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Test rupt7 Description sheet

Test name
rupt7

Calculation type
MECHANICS ELASTIC

Finite element type
QUA8 DIME 2

Topic
STRESS INTENSITY FACTOR FOR AN OBLIQUE EDGE CRACK IN A PLATE SUBJECTED TO UNIFORM TENSILE STRESS

The structure is a finite plate enclosing an oblique edge crack . This structure is subject to uniform tensile stress. The CASTEM solution for the stress intensity factors K is given with the displacements method in 3 nodes at the crack tip (including the mid-node) from the local displacements wich are computed in the local basis () for the crack by the relationship :

c and d are the coefficients of the line :
a and b are the coefficients of the line :

u : is the displacement following the ox axis.
v : is the displacement following the oy axis .
r : is the distance between the point, where K is computed, and the local origin.
This CASTEM K value for the stress intensity factor is compared to the result obtained with the analytical method .

Reference CASTEM
Test NAFEMS : D.P.Rooke and D. J. Cartwright in Copendium of Intensity Factors HMSO (1976) : rupt7 Modélisation des structures élastiques dans CASTEM 2000.
Version
97' customer version
Model description

Test rupt7 Results

RESULTS

CASTEM FIGURES

*           Test Rupt7.dgibi: Jeux de données         *
*           ---------------------------------         *
*                                                     *
*******************************************************
*****   CALCUL DU FACTEUR D'INTENSITE DE          *****
*****   CONTRAINTES D'UNE PLAQUE AVEC FISSURE     *****
*****   RECTILIGNE INCLINEE DEBOUCHANTE CHARGEE   *****
*****   EN TRACTION UNIFORME                      *****
*****   LARGEUR DE LA PLAQUE : 100 mm ;           *****
*****   LONGUEUR   : 250 mm                       *****
*****   LONGUEUR DE FISSURE : 50 mm   ;           *****
*****   ANGLE / OX : 45°                          *****
*****   CONTRAINTE UNIFORME : 10 daN/mm2          *****
*******************************************************
opti dime 2 elem qua8 mode plan defo echo 0;
*--------------DEFINITION DU MAILLAGE-----------------;

p0=0. 0.;p11=100. 0.;p12=p11 plus (0. 0.);p2=250. 0.;
p3=0. 100.;
a=50. * (cos 45);
pf=p11 plus (a a);
p4=(p11 plus (a 0.)) plus (0. 100.);p5=250. 100.;
i=0; pd=prog 1. 2.5 5. 9. 13. 19. 25. ;
cim1=pf;cim2=pf;cim3=pf;cim41=pf;cim42=pf;
plfim1=pf;plfim2=pf;pim2=pf;
repeter bbbb 7;
   i= i + 1;v=extr pd i;
   pi1=pf moin (v 0.);pi2=pf plus (0. v);
   pi3=pf plus (v 0.);pi4=pf moin (0. v);
   ci4=cerc  2 pi4 pf pi1;
   plfi1=ci4 poin 2;plfi2=plfi1 plus (0. 0.);
   n=2;si ((i ega 6 ) ou (i ega 7));n=4;finsi;
   m = n / 2;
   ci1=cerc  n pi1 pf pi2;
   ci2=cerc  n pi2 pf pi3;
   ci3=cerc  n pi3 pf pi4;
   ci41=cerc  m pi4 pf plfi1;
   ci42=cerc  m plfi2 pf pi1;
   sui1=cout cim1 ci1;
   sui2=cout cim2 ci2;
   sui3=cout cim3 ci3;
   sui41=cout cim41 ci41;
   sui42=cout cim42 ci42;
   elim 0.001 (sui42 et sui1);
   elim 0.001 (sui1 et sui2 et sui3 et sui41);
   lfi1=(cont sui41) comp plfim1 plfi1;
   lfi2=(cont sui42) comp plfim2 plfi2;
   lhi=pim2 d 1 pi2;
   cim1=ci1;cim2=ci2;cim3=ci3;cim41=ci41;cim42=ci42;
   plfim1=plfi1;plfim2=plfi2;pim2=pi2;
   sut1=sui42 et sui1;
   sut2=sui2 et sui3 et sui41;
   si (i ega 1);
     sutt1=sut1;sutt2=sut2;lf1=lfi1;lf2=lfi2;lh=lhi;
   sinon;
     sutt1=sutt1 et sut1;sutt2=sutt2 et sut2;
     lf1=lf1 et lfi1;lf2=lf2 et lfi2;lh=lh et lhi;
   finsi;
fin bbbb;
lg=p0 d 5 p3;ld=p2 d 5 p5;
lhc= p4 d 2 pi2;
lig1=lg et (p3 d 8 p4) et lhc;
lig1=lig1 et (inve ci1) et ci42;
lig1=lig1 et (plfi2 d 2 p12 d 5 p0);
su1=surf lig1 plan;obj1=su1 et sutt1;elim 0.001 obj1;
lig2=ld et (p5 d 8 p4) et lhc;
lig2=lig2 et ci2 et ci3 et ci41;
lig2=lig2 et (plfi1 d 2 p11 d 8 p2);
su2=surf lig2 plan;obj2=su2 et sutt2;elim 0.001 obj2;
perobj1 = inve ( cont sutt1);
perobj2 = inve ( cont sutt2);
lifis1 = perobj1 comp pf plfi2;
lifis2 = perobj2 comp pf plfi1;

obj=obj1 et obj2;

*-----------DEFINITION DU MATERIAU-------------------;
objaf = modl obj MECANIQUE ELASTIQUE ISOTROPE;
mat=mate objaf youn 20000 nu 0.3;
*-----------MATRICE DE RIGIDITE----------------------;
rig=rigi mat objaf;
*-----------DEFINITION DU CHARGEMENT-----------------;
cdl=(bloq ux uy p0) et (bloq uy p2);
f1=pres mass objaf lg -10.;
f2=pres mass objaf ld -10.;
*-----------RESOLUTION ET CONTRAINTES----------------;
u=reso (rig et cdl) (f1 et f2);
sig=sigm mat objaf u;
*----CALCUL DU FACTEUR D'INTENSITE DE CONTRAINTES----;
*----APPEL DE LA PROCEDURE SIF  ---------------------;
SUPTAB = TABLE;
SUPTAB.'MODMIXTE' = VRAI;
SUPTAB.'LIFIS1' = lifis2;
SUPTAB.'LIFIS2' = lifis1;
SUPTAB.'FRTFISS' = pf;

SIF SUPTAB MAT U;
K1MOY = SUPTAB.K1;
K2MOY = SUPTAB.K2;

*-----DEPOUILLEMENT  ---------------------------------;


mess '***********************';
mess '* SOLUTION ANALYTIQUE *';
mess '***********************';
mess 'K1th = 150.37 ';
EC1 = (K1MOY - 150.37) / K1MOY * 100;
mess 'ECART entre K1MOY et K1TH : 'EC1'%';
mess 'K2th =  -71.4 ';
EC2 = (K2MOY + 71.4) / K2MOY * 100;
mess'ECART entre K2MOY et K2TH : 'EC2'%';

*-----------------------------------------------------;
*------------------MESSAGE D'ERREUR-------------------*
SI ((EC1 < 0.6) ET (EC2 < 0.25));
  MESS 'ERR   0';
  ERRE  0;
SINON;
  MESS 'ERR   5';
  ERRE  5;
FINSI;
*-----------------------------------------------------*
fin;



ven 3 nov 04:14:58 NFT 2000