Civil Engineering Reference
In-Depth Information
Substituting Eq. (4.51) back into Eq. (4.50) and solving for the restoring force
F
, the moments
m
1
and
m
2
at the RHs and the shear force
V
gives:
<
=
<
=
F
m
1
m
2
V
153
:
0009 kN
−229
:
5014 kN-m
−229
:
5014 kN-m
153
:
0009 kN
ð4
:
52Þ
=
:
;
:
;
Then, the rotations
θ
1
and
θ
2
at the column ends and the shear deformation
τ
of the structure can
be obtained:
θ
1
=
θ
2
= −0
:
0162 rad,
τ
=0
:
0117 m
ð4
:
53Þ
Comparing the demands in Eqs. (4.52) and (4.53) with the corresponding capacities in
Tables 4.1 and 4.2, we can find that the strength of SH starts to deteriorate. However, due to
the same primary curve parameters of the two RHs shown in Table 4.1 and the assumption of
the procedure of dividing the total primary curve to the bending and shear ones for the SDOF
system in previous section, the RHs and SH should behave on the same type of branches at the
same force level. Therefore, assume that the RHs and the SH behave on the corresponding
softening branches and the moments
m
1
and
m
2
and shear force
V
can be expressed in the terms
of
θ
1
00
,
θ
2
00
and
τ
00
according to Eq. (4.5):
m
1
= −407
:
02−1
:
1×10
4
<
θ
0
1
m
2
= −407
:
02−1
:
1×10
4
θ
00
ð4
:
54Þ
:
V
= 390
:
67−3
:
3×10
4
τ
00
Therefore, Eq. (4.50) becomes:
0.6391
−
0.9586
−
0.9586
0.6391
0.06
F
′′
−
0.9586
1.9587
−
0.11
0.9171
−
0.9586
−θ
−
407.02
1
5
×
10
=
′′
−
0.9586
0.9171
1.9587
−
0.11
−
0.9586
−θ
−
407.02
2
′′
0.6391
−
0.9586
−
0.9586
0.6391
−
0.33
−τ
390.67
ð4
:
55Þ
Solving for the plastic deformations
θ
1
00
,
θ
2
00
and
τ
00
in Eq. (4.55) gives:
<
=
<
=
θ
0
1
θ
0
2
τ
00
−0
:
0167 rad
−0
:
0167 rad
0
:
0075m
=
ð4
:
56Þ
:
;
:
;
Substituting Eq. (4.56) back into Eq. (4.35) and solving for the restoring force
F
, the moments
m
1
and
m
2
at the RHs and the shear force
V
gives:
<
=
<
=
F
m
1
m
2
V
145
:
25 kN
−217
:
88 kN-m
−217
:
88 kN-m
145
:
25 kN
ð4
:
57Þ
=
:
;
:
;
