Spring Rate Calculator (Helical Compression Spring)
By Santaji GadepatilinUpdated November 3, 2026
The stiffness of a coil spring comes from twisting its wire, so it depends strongly on the wire diameter and the coil diameter. This spring rate calculator uses k = G d⁴ ÷ (8 D³ n) for a helical compression or extension spring, then finds the deflection under a load and the corrected shear stress in the wire using the Wahl factor.
How to use the spring rate calculator
Enter the wire diameter, the mean coil diameter, the number of active coils and the wire material’s shear modulus, plus a load to check. The calculator returns the spring rate in N/mm, the deflection under that load, the spring index and the Wahl-corrected shear stress in the wire. Compare the stress with the wire’s allowable value (often 40–50% of its tensile strength for static loads).
Spring rate formula
k = G d⁴ ÷ (8 D³ n) · x = F ÷ k
τ = Kw × 8 F D ÷ (π d³) · Kw = (4C − 1) ÷ (4C − 4) + 0.615 ÷ C
Example
Music wire (G = 79.3 GPa), d = 2 mm, D = 20 mm, 10 active coils, loaded with 20 N:
- k = 79.3 × 10⁹ × 0.002⁴ ÷ (8 × 0.02³ × 10) = 1.982 N/mm
- Deflection = 20 ÷ 1.982 = 10.09 mm
- Index C = 20 ÷ 2 = 10; Wahl factor Kw = 1.1448
- τ = 1.1448 × 8 × 20 × 0.02 ÷ (π × 0.002³) = 145.8 MPa
Wahl correction factor
| Spring index C | Kw |
|---|---|
| 4 | 1.404 |
| 5 | 1.310 |
| 6 | 1.252 |
| 8 | 1.184 |
| 10 | 1.145 |
| 12 | 1.119 |
| 14 | 1.102 |
| 16 | 1.088 |
Spring rate for steel springs with 10 active coils
| Wire d | Mean coil D | Rate k |
|---|---|---|
| 1 mm | 10 mm | 0.99 N/mm |
| 1.5 mm | 12 mm | 2.90 N/mm |
| 2 mm | 16 mm | 3.87 N/mm |
| 2 mm | 20 mm | 1.98 N/mm |
| 3 mm | 25 mm | 5.14 N/mm |
| 4 mm | 32 mm | 7.74 N/mm |
| 5 mm | 40 mm | 9.68 N/mm |
| 6 mm | 50 mm | 10.28 N/mm |
Notes
- Springs in series add like resistors in parallel (1/k = 1/k₁ + 1/k₂); springs in parallel add directly.
- The natural frequency of the spring with a mass on it follows from the spring-mass frequency calculator.
Frequently asked questions
What is the formula for spring rate?
k = G d⁴ ÷ (8 D³ n), where G is the shear modulus of the wire, d the wire diameter, D the mean coil diameter and n the number of active coils.
How do wire and coil diameter affect stiffness?
Stiffness rises with the fourth power of wire diameter and falls with the cube of coil diameter. A 10% thicker wire makes the spring 46% stiffer; a 10% larger coil makes it 25% softer.
What is the Wahl factor?
A correction, K_w = (4C − 1)/(4C − 4) + 0.615/C, for the extra stress on the inside of the coil caused by wire curvature and direct shear. It matters most for tightly wound springs with a small index C.
What spring index is recommended?
C = D/d between about 4 and 12. Below 4 the spring is hard to coil; above 12 it tends to buckle and tangle.