Normal force in Linear Elastic Solver

Numerical methods and mathematical models of Elmer
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Matthew
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Normal force in Linear Elastic Solver

Post by Matthew »

Hello everyone,

I just want to ask regarding the normal force for boundary condition in linear elastic solver. What is the unit of this force if assuming the SI units of all variables i.e. stress is in Pascal.

Is this force is in unit Newton (N) or in other units such as N/m (force per unit length) and N/m^2 (force per unit area). If so, do I need to convert my pressure value into force before applying this to my model?

Or is it in the unit of pressure (Pascal or N/m^2)?

Thank you very much,
Jamil
raback
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Re: Normal force in Linear Elastic Solver

Post by raback »

Hi

The unit is that of pressure i.e. Pa if you're working in SI units. The name may be somewhat misleading.

Generally all BCs and body forces are distributed ones, and you must separately request if you want them normalized to some integral value over the area/volume.

-Peter
Matthew
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Re: Normal force in Linear Elastic Solver

Post by Matthew »

Thanks Peter,

I tried to run a simple simulation by using a coupling of heat equation and linear elastic solver.

I have a "donut" shape geometry shown in the figure below.

I want to apply a normal force at the inner boundary, so that the displacement at the boundary can be free. This normal force must balance with the stress tensor, as given below (taken from Elmers Model Manual):

tau = Cijkl*Ekl - Bij*(T-To) = P (1)

where:
Cijkl = elastic modulus
Ekl = strain tensor
Bij = heat expansion tensor
T = temperature field
To = reference temperature
P = normal force

To determine P at displacement zero, P = -Bij*(T-To).

I have the following values of parameters:
Young modulus = 584 Pa
Poisson's ratio = 0.35
Bij = 1 (unit I supposed to be Pa/K ?)
To = 0
T = 1330 K

So I get the value P = -1330 Pa.

As I mentioned earlier, I also coupled the linear elastic solver with heat equation solver, to see the thermal expansion. The parameters value for heat equation solver are as follows:

Density = 1 (kg/m^3)
Heat capacity = 1/3244 (J/K)
Heat conductivity = 3.6e-12 (Watts/m.K)
Heat source = 0 (set as zero for now)

The outer boundary is set as fix at zero displacement and temperature fixed at 1330 K. The initial conditions are displacement = zero and temperature = 1330 K.

Expected results: The inner circle will not expand as the pressure applied in the inner boundary balances the stress in the material.

However, I got the following result (in the second figure) in which the inner circle expands greatly. Any mistake that I did? I suspect it came from the application of inner pressure (P) and also the definition of heat expansion coefficient Bij.

I also uploaded the .sif files and other related files (mesh etc) for you to see.

Please provide your suggestion and guidance. Thank you.

-Jamil-
Attachments
mesh.boundary
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case.sif
(3.94 KiB) Downloaded 286 times
figures.png
figures.png (184.8 KiB) Viewed 4342 times
Matthew
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Re: Normal force in Linear Elastic Solver

Post by Matthew »

One more question:

Is it necessary to write the Coordinate System String to Cartesian 2D in solving 2D simulation? How to change this in the ELMER GUI (or is it automatically recognised as 2D depending on the geometry used?)?

Jamil
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Re: Normal force in Linear Elastic Solver

Post by raback »

Hi

The dimension follows the mesh dimension automatically. 2D/3D is actually redundant.

-Peter
Matthew
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Re: Normal force in Linear Elastic Solver

Post by Matthew »

Hi Peter,

Can you help me with the problem I asked about the "donut" model? Any suggestion would be appreciated.

Best regards,
Jamil
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