Optimizing Tolerances for Chain Link and Bolt Connections: A 3 Sigma Approach

[f00drunner]

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.

 

[Bystander]

"n" transitions between the chain link and the bolt.

Google "quadrature."

 

[f00drunner]

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.

 

[bsheikho]

So it seems your worst case scenario shouldn't exceed 0.4mm, so you would need to assume ALL the links are not perfect.

 

[f00drunner]

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.

 

[bsheikho]

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.

 

[f00drunner]

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.

 

[bsheikho]

I guess each part would be manufactured differently, so each would have its own tolerance. But what you said makes a lot of sense, the hole with the positive tolerance, the bolt with the negative tolerance.

 

[Nidum]

http://www.mitcalc.com/doc/tolanalysis1d/help/en/tolanalysis1d.htm

 

[Bandit127]

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%) into 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.

 

[f00drunner]

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, shouldn't the 3sigma compared to a worst-case approach widen the tolerances?