Calculate total deflection -two close-coiled helical springs:

Two close-coiled helical springs are compressed among two parallel plates by a load of 1000 N. The springs contain a wire diameter of 10 mm and radii of the coils are 50 mm and 75 mm. Each of spring has 10 coils and of the similar initial length. If the smaller spring is placed inside the larger one, calculate following

a. the total deflection, and

b. the maximum stress in each spring.

Take G = 40 GPa.

Solution

In this type of case, the springs are associated in parallel.

Δ₁ = Δ₂

Δ= 64 W R³ n / G d ⁴

 (64 W₁ × 50 × n ) / Gd ⁴ =( 64 W₂ × 75  × n) /Gd ⁴

∴          W₁ = 3.375 W₂

W₁ + W₂ = 1000

⇒         3.375 W₂ + W₂ = 1000

∴          W₂ = 229 N

∴          W₁ = 1000 - 229 = 771 N

Δ₁ = Δ₂ =64 × 771 × 50³ × 10 / (40 × 10³ × 10⁴) = 6.168 mm

τ _max1 = 16 W R / π d ³ = 16 × 771 × 50/ π × 10³  = 196.3 N/mm²

τ _{max 2} = 16 W R / π d ³ = 16 × 229 × 75 / π × 10³ = 87.5 N/mm²