Shaft Diameter from Allowable Torsional Shear

The torsion formula for a round shaft says the shear stress at radius $r$ is $\tau = Tr/J$. For a solid round section the polar second moment of area is $J = \pi d^4/32$, and the stress is worst at the surface, where $r = d/2$. Substitute both and the $d$s collapse:

\[ \tau_{max} = \frac{T(d/2)}{\pi d^4/32} = \frac{16T}{\pi d^3} \qquad \Longrightarrow \qquad d = \sqrt[3]{\frac{16T}{\pi\tau}} \]

Torsional shear varies linearly from zero on the axis to maximum at the surface, which tells you immediately that the core of a solid shaft is nearly idle. That is the argument for hollow shafts: removing the inner half of the diameter takes away only about 6% of the torsional strength while removing 25% of the mass, and it is why driveshafts and aircraft transmission shafts are tubes.

The cube root is generous

Because diameter enters cubed, torque capacity is extremely sensitive to size — and size is correspondingly insensitive to torque. Eight times the torque needs only twice the diameter. A 10% increase in diameter buys a 33% increase in torque capacity. This is why shafts rarely look as heavy as the loads they carry would suggest, and why "go up one size" is such an effective piece of shop advice.

What this equation leaves out, which is most of it

Almost no real shaft carries steady torsion alone. A shaft with a pulley, a gear or a sprocket on it is also a beam, carrying bending from the belt pull or the tooth loads. And because the shaft turns, a point on its surface moves from tension to compression and back on every revolution: the bending stress is fully reversed, at the shaft speed, and the shaft is in high-cycle fatigue rather than static loading. A shaft at 1,750 rpm accumulates a million cycles in under ten hours.

The classical way to combine them is an equivalent torque. On the maximum-shear-stress (Tresca) theory, $T_e = \sqrt{M^2 + T^2}$, and the same equation is then used with $T_e$ in place of $T$. Modern practice follows a fatigue method instead — the ASME shaft equation, or a Soderberg/Goodman construction with the mean and alternating components treated separately, and with stress-concentration and surface-finish factors applied. Either way, the number from this page is a floor, not an answer.

Two more subtractions

A keyway removes material and, more importantly, concentrates stress at its corners, with a stress concentration factor around 2 to 3 depending on the fillet radius. The common allowance is to size a keyed shaft 5–10% larger than the plain calculation, and to insist on a properly radiused keyway rather than a sharp-cornered one. Every shoulder, snap-ring groove and cross-hole does something similar.

And stiffness can govern instead of strength. A shaft that is strong enough may still wind up too far under load, which upsets gear mesh alignment and timing. The traditional limit is about one degree of twist per 20 diameters of length, from $\phi = TL/(GJ)$. Lateral stiffness matters too: the shaft's first bending critical speed must sit well clear of the running speed, usually by 20% or more.

Choosing the allowable

An allowable shear stress is a design decision, not a material property. It comes from the material's yield or endurance strength divided by a factor of safety, reduced by whatever the fatigue analysis demands, and it is what codes and standards exist to pin down for a given application. This page will use whatever number you give it, and the honesty of the answer is entirely the honesty of that number.