Civil Engineering Reference
In-Depth Information
36.43
◦
into Equation (5.33) we obtain the required longitudinal steel:
Substituting this
α
r
=
43
◦
)
ρ
f
y
=
σ
+
τ
t
(tan 36
.
(5.40)
τ
t
in the second term is multiplied by tan 36.43
◦
<
Since
1,
ρ
f
y
will be less than that if
45
◦
, thus resulting in a more economical design. However, when
α
r
=
α
r
is designed to be
less than 45
◦
, the crack widths are expected to be larger.
5.1.5.3 Solutions
The fixed angle
α
1
and the principal applied stresses,
σ
1
and
σ
2
, for the RC 2-D element can be
calculated directly from the applied membrane stresses
σ
,
σ
t
and
τ
t
. From Equation (4.25)
in Chapter 4:
(
σ
−
σ
t
)
2
2
.
13
−
(
−
2
.
13)
cot 2
α
1
=
=
=
0
.
577
τ
t
2(3
.
69)
60
◦
,
30
◦
.
2
α
1
=
and
α
1
=
From Equations (4.28) and (4.29) in Chapter 4:
σ
−
σ
t
2
2
2
2
σ
+
σ
t
2
2
.
13
−
2
.
13
.
13
+
2
.
13
2
σ
1
=
+
+
τ
t
=
+
+
3
.
69
2
2
2
=
0
+
4
.
26
=
4
.
26 MPa
σ
−
σ
t
2
2
σ
+
σ
t
2
2
σ
2
=
−
+
τ
=
0
−
4
.
26
=−
4
.
26 MPa
.
t
The Mohr circle for the applied stresses is shown in Figure 5.3(b). Indicated in this circle
are the angle 2
60
◦
,
4.26 MPa.
The reinforcement can be designed according to the basic equations of the equilibrium
(plasticity) truss model (Equations 5.33 and 5.34). Since
α
1
=
σ
1
=
4.26 MPa, and
σ
2
=−
ρ
f
y
=
ρ
t
f
ty
, the rotating angle
α
r
can be found by equating these two equations:
σ
+
τ
t
tan
α
r
=
σ
t
+
τ
t
cot
α
r
τ
t
(tan
α
r
−
cot
α
r
)
=−
(
σ
−
σ
t
)
Noticing that (tan
α
r
−
cot
α
r
) is equal to
−
2cot2
α
r
, the angle
α
r
is expressed as:
α
r
=
σ
−
σ
t
2
.
−
−
.
2
13
(
2
13)
=
=
.
cot 2
0
577
τ
lt
2(3
.
69)
60
◦
, and
30
◦
. Also, tan
2
α
r
=
α
r
=
α
r
=
0.577 and cot
α
r
=
1.732
Substituting
α
r
back into Equations (5.33) and (5.34), we have
ρ
f
y
=
σ
+
τ
t
tan
α
r
=
2
.
13
+
3
.
69(0
.
577)
=
4
.
26 MPa
ρ
t
f
ty
=
σ
t
+
τ
t
cot
α
r
=−
2
.
13
+
3
.
69(1
.
732)
=
4
.
26 MPa
4
26
413
=
.
ρ
=
ρ
t
=
ρ
=
0
.
0103
