How shaft diameter is calculated
A shaft carries torque from the power it transmits and bending from gears, pulleys and belt pull. The two stresses combine through the von Mises criterion to give the smallest diameter that won't yield. A rotating shaft sees fully reversed bending every revolution, though, so fatigue usually sets the final size.
- Torque:
T = P/ω = 9549·P[kW] / n[rpm]N·m - Yield:
d = [ 16·n/(π·Sy) · √(4M² + 3T²) ]^⅓ - Fatigue (DE-Goodman):
d = [ 16·n/π · (2·Kf·M/Se + √3·Kfs·T/Sut) ]^⅓
Worked example: motor shaft
A 7.5 kW motor at 1450 rpm transmits 49.4 N·m. With 120 N·m of bending, 1045 cold-drawn steel and a safety factor of 2, yield needs 17.0 mm, so the standard 17 mm size. Add an end-mill keyseat and check fatigue, and the shaft needs 30.3 mm, so a 32 mm standard size: that's why motor shafts look generous.
Good to know
- Keyseats, shoulders and retaining-ring grooves concentrate stress. Size the shaft at those sections, not in the plain middle.
- Deflection at gears and bearings, and the first critical speed, can govern long or fast shafts even when stress is fine.
- Aluminum has no true endurance limit, so its fatigue strength keeps falling with cycles.
- Pro adds the fatigue diameter with surface, size and stress-raiser factors, and the angle of twist.