Moment of Inertia of a Circle

Moment of Inertia of a Circle — Formula and Worked Example

A solid circular section has the same second moment about every centroidal axis, because it is symmetric in every direction. That makes it the natural choice for shafts, which are loaded in bending from continually changing directions as they rotate.

Formulas

Because Iₓ = Iₖ, the polar moment is simply J = 2I. J is what you need for torsion (τ = Tr/J); I is what you need for bending (σ = My/I). Mixing them up is one of the most common errors in shaft design.

Worked Example — A 50 mm diameter solid circle

Values computed by the calculator above. Change any dimension and every property updates live.

What is the moment of inertia of a circle?

For a solid circle of diameter d, I = πd⁴/64 about any axis through the centre. In terms of radius that is πr⁴/4.

What is the difference between I and J for a circle?

I is the second moment about a diameter and governs bending. J is the polar second moment about the axis through the centre and governs torsion. For a circle J = Ix + Iy = 2I = πd⁴/32.

Why are shafts circular?

A rotating shaft is bent from every direction in turn, and a circle is the only shape whose second moment is identical about every centroidal axis. It also has no stress-concentrating corners and is straightforward to machine and seal.

Where πd⁴/64 Comes From

It is easiest to find the polar moment first. Using an annular element of radius ρ and thickness dρ, dA = 2πρ dρ, so

J = ∫₀^r ρ²(2πρ) dρ = 2π[ρ⁴/4]₀^r = πr⁴/2 = πd⁴/32

By symmetry Iₓ = Iₖ, and since J = Iₓ + Iₖ it follows immediately that each is half of J, giving I = πd⁴/64. That symmetry argument is quicker than integrating the bending case directly.

Common Shaft Diameters

Note how fast these grow: going from 25 mm to 50 mm multiplies the area by 4 but I and J by 16, because they scale with d⁴. Doubling a shaft diameter is a very expensive way to add mass and a very cheap way to add stiffness.

Moment of Inertia of a Circle About a Tangent

A tangent is a line touching the edge, parallel to a diameter and a distance r from the centre. The parallel-axis theorem adds A·r² to the centroidal value:

I_tangent = πr⁴/4 + (πr²)(r²) = 5πr⁴/4 = 5πd⁴/64

That is five times the value about a diameter. For the 50 mm circle above it is 1,533,981 mm⁴.

Semicircle, Quarter Circle and Other Circular Sections

The centroid of a semicircle is 4r/3π from the flat edge, and of a quarter circle 4r/3π from each straight edge. The centroidal values are the edge values minus the parallel-axis term A·(4r/3π)²; forgetting that term is the usual error in composite sections built from half-rounds.

Worked Example in Inches — a 2 in diameter round bar

With d = 2 in: I = π(2)⁴/64 = 0.7854 in⁴, J = π(2)⁴/32 = 1.5708 in⁴, S = π(2)³/32 = 0.7854 in³ and r = d/4 = 0.5 in. Switch the calculator to Imperial and it works in inches throughout.

Is the moment of inertia of a circle the same about x and y?

Yes. A circle is symmetric about every diameter, so I is the same about every axis through the centre: Iₓ = Iₖ = πd⁴/64, and the product of inertia is zero. That is why the polar moment is exactly twice it.

Why is it πd⁴/64 and not πr⁴/4?

They are the same number. Substituting r = d/2 gives π(d/2)⁴/4 = πd⁴/64. Engineers usually quote the diameter form because shafts and bars are specified by diameter.

What is J for a circle?

J = πd⁴/32 = πr⁴/2, the polar moment about the axis through the centre. It is used in the torsion formula τ = Tr/J; try it in the shaft torsion simulator.

Other Cross-Sections

The same calculator handles all eight standard sections:

• I-Beam
• Channel
• Angle
• T-Section
• Rectangle
• Circle (this page)
• Hollow Circle
• Hollow Rect

For the full section-property reference and the parallel-axis theorem, use the moment of inertia calculator for any cross-section.

Updated 4 October 2026