How this calculation works
A rectangular shaft key carries the tangential force needed to transmit torque between a shaft and hub. Two different projected areas matter: one for key shear and another for bearing compression. This page keeps those average stresses separate.
F = 2T/d; τkey = F/(bL); σbearing = F/[(h/2)L]A worked example
For 100 N m on a 40 mm shaft, tangential force is 5000 N. With a 12 mm wide, 8 mm high key and 40 mm effective length, average shear is 10.4167 MPa and average bearing stress is 31.25 MPa.
Good to know
One rectangular key and uniform load distribution only. Edge loading, fatigue, keyway stress concentrations, hub strength and shaft weakening are absent. The page does not supply a standard key size or an allowable material stress.
Reference: RoyMech — key shear and bearing stress (opens in a new tab) ↗
What this model answers
A rectangular shaft key carries the tangential force needed to transmit torque between a shaft and hub.
Two different projected areas matter: one for key shear and another for bearing compression. This page keeps those average stresses separate.
The Hollow shaft torsion stress and twist calculator checks the shaft’s nominal pure-torque stress and twist. A key’s shear and bearing areas require their own checks.
Keep the input conventions consistent
Use the effective straight engagement length rather than automatically taking the full manufactured key length.
The simplified bearing area assumes half of the key height is engaged in the loaded face. Actual chamfers and keyway depths can change that assumption.
Read the result carefully
A longer key reduces both average stresses proportionally in the uniform-load model.
Increasing width reduces shear stress, while increasing height reduces the assumed bearing stress. Neither change proves that the weakened shaft section has adequate strength.
A key transfers torque through forces acting at the shaft radius. The same force can produce distinct average shear and bearing stresses because the resisting areas differ. Key engagement and usable contact length must follow the entered model; a longer physical key is not automatically fully effective at both ends.
Work through the numbers
For 100 N m on a 40 mm shaft, tangential force is 5000 N.
With a 12 mm wide, 8 mm high key and 40 mm effective length, average shear is 10.4167 MPa and average bearing stress is 31.25 MPa.
Where the calculation stops
One rectangular key and uniform load distribution only.
Edge loading, fatigue, keyway stress concentrations, hub strength and shaft weakening are absent. The page does not supply a standard key size or an allowable material stress.
A few questions, answered
Why is bearing area based on half the height?
That is this simplified model’s assumed engagement depth; real shaft and hub keyways may require their actual contact depths.
What should I change in the comparison?
Double effective engagement length and compare both average stresses.