Springs
Springs store energy and provide controlled force. Learn helical compression and extension springs, spring rate, shear stress with the Wahl factor, spring index, buckling and surge, springs in series/parallel, and leaf and torsion springs.
Valve springs, suspension coils, pen clickers, clutch springs and safety valves all rely on springs to push back with a predictable force. A helical spring is really a coiled torsion bar: the wire twists as the spring compresses.
Design means getting the right stiffness and force range while keeping wire stress below fatigue limits and avoiding buckling or resonance.
An analogy
A spring is a mechanical battery: compress it and it stores energy, release it and it gives the energy back.
Spring rate and Hooke's law
F = kx. Rate depends strongly on wire diameter (d⁴) and coil diameter (D³): a thicker wire or smaller coil makes a much stiffer spring. More active coils make it softer.
Stress and spring index
Wire stress is torsional shear plus direct shear and curvature effects, captured by the Wahl factor. Keep the spring index C between about 4 and 12: low C is hard to coil and highly stressed; high C tangles and buckles.
Ends, solid length and buckling
Squared-and-ground ends sit flat; each end adds an inactive coil. Solid height (all coils touching) must not be reached in service. Long, slender compression springs buckle — guide them in a bore or over a rod when free length > ~4D.
Fatigue and surge
Cyclic springs (engine valves) are designed on torsional fatigue with Goodman/Zimmerli data; shot peening adds ~20–30 % fatigue strength. Springs have their own natural frequency f = (d/2πnD²)√(G/32ρ)·… (surge); it must be far above operating frequencies to avoid valve float.
Formulas
- Spring rate:
k = G d⁴ / (8 D³ n) - Shear stress:
τ = K_w · 8 F D / (π d³) - Wahl factor:
K_w = (4C − 1)/(4C − 4) + 0.615/C - Series / parallel:
1/k = Σ1/k_i (series), k = Σk_i (parallel) - Energy:
U = ½ k x²
Worked example: Spring rate and stress
Given: Steel G = 80 GPa, d = 4 mm, D = 32 mm, n = 10; F = 200 N
Find: k, deflection, stress.
- k = 80 000 × 4⁴/(8 × 32³ × 10) = 7.81 N/mm
- x = 200/7.81 = 25.6 mm
- C = 8 → K_w = 1.184
- τ = 1.184 × 8 × 200 × 32/(π × 4³) = 302 MPa
Answer: 7.81 N/mm, 25.6 mm, ≈ 302 MPa
Worked example: Energy stored
Given: k = 7.81 N/mm, x = 25.6 mm
Find: Energy.
- U = ½ × 7.81 × 25.6² = 2560 N·mm
Answer: 2.56 J
Common mistakes
- Wrong: Forgetting the Wahl (curvature) factor. — Right: Apply K_w — inner-fibre stress is 10–30 % higher.
- Wrong: Allowing the spring to go solid. — Right: Keep a clash allowance (≈ 10–15 % of working deflection).
Applications
- Automotive: Coil suspension springs, valve springs, clutch diaphragm springs.
- Safety devices: Pressure-relief valves, spring-applied brakes.
- Consumer products: Switches, mattresses, retractable pens.
Key takeaways
- k = Gd⁴/(8D³n).
- τ = K_w 8FD/(πd³).
- Keep C ≈ 4–12; prevent solid and buckling.
- Series adds compliance, parallel adds stiffness.
Going further
- Variable-rate springs: Variable pitch or conical springs give progressive rates in suspensions.
- Belleville washers: Disc springs give very high force in small space and adjustable characteristics by stacking.
- Gas springs: Nitrogen-filled struts provide nearly constant force (tailgates, office chairs).