2.3 Varying Material and Geometry Parameters
21
node 4:
E
X
−
A 3 + A 4
2
u 3 +
A 3 + A 4
2
+
A 4 + A 5
2
u 4 −
A 4 + A 5
2
u 5
= 0 .
(2.42)
The consideration of the displacement boundary conditions according to Eq. (2.30),
i.e. u 1 = 0 and u 5 = u 0 , in the three last equations and rearranging the set of equations
in matrix form results in:
E
2X
⎡
⎢
⎣
A 1 + 2 A 2 + A 3 −(A 2 + A 3 )
0
−(A 2 + A 3 ) A 2 + 2 A 3 + A 4 −(A 3 + A 4 )
0
−(A 3 + A 4 ) A 3 + 2 A 4 + A 5
⎤
⎥
⎦
⎡
⎢
⎣
u 2
u 3
u 4
⎤
⎥
⎦ =
⎡
⎢
⎣
0
0
E
2X
(A 4 + A 5 )u 0
⎤
⎥
⎦ .
(2.43)
The same system of equations would be obtained based on a finite element approach
with linear elements, see [8].
If the boundary condition at the right-hand end is given as a single force F 0 , one
may follow first of all the idea based on a fictitious node n + 1 outside the structure
as shown in the previous section in Fig. 2.5. The evaluation of node 5 based on the
general scheme (2.39) gives
node 5:
E
X
−
A 4 + A 5
2
u 4 +
A 4 + A 5
2
+
A 5 + A 6
2
u 5 −
A 5 + A 6
2
u 6
= 0 .
(2.44)
If the centered difference scheme from Table 1.1 is used, the gradient at the boundary
node together with the force equilibrium, i.e. N 5 = F 5 , reads as
du
dX
5
=
u 6 − u 4
2X
=
F 0
E A 5
.
(2.45)
The last equation can be rearranged for u 6 and this relationship can be introduced
into the evaluation of node 5 according to Eq. (2.44). This gives finally an expression
for the displacement u 5 in the form:
u 5 = u 4 +
2F 0 X
E
A 5 + A 6
A 5 (A 4 + 2 A 5 + A 6 )
.
(2.46)
The last equation still contains the fictitious area A 6 and might be not the best
approach to solve the problem.