Question

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.\

Answer 1

Solution :

Given :

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

  = 0.188 cm

R = mean radius of coil $=\frac{0.85}{2}$

                                      = 1.08 cm

G = modulus of rigidity of carbon steel wire = $79 \ GN/m^2$

L = length of wire of spring = 2 inch

∴ L = 5.08 cm

k = spring constant

k = axial load/ axial deflection

 $k= \frac{Gd^4}{64R^3}$ ..................(1)

This is taken from the Torsion equation,

$\frac{T}{I_P}=\frac{\tau_{max}}{d/2}=\frac{G \theta}{2 \pi R}$

$\frac{WR}{\frac{\pi}{32} d^4} = \frac{\tau_{max}}{d/2} =\frac{G \theta}{2 \pi R}$

∴ From equation (1)

$k =\frac{79 \times 10^9 \times (0.188)^4 \times 10^{-8}}{64 \times (1.08)^3 \times 10^{-6}}$

$k = 11334 \ N/m$

So the axial compression is $\delta =0.618 \ inch$ = 1.57 cm

Axial load required W= k x δ

                                   = 11334 x 1.57

                                   = 177.95 N