How a Shaft Actually Carries Torque

Why this matters

Shafts almost never break from the torque they carry. They break from something else entirely, at a place where somebody cut, stamped, pressed or corroded the one part of the shaft that was doing the work. Understanding where torsional stress actually lives makes both of those facts obvious, and it kills a whole family of bad repairs: the heavier shaft that fixes nothing, the polished-out crack, the weld build-up that puts a fresh crack-starter exactly where the old one was.

This article is mostly a list of things that do not carry the torque. That turns out to be the useful list.

Before you handle it

If you are removing a shaft, isolate and lock out under 29 CFR 1910.147, confirm the wheel has stopped rather than assuming the contactor drop stopped it, and restrain a spring-loaded motor base before belts come off, because the tensioned base is stored energy under that standard. A puller under load is stored energy too: stay out of line with the screw, use a shield or restraint, and relieve the load before releasing. A fan shaft with a wheel on it is heavy and awkward; rig it or use two people rather than manhandling it out of a housing, because a dropped shaft breaks feet.

A broken shaft has fracture edges that will open a hand, so handle the pieces on a bench with cut-resistant gloves. Do not grind or wire-brush a fracture face before it has been looked at: the surface is the evidence, and abrasive work on it is also a respirable dust exposure that wants local exhaust. If anyone proposes welding a repair, that is a metallurgical decision before it is a welding decision, and the welding fume is its own hazard route: on stainless or chrome-bearing alloy the arc generates hexavalent chromium, which needs local exhaust ventilation and respiratory protection under a written program per 29 CFR 1910.134 alongside the chromium(VI) standard at 29 CFR 1910.1026. Eye protection and gloves do not address an inhalation route.

The stress the shaft actually sees

Torque applied to a round shaft produces shear stress that varies linearly with radius: zero at the centreline and maximum at the outside surface. For a solid round shaft the maximum is

tau = 16 T / (pi d cubed)

and the polar second moment of area, the geometric property that resists twist, is J = pi d to the fourth over 32.

Two consequences follow, and they are the whole article.

The outer material does nearly all of the work. Integrate the contribution to J from 0.8 of the radius out to the surface and you get 1 minus 0.8 to the fourth, which is 0.59. The outer fifth of the radius carries 59 percent of the shaft's torsional capacity. The inner half of the radius, which is a quarter of the cross-sectional area, contributes about 6 percent.

Anything that interrupts the surface interrupts the load path where it is densest. A keyway, a step, a snap-ring groove, a cross-hole, a stamped identification number, a corrosion pit. None of those would matter much if the stress were spread evenly through the section. It is not.

What does not carry the torque

The core. Covered above. This is why a hollow shaft costs so little: with an inner diameter half the outer, J falls to 1 minus 0.5 to the fourth, which is 94 percent, while cross-sectional area and therefore weight fall to 75 percent. You give up 6 percent of the torsional capacity to remove a quarter of the mass. Every driveshaft you have ever seen is hollow for exactly this reason.

The setscrew. A setscrew holds a hub axially and keeps it from creeping. It transmits torque only through friction and a small point indentation, a fraction of what a key of the same size joint transmits, and once it has dimpled the shaft it has also created a stress raiser on the surface that matters. Treating a setscrew as a torque device is how sheaves end up spinning on shafts.

A clearance-fit hub. If there is measurable clearance between hub bore and shaft, the fit is carrying nothing and the key is carrying all of it. That distinction shows up as a wallowed keyway rather than as a broken shaft, and a sibling article covers it.

Paint, plating, oxide and surface coatings. They contribute no strength. What they can do is hide the surface condition that decides the shaft's fatigue life, which is why the first step in inspecting a suspect shaft is cleaning it enough to see the metal.

The material's strength grade, on its own. Doubling the yield strength doubles the static torque the shaft could survive, and static torque is almost never what breaks it. Fatigue strength does scale with tensile strength, but the surface factor and the stress concentration factor scale against you at the same time: a higher-strength steel is more notch sensitive, so it gets less benefit from its own strength at a sharp corner than a milder steel does.

The nominal stress you computed. This is the important one. The formula above gives the stress in a plain round section with nothing cut into it. At any geometric feature the real stress is that number times a concentration factor. Published factors for a keyway in torsion cluster around 2 to 3, and they depend strongly on the fillet radius at the keyway corner and on how the keyway is ended. For a shoulder step in torsion, the factor falls steeply as the fillet radius grows relative to the shaft diameter: going from a radius of 2 percent of the diameter to 10 percent takes it from roughly 2.5 down to about 1.5 at a typical step ratio, and the value also depends on how big the step is. The sharp corner is the variable, not the feature. A generous radius costs nothing to specify and buys most of the difference.

The worked case: a shaft that broke at 109 psi

A belt-driven fan shaft, 1.500 in diameter, driven by a 2 hp motor at 1750 rpm, sheave pressed on and overhung 2.0 in beyond the bearing. It broke at the edge of the sheave hub after about four years.

