Civil Engineering Reference
In-Depth Information
The applied stresses at the level of first yield are:
σ
=
m
σ
1
=
0
.
5(3
.
65)
=
1
.
82 MPa
σ
t
=
m
t
σ
1
=−
0
.
5(3
.
65)
=−
1
.
82 MPa
τ
t
=
m
t
σ
1
=
0
.
866(3
.
65)
=
3
.
16 MPa
≈
3
.
17 MPa OK
.
These calculations show that the stress and strain conditions analyzed by the softened truss
model are almost identical to those obtained by the Mohr compatibility truss model in Figure
5.10, Example Problem 5.2, Section 5.3.4. Therefore, the Mohr circles for the stresses and
strains calculated by the RA-STM in this problem will not be given here.
2. Select
0.0005 (after yielding of steel in both directions)
This is a point after the yielding of both the longitudinal and the transverse steel. The stress
condition of the 2-D element is shown in Figure 5.3(a)-(d). The applied principal tensile stress
must be
ε
d
=−
4.26 MPa. The strain condition, however, will be determined here. After several
cycles of the trial-and-error process we assume:
σ
1
=
ε
r
=
.
σ
1
=
.
0
0108
and
4
26 MPa
0
.
9
0
.
9
Equation 8
ζ
=
√
1
ε
r
=
√
1
0108)
=
0
.
329
+
+
600(0
.
600
ε
d
ζε
o
=
−
0
.
0005
002)
=
0
.
760
<
1
Ascending branch
0
.
329(
−
0
.
2
ε
d
ζε
o
2
ε
d
ζε
o
f
c
Equation
7
a
σ
d
=
ζ
−
760)
2
]
=
.
−
.
.
−
.
=−
.
0
329(
27
6)[2(0
760)
(0
8
54 MPa
Solve the longitudinal steel strain
ε
(assume yielding of steel)
ε
r
−
ε
d
−
σ
d
0
.
0108
+
0
.
0005
10
−
3
/MPa
=
=
1
.
323
×
8
.
54
m
σ
1
−
ρ
f
−
ρ
p
f
p
ε
=
ε
r
+
ε
r
−
ε
d
−
σ
d
Equation 13
P
10
−
3
=
0
.
0108
+
1
.
323
×
(0
.
5(4
.
26)
−
4
.
26
−
0)
=
0
.
00798
>
0
.
00207 (
ε
y
)OK
Solve the transverse steel strain
ε
t
(assume yielding of steel)
m
t
σ
1
−
ρ
t
f
t
−
ρ
tp
f
tp
=
ε
r
+
ε
r
−
ε
d
−
σ
d
ε
t
Equation 14
P
10
−
3
=
0
.
0108
+
1
.
323
×
(
−
0
.
5(4
.
26)
−
4
.
26
−
0)
=
0
.
00234
>
0
.
00207 (
ε
y
)
,
OK
,
shortly after yielding
