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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?
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%
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