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Substitute these relationships into Eq. (7.71a)
*
0
2
6
!
L
3
1
0
EI
2
]
k
u
=
[026
ξ
12
ξ
d
ξ
ξ
*
"
12
ξ
2
00 0 0
04 6 8
0612 18
0818
14
5
EI
L
3
=
(3)
ξ
ξ
1
/
6
p
0
L
1
0
(
2
1
/
12
p
u
=
1
−
ξ)
d
ξ
=
p
0
L
(4)
3
ξ
1
/
20
4
ξ
1
/
30
Turn now to the boundary relation
R
of Eq. (7.70). Recall that the terms in
R
are taken from
]
0
[
−
V
δw
−
M
δθ
on
S
u
.
The
S
u
boundary conditions are
w(
0
)
=
0
,
θ(
0
)
=
0
,
and
w(
L
)
=
0
,
and, hence, only those terms corresponding to
δw(
0
)
,
δθ(
0
)
,
and
δw(
L
)
are retained in
R
.
Thus,
R
becomes
EI
N
u
(
)
N
u
(
N
u
(
N
u
(
N
u
N
u
(
R
=
L
)
L
)
−
0
)
0
)
+
(
0
)
0
*
1
1
1
1
0
0
0
0
1
0
0
0
!
EI
L
3
=
[0
0
6
24]
−
[0060]
+
[0200]
*
"
02624
00624
00624
00624
EI
L
3
=
(5)
R
T
The system of equations for
w
,
i.e., [
k
u
+
(
R
+
)
]
w
−
p
u
=
0
,
becomes
026 4
2 4 12 32
6 2 4 8
24
w
1
w
2
w
3
w
4
1
/
6
=
EI
L
3
1
/
12
p
0
L
1
/
20
(6)
32
48
76
.
8
1
/
30
k
w
=
p
u
which has the solution
−
0
.
4167
p
0
L
4
120
EI
4
.
0000
=
w
(7)
−
5
.
2916
1
.
8229
As a second case, use a trial function with five unknown parameters
2
3
4
][
5
]
T
w(
)
=
ξξ
ξ
ξ
w
w
w
w
w
x
[1
(8)
1
2
3
4
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