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Applied load shall elongate the spring:
Design a close coiled helical spring which shall deflect 100 mm under a load of 600 N. The radius of the coil is 6 times the wire diameter. The maximum shear stress is not to exceed 8 MPa. G = 80 GPa. What suddenly applied load shall elongate the spring by 100 mm?
Solution
Δ = 100 mm, W = 600 N, R = 6d, τmax = 80 N/mm2 ,
G = 90 GPa = 80 × 103 N/mm2
τmax = 16 W R / π d3
⇒ 80 = 16 × 600 × (6d )/ π d 3
∴ d = 15.14 mm
R = 6d = 90.84 mm
Δ= 64 W R3 n / Gd 4
⇒ 100 =64 × 600 × 90.843 × n/(80 × 103 ) (15.14)4
= n = 14.6
A suddenly applied load shall cause double the stress caused by same load if applied gradually. Therefore, in this case if same stress is to be induced in spring wire cross-section by suddenly applied load it shall be W/2 = 600/2 = 300 N .
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