Torsion: From Geometry to Shaft Design

Axial loading gave you a very comfortable answer. Pull a bar along its own axis and the stress is load over area, the same at every point of the section, and the strain is the same everywhere too. Torsion is the next load case, and almost none of that comfort survives. Here is the question. A round shaft carries a torque T. How is that torque shared out across the cross section, and how much can the shaft carry before the material gives way? We will answer it the way it should be answered, by looking at what the twist does to the material and letting the algebra follow. So here is the shaft. A solid circular bar, with its axis running along z, clamped at the left end and free at the right. Notice that the section is a circle. That matters more than you might think: a square shaft under torque warps out of plane, and everything we are about to do would fail for it. Now hang a torque T on the free end, and rule one straight line along the surface, running from a point A at the fixed end to a point B at the far end. Apply the torque, and the shaft twists. The far end rotates through an angle we will call phi. Watch the cross sections while it turns. Each one rotates in its own plane, and because it is a circle it lands straight back onto itself. Nothing warps out of plane, nothing changes diameter, and no section slides along the axis. Plane sections stay plane. Now paint four radii onto the end face. They swing round together, all through the same angle, and every one of them is still perfectly straight. The section turns like a rigid disc on a hub. The line along the surface is the one thing that does not survive. Here is where it started, in gray, and here is where it has gone. B has been carried round the rim while A stayed put, so the line has tilted over, and that tilt is a shear angle. Notice that the line has barely changed length. There is no stretching here, only shearing. So, three facts, and every one of them came out of the picture rather than out of a material property. Sections stay plane and circular. Each section rotates rigidly, so radii stay straight. And a line along the surface tilts through an angle, which we will call gamma. That last one is the one we can use, because gamma is a strain. Put a number on it, and the whole of torsion falls out.

Now put a number on that tilt. Two pictures. On the left we are looking straight down the axis at the far face of the shaft. On the right I have taken the cylindrical surface at some radius rho, slit it along a line and unrolled it flat, which turns the whole question into plane trigonometry. Start on the left. The whole section has rotated through the angle phi. A material point P sitting at radius rho has been carried round to P prime, and the distance it has travelled is the arc rho phi. Radius times angle, with the angle measured in radians. And there is the key fact already. The distance a point moves is proportional to how far out it sits. Bring the point in toward the axis and it travels less. Push it out to the surface and it travels the most. Settle it back somewhere in between, and remember that a point sitting exactly on the axis does not move at all. Now the right hand picture. The line A B was ruled straight along the shaft, parallel to the axis, and its length is L. A sits at the fixed end and does not move. B sits on the far face, and B has just been carried sideways by exactly that same distance, rho phi. So the line has swung round to A B prime, and the angle at A between where it was and where it is now is the shear strain, gamma. Its tangent is that sideways movement divided by the length: rho phi over L. Torsional strains are small, well under a degree in any shaft you would actually use, so the tangent is the angle to any accuracy that matters. Shear strain is rho phi over L. Now look at what is fixed in that expression. Phi belongs to the whole shaft. So does L. Both of them are the same number for every single point in this cross section. The only thing that varies from point to point is rho. So shear strain is proportional to radius, and nothing but geometry went into that. Watch it happen. At the axis, where rho is nearly zero, the strain almost vanishes. Halfway out it is half of its largest value. And at the outer surface, at radius c, it reaches its maximum, c phi over L. Divide the two and the shaft's own dimensions cancel out. Gamma is rho over c times gamma max. A straight line, zero at the centre and largest at the surface. That is the strain distribution, and no material property has appeared yet. One last way to picture it. Think of the shaft as a nest of thin tubes, one inside the next. Each tube shears a little relative to its neighbour, and the further out you go the more of that sliding has piled up. The skin does the most work. The axis does none.

