Correction Factor

Consider a helical torsion spring is made from a 0.5 cm diameter wire. If the mean spring diameter is 50 mm, what is the correction factor?

Expand Hint
Correction factor:
$$$K_i=\frac{(4C^2-C-1)}{[4C(C-1)]}$$$
Hint 2
$$$C=\frac{D}{d}$$$
where $$D$$ is the mean spring diameter and $$d$$ is the wire diameter.
Correction factor:
$$$K_i=\frac{(4C^2-C-1)}{[4C(C-1)]}$$$
where $$C=D/d$$ , $$D$$ is the mean spring diameter, and $$d$$ is the wire diameter.

First, let’s solve for $$C$$ :
$$$C=\frac{50mm}{5mm}=10$$$
Thus,
$$$K_i=\frac{(4(10)^2-(10)-1)}{[4(10)(10-1)]}=\frac{(400-10-1)}{[40(9)]}=\frac{389}{360}=1.08$$$
1.08