Answer to Question #118386 in Physics for Ariyo Emmanuel

Question

(2) A solid shaft of 10cm diameter transmits 74 kW at 150 rev/min. Calculate (a) the torque on the shaft, (b) the maximum shear stress developed, (c) the angle of twist in a length of 1.50m, and (d) the shear stress at a radius of 3cm. Take G = 8 MN/cm2

Answer 1

(a) The torque is

$$T=\frac{P}{2\pi n}=\frac{74000}{2\pi 150/60}=4710\text{ N}\cdot\text{m}.$$

(b) The maximum shear stress developed

$$\tau=\frac{T}{J}R=\frac{32T}{\pi d^2}R=\frac{32\cdot4710}{0.1^2\pi}0.05=240\cdot10^3\text{ Pa}.$$

(c) The angle of twist in a length of 1.50 m

$$\frac{T}{J}=\frac{G\phi}{L}\rightarrow\phi=\frac{TL}{GJ}=\frac{PL}{2(n/60)\pi^2 d^4G}=\\ \space\\ =\frac{74000\cdot1.5}{2(150/60)\pi^2 0.1^4\cdot8\cdot10^{10}}=2.81\cdot10^{-4}\text{ rad},$$

or 0.016°.

(d) The shear stress at a radius of 3 cm is

$$\tau=\frac{32T}{\pi d^2}r=\frac{32\cdot4710}{0.1^2\pi}0.03=144000\text{ Pa}.$$