Determine the angle of twist - solid steel shaft, Mechanical Engineering

Determine the angle of twist - solid steel shaft:

A solid steel shaft 6 m long is fixed at each of the end. A torque of 3 kN-m is applied to the shaft at a section 2 m from one end. What is the fixing torques developing on the ends of the bar? If the shaft diameter is equal to 40 mm, what are the maximum stresses in two positions? Also determine the angle of twist where torque is applied.

Take E = 80 GPa.

Solution

E = 80 GPa = 0.8 × 1011 N/mm2

T = 3 kN-m

d = 40 mm = 0.04 m

J =  (π /32 )d 4 = (π /32)× (0.04)4  = 2.5 × 10- 7 m4

R = 40/2 = 20 mm = 0.02 m4

Z P        = J/ R = 1.26 × 10- 5 m3

910_Determine the angle of twist.png

Portion AC

T1 = torque

θ1 = (T1 l1 / GJ ) = 2T1 / G J

τmax      = (T1 /J ) R =  T1 / Z P

Portion CB

θ2 = T2  l2/ GJ  = 4T2 / GJ

tmax = (T2 /J ) R =  T2 / Z P

∴          T1  + T2  = 3

T2  = (3 - T1 )

θ1 = θ2                                          ---------- (1)

⇒         2T1/ GJ = 4T2/GJ

T2  = 2T2               -----------------(2)

 (1) and (2),

2T2  + T2  = 3

⇒ T2  = 1 kN-m --------- (3)

T1  = 2T2  = 2 × 1 = 2 kN-m     -------- (4)

Maximum Stresses

Portion AC

τmax = (2 × 103) / (1.26 × 10- 5) = 159 × 106 N/m2  = 159 N/mm2

 Portion CB

τmax = (1 × 103) / (1.26 × 10- 5)

 = 80 × 106 N/m2  = 80 N/mm2

 Angle of twist :

θ= 2T1 /GJ

=(2 × (2 × 103 )) / ((0.8 × 1011 ) (2.5 × 10- 7 )) = 2 × 10-1 radian = 0.2 radians

Posted Date: 1/21/2013 5:11:03 AM | Location : United States







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