Compute the diameter of a solid shaft:
Compute the diameter of a solid shaft transmitting 150 kW at 25 rpm, if the maximum shear stress in the shaft is not to exceed 70 MPa. Compare it with the shaft delivering similar power at 25000 rpm.
Solution
Power transmitted, P = 150 kW = 150 × 10^{3} W
Number of revolutions, N = 25 rpm
Since, 1 Pa = 1 N/m^{2} and 1 Mega Pascal = 10^{6} N/m^{2 }= 1 N/mm^{2}
Then, maximum shear stress, τ_{m} = 70 MPa = 70 N/mm^{2}
Let T be the torque transmitted in N m.
P = 2πNT /60
150 ×10^{3} = (2π × 25/60 ) × T
T = 57.28 × 10^{3} N m
T = 57.28 × 10^{6} N mm
T/ J= τ_{m} /(d/2)
d = 160.9 mm
If N = 25000 rpm, then P = 2πNT /60
150 ×10^{3} = (2π × 25000 × T )/60
T = 57.28 N m = 57.28 ×10^{3 }N mm
T/ J= τ_{m}/(d/2)
57.28 × 10^{3} / (π/32) d ^{4} = 70/(d/2)
d = 16.09 mm
It is seen from this example that the size of the shaft is decreased very much if the power is transmitted at high speed. That is the cause for the modern tendency to utilize high speed machines, those results in considerable saving in the material cost.