Moment of Inertia of a Circle
I = πd⁴/64 • Polar Moment J & Section Modulus
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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
| Quantity | Expression |
|---|---|
| Area | A = πd² / 4 |
| Second moment (any centroidal axis) | I = πd⁴ / 64 |
| Polar second moment | J = Iₓ + Iₖ = πd⁴ / 32 |
| Section modulus | S = πd³ / 32 |
| Radius of gyration | r = d / 4 |
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
| Quantity | Value |
|---|---|
| d (diameter) | 50 mm |
| A | 1,963 mm² |
| I = πd⁴/64 | 306.8 × 10³ mm⁴ |
| J = πd⁴/32 | 613.6 × 10³ mm⁴ |
| S = πd³/32 | 12.3 × 10³ mm³ |
| r = d/4 | 12.5 mm |
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 = ∫₀r ρ²(2πρ) dρ = 2π[ρ⁴/4]₀r = 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
| d (mm) | A (mm²) | I (mm⁴) | S (mm³) | J (mm⁴) |
|---|---|---|---|---|
| 10 | 79 | 491 | 98 | 982 |
| 16 | 201 | 3,217 | 402 | 6,434 |
| 20 | 314 | 7,854 | 785 | 15,708 |
| 25 | 491 | 19,175 | 1,534 | 38,350 |
| 30 | 707 | 39,761 | 2,651 | 79,522 |
| 40 | 1,257 | 125,664 | 6,283 | 251,327 |
| 50 | 1,963 | 306,796 | 12,272 | 613,592 |
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.
What is the moment of inertia of a semicircle or quarter circle?
About its own centroidal axis a semicircle of radius r has I ≈ 0.1098r⁴, with the centroid 4r/3π from the flat edge. About the flat edge itself it is πr⁴/8. The difference is the parallel-axis transfer term, and forgetting it is the usual source of error in composite sections built from half-rounds.
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, see the Moment of Inertia Calculator hub.