Civil Engineering Reference
In-Depth Information
100
80
60
40
20
0
0
5
10
15
20
25
30
Displacement (cm)
Figure 7.7
Pushover curve of the one-story frame with update on geometric nonlinearity.
At this point,
q
1− 155
:
2
=
600
2
m
yc
,
1
= 100 ×
ð
Þ
=96
:
60 kNm
ð7
:
74aÞ
q
1− 244
:
8
=
600
2
m
yc
,
3
= 100 ×
ð
Þ
=91
:
30 kNm
ð7
:
74bÞ
The calculation process continues, and the final pushover curve is shown in Figure 7.7.
Example 7.3 Pushover of One-Story Frame with No Update on Geometric Nonlinearity
Consider again the one-story one-bay moment-resisting frame as shown in Figure 7.6 with
three DOFs (
n
= 3) and six PHLs (
q
= 6). Let
I
c
=
I
b
=
I
and
L
c
=
L
b
=
L
. Assume again
that only a lateral force of
F
o
is applied at the horizontal degree of freedom
x
1
, this gives
F
1
=
F
o
and
F
2
=
F
3
= 0. It then follows that the global stiffness matrices of the FAM for this
one-story frame are given in Eq. (7.52), and the governing equation is given in Eq. (7.53).
For the numerical illustration, again let
E
= 200 GPa,
I
b
=
I
c
=
I
=20×10
6
mm
4
,
L
b
=
L
c
=
L
= 4 m, and
P
= 200 kN. Now assume that geometric nonlinearity is not updated as the
new axial force is calculated (i.e. the global stiffness matrices remain constant due to the initial
load of
P
= 200 kN), then Eq. (7.53) becomes
1379
.
1479
.
1479
.
1479
.
1479
.
1479
.
1479
.
0
0
x
F
o
1
x
0
1479
.
7892
.
2000
2027
.
3892
.
0
0
4000
2000
2
1479
.
2000
7892
.
0
0
2027
.
3892
.
2000
4000
x
0
3
θ
′
1479
.
2027
.
0
3892
.
2027
.
0
0
0
0
−
m
1
1
1479
.
3892
.
0
2027
.
3892
.
0
0
0
0
−
θ ′
=
m
2
2
−
θ ′
1479
.
0
2027
.
0
0
3892
.
2027
.
0
0
m
3
3
θ
′
1479
.
0
3892
.
0
0
2027
.
3892
.
0
0
−
m
4
4
θ
′
0
4000
2000
0
0
0
0
4000
2000
−
m
5
5
0
2000
4000
0
0
0
0
2000
4000
−
θ ′
m
6
6
ð7
:
75Þ
