Civil Engineering Reference
In-Depth Information
Calculate the longitudinal steel stress
f
:
f
cr
f
y
1
.
5
2
1
.
5
1
ρ
1
059
446
.
Equation 18
e
B
=
=
=
0
.
0175
0
.
01789
.
5
f
y
=
Equation 18
d
(0
.
93
−
2
B
)
f
y
=
(0
.
93
−
2
×
0
.
0175) 446
.
5
=
399
.
6MPa
ε
y
=
f
y
/
E
s
=
399
.
6
/
200 000
=
0
.
00200
<
ε
=
¯
0
.
004534
Notice that the longitudinal steel has yielded.
Equation
18
b
f
=
(0
.
91
−
2
B
)
f
y
+
(0
.
02
+
0
.
25
B
)
E
¯
ε
=
(0
.
91
−
2
×
0
.
0175) 446
.
5
+
(0
.
02
+
0
.
25
×
0
.
0175) 200 000 (0
.
004534)
=
390
.
7
+
22
.
10
=
412
.
8MPa
Calculate the transverse steel stress
f
t
:
f
cr
f
ty
1
.
5
2
1
.
5
1
ρ
t
1
059
462
.
Equation 19
e
B
=
=
=
0
.
0249
0
.
01193
.
6
f
y
=
Equation 19
d
(0
.
93
−
2
B
)
f
ty
=
(0
.
93
−
2
×
0
.
0249) 462
.
6
=
407
.
2MPa
ε
y
=
f
y
/
E
s
=
407
.
2
/
192 400
=
0
.
00212
<
ε
t
=
¯
0
.
006677
Notice that the transverse steel strain has also yielded and its strain (¯
ε
t
) is greater than the
longitudinal steel strain (¯
ε
).
Equation 19
b
f
t
=
(0
.
91
−
2
B
)
f
ty
+
(0
.
02
+
0
.
25B)
E
t
¯
ε
t
=
(0
.
91
−
2
×
0
.
0249) 462
.
6
+
(0
.
02
+
0
.
25
×
0
.
0249) 192 400 (0
.
006677)
=
397
.
9
+
33
.
70
=
431
.
6MPa
Check equilibrium equations
20
and
21
:
(
ρ
f
+
ρ
t
f
t
)
1
=
0
.
01789(412
.
5)
+
0
.
01193(431
.
6)
=
7
.
385
+
5
.
149
=
12
.
53 MPa
(
ρ
f
−
ρ
t
f
t
)
1
=
0
.
01789(412
.
5)
−
0
.
01193(431
.
6)
=
7
.
385
−
5
.
149
=
2
.
236 MPa
−
σ
2
c
1
c
Equation
20
(
ρ
f
+
ρ
t
f
t
)
2
=
(
σ
+
σ
t
)
+
σ
=
(0
+
0)
−
(0
.
2785
−
12
.
81)
=
12
.
53 MPa
≈
12
.
53 MPa OK
