6.3 Answers for Problems from Chap. 3
135
Table 6.2 Values of the inner bending moment at the grid points
Grid point
Coordinate X
M Y (X )
1
0
3F0 L
2
2
L
4
F 0 L
3
L
2
F0 L
2
4
3L
4
F0 L
4
5
L
0
3.19 Finite difference approximation of a stepped cantilevered Euler–Bernoulli
beam with two single forces based on five domain nodes
The internal bending moment distribution is obtained as
M y (X ) = F 0
L
2
− X
+ F 0 (L − X ) for 0 ≤ X ≤
L
2
,
(6.127)
M Y (X ) = F 0 (L − X ) for
L
2
≤ X ≤ L ,
(6.128)
whereas the values at the five grid points are collected in Table 6.2.
Equation (3.27) must be evaluated for four different grid points since we have
four unknown displacements (u 2 , . . . , u 5 ) to determine. Considering grid points i =
2, . . . , 5 introduces the fictitious node 6 at the right-hand side. However, this node
cannot be eliminated based on the moment relation since we used this relationship
already for node 5. Furthermore, the shear force relation would introduce a further
fictitious node. Thus, it is recommended to state Eq. (3.27) for the grid points i =
1, . . . , 4:
node 1:
E2I Y
X 2 (u 2 − 2u 1 + u 0 ) = −
3F 0 L
2
,
(6.129)
node 2:
E2I Y
X 2 (u 3 − 2u 2 + u 1 ) = −F 0 L ,
(6.130)
node 3:
E
2I Y +I Y
2
X 2 (u 4 − 2u 3 + u 2 ) = −
F 0 L
2
,
(6.131)
node 4:
E I Y
X 2 (u 5 − 2u 4 + u 3 ) = −
F 0 L
4
,
(6.132)
or in matrix notation under consideration of the conditions at the left-hand boundary,
i.e., u 1 = 0 and u 0 = u 2 :
135
Table 6.2 Values of the inner bending moment at the grid points
Grid point
Coordinate X
M Y (X )
1
0
3F0 L
2
2
L
4
F 0 L
3
L
2
F0 L
2
4
3L
4
F0 L
4
5
L
0
3.19 Finite difference approximation of a stepped cantilevered Euler–Bernoulli
beam with two single forces based on five domain nodes
The internal bending moment distribution is obtained as
M y (X ) = F 0
L
2
− X
+ F 0 (L − X ) for 0 ≤ X ≤
L
2
,
(6.127)
M Y (X ) = F 0 (L − X ) for
L
2
≤ X ≤ L ,
(6.128)
whereas the values at the five grid points are collected in Table 6.2.
Equation (3.27) must be evaluated for four different grid points since we have
four unknown displacements (u 2 , . . . , u 5 ) to determine. Considering grid points i =
2, . . . , 5 introduces the fictitious node 6 at the right-hand side. However, this node
cannot be eliminated based on the moment relation since we used this relationship
already for node 5. Furthermore, the shear force relation would introduce a further
fictitious node. Thus, it is recommended to state Eq. (3.27) for the grid points i =
1, . . . , 4:
node 1:
E2I Y
X 2 (u 2 − 2u 1 + u 0 ) = −
3F 0 L
2
,
(6.129)
node 2:
E2I Y
X 2 (u 3 − 2u 2 + u 1 ) = −F 0 L ,
(6.130)
node 3:
E
2I Y +I Y
2
X 2 (u 4 − 2u 3 + u 2 ) = −
F 0 L
2
,
(6.131)
node 4:
E I Y
X 2 (u 5 − 2u 4 + u 3 ) = −
F 0 L
4
,
(6.132)
or in matrix notation under consideration of the conditions at the left-hand boundary,
i.e., u 1 = 0 and u 0 = u 2 :