Brzev ch13 ism

13-1
Chapter 13 - Solutions
__________________________________________________________________________
First, check whether the wall meets criteria related to the Empirical Method for bearing wall
design (A23.3 Cl.14.2), as follows
a) Solid rectangular cross section constant along the wall height.
b), c) Wall is loaded concentrically.
d) The wall is supported against lateral displacement at the top and bottom.
In conclusion, the wall meets the above criteria and so the Empirical method can be used in the
design.
1. Determine the wall thickness.
Wall thickness is restricted based on A23.3 Cl.14.1.7.1, as follows:
Therefore, the 180 mm value governs, however use
2. Check the bearing resistance.
13.1.
hu 4500mm=
t
hu
25
------ 4500
25
------------ 180mm= =≥
t 150mm≥
t 200mm=
Copyright © 2006 Pearson Education Canada Inc.

13-2
The contact area is
Bearing resistance (A23.3 Cl.10.8.1) can be determined as
[12.28]
Since
okay
the wall bearing resistance is adequate.
3. Determine the factored axial load resistance.
• Determine the gross area of the wall section .
Determine according to A23.3 Cl.14.1.3.1.
i) bearing width
ii) will be determined by drawing the lines sloping downward from each side of the bearing
area - the slope is 2:1. Note that the value is limited by the intersection with the lines corre-
sponding to the adjacent point loads.
(based on the intersecting points)
iii)
The smallest value governs, hence
A1( )
A1 200mm 200mm× 40000mm
2
= =
Br 0.85φcfc′A1=
0.85 0.65 25MPa 40000mm
2
××× 552kN==
Br 552kN Pf 300kN=>=
Ag( )
lb
a 200mm=
lb a 2 9t×+ 200 2 9 200××+ 3800mm= = =
lb
lb
lb 2500mm=
lb s≤ 2500mm=
lb 2500mm=

13-3
Let us determine the gross area
• Determine the factored axial load resistance.
(pin supported wall)
Use [3.7]
(A23.3 Eq.14.1) [13.13]
Since
okay
the wall is adequate for the given loads.
4. Determine the distributed and concentrated wall reinforcement.
• Distributed horizontal reinforcement
i) Find the area of reinforcement.
Minimum required area of horizontal reinforcement (A23.3 Cl.14.1.8.6) can be determined as
ii) Determine the required bar spacing.
(15M bars, Table A.1)
[3.29]
Find the maximum permitted bar spacing according to A23.3 Cl.14.1.8.4.
In this case,
Horizontal reinforcement: 15M@500.
iv) Check whether one layer of reinforcement is adequate.
One layer is adequate when (A23.3 Cl.14.1.8.3). Since
Ag lb t× 2500mm 200mm× 500000mm
2
= = =
k 1.0=
α1 0.8≅
Pr
2
3
---α1φcfc′Ag 1
khu
32t
---------
2
–=
2
3
--- 0.8 0.65 25( MPa) 500 10
3
mm
2
×( ) 1
1 4500mm×
32 200×
-------------------------------
2
– 2191kN=××××=
Pr 2191kN Pf 300kN=>=
Ag 1000mm t× 1000mm 200mm× 200 10
3
mm
2
×= = =
Ahmin 0.002Ag 0.002 200 10
3
mm
2
×( ) 400mm
2
m⁄= = =
Ab 200mm
2
=
s Ab
1000
As
------------≤ 200mm
2 1000
400mm
2
m⁄
------------------------------× 500mm= =
s 500mm=
smax
3t 3 200 600mm=×=
500mm governs←
≤
s smax 500mm= =
t 210mm<

13-4
okay
• Distributed vertical reinforcement
i) Determine the area of vertical reinforcement.
Minimum area of vertical reinforcement (A23.3 Cl.14.1.8.5):
[13.1]
ii) Required bar spacing
[3.29]
iii) Check whether the bar spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).
Since
use
Vertical reinforcement: 15M@500
iv) Concentrated reinforcement (A23.3 Cl.14.1.8.8): 2-15M bars at each end
5. Sketch a design summary.
t 200mm 210mm<=
Ag 1000mm t× 1000mm 200mm× 200 10
3
mm
2
×= = =
Avmin 0.0015Ag 0.0015 200 10
3
mm
2
×( ) 300mm
2
m⁄= = =
Ab 200mm
2
=
s Ab
1000
As
------------ 200mm
2 1000mm
300mm
2
m⁄
------------------------------ 666mm= =≤
smax
3t 3 200 600mm=×=
500mm governs←
≤
s 666mm smax> 500mm= =
s 500mm=

