# 47a where upon use of 545 and 546 n12 n12

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x1 (1 ; ) + x2 y y1 y2 x = x1 (1 ; ) + x3 y y1 y3 : The Jacobian determinant of the transformation can vanish depending on the location of Node 5. The analysis may be simpli ed by constructing the transformation in two steps. In the rst step, we use a linear transformation to map an arbitrary element onto a canonical element having vertices at (0 0), (1 0), and (0 1) but with one curved side. In the second step, we remove the curved side using the quadratic transformation (5.4.7). The linear mapping of the rst step has a constant Jacobian determinant and, therefore, cannot a ect the invertibility of the system. Thus, it su ces to consider the second step of the transformation as shown in Figure 5.4.5. Setting (x1 y1) = (0 0), (x2 y2) = (1 0), and (x3 y3) = (0 1) in (5.4.7a) yields x( y( Using (5.4.7c-e) Calculating the Jacobian ) 1 ^2 0 ^2 x5 ^ 2 ) = 0 N2 + 1 N3 + y5 N5 : x( y( )= ) J ( )= x x yy e (1 ; 2 ) + 4 (1 ; 2 ) ;+ = 1;22 + 44x5 y5 x5 : y5 ;2 + 4x5 1 ; 2 + 4y5 5.4. Coordinate Transformations y 3 1 0 1 0 21 11111111 00000000 11111111 00000000 11111111 00000000 1 0 11111111 00000000 1 0 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 η 3 1 0 1 0 6 1 0 1 0 5 1 0 1 0 1 0 1 0 1 0 1 0 5 6 1 0 1 0 1 0 1 0 1 0x 1 0 1 0 1 0 1 4 2 1 4 1 0ξ 1 0 2 Figure 5.4.5: Quadratic mapping of a right triangle having one curved side. The shaded region indicates where Node 5 can be placed without introducing a singularity in the mapping. we nd the determinant as det(Je( ) = 1 + (4x5 ; 2) + (4y5 ; 2) : The Jacobian determinant is a linear function of and thus, as with Example 5.4.1, we need only ensure that it has the same sign at each of its three vertices. We have det(Je(0 0)) = 1 det(Je(0 1)) = 4x5 ; 1 det(Je(1 0)) = 4y5 ; 1: Hence, the Jacobian determinant will not vanish and the mapping will be invertible when x5 > 1=4 and y5 > 1=4 (cf. Problem 2 at the end of this section). This region is shown shaded on the triangle of Figure 5.4.5. Problems 1. Consider the second-order serendipity shape functions of Problem 4.3.1...
View Full Document

## This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online