Civil Engineering Reference
In-Depth Information
Tracing the peak of
σ
d
/
f
c
(
α
r
=
30
◦
and
ε
t
=
ε
y
)
ε
r
Table 5.5
σ
d
/
f
c
ζ
ε
d
ζε
o
ε
d
Eq. (5-156)
Eq. 8
Eq. 7
a
Eq. 7
b
−
0.00050
0.00978
0.34342
0.72797
<
1
0.31801
−
0.00060
0.01008
0.33901
0.88493
<
1
0.33452
−
0.00064
0.01020
0.33729
0.94874
<
1
0.33640
−
0.00065
0.01023
0.33686
0.96478
<
1
0.33644 peak
−
0.00066
0.01026
0.33644
0.98086
<
1
0.33632
−
0.00070
0.01038
0.33476
1.04552
>
1
0.33473
−
0.00080
0.01068
0.33067
1.20966
>
1
0.33007
The coordinate of
ρ
=
ρ
t
=
0.0112 are given as point C in Figure 5.28. It is a point on the
α
r
from 0
◦
to 45
◦
and repeating the
same procedures we have the whole inner boundary curve
I
. Similarly, the inner boundary
curve
I
t
can be obtained using Equation (5.155) and varying the angle
30
◦
. By varying the angle
inner boundary
I
with
α
r
=
α
r
from 45
◦
to 60
◦
.On
this boundary curve
ε
t
>ε
y
.
It is interesting to point out that the design of reinforcement using the equilibrium (plasticity)
truss model in Example Problem 5.1, Section 5.1.5, resulted in the steel percentage of
ε
=
ε
y
and
ρ
=
ρ
t
=
0.0103. This pair of
ρ
and
ρ
t
is plotted as point E in Figure 5.28 for a concrete strength of
f
c
=
413 MPa (60 000 psi). Point E is
located in region (1), just within the inner boundary
I
, meaning that the 27.56 MPa (4000 psi)
concrete did not crush when both the longitudinal and the transverse steel yielded. If a lower
strength of concrete (say, 20 MPa or 3000 psi) is used, point E would be located in region (2)
outside the inner boundary
I
. This situation means that the concrete would crush before the
yielding of the transverse steel, and the assumption of the yielding of all the steel (
f
=
27.56 MPa (4000 psi) and a steel yield strength of
f
y
=
f
t
=
f
y
) could not be ensured.
The outer boundary curves,
O
and
O
t
, in Figure 5.28 can be obtained by the preceding
four-step procedures with one exception in step 3. In the case of
O
t
, the longitudinal steel stress
will yield,
f
=
f
y
, but the transverse steel stress will not,
f
t
<
f
y
. Hence, before the calculation
ρ
t
by Equation (5.154) the pre-yield transverse steel stress
f
t
has to be calculated from the transverse steel strain
of transverse steel percentage
ε
t
by the strain compatibility condition,
ε
t
=
ε
r
−
ε
+
ε
d
.Thestrain
ε
is, of course, equal to the yield stress
ε
y
, and the strains
ε
d
and
ε
r
are obtained in step 2 after the peak of the normalized concrete stress is located.
5.5 Concluding Remarks
The rotating angle softened truss model (RA-STM) for shear has the following characteristics
when compared with the fixed angle theory discussed in Chapter 6.
1. The direction of cracks is assumed to follow the principal
r
−
d
coordinate of the concrete
element. Consequently, the concrete shear stress
d
coordinate must vanish
and the 'contribution of concrete' (
V
c
) must be zero. Chapter 5 clearly shows the elegance
and the purity of the rotating angle theory.
τ
rd
in the
r
−
