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/ σb
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