Torsion

Every stress calculation so far divided a force by an area, which quietly assumed that every square millimetre of the section worked equally hard. Twist a circular shaft and that stops being true: the material near the axis barely moves at all, and where the metal sits becomes a design variable in its own right.

What twisting does to the geometry

Take a straight shaft of circular section, fix one end, and apply a torque $T$ about the axis at the other. Scribe a line along the surface before loading and it becomes a helix; scribe radial lines on the end face and they stay straight.

Two observations, both confirmed by experiment and both true only for circular sections, carry the entire derivation. Plane cross sections remain plane and do not warp out of their own plane. Radii remain straight, so each cross section rotates as a rigid disc relative to its neighbours. The reason both hold is symmetry: a circular section looks the same after any rotation about the axis, so there is no preferred direction in which warping could occur. A square shaft has corners, has preferred directions, and does warp, which is why the last section of this lesson has to walk the result back.

Given those two facts, consider a point at radius $r$ on a shaft of length $L$ whose far end has rotated through an angle $θ$ radians. A line originally along the axis has been dragged sideways by an arc length $r θ$ over a length $L$ , so the right angle between the axial line and the radius has closed by

γ = \frac{r θ}{L}

That is a pure shear strain, and it is proportional to $r$ . On the axis it is zero. At the surface it is largest. The material in the middle of a solid shaft is doing almost nothing.

From strain to the torsion formula

Within the elastic range, shear stress is proportional to shear strain through the shear modulus $G$ , so

τ = G γ = \frac{G θ}{L} r

The stress rises linearly from zero on the axis to a maximum at the outer surface. $G$ is not an independent property for an isotropic material: it is fixed by $E$ and Poisson's ratio through $G = E / [2 (1 + ν)]$ . For steel with $E = 210$ GPa and $ν = 0.3$ that gives $G = 80.8$ GPa, and the tabulated value for structural steel is $81$ GPa. Aluminium comes out near $26$ GPa.

Now impose equilibrium. The shear stress on an annular ring of radius $r$ and thickness $d r$ acts on an area $2 π r d r$ and has a moment arm $r$ about the axis, so it contributes $τ \cdot 2 π r d r \cdot r$ to the torque. Substituting for $τ$ and integrating,

T = \frac{G θ}{L} \int_{0}^{R} 2 π r^{3} d r = \frac{G θ}{L} J

where the integral has been given a name:

J = \int r^{2} d A

is the polar second moment of area. Eliminating $G θ / L$ between the last two results gives the torsion formula, which is normally written as a chain of three equal ratios:

\frac{T}{J} = \frac{τ}{r} = \frac{G θ}{L}

The first equality gives stress from torque, the second gives the twist. Notice what $J$ has done: it is the only place the shape of the section enters, and because $r$ appears squared, material far from the axis counts far more than material near it. That is the whole of the shape argument, and it recurs in the bending lesson with $J$ replaced by a different second moment.

The polar second moment

For a solid circle of radius $R$ , the integral above evaluates to $π R^{4} / 2$ , or in terms of diameter

J = \frac{π d^{4}}{32}

For a hollow circular section the integral simply runs from the inner radius outward, so $J = π (D^{4} - d^{4}) / 32$ . Nothing else in the derivation changes, because a hollow shaft satisfies the same two geometric observations.

The fourth power is worth pausing on. Doubling the diameter of a solid shaft multiplies its torsional stiffness and its torque capacity at a given stress by sixteen and by eight respectively, while multiplying its weight by four.

Example. A solid steel shaft of $50$ mm diameter carries a torque of $1.2$ kN m over a length of $1.5$ m. Find the maximum shear stress and the angle of twist, taking $G = 81$ GPa.

$J = π \times 50^{4} / 32 = 613 592$ mm⁴. The maximum stress is at the surface, $r = 25$ mm:

τ = \frac{T r}{J} = \frac{1.2 \times 10^{6} \times 25}{613 592} = 48.9 MPa

The twist is $θ = T L / G J = (1.2 \times 10^{6} \times 1500) / (81 000 \times 613 592) = 0.0362$ rad, which is $2.08$ degrees. A rule of thumb for power transmission shafts is to keep the twist under one degree per metre, and this one is at $1.38$ degrees per metre, so it would be sized on stiffness rather than strength.

Now you. A solid steel shaft of $60$ mm diameter carries $2$ kN m over $2$ m. Find the maximum shear stress and the twist in degrees.

Answer

$J = π \times 60^{4} / 32 = 1.272 \times 10^{6}$ mm⁴. Then $τ = 2 \times 10^{6} \times 30 / 1.272 \times 10^{6} = 47.2$ MPa, and $θ = 2 \times 10^{6} \times 2000 / (81 000 \times 1.272 \times 10^{6}) = 0.0388$ rad, or $2.22$ degrees.

