Force required for press fit 1

[pmainland]

Hello,
I am looking for some help in getting a ball park figure for a press fit. We are looking at high force linear servo motor from Rockwell to do the job. My problem is I'm not sure how much force is required. I have tried a few formulas I found on the web but the resulting answer was out of the ball park. I left the drawings at work but from what I can remember the details are:

Pin: .395" dia (solid), 303/304 S.S. .25 long

Mating hole: 12L14 steel, 52RC.

Maximum interference is .0018"

Any help would be greatly appreciated!!

 

[electricpete]

There is a link to an on-line calculator here
faq404-1230

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[pmainland]

electricpete,
I've tried to use the interference fit calculator at that site but I must be doing something wrong or using the wrong calculator. I do not see a text box to imput the size of the hole the pin is being inserted into.

 

[GregLocock]

Well they've certainly stuffed that up. You have to log in to change the pin diameter, d.

That is a lot of interference you are using

Cheers

Greg Locock

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

 

[pmainland]

It appears you must own a copy of Advanced engineering design to get the answer key to access full calculator use. Thanks anyways. My engineering came up with the .0018 max interference, From what I remember the min was .001.

Thanks!

 

[electricpete]

Sorry, I didn't realize that link didn't work.

Here is my quick calc. Worth what you paid for it (0) in readability and accuracy. If you see any error, let me know. Note for the geomtery of cylinder within a cylinder I had to assume OD for the outer cyliner... I assumed 2" OD

> P := delta*E/ (2*rb*rc^2) * (rc^2-rb^2);
> nu:=0.33;
> delta:=0.0018/2*inch;
delta := .0009000000000 inch
> rb:=0.395/2*inch;
rb := .1975000000 inch
> rc:=1*inch; # ASSUME TWO INCH O.D!
> L:=0.25*inch;
> E:=30e6*lbf/inch^2;

> P;
65688.18040 lbf/inch^2

> mu:=0.15;
> F:=mu*2*pi*rb*L*P;
F := 3056.789042 lbf
> P := delta*E/ (2*rb*rc^2) * (rc^2-rb^2);
> nu:=0.33;
> delta:=0.0018/2*inch;
delta := .0009000000000 inch
> rb:=0.395/2*inch;
rb := .1975000000 inch
> rc:=1*inch; # ASSUME TWO INCH O.D!
> L:=0.25*inch;
> E:=30e6*lbf/inch^2;

> P;
65688.18040 lbf/inch^2

> mu:=0.15;
> F:=mu*2*pi*rb*L*P;
F := 3056.789042 lbf
> P := delta*E/ (2*rb*rc^2) * (rc^2-rb^2);
> nu:=0.33;
> delta:=0.0018/2*inch;
delta := .0009000000000 inch
> rb:=0.395/2*inch;
rb := .1975000000 inch
> rc:=1*inch; # ASSUME TWO INCH O.D!
> L:=0.25*inch;
> E:=30e6*lbf/inch^2;

> P;
65688.18040 lbf/inch^2

> mu:=0.15;
> F:=mu*2*pi*rb*L*P;
F := 3056.789042 lbf
> P := delta*E/ (2*rb*rc^2) * (rc^2-rb^2);
> nu:=0.33;
> delta:=0.0018/2*inch;
delta := .0009000000000 inch
> rb:=0.395/2*inch;
rb := .1975000000 inch
> rc:=1*inch; # ASSUME TWO INCH O.D!
> L:=0.25*inch;
> E:=30e6*lbf/inch^2;

> P;
65688.18040 lbf/inch^2

> mu:=0.15;
> F:=mu*2*pi*rb*L*P;
F := 3056.789042 lbf


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[electricpete]

Whoops. It looks like everything after
F := 3056.789042 lbf
is a repeat.

Also, sorry for all those decimal places... obviously not intended to convey the precision.

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[GregLocock]

I get 2 tons,

Hands up everyone who thinks 1.5 or 2 tons is a reasonable force to apply to a dowel 0.4" OD by 1/4" long

Not me.

You'll yield the SS pin.

Cheers

Greg Locock

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

 

[random88]

The Machinery's handbook has the formulas you need.
(p 657 in the 27th edition).

