Civil Engineering Reference
In-Depth Information
Assume that the plastic hinges for the beams and columns take the following moment versus
plastic rotation relationships:
θ
0
i
=0
m
i
=
F
o
L
=
2+ 2
EI
=
L
m
i
≤
F
o
L
=
2
m
i
>
F
o
L
=
2
,
if
then
i
=1,…,8
ð2
:
127aÞ
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
:
127bÞ
where
i
=1,…, 8 denotes PHLs #1 to #8 in the columns, and
i
=9,…, 12 denotes PHLs #9 to
#12 in the beams.
The structure is first assumed to be linear, this gives
Θ
00
=
0
. Then using the first matrix equa-
tion of Eq. (2.126) gives
<
=
<
=
2
4
3
5
48
EI
=
L
3
−24
EI
=
L
3
−6
EI
=
L
2
−6
EI
=
L
2
x
1
x
2
x
3
x
4
x
5
x
6
F
o
F
o
0
0
0
0
0
0
−24
EI
=
L
3
24
EI
=
L
3
6
EI
=
L
2
6
EI
=
L
2
6
EI
=
L
2
6
EI
=
L
2
6
EI
=
L
2
0
12
EI
=
L
2
EI
=
L
2
EI
=
L
0
=
6
EI
=
L
2
0
2
EI
=
L
12
EI
=
L
0
2
EI
=
L
:
;
:
;
−6
EI
=
L
2
6
EI
=
L
2
2
EI
=
L
0
8
EI
=
L
2
EI
=
L
−6
EI
=
L
2
6
EI
=
L
2
0
2
EI
=
L
2
EI
=
L
8
EI
=
L
ð2
:
128Þ
Solving for the displacements
x
gives
<
:
=
;
<
:
=
;
x
1
x
2
x
3
x
4
x
5
x
6
2
F
o
L
3
=
15
EI
=
4
EI
−
F
o
L
2
F
o
L
3
=
10
EI
=
ð2
:
129Þ
−
F
o
L
2
=
10
EI
−
F
o
L
2
=
20
EI
−
F
o
L
2
=
20
EI
Then substituting Eq. (2.129) into the second matrix equation of Eq. (2.126), the moments are
calculated as
<
=
<
=
<
=
<
=
m
1
m
2
m
3
m
4
m
5
m
6
m
7
m
8
m
9
m
10
m
11
m
12
0
:
6
0
:
4
0
:
2
0
:
3
0
:
6
0
:
4
0
:
2
0
:
3
−0
:
6
−0
:
6
−0
:
3
−0
:
3
m
y
1
m
y
2
m
y
3
m
y
4
m
y
5
m
y
6
m
y
7
m
y
8
m
y
9
m
y
10
m
y
11
m
y
12
0
:
5
0
:
5
0
:
5
0
:
5
0
:
5
0
:
5
0
:
5
0
:
5
−0
:
4
−0
:
4
−0
:
4
−0
:
4
=
F
o
L
,
=
F
o
L
ð2
:
130Þ
:
;
:
;
:
;
:
;
