.
Writing $\frac{\pi(D - d)^4}{32}$ for a hollow shaft. Subtract the fourth powers, not the diameters.
Mixing up area moment (m⁴) and mass moment (kg m²).

The radius of gyration gives a single length summary of either kind of moment, and the rotation hub lists more turning formulas.

Frequently asked questions

What is the formula for polar moment of inertia?

For a solid circular section J = πd⁴ / 32 = πr⁴ / 2. For a hollow circular section J = π(D⁴ − d⁴) / 32.

What is the unit of polar moment of inertia?

For the area quantity used in shafts it is m⁴, often quoted in mm⁴. The mass version has unit kg m².

How is polar moment related to Iₓ and Iᵧ?

By the perpendicular axis theorem, J = Iₓ + Iᵧ for axes in the plane of the section through its centroid.

Why are hollow shafts used?

Most of J comes from material far from the centre. A hollow shaft keeps that outer material and removes the inner part, saving weight while losing little stiffness.