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/mm². 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/mm², solid length = 45 mm, d = ?,

R = ?, n = ?

Use Eq. (10) for stiffness

k = W/ Δ          = G d ⁴/            64 R³ n

or         1 = (40 × 10³ ) d ⁴ / 64 R³ n

∴ R³ n/ d ⁴     = 625             ---------- (1)

Use Eq. (3) for maximum shearing stress with K = 1

τ_max      = 16 W R  /π d³

or         150 =16 × 50 × R /π d ³

∴          R / d ³  = 0.6                          ------------- (2)

or         R = 0.6 d³

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/d⁴)  = 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