Answer
As a wave spring is compressed from its free height towards its solid height, the waves flatten, which naturally causes the mean diameter to expand. This expansion can be approximated by the formula $D_{expanded} = \sqrt{D_{free}^2 + (h^2 / \pi^2)}$, where $h$ is the height of the wave. If the spring is installed in a tight bore, this radial expansion can lead to 'binding' or friction against the bore wall, which artificially increases the spring rate and causes unpredictable hysteresis. Designers must ensure that the clearance between the Outer Diameter (OD) and the bore, or the Inner Diameter (ID) and the shaft, is sufficient to accommodate this expansion at maximum deflection. In precision medical devices, the use of a 'shim-end' wave spring can mitigate this by providing a flat contact surface that stabilizes the expansion.