 

[israelkk]

electricpete

mu=0.15 is too optimistic. You have to take into account that it may even go as high as 0.5. Add to this s factor of safety to cover all the bases.

 

[dimjim]

You might be able to save some force because the
leading radius would reduce the .25 length and
probably push some of the material down as it is
penetrating the material as well as force the
high points of the surface finish into the valleys
of the surface finish. Would the hole also have
a chamfer on it to start the pin?

 

[GregLocock]

It's a bit of a crazy shape for a shear locator.

By the time you've squooshed (technical term) the soft ss pin into the hole any pretence at radial accuracy is long gone, I reckon. The swarf from the pin will hold the two surfaces apart as well.


Cheers

Greg Locock

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

 

[Tmoose]

I reckon A press fit that generates swarf, flotsam, or jetsam may not be a press fit any longer. A little while ago a normally excellent supplier let the "new guy" install (15) press fit thick walled 4 inch OD Drill bushings that skived the bore during installation. They were out of position, tipped, and the IDs were distorted up to 0.015 inch.

Thermal installation, or if your application allows it, a more modest fit with Loctite may be more friendly and even more accurate.

 

[RPstress]

I get 2 to 4 tons.

 

[pmainland]

I just wanted to say thank you to everyone who took the time to respond to my post. I guess I did not remember the drawings as good as I thought. No parts are hardened. The pin material is 12L14. The housing (hole) material is 1215. The pin diameter is .3940 +/- .0003 X .150" long and the hole dia is .3933 +/- .0002. It should be a max interference of .0012 and min of .0002. I used the formula from the Machinerys handbook and came up with 414 lbs force. I used 850 as the pressure factor. Again, Thanks for the help!

 

[electricpete]

I looked at Machinery's Handbook, but I'm not familiar with that approach. The table doesn't go below 1" diameter... did you extrpolate?

Using the high end interference of 0.0012", I only came up with < 200 pounds. Did I do it wrong? Here is my work:

The approximate ultimate pressure in tons can be determined by the use of the following formula in conjunction with the accompanying table of Pressure Factors for Forced Fits. Assuming that A = area of surface in contact in “fit”; a = total allowance in inches; P = ultimate pressure required, in tons; F = pressure factor based upon assumption that the diameter of the hub is twice the diameter of the bore, that the shaft is of machine steel, and that the hub is of cast iron
P=A*a*F/2

> A:=2*3.14159*(0.395/2)*inch*0.15*inch;
A := .1861392075 inch^2
> a:=0.0012*inch;
> F:=850*ton/inch^3;
> P:=A*a*F/2;
P := .09493099585 ton
> P*2000*lbf/ton;
P:= 189.8619917 lbf


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[electricpete]

Just to clarify, my calc directly above was based on Machinery's Handbook... did I misapply that? How did you apply it to come up with 400 pounds?

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[pmainland]

oops sorry about the typo, I came up with an answer very close to yours.

 

[electricpete]

OK, next question. Repeat my previous calc (13 May 07 23:21) trying to make it look like the Machinery Handbook Assumptions (Router=2*Rinner) still gives F~1400 lbf, which is much higher than Machinery Handbook result 200 lbf. Why?

Here is the calc I did to try to recreate Machinery Handbook assumptions: Use hub OD=2*hub ID. For cast iron, use conservatively low E=25e6. Also use conservatively low mu=0.15.

P := delta*E/ (2*rb*rc^2) * (rc^2-rb^2);
> delta:=0.0018/2*inch;
delta := .0009000000000 inch
> rb:=0.395/2*inch;
rb := .1975000000 inch
> rc:=0.395*inch; # twice the inner radius
rc := .395 inch
> L:=0.15*inch;
> E:=25E6*lbf/inch^2;
>
> P;

P=42721.51898*lbf/inch^2

> mu:=0.15;
> F:=mu*2*pi*rb*L*P;
F := 1192.823461 lbf

So why does Machinery Handbook give ~ 200lbf and the other calc gives 1400lbf when I tried to recreate the assumptions stated above for Machinery Handbook?

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[GregLocock]

I get 1625 lbf, for the same calculation, I think, assuming a steel pin and cast housing.

Cheers

Greg Locock

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