How this instrument works
The torsion constant J measures how strongly a shaft's cross-section resists twisting about its own axis — the rotational counterpart to the area moment of inertia used in bending. For a circular section, solid or hollow, J is found by integrating r² over every scrap of area in the cross-section: each ring of material at radius r contributes area 2πr·dr and torsional resistance r²·(2πr·dr), and the whole disc integrates out to πD⁴⁄32. A hollow shaft simply subtracts the inner disc that was never there.
That fourth-power dependence on diameter is why torsion constant grows so fast with size and so slowly with wall thickness. Shear stress under torque is τ = Tr⁄J, rising in a straight line from zero at the centreline to a maximum at the outer surface — material near the axis sits at low stress and does little structural work, while material at the rim does the most of it. Boring out the centre of a solid bar, as the hollow term in this formula does, removes the least useful material first, which is exactly why marine propeller shafts, aircraft torque tubes, and automotive half-shafts are so often built as tubes rather than solid rounds.
The formula only applies to circular cross-sections, and that limit is worth respecting. For a square, rectangular, or I-shaped bar, the torsion constant used in the twist equation is smaller than the polar moment of inertia and has no closed algebraic form — engineers pull it from tables or run a numerical solution such as the membrane analogy or a finite-element model instead. Confusing the two is a genuine and common error: the polar moment of inertia and the torsion constant coincide only for round shafts, which is precisely the case this instrument covers.
- Enter the shaft's Outer diameter — the full diameter measured across the tube or bar.
- Enter the Inner diameter (0 for solid shaft) — the bore diameter if the shaft is hollow, or leave it at 0 for a solid round.
- Read Torsion constant, m⁴ (J) — the shape factor used in τ = Tr⁄J for stress and θ = TL⁄(GJ) for angle of twist.
- Try Din = 0 first to see the solid-shaft baseline, then raise it toward Dout to see how little stiffness a thin wall actually gives up.
Worked example — a 50 mm hollow steel drive shaft
Take a hollow steel tube with a 50 mm outer diameter and a 40 mm inner diameter — a 5 mm wall all the way round, a common propeller-shaft or torque-tube proportion. In metres, D_out⁴ = 0.05⁴ = 6.25×10⁻⁶ m⁴ and D_in⁴ = 0.04⁴ = 2.56×10⁻⁶ m⁴. The formula gives J = π(6.25×10⁻⁶ − 2.56×10⁻⁶) ⁄ 32 = π(3.69×10⁻⁶) ⁄ 32 = 3.6226×10⁻⁷ m⁴, or 362,265 mm⁴ if the smaller unit reads easier.
For comparison, a solid 50 mm round of the same outer diameter works out to J = π(0.05⁴) ⁄ 32 = 6.1359×10⁻⁷ m⁴ — about 70% more than the tube. But the solid bar carries nearly three times the cross-sectional area (1.963×10⁻³ m² against 7.069×10⁻⁴ m² for the tube), so the hollow shaft delivers 59% of the solid bar's torsional stiffness for 36% of the material — the quantitative reason hollow shafts show up wherever weight matters as much as strength.
Questions
Why does a hollow shaft carry almost as much torque as a solid one?
Because shear stress from torque rises in a straight line from zero at the shaft's centreline to a maximum at its outer surface (τ = Tr⁄J), so material near the axis is barely stressed and contributes little strength for its weight. Boring it out removes mostly idle material: the 50/40 mm tube above keeps 59% of a solid 50 mm shaft's torsion constant while using only 36% of its cross-sectional area.
Is the torsion constant the same thing as the polar moment of inertia?
Only for circular cross-sections, which is the one case where the two happen to coincide — π(D_out⁴ − D_in⁴)⁄32 is both at once. For any non-circular section such as a square, rectangle, or I-beam, the torsion constant used in the twist equation is smaller than the polar moment of inertia and has no simple closed formula; it comes from tables or a numerical solution such as the membrane analogy or finite elements.
What do I enter for a solid shaft?
Leave the Inner diameter (0 for solid shaft) field at 0. The formula then reduces to J = πD_out⁴⁄32, the standard solid-round torsion constant, since subtracting a zero-diameter hole changes nothing about the result.
Where does J get used after I calculate it?
In the two standard torsion equations for a circular shaft: shear stress τ = Tr⁄J, where r is usually the outer radius and T the applied torque, and angle of twist θ = TL⁄(GJ), where L is the shaft length and G the material's shear modulus. J itself depends only on geometry — you still need T, G, and L to get a stress or a twist angle out of it.
Does this formula work for square or rectangular shafts?
No. It is exact only for solid or hollow circular cross-sections. A square bar of side a has a torsion constant of roughly 0.1406a⁴, not πa⁴⁄32, because torque-induced shear no longer varies smoothly with radius once corners are involved; rectangular and other open sections need their own formulas or a numerical solve.
Why does J scale with the fourth power of diameter?
Because torsional resistance is the integral of r² over the cross-sectional area, and the area itself scales with r²; multiplying those two contributions together gives a dependence on the fourth power of the radius, and so of the diameter. It is the same reason bending stiffness, the area moment of inertia, also scales as the fourth power of a beam's depth.