Civil Engineering Reference
In-Depth Information
Mild steel
ε
s
≤
ε
y
f
s
=
E
s
¯
ε
s
¯
18
a
,
19
a
ε
s
>ε
y
f
s
=
(0
.
91
−
2
B
)
f
y
+
(0
.
02
+
0
.
25
B
)
E
s
¯
ε
s
¯
18
b
,
19
b
f
s
=
f
p
−
E
s
(¯
ε
p
−
ε
s
)
¯
ε
s
<
¯
ε
p
¯
18
c
,
19
c
ε
y
=
f
y
/
f
y
=
E
s
(0
.
93
−
2
B
)
f
y
18
d
,
19
d
f
cr
f
y
1
.
5
31
f
c
(
MPa
) and
1
ρ
B
=
f
cr
=
0
.
ρ
≥
0
.
15%
18
e
,
19
e
Equations
to
are intended for the longitudinal steel when the subscripts s are
18
a
18
e
replaced by the subscript
, and Eqs.
19
a
to
19
e
are intended for the transverse steel when
the subscripts
s
are replaced by the subscript
t
.
6.1.11.2 Solution Algorithm
The 19 equations, Eqs.
1
to
19
will be used to solve 19 unknown variables. These 19
equations involve 22 variables, namely, 8 stresses (
c
c
c
σ
,
σ
t
,
τ
t
,
σ
1
,
σ
2
,
τ
12
,
f
,
f
t
), 10 strains
(
ε
,
ε
t
,
γ
t
,
ε
1
,
ε
2
,
γ
12
,¯
ε
1
,¯
ε
2
,¯
ε
,¯
ε
t
), the deviation angle
β
, the softening coefficients
ζ
, and
the two Hsu/Zhu ratios
ν
12
and
ν
21
. In the case of pure shear, the two applied normal stresses
σ
=
0 and
σ
t
=
0. When a value of
ε
2
is selected, the remaining 19 unknown variables can be
solved by the 19 equations.
To develop an efficient solution algorithm, the first two equilibrium equations 1 and 2
are summed and subtracted to obtain the following two equations, which are used as the
convergence criteria for the solution procedure (Hsu and Zhu, 2002):
−
σ
2
c
1
c
ρ
f
+
ρ
t
f
t
=
(
σ
+
σ
t
)
+
σ
20
−
σ
2
cos 2
c
1
c
c
12
sin 2
ρ
f
−
ρ
t
f
t
=
(
σ
−
σ
t
)
−
σ
α
1
+
2
τ
α
1
21
ε
2
The iterative procedure of SMM is illustrated by the flow chart in Figure 6.23. First select a
γ
12
. After completing the DO-loops in the flow chart,
Figure 6.23, using Equations
4
,
5
, and
7
-
21
, the shear stress
ε
1
and
value and assume two values for
τ
t
and the shear strain
γ
t
for a selected value of
ε
2
are calculated from Equations
3
and
6
, respectively. By selecting a
series of
ε
2
values, it is possible to plot the
τ
t
−
γ
t
curves. Curves relating any two variables
can similarly be plotted.
6.1.12 Example Problem 6.1
6.1.12.1 Problem Statement
45
◦
, has been tested by Chintrakarn
(2001). This 2-D element was reinforced with 2#6 mild steel bars at 189 mm spacing in the
longitudinal direction and 2#2 mild steel bars at 189 mm spacing in the transverse direction.
The properties of the longitudinal steel are:
A RC 2-D element M3, as shown in Figure 6.2(b) with
α
1
=
ρ
=
1.70%,
f
y
=
425.4 MPa, and
E
s
=
212 700
MPa. For transverse steel:
ρ
t
=
0.19%,
f
ty
=
457.9 MPa, and
E
s
=
184 600 MPa. The
