Find the maximum shear stress - stepped circular bar:
A stepped circular bar ABC 4 m long fixed at one of the end 'A' is subjected to a torque of 10 kN-m. If the portion AB is solid shaft of 100 mm diameter & the portion BC is a hollow with an external diameter 100 mm and internal diameter 80 mm. discover the maximum shear stress & maximum angle of twist.
AB = 2 m; BC = 2 m; G = 80 kN/mm^{2}.
Solution
G = 80 kN/mm^{2} = 0.8 × 10^{11} N/mm^{2}
Figure
Portion BC
l = 2 m, d_{1} = 100 mm, d_{2} = 80 mm, T = 180 kN-m = 10 × 10^{3} N-m
= π (0.1^{4} - 0.08^{4} ) = 5.8 × 10^{- 6} m^{4}
R = 100 = 50 mm = 0.05 m
T / J = τ_{max}/ R = G θ/ l
τ_{max} = (T/J) × R = (10× 10^{3} )(5.8 × 10^{- 6} ) × (0.05)
= 86.2 × 10^{6} N/mm^{2} = 86.2 N/mm^{2}
Portion AB
Solid T = 10 kN-m
d = 100 mm
J = ( π / 32 )(0.1)^{4 } = 9.8 × 10^{- 6 }m4
R = d /2 = 50 mm = 0.05 m
τ _{max} = T R / d J = ((10 × 10 ^{3}) (2) / (9.8 × 10^{- 6} ) )(0.05) = 51 × 10^{6} N/m^{2} = 51 N/mm^{2}
θ= Tl / GJ = (10 × 10^{3}) (2) / ((0.8 × 10^{11}) (9.8 × 10^{- 6}))
= 2.55 × 10^{- 2} radians
∴ Maximum shear stress = 86.2 N/mm^{2}
θ_{A} = 0
θ_{B} = 4.3 × 10^{- 2} radians
θ_{C } = 4.3 × 10^{- 2} + 2.55 × 10^{- 6} = 6.85 × 10^{ 6} radians