Strain is geometry. Stress needs the material, and for that we need one constitutive law. Below the elastic limit, shear stress and shear strain are proportional. Tau equals G gamma, where G is the shear modulus, the shear counterpart of Young's modulus. Substitute the strain we just derived. Tau is G rho phi over L. G is a material constant, phi and L belong to the whole shaft, so once again the only quantity that varies across the section is rho. Shear stress grows linearly with radius. So here is the stress at one point, a distance rho out from the axis. It acts tangentially, at right angles to the radius, and its size follows that straight line rule. Move the point outward and the arrow grows in exact proportion. Take it right out to the outer surface, at radius c, and the stress reaches its largest value there. Every point on that rim sits at the same radius, so the entire outer skin is at the maximum together. And here is the whole distribution across a radius. Nothing at the axis, rising straight out to tau max at the surface. Written down, that is tau equals rho over c times tau max. Now collect what all of that stress adds up to. Take a small patch of area, d A, sitting at radius rho. The force on that patch is stress times area, tau d A. That force acts tangentially, at a distance rho from the axis, so its moment about the axis is rho times tau d A. That is the torque this one little patch carries. The total torque is the sum over every patch in the section, which is the integral of rho tau d A taken over the area. Now substitute the stress distribution. Tau max over c is the same number everywhere in this section, so it comes straight outside the integral, and what is left behind is the integral of rho squared d A. Stop and look at that integral, because it is the whole point. There is no load in it and no material property in it. It contains nothing but the shape of the cross section, and where that shape puts its area. That integral is the polar second moment of area, J. So work it out for a solid circle. The natural element is a thin ring, because every point on a ring sits at the same radius rho. Give it a thickness d rho, and its area is its circumference times that thickness, two pi rho d rho. Substitute that in and the double integral collapses to an ordinary one in a single variable, running from zero at the axis out to c at the surface. Sweep the ring outward and you are sweeping through exactly that. Integrate rho cubed, and J comes out as pi c to the fourth over two. Notice the fourth power. Doubling the radius multiplies J by sixteen. Put J back into the torque relation and rearrange. The shear stress at radius rho is T rho over J. That is the torsion formula, and it is the exact counterpart of stress equals load over area. And phi came along for the ride. Thread the same substitution back through the strain we derived, and the angle of twist is T L over G J. Strength and stiffness both hang on the same integral. So the whole of torsion sits in J, and J is a statement about where the material is. Which raises exactly one design question. Where should you put it?

So where should the material be? Here is the cross section of a solid steel shaft, thirty millimetres in radius, carrying a torque about its own axis. Go back to what J was actually measuring, the integral of rho squared d A. Every scrap of area is weighted by the square of its distance from the axis. So it is worth asking what the middle of this bar is earning. Shade in everything inside half the radius, everything within fifteen millimetres of the axis. By area, that core is a quarter of the whole bar. A quarter of the weight you carry around, and a quarter of the steel you paid for. But J for that core alone is pi times fifteen to the fourth over two, and fifteen to the fourth is one sixteenth of thirty to the fourth. So a quarter of the material is contributing about six percent of the torsional stiffness. The middle of a shaft is very close to dead weight. So take it out, and put it where the rho squared can get at it. Here is a tube, outer radius fifty millimetres, inner radius forty. Its area is pi times fifty squared minus forty squared, which is pi times nine hundred, and that is exactly the area of the solid bar. Same steel per metre, same weight. Now put numbers against them. Both sections have an area of about two thousand eight hundred and thirty square millimetres, so a metre of each weighs the same. J for the solid bar is pi over two times thirty to the fourth, about one point two seven million millimetres to the fourth. For the tube it is pi over two times fifty to the fourth minus forty to the fourth, and that comes to five point eight million. Four and a half times as much, for the same weight of steel. That is stiffness. Strength is a slightly different sum, because the torsion formula puts the peak stress right at the outer surface, where rho equals c. The torque you can carry at a given allowable stress is tau times J over c, and the tube has the bigger c. Work it out. Forty two thousand cubic millimetres for the solid bar, one hundred and sixteen thousand for the tube. Two point seven times the torque, at exactly the same weight. That is why a drive shaft is a tube. There are limits, of course. Make the wall too thin and the tube buckles, or dents, long before the material anywhere near yields, and a hollow shaft costs more to make and more to join. But the trend is not subtle, and it is why almost every drive shaft, every bicycle frame and every aircraft control tube you will ever meet is hollow. So, the whole argument in four lines. Geometry alone gave us shear strain proportional to radius. An elastic material turned that into shear stress proportional to radius. Integrating the moment of that stress over the area gave the torque, and dragged the polar second moment out into the open as an integral. And because that integral weights area by rho squared, metal near the axis earns almost nothing. And all of it came out of one picture. A line ruled along the surface of a bar, and what happened to that line when you twisted it.