13-5
_________________________________________________________________________
First, check whether the wall meets criteria related to the Empirical Method for bearing wall
design (A23.3 Cl.14.2), as follows
a) Solid rectangular cross section constant along the wall height
b), c) The wall is concentrically loaded
d) The wall is supported against lateral displacement at the top and bottom
In conclusion, the wall meets the above criteria and the Empirical Method can be used in this
design.
1. Determine the wall thickness.
(A23.3 Cl.14.1.7.1)
Since
use .
2. Determine the factored axial load.
(NBC 2005 Table 4.1.3.2)
13.2.
hu 4000mm=
t
hu
25
------
4000mm
25
---------------------- 160mm= =≥
t 150mm≥
t 160mm=
DL 80kN m⁄=
LL 40kN= m⁄
wf 1.25DL 1.5LL+=
1.25 80kN m⁄ 1.5 40
kN
m
-------×+× 160kN m⁄==

13-6
Find the gross area of the wall section (consider a strip of unit length ).
[3.7]
(A23.3 Eq.14.1) [13.13]
Since
okay
The wall axial resistance is adequate for this design.
• Distributed horizontal reinforcement.
i) Find the minimum horizontal reinforcement area (A23.3 Cl.14.1.8.6).
ii) Find the required bar spacing.
[3.29]
iii) Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)
Use
Horizontal reinforcement: 15M@480
One layer is adequate when , so
Ag( ) lb 1000mm=
Ag lb t× 1000mm 160mm× 160 10
3
mm
2
×= = =
k 1.0=
α1 0.8≅
Pr
2
3
---α1φcfc′Ag 1
khu
32t
---------
2
–=
2
3
--- 0.8 0.65 25MPa 160 10
3
mm
2
×( )×××× 1
1 4000mm×
32 160mm×
-------------------------------
2
– 540= kN m⁄=
Pr 540.3kN m⁄ wf> 160kN m⁄= =
Ag 1000 t× 1000 160× 160 10
3
mm
2
×= = =
Ahmin 0.002Ag 0.002 160 10
3
mm
2
×( ) 320mm
2
m⁄= = =
Ab 200mm
2
=
s Ab
1000
As
------------≤ 200mm
2 1000mm
320mm
2
m⁄
------------------------------ 625mm==
smax
3t 3 160 480mm governs←=×=
500mm
≤
s 480mm=
t 210mm<

13-7
okay
i) Find the minimum required area (A23.3 Cl.14.1.8.5).
(same as Step 4)
[3.29]
Since
Use
iv) Concentrated reinforcement at wall ends (A23.3 Cl.14.1.8.8): 2-15M bars at each end
t 160mm 210mm<=
Ag 160 10
3
mm
2
×=
Avmin 0.0015Ag 0.0015 160 10
3
mm
2
×( ) 240mm
2
m⁄= = =
Ab 200mm
2
=
s Ab
1000
As
------------ 200mm
2 1000mm
240mm
2
m⁄
------------------------------× 833mm= =×≤
smax
3t 3 160× 480mm governs←= =
500mm
≤
s 833mm smax 480mm=>=
s 480mm=

13-8
__________________________________________________________________________
1. Find the shear force and bending moment distribution in the wall.
Load analysis (NBC 2005 Table 4.1.3.2):
Find the average pressure:
Use the equivalent wall model loaded by a uniform pressure of 60 kPa (see the sketch below).
Note that very similar results would be obtained by using the “actual” pressure distribution.
13.3.
p1f 1.5p1 1.5 20× 30kPa= = =
p2f 1.5p2 1.5 60× 90kPa= = =
pof
p1f p2f+
2
--------------------
30 90+
2
------------------ 60kPa= = =
Mf
pof l( )
2
8
----------------
60kPa( ) 4m( )
2
8
------------------------------------- 120kNm m⁄= = =
Vf
pof l×
2
--------------- 60kPa 4m×
2
------------------------------- 120kN m⁄= = =

