?? plate1.inp
字號:
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;
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