Torsion of cylindrical bars

Grip a straight steel rod at one end, twist the other, and something invisible happens inside the metal long before it snaps: every internal fibre is being sheared, more violently the farther it sits from the centre. This is torsion — the twisting of a circular shaft by equal and opposite end torques — and it is the opening idea of the "Torsion and Flexural Loading" unit in ME10001 Mechanics. Drive shafts, propeller shafts, axles, and the screwdriver in your hand all live or die by this one relationship between an applied torque, the shaft's geometry, and the shear stress and twist it produces. This article builds the torsion formula from three simple assumptions about how a circular shaft deforms, shows how the polar moment of inertia $J$ packages the cross-section's geometry into a single number, extends the idea to hollow shafts, multi-segment superposition, statically indeterminate shafts, and power transmission, and works two fully-solved numerical examples so the arithmetic is never a mystery.

The core insight

In a circular shaft under torque, shear stress and angle of twist both scale linearly with radius and torque through $\tau = T\rho/J$ and $\phi = TL/JG$ , where $J$ is the polar moment of inertia of the cross-section.

The core idea

Three assumptions carry the whole theory. For a straight circular shaft (solid or hollow) twisted by end torques: (1) plane cross-sections remain plane and simply rotate as rigid discs about the shaft axis — they do not warp; (2) radii of the cross-section stay straight during twisting; (3) shear strain $\gamma$ varies linearly with radial distance $\rho$ from the axis, zero at the centre and maximum at the outer surface. These are purely kinematic statements about how the shaft deforms — no material property is needed yet.

Hooke's law turns strain into stress. Since $\tau = G\gamma$ for a linearly elastic material, and $\gamma$ is linear in $\rho$ , shear stress $\tau$ must also be linear in $\rho$ . Matching this stress distribution to the requirement that it be statically equivalent to the internal torque $T$ (integrating $\tau \rho$ over the area) gives the torsion formula:

\tau = \frac{T\rho}{J}, \qquad \tau_{max} = \frac{Tc}{J}

where: $T$ is the internal torque carried by the section, $\rho$ is the radial distance from the axis to the point of interest, $c$ is the outer radius of the shaft, and $J$ is the polar moment of inertia of the cross-section. Shear stress is exactly zero at the centre and rises linearly to a maximum $\tau_{max}$ at the outer fibre — it is never uniform across a solid section.

$J$ packages the geometry. For circular cross-sections:

J = \frac{\pi c^4}{2} \ \text{(solid shaft)}, \qquad J = \frac{\pi\left(c_o^4 - c_i^4\right)}{2} \ \text{(hollow shaft)}

where: $c_o$ and $c_i$ are the outer and inner radii of a hollow (tubular) shaft. A hollow shaft removes material near the low-stress centre, so it delivers a much better strength-to-weight ratio than a solid shaft of the same outer radius — this is why drive shafts and aircraft components are often tubular.

\phi = \frac{TL}{JG}

where: $\phi$ is the angle of twist in radians between two cross-sections a distance $L$ apart, and $G$ is the shear modulus of the material. Unlike $\tau$ , the twist $\phi$ does not depend on $\rho$ — it is a single number describing the relative rotation of the whole end section.

Superposition for multi-segment shafts. If a shaft has several segments with different $T$ , $J$ , or $G$ , the total twist between the outermost ends is the algebraic sum of the segment twists, $\phi_{total} = \sum T_iL_i/(J_iG_i)$ , with a consistent sign convention for torque direction fixed first.

Statically indeterminate shafts. When a shaft is torqued between two fixed supports (both ends built into rigid walls), equilibrium alone gives one equation but there are two unknown reactive torques — the problem is indeterminate. The missing equation is a compatibility condition: since both ends are fixed, the net twist across the whole shaft between the supports must be zero. Solving equilibrium and compatibility together fixes both reactions.

