force for press fit ? 1

[mrkoko]

I'm trying to calculate the for required to press a 1/2 pin in to 1/2 hole... total interference will be 0.023" and the thickness of the receiving parts is 0.394. I was using the formula out of the machinery handbook P=(AxaxF)/2 where P is in tons, A is the area in contact between pin and holes surfaces, 'a' is the interference and F is some force factor. My problem is the handbook only gives F values for holes 1inch and over. Does anyone know what F factor I can use to give me the results I am looking for.

 

[eromlignod]

That's a gigantic amount of interference! The pin will probably shave out the inside of the hole. Are you sure you don't mean 0.0023"?

Don
Kansas City

 

[Cheddarcaveman]

0.023" interference on a 1/2" part is HUGE. A standard interference would be less than 0.002" so I suspect you've got a point in the wrong place!

 

[desertfox]

Hi mrkoko

I agree with the others that there is to much interference.
But have a look at this site anyway:-

http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm#Press_fits

regards

desertfox

 

[GregLocock]

F is the friction coefficient 0.15 to 0.3 typically.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

Fhas gotta be more than the friction co-efficient ... if the solution is in tons (real imperial 2240 lbs tons) and the inputs are linear dimensions (area, interference), then there's a bunch of other stuff in F other than just friction.

agree with the other posters about the interference ... if you really have to do this, consider how you're going to support the workpiece (to avoid damaging it)

 

[Twoballcane]

It sounds like you may have to heat the female and chill the male to overcome the interference.

Tobalcane
"If you avoid failure, you also avoid success."

 

[byrdj]

for steel, the delta T would need to be 2,000F

 

[vc66]

The chart in Machinery's Handbook is for a machined Steel Pin and an Iron Hub. Assuming this is your case, then you may want to put the values into an Excel spreadsheet, and extrapolate what a conservative number would be for your 1/2 inch case. I'm not sure about this, though, because below 1 inch the pressure factor may become distorted. I would check my calculations again as to the interference necessary--.023 is a huge interference, as has been stated. Otherwise you best best is a shrink fit.

V

Mechanical Engineer
"When I am working on a problem, I do not think of beauty, but when I've finished, if the solution is not beautiful, I know it is wrong."

- R. Buckminster Fuller

 

[sreid]

Even if you can shrink this together, the outer part will fail in tension before you reach room temperature.

 

[GregLocock]

Sorry I misread the formula


Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[mrkoko]

Thanks for all the input. I probably should have mentioned this before but it is a stud I'm looking to press in, in a production environment. The stud has a splined surface of sorts on the portion of the stud which is being pressed. This makes the contact surface area which is contacting the receiving hole much less, so I think the 0.023 press should be achievable. When I estimate the F valve for the 1/2 dia pin I get a required force to press stud of about 1600 lbs.

 

[eromlignod]

Ah, that's a different story. Now you're cold forming!

This may sound like a cop-out, but if I were you, I'd find a shop with a hand-crank hydraulic press (lots of machine shops have them)

http://www.machinetooldistributor.com/Images/Dake_hydraulic_press.jpg

and press one in by hand and read the pressure gauge. You can calculate until you're blue in the face and still not get as accurate an estimate as this.

Don
Kansas City

 

[hydtools]

The spline of sorts is probably a straight knurl. Like eromlignod suggests, measure the force to insert the pin.

Provide some mechanical stop so the pin is not inserted too far.

Ted

 

[Cockroach]

I think that "splined surface" you are referring to is a fluted area for acceptance of material galled from the ID of your hub during the pressing operation.

I was skeptical of the 0.023 inch interference fit until reading your last post. Typically these fluted areas are used on shafts for torque transmission. They add a little to the cost of the piece, but are effective as an anti-slip device in torque applications.

I've used Knurl before to buildup the shaft for a greater press fit, it does sort of the same thing but not as effective as the fluted barrel type like you have.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada