Torsion

Torsion is twisting. Learn shear stress τ = Tr/J, angle of twist θ = TL/GJ, power transmission P = Tω, solid vs hollow shafts, and designing shafts for strength and stiffness.

Every motor, engine and turbine transmits power through a rotating shaft. The shaft carries a torque that twists it, creating shear stress that is zero at the centre and maximum at the surface.

Shaft design checks two things: the stress must stay below allowable, and the twist must be small enough for accurate, vibration-free operation.

An analogy

Wringing out a towel: your hands twist opposite ways and the fibres on the outside spiral the most. That spiral is shear strain.

Torque and power

Power = torque × angular speed. At a given power, a slow shaft carries a large torque and needs to be thick; a fast one can be slim. That's why gearboxes put the thick shafts on the slow output side.

Stress distribution

Shear stress rises linearly from zero at the centre to maximum at the surface. The material near the centre does little work — so hollow shafts save weight with small strength loss.

Stiffness: angle of twist

θ = TL/(GJ). Machine-tool spindles and long drive lines limit twist to around 0.25–1° per metre to avoid positioning error and torsional vibration.

Combined bending and torsion; non-circular sections

Real shafts also bend from gear and belt loads. The equivalent (von Mises) stress σ_eq = √(σ² + 3τ²) or ASME shaft equations are used. Non-circular sections warp; open thin-walled sections (channels, angles) are very weak in torsion compared with closed tubes.

Formulas

• Torsion equation: T/J = τ/r = Gθ/L
• Maximum shear (solid shaft): τ_max = 16T / (π d³)
• Power: P = T · ω = 2πNT / 60
• Hollow shaft J: J = π(D⁴ − d⁴) / 32

Worked example: Motor shaft sizing

Given: P = 20 kW at 1440 rpm; τ_allow = 40 MPa

Find: Torque and minimum diameter.

• T = 20 000/(2π × 1440/60) = 132.6 N·m
• d = ∛(16T/πτ) = ∛(16 × 132 600/(π × 40)) = 25.7 mm
• Choose Ø28 or Ø30 mm (keyway weakens it)

Answer: T = 132.6 N·m, d ≥ 25.7 mm

Worked example: Twist of the chosen shaft

Given: Ø30 mm, T = 132.6 N·m, G = 80 GPa; L = 1 m

Find: Stress and twist.

• J = π × 30⁴/32 = 79 522 mm⁴
• τ = 16 × 132 600/(π × 30³) = 25.0 MPa
• θ = TL/GJ = 132 600 × 1000/(80 000 × 79 522) = 0.0208 rad = 1.19°

Answer: 25 MPa, 1.19° per metre

Common mistakes

• Wrong: Using rpm directly as ω. — Right: ω = 2πN/60 rad/s.
• Wrong: Ignoring keyways and shoulders. — Right: Apply stress-concentration factors (keyway K_ts ≈ 2–3 in fatigue).

Applications

• Drive shafts: Automotive propeller shafts are thin-wall tubes: high J, low mass, high critical speed.
• Torsion bars: Some vehicle suspensions and anti-roll bars use shaft twist as a spring.
• Torque sensors: Strain gauges at 45° on a shaft measure torque via shear strain.

Key takeaways

• τ = Tr/J; solid: 16T/πd³.
• P = Tω: slow shafts need more torque capacity.
• θ = TL/GJ; check stiffness too.
• Hollow shafts save weight efficiently.

Going further

• Critical speed: Shafts whirl violently at their natural bending frequency; design speed is kept well away from it (Dunkerley/Rayleigh methods).
• Torsional vibration: Engine crankshafts use dampers because firing pulses excite torsional resonances.
• ASME shaft design: The DE-Goodman equation combines alternating bending with steady torque and fatigue factors to find shaft diameter.