I have estimated a B10 life for a bearing based on my operating conditions. What I'm trying to do is determine how many failures/hour I can expect based on that B10 life. Assuming that bearings tend to fail according to a known distribution, there ought to be a simple way to convert B10 to failure rate in failures/hour. I just can't find it.
B10 is defined as 100 million
revolutions. You have to decide
how many revolutions per hour,
how many hours per day, and
how many days per year to calculate
the number of revolutions per
year. Then divide the 100 million
revolutions by the total revolutions
per year to get the amount of years.
Maybe an example will help to clarify what I'm trying to do. Let's say I have 100 bearings which, for my operating conditions, have a B10 life of 10,000 hours. Now I put all of those bearings to test and after 10,000 hours, 90 bearings are still operating and 10 have failed. If I want to determine the failure rate of my bearings in failures/hour, I know that I have 10 failures, but I need to know the total number of operating hours for all 100 bearings. Now I know that 90 of the bearings have 10,000 hours, but 10 failed before that. How many hours do I attribute to the 10 bearings that failed before 10,000 hours. Assuming that bearings tend to fail according to a distribution (normal, exponential, etc.) I should be able to assign a number of hours to the failed bearings based on the failure distribution. This is where I get stuck. This must be done often for estimation of MTBF, so I assume there is a straightforward way to do it.
I found what I needed. I found a report from SKF indicating that rolling element bearing failures tend to follow a Weibull distribution with a slope of 1.11. Given this distribution, the MTTF is approximately 7 times the B10 life. The report is titled, "Reliability and Life: An Introduction to Reliability Terminology, Modeling, Prediction, and MTBF." It can be obtained at
You might want to check out a Weibull
distribution slope curve. It is 1.0 for 0
to 50, 1.5 for 50 to 70, 2 for 70 to 80,
3.5 for 80 to 90, and 8 for 90 to 100.
I think the mean would be nearer to
8 times the b10 life.
Counting the area under the slope at
each 10 percent interval yields
.5,1.5,2.5,3.5,4.5,6.5,7,9,12,17.
SKF should be a good source for
an approximation but it does look
a little low to me.