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How does the mean diameter ($D_m$) sensitivity impact the fatigue life of a wave spring?

2026-06-16 FAQ

The fatigue life of a wave spring is inversely related to the stress range $\Delta \sigma$ experienced during cycling. Since $\sigma \propto D_m / (b t^2 N_w^2)$, the mean diameter plays a significant role. Increasing $D_m$ for a fixed load $P$ actually increases the bending moment and the resulting stress. Using the Goodman relation, $\frac{\sigma_a}{S_e...

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The fatigue life of a wave spring is inversely related to the stress range $\Delta \sigma$ experienced during cycling. Since $\sigma \propto D_m / (b t^2 N_w^2)$, the mean diameter plays a significant role. Increasing $D_m$ for a fixed load $P$ actually increases the bending moment and the resulting stress. Using the Goodman relation, $\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_u} = 1$ (where $\sigma_a$ is alternating stress, $\sigma_m$ is mean stress, $S_e$ is endurance limit, and $S_u$ is ultimate tensile strength), engineers must optimize the diameter to minimize the stress range. In subsea valves where long-term reliability is paramount, a larger number of waves $N_w$ is often preferred over a larger $D_m$ to reduce the stress per wave and extend the cycle life into the $10^6$ range.

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