Biomedical Engineering Reference
In-Depth Information
independent. A solution for this dilemma is to eliminate
λ
3
from the shape functions by
using:
λ
3
=
1
−
λ
1
−
λ
2
.
(17.25)
A more obvious way of solving the problem is to substitute Eq. (
17.18
) into the equa-
tions for the shape functions, thus eliminating all triangular coordinates and directly
determine
∂
N
i
/∂
x
and
∂
N
i
/∂
y
.
•
The second issue is the difference in integration limits which have to correspond with
a triangle. For the square element in Section
17.2
the domain for integration is simple,
meaning that the surface integral can be split into two successive single integrals with
ξ
and
η
as variables. The integration limits are
−
1 to +1. For the triangle this is more
complicated and the limits of integration now involve the coordinate itself. This item
will be discussed shortly in Section
17.5
.
17.4
Lagrangian and Serendipity elements
In principle higher-order elements are more accurate than the linear ones discussed
so far. However, the computation of element coefficient matrices and element
arrays is more expensive for higher-order elements, and the cost-effectiveness
depends on the particular problem investigated. Cost-effectiveness in the sense
that there is a trade-off between the accuracy, using a smaller number of
higher-order elements, versus using more linear elements.
Higher-order elements can systematically be derived from Lagrange polynomi-
als. A (one-dimensional) set of Lagrange polynomials on an element with domain
ξ
1
≤
ξ
≤
ξ
n
is defined by
b
=
1,
b
=
a
(
ξ
−
ξ
b
)
b
=
1,
b
=
a
(
l
n
−
1
(
ξ
)
=
(17.26)
a
ξ
a
−
ξ
b
)
(
ξ
−
ξ
1
)
...
(
ξ
−
ξ
a
−
1
)(
ξ
−
ξ
a
+
1
)
...
(
ξ
−
ξ
n
)
=
,
ξ
a
−
ξ
1
)
...
ξ
a
−
ξ
a
−
1
)(
ξ
a
−
ξ
a
+
1
)
...
ξ
a
−
ξ
n
)
(
(
(
with
n
the number of nodes of the element and with
a
=
1, 2,
...
n
referring to a
node number. Notice that the above polynomial is of the order (
n
−
1).
For instance, first-order (linear) polynomials are found for
n
=
2, hence
ξ
−
ξ
2
ξ
1
−
ξ
2
,
l
1
=
(17.27)
and
ξ
−
ξ
1
ξ
2
−
ξ
1
.
l
2
=
(17.28)