13-9
2. Design the wall for flexure.
i) Find the effective depth .
Use 20M bars
cover (interior exposure, Table A.2)
(wall thickness)
ii) Find the required moment resistance.
Set
iii) Find the required area of vertical tension reinforcement.
Design the wall for flexure based on a strip of unit width .
Find the reinforcement area - use the direct method.
[5.4]
iv) Select the amount of reinforcement in terms of size and spacing.
Use 20M bars see Table A.1
[3.29]
v) Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)
Since
Use .
vi) Confirm that the CSA A23.3 maximum tension reinforcement (Cl.10.5.2) requirement has
been satisfied.
Actual reinforcement area:
d
db 20mm=( )
20mm=
t 300mm=
d t cover–
db
2
-----– 300 20–
20
2
------– 270mm= = =
Mr Mf 240kNm m⁄= =
b 1000mm=
As 0.0015fc′b d d
2 3.85Mr
fc′b
------------------––=
0.0015 25MPa 1000mm 270 270( )
2 3.85 120 10
6
×( )×
25MPa( ) 1000mm( )
---------------------------------------------------–– 1377mm
2
m⁄=××=
Ab 300mm
2
=( )
s Ab
1000
As
------------ 300mm
2
( )=×
1000mm
1377mm
2
------------------------ 218mm=≤
smax
3t 3 300 900mm=×=
500mm governs←
≤
s 218mm smax 500mm=<=
s 200mm=

13-10
Reinforcement ratio:
[3.1]
Balanced reinforcement ratio:
Since
okay
vii) Check the CSA A23.3 minimum reinforcement requirement (Cl.14.1.8.5).
[13.1]
Since
okay
The vertical reinforcement is adequate.
3. Design the wall for shear.
Design shear force is
a) Concrete shear resistance .
Find the effective shear depth :
So,
(larger value governs)
Set (unit strip)
Find (A23.3 Cl.11.3.6.3b).
(A23.3 Eq.11.9) [6.13]
As Ab
1000
s
------------ 300mm
2
( )
1000mm
200mm
---------------------- 1500mm
2
m⁄===
ρ
As
bd
------
1500mm
2
1000mm 270mm×
----------------------------------------------- 0.0055= = =
ρb 0.022= fc′ 25MPa Table A.4,=( )
ρ 0.0055 ρb 0.022=<=
Ag 1000mm t× 1000mm 300mm× 300 10
3
mm
2
×= = =
Avmin 0.0015Ag=
0.0015 300 10
3
mm
2
×( ) 450mm
2
m⁄==
As 1500mm
2
= m⁄ 450mm
2
m⁄>
Vf 120kN m⁄=
Vc( )
dv( )
dv
0.9d
0.72t
0.9 270mm×
0.72 300mm×
243mm
216mm
= = =
dv 243mm 240mm≅=
bw 1000mm=
β
β
230
1000 dv+
-----------------------=

13-11
Find .
(A23.3 Eq.11.6) [6.12]
b) Is shear reinforcement required?
According to A23.3 Cl.11.2.8.1, shear reinforcement is not required when
Since
it follows that the shear reinforcement is not required, however CSA A23.3 Code requires mini-
mum provision of horizontal reinforcement in walls.
c) Determine the minimum distributed horizontal reinforcement.
i) Reinforcement area (A23.3 Cl.14.1.8.6)
20M bars:
[3.29]
iii) Check whether the spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).
Since
okay
According to A23.3 Cl.14.1.8.3, one layer of reinforcement is adequate when the wall thickness is
less than 210 mm, that is, . In this case, . Therefore, two layers of rein-
230
1000 240+
--------------------------- 0.185 0.18≅==
Vc
Vc φcλβ fc′bwdv=
0.65 1.0( ) 0.18( ) 25MPa 1000mm( ) 240mm( )= 140kN m⁄=
Vf Vc≤
Vf 120kN m⁄= Vc 140kN m⁄=<
Ag 1000mm t× 1000mm 300mm× 300 10
3
mm
2
×= = =
Ahmin 0.002Ag=
0.002 300 10
3
mm
2
×( ) 600mm
2
m⁄==
Ab 300mm
2
, Table A.1=( )
s Ab
1000
As
------------ 300mm
2 1000mm
600mm
2
----------------------× 500mm= =≤
s 500mm=
smax
3t 3 300 900mm=×=
500mm governs←
≤
s smax 500mm= =
t 210mm< t 300mm=

13-12
forcement are required. Instead of using 20M bars for horizontal reinforcement, it would be bet-
ter to use 15M bars.
For 2-15M bars at 500 mm spacing, the area is
Since
Use 2 layers of horizontal reinforcement, that is, 15M@500.
Note that one layer of vertical reinforcement has been specified at the exterior wall face
(15M@500) in order to satisfy the requirement for 2 layers of reinforcement. The main vertical
tension reinforcement for this wall is located at the interior face.
Ab 2 200 400mm
2
=×=
As Ab
1000
s
------------ 400mm
2
( )
1000mm
500mm
----------------------× 800mm
2
m⁄= = =
As 800mm
2
= m⁄ 600mm
2
m⁄>

