Compression spring rate
Work out the rate of a helical compression spring from its wire diameter, mean coil diameter, active coils and material, then check the shear stress and deflection at your working load. The Wahl factor is applied because a spring is a curved beam, and the inside of the coil carries more stress than a straight-wire calculation suggests.
| Quantity | Value | Basis |
|---|---|---|
| Mean coil diameter | 20.00 mm | outside 22.00 mm − wire |
| Spring index C | 10.00 | D / d — 4 to 12 is the practical band |
| Wahl factor | 1.1448 | curvature and direct shear correction |
| Rate | 1.982 N/mm | G d⁴ / (8 D³ N) |
| Load at 10 mm | 19.8 N | rate × deflection |
| Deflection at load | 25.22 mm | 50 N ÷ 1.982 |
| Shear stress at load | 364.4 MPa | Kw · 8 F D / (π d³) |
| Solid height | 24.0 mm | d (N + 2) |
| Wire length | 0.754 m | ≈ π D (N + 2) |
Keep the working stress below roughly 45% of the material’s torsional yield for a spring that has to survive millions of cycles.
k = G d⁴ / (8 D³ N) · N = active coils only
C = D / d · practical band 4–12
Kw = (4C−1)/(4C−4) + 0.615/C · curvature + direct shear
τ = Kw · 8 F D / (π d³) · at the working load
δ = F / k · linear within the elastic range
A clear path from inputs to results.
A compression spring's rate is fixed by four numbers: wire diameter, mean coil diameter, active coil count and the shear modulus of the material. Because the wire diameter enters to the fourth power and the coil diameter to the third, small changes in wire move the rate far more than anything else. This calculator gives the rate, then checks the shear stress and deflection at your working load with the Wahl correction applied — the inside of a coil carries more stress than a straight-wire calculation implies.
How to use
- Enter the wire diameter.
- Enter the mean coil diameter — outside diameter minus one wire diameter.
- Count only the active coils; the two end coils that sit on the seat do not deflect.
- Pick the wire material, which sets the shear modulus.
- Enter the working load you expect the spring to carry.
- Read the rate, the stress at that load, the deflection and the solid height.
The formula
Reference: Wahl (1963) coil spring correction; Shigley's Mechanical Engineering Design ch. 10; ASTM A228 / A231 material shear moduli.
Worked example
3 mm wire, 24 mm mean diameter, 8 active coils, chrome-vanadium, 120 N
- Wire diameter 3 mm.
- Mean coil diameter 24 mm.
- 8 active coils.
- Chrome-vanadium wire, G = 77,200 MPa.
- Working load 120 N.
- 1
Rate: 77,200 × 3⁴ / (8 × 24³ × 8) = 6,253,200 / 884,736 = 7.068 N/mm.
- 2
Spring index: C = 24 / 3 = 8.0.
- 3
Wahl factor: (4×8 − 1)/(4×8 − 4) + 0.615/8 = 1.125 + 0.077 = 1.184.
- 4
Stress at 120 N: 1.184 × 8 × 120 × 24 / (π × 3³) = 321.6 MPa.
- 5
Deflection: 120 / 7.068 = 16.98 mm. Solid height: 3 × (8 + 2) = 30 mm.
- 6
At 25 mm of compression the spring carries 25 × 7.068 = 176.7 N.
Rate 7.068 N/mm · index 8.0 · Wahl 1.184 · 321.6 MPa at 120 N · 16.98 mm deflection · 30 mm solid.
Interpreting the result
The spring sits at index 8, comfortably inside the 4–12 band where a spring can be wound and will not surge badly. At 321.6 MPa it is working but not at the limit: music wire at that diameter typically tolerates a torsional yield around 1,400–1,600 MPa, so this is roughly 20–23% of yield — good for a long-life spring. Push the wire down to 2 mm at the same coil diameter and the rate collapses to 1.4 N/mm while the stress at 120 N climbs past 1,000 MPa, which is why wire diameter is the first thing to change and the last thing to guess.
Assumptions & limitations
- Rate applies to the linear, elastic range only. A spring that compresses toward solid becomes progressively stiffer.
- Active coils exclude the two end coils, which is correct for squared and ground ends. Plain ends add no dead coil; open ends change the count.
- The mean coil diameter is used, not the outside diameter. Using the outside diameter overstates the rate by roughly 10% at index 8.
- Wahl correction accounts for curvature and direct shear. It does not account for set, relaxation or fatigue damage.
- Shear modulus is taken as constant with temperature. Above roughly 120 °C for steel, it is not.
Frequently asked questions
Why does wire diameter matter so much?
Rate scales with d⁴. Doubling the wire diameter multiplies the rate by sixteen at the same coil diameter, while also raising the stress margin dramatically. It is the single most powerful design lever, and the reason a small manufacturing tolerance on wire has a large effect on rate.
What is the Wahl factor for?
A spring is a curved beam, so the stress on the inside of the coil is higher than the average. Wahl's factor combines that curvature effect with the direct shear carried by the wire. Ignoring it understates the peak stress by 10–20% at typical indices.
Does the number of active coils change the rate linearly?
Yes — rate is inversely proportional to active coil count. Ten active coils give half the rate of five. That makes coil count the easiest way to trim a rate after the wire and diameter are fixed.
What is a good spring index?
Between 4 and 12. Below 4 the spring is hard to wind and the stress concentration is severe. Above 12 the spring is floppy, prone to buckling and surge, and the wire is easily over-stressed. Index 7 to 9 is the sweet spot for most designs.
How do I set the installed length?
Work backwards from the solid height. Installed length minus solid height is the total travel available; divide the working load by the rate to get the deflection, and check that deflection plus preload stays inside that travel. A spring should never reach solid under normal load.
Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.