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

σ_b = 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