Knowledge Answer

How does radial expansion affect the clearance fit of a wave spring in a bore during compression?

2026-06-16 FAQ

As a wave spring is compressed from its free height to a working height, the waves flatten, causing an increase in the outer diameter ($O.D.$). This radial expansion $\Delta D$ can be approximated by the formula $\Delta D = 0.02 \cdot \frac{(W_f - W_h) \cdot N^2}{D_m}$, where $W_f$ is free height and $W_h$ is work height. If the spring is housed in a bore...

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As a wave spring is compressed from its free height to a working height, the waves flatten, causing an increase in the outer diameter ($O.D.$). This radial expansion $\Delta D$ can be approximated by the formula $\Delta D = 0.02 \cdot \frac{(W_f - W_h) \cdot N^2}{D_m}$, where $W_f$ is free height and $W_h$ is work height. If the spring is housed in a bore with insufficient clearance, it will bind against the wall, causing a non-linear spike in the spring rate and localized wear. Designers must ensure that $Bore_{min} > O.D._{max} + \Delta D$. In high-temperature environments using $17-7PH$ stainless steel, the thermal expansion coefficient $\alpha$ must also be added to the radial expansion calculation to prevent catastrophic interference at the operating temperature $T_{op}$.

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