Determine mean coil radius - close coiled helical spring, Mechanical Engineering

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, τmax = 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

τmax      = 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

Posted Date: 1/21/2013 1:09:34 AM | Location : United States







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