Power transmission. A rotating shaft carrying torque $T$ at angular speed $\omega$ transmits power

P = T\omega = 2\pi n T

where: $\omega$ is in rad/s, $n$ is the rotational speed in rev/s, and $P$ is in watts. This lets an engineer start from a required power and speed, pick an allowable shear stress, and back out the shaft diameter needed.

Figure

Cross section of a solid circular shaft showing shear stress zero at the centre and rising linearly to a maximum at the outer radius

Derivation ADeriving the torsion formula from kinematics and equilibrium

The goal is to combine the three kinematic assumptions of circular-shaft torsion (plane sections rotate rigidly, radii stay straight, shear strain varies linearly with radial distance) with Hooke's law and equilibrium of internal torque to arrive at $\tau = T\rho/J$ , showing exactly why $J$ must be the polar moment of inertia and nowhere else does the shape of the cross-section enter the formula.

Step 1. Express the linear strain assumption. Because plane sections rotate rigidly about the axis, a fibre at radius $\rho$ over a length $L$ twists through angle $\phi$ , and geometry gives the shear strain there as $\gamma(\rho) = \rho\phi/L$ . This is linear in $\rho$ : zero at the centre, maximum at the outer radius $c$ , where $\gamma_{max} = c\phi/L$ . Dividing, $\gamma(\rho) = (\rho/c)\gamma_{max}$ .

Step 2. Apply Hooke's law for shear, $\tau = G\gamma$ , to convert the strain distribution into a stress distribution. Since $\gamma$ is linear in $\rho$ and $G$ is constant across the cross-section, $\tau(\rho) = (\rho/c)\tau_{max}$ — shear stress is also linear in $\rho$ , zero at the centre and $\tau_{max}$ at the outer fibre.

Step 3. Enforce equilibrium: the internal torque $T$ carried by the section must equal the resultant moment of this stress distribution about the axis. Each elemental area $dA$ at radius $\rho$ carries force $\tau(\rho)\,dA$ acting tangentially, contributing moment $\rho\,\tau(\rho)\,dA$ . Summing (integrating) over the whole area, $T = \int \rho\,\tau(\rho)\,dA = \int \rho \cdot (\rho/c)\tau_{max}\,dA = \dfrac{\tau_{max}}{c}\int \rho^2\,dA$ .

Step 4. Recognise the integral. $\int \rho^2\,dA$ is by definition the polar moment of inertia $J$ of the cross-section about its centroidal axis — a purely geometric property that does not depend on the material or the torque. So $T = \dfrac{\tau_{max}}{c}J$ , which rearranges to $\tau_{max} = Tc/J$ , and combining with Step 2's ratio gives the general form at any radius, $\tau(\rho) = T\rho/J$ — the torsion formula. The same substitution, run through $\phi = \rho$ -independent algebra using $\gamma_{max} = c\phi/L$ and $\tau_{max} = G\gamma_{max}$ , yields $\phi = TL/(JG)$ . Result: $\tau = T\rho/J$ with $\tau_{max} = Tc/J$ , and $\phi = TL/(JG)$ , exactly matching the formulas quoted above.

Worked example

A solid circular shaft of diameter $d = 50\ \text{mm}$ carries a torque $T = 1.2\ \text{kN}\,\cdot\,\text{m}$ . Find the maximum shear stress and the angle of twist over a length $L = 2\ \text{m}$ , taking $G = 80\ \text{GPa}$ .

Step 1. Convert diameter to radius: $c = d/2 = 25\ \text{mm} = 0.025\ \text{m}$ . This is the single most common slip in torsion problems — plugging the diameter straight into the formula instead of the radius — so it is worth doing explicitly.

