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/mm^{2} = 0.8 × 1011 N/mm^{2 }
Portion CD
T =+ 2 kN-m = + 2 × 10^{3} N-m , l = 1 m
d = 60 mm = 0.06 m
J = (π /32 )d ^{4} = π (0.06)^{4} = 1.3 × 10^{- 6} m^{4}
R = d/2 = 60 /2= 30 = 0.03 m
T / J = τ_{max}/ R = G θ/ l
τ _{max} = (T/ J) × R = ((2 × 10^{3} ) /(1.3 × 10^{- 6} )) × 0.03 = 46.2 × 10^{6} N/mm^{2}
θ= Tl / GJ = ((+ 2 × 10^{3} ) × 1 )/((1.3 × 10^{- 6} ) (0.8 × 10^{11} ) )= + 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) × 0.075^{4} = 1.3 × 10^{- 6} m^{4}
R = d/ 2 = 0.0375 m
τ _{max} = (T / J) × R = (1 × 10^{3} /3.1 × 10^{- 6} )× 0.0375 = 12.1 N/mm^{2}
θ= Tl / GJ = -( (1 × 10^{3} ) × 2) / ((0.8 × 10^{11} ) (3.1 × 10^{- 6} ))
= - 0.8 × 10^{- 2} radians
Portion AB
l = 1 m
T = + 2 - 3 + 4 = + 3 kN-m
d_{1} = 100 mm = 0.1 m
d_{2} = 80 mm = 0.08 m
J = (π /32) (0.1^{4} - 0.08^{4 }) = 5.8 × 10^{- 6} m^{4}
R = d_{1}/2 = 50 mm = 0.05 m
τ _{max} = (T/J) × R = (3 × 10^{3} ) (0.05) /(5.8 × 10^{- 6} ) = 259 × 10 N/m = 25.9 N/mm
θ= Tl/ GJ =((+ 3 × 10^{3} ) × (1)) / ( (0.8 × 10^{11} ) (5.8 × 10^{- 6 }))
= + 0.65 × 10^{- 2} radians
∴ Maximum shear stress = 46.2 N/mm^{2}
θ_{A} = 0
θ_{B} = 1.9 × 10^{- 2} radians
θ_{C} = + 1.9 × 10^{- 2} - 0.8 × 10^{- 2} = 1.1 × 10^{- 2 }radians
θ_{D} = 1.1 × 10^{- 2} + 0.65 × 10^{- 2} = 1.75 × 10^{- 2} radians
∴Angle of twist at free end = 1.75 × 10^{- 2} radians