Answer to Question #151944 in Mechanical Engineering for Banda

Question

A hollow shaft is 225mm out side diameter and 150mm inside diameter . calculate,
1. the maximum power this shaft can transmit at 150rev/min if the maximum shear stress in not to exceed 70MN/m^2....?
2. The diameter of a solid shaft of the same material which would transmit the same maximum power at the same speed with the same stress

Answer 1

$J=\frac{\pi}{2}(r_2^4-r_1^4)=\frac{\pi}{2}(0.1125^4-0.075^4)=0.0002(m^4)$

$T=\frac{J\tau}{r_2}=\frac{0.0002\cdot70\cdot10^6}{0.1125}=124.444(kN\cdot m)$

1) At $110 (rev/min)$ the power generated is

$124444\cdot 2\cdot3.14\cdot \frac{110}{60}=1.43(MW)$ . Answer

2) $J=\frac{\pi r^4}{2}=\frac{3.14 r^4}{2}=1.57r^4$

$\frac{J\tau}{r_2}\cdot2\pi\nu= \frac{1.57r^4\tau}{r}\cdot2\pi\nu$

$\frac{J}{r_2}= \frac{1.57r^4}{r}\to r=(\frac{J}{1.57r_2})^{1/3}=(\frac{0.0002}{1.57\cdot 0.1125})^{1/3}=0.104(m)$

$=(\frac{1430000}{2\cdot3.14\cdot(110/60)\cdot1.57\cdot40\cdot10^6})^{1/3}=0.1255(m)$

$d=2\cdot0.104=0.208(m)$ . Answer