Civil Engineering Reference
In-Depth Information
ε
2
ζε
o
=
¯
−
0
.
00300
Equation 13
b
0024)
=
11
.
63
>
1 descending branch
0
.
1075(
−
0
.
1
2
(¯
ε
2
/ζ ε
o
)
−
1
2
f
c
σ
=
ζ
−
(4
/ζ
)
−
1
1
2
11
.
63
−
1
=
0
.
1075 (
−
48
.
1)
−
(4
/
0
.
1075)
−
1
=
0
.
1075 (
−
48
.
1) (0
.
9138)
=−
4
.
725 MPa
Equation 16
b
ε
1
=
¯
0
.
21716
>ε
cr
=
0
.
00008 descending branch
31
f
c
(
MPa
)
31
√
48
f
cr
=
.
=
.
.
=
.
0
0
1
2
150 MPa
f
cr
ε
cr
¯
0
.
4
150
0
0
.
4
.
00008
σ
1
=
=
.
=
.
2
0
2285 MPa
ε
1
0
.
021716
c
c
2
σ
1
−
σ
0
.
2285
+
4
.
725
c
Equation
17
τ
12
=
ε
1
−
ε
2
)
γ
12
=
00300)
(0
.
01672)
2(
2(0
.
027416
+
0
.
=
81
.
42(0
.
01672)
=
1
.
361 MPa
Calculate longitudinal steel stress
f
:
f
cr
f
y
1
.
5
2
1
.
5
.
1
ρ
1
150
425
=
=
=
.
Equation 18
e
B
0
0211
0
.
0170
.
4
f
y
=
Equation
18
d
(0
.
93
−
2
B
)
f
y
=
(0
.
93
−
2
×
0
.
0211)(425
.
4)
=
377
.
7MPa
ε
y
=
f
y
/
E
s
=
377
.
7
/
212
,
700
=
0
.
00178
>
ε
=
¯
0
.
000998
Notice that longitudinal steel did not yield.
Equation
18
a
f
=
E
s
¯
ε
=
212 700 (0
.
000998)
=
212
.
3MPa
Calculate transverse steel stress
f
t
:
f
cr
f
ty
1
.
5
2
1
.
5
1
ρ
t
1
150
457
.
Equation 19
e
B
=
=
=
0
.
1693
0
.
0019
.
9
f
y
=
.
−
2
B
)
f
ty
=
.
−
×
.
.
=
.
Equation 19
d
(0
93
(0
93
2
0
1693)(457
9)
270
8MPa
ε
y
=
f
y
/
E
s
=
270
.
8
/
184
,
600
=
0
.
00147
<
¯
ε
t
=
0
.
01772
Notice that the transverse steel reached into the strain-hardening range.
Equation
19
b
f
t
=
(0
.
91
−
2
B
)
f
ty
+
(0
.
02
+
0
.
25B)
E
s
¯
ε
t
=
(0
.
91
−
2
×
0
.
1693) (457
.
9)
+
(0
.
02
+
0
.
25
×
0
.
1693) (184 600) (0
.
01772)
=
261
.
6
+
203
.
9
=
465
.
5MPa
.
