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How is the theoretical spring rate (k) for a multi-turn Crest-to-Crest wave spring derived considering the serial arrangement of waves?

2026-06-16 FAQ

The spring rate for a multi-turn Crest-to-Crest wave spring is calculated by treating each turn as a series of individual waves. For a spring with $N$ turns and $n$ waves per turn, the total number of active waves is $Z = n \cdot N$. Using the standard deflection formula for a curved beam, the rate is $k = \frac{E \cdot b \cdot t^3 \cdot n^4}{I_{coeff} \c...

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The spring rate for a multi-turn Crest-to-Crest wave spring is calculated by treating each turn as a series of individual waves. For a spring with $N$ turns and $n$ waves per turn, the total number of active waves is $Z = n \cdot N$. Using the standard deflection formula for a curved beam, the rate is $k = \frac{E \cdot b \cdot t^3 \cdot n^4}{I_{coeff} \cdot D_m^3 \cdot N}$, where $E$ is the Young's Modulus, $b$ is the radial wall, $t$ is the material thickness, and $D_m$ is the mean diameter. The factor $1.15$ or similar empirical constants ($I_{coeff}$) are used to account for the actual boundary conditions at the contact points. In a multi-turn configuration, the total deflection $f_{total}$ is the sum of the deflections of each turn. Because the load $P$ is constant throughout the serial stack, the total rate is inversely proportional to the number of turns $N$. For precision aerospace applications, the operating stress $\sigma$ must be checked using $\sigma = \frac{3 \cdot π \cdot P \cdot D_m}{4 \cdot b \cdot t^2 \cdot n^2}$, ensuring it remains below the yield strength at the operating temperature.

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