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the extended Ritz method gives
p
0
L
4
90
EI
(
2
3
5
6
w
=
3
ξ
−
5
ξ
+
3
ξ
−
ξ
)
2
3
5
6
=
0
.
2304
ξ
−
0
.
384
ξ
+
0
.
2304
ξ
−
0
.
0768
ξ
which is equal to the exact solution.
7.18 Prove that the stationary value of the functional
(
u
)
=
L
(
uLu
−
2
u f
)
dx
is equivalent to the solution of the differential equation
Lu
−
f
=
0
.
Hint:
Introduce [Appendix I, Eq. (I.3)]
u
=
u
+
η.
Then
(
u
)
=
[
(
u
+
η)
L
(
u
+
η)
−
2
(
u
+
η)
f
]
dx
L
=
0
=
Set
d
()
d
=
().
0
d
d
d
d
If
L
is linear,
L
(
u
+
η)
=
L
(
u
+
η)
. First, carry out the differentiation and
then set
=
0
.
L
(
uL
η
+
η
Lu
−
2
η
f
)
dx
=
0
If
L
is self-adjoint,
L
2
(
−
)η
=
Lu
f
dx
0 and the fundamental lemma of Appendix
I implies that
Lu
−
f
=
0
.
7.19 Prove that the stationary value of the functional
V
(
u
T
Lu
2
u
T
f
(
u
)
=
−
)
dV
is equivalent to the solution of the differential equation
Lu
−
f
=
0
.
Hint:
Follow the procedure outlined in the hint of Problem 7.18.
7.20 Find the response of the beam of Fig. P7.20 using the simple Galerkin's method and
the extended Galerkin's method. Compare the results with the exact solution. The
applied distributed load should be represented by a second order polynomial.
Hint:
See Problem 7.17 for the loading distribution.
FIGURE P7.20
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