Fewpal
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A solid shaft transmits power at the rate of 2000KW at the speed of 60RPM. If the safe allowable stress is 80MN/m2, find the minimum diameter of the shaft.

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P = 2000KW = 2000 × 103

N = 60RPM 

τ = 80MN/m2 = 80 × 106 N/m

d = ?

Using the relation: P = 2π.N.Tmax/60 watts 

2000 X 103 = 2π.60.Tmax/60 

Tmax = 318309.88 N–m

Now using the relation Tmax = (π/16) τmax.D3 

318309.88 = (π/16) × 80 × 106D3 

D = 0.2726 m or D = 272.63 mm

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