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The terms in brackets are the virtual work expressions
]
0
[
−
δ
V
w
−
V
δw
−
δ
M
θ
−
M
δθ
on that portion of the boundary where the displacement boundary conditions occur, i.e.,
on
S
u
.
N
u
Introduce the trial solution
w(
x
)
=
w
,
and note that
w
T
N
u
δw
=
N
u
δ
w
=
δ
Then
w
T
EI
L
0
N
T
u
L
N
u
N
u
−
N
T
u
N
u
−
N
u
N
u
on
S
u
N
u
+
N
T
u
N
u
dx
=
δ
−
δ
W
+
w
0
p
z
dx
L
N
u
−
=
0
0
Let
EI
N
u
N
u
L
N
u
−
N
u
R
=
(7.70)
0
which corresponds to the term [
−
V
δw
−
M
δθ
]
0
on
S
u
.
Then
w
T
EI
L
0
L
!
"
=
N
T
u
N
u
dx
R
T
=
δ
N
u
p
z
dx
δ
W
+
(
R
+
)
w
−
0
(7.71a)
0
k
u
p
u
Thus, the free parameters
w
can be found from
R
T
+
(
+
)
−
=
[
k
u
R
]
w
p
u
0
(7.71b)
EXAMPLE 7.9 Beam with Linearly Varying Loading
Consider again the beam of Fig. 7.1. This time we will employ a trial function that does not
necessarily satisfy the displacement boundary conditions.
We begin with a polynomial trial solution in the form
w
1
w
2
w
3
w
4
2
3
4
]
w(
x
)
=
N
u
w
=
[
ξξ
ξ
ξ
(1)
Although it is not necessary that any boundary conditions be satisfied, note that the con-
dition
w(
0
)
=
0isfulfilled by (1). With
d
ξ
=
dx
/
L
the needed derivatives become
1
L
[1
N
u
=
2
3
]
2
ξ
3
ξ
4
ξ
1
L
2
[026
N
u
=
2
]
ξ
12
ξ
(2)
1
L
3
[00624
N
u
=
ξ
]
which are obtained using
d
N
u
dx
d
N
u
d
d
ξ
dx
N
u
=
=
ξ
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