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Effects of Thermal Expansion on Preload of on Composite Cylinders 1

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brady1

brady1

Mechanical
Mar 19, 2008
14
I have graphite cylinder fitted into an aluminum cylinder. The aluminum cylinder is first placed over the graphite cylinder with a 0.05-0.1mm gap, heated uniformly to 400 degrees and stretched longitudinally . Upon cooling the Aluminum cylinder shrinks plastically resulting in a 10 MPA contact pressure at the contacting interface. The dimensions are listed below:

Aluminum Cylinder/Jacket

OD: 60mm
ID: 57mm
CTE: 24.5E-9
E: 71 MPA

Graphite Core

OD: 57mm
ID: 11mm
CTE: 8.9E-6
E 14.5 MPA

How do I determine the effects on the pre-load for a 166 deg uniform temperature change. I reasoned that only some preload will be lost given an initial 10 MPA pre-stress.Note that aluminum expands about 3 times faster than graphite.

 
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I have a shrink fit program that uses as references to "Theory and Design of Pressure Vessels by John F. Harvey, Van Nostrand Reinhold Company, 1985. Additional formula and examples based on Mechanical Engineering Design by J.E. Shigley, second addition 1972. It appears to use use Harvey as the primary reference in respect to temperature effects.

It will calculate with an initial clearance and the loss interference pressure due to temperature or spin giving negative values. It uses a logarithmic temperature distribution in the cylinders.

The sad part is that I don't have it running now. It is a DOS program that runs on Symphony and the program is on 51/2" floppy and they want $75.00 to put it on a disk or 3 1/4" floppy.
 
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brady1

Okay the formula your using calculates the hoop stress in a cylinder for a given internal pressure, a thin walled cylinder is considered to have uniform hoop stress and so the formula you have used will give you the internal pressure to put the whole cylinder into its yield state

p= 2*(2/28.5)*100= 14Mpa using your figures not 7Mpa

further the wall thickness of the aluminium is 1.5mm not 2mm
60-57 =3.

Now forget the heating and shrinking bit and consider that you require to put the aluminium into it yield state by virtue of an interference fit so we need 14Mpa at the interface of aluminium and graphite, now to calculate the intereference using the formula I posted yesterday:-

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

you will find this formula on page 74

see page 72 and 73,74 of this reference:-

materials+by+william+j+nash&source=bl&ots=qIPFARgcAu&si
g=qKrZARq3KtmrCfOeFk0LreW4Yh4&hl=en&ei=m3W_S8mbFZ260gS
03P2ZCQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0C
DAQ6AEwCQ#v=onepage&q&f=false


According to me E for aluminium = 70Gpa not 70Mpa as stated in your original post and E for graphite 14.5Gpa not 14.5Mpa
now:-

14*28.5^2/(70*10^3*2) + 14*28.5^2/(14.5*10^3*23)=0.115315mm

that figure is the radial interference between the graphite and aluminium components needed to put the aluminium into its yield stress, any increase on this interference will cause the aluminium to exceed its yield stress, go further and you will fracture the aluminium.
Now for the temperature rise, any stresses caused by the temperature rise can simply be added algerbraically to the mechanical stresses induced by the interference.
However aluminium will expand at a much greater rate than the graphite and therefore with the interference as calculated the composite will fall apart.
The residual stress you refer to is the yield stress in tension which in this case doesn't help you.
When I first mentioned residual stress, I had a compressive residual stress in mind but clearly from your posts you don't have that.

desertfox
 
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What is the aluminum alloy? Any concern that 400C is in the annealing range?

desertfox, you are correct that the aluminum sleeve will reach yield strength as it shrinks tending to go more than your calculated interference based on reaching yield. It should not fail until the elongation reaches, what?, 10-30%, depending on alloy and condition.

Ted
 
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It is a Russian grade aluminum alloy AMg2M-AMg6M. The yield strength lies in the range 100-170MPa.
 
