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=
a hinge at
b
,
M
b
=
0
p
2
and
v
2
=
θ
b
would be the unknown variable. From Eq. (5.101)
v
2
=−
k
−
1
22
k
21
v
1
(5.102)
so that forces
p
1
are related to the displacements
v
1
,by
p
1
=
k
11
−
k
12
k
−
1
22
k
21
v
1
(5.103)
This serves as a set of springs for which forces have known dependencies (spring rates) on
displacements. The displacement dependent forces of Eq. (5.103) are readily incorporated
into the analysis. That is, the effect of
p
1
can be assembled into the global stiffness matrix
K
using
=
k
11
22
k
21
.
−
k
12
k
−
1
p
1
···
0
v
1
···
0
0
··············· · ···
0
(5.104)
.
0
for the
i
th element. After the global displacements have been computed, the variables, e.g.,
reactions,
v
2
can be determined using Eq. (5.102).
There are a number of other effects that can be accounted for in a similar fashion. For
example, it may be desirable to model a joint as having a certain degree of flexibility. As
formulated above, an appropriate dependency between forces and displacements at the
joint can be incorporated into the displacement method analysis.
5.3.14 Rigid Elements
Rigid elements can be the source of numerical instabilities in an analysis. For example,
for a beam element that is rigid against axial deformation,
EA
approaches infinity and
the determinant
Thus, a rigid element, which implies that some of its dis-
placements will be the same, i.e., some displacements are dependent, can lead to a sin-
gular stiffness matrix. This singularity is different from the singularity property exhib-
ited by all element stiffness matrices (Chapter 4, Section 4.3). As is to be illustrated in
Example 5.7, this problem can be corrected by taking into account the dependency of cer-
tain variables. For example, if a bar element, which extends from
x
(
K
)
→
0
.
=
a
to
x
=
b
, is rigid
in the axial direction, set
u
b
=
u
a
.
Then certain rows and columns of the stiffness matrix
should be reorganized.
EXAMPLE 5.7 Frame with Rigid Members and Distributed Loading
Return to the frame of Figs. 5.21 and 5.22, and suppose that a distributed load is placed
between nodes
a
and
b
as shown in Fig. 5.28. Furthermore, assume that members 2 and 3
cannot extend and compress, i.e., they are rigid with respect to axial deformation. Other than
these two characteristics, the material and geometric properties are the same as employed
in Example 5.5. The response due to two loading cases is sought.
Loading Case 1:
p
0
=
2
.
0 kN/m (Fig. 5.28)
Loading Case 2:
P
Z
=
20 kN,
P
X
=
10 kN, the same as in Example 5.5
p
i
0
,
=
k
i
p
i
The element stiffness matrix of Eq. (5.94) applies and can be inserted in
v
i
−
where
p
i
0
.
accounts for the distributed loading applied to element
i
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