plate1.inp

计算一般结构受载求位移

example: infinite plate with a central circular opening
COODINATE DATA=
NUMNP=27;
   X=A, 1.0000, 1;  A, 2.5000, 2;  A, 5.0000, F, 3, 3, 5;
     A, 0.9807, 4;  A, 2.4520, 5;  A, 0.9239, 7;  A, 2.3097, 8;
     A, 0.8315,10;  A, 2.0787,11;  A, 0.7071,13;  A, 1.7678,14;
     A, 0.5556,16;  A, 1.3889,17;  A, 3.3409,18;  A, 0.3827,19;
     A, 0.9567,20;  A, 2.0711,21;  A, 0.1951,22;  A, 0.4877,23;
     A, 0.9946,24;  A, 0.0000,25,26,27;       E;
   Y=A, 0.0000, 1, 2, 3;           A, 0.1951, 4;  A, 0.4877, 5;
     A, 0.9946, 6;  A, 0.3827, 7;  A, 0.9567, 8;  A, 2.0711, 9;
     A, 0.5556,10;  A, 1.3889,11;  A, 3.3409,12;  A, 0.7071,13;
     A, 1.7678,14;  A, 5.0000, F,15, 3, 5;        A, 0.8315,16;
     A, 2.0787,17;  A, 0.9239,19;  A, 2.3097,20;  A, 0.9808,22;
     A, 2.4520,23;  A, 1.0000,25;  A, 2.5000,26;  E;
BLOCK EXPRESSION DATA=
   A,  3, 6, 9, 7, 4, 1;   A,  9,12,15,13,10, 7;
   A, 15,18,21,19,16,13;   A, 21,24,27,25,22,19;  E;
POINTS DISTRIBUTION DATA=
   A,3,11,2,10,3,3,2; F,1,1,4;  E;
MATERIAL PROPERTIES=
   3, 1, 1.0E8, 1.0E8, 1.0, 1.0, 1.0;
   1, 1, 1.0E3, 0.300, 1.0, 1.0, 1.0;   E;
MATERIAL PROPERTY CITING NUMBER OF INTERFACES=
   1,    E;
MATERIAL PROPERTY CITING NUMBER OF BLOCKS=
   2,    E;
SEGMENT BOUNDARY CONDITION=
   A, 01, 4, 3;   A, 10, 1, 6;   E;
POINT BOUNDARY CONDITION=
   E;
LOADING DATA=
   LDMDL=1,  LDCS=1,
      1, A, 1.0, 0.0,  1,1,0,  1,2,0,  2,1,0,  2,2,0,  E;
      1.0,    E;
GRAPH DATA=
 1, 1.0E2;  E;

   A, 01, 4, 0.0, 1.0; 4, 0.0, 2.0; 4, 0.0, 3.0; 4, 0.0, 4.0; 4, 0.0, 5.0;
   A, 10, 1, 1.0, 0.0; 1, 2.0, 0.0; 1, 3.0, 0.0; 1, 4.0, 0.0; 1, 5.0, 0.0;

   A,  3, 6, 9, 7, 4, 1;   A,  9,12,15,13,10, 7;
   A, 15,18,21,19,16,13;   A, 21,24,27,25,22,19;  E;

   A,  3, 9,15,13, 7, 1;   A, 15,21,27,25,19,13;  E;