Determine the axial extension along a load:
An open coiled helical spring is created having 10 turns of a mean radius of 60 mm. The wire diameter is 10 mm and coils make an angle of 38^{o} along a plane perpendicular to the axis of the coil. Determine
(1) the axial extension along a load of 200 N,
(2) the angle of rotation of free end.
E = 200 GPa, G = 80 GPa.
Solution
n = 10, R = 60 mm, d = 10 mm, α = 38, E = 200 GPa, G = 80 GPa
(a) Axial Extension
Δ= 64 W R^{3} sec α/d^{4} ( cos2 α/G +2 sin ^{2} α /E)
=64 (200) (60)^{3} (10) sec 30 / 10^{4 }((cos^{2} 30/80 × 10^{3)}) + (2 sin ^{2} 30 /200 × 10^{3 })) = 36.7 mm
(a) Angle of Rotation at Free End
φ= 64 W R^{3} sin α/d^{4 }((1/G - 2/E))
=64 (200) (60)^{3} (10) sec 30/10^{4 }( 1/^{ }(80 × 10^{3}) - 2/ (200 × 10^{3})) = 0.0567 radian