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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/mm2 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
σb = 150 N/mm2
U = 125000 N-mm
n =?
R0 =?
E = 200 × 103 N/mm2
σ = 3Wl /2 nbt2
⇒ 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 × 103 ) (100) (12)3 = 30.6 mm
U = (½) W Δ
⇒ 125000 = 1029 n × (30.6/2)
∴ n = 79; 8
R0 = l 2 / 8 Δ = (1400)2 /8 × 30.6 = 8007 mm
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