Determine the maximum permissible axial load:
In an open coiled spring of 10 coils the stresses because of bending and twisting are 120 N/mm^{2} and 150 N/mm^{2} 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 × 10^{3} N/mm^{2} , τ = 120 N/mm^{2},
G = 80 GPa = 80 × 10^{3} N/mm^{2} , W = ?, d = ?
σ_{ b }= 32 W R sin α / π d ^{3}= 120 N/mm^{2}
τ= 16 W R cos α/ π d ^{3} =150 N/mm^{2}
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; cos^{2} α= 0.86 , sec α = 1.075
Δ= 64 W R^{3} sec α/d^{4} (cos^{2} α/G+ 2 sin ^{2} α/E)
⇒ 20 = (64 W (5d )^{3} (10)/ d ^{4} × 0.93)[ 0.86/ (80 × 10^{3}) +((2 × 0.14)/ (200 × 10^{3})]
∴ 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