CPK for destructive test (Ball shear)
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 This topic has 7 replies, 7 voices, and was last updated 20 years, 2 months ago by Pipkin.

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September 16, 2002 at 5:17 pm #30348
Miguel CamargoParticipant@MiguelCamargo Include @MiguelCamargo in your post and this person will
be notified via email.Hi folks,
We have a CPK metric for a specific destructive test (ball shear). The CPK level is around 1.00, being mainly affected by the dispersion (stdev). However, looking at the data distribution we can see that there is a lot of variation due to some units with a extremely good result. As a way to eliminate this problem, we’d been thinking on changing this destructive test to a nondestructive test. However, we need to figure out a way for calculating the CPK under this new scenario.
Any idea???
0September 16, 2002 at 7:44 pm #78964Destructive test results often follow a skewedright distribution, having a long “tail” to the right, as you described for your situation.
There are several methods available for estimating the Cpk index for nonnormally distributed output distributions. You could use one of these instead of attempting to develop a new nondestructive test (whose results might also follow a nonnormal distribution).
In one such method, you plot the test results on normal probability paper and then fit a curve through the plotted points. From this curve, you estimate the .135 percentile (the value below which lies .135 percent of the cumulative distribution), the 50 percentile (the median), and the 99.865 percentile. With these values, the formula for Cpk is:
Cpk = Minimum[ (50 percentile – LSL) / (50 percentile – .135 percentile),
(USL – 50 percentile) / (99.865 percentile – 50 percentile)]
With destructive testing, I believe you would have only one specification, in your case, just a LSL. Thus, the above equation simplifies to:
Cpk = Cpl = (50 percentile – LSL) / (50 percentile – .135 percentile)
I took this procedure from the book Measuring Process Capability by DR Bothe. It is also mentioned by the ISO TC 69 on applications for statistical methods in document N13 by working group 6 of subcommitte 4.
I hope this helps.0September 16, 2002 at 10:00 pm #78968
Miguel.CamargoParticipant@Miguel.Camargo Include @Miguel.Camargo in your post and this person will
be notified via email.Hi David,
Definitely, your proposal sounds much better. Could you please tell me where can I get the ISO standard or sending me any related information (presentation, explanation, web page, etc) about this formulas?
My email address is [email protected]
Thanks a lot
0September 30, 2002 at 12:23 pm #79313
Sonja RousseauMember@SonjaRousseau Include @SonjaRousseau in your post and this person will
be notified via email.Look at u charts, I have used these in the past for go/nogo and they work very well for this type of test
0September 30, 2002 at 2:31 pm #79323Miguel,
You are measuring the output of a process what key factors are involved in the process ( welding amps, pressure. etc) do you monitor the input variables?
Do you monitor the thickness ofthe material?
Performing a Cpk is very straight forward. However, if you have a variability problem I’d obtain data on the key process input variables then run a regression analysis or multi vari study to see what influences the output variable.
Do you maintain process control charts to determine if this process is stable over time? If not do that first.
Is the process centered ?
The purpose for the capability requirement is that the customers wants assurcance that the process is capable of delivering conforming product X% of the time. What does your chartage tell you about the process ?:
0October 2, 2002 at 9:04 pm #79379
Miguel CamargoParticipant@MiguelCamargo Include @MiguelCamargo in your post and this person will
be notified via email.Hi folks,
As Dave already said, “Destructive test results often follow a skewedright distribution, having a long “tail” to the right”. Actually this is the current situation.
On tis moment, I am not focused on the root causes of that variation. Basically, I would like to validate the method we’d been using for calculating CPK prior to go to that level, most of the books have two options:
1) Transform data for achieving normality: This is not fine for me, because I am calculating CPK (LSL only) for a destructive test. Test data follow a skewedright distribution, having a long “tail” to the right, so that even normalizing the data the stdev is huge resulting on a poor CPK (around 1).
2) Utilize percentiles of the distribution instead of the mean and multiples of sigma: Using this method the CPK was much better. However, it seems like there is not any accepted standard/formula for calculating CPK using this method. As you can figure out, this is critical due to the results could be significantly different, as well as would be hard to convince to my customers if we don’t have a commonly accepted reference, therefore a ISO document about this would be excellent.
Option A:
Cpk = Cpl = (Median – LSL) / (Median – .005 percentile)
Option B:
Cpk = Cpl = (Median – LSL) / (Median – .135 percentile)0October 3, 2002 at 1:25 pm #79395
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.I don’t know an ISO document for this, but may be the common sense can convince your customer:
The formula Cpk = (average – LSL) / (3 sigma) is world wide accepted, when it is a normal distribution (provided that CPL is what applies). The concept behind the formula is to compare the length of the distance between the half of the distribution to the specification limit with the width of the half of the distribution that goes to thiat side. Because the full width is taken as 6 sigma. If you want to use the Cpk value for an estimation of how many parts will fall beyond the specification limit, the formula will not work for non normal distributions. However, if you use percentiles instead of sigmas, for any distribution shape same Cpk values will lead to same PPMs.
Translating the previous formula, valid only for normal distributions, to percentiles valid for any distribution:
Cpk = (50% percentile – LSL) / (50% percentile – 0.135% percentile).
Applied to a normal distribution, both formulas will give the same result. And applied on a non – normal distribution, there will be as many parts below LSL as in a normal distributuion with the same Cpk (CPL).
Hope this helps.0October 4, 2002 at 5:21 am #79446HI,
BESIDES NORMALIZING & CLEMENT’S METHOD. DO YOU KNOW ANY ADDITIONAL METHOD FOR HANDLING NONNORMAL DISTRIBUTIONS?
BR,
0 
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