Static simulation: obtaining displacements due to elastic and plastic deformations

Simulation Generality
1

Hello

After several searches I discovered this forum which may be able to help me:

In the static simulation on SW, I apply a force to an assembly. I have elastic AND plastic deformation.

I have created my displacement path and I get a "global" displacement. However, I would like to be able to break down this displacement into two parts: the one that is due to elastic deformation (and therefore will return to 0 once the force is cancelled), and the one that is due to plastic deformation. Ideally, I would also like to be able to display a displacement trace after the effort is canceled, and therefore visualize the displacements after the effort returns to 0, in order to do a before/after.

I thank in advance all those who will take the time to help me, I hope my question is clear enough!

Have a good day:)

2

Hello

Which simulation version do you have?

For Solidworks premium and Standard Simulation, there is only linear statics. beyond that, the results are wrong...

from memory, from Simulation pro. There is a nonlinear static module

3

Hello

I'm currently in standard simulation only. But tomorrow we receive a premium license precisely to have good results once the elastic limit of the material is exceeded. (I think you're confusing Simul Pro and Premium, Premium being the maximum)

But in any case I would like to be able to compare the results obtained between linear statics (standard simulation) and non-linear statics (premium simulation).

 

Thank you for your quick response!

4

Hello Mezlo4763

I don't think Icome is confusing because the names in SW are absurd.

For example, I have the premium version which includes CAD and static simulation.

To get the version with dynamic simulation I would have to switch to the PRO version.

Solidworks uses the word "premium" several times to refer to different realities, hence the impression of confusion.

@tous if one day you want to have a little fun, take the basic SW salesman (a new one is even funnier) and have you explain the differences between the premium which is not the premium of the premium, which he is talking about, when he talks about the premium (aspirin not provided).

Have a good day premium hurry of the day

2 Likes
5

Hello

 

ahah okay I'll take note ;)

But I just checked we will have the non-linear static option (we buy the license only for that....)

Thank you

6

Hello

I'm revisiting the subject, I still haven't managed to get only the plastic deformations after putting the force back to 0 during my simulation.

thank you in advance:)

7

Hello Mezlo4763

I have the impression that there is a big confusion of vocabulary (or else I have never understood anything and from the beginning to the simulation, which is possible).

Let me explain! ;-)

Simulation with Solidworks is only done in the elastic domain and never in the plastic domain.

In order for us to agree on the vocabulary, it is necessary to know that plastic means that there has been a definitive deformation of the material: like, when we have bent a sheet metal, when we have done a stamping.

The criterion of ruin cannot be determined since the yield strength Re is the ultimate limit not to be exceeded. (Just look at the variations in the results in the fracture tests on parts that are not specimens.)

However, the actual stresses experienced may be higher than the calculated stress , in particular due to stress concentrations (notches, holes, chamfers, fillets, etc.).

As mentioned earlier, yield strength (R) is often an ultimate limit that should not be exceeded for in-service parts. However, the actual stress on the parts may be higher than the stress calculated; Indeed, in operation the force can be higher than expected, and as far as the parts are concerned, variations in shape (notches, holes, fillets, etc.) lead to stress concentrations. In all cases, the nominal stress, if it corresponds to a static equilibrium condition, must remain below the yield strength.

To take into account these unforeseen phenomena, a practical limit value Rp is used, which is less than Re. To arrive at a correct value, a safety factor (S) must be used. Depending on the area of use of the part, this is why the Re is divided by S.

If the yield limit is exceeded with the safety factor, the system is considered to be defective and that this is in some way the criterion of failure.

Kind regards

 

2 Likes
8

Hello

It's also very possible that it's me who doesn't understand anything;)

In any case, I agree with you on the term plastic deformation.

However, in what is now called the premium  simulation pack (according to my salesman haha), you can do non-linear static simulation. I understand by this that we are after the elastic limit Re, right?

For the rest of your answer: my product that I want to simulate will only be used once, on a static test bench. And to validate my test, I have the right to have plastic deformation. Measuring at the point where the force is exerted, I must have a displacement after returning to zero that is less than 25% of the maximum displacement observed during the test.

Example: on my test bench, I apply my effort. I measure a displacement of 10mm, then I come back to zero force but I only come back to 2mm from the starting point.

So I did indeed have plastic deformation, but my test is validated because 2mm represents 20% of the total displacement measured.

So what I'm looking to see with simulation is this shift to the point of application when my effort is canceled.

There you go, maybe I'm sinking :p

Thank you:)

9

Good evening

No, you're not sinking because we're on this forum to help each other.

We're cool to keep you  from sinking in some way.

I see where the possible confusion is but I have to prepare an answer for you because it's not easy.

In the meantime, what I understand is in your test, you have a part of the whole that deforms beyond the elastic (goes into the plastic) while another part that has remained in the elastic domain returns to its initial shape.

Typically you embed a tube in a wall that under its own weight flexes slightly according to the length. But a bully not there and hangs from the tube which bends squarely.

If you watch the film in slow motion  (ON slowmotion) you will see that the tube curves until you exceed the maximum concentration point close to the wall.
The bully having left, you see that the tube came back straight shortly after the fatal 90° fold. This means that you may have localized and more or less significant plastic deformity. And at the same time another part that remains strictly in the elastic domain and returns to its initial shape.

It is from this point of view that we must understand why  we invoke non-linearity :-)

Always if I understand your problem ;-)  ;-)     When you write (( I have the right to have plastic deformation. Measuring at the point where the force is exerted, I must have a displacement after returning to zero that is less than 25% of the maximum displacement observed during the test.  This means that if I take the example of the tube: you would be entitled to a permanent deformation of 6% at the wall level and that the rest of the tube regains its shape at natural bending under its own natural weight. The only problem is that your part is partially or totally screwed up in the sense that the next time you try, you will have a new plastic deformation and that we hope will be controlled, otherwise you will go straight to total ruin.

I'm good there! hein say ;-)

Kind regards

 

 

10

Good evening

I will try to explain to you simply without giving a lecture on the subject.
In italics appear portions of text copied for the most part in the SW help, the quotations are taken from it unless otherwise indicated. I cut parts of sentences to shorten the text and make it coherent.

In a few words: Non-linear does not mean that you move into the plastic field.

Here is potentially the area of "possible confusion" between Linear, nonlinear and plastic deformation.
This means that you will never be able to reproduce a plastic deformation through simulation - even if it is non-linear - at least with the SolidWorks simulation modules, regardless of the standard or professional level.

To understand this, we have to start from a definition of non-linear.

The difference between the two possibilities is shown in this simple diagram

Quotation:  Linear static analysis, on the other hand, assumes that the relationship between loads and the induced response is linear. For example, if we double the intensity of the loads, the induced response is linear. Example: if you double the intensity of the loads, the response (displacements, deformations, stresses, reaction forces, etc.) will also be doubled.

Linear analysis is based on static and linearity assumptions and is, therefore, valid as long as these assumptions are. When one (or more) of these assumptions is no longer true, linear analysis produces erroneous predictions and nonlinear analyses should be used to model nonlinearities.

....... The rest and the full proposal can be read in the attached PDF document. ....

To conclude provisionally on the choice of this or that version, we can remember that the first level of version delivered in coherence with the standard version only allows linear simulation.
If you want to do nonlinear static simulations, you need the PRO version.

Kind regards


statique_lineaire_ou_non_lineaire.pdf