13-13
_________________________________________________________________________
a) Design the wall for the given reinforcement.
1. Determine the design bending moments and shear forces in the wall.
Load analysis (NBC 2005 Table 4.1.3.2):
Find the sum of moments about the support B as
So, the reaction at point A is equal to
The resultant of lateral load can be calculated as (see the sketch above)
Find the reaction at point B from the equilibrium of lateral forces as
or
13.4.
pf 1.5 30×
kN
m
2
------- 45
kN
m
2
-------= =
ΣMB
45
kN
m
2
------- 4m×
2
---------------------------- 4m
3
-------- A 3m( )– 0= =
A 40kN=
R
45
kN
m
2
------- 4m×
2
---------------------------------- 90kN==
ΣH A B R 0=–+=

13-14
Sketch the shear force diagram.
The point of zero shear (located at a distance from the top of the wall) can be determined from
the equilibrium of lateral forces on a free-body diagram of the top portion of the wall (see the
sketch above):
Equation of equilibrium of horizontal forces for the free-body diagram:
or
The distance can be found from the above equation as
Develop the bending moment diagram.
The following critical moment values need to be determined:
• Bending moment at support A (negative):
• Maximum bending moment at the distance from the top of the wall:
B R A– 90= kN 40kN 50kN=–=
x
px
pf
4
---- x 11.25x=×=
ΣH
px x
2
×
2
----------------- A 0=–=
11.25x
2
2
------------------- 40kN=
x
x 2.667m=
M
11.25(
kN
m
2
------- 1m( )
2
×
6
-------------------------------------------------– 1.87kNm 2kNm–≅–= =
x 2.67m=

13-15
Find the effective depth .
Estimate (reinforcement located in the middle of the wall - 1 layer).
Section at the midspan (maximum positive bending moment)
Find the required moment resistance.
Set
Find the required area of vertical tension reinforcement.
[5.4]
Select the amount of reinforcement in terms of size and spacing.
[3.29]
M
11.25
kN
m
2
------- 2.67m( )×
2
6
--------------------------------------------------– 40kN( ) 1.67m( )+ 53kNm= =
d
d 100mm=
Mr Mf 53kNm m⁄= =
b 1000mm=
As 0.0015fc′b d d
2 3.85Mr
fc′b
------------------––=
0.0015 25MPa( ) 1000mm( ) 100mm 100mm
2 3.85 53 10
6
×( )
25MPa 1000mm( )
----------------------------------------------––=
2142mm
2
m⁄=
Ab 300mm
2
=
s Ab
1000
As
------------ 300mm
2
( )=×
1000mm
2142mm
2
------------------------ 140mm=≤

13-16
iii) Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)
Since
Use .
Confirm that the CSA A23.3 maximum tension reinforcement requirement (Cl.10.5.2) has been
satisfied.
[3.1]
Since
This amount of reinforcement is acceptable per CSA A23.3, however it is considered to be large
as it is very close to the balanced reinforcement.
Check the CSA A23.3 minimum reinforcement requirement (Cl.14.1.8.5).
[13.1]
Since
okay
Section at the support A (negative bending moment)
Set
[5.4]
smax
3t 3 200 600mm=×=
500mm governs←
≤
s 140mm=
ρ
As
bd
------
2142mm
2
1000mm 100mm×
----------------------------------------------- 0.0214= = =
ρb 0.022= fc′ 25MPa Table A.4,=( )
ρ 0.0214 ρb 0.022=<=
Ag 1000mm t× 1000mm 200mm× 200 10
3
mm
2
×= = =
Avmin 0.0015Ag=
0.0015 200 10
3
mm
2
×( ) 300mm
2
m⁄==
As 2143mm
2
= m⁄ 300mm
2
m⁄>
Mr Mf 2kNm m⁄= =
As 0.0015fc′b d d
2 3.85Mr
fc′b
------------------––=

