Civil Engineering Reference
In-Depth Information
0
:
49
1
:
00
= 3105
:
4kg
25000
0
1
M
Φ
1
=0
:
49 1
:
00
M
1
=
Φ
f
g
0
25000
0
:
49
1
:
00
= 110110 N
=
m
761030
−330880
T
K
1
=
Φ
1
K
Φ
1
=0
:
49 1
:
00
f
g
−330880
251470
0
:
49
1
:
00
= 739
:
7 N-s
=
m
1163
:
7
−525
:
7
T
C
1
=
Φ
1
C
Φ
1
=0
:
49 1
:
00
f
g
−525
:
7
854
:
1
1
=
K
1
M
1
=
110110
2
ω
3105
:
4
=35
:
46
2
ω
1
M
1
=
739
:
7
C
1
ς
1
=
2×5
:
95 × 3105
:
4
=0
:
02
γ
1
=
Ml
M
1
=1
:
2
The NPAS were performed and through using the graphic method and Eqs. (8.86) and
(8.87), the yield moment is satisfied by
Ω
1
,
y
=
Γ
1
V
1
,
y
M
1
=
1
:
2 × 16000
3105
:
4
=6
:
18N-m
ð8
:
91Þ
Thus, governing equations of the 1st modal system in the FAM is written as
Y
1
ðÞ+0
:
24
Y
1
ðÞ+35
:
46
Y
1
ðÞ= −1
:
2
x
g
ðÞ+35
:
46
Y
0
1
ðÞ
ð8
:
92Þ
a
s
,
1
ðÞ=35
:
46
Y
1
ðÞ−35
:
46
θ
0
1
ðÞ
ð8
:
93Þ
Ω
1
ðÞ=35
:
46
Y
1
ðÞ−35
:
46
θ
0
1
ðÞ
ð8
:
94Þ
<
Ω
1
ðÞ≥ 6
:
18
1
35
:
46
35
:
46
Y
1
ðÞ−
Ω
y
,
1
θ
0
1
ðÞ=
ð8
:
95Þ
:
0
Ω
1
ðÞ<6
:
18
Perform the state space formulation presented in Chapter 2.2 on the 1st modal system. Dis-
placement and acceleration time history of each floor, and plastic rotation in the 1st modal coor-
dinate time history is shown in Figure 8.10.
(2) 2nd modal system
According to Eq. (8.49), there are
= 12824 kg
25000
0
−2
:
03
1
:
00
2
M
Φ
2
=
M
2
=
Φ
f
−2
:
03 1
:
00
g
0
25000
2
:
03
1
:
00
= 4738900N
=
m
761030
−330880
T
K
2
=
Φ
2
K
Φ
2
=
−2
:
03 1
:
00
f
g
−330880
251470
