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For each case, the values of the displacements at the boundaries are
A. The same as D.
B.
w(
0
)
=
0
(satisfied by the trial solution)
0035
p
0
L
3
EI
w
(
0
)
=−
0
.
=
0
,
(18)
000955
p
0
L
4
EI
w(
1
)
=
0
.
=
0
,
01805
p
0
L
4
EI
C.
w(
0
)
=−
0
.
=
0
,
0416
p
0
L
3
EI
w
(
)
=
.
=
0
0
0
,
(19)
01
p
0
L
4
EI
w(
1
)
=
0
.
=
0
,
D.
w
(
w(
0
)
=
0
,
0
)
=
0
,
w(
1
)
=
0
(20)
The displacement, moment, and shear force are compared in Fig. 7.5. Note that case B,
which employed a trial function that satisfied one of the displacement boundary conditions,
led to better results than case C, whose trial function satisfied no displacement boundary
conditions. The higher order polynomial of case D was the best of all, since it gave the exact
solution.
Extended Galerkin's Method
Recall from Section 7.4.2 that Galerkin's method requires that the trial solution satisfy
all boundary conditions, i.e., both displacement and force boundary conditions. These
constraints can be relaxed by including these conditions in the global representation of the
fundamental equations. We begin with the global form of the equations of equilibrium,
along with the displacement and force boundary conditions. For a continuum
S
u
δ
u
T
D
T
σ
u
T
p
T
−
V
δ
(
+
p
V
)
+
S
p
δ
(
−
)
−
(
−
)
=
dV
p
p
dS
u
u
dS
0
(7.72)
and for a beam
L
0
δw(
i
v
]
0
x
)
[
EI
w
(
x
)
−
p
z
(
x
)
]
dx
+
[
δw(
V
−
V
)
+
δθ(
M
−
M
)
on
S
p
]
0
=
−
[
δ
M
(θ
−
θ)
+
δ
V
(w
−
w)
0
(7.73)
on
S
u
A
ssume that all applied boundary forces and displacements are zero, i.e.,
V
=
M
=
θ
=
w
=
0
.
Also, since
d
dx
w
w
θ
=−
M
=−
EI
V
=−
EI
Eq. (7.73) can be written as
L
0
δw(
L
0
δw(
i
v
w
(
))
+
(
−
δw
)(
−
w
)
]
0
x
)
[
EI
w
(
x
)
]
dx
−
x
)
p
z
(
x
)
dx
+
[
δw(
−
EI
x
EI
on
S
p
(
−
w
)(
−
δw
)
+
w(
−
δw
)
]
0
=
−
[
EI
EI
0
(7.74)
on
S
u
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