13-17
[3.29]
(A23.3 Cl.14.1.8.4, same as for other section)
Since
Use .
Find the actual reinforcement area.
[13.1]
Since
okay
Vertical reinforcement: 15M@500.
Design shear force:
• Concrete shear resistance .
So,
As 0.0015 25( ) 1000( ) 100 100
2 3.85 2 10
6
×( )
25 1000( )
---------------------------------–– 58mm
2
m⁄==
Ab 200mm
2
=
s Ab
1000
As
------------ 200mm
2
( )=×
1000mm
58mm
2
---------------------- 3448mm=≤
smax 500mm=
s 500mm=
As Ab
1000
s
------------ 200mm
2
( )
1000mm
500mm
----------------------× 400mm
2
m⁄= = =
Ag 1000mm t× 1000mm 200mm× 200 10
3
mm
2
×= = =
Avmin 0.0015Ag=
0.0015 200 10
3
mm
2
×( ) 300mm
2
m⁄==
As 400mm
2
= m⁄ 300mm
2
m⁄>
Vf 50kN m⁄=
Vc( )
dv( )
dv
0.9d
0.72t
0.9 100mm×
0.72 200mm×
90mm
144mm
= = =
dv 140mm≅

13-18
Set (unit strip)
Find (A23.3 Cl.11.3.6.3b).
(A23.3 Eq.11.9) [6.13]
Find .
(A23.3 Eq.11.6) [6.12]
• Is shear reinforcement required?
Since
• Determine the minimum distributed horizontal reinforcement.
15M bars
[3.29]
Since
bw 1000mm=
β
β
230
1000 dv+
-----------------------=
230
1000 140+
--------------------------- 0.2==
Vc
0.65 1.0( ) 0.2( ) 25MPa 1000mm( ) 140mm( )= 91kN m⁄=
Vf Vc≤
Vf 50kN m⁄= Vc 91kN m⁄=<
Ag 1000mm t× 1000mm 200mm× 200 10
3
mm
2
×= = =
Ahmin 0.002Ag=
0.002 200 10
3
mm
2
×( ) 400mm
2
m⁄==
Ab 200mm
2
, Table A.1=( )
s Ab
1000
As
------------ 200mm
2 1000mm
400mm
2
----------------------× 500mm= =≤
s 500mm=
smax
3t 3 200 600mm=×=
500mm governs←
≤

13-19
okay
b) Comment on the effectiveness of placing the reinforcement in the middle of the wall.
It is convenient to place vertical reinforcement in the middle of the wall; in this way, the wall is
able to resist both positive and negative bending moments without adjusting the location of rein-
forcing steel to suit the moment diagram. This is a good approach to save labour cost on site, as
well as to reduce complexities of reinforcement installation and minimize chances for field errors.
c) Does this reinforcement strategy represent the most cost-effective solution?
The required amount of vertical reinforcement placed in the middle of the wall is very large (very
close to the balanced reinforcement). In order to reduce the amount of reinforcement, an alterna-
tive solution can be considered wherein the reinforcement is placed close to the inside wall face.
This will result in an increased effective depth which will in turn result in a reduced amount of
vertical reinforcement.
When the reinforcement is moved to the inside face of the wall, the effective depth can be esti-
mated as
Consider the section with the maximum bending moment, where
Find the required area of vertical tension reinforcement- use the direct method.
s smax 500mm= =
d 200mm 35mm–≅ 165mm=
Mr Mf 53kNm m⁄= =

13-20
[5.4]
Use 15M bars ( , see Table A.1)
[3.29]
Use 15M@200.
This alternative solution results in significantly smaller amount of vertical reinforcement - there is
a 50% reduction in the total reinforcement area (1009 mm2
versus 2142 mm2
).
Vertical wall elevation and the reinforcement arrangement is presented below.
As 0.0015fc′b d d
2 3.85Mr
fc′b
------------------––=
0.0015 25MPa( ) 1000mm( ) 165 165mm
2 3.85 53 10
6
×( )
25MPa 1000mm( )
----------------------------------------------––=
1009mm
2
m⁄=
Ab 200mm
2
=
s Ab
1000
As
------------ 200mm
2
( )=×
1000mm
1009mm
2
------------------------ 198mm=≤

13-21
_________________________________________________________________________
a) Draw the bending moment diagram for the wall and the footing.
Gravity load effects
Factored axial load (NBC 2005 Table 4.1.3.2)
Eccentricity
Factored bending moment:
Axial forces in the slabs (see the sketch above):
Bending moment diagram is shown on the sketch below.
Lateral load effects
Factored soil pressure (NBC 2005 Table 4.1.3.2)
13.5.
Pf 1.25P 1.25 150 188kN m⁄=×==
e
1m
3
-------- 0.25m
2
---------------– 0.2m= =
Mf Pf e× 188=
kN
m
------- 0.2m× 38
kNm
m
------------= =
T C
Mf
h
-------
38kNm m⁄
3m
--------------------------- 12.5kN m⁄= = = =
pf 1.5 20kPa× 30kPa= =

