How do interference fits change as temperature increases?

[GrantR]

Hello,

We are going to be assembling a hardened steal bushing into a A380 aluminum component. The interference fit is size on size. The interference is created by pressing this bushing into the aluminum component.

Recently there has been some concern that at elevated temperatures, the interference would drop to the point that it is almost negligable (about 1 micron).

I know the thermal coefficients of expansion of the aluminum and steel, and I have the interference values at room temperature (the inside diameter of the aluminum component, and the outside diameter of the bushing).

Could someone tell me either the formula to calculate the change in interference as a function of temperature or maybe a website if any that would have this information.

Thanks for any response.

Sincerely.

GrantR

 

[ivymike]

um, isn't this a matter of calculating the change in diameter of each, then subtracting one from the other (assuming uniform temp)?

 

[Cockroach]

I would think so IvyMike. Simply compute the new diameters based on steel changes at that temperature and apply the same computational model used to do interference fits at ambient temperatures.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[GrantR]

Yes I am assuming that this is all that is to it. But all of the literature that I have refers to the linear thermal coefficient of expansion, on straight/planar items.

How would you calculate the change in Diameter of a hole and of the outside diameter of the bushing.

My first thought was that you could "un-roll" the diameter, in other words take the circumference of the hole or bushing and calculate how much that grows and then relate this back to the change in diameter.(circumference = pi()*dia) But I don't think this is that simple, because the internal diameter is not free to expand because of the surrounding material.

This may be basic mechanics but the solution evades me at the moment.

Any help is appreciated.

Thanks.

 

[metengr]

The equation for determining the increase or decrease of a solid or thru bore (OD or ID) shaft diameter is based on the LCT equation. In this case length (L) can be treated as shaft diameter;

delta diameter = thermal coefficient of linear expansion x original diameter x delta T

 

[dmalicky]

Some assumptions usually made in differential thermal expansion problems:
1. The temp of each part is ~uniform throughout
2. The temps of each part vary together with time; i.e., one does not heat up or cool down dramatically before the other. They need not reach the same steady state temp though.
3. (optional): The temp of 1 part is the same as the other at any point in time.

If assumption #1 is true, then the "linear" coef. of thermal expansion (CTE) and DeltaX = CTE * L * DeltaT apply to any dimension on the material: linear, diameter, diagonal, whatever. The classic explanation is this: Draw a circle on a sheet of (stress-relieved) metal. Now heat the part up so its temp is uniform thruout (internal stresses are thus still zero). Now magically cut out that circle with zero kerf. Both the OD of the circle and the ID of the hole will be the same: they fit before, so they must fit after. Both ODs and IDs expand the same amount with temp. Your "unrolling" idea will actually work as the Pi term will cancel out--provided assumption #1 is true. But if, e.g., the heat is coming from the ID of the hole and the outside surface of the aluminum is, say, water cooled, then as you say, the material surrounding the hole will indeed constrain its expansion. If Assumption #1 isn't true then you need data on the temp distributions, and maybe FEA to solve it. If it is sortof true then some people take an average temp to get in the ball park.

Assumption #2 can get you into trouble if not true, but can be modelled by looking at snapshots in time.

If all 3 of the assumptions are true, the equations simplify to
DeltaFit = (CTEofOuterPart - CTEofInnerPart) * NominalDiameter * DeltaT

David Malicky

 

[GregLocock]

Shigley goes through an example of shrink fitting which is directly analagous to this case, if a simple calculation ignoring the precompression is not sufficient. I think you'll find the simple calcualtion applies though.


When metals expand due to temperature all the molecules move further apart, it is analogous to magnifying the structure. As such all linear dimensions increase in the same proportion.


Cheers

Greg Locock

 

[desertfox]

Hi GrantR

I believe it is just a case of using the existing diameters and multipling them by the coefficient of expansion and subtracting them, merely use the diameters as lengths.
If you know the interference you can actually workout the temperature at which there is zero intereference between the parts and then you know your max temperature your joint can go to.
A word of caution though the coefficient of expansion can vary with temperatue, so if you use the values which most text books quote for materials then you will find that these expansion coefficients are for materials at room temperature. Also note the comments of the other posts regarding uniform heating, your joint could fail sooner if heating of joint is not uniform.

regards desertfox

 

[GrantR]

Thank you all for your responses!

 

[MintJulep]

But what diameter to use?

Press-fit a rubber cylinder into a steel plate and the rubber will compress, but the steel will not expand.

Press a steel cylinder into a rubber sheet and the rubber will expand, but the steel will not compress.

Two objects of similar E will share the deformation.

Seems that you need to determine the interface diameter first.

 

[desertfox]

Hi GrantR

On reading your post for the second time I found, I missed the line where you stated that the intereference was size on size. If this is the case then yes you are in trouble with temperature rise, you need to increase the intereference so that the steel shaft is larger than the alumium bored part. If you have zero intereference (ie size on size) then you generate no stresses in the components at there interface. Therefore with a size and size fit and uniform heating your joint will fail with a 1 degree temperature rise above that which the joint was made.
Have a look at THREAD301-45683.

regards desertfox

 

[IFRs]

If the steel part is larger in diameter than the hole in the alumimum part, you will scrape off the aluminum when the parts are pressed together. The aluminum part will be larger than you think and the calculations will show interference when there is not. I would heat up the aluminum part so it slides nicely over the steel part, then the aluminum ID will be what you bored it out to.

 

[sreid]

I agree with IFRs that shrink fitting will work better than pressing. It can be tricky doing interferance fits between aluminum and steel. You don't want to lose the interferance at elevated temperature but you can't have too much interferance or the aluminum will crack. And the stresses get worse at colder temperatures.

 

[unclesyd]

I don’t have the particulars but we used a hardened spring bushing to handle some of the same conditions as you post. You might want to take a look at the following.

http://www.connexusa.com

 

[XELR8]

The change in radius as a function of temperature will be:

dR = alpha*Ro*dTemp

dR = change in radius
alpha = thermal coefficeint
Ro = original diameter
dTemp = final - inital temp

The stresses associated with these changes will need to evaluated at various temperatures or your specific application.

 

[XELR8]

Correction from previous post:

Ro = original radius.

 

[diamondjim]

I am with Ivymik's analysis.
Yes, you will loose preload as
the temperature increases.
Conversely, you will increase the
preload or interference fit based
on the difference in diameters
of the two parts effected.