Civil Engineering Reference
In-Depth Information
3875
f
c
=
properties of concrete are:
f
c
=
48.1 MPa,
ε
o
=
0.0024, and
E
c
=
26 875 MPa.
ρ
ρ
t
ratio of 9.0, and thus a very large
value of 16.7
◦
Since panel M3 has a very large
β
at
peak load, it is a good example to illustrate the power of SMM.
Panel M3 was subjected to a principal tensile stress
σ
2
of equal magnitude in the 1-2 coordinate. These applied stresses resulted in a pure shear stress
in the
σ
1
, and a principal compressive stress
−
t
coordinate and during its loading history: i.e.
σ
=
0MPa,
σ
t
=
0 MPa and an
applied shear stress
τ
t
that increased until failure. Analyze the behavior of this 2-D element M3
by the softened membrane model (SMM). Particular interest is placed in the stress and strain
conditions at the first yield of steel, at the peak load stage, and in the descending branch when
the selected concrete compressive strain
ε
2
=−
0.003. The computer program was terminated
at
0.008.
In order to use the SMM, the Poisson effect must be taken into account by using Equations
7 - 12 . The constitutive laws of concrete in compression, in tension and in shear were
calculated by Equations 13 - 17 , while the constitutive laws of smeared mild steel were
calculated by Equations 18
a
- 18
e
and 19
a
- 19
e
.
|
ε
2
|
>
|
ε
lim
|
, where the limiting concrete compressive strain
ε
lim
is taken as
−
6.1.12.2 Solution
The calculations according to the flow chart in Figure 6.23 are best done by computer, because
the procedure involves a nested DO-loop. The computer-calculated results are recorded in
Table 6.1 for three critical points: point 1 at first yield; point 2 at the peak strength; and point
3 in the descending branch. Since the SMM method is most powerful when applied to the
descending branch, we will calculate all the stresses and strains at point 3 in a step-by-step
manner.
Select
ε
2
=−
0.00300. At the final cycle of the calculation:
Assume
γ
12
=
0
.
01672
Assume
ε
1
=
0
.
027416
45
◦
,
α
1
=
because the 2-D element is subjected to pure shear
.
α
1
−
γ
1
2
ε
=
ε
1
cos
2
α
1
+
ε
2
sin
2
Equation
4
2sin
α
1
cos
α
1
=
0
.
027416(0
.
5)
+
(
−
0
.
00300)(0
.
5)
−
(0
.
01672)(0
.
5)
=
0
.
003848
α
1
+
γ
1
2
=
ε
1
sin
2
α
1
+
ε
2
cos
2
Equation
5
ε
t
2sin
α
1
cos
α
1
=
0
.
027416(0
.
5)
+
(
−
0
.
00300)(0
.
5)
+
(0
.
01672)(0
.
5)
=
0
.
020568
Equation
7
b
ν
12
=
1
.
9
since
ε
t
>ε
y
Equation 8
ν
21
=
0
after cracking
1
ν
12
Equation
9
ε
1
=
¯
−
ν
12
ν
21
ε
1
+
−
ν
12
ν
21
ε
2
=
ε
1
+
ν
12
ε
2
1
1
=
0
.
027416
+
1
.
9(
−
0
.
00300)
=
0
.
021716
ν
21
1
Equation
ε
2
=
¯
1
−
ν
12
ν
21
ε
1
+
1
−
ν
12
ν
21
ε
2
=
ε
2
=−
0
.
00300
10
