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PLC • VFD • HMI • Control Panels • In

18/08/2026

22 kW motor at 960 RPM — what is the shaft torque? And what diameter shaft does it need? Two calculations in one worked example.

SHAFT TORQUE & POWER FORMULAS:

T (N·m) = P (kW) × 9,550 / N (RPM)
P (kW) = T (N·m) × N (RPM) / 9,550
d (m) = ∛(16T / π × τ_allow)

GIVEN:
P = 22 kW | N = 960 RPM | Material: mild steel (τ_allow = 40 MPa)

STEP 1 — Shaft Torque:
T = 22 × 9,550 / 960 = 210,100 / 960 = 218.9 N·m

T = 219 N·m

STEP 2 — Shaft Diameter:
d = ∛(16 × 219 / (π × 40,000,000))
d = ∛(3,504 / 125,663,706)
d = ∛(2.789 × 10⁻⁵)
d = 0.0306 m = 30.6 mm

Select d = 35 mm standard shaft (round UP — never down)

CONSTANT 9,550 EXPLAINED:
9,550 = 60,000 / (2π) = unit conversion from kW and RPM to N·m
For HP and RPM: T (lb·ft) = HP × 5,252 / N

CROSS-CHECK:
Power back-calculated: P = 219 × 960 / 9,550 = 22.03 kW ✓

Save this — the most fundamental relationship in mechanical power transmission.

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