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where
w
.
.
a
k
aa
k
ab
0
θ
a
...... .
......
. ......
w
.
k
bb
.
b
K
=
V
=
k
ba
k
bb
+
k
bc
θ
b
...... .
......
. ......
w
.
.
c
k
cb
k
cc
0
θ
c
−
−
−
12
6
12
6
0
0
−
64 62 00
EI
−
26 40
−
12
−
6
=
(9)
−
62 08 62
00
3
−
26 26
00
−
62 64
The global loading vector
P
0
is assembled in a fashion similar to the global stiffness matrix.
Use Eq. (5.99b)
2
P
0
j
p
i
j
=
i
=
1
Thus,
51
−
51
−
P
0
a
P
0
b
P
0
c
0
p
a
p
b
+
8
8
60
4
9
2
p
0
120
p
0
120
39
+
21
P
0
p
b
=
=
=
=
(10)
7
−
3
p
c
9
2
P
0
The boundary conditions
w
=
θ
=
w
=
0 imposed on
KV
=
give the nonsingular
a
a
c
system
24
0
−
6
w
b
θ
b
θ
c
60
4
2
EI
120
p
0
=
08
2
(11)
3
−
62
4
Here the rows corresponding to the unknown reactions (resulting from the pr
es
cribed
displacements
w
a
,
θ
a
,
and
w
c
) are ignored. Thus, (11) is equivalent to
K
22
V
y
=
P
of Eq.
(5.73). The solution to these relations is
w
4
.
5
b
3
p
0
=
θ
−
1
.
5
(12)
b
120
EI
θ
8
.
0
c
The internal forces are computed using
k
i
v
i
p
i
0
p
i
−
=
(13)
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