Torsional Shear Stress (τ = Tr/J)

Also known as shaft torsion stress · Tr/J

τ=TrJ\tau = \frac{T r}{J}

Worked example: 1 kN·m on a 50 mm shaft → 40.74 MPa at the surface — press Try an example to run it live, then adjust anything.

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Torsional Shear Stress (τ = Tr/J) explained

TrτJ

Twist a round shaft and every cross-section rotates rigidly relative to its neighbour, so the shear strain — and therefore the shear stress — grows linearly from zero at the centre to a maximum at the outer surface. That is the whole content of τ = Tr/J. Coulomb established the linear torque-twist behaviour experimentally in 1784 while building his torsion balance, and the modern form followed from Navier's generation. A 1 kN·m torque on a 50 mm shaft (J = 6.136 × 10⁻⁷ m⁴, r = 0.025 m) produces τ = 1000 × 0.025 ÷ 6.136 × 10⁻⁷ = 40.7 MPa at the surface.

Because only the surface matters, hollow shafts are the efficient answer, and any surface defect is disproportionately dangerous — a machining groove, a keyway or a sharp fillet at a shoulder is exactly where the stress is already highest, which is why fatigue failures of drive shafts nearly always start at a step or a keyway corner. Enter J in m⁴ and T in newton-metres; also remember that a shaft carrying torque plus bending must be checked on the combined stress, not each separately, and that ductile shafts fail on a plane perpendicular to the axis while brittle ones (cast iron, chalk) break on a 45° helix, following the principal tension.

Torsional Shear Stress (τ = Tr/J) formula

τ=TrJ\tau = \frac{T r}{J}
Where
  • τ\tau= Torsional shear stress (kPa)
  • TT= Applied torque (N·m)
  • rr= Radius to the point (m)
  • JJ= Polar moment of inertia (mm⁴)

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