Sizing a shaft from the power it carries

Shafts are almost never specified by torque directly. They are specified by the power they transmit and the speed they run at, and the torque follows from mechanics: power is torque times angular velocity, $P = T ω$ , with $ω = 2 π N / 60$ for a speed of $N$ revolutions per minute.

Combining that with the torsion formula gives the design equation. For a solid shaft, $τ_{\max} = T (d / 2) / (π d^{4} / 32) = 16 T / (π d^{3})$ , so

d^{3} = \frac{16 T}{π τ_{allow}}

Example. A solid steel shaft transmits $30$ kW at $600$ rpm. The allowable shear stress is $50$ MPa. What diameter is needed?

$ω = 2 π \times 600 / 60 = 62.83$ rad s⁻¹, so $T = 30 000 / 62.83 = 477.5$ N m, which is $4.775 \times 10^{5}$ N mm. Then $d^{3} = 16 \times 4.775 \times 10^{5} / (π \times 50) = 48 650$ mm³, giving $d = 36.5$ mm. A $40$ mm shaft would be specified, since that is a stock size and the next one down is not.

Now you. A solid steel shaft transmits $20$ kW at $300$ rpm with an allowable shear stress of $45$ MPa. What diameter is needed?

Answer

$ω = 31.42$ rad s⁻¹ and $T = 20 000 / 31.42 = 636.6$ N m. Then $d^{3} = 16 \times 6.366 \times 10^{5} / (π \times 45) = 72 050$ mm³ and $d = 41.6$ mm, so a $45$ mm shaft.

Note that halving the speed doubles the torque for the same power, so slow machinery needs fat shafts. This is why a ship's propeller shaft turning at 100 rpm is enormous and a turbine shaft delivering the same power at 3000 rpm is not.

Why shafts are hollow

The linear stress distribution says the core of a solid shaft is nearly idle, and the arithmetic says how nearly.

Example. Compare a solid steel shaft of $60$ mm diameter with a hollow one of $60$ mm outside and $40$ mm bore, on the basis of torque carried per unit of material.

Solid: $J = 1.272 \times 10^{6}$ mm⁴ and area $2827$ mm². The torque at a given surface stress $τ$ is $T = τ J / R = τ \times 1.272 \times 10^{6} / 30 = 42 410 τ$ , so torque per unit area is $15.0 τ$ .

Hollow: $J = π (60^{4} - 40^{4}) / 32 = 1.021 \times 10^{6}$ mm⁴ and area $1571$ mm². The torque is $τ \times 1.021 \times 10^{6} / 30 = 34 030 τ$ , so torque per unit area is $21.7 τ$ .

Boring out the middle removes $44$ per cent of the material and only $20$ per cent of the torsional strength, so the hollow shaft is $44$ per cent more efficient per kilogram. This is why drive shafts, bicycle axles and aircraft control tubes are tubes.

Now you. Repeat for a hollow shaft of $60$ mm outside and $45$ mm bore.

Answer

$J = π (60^{4} - 45^{4}) / 32 = 8.698 \times 10^{5}$ mm⁴ and the area is $1237$ mm². The torque is $τ \times 8.698 \times 10^{5} / 30 = 28 990 τ$ and the torque per unit area is $23.4 τ$ , which is $56$ per cent better than the solid shaft.

The limit is not efficiency but local buckling: as the wall gets thinner relative to the diameter, the tube stops failing by shear yielding and starts failing by the wall wrinkling into diagonal folds, which is the same instability the buckling lesson meets in a different guise.

The circular assumption, and what breaks without it

Every result in this lesson rests on cross sections staying plane, and that is a property of the circle. Load a square bar in torsion and the corners of the section pull out of plane in a saddle shape, so the elementary derivation fails at its first step. The maximum shear stress in a rectangular section is not even at the corner furthest from the centre; it is at the midpoint of the longer side, and the corners carry no shear stress at all. Solutions for these cases exist, they come from a different route entirely, Saint-Venant's torsion function, and they are always quoted as tabulated coefficients rather than as a single formula.

The difference that matters most in structures is between closed and open thin-walled sections. A closed tube resists torque by a shear flow that circulates continuously around the wall, and it is very stiff. Slit that tube along its length and the shear flow can no longer circulate, so the section resists only by shearing through its own thickness, and its torsional stiffness collapses by a factor of hundreds for the same amount of material. A rolled steel I-beam is an open section and is therefore hopeless in torsion, which is exactly why a beam carrying an eccentric load has to be checked for twisting separately, and why box girders are used wherever torsion matters, in a curved bridge deck or a crane runway.

Torsion has done its main job for this course, which is to establish that the second moment of the area, not the area, is what a section offers against a moment. Bending applies the same idea in the direction that beams actually work, and the result decides the shape of nearly every structural member ever rolled.