Hooke's Law
Hooke's Law states that within the elastic limit, the force a spring exerts is directly proportional to its deflection. The spring constant k (stiffness) is the proportionality constant.
Core Formula
F = k × x
k = F / x (N/m)
x = F / k (m)
F = force (N)
k = spring constant (N/m)
x = deflection from natural length (m)Worked Example
Suspension spring: k=25,000 N/m, load=500 N
x = F/k = 500/25,000 = 0.02 m = 20 mm deflection
Spring test: 150 N force → 30 mm deflection
k = 150/0.03 = 5,000 N/mSprings in Combination
Parallel (same deflection, forces add):
k_total = k₁ + k₂ + k₃
Series (same force, deflections add):
1/k_total = 1/k₁ + 1/k₂ + 1/k₃
Two equal springs parallel: k_total = 2k
Two equal springs series: k_total = k/2Natural Frequency
ω_n = √(k/m) (rad/s)
f_n = ω_n / 2π = (1/2π)√(k/m) (Hz)
Car suspension: k=50,000 N/m, m=400 kg
f_n = (1/2π)√(50000/400) = 1.78 Hz (comfortable ride range)Calculate spring force: Free Spring Constant Calculator
Hooke's Law and Spring Constant
F = k × x, where F = force (N), k = spring constant / stiffness (N/m), x = displacement from equilibrium (m). A higher k means stiffer spring: less deflection for the same force. Rearranging: k = F/x. Example: a spring deflects 50 mm (0.05 m) under a 100 N load: k = 100/0.05 = 2,000 N/m = 2 kN/m. Springs in series (softer combination): 1/k_total = 1/k₁ + 1/k₂. Springs in parallel (stiffer combination): k_total = k₁ + k₂.
Applications
- Automotive suspension: Spring rates of 15–30 kN/m (comfort car) to 60–200 kN/m (sports car). Corner weight (load per spring) determines required spring rate for target ride height.
- Industrial machinery: Vibration isolation springs sized so natural frequency (f = (1/2π)√(k/m)) is well below excitation frequency.
- Return springs in valves: Must provide sufficient closing force at maximum lift against fluid pressure. Rate must not cause pressure oscillation (flutter).
- Weighing scales: Linear spring deflection directly indicates weight. Precision scales use very low-rate springs for sensitivity.
Frequently Asked Questions
What is the natural frequency of a spring-mass system?
f_n = (1/2π) × √(k/m) Hz, or ω_n = √(k/m) rad/s. A car with spring rate 20 kN/m and sprung mass 350 kg: f_n = (1/2π) × √(20000/350) = (1/2π) × 7.56 = 1.20 Hz — typical for ride comfort. Resonance occurs when excitation frequency matches f_n (e.g., road roughness wavelength at vehicle speed). Dampers (shock absorbers) reduce resonance amplitude.
What is the elastic limit and why does it matter?
Hooke's Law applies only within the elastic (proportional) limit — below the yield stress of the spring material. Beyond this limit, permanent deformation (set) occurs and the spring no longer returns to its original length. Spring design ensures maximum working load produces stress below the fatigue limit (for cyclic loading) or below ~40–60% of yield strength (for static loads). Springs that are over-compressed and permanently deformed are said to have "taken a set."
How is spring constant related to material and geometry?
For a helical coil spring: k = Gd⁴ / (8D³n), where G = shear modulus of material, d = wire diameter, D = mean coil diameter, n = number of active coils. Stiffening options: use stiffer material (higher G), thicker wire (d⁴ effect is powerful), reduce coil diameter, reduce coil count. Leaf springs: k ∝ Ebt³ / L³ where b = width, t = thickness, L = span — thickness is dominant (t³).