Effects of Thermal Expansion on Preload of on Composite Cylinders 1

• Apr 7, 2010

• #2

GregLocock

Automotive

Apr 10, 2001

23,759

"plastically"?

Anyway look up shrink fit in Shigley. Lame's equation.

Cheers

Greg Locock


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• Apr 7, 2010

• Thread starter
• #3

brady1

Mechanical

Mar 19, 2008

14

Thank you for your prompt reply GregLocock

Yes, it experiences plastic deformation. There is actually a 0.05mm- 0.1mm gap between the aluminum and graphite before the Aluminum is heated to 400 degrees and stretched. Upon cooling, the aluminum cylinder`s inner diameter does regain it initial radius.

In shrink fits, the inner diameter of the outer shell is normally smaller than the outer diameter of the inner cylinder.

In the case above, the aluminum starts off with a larger diameter and is stretched longitudinal by applying heat such that it does not regain it original diameter upon cooling. Because of this, the hoop stress is basically equal to the yield strength of the material. The prestress/preload is found by the equation:

P = (t/Ri)*theta

P=preload/contact pressure/shrinkage pressure
t=wall thickness of aluminum
Ri= inner radius of aluminum
theta= yield strength

This information is taken straight from the original designers/manufacturers of the part under question.

I have checked Shigley, but still not sure how to determine the change in preload as the temperature rises.Also note that these are 2 thick walled cylinders.

 

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• Apr 8, 2010

• #4

GregLocock

Automotive

Apr 10, 2001

23,759

Lame's equation admittedly only applies if the stresses are under the elastic limit.

Sorry, I don't off hand know how to handle plastic deformation in this case - often energy methods work well.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

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• Apr 8, 2010

• #5

hydtools

Mechanical

Apr 4, 2007

3,265

That is an interesting method for creating a shrink fit. Stretching the outer tube reduces the tube inside diameter, I assume until it just touches the core OD. Then release the stretching force and allow the aluminum tube to cool and shrink. I would expect applying the thermal expansion equations would lead to the solution for reduced shrink fit stress. I would expect that heating the assembly to 400 degrees would return the aluminum to its diameter before cooling. The graphite core will have increased in diameter at the same time so there will still be some interference.

The method allows for a free fit assembly at room temperature. Interesting.

Ted

 

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• Apr 8, 2010

• Thread starter
• #6

brady1

Mechanical

Mar 19, 2008

14

Thank again for your replies,

Yes, it is very interesting indeed and was accomplished. This part came from overseas actually. This is how they created the shrink fit.

The basically stretched and heated the aluminum to a point where it would shrink back to a diameter smaller than that of the graphite if one could magically make the graphite disappear. So in essence, it actually would not go back to its initial diameter which exceeded that of the graphite by 0.1mm.

This is evidence that it experienced plastic deformation to the point where it wants to shrink back to a smaller diameter instead of its original larger diameter. It was important to ensure thermal contact between the two cylinder to reduce the thermal resistance.

The temperatures that it is expected to reach during operation is about half the 400 degrees it was heated to achieve the fit and hence contact pressure.

so in that instance, I do not think the two will lose contact even though the aluminum expands 3 time faster than the graphite, but there will be some lost of preload. That is what I want to determine.

Thank again

 

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• Apr 8, 2010

• #7

Wrenchbender

Mechanical

Dec 16, 2008

140

Brady, I think your second sentence is not really relevant. Who cares what the processing path, poisson stretch etc. was? You have this assembly in equilibrium, at room ambient, with a known interface pressure! The nominal geometry, as well as CTE's and elastic mods are also known. I would try -
a. Check published data for CTEs as a function of T. (Maybe you will luck out and it can be assumed constant over the temperature range. Otherwise find some empirical eqn.)
b. From the known interface pressure, back calculate the diametral interference that would produce this.
c. Raise T by 166 degC for a new diametral interference.
d. Calc a new interface pressure.

(Hope that was good advice. I haven't tried it, but I believe all the necessary material properties are known.)

