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Solid shaft 300 kW, 120 rpm with twist limit

Short answer

Mean torque 23.87 kN·m, maximum 31.04 kN·m. Strength criterion gives diameter 116.5 mm, stiffness 18.2 mm; the larger governs: required diameter ≈116.5 mm.

A solid shaft is to transmit 300 kW at 120 rpm. Determine the diameter if the allowable shear stress is 100 N/mm2100\ \text{N/mm}^2 and the allowable angle of twist is 30∘30^\circ per diameter length of the shaft. Assume that the maximum torque is 1.3 times the mean torque. Take G=105 MPaG = 10^5\ \text{MPa}.

Data: P=300P = 300 kW, N=120N = 120 rpm, τallow=100 N/mm2\tau_{allow} = 100\ \text{N/mm}^2, G=105 N/mm2G = 10^5\ \text{N/mm}^2, allowable twist 30∘30^\circ per diameter length (taken as stated, L=dL = d), Tmax=1.3 TmeanT_{max} = 1.3\,T_{mean}.

Torques

Tmean=60P2πN=60(300×103)2π(120)=23873.2 N⋅m,Tmax=1.3Tmean=31035.2 N⋅mT_{mean} = \frac{60P}{2\pi N} = \frac{60(300\times10^3)}{2\pi(120)} = 23873.2\ \text{N·m}, \qquad T_{max} = 1.3T_{mean} = 31035.2\ \text{N·m}

Strength

d3=16Tmaxπτ=16(3.1035×107)π(100)=1.5806×106 mm3⇒d1=116.49 mmd^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(3.1035\times10^{7})}{\pi(100)} = 1.5806\times10^{6}\ \text{mm}^3 \Rightarrow d_1 = 116.49\ \text{mm}

Stiffness

θ=TmaxLGJ=32 Tmax dπGd4=32TmaxπGd3,θ=30∘=0.5236 rad\theta = \frac{T_{max}L}{GJ} = \frac{32\,T_{max}\,d}{\pi G d^4} = \frac{32T_{max}}{\pi G d^3}, \quad \theta = 30^\circ = 0.5236\ \text{rad} d3=32(3.1035×107)π(105)(0.5236)=6.0375×103 mm3⇒d2=18.21 mmd^3 = \frac{32(3.1035\times10^{7})}{\pi(10^5)(0.5236)} = 6.0375\times10^{3}\ \text{mm}^3 \Rightarrow d_2 = 18.21\ \text{mm}

Design

The larger value governs, so strength controls the design: d=116.49d = 116.49 mm.

Answer: diameter of the shaft ≈116.5 mm\approx 116.5\ \text{mm} (adopt 120 mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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