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/mm^{2}.
G = 80 GPa.
Solution
For spring in series, W is similar.
Δ = Δ_{1} + Δ_{2} --------- (1)
1/ K = 1/ K_{1} + 1/ K_{2} --------- (2)
Δ= 64 W R^{3} n / Gd ^{4}
Δ1 =64 W × 15^{3} × 10 / (80 × 10^{3 }× 2.54)
W/ Δ1 = K1 = 1.45 N/mm ---------- (3)
K = 750 N/m = 0.75 N/mm ------- (4)
Δ_{2} =64 W × 15^{3} × 10/80 × 10^{3} × d ^{4}
W/ Δ_{2} = K_{2} = d^{4} (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 d^{3}
⇒ 200 = 16 W × 15 /π (2.5)^{3}
∴ W = 40.9 N
Total extension, Δ= W /K = 40.9/0.75 = 54.5 mm