Find out the power - hollow shaft:
A hollow shaft of internal diameter 400 mm & external diameter 450 mm is needed to transmit power at 120 rpm. Find out the power it may transmit, if the shear stress is not to exceed 50 N/mm^{2} & the maximum torque exceeds the mean by 30%.
Solution
do = 450 mm, di = 400 mm, N = 120 rpm
τ_{max} = 50 N/mm^{2} = 50 × 106 N/m^{2}
T_{max} = τ_{max} × Z P
= (π /32 )(450^{4} - 400^{4} ) = 1.51 × 10^{9} mm^{4}
R = do /2 = 450 = 225 mm
Z _{P} = J/ R = 1.51 × 10^{9} /225 mm^{3} = 6.7 × 10^{- 3} m^{3}
τ_{max} = (50 × 10 ) (6.7 × 10) = 335 × 10 N-m
τ_{max } = 1.3 × T
⇒ 335 × 103 = 1.3 T
∴ T = 258 × 10^{3} N-m
∴ Power, P = 2π NT/60 = 2π × 120 × 258 × 10^{3 }/60
= 3242 × 103 watts = 3242 kW