Library / Foundations: Physics / FPH-65
Foundations: Physics
Equivalent stiffness of two springs in series
Single spring stiffness that behaves like two springs joined end to end.
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- FPH-65
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- Foundations: Physics
- Result
- keq [N/m]
Springs in series carry the same force and their extensions add, so their compliances (1/k) add. For two springs, k_eq = k₁k₂/(k₁ + k₂), always smaller than the softer spring. Use the parallel page when the springs sit side by side and share the load.
Single spring stiffness that behaves like two springs joined end to end. Both carry the same force, their extensions add, so the combination is softer than either one.
Calculator
- Softer of the two400 N/m
- Total compliance, 1/k_eq0.004167 m/N
- Natural frequency with a 1 kg mass2.466 Hz
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The pair acts like one 240 N/m spring, softer than both 400 and 600 N/m.
- The softer spring dominates: the combination is always below the smaller stiffness, and is half of it when the two are equal.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N/m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k1 Stiffness of spring 1 | 1 to 1,000,000 N/m | 0.9983 | 599.6 | 598.6 | +10% | +5.77 % |
| k2 Stiffness of spring 2 | 1 to 1,000,000 N/m | 0.9975 | 399.8 | 398.8 | +10% | +3.77 % |
Sweep of Stiffness of spring 1
| Stiffness of spring 1 (N/m) | Equivalent stiffness (N/m) |
|---|---|
| 1 | 0.9983 |
| 5.623 | 5.571 |
| 31.62 | 30.04 |
| 177.8 | 137.2 |
| 1,000 | 375 |
| 5,623 | 542.2 |
| 31,620 | 588.8 |
| 177,800 | 598 |
| 1,000,000 | 599.6 |
Sweep one variable
Variables and how each one contributes
Equivalent stiffness
What the equation gives you. Everything else feeds it.
Stiffness of spring 1
The stiffness of spring 1, in N/m. Raises the equivalent stiffness, but by less than 10 % because the softer spring limits the result. Raising it by 10 % (to 440) moves the result by +5.77 %, from 240 to 253.8 N/m.
Stiffness of spring 2
The stiffness of spring 2, in N/m. Raises the equivalent stiffness, but by less than 10 % because the softer spring limits the result. Raising it by 10 % (to 660) moves the result by +3.77 %, from 240 to 249.1 N/m.
Worked example with the values above
- Compliances: 1/k₁ + 1/k₂ = 1/400 + 1/600 = 0.00416667 m/N.
- Invert: k_eq = 1 / 0.00416667.
- k_eq = 240 N/m.
Assumptions and limits
- Massless springs that obey Hooke’s law, joined end to end.
- Both springs carry the same force.
Common mistakes
- Adding the stiffnesses (that is the parallel case).
- Using the series formula for springs mounted side by side.
- Forgetting that three or more springs need 1/k_eq = Σ 1/kᵢ.
Statistics and uncertainty
The result is dominated by the softer spring, so its tolerance matters most. If one spring is ten times stiffer than the other, the combination is within 10 % of the soft one, and the stiff spring’s tolerance has little effect.
Where it is used
- Combining a suspension spring with a flexible mount.
- Finding the natural frequency of a mass on two springs in line.
Maths you need
- Hooke’s law: F = kx
- Reciprocals: adding compliances