6.3 Answers for Problems from Chap. 3
133
u Z
L
2
= −
q 0 L
4
15E I 0
.
(6.112)
relative error =
−
31
2048
+
1
15
−
1
15
× 100 = −77.295% .
(6.113)
3.17 Finite difference approximation of a simply supported beam with varying
bending stiffness—constant bending moment
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 = M 0 .
(6.114)
The following finite difference scheme can be derived:
u i+1 − 2u i + u i−1 = X
2
× M 0 ×
1 +
2X i
L
− 1
2
E I 0
.
(6.115)
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
⎤
⎥
⎥
⎦ =
M 0 L
2
E I 0
⎡
⎢
⎢
⎣
5
64
1
16
5
64
⎤
⎥
⎥
⎦ .
(6.116)
Solution:
u 2 = −
7M 0 L
2
64E I 0
, u 3 = −
9M 0 L
2
64E I 0
, u 4 = −
7M 0 L
2
64E I 0
.
(6.117)
Analytical solution:
u Z (X ) =
2M 0
E I 0 L 2
1
2
X
2 L
2
−
1
3
X
3 L +
1
6
X
4
+ c 1 X + C 2 .
(6.118)
Consideration of the boundary conditions gives c 2 = 0 and c 1 = −
2M 0 L
E I 0
. Based on
the analytical solution, the displacement in the middle of the beam is obtained as:
u Z
L
2
= −
31M 0 L
2
48E I 0
.
(6.119)
relative error =
−
9
64
+
31
48
−
31
48
× 100 = −78.226% .
(6.120)