Multiple subcases, no effect of first subcase to last subcase

Multiple subcases, no effect of first subcase to last subcase

Anonymous
Not applicable
2,046 Views
3 Replies
Message 1 of 4

Multiple subcases, no effect of first subcase to last subcase

Anonymous
Not applicable

Hello!

 

Looked a litte around but not clearly sure how the subcases are working together.

 

I have totally 2 subcases and use Nastran In-CAD 2019.

 

I give in the first subcase a force outside and in the second subcase a force inside.

If I run the subcases alone (deactivating each time the other one) it works without problems.


If I run the analysis with both subcases i get the same results like just in subcase 2.

Looks like the first subcase gives no information to the second subcase.

 

I read that in nonlinear static analysis the results from subcase 1 goes automatically to subcase 2. (Take a look at screenshot)

 

On animating the the last increment of subcase 2  i just see the movement from subcase 2 and no movement of subcase 1.

 

When the second subcase starts I see no deformation of subcase 1 in it.

 

Need I special settings  like result file of subcase 1 as input in subcase 2 ?

 

Thanks for help!

Subcase.png

0 Likes
Accepted solutions (2)
2,047 Views
3 Replies
Replies (3)
Message 2 of 4

John_Holtz
Autodesk Support
Autodesk Support
Accepted solution

Hi @Anonymous

 

I see a couple of possible problems. Let me start with your comment "On animating the last increment of subcase 2  i just see the movement from subcase 2 and no movement of subcase 1."

 

When you animate the results of one step, the animation starts at 0 load, linearly increments the result of the one step to the full value, and then back to zero. What you want to do is animate all of the step. See the article "Animation does not match the individual results of an In-CAD analysis" for more details.

 

The second issue is that the load in subcase 1 is removed in subcase 2, unless the load is assigned to both subcases. If the load in subcase 1 is removed, then it is possible that the final result (at the end of subcase 2) would match the results when you run the analysis with just the subcase 2 load. (If the load in subcase 1 does not create any nonlinear effects that change the behavior in subcase 2, then ramping load 1 up to the maximum in subcase 1 and ramping load 1 back to 0 in subcase 2 would give the same result as running the analysis with just load 2 in subcase 2.)

 

If you edit load 1 in subcase 1, you should see the options to assign the load to multiple subcases on the left side of the dialog. See this article for details on how the loads change from one subcase to another: "How do subcases function in a non-linear solution type in Nastran".

 

Let us know if you have any follow-up questions. It can be confusing at first. Smiley Happy.

 

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided, indicate the version of Inventor Nastran you are using.
If the issue is related to a model, attach the model! See What files to provide when the model is needed.
0 Likes
Message 3 of 4

Anonymous
Not applicable

Hello !

At first thanks for explaining! It helped a little to understand. But its not totally clear now in my case.

 

I tried your solution with copy the Load from Subcase 1 in Subcase 2.  The analysys worked but in Subcase 2 the upper side is fixed so the force will not work on this. (Look at the picture)

 

I want to give a force from the upper side when the lower side is fixed (Subcase 1). I want to keep the result and then giving a force

from the lower side while the upper side is fixed(Subcase 2).

 

Is there a solution two  simulate this case ? Otherwise one option is to use both forces in one (Sub)case but then I need smart boundary conditions.

 

Hope you get a better view now of my problem.


Greetings and thanks a lot for helping!

 

 

Subcase1-2.png

0 Likes
Message 4 of 4

Anonymous
Not applicable
Accepted solution

Hello!

After a while I tried a solution that works good.  My problem was the boundary conditions when i use the force in the first subcase and then in the second subcase.  But I fixed this problem. I used a movement as a force and not directly a force in newton. So I can use the bondary conditions in both subcases.

0 Likes