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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
(θ
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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