Discover the angle of twist:
Discover the diameter of the shaft needed to transmit 60 kW at 150 r.p.m., if the maximum torque is possible to exceed the mean torque by 25% for a maximum permissible shear stress of 60 N/mm^{2}. Discover also the angle of twist for a length of 2.5 metres.
Take G = 8 × 104 N/mm^{2}.
Solution
Here, P = 60 kW = 60 × 10^{3} W
N = 150 rpm, T_{max }= 1.25 T_{mean}, τ_{m} = 60 N/mm^{2}
l = 2.5 m = 2.5 × 103 mm^{2}
We know,
P = 2πNT /60
60 ×10^{3} = (2π ×150 × T)/60
T = 3819.7 N m = 3.82 ×10^{6} N mm, T is the mean torque
T_{max} = 1.25 T_{mean} = 1.25 × 3.8197 × 10^{6} = 4.746 ×10^{6} N mm
T _{max} = (π/16 ) τ_{m} d ^{3}
4.7746 × 10^{6 } = ( π /16 )× 60 × d ^{3}
d ^{3 } = 4.7746 × 10^{6}× 16 / π× 60
∴ d = 74 mm
T / J = Gθ/ l
4.7746 ×106 / (π/32) × (74)^{4} = (8 ×10^{4}/2500 )× θ
θ = 0.0507 radians
θ = 2° 54′