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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

Hello,

I'm currently doing some tests as part of my master thesis.

My problem is that I have a cube on which I imposed constraints on three face in their normal direction in order to be able to impose a displacement on the top face and have result that I can compared with analytical result (by computing the Hooks matrix).

Since my cube is fully constrained and I only impose a displacement in the Z direction, I should have zero strain in the others. But this is not the case, I have strain in the X and Y direction in the order of 10^-3

Where does this error can from ?

Thank you for you time.

Ilan

*Last edited by IlanRossler (2022-03-07 14:58:39)*

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**jonas loenartz****Member**- Registered: 2021-10-01
- Posts: 51

Hello,

The elongation in z-direction leads to lateral contraction of the cube, and this should result in a strain-component in the x- and y-direction.

edit: strain-component

*Last edited by jonas loenartz (2022-02-25 15:25:03)*

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

Hello, thank you for your answer.

Normally it shouldn't happen if I have constrained three face of the cube in their normal direction. I did that to avoid lateral contraction. So i'm excepting no stress and strain in the x and y direction.

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**jonas loenartz****Member**- Registered: 2021-10-01
- Posts: 51

Based on what I can deduct from your command-file, the faces opposite can still move freely towards the constrained faces.

Jonas

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,829

hello

1- it would be interesting to have the mesh

2- what happens with an isotropic material ?

jean pierre aubry

consider reading my book

freely available here https://framabook.org/beginning-with-code_aster/

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

Dear jean pierre aubry,

I have the same error (strain in X and Y non zero) if I use an isentropic material.

I've attached my mesh.

Dear Jonas,

But my cube is fully constrained no ? By imposing zero displacement I've locked all DDL of my cube no ?

Sorry if some of my questions/answers seems obvious, I'm still learning.

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,829

fully constrained????

are you sure

what does it should mean in your idea,

*Last edited by jeanpierreaubry (2022-02-25 16:56:48)*

consider reading my book

freely available here https://framabook.org/beginning-with-code_aster/

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,829

the deformed shape i get!!

consider reading my book

freely available here https://framabook.org/beginning-with-code_aster/

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

Ok maybe I'll start over because I think I got lost along the way sorry.

What I want is to impose a displacement on the top face of this cube and apply some constraints that would allow me to calculate the stress analytically with the help of the Hooke matrix easily since I know the strain and check if the result given by code_aster match what I've computed myself. (photo of the matrix attached).

What I first did was to use LIASION = ENCASTRE on the bottom face and just impose a displacement on the top face. But I was advised that this was not the right way to do it as it made the deformation of the cube ''unrealistic''. I was then advised to constraint three of the face of the cube like I explained in the first post

What should I do in order to achieve what I want ?

Thank you for your help.

*Last edited by IlanRossler (2022-02-25 17:26:03)*

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**jonas loenartz****Member**- Registered: 2021-10-01
- Posts: 51

Ok, what I think you are trying to achieve, is achievable by blocking DX for the 'frontface' and DY for the 'rightface' as well.

```
load = AFFE_CHAR_MECA(identifier='6:1',
DDL_IMPO=(_F(DZ=5.0,
GROUP_MA=('topface', )),
_F(DX=0.0,
GROUP_MA=('rearface', 'frontface')),
_F(DZ=0.0,
GROUP_MA=('bottomface', )),
_F(DY=0.0,
GROUP_MA=('leftface', 'rightface'))),
MODELE=model)
```

Jonas

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

jonas loenartz wrote:

Ok, what I think you are trying to achieve, is achievable by blocking DX for the 'frontface' and DY for the 'rightface' as well.

`load = AFFE_CHAR_MECA(identifier='6:1', DDL_IMPO=(_F(DZ=5.0, GROUP_MA=('topface', )), _F(DX=0.0, GROUP_MA=('rearface', 'frontface')), _F(DZ=0.0, GROUP_MA=('bottomface', )), _F(DY=0.0, GROUP_MA=('leftface', 'rightface'))), MODELE=model)`

Jonas

By doing that I've strain in the order of 10^-16 and 10^-18 in the X and Y directions. Why is this value not zero ? Because of the precision of code_aster ?

Also if I may ask, can you explain me why doing what you told me solved my problem ?

*Last edited by IlanRossler (2022-02-25 17:34:18)*

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,829

jonas is right

but the i doubt the coefficients you give for the elastic properties of the material do not match

i do not think you can have three poisson ratio with the same value on a material that has 3 different E modulus, am i right?

consider reading my book

freely available here https://framabook.org/beginning-with-code_aster/

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

jeanpierreaubry wrote:

jonas is right

but the i doubt the coefficients you give for the elastic properties of the material do not match

i do not think you can have three poisson ratio with the same value on a material that has 3 different E modulus, am i right?

Yes I'm aware that my coefficient are not right because I have yet to determine the real ones from experimental results. It shouldn't matter though because I use the same coefficient in my analytical calculation no ?

