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 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