Civil Engineering Reference
In-Depth Information
There are 9 equations and the skyline storage is 39
Node Displacement(s)
1 0.0000E+00 0.0000E+00
2 -0.1042E-02 0.0000E+00
3 0.5333E-03 -0.6562E-04
4 0.9500E-03 -0.3046E-02
5 -0.1425E-02 -0.2981E-02
6 -0.1692E-02 -0.7263E-02
Element Actions
1 0.2667E+02 -0.2000E+02 -0.2667E+02 0.2000E+02
Axial force = -0.3333E+02
2 -0.2667E+02 0.0000E+00 0.2667E+02 0.0000E+00
Axial force = 0.2667E+02
3 -0.2083E+02 0.0000E+00 0.2083E+02 0.0000E+00
Axial force = 0.2083E+02
4 -0.5833E+01 0.4375E+01 0.5833E+01 -0.4375E+01
Axial force = 0.7292E+01
5 0.0000E+00 -0.4375E+01 0.0000E+00 0.4375E+01
Axial force = -0.4375E+01
6 0.7500E+01 0.5625E+01 -0.7500E+01 -0.5625E+01
Axial force = -0.9375E+01
7 0.1917E+02 0.0000E+00 -0.1917E+02 0.0000E+00
Axial force = -0.1917E+02
8 0.0000E+00 0.4375E+01 0.0000E+00 -0.4375E+01
Axial force = -0.4375E+01
9 -0.1333E+02 0.1000E+02 0.1333E+02 -0.1000E+02
Axial force = 0.1667E+02
10 0.1333E+02 0.0000E+00 -0.1333E+02 0.0000E+00
Axial force = -0.1333E+02
Figure 4.8
Results from first Program 4.2 example
Nodal coords (m)
z
Node xyz
1 0 0 0
2 1.25 3 0
3 3.5 2 0
4 4 1 0
5 2 1.5 3
Q
5
y
Load components (kN)
Q
x
= 20.0
Q
y
= -20.0
Q
z
= 30.0
EA = 5
×
10
5
kN
2
4
3
2
1
3
4
x
1
Figure 4.9
Mesh and data for second Program 4.2 example (
Continued on page 121
)
