Compression spring calculator: spring rate, force, stress and solid length
This compression spring calculator uses the standard formulas for cold-formed helical compression springs: spring rate R = G × d⁴ ÷ (8 × Dm³ × n). A spring of 2 mm patented spring steel wire (class C/D) with 20 mm outer diameter and 8 active coils has a rate of 3.494 N/mm, so 10 mm of deflection takes 34.94 N.
Example: Material: patented spring steel wire, class C/D · Wire diameter d 2 mm · Outer diameter De 20 mm · Active coils n 8 · Free length L0 (optional) 50 mm · Deflection s (optional) 10 mm → Spring rate: R = 3.494 N/mm. Source: Federnfabrik Schmid – Normfedern catalogue, “Druckfederberechnung / Compression spring calculation”: formulas 3.1–3.10 (R = G·d⁴/(8·Dm³·n), τ = 8·Dm·F/(π·d³), τc zul ≦ 0.56·Rm, nt = n + 2, Lc ≦ nt·d), materials table with G and Rm (read 09.10.2026). Updated: .
How it is calculated
Spring rate R = G × d⁴ ÷ (8 × Dm³ × n). G is the shear modulus of the wire, d the wire diameter, Dm the mean coil diameter (outer diameter minus wire) and n the number of active coils. The force at a deflection s is F = R × s, the shear stress τ = 8 × Dm × F ÷ (π × d³). Federnfabrik Schmid gives these formulas for cold-formed compression springs with round wire up to 17 mm, more than 2 active coils and a spring index Dm ÷ d of 4 to 20.
Examples (spring steel, G = 81,500 N/mm²)
| Wire d | Outer De | Coils n | Rate R |
|---|---|---|---|
| 1 mm | 10 mm | 10 | 1.397 N/mm |
| 2 mm | 20 mm | 8 | 3.494 N/mm |
| 3 mm | 30 mm | 6 | 6.987 N/mm |
| 4 mm | 40 mm | 6 | 9.316 N/mm |
| 5 mm | 40 mm | 8 | 18.563 N/mm |
The wire diameter counts to the fourth power: with the same Dm and n, double the wire gives sixteen times the rate. The same spring in stainless steel 1.4310 (G = 70,000 N/mm²) only reaches 3.001 N/mm.
Solid length and stress check
With one closed coil at each end the total coils are nt = n + 2, and with ground ends the solid length is at most nt × d – 20 mm for the example. With a free length of 50 mm the spring can be compressed 30 mm, which takes 104.81 N and gives τc ≈ 601 N/mm². Schmid allows τc up to 0.56 × Rm; for class C/D wire of 2 mm, Rm = 2220 − 820 × lg d = 1973 N/mm², so the limit is 1105 N/mm² – the spring is safe at solid length. For Hooke’s law in general see the spring constant calculator; force units are in the force calculator.
Frequently asked questions
How do you calculate the spring rate of a compression spring?
R = G × d⁴ ÷ (8 × Dm³ × n). For d = 2 mm, Dm = 18 mm, n = 8 and G = 81,500 N/mm²: 3.494 N/mm.
What is the mean diameter?
Outer diameter minus wire diameter: 20 − 2 = 18 mm. It is the diameter through the wire centre.
How much force does the spring need for 10 mm?
Rate times deflection: 3.494 N/mm × 10 mm = 34.94 N.
How do I calculate the solid length?
With one closed coil per end, nt = n + 2; ground ends give at most nt × d. 10 coils × 2 mm = 20 mm.
When is the spring overstressed?
When the shear stress at solid length exceeds 0.56 × Rm. For 2 mm class C/D wire the limit is about 1105 N/mm².
Which shear modulus should I use?
Schmid lists 81,500 N/mm² for patented spring steel, 79,500 for oil-tempered wire and 70,000 for stainless 1.4310.
Sources and legal basis
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