Discover the torque on the shaft:
A steel shaft transmits 105 kW at 160 rpm. If the shaft is 100 mm diameter, discover the torque on the shaft and the maximum shear stress induced. Discover also the twist of the shaft in a length of 6 m.
Take G = 8 × 104 N/mm^{2}.
Solution
P = 105 kW = 105 × 10^{3} W
N = 160 rpm
d = 100 mm
l = 6 m = 6000 mm
G = 8 × 10^{4} N/mm^{2}
We know, P = 2πNT /60
105 × 10^{ 3} = (2π × 160 /60 )× T
T = 6266 N m = 6.266 ×10^{6} N mm
Note T = ( π/16) τ_{m}(100)^{3}
6.266 × 10^{6} = π/ 16 τ_{m} (100)^{3}
τ_{m} = 31.19 N/mm^{2}
T / J = Gθ/ l or θ = Tl/ GJ
J = (π/32 )d ^{4 } = (π/32) × 10^{8 } = 9.82 × 10^{6 } mm4
θ= 6.266 × 10^{6} × 6 × 10^{3} / (8 × 10^{4} × 9.82 × 10^{6})
θ = 0.04786 radian
or
θ= 0.04786 × (180/ π) deg . = 2.74 deg . or 2o 44.4′