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Equivalent Spring Stiffness Calculator
Calculate combined stiffness for springs in series or parallel
Calculation Parameters
Based on ISO 26909 and Hooke's Law principles
Calculation Results
Equivalent Stiffness:
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Total Compliance:
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Force Distribution:
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Deflection Distribution:
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System Analysis:
How the Calculator Works
Springs in Parallel
When springs are arranged side-by-side (parallel), they share the load equally:k_eq = k₁ + k₂ + k₃ + ... + k_n
Characteristics:
- Total stiffness increases
- Same deflection for all springs
- Force is distributed among springs
- Used to increase load capacity
Springs in Series
When springs are connected end-to-end (series), they experience the same force:1/k_eq = 1/k₁ + 1/k₂ + 1/k₃ + ... + 1/k_n
Characteristics:
- Total stiffness decreases
- Same force through all springs
- Total deflection is sum of individual deflections
- Used to increase working range
Mixed Configurations
Complex arrangements combine series and parallel connections:- Calculate parallel groups first
- Then calculate series combinations
- Work from inside out for nested configurations
Spring Types and Applications
- Compression Springs: Most common, resist compressive forces
- Extension Springs: Resist tensile forces, have initial tension
- Torsion Springs: Resist rotational forces, k in N·m/rad
- Disc Springs: High load capacity in small space, non-linear
Important Considerations
- Spring rate may vary with deflection (non-linear springs)
- Consider coil binding in compression springs
- Account for initial tension in extension springs
- Temperature affects spring stiffness
- Fatigue life depends on stress range
Practical Applications
- Vibration Isolation: Series springs for lower frequency
- Load Sharing: Parallel springs for heavy loads
- Fine Tuning: Mixed configurations for specific characteristics
- Redundancy: Multiple springs for safety
📘 Spring Stiffness Calculator
Calculates equivalent stiffness of multiple springs in series, parallel, or mixed configurations.
Parallel: k = k₁ + k₂ + ... | Series: 1/k = 1/k₁ + 1/k₂ + ...
💼 Applications
- Compressor Vibration Isolation: Required fn = 5 Hz, mass 1200 kg. Need k = 118 kN/m. Solution: 4 springs parallel × 29.5 kN/m each.
- Instrument Suspension: Have 5000 N/m springs, need 2000 N/m. Solution: 2 in series → k = 2500 N/m. Add adjustment for fine-tuning.
- Two-Stage Isolation: Upper: 4 springs × 10000 N/m parallel = 40 kN/m. Lower: 4 × 8000 N/m = 32 kN/m. Stages in series → effective ~18 kN/m.
- Emergency Replacement: Broken spring 12000 N/m. Only 6000 N/m available. Solution: 2 parallel = 12000 N/m ✓
Spring Formula:
Helical spring: k = Gd⁴ / (8D³n) where G = shear modulus (80 GPa steel), d = wire Ø, D = mean coil Ø, n = active coilsCategories: