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A very simple test, however, gave unexpected result

Posted 2 set 2018, 15:06 GMT-4 Structural & Acoustics, General Version 5.3a 8 Replies

Chi Zhang
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Hi,

I set up a simple test case to understand COMSOL and FE simulation, however, could not understand the result.

2D, linear elastic material in solid mechanics interface, stationary study step. a block of aluminium. Initial location of this block is (0,0)

two study steps:
1st step, move the block for 1 meter with "prescribed displacement". location of the block is (1,0);
2nd step, set the result of the first step as initial "values of dependent variables", and remove all the boundary conditions.

to my surprise, the result of the 2nd step showed that the block moved back to (0,0) instead of staying at (1,0).

Could someone explain me the drving force of this movement? Is there any way to keep the block stay at (1,0) without any boundary condition except for "free"?

the .mph file has been attached.

best, Chi

Attachments:

8 Replies Last Post 11 set 2018, 16:50 GMT-4
Edgar J. Kaiser Certified Consultant
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Posted: 6 years ago 4 set 2018, 06:03 GMT-4

Chi,

the second step is solved for a new set of dependent variables (u2, v2). That is why the first step doesn't matter there.

Cheers Edgar

-------------------
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Chi, the second step is solved for a new set of dependent variables (u2, v2). That is why the first step doesn't matter there. Cheers Edgar
Chi Zhang
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Posted: 6 years ago 10 set 2018, 10:51 GMT-4
Updated: 6 years ago 10 set 2018, 12:03 GMT-4

Chi,

the second step is solved for a new set of dependent variables (u2, v2). That is why the first step doesn't matter there.

Cheers Edgar

Dear Edgar,

thanks for your reply. I feel that what you said is important. However, after trying to set the dependent variables of interface 2 as (u, v) instead of the default (u2, v2), the expected result (result of the 2nd step staying at (1,0)) was still not shown.

>Chi, > >the second step is solved for a new set of dependent variables (u2, v2). That is why the first step doesn't matter there. > >Cheers >Edgar Dear Edgar, thanks for your reply. I feel that what you said is important. However, after trying to set the dependent variables of interface 2 as (u, v) instead of the default (u2, v2), the expected result (result of the 2nd step staying at (1,0)) was still not shown.
Edgar J. Kaiser Certified Consultant
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Posted: 6 years ago 10 set 2018, 13:30 GMT-4

It is still a different set of variables, I guess. solid.u and solid2.u. Not sure what other effects you might see by choosing the identical specifiers, not advisable in gerneral to my experience.

Can you describe what the intention behind these tests is. There may be other ways to do it.

-------------------
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
It is still a different set of variables, I guess. solid.u and solid2.u. Not sure what other effects you might see by choosing the identical specifiers, not advisable in gerneral to my experience. Can you describe what the intention behind these tests is. There may be other ways to do it.
Chi Zhang
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Posted: 6 years ago 10 set 2018, 13:53 GMT-4

It is still a different set of variables, I guess. solid.u and solid2.u. Not sure what other effects you might see by choosing the identical specifiers, not advisable in gerneral to my experience.

Can you describe what the intention behind these tests is. There may be other ways to do it.

Hi Edgar,

thanks for your time!

According to COMSOL reference manual (chapter "Predefined and Built-In Variables") and my test, "the dependent variables (field variables) are unique within a model". In fact, "solid.u" does not exist.

In Abaqus, one could "deactivate" some boundary conditions, which means if one in the first step apply a displacement and in the next step "deactivate" it, the bulk will stay at the displaced location. I would like to reproduce this "deactivation" function in COMSOL.

As I said, in COMSOL, in step 2 where I remove all the boundary condition, the bulk material to my surprise come back. I want very much to know the driving force of the coming-back movement (when I do not apply any boudary condition), because I feel that this is important to understand the logic behind COMSOL and finite element simulation.

best, Chi

>It is still a different set of variables, I guess. solid.u and solid2.u. Not sure what other effects you might see by choosing the identical specifiers, not advisable in gerneral to my experience. > >Can you describe what the intention behind these tests is. There may be other ways to do it. Hi Edgar, thanks for your time! According to COMSOL reference manual (chapter "Predefined and Built-In Variables") and my test, "the dependent variables (field variables) are unique within a model". In fact, "solid.u" does not exist. In Abaqus, one could "deactivate" some boundary conditions, which means if one in the first step apply a displacement and in the next step "deactivate" it, the bulk will stay at the displaced location. I would like to reproduce this "deactivation" function in COMSOL. As I said, in COMSOL, in step 2 where I remove all the boundary condition, the bulk material to my surprise come back. I want very much to know the driving force of the coming-back movement (when I do not apply any boudary condition), because I feel that this is important to understand the logic behind COMSOL and finite element simulation. best, Chi
Jeff Hiller COMSOL Employee

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Posted: 6 years ago 10 set 2018, 13:57 GMT-4

Hi Chi,

In a stationary study, the "initial values" are used as the initial guess for the solver when solving nonlinear models (See Introduction to COMSOL Multiphysis manaual, page 71). They do not have the physical meaning you think they do.

The solution to your step two is a null displacement because there are no loads applied, an no other reason for your structure to deform.

Jeff

-------------------
Jeff Hiller
Hi Chi, In a stationary study, the "initial values" are used as the initial guess for the solver when solving nonlinear models (See Introduction to COMSOL Multiphysis manaual, page 71). They do not have the physical meaning you think they do. The solution to your step two is a null displacement because there are no loads applied, an no other reason for your structure to deform. Jeff
Chi Zhang
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Posted: 6 years ago 10 set 2018, 14:09 GMT-4

the result of the 2nd step showed that the block moved back to (0,0) instead of staying at (1,0).

Hi Jeff,

thanks!

Now I set all "initial values" as default ("0"), as you said they are initial guess which do not have physical meanings.

I apologize that I do not undersand the second paragraph of your post.

  1. what is the driving force of the block coming back in step 2 (when I do not apply any boudary condition)?
  2. how to keep the block stay at (1,0) instead of moving back to (0,0) after step 2?

best, Chi

>the result of the 2nd step showed that the block moved back to (0,0) instead of staying at (1,0). Hi Jeff, thanks! Now I set all "initial values" as default ("0"), as you said they are initial guess which do not have physical meanings. I apologize that I do not undersand the second paragraph of your post. 1. what is the driving force of the block coming back in step 2 (when I do not apply any boudary condition)? 2. how to keep the block stay at (1,0) instead of moving back to (0,0) after step 2? best, Chi
Jeff Hiller COMSOL Employee

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Posted: 6 years ago 10 set 2018, 14:27 GMT-4

In step 2, the block does not "come back" from anywhere. It's a stationary study, and again in a stationary study the initial value field only provides an initial guess for the solver.

It sounds as if you mean to do a transient step as your second step perhaps.

Jeff

-------------------
Jeff Hiller
In step 2, the block does not "come back" from anywhere. It's a stationary study, and again in a stationary study the initial value field only provides an initial guess for the solver. It sounds as if you mean to do a transient step as your second step perhaps. Jeff
Chi Zhang
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Posted: 6 years ago 11 set 2018, 16:50 GMT-4

In step 2, the block does not "come back" from anywhere. It's a stationary study, and again in a stationary study the initial value field only provides an initial guess for the solver.

It sounds as if you mean to do a transient step as your second step perhaps.

Jeff

Thanks.

now I understand that currently it is a stationary study, and what I actually need to simulate this rigid body movement is transient study

>In step 2, the block does not "come back" from anywhere. It's a stationary study, and again in a stationary study the initial value field only provides an initial guess for the solver. > >It sounds as if you mean to do a transient step as your second step perhaps. > >Jeff Thanks. now I understand that currently it is a stationary study, and what I actually need to simulate this rigid body movement is transient study

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