example: cantilevered beam idealized by one block
COODINATE DATA=
   NUMNP=4;
     A, X=0.0, 1,3;   A, 12.0, 2,4;   E;
     A, Y=0.0, 1,2;   A,  2.0, 3,4;   E;
BLOCK EXPRESSION DATA=
   A, 1,2,4,3;  E;
POINTS DISTRIBUTION DATA=
   A,11,7,10,6,3,3,2; 1;  E;
MATERIAL PROPERTIES=
   3, 1, 1.0E5, 1.000000E5, 1.0, 1.0, 1.0;
   1, 1, 1.0E0, 0.33333333, 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, 1, 4;    E;
POINT BOUNDARY CONDITION=
   A, 00, 1, 0.0, 0.0,  E;
LOADING DATA=
   LDMDL=1,  LDCS=1,
      1, A, 0.0, -3.0,  1,2,0;  E;
      1.0,    E;
GRAPH DATA =
   1, 1.0E-4, E;

   A, 00, 1, 0.0, 0.0,
   A, 01, 1, 0.0, 0.5; 1, 0.0, 1.0; 1, 0.0, 1.5; 1, 0.0, 2.0;    E;
   A, 00, 1, 4;  E;
   0, A, 0.0, -3.0,  1,12.0,2.0; 1,12.0,0.0;  E;