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Substitution of these expressions into Eq. (11.76), with N 0
=−
P , leads to
v T [ EI G T k EI G
P G T k P G ] v
δ
W
= δ
+
=
0
(2)
with
2
3
4
2
3
3
2
12
5
2
ξ
6
ξ
12
ξ
1
L 2 1
1
1
L
N u N u Ld
3
4
5
k P
=
ξ =
2
ξ
6
ξ
12
ξ
Ld
ξ =
1
2
6
5
(3)
2
0
0
4
5
6
2
ξ
6
ξ
12
ξ
2
5
12
7
1
2
00 8
00 6
0024
0024
ξ
1
L 4 1
1
1
L 3
N u N i u
3
k EI
=
Ld
ξ =
0024
ξ
Ld
ξ =
(4)
0
0
4
0024
ξ
/
5
0857 L 3
This gives 1
.
8 EIL
P 0
.
=
0or
1
.
8
EI
L 2 =
21 EI
L 2
P cr
=
(5)
0
.
0857
Consider a two-term assumed displacement that satisfies all the boundary conditions.
2
3
4
3
4
5
w = (
3
ξ
/
2
5
ξ
/
2
+ ξ
)w 1 + (
4
ξ
/
3
7
ξ
/
3
+ ξ
)w 2
(6)
For this assumed displacement,
3
/
2
0
w 1
w
G
/
/
5
24
3
v
=
=
(7)
1
7
/
3
2
0
1
=
ξ
2
ξ
3
ξ
4
ξ
5 ]
N u
[
N u =
[ 2
ξ
3
ξ
2
4
ξ
3
5
ξ
4 ]
/
L
N u =
2
3 ]
L 2
[26
ξ
12
ξ
20
ξ
/
(8)
N u =
2 ]
L 3
[0624
ξ
60
ξ
/
N i u =
L 4
[0024120
ξ
]
/
2
PL 2
With these expressions and
λ = ε
=
/
EI , Eq. (11.76) leads to
1
0
0
0
.
80
0
.
80
.
0857
0
.
0429
λ
=
(9)
0
.
80
0
.
6095
0
.
0429
0
.
0254
λ
2
λ +
.
=
.
or
8714
1359
2
0
Then
35 EI
L 2
P cr =
20
.
(10)
L 2
as compared to the exact solution of P cr
=
20
.
19 EI
/
.
Note that Galerkin's method appears
to provide an upper bound to the exact solution.
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