13-22
Bending moment diagram is shown on the sketch below.
Combined gravity and lateral loads
Shear force diagram is obtained using the principle of superposition, by combining the corre-
sponding shear force values for gravity and lateral loads (see below).
Maximum bending moment develops at the point of zero shear (distance 1.08 m from the top of
the wall). The corresponding value can be obtained as follows
Mf
pf h
2
×
8
----------------
30kPa( ) 3m( )
2
×
8
------------------------------------------- 33.8kNm m⁄= = =
Mf 32.5kN m⁄( ) 1.08m( ) 30kN m⁄( )
1.08m( )
2
2
---------------------- 17.6kNm m⁄=×–×=

13-23
Bending moment diagram due to combined gravity and lateral loads is shown below.
Find the inflection point located at a distance from the base of the wall (see the sketch above).
Bending moment at a distance from the base of the wall is equal to zero, that is,
or
The solution of this quadratic equation is
b) Design the flexural reinforcement for the footing and the exterior wall face.
Use 20M bars
Footing:
cover (Table A.2)
(footing thickness)
Wall:
cover (exterior exposure, Table A.2)
(wall thickness)
x
Mx x
Mx 38– 57.5 x( ) 30
x( )
2
2
----------–+ 0= =
15x
2
57.5x– 38+ 0=
x 0.85m=
d
db 20mm=( )
75mm=
t 300mm=
d t cover–
db
2
-----– 300 75–
20
2
------– 215mm= = =
50mm=
t 250mm=

13-24
Use
for the design (smaller value).
Set
[5.4]
[3.29]
Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)
Since
Use .
satisfied.
d t cover–
db
2
-----– 250 50–
20
2
------– 190mm= = =
d 190mm=
Mr Mf 38kNm m⁄= =
b 1000mm=
As 0.0015fc′b d d
2 3.85Mr
fc′b
------------------––=
0.0015 25MPa 1000mm 190 190( )
2 3.85 38 10
6
×( )×
25MPa( ) 1000mm( )
---------------------------------------------------–– 603mm
2
m⁄=××=
Ab 300mm
2
=( )
s Ab
1000
As
------------ 300mm
2
( )=×
1000mm
603mm
2
---------------------- 497mm 450mm≅=≤
smax
3t 3 250 750mm=×=
500mm governs←
≤
s 450mm=
As Ab
1000
s
------------ 300mm
2
( )
1000mm
450mm
---------------------- 667mm
2
m⁄===

13-25
[3.1]
Since
okay
[13.1]
Since
okay
Determine the reinforcement length for the wall (above the footing-wall interface).
Development length for 20M bars in this example (bottom bars, regular uncoated bars, normal-
density concrete) can be determined from Table A.5 as
A23.3 Cl.12.11.3 (as applied to continuous beams) states that (see Section 9.7.8)
(A23.3 Eq.12-6) [9.8]
where
(greater value governs)
So,
Since
okay
Therefore, the total wall reinforcement length above the wall-footing interface can be obtained as
ρ
As
bd
------
667mm
2
1000mm 190mm×
----------------------------------------------- 0.0035= = =
ρb 0.022= fc′ 25MPa Table A.4,=( )
ρ 0.0035 ρb 0.022=<=
Ag 1000mm t× 1000mm 250mm× 250 10
3
mm
2
×= = =
Avmin 0.0015Ag=
0.0015 250 10
3
mm
2
×( ) 375mm
2
m⁄==
As 667mm
2
= m⁄ 375mm
2
m⁄>
ld 575mm=
ld
Mr
Vf
------- la+≤
la
d
12db
190mm
12 20mm 240mm=×
= =
la 240mm=
ld
38kNm
57.5kN
------------------- 0.24m+ 0.66m 0.24m+ 0.9= m 900mm==≤
ld 575mm 900mm<=

