134
6 Answers to Supplementary Problems
Table 6.1 Values of the distributed load at the grid points
Grid point
Coordinate X
q Z (X )
1
0
−αq 0
2
L
4
−
3
4 α +
1
4 β
q 0
3
L
2
−
1
2 α +
1
2 β
q 0
4
3L
4
−
1
4 α +
3
4 β
q 0
5
L
−βq 0
3.18 Finite difference approximation of a simply supported beam loaded by a
linearly distributed load
The function of the distributed load can be expressed as
q Z (X ) = −q 0
α + (β − α)
X
L
,
(6.121)
whereas the values at the five grid points are collected in Table 6.1.
Evaluation of the finite difference approximation of the fourth-order differential
equation according to Eq. (3.9) at the inner nodes i = 2, . . . , 4 and consideration of
the boundary conditions gives:
node 2:
E I Y
X 3 (5u 2 − 4u 3 + u 4 ) = −X
3
4
α +
1
4
β
q 0 ,
(6.122)
node 3:
E I Y
X 3 (−4u 2 + 6u 3 − 4u 4 ) = −X
1
2
α +
1
2
β
q 0 ,
(6.123)
node 4:
E I Y
X 3 (u 2 − 4u 3 + 5u 4 ) = −X
1
4
α +
3
4
β
q 0 .
(6.124)
or in matrix notation with X = L/4:
⎡
⎣
5 −4 1
−4 6 −4
1 −4 5
⎤
⎦
⎡
⎣
u 2
u 3
u 4
⎤
⎦ = −
q 0 L
4
256E I Y
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
3
4
α +
1
4
β
1
2
α +
1
2
β
1
4
α +
3
4
β
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
.
(6.125)
The solution of this linear system of equations gives the unknown nodal values as:
⎡
⎣
u 2
u 3
u 4
⎤
⎦ = −
q 0 L
4
4096E I Y
⎡
⎣
21α + 19β
28α + 28β
19α + 21β
⎤
⎦ .
(6.126)
6 Answers to Supplementary Problems
Table 6.1 Values of the distributed load at the grid points
Grid point
Coordinate X
q Z (X )
1
0
−αq 0
2
L
4
−
3
4 α +
1
4 β
q 0
3
L
2
−
1
2 α +
1
2 β
q 0
4
3L
4
−
1
4 α +
3
4 β
q 0
5
L
−βq 0
3.18 Finite difference approximation of a simply supported beam loaded by a
linearly distributed load
The function of the distributed load can be expressed as
q Z (X ) = −q 0
α + (β − α)
X
L
,
(6.121)
whereas the values at the five grid points are collected in Table 6.1.
Evaluation of the finite difference approximation of the fourth-order differential
equation according to Eq. (3.9) at the inner nodes i = 2, . . . , 4 and consideration of
the boundary conditions gives:
node 2:
E I Y
X 3 (5u 2 − 4u 3 + u 4 ) = −X
3
4
α +
1
4
β
q 0 ,
(6.122)
node 3:
E I Y
X 3 (−4u 2 + 6u 3 − 4u 4 ) = −X
1
2
α +
1
2
β
q 0 ,
(6.123)
node 4:
E I Y
X 3 (u 2 − 4u 3 + 5u 4 ) = −X
1
4
α +
3
4
β
q 0 .
(6.124)
or in matrix notation with X = L/4:
⎡
⎣
5 −4 1
−4 6 −4
1 −4 5
⎤
⎦
⎡
⎣
u 2
u 3
u 4
⎤
⎦ = −
q 0 L
4
256E I Y
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
3
4
α +
1
4
β
1
2
α +
1
2
β
1
4
α +
3
4
β
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
.
(6.125)
The solution of this linear system of equations gives the unknown nodal values as:
⎡
⎣
u 2
u 3
u 4
⎤
⎦ = −
q 0 L
4
4096E I Y
⎡
⎣
21α + 19β
28α + 28β
19α + 21β
⎤
⎦ .
(6.126)
