How this instrument works
The angle of twist, θ, is how far one end of a shaft rotates relative to the other while torque runs through it — a wind-up, not a bend. The formula θ = TL ⁄ (GJ) puts torque and length on top because doubling either one doubles the rotation: twice the twisting load, or twice the distance it acts over, both double θ. Shear modulus G and polar moment of inertia J sit underneath because they measure resistance — G is how stiffly the material itself resists distorting under shear, and J is how much of the cross-section's area sits far from the centre, where it does the most work fighting rotation.
That denominator, GJ, is called torsional rigidity, and the formula behind it rests on one geometric assumption: a flat cross-section stays flat and simply rotates as you move along the shaft. That is true for solid or hollow circular shafts and false for almost any other shape. Barré de Saint-Venant worked out the general torsion problem in 1855 and showed that square, rectangular and I-beam sections warp out of their own plane under twist — which is why GJ only ever stands alone for round bars, while other shapes need a separate torsion constant in its place. Inside a circular shaft, shear stress climbs in a straight line from zero at the centre to a maximum at the outer surface, τ = Tr ⁄ J, and Hooke's law in shear, τ = Gγ, is what ties that stress back to the strain that adds up into θ.
The formula assumes G and J stay constant along the shaft and that the material never leaves its linear-elastic range. A stepped shaft, one whose diameter changes partway along, needs its twist worked out segment by segment and summed — θ = ΣTᵢLᵢ ⁄ (GᵢJᵢ) — because that kind of superposition only holds while everything stays linear. Push the shear stress past the material's yield point and the relationship curves: plastic torsion follows Nadai's sand-heap analogy rather than a straight-line ratio, and past that point this formula, like the calculator built on it, is no longer the right tool.
- Enter the twisting load into Applied torque — newton-metres by default, or switch to ft·lb or in·lb if that's how your torque wrench reads.
- Enter the span the twist builds up over into Shaft length, the distance between the two points whose rotation you're comparing.
- Enter the material's stiffness into Shear modulus — structural steel sits near 79,300 MPa, aluminium nearer 26,000 MPa — in kPa or MPa.
- Enter the cross-section's resistance to twisting into Polar moment of inertia, m⁴; for a solid round shaft, J = πd⁴⁄32.
- Read Angle of twist in degrees, or switch its unit to radians if you're feeding the result into a further calculation.
Worked example — a 50 N·m torque on a 1 m steel shaft
A machine shop is checking a 1 m length of solid steel shaft, shear modulus 79.3 GPa, with a polar moment of inertia of 1.5 × 10⁻⁶ m⁴ — equivalent to a solid round cross-section about 62.5 mm across. A 50 N·m torque is applied at one end. Enter 50 into Applied torque, 1 into Shaft length, 79300 into Shear modulus (its MPa unit), and 0.0000015 into Polar moment of inertia, m⁴.
The formula gives θ = 50 × 1 ⁄ (79,300,000,000 × 0.0000015) = 50 ⁄ 118,950 = 0.00042034468264 rad, which the readout converts to 0.0241°. That is a wind-up too small to see — roughly the angular width of a coin viewed from 40 metres away — which is exactly the point of stiff structural shafting: it should carry torque without visibly moving.
Scale matters more than it looks. At this torque and cross-section, the shaft would need to run more than 41 metres before twist reached a full degree. A common shaft-design guideline caps torsional deflection at roughly 1° over a length of twenty diameters — 1.25 m for this shaft — and at that span the same torque produces only about 0.03°, near 3% of that conventional limit, meaning geometry this stiff sits nowhere near where twist becomes a real design concern.
Questions
Does this formula work for any shaft cross-section?
No — only for solid or hollow circular shafts. The derivation assumes a flat cross-section stays flat and simply rotates as you move along the shaft, which only holds for round bars. Barré de Saint-Venant showed in 1855 that square, rectangular and I-beam sections warp out of plane under torque, so GJ has to be replaced with a separate torsion constant for anything that isn't circular.
Why do torque and length multiply while shear modulus and polar moment divide?
Torque and length are what drive the twist — apply more torque, or let it act over a longer shaft, and rotation grows in direct proportion, so both sit on top. Shear modulus and polar moment measure resistance — a stiffer material or a beefier cross-section fights the twist harder, so both sit on the bottom, shrinking θ as either one grows.
What if the shaft changes diameter or material partway along its length?
Split it into segments and add the results: θ_total = ΣTᵢLᵢ ⁄ (GᵢJᵢ), one term per section using that section's own torque, length, shear modulus and polar moment. This works because the underlying relationship is linear — it breaks down only once a segment is stressed past its own elastic limit.
How is polar moment of inertia different from the moment of inertia used for bending?
Polar moment of inertia, J, measures resistance to twisting about the shaft's own axis; area moment of inertia, I, measures resistance to bending about a perpendicular axis instead. For a solid or hollow circular section the two are related simply, J = 2I, a consequence of the perpendicular axis theorem — but that shortcut only holds for circles.
Is there a practical limit on how much twist a shaft should have?
Function dictates it more than any universal number, but a widely used rule of thumb for general machinery shafting caps torsional deflection at about 1° over a length of twenty diameters. Precision equipment — printing rollers, machine-tool spindles, robot arms — often needs far tighter limits, since even a fraction of a degree can throw off registration or positioning.
Does θ = TL/(GJ) still hold once the shaft yields?
No. The formula comes straight from Hooke's law in shear, τ = Gγ, which only holds up to the material's shear yield point. Push torque past that and stress redistributes across the section — the classic treatment is Nadai's sand-heap analogy for fully plastic torsion — and rotation grows faster than this linear formula predicts.