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.
Expand Hint
The spring force:
$$$F=kx$$$
where
$$k$$
is the spring constant, and
$$x$$
is the spring’s displacement.
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
Time Analysis
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