Civil Engineering Reference
In-Depth Information
x
=
1
h
0
2
M
f
cm
bh
01
−
1
−
=
1
1
10
6
2
×
194
×
−
−
×
565 = 113 mm
13
.
5
×
250
×
565
2
Bending moment resisted by prestressied reinforcements, Δ
M
,is
A
s
f
y
h
0
−
x
2
Δ
M
=
M
max
−
565
=7
.
2
113
2
10
6
−
10
7
kN
= 194
×
1140
×
210
×
−
×
·
m
Thus,
10
7
Δ
M
f
py
h
+
a
p
−
7
.
2
×
= 435 mm
2
A
p
=
=
600 + 30
x
2
113
2
290
×
−
Choose 2 20,
A
p
= 628 mm
2
.
e. Determine the tension control stress and calculate loss of prestress. According to Table
3.4, adopt the tension control stress as
σ
con
=0
.
85
f
pyk
=0
.
85
×
315 = 268 MPa
Calculate the loss of prestress. Two ends of the prestressed reinforcements are anchored
by welding, so
σ
l
1
=0.
Calculate the loss of prestress caused by friction at lower supporting points,
σ
12
.Known
from Fig. 3.29, the angle between diagonal brace and longitudinal axis is 0.565 rad, assume
the friction coecient
μ
=0
.
25. Then substituting them into Eq. (3.43) gives
e
−μθ
) = 268(1
e
−
0
.
25
×
0
.
565
)=35
.
3MPa
σ
l
2
=
σ
con
(1
−
−
The loss of prestress caused by stress relaxation can be obtained by Eq. (3.44).
σ
l
4
=0
.
05
σ
con
=13
.
4MPa
Hence,
σ
l
=
σ
l
2
+
σ
l
4
=35
.
3+13
.
4=48
.
7MPa
f. Calculate prestress internal forces.
10
5
N
N
p
=
σ
con
A
p
= 268
×
628 = 2
.
2
×
h
2
+
a
p
=2
.
2
600
2
+30
=7
.
26
10
6
×
10
7
N
M
p
=
N
p
×
×
·
mm
g. Check the normal sectional strength of original beam according to eccentric compressive
members. External bending moment resisted by beam is
M
=
M
max
−
M
p
= 194
−
72
.
6 = 121
.
4kN
·
m
10
5
e
0
=
M
N
=
121
.
4
×
= 552 mm
>
0
.
3
h
0
2
.
2
×
10
5
Thus,
e
a
=0
,
i
=
e
0
+
e
a
= 552 mm
