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Open coiled helical spring:
In an open coiled helical spring along with α = 38o, if the inclination of the coil is ignored, determine the percentage by which the axial extension is under estimated.
E = 200 GPa, G = 80 GPa.
Solution
Δ= (64 W R3 n / d 4 cos α)( (cos2 α/G)+(2 sin 2 α /E))
= K/ cos α[cos2 α/G +2 sin 2 α/E]
α= 38o , Δ = K × (0.115/0.866) × 10- 4 = K × 0.0147
If α is neglected,
α= 0, Δ2 = K × 0.0125
% Error = ((Δ1 - Δ2 ) / Δ1 )× 100
= (0.0147 - 0.0125/0.0147) × 100 = 14.96%
inertia force?
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