Civil Engineering Reference
In-Depth Information
m
5
+
m
6
L
b
= 200−
−
90
:
00
−
89
:
26
4
P
2
=
P
−
= 244
:
8kN
ð7
:
91bÞ
Equation (7.54) is now used to ensure PHLs #1 and #3 continue to yield and PHL #5 has
reached yielding:
2
2
155
:
2
600
102
:
86
100
PHL#1
:
+
=1
:
12
ð7
:
92aÞ
2
2
244
:
8
600
99
:
33
100
PHL#3
:
+
=1
:
15
ð7
:
92bÞ
90
90
PHL#5
:
=1
:
00
ð7
:
92cÞ
which indicates that PHLs #1, #3, and #5 have reached or exceeded yielding at a lateral force of
F
o
= 84.17 kN that causes a lateral displacement of
x
1
= 11.19 cm.
At this point,
q
1− 155
:
2
=
600
2
m
yc
,
1
= 100 ×
ð
Þ
=96
:
60 kNm
ð7
:
93aÞ
q
1− 244
:
8
=
600
2
m
yc
,
3
= 100 ×
ð
Þ
=91
:
30 kNm
ð7
:
93bÞ
The calculation process continues, and the final pushover curve is shown in Figure 7.8. The
pushover curve that was obtained in Example 7.2 with updates to geometric nonlinearity
(see Figure 7.7) is also shown in Figure 7.8 as a comparison.
Figure 7.8 shows that the difference between whether or not to update the geometric non-
linearity at every step due to changes in column axial force on the pushover curve is negligible.
One assumption often made among various software packages is that the effect of changing
axial force on the geometric stiffness matrices is ignored, and therefore the geometric
100
80
60
40
Geo update
No update
20
0
0
5
10
15
20
25
30
Displacement (cm)
Figure 7.8
Comparison of pushover curves on geometric nonlinearity with and without updates.
