Discover the angle of twist, Mechanical Engineering

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.


Here,   P = 60 kW = 60 × 103 W

N = 150 rpm,   Tmax = 1.25 Tmean,  τm  = 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

Tmax = 1.25 Tmean = 1.25 × 3.8197 × 106 = 4.746 ×106 N mm

T max =  (π/16 ) τm        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′

Posted Date: 1/21/2013 2:46:10 AM | Location : United States

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