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 = ?, τ_max  = ?

(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

Posted Date: 1/21/2013 1:18:06 AM | Location : United States

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