Compression Spring Rate
Calculate the spring rate (stiffness) of a helical compression spring from wire diameter, mean coil diameter, number of active coils and shear modulus. Optionally enter a load to get deflection and wire shear stress.
Find the Stiffness of a Helical Compression Spring
Compression Spring Rate Formula
Where:
- \(k\) — spring rate, or stiffness (N/mm)
- \(G\) — shear modulus of the wire material (N/mm², i.e. MPa)
- \(d\) — wire diameter (mm)
- \(N_a\) — number of active coils
- \(D\) — mean coil diameter (mm)
This formula applies to a close-coiled helical spring made of round wire, where the load is carried primarily by torsion of the wire. It shows that stiffness grows very rapidly with wire diameter (to the fourth power) and falls with the cube of the coil diameter and with the number of active coils.
Compression Spring Rate – Calculation Example
Example: a music-wire spring with a wire diameter of 3 mm, a mean coil diameter of 25 mm, 10 active coils and a shear modulus of 79.3 GPa (79,300 N/mm²), loaded with 250 N.
\[ k = \frac{79\,300 \times 3^{4}}{8 \times 10 \times 25^{3}} = \frac{79\,300 \times 81}{1\,250\,000} \approx 5.139 \ \text{N/mm} \]
\[ f = \frac{F}{k} = \frac{250}{5.139} \approx 48.65 \ \text{mm} \]
\[ \tau = K_w \, \frac{8 F D}{\pi d^{3}} = 1.176 \times \frac{8 \times 250 \times 25}{\pi \times 27} \approx 693 \ \text{MPa} \]
Result: the spring rate is k ≈ 5.139 N/mm (29.34 lbf/in). Under a 250 N load the spring deflects about 48.65 mm and the shear stress in the wire is roughly 693 MPa (≈ 100 ksi).
How to Use the Calculator
- Enter the Wire diameter (d) of the round wire and select mm or in.
- Enter the Mean coil diameter (D), measured to the centre of the wire (outer diameter minus one wire diameter), with its unit.
- Enter the Number of active coils (Na). For closed and ground ends subtract 2 from the total coils; for closed unground ends subtract 3; for plain ends count all coils.
- Enter the Shear modulus (G) of the wire material and pick its unit (GPa, Mpsi or psi). Typical values for common spring wires are listed in the field description.
- Optionally enter an Applied axial load (F) in N, lbf or kgf. The calculator then also reports the deflection (from Hooke's law, \(f = F/k\)) and the shear stress in the wire, including the Wahl curvature correction factor.
- Read the spring rate in N/mm, lbf/in and N/m in the results table, together with any warnings about the spring index or stress level.
What Is Spring Rate?
Spring rate, also called spring constant or stiffness, is the force required to compress (or extend) a spring by one unit of length. A spring with a rate of 5 N/mm needs 5 newtons of force for every millimetre of compression — 10 N compresses it 2 mm, 20 N compresses it 4 mm, and so on, as long as the spring stays in its elastic range. Spring rate is the single most important number describing a compression spring, because it determines how the spring will behave in a mechanism: how much force it exerts at a given displacement and how far it will sit under a given load.
Although a compression spring looks like it is being squeezed, the wire itself is loaded almost entirely in torsion. Each cross-section of the wire twists slightly as the load is applied, and this twisting is what stores the energy. That is why the formula uses the shear modulus \(G\) rather than Young's modulus, and why it has the same \(d^4 / D^3\) character as a torsion formula — the same physics covered in more detail by our shaft torsion calculator.
The strong dependence on dimensions matters in practice. Doubling the wire diameter makes the spring 16 times stiffer; doubling the coil diameter makes it 8 times softer; doubling the number of coils makes it twice as soft. This is why spring manufacturers quote wire and coil diameters to very tight tolerances — small dimensional errors produce noticeable rate changes. If you are checking whether a spring is strong enough as a component, the wire stress result also serves as a quick sanity check, similar in spirit to verifying section modulus in beam design.
Interpreting the Results
The main result, \(k\), is given in N/mm and lbf/in. If you entered a load, the deflection \(f = F/k\) tells you how far the spring shortens under that load, and the shear stress tells you whether the wire is working comfortably within its material limits. For common spring steels, keep the corrected shear stress well below about 45–50 % of the wire's tensile strength for static loads, and lower still for cyclic service. The calculator flags results that approach roughly 1000 MPa, a typical threshold for common spring wires.
