Question 1

How do I calculate the required length of a shaft key?

Accepted Answer

Find the tangential force at the shaft surface F = 2T/d (the force acts at radius d/2, hence the factor of 2). Then size the key for two failure modes: shear across the width, required L_shear = n·F/(b·S_sy) with the distortion-energy shear allowable S_sy = 0.577·S_y; and bearing (crushing) on the engaged half-height h/2, required L_bearing = 2·n·F/(h·S_y). Take L = max(L_shear, L_bearing) and round up to a standard length. For a square key the bearing check always governs.

Question 2

Why does bearing govern a square key instead of shear?

Accepted Answer

With the common convention — shear allowable 0.577·S_y and bearing allowable S_y on only half the key height — the ratio L_bearing/L_shear for a square key (b = h) is exactly 2·0.577 = 1.154. So the bearing (crushing) length is always 15.4% longer than the shear length and therefore governs. A square key is 'equally strong in shear and crushing' only if the bearing allowable is also taken as 0.577·S_y.

Question 3

How is the key cross-section b×h chosen?

Accepted Answer

The width b and height h are not free — they are fixed by the shaft diameter. This calculator looks the shaft diameter up in the standard band table: DIN 6885-1 (≡ ISO/R 773) for metric keys (e.g. a 22–30 mm shaft → 8×7 mm key) or ASME/ANSI B17.1 for inch keys (e.g. a 7/8–1¼ in shaft → 1/4 in square key). You can override b and h manually if you are using a non-standard key.

Question 4

Should the bearing allowable be S_y or S_y/2?

Accepted Answer

Shigley's Mechanical Engineering Design uses the yield strength S_y directly as the bearing (crushing) allowable, applying the factor of safety n separately; this is the default here and reproduces the textbook worked examples exactly. Some machine-design texts are more conservative and use S_y/2 as the allowable bearing stress, which roughly doubles the required bearing length. The tool exposes a labeled {S_y | S_y/2} toggle so you can pick the basis your standard or instructor requires.

Question 5

What if the required key is very long?

Accepted Answer

If the standard length exceeds about 1.5 times the shaft diameter the key load distributes very non-uniformly along its length, so simply specifying a longer key does not keep adding capacity. The fix is to use two keys 180° apart, a larger key cross-section or shaft, or a splined connection — not to divide the torque between two keys (load sharing is uneven). The tool flags this with a long-key warning.

Key & KeywayDesign Calculator

Key & Keyway
Design Calculator