Determine Mean coil radius - close coiled helical spring:
A close coiled helical spring contains a stiffness of 1 kN/m in compression to a maximum load of 50 N and a maximum shearing stress of 150 N/mm^{2}. The solid length of the spring is equal to 45 mm. Resolve out the wire diameter, mean coil radius, and number of coils. Take G = 40 GPa.
Solution
W = 50 N, k = 1 kN/m = 1 N/mm, τ_{max} = 150 N/mm^{2}, solid length = 45 mm, d = ?,
R = ?, n = ?
Use Eq. (10) for stiffness
k = W/ Δ = G d ^{4}/ 64 R^{3} n
or 1 = (40 × 10^{3} ) d ^{4 }/ 64 R^{3} n
∴ R^{3} n/ d ^{4} = 625 ---------- (1)
Use Eq. (3) for maximum shearing stress with K = 1
τ_{max} = 16 W R /π d^{3}
or 150 =16 × 50 × R /π d ^{3}
∴ R / d ^{3 } = 0.6 ------------- (2)
or R = 0.6 d^{3}
Note solid length of spring is length while all coils are touching each other, so
nd = 45
n = 45 /d ---------- (3)
Use R from Eq. (2) and n from Eq. (3) in Eq. (i)
(0.6 d ^{3} )^{3} ( 45/d)(1/d^{4}) = 625
Or d ^{4} = 64.3
∴ d = 2.83 mm
From Eq. (3)
n = 45 /2.83 = 15.9 (say 16)
R = 0.6 d^{ 3} = 0.6 (2.83)^{3} = 13.6 mm