Torsion of Shafts and Springs

Shafts carry power by twisting. The torsion equation gives their shear stress and angle of twist; the same ideas explain the stiffness of helical springs, whose wire is also loaded in torsion.

The torsion equation

For a circular shaft:

T/J = τ/r = Gθ/L

• τ grows linearly from zero at the centre to its maximum at the surface.
• Polar moment of inertia: solid J = πd⁴/32; hollow J = π(D⁴ − d⁴)/32.
• Maximum shear stress: solid shaft τmax = 16T/(πd³).
• Torsional stiffness = T/θ = GJ/L.

Power: P = 2πNT/60 (N in rpm, T in N·m, P in W).

A hollow shaft is stronger and stiffer than a solid shaft of the same weight, because material near the centre carries little stress.

Shafts in series and parallel

• Series (end to end): same torque, angles of twist add: θ = Σ TLᵢ/(GᵢJᵢ).
• Parallel (for example a composite shaft, or a shaft fixed at both ends with a torque in between): same angle of twist, torques add.

Combined bending and torsion

A shaft with bending moment M and torque T has:

• Equivalent torque Te = √(M² + T²) → τmax = 16Te/(πd³)
• Equivalent moment Me = ½(M + √(M² + T²)) → σmax = 32Me/(πd³)

Close-coiled helical springs

For a spring of mean coil diameter D (radius R), wire diameter d and n active turns under axial load W:

• Wire shear stress τ = 8WD/(πd³) (times a Wahl factor for curvature).
• Deflection δ = 8WD³n/(Gd⁴) = 64WR³n/(Gd⁴).
• Stiffness k = Gd⁴/(8D³n).

Springs in series: 1/k = 1/k₁ + 1/k₂. In parallel: k = k₁ + k₂. Cutting a spring in half doubles the stiffness of each half.

Worked example

A solid shaft transmits 40 kW at 400 rpm with τallow = 50 MPa. Find the minimum diameter.

• T = 40000 × 60/(2π × 400) = 954.9 N·m.
• d = (16T/(πτ))^(1/3) = (16 × 954.9 × 10³/(π × 50))^(1/3) = 46.0 mm.

Practice