1
$\begingroup$

a description for virbation monitoring of industrial bearings states:

peak amplitues at 1 x rpm

should be an indicator for a fault in my case. Can somebody explain me what I have to understand under 1 x rpm. My chart (FFT) shows Amplitude in m/s^2 on the y-axis and Frquency in [Hz] on the x-axis.

Do I need to know how much rotations my bearings do in one minute? If somebody can give an example, also was 2 x rpm mean and how I calculate this value?

Thanks in advance!

$\endgroup$
1
  • $\begingroup$ Rpm means revolutions per minute. Then 2 x rpm means twice the initial rpm. $\endgroup$
    – Solar Mike
    Jun 11, 2018 at 21:23

1 Answer 1

4
$\begingroup$

RPM = revolutions per minute.

In the context of vibration monitoring, 1 x rpm means a vibration frequency that is the same as the rotation speed (one vibration oscillation per revolution of the rotor), 2 x rpm means a frequency 2 x the rpm (two oscillations per revolution), etc.

Because of nonlinear effects in the mechanical system, vibrations can be excited at different multiples of the rotation speed (1x, 2x, 3x, etc) and even at fractional multiples (for example 0.5x). However the "synchronous" vibration at 1x RPM usually has the biggest amplitude.

In fact some faults in mechanical bearings can excite vibrations at a frequency which doesn't have any "obvious" relation to the rotor speed - for example problems with the bearing lubrication system can excite so-called "oil whip" vibrations at about 0.9 x the rotor speed.

$\endgroup$
2
  • $\begingroup$ thanks for your explanation. So it is important to know how much my rotation speed is to know where 1 x rpm is in the FFT-Chart. Is there an explanation about the relationship between FFT-Chart and Vibration patterns? What is the reason or explanation that works? $\endgroup$ Jun 12, 2018 at 7:19
  • $\begingroup$ You haven't explained very well what your "chart" actually shows, so it's hard to guess how you should use the data. $\endgroup$
    – alephzero
    Jun 12, 2018 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.