Page 548 - Aircraft Stuctures for Engineering Student

12.8 Finite element method for continuum structures 529

displacement and force vectors are

, (12.95)

As in the case of the triangular element we select a displacement function which
satisfies the total of eight degrees of freedom of the nodes of the element; again this
displacement function will be in the form of a polynomial with a maximum of eight
coefficients. Thus
u(x, v) = a1 + a2x + a3y + a4xy 1 (12.96)
v(x,y) = a5 + (Y6x f a7y f a8Xy
The constant terms, al and a5? are required, as before, to represent the in-plane rigid
body motion of the element while the two pairs of linear terms enable states of con-
stant strain to be represented throughout the element. Further, the inclusion of the xy
terms results in both the u(x, y) and v(x, y) displacements having the same algebraic
form so that the element behaves in exactly the same way in the x direction as it does
in the y direction.
Writing Eqs (12.96) in matrix form gives

{ ::::;;} = [ 000 0 lxyxy (12.97)
1
1xyxyoooo

ff7
or

(12.98)

Now substituting the coordinates and values of displacement at each node we obtain
-1 xi yi 0 0 0 0
0 0 0 1 xi yi Xi-vi
1 xj Yj 0 0 0 0
0 0 0 1 xj yj xjyj (12.99)
0 0 0 0 b
xk Yk
0 0 0 xk Yk XkYk
1 XI Yi 0 0 0 0
0 0 0 1 XI Yl -w4