Knowledge Answer

How does the mean diameter (Dm) variation influence the radial expansion of a wave spring during compression?

2026-06-16 FAQ

As a wave spring is compressed toward its solid height, the waves flatten, causing the mean diameter $D_m$ to expand. This radial expansion $\Delta D$ can be approximated by $\Delta D = 0.02 \cdot \frac{(L_0 - L_1)^2}{D_m \cdot Z}$, where $L_0$ is the free height and $L_1$ is the work height. In precision bore installations, this expansion must be account...

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As a wave spring is compressed toward its solid height, the waves flatten, causing the mean diameter $D_m$ to expand. This radial expansion $\Delta D$ can be approximated by $\Delta D = 0.02 \cdot \frac{(L_0 - L_1)^2}{D_m \cdot Z}$, where $L_0$ is the free height and $L_1$ is the work height. In precision bore installations, this expansion must be accounted for to prevent binding against the housing. If the clearance is insufficient, the resulting radial friction will artificially increase the measured spring rate and may lead to premature fatigue failure due to localized stress concentrations.

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