The Finite Element Method - Mechanics of Structures: Variational and Computational Methods - page 407

For the isoparametric element, the geometry, i.e., the coordinate relations, is defined by the
interpolation functions for the displacement.

4

1
4 [

x = N i x i = ( 1 − ξ)( 1 − η) x 1 + ( 1 + ξ)( 1 − η) x 2

i = 1

+ ( 1 + ξ)( 1 + η) x 3 + ( 1 − ξ)( 1 + η) x 4 ]

(6.132)

4

1
4 [

y = N i y i = ( 1 − ξ)( 1 − η) y 1 + ( 1 + ξ)( 1 − η) y 2

i = 1

+ ( 1 + ξ)( 1 + η) y 3 + ( 1 − ξ)( 1 + η) y 4 ]

ξ =

For example, the line 1 in the master element is transformed to

η) = (

1 − η)

2

x 2 + (

1 + η)

2

x ( 1 , x 3

x 2 + x 3 x 3 − x 2

= + η

2 2

η) = (

1 − η)

2

+ (

1 + η)

2

y ( 1 , y 2 y 3

y 2 + y 3 y 3 − y 2

= + η

2 2

c in the quadri-
lateral element. If the above transformation relations are invertible, line b −

1 in the master element is transformed to a line b − Thus, the line ξ =

c can also be

1 in the master element. Similarly, lines c − d ,d − a , and a − b

transformed to ξ = can

be tranformed to η = 1 , ξ =− 1 , and η =− 1 in the master element. Thus, the complete

geometry can be transformed.
Using the coordinate transformation, the derivatives in the global coordinate system,
where the slave element resides, can also be transformed to those in the master element. In
the calculation of the element stiffness matrix and loading vector for the slave element, the
derivatives

∂/∂ x, ∂/∂ y are required. From the chain rule of differentiation,

∂
∂ x = ∂

∂ξ
∂ x + ∂

∂η
∂ ∂ξ ∂η x

(6.133)

∂
∂ y = ∂

∂ξ
∂ y + ∂

∂η
∂

∂ξ ∂η y

∂ξ
∂

∂ξ
∂

∂η
∂

∂η
∂

To evaluate ∂/∂ x and ∂/∂ y, the derivatives

x , y ,

and are needed. However, the

x y

explicit relationship in Fig. 6.39a are difficult to obtain. This

differs from the case of Fig. 6.10 where, for a rectangular element with the origin at the
lower left corner,

ξ = ξ( x, y ) and η = η( x, y )

ξ = x / a and η = y / b . To obtain the desired derivatives, first establish

∂ξ = ∂

∂

x ∂ ∂ξ + ∂

x

y ∂

y
∂ξ ∂ ∂

∂η = ∂

∂

x ∂

∂η + ∂

y ∂

x y

∂η

(6.134)

∂ ∂

or, in matrix notation,

∂ξ
∂
∂η

∂ x
∂ξ

∂ x
∂
∂

J ∂ x
∂
∂

∂ y

∂ξ

= = (6.135a)

∂ x

∂η

∂ y

∂η

y y