13-26
c) Design the vertical and horizontal reinforcement at the interior wall face.
cover (interior exposure, Table A.2)
(wall thickness)
Set
[5.4]
Use 15M bars ( , see Table A.1)
[3.29]
Find the maximum permitted bar spacing (A23.3 Cl.14.1.8.4)
Since
Use .
l 0.85m 0.575m+ 1.425m 1.5m≅= =
d
20mm=
t 250mm=
d t cover–
db
2
-----– 250 20–
20
2
------– 220mm= = =
Mr Mf 18kNm m⁄≅=
b 1000mm=
As 0.0015fc′b d d
2 3.85Mr
fc′b
------------------––=
0.0015 25MPa 1000mm 220 220( )
2 3.85 18 10
6
×( )×
25MPa( ) 1000mm( )
---------------------------------------------------–– 240mm
2
m⁄=××=
Ab 200mm
2
=
s Ab
1000
As
------------ 200mm
2
( )=×
1000mm
240mm
2
---------------------- 833mm=≤
smax
3t 3 250 750mm=×=
500mm governs←
≤
s 500mm=

13-27
satisfied.
[3.1]
Since
okay
[13.1]
Since
okay
Design shear force:
• Concrete shear resistance .
So,
Set (unit strip)
Find (A23.3 Cl.11.3.6.3b).
As Ab
1000
s
------------ 200mm
2
( )
1000mm
500mm
---------------------- 400mm
2
m⁄===
ρ
As
bd
------ 400mm
2
1000mm 220mm×
----------------------------------------------- 0.0018= = =
ρb 0.022= fc′ 25MPa Table A.4,=( )
ρ 0.0018 ρb 0.022=<=
Ag 1000mm t× 1000mm 250mm× 250 10
3
mm
2
×= = =
Avmin 0.0015Ag=
0.0015 250 10
3
mm
2
×( ) 375mm
2
m⁄==
As 400mm
2
= m⁄ 375mm
2
m⁄>
Vf 57.5kN m⁄=
Vc( )
dv( )
dv
0.9d
0.72t
0.9 220mm×
0.72 250mm×
198mm
180mm
= = =
dv 200mm≅
bw 1000mm=
β

13-28
(A23.3 Eq.11.9) [6.13]
Find .
(A23.3 Eq.11.6) [6.12]
• Is shear reinforcement required?
Since
• Determine the minimum distributed horizontal reinforcement.
15M bars
[3.29]
Since
Use .
β
230
1000 dv+
-----------------------=
230
1000 200+
--------------------------- 0.19==
Vc
0.65 1.0( ) 0.19( ) 25MPa 1000mm( ) 200mm( )= 123.5kN m⁄=
Vf Vc≤
Vf 57.5kN m⁄= Vc 123.5kN m⁄=<
Ag 1000mm t× 1000mm 250mm× 250 10
3
mm
2
×= = =
Ahmin 0.002Ag=
0.002 250 10
3
mm
2
×( ) 500mm
2
m⁄==
Ab 200mm
2
, Table A.1=( )
s Ab
1000
As
------------ 200mm
2 1000mm
500mm
2
----------------------× 400mm= =≤
s 400mm=
smax
3t 3 250 750mm=×=
500mm governs←
≤
s 400mm=

13-30
______________________________________________________________________
• Determine the gross area of the wall section .
Determine according to A23.3 Cl.14.1.3.1.
i) bearing width
ii) will be determined by drawing the lines sloping downward from each side of the bearing
area - the slope is 2:1. Note that the value is limited by the intersection with the lines corre-
sponding to the adjacent point loads.
(based on the intersecting points, see the sketch above)
iii)
The smallest value governs, hence
Let us determine the gross area
13.6.
Ag( )
lb
a 300mm=
lb a 2 9t×+ 300 2 9 300××+ 5700mm= = =
lb
lb
lb 1200mm=
lb s≤ 1200mm=
lb 1200mm=

13-31
• Determine the factored axial load resistance
Use [3.7]
(A23.3 Eq.14.1) [13.13]
Since
okay
the wall is adequate for the given loads.
• Check whether one layer of reinforcement is adequate.
One layer is adequate when (A23.3 Cl.14.1.8.3). Since
use 2 layers of reinforcement
• Distributed horizontal reinforcement
i) Find the area of reinforcement.
Minimum required area of horizontal reinforcement (A23.3 Cl.14.1.8.6) can be determined as
Use 2 layers
[3.29]
Find the maximum permitted bar spacing according to A23.3 Cl.14.1.8.4.
Since
Ag lb t× 1200mm 300mm× 360000mm
2
= = =
k 1.0=
α1 0.8≅
Pr
2
3
---α1φcfc′Ag 1
khu
32t
---------
2
–=
2
3
--- 0.8 0.65 25( MPa) 360 10
3
mm
2
×( ) 1
1 4000mm×
32 300×
-------------------------------
2
– 2578kN=××××=
Pr 2578kN Pf 500kN=>=
t 210mm<
t 300mm 210mm>=
Ag 1000mm t× 1000mm 300mm× 300 10
3
mm
2
×= = =
Ahmin 0.002Ag 0.002 300 10
3
mm
2
×( ) 600mm
2
m⁄= = =
Ab 200mm
2
=
s Ab
1000
As
------------≤ 2 200× mm
2
( )
1000
600mm
2
m⁄
------------------------------× 666mm= =
smax
3t 3 300 900mm=×=
500mm governs←
≤

