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# 44 hierarchical edge and interior shape functions n42

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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. Three-Dimensional 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 higher-order 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 two-dimensional 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. Three-Dimensional Shape Functions 25 monomial terms (cf., e.g., B...
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