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hi hydtools

That was very clever of you to find that link.

This is interesting problem and I take your point about it shrinking further than the intereference calculated by my formula.
However if we take poissons ratio to be 0.3, although I know it changes depending on temperature and consider the increasing length as being 0.006, as given in your link then the change in diameter is only 6.84*10^-4 which again wouldn't give enough interference.
Okay if it works I must be wrong so what am I missing (grin)

desertfox
 
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Hu hytools

I really do not think that the composites will fall apart like hytool said. It makes sense that to relax the shrinkage pressure, one would have to raise the temperature to at least 400 C as used in creating the fit.

In designing this composite, the importance of a very low contact resistance was taken into account. In operation the device sees a maximum temperature rise of only 160 C compared to its 380 C rise during manufacturing. Any ideas on this? Does this convince you that the two part will stay in contact.

I am not discounting the fact that they can come apart though.
 
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Hi brady1

Well just looking at my figures particularly the last calculations, which if I doubled the radial interference to get the diameteral interference, which is 0.23mm and compare it with the expansion of the aluminium cylinder for a 160 deg rise which is 0.223mm I would say its on the borderline of seperating, whilst I haven't taken into account the extra interference due to Poisson's ratio, which very small compared to the previously calculated figure I suppose it might help in keeping the composite together. I did think at first, which was my mistake that it was going upto 400 degrees in service.
If your asking why the aluminium was at 400 degrees for manufacturing then thats fairly easy to answer ie:-
hot working will prevent work harding, lowers forces required to stretch tube, refines grain structure to name but a few.

desertfox
 
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Is anyone of you familiar with zone-normalized deformation ?
 
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Hi brady1

Do you mean strains beyond yield into the plastic zone?
If you do just looking at a text book, your aluminium cylinder won't have any residual stress because all the wall section as been put into yield.

desertfox
 
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brady1,
I think you are pulling that phrase from the above linked Russian report.
I think the authors have incorrectly called the annealing of the aluminum as it is stretched 'normalizing'. Normalizing is used only with ferrous metals. The authors may have more properly coined the phrase as zone-annealing deformation. They heated to an annealing temperture the aluminum tube in zones as the as it was stretched in order to soften the cold working due to stretching.

Ted
 
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Hi,

Thank you desert fox and hytools.

So when the author says that the aluminum shrink plastically. That does not mean that the aluminum was deformed, does it.

If the aluminum was annealed, does that means that its yield strength was decreased to the 100Mpa yield strength that they used to find the shrinkage pressure.

My material science knowledge is a little rusty.

brady
 
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What is meant by shrinking plastically is that the aluminum reaches its yield strength as it shrinks around the graphite and continues to remain in the plastic region as it further cools and shrinks. Once the aluminum enters its plastic region it shrinks without significant increase in stress.

The author initially deformed the aluminum tube by stretching it.

Yes, when the aluminum is annealed its strength is reduced. Strength increased by cold working, the stretching, or aging is reduced by annealling. We do not know what the original condition (strength) of the tube was before being stretched.

Ted
 
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This is not a very good job but it's the best I can do with what I have.

The syntax shown is for Symphony or Lotus 123. This is the Thermal Stress part of the program I mentioned in a previous. I hope you can glean some information from it.
All units are US standard. There maybe an error on page A-8 for the C2 value he uses the B value for the C cal lout and caries through with it.






 
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Hi unclesyd

Only downloads thermal2 and thermal5 have anything to read all the rest are blank including shrink1 and shrink2.

desertfox
 
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I think it's working now. I click on the icon.

On page A-2 the last line on the shrink fit part is missing.

Axial = (F1)= 3082 psi

 
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Hi unclesyd

I did get to see the other files you posted, however as far as I can see all these examples are interference fits within the elastic limits of the materials unless I am missing something, whereas the OP as completely yielded the outer cylinder in his case so what effect will that have on any future expansion due to temperature rise?

desertfox
 
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