Learn: Circular-shaft torsional shear
Twisting a circular shaft creates shear stress that increases with distance from its centre. The formula gives the stress at the particular radius you choose, not necessarily the maximum surface stress.
What the formula is saying
In elastic circular-shaft torsion, the shear strain varies linearly with radius. J describes the section’s polar spread of area, so τ grows with torque and radius but decreases with J.
Read the symbols in plain language
- T
- Torque
Torque. The elastic stress grows linearly as the evaluated point moves away from the centre.
N·mUse N·m as the base unit shown here. Torque is N·m, radius is m and J is m⁴. Their combination gives N/m² or Pa. Convert kN·m to N·m before substitution.
- r
- Radius to point
Radius to point. The elastic stress grows linearly as the evaluated point moves away from the centre.
mMetres measure length; 1 m = 1000 mm.
- J
- Polar second moment
Area-weighted squared distance from the shaft centre: the sum of the two perpendicular centroidal second moments.
m⁴The fourth power of metres is used for a second moment of area; 1 m⁴ = 10¹² mm⁴.
- τ
- Result to find
Circular-shaft torsional shear. A larger polar spread of area resists the same torque with less shear stress.
Pa
Sort out the units first
Torque is N·m, radius is m and J is m⁴. Their combination gives N/m² or Pa. Convert kN·m to N·m before substitution.
Assumptions before calculating
Treat the material as homogeneous and linearly elastic, with small strains. Use properties for the actual temperature and loading direction; the calculation is a model of behaviour before yielding, not a failure test.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find τ and explain the result in the stated output unit.
- T · Torque
- 5000 N·m
- r · Radius to point
- 0.05 m
- J · Polar second moment
- 0.00000982 m⁴
Multiply torque by the chosen radius
The elastic stress grows linearly as the evaluated point moves away from the centre.
(5000) × (0.05) = 250 N·m²Divide by polar second moment
A larger polar spread of area resists the same torque with less shear stress.
(250) ÷ (0.00000982) ≈ 25458248.47 Pa
Does this worked answer make sense?
Doubling torque doubles the stress. For a solid shaft, the stress is zero at the centre and largest at the outer radius.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- T · Torque
- 2000 N·m
- r · Radius to point
- 0.04 m
- J · Polar second moment
- 0.00000402 m⁴
Multiply torque by the chosen radius
The elastic stress grows linearly as the evaluated point moves away from the centre.
(2000) × (0.04) = 80 N·m²Divide by polar second moment
A larger polar spread of area resists the same torque with less shear stress.
(80) ÷ (0.00000402) ≈ 19900497.51 Pa
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- T · Torque
- 3000 N·m
- r · Radius to point
- 0.03 m
- J · Polar second moment
- 0.00000127 m⁴
Find: Learn: Circular-shaft torsional shear
A hint, not the answer
In elastic circular-shaft torsion, the shear strain varies linearly with radius. J describes the section’s polar spread of area, so τ grows with torque and radius but decreases with J.
Torque is N·m, radius is m and J is m⁴. Their combination gives N/m² or Pa. Convert kN·m to N·m before substitution.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Multiply torque by the chosen radius
The elastic stress grows linearly as the evaluated point moves away from the centre.
(3000) × (0.03) = 90 N·m²Divide by polar second moment
A larger polar spread of area resists the same torque with less shear stress.
(90) ÷ (0.00000127) ≈ 70866141.73 Pa
Avoid the common trap
Do not use diameter instead of radius. Do not substitute bending I for polar J. A centre-point value of zero does not mean the outer surface is unstressed.
When this method applies — and when it does not
Use a straight circular solid or hollow shaft under Saint-Venant torsion, away from end effects. The chosen radius must lie within the material, which cannot be checked from J alone. Noncircular shafts require their own torsion solution.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Circular-shaft torsional shear. Use the relevant elastic-response, equilibrium, constitutive-relations or shear-and-torsion module; the lesson states its particular sign convention.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.