Torsion in circular shafts

Engineering · Second-year engineering · about 10 min

Drive shafts, axles and bolts being tightened all carry torque, a twisting moment. Torsion theory tells you how much shear stress that creates inside the shaft and how far the shaft twists, which is exactly what you need to choose a safe diameter.

The key idea

For a solid circular shaft of diameter $d$ carrying torque $T$ :

\tau = \frac{T\,r}{J} \qquad J = \frac{\pi d^4}{32} \qquad \theta = \frac{T\,L}{G\,J}

The shear stress $\tau$ is largest at the outer surface ( $r = d/2$ ), and the angle of twist $\theta$ (in radians) grows with length $L$ and falls with the shear modulus $G$ .

A solid steel cylinder drawn in a slight three-quarter view, lying horizontally, so its round free end can be seen. Its left end is fixed into a hatched wall. A curved dark red arrow around the free right end shows the torque. A straight thin line runs along the top of the cylinder from the wall to the right end. A second thin line starts at the same point at the wall and ends further round the free end. Mark the angle between the two lines at the free end. A dimension for the length runs under the cylinder and a small dimension shows the diameter across the free end. — Torque twists the shaft, and the twist grows along its length.

Try it yourself

Change the torque and diameter and watch the stress and twist respond.

Why the stress is biggest at the surface

When a shaft twists, each cross-section rotates a little relative to the next. Material further from the centre moves further, so it is strained more, and the shear stress rises linearly from zero at the centre to its maximum at the outer edge. That's also why hollow shafts are efficient: the material near the centre does very little work.

A circular cross-section of a solid shaft, seen face on, in light steel grey. A thin radius line runs from the centre O straight up to the top edge. Along that line, draw five small dark red arrows at right angles to the radius, all pointing right, with their lengths growing evenly from nothing at the centre to the longest at the edge. — Shear stress rises in a straight line from zero at the centre to its maximum at the surface.

The polar moment of area

$J$ measures how well a cross-section resists twisting.

Solid shaft: $J = \dfrac{\pi d^4}{32}$
Hollow shaft: $J = \dfrac{\pi (D^4 - d^4)}{32}$

Because of the fourth power, doubling the diameter makes the shaft sixteen times stiffer in torsion.

Two circular cross-sections side by side, seen face on. Left: a solid light steel grey circle with a horizontal diameter dimension. Right: a ring of light steel grey with an empty white centre, with its outer diameter dimension above it and its inner diameter dimension across the hole. — Solid and hollow shafts, seen end on.

Worked example: stress and twist in a steel shaft

A solid steel shaft of diameter $40$ mm and length $1.5$ m carries a torque of $500$ N·m. Take $G = 80$ GPa. Find the maximum shear stress and the angle of twist.

Show the solution

Polar moment of area:

J = \frac{\pi (0.04)^4}{32} = 2.513 \times 10^{-7} \text{ m}^4

Maximum shear stress at $r = 0.02$ m:

\tau_{max} = \frac{T r}{J} = \frac{500 \times 0.02}{2.513 \times 10^{-7}} = 39.8 \text{ MPa}

Angle of twist:

\theta = \frac{T L}{G J} = \frac{500 \times 1.5}{80 \times 10^{9} \times 2.513 \times 10^{-7}} = 0.0373 \text{ rad} = 2.14^\circ

Common mistakes

Using the diameter instead of the radius in the stress formula

$\tau = \frac{Tr}{J}$ uses the radius, $r = d/2$ .

Mixing millimetres and metres

Put everything in metres and newtons, and the answer comes out in pascals. A $40$ mm diameter is $0.04$ m.

Leaving the angle of twist in radians when degrees are asked for

Multiply by $\frac{180}{\pi}$ : $0.0373$ rad is $2.14^\circ$ .

Check your understanding

1. Where is the shear stress largest in a twisted solid shaft?

2. If you double a solid shaft's diameter, its polar moment of area J becomes...

3. Doubling the length of a shaft under the same torque does what to its angle of twist?

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