Civil Engineering Reference
In-Depth Information
1
2
p
c
t
d
+
1
2
(3987)(96)
t
d
(96)
2
Equation [22]
A
o
=
A
c
−
=
975 000
−
+
793 000 mm
2
=
Equation [23]
p
o
=
p
c
−
4
t
d
=
3987
−
4(96)
=
3603 mm
5031 mm
2
A
=
13(387)
=
790 mm
2
A
p
=
8(98
.
7)
=
Solve
ε
A
o
(
−
ε
d
)(
−
σ
d
)
Equation [20]
ε
=
ε
d
+
(
A
f
+
A
p
f
p
)
793 000(0
.
0005)(13
.
51)
=−
0
.
0005
+
(5031
f
+
790
f
p
)
Assume elastic range
10
3
Equation [15
a
]
f
=
E
s
ε
=
200
×
ε
before yielding
10
3
(0
Equation [17
a
]
f
p
=
E
ps
(
ε
dec
+
ε
)
=
200
×
.
005
+
ε
) before elastic limit
10
−
6
Then (
ε
+
0
.
0005)(
ε
+
0
.
000678)
=
4
.
60
×
ε
=
0
.
00156
ε
y
=
0
.
00207
>
0
.
00156 O.K. for mild steel
ε
ps
at 0
.
7
f
pu
−
ε
dec
=
0
.
00652
−
0
.
005
=
0
.
00152
≈
0
.
00156
OK for prestressed steel
Solve
ε
t
A
o
s
(
−
ε
d
)(
−
σ
d
)
Equation [21]
ε
t
=
ε
d
+
p
o
(
A
t
f
t
+
A
tp
f
tp
)
793 000(203)(0
.
0005)(13
.
51)
=−
0
.
0005
+
(3603)(284
f
t
+
0)
1
.
063
f
t
ε
t
+
0
.
0005
=
Assume elastic range
10
3
Equation [16
a
]
f
t
=
E
s
ε
t
=
200
×
ε
t
before yielding
t
10
−
6
Then
ε
+
0
.
0005
ε
t
−
5
.
32
×
=
0
ε
t
=
0
.
00207
=
ε
y
(transverse steel just yielded)
Check
ε
r
Equation [24]
ε
r
=
ε
+
ε
t
−
ε
d
=
0
.
00156
+
0
.
00207
+
0
.
0005
=
0
.
00413
≈
0
.
00414 assumed OK
