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beam of Fig. 5.5, we simply note that
w
=
0 and
θ
=
0 since the left end is fixed. The
a
a
conditions at the simply supported right end are
w
=
0 and
M
x
=
L
=
0. These are applied
x
=
L
to the first and fourth rows of (3). Equation (3), that is,
z
x
=
L
=
Uz
a
,
can be written as
.
.
0
w
=
0
U
U
w
w
=
0
w
V
w
M
.
.
V
θ
=
0
···
···
···
···
.
.
=
(4)
V
···
···
···
···
.
.
M
0
M
=
0
U
MV
U
MM
M
.
.
1
0
0
1
1
x
=
L
x
=
L
x
=
0
w
=
θ
=
Cancel columns 1 and 2 because
0
0
0
θ
L
and
V
x
=
L
are unknown
where
U
kj
and the loading vector components are the elements of
U
of (3). The equations
w
Ignore rows 2 and 3 because
x
=
0 are used to compute
M
a
and
V
a
,
the remaining unknown initial
parameters. Thus from (4)
M
=
0 and
M
x
=
L
=
x
=
L
0
0
U
w
V
V
a
M
a
w
M
0
0
U
w
M
L
=
=
+
(5)
U
MV
U
MM
x
=
x
=
L
x
=
L
or
0
L
0
=
V
a
U
w
V
+
M
a
U
w
M
+
w
(6)
M
L
=
+
+
0
V
a
U
MV
M
a
U
MM
Equations (5) and (6) are solved for
M
a
,V
a
,
giving
M
0
U
w
M
0
U
MM
V
a
=
(
−
w
)
|
/
∇
x
=
L
(7)
0
U
MV
M
0
U
w
V
M
a
=
(w
−
)
|
/
∇
x
=
L
where
U
ij
and the loading components are the coefficients of the overall transfer matrix
U
and
∇=
(
U
w
V
U
MM
−
U
w
M
U
MV
)
|
(8)
x
=
L
For the beam of Fig. 5.5, (3) gives
.
3
EI
2
p
0
4
EI
4
3
−
2
8
15
1
−
2
−
EI
.
3
2
2
EI
2
EI
−
p
0
01
EI
0
.
−
00
1
p
0
U
=
(9)
.
2
−
4
p
0
00 2
1
3
.
···
···
···
···
···
0
.
00
0
1
4
EI
2
8
15
p
0
and
M
L
=
−
4
p
0
L
Introduce the boundary conditions to (9) or use (7) with
w
=
to
3
find the initial parameters
V
a
and
M
a
. Thus,
−
−
V
a
M
a
3
EI
2
4
4
3
−
2
8
15
p
0
EI
EI
=
(10)
2
4
p
0
2
1
3
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