Squeeze Pin Connection

[SergD]

I am trying to figure out how to analyze a squeeze pin connection. I have a .935" Dia. pin that goes in a .938" Dia. hole. The pin then gets compressed by a press with a 88,350 lbs of force to create a interference fit. Both the shaft and pins are made of SS316. I want to know how much of an interference fit this process creates. Any suggestions are greatly appreciated.

 

[btrueblood]

Squeeze pin? Sounds more like a cold-set rivet?

 

[EdStainless]

You know that in 316SS you can't press this pin in, right?
You should be able to do it with about a 500F temp differential.
And are you really asking how much of the interference will be on each part?
Since both material have the same strength and modulus this is fairly straight forward.
They will each deform the same amount.
The machinery handbook has a standard method for calculating this.
How did you calculate the stress? You had to know the sizes in order to do that....


= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[GregLocock]

"I have a .935" Dia. pin that goes in a .938" Dia. hole." Why does a pin with a 3 thou clearance need to be pressed in? Is it very long, or bent?

Cheers

Greg Locock


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[SergD]

Thanks for the responses. Here are some clarifications:

Not sure if cold set rivet is the correct term either. I see the similarities but the pin has no heads that get formed with the pressing process. The whole body of the pin expands.

You know that in 316SS you can't press this pin in, right? Could someone expand on this? This is a process we use often.
And are you really asking how much of the interference will be on each part? Yes.
The machinery handbook has a standard method for calculating this. Could someone refer me to a specific section?
How did you calculate the stress? The force of 88,350 lbs is just based on the current process (it's based on the pressure setting the press).

The reason for pressing the pin is to hold the components together. The pin is placed thru a shaft, and disc, and then it gets pressed to expand it to create an interference fit.

The way I'm thinking I can analyze this problem is by breaking the process into two problems. First, I would consider the squeezing of the pin as a open die forging operation which would give me the diameter of the pin after compressing it. Then I would consider the interference fit of this new diameter pin and the shaft and disc. Seems a bit simplistic, but that's the only way I can think of solving this problem right now.

The whole reason why I'm looking into this is because I have a request to have the material of the pin be SS 17-4. So I need to figure out how the current process works, and see if the press we have has the ability of compressing the 17-4 pin to the same interference fit as our current design.

 

[GregLocock]

Ah sorry, I thought you were pressing the pin in, not setting it once in place.

"The way I'm thinking I can analyze this problem is by breaking the process into two problems. First, I would consider the squeezing of the pin as a open die forging operation which would give me the diameter of the pin after compressing it. Then I would consider the interference fit of this new diameter pin and the shaft and disc. Seems a bit simplistic, but that's the only way I can think of solving this problem right now."

I think that'll give the wrong answer. The second bit seems all wrong. Lame's equation doesn't help much as it is elastic, not plastic.



Cheers

Greg Locock


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[GregLocock]

Here'a relevant equation

Forging Force Requirement:

The forging force, F, required to forge material by impression – die forging operation can be determined by the relation

F = k . s f . A

where k is a constant (whose value can be taken from Table 2.1 s f is the flow stress of material at the forging temperature, and A is the projected area of the forging including the flash.

In hot forging of most non – ferrous metals and alloys, the forging pressure is generally in the range of 500 MPa to 1000 MPa.

http://nptel.ac.in/courses/112107144/Metal Forming & Powder metallurgy/lecture2/lecture2.htm

Cheers

Greg Locock


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[BrianE22]

A couple of simultaneous equations - the first uses the same pressure on the OD of the pin and the ID of the hole. The second has the radial displacement of the pin equal to the radial displacement (opposite sign) of the hole. This ignores the parts other than the pin and hole.

 

[Screwman1]

You aren't going to end up with any interference- the pin cannot expand beyond the boundary of the hole to create interference.

 

[Nescius]

You aren't going to end up with any interference- the pin cannot expand beyond the boundary of the hole to create interference.

The hole material is not infinitely stiff. Picture it like this:

Steel pin. Rubber block with hole. The fit is free. Now, upset the pin. Really smash it so it gets much fatter. The rubber hole will now be stretched around the pin. Interference fit achieved.

Or think of it like this:

Both the pin and the hole will be subject to forces that result in deformation. The forces are removed and there's some springback, blah, blah.. It is mathematically irrelevant that the pin is in the hole at the time. You could substitute another mechanism to mimic the force on the inside of the hole as well as substitute another mechanism to apply force to the outside of the pin as you upset it. If you could then superimpose the pin onto the hole, you'll find that there may be interference, depending on the properties of the pin and hole materials.

 

[Tmoose]

Have you thought about how much retention force you need ?