Helical Spring Rate

Helical spring rate: k = G·d⁴ / (8·D³·n)

For a round-wire helical spring, whether it works in compression or extension, the stiffness (spring rate) comes out as k = G·d⁴ / (8·D³·n). Here G is the shear modulus of the wire material (around 79–81 GPa for music wire and most carbon-steel springs, closer to 73 GPa for stainless 302), d is the wire diameter, D is the mean coil diameter (outer minus the wire), and n counts the active coils. Force and deflection then obey Hooke's law: F = k·x. Run the numbers for d = 2 mm, D = 20 mm, n = 10 and G = 79 GPa and you land on k ≈ 1.98 N/mm, so a 10 N load squeezes the spring roughly 5 mm. Designers tend to keep the spring index C = D/d between 4 and 12 so the part is easy to make, and the Wahl correction factor Kw handles the stress concentration when you size the wire against fatigue.

Applications

You find these springs in automotive suspensions (coil-over springs and McPherson struts) and in the valve springs of internal-combustion engines. They also turn up in mechanical scales, in retractable pens and click mechanisms, in mousetraps and toys, in railway truck springs, in pressure-relief valves, in garage-door counterbalance assemblies and in mattresses. Mechanical watch mainsprings work the same way too, though those are spiral rather than helical.

FAQ

Why does d appear to the 4th power? A helical spring loads its wire mostly in torsion, and the torsional stiffness of a round bar scales with the polar moment of inertia, which goes as d⁴.

Active vs total coils? Only the active coils n actually flex. End coils that have been ground or closed sit idle, so a typical compression spring ends up with n_active = n_total − 2.

What happens beyond the elastic limit? The spring takes a permanent set. Its free length shrinks and k drops a little. To avoid that, designers keep the corrected shear stress (Kw·τ) under the material's allowable for the fatigue life they're after.