132
6 Answers to Supplementary Problems
Special case k = 4, E I Y = 1 and L = 1:
u 3 = u
L
2
= −0.0224451F 0 .
(6.105)
Relative error (analytical solution cf. [1]):
relative error =
−0.0224451F 0 + 0.0200233F 0
−0.0200233F 0
× 100 = 12.094% . (6.106)
3.16 Finite difference approximation of a simply supported beam with varying
bending stiffness—constant distributed load
With the bending stiffness and bending moment distribution, the following differential equation can be written:
E I 0
1 +
2X
L
− 1
2 ×
d
2 u Z (X )
dX 2 =
q 0 X
2
(L − X ) .
(6.107)
The following finite difference scheme can be derived:
u i+1 − 2u i + u i−1 = X
2
×
q 0 X i
2
(L − X i ) ×
1 +
2X i
L
− 1
2
E I 0
.
(6.108)
Under consideration of the boundary conditions, i.e. u 1 = u 5 = 0, the following
system of equations is obtained:
⎡
⎢
⎢
⎣
−2 1 0
1 −2 1
0 1 −2
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
u 2
u 3
u 4
⎤
⎥
⎥
⎦ =
q 0 L
4
E I 0
⎡
⎢
⎢
⎣
15
2048
1
128
15
2048
⎤
⎥
⎥
⎦ .
(6.109)
Solution:
u 2 = −
23q 0 L
4
2048E I 0
, u 3 = −
31q 0 L
4
2048E I 0
, u 4 = −
23q 0 L
4
2048E I 0
.
(6.110)
Analytical solution:
u Z (X ) = −
q 0
E I 0 L 2
1
15
X
6
−
1
12
X
4 L
2
−
1
6
X
3 L
3
+ c 1 X + C 2 .
(6.111)
Consideration of the boundary conditions gives c 2 = 0 and c 1 = −
11q 0 L
3
60E I 0
. Based on
the analytical solution, the displacement in the middle of the beam is obtained as: