Circular Shaft Torsion

τ = \frac{T ρ}{J}, θ = \frac{T L}{G J}

For circular shafts, shear stress varies linearly with radius. J is the polar moment of inertia (π/32)d⁴ for a solid round. Angle of twist is analogous to PL/AE.

$T$ (N·m): Torque
$J$ (m⁴): Polar moment of inertia
$ρ$ (m): Radial coordinate

Step-by-step solved example

Solid steel shaft d = 40 mm, T = 400 N·m. Max shear stress?

Circular shaft: τ = Tr/J, twist θ = TL/GJ.

1. J
J = π/32 × 40⁴ = 2.513×10⁵ mm⁴.
2. τ_max
τ = T r / J = 400e3 N·mm × 20 / 2.513e5 = 31.8 MPa.
$τ_{m a x} = 31.8 MPa$

Answer: 31.8 MPa

10 practice questions

0/10 correct

1.Solid circular J is

2.Hollow shaft vs solid of same outer d: τ_max is

3.Power P related to torque and ω by

4.Angle of twist θ doubles if L doubles (same T, G, J)

5.τ = 0 at the center of a solid round shaft because

6.G for steel is about

7.A 2 kN force on a 150 mm wrench: torque is

8.Polar J of a thin tube ≈

9.Noncircular shafts

10.d = 20 mm solid, T = 50 N·m. τ_max ≈