How this instrument works
A helical spring stores energy mainly by torsion: the axial force pulling or pushing on the coil acts at a lever arm equal to the coil radius, so each turn of wire twists like a short torsion bar rather than bending. That is why the shear modulus G, the material's resistance to angular distortion, appears in the formula instead of Young's modulus, which describes resistance to straight-line stretching. The diameter d is raised to the fourth power because a round wire's resistance to twisting — its polar moment of inertia, J = πd⁴⁄32 — grows with the fourth power of diameter, so a modest change in gauge shifts the rate dramatically.
The coil diameter D is cubed because it works against the spring twice over: it sets the torque arm the axial force acts through, and it sets how much wire length each coil contributes. Both effects scale with D, and multiplying them together gives the cube. The number of active coils n sits in the denominator because coils behave like springs wired in series — each one adds more wire that must twist under the same load, so total deflection grows with n and the rate falls in inverse proportion, the same arithmetic that makes series resistors conduct less than any one of them alone.
The identity assumes a thin wire wound at a small helix angle, so the load is carried almost entirely as torsion with only a negligible bending and direct-shear component — true for the pitch angles, typically under 10 degrees, seen in ordinary compression and extension springs. It also says nothing about failure: a spring's rate can be calculated exactly here while the material is one overload away from yielding, because that check needs the Wahl stress-concentration factor and an allowable shear stress, a separate calculation from the stiffness this formula returns.
- Enter the shear modulus G of the wire material — the unit menu offers kPa or MPa; common spring steel is about 79,300 MPa (79.3 GPa).
- Enter the wire diameter d, the thickness of that strand, in metres — 3 mm is 0.003 m.
- Enter the coil diameter D measured to the wire's centreline (the mean diameter), not the outer diameter of the coil.
- Enter the number of active coils n, the turns that actually twist and deflect under load, excluding any flattened end coils.
- Read the spring rate k in newtons per metre — the force needed to stretch or compress the coil by one metre.
Worked example — a 3 mm steel wire coil, 25 mm across, 10 turns
Take a spring wound from steel wire with a shear modulus of 79.3 GPa, or 79,300,000,000 Pa, using wire 3 mm thick — 0.003 m — coiled to a mean diameter of 25 mm, or 0.025 m, with 10 active coils. Raise the wire diameter to the fourth power: 0.003⁴ = 0.000000000081 m⁴, or 8.1 × 10⁻¹¹ m⁴. Multiply that by the shear modulus: 79,300,000,000 × 8.1 × 10⁻¹¹ = 6.4233.
Now the denominator: cube the coil diameter, 0.025³ = 0.000015625 m³, then multiply by 8 and by the 10 active coils: 8 × 0.000015625 × 10 = 0.00125. Divide the two results: k = 6.4233 ⁄ 0.00125 = 5138.64 N/m. That is the force, in newtons, needed to stretch or compress this spring by a full metre — about 5.14 newtons per millimetre of travel, a realistic rate for a compact mechanical return spring, and exactly the figure an engineer would target by choosing this wire size, coil size, and turn count.
Questions
Why does spring rate depend on shear modulus rather than Young's modulus?
Because a helical spring stores energy mainly by twisting the wire, not stretching it — each coil behaves like a short torsion bar loaded by the axial force acting on the coil radius. Torsional stiffness is governed by the shear modulus G, the resistance to angular distortion, while Young's modulus E describes resistance to straight-line stretching instead. Using E here would overstate steel's stiffness by roughly two and a half times, since E is about 200 GPa against G near 79 GPa for the same steel.
What is the difference between active coils and total coils?
Active coils are the turns free to twist and contribute to deflection; total coils include the end turns, which are often ground flat so they sit flush against the mounting seat and barely flex. A spring with 12 total coils might have only 10 active ones. Using the total count instead of the active count in this formula understates the true spring rate, sometimes by 10 to 20 percent, which is why spring catalogs list both figures separately.
Does coil diameter mean the inside, outside, or mean diameter?
Mean diameter — measured centreline to centreline across the coil, equal to the outer diameter minus one wire diameter. Entering the outer diameter instead makes D too large, and because D is cubed in the formula, that overstatement can understate the calculated rate by 30 percent or more for a thick strand. Callipers on a real spring usually read the outer diameter, so subtract that diameter before typing in D.
Why is the wire diameter raised to the fourth power?
Because a round wire's resistance to twisting, its polar moment of inertia, scales with the fourth power of its diameter: J = πd⁴⁄32. Small changes in that dimension therefore have an outsized effect on the result — increasing d by 25 percent very roughly doubles the spring rate with everything else fixed, which is why spring designers usually adjust gauge before they adjust the coil count.
Can this formula predict when the spring will break?
No — it gives stiffness, not strength. Whether a spring survives a given load depends on the shear stress in the wire, which needs the Wahl correction factor to account for stress concentration on the inside of each coil, checked against the material's allowable shear stress. Two springs can share the exact rate calculated here while one sits safely inside its elastic limit and the other is already yielding.
What happens to the rate if I add more coils?
It drops. Coils behave like springs connected in series: each additional active coil adds wire length that must twist under the same force, so total deflection grows and the rate falls in inverse proportion to the coil count. Doubling the coil count from 10 to 20 in the worked example halves the rate from 5138.64 N/m to about 2569.32 N/m, with the spring's other dimensions unchanged.