Saturday, July 6, 2019

INITIAL STAGES OF BEARING FAILURE PRODUCES HIGH FREQUENCY VIBRATIONS? HOW?

To understand this, let's take an example of a simple pendulum.
We know that time period of oscillation of a simple pendulum is given as,
T=2π√(L/g)
Where T is the time period of oscillation
L is the length of the pendulum
g is the acceleration due to gravity
From the equation, it is clear that the time period is proportional to the length of the pendulum.

From the figure, it can be seen that for the same angle or force, depending on the length, the amplitude or oscillation length changes. The amplitude is directly proportional to the length of the pendulum which in turn is proportional to the time period of oscillation.
i.e.; for a shorter amplitude or length, the time period will be less and for larger amplitude and length, the time period will be more.
The frequency is reciprocal of the time period, 1/T.
Hence, for small amplitudes, the time period will be less and frequency will be more.

Now, coming back to the bearings, initial failures are microscopic, the damage gap of which will be in nanometers or microns (just like a shorter length pendulum) which produces small amplitude vibrations. This, in turn, generates very high frequencies. As the failure progress, the damage becomes macroscopic, with the damage gap visible to the naked eye, producing high amplitude vibrations at lower frequencies.

In a more detailed way, as the damages on bearing are microscopic, the forcing component (ball, or races) will pass through the defect quickly, making the contact time small and frequencies high. As the damage grows, there will be more contact time and hence frequencies becomes lower. When the damage is severe, the contact time will be more and impacts are produced, leading to raised noise floor in vibration spectrum.

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