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/mm², 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/mm²

b = 12 t

E = 200 GPa = 200 × 103 N/mm²

Δ= 3W l ³ / 8 nb t ³ E

⇒         50 =3 × 8000 × 800³ / 8 × n (12 t ) t ³ (200 × 10³ )

∴          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 × 7³  = 200

∴          n = 5.8 ;  6

Actual deflection under the proof load,

Δ₀ =      (3 (8000) (800)³ )/ (8 × 6 × 84 × 7³  × 200 × 10) = 44.4 mm

Initial radius of curvature,

R₀  = l ² / 8 Δ₀ =(800)²/(8 × 44.4) = 1802 mm