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6.5.4
h-Convergence and p-Convergence
The most common method for achieving an improved approximate solution is by succes-
sively using finer meshes of elements. This approach is usually called
h-convergence
. Alter-
natively, the introduction of higher order trial functions can lead to a more accurate solution
for a constant set of mesh divisions (
p-convergence
). Although highly problem dependent,
often the use of higher order trial functions results in an improved solution at less computa-
tional expense than the use of mesh refinement [Szabo and Babuska, 1991]. A combination
of h- and p-convergence is often considered as the most favorable approach [Oden, 1990].
Much of the work in this chapter utilizes interpolation functions, wherein the unknown
parameters
u
(Eq. 6.13) are replaced by nodal values
v
. A drawback to the use
of interpolation functions is that the introduction of higher order trial functions involves a
completely new element matrix. On the other hand, if a trial function expressed in terms of
the generalized displacements
u
of
u
=
N
u
u
is employed, the effect of successive higher order trial func-
tions can be additive. Such trial functions, which are called
hierarchical
, are illustrated below.
For the two-dimensional plane problem presented earlier in this chapter, the shape func-
tion is obtained from (Eq. 6.11)
=
+
ξ
+
ξη
+
u
x
u
1
u
2
u
3
u
4
η
(6.51)
where
,
4 are the unknown constants (generalized displacements). In this
case, the number of unknown constants is equal to the number of nodes; consequently, it
is convenient to express the shape function in terms of the nodal displacements. However,
more terms can be included in the shape function. Instead of using Eq. (6.51), let the shape
function
u
x
be expressed as
u
x
u
i
,i
=
1
,
2
,
...
2
2
=
+
ξ
+
η
+
ξη
−
−
u
1
u
2
u
3
u
4
u
5
ξ
u
6
η
u
1
u
2
.
2
2
]
=
[1
ξηξη
−
ξ
−
η
u
6
N
u
2
]
u
1
=
[
N
u
1
(6.52)
u
2
where
2
2
]
N
u
1
=
[1
ξηξη
]
,
N
u
2
=
[
−
ξ
−
η
u
4
]
T
,
u
6
]
T
u
1
=
[
u
1
u
2
u
3
u
2
=
[
u
5
Substitute the nodal coordinates into Eq. (6.52), giving
u
1
u
x
1
u
x
2
u
x
3
u
x
4
1000
0
0
u
2
.
=
1100
−
10
1111
−
1
−
1
1010
0
−
1
u
6
u
1
[
N
u
1
N
u
2
]
=
v
x
(6.53)
u
2
in which
1000
1100
1111
1010
00
−
10
N
u
1
=
N
u
2
=
−
1
−
1
0
−
1
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