2.4 Solved Problems
31
Fig. 2.12 Finite difference discretization of the stepped rod
node 2:
E
2 (−2 Au 1 + (2 A + 2 A)u 2 − 2 Au 3 ) = 0 ,
(2.86)
node 3:
E
X 2 (−2 Au 2 + (2 A + A)u 3 − Au 4 ) = 0 ,
(2.87)
node 4:
E
X 2 (−Au 3 + (A + A)u 4 − Au 5 ) = 0 ,
(2.88)
node 5:
E
X 2 (−Au 4 + (A + A)u 5 − Au 6 ) = 0 .
(2.89)
The horizontal displacement is zero at the left-hand end and it can be immediately
concluded that u 1 = 0. Equation (2.89) contains still the displacement of the fictitious
node 6 and a balance between the internal normal force (N X ) and the external load
(F 0 ) at node 5 allows to derive a relationship to eliminate u 6 (see Example 2.1 for
details):
u 6 = u 4 +
2X F 0
E 5 A 5
.
(2.90)
Introducing the last relation in Eq. (2.89), the following matrix scheme can be stated:
⎡
⎢
⎢
⎣
4 A −2 A 0
0
−2 A 3A −A 0
0 −A 2 A −A
0
0 −2 A 2 A
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
u 2
u 3
u 4
u 5
⎤
⎥
⎥
⎦ =
X F 0
E
⎡
⎢
⎢
⎣
0
0
0
2
⎤
⎥
⎥
⎦ ,
(2.91)
or
⎡
⎢
⎢
⎣
4 −2 0 0
−2 3 −1 0
0 −1 2 −1
0 0 −2 2
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
u 2
u 3
u 4
u 5
⎤
⎥
⎥
⎦ =
X F 0
E A
⎡
⎢
⎢
⎣
0
0
0
2
⎤
⎥
⎥
⎦ .
(2.92)