Civil Engineering Reference
In-Depth Information
These applied stresses at first yield are recorded in terms of the Mohr circle in Figure
5.10(c). Accordingly, the stresses in the steel are:
f
=
0
.
857 (482)
=
413 MPa
f
t
=
0
.
857 (259)
=
222 MPa
ρ
f
=
0
.
857 (4
.
97)
=
4
.
26 MPa
ρ
t
f
t
=
.
.
=
.
0
857 (2
67)
2
29 MPa
These stresses for the steel are recorded in Figure 5.10(e). From the applied stresses and the
stresses in the steel we can calculate the stresses in the concrete element at first yield:
σ
−
ρ
f
=
1
.
82
−
4
.
26
=−
2
.
44 MPa
σ
t
−
ρ
t
f
t
=−
1
.
82
−
2
.
29
=−
4
.
11 MPa
σ
d
=
0
.
857 (
−
7
.
64)
=−
6
.
55 MPa
These stresses for the concrete are used to plot the Mohr circle in Figure 5.10(d). Finally,
the strains are calculated as follows:
σ
d
E
c
=
−
6
.
55
10
−
3
Equation (5
.
75)
ε
d
=
800
=−
0
.
265
×
24
,
f
E
s
=
413
200
10
−
3
Equation (5
.
73)
ε
=
000
=
2
.
07
×
,
f
t
E
s
=
222
200
10
−
3
Equation (5
.
74)
ε
t
=
000
=
1
.
11
×
,
10
−
3
10
−
3
Equation (5
.
71)
ε
r
=
ε
+
ε
t
−
ε
d
=
(2
.
07
+
1
.
11
+
0
.
265)
×
=
3
.
45
×
γ
t
2
=
Equation (5
.
72)
(
ε
r
−
ε
d
)sin
α
r
cos
α
r
10
−
3
(0
10
−
3
=
(3
.
45
+
0
.
265)
×
.
4831)
=
1
.
795
×
These strains are given in terms of the Mohr's strain circle in Figure 5.10(b). It should be
observed that the
α
r
value of 37.54
◦
for the strain condition is identical to that for the stress
condition in the concrete element, because the same
r-d
principal coordinate is used for both
stresses and strains.
Figure 5.10(b)-(d) shows the stresses and strains of the 2-D element at the first yield of
longitudinal steel. The set of externally applied stresses at this load stage is:
σ
=
1.82 MPa
(tension),
3.16 MPa. This set of applied stresses
at first yield is 0.857 times the set of applied stresses shown in Figure 5.10(a). The principal
stresses,
σ
t
=−
1.82 MPa (compression) and
τ
t
=
σ
1
and
σ
2
, in Figure 5.10(c) are calculated as follows:
1
(
σ
−
σ
t
2
2
σ
+
σ
t
2
1
.
82
−
1
.
82
.
82
+
1
.
82
σ
1
=
+
)
2
+
τ
2
=
+
+
.
16
2
3
t
2
2
=
0
+
3
.
65
=
3
.
65 MPa
σ
2
=
0
−
3
.
65
=−
3
.
65 MPa
