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These are the unassambled displacement and force vectors. The applied loadings are
M
b
M
c
P
=
(2)
We begin the analysis by developing the fundamental equations. Use Sign Convention
2. For a single bar element, the flexibility matrix is given by Chapter 4, Eq. (4.36a) as
i
4
i
M
a
M
b
i
θ
12
EI
−
2
a
=
(3)
θ
−
24
b
v
i
R
f
i
p
i
R
=
Place the element matrices into a single unassembled global flexibility matrix
f
such that
v
=
fp
where
2
.
.
4
−
0
0
.
······ · ······ · ······
.
24
.
−
2
.
12
EI
4
−
f
=
0
0
(4)
.
−
24
.
...
.
...
.
...
.
.
4
−
2
0
0
.
.
−
24
As the next step, we will establish
b
0
and
b
1
. Although numerous choices are possible,
we select the reactions at
b
and
c
as the redundants. Note that the beam would be statically
determinate if the supports at
b
an
d
c
were removed. Matrices
b
0
and
b
1
are found from
the relationship [Eq. (5.125)]
p
=
b
0
P
+
b
1
P
x
with
R
b
R
c
P
x
=
(5)
To find
b
0
, set
P
x
equal to zero. This means that
p
=
b
0
P
or
M
a
M
b
M
b
M
c
M
c
M
d
b
0
M
b
M
c
=
(6)
Let
M
b
1 to form the
second column. Use the configurations (Sign Convention 2) of Fig. 5.32 and summation of
moments to calculate the forces
p
which correspond to the first column of
b
0
. The second
=
1
, M
c
=
0 to form the first column of
b
0
, and let
M
b
=
0
, M
c
=
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