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Unformatted text preview: ). 4.5.1 Lagrangian Shape Functions on Tetrahedra
Let us begin with a linear shape function on a tetrahedron. We introduce four nodes
numbered (for convenience) as 1 to 4 at the vertices of the element (Figure 4.5.1). Imposing the usual Lagrangian conditions that Nj (xk yk zk ) = jk , j k = 1 2 3 4, gives 4.4. ThreeDimensional Shape Functions 23 the shape functions as
1
04 (0,0,0,1)
1
0 (ζ1,ζ2,ζ 3ζ4)
, P 1
03 (0,0,1,0)
1
0 1
0 1
10
(1,0,0,0) 1
0
1
02 (0,1,0,0) Figure 4.5.1: Node placement for linear shape functions on a tetrahedron and de nition
of tetrahedral coordinates.
m
Nj (x y z) = Dk lC (x y z) where jklm (j k l m) a permutation of 1 2 3 4 2
1x
6 1 xk
Dk l m(x y z) = det 6 1 x
4
l
1 xm y
yk
yl
ym (4.5.1a) 3 z
zk 7
7
zl 5
zm (4.5.1b) 2
3
1 xj yj zj
6
7
Cj k l m = det 6 1 xk yk zk 7 :
(4.5.1c)
4 1 xl yl zl 5
1 xm ym zm
Placing nodes at the vertices produces a linear shape function on each face that is uniquely
determined by its values at the three vertices on the face. This guarantees continuity of
bases constructed from the shape functions. The restriction of U to element e is
4
X
U (x y z) = cj Nj (x y z):
(4.5.2)
j =1 As in two dimensions, we may construct higherorder polynomial interpolants by
either mapping to a canonical element or by introducing \tetrahedral coordinates." Focusing on the latter approach, let
j = Nj (x y z) j=1 2 3 4 (4.5.3a) 24 Finite Element Approximation
11
00
4
11
00
11
00 ζ
11
004 (0,0,1)
11
00
11
00 z 1111
0000 η
1111
0000
1
03 (0,1,0)
1111
0000
1
0 1
03
1
0
1
0 1
11
00
11
00
11
00 11
00
11
00
11
002 y 11
00
11
00
11
00 1 (0,0,0) x 11
00
11
00
11
00 ξ 2 (1,0,0) Figure 4.5.2: Transformation of an arbitrary tetrahedron to a right, unit canonical tetrahedron.
and regard j , j = 1 2 3 4, as forming a redundant coordinate system on a tetrahedron.
The coordinates of a point P located at ( 1 2 3 4) are (Figure 4.5.1)
VP 234
VP 134
VP 124
VP 123
(4.5.3b)
2= V
3= V
4= V
1= V
1234
1234
1234
1234
where Vijkl is the volume of the tetrahedron with vertices at i, j , k, and l. Hence, the
coordinates of Vertex 1 are (1 0 0 0), those of Vertex 2 are (0 1 0 0), etc. The plane
= 0 is the plane A234 opposite to vertex 1, etc. The transformation from physical to
tetrahedral coordinates is
232
32 3
x
x1 x2 x3 x4
1
6 y 7 6 y1 y2 y3 y4 7 6 2 7
6 7=6
(4.5.4)
4 z 5 4 z1 z2 z3 z4 7 6 3 7 :
54 5
1
1111
4
The coordinate system is redundant as expressed by the last equation.
The transformation of an arbitrary tetrahedron to a right, unit canonical tetrahedron
(Figure 4.5.2) follows the same lines, and we may de ne it as
= N2 (x y z) = N3 (x y z) = N4 (x y z): (4.5.5) The face A134 (Figure 4.5.2) is mapped to the plane = 0, the face A124 is mapped to
= 0, and A123 is mapped to = 0. In analogy with the twodimensional situation, this
transformation is really the same as the mapping (4.5.3) to tetrahedral coordinates.
A complete polynomial of degree p in three dimensions has np = (p + 1)(p + 2)(p + 3)
6 (4.5.6) 4.4. ThreeDimensional Shape Functions 25 monomial terms (cf., e.g., B...
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This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.
 Spring '14
 JosephE.Flaherty
 The Land

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