Torsional Shear Stress⚠ unverified
Mechanical / Shafts · Shear stress in a shaft under torsion
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| T | T | N*m | 500.0 | Torque |
| r | r | m | 0.025 | Radius |
| J | J | m^4 | 6e-08 | Polar moment of area |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| tau | τ | Pa | Shear stress |
The science & history
Understanding the Parameters
-
Torque $T$ — the twisting moment carried by the shaft. Stress is directly proportional to it; the peak operating torque sets the maximum stress.
-
Radius $r$ — the through-section coordinate. Shear stress is zero at the centre and grows linearly to a maximum at the outer surface ($r = d/2$), which is why torsional failures start at the surface and why surface finish and stress raisers (keyways, fillets) matter so much. Use $r = d/2$ for the governing peak stress.
-
Polar moment $J$ — the section's torsional resistance (Polar Moment solid shaft, Polar Moment Hollow). Because $J \propto d^4$, the stress falls as $1/d^3$ when the surface radius is used — the same steep diameter dependence seen throughout shaft design.
Derivation (Approaching a Proof)
The derivation rests on one kinematic assumption for circular shafts: plane cross-sections remain plane and rotate rigidly about the axis (no warping — a special property of the circle).
-
Kinematics. Under a twist, a section at distance $x$ rotates by angle $\phi(x)$. A fibre at radius $r$ shears by an angle $\gamma = r\,\dfrac{\mathrm{d}\phi}{\mathrm{d}x}$ — strain grows linearly with radius.
-
Constitutive law. For a linear-elastic material, $\tau = G\gamma = G r\,\dfrac{\mathrm{d}\phi}{\mathrm{d}x}$, so shear stress is also linear in $r$: $\tau(r) = \tau_{\max}\, r/R$.
-
Equilibrium. The internal shear stresses must sum to the applied torque. Each area element contributes a moment $\tau\, r\, \mathrm{d}A$ about the axis: $$T = \int_A \tau\, r\, \mathrm{d}A = \frac{\tau_{\max}}{R}\int_A r^2\, \mathrm{d}A = \frac{\tau_{\max}}{R}\,J,$$ using $J \equiv \int_A r^2\,\mathrm{d}A$ by definition.
-
Solve. Rearranging, $\tau_{\max} = TR/J$, and for a general radius $\tau = Tr/J$.
The whole result flows from the linear strain distribution (kinematics) combined with the definition of the polar moment (equilibrium) — no empirical constant enters.
Dimensional check. $[\tau] = \dfrac{\text{N}\cdot\text{m}\cdot\text{m}}{\text{m}^4} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$. ✓
History and Development
The torsion formula descends from Coulomb's 1784 torsion experiments and Saint-Venant's rigorous 1850s theory of torsion, which showed that only circular sections twist without warping (so the elementary formula is exact for them). It is foundational to machine design — power transmission shafts, axles, torsion bars — and combines with bending via the von Mises criterion for real shafts carrying both (see Combined Stress Shaft).
Related Concepts: Polar Moment solid shaft, Polar Moment Hollow, Shaft Diameter torsion, Combined Stress Shaft, Shear Stress, Keyway Stress Reduction
Notes: Maximum at the outer surface ($r = d/2$). For hollow shafts use $J$ from Polar Moment Hollow with $r = d_o/2$. Non-circular sections warp — use Saint-Venant's torsion constant, not $J$. Apply a stress-concentration factor at keyways/fillets (Keyway Stress Reduction).