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Relative displacements of 2 grids: state of the art?

Stéphane Combet

Hello,

I have a model in which I am interested in relative displacements between couples of grids.

What is the best method to do this today?

I mean I was used to create SPOINT and MPC for this, but is there a more modern (and easier) way to perform that?

Thank you!

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Rajashri_Saha

Hi Stephane,

You can calculate relative displacement while post processing in Hyperview.

See the below image. 1st plot the displacement contour. In measure group you need to select relative displacement as result type and then need to select the nodes.

Thanks

Rajashri

Stéphane Combet

Thank you for your answer Rajashri.

I don't understand how I get relative displacement there.

It just displays global displacement values.

I would like to set a node considered as a reference, and to have displacements relative to this node.

With the values displayed, I could eventually calculate by myself

relative disp = ((x1-x2)²+(y1-y2)²+(z1-z2)²)^(1/2)

but my question is "is there a way to have it directly?"

Maybe I didn't get correctly what you propose?

Adriano_Koga

in post-process, Relative displacement or incremental distance measures should do the work.

On the solver side, depending on the stiffness of your model, you could add a small spring element (CELAS) into your model, with unit stiffness (1.0N/mm). Then if you request the element force for this spring, you will get indirectly the relative displacement / elongation of your spring.

Of course, this assuming that 1N/mm stiffness will not change so much your model.

Rajashri_Saha

Thank you for your answer Rajashri.

I don't understand how I get relative displacement there.

It just displays global displacement values.

I would like to set a node considered as a reference, and to have displacements relative to this node.

With the values displayed, I could eventually calculate by myself

relative disp = ((x1-x2)²+(y1-y2)²+(z1-z2)²)^(1/2)

but my question is "is there a way to have it directly?"

Maybe I didn't get correctly what you propose?

In Relative displacement as result you need to do subtraction again as it is giving the nodal displacement value of individual node. You can use incremental distance instead. See the below image for this.

Thanks

Rajashri