Script to calculate a deformation with a rotating force vector.

Hello Kees,

What the Doc says is:

PAR_LOT =
Way the commands are treated:

'OUI': (default); superviser analyses all the commands before asking the execution.

'NON': after analysing one command the supervisor asks for it's execution and then analyses (and executes) the next command (treatment command by command).

I don't understand very well neither why if all the commands are analysed before execution DEFI_LIST_REEL returns an error...

I just installed CA 10.1 as I thought that perhaps the command was not supported by the version that I use.

Now I will continue with the tests...

Thanks again for your great help!

Hello,

My script is now working OK.
As the Force is not applied directly near the weak part of the structure, to limit the size of the mesh, I have tried to use: LIAISON_SOLIDE as described in:

www.caelinux.org/wiki/index.php/Contrib:.../10_1_x_cases/torque

I apply a force to a single node with:

[code:1]AFFE_CHAR_MECA(MODELE=MODE,
FORCE_NODALE=_F(GROUP_NO=('F_Point',),
FX=AMPLITUDE,),);[/code:1]

The node F_Point is included in the group of nodes 'Top_F':

[code:1]LIA=AFFE_CHAR_MECA(MODELE=MODE,
LIAISON_SOLIDE=_F(GROUP_NO='Top_F',),);[/code:1]

It seems that the AMPLITUDE force applied to 'F_Point' is then distributed on the other nodes of Top_F. The effective charge applied to 'F_Point' seems to be AMPLITUDE/Surface('Top_F').
Should it be like that?

Here it is the CA script.


[code:1]
DEBUT(PAR_LOT='NON'); # needed to avoid an error at DEFI_LIST_REEL

import Numeric

AMPLITUDE=1; # amplitude in N
PI=Numeric.pi; # Pi

# divided into tsteps
step = 45; ## step in degrees
tend = 360; ## end angle

print 'step: ',step
print 'tend: ',tend

time=DEFI_LIST_REEL(DEBUT=0.0,
INTERVALLE=_F(JUSQU_A=tend-step,PAS=step,),
INFO=2,TITRE='Angle',);

lenTime = len(time.Valeurs());

#Cos values for angles from 0 to 'tend'-'step' in steps of 'step'
fcs = DEFI_LIST_REEL(INFO=2,TITRE='cos ',VALE = [(Numeric.cos(ii*step*PI/180.0)) for ii in range(0,lenTime)], )

#Sin values for angles from 0 to 'tend'-'step' in steps of 'step'
fsn = DEFI_LIST_REEL(INFO=2,TITRE='sin ',VALE = [(Numeric.sin(ii*step*PI/180.0)) for ii in range(0,lenTime)], )

rampx = DEFI_FONCTION(NOM_PARA='INST',
VALE_PARA=time,
VALE_FONC=fcs,
PROL_DROITE = 'CONSTANT',
PROL_GAUCHE = 'CONSTANT',
INFO=2, # nice to check the values
TITRE='rampx',);

rampy = DEFI_FONCTION(NOM_PARA='INST',
VALE_PARA=time,
VALE_FONC=fsn,
PROL_DROITE = 'CONSTANT',
PROL_GAUCHE = 'CONSTANT',
INFO=2, # nice to check the values
TITRE='rampy',);

MA=DEFI_MATERIAU(ELAS=_F(E=100e3,
NU=0.300,),); # E in MPa

MESH=LIRE_MAILLAGE(FORMAT='MED',);

MAIL=CREA_MAILLAGE(MAILLAGE=MESH,
CREA_POI1=_F(NOM_GROUP_MA='F_Point',
GROUP_NO='F_Point',),);

MODE=AFFE_MODELE(MAILLAGE=MAIL,
AFFE=(_F(TOUT='OUI',
PHENOMENE='MECANIQUE',
MODELISATION='3D',),
_F(GROUP_MA='F_Point',
PHENOMENE='MECANIQUE',
MODELISATION='DIS_TR',),),);

