The Finite Element Method - Mechanics of Structures: Variational and Computational Methods - page 380

of Fig. 6.27c. For this case, Eq. (6.65c) becomes

u (ξ , η) = N 1 ξ N 1 η v 1 + N 2 ξ N 1 η v 5 + N 3 ξ N 1 η v 2 + N 1 ξ N 2 η v 8 + N 2 ξ N 2 η v 9 + N 3 ξ N 2 η v 6

+ N 1 ξ N 3 η v + N 2 ξ N 3 η v + N 3 ξ N 3 η v = Nv (6.65d)

4 7 3

In matrix notation, Eq. (6.65c) appears as

v 1 v 4 v 7 N 1 η

N 2 η
N 3 η

=

N T
ξ

u (ξ , η) = [ N 1 ξ N 2 ξ N 2 ξ ] v 2 v 5 v 8 RN η (6.65e)

v 3 v 6 v 9

and Eq. (6.65d) becomes

v v v N 1 η

N 2 η
N 3 η

1 8 4

=

N T
ξ

u (ξ , η) = [ N 1 ξ N 2 ξ N 3 ξ ] v v v RN η . (6.65f)

5 9 7

v v v

2 6 3

This is referred to a biquadratic interpolation . The two-dimensional Lagrangian interpolation
function contains a complete order m polynomial plus some individual terms of higher
order. Since the convergence of the finite element process can be shown to be related to
the highest order complete polynomial, it can be useful to eliminate some terms, e.g., those
corresponding to the inner nodes, so that the interpolation functions depend only on corner
and boundary points.

EXAMPLE 6.7 Shape Functions of a Bilinear Element
For the bilinear element in Fig. 6.10, construct the shape functions corresponding to u x , the
displacement in the x direction, directly from the Lagrangian interpolation polynomials.
The bilinear polynomial of Eq. (6.11)

= + ξ + ξη +

u x u 1 u 2 u 3 u 4 η (1)

This polynomial is, according to the Pascal triangle, not complete in the sense of not in-
cluding all of the terms of order 2 and below. On the other hand, the element is complete
because (1) satisfies all the completeness conditions given in Section 6.5.2. Alternatively, a
shape function similar in form to that of Eq. (6.65f) can be assembled for the displacement u x .

u x = N 1 ξ N 1 η u x 1 + N 2 ξ N 1 η u x 2 + N 2 ξ N 2 η u x 3 + N 1 ξ N 2 η u x 4 (2)

or

] u x 1

N 1 η
N 2 η

u x 4

u x = [ N 1 ξ N 2 ξ (3)

u x 2 u x 3

where N i ξ and N i η ,i = 1 , 2 are the Lagrangian polynomials of Eq. (6.64) with m = 2. Thus,

with ξ 1 = 0 and ξ 2 = 1 (and η 1 = 0 , η 2 = 1 ) ,

N 1 ξ = ξ −

1

ξ = ξ −

0

1 =− (ξ −

1 ) = ( 1 − ξ) ,

0 = ξ

2

− −

0 1

(4)

N 1 η = η −

1

η = η −

0

1 =− (η −

1 ) = ( 1 − η) ,

0 = η

2

0 − 1 −