Torsional Stress
Shear stress and angle of twist in a solid or hollow shaft.
Occurs at the outer surface. Shear stress varies linearly from zero at the centre to this maximum at the rim.
Over 1 m. Long drive shafts are often limited by twist rather than stress, since excessive wind-up causes vibration and timing errors.
Cross-sectional area 1257 mm². Because J depends on the fourth power of diameter, material near the surface does almost all the work — which is why hollow shafts are so efficient.
Removing the core of a shaft costs very little strength: the inner half of the radius contributes only about 6% of J, because the fourth-power relationship weights the outside so heavily. That is why drive shafts, bicycle frames, and scaffolding are tubes rather than bars.
Twisting a shaft produces shear stress τ = Tr ÷ J, greatest at the outer surface. A solid 40 mm steel shaft carrying 500 N·m sees about 39.8 MPa and twists 1.43° over a metre.
Stress grows outward from the centre
When a shaft twists, each cross-section rotates slightly relative to the next. Material at the centre barely moves relative to its neighbour and carries almost no stress; material at the surface moves the most and carries the maximum. Shear stress varies linearly from zero at the axis to its peak at the rim.
That distribution is the key to everything practical about torsion. The core of a solid shaft contributes very little, which is why drive shafts, bicycle frames, and scaffolding poles are tubes. Removing the inner half of the radius costs only about 6% of the torsional stiffness while removing 25% of the material.
Two independent limits
A shaft can be governed by strength — the peak shear stress must stay below what the material tolerates — or by stiffness, meaning the total angle of twist must stay within an acceptable limit. Long drive shafts are frequently limited by twist rather than stress, because excessive wind-up causes vibration, timing errors in machinery, and poor response.
J is the polar second moment of area, r the outer radius, G the shear modulus, and θ the angle of twist in radians. Steel G ≈ 80 GPa, aluminium ≈ 26 GPa.
Worked example: 40 mm steel shaft, 500 N·m
Section property first, then stress and twist:
- 1 Compute the polar second moment. J = π × 0.04⁴ ÷ 32 = 2.513 × 10⁻⁷ m⁴.
- 2 Take the outer radius. r = D ÷ 2 = 0.02 m — the point of maximum stress.
- 3 Apply τ = Tr/J. 500 × 0.02 ÷ 2.513e−7 ≈ 39.8 MPa.
- 4 Compare with the allowable. Against a 100 MPa allowable that is 40% utilisation, so the shaft is comfortable on strength.
- 5 Compute the twist. θ = TL/(GJ) = 500 × 1 ÷ (80 GPa × 2.513e−7) = 0.0249 rad.
- 6 Convert to degrees. 0.0249 × 180/π = 1.425° over one metre.
Why hollow shafts win
A 40 mm outer diameter shaft, comparing solid against increasingly thin-walled tubes.
| Bore | J relative to solid | Material relative to solid |
|---|---|---|
| Solid | 100% | 100% |
| 10 mm | 99.6% | 93.8% |
| 20 mm | 93.8% | 75.0% |
| 30 mm | 68.4% | 43.8% |
| 35 mm | 41.4% | 23.4% |
Where the simple theory applies
These formulas hold for circular sections — solid or hollow — loaded within the elastic range. Circular shafts are special because their cross-sections stay flat when twisted. Non-circular sections warp out of plane, which makes their torsion analysis considerably harder and means a square bar does not follow τ = Tr/J.
Two practical cautions. Torsional failure often starts at a stress concentration rather than in the plain shaft: a keyway, a shoulder, a cross-hole, or a sharp fillet can multiply local stress several times over, which is why fillet radii on stepped shafts matter so much.
And shafts in machinery rarely see steady torque. Fluctuating or reversing loads mean fatigue governs rather than static strength, and the allowable stress for millions of cycles is far below the static shear strength. A shaft that passes this static check can still fail in service if the loading cycles.