Civil Engineering Reference
In-Depth Information
Tracing the peak of
σ
d
/
f
c
(
α
r
=
45
◦
)
ζ
Table 5.4
σ
d
/
f
c
ε
d
ζε
o
ε
d
Eq. (5-148)
Eq. 7
a
Eq. 7
b
−
0.00050
0.46266
0.54035
<
1
0.36493
−
0.00060
0.45904
0.65354
<
1
0.40392
−
0.00070
0.45550
0.76839
<
1
0.43106
−
0.00080
0.45204
0.88488
<
1
0.44605
−
0.00086
0.45000
0.95556
<
1
0.44911
−
0.00087
0.44966
0.96739
<
1
0.44918 peak
−
0.00088
0.44933
0.97924
<
1
0.44914
−
0.00090
0.44866
1.00298
>
1
0.44866
−
0.00100
0.44535
1.12271
>
1
0.44480
In Table 5.4 the peak strain
ε
o
of nonsoftened concrete is taken as
−
0.002 in the calculation
of the ratio
ε
d
/ζ ε
o
. It should be recalled that the concrete strain
ε
d
is in the ascending branch
when
ε
d
/ζ ε
o
<
1 and is in the descending branch when
ε
d
/ζ ε
o
>
1. Table 5.4 illustrates that
f
c
peaks at a value of 0.44918
the normalized concrete stress
σ
d
/
≈
0.4492 at a concrete strain
ε
d
of 0.00087. The corresponding
ε
d
/ζ ε
o
value of 0.96739 is in the ascending branch but is
f
c
.
very close to the peak stress of
ζ
0.4492
f
c
and
f
s
=
Substituting
σ
d
=
f
y
into Equation (5.146) we obtain the total percentage
of steel for the balanced condition:
m
+
4492
f
c
f
y
ρ
+
ρ
t
=
−
0
.
2
m
t
+
m
t
(5.149)
2
m
t
In the case of Example Problem 5.4, Section 5.4.6, where
m
=
0.5,
m
t
=−
0.5,
m
t
=
0.866,
f
c
=
27.56 MPa (4000 psi) and
f
y
=
413 MPa (60 000 psi), the total percentage of
steel
ρ
+
ρ
t
is
0
ρ
+
ρ
t
=
−
0
.
4492(
−
27
.
56)
.
5
+
2(0
.
866)
−
0
.
5
=
0
.
0300 or 3
.
00%
413
2(0
.
866)
3.73, the percentages of
the steel in the longitudinal and the transverse directions for the balanced condition are:
Solving this equation with the equal strain condition of
ρ
/ρ
t
=
3
.
73
ρ
=
73
(0
.
0300)
=
2
.
36%
1
+
3
.
1
ρ
t
=
.
=
.
.
73
(0
0300)
0
633%
+
.
1
3
The balanced point B, having a coordinate of
ρ
=
2.36% and
ρ
t
=
0.633%, lies on the
dotted line OA in Figure 5.28.