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,829

considering the learning level where you are

i would first try to solve the problem with an isotropic material

1e-15 is zero for all practical cases with any numerical software

consider reading my book

freely available here https://framabook.org/beginning-with-code_aster/

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**jonas loenartz****Member**- Registered: 2021-10-01
- Posts: 51

By doing that I've strain in the order of 10^-16 and 10^-18 in the X and Y directions. Why is this value not zero ? Because of the precision of code_aster ?

Also if I may ask, can you explain me why doing what you told me solved my problem ?

Yes, it is 0, but due to the precision (not the precision of code_aster, but numerical precision of a computer in general) you have the very small residual value.

And for your second question, you blocked only the DOFs on the faces you specified in the direction you specified. So contraction was still possible, because the other nodes were able to move towards the constrained faces. By blocking the opposite faces as well, you locked the width of the cube.

I can really recommend to you working through Mr. Aubrys book. It helped me a lot, getting started with code_aster. But I had some experience with FEM before.

*Last edited by jonas loenartz (2022-02-25 17:49:58)*

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

jeanpierreaubry wrote:

considering the learning level where you are

i would first try to solve the problem with an isotropic material1e-15 is zero for all practical cases with any numerical software

Ok I will do that.

*Last edited by IlanRossler (2022-02-25 17:47:35)*

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

Thank you both jonas loenartz and jeanpierreaubry for taking the time to answer my questions even though they weren't always clear. It helps me a lot and I will start over with your advices in mind.

Sincerely,

Ilan Rossler

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

jeanpierreaubry wrote:

considering the learning level where you are

i would first try to solve the problem with an isotropic material1e-15 is zero for all practical cases with any numerical software

I did try with an isotropic material but I didn't manage to match the analytical solution with the one from the result of the simulation. I think this may come from a misunderstanding of the computed field.

This is the result I got from the simulation :

SIGM_ELNO = 8.27206e+7, 8.27206e+7, 1.47059e+8, -3.33648e-7, -8.66086e-7, 7.05369e-8

EPSI_ELNO = 1.03456e-16, -5.20126e-17, 0.025, -1.29646e-16, -3.36536e-16, 2.74086e-17

But this raises two questions :

- Do I compute the correct field ?

- If so, why doesn't it seems correct when I use Hooke's law saying that : E = stress/strain. (I used 3500e6 as E)

Ilan

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,829

i can see some strange things in your model

if we are in N, m unit system

E_L=3500000000 N/m2 is a normal value although quite low

then the cube sides are 200 m long, you need a rather big test bed to perform the traction test!!

if we are in N, mm unit system

the cube sides are 200 mm long

and E_L=3500000000 N/mm2 = 3.5E9 N/mm2 depicts a material unknown on earth (16666 times stiffer than steel!)

nevertheless it is a sound engineering practice not to use crazy values for the physical parameters!

this being said using

PLAani=DEFI_MATERIAU(ELAS=_F(E=3500000000.0,NU=0.36,),);

and your initial boundary conditions

i get the following results

EPSI_ELNO, EPZZ = 0.025; SIGM_ELNO, SIZZ = 8.75E7 in agreement with E value

EPSI_ELNO, EPXX = 0.009 in agreement with poisson ration NU

*Last edited by jeanpierreaubry (2022-02-26 11:32:19)*

consider reading my book

freely available here https://framabook.org/beginning-with-code_aster/

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

Dear jeanpierreaubry,

Thank you for your answer and your advices I've indeed obtained the same results.

I'm now trying to do as I initially planned and to use an anisotropic material with the following properties (I know they are unrealistic but I use those to compare to the isentropic case).

PLAani = DEFI_MATERIAU(

ELAS_ORTH=_F(

E_L=3500000000.0,

E_N=3000000000.0,

E_T=3500000000.0,

G_LN=1287000000.0,

G_LT=1287000000.0,

G_TN=1287000000.0,

NU_LN=0.36,

NU_LT=0.36,

NU_TN=0.36

)

)

Next I'm trying to see if I can matched code_aster results (with MATLAB) by resolving the equation with the Hooke matrix that I posted before. Sadly I don't get the same result from my computation and from code_aster.

For example I have from code_aster that

SIXX= -1.6842e-6 SIYY = -1.42045e-6 SIZZ = 7.5e+7

EPXX = -0.00771429 EPYY = -0.00771429 EPZZ = 0.025

But based on the equation I know that

EPXX = 1/E_L * SIXX - NU_LT/E_L * SIYY - NU_LN/E_N * SIZZ

And that is equal to -0.009 and I don't understand the source of the mismatch.

Do I use the wrong formula or am I missing something in the simulation ?

*Last edited by IlanRossler (2022-03-01 19:39:02)*

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,829

my textbook does not give the same hook matrix

consider reading my book

freely available here https://framabook.org/beginning-with-code_aster/

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**IlanRossler****Member**- Registered: 2022-01-28
- Posts: 22

Ok I've understand my mistake. In my calculation I've written that v_31 = v_13 but since E_3 != E_1 it is not true.

I think this close the subject.

I want again to thank you both for taking the time to explain everything to me, I'll definitely go through your book.

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