Civil Engineering Reference
In-Depth Information
<
=
<
=
2
4
3
5
125
:
93
−62
:
97
0
0
−8
:
14
−8
:
14
−0
:
03
−0
:
06
x
3
x
4
x
5
x
6
F
1
F
2
0
0
0
0
−62
:
97
62
:
97
8
:
14
8
:
14
8
:
14
8
:
14
0
8
:
14
36
:
06 1
:
75
8
:
14
0
×10
6
=
ð5
:
78Þ
0
8
:
14
1
:
75 36
:
06
0
8
:
14
:
;
:
;
−8
:
14
8
:
14
8
:
14
0
19
:
78
1
:
75
−8
:
14
8
:
14
0
8
:
14
1
:
75
19
:
78
The solution of Eq. (5.78) is
<
=
<
=
−0
:
03 m
−0
:
06 m
0
:
0114 rad
0
:
0114 rad
0
:
0070 rad
0
:
0070 rad
x
1
x
2
x
3
x
4
x
5
x
6
=
N
F
1
F
2
−114470
−1588900
ð5
:
79Þ
=
and
:
;
:
;
Then, the moments and axial resistant force are expressed by
2
4
3
5
8
<
:
9
=
;
8
<
:
9
=
;
m
1
m
2
m
3
m
4
m
5
m
6
m
7
m
8
m
9
m
10
m
11
m
12
P
1
P
2
8
:
14
0
8
:
14
0
0
0
−
151370 N-m
−58543 N-m
−1309 N-m
−36903 N-m
−151370 N-m
−58543 N-m
−1309 N-m
−
8
:
14
0
16
:
28
0
0
0
−8
:
14
8
:
14 16
:
28
0
8
:
14
0
−8
:
14
8
:
14
8
:
14
0
16
:
28
0
<
:
=
;
<
:
=
;
x
1
x
2
x
3
x
4
x
5
x
6
8
:
14
0
0
8
:
14
0
0
03
−0
:
06
0
:
0114
0
−
0
:
8
:
14
0
0
16
:
28
0
0
−8
:
14
8
:
14
16
:
28
8
:
14
0
0
=
K
0
T
×10
6
=
=
−
8
:
14
8
:
14
0
8
:
14
0
16
:
28
0114
0
:
0070
0
:
0070
:
36903 N-m
59852 N-m
59852 N-m
36903 N-m
36903 N-m
−1563400 N
−1563400 N
0
0
3
:
50
1
:
75
0
0
0
1
:
75
3
:
50
0
0
0
0
0
0
3
:
50
1
:
75
0
0
0
0
1
:
75
3
:
50
52
:
11
52
:
11
0
0
0
0
−52
:
11
0
0
0
0
0
ð5
:
80Þ
Comparing moments demands in Eq. (5.80) with the corresponding yield moments shows
that the moments at PHL #1, #5, #9, and #10 have exceeded their corresponding yield moment.
For other PHLs, their moment demands are less than the corresponding yield moment, so the
assumption that these four PHLs remain elastic is correct. In addition, the axial force of the
brace has also exceeded their maximum capacity in compression, so the SH#7 has stepped into
its
buckling
stage,
region A-B. Therefore,
there
are
the
following
relations:
θ
00
2
=
θ
00
3
=
θ
00
4
=
θ
00
6
=
θ
00
7
=
θ
00
8
=
θ
00
11
=
θ
00
12
=
δ
00
2
=0
The moments at the PHs and axial resistant force at the SHs are expressed by
m
=
m
c
m
c
m
b
1
m
b
2
F
1
F
2
½
ð5
:
81Þ
in which,
m
c
=
½
−
130000
−
130000 1110208111
N-m,
m
b
1
= 50000 50000
½
N-m and
m
b
2
=
½
−8111 −8111
N-m
Finally, the governing equation set is satisfied by
