Knowledge Answer

How can 'Resonance' cause a spiral retaining ring to fail in a high-vibration environment, and how is the 'Natural Frequency' calculated?

2026-06-16 FAQ

If the frequency of system vibration matches the natural frequency of the retaining ring, the ring can undergo high-amplitude oscillations, leading to a loss of cling force. The fundamental natural frequency $f_n$ for a circular ring is $f_n = _x000c_rac{k}{2 ext{π}} ext{sqrt} _x000c_rac{E I}{m R^3}$. In a resonant state, the ring 'dances' in the groove,...

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If the frequency of system vibration matches the natural frequency of the retaining ring, the ring can undergo high-amplitude oscillations, leading to a loss of cling force. The fundamental natural frequency $f_n$ for a circular ring is $f_n = _x000c_rac{k}{2 ext{π}} ext{sqrt} _x000c_rac{E I}{m R^3}$. In a resonant state, the ring 'dances' in the groove, causing rapid fretting wear or even 'jumping' out of the groove entirely. This is common in reciprocating compressors. To solve this, engineers 'tune' the ring by changing its mass (thickness) or its stiffness (width), or by using a multi-turn ring which has higher internal damping due to inter-turn friction.

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