spring does not work

[ivo_grigoliZZ6J2]

Hello everyone

I am calculating an assembly with a spring in Inventor Nastran for the first time. In order to better understand the exact procedure, I have greatly simplified the structure for the first step. Two plates move towards each other with a forced movement, with a spring mounted between them (see attached images)

The calculation runs without an error message. However, when I read out the bearing forces, all values are 0, which is of course not correct.

I'm sure it's a stupid mistake on my part and I'm grateful for any tips.

Thank you very much

Inventor Nastran Version 2024

[John_Holtz]

Hi @ivo_grigoliZZ6J2 . Welcome to the Inventor Nastran forum.

Please attach your model so that someone can look at it and get the answers to question we have. See What files to provide when the model is needed - Autodesk Community.

For example,

1. How is the spring connected to the faces of the model? (Loads are transferred only when the nodes on the elements are connected together.)
2. What result is a "bearing force"?
3. What type of contact is defined between the parts?
4. Where does the load go? (I assume you see the load appear in the constraint reaction, so it is transmitted through the model somehow. That should indicate why you are not seeing what you expect to see in the spring.)

John

[ivo_grigoliZZ6J2]

Good morning @John_Holtz, thanks for the input

The 3D-File is attached.

In the meantime, I have found out why, but I don't quite understand it yet.

But first to explain the model:

One plate is mounted on the bolts with "sliding/not separated" and moves towards the second, fixed plate by "forced displacement". The bolts are rigidly connected to the fixed plate.

So far, everything works. Now I want to fit a spring between the two plates. This spring has a factor K3 (as it is mounted in the global z-direction) of 10.

To connect the spring, I have defined the two contact surfaces as rigid. In this way, the spring connects to the respective center point of the surfaces.

Now to the point where it becomes unclear:

Since the rigid surface should / may move, I have released all degrees of freedom of the rigid connection (see pictures).

The reaction forces on the fixed plate are then zero. However, if I fix the degrees of freedom of the rigid connection, there are also reaction forces.

My question is when and wich of the degrees of freedom of the rigid connection should be released.

I hope my request is clear so far.

Many thanks in advance for your input.

[John_Holtz · Accepted solution]

Interesting. You post is the third one in the last week where the answer is related to the degrees of freedom on the rigid body connectors! (Before that, the subject rarely comes up.)

Thanks for the description. I have not look at your model because I am using Inventor 2023 at this time and it will not read your 2024 model. But I can clarify your understanding based on your description.

• The checkboxes for the rigid body connectors (Tx, Ty, Tz, Rx, Ry, Rz) are not like the checkboxes for a constraint. The checkboxes for the rigid connector would be better labeled as Fx, Fy, Fz, Mx, My, Mz because they indicate what type of force or moment is transferred between the rigid connector and the model. (To programmers, translation in X and forces in X are closely related, and so on.)
• As a minimum, you want to transmit forces in the Z direction from the spring to the rigid connector to the solids. Therefore, Tz needs to be checked. You may also need Tx and Ty to prevent the rigid connectors and springs from moving freely in the X and Y direction. In other words, you need Fx and Fy transmitted between the solid and rigid connectors to prevent the spring/rigid body combination from moving freely in X and Y.
• Because you are using solid elements, they do not transmit moments. Whether Rx Ry Rz is checked or unchecked won't make a difference.

See this article for more details: Difference between rigid and interpolated type of rigid body connectors in Inventor Nastran.

John

[ivo_grigoliZZ6J2]

Many thanks for the detailed answer. The labeling is indeed a bit confusing, but now I understand it.

Thank you for your help

Ivo

[marchand_dupreez]

Hi,

I am also finding it impossible to be able to use such a basic function, I want to be able to model a simple bellow to counteract thermal expansion, so it take significatnt effort to get the thermal expansion model right and then not to be able to select a simple face (pressure surface) and a spring stiffness. Why is this such a schlep? Not trying to be diminishing, but competitor tools are so much simpler and easier to use with repetative success, everytime I try to use this spring connector It's like an intense negotiation just to get it to work. Please advise how can I apply a spring stiffness to a simple face of a solid body? Thanks!

[John_Holtz]

Hi @marchand_dupreez . Welcome to the Inventor Nastran forum.

You may want to create a new post because your model and issue is likely different than the original questions from ivo_grigoliZZ6J2.

It sounds like you are trying to attach a 1-point spring to a surface; that is, a point to an area. Inventor does not have the intelligence to know how to distribute the point reaction in the spring to an entire surface. Probably what you want to do is apply a connector to the face ("Connector > Rigid Body", either "Rigid" or "Interpolation") and use the center point of the rigid body connector as the attachment point of the spring. (I'm guessing you will do the same thing on the other end of the bellows.)

Another common mistake is not applying all 6 spring stiffness values, or applying them in the wrong direction.

1. The stiffness values 1, 2, 3 correspond to the directions defined by the coordinate system chosen on the spring dialog. (By default, 1-2-3 are aligned to the global axes X-Y-Z, but you can choose your own coordinate system to represent springs at different angles and so on.) The K1 stiffness is not necessarily the axial direction like in other software.
2. By leaving any stiffness values blank, you are telling the analysis the spring has 0 stiffness in that direction. That is a problem because it implies a force or moment of zero will  create an infinite displacement or rotation, and mathematics does not like to calculate 0/0.

Feel free to provide additional details or a model so that we can understand your issue more clearly.

John