MATE=AFFE_MATERIAU(MAILLAGE=MAIL,
AFFE=_F(TOUT='OUI',
MATER=MA,),);

CARA=AFFE_CARA_ELEM(MODELE=MODE,
DISCRET=_F(CARA='K_TR_D_N',
GROUP_MA='F_Point',
VALE=(1.0,1.0,1.0,1.0,1.0,1.0,),),);

# Boundary conditions: Fixed side
ENCAS=AFFE_CHAR_MECA(MODELE=MODE,
DDL_IMPO=_F(GROUP_MA='Base',
DX=0,
DY=0,
DZ=0,),);

# X Component of the rotating Force Vector
F_x=AFFE_CHAR_MECA(MODELE=MODE,
FORCE_NODALE=_F(GROUP_NO=('F_Point',),
FX=AMPLITUDE,),);

# Y Component of the rotating Force Vector
F_y=AFFE_CHAR_MECA(MODELE=MODE,
FORCE_NODALE=_F(GROUP_NO=('F_Point',),
FY=AMPLITUDE,),);

# Top_F containts F_Point node (point wher F is applied) and the top Face nodes
LIA=AFFE_CHAR_MECA(MODELE=MODE,
LIAISON_SOLIDE=_F(GROUP_NO='Top_F',),);

RESU=MECA_STATIQUE(MODELE=MODE,
CHAM_MATER=MATE,
LIST_INST=time,
INFO=2,
CARA_ELEM=CARA,
EXCIT=(_F(CHARGE=ENCAS),
_F(CHARGE=F_x,FONC_MULT=rampx,),
_F(CHARGE=F_y,FONC_MULT=rampy,),
_F(CHARGE=LIA),),
OPTION='SIEF_ELGA_DEPL',);

RESU=CALC_ELEM(reuse =RESU,
MODELE=MODE,
CHAM_MATER=MATE,
RESULTAT=RESU,
OPTION=('SIGM_ELNO_DEPL','EQUI_ELNO_SIGM',),
EXCIT=(_F(CHARGE=ENCAS),
_F(CHARGE=F_x,FONC_MULT=rampx,),
_F(CHARGE=F_y,FONC_MULT=rampy,),
_F(CHARGE=LIA),),);

RESU=CALC_NO(reuse =RESU,
RESULTAT=RESU,
OPTION=('SIGM_NOEU_DEPL','EQUI_NOEU_SIGM',),);

# define table TB_disp
TB_Z=POST_RELEVE_T(ACTION=(_F(OPERATION='EXTRACTION',
INTITULE='Displacements',
RESULTAT=RESU,
NOM_CHAM='DEPL',
TOUT_ORDRE='OUI',
GROUP_NO='POS',
NOM_CMP=('DZ',),),),
TITRE='DZ',);

# print displacements to file
IMPR_TABLE(TABLE=TB_Z,
FORMAT='TABLEAU',
UNITE=8,
SEPARATEUR=' * ',
TITRE='Displacements Z at POS',);

IMPR_RESU(FORMAT='MED',
UNITE=80,
RESU=_F(MAILLAGE=MAIL,
RESULTAT=RESU,
NOM_CHAM=('SIGM_NOEU_DEPL','EQUI_NOEU_SIGM','DEPL',),),);

FIN();
[/code:1]<br /><br />Post edited by: Patufet, at: 2010/03/20 18:32

Claus,

thank you for your reply. It seems to me than more than the number of nodes it is the surface itself. To obtain the expected displacement I had to apply the force:

AMPLITUDE/Surface(Top_F) instead of AMPLITUDE.

In my test I applied 1N and expected a displacement of 0.2 mm. I obtained instead a displacement of about 0.128 mm, the surface of Top_F being 0.8x0.8=0.64 mm^2.

The number of nodes of Top_F (excluding F_Point) was 225 in my test.

Regards.<br /><br />Post edited by: Patufet, at: 2010/03/20 22:09