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Unformatted text preview: ear polynomials.) The equation of the line connecting Nodes 1 and 3 of the triangular element shown on the left of Figure 4.2.4 is N21 = 0. Likewise, the equation of a line passing through Node 2 and parallel to the line passing through Nodes 1 and 3 is N21 = 1. Thus, to map the line N21 = 0 onto the line = 0 in the canonical plane, we should set = N21 (x y). Similarly, the line joining Nodes 1 and 2 satis es the equation N31 = 0. We would like this line to become the line = 0 in the transformed plane, so our mapping must be = N31 (x y). Therefore, using (4.2.2) 2 3 2 3 1xy 1xy det 4 1 x1 y1 5 det 4 1 x1 y1 5 1 x3 y3 1 x2 y2 3 3 : (4.2.6) = N21 (x y) = 2 = N31 (x y) = 2 1 x2 y2 1 x3 y3 det 4 1 x1 y1 5 det 4 1 x1 y1 5 1 x3 y3 1 x2 y2 As a check, evaluate the determinants and verify that (x1 y1) ! (0 0), (x2 y2) ! (1 0), and (x3 y3) ! (0 1). Polynomials may now be developed on the canonical triangle to simplify the algebraic 4.2. Lagrange Shape Functions on Triangles 7 1 0 1 03 1 0 11 00 11 00 11 00 8 90 1 1 N 1= 0 2 1 0 1 0 1 0 1 0 1 0 11 0N 1= 1/3 1 0 10 0 N 1= 0 1 1 1 0 1 06 1 0 1 0 1 0 1 50 1 02 1 0 1 0 N 1= 2/3 1 4 1 0 1 0 1 0 1 0 1 07 1 0 Figure 4.2.5: Geometry of a triangular nite element for a cubic polynomial Lagrange approximation. complexity and subsequently transformed back to the physical element. 2. Transformation using triangular coordinates. A simple procedure for constructing Lagrangian approximations involves the use of a redundant coordinate system. The construction may be described in general terms, but an example su ces to illustrate the procedure. Thus, consider the construction of a cubic approximation on the triangular element shown in Figure 4.2.5. The vertex nodes are numbered 1, 2, and 3 edge nodes are numbered 4 to 9 and the centroid is numbered as Node 10. Observe that the line N11 = 0 passes through Nodes 2, 6, 7, and 3 the line N11 = 1=3 passes through Nodes 5, 10, and 8 and the line N11 = 2=3 passes through Nodes 4 and 9. Since N13 must vanish at Nodes 2 - 10 and be a cubic polynomial, it must have the form N13 (x y) = N11 (N11 ; 1=3)(N11 ; 2=3) where the constant is determined by normalizing N13 (x1 y1) = 1. Since N11(x1 y1) = 1, we nd = 9=2 and N13(x y) = 9 N11 (N11 ; 1=3)(N11 ; 2=3): 2 The shape function for an edge node is constructed in a similar manner. For example, in order to obtain N43 we observe that the line N21 = 0 passes through Nodes 1, 9, 8, and 3 8 Finite Element Approximation the line N11 = 0 passes through Nodes 2, 6, 7, and 3 and the line N11 = 1=3 passes through Nodes 5, 10, and 8. Thus, N43 must have the form N43(x y) = N11N21 (N11 ; 1=3): Normalizing N43 (x4 y4) = 1 gives Hence, = 27=2 and 21 N43 (x4 y4) = 3 3 ( 2 ; 1 ): 33 N43 (x y) = 27 N11 N21 (N11 ; 1=3): 2 3 must vanish on the boundary of the triangle and is, Finally, the shape function N10 thus, determined as 3 N10(x y) = 27N11N21 N31 : 3 The cubic shape functions N13 , N43 , and N10 are shown in Figure 4.2.6. The three linear shape functions Nj1 , j = 1 2 3, can be regarded as a redundant coordinate system...
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