Tolerance management problem

  • Thread starter f00drunner
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  • #1
My problem is as following. I have a chain with "n" transitions between the chain link and the bolt. Now the whole chain rests without strain and you pull it, it elongates "x" millimeters.

I have to calculate the maximum accepted play of each transition between a bolt and a chain link in relation to "n" and "xmax" with a probability of 3 sigma.

So for example I have a chain of 15 transitions, the chain is allowed to elongate 0,4 mm, how much play is the maximum for each borehole/bolt.

I am honestly lost at this. I tried googling it for honestly two hours, but it seems like I have no idea what I am even looking for.
 

Answers and Replies

  • #2
Bystander
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"n" transitions between the chain link and the bolt.
Google "quadrature."
 
  • #3
Residual sum of squares being the one?
Sorry if I seem dumb or not proactive enough, but I am honestly at a loss, tried reading 10 page articles and much more, but I just don't seem to grasp the full problem.
 
  • #4
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So it seems your worst case scenario shouldn't exceed 0.4mm, so you would need to assume ALL the links are not perfect.
 
  • #5
Exactly, that is also what I already have written down here. Problem is that I also need the solution for a probability of 3ς (=99,7%), which is a problem I have trouble solving.

Edit: I just reread my posts and saw that I didnt really elaborate on the kicker of the problem.
I am all focused on tolerances. Let's say the bolt has a diameter of 10-0,4mm and the hole a diameter of 10+0,4mm, the play ranges from 0...0,8mm. Both dimensions are gauss-distributed around the middle of the tolerance (so the bolt around 10,2mm). Now I feel like the problem is a lot clearer.
 
Last edited:
  • #6
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I'm trying to learn from your example myself. Couldn't you define your 3 sigma point (or 3 standard deviations away from the perfect) as being your worst but acceptable tolerance. And your mean as the perfect scenario.
 
  • #7
That was what I was also trying to do. What really confuses me is that I have two tolerances per play that interact and have nothing to do with each other.
 
  • #8
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I guess each part would be manufactured differently, so each would have its own tolerance. But what you said makes alot of sense, the hole with the positive tolerance, the bolt with the negative tolerance.
 
  • #10
Bandit127
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How many degrees of freedom do you have in the bolt/link system? n
Your tolerance is 0.4 mm.
You tolerance per degree of freedom is 0.4/n. I will call this x for simplicity.

You have to fit 3 SD (99.7%) in to x. So 1 SD = x/3. That is your tolerance. You can apply this symmetrically for bolt and link.

Introduce a manufacturing system (production plus measurement) with a CP of at least 1.
 
  • #11
So what you are saying is, that If I want my chain to elongate less than 0,4mm over 15 links, I get an "x" (play per link) of 0,026mm.
So if i now want to achieve 3 sigma, i have to divide these 0,026mm by three?

From my understanding, shouldnt the 3sigma compared to a worst-case approach widen the tolerances?
 

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