Civil Engineering Reference
In-Depth Information
g. Calculate the internal forces caused by prestress and its effect. As to the beams
retrofitted by exposed prestressed reinforcements, the section strength should be checked
considering the beam as eccentric compressive member.
h. Based on
σ
con
, calculate the prestress internal forces
10
5
N
N
p
=
σ
con
A
p
= 425
×
760 = 3
.
23
×
10
5
×
M
p
=
N
p
(
y
0
+
a
p
)=3
.
23
×
(442
.
9 + 15)
10
8
N
=1
.
48
×
·
mm = 148 kN
·
m
i. Check the normal section bearing capacity of original beam, considering it as an eccen-
tric compressive member.
External bending moment
M
acted on the beam is
M
=
M
max
−
M
p
= 557
.
2
−
148 = 409
.
2kN
·
m
10
5
M
N
409
.
2
×
e
0
=
=
= 1267 mm
>
0
.
3
h
0
3
.
23
×
10
5
Then
e
a
=0
,
i
=
e
0
+
e
a
= 1267 mm
Assume
η
=1,
ηe
i
= 1267
>
0
.
3
h
0
, check it as large eccentric compression member firstly.
Tentatively calculate the value of
x
(taking no account of the compressive constructional
reinforcements) according to Eq. (3.25)
x
=
N
p
+
f
y
A
s
f
cm
b
i
3
.
23
×
10
5
+ 2281
×
310
=
13
.
5
×
400
10
6
5400
=
1
.
03
×
= 190 mm
>h
i
= 100 mm
This shows that neutral axis goes through the flange of the section.
x
needs to be recal-
culated:
10
6
−
f
cm
(
b
i
−
b
)
h
i
10
6
−
x
=
1
.
03
×
=
1
.
03
×
13
.
5
×
(400
−
200)
×
100
f
cm
b
13
.
5
×
200
= 280 mm
<ξ
b
h
0
=0
.
53
×
640 = 339 mm
(belongs to large eccentric compression members).
e
=
ηe
i
+
y
0
−
a
s
= 1267 + 442
.
9
−
60 = 1650 mm
10
5
×
10
8
N
Ne
=3
.
23
×
1650 = 5
.
33
×
·
mm
h
0
−
f
cm
bx
h
0
−
+
f
cm
(
b
i
−
h
2
x
2
b
)
h
i
640
+13
.
5
640
280
2
100
2
=13
.
5
×
200
×
280
×
−
×
(400
−
200)
×
100
×
−
10
8
N
10
8
N
=5
.
37
×
·
mm
>Ne
=5
.
33
×
·
mm
= 760 mm
2
, which meets the
The calculation indicates that it's appropriate to adopt
A
p
bearing capacity demand.
