The Finite Element Method - Mechanics of Structures: Variational and Computational Methods - page 344

corners, e.g., Eq. (6.14), are assembled,

= = N ux

u x 1
u x 2
u x 3
u x 4

1000
1100
1111
1001

u 1

u 2

v x = u x (6.15)

u 3

u 4

4 matrix N ux is readily inverted.

The 4 ×

1000

=

− 1100

1

N − 1
ux

= G x (6.16)

− 11 − 1

− 1001

Then

= N − 1 u x

ux v x

= G x v x

and, from Eq. (6.11),

u x (ξ , η) = N ux G x v x = N x v x (6.17)

with

1000

=

− 1100

1

[1 ξξηη ] [ ( 1 − ξ)( 1 − η) ξ( 1 − η) ηξ η( 1 − ξ) ]

− 11 − 1

(6.18)

− 1001

G x = N − 1

ux

N ux N x

Thus, in Eq. (6.17) the assumed displacement u x has been expressed in terms of the unknown
nodal displacements.
The same manipulations for u y lead to similar relationships. For the y coordinate, N uy G y =

N y can be found, where N y (ξ , η) = N x (ξ , η). If the displacements u x and u y are placed to-

gether,

u x
u y

N x 0
0N y

v x
v y

Nv i N i v i u i u = = = = Nv = = (6.19a)

where superscript i has been included to indicate that this is the i th element, or

u x 1
u x 2
u x 3
u x 4
···
u y 1
u y 2
u y 3
u y 4

u x
u y

(

1 − ξ)( 1 − η) ξ( 1 − η) ξη η( 1 − ξ) 0

=

0 ( 1 − ξ)( 1 − η) ξ( 1 − η) ξη η( 1 − ξ)

v i
(6.19b)

u N