Moment of Inertia of a Hollow Circle
Tubes & Pipes • I, Polar Moment J & Section Modulus
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Moment of Inertia of a Hollow Circle — Formula and Worked Example
A tube removes material from where it contributes least — near the neutral axis — and keeps it where the r² weighting matters most. That is why a hollow shaft is dramatically stiffer per unit mass than a solid one of the same weight.
Formulas
| Quantity | Expression |
|---|---|
| Area | A = π(D² − d²) / 4 |
| Second moment (any centroidal axis) | I = π(D⁴ − d⁴) / 64 |
| Polar second moment | J = π(D⁴ − d⁴) / 32 |
| Section modulus | S = 2I / D |
| Radius of gyration | r = √(D² + d²) / 4 |
The fourth-power terms subtract, so removing the middle costs surprisingly little stiffness. A 60/40 mm tube keeps about 80% of the I of a 60 mm solid bar while using only 56% of the material. Setting d = 0 correctly recovers the solid-circle formulas.
Worked Example — A 60 mm outside / 40 mm inside tube
| Quantity | Value |
|---|---|
| D (outside) | 60 mm |
| d (inside) | 40 mm |
| A | 1,571 mm² |
| I | 510.5 × 10³ mm⁴ |
| J = 2I | 1,021.0 × 10³ mm⁴ |
| S = 2I/D | 17.0 × 10³ mm³ |
| I retained vs 60 mm solid | 80% |
| material used vs solid | 56% |
Values computed by the calculator above. Change any dimension and every property updates live.
What is the moment of inertia of a hollow tube?
I = π(D⁴ − d⁴)/64, where D is the outside diameter and d the inside diameter. The polar moment is twice that, J = π(D⁴ − d⁴)/32.
Why is a hollow shaft more efficient than a solid one?
Second moment weights area by the square of its distance from the axis, so material near the centre barely contributes. Removing it saves a lot of mass for very little loss of I or J. A 60/40 mm tube retains roughly 80% of the stiffness of a 60 mm solid bar with 56% of the material.
Can I use these formulas for a thin-walled tube?
Yes, they are exact for any wall thickness. For a very thin wall the approximation I ≈ πr³t (with r the mean radius and t the wall) is often quicker and is accurate to within a few percent once D/t exceeds about 20.
Why Removing the Core Costs So Little
Because I weights area by the square of its distance from the axis, material near the centre contributes almost nothing. Boring out a 60 mm bar, holding the outside diameter constant:
| Bore (mm) | A (mm²) | A kept | I (mm⁴) | I kept | Stiffness per unit mass |
|---|---|---|---|---|---|
| solid | 2,827 | 100% | 636,173 | 100% | 1.00 |
| 20 | 2,513 | 89% | 628,319 | 99% | 1.11 |
| 30 | 2,121 | 75% | 596,412 | 94% | 1.25 |
| 40 | 1,571 | 56% | 510,509 | 80% | 1.44 |
| 45 | 1,237 | 44% | 434,884 | 68% | 1.56 |
| 50 | 864 | 31% | 329,376 | 52% | 1.69 |
The last column is the point: at a 40 mm bore the tube keeps 80% of the stiffness with 56% of the material, so stiffness per unit mass is 1.44× that of the solid bar. Push to a 50 mm bore and it reaches 1.69×. The limit is set by local buckling and by wall thickness practicalities, not by the maths.
Thin-Wall Approximation
When the wall is thin relative to the radius, I ≈ πr³t is often quicker, where r is the mean radius and t the wall thickness. For the 60/40 tube above, r = 25 mm and t = 10 mm gives π(25)³(10) = 490,900 mm⁴ against the exact 510,509 mm⁴ — about 4% low. The approximation tightens rapidly as D/t grows and is within 1% once D/t exceeds roughly 20.
Why does a bicycle frame use tubes rather than solid rods?
For the same reason: a frame must resist bending and torsion, both of which depend on material being far from the axis. A tube delivers nearly the same I and J as a solid rod of the same outside diameter at roughly half the mass. The trade-off is that thin walls become vulnerable to local denting and buckling, which is why tube walls are butted — thicker at the ends where stresses concentrate.
Other Cross-Sections
The same calculator handles all eight standard sections:
For the full section-property reference and the parallel-axis theorem, see the Moment of Inertia Calculator hub.