Calculate spring constant (rate) for helical compression and extension springs. Essential for mechanical design and engineering.
Spring rate (stiffness, constant) k defines the relationship between applied force and resulting deflection for a linear elastic spring. For helical compression and extension springs with round wire, it is derived from the fundamental torsion of a rod.
Basic Formula (Round Wire Helical Spring):
k = \frac{G \cdot d^4}{8 \cdot D^3 \cdot N}
Derivation: The torque on the wire is F·D/2, the torsional deflection per coil is (16·F·D²)/(π·G·d⁴), summing over N coils gives total deflection, leading to the formula above.
Spring Index (C = D/d)
Recommended range: 4 ≤ C ≤ 12. Low C causes high curvature stress; high C may lead to buckling. The calculator displays the index in the results.
| Material | G (MPa) | G (psi) | Notes |
|---|---|---|---|
| Music wire (ASTM A228) | 79,300 | 11.5×10⁶ | High strength, general use |
| Stainless steel (302/304) | 69,000 | 10.0×10⁶ | Corrosion resistant |
| Phosphor bronze (ASTM B159) | 27,500 | 4.0×10⁶ | Electrical/conductive, corrosion resistant |
| Beryllium copper | 48,300 | 7.0×10⁶ | High conductivity, high strength |
| Inconel 600 | 75,800 | 11.0×10⁶ | High temperature, corrosion resistant |
End Conditions & Active Coils:
Stress & Wahl Correction: The formula above gives stiffness, but to avoid failure, check maximum stress using the Wahl factor Kw. The maximum shear stress τmax = Kw·(8·F·D)/(π·d³) should be below material’s allowable stress. The calculator provides τmax based on the force at entered deflection.
Buckling: For compression springs, if free length / mean diameter > 4, buckling may occur under load. Use guides or reduce slenderness.
Set Point & Fatigue: Overstressing can cause permanent set (loss of length). For cyclic loading, design for infinite life by keeping stress below endurance limit.
Design a compression spring for a valve: Required force 50 N at 10 mm compression. Choose material: music wire (G=79,300 MPa). Try d=2.5 mm, D=20 mm, N=8 → k ≈ 6.05 N/mm. At 10 mm, force = 60.5 N (slightly high). Adjust N to 9 → k ≈ 5.38 N/mm → force ≈ 53.8 N, close. Fine-tune dimensions to meet exact force and stress limits.
Calculator Features (Enhanced):