Computer Aided Engineering Design

326 COMPUTER AIDED ENGINEERING DESIGN

Note that the interpolation above is equally biased along the x and y directions. To determine the
coefficients a 0 ,a 1 and a 2 , we need to solve

ui = a 0 + a 1 xi + a 2 yi
uj = a 0 + a 1 xj + a 2 yj
uk = a 0 + a 1 xk + a 2 yk (11.10b)
to get

a

uxy xy u xy yx u xy xy

xxyy xxyy

ijkkjjk i k i k ij ji

jik j k jj i

0 =

( – ) + ( – ) + ( – )

( – )( – ) – ( – )( – )

a

uy y u y y uy y

xxyy xxyy

ijk j k i k ij

jik j k jj i

1 =

( – ) + ( – ) + ( – )

( – )( – ) – ( – )( – ) (11.10c)

a

ux x u x x u x x

xxyy xxyy

i k jjikkji

jik j k jj i

2 =

( – ) + ( – ) + ( – )

( – )( – ) – ( – )( – )

and thus

u

xy y x y y x y y

xy y x y y x y yu

xy y xy y x y y

x

j k j kkj

ijk j k i k ij

i

i kki k i

i

=

( – ) + ( – ) + ( – )

( – ) + ( – ) + ( – ) +

( – ) + ( – ) + ( – )

(yyy xyy xyyj – ) + k j( – ) + k i k( – )ijuj

+

( – ) + ( – ) + ( – )

( – ) + ( – ) + ( – )

xy y xy y xy y

xy y x y y x y y

u

ij j i i j

ij k j k i k ij

k (11.10d)

or u
A
A
u

A

A

u

A

A

i i u N xyu N xyu N xyu

j

j

k

= + + k = iij j( , ) + ( , ) + kk( , ) (11.10e)

whereAi,Aj and Ak are the triangular areas shown in Figure 11.12 and A is the area of the triangular
element (A = Ai +Aj + Ak). Note that for P in the interior of the triangle, the shape functions Ni(x,y)

Figure 11.12 A triangular element

y

x

i

j

ui, fix

vi, fiy

Ak

uj, fjx

vj, fjy

Ai

Aj P

k uk, fkx

vk, fky