Shaft Torsion Calculator

Calculate torsional shear stress, angle of twist, and polar moment of inertia for shafts.

Steel=80, Al=26, Cu=45
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Torsion Produces Shear Stress and Twist

A torque applied to a circular shaft creates shear stress that increases from zero at the center to a maximum at the outer surface. For elastic Saint-Venant torsion of a circular shaft, τ(r)=Tr/J and θ=TL/(JG). The polar second moment J describes geometric resistance to torsion: J=πd4/32 for a solid circular shaft and π(D4−d4)/32 for a hollow shaft.

The fourth-power dependence makes shaft diameter extremely influential. These formulas are intended for circular shafts in linear elastic torsion; noncircular sections warp and require different torsion constants and stress distributions.

τmax=Tc/J,   θ=TL/(JG)
SymbolMeaningWhy it appears / units
TApplied torqueN·m.
JPolar second momentm⁴; geometric torsional stiffness term.
GShear modulusPa.
cOuter radiusm; location of maximum shear stress.

Strength and stiffness are separate checks. A shaft can remain below allowable shear stress yet twist too much for alignment or control requirements. Hollow shafts can be efficient because material far from the center contributes strongly to J.

Torsional shear stress should increase linearly from the center to the outer radius in a circular shaft. The maximum is τmax=Tr/J. Because J depends strongly on diameter, a small diameter change can produce a large change in stress and twist.

Worked Examples

Example 1: Solid steel shaft: T=500N·m, d=50mm, L=1m, G=80GPa
J=πd⁴/32=613,592mm⁴, c=25mm
Result: τ=500e6×25/613592=20.4 MPa
Low torsional stress, mild steel τ_y≈145 MPa
Example 2: Power shaft: T=2000N·m, d=80mm
J=πd⁴/32=4.02×10⁶mm⁴
Result: τ=2000e6×40/4.02e6=19.9 MPa
High torque, larger shaft
Example 3: Solid shaft stress
d=40mm, T=500N·m
Result: τmax≈39.8MPa
The diameter enters through d³ in the simplified solid-shaft stress expression.
Example 4: Doubling diameter
same torque, solid circular shaft
Result: maximum stress becomes 1/8
For a solid shaft, τmax is proportional to 1/d³.

Common Mistakes

⚠️
Using the area moment I instead of polar moment J

Bending uses I; circular-shaft torsion uses J.

⚠️
Mixing millimeters and meters

J contains length to the fourth power, so unit mistakes can change results by huge factors.

⚠️
Applying circular-shaft formulas to arbitrary sections

Rectangular and open sections warp and need different torsion constants and stress relations.

Frequently Asked Questions

Hollow vs solid shaft efficiency?
For same weight, hollow shaft resists more torsion. J∝d⁴; moving material far from center increases J dramatically. Hollow shaft: 15-25% more torque capacity per unit weight.
Power-torque relationship?
P = Tω = T×2πn/60 where n is RPM. A motor delivering 10 kW at 1500 RPM: T = 10000/(2π×1500/60) = 63.7 N·m. Calculate torque from nameplate power and speed, then size shaft.
Why is shear stress zero at the shaft center?
In elastic circular-shaft torsion, τ=Tr/J. At radius r=0, the local shear stress is zero and then rises linearly toward the surface.
Why are hollow shafts efficient in torsion?
Material farther from the center contributes strongly to polar moment J, so removing low-radius material can save weight while retaining much torsional stiffness.
What controls angle of twist?
Twist increases with torque T and length L and decreases with polar moment J and shear modulus G.
Is shear modulus the same as Young’s modulus?
No. For isotropic linear elastic materials they are related by G=E/[2(1+ν)], where ν is Poisson’s ratio.
Which polar moment of inertia belongs in the torsion formula?
For a solid circular shaft, J=πd4/32; for a hollow circular shaft, J=π(D4−d4)/32. The ordinary area moment I used in beam bending is not interchangeable with J. Noncircular sections require different torsion treatment, and stress concentrations near keyways or shoulders are not represented by the elementary circular-shaft formula.

Formula Explorer connections

Interpretation: This formula is the rotational counterpart of linear mechanics, relating angle, angular motion, torque, inertia or rotational energy. Assumption: Define the rotation axis and sign convention. Rigid-body behavior, no slipping, steady rotation or negligible bearing losses may be assumed.

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