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**dbpatankar****Member**- From: Roorkee, Uttarakhand, India
- Registered: 2010-05-22
- Posts: 172

Hello,

I am trying to simulate a simple friction behaviour using DIS_CHOC. The problem is same as given in SDND100 test case with only difference being that its solution is expected through DYNA_NON_LINE and not through DYNA_VIBRA/TRANS_HARM. However the effect of friction is not seen in the solution. I tried with various orientations given to POI element without any effect on the solution. Can someone have a look and see the issue. I am attaching the mail and comm files.

I need to use DYNA_NON_LINE because I am going to use it in a bigger sized problem involving friction.

Moreover, here are a few quick issues I need help on in case you are not able to look into the problem in detail :

1) In DEFI_MATERIAU/DIS_CONTACT : K_n is normal stiffness which is used only for getting the normal force F_n. In case there is no shock, JEU=0. Then how is F_n calculated? Is it just equal to K_n?

2) How local coordinate axes for POI element should be oriented in case of friction? My understanding is that the local X-axis is the direction of shock ( Direction of K_n) and local Y-axis (K_tan) is direction of friction. Is this correct?

3) If you manage to look into the comm files, do let me know if I am on right track or completely off the track with some hints to bring me on track.

Thanks.

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,569

hello

my first answer would be

do not use discrete point element for that

but discrete line element with a very short length such that orientation is easy to define

in addition

a POI1 element with a stiffness is not the same thing as a line elemnt with 2 nodes

NOT AT ALL!!

jean pierre aubry

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**dbpatankar****Member**- From: Roorkee, Uttarakhand, India
- Registered: 2010-05-22
- Posts: 172

jeanpierreaubry wrote:

hello

my first answer would be

do not use discrete point element for that

but discrete line element with a very short length such that orientation is easy to define

Had tried that as well, without much different result. Posting below my mesh and comm file which has friction element modelled as a SEG2 element with length 0.1.

1) Mesh .mail file

```
COOR_3D
N1 0.0 0.0 0.0
N2 0.0 0.0 0.1
FINSF
POI1
%E1 N1
E2 N2
FINSF
SEG2
E3 N1 N2
FINSF
GROUP_NO NOM=FIX
N1
FINSF
GROUP_NO NOM=LOAD
N2
FINSF
%GROUP_MA NOM=FIXE
%E1
%FINSF
GROUP_MA NOM=LOADE
E2
FINSF
GROUP_MA NOM=BAR
E3
FINSF
FIN
```

