Civil Engineering Reference
In-Depth Information
Calculation at the peak point 3 using FA-STM:
Select ¯
ε
2
=−
0
.
00067
Assume ¯
ε
1
=
0
.
01188
Assume
γ
12
=
0
.
002143
45
◦
,
α
1
=
because the 2-D element is subjected to pure shear
.
α
1
−
γ
1
2
ε
1
cos
2
ε
2
sin
2
Equation
11
¯
ε
=
¯
α
1
+
¯
2sin
α
1
cos
α
1
=
0
.
01188(0
.
5)
+
(
−
0
.
00067) (0
.
5)
−
(0
.
002143) (0
.
5)
=
0
.
004534
α
1
+
γ
1
2
ε
1
sin
2
ε
2
cos
2
Equation
12
ε
t
¯
=
¯
α
1
+
¯
2sin
α
1
cos
α
1
=
0
.
01188(0
.
5)
+
(
−
0
.
00067) (0
.
5)
+
(0
.
002143) (0
.
5)
=
0
.
006677
2
tan
−
1
2
tan
−
1
1
γ
12
1
0
.
002143
845
◦
Equation 15
β
=
=
=
4
.
(¯
ε
1
−
¯
ε
2
)
(0
.
01188
+
0
.
00067)
5
9
1
8
f
c
≤
.
1
|
β
|
24
◦
Equation 14
ζ
=
0
.
√
1
−
+
400¯
ε
1
5
9
1
845
◦
24
◦
8
√
44
.
1
4
.
=
1
≤
0
.
√
1
−
+
400(0
.
01188)
.
=
(0
.
8734) (0
.
4170) (0
.
7981)
=
0
.
2906
ε
2
ζε
o
=
¯
−
0
.
00067
Equation
13
b
00235)
=
0
.
981
<
1 ascending branch
0
.
2906(
−
0
.
2
¯
2
¯
ε
2
ζε
o
ε
2
ζε
o
c
2
f
c
σ
=
ζ
−
1)
2 (0
981)
2
=
0
.
2906(
−
44
.
.
981)
−
(0
.
=
0
.
2906(
−
44
.
1) (0
.
9996)
=−
12
.
81 MPa
Equation
16
b
ε
1
=
¯
0
.
01188
>ε
cr
=
0
.
00008 descending branch
31
f
c
(
MPa
)
31
√
44
f
cr
=
0
.
=
0
.
.
1
=
2
.
059 MPa
f
cr
ε
cr
¯
0
.
4
059
0
0
.
4
.
00008
c
1
σ
=
=
2
.
=
0
.
2785 MPa
ε
1
0
.
01188
c
1
c
2
σ
−
σ
0
.
2785
+
12
.
81
c
12
Equation
17
τ
=
ε
2
)
γ
12
=
00067)
(0
.
002143)
ε
1
−
.
+
.
2(¯
¯
2(0
01188
0
=
521
.
58(0
.
002143)
=
1
.
1177 MPa
