Civil Engineering Reference
In-Depth Information
=0
:
02 = 2
%
ζ
=
C
=
2
ω
M
=0
:
251
=
2×
p
39
:
48
×1
:
0
ð3
:
104eÞ
m
p
=
d
y
×3
EI
=
L
2
=0
:
10 × 394
:
8=39
:
48 kNm
ð3
:
104fÞ
Δθ
00
=0
m
k
+1
=
m
k
+0
:
08
K
00
Δθ
00
m
k
+1
≤ 39
:
48
m
k
+1
>39
:
48
,
if
then
ð3
:
104gÞ
The recursive equation as given in Eq. (3.57) becomes
=
x
k
x
k
+
a
k
+
x
0
k
x
k
+1
x
k
+1
0
:
998028 0
:
009981
−0
:
39403 0
:
995520
−0
:
000100
−0
:
009955
0
:
003940
0
:
39303
ð3
:
105Þ
and the governing equations of the FAM in Eqs. (3.59) and (3.60) are
Þ
Δθ
00
= 394
:
8
Þ
θ
0
k
m
k
+1
+ 3948
ð
ð
Þ
x
k
+1
− 3948
ð
ð3
:
106aÞ
x
0
k
+1
=10
θ
0
k
+
Δθ
00
ð3
:
106bÞ
Finally, the absolute acceleration response can be calculated using Eq. (3.62) as
y
k
+1
= −0
:
251
x
k
+1
−39
:
48
x
k
+1
−
x
0
k
+1
ð3
:
107Þ
Assume that the earthquake ground acceleration record is the 1994 Northridge earthquake
record as shown in Figure 3.3, and the initial displacement and velocity are both set equal to
zero (i.e.
x
0
=
x
0
= 0). It follows that the displacement, velocity, acceleration, moment, and plas-
tic rotation response histories can be evaluated using Eqs. (3.105) through Eq. (3.107), and the
results are presented in Figure 3.7. The responses are plotted along with the linear responses
from Example 3.1 as comparisons (note that both the present example and Example 3.1 use
the same period of 1.0 s, damping of 2%, and the same Northridge earthquake ground motion
as the input). In addition, the initial yielding moment and the backbone curve are plotted in the
figure as appropriate.
Finally, assume that the SDOF column has properties of
C
= 0.419 kN s/m,
I
= 5.85 m
4
,
d
y
= 5 cm, and a bilinear moment versus plastic rotation relationship with a kinematic softening
ratio of −4%. This gives the following structural properties:
3
=17
:
55 kN
=
m
K
=3
EI
=
L
3
= 3 × 1000 × 5
:
85
=
1ðÞ
ð3
:
108aÞ
2
= 175
:
5kN
K
0
=3
EI
=
L
2
= 3 × 1000 × 5
:
85
=
1ðÞ
ð3
:
108bÞ
K
00
=3
EI
=
L
= 3 × 1000 × 5
:
85
=
10 = 1,755 kNm
ð3
:
108cÞ
p
M
=
K
p
1
=
17
:
55
T
=2
π
=2
π
=1
:
5s
ð3
:
108dÞ
=0
:
05 = 5
%
ζ
=
C
=
2
ω
M
=0
:
419
=
2×
p
17
:
55
×1
:
0
ð3
:
108eÞ
m
p
=
d
y
×3
EI
=
L
2
=0
:
05 × 175
:
5=8
:
775 kNm
ð3
:
108fÞ
