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Mechanics · Shafts and torsion

Solid or hollow shaft torsion

Evaluate the elastic response of a circular shaft under constant torque, with zero or non-zero inner diameter.

Created v1.0

By Thibaut Grzelak, Mechanical Analysis Engineer

Shafts and torsion
01Calculation inputsτmax = T(D/2)/J ; θ = TL/(GJ)
Circular shaft under torqueT is transmitted torque, L is loaded length, D is outer diameter and d is the optional inner diameter.TLDd
J = π(D⁴ − d⁴) / 32τmax = T·D / (2J)
Input identificationT is transmitted torque, L is loaded length, D is outer diameter and d is the optional inner diameter.

Results are up to date.

03

Results

Metric
Polar second moment
1,021,018mm⁴

Circular-section geometric torsion property.

Maximum shear stress
58.7649MPa

Shear stress at the outer surface.

Angle of twist
1.402909°

Relative rotation between shaft ends.

Torsional stiffness
81,681.41N·m/rad

Ratio T/θ = GJ/L.

View calculation detailsFormulas, SI conversion, numerical substitution and resultsOpen ↓

Calculation note

Full calculation details

Solid or hollow circular homogeneous prismatic shaft. Constant torque and linear-elastic behaviour. Plane sections remain plane and stress concentrations are excluded. The formula does not directly apply to non-circular sections.

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01Formulas and symbols

Formulas used

Jπ(D⁴ − d⁴)/32Polar second moment
τmaxT(D/2)/JMaximum shear stress
θTL/(GJ)Angle of twist
ktGJ/LTorsional stiffness
02Assumptions and limits

Scope of validity

  • Solid or hollow circular homogeneous prismatic shaft.
  • Constant torque and linear-elastic behaviour.
  • Plane sections remain plane and stress concentrations are excluded.
  • The formula does not directly apply to non-circular sections.
03Validation example

Reference numerical case

  1. T = 2 kN·m, D = 60 mm, d = 40 mm, L = 1 m and G = 80 GPa.
  2. J = 1,021,017.6 mm⁴.
  3. τmax = 58.7649 MPa.
  4. θ = 1.40291° and kt = 81,681.4 N·m/rad.
04Frequently asked questions

Questions about the calculation

Why is the formula limited to circular sections?

A circular section avoids the complex warping of non-circular sections. Rectangles and open profiles require a dedicated torsion constant.

Which shear modulus G should be used?

Use the material shear modulus at the relevant temperature. For an isotropic material, G = E/[2(1+ν)].

Does torque sign change the maximum stress?

It changes stress and rotation direction. The displayed maximum values use the torque magnitude.

05References

Technical references

  1. Gere & Goodno, Mechanics of Materials, torsion of circular shafts.
  2. Roark’s Formulas for Stress and Strain, 9th edition.

Calculation path and sources

A strength check should connect loads, section properties, stresses and relevant failure modes.

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External technical sources