Page 546 - Aircraft Stuctures for Engineering Student

12.8 Finite element method for continuum structures 527

Then, from Eqs (iii), (iv) and (v)
2u3 - u1 - u2
a3 =
4
Substituting for al, a2 and a3 in the first of Eqs (12.82) gives

-
u=ul+(7)x+( u1 2u3 - 4 UI - 2.42 )Y
u2
or
u= (1-,-++ (;-$)"f;u3 V
X
(vii)
Similarly

(viii)

Now from Eqs (12.88)

du u1 u2
&, = - = -- + -
ax 4 4

and

Hence

- du
-
dX
av
- k (ix)
dY 4 v2
dtl av
-+- -1 -1 -1 1 2 0 u3
-ay ax -

Also

Hence

-a -b a -b 0 2b
-b -U b -U 0 2a 1
--c -c --c c 2c 0