Civil Engineering Reference
In-Depth Information
Example 4.11
The calculated ratio in Example 4.8 is ρ
f
= 1.65%. Repeat the example
for the reinforcement ratios of ρ
f
=
1.0% and ρ
f
=
2.0%. Assume that
the effective depth of the beam,
d,
is unknown but the width of the
beam is unchanged (
b
= 16.0 in.).
Solution:
ω
b
= 0.1667
ρ
f
= 1.0%
ρ
f
= 1.65%
ρ
f
=
2.0%
ω
f
=
0.150
ω
f
=
0.2475
ω
f
=
0.300
FRP rupture
Concrete
crushing
Concrete
crushing
f
= 1.00
f
= 0.798
f
= 0.713
—
ω′
f
=
0.1975
ω′
f
=
0.2139
bd
2
= 9978 in.
3
bd
2
= 7736 in.
3
bd
2
= 7218 in.
3
d
= 25.0 in.
d
= 22.0 in.
d
= 21.2 in.
A
f
= 4.0 in.
2
A
f
= 5.81 in.
2
A
f
= 6.80 in.
2
Use 4#9
Use 5#10
Use 6#10
4.6 STRENGTH-REDUCTION FACTORS FOR FLEXURE
4.6.1 ACI 440.1R-06 approach
The relationship suggested by ACI 440.1R-06 for the strength-reduction
factor shown in Figure 4.9 is given in the following equation:
ρ≤ρ
0.55
for
f
b
ρ
ρ
f
b
(4.46)
φ=
0.30.25
+
ρ<ρ<
1.4
ρ
for
b
f
b
0.65
for
ρ≥
1.4
ρ
f
b
φ
0.65
ρ
f
φ =0.30+0.25
0.55
ρ
b
Transition
Concrete
Crushing
FRP
Rupture
ρ
f
ρ
b
1.4ρ
b
Figure 4.9
Strength-reduction factor as per ACI 440.1R-06.
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