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Steel shaft 200 hp: torque, shear, twist

Short answer

Torque = 9.498 kN·m, maximum shear stress = 36.34 MPa, angle of twist over 5 m = 0.0367 rad (2.10°).

A steel shaft transmits 200 horse power at 150 rpm. If the shaft is 110 mm in diameter, find the torque in the shaft and the maximum shear stress developed. Also, determine the angle of twist for the shaft in the length 5 m. Take G=90 GN/m2G = 90\ \text{GN/m}^2. (1 hp = 746 watt)

Data: P=200P = 200 hp =200×746=149200= 200\times746 = 149200 W, N=150N = 150 rpm, d=110d = 110 mm, L=5L = 5 m, G=90 GN/m2=90×103 N/mm2G = 90\ \text{GN/m}^2 = 90\times10^3\ \text{N/mm}^2.

Torque

T=60P2πN=60(149200)2π(150)=9498.4 N⋅m=9.4984×106 N⋅mmT = \frac{60P}{2\pi N} = \frac{60(149200)}{2\pi(150)} = 9498.4\ \text{N·m} = 9.4984\times10^{6}\ \text{N·mm}

Maximum shear stress

τmax=16Tπd3=16(9.4984×106)π(110)3=36.34 N/mm2\tau_{max} = \frac{16T}{\pi d^3} = \frac{16(9.4984\times10^{6})}{\pi(110)^3} = 36.34\ \text{N/mm}^2

Angle of twist in 5 m

J=π32(110)4=1.4374×107 mm4J = \frac{\pi}{32}(110)^4 = 1.4374\times10^{7}\ \text{mm}^4 φ=TLGJ=(9.4984×106)(5000)(90×103)(1.4374×107)=0.03671 rad=2.103∘\varphi = \frac{TL}{GJ} = \frac{(9.4984\times10^{6})(5000)}{(90\times10^3)(1.4374\times10^{7})} = 0.03671\ \text{rad} = 2.103^\circ

Answer: T=9.498T = 9.498 kN·m; τmax=36.34\tau_{max} = 36.34 MPa; φ=0.0367\varphi = 0.0367 rad =2.10∘= 2.10^\circ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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