need press fit formula 3

[marmon]

Im looking for a formula which gives you the tonnage needed to remove a press fit, which factors diameter length of press fit, the press itself in thousandths of an inch, and maybe wall thickness of the mating peices.

our problem is removing a .005" fit 21 inches long. the shaft is 7" in dia and the wall thickness has been machined down to1.25" wall.

 

[metengr]

Why don't you use heat to break the interference fit? Heat the component that is joined to the shaft and keep the shaft as cool as possible during the application of heat.

 

[marmon]

well its a roll, w a journal pressin 21" so only the end is accessable. Also that would require huge amounts of heat to expand .005-,006 thou

 

[metengr]

Not necessarily. Using a linear coefficient of expansion of 6.7 X 10^-6 in/in deg F for the roll (assuming it is steel) you can calculate the temperature needed to expand the roll diameter to break the fit.

Otherwise, you need to hollow out the solid shaft even further to make it thin enough so that it will collapse.

 

[desertfox]

Hi marmon

try this formula:-


Here are the formula you need:-



Pa= f*3.142*d*L*Pc

where f=friction coefficient
Pc=contact pressure between the two members
d= nominal shaft dia
L=length of external member.
Pa= axial force required to interference fit
to calculate Pc for a given interference use the formula:-

Pc=x/[Dc*[((Dc^2+Di^2)/(Ei(Dc^2-Di^2))+....................
((Do^2+Dc^2)/(Eo*(Do^2-Dc^2))-((Ui/Ei)+
Ui/Eo))]



where x = total interference
Dc=dia of shaft
Di = dia of inner member(this is zero for solid shaft)
Do= outside dia of collar
Uo=poissons ratio for outer member
Ui=poissons ratio for inner member
Eo=modulus of elasticity for outer member
Ei=modulus of elasticity for inner member

This formula for Pc will simplify if the materials are the same.
Hope this helps

regards desertfox

 

[GregLocock]

faq404-1230



Cheers

Greg Locock

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

 

[GregLocock]

260 tons, assuming both parts are steel, mu is 0.3

Coo.

I'd be thinking about heat.



Cheers

Greg Locock

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

 

[marmon]

we ended up boring a 4" hole in the shaft then lancing @ 2 points full length almost completely through the shaft and it still took 150 ton to get it out. the last shaft must have been put in in the 50's

Thanks for the help

 

[marmon]

Greg:
using the calculator i get some unlikely results
for the Modulus of elasticity is stell not .029GPA?
also for the diametrical interferance between the parts "delta"/d
is this divded once to get "delta" then again in the equation?
"delta"/d?
last but not least most lineal units were in millimeters, but "d" shows its units in meters is this correct for the equation? im getting answers much lower than inticipated

 

[marmon]

never mind the .029, my book had an error

 

[desertfox]

Hi marmon

Using the formula I suplied in my last post I calculated you need at least 199.4 Tons to remove the shaft assuming
friction coefficient was 0.3 and also 0.3 for poissons ratio.

Regards

desertfox

 

[desertfox]

Hi marmon

sorry should read 195.44 Tons not 199.44 Tons


regards

desertfox

 

[marmon]

desert fox.

My memory recalls this steel as having a friction co-efficient of .8 dry!!!.

This is a handy formula for me as i work at a machine shop
just had trouble finding one THANK YOU
for total interferance fit did you use diametrical? of .006"? Im not at the office till tomorrow and have no good info were i am for the imperial young's modulus value for this steel. internets giving me conflicting numbers.
I assume the units you use is psi? also everything linear in inches?
Very interested to see the number i come up with, seeing as you 2 had varying results, with a ,8 friction co-efficient im guessing over 300 ton, i will also like to see what the rating was before we bored out the 4" hole.
The machinists at the shop use a rule of thumb that after .006 thou consider it one piece, i see why now
Thanks for the help

 

[desertfox]

Hi marmon

You can use imperial units throughout with the formula I have posted, were metric over here so I just converted to imperial units at the end. I have never used 0.8 as a coefficient of friction for steel on steel in fact the highest I have ever used is 0.4. However that dosen't mean 0.8 isen't a valid number. For the interference I used .005"
as given in your original post.
Finally I didn't get varying results I just wrote the answer down wrong the first time.
If you want to calculate the force at a friction coeficient of 0.8 then just multiply the 195.44 Tons * 0.8/0.3=521 Tons.

regards

desertfox

 

[notTiger]

Hi

Need some help in designing a press - interference fit

We have a 6-4 Titanium collar (OD 22.2mm ID TO BE DETERMINED) which we currently connect by loctite to a:

14.1mm OD,
13.1mm ID
304 Stainless tube (not annealed)
6mm insertion of tube into the titanium

Despite our best efforts the loctite works 95% of the time only!

I wonder if someone could help us design the ID of aperture in the titanium part- for a press fit

The press force available would be that normally available in workshops only

We out source production and want to make some best guess prototypes, and as you know titanium is expensive+

Cheers Simon

 

[desertfox]

Hi notTiger

We need more information:- Modulus of Elasticity for both materials, service loads and whether the joint is subject to temperature.

regards desertfox

 

[notTiger]

Hi DesertFox _ thanks.....

http://www.matweb.com/search/SpecificMaterial.asp?bassnum=MTP644

- Gives:

Titanium 6-4
Tensile Strength, Ultimate 900 MPa 131000 psi
Tensile Strength, Yield 830 MPa 120000 psi
Modulus of Elasticity 114 GPa 16500 ksi Average of tension and compression

http://www.matweb.com/search/SpecificMaterial.asp?bassnum=MQ304A

- Gives:

SS 304
Tensile Strength, Ultimate 505 MPa 73200 psi
Tensile Strength, Yield 215 MPa 31200 psi at 0.2% offset
Modulus of Elasticity 193 - 200 GPa 28000 - 29000 ksi

The force it needs to survive is 2x what a person can acheive with a lever distance of 150mm acting on a 5mm cam tool - best way to visualise this is visit our website

Tool (07) acts in hole (02) and is turned 90* to release the mechanism and allow putter shaft adjustment - It is a cam-operated clutch mechanism I designed

Cheers

Simon
(not an engineer so stumbling in the dark here a bit!)

 

[notTiger]

Oops here is the web link

Sorry!

http://www.pukugolf.com/technology.overview.htm

 

[desertfox]

Hi notTiger

From my formula above I estimate that an interference of 0.1mm between mating parts would require a force of 6701N
or 1503lbf to move it.
Basically you would need this force to assemble it.
If you use the above formula I posted earlier you can get various for different fits.
Note last line of the formula should read Uo/Eo not Ui/Eo

regards desertfox

 

[notTiger]

Thanks DF

My little company has about a dozen patents in process - mostly in fasteners, and my extreme lack of engineering knowledge has been both good (look at thing diferrently!) and bad (no idea on the maths etc!)

Cheers

Simon