Civil Engineering Reference
In-Depth Information
When transverse reinforcement goes across doors or windows, it is better to bend the rebar
perpendicular to the wall surface into a straight hook along edges of holes and then anchor
it. Holes with penetrating S-shape reinforcement in the wall must be drilled by machine. It
is better to drill holes with penetrating reinforcement in a slab by machine.
2. Compressive capacity calculation for retrofitted wall with reinforcement mat layer
Wall body retrofitted with reinforcement mat layer becomes composite masonry and its
compressive capacity of normal section can be calculated by Eqs. (4.2)
∼
(4.5).
3. Shear capacity calculation for retrofitted wall with reinforcement mat layer
There are many factors influencing shear capacity for retrofitted walls with layers, such as
compressive stress of upper wall, thickness and shear strength of mortar layer, reinforcement
layout quantity and strength of layer, thickness and shear strength of the original wall. In
reference to relative test results, shear capacity for a retrofitted wall with reinforcement mat
layer can be checked by the following formula:
(
f
vz
+0
.
7
σ
0
)
A
k
1
.
9
V
k
(4.7)
where
V
k
is the shear of the
k
th wall;
σ
0
is average compressive stress of wall section in
half of story height;
A
k
is section area of the
k
th wall in half of story height (area of doors
and windows should be deduced);
f
vz
is equivalent shear strength of original brick wall after
retrofitting (conversion shear strength for short) and lower value based on the two following
formulas should be adopted according to different repairing and retrofitting conditions, when
controlled by layer mortar strength:
nt
1
t
m
f
v
1
+
2
3
f
v
+
0
.
03
nA
sv
1
√
st
m
f
vz
=
f
y
(4.8)
when controlled by reinforcement strength:
0
.
4
nt
1
t
m
f
v
1
+0
.
26
f
v
+
0
.
03
nA
sv
1
f
vz
=
√
st
m
f
y
(4.9)
where
t
1
is thickness (mm) of cement mortar layer or mortar layer with reinforcement mat;
t
m
is thickness (mm) of original brick wall;
n
is the number of retrofitted layers of one wall;
f
v
1
is mortar shear strength (N/mm
2
) of layer and can be calculated by
f
v
1
=1
.
4
√
M
,
M
is mortar strength grade of layer;
f
v
is design value of brick masonry shear strength along
full-length cracks,
f
v
= 0 will be adopted for cracking wall without repairing cracks;
A
sv
1
is
section area of single reinforcement;
f
y
is design value of reinforcement strength (MPa);
s
is
spacing (mm) of reinforcements in reinforcement mat, and the unit of
√
s
in the formula is
still mm.
Example 4.2
Consider a four-floor brick house with walls between windows. The
wall width is 2.7 m and thickness 0.24 m. Its brick strength grade is MU7.5 and mortar
strength grade M2.5. Average compressive stress of the wall in half height of the first floor
is 0.39 MPa. Design value of seismic shear force the wall is
V
k
= 180 kN. Determine seismic
strength of the wall; if its strength is deficient, retrofit the wall by cement mortar layer with
reinforcement mat and check again.
a. Check seismic strength of original wall
f
v
=0
.
09 MPa can be obtained as wall shear strength along full-length cracks in
Code for
Design of Masonry Structures. ξ
N
=1
.
426 can be got as normal stress effect coecient of
masonry strength in
Code for Seismic Design of Buildings
. Masonry seismic shear strength
is
f
vE
=
ξ
N
f
v
=1
.
426
×
0
.
09 = 0
.
128 MPa
f
vE
·
A/γ
RE
=0
.
128
×
2700
×
240
/
1
.
0 = 82944 N
<V
k
