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!!
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
FAQ404-1230: Press Fits - quick calculator
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RE: Force required for press fit
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
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
Thanks!
RE: Force required for press fit
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
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
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
(p 657 in the 27th edition).
RE: Force required for press fit
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
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
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
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
RE: Force required for press fit
RE: Force required for press fit
Using the high end interference of 0.0012", I only came up with < 200 pounds. Did I do it wrong? Here is my work:
> 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
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RE: Force required for press fit
RE: Force required for press fit
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
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
http://www.tribology-abc.com/sub23.htm
RE: Force required for press fit
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
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
Cheers
Greg Locock
Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.