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Determine the mean radius of an open coiled spring:
Determine the mean radius of an open coiled spring of helix angle of 38o, to give a vertical displacement of 20 mm and an angular rotation of 0.02 radian at free end under an axial load of 30 N. The material available is equal to 6 mm diameter steel bar. Take E = 200 GPa, G = 80 GPa.
Solution
α = 30o, Δ = 20 mm, φ = 0.02 radian, W = 30 N, E = 200 GPa, G = 80 GPa
Δ= (64 W R3 sec α /d4 ) + (cos2 α /G + 2 sin 2 α /E) -------- (1)
φ= (64 W R3 sin α/d4 )((1/G -2/E) ------------- (2)
Eq. (2) divided by Eq. (1), we obtain
1000 = R (0.115)/( 0.433 × 0.025)
∴ R = 94 mm
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