 

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• Apr 8, 2010

• Thread starter
• #8

brady1

Mechanical

Mar 19, 2008

14

Thank you Bestwrench , I have actually calculated the diametrical interference. The CTE can be assumed constant. But I am still not sure how to accomplish C and d from your advice. How do i account for the interface pressure which is trying to avoid thermal expansion. Also, if i assume no preload and raise the temperture, there will be a gap between the two piece because the aluminum is on the outside.

Can you elaborate a little on c and d.

Thank you

 

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• Apr 8, 2010

• #9

rb1957

Aerospace

Apr 15, 2005

15,993

if you've calculated the diametrical interference, then you know the ID of the sleeve, ie the real ID if it wasn't stretched over the tube. wouldn't this be the starting poiint for heating the tube to 166deg ? you know C so you can calc the expansion of the sleeve. and so get the new un-stretched ID; if less than the tube OD then there'll still be some strectch, some preload, in the sleeve.

no ?

 

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• Apr 8, 2010

• #10

hydtools

Mechanical

Apr 4, 2007

3,265

Calculate the shrink of the aluminum as it cools from 400C of the contact diameter 57mm. I calculate the reduction in diameter to be 0,5244mm. That would be the approximate interference ignoring contraction of the graphite due to compression of its diameter and expansion of the aluminum due to the interference pressure.

Calculate the increase in diameter due to increasing temperature. As long as the increase is less than 0,5244mm the parts will remain in contact.

Ted

 

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• Apr 8, 2010

• Thread starter
• #11

brady1

Mechanical

Mar 19, 2008

14

Thank you guys. I think I know the step to arrive at the new interface pressure.

Given that I can prove that they will stay in contact. This means that the change in their radius will be equal. I can find the diametrical interference for a 166C temperature rise. Using those new diameters,I can equate the formulas for deformation in the aluminum and the graphite to find the new interface pressure. Is that right? I am sure it is.

The last question is:

Do i also apply a temperature rise to the graphite as well and use its expanded diameter OD along with the expanded diameter for the aluminum? Or do I just apply the temperature rise to the aluminum ?

 

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• Apr 8, 2010

• #12

rb1957

Aerospace

Apr 15, 2005

15,993

"This means that the change in their radius will be equal." i don't think this is quite right ... the graphite tube is going to be extremely stable with temperature (as shown by it's C). but you've calc'd the ID of the sleeve having been stretched and heated and thoroughly deformed, and this is smaller than the tube OD, hence the preload between the surfaces.

now if you heat the combined tube and sleeve, the sleeve ID will expand more than the graphite tube OD, but the sleeve expansion is based on it's unstretched ID; and so long as this is thermally expanded ID is less than the tube OD (expanded if you want, but there'll be Very little difference) then there'll be interference, and you'll be able to calculate the contact stress.

no ?

 

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• Apr 8, 2010

• Thread starter
• #13

brady1

Mechanical

Mar 19, 2008

14

I understand what you are saying rb1957. So given that I can find the total deformation. I am not sure what radii to use for the OD of the graphite and ID of the aluminum, i.e, the dimension after they have been assembled. do i used the 57mm as the common interface diameter.

How do i find the final contact radius after assembly ?
I know that the graphite OD will get smaller due to the contact pressure.

 

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• Apr 8, 2010

• #14

hydtools

Mechanical

Apr 4, 2007

3,265

Knowing contact pressure you can calculate the final assembled interface diameter.

Then calculate the thermal change in diameter for the graphite and for the aluminum. The hot outside diameter of the graphite will be larger than the hot inside diameter of the aluminum by the amount of remaining interference.

Ted

 

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• Apr 8, 2010

• #15

Compositepro

Chemical

Oct 22, 2003

7,822

How could you possibly "know" the contact pressure? That in itself must be calculated number.

