Find the maximum shear stress at free end, Mechanical Engineering

Find the maximum shear stress at free end:

A stepped shaft ABCD, with A is fixed end and D is free end. AB = 1 m, BC = 2 m, CD = 1 m. AB is a hollow shaft of 100 mm outer diameter and 80 mm internal diameter. BC is solid shaft of 75 mm diameter CD is a solid shaft of 69 mm diameter. The torque applied at end D is 2 kN-m clockwise at C is 3 kN-m anticlockwise at B is 4 kN-m clockwise. Take G = 80 kN/mm2. Find the maximum shear stress and the angle of twist at free end.

Solution

G = 80 kN/mm2  = 0.8 × 1011 N/mm2    

520_Find the maximum shear stress at free end.png

Portion CD

T =+ 2 kN-m = + 2 × 103 N-m ,            l = 1 m

d = 60 mm = 0.06 m

J =  (π /32 )d 4  = π (0.06)4  = 1.3 × 10- 6 m4

R = d/2 = 60 /2= 30 = 0.03 m

T / J = τmax/ R = G θ/ l

τ max     = (T/ J) × R = ((2 × 103 ) /(1.3 × 10- 6 )) × 0.03 = 46.2 × 106 N/mm2

θ= Tl / GJ   = ((+ 2 × 103 ) × 1 )/((1.3 × 10- 6 ) (0.8 × 1011 ) )= + 1.9 × 10- 2 radians

Portion BC

l = 2 m,            T = + 2 - 3 = - 1 kN-m

d = 75 mm = 0.075 m

J =  (π /32 )d 4  = (π/32) × 0.0754  = 1.3 × 10- 6 m4

R = d/ 2 = 0.0375 m

τ max     = (T / J) × R = (1 × 103 /3.1 × 10- 6 )× 0.0375 = 12.1 N/mm2

θ= Tl / GJ   = -( (1 × 103 ) × 2) / ((0.8 × 1011 ) (3.1 × 10- 6 ))

= - 0.8 × 10- 2 radians

Portion AB

l = 1 m

T = + 2 - 3 + 4 = + 3 kN-m

d1 = 100 mm = 0.1 m

d2 = 80 mm = 0.08 m

 J =  (π /32) (0.14  - 0.084 ) = 5.8 × 10- 6 m4

R = d1/2 = 50 mm = 0.05 m

τ max     = (T/J)  × R =   (3 × 103 ) (0.05) /(5.8 × 10- 6 ) = 259 × 10  N/m   = 25.9 N/mm

θ= Tl/ GJ   =((+ 3 × 103 ) × (1)) / ( (0.8 × 1011 ) (5.8 × 10- 6 ))

= + 0.65 × 10- 2 radians

∴          Maximum shear stress = 46.2 N/mm2

θA = 0

θB  = 1.9 × 10- 2 radians

θC  = + 1.9 × 10- 2  - 0.8 × 10- 2  = 1.1 × 10- 2 radians

θD  = 1.1 × 10- 2  + 0.65 × 10- 2  = 1.75 × 10- 2 radians

∴Angle of twist at free end = 1.75 × 10- 2 radians

Posted Date: 1/21/2013 5:04:09 AM | Location : United States







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