Determine the greatest load carried by the composite springs:

A composite spring contain two close-coiled helical spring associated in series, each spring has 10 coils at a mean radius of 15 mm. Determine the diameter of one if the other is 2.5 mm and the stiffness of the composite spring is 750 N/mm. Determine the greatest load that may be carried by the composite springs, and the corresponding extension, for a maximum shearing stress of 200 N/mm².

G = 80 GPa.

Solution

For spring in series, W is similar.

 Δ = Δ₁ + Δ₂       --------- (1)

1/ K  =  1/ K₁  +   1/ K₂                                 --------- (2)

Δ= 64 W R³ n / Gd ⁴

Δ1 =64 W × 15³ × 10 / (80 × 10³ × 2.54)

W/ Δ1 = K1 = 1.45 N/mm       ---------- (3)

 K = 750 N/m = 0.75 N/mm ------- (4)

Δ₂ =64 W × 15³ × 10/80 × 10³ × d ⁴

W/ Δ₂ = K₂  = d⁴ (0.037)                             ----------- (5)

By using (2), (3), (4) and (5), we obtain

1/0.75  = (1/1.45 )+1/ (0.037 × d ⁴)

∴          d = 2.55 mm

τ_max      = 16 W R /π d³

⇒         200 = 16 W × 15 /π (2.5)³

∴          W = 40.9 N

 Total extension, Δ=    W /K = 40.9/0.75 = 54.5 mm