Civil Engineering Reference
In-Depth Information
Therefore, in summary,
2
θ
1
θ
2
θ
3
θ
4
θ
5
θ
6
m
2
F
L
3
F
L
6
EI
o
o
1
0
m
F
L
3
o
3
2
x
F
L
3
EI
1
o
2
F
L
6
EI
m
2
F
L
3
o
2
o
3
x
=
−
F
L
3
EI
,
=
,
=
ð2
:
162Þ
2
o
0
m
F
L
3
o
2
4
x
−
F
L
3
EI
3
o
2
−
5
F
L
18
EI
m
−
F
L
3
o
o
5
2
−
5
F
L
18
EI
m
−
F
L
3
o
o
6
which are the same as those obtained in Eq. (2.76) of Example 2.6.
Example 2.15 Nonlinear Solution to the Two-Story Frame with Static Condensation
Consider the two-story moment-resisting frame used in Example 2.12 with the same applied
loads, i.e.
F
1
=
F
2
=
F
o
and
F
3
=
F
4
=
F
5
=
F
6
= 0. Assume that the moment versus plastic rota-
tion relationships are:
θ
0
i
=0
m
i
=
F
o
L
=
2+ 2
EI
=
L
m
i
≤
F
o
L
=
2
if
m
i
>
F
o
L
=
2
,
then
i
=1,…,8
ð2
:
163aÞ
Þ
θ
0
i
ð
θ
0
i
=0
m
i
=2
F
o
L
=
5+
EI
=
ð Þ
θ
0
i
m
i
≤ 2
F
o
L
=
5
m
i
>2
F
o
L
=
5
,
if
then
i
=9,…,12
ð2
:
163bÞ
The stiffness matrices were presented in Eq. (2.125). Recognizing
F
3
=
F
4
=
F
5
=
F
6
=0,
static condensation can be performed with the matrices and vectors being partitioned as follows:
x
F
o
1
x
F
o
2
x
x
F
0
ad
d
3
=
=
ð2
:
164aÞ
,
x
x
0
0
r
4
x
0
5
x
0
6
3
3
2
2
48
EI
L
−
24
EI
L
0
0
−
6
EI
L
−
6
EI
L
3
3
2
2
2
2
−
24
EI
L
24
EI
L
6
EI
L
6
EI
L
6
EI
L
6
EI
L
2
K
K
0
6
EI
L
12
EI
L
2
EI
L
2
EI
L
0
dd
dr
=
2
K
K
0
6
EI
L
2
EI
L
12
EI
L
0
2
EI
L
rd
rr
2
2
−
6
EI
L
6
EI
L
2
EI
L
0
8
EI
L
2
EI
L
2
2
−
6
EI
L
6
EI
L
0
2
EI
L
2
EI
L
8
EI
L
ð2
:
164bÞ
