Civil Engineering Reference
In-Depth Information
2
2
236
:
4
600
92
:
85
100
PHL#3
:
+
=1
:
02
ð7
:
86bÞ
which indicates that both PHLs #1 and #3 have reached or exceeded yielding at a lateral force of
F
o
= 75.51 kN that causes a lateral displacement of
x
1
= 8.19 cm.
The analysis continues with both PHLs #1 and #3 yielded. The lateral force is now applied up
to
F
o
= 84.17 kN. Using the same stiffness matrix in Eq. (7.75) by assuming no update to the
geometric nonlinearity even when the axial compressive force in the column changes, extract-
ing Rows 1, 2, 3, 4, and 6, while making use of Eq. (7.55a), gives:
<
=
<
=
2
4
3
5
1379
:
9 1479
:
9 1479
:
9 1479
:
9 1479
:
9
1479
:
9 7892
:
2
x
1
x
2
x
3
−
θ
0
1
−
θ
0
3
84
:
17
0
0
96
:
21
91
:
91
2000
2027
:
30
1479
:
9
2000
7892
:
2
0
2027
:
3
=
ð7
:
87Þ
:
;
:
;
1479
:
9 2027
:
3
4692
:
20
0
1479
:
9
0
2027
:
3
0
4692
:
2
where
q
1− 163
:
6
=
600
2
m
yc
,
1
= 100 ×
ð
Þ
=96
:
21 kNm
ð7
:
88aÞ
q
1− 236
:
4
=
600
2
m
yc
,
3
= 100 ×
ð
Þ
=91
:
91 kNm
ð7
:
88bÞ
Solving for the displacements at the DOFs and plastic rotations in Eq. (7.87) gives
<
=
<
=
x
1
x
2
x
3
−
θ
0
1
−
θ
0
3
0
:
1119
−0
:
0151
−0
:
0148
−0
:
0083
−0
:
0093
=
ð7
:
89Þ
:
;
:
;
Then, substituting Eq. (7.89) back into the last six equations of Eq. (7.75) gives the moments
<
=
<
=
2
4
3
5
<
=
m
1
m
2
m
3
m
4
m
5
m
6
1479
:
9 2027
:
3
0
3892
:
20
102
:
81
90
:
00
99
:
37
89
:
26
−90
:
00
−89
:
26
0
:
1119
−0
:
0151
−0
:
0148
−0
:
0083
−0
:
0093
1479
:
9 3892
:
2
0
2027
:
30
1479
:
9
0
2027
:
3
0
3892
:
2
=
=
ð7
:
90Þ
1479
:
9
0
3892
:
2
0
2027
:
3
:
;
:
;
:
;
0
4000
2000
0
0
0
2000
4000
0
0
and the column axial force are updated as
P
1
=
P
+
m
5
+
m
6
L
b
= 200 +
−
90
:
00
−
89
:
26
4
= 155
:
2kN
ð7
:
91aÞ
