Determine the maximum permissible axial load:

In an open coiled spring of 10 coils the stresses because of bending and twisting are 120 N/mm² and 150 N/mm² respectively while the spring is loaded axially. Supposing the mean radius of the coil is 5 times the wire diameter, determine the maximum permissible axial load and the wire diameter for a maximum extension of 20 mm.

E = 200 GPa, G = 80 GPa.

Solution

n = 10, R = 5d, Δ = 20 mm, E = 200 GPa = 200 × 10³ N/mm² , τ = 120 N/mm²,

G = 80 GPa = 80 × 10³ N/mm² , W = ?,  d = ?

σ _b = 32 W R sin α / π d ³= 120 N/mm²

τ= 16 W R cos α/ π d ³ =150 N/mm²

Taking ratio of τ/ σ_b

2 tan α=  120 / 150

tan α= 0.4

sin α= 0.37;  sin ² α = 0.14

cos α= 0.93;  cos² α= 0.86 , sec α = 1.075

Δ= 64 W R³ sec α/d⁴    (cos² α/G+ 2 sin ² α/E)

⇒         20 = (64 W (5d )³ (10)/ d ⁴ × 0.93)[ 0.86/ (80 × 10³) +((2 × 0.14)/ (200 × 10³)]

∴          W / d  = 19.99             ------------ (1)

              32 × W × (5d ) × 0.37/ π d ³  = 120

⇒         W  / d ² = 6.37                           ---------- (2)

Eq. (1) divided by Eq. (2), we obtain

d = 3.0 m

W = 19.99 × 3 = 1145.43 N

Posted Date: 1/21/2013 1:27:41 AM | Location : United States

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