Civil Engineering Reference
In-Depth Information
According to Eq. (3.55), the area of the new adding reinforcement can be computed by
(
ϕ
=0
.
935):
=
N
0
.
8
f
y
1
A
s
1
f
y
A
s
−
ϕ
−
f
c
A
c
−
0
.
8
f
c
1
A
c
1
=
3
.
6
10
6
0
.
935
×
−
10
×
400
×
550
−
310
×
2036
−
0
.
8
×
12
.
5
550)
0
.
8
×
(500
×
650
−
400
×
×
310
=
1
.
02
10
6
248
<
0
10
6
−
×
1
.
05
×
The above-mentioned calculation explains that only using the shotcrete around the column
can satisfy the request for retrofitting. So, only 4
14 longitudinal constitutional rebars
are needed.
Example 3.7
In a five-story factory that is a frame structure, lifting large components
damaged the frame form, which induced the second floor to tilt seriously. Some columns
need to be retrofitted. Calculations of retrofitting a side column are given:
a. Design data. The section size of this central column is 400 mm
600 mm. The
concrete strength grade is C20. The story height
H
=5
.
0 m. The original design external
forces are
N
0
×
=
A
s
= 600 kN,
M
0
= 360 kN
·
m. The reinforcement is 4
14 (
A
s
= 1256
mm
2
). The extra design moment due to tilt is Δ
M
=50kN
m.
b. Retrofitting method. Single-side thickening is applied because the moment is unidirec-
tional. First, coarsen tensile concrete surface of original column and expose 80 mm stirrups.
Then weld U-shape stirrups on original stirrups and weld the longitudinal rebar to original
stirrups by short bars. Finally, spray 50 mm C25 fine concrete.
c. Computing
h
01
,
η
and
e
. From Fig. 3.40, the effective height of the retrofitted section
can be computed by:
·
h
01
= 600
−
10 = 590 mm
l
h
=
1
.
0
×
5
=7
.
69
<
8
η
=1)
0
.
65
10
3
600
e
0
=
M
N
360
×
=
= 600 mm
>
0
.
3
h
01
(large eccentric)
10
3
600
50
×
e
a
=
=83
.
3mm (
e
i
=
e
0
+
e
a
= 683
.
3 mm)
So,
h
2
e
=
ηe
i
+
h
01
−
= 683
.
3 + 590
−
325 = 948
.
3mm
80
35 15 35
welding
U-shaped stirrups
new adding reinforcement
50
600
Fig. 3.40
Retrofit of column.