2) .comm file

```
mesh = LIRE_MAILLAGE(FORMAT='ASTER',
UNITE=20)
model = AFFE_MODELE(AFFE=_F(MODELISATION=('DIS_T', ),
PHENOMENE='MECANIQUE',
TOUT='OUI'),
MAILLAGE=mesh)
elemprop = AFFE_CARA_ELEM(DISCRET=(_F(CARA='M_T_D_N',
GROUP_MA=('LOADE', ),
VALE=1.0),
_F(CARA='A_T_D_N',
GROUP_MA=('LOADE', ),
VALE=(0.04, 0.0, 0.0)),
_F(CARA='K_T_D_N',
GROUP_MA=('LOADE', ),
VALE=(1.0, 0.0, 0.0)),
_F(CARA='M_T_D_L',
GROUP_MA=('BAR', ),
VALE=0.01),
_F(CARA='K_T_D_L',
GROUP_MA=('BAR', ),
VALE=(0.01, 0.0, 0.0))),
MODELE=model)
BARM = DEFI_MATERIAU(ELAS=_F(E=1.0,
NU=0.1,
RHO=0.01))
FRICM = DEFI_MATERIAU(DIS_CONTACT=_F(AMOR_NOR=0.0,
AMOR_TAN=0.0,
COULOMB=0.4,
DIST_1=0.05,
DIST_2=0.05,
JEU=0.0,
RIGI_NOR=1.0,
RIGI_TAN=10000.0))
fieldmat = AFFE_MATERIAU(AFFE=_F(GROUP_MA=('BAR', ),
MATER=(FRICM, )),
MODELE=model)
SINF = FORMULE(NOM_PARA=('INST', ),
VALE='1.0*sin(0.4*INST)')
listr = DEFI_LIST_REEL(DEBUT=0.0,
INTERVALLE=_F(JUSQU_A=40.0,
PAS=0.01))
times = DEFI_LIST_INST(DEFI_LIST=_F(LIST_INST=listr),
METHODE='AUTO')
BC = AFFE_CHAR_MECA(DDL_IMPO=(_F(DY=0.0,
DZ=0.0,
TOUT='OUI'),
_F(GROUP_NO=('FIX', ),
LIAISON='ENCASTRE')),
MODELE=model)
load0 = AFFE_CHAR_MECA(FORCE_NODALE=_F(FX=1.0,
GROUP_NO=('LOAD', )),
MODELE=model)
resnonl = DYNA_NON_LINE(CARA_ELEM=elemprop,
CHAM_MATER=fieldmat,
COMPORTEMENT=_F(DEFORMATION='PETIT',
GROUP_MA=('BAR', ),
RELATION='DIS_CHOC'),
CONVERGENCE=_F(RESI_GLOB_RELA=1e-06),
EXCIT=(_F(CHARGE=BC),
_F(CHARGE=load0,
FONC_MULT=SINF)),
INCREMENT=_F(LIST_INST=listr),
METHODE='NEWTON',
MODELE=model,
NEWTON=_F(MATRICE='ELASTIQUE',
PREDICTION='ELASTIQUE',
REAC_ITER=100),
SCHEMA_TEMPS=_F(FORMULATION='DEPLACEMENT',
SCHEMA='NEWMARK'))
DX = RECU_FONCTION(GROUP_NO=('LOAD', ),
NOM_CHAM='DEPL',
NOM_CMP='DX',
RESULTAT=resnonl)
IMPR_FONCTION(COURBE=_F(COULEUR=1,
FONCTION=DX,
MARQUEUR=0,
STYLE=1),
FORMAT='XMGRACE',
PILOTE='POSTSCRIPT',
UNITE=29)
```

Do you see any problem in these files?

in addition

a POI1 element with a stiffness is not the same thing as a line elemnt with 2 nodes

NOT AT ALL!!

Can you please elaborate? My impression is that a POI1 element with Kx = 1 (in K_T_D_N) is equivalent to SEG2 Discrete element with Kx = 1 (in K_T_D_L).

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**jeanpierreaubry****Guru**- From: nantes (france)
- Registered: 2009-03-12
- Posts: 3,569

dbpatankar wrote:

Can you please elaborate? My impression is that a POI1 element with Kx = 1 (in K_T_D_N) is equivalent to SEG2 Discrete element with Kx = 1 (in K_T_D_L).

a prime from the secnd edition of my book

Using 0 length line element

In the example in chapter ?? we used K_TR_D_L elements with a length of $10 mm$, the problem could have been solved using a length of 0 mm i.e. the two nodes of the elements would be sitting at exactly the same place.We do not recommend doing so for the following reasons :

-at mesh times the two nodes may be merged into a single one and the line element will be discarded raising a run time error, this will happen in Gmsh if we use Coherence at some place in the .geo file ;

-as the length of the line is 0 its orientation is somewhat difficult to know unless it is forced with the keyword ORIENTATION with CARA=VECT_Y;

-in true life the joint between two elements is having a physical length, better use it!In the same manner, using a single node, POI1 element with K_TR_D_N properties to model a joint is really calling for trouble!

The problem seems sane and should be solved without any warning or error but the results will not be the expected one.

With this we do not have an element with one or more relaxed DOF at one end but a fixed joint between two elements with a node carrying some stiffness at the same place, which is not the same thing!

if you have one element with only one node how can you see a different behavior at each end of the element which is what should happen in the case of friction

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