Question

What is the speed of 75 mm shaft transmitted 69.5 kW if stress is not to exceed 23.2 Mpa

Answer 1

Given:

• Diameter of the shaft, D= 75 mm= 0.075 m
• Power, P= 69.5 KW= 69.5×1000 W= 69500 W
• Stress,
  7
  ​​​​​​​= 23.4 MPa= 23.4×10^6 Pa

To find:

• Speed (N)

Note:

• Polar moment of inertia of circular shaft, J= πD^4/32, where D= Diameter of the shaft in m
• 1 Pa= 1 NN/m^2

Solution:

We know that, Power, P= 2πNT/60, where N= Speed in rpm, T= Torque in Nm.

We know that, by Torsion equation, Torque, T=
7
J/R, where
7
= Stress in N/m^2, J= Polar moment of inertia in m^4, R= Radius of the shaft in m

• T=  
  TJ R
  
• T=  
  T(TD4/32) D/2
  
• T=. $\frac{\tau(\pi D^3) }{16}$
• T= .$\frac{23.2(10^6)(\pi)( 0.075^3) }{16}$
• T= Nm
• Torque, T= 1921.771 Nm

​​​​​​​As said before, Power, P= 2πNT/60

• P=  $\frac{2\pi NT}{60}$
• 69500= $\frac{2\pi N(1921.771)}{60}$
• N=  $\frac{(69500)(60)}{2\pi (1921.771)}$
• N=  345.35 rpm
• Speed, N= 345.35 rpm

​​​​​​​Answer:

• Speed of the shaft= 345.35 rpm