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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 38o 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 R3 sec α/d4 ( cos2 α/G +2 sin 2 α /E)
=64 (200) (60)3 (10) sec 30 / 104 ((cos2 30/80 × 103)) + (2 sin 2 30 /200 × 103 )) = 36.7 mm
(a) Angle of Rotation at Free End
φ= 64 W R3 sin α/d4 ((1/G - 2/E))
=64 (200) (60)3 (10) sec 30/104 ( 1/ (80 × 103) - 2/ (200 × 103)) = 0.0567 radian
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