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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 Spring '14
 JosephE.Flaherty
 Polynomials, The Land, Quadratic equation, NJ, Degree of a polynomial, shape functions

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