A carbon steel helical compression spring with squared and ground ends has an outside diameter of 0.85 in and a wire diameter of 0.074 in. The free length is 2 in, and the solid length is 0.618 in. Find the spring constant and the load required to compress the spring to the solid length.\

Solution :

Given :

d = diameter of the wire of the spring = 0.074 in

  = 0.188 cm

R = mean radius of coil

                                      = 1.08 cm

G = modulus of rigidity of carbon steel wire =

L = length of wire of spring = 2 inch

∴ L = 5.08 cm

k = spring constant

k = axial load/ axial deflection

  ..................(1)

This is taken from the Torsion equation,

∴ From equation (1)

So the axial compression is = 1.57 cm

Axial load required W= k x δ

                                   = 11334 x 1.57

                                   = 177.95 N

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answer:

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Explanation: