Civil Engineering Reference
In-Depth Information
c
,and
L
1
elongated to
L
1
+Δ
L
1
, horizontal length of which is
L
1
−
Δ
L
1
. According to
geometrical relationship and original beam's opposite deflection caused by prestress,
f
)
2
=(
L
1
+Δ
L
1
)
2
−
(
L
1
−
Δ
L
2
)
2
(Δ
H
+
H
−
f
a
d
H
b
c
c
Δ
L
2
Δ
H
−
f
L
1
L
1
L
2
L
2
Fig. 3.21
Stretching curved prestressed reinforcement retrofitted beam at two points.
Develop the right item in the above formula, neglect higher order micro contents of
Δ
L
1
L
1
,
Δ
L
2
,andas
L
1
−
=
H
2
,thenget
f
)
2
=
H
2
+2
L
1
Δ
L
1
+2
L
1
(Δ
H
+
H
−
Δ
L
2
So the vertical top bracing value Δ
H
can be obtained by the formula:
H
+
H
2
+2(
L
1
Δ
L
1
+
L
1
Δ
L
2
)
Δ
H
=
f
−
(3
.
37)
where
L
1
and
L
1
are oblique reinforcement's initial length and its horizontal length, respec-
tively; Δ
L
1
is retrofitted reinforcement's deformation at the segment
L
1
:
σ
con
E
s
Δ
L
1
=
L
1
+
a
Δ
L
2
is retrofitted reinforcement's deformation at the segment
L
2
(horizontal segment/2);
H
is initial kink length of prestressed reinforcement (Fig. 3.21);
f
is opposite deflection
caused by prestress internal force. When opposite deflection is calculated, beam stiffness is
computed as Eq. (3.23). The factor that f influences on Δ
H
can be neglected if prestress is
small.
b. One-point vertical stretching curved reinforcement.
Fig. 3.22 shows vertical stretching curved prestressed reinforcement at one point to retrofit
beam. The difference from two-point stretching method lies on
L
2
=0
,
Δ
L
2
=0. Conse-
quently, calculation formula of stretching value can be obtained by substituting Δ
L
2
=0
into Eq. (3.37), and ordering Δ
L
=2Δ
L
1
,whichis
H
+
H
2
+
L
1
Δ
L
1
Δ
H
=
f
−
(3
.
38)
f
a
b
c
H
c
Δ
H
−
f
L
1
L
1
Fig. 3.22
Stretching curved prestressed reinforcement retrofitted beam at one point.
c. Vertical stretching straight-line reinforcement.
Fig. 3.23 shows stretching straight line prestressed reinforcement by one point and two
points method to retrofit beam. The difference from curved prestressed reinforcement is
L
1
=
L
1
,H
= 0. Substituting them into Eq. (3.37) and adopting
L
as the whole length
