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    

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 = τ_max/ 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

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