Page 550 - Aircraft Stuctures for Engineering Student

12.8 Finite element method for continuum structures 531

whence, from either of Eqs (v) or (vi)
a' = -0.000125 (viii)
Adding Eqs (i) and (ii)

CY1 - a3 = 0.002
Adding Eqs (iii) and (iv)
CX~ + a3 = -0.0015
Then adding Eqs (ix) and (x)
(~1 = 0.00025
and, from either of Eqs (ix) or (x)

a3 = -0.00175 (xii)
The second of Eqs (12.96) is used to determine a5, Q6, a7, (Yg in an identical manner to
the above. Thus
a5 = -0.001

06 = 0.00025
a7 = 0.002
a8 = -0.00025

Now substituting for q, CX~, . . . , Qg in Eqs (12.96)
ui = 0.00025 - 0.000125~ - 0.00175~ - 0.000625~~
and
~j = -0.001 + 0.00025~ + 0.002~ - 0.00025~~
Then, from Eqs (12.88)
au
Ex=-- - -0.000125 - 0.000625j~
ax
av
E,. = - = 0.002 - 0.00025~
aY
au av
yxy = - + - = -0.0015 - 0.000625~ - 0.00025~
ay ax
Therefore, at the centre of the element (x = 0, y = 0)

E, = -0.000125
E). = 0.002
rx,, = -0.0015
so that, from Eqs (12.92)
E
2ooooo (-0.000125 + (0.3 x 0.002))
= -(E.y + V&J = -
1-9 1-0.3'