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N
u
δθ
=−
δw
=
An approximate displacement of the form
w(
x
)
=
w
is chosen. Then
N
u
δ
δw
(
N
u
(
)δ
=−
w
,
x
)
=
x
w
,
and
δ
W
=
0 becomes
L
0
(δ
L
0
δ
w
T
N
u
V
M
L
0
w
T
N
T
u
EI
N
u
w
T
N
u
N
u
−
δ
W
=
w
)
dx
−
p
z
dx
−
δ
−
EI
L
0
L
!
"
=
+
N
u
V
N
u
M
L
N
T
u
N
u
(
N
u
(
(
)
)
−
)
p
z
(
)
−
x
x
dx
w
x
x
dx
0
=
δ
w
T
0
0
k
u
p
u
(7.45)
From this form, it is apparent that
k
u
is inherently symmetric. Finally, we obtain the system
of equations
k
u
w
=
p
u
(7.46)
for the free parameters
w
.
EXAMPLE 7.5 Beam with Linearly Varying Loading
Return to the fixed-hinged beam of Fig. 7.1. The geometrical (displacement) boundary
conditions on
S
u
are
)
=
w
(
w(
0
)
=
0
,
−
θ(
0
0
)
=
0
,
w(
L
)
=
0
(1)
and the static boundary condition is
M
(
L
)
=
0
(2)
on
S
p
.
We will begin with a two-parameter approximate deflection
]
w
1
w(ξ)
=
N
u
1
(ξ )w
1
+
N
u
2
(ξ )w
2
=
[
N
u
1
(ξ )
N
u
2
(ξ )
=
N
u
w
(3)
w
2
Since the variational principle of Eq. (7.45) contains second order derivatives, it is necessary
to select an approximate deflection containing at least a third order polynomial. Choose
2
3
N
u
1
(ξ )
=
a
1
+
a
2
ξ
+
a
3
ξ
+
a
4
ξ
(4)
2
3
4
N
u
2
(ξ )
=
b
1
+
b
2
ξ
+
b
3
ξ
+
b
4
ξ
+
b
5
ξ
The approximation (3) must satisfy the geometrical boundary conditions, and the con-
stants
a
i
,i
=
1
,
2
,
3
,
4 and
b
i
,i
=
1
,
2
,
3
,
4
,
5 will be chosen to accomplish this. Thus,
w(
0
)
=
0
requires that
a
1
=
b
1
=
0
w
(
)
=
=
=
0
0
requires that
a
2
b
2
0
(5)
w(
)
=
+
=
L
0
requires that
a
3
a
4
0
b
3
+
b
4
+
b
5
=
0
We can now select any values for
a
i
,b
i
that satisfy these relations, e.g.,
a
3
=−
a
4
=
1
,
b
3
=
b
5
=
1
,
b
4
=−
2
(6)
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