Discover the angle of twist - uniformly varying shaft:

A uniformly varying shaft of length 2 m is subjected to a torque of 2 kN-m. The diameter at the ends shall be 60 mm and 120 mm. If G = 80 kN/mm2, discover the angle of twist. If the average diameter is utilized in the estimate of angle of twist, what is the percentage error?

Solution

d₁ = 60  mm;   d₂ = 120 mm

R₁ = 30 mm;    R₂ = 60 mm

T = 2 kN-m = 2 × 103 N-m,     l = 2 m

G = 80 kN/mm₂ = 0.8 × 1011 N/m₂

=2 × (2 × 10)³ × 2/ (3π× 0.8 × 10¹¹) (( 0.03²  + 0.03 × 0.06 + 0.06² )/ (0.03³ × 0.06³))

= 0.011 radians

If average diameter is used :

d = 60 + 120 /2 = 90 mm

J =  (π/32) d ⁴  = π (0.09)⁴  = 6.44 × 10^{- 6} m4

∴          θ₂ = Tl / GJ =   ((2 × 10³ ) (2) )/((0.8 × 10¹¹ ) (6.44 × 10^{- 6} )) = 0.008 radians

% error = ((θ₁ - θ₂ )/ θ₁ ) × 100

=( (0.011 - 0.008) /0.011 )× 100 = 27.3%

Posted Date: 1/21/2013 4:58:13 AM | Location : United States

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