## Angle of twist is zero on the shaft, Mechanical Engineering

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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

⇒  TA  - 8 + 10 + TD  = 0

∴          TA  + TD  = - 2

G = 80 GPa = 0.8 × 1011 N/m2              -------------- (1)

d = 100 mm = 0.1 m

J =  (π /32)d 4 = (π/32) × (0.1)4  = 9.8 × 10- 6 m4

R = d /2= 100/2 = 0.05 m ,   ZP = 1.96 × 10- 4 m3 Figure (a) Figure (b) : Line Diagram

Portion AB

θ 1 = (TA . l )/GJ = (TA × 3 )/GJ                         ---------- (2)

Portion BC

θ2 = ((TA  - 8) × 5 )/     GJ               -------- (3)

Portion CD

θ3 = ((TA - 8 + 10) × 2)/ GJ = ((TA + 2) × 2)/ GJ                      ---------- (4)

Angle of twist at D co.r.t A = 0

(TA × 3 )/ G J + ((TA  - 8) × 5 )/ GJ + ((TA  + 2) × 2)/ GJ = 0

∴          TA  = 3.6 kN-m                                       ----------- (5)

From Eqs. (1) and (5),

3.6 + TD = - 2

TD = - 5.6 kN-m = 5.6 kN-m    (  )

Angle of Twist

θ1 = ((3.6 × 103 ) × 3) / ((0.8 × 1011 ) (9.8 × 10- 6 ))

= 1.4 × 10- 2 radians

θ2 =      ((3.6 - 8) × 103 × 5 )/(0.8 × 1011 ) (9.8 × 10- 6 )= - 2.8 × 10- 2 radians

θ3 =      (3.6 + 2) × 103  × 2 /((0.8 × 1011 ) (9.8 × 10- 6 ))= 1.4 × 10- 2 radians

θAB  = θ1 = 1.4 × 10- 2 radians

θ AC  = θ1 + θ2  = 1.4 × 10- 2 - 2.8 × 10- 2 = - 1.4 × 10- 2 radians

θAD  = θAC  + θ3 = - 1.4 × 10- 2

Shear Stress

+ 1.4 × 10- 2  = 0

τ max     = (T/ J) . R =  T/ Z P

Portion AB

τmax = 3.6 × 103 / (1.96 × 10- 4) = 18.4 × 106 N/m2  = 18.4 N/mm2

Portion BC

τ max = (3.6 - 8) × 103/ (1.96 × 10- 4)

= - 22.4 × 106 N/m2  = - 22.4 N/mm2

Portion CD

τ max      = (3.6 + 2) × 103/1.96 × 10- 6 = 28.6 × 106 N/m2  = 28.6 N/mm2

Zero Angle of Twist

It occurs between B and C

θ1 = 1.4 × 10- 2 radians  ∴ x = 2.5 m from B.

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