John Berner
Applications Research Inc.
Golden Valley, Minn.


Selecting a set of bearings to withstand anticipated loads is a common design task with powertransmission equipment. Another deals with finding a cumulative component reliability that exceeds the standard 90% at a specified number of cycles or hours.
A good way to begin a reliability study defines what constitutes failure. Then identify the most significant failure modes for the machine.
A next step analyzes test data for each failure mode to establish the reliability of each as a function of time, cycles, or some other quantifiable measurement. In other words, at time t:
R_{1}= f_{1}(t),
R_{2}= f_{2}(t)
and so on. Then total system reliability becomes:
R_{system}= R_{1} R_{2}R_{3}
and so on at time t.
As a general rule, evaluate test data in terms of its Weibull distribution because it is rare that any set of field or lab data will fall clearly into a particular standard distribution, such as exponential, binomial, or normal. In addition, a particular failure mode may not occur until something else has worn to a certain level. In other words, a failure mode may be, at least partially, a secondary failure. These usually show up as an unreasonably high Weibull slope number. It can generally be corrected by applying Weibull's third parameter, or offset, from the origin. This is a fairly straightforward calculation for most any reasonably capable Weibull software routine.
Bearings present a unique problem in that they are typically specified by the manufacturer in terms of a dynamic load rating. For ball bearings, this is the load at which about 10% of them fail when rotated to 1 million revolutions of the outer race. Roller bearings are specified using a similar system, but the 10% failure corresponds to 3,000 hr at 500 rpm, or 90 million revolutions. The machine's rpm can then be converted to hours.
For example, consider an individual ball bearing with a Basic Dynamic Capacity, C_{BD}, rating of 3,000 lb. It is to carry an effective load, P_{eff}, of 400 lb at 2,500 rpm. Since wear damage varies as the cube of the load, the number of revolutions corresponding to 10% (B10) failure will be
At 2,500 rpm, this corresponds to
Where L_{10h} = bearing life, hr , and = bearing speed, rpm
This is the first 10% failure point.
This is all well and good for finding one bearing's 10% failure hours (or revolutions). However, designers should be interested in the complete system which probably includes more than one bearing. The reliability of agricultural hardware or equipment is typically specified by the marketing group as something like "No more than 1% failure (99% reliability) through the first year and no more than 10% failure (90% reliability) through the fifth year." Many consumer products are subject to similar requirements, particularly in the first year.
To come up with a system that exceeds 99% reliability through the first year and 90% through the fifth, all criticalcomponent reliabilities must exceed those figures by a significant amount. Bearings are typically included in the category of critical components.
One reliability indicator comes from Weibull plots. For example, several research projects tend to show a range of variability in the value of Beta, the Weibull shape (slope) factor for bearings. It is also reasonable to assume that the load level on the bearing or perhaps even the ratio of the Basic Dynamic Capacity to the applied load can affect the Weibull shape factor. For instance, a light load might produce a shape factor close to 1.0. A heavy load might increase it to 4.0 or more.
So a reasonable starting value of Weibull Beta or shape factor for most agriculturalequipment (including lawn and garden) bearings is around 2.0. There are good reasons for the assumptions but they are involved and beyond the scope of this article.
To continue, recall that a straight line is defined either by two points, or one point and the line's slope. For example, Weibull cumulative density function and its linearized version are:
where F(t) = fraction of population failed, = characteristic life, and = slope or shape factor. Taking the natural log of both sides gives:
the standard slopeintercept form for a straight line for which is the slope.
Referring back to the earlier example, we know one of the points to be
We also know that = 2.0, so
and
Also, R(t) = 1F(t)
so
Suppose the most significant failure modes for a particular mechanism involve four bearings at critical locations. Using the method above, we've calculated the other three bearing characteristic lives for this application. We have:
We want the reliability at 1,000 hr for all bearings considered as a single system. Therefore:
Similarly, R_{2}(1,000) = 97.30%, R_{3}(1,000) = 98.23%, and R_{4}(1,000) = 99.04%
R_{system}(1,000) = R_{1}R_{2}R_{3}R_{4 }=93.41%
Another way to accomplish the above is with statistical analysis software with Weibull plot functions that include a Multiple Mode calculation routine.