04 // Interactive Formula Compendium // ENGINEERINGτ_max = (T · r) / J, θ = (T · L) / (G · J)
Circular Shaft Torsion & Polar Shear Stress Formula
The torsion formula τ = Tr/J relates applied torque T and outer shaft radius r to peak surface shear stress via polar moment of inertia J.
Formula Reference
Main Formula
τ_max = (T · r) / J, θ = (T · L) / (G · J)
Rearranged Forms
J_solid = (π · d⁴) / 32
P = T · ω
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| tau | Peak Surface Shear Stress | Pa |
| T | Applied Torque | N·m |
| r | Outer Shaft Radius | m |
| J | Polar Moment of Inertia | m⁴ |
Inputs & Parameters
[N·m]
[m]
[m⁴]
Calculated Result (Peak Surface Shear Stress)
tau =4.88926e+7Pa
Formula Notes & Background
- Axisymmetric circular cross-sections do not warp under pure torsion; shear strain varies linearly with radial distance γ(ρ) = ρ·(dθ/dx).
- For a hollow shaft of outer diameter D and inner diameter d, J = (π/32)(D⁴ − d⁴).
Step-by-Step Examples
#1: Solid 50 mm diameter steel drive shaft (r = 0.025 m, J = π(0.05)⁴/32 = 6.136×10⁻⁷ m⁴) transmitting T = 1,200 N·m.
T = 1,200 N·mr = 0.025 mJ = 6.136×10⁻⁷ m⁴
τ_max = (1,200 × 0.025) / 6.136×10⁻⁷ = 30 / 6.136×10⁻⁷
→ τ_max = 48.89 MPa
#2: Angle of twist over L = 1.5 m for the same shaft with shear modulus G = 79.3 GPa.
L = 1.5 mG = 7.93×10¹⁰ Pa
θ = (1,200 × 1.5) / (7.93×10¹⁰ × 6.136×10⁻⁷) = 1800 / 48,658
→ θ = 0.0370 rad (2.12°)
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Frequently Asked Questions
Why are automotive drive shafts hollow tubes rather than solid bars?+
Core material near ρ = 0 carries almost zero shear stress while adding rotational inertia and weight; a hollow tube achieves nearly identical J at a fraction of the mass.
Does τ = Tr/J apply to square or rectangular shafts?+
No — non-circular shafts warp out of plane and develop peak shear stress at the midpoints of the widest flat faces (requiring Saint-Venant torsion constants).
How do I convert motor power (kW) and RPM into shaft torque T?+
Use T = P / ω = (1000 · P_kW × 60) / (2π · RPM) ≈ 9549.3 × P_kW / RPM.