AboutAaron Overbeek Expertise I am a Six Sigma Master Black Belt and Lean Champion. My expertise ranges from tacticle (local) implementation to strategic (global) implementation of Lean Six Sigma. Also knowledgable in Best Manufacturing Practices, Quality, Supply Chain, Engineering, and EH&S. I am not a consultant and do not mind follow up questions.
Experience Bachelors of Science in Manufacturing Engineering from the University of Michigan. Lean Champion and Six Sigma Master Black Belt. Member of the American Society of Safety Engineers, Society of Manufacturing Engineers, and ISO audetation.
Director of Operational Excellence and Strategic Sourcing for multi-billion dollar corporation.
I'm a product design engineering student who currently working on a thesis of designing tolerance of rolled form and welded door sash car assembly. I wanted to set a new tolerance on a few areas of the door frame which I found are the critical areas and always out of the tolerance range.
My problem here is on how to develop a new tolerance after I had known the standard deviation from the capability histogram. PC and PCI also had already calculated. I mean in terms of what equation to use to develop a new tolerance and does the capability histogram must be at the center of the nominal value in order to use the standard deviation in any of equation related? (because the data I collected is within the tolerance spec. but slightly shifted to the left or right from the nominal value)
Lastly, is it possible to do stack up tolerance on the door sash on a joint part (two parts) where the joint are welded and how? Sorry to ask but I'm totally new in this area (SPC method) and still learning. Your patience and guidance are much appreciated.
ANSWER: A primary concern is to determine how wide the tolerances may be without affecting other factors or the outcome of a process. This can be by the use of scientific principles, engineering knowledge, and professional experience. Experimental investigation is very useful to investigate the effects of tolerances: Design of experiments, formal engineering evaluations, etc.
A good set of engineering tolerances in a specification, by itself, does not imply that compliance with those tolerances will be achieved. Actual production of any product (or operation of any system) involves some inherent variation of input and output. Measurement error and statistical uncertainty are also present in all measurements. With a normal distribution, the tails of measured values may extend well beyond plus and minus three standard deviations from the process average. One, or both, tails might extend beyond the specified tolerance.
If the amount of variation is larger than the difference between the upper spec limit minus the lower spec limit, our product or service output will always produce defects, it will not be capable of meeting the customer or process output requirements.
As you have learned, variation exists in everything. There will always be variability in every process output, you can’t eliminate it completely, but you can minimize it and control it. You can tolerate variability if the variability is relatively small compared to the requirements and the process demonstrates long-term stability, in other words the variability is predictable, and the process performance is on target meaning the average value is near the middle value of the requirements.
The output from a process is either: capable or not capable, centered or not centered. The degree of capability and/or centering determines the number of defects generated. If the process is not capable, you must find a way to reduce the variation.
And if it is not centered, it is obvious that you must find a way to shift the performance. But what do you do if it is both incapable and not centered? It depends, but most of the time you must minimize and get control of the variation first, this is because high variation creates high uncertainty, you can’t be sure if your efforts to move the average are valid or not. Of course, if is just a simple adjustment to shift the average to where you want it, you would do that before addressing the variation.
You have two courses of action here. You can change the process to have your normal distribution centered closer to the mean, or you can increase the tolerances.
To shift the normal distribution curve: Multi-Vari Charts are intended to be used as a passive study, but later in the process they can be used as a graphical representation where factors were intentionally changed. The only caveat with using MINITABTM to graph the data is that the data must be balanced. Each source of variation must have the same number of data points across time.
Typically, we start with a data collection sheet that makes sense based on our knowledge of the process. Then follow the steps.
If we only see minor variation in the sample, it is time to go back and collect additional data. When your data collection represents at least 80% of the variation within the process then you should have enough information to evaluate the graph.
Remember for a Multi-Vari Analysis to work the output must be Continuous and the sources of variation Discrete.
Feel free to follow-up.
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QUESTION: Thanx for the long explanation you gave. It seems I understand it better than before. But sir, what if the distribution is already normal but not centered (but it is almost near to the mean value), can we still proceed with the analysis and use the standard deviation found to generate a new tolerance?
Again, after we had found the new tolerance is it possible to do tolerance stack up on the welded area? and what method or equation related to use? Sorry for taking your time to answer this question. Thank you.
Answer Stacking two components that are assembled together linearly, and the assembly length is 40 +/- 1 mm. We need to find the tolerance for each component so that Cp = 2.0, and Cpk>1.5
Stacking Tolerance Assumptions: Both components have the same symmetrical tolerance around their nominal value, and both nominal values sum. The manufacturing process of both components have the same variation and meet the capability requirement of Cpk not lower than 1.33. The assembly process adds no extra variation on the total length (i.e L=L1+L2)
Because Cp=2 and tolerance range=2mm, then S=1/6mm for the assembly. Also, because Cpk>1.5 the average must be at least 4.5 S = 0.75mm away from the closest limit, so the maximum shift of the average from the target value (40) is 0.25mm.
To assure that the assembly average will not be more than 0.25mm away from the target, the average of the components must not be more than 0.125mm away from the target.
S^2 = S1^2 + S2^2= 2 x S1^2 ==> S1 = S/sqrt(2) = 0.11785mm (this is the standard deviation of the manufacturing process of the components). The average of the manufacturing process of the components shall be at least 4 x S1 = 0.47140mm away from the specification limit to assure a Cpk>1.33.
If we take "Distance form the specification limit to the average" = 0.47140mm and "Distance form the average to the target" = 0.125mm; then "Distance from the specification limit to the target" = 0.125mm + 0.47140mm = 0.59640mm (let's say 0.6mm?)
So a tolerance of +/-0.6 for the components will assure a Cp=2 and Cpk>1.5 for the assembly, if the components are manufactured with a Cpk>1.33.
Note however that you said Cp=2, and not Cp>2, that means that S1=0.11785mm and not smaller. If S1 was improved (reduced), you could offset more and more the average of the components maintaining a Cpk>1.33, In the limit, with a very low variation in the components (S1) you could keep a Cpk>1.33 withh all the distribution very close to the specification limit (let's say +0.6mm) and the sum of components will be around 40 +1.2mm (i.e. out of tolerance)
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