Civil Engineering Reference
In-Depth Information
Substituting Eq. (4.82) back in Eq. (4.77) and solving for the moments at the RHs and shear
forces due to the SHs gives:
<
=
<
=
m
1
m
2
m
3
m
4
m
5
m
6
V
7
V
8
−152
:
58 kN-m
−72
:
42 kN-m
−152
:
58 kN-m
−72
:
42 kN-m
72
:
42 kN-m
72
:
42 kN-m
75 kN
75 kN
=
ð4
:
83Þ
:
;
:
;
Comparing again the demands in Eq. (4.83) with the corresponding capacities listed in
Tables 4.1, 4.2, and 4.3, the assumption that the locations of RHs #1, #3, #5, #6 and SHs
#7, #8 are in the plastic domain while the locations of RHs #2 and #4 remain elastic is correct.
In summary, there are
<
=
<
=
<
=
<
=
θ
0
1
θ
0
2
θ
0
3
θ
0
4
θ
0
5
θ
0
6
τ
0
7
τ
0
8
−0
:
0007 rad
0 rad
−0
:
0007 rad
0 rad
0
:
0003 rad
0
:
0003 rad
0
:
0003m
0
:
0003m
m
1
m
2
m
3
m
4
m
5
m
6
V
7
V
8
−152
:
58 kN-m
−72
:
42 kN-m
−152
:
58 kN-m
−72
:
42 kN-m
72
:
42 kN-m
72
:
42 kN-m
75 kN
75 kN
<
=
<
=
x
1
x
2
x
3
0
:
0048m
0
:
0015m
0
:
0015m
=
,
=
,
=
:
;
:
;
:
;
:
;
:
;
:
;
ð4
:
84Þ
Note that the joint equilibriums are satisfied when
m
2
= −
m
5
and
m
4
= −
m
6
and the force equi-
libriums of the columns are satisfied when
V
7
= −(
m
1
+
m
2
)/
L
c
and
V
8
= −(
m
3
+
m
4
)/
L
c
.
Figure 4.12 summarizes these results.
0.0048
72.42
72.42
0.0003
F
1
0.0003
72.42
72.42
0.0015
0.0015
75
75
0.0003
0.0003
75
75
0.0007
0.0007
152.58
152.58
Figure 4.12
Graphical illustration of the response of nonlinear one-story one-bay RC frame.
