Knowledge Answer

How does the number of active waves and turns influence the theoretical spring rate in Crest-to-Crest wave springs?

2026-06-16 FAQ

In Crest-to-Crest wave springs, the spring rate $K$ is inversely proportional to the number of turns $n$ and directly proportional to the fourth power of the number of waves $N$ per turn. The governing equation for the load $P$ is $P = \frac{E \cdot b \cdot t^3 imes N^4 imes f}{1.59 \cdot D_m^3 imes n}$, where $E$ is the Modulus of Elasticity, $b$ is the...

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In Crest-to-Crest wave springs, the spring rate $K$ is inversely proportional to the number of turns $n$ and directly proportional to the fourth power of the number of waves $N$ per turn. The governing equation for the load $P$ is $P = \frac{E \cdot b \cdot t^3 imes N^4 imes f}{1.59 \cdot D_m^3 imes n}$, where $E$ is the Modulus of Elasticity, $b$ is the radial wall, $t$ is the material thickness, $f$ is the deflection, and $D_m$ is the mean diameter. Increasing the number of turns $n$ effectively adds springs in series, reducing the overall rate, while increasing the wave count $N$ stiffens the spring exponentially. Engineers must balance $N$ to avoid 'bottoming out' or exceeding the material's elastic limit at the wave peaks.

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