Applied load shall elongate the spring, Mechanical Engineering

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?


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 .

Posted Date: 1/21/2013 1:14:22 AM | Location : United States

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