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-1.16
-1.16
-0.49
-0.49
12
5
7
4
6
9
3
A
8
1.41
0.86
0.86
1.41
-1.81
-1.81
-0.79
-0.56
-0.56
1
9
4
5
6
8
2
3
7
10
12
B
11
0.44
0.37
0.37
1.37
1.37
1.74
-2.20
-1.36
-1.09
-0.9
-0.56
-0.18
1 2 3 4
5
6
8
13
15
7
9 10 11 12
C
16 17 18
19 20 21 22 23
14
0.69
0.97
1.65
2.08
FIGURE 6.9
Shear force distributions for mesh refinements A, B, and C, using the material law
V
w
.
=−
EI
6.4
Finite Element Method for Plane Problems
A two-dimensional structure provides the opportunity to describe the finite element method
for problems of reasonably general geometry. We begin by establishing a 4-node rectangu-
lar element (Melosh, 1963), which will be used here as a fundamental building block for a
multi-degree-of-freedom model of a structure.
6.4.1 Rectangular Element
Consider the rectangular element with four nodes and eight degrees of freedom shown
in Fig. 6.10. To describe the displacement field in this two-dimensional element, assume
dispacements are in the
x, y
plane, i.e., this is an in-plane problem, and represent the
displacement
u
with the two rectilinear components
u
x
,
u
y
, i.e.,
u
x
(ξ
,
η)
u
=
(6.10)
u
y
(ξ
,
η)
Introduce a trial displacement solution (the trial function)
u
with eight unknowns, i.e., with
eight degrees-of-freedom (DOF) (coordinates). Assume that a polynomial can adequately
represent each of the displacements
u
x
and
u
y
. For the displacement in the
x
direction,
choose the bilinear polynomial
u
1
η
=
1
ξξηη
u
2
N
ux
=
+
ξ
+
ξη
+
u
x
(ξ
,
η)
=
u
x
u
1
u
2
u
3
u
4
(6.11)
u
3
u
4
where the origin of the coordinates
ξ
,
η
is at node 1. This polynomial contains the four
unknowns,
u
1
,
u
2
,
u
3
,
u
4
. Similarly, the displacement in the
y
direction is chosen to have the
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