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Thus,
k
u
w
=
p
u
(7.65b)
as is given in Eq. (7.59).
EXAMPLE 7.7 Beam with Linearly Varying Loading
Consider again the fixed-hinged beam of Fig. 7.1.
Use the single parameter polynomial approximation that was derived in Example 7.1
(
m
=
1
)
2
3
4
w
=
N
u
)w
1
(1)
As noted in Example 7.1, this satisfies all of the boundary conditions. Galerkin's relationship
of Eq. (7.65) in terms of
w
=
(
3
ξ
−
5
ξ
+
2
ξ
ξ
appears as
L
EI
N
u
N
i
u
dx
p
0
L
=
N
u
dx
−
0
(
−
ξ)
w
1
0
w
T
δ
(2)
0
k
u
p
u
Introducing
N
i
u
=
/
L
4
.
48
1
0
δw
1
0
δw
48
L
4
EI
2
3
4
2
3
4
(
3
ξ
−
5
ξ
+
2
ξ
)w
1
Ld
ξ
−
1
p
0
(
1
−
ξ)(
3
ξ
−
5
ξ
+
2
ξ
)
Ld
ξ
=
0
(3)
1
leads to
p
0
L
4
108
EI
48
EI
L
3
3
20
w
1
15
p
0
L
·
=
or
w
=
1
1
Finally, the approximate deflection is
p
0
L
4
108
EI
(
2
3
4
w
=
3
ξ
−
5
ξ
+
2
ξ
)
(4)
Let's continue this problem by using an assumed deflection with two terms
(
m
=
2
).
Introduce
2
3
4
2
3
4
5
w
=
w
(
3
ξ
−
5
ξ
+
2
ξ
)
+
w
(ξ
+
ξ
−
4
ξ
+
2
ξ
)
1
2
(5)
N
u
1
N
u
2
Note that all boundary conditions are still satisfied. We proceed to establish Galerkin's
condition (2) with
N
u
=
[
N
u
1
N
u
2
]
,
w
=
[
w
1
w
2
]
T
.
The matrices
k
u
and
p
u
are found to be
3
2
3
4
ξ
−
5
ξ
+
2
ξ
L
4
1
[48
−
96
+
240
ξ
]
Ld
ξ
EI
2
3
4
5
ξ
+
ξ
−
4
ξ
+
2
ξ
k
u
=
N
i
u
0
N
u
L
3
7
.
25
.
6
EI
=
(6)
5
.
65
.
3714
p
0
L
0
.
06
p
u
=
(7)
0
.
0476
Finally,
k
u
w
=
p
u
gives
w
0
p
0
L
4
EI
.
0125
1
w
=
=
(8)
w
−
0
.
00417
2
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