Calculate the radius of plates:
A laminated steel spring, simply supported at the ends & centrally loaded, along a span of 0.8 m; is needed to carry a proof load of 8 kN; and the central deflection is not to exceed 50 mm; the bending stress should not exceed 400 kN/mm^{2}, plates are available in multiples of 1 mm for thickness and 4 mm for width. Calculate appropriate values of width, thickness and number of plates, & calculate the radius to which the plates should be made. Consider width = 12 × thickness.
E = 200 GPa.
Solution
Simply Supported Spring l = 0.8 m
W = 8 kN = 800 N
Δ = 50 mm
σ_{b} = 400 N/mm^{2}
b = 12 t
E = 200 GPa = 200 × 103 N/mm^{2}
Δ= 3W l ^{3} / 8 nb t ^{3} E
⇒ 50 =3 × 8000 × 800^{3} / 8 × n (12 t ) t ^{3 }(200 × 10^{3} )
∴ nt ^{4} = 12,800 ---------- (1)
σ= 3Wl /2 nb t^{ 2}
⇒ 400 = 3 × 8000 × 800 / 2 n (12 t ) t ^{2}
∴ nt ^{3} = 2, 000 ----------- (2)
From Eq. (1) to Eq. (2), we obtain
⇒ t = 6.4 mm = 7 mm
b = 12 t = 12 × 7 = 84 mm
nt ^{3} = 2, 000
⇒ n × 7^{3} = 200
∴ n = 5.8 ; 6
Actual deflection under the proof load,
Δ_{0} = (3 (8000) (800)^{3 })/ (8 × 6 × 84 × 7^{3} × 200 × 10) = 44.4 mm
Initial radius of curvature,
R_{0} = l ^{2 }/ 8 Δ_{0} =(800)^{2}/(8 × 44.4) = 1802 mm