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⚙️ Spring Rate Calculator

Calculate the spring constant (rate) of a helical compression spring.

⚙️ Spring Rate Calculator

Calculate the spring constant (rate) of a helical compression spring.

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Calculation Result

Spring Rate (k)- lbf/in
Mean Diameter (D): - in

Spring Geometry

D_o = 20 mm
d = 2 mm

Calculator Description

The spring rate (k) expresses how "stiff" a coil spring is — the deflection (in m) produced per newton of applied load. Helical compression springs are used for cushioning, return force, load support and vibration isolation, and their behavior is set by the material shear modulus and the coil geometry.

What this calculator finds

Given the wire diameter d, mean coil diameter D, number of active coils n and the material shear modulus G, the calculator returns the spring rate k, letting you predict deflection under a design load or the stiffness you need.

Why it matters

  • Sizing stiffness for valve springs, suspensions, clutches and other machine parts
  • Choosing coil count and diameters to meet a target force–deflection curve
  • Feeding the pre-step input for stress and fatigue-life calculations

Formula

Spring Rate Formula (round-wire helical spring)

This is the standard closed-form relation for a round-wire spring in the linear elastic range, without curvature (Wahl) correction.

k=Gd48D3nk = \dfrac{G d^4}{8 D^3 n}
  • kSpring rate [N/m or lb/in]
  • GShear modulus [Pa or psi]
  • dWire diameter [m or in]
  • DMean coil diameter [m or in]
  • nNumber of active coils

How the formula works

  • k scales with d⁴, so a small increase in wire diameter sharply raises stiffness.
  • k is inversely proportional to D³, so a larger coil diameter makes the spring more flexible.
  • More active coils n lower the stiffness (inverse proportionality).
  • A larger shear modulus G (steel > copper) gives a stiffer spring.

Worked example

For a steel spring (G ≈ 79.3×10⁹ Pa) with wire diameter d = 2 mm, mean diameter D = 20 mm and n = 10 active coils: with d⁴ = 1.6×10⁻¹¹ and D³ = 8×10⁻⁶, k = 79.3×10⁹ × 1.6×10⁻¹¹ / (8 × 8×10⁻⁶ × 10) ≈ 1980 N/m (≈ 11.3 lb/in).

Useful Tips

Practical tips

  • To change stiffness a lot, adjusting wire diameter d is far more effective than changing D.
  • Active coils differ by end-fixing style; subtract the inactive (dead) coils from the total.
  • For precision design, apply the Wahl correction factor to account for curvature and stress concentration.

Limitations & cautions

  • The formula is valid only in the linear elastic range, before yielding; permanent set invalidates it.
  • Very small or very large spring index (D/d) introduces stress concentration not captured here.
  • Rising temperature lowers G and therefore the stiffness as well.