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FIGURE 8.6
Fixed, simply supported beam with ramp loading.
Use a simple central difference formula from Table 8.1 (Eq. 8.2)
i
v
h 4
w
= (
1
/
)(w i 2
4
w i 1 +
6
w i
4
w i + 1 + w i + 2 )
(3)
i
For p i
of (2), we simply use the value of the distributed load at node i
.
From (2) and (3), we
obtain the system of equations
p 0 h 4
i
=
1:
w 1
4
w 0 +
6
w 1
4
w 2 + w 3
= (
4
/
5
)(
)/
EI
p 0 h 4
i
=
2:
w 0
4
w 1 +
6
w 2
4
w 3 + w 4
= (
3
/
5
)(
)/
EI
(4)
p 0 h 4
i
=
3:
w 1
4
w 2 +
6
w 3
4
w 4 + w 5
= (
2
/
5
)(
)/
EI
p 0 h 4
i
=
4:
w 2
4
w 3 +
6
w 4
4
w 5 + w 6 = (
1
/
5
)(
)/
EI
The boundary conditions will reduce the number of unknowns in (4). Since the displace-
ments are zero at the ends,
w 0 = w 5 =
0
.
The displacement
w 1 at the fictitious node can be
w 0 =
identified using the slope condition
0
0 =
0
).
The simple central difference formula
w 0
for
is
w 0 = (
1
/
2 h
)( w 1
+ w
)
(5)
1
w 0 =
The condition
0 gives
w 1 = w 1
(6)
w 5 =
From the moment boundary condition
0 and Eq. (8.4), we find
w 5 =
h 2
0
= (
1
/
)(w
2
w
+ w
)
or
w
=− w
(7)
4
5
6
6
4
Physically, conditions (6) and (7) would appear as in Fig. 8.7.
Equation (4) reduces to
=
7
410
w
4
3
2
1
1
h 4
EI
p 0
5
46
41
w
2
(8)
1
46
4
w
3
01
45
w
4
Although this matrix is symmetric, in general finite difference matrices are not. The solution
to (8) is
w 1
w
0
.
001243
=
p 0 L 4
EI
0
.
002553
2
(9)
w
0
.
002793
3
w
0
.
001788
4
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