📄 testb.out
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< M A T L A B >
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Version 6.5.0.180913a Release 13
Jun 18 2002
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*********************************************************************
* FEMjr v 2.0 - Finite Element Method junior - An FEM framework *
*********************************************************************
Enter name of input data file:
inputfile =
testb.dat
<<<<---- Data File Echo ---->>>>
TITLE - Three Bar Truss (from Allaire pg. 117)
NODES - # of nodes: 3; # of dimensions: 2
# ID x y z
1 1 0 150
2 2 0 0
3 3 260 150
ELEMENTS - # of elements: 3
# ID elemType propID connectivity
3 1 truss2d 1 1 3
3 2 truss2d 1 2 1
3 3 truss2d 2 2 3
PROPERTIES - # of properties: 2
# ID propType properties
1 1 MatTruss 69 200
2 2 MatTruss 207 100
CONSTRAINTS - # of constraints: 3
# nodeID dof value
1 1 1 0
2 1 2 0
3 2 1 0
LOADS - # of loads: 1
# nodeID dof value
1 3 2 -0.4
Stiffness Matrix and Load Vector after assembly
kglobal =
53.0769 0 0 0 -53.0769 0
0 92.0000 0 -92.0000 0 0
0 0 51.7404 29.8502 -51.7404 -29.8502
0 -92.0000 29.8502 109.2213 -29.8502 -17.2213
-53.0769 0 -51.7404 -29.8502 104.8173 29.8502
0 0 -29.8502 -17.2213 29.8502 17.2213
load_vector =
0
0
0
0
0
0
Load Vector after assembly of point loads
load_vector =
0
0
0
0
0
-0.4000
Constrained Stiffness Matrix and Load Vector
STIFFNESS MATRIX
r/c 1 2 3 4 5 6 7 8
1 53 0 0 0 0 0
2 0 92 0 0 0 0
3 0 0 52 0 0 0
4 0 0 0 1.1e+02 -30 -17
5 0 0 0 -30 1e+02 30
6 0 0 0 -17 30 17
LOAD VECTOR
Row Value
1 0
2 0
3 0
4 0
5 0
6 -0.4
<<<<---- Solution Results ---->>>>
SOLUTION VECTOR
dof# value
1 0
2 0
3 0
4 -0.0043478
5 0.013063
6 -0.050217
Strain Energy (assume homogeneous displacement BCs) = 0.010043
ELEMENTAL POSTPROCESSING
ID f1_x f1_y f2_x f2_y magnitude stress strain energy
1 -0.69333 0 0.69333 0 0.69333 0.0034667 0.0045284
ID f1_x f1_y f2_x f2_y magnitude stress strain energy
2 0 -0.4 0 0.4 0.4 0.002 0.00086957
ID f1_x f1_y f2_x f2_y magnitude stress strain energy
3 0.69333 0.4 -0.69333 -0.4 0.80044 -0.0080044 0.0046454
Strain Energy = 0.010043
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