Civil Engineering Reference
In-Depth Information
Since
U
1
=
0
, we remove the first row and column to obtain
=
4
k
−
3
k
0
U
2
U
4
−
F
F
3
2
F
−
3
k
5
k
−
2
k
0
−
2
k
2
k
as the system of equations governing displacements
U
2
and
U
4
and the unknown nodal
force
F
3
. This last set of equations clearly shows that we cannot simply strike out the row
and column corresponding to the
nonzero
specified displacement
because it appears in
the equations governing the active displacements. To illustrate a general procedure, we
rewrite the last matrix equation as
=
−
3
k
−
2
k
U
2
U
4
5
k
F
3
−
F
2
F
−
3
k
4
k
0
−
2
k
0
2
k
Next, we formally partition the stiffness matrix and write
=
−
3
k
−
2
k
U
2
U
4
[
K
]
K
U
]
[
K
U
]
K
UU
]
{}
{
U
}
{
F
}
{
F
U
}
5
k
−
=
3
k
4
k
0
−
2
k
0
2
k
with
[
K
]
=
[
5
k
]
[
K
U
]
=
[
−
3
k
−
2
k
]
−
3
k
−
[
K
U
]
=
[
K
U
]
T
=
2
k
4
k
0
0
k
[
K
UU
]
=
{} = {}
U
2
U
4
{
U
} =
{
F
} = {
F
3
}
−
F
2
F
{
F
U
} =
From the second “row” of the partitioned matrix equations, we have
[
K
U
]
{
U
}={
F
U
}
and this can be solved for the unknown displacements to obtain
{}+
[
K
UU
]
[
K
UU
]
−
1
(
{
U
}=
{
F
}−
[
K
U
]
{}
)
provided that
[
K
UU
]
−
1
exists. Since the constraints have been applied correctly, this
inverse does exist and is given by
1
4
k
0
[
K
UU
]
−
1
=
1
2
k
0
