Thermal Expansion and Poisson ratio

[DanielHuber]

If I simulate the thermal expansion of a square sheet with Abaqus, I receive an expansion that is too large by the Poisson number.
E.g. a simple set up:
A 1x1 square sheet, with a thermal expansion coefficient of 1E-6 and a temperature increase of 1000 degree should expand by 1E-3. Instead I receive an expansion of 1.3E-3 if the Poisson ratio is set to 0.3.

Can anybody give some enlightenment? Or am I fooling me myself???
cheers, Daniel

 

[corus]

..should increase by 1e-3, instead I receive an expansion of 1e-3? Have I misread this somewhere?

 

[DanielHuber]

note that the expansion is: 1.3e-3, not 1e-3!!!!

I think I found the reason for the extensive expansion. In a 2dim Model, Abaqus seems to keep the third dimension constant! Due to the thermal expansion, the 3.dim would expand by 0.001xlength. With a length of 1 this would give 0.001. Keeping it constant results in compressing it by 0.001. With a Poisson ratio of 0.3 this results in an additional expansion along the first dimension by 0.0003, exactly what I observe.
Now, I am new to Abaqus and you may help me understand, where the constant third dimension is defined. The only thing I found is in the "Edit Section" dialog, where there is an entry "Plain strain/stress thickness". But I did not check this!
cheers, Daniel

 

[corus]

My fault. The problem with 2D thermal stresses is that you need generalised plane strain where free expansion out of plane is allowed. You have defined plane strain, with zero strain. Change the element type, define a reference point, and restrain that so that only the rotational freedoms of the plane are restrained. For that you need to manually edit the inp file as you can' do it in CAE.

 

[DanielHuber]

I finally succeeded,thank's a lot for your help.
One last question. I did the thermal expansion because I did not succeed in modeling an interference fit. It works now, but it looks like a rather clumsy way for doing it. Is there not a more elegant solution? And why is it so difficult to model an interference fit directly?
cheers, Daniel

 

[corus]

Two ways of modelling interference is to use thermal expansion as you have done, and secondly to impose an interference fit directly in the interaction module. There's an option in the contact set up to include this.

 

[DanielHuber]

I tried the direct way first, but had troubles with convergence. That is why I tried thermal expansion. Can I find an example of an interference fit somewhere?
cheers, Daniel

 

[Ashu28]

Hello Daniel,
I am doing interference fit on a model using ABAQUS CAE and have been successful with it. However, now I want to try using temperature approach that you mentioned.
Can you please tell me how did u go about it? I have changed the material properties to include the thermal coefficient but am not sure about the boundary conditions and loading.

Regards

 

[DanielHuber]

Hi Ashu28,
unfortunately I do no more have these calculations I. But I prepared an example, OneCyl.cae, for you. I try to upload it.
OneCyl is a simple cylinder that get heated from 0 to 100 degree and threfore expands.
Points to pay attention:
-Element type: Coupled Temperature Displacement
-Analysis type: Generalized plain strain
-Temperatures can be specified in boundary conditions.
- Convergence is difficult, therefore use damping e.g.: Automatic stabilization (see under "step")

I then made a second example, TwoCyl, with two concentric cylinder. The inner cylinder is then heated and expands, creating an interference fit. On a first view this seems to work, but the grids on both cylinders interpenetrate. Beeing pretty new to Abaqus, I did not yet find out what is wrong here. This itches, because last time this did not happen.

I would like to learn how to make an interference fit. Could you please send me an example.

cheers, Daniel

 

[emadomrani]

Hello everybody

I have problem to simulate cooling a a part
my part include 2 section, material of top section is Aluminum and Material of bottom section is Copper and these part are welded together
the initial temperature of this part is 433K,
This part quench during 1 second and final temperature is 298K

Please help me to simulate this problem

 

[DanielHuber]

Hi Ashu28,
I just realized that uploading more than one file does not work.
Therefore, I zipped everything into one file.
Daniel