Torsional stress. Torque is 5252 x hp / rpm in lb-ft, so 5252 x 2 / 1750 gives 6.00 lb-ft, which is 72 lb-in.

tau = 16 x 72 / (pi x 1.5 cubed) = 1152 / 10.60 = 109 psi

Shear yield for a medium-carbon shafting steel is roughly 0.58 of its tensile yield, so for a yield around 45 ksi that is about 26 ksi. The shaft is at 0.4 percent of it. Even a hard jam at three times motor rated torque puts it at about 326 psi, which is 1.3 percent. Torque was never within two orders of magnitude of a problem, and nothing about a bigger or better shaft would have helped that number.

Bending stress at nameplate duty. Second moment of area I = pi x 1.5 to the fourth over 64, which is 0.2485 in to the fourth, and c is 0.75 in. Belt pull at the drive table figure is 150 lbf, overhung 2.0 in, so the bending moment at the bearing is 300 lb-in.

sigma = M c / I = 300 x 0.75 / 0.2485 = 905 psi

That stress is fully reversed, because the belt pull direction is fixed in space and the shaft rotates inside it. Every revolution is a complete tension-compression cycle, so at 1750 rpm the shaft accumulates about 2.5 million cycles a day.

905 psi is still nowhere near an endurance limit, which for a medium-carbon steel in a decent surface condition is in the tens of thousands of psi. At nameplate duty this shaft could not have broken, and that conclusion is the finding, not a dead end. It means the load that broke it is not the load on the nameplate.

What was actually there. Two things, found on inspection. The drive had been chased for a squeal over several years and measured at about three times the drive manufacturer's tension figure. And under the sheave hub, revealed when it came off, the shaft was pitted with corrosion and fretted at the hub edge, with the characteristic reddish-brown powder.

Redo it. Belt pull about 450 lbf, moment 900 lb-in:

sigma = 900 x 0.75 / 0.2485 = 2,716 psi

Fretting at the edge of a press fit is not a geometric notch and it does not have a clean concentration factor. Published fatigue strength reductions for fretting at a press-fit edge are commonly quoted in the 2 to 5 range and depend on contact pressure, slip amplitude and the materials, so this is a range you reason with rather than a spec you apply. At a factor of 4, the local stress amplitude is about 10.9 ksi. Meanwhile the corrosion pitting cuts the endurance limit of a medium-carbon steel to well under half its polished value.

That is the honest conclusion available from a field calculation: the shaft was not remotely at risk at rated duty, and the combination of a drive at triple tension with a corroded, fretted press-fit edge moves the amplitude and the endurance limit into the same range. It does not prove the failure to a decimal place, and it should not be written up as if it did. It does tell you what to change.

Reading which load broke it

The fracture face separates the two cases without any arithmetic, and it is worth knowing because it settles arguments.

A rotating-bending fatigue fracture is flat and perpendicular to the shaft axis. It has a smooth, often beach-marked region where the crack grew slowly, and a rougher final region where the remaining section let go at once. The ratio of the two matters: a small final zone means the nominal stress was low and the crack had to grow a long way before the shaft could not hold, which points at a stress raiser rather than an overload. A large final zone means the shaft was genuinely close to its limit.

A torsional failure on a ductile shaft looks nothing like that. A static torsional overload gives a flat, transverse face with visible twisting and smeared material. A torsional fatigue crack tends to run at about 45 degrees to the axis, following the plane of maximum tensile stress, which produces a helical fracture.

If you find a flat transverse fatigue face and someone is telling you the shaft was overloaded in torque, the fracture disagrees with them.

Where this changes what you do

Specify the radius, not the diameter. On a stepped shaft, a generous fillet at the shoulder buys more fatigue life than a size increase does, and it is free at manufacture. A sharp corner is a defect that looks like a drawing.

Do not polish out a crack. Removing visible cracking on a rotating shaft does not remove the sub-surface crack front, and grinding removes any compressively stressed skin from rolling or peening while it is at it. That skin is on the surface, which is where all the stress is.

Weld only with the metallurgy settled. Welding on a shaft that sees reversed bending puts a heat-affected zone with different properties, plus residual tensile stress, exactly at the surface where fatigue starts. Where the material or its heat treatment is unknown, the honest answer to the customer is a new shaft.

Fix the tension before you fix the shaft. In the case above, replacing the shaft without measuring the drive gives you the same failure on a fresh part, on the same clock.

Address the fretted joint, not just the fretting. A press fit that has fretted has been micro-slipping, which means the fit is not tight enough for the load reversals it sees. Cleaning it up and reassembling to the same fit reproduces it.

References

• Trade-standard mechanical design references for torsional stress distribution, polar second moment of area, and published stress concentration factors, which are geometry-specific and vary with fillet radius and step ratio
• ISO 15243 and general failure-analysis practice for the interpretation of fatigue fracture surfaces, beach marks and final fracture zones
• 29 CFR 1910.147 (hazardous energy, stored energy in a tensioned drive and a loaded puller); 29 CFR 1910.134 and 29 CFR 1910.1026 (respiratory protection program; chromium(VI) in welding fume)
• Drive manufacturer documentation for tension-versus-deflection values, since drive over-tension is the load that most often is not in the calculation
• See related: What a Keyway Does and How It Fails; Why Shaft Deflection Matters More Than Shaft Strength