Shear Stress in a Parallel Key

A key is a small rectangular bar sitting half in a slot in the shaft and half in a slot in the hub, and its only job is to stop the two from turning relative to one another. The torque arrives at the shaft surface as a tangential force, $F = 2T/d$, and the key resists it across a shear plane whose area is its width times its length. So:

\[ \tau = \frac{F}{wL} = \frac{2T}{d\,w\,L} \]

The shear plane is the one lying flush with the shaft surface, splitting the key into its shaft half and its hub half. That is why the width appears and not the height — the key is being sliced along its length, not crushed flat.

The second check, which is often the one that governs

A key can also fail in crushing (bearing) on its side faces, where the hub pushes against it. The bearing area is the key length times only the half of the height standing proud of the shaft, so for a key sunk half its depth:

\[ \sigma_{crush} = \frac{4T}{d\,h\,L} \]

Compare the two. For a square key, $h = w$, so the crushing stress is exactly twice the shear stress. Whether that matters depends on the material: a ductile steel's allowable compressive stress is roughly twice its allowable shear, so for a square key in steel the two checks come out close together and either can govern. For a flat (rectangular) key, where $h < w$, crushing is more likely to be the limit. Check both, always.

The key is meant to break

This is the design intent and it is worth stating plainly: the key is normally specified in a material softer than both the shaft and the hub, so that an overload shears the key. A sheared key is a part you drive out with a brass punch and replace in ten minutes. A twisted shaft or a split hub is a rebuild. Specifying a key harder than the shaft — which happens when someone substitutes whatever bar stock is on the shelf — inverts the intended failure order and turns a cheap failure into an expensive one.

Length has diminishing returns

Doubling a key's length does not double its capacity in practice. The shaft winds up elastically along the key's length, so the end where the torque enters carries far more than its share and the far end carries almost nothing. Beyond roughly 1.5 shaft diameters of key length the extra material contributes very little. When one key is genuinely not enough, the answers are two keys at 180°, a splined hub (which is effectively many small keys cut integrally, and spreads the load properly), or a keyless friction fit — a taper lock, a shrink disc, or a polygon connection.

Standard proportions, and why they exist

Key sizes are standardised against shaft diameter — the width is conventionally about a quarter of the shaft diameter and the height a quarter or a sixth, with the exact series given in the public dimensional standards (ANSI B17.1 and ISO/R 773). Those proportions are not arbitrary: they are chosen so that a key of the standard size, in the standard material, is a little weaker than the shaft it sits in. If your calculation says a standard key is badly overstressed for the torque you are passing, the real conclusion is usually that the shaft is undersized too.