The calculator also reports the spring index \(C = D/d\). Values between about 4 and 12 are considered manufacturable and well-behaved: below 4 the coiling stresses are severe and the spring is hard to wind; above 15 the coil is flimsy and may buckle, tangle or behave inconsistently. The example spring above has \(C = 25/3 \approx 8.3\), squarely in the good range.
Assumptions and Practical Tips
- Close-coiled round wire: the formula assumes a helical spring of round wire with closely spaced coils, loaded axially, with negligible direct shear and bending effects.
- Linear elastic behaviour: the rate is constant only until the wire yields or the coils reach solid height. Once coils touch, the effective rate rises sharply.
- Count active coils correctly: end coils that are closed (touching) barely deflect. Using the total coil count instead of the active count is the most common source of error — it can underestimate the rate by 20–30 % on short springs.
- Check deflection against available travel: the calculated deflection must fit between the free length and the solid height of the spring; otherwise the spring bottoms out.
- Use the right shear modulus: stainless steel and non-ferrous wires are noticeably softer than music wire (≈ 69 GPa vs ≈ 79.3 GPa), which lowers the rate accordingly.
- Temperature and service: at elevated temperatures springs relax and lose force over time; the calculated rate applies to the wire at room temperature.
When sizing the wire for a given load, remember that stress rises with \(1/d^3\) while rate rises with \(d^4\), so there is a direct trade-off between stiffness and stress — a thicker wire gives a stiffer, lower-stressed spring but a much higher force for the same travel. For pressure-containing parts a similar wall-thickness logic applies, as covered in our pipe wall thickness calculator.
Frequently Asked Questions
How do I convert spring rate between N/mm and lbf/in?
1 N/mm equals about 5.710 lbf/in, and 1 lbf/in equals about 0.1751 N/mm. So multiply a metric rate by 5.71 to get lbf/in, or divide by 5.71 to go the other way. If you work in SI base units, remember that 1 N/mm is the same as 1000 N/m — a common slip when comparing with textbook values quoted in N/m.
How can I measure the spring rate of a real spring?
Compress the spring between flat plates and record its length at two known loads. The rate is the change in force divided by the change in length: \( k = \Delta F / \Delta L \). Use two points well apart (e.g. 10% and 60% of travel) and measure free length first. A bathroom-scale and caliper setup is accurate enough for most hobby checks. A single load/free-length measurement is not enough, because springs often have an initial settle region.
How do I combine several springs — does the rate add up?
Springs in parallel (stacked so they compress the same distance) add directly: \( k = k_1 + k_2 \). Springs in series (end to end, each carrying the same load) add as reciprocals: \( 1/k = 1/k_1 + 1/k_2 \), so the combination is always softer than the softest spring. This is why nesting two springs inside each other roughly doubles the effective rate.
Does this formula work for extension (tension) springs?
Yes, the rate formula \( k = Gd^4 / (8N_a D^3) \) applies to close-coiled helical extension springs too, since the wire is loaded in torsion the same way. Note that an extension spring with closed initial tension may show a higher apparent rate over the first few millimetres of extension until the coils separate. For torsion springs (loaded by twisting, not axial force) the formula is different — use a torsion spring calculator.
What is a typical spring rate value?
There is no single 'good' value — rates range from under 0.1 N/mm for delicate instrument springs to over 1000 N/mm for heavy automotive suspension springs. Typical passenger-car suspension springs are around 15–70 N/mm (85–400 lbf/in); small ballpoint pen and valve springs are often 0.1–5 N/mm. The right rate is set by how much force you need over the available travel, not by any universal benchmark.
Why does my measured spring rate differ from the calculated one?
The most common causes are using total coils instead of active coils, measuring outer diameter rather than mean diameter, and material substitution — stainless steel is about 13% softer than music wire at the same dimensions. Manufacturing tolerances also matter: because rate varies with \( d^4 \), a wire just 2% undersized makes the spring about 8% softer. Also check that the spring was not compressed past its linear range or to solid height during the measurement.
How do I find the maximum load a compression spring can take?
Two limits apply, and the lower one wins. First, the stress limit: keep the corrected shear stress (which the calculator shows) below roughly 45–50% of the wire's tensile strength for static loads, and lower for cyclic duty. Second, the geometric limit: the spring must not compress to solid height. Multiply the available travel (free length minus solid height) by the rate to get the force at solid height, and stay safely below it — typically at 85% of maximum deflection for cyclic service.