Civil Engineering Reference
In-Depth Information
2
4
3
5
448,000
−244,900
39,160
−3,716
346
:
0
−29
:
4
−244,900
432,000
−260,200
38,140
−3,551
305
:
4
39,160
−260,200
439,900
−248,000
36,170
−3,106
K
e
=
kN
=
m
−3,716
38,140
−248,000
376,600
−187,800
24,430
346
:
0
−3,551
36,170
−187,800
278,200
−123,400
−29
:
4
305
:
4
−3,106
24,430
−123,400
101,800
ð7
:
176Þ
Let the mass be 200.0 Mg on each floor, giving a total mass of 1,200 Mg for the entire frame.
The diagonal nonzero mass matrix
M
dd
can be written as
2
3
000000
0 200 0 0 0 0
000000
000000
0 0 0 0 200 0
000000
4
5
M
dd
= 200
:
0×
I
=
Mg
ð7
:
177Þ
Using the condensed elastic stiffness matrix
K
e
in Eq. (7.176) and the diagonal
M
dd
mass matrix
in Eq. (7.177), the 6 periods of vibration are calculated and summarized in Figure 7.27. The
damping is assumed to be 2% in all six modes of vibration. This gives a damping matrix of
2
4
3
5
−4
:
2
−4
:
6
−2
:
0
−1
:
3
361
−114
−114
329
−130
−9
:
2
−6
:
2
−3
:
4
−4
:
2
−130
327
−129
−7
:
2
−7
:
0
C
dd
=
kN s
=
m
ð7
:
178Þ
−4
:
6
−9
:
2
−9
:
0
−129
302
−110
−2
:
0
−6
:
2
−7
:
2
−110
259
−99
:
2
−1
:
3
−3
:
4
−7
:
0
−9
:
0
−99
:
2
150
Assume that all 84 plastic hinges exhibit elastic-plastic behavior with moment capacity of
the
i
th plastic hinge,
m
c
,
i
, calculated as
m
c
,
i
=
f
y
×
Z
i
i
=1,…,84
ð7
:
179Þ
where
f
y
is the yield stress of steel and
Z
i
is the plastic section modulus of the
i
th plastic hinge.
This gives
(
m
i
ðÞ≤
m
c
,
i
m
i
ðÞ>
m
c
,
i
Δθ
0
i
ðÞ=0
m
i
ðÞ=
m
c
,
i
if
,
then
i
=1,…,84
ð7
:
180Þ
By subjecting the 6-story frame to the 1995 Kobe earthquake ground motion as shown in
Figure 3.2, Figures 7.28 to 7.30 show the displacement, velocity, and absolute acceleration
responses, respectively, of selected floors. In addition, the same responses but for the case
without considering any geometric nonlinearity (see Example 3.9) are presented in the figures
for comparisons.
