3.4 Solved Problems
65
Fig. 3.20 Modeling
approach to consider a single
force in the fourth-order
differential equation based
on n grid points
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = 0 ,
(3.122)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = 0 ,
(3.123)
node 4:
E I Y
X 3 (u 6 − 4u 5 + 6u 4 − 4u 3 + u 2 ) = 0 ,
(3.124)
node 5:
E I Y
X 3 (u 7 − 4u 6 + 6u 5 − 4u 4 + u 3 ) = 0 ,
(3.125)
node 6:
E I Y
X 3 (u 8 − 4u 7 + 6u 6 − 4u 5 + u 4 ) = 0 ,
(3.126)
node 7:
E I Y
X 3 (u 9 − 4u 8 + 6u 7 − 4u 6 + u 5 ) = −F 0 .
(3.127)
It should be noted here that Eqs. (3.124)–(3.125), i.e., the equations with the gray
background, are not affected by any boundary or fictitious nodes. These equations will
help us later to construct a scheme for a larger number of nodes (n > 6). The vertical
displacement is zero at left-hand ends and it can be immediately concluded that
u 1 = 0. Using the condition for the rotation at node 1 (
du
dX
1
= 0) and the conditions
for the internal bending moment ( M Y | 7 = 0) and the shear force ( Q Z | 7 = −F 0 )
at node 7 (see example Problem 3.1 for details of the derivations), one can deduct
the following additional condition based on centered difference schemes for the
derivatives: u 0 = u 2 , u 6 = 2u 5 − u 4 , and u 7 = 4u 5 − 4u 4 + u 3 + 0.
Thus, Eqs. (3.122)–(3.127) can be rearranged to give under consideration of the
values of the boundary and fictitious nodes
node 2: 7u 2 − 4u 3 + u 4 + 0 + 0 + 0 = 0 ,
(3.128)
node 3: − 4u 2 + 6u 3 − 4u 4 + u 5 + 0 + 0 = 0 ,
(3.129)
node 4: u 2 − 4u 3 + 6u 4 − 4u 5 + u 6 + 0 = 0 ,
(3.130)
node 5: 0 + u 3 − 4u 4 + 6u 5 − 4u 6 + u 7 = 0 ,
(3.131)