Civil Engineering Reference
In-Depth Information
E
,
I
b
,
L
b
x
3
#5
#6
x
2
3
x
1
#2
#4
M
=
5
0 Mg
dd
1
2
C
=
2
0 Mg s/m
dd
E
,
I
c
,
L
c
E
,
I
c
,
L
c
K
=
315 kN/m
PH
L #1
#3
Figure 3.9
One-story one-bay moment-resisting frame as a SDOF system.
Example 3.7 One-story Moment-Resisting Frame
Consider a one-story one-bay moment-resisting frame as shown in Figure 3.9 with mem-
bers assumed to be axially rigid. This frame has a total of 3 DOFs (i.e.
n
=3)and6PHLs
(i.e.
q
= 6) as labeled in the figure. The stiffness matrices
K
,
K
0
, and
K
00
were presented in
Eq. (2.54) of Example 2.3. Assume that the frame can be reduced to a SDOF system with a
mass of
M
dd
= 5.0 Mg and a damping
C
dd
= 2.0 Mg s/m at the DOF #1, while the
mass moment of inertia at DOFs #2 and #3 are zero. Therefore, static condensation can
be applied to eliminate the DOFs for
x
2
and
x
3
, and the condensed stiffness matrices
K
,
K
0
, and
K
00
were presented in Eq. (2.153) of Example 2.14 for the special case where
I
b
=
I
c
=
I
and
L
b
=
L
c
=
L
.
Assume that the one-story frame has member properties of
E
= 200.0 GPa,
I
b
=
I
c
=
6.0 × 10
6
mm
4
, and
L
b
=
L
c
= 4.0 m. Then, according to Eq. (2.153),
the condensed
stiffness matrices become
K
=
84 × 200 × 3
5ðÞ
= 315kN
=
m
ð3
:
142aÞ
3
K
0
=
K
0
d
−
K
dr
K
−1
rr
K
r
= 360 270 360 270
½
−270
−270
kN
ð3
:
142bÞ
2
4
3
5
1,040
280
40
80
−280
−80
280
560
80
160
−560
−160
40
80
1, 040
280
−80
−280
K
00
=
kNm
=
rad
ð3
:
142cÞ
80
160
280
560
−160
−560
−280
−560
−80
−160
560
160
−80
−160
−280
−560
160
560
The period of vibration
T
and the damping ratio
ζ
can also be calculated:
p
5
:
0
=
315
p
M
dd
=
K
=2
π
T
=2
π
=0
:
792s
ð3
:
143aÞ
C
2
:
0
2
p
KM
dd
p
ζ
=
=
=0
:
0252 = 2
:
52
%
ð3
:
143bÞ
315 × 5
:
0
2
