## Free Length Spring

Consider a free length spring is compressed to 50 mm with a 1 N applied force. If the 55 mm diameter helical compression spring has a 3 mm wire diameter and 15 coils, what is the spring’s free length in mm? Assume the shear modulus of elasticity is 300 N/mm^2.

##
__
__**Hint**

**Hint**

The spring force:

$$$F=kx$$$

where
$$k$$
is the spring constant, and
$$x$$
is the spring’s displacement.

##
__
__**Hint 2**

**Hint 2**

$$$k=\frac{d^4G}{8D^3N}$$$

where
$$G$$
is the shear modulus of elasticity,
$$d$$
is the wire diameter,
$$D$$
is the mean spring diameter, and
$$N$$
is the number of active coils.

A spring’s deflection and force are related by
$$F=kx$$
where the spring rate (spring constant)
$$k$$
is given by:

$$$k=\frac{d^4G}{8D^3N}$$$

where
$$G$$
is the shear modulus of elasticity,
$$d$$
is the wire diameter,
$$D$$
is the mean spring diameter, and
$$N$$
is the number of active coils.

Combining the two equations:

$$$\frac{F}{x}=\frac{d^4G}{8D^3N}$$$

Solving for displacement:

$$$x=\frac{8(55mm)^3(15)(1N)}{(3mm)^4(300N/mm^2)}=\frac{19,965,000mm^5}{24,300mm^4}=821.6\:mm$$$

Recall that displacement is:

$$$x=l_f-l_o$$$

where
$$l_f$$
is the final length, and
$$l_o$$
is the original length.

Because the spring is compressed, displacement is negative:

$$$-821.6mm=55mm-l_o$$$

$$$l_o=55mm+821.6mm=876.6\:mm$$$

876.6 mm