Civil Engineering Reference
In-Depth Information
The tensile strain
ε
r
can now be checked:
Equation
15
ε
r
=
ε
+
ε
t
−
ε
d
=
0
.
00798
+
0
.
00234
+
0
.
0005
=
0
.
01082
≈
0
.
0108 (assumed) OK
Check
σ
1
:
Equation
9
a
f
=
413 MPa (extensively yielded since
ε
=
0
.
0079
0
.
00207)
Equation 10
a
f
t
=
413 MPa (shortly after yielding since
ε
t
=
0
.
00234)
ρ
f
=
0
.
0103(413)
=
4
.
26 MPa
ρ
t
f
t
=
0
.
0103(413)
=
4
.
26 MPa
=
ρ
t
f
t
)
+
ρ
f
)
=
.
.
−
.
.
=
B
m
(
m
t
(
0
5(4
26)
0
5(4
26)
0
C
=
(
ρ
f
)(
ρ
t
f
t
)
=
(4
.
26)(4
.
26)
=
18
.
15 MPa
B
4
SC
15)
B
2
(0)
2
1
2
S
1
σ
1
=
±
−
=
−
±
−
−
.
Equation 17
0
4(
1)(18
2(
−
1)
=−
0
.
5(
−
8
.
52)
=
4
.
26 MPa
=
4
.
26 MPa (assumed)
OK
ε
d
,
ε
r
,
ε
, and
ε
t
and the stresses,
σ
d
and
σ
1
, are calculated from the
Now that the strains
trial-and-error process, we can determine the angle
α
r
, the shear stress
τ
t
and the shear strain
γ
t
:
ε
t
−
ε
d
ε
−
ε
d
=
0
.
00234
+
0
.
0005
Equation 16 tan
2
α
r
=
0005
=
0
.
334
0
.
00798
+
0
.
tan
α
r
=
0
.
577
30
◦
60
◦
α
r
=
2
α
r
=
τ
t
=
−
σ
d
)sin
α
r
cos
α
r
=
.
.
.
Equation 3
(
(8
54)(0
500)(0
866)
=
3
.
69 MPa
Equation 6
γ
t
/
2
=
(
ε
r
−
ε
d
)sin
α
r
cos
α
r
=
(0
.
0108
+
0
.
0005)(0
.
500)(0
.
866)
=
0
.
00489
The applied stresses at the level near ultimate where both steel have yielded are:
σ
=
m
σ
1
=
0
.
5(4
.
26)
=
2
.
13 MPa
σ
t
=
m
t
σ
1
=−
0
.
5(4
.
26)
=−
2
.
13 MPa
τ
t
=
m
t
σ
1
=
0
.
866(4
.
26)
=
3
.
69 MPa
