## Large Diameter Interference FIts: Calculating Removal Load

## Large Diameter Interference FIts: Calculating Removal Load

(OP)

Hello. I have 2 large diameter cylindrical shells that are "rabbetted" together. I am trying to determine the approximate load required to pull them apart without much success. The parameters are as follows (all length dimensions in inches) and result in a .005" (diametric) interference:

OD2 = 40.525

ID2 = 40.175

t2 = .350

OD1 = 40.180

ID1 = 39.98

t1 = .200

The axial engagement length of the fit is:

L = .125

The materials are the same for inner & outer shell:

E = 221 GPa

I have tried a couple of approaches that yield very different numbers. I first tried just considering the force required to deflect the inner shell by the interference value (radially .0025") and then using this "normal force" to determine the frictional load to overcome. I used a formula in Roarks to determine this force. This requires a significant amount of force compared to using a "shaft/hub" press fit calculator such as the one at http://www.tribology-abc.com/sub23.htm. The difference in answers is from about 1200 lbf to 20 lbf, respectively.

I think a couple of things are happening here, but could use some guidance:

1) I believe that the Roarks approach might be overly conservative since it does not consider the outer shell and that it will expand a little to accommodate the inner shell.

2) For a shaft/hub equation, I am guessing that they might not be too accurate for "shafts/hubs" as large as the ones I am looking at. But maybe someone could comment on this?

**Outer Shell**(female):OD2 = 40.525

ID2 = 40.175

t2 = .350

**Inner Shell**(male):OD1 = 40.180

ID1 = 39.98

t1 = .200

The axial engagement length of the fit is:

L = .125

The materials are the same for inner & outer shell:

*v*= 0.3E = 221 GPa

I have tried a couple of approaches that yield very different numbers. I first tried just considering the force required to deflect the inner shell by the interference value (radially .0025") and then using this "normal force" to determine the frictional load to overcome. I used a formula in Roarks to determine this force. This requires a significant amount of force compared to using a "shaft/hub" press fit calculator such as the one at http://www.tribology-abc.com/sub23.htm. The difference in answers is from about 1200 lbf to 20 lbf, respectively.

I think a couple of things are happening here, but could use some guidance:

1) I believe that the Roarks approach might be overly conservative since it does not consider the outer shell and that it will expand a little to accommodate the inner shell.

2) For a shaft/hub equation, I am guessing that they might not be too accurate for "shafts/hubs" as large as the ones I am looking at. But maybe someone could comment on this?

NX7.5.5.4 - Teamcenter 8

ANSYS Workbench 14.5

## RE: Large Diameter Interference FIts: Calculating Removal Load

The biggest contributors to uncertainty is the friction and whether any corrosion has increased the amount of interference between the parts and with it the pre-load.

## RE: Large Diameter Interference FIts: Calculating Removal Load

Actually assembling a joint as described requires a chamfer or special tooling. A 45 deg chamfer won't do. An appropriate chamfer would take up ~half the specified length of engagement.

Even with no chamfer and some kind of magical special tooling to assemble the joint, I'd tend to believe it would take only 20 pounds of force to knock it apart. Or it would separate under gentle handling.

I sure hope this is an academic exercise; I hate to see such, er, used food, in actual product designs. No offense intended.

Mike Halloran

Pembroke Pines, FL, USA

## RE: Large Diameter Interference FIts: Calculating Removal Load

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

## RE: Large Diameter Interference FIts: Calculating Removal Load

i'd look up the irish stress theoretican ... timoshenko, probably plates and shells.

Quando Omni Flunkus Moritati

## RE: Large Diameter Interference FIts: Calculating Removal Load

@GregLocock: No CAD here...that's all natural talent ;) I listed the t's for convenience ... guess I need to work on my definition of convenient.

@ Dave & RB: Thanks for that. There's no relative heating for the fit. I am a contractor for a major aviation company and it appears they do these kinds of fits all the time. Unfortunately we are discouraged from asking them questions, so I am left to fend for myself. I thought that maybe someone here had done this before and had a nice formula obviating the need to re-derive something. I am all for derivations, but this seemed like it may have been a common issue. Maybe not though.

Thanks for the replies :)

NX7.5.5.4 - Teamcenter 8

ANSYS Workbench 14.5

## RE: Large Diameter Interference FIts: Calculating Removal Load

i'm assuming the SRM/MM doesn't go into any detail ... possibly just "sent the ass'y back to us for maintenance". I suspect that they have a proprietary process for disassembly (without wrecking the parts) and that's why they're not talking to you ...

go carefully ...

40" diameter ... axle or engine shaft ?

Quando Omni Flunkus Moritati

## RE: Large Diameter Interference FIts: Calculating Removal Load

Thanks again.

NX7.5.5.4 - Teamcenter 8

ANSYS Workbench 14.5

## RE: Large Diameter Interference FIts: Calculating Removal Load

Quando Omni Flunkus Moritati

## RE: Large Diameter Interference FIts: Calculating Removal Load

.005 = XdegF*13ppm/degF*40;

X = .005/40/(13/1000000)degF

## RE: Large Diameter Interference FIts: Calculating Removal Load

One tap with a hammer should release it.

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?