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consideration, the second and fourth rows of the stiffness matrix of Eq. (4.12) become
M
a
k
22
k
24
θ
4
EI
2
EI
θ
a
a
=
=
(4.20)
2
EI
4
EI
θ
b
M
b
k
42
k
44
θ
b
k
R
k
R
Define the
reduced force
and
displacement vectors
M
a
M
b
θ
a
p
R
=
and
v
R
=
(4.21)
θ
b
Then Eq. (4.20) becomes
p
R
=
k
R
v
R
,
where
k
R
is the
reduced stiffness matrix
. We wish to find
f
of
v
R
=
fp
R
.
Since
k
R
is nonsingular, we can obtain the flexibility matrix by inversion.
Thus,
k
44
−
k
24
k
22
−
1
1
−
k
42
k
22
EI
/
−
/
k
24
3
1
6
k
−
1
R
f
=
=
=
k
24
k
42
=
(4.22)
k
42
k
44
−
1
/
61
/
3
k
22
k
44
−
The relationship between the stiffness and flexibility matrices can be illustrated fur-
ther by the following two cases which are examples of constrained beam elements. The
flexibility matrix depends upon which variables are chosen as independent variables.
Case 1
Suppose the shear force
V
b
and the moment
M
b
are chosen to be independent variables,
and it is desired to eliminate or to suppress the displacements associated with rigid body
motion. From Eq. (4.17b),
w
θ
U
vv
I
]
v
a
v
b
v
=
=
v
b
−
U
vv
v
a
=
[
−
=
gv
(4.23a)
with
−
1
10
g
=
(4.23b)
0
−
101
Recall from the derivation of Section 4.2.2 that the rigid body displacements are
w
a
−
θ
a
and
θ
a
. Physically, this can be considered to correspond to a beam with the right end
free and the left end cantilevered (Fig. 4.4), where the cantilevered end can translate and
rotate. Remove the rigid body displacement (
w
a
−
θ
a
,
θ
a
)from
w
b
and
θ
b
, thereby defining
new displacement variables, say
w
R
and
θ
R
, which, according to Eq. (4.4), are identical to
w
and
θ
, respectively.
w
R
=
w
b
−
(w
a
+
θ
a
)
=
w
(4.24)
θ
R
=
θ
b
−
θ
a
=
θ
Define
w
R
θ
R
v
R
=
(4.25a)
Set
V
R
=
V
b
,M
R
=
M
b
or
V
b
M
b
p
R
=
p
b
=
(4.25b)
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