Civil Engineering Reference
In-Depth Information
Force equilibrium:
0.85
fb cAf
β= +
Af
(5.36)
c
1
s
s
f
fe
0.85
fbaAf
f
−
c
s
s
A
=
(5.37)
f
fe
f
f
a
d
f
f
c
s
ρ=
0.85
−ρ
(5.38)
f
s
fe
fe
Moment equilibrium:
a
a
MMAf d
=φ =φ −+φΨ
Af
d
−
(5.39)
u
n
ss
f
f
fe
f
2
2
M
fbd
f
f
d
d
a
f
f
d
d
a
u
s
fe
=ρ −−Ψρ
d
d
−
(5.40)
s
f
f
f
2
2
2
φ
2
2
c
f
cf
c
f
Substituting Equation (5.38) into Equation (5.40) and rearranging the terms,
3
2
a
d
(
)
a
d
d
d
a
d
0.208
ˆ
0.208
ˆ
−+
2
Q
+
2.77
Q
+
Q
β
1
2
1
1
f
f
f
f
β
d
d
d
2.77
ˆ
1
+
Q
Ψ− =
0
(5.41)
f
1
d
f
f
where
87
d
d
ˆ
Q
=ρ
when U.S. customary units are used
(5.42a)
1
s
f
c
f
600
d
d
ˆ
Q
=ρ
when S.I. units are used
(5.42b)
1
s
f
c
f
M
fbd
d
d
ˆ
u
Q
=
φ
−Ψ−
Q
(5.43)
2
1
f
2
c
f
f
Now we need to determine at which level of FRP and steel reinforcement the
mode of failure switches from ductile crushing to brittle crushing. For that, we need
to derive a balanced FRP ratio at which concrete extreme-compression fiber reaches
0.003 at the same time that ε
s
= ε
y.
This is called ρ
f
bal
, and it is derived by Rasheed
and Pervais (2003) as follows:
ε
=
ε+ε
0.003
0.003
cu
bal
cu
y
bal
c
=
d
(5.44)
c
d
+ε
y
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