Determine the mean radius of an open coiled spring:
Determine the mean radius of an open coiled spring of helix angle of 38^{o}, 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 = 30^{o}, Δ = 20 mm, φ = 0.02 radian, W = 30 N, E = 200 GPa, G = 80 GPa
Δ= (64 W R^{3} sec α /d^{4} ) + (cos^{2} α /G + 2 sin ^{2} α /E) -------- (1)
φ= (64 W R^{3} sin α/d^{4} )((1/G -2/E) ------------- (2)
Eq. (2) divided by Eq. (1), we obtain
1000 = R (0.115)/( 0.433 × 0.025)
∴ R = 94 mm