6.3 Answers for Problems from Chap. 3
131
E I Y
d
2 u
dX 2 = −M Y → (E I Y ) i
u i+1 − 2u i + u i−1
X 2
= −M Z ,i .
(6.97)
→ M Y,2 = M Y,4 = +
F 0 L
8
, M Y,3 = +
F 0 L
4
.
(6.98)
The boundary values, i.e. M Y,1 = M Y,5 = 0 result from the BC or from a forward
(i = 1) or backward (i = 5) scheme.
FD shear force distribution:
E I Y
d
3 u
dX 3 = −Q Z → (E I Y ) i
u i+2 − 2u i+1 + 2u i−1 − u i−2
2X 3
= −Q Z ,i . (6.99)
Q Z ,3 = −
64E I Y
2L 3
u 5
0
−2u 4 + 2u 2 − u 1
0
= 0 .
(6.100)
The boundary values, i.e. Q Z ,1 = 2F 0 and Q Z ,5 = −2F 0 result from a forward (i =
1) or backward (i = 5) scheme, respectively. The values Q Z ,3 and Q Z ,4 cannot be
calculated based on the actual subdivision.
Analytical solution (0 ≤ X ≤
L
2
):
u(X ) =
F 0
48E I Y
3L
2 X − 4X
3
, M Y (X ) = +
F 0 X
2
, Q y (X ) = +
F 0
2
. (6.101)
3.15 Finite difference approximation of a simply supported beam on elastic
foundation based on five domain nodes
The single force F 0 is understood as the integral value of a distributed load q 0 , which
is acting over a length of X .
Finite difference approximation of the partial differential equation:
E I Y ×
u i+2 − 4u i+1 + 6u i − 4u i−1 + u i−2
X 3
+ ku i X = R i .
(6.102)
Linear system of equations:
⎡
⎢
⎢
⎣
5 +
kX
4
E I Y
−4
1
−4
6+
kX
4
E I Y
−4
1
−4
5+
kX
4
E I Y
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
u 2
u 3
u 4
⎤
⎥
⎥
⎦ =
⎡
⎢
⎢
⎣
0
−
F 0 X
3
E I Y
0
⎤
⎥
⎥
⎦ .
(6.103)
Solution:
u 3 = u
L
2
= −
6E I Y +
kL
4
256
L
3 F 0
64
4(E I Y ) 2 +
3E I Y kL 4
64
+
k 2 L 8
65536
.
(6.104)