Polar Moment Hollow⚠ unverified
Mechanical / Shafts · Compute the polar moment of inertia for a hollow circular shaft
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| do | do | m | 1.0 | Outer diameter |
| di | di | m | 1.0 | Inner (bore) diameter |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | J | m^4 | Polar moment of inertia, in metres^4 (m^4) |
The science & history
Understanding the Parameters
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Outer diameter $d_o$ — dominates $J$ through its fourth power. Torsional capacity is set almost entirely by how far the outermost material sits from the axis, so a modest increase in outer diameter is far more effective than adding wall thickness inward.
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Inner diameter $d_i$ — subtracts the bore's contribution, also as a fourth power. Because $d_i^4$ grows slowly until $d_i$ is a large fraction of $d_o$, a generous bore removes a lot of weight (area $\propto d_i^2$) while sacrificing comparatively little $J$ ($\propto d_i^4$). That mismatch is the whole efficiency argument for hollow shafts.
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Strength-to-weight — a tube with $d_i/d_o = 0.7$ removes ~49 % of the area but only ~24 % of the $J$: nearly the same torsional strength at half the weight.
Derivation (Approaching a Proof)
The polar moment is additive over area, so the hollow section is simply the solid outer circle with the inner circle removed. Using the solid-shaft result $J_{\text{solid}} = \pi d^4/32$ (see Polar Moment solid shaft) for each diameter:
$$J = \int_A r^2\,\mathrm{d}A = J_{\text{outer disc}} - J_{\text{bore disc}} = \frac{\pi d_o^4}{32} - \frac{\pi d_i^4}{32} = \frac{\pi (d_o^4 - d_i^4)}{32}.$$
Equivalently, integrate the ring element $\mathrm{d}A = 2\pi r\,\mathrm{d}r$ directly between the inner and outer radii $R_i = d_i/2$ and $R_o = d_o/2$:
$$J = 2\pi\int_{R_i}^{R_o} r^3\,\mathrm{d}r = \frac{\pi}{2}\left(R_o^4 - R_i^4\right) = \frac{\pi(d_o^4 - d_i^4)}{32}.$$
Both routes give the same result because the second moment of area superposes: the material that isn't there (the bore) simply doesn't contribute to the integral. Setting $d_i = 0$ recovers the solid shaft.
Dimensional check. $[J] = \text{m}^4$. ✓
History and Development
The hollow-shaft result follows directly from the same 19th-century torsion theory (Coulomb, Saint-Venant) as the solid case, via the additivity of the area integral. Its practical importance grew with weight-critical machinery — automotive and aircraft drive shafts, propeller shafts, and bicycle frames — where the torsional-strength-per-weight advantage of tubes is decisive. Tabulated in Shigley and Machinery's Handbook.
Related Concepts: Polar Moment solid shaft, Torsional Shear Stress, Shaft Diameter torsion, Rectangular Moment of Inertia, Critical Speed Shaft
Notes: Hollow round section. Set $d_i = 0$ to recover the solid shaft. Torsional stress in a tube is still $\tau = Tr/J$ with $r = d_o/2$ at the outer surface (see Torsional Shear Stress).