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Force required for press fit

Force required for press fit

Force required for press fit

(OP)
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!!

RE: Force required for press fit

There is a link to an on-line calculator here
FAQ404-1230: Press Fits - quick calculator

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RE: Force required for press fit

(OP)
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.

RE: Force required for press fit

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: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Force required for press fit

(OP)
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!

RE: Force required for press fit

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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RE: Force required for press fit

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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RE: Force required for press fit

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: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Force required for press fit

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

RE: Force required for press fit

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.

RE: Force required for press fit

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?

RE: Force required for press fit

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: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Force required for press fit

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.

RE: Force required for press fit

I get 2 to 4 tons.

RE: Force required for press fit

(OP)
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!

RE: Force required for press fit

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:

Quote:

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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RE: Force required for press fit

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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RE: Force required for press fit

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

RE: Force required for press fit

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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RE: Force required for press fit

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

Cheers

Greg Locock

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

RE: Force required for press fit

From Unclesyd's reference:
The interference should be corrected for the effect of surface roughness. The flattening of roughness peaks by mounting under high pressure is generally estimated at 10% of the Rz-value where Rz=6Ra (ISO 6336-2). The correction at the diametrical interference thus becomes 0.1·2·(Rz1+Rz2). If the surface roughness of the shaft Ra1=0.5?m and that of the bore Ra2=1?m then d?=0.1·2·6·(0.5+1)=1.8?m and ?eff=?m-d? where ?eff=effective value of diametrical interference, ?m=measured value of diametrical interference and d?=correction for surface roughness.

Even so, it would not account for that amount of difference.

RE: Force required for press fit

electricpete your equation ignores the  Poissons ratio correction. If I put nu=0.0 into my calculation I get 1190 lbf

Cheers

Greg Locock

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

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