6.2 Answers for Problems from Chap. 2
123
Fig. 6.5 Finite difference
discretization of the
bi-material rod
u II (X ) =
1
k II
−
k II
2(k I + k II )
p 0 L +
k I k II
k I + k II
u 0
L
X
+
k II
k I + k II
p 0 L
2
+
k II (k II − k I )
k I + k II
u 0
.
(6.47)
Assigning the specific values k I = 2k II = 1, L I = L II = 1, p 0 = 1, and u 0 = 1,
Eqs. (6.46) and (6.47) can be simplified to the following expression:
u I (X ) =
7X
6
−
X
2
2
,
(6.48)
u II (X ) =
X
3
+
1
3
.
(6.49)
The finite difference discretization of the bi-material rod is shown in Fig. 6.5 for five
domain nodes.
The evaluation of the finite difference approximation according to Eq. (2.56) at
the inner nodes i = 2, . . . , 5 gives:
node 2:
1
X
(0 + (E A) I (u 3 − 2u 2 + u 1 )) = −p 0 ,
(6.50)
node 3:
1
X
(E A) II − (E A) I
4
(u 4 − u 2 ) +
(E A) I + (E A) II
2
(u 4 − 2u 3 + u 2 )
=
− p 0
2
= −R ,
(6.51)
node 4:
1
X
(0 + (E A) II (u 5 − 2u 4 + u 3 )) = 0 ,
(6.52)
or in matrix notation under consideration of the boundary conditions, i.e., u 1 = 0
and u 5 = u 0 :
⎡
⎢
⎢
⎢
⎢
⎢
⎣
−2
1
0
3(E A) I +(E A) II
4
−((E A) I + (E A) II )
(E A) I +3(E A) II
4
0
1
−2
⎤
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎣
u 2
u 3
u 4
⎤
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
−
p 0
2
(E A) I
−
p 0
2
2
−u 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
. (6.53)
123
Fig. 6.5 Finite difference
discretization of the
bi-material rod
u II (X ) =
1
k II
−
k II
2(k I + k II )
p 0 L +
k I k II
k I + k II
u 0
L
X
+
k II
k I + k II
p 0 L
2
+
k II (k II − k I )
k I + k II
u 0
.
(6.47)
Assigning the specific values k I = 2k II = 1, L I = L II = 1, p 0 = 1, and u 0 = 1,
Eqs. (6.46) and (6.47) can be simplified to the following expression:
u I (X ) =
7X
6
−
X
2
2
,
(6.48)
u II (X ) =
X
3
+
1
3
.
(6.49)
The finite difference discretization of the bi-material rod is shown in Fig. 6.5 for five
domain nodes.
The evaluation of the finite difference approximation according to Eq. (2.56) at
the inner nodes i = 2, . . . , 5 gives:
node 2:
1
X
(0 + (E A) I (u 3 − 2u 2 + u 1 )) = −p 0 ,
(6.50)
node 3:
1
X
(E A) II − (E A) I
4
(u 4 − u 2 ) +
(E A) I + (E A) II
2
(u 4 − 2u 3 + u 2 )
=
− p 0
2
= −R ,
(6.51)
node 4:
1
X
(0 + (E A) II (u 5 − 2u 4 + u 3 )) = 0 ,
(6.52)
or in matrix notation under consideration of the boundary conditions, i.e., u 1 = 0
and u 5 = u 0 :
⎡
⎢
⎢
⎢
⎢
⎢
⎣
−2
1
0
3(E A) I +(E A) II
4
−((E A) I + (E A) II )
(E A) I +3(E A) II
4
0
1
−2
⎤
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎣
u 2
u 3
u 4
⎤
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
−
p 0
2
(E A) I
−
p 0
2
2
−u 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
. (6.53)
