# 53 to tetrahedral coordinates a complete polynomial

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Unformatted text preview: renner and Scott 3], Section 3.6). With p = 2, we have n2 = 10 monomial terms and we can determine Lagrangian shape functions by placing nodes at the four vertices and at the midpoints of the six edges (Figure 4.5.3). With p = 3, we have n3 = 20 and we can specify shape functions by placing a node at each of the four vertices, two nodes on each of the six edges, and one node on each of the four faces (Figure 4.5.3). Higher degree polynomials also have nodes in the element's interior. In general there is 1 node at each vertex, p ; 1 nodes on each edge, (p ; 1)(p ; 2)=2 nodes on each face, and (p ; 1)(p ; 2)(p ; 3)=6 nodes in the interior. 4 1 0 1 0 1 0 8 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 09 1 0 1 0 1 0 1 0 5 1 010 1 0 1 0 1 0 1 0 1 0 1 0 1 06 1 0 1 03 1 0 1 0 2 1 0 11 00 1 0 11 00 1 0 11 00 1 0 11 00 11 00 11 00 11 11 00 00 11 00 1 0 11 11 00 00 11 00 11 00 1 0 1 0 1 0 1 0 1 0 1 0 1 0 11 00 1 11 0 00 1 0 1 11 11 0 00 00 11 00 1 0 11 00 1 0 Figure 4.5.3: Node placement for quadratic (left) and cubic (right) interpolants on tetrahedra. Example 4.5.1. The quadratic shape function N12 associated with vertex Node 1 of a tetrahedron (Figure 4.5.3, left) is required to vanish at all nodes but Node 1. The plane 1 = 0 passes through face A234 and, hence, Nodes 2, 3, 4, 6, 9, 10. Likewise, the plane 2 1 = 1=2 passes through Nodes 5, 7 (not shown), and 8. Thus, N1 must have the form N12 ( 1 2 3 4) = 1 ( 1 ; 1=2): Since N12 = 1 at Node 1 ( 1 = 1), we nd = 2 and N12 ( 1 2 3 4 ) = 2 1 ( 1 ; 1=2): Similarly, the shape function N52 associated with edge Node 5 (Figure 4.5.3, left) is required to vanish on the planes 1 = 0 (Nodes 2, 3, 4, 6, 9, 10) and 2 = 0 (Nodes 1, 3, 4, 7, 8, 10) and have unit value at Node 5 ( 1 = 2 = 1=2). Thus, it must be N52 ( 1 2 3 4) = 4 1 2: 26 Finite Element Approximation 1,1,2 2,1,2 1 0 1 0 1 0 1 0 1 0 1 0 ζ 1 01,2,2 1 0 1 0 1 0 1 02,2,2 1 0 111 000η 111 000 1 0 111 000 1 0 111 000 ξ 000 1 0 111 1,1,1 1 0 1 0 1 0 2,1,1 1 0 1 0 1 0 1 0 1 0 1 0 1,2,1 2,2,1 1 0 1 0 11 00 1 0 1 0 1 0 1 0 11 00 1 0 1 0 11 00 11 00 1 11 0 00 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 11 00 1 0 11 00 1 0 1 0 1 0 1 0 11 00 1 0 11 00 1 0 1 0 1 0 1 0 11 00 1 0 11 00 1 0 1 0 1 0 1 0 11 00 1 0 1 0 1 0 1 0 11 00 1 0 1 0 1 0 11 00 1 0 1 0 Figure 4.5.4: Node placement for a trilinear (left) and tri-quadratic (right) polynomial interpolants on a cube. 4.5.2 Lagrangian Shape Functions on Cubes In order to construct a trilinear approximation on the canonical cube f j;1 1g, we place eight nodes numbered (i j k), i j k = 1 2, at its vertices (Figure 4.5.4). The shape function associated with Node (i j k) is taken as Ni j k( ) = Ni( )Nj ( )Nk ( ) (4.5.7a) where Ni( ), i = 1 2, are the hat function (4.3.1d,e). The restriction of U to this element has the form 222 XXX U( )= ci j k Ni j k( ) (4.5.7b) i=1 j =1 k=1 Once again, ci j k = Ui j k = U ( i j k ). The placement of nodes at the vertices produces bilinear shape functions on each...
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