Civil Engineering Reference
In-Depth Information
In the case of
ε
=
ε
y
for mild steel, the compatibility equation for
ε
, Equation (5.97) or
4 , becomes:
1
cos
2
ε
r
=−
ε
d
tan
2
α
r
+
ε
y
(5.155)
α
r
Similarly, in the case of
ε
t
=
ε
y
, the compatibility equation for
ε
t
, Equation (5.98) or 5 ,
give:
1
sin
2
ε
r
=−
ε
d
cot
2
α
r
+
ε
y
(5.156)
α
r
ε
r
in Equations (5.155) or (5.156) can be used in conjunction with
Equation
8
to calculate the softening coefficient of concrete
The tensile strain
. Then the softened stress-strain
relationship of Eqs.
7
a
and
7
b
will be used to trace and to locate the peak of the concrete
stress. The procedures to plot the inner boundary curves,
I
and
I
t
, are summarized as follows:
ζ
Step 1: Select an angle
α
r
and insert it into Equations (5.153)-(5-156).
Step 2: By increasing the strain
ε
d
incrementally and calculating the softening coefficient
ζ
by
Equation 8 in conjunction with the strain
ε
r
from Equation (5.155) or (5.156), we can trace
and locate the peak of the normalized concrete stress
f
c
by the stress-strain relationship
of Equations 7
a
or 7
b
, similar to the process in Table 5.4.
Step 3: Substituting the peak value of
σ
d
/
f
y
into Equations (5.153) and (5.154)
we obtain the percentages of the longitudinal and the transverse steel
σ
d
and
f
=
f
t
=
ρ
and
ρ
t
respectively,
to locate one point on the inner boundary curves for the selected angle
α
r
.
Step 4: Select the angle
α
r
incrementally and repeat steps 1-3; we can generate a series of
points to plot the inner boundary curves.
30
◦
, for example, and locate a point on the inner boundary curve
I
, where
Take
α
r
=
ε
t
=
ε
y
and
ε
>ε
y
:
1
sin
2
30
◦
=−
ε
d
(3)
=−
ε
d
cot
2
30
◦
+
ε
y
Equation (5
.
156)
ε
r
+
0
.
00207(4)
=
3(
−
ε
d
)
+
0
.
00828
Now we can find the peak of the concrete stress as shown in the following Table 5.5.
Since the peak of concrete stress occurs at
f
c
=
0.33644, the longitudinal steel and the
transverse steel can be calculated from Equations (5.153) and (5.154) as:
σ
d
/
m
m
t
sin
2
30
◦
3364
f
c
f
y
−
0
.
sin 30
◦
cos 30
◦
+
ρ
=
0
5)
2
−
0
.
3364(27
.
56)
.
5
=
866
(0
.
5)(0
.
866)
+
(0
.
=
0
.
0112
413
0
.
m
t
m
t
cos
2
30
◦
3364
f
c
f
y
−
0
.
sin 30
◦
cos 30
◦
+
ρ
t
=
−
866)
2
−
0
.
3364(27
.
56)
0
.
5
=
866
(0
.
5)(0
.
866)
+
(0
.
=
0
.
0112
.
413
0
