Compression Spring Calculator
Spring rate, load, shear stress with the Wahl correction, solid height and buckling check from wire diameter, mean diameter, active coils and material
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Spring rate, load, Wahl-corrected shear stress, solid height and a buckling check for a helical compression spring from wire diameter, mean diameter, active coils and material, with a sketch and a load–deflection line.
How to use
- Enter "Wire diameter d", "Mean diameter D" (or switch "Diameter entered" to "Outside diameter") and "Active coils Na".
- Choose the "End type" and enter the "Free length L0"; leave it empty to skip the solid-height and buckling checks.
- Pick a "Material"; the "Shear modulus G" is filled with a representative value you can edit.
- Under "Working condition" choose "Deflection given" or "Load given", enter the value and read the "Spring rate k", stresses and warnings.
How it works
- k = G·d⁴ ÷ (8·D³·Na), F = k·x, τ = Kw × 8·F·D ÷ (π·d³), Wahl factor Kw = (4C − 1)/(4C − 4) + 0.615/C, spring index C = D/d. Total coils and solid height follow the end type (Shigley Table 10-1).
- Tensile strength is estimated as Rm = A ÷ dᵐ (Shigley Table 10-4); the static shear limit is 45 % of Rm for cold-drawn carbon steel, 50 % hardened and tempered, 35 % stainless and nonferrous. The buckling limit of the free length is 2.63·D/α.
- All values are estimates; fatigue life and set are not covered.
Questions and answers
- How is the spring rate calculated?
- k = G·d⁴ ÷ (8·D³·Na). Music wire (G = 81 GPa), 2 mm wire, 20 mm mean diameter and 10 active coils give k = 81000 × 16 ÷ (8 × 8000 × 10) = 2.025 N/mm.
- Why multiply by the Wahl factor?
- Stress on the inside of a coil is higher because of the curvature. The Wahl factor corrects for this and for direct shear: about 1.145 at a spring index of 10 and 1.25 at 6.
- What spring index should I aim for?
- Usually 4 to 12. Below 4 the spring is hard to coil; above 12 the coils tangle and dimensions scatter.
- When does a compression spring buckle?
- When the free length exceeds 2.63·D/α: about 5.26 mean diameters with both ends fixed and parallel (α = 0.5), about 1.3 with one end free (α = 2). Longer springs need a guide rod or tube.