 

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• Apr 8, 2010

• #16

rb1957

Aerospace

Apr 15, 2005

15,993

i guess i assumed (yeah, i know) that you had followed the ID of the sleeve through it's tortuous manufacturing process, in order to determine the diametrical interference. Possibly you're inferring the ID based on the interference stress. However you've done it, you know (as posted) the diametrical interfernce as manufactured, so you know the ID the sleeve wants to be at (post torture), and this is the diamter that now expanding when you heat the combination again, to 166deg. The Al sleeve will expand much more than the graphite tube, but the Al ID may (or may not) be in contact with the tube OD.

it does seem to be a very complex manufacturing process, it must've been fun developing it ! many pieces for the "rogues galley" i suspect.

 

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• Apr 8, 2010

1
• #17

desertfox

Mechanical

Dec 29, 2002

5,560

Hi brady1

Please check your data in your original post ie E = 70Gpa for aluminium and E=14.5Gpa for graphite, according to the sites I have looked at and also CTE expansion for aluminium is 23.6 * 10^-6.
I used this formula as you to obtain the interference fit between the two components and using your interface pressure:-

p*r^2/(E1*t) + p*r^2/(E2*t) = interference
radial

p= 10Mpa r= common interface radius ie 57/2 (your case)
E1=elastic M of aluminium E2= elastic M graphite

This gives an intereference of 0.0658mm which isn't enough intereference for the two parts to stay together for the temperature rise you quoted unless my figures are incorrect hence the request to check your data.
Finally the formula is for thin cylinders and not thick cylinders although your aluminium cylinder is thin, the graphite one is classed as thick but I believe its good enough for an estimate.
My final comment is a question, after cooling of the aluminium are there any residual stressses in it because they need to be taken into account, if they are present or at least I believe so.

desertfox

 

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• Apr 9, 2010

• Thread starter
• #18

brady1

Mechanical

Mar 19, 2008

14

Compositepro the initial contact pressure is was found based on the fact that the aluminum experienced plastic deformation during processing.During processing, it is known that the relative lengthening is three times larger that the permanent deformation at the yield strength. Therefore, the final tensile strength in the aluminum is practically equal to the yield strength. The pressure applied to the graphite is found using a known equation where:

P = (a/R)* hoop stress

This pressure is due to the aluminum attempting to shrink during cool, but cant because of the presence of the graphite.

remember that the hoop/circumferential stress is equal to the yield strength

desertfox there are residual stresses after cooling the aluminum like i have stated in this posted and those above. The aluminum want to shrink by an amount equal to 400C*24.5*10^-6*(57)= 0.5586mm.

As long as the operation temperature stays below 400C, the part should remain in contact like hytools suggested. I believe they will.

I have the device and know its final parameters and can calculate the interference.

Thank you

 

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• Apr 9, 2010

• #19

desertfox

Mechanical

Dec 29, 2002

5,560

brady1

If you heat up the aluminium it wants to expand not shrink as stated in your last post.
Also notice you gave GTE 24.5*10^-9 in your original post now in your last post its 24.5*10^-6 which I mentioned in my previous post.
Finally please upload your intereference calculation which you state is 10MPa, because my calulation shows the radial intereference to by only 0.0658mm radially based on your 10MPa.

desertfox

 

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• Apr 9, 2010

• Thread starter
• #20

brady1

Mechanical

Mar 19, 2008

14

The CTE is actually 24.5*10^-6 .My mistake. The yield strength for this aluminum is between 100-170Mpa.according to the manufacturer`s the aluminum underwent deformation such that its final stress is equal to its yield strength. I used 100Mpa to be conservative.

The manufacture went on to calculate the contact pressure using the formula:

theta(tangential)=(p*di)/2t

and p= (2t*theta(tangential))/di
p= 2*(2mm/28.5mm)*100Mpa=7Mpa

so use 7Mpa instead of 10Mpa

where:

theta(tangential) is that tangential stress and equal to the yield stress. In that case i used 100Mpa.

P= contact of radial pressure

di= internal radius of sleeve

t= wall thickness

this formula is stated in shigley as the average tangential stress and is valid for thick and thin walled cylinders.

You can use this information and that provided above to find and analogous diametrical interference.

Also in the manufacturing process, it was heated and stretched over the graphite. Upon cooling, it shrunk imparting a pressure(p) that can be found using the formula that i stated earlier.

 

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