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/mm² & the maximum torque exceeds the mean by 30%.

Solution

do = 450 mm,  di = 400 mm,   N = 120 rpm

τ_max = 50 N/mm² = 50 × 106 N/m²

T_max  = τ_max  × Z P

=  (π /32 )(450⁴  - 400⁴ ) = 1.51 × 10⁹ mm⁴

R = do /2 = 450 = 225 mm

Z _P        = J/ R = 1.51 × 10⁹ /225 mm³ = 6.7 × 10^{- 3}  m³

τ_max  = (50 × 10 ) (6.7 × 10) = 335 × 10  N-m

τ_max  = 1.3 × T

⇒         335 × 103  = 1.3 T

∴          T = 258 × 10³ N-m

∴          Power, P = 2π NT/60 = 2π × 120 × 258 × 10³ /60

                         = 3242 × 103 watts = 3242 kW