Hong Life CalculatorsUnderstanding the helical compression spring calculator — concept and principle
A helical compression spring stores energy by twisting its wire, so its stiffness and stress depend on the wire diameter, the coil diameter and the number of working coils. This calculator gives the spring rate, the deflection under your load, the corrected maximum shear stress in the wire and the spring index.
How it is calculated
- Spring rate: k = G d⁴ / (8 D³ n), with G in MPa (GPa × 1,000), d and D in mm, giving N/mm.
- Deflection: δ = F / k.
- Spring index: C = D / d.
- Wahl factor: K_w = (4C − 1)/(4C − 4) + 0.615/C.
- Maximum shear stress: τ = K_w × 8 F D / (π d³), in MPa.
The principle
Each coil acts like a short bar in torsion: the axial force F creates a torque F·D/2 on the wire. Applying torsion theory to the full length of wire, about π D n, gives the rate formula, which is why stiffness rises with the fourth power of wire diameter and falls with the cube of coil diameter. The basic stress 8FD/(πd³) is the torsional stress on a straight bar. A. M. Wahl showed that wire curvature concentrates stress on the inside of the coil and that direct shear adds to it; his factor K_w combines both effects.
Worked example
- Inputs: d = 2 mm, D = 16 mm, 8 active coils, G = 79 GPa, F = 100 N.
- k = 79,000 × 2⁴ / (8 × 16³ × 8) = 1,264,000 / 262,144 = 4.822 N/mm.
- δ = 100 / 4.822 = 20.74 mm.
- C = 16/2 = 8; K_w = 31/28 + 0.615/8 = 1.1071 + 0.0769 = 1.184.
- Uncorrected stress = 8 × 100 × 16 / (π × 8) = 509.3 MPa; τ = 1.184 × 509.3 = 603.0 MPa.
| Mean coil diameter (mm) | Spring rate (N/mm) |
|---|---|
| 4 | 308.59375 |
| 8 | 38.57421875 |
| 12 | 11.42939815 |
| 16 | 4.821777344 |
| 20 | 2.46875 |
| 24 | 1.428674769 |
| 32 | 0.602722168 |
How to read the result
Compare τ with the allowable shear stress of the wire. Allowable values are a fraction of the wire's tensile strength, which itself rises as wire gets thinner; for static loading Shigley's tables give roughly 35–50% of tensile strength depending on material, and fatigue loading allows less. A spring index between about 4 and 12 is easy to manufacture; below that, coiling is hard, above it the spring tends to tangle and buckle. Check that the deflection fits: with squared and ground ends the solid length is about d × (n + 2) = 20 mm here, so the free length must be at least 20 + 20.7 mm plus clearance.
Common mistakes
- Entering the outside diameter instead of the mean diameter (D = outside − d).
- Counting total coils as active coils; closed ends are inactive.
- Using GPa without converting, which makes k 1,000 times too small.
- Ignoring buckling of long, slender springs.
Frequently asked questions
What shear modulus should I use?
About 79 GPa for common carbon spring steels; stainless spring wire is lower, around 69–70 GPa.
How do I make the spring softer?
Add active coils, use thinner wire or a larger coil diameter; thinner wire changes the rate most.
What if the spring is compressed solid?
Stress at solid length exceeds the working stress; keep it below the allowable to avoid permanent set.
Sources: Wahl, A. M., Mechanical Springs, McGraw-Hill · Budynas & Nisbett, Shigley's Mechanical Engineering Design, McGraw-Hill · EN 13906-1, Cylindrical helical springs made from round wire and bar — Calculation and design — Part 1: Compression springs