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/mm². Discover also the angle of twist for a length of 2.5 metres.

Take G = 8 × 104 N/mm².

Solution

Here,   P = 60 kW = 60 × 10³ W

N = 150 rpm,   T_max = 1.25 T_mean,  τ_m  = 60 N/mm²

l = 2.5 m = 2.5 × 103 mm²

 We know,

P = 2πNT /60

60 ×10³  = (2π ×150 × T)/60

T = 3819.7 N m = 3.82 ×10⁶ N mm, T is the mean torque

T_max = 1.25 T_mean = 1.25 × 3.8197 × 10⁶ = 4.746 ×10⁶ N mm

T _max =  (π/16 ) τ_m        d ³                   

4.7746 × 10⁶  = ( π /16 )× 60 × d ³

d ³  = 4.7746 × 10⁶× 16 / π× 60

∴          d = 74 mm

T / J = Gθ/ l

4.7746 ×106 / (π/32) × (74)⁴  =           (8 ×10⁴/2500 )× θ

θ = 0.0507 radians

θ = 2° 54′