Shaft Torsion

How to Use the Shaft Torsion Calculator

1. Enter the Applied torque (T) and select its unit — N·m, kN·m, lb·ft or lb·in.
2. Choose the Shaft type: solid circular shaft or hollow circular shaft (tube). For a hollow shaft, also enter the Inner diameter (d), which must be smaller than the outer diameter.
3. Enter the Outer diameter (D) and the Shaft length (L) with their units (mm, cm, m, in or ft).
4. Enter the Shear modulus (G) of the material — steel ≈ 80 GPa, aluminum ≈ 27 GPa, copper ≈ 45 GPa — and choose the unit (GPa, MPa, psi or ksi).
5. Read the results: the maximum shear stress (τmax) in MPa, ksi and psi, and the angle of twist (φ) in degrees, radians, degrees per metre and degrees per foot.

What Is Shaft Torsion?

When a torque is applied along the axis of a shaft — as happens in every transmission shaft, drive line, drill string or propeller shaft — the material is twisted. Each cross-section rotates relative to the neighbouring one, and this rotation is resisted by shear stresses distributed over the cross-section. For circular shafts (solid or with a concentric bore) the analysis is exact and simple: the shear stress varies linearly from zero at the centre to a maximum at the outer surface, which is why shafts and tubes fail or yield first on their outside.

Two quantities matter in a torsion check. The first is the maximum shear stress, \(\tau_{max} = Tc/J\): it must stay below the allowable shear stress of the material. For ductile steels, the yield strength in shear is roughly 0.5–0.6 times the tensile yield strength, and a suitable safety factor is applied on top. The second is the angle of twist, \(\varphi = TL/JG\): even a shaft that is strong enough may twist too much, causing misalignment, vibration, wind-up in control systems or positional error in precision drives.

The geometry term \(J\), the polar moment of inertia, scales with the fourth power of the diameter. Doubling the diameter of a solid shaft reduces the stress and the twist by a factor of 16. This is why hollow shafts are so effective: removing the lightly stressed material near the centre costs very little stiffness but saves a great deal of weight, which is why drive shafts, roll shafts and many aerospace parts are tubes.

When interpreting the result, compare \(\tau_{max}\) with the material's allowable shear stress, and compare \(\varphi\) with your twist limit. As a traditional rule of thumb, machine shafts are limited to about 1° of twist over a length of 20 diameters; precision and long-line shafting often needs much tighter limits.

Assumptions, Limitations and Practical Tips

• Circular cross-sections only. The formulas are exact for solid or hollow circular shafts with a concentric bore. Non-circular sections (rectangular bars, keyways, splines) twist very differently and need separate methods.
• Uniform shaft, constant torque. The calculation assumes a constant cross-section, a single material and a torque that does not change along the length. For stepped shafts or several torques, compute each segment separately and add the twist angles.
• Linear elastic behaviour. Stresses must remain below the elastic limit; the result does not predict yielding or plastic collapse.
• Keyways, holes and shoulders concentrate stress. A real shaft with a keyseat or a shoulder may see local stresses 1.5–3 times the nominal value, so allow for stress concentration factors in critical designs.
• Check both criteria. A short, stout shaft is usually governed by stress; a long, slender shaft is often governed by twist. Also remember that a long shaft can fail by buckling or whirling — a related check is our Euler buckling load calculator.
• For shafts that also carry bending loads (gears, belts, pulleys), superimpose the bending stresses from a beam analysis — see the simply supported beam with center point load and uniformly loaded beam calculators.