Torsion of circular shafts

A torque is a moment about a member's own axis. Drive shafts in pumps and turbines, the shafts of motors that drive conveyors and lift gates, drilled piles twisted into the ground, and steel tubes in sign structures all carry torque. In buildings and bridges, torsion also arises when loads act off the shear centre of a beam, as in a spandrel beam that supports a floor slab on one side only.

This lesson develops the theory for circular shafts, solid or hollow, where the results are exact within the elastic range and remarkably simple. It then shows how to transmit power, how to handle shafts with several segments or redundant supports, and how thin-walled closed sections resist torque.

How a circular shaft deforms

Draw a grid of lines on the surface of a rubber rod and twist it. The circles around the rod stay circles and stay the same size, and the cross-sections stay flat and simply rotate relative to one another. Straight lines along the rod become helices. These observations, which can be proved by symmetry for any circular shaft, lead to the key geometric result.

Consider a shaft of length \(L\) whose end rotates by an angle \(\phi\) (the angle of twist, in radians) relative to the other end. A line on the surface at radius \(c\) tilts by an angle \(\gamma\), which is the shear strain. For a small angle, the arc length at the end is \(c\,\phi = L\,\gamma\), so the shear strain at the surface is \(\gamma_{\max} = c\,\phi/L\). At any interior radius \(\rho\) (rho), the same reasoning gives

Shear strain grows linearly from zero at the centre to a maximum at the outer surface. If the material is linear elastic, \(\tau = G\gamma\), so the shear stress also varies linearly with radius.

The torsion formula

Equilibrium connects the stress to the torque. A ring of area \(dA\) at radius \(\rho\) carries a shear force \(\tau\,dA\), which has a moment \(\rho\,\tau\,dA\) about the axis. Adding these up over the section must give the internal torque \(T\). With \(\tau = \tau_{\max}\,\rho/c\):

The integral \(J\) is the polar moment of inertia of the cross-section. Rearranging gives the torsion formula:

For a solid circular section of radius \(c\) (diameter \(d\)), and for a hollow tube with outer and inner radii \(c_o\) and \(c_i\):

Because \(J\) depends on the fourth power of the radius, material far from the centre is very effective. The material near the centre carries little stress and contributes little to \(J\), which is why hollow shafts are lighter for the same strength. Long drive shafts in vehicles and the tubular members of sign gantries take advantage of this.

Shear stresses on the cross-section are accompanied, by equilibrium of a small element, by equal shear stresses on longitudinal planes. That has practical consequences. A wooden shaft, weak in shear along the grain, splits lengthwise. A brittle material such as chalk or cast iron fails on a helix at 45° to the axis, because the 45° planes carry the maximum tensile stress, as lesson 8 will show. Twist a piece of chalk and you will see the helical fracture.

Angle of twist

Combining \(\gamma = \rho\phi/L\), \(\tau = G\gamma\) and \(\tau = T\rho/J\) gives the angle of twist for a uniform shaft carrying a constant internal torque:

The product \(GJ\) is the torsional rigidity, the torsional counterpart of the axial rigidity \(AE\). The angle is in radians. Limiting twist matters for machinery, where too much twist causes vibration and misalignment, and for structures, where twist of a beam can crack finishes or damage connections. A common rule for line shafting is a twist of no more than about 1° over a length of 20 diameters, although specific limits depend on the application.

For shafts with several segments, or with several gears applying torques along their length, use the same bookkeeping as for axial bars: find the internal torque in each segment with the method of sections, then add the segment twists, \(\phi = \sum T_iL_i/(G_iJ_i)\). Use the right-hand rule for signs: a torque whose vector points out of the cut face is positive.

Power transmission

Shafts usually exist to transmit power. Power is torque times angular speed:

With \(T\) in N·m and \(\omega\) in rad/s, \(P\) is in watts. Rotational speed is often given in revolutions per minute (rpm), so \(\omega = 2\pi n/60\). At a given power, the torque is inversely proportional to speed. That is why a gearbox that slows a motor down to drive a heavy load needs a much thicker output shaft than input shaft.

Statically indeterminate shafts

A shaft fixed at both ends and loaded by a torque between them is statically indeterminate, exactly like the axial bar between two walls in lesson 2. The method is identical: equilibrium of torques, a compatibility condition that the total twist between the fixed ends is zero, and the torque–twist relation \(\phi = TL/(GJ)\).

Compound shafts, such as a steel core inside a bonded aluminum sleeve, are also indeterminate. Both parts twist through the same angle, so the torque divides in proportion to their torsional rigidities \(GJ\), just as load divided in proportion to \(AE\) in the composite post.

Thin-walled closed sections and noncircular shafts

For noncircular sections, the cross-sections warp out of plane when twisted, and the simple formulas no longer apply. Solid rectangular bars, for example, have their maximum shear stress at the middle of the long side, not at the corners, and their stiffness is much lower than a circle of equal area. Open sections such as I-beams and channels are especially weak and flexible in torsion, which is why they are poor choices for members that must resist twisting.

