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
d_{1} = 60 mm; d_{2} = 120 mm
R_{1} = 30 mm; R_{2} = 60 mm
T = 2 kN-m = 2 × 103 N-m, l = 2 m
G = 80 kN/mm_{2} = 0.8 × 1011 N/m_{2}
=2 × (2 × 10)^{3} × 2/ (3π× 0.8 × 10^{11}) (( 0.03^{2} + 0.03 × 0.06 + 0.06^{2} )/ (0.03^{3} × 0.06^{3}))
= 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
∴ θ_{2} = Tl / GJ = ((2 × 10^{3} ) (2) )/((0.8 × 10^{11} ) (6.44 × 10^{- 6} )) = 0.008 radians
% error = ((θ_{1} - θ_{2} )/ θ_{1} ) × 100
=( (0.011 - 0.008) /0.011 )× 100 = 27.3%