Angle of twist is zero on the shaft:
A shaft 10 m long and 100 mm diameter during its length is fixed at the ends and is subjected to two opposite torques of 8 kN-m and 10 kN-m at a distance of 3 m and 8 m from one end. Discover the fixing torques at the ends. What is the maximum angle of twist and the maximum shear stress in the shaft? At what section the angle of twist is zero on the shaft?
Take E = 80 GPa.
Solution
Σ T = 0
⇒ T_{A} - 8 + 10 + T_{D} = 0
∴ T_{A} + T_{D} = - 2
G = 80 GPa = 0.8 × 10^{11} N/m^{2} -------------- (1)
d = 100 mm = 0.1 m
J = (π /32)d ^{4} = (π/32) × (0.1)^{4} = 9.8 × 10^{- 6} m^{4}
R = d /2= 100/2 = 0.05 m , Z_{P} = 1.96 × 10^{- 4} m^{3}
Figure (a)
Figure (b) : Line Diagram
Portion AB
θ_{ 1} = (T_{A} . l )/GJ = (T_{A} × 3 )/GJ ---------- (2)
Portion BC
θ_{2} = ((T_{A} - 8) × 5 )/ GJ -------- (3)
Portion CD
θ_{3 }= ((T_{A} - 8 + 10) × 2)/ GJ = ((T_{A} + 2) × 2)/ GJ ---------- (4)
Angle of twist at D co.r.t A = 0
(T_{A} × 3 )/ G J + ((T_{A} - 8) × 5 )/ GJ + ((T_{A} + 2) × 2)/ GJ = 0
∴ T_{A} = 3.6 kN-m ----------- (5)
From Eqs. (1) and (5),
3.6 + T_{D} = - 2
T_{D} = - 5.6 kN-m = 5.6 kN-m ( )
Angle of Twist
θ_{1} = ((3.6 × 10^{3} ) × 3) / ((0.8 × 10^{11} ) (9.8 × 10^{- 6} ))
= 1.4 × 10^{- 2} radians
θ_{2} = ((3.6 - 8) × 10^{3} × 5 )/(0.8 × 10^{11} ) (9.8 × 10^{- 6} )= - 2.8 × 10^{- 2} radians
θ_{3} = (3.6 + 2) × 10^{3} × 2 /((0.8 × 10^{11} ) (9.8 × 10^{- 6} ))= 1.4 × 10^{- 2} radians
θ_{AB} = θ_{1 }= 1.4 × 10^{- 2} radians
θ _{AC} = θ_{1} + θ_{2} = 1.4 × 10^{- 2} - 2.8 × 10^{- 2} = - 1.4 × 10^{- 2} radians
θ_{AD } = θ_{AC } + θ_{3} = - 1.4 × 10^{- 2}
Shear Stress
+ 1.4 × 10^{- 2} = 0
τ _{max} = (T/ J) . R = T/ Z_{ P}
Portion AB
τ_{max }= 3.6 × 10^{3} / (1.96 × 10^{- 4}) = 18.4 × 10^{6} N/m^{2 } = 18.4 N/mm^{2}
Portion BC
τ _{max }= (3.6 - 8) × 10^{3}/ (1.96 × 10^{- 4})
= - 22.4 × 10^{6} N/m^{2} = - 22.4 N/mm^{2}
Portion CD
τ _{max} = (3.6 + 2) × 10^{3}/1.96 × 10^{- 6 }= 28.6 × 10^{6} N/m^{2} = 28.6 N/mm^{2}
Zero Angle of Twist
It occurs between B and C
θ_{1} = 1.4 × 10^{- 2} radians
∴ x = 2.5 m from B.