Torsion of circular shafts

By the end of this topic you will be able to

Calculate the polar second moment of area J for solid and hollow circular sections.
Use the torsion formula to find the maximum shear stress in a shaft.
Calculate the angle of twist and explain why hollow shafts are efficient.

Twisting a shaft

A shaft that transmits power, such as the drive shaft of a car or the spindle of a pump, is loaded by a torque $T$ (a twisting moment). Cut the shaft anywhere and the internal shear stresses on the cut face must add up to the torque.

For a circular shaft (solid or hollow), plane sections remain plane and radii remain straight. That gives a strain that grows in proportion to the distance $\rho$ from the axis, and hence shear stress that grows linearly:

\tau = \frac{T\rho}{J}

The stress is zero on the axis and greatest at the outer surface, $\rho = r$ :

\tau_{\text{max}} = \frac{Tr}{J}

Polar moment of area

$J$ measures how the area is spread around the axis:

J_{\text{solid}} = \frac{\pi d^4}{32}, \qquad J_{\text{hollow}} = \frac{\pi}{32}\left(d_o^4 - d_i^4\right)

Angle of twist

The two ends of a shaft of length $L$ rotate relative to each other through

\phi = \frac{TL}{GJ} \quad \text{(radians)}

where $G$ is the shear modulus (about 80 GPa for steel). The product $GJ$ is the torsional stiffness.

Why hollow shafts?

The material near the axis is lightly stressed and contributes little to $J$ . Removing it saves weight with little loss of strength, so hollow shafts are common in vehicles and aircraft. Try it in the model: set the inner-to-outer ratio to 0.6 and compare.

Interactive modelApply a torque and compare solid and hollow shafts.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example

A solid steel shaft, 50 mm in diameter and 1 m long, transmits a torque of 2 kN·m. $G = 80$ GPa. Find the maximum shear stress and the angle of twist.

Polar moment. $J = \pi\times50^4/32 = 613\,600$ mm⁴.

Stress. $\tau_{\text{max}} = \dfrac{T r}{J} = \dfrac{2\times10^6\times25}{613\,600} = 81.5$ MPa.

Twist. $\phi = \dfrac{TL}{GJ} = \dfrac{2\times10^6\times1000}{80\,000\times613\,600} = 0.0407$ rad $= 2.33^\circ$ .

Hollow alternative. With the same outer diameter and a bore of 30 mm, $J = \pi(50^4 - 30^4)/32 = 534\,100$ mm⁴: the stress rises to 93.6 MPa, only 15% higher, while the cross-section area falls by 36%.

Practice problems

Find $J$ for a solid shaft 40 mm in diameter.

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$J = \pi\times40^4/32 = 251\,300$ mm⁴.
A 40 mm shaft carries 1.0 kN·m. Find the maximum shear stress.

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$\tau = 1.0\times10^6\times20/251\,300 = 79.6$ MPa.
A 40 mm steel shaft ( $G = 80$ GPa) is 1.5 m long and carries 1.0 kN·m. Find the angle of twist.

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$\phi = 1.0\times10^6\times1500/(80\,000\times251\,300) = 0.0746$ rad $= 4.27^\circ$ .
Doubling the diameter of a shaft: by what factor does the maximum stress fall for the same torque?

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$J$ grows by $2^4 = 16$ and $r$ by 2, so $\tau = Tr/J$ falls by a factor of 8.

Key formulae

Polar moment, solid shaft: $J = \frac{\pi d^4}{32}$
Polar moment, hollow shaft: $J = \frac{\pi}{32}\left(d_o^4 - d_i^4\right)$
Torsion formula: $\tau = \frac{T\rho}{J}, \quad \tau_{\text{max}} = \frac{Tr}{J}$
Angle of twist: $\phi = \frac{TL}{GJ}$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (45 min). 10 min: demonstration (twist a rubber eraser with ink lines on it, or a chalk stick until it fails at 45°); 15 min: derive linear distribution; 10 min: worked example; 10 min: hollow shafts with the model.

Activity. Challenge: find the inner/outer ratio that keeps the stress within 20% of the solid shaft while saving the most weight.

Discussion. Why does chalk twisted to failure crack along a spiral?

Assessment idea. Compare a solid and a hollow shaft of equal mass.