Determine the number of leaves and initial curvature:
A leaf spring with a span of 1.40 m consist width and thickness of leaves to be 100 mm and 12 mm respectively. The maximum bending stress is equal to 150 N/mm^{2} and the spring should absorb 125000 N-mm when straightened. Determine the number of leaves and initial curvature. Take E = 200 GPa.
Solution
d = 1400 mm
b = 100 mm
t = 12 mm
σ_{b} = 150 N/mm^{2}
U = 125000 N-mm
n =?
R_{0 }=?
E = 200 × 103 N/mm^{2}
σ = 3Wl /2 nbt^{2}
⇒ 150 = 3W × 1400 / 2 n (100) (12)^{2}
∴ W = 1029 n --------------- (1)
Δ= 3W l ^{3} / 8 nE b t^{ 3} = 3 (1029 n) (1400)^{3} / 8 n (200 × 10^{3} ) (100) (12)^{3} = 30.6 mm
U = (½) W Δ
⇒ 125000 = 1029 n × (30.6/2)
∴ n = 79; 8
R_{0} = l ^{2 }/ 8 Δ = (1400)^{2} /8 × 30.6 = 8007 mm