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Hollow shaft 30 kW at 240 rpm: diameters

Short answer

Torque T=60P/(2πN)=1.194 kN·m. Using T=(π/16)τD³(1-k⁴) with k=0.5, τ=100 MPa, gives D=40.2 mm, d=20.1 mm.

Determine the internal and external diameters of a hollow shaft to transmit a power of 30 kW at 240 rpm if the shear stress must not exceed 100 MPa. The outside diameter is to be twice the inside diameter.

Data: P=30P = 30 kW, N=240N = 240 rpm, τmax=100 MPa\tau_{max} = 100\ \text{MPa}, D=2dD = 2d (so k=d/D=0.5k = d/D = 0.5 and 1−k4=0.93751-k^4 = 0.9375).

Torque

T=60P2πN=60(30×103)2π(240)=1193.7 N⋅m=1.1937×106 N⋅mmT = \frac{60P}{2\pi N} = \frac{60(30\times10^3)}{2\pi(240)} = 1193.7\ \text{N·m} = 1.1937\times10^{6}\ \text{N·mm}

Diameters

T=π16τ D3(1−k4)⇒D3=16Tπτ(1−k4)=16(1.1937×106)π(100)(0.9375)=6.4846×104 mm3T = \frac{\pi}{16}\tau\,D^3(1-k^4) \Rightarrow D^3 = \frac{16T}{\pi\tau(1-k^4)} = \frac{16(1.1937\times10^{6})}{\pi(100)(0.9375)} = 6.4846\times10^{4}\ \text{mm}^3 D=40.18 mm,d=D2=20.09 mmD = 40.18\ \text{mm}, \qquad d = \frac{D}{2} = 20.09\ \text{mm}

Answer: external diameter ≈40.2\approx 40.2 mm, internal diameter ≈20.1\approx 20.1 mm. (In practice adopt 41 mm and 20.5 mm.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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