Thin-walled closed tubes of any shape, such as rectangular hollow structural sections and box girders, are an important exception. They resist torque efficiently through a constant shear flow \(q = \tau\,t\) around the wall. Equilibrium gives Bredt's formula:

where \(A_m\) is the area enclosed by the centreline of the wall and \(t\) is the local wall thickness. The highest stress occurs where the wall is thinnest. Box girders in curved highway bridges rely on this closed-section behaviour to carry the large torques produced by the curvature.

Designing a shaft with code

In practice a shaft must satisfy both a stress limit and a twist limit, and the governing one depends on its length. The program below sizes a solid steel shaft for a given power and speed against both criteria and reports which one controls.

# Size a solid steel shaft for power, stress and twist limits
import math

G = 77000.0                 # MPa (N/mm^2)
tau_allow = 55.0            # MPa
def size_shaft(power_kW, rpm, length_m, twist_limit_deg_per_m=0.25):
    T = power_kW * 1e6 / (2 * math.pi * rpm / 60)          # N*mm
    d_stress = (16 * T / (math.pi * tau_allow)) ** (1 / 3)
    phi_allow = math.radians(twist_limit_deg_per_m * length_m)
    # phi = T L / (G J) with J = pi d^4 / 32  ->  d^4 = 32 T L / (pi G phi)
    d_twist = (32 * T * length_m * 1000 / (math.pi * G * phi_allow)) ** 0.25
    d = max(d_stress, d_twist)
    rule = "stress" if d_stress >= d_twist else "twist"
    print(f"{power_kW:5.0f} kW at {rpm:5.0f} rpm: T = {T/1e6:6.3f} kN*m  "
          f"d(stress) = {d_stress:5.1f} mm  d(twist) = {d_twist:5.1f} mm  -> {rule} governs, use {math.ceil(d/5)*5} mm")

size_shaft(45, 1750, 1.5)
size_shaft(45, 300, 1.5)
size_shaft(200, 900, 4.0)
print("Same shafts with a looser limit of 2.0 degrees per m:")
size_shaft(45, 1750, 1.5, 2.0)
size_shaft(45, 300, 1.5, 2.0)
size_shaft(200, 900, 4.0, 2.0)

   45 kW at  1750 rpm: T =  0.246 kN*m  d(stress) =  28.3 mm  d(twist) =  52.2 mm  -> twist governs, use 55 mm
   45 kW at   300 rpm: T =  1.432 kN*m  d(stress) =  51.0 mm  d(twist) =  81.2 mm  -> twist governs, use 85 mm
  200 kW at   900 rpm: T =  2.122 kN*m  d(stress) =  58.1 mm  d(twist) =  89.6 mm  -> twist governs, use 90 mm
Same shafts with a looser limit of 2.0 degrees per m:
   45 kW at  1750 rpm: T =  0.246 kN*m  d(stress) =  28.3 mm  d(twist) =  31.1 mm  -> twist governs, use 35 mm
   45 kW at   300 rpm: T =  1.432 kN*m  d(stress) =  51.0 mm  d(twist) =  48.3 mm  -> stress governs, use 55 mm
  200 kW at   900 rpm: T =  2.122 kN*m  d(stress) =  58.1 mm  d(twist) =  53.3 mm  -> stress governs, use 60 mm

With the strict limit of 0.25° per metre, typical of precision line shafting, twist governs every case. Because the limit is stated per metre, the shaft length cancels out of the twist criterion. With a looser limit of 2° per metre, the required twist diameters shrink by a factor of \(8^{1/4} \approx 1.68\), and stress governs the two high-torque shafts. Changing the inputs and watching which criterion controls is a quick way to build engineering judgment.

Keyways, shoulders and fatigue

Real shafts are not smooth cylinders. Gears and pulleys are fixed with keys that sit in keyways cut into the shaft, bearings sit against shoulders where the diameter changes, and retaining rings sit in grooves. Each of these raises the local shear stress above \(Tc/J\) by a stress concentration factor, just as holes did for axial members in lesson 2. A shoulder with a generous fillet radius may have a factor only slightly above 1, while a sharp-cornered keyway can approach 2 or more.

Rotating shafts that also carry bending from belts or gears see a full reversal of bending stress every revolution, so a shaft turning at 1,750 rpm accumulates millions of cycles in a few days. Fatigue, not static strength, therefore governs most machine shafts, and the cracks almost always start at keyways, fillets and grooves. The allowable stress of 55 MPa used in the examples above is deliberately conservative to allow for these effects in a preliminary design. A full shaft design combines torsion with bending, as lesson 7 shows, and then checks fatigue at each discontinuity.