esys-escript

get velocity and set force at tagged boundary nodes

Asked by Hongyang Cheng

Hello,

After solving the ODE (rho*a = sigma), is it possible to get current nodal velocity (v = v+ a*dt) on the boundary? I also need to update boundary force at the corresponding nodes for solving the ODE in the next time step.

I looked into [1,2] and used PhysicalNames for tagging the boundaries. ReadGmash() seems to work for ver.4.0, as "domain.getTag(tag_name)" returns the correct tag id. However, a.isTagged returns False, I am guessing the Data on FunctionSpace does not inherit the tags from the domain. Does anyone know how to use tagging to export and import data?

[1] https://answers.launchpad.net/escript-finley/+question/243686
[2] https://answers.launchpad.net/escript-finley/+question/247567

Cheers

Hongyang

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Lutz Gross
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Launchpad Janitor (janitor) said : 		#1

This question was expired because it remained in the 'Open' state without activity for the last 15 days.

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Lutz Gross (l-gross) said : 		#2

Your statement "nodal velocity on the boundary" is not clear?
Do you mean how to interpolate the nodal velocity to a boundary element (that would be on the quadrature point)?
Do you want to have a py-list of the values at the nodes on a particular set of nodes located at portion of the domain boundary tagged with a specific name?

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Hongyang Cheng (alex-cheng) said : 		#3

Hi, Lutz

Thanks for your reply and happy new year. I meant the second. I wanted to have a py-list of velocities at the nodes located at the boundary of Solution FunctionSpace. Just for the record, I have used getReferenceIDFromDataPointNo() and getTagFromDataPointNo() to do it.

Since you mentioned boundary elements, I am curious about how traction on boundary elements is converted to nodal force. If I use four-point quadrature tri6 elements to define a rectangle domain and set isotropic confining pressure, the nodal force at a mid node will be double of that at a corner node. Could you please tell me why nodal force is not uniformly distributed?

Hongyang

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Lutz Gross (l-gross) said : 		#4

Look's like you solved the problem.

Second question: are you referring to a boundary condition in the form sigma_{ij}n_j = p_0*n_i with a constant pressure p_0?

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Hongyang Cheng (alex-cheng) said : 		#5

Yes, that's correct. In this case, the only coefficient assigned to the PDE should be "y" and nodal force was checked by getting the right hand side of the PDE.

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Best Lutz Gross (l-gross) said : 		#6

If you look at the lower left corner of a rectangular domain then y=[-p0,0] for elements on the vertical boundary and y=[0,-p0] on the element on the horizontal boundary. If you look at a node on the boundary it gets contributions from two neighboring elements which are in essence [-p0,0] on the vertical boundary and [0,-p0] on the horizontal boundary. This gives you node forces of about [-2p0,0] and [0,-2p0] except for the corner where you get [-p0,-p0] due to the different orientation of the surface force on the neighboring elements.
Hope this makes sense ...

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Hongyang Cheng (alex-cheng) said : 		#7

Thanks Lutz Gross, that solved my question.