The maximum stress in each material:
A copper tube of external diameter 60 mm and internal diameter 40 mm is closely fitted to a steel rod of 40 mm diameter to compose a composite shaft. If a torque of 6 kN-m is to be resisted by the shaft, discover the maximum stress in each material and the angle of twist in 2 m length.
Take G = 80 kN/mm for steel, and
G = 40 kN/mm^{2} for copper.
Solution
Copper Tube
G = 40 kN/mm^{2 } = 0.4 × 1011 N/m2 , l = 2 m
d_{1} = 60 mm = 0.06 m
d_{2} = 40 mm = 0.04 m
TC = torque resisted by copper tube
θ= Tl/ GJ = (TC ) (2) /((0.4 × 10^{11} ) (1 × 10^{- 6 })) = 2 T_{C} × 10^{- 5} radians ---------- (1)
τ_{max} = (T/J )× R = (1 × 10^{- 6} ) × (0.03) = 0.03 TC × 10 N/m
= 0.03 T_{C} N/mm^{2} ------------ (2)
Figure
G = 80 kN/mm^{2} = 0.8 × 10^{11} N/m^{2} , l = 2 m
d = 40 mm
J = (π/32) d^{ 4} = (π /32)× 0.044 = 2.5 × 10^{- 7} m^{4}
R = 20 mm = 0.02 m
θ= Tl / GJ = (T_{S }) (2) / ((0.8 × 10^{11} ) (2.5 × 10^{- 7} )) = T_{S} × 10^{- 4} radians --------- (3)
τ_{max} =(T/ J) × R = Ts / ((2.5 × 10- 7 ) (0.02)) = 0.008 T_{S} × 10^{7 } N/m
= 0.08 T_{S} N/mm^{2} --------- (4)
In composite shafts, θ_{C} = θ_{S}
⇒ 5 T_{C} × 10^{- 5} = T_{S} × 10^{- 4}
∴ T_{C} = 2T_{S} -------- (5)
T_{S} + T_{C} = T
⇒ T_{S} = 2T_{S} = 6
∴ T_{S} = 2 kN-m T_{C} = 4 kN-m
∴ Angle of twist, θ= (2 × 10^{3} ) × 10^{- 4 }= 0.2 radians
Maximum Shear Stress
For copper shaft : For steel shaft :
τ_{max} = 0.03 × 4 × 10^{3} = 120 N/m^{2}
τ_{max} = 0.08 × 2 × 10^{3} = 160 N/m^{2}