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In-Depth Information
N
2
i
∑
∑
∆
=
0
m
θ
(44)
pjh ph
j p
i
=
1
h
=
1
i
(
)
+
M
=
0 5
.
f BJd Jd
3
−
d
'
f Bd d
(
−
d
')
ρ
;
y
c
y
d
d
'
Zou and Chan (2005) also adopted the follow-
ing assumptions for relating the plastic rotation
to the moment and reinforcements in the beam
and column.
They assumed a bilinear M-θ relation similar to
the one shown in Figure (7); accordingly a linear
relation, as in Eq.(45), can be found between
θ
p
, the
plastic rotation, and the moment in excess of M
y
.
2
(
)
(
)
2
J
=
ρ
+
ρ
'
n
+
2
ρ
+
ρ
'
n
− +
ρ
ρ
'
n
sc
sc
sc
(46)
2
(
)
f BD
n
2
f Bd d
−
d
'
D Jd
J
ρ
⋅
y
y
M
=
⋅
−
+
⋅
;
y
2
2
3
1
−
J
2
d
1
−
J
sc
2
F
f Bd
F
f Bd
D
d
F
f Bd
J
= −
2
ρ
+
X
n
+
2
ρ
+
X
n
2
+
2
n
ρ
+
X
sc
sc
sc
y
y
y
(47)
M M
M
−
−
U
y
θ
=
θ
(45)
p
P
M
in which
n
sc
is the ratio of modules of elasticity of
steel,
Es
and concrete
Ec
.
ρ
and
ρ'
are the ratios
of reinforcements in tensile and compression zone
of concrete beam. Zou and Chan (2005) assumed
that
ρ
and
ρ'
are linearly dependent to each other.
They were assumed equal in columns.
U
y
Zou and Chan assumed
M
=
1 1
.
M
and
U
y
could write
θ
p
as a function of
M
Zou and Chan (2005) used an explicit rela-
tionship between M
y
and reinforcement ratios
ρ
and
ρ'
for beams, Eq.(46), and beam-columns,
Eq.(47), as follows:
They expressed the relationship between plastic
rotation and reinforcement ratios by second order
Taylor series expansion as follows:
0
2
0
+
∂
θ ρ
ρ
(
)
(
∂
θ ρ
ρ
(
)
(
1
2
0
p
0
p
0 2
θ ρ
( )
=
θ ρ
(
)
ρ
−
ρ
)
+
ρ
−
ρ
)
p
p
2
∂
∂
(48)
Figure 7. A bilinear M-θ relation
To find the derivatives of
θ
p
with respect to
design variables it is noted that Eqs.(46 and 47
provide explicit relation between M and
ρ.
There-
fore, using Eq.(45)
θ
p
can be expressed in terms
of
ρ
and the derivatives can be easily obtained.
Substituting
θ
p
from Eq.(48) into Eq.(44) and
utilizing chain rule in differentiating, produces an
explicit relation between pushover displacement
and reinforcement ratios.
N
∂
∆
(
ρ
0
)
(
N
∂
2
∆
(
ρ
0
)
1
2
i
i
∑
∑
0
j
i
0
j
i
0 2
∆
(
ρ
)
=
∆
(
ρ
)
+
ρ
−
ρ
)
+
(
ρ
−
ρ
)
j
i
j
i
i
i
2
i
i
∂
ρ
∂
ρ
i
=
1
i
=
1
i
i
(49)
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