A shaft that survives its torque still has to hand that torque to something — a gear, a sprocket, a coupling hub. The usual handover is a parallel key: a rectangular bar of steel sitting half in a slot in the shaft and half in a slot in the hub, sheared across its width by the drive.

Its check is $\tau = \dfrac{2T}{d\,w\,L}$ — tau equals two T over d w L. $\tau$ is the average shear stress in the key, in N/mm²; $T$ is the torque passing through it, in N·mm; $d$ is the shaft diameter at the keyway; $w$ is the key width, the dimension crossing the shear plane; and $L$ is the key length, running along the shaft. Width and length are different letters doing genuinely different jobs, and mixing them up is the setup mistake this lesson exists to prevent.

The 2 is not decoration. The torque arrives at the key as a tangential force at the shaft SURFACE, $F = \dfrac{2T}{d}$, because the arm is the radius and the relation is written in terms of the diameter. Divide that force by the key's shear area $w \times L$ and you have the whole thing. Drop the 2 and you report half the stress you actually have.

And now the design intent, which is the real lesson. A key is meant to be the weakest part of the drive — softer than the shaft and softer than the hub — so that an overload shears a two-dollar bar you can drive out with a punch, instead of a shaft that needs the machine stripped. It is a mechanical fuse. Fit a key harder than the shaft and you have inverted that on purpose, turning a cheap failure into an expensive one. Two footnotes for later: shear is only half the check (a key also crushes on its side faces), and past about 1.5 shaft diameters of length the extra key is decorative, because the shaft's own wind-up stops loading it evenly.