The most popular coordinate transformations are

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: isoparametric, and superparametric when their polynomial degree is, respectively, lower than, equal to, and greater than that used for the trial function. As we have seen in Chapter 4, the transformations use the same shape functions as the nite element solutions. We illustrated linear (Section 4.2) and bilinear (Section 4.3) transformations for, respectively, mapping triangles and quadrilaterals to canonical elements. We have two tasks in front of us: (i) determining whether higher-degree piecewise polynomial mappings can be used to advantage and (ii) ensuring that these transformations will be nonsingular. Example 5.4.1. Recall the bilinear transformation of a 2 2 canonical square to a quadrilateral that was introduced in Section 4.3 (Figure 5.4.1) y 2,2 (x 22 22) ,y 1 0 1 0 (x 12,y12 ) η 1,2 1 0 1 0 1,2 1 0 1 0 2,2 1 0 1 0 1 h2 1 02,1 1 0 α 12 1 0 1 0 1,1 (x ξ (x 21,y21 ) h1 ,y ) 11 11 x 1 1,1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 2,1 1 1 0 1 0 1 Figure 5.4.1: Bilinear mapping of a quadrilateral to a 2 2 square. x( y( 2 2 ) = X X xij N ( ) yij i j i=1 j =1 ) (5.4.2a) i j=1 2 (5.4.2b) where Ni j ( ) = Ni ( )Nj ( ) and Ni ( ) = (1 ; )=2 if i = 1 : (1 + )=2 if i = 2 (5.4.2c) The vertices of the square (;1 ;1), (1 ;1), (;1 1), (1 1) are mapped to the vertices of the quadrilateral (x1 1 y1 1), (x2 1 y2 1), (x1 2 y1 2), (x2 2 y2 2). The bilinear transformation is linear along each edge, so the quadrilateral element has straight sides. 5.4. Coordinate Transformations 17 Di erentiating (5.4.2a) while using (5.4.2b,c) x = x21 ; x11 N1 ( ) + x22 ; x12 N2( ) 2 2 y = y21 ; y11 N1( ) + y22 ; y12 N2 ( ) 2 2 x = x12 ; x11 N1( ) + x22 ; x21 N2( ) 2 2 y12 ; y11 N ( ) + y22 ; y21 N ( ): y= 2 1 2 2 Substituting these formulas into (5.4.1) and evaluating the determinant reveals that the quadratic terms cancel hence, the determinant of Je is a linear function of and rather than a bilinear function. Therefore, it su ces to check that det(Je) has the same sign at each of the four vertices. For example, det(Je(;1 ;1)) = x (;1 ;1)y (;1 ;1) ; x (;1 ;1)y (;1 ;1) or det(Je(;1 ;1)) = (x21 ; x11 )(y12 ; y11)...
View Full Document

Ask a homework question - tutors are online