Step 2. Compute the polar moment of inertia for the solid section:

J = \frac{\pi c^4}{2} = \frac{\pi (0.025)^4}{2} = 6.136\times10^{-7}\ \text{m}^4

Step 3. Apply the torsion formula at the outer fibre:

\tau_{max} = \frac{Tc}{J} = \frac{(1200)(0.025)}{6.136\times10^{-7}} = 48.9\times10^{6}\ \text{Pa} = 48.9\ \text{MPa}

Step 4. Apply the angle-of-twist formula over the given length:

\phi = \frac{TL}{JG} = \frac{(1200)(2)}{(6.136\times10^{-7})(80\times10^{9})} = 0.0489\ \text{rad} \approx 2.80^\circ

The maximum shear stress is 48.9 MPa at the outer surface, and the shaft twists by 0.0489 rad, about 2.80 degrees, over the 2 m length.

Competitive stretch

Now make the shaft hollow instead: outer diameter $100\ \text{mm}$ , inner diameter $60\ \text{mm}$ , transmitting power at $n = 10\ \text{rev/s}$ with allowable shear stress $\tau_{allow} = 60\ \text{MPa}$ ( $G = 80\ \text{GPa}$ ). With $c_o = 0.05\ \text{m}$ , $c_i = 0.03\ \text{m}$ : $J = \pi(c_o^4 - c_i^4)/2 = 8.545\times10^{-6}\ \text{m}^4$ . Rearranging $\tau_{max} = Tc_o/J$ gives the allowable torque $T_{allow} = \tau_{allow}J/c_o = 10{,}254\ \text{N}\,\cdot\,\text{m}$ . With $\omega = 2\pi n = 62.83\ \text{rad/s}$ , the maximum power is $P = T\omega \approx 644\ \text{kW}$ — the same torsion machinery, now sized for a design problem instead of an analysis problem.

Common traps

Using the outer diameter instead of the outer radius $c$ in $\tau_{max} = Tc/J$ — always halve the diameter before substituting.
Assuming shear stress is uniform across a solid section; it is zero at the centre and rises linearly to a maximum only at the outer fibre.
Confusing $J$ (polar moment of inertia, for torsion) with $I$ (rectangular moment of inertia, for bending) — they are different area properties computed differently and are not interchangeable.
In indeterminate-shaft problems, writing equilibrium alone and stopping — the extra unknown reactive torque needs the compatibility equation (net twist between the fixed supports equals zero) to close the problem.

Ask the tutor: Want a walkthrough of how to set up and solve a statically indeterminate shaft, or how to combine superposition across segments with different torque, diameter, and material — ask the AI tutor to work one through with you step by step.

Why does J use rho squared while the torsion formula itself only has rho to the first power

They answer different questions. The stress at a point, $\tau = T\rho/J$ , is linear in $\rho$ because shear strain grows in direct proportion to distance from the axis. But $J = \int \rho^2\,dA$ shows up when summing the moment of that stress over the area to match the internal torque: each stress element already carries one factor of $\rho$ , and computing its moment about the axis multiplies by another $\rho$ , giving $\rho^2$ inside the integral. The squared term in $J$ is a bookkeeping consequence of integrating a moment, not a contradiction of the linear stress distribution.

Does the torsion formula work for a shaft with a square or rectangular cross section

No, not directly. The whole derivation rests on radii staying straight and cross-sections rotating as rigid discs, which is only true for circular (solid or hollow) sections by symmetry. Non-circular sections warp out of their own plane when twisted, so the simple $\tau = T\rho/J$ and $\phi = TL/JG$ formulas do not apply; those shapes need the separate, more advanced theory of torsion of non-circular sections, which is outside this topic's scope.

If I have a shaft made of two different materials welded end to end, how do I find the total twist

Treat each segment separately and add. Compute $\phi_i = T_iL_i/(J_iG_i)$ for each segment using its own torque, its own $J$ (which depends on its diameter), and its own shear modulus $G$ , then sum the segment twists algebraically using a single consistent sign convention for torque direction. This is exactly the superposition idea used for shafts of varying cross-section or torque, extended to varying material as well — nothing new is needed beyond applying the same formula segment by segment.

Torsion of cylindrical bars

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads