Determine equivalent maximum shear stress in spring:
A close coiled helical spring contains a stiffness of 10 N/mm. Its length while fully compressed, with adjacent coils touching each other is 400 mm. G = 80 GPa.
(a) Find out the wire diameter and the mean coil radius, if their ratio is 0.02.
(b) If the gap among any two adjacent coils is 2 mm, what maximum load may be applied before the adjacent coils touch?
(c) What is the equivalent maximum shear stress in spring?
Solution
k = 10 N/mm, nd = 400 mm, G = 80 GPa = 80 × 103 N/mm2 , d = ?, R = ?,
d /R= 0.02 ∴R = 5d , Gap = 2 mm, W = ?, τmax = ?
(a) k = W / Δ = G d 4/64 R3 n
G d 4
∴ W/ Δ = G d 4/64 R3 n
or 10 =80 × 103 × d 4 / 64 (5d )3 . (400/d)
∴ d = 20 mm
R = 5d = 5 × 20 = 100 mm
n = 400/ d = 400 /20= 20
(b) Gap between adjacent coils = 2 mm
Total gap = 2n = 2 × 20 = 40 mm, i.e. spring ought to deflect 40 mm if coils are to touch each other.
Δ= 64 W R3 n/ Gd 4
∴ 40 = 64 × W × 1003 × 20 / 80 × 103 (20)4
or W = 400 N
∴ τ max = 16 W R / π d 3 = 16 × 400 × 100 / π (20)3 = 25.5 N/mm