Civil Engineering Reference
In-Depth Information
The reader is urged to utilize these reaction force components and
check the equilibrium conditions of the structure.
Step 9.
Compute strain and stress in each element. The major computational
task completed in Step 7 provides the displacement components of
each node in the global coordinate system. With this information and
the known constrained displacements, the displacements of each
element in its element coordinate system can be obtained; hence, the
strain and stress in each element can be computed.
For element 2, for example, we have
u
(2
1
=
U
1
cos
2
+
U
2
sin
2
=
0
04618 )
√
2
u
(2
2
=
U
7
cos
2
+
U
8
sin
2
=
(
−
0
.
01600
+
0
.
/
2
=
0
.
02134
The axial strain in element 2 is then
u
(2
2
u
(2
1
L
(2)
−
0
02133
40
√
2
.
(2)
771(10
−
4
)
ε
=
=
=
3
.
and corresponding axial stress is
(2)
(2)
=
E
ε
=
3771 psi
The results for element 2 are presented as an example only. In finite
element software, the results for each element are available and
can be examined as desired by the user of the software
(postprocessing).
Results for each of the eight elements are shown in Table 3.5; and
per the usual sign convention, positive values indicate tensile stress
while negative values correspond to compressive stress. In obtaining
the computed results for this example, we used a spreadsheet program
to invert the stiffness matrix, MATLAB to solve via matrix inversion,
and a popular finite element software package. The solutions resulting
from each procedure are identical.
Table 3.5
Results for the Eight Elements
Element
Strain
Stress, psi
5.33(10
−
4
)
1
5333
3.77(10
−
4
)
2
3771
−
4.0(10
−
4
)
3
−
4000
1.33(10
−
4
)
4
1333
5.33(10
−
4
)
5
5333
−
5.67(10
−
4
)
6
−
5657
2.67(10
−
4
)
7
2667
4.00(10
−
4
)
8
4000
