Computer Aided Engineering Design

332 COMPUTER AIDED ENGINEERING DESIGN

u = a 0 + a 1 x+a 2 y+a 3 xy (11.11a)

Note again that the polynomial basis [1 xyxy] is chosen such that there is no relative bias along the
x and y directions. Alternatively, to ease calculations, the four-node element may be mapped onto a
square shown on the right in Figure 11.15 with local coordinates (ξ,η). We can first determine the
mapping between the global and local coordinate systems. To map the xcoordinate of P on the square,
we may use

x = c 0 + c 1 ξ+c 2 η+c 3 ξη (11.11b)

employing the same reasoning that we need to map four nodes for which we need four unknowns and
the basis of four polynomial functions (as chosen above) should not have any bias towards any local
coordinate. Also the x coordinate of P would depend only on the x nodal coordinates, i.e. x 1 ,x 2 ,x 3 and
x 4. Employing the conditions in Eq. (11.11b) at four nodes, we have

x 1 = c 0 – c 1 – c 2 + c 3

x 2 = c 0 + c 1 – c 2 – c 3 (11.11c)

x 3 = c 0 + c 1 + c 2 + c 3

x 4 = c 0 – c 1 + c 2 – c 3

Solving Eqs. (11.11c) gives

c 0 =^14 (x 1 + x 2 + x 3 + x 4 ) c 1 =^14 (–x 1 + x 2 + x 3 – x 4 ) (11.11d)

c 2 =^14 (–x 1 – x 2 + x 3 + x 4 ) c 3 =^14 (x 1 – x 2 + x 3 – x 4 )

Substituting the coefficients in Eq. (11.11b) and rearranging, we get

x =^1

4

(1 – ξ)(1 – η)x 1 +^1

4

(1 + ξ)(1 – η)x 2 +^1

4

(1 + ξ)(1 + η)x 3 +^1

4

(1 – ξ)(1 + η)x 4 (11.11e)

or x = N 1 (ξ,η)x 1 + N 2 (ξ,η)x 2 + N 3 (ξ,η)x 3 + N 4 (ξ,η)x 4 (11.11f)

whereN 1 (ξ,η),... , N 4 (ξ,η) are the shape functions. Note that within the square, all shape functions
are greater than 0. At node 1, N 1 (–1, –1) = 1 while all the other functions are zero. Likewise, at node
2 only N 2 (1, –1) = 1, and so on. To map the y coordinate of P on the square, we can proceed in a
manner similar to steps given in Eqs. (11.11b)-(11.11f) to get

y = N 1 (ξ,η)y 1 + N 2 (ξ,η)y 2 + N 3 (ξ,η)y 3 + N 4 (ξ,η)y 4 (11.11g)
Eqs. (11.11f) and (11.11g), therefore, generically interpolate any nodal information onto a point
interior to the quadrilateral element. We can use identical functions to interpolate the u and v
displacements at point P, that is,

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

v = N 1 (ξ,η)v 1 + N 2 (ξ,η)v 2 + N 3 (ξ,η)v 3 + N 4 (ξ,η)v 4

which can be combined in the matrix form as

u NN N N

NNN N

uu u uT

v

vvvv

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣

⎢

⎤

⎦

= ⎥

0000

00 0 0

[ ] =

12 3 4

1234

11 2 2 3 3 4 4 Nu (11.11h)