Civil Engineering Reference
In-Depth Information
It turns out that it is much better to use an
element-oriented
approach. For
every element
T
∈
T
h
, we find the additive contribution from (8.1) to the stiffness
matrix. If every element contains exactly
s
nodes, this requires finding an
s
×
s
submatrix. We transform the triangle
T
under consideration to the reference triangle
T
ref
. Let
F
:
T
ref
→
T, ξ
→
Bξ
+
x
0
be the corresponding linear mapping. Then
the contribution of
T
is given by the integral
µ(T )
µ(T
ref
)
a
kl
(B
−
1
)
k
k
(B
−
1
)
l
l
∂
k
N
i
∂
l
N
j
dξ.
(
8
.
2
)
T
ref
k,l
k
,l
Here
µ(T )
is the area of
T
. After transformation, every function in the nodal basis
coincides with one of the normed
shape functions N
1
,N
2
,...,N
s
on the reference
triangle. These are listed in Table 4 for linear and quadratic elements.
6
For the
model problem 4.3, using a right triangle
T
(with right angle at point number 1),
we get
1
1
2
(u
1
−
u
3
)
2
,
where
u
i
is the coefficient of
u
in the
N
i
expansion. This gives
2
(u
1
−
u
2
)
2
a(u, u)
|
T
=
+
2
−
1
−
1
1
2
a(ψ
i
,ψ
j
)
|
T
=
−
11
−
1
1
for the stiffness matrix on the element level. For linear elements, it is also easy
to find the so-called
mass matrix
whose elements are
(ψ
i
,ψ
j
)
0
,T
. For an arbitrary
triangle,
211
121
112
.
µ(T )
12
(ψ
i
,ψ
j
)
0
,T
=
(
8
.
3
)
For differential equations with variable coefficients, the evaluation of the inte-
grals (8.2) is usually accomplished using a Gaussian quadrature formula for multiple
6
To avoid indices, in Table 4 we have written
ξ
and
η
instead of
ξ
1
and
ξ
2
.
In addition, we note that for the quadratic triangular elements, the basis functions
N
1
,N
2
and
N
3
can be replaced by the corresponding nodal basis functions of linear elements.
The coefficients in the expansion
i
=
1
z
i
N
i
then have a different meaning:
z
1
,z
2
and
z
3
are still the values at the vertices, but
z
4
,z
5
and
z
6
become the deviations at the midpoints
of the sides from the linear function which interpolates at the vertices.
This basis is not a purely nodal basis, although the correspondence is very simple.
However, it has two advantages: we get simpler integrands in (8.2), and the condition
number of the system matrix is generally lower (cf. hierarchical bases).
Search WWH ::
Custom Search