WARNING: The displacements and/or stresses at intermediate points are generated linearly using the coordinates of the nodes. Consequently the automatic generation of the boundary conditions can not be performed if the boundary is curved. However it can be used for the case of constant evolution. Plate in traction, symmetry conditions. ************************************************** General constants : Number of elements........ 8 Number of internal points. 1 Elasticity modulus........ 2100000. Poisson coefficient....... .200 Number of Gauss points.... 4 Type of problem........... Plane Stress node x coor y coor 1 0.0000 0.0000 2 1.5000 0.0000 3 3.0000 0.0000 4 3.0000 1.5000 5 3.0000 3.0000 6 1.5000 3.0000 7 0.0000 3.0000 8 0.0000 1.5000 9 0.0000 0.0000 node code ul vl ug vg 1 16 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 2 15 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 3 4 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 4 1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 5 1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 6 1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 7 6 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 8 15 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 9 16 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 node code nsb tsb nsa tsa 1 16 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 2 15 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 3 4 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 4 1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 5 1 0.0000E+00 0.0000E+00 0.1000E+04 0.0000E+00 6 1 0.1000E+04 0.0000E+00 0.1000E+04 0.0000E+00 7 6 0.1000E+04 0.0000E+00 0.0000E+00 0.0000E+00 8 15 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 9 16 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 Plate in traction, symmetry conditions. ************************************************** Solution on the boundary: boundary points node xdisplacement ydisplacement 1 0.0000000E+00 -0.0000000E+00 2 -0.1785776E-03 -0.0000000E+00 3 -0.3571440E-03 -0.0000000E+00 4 -0.3571587E-03 0.7142944E-03 5 -0.3571416E-03 0.1428567E-02 6 -0.1785758E-03 0.1428579E-02 7 -0.0000000E+00 0.1428573E-02 8 0.0000000E+00 0.7142969E-03 node nor stress bf nor stress af tan stress bf tan stress af 1 -0.2708934E-01 0.1000017E+04 0.0000000E+00 0.0000000E+00 2 0.9999803E+03 0.9999803E+03 0.0000000E+00 0.0000000E+00 3 0.1000021E+04 0.0000000E+00 0.0000000E+00 0.0000000E+00 4 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00 5 0.0000000E+00 0.1000000E+04 0.0000000E+00 0.0000000E+00 6 0.1000000E+04 0.1000000E+04 0.0000000E+00 0.0000000E+00 7 0.1000000E+04 -0.2291476E-01 0.0000000E+00 0.0000000E+00 8 0.2391433E-01 0.2391433E-01 0.0000000E+00 0.0000000E+00 Solution in the domain: internal points x coor y coor xdisplacement ydisplacement 1.000 1.000 -0.11906780E-03 0.47612139E-03 x coor y coor stress xx stress yy stress xy 1.000 1.000 0.00000000E+00 0.00000000E+00 0.00000000E+00