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