13-32
Use
Horizontal reinforcement: 15M@500 (2 layers).
i) Determine the area of vertical reinforcement.
Minimum area of vertical reinforcement (A23.3 Cl.14.1.8.5):
[13.1]
Use 2 layers
[3.29]
Since
use
Vertical reinforcement: 10M@400 (2 layers)
• Concentrated reinforcement (A23.3 Cl.14.1.8.8): 2-15M bars at each end
s smax 500mm= =
Ag 1000mm t× 1000mm 300mm× 300 10
3
mm
2
×= = =
Avmin 0.0015Ag 0.0015 300 10
3
mm
2
×( ) 450mm
2
m⁄= = =
Ab 100mm
2
=
s Ab
1000
As
------------ 2 100× mm
2
( )
1000mm
450mm
2
m⁄
------------------------------ 444mm= =≤
smax
3t 3 300 900mm=×=
500mm governs←
≤
s 444mm smax< 500mm= =
s 400mm=

13-33
_______________________________________________________________________
1. Determine the design axial loads, bending moments, and shear forces.
Wind load:
Factored wind load (NBC 2005 Table 4.1.3.2):
Find the factored bending moment.
Find the factored shear force.
Find the factored axial load at the base of the wall.
13.7.
w 150kN m⁄=
wf 1.4 w× 1.4 150× 210kN m⁄= = =
Mf
wf l×
2
2
----------------
210kN m⁄( ) 7.0m( )
2
2
-------------------------------------------------- 5145kNm= = =
Vf wf l× 210
kN
m
------- 7.0m( ) 1470kN= ==
Pf 1.25P2 1.25P1+ 1.25 700× 1.25 500×+ 1500kN= = =

13-34
Develop the bending moment, shear force, and axial force diagrams.
The critical section is at the base of the wall, where
Design shear force:
Wall length:
Wall thickness:
The effective shear depth is
Find (A23.3 Cl.11.3.6.3), assuming a minimum shear reinforcement:
Find
(A23.3 Eq. 11.6) [6.12]
Since
it follows that the shear reinforcement is required.
Find the required spacing of shear reinforcement
Mf 5145kNm 5100kNm≅=
Vf 1470kN= 1500kN≅
Pf 1500kN=
Vf 1500kN=
lw 5000mm=
t 200mm=
dv( )
dv 0.8 lw× 0.8 5000mm( )× 4000mm= = =
β
β 0.18=
Vc
Vc φcλβ fc′tdv=
0.65 1.0 0.18 30MPa 200mm 4000mm 512.7kN 510kN≅=××××=
Vf 1500kN Vc 510kN=>=

13-35
where
[6.9]
Use 15M bars, hence
Find (A23.3 cl.11.3.6.3) as
The spacing of horizontal reinforcement can be determined from equation (6.9) as
So,
Check whether the spacing is within the CSA A23.3 prescribed limits (Cl.14.1.8.4).
Since
okay
Check whether one layer of reinforcement is adequate (A23.3 Cl.14.1.8.3).
Since one layer of reinforcement is adequate.
3. Design the wall for flexure and axial load.
Design axial and flexural load effects:
• Determine the minimum vertical reinforcement (A23.3 Cl.14.1.8.5).
Vs Vf Vc– 1500kN 510kN– 990kN= =≥
Vs
φsAhfydv θcot
s
----------------------------------=
Ah 200mm
2
=
θ
θ 35°= θcot 1.43=
s
φsAhfydv θcot
Vs
----------------------------------=
0.85 200mm
2
( )× 400MPa( ) 4000mm( )1.43
990 10
3
N×
------------------------------------------------------------------------------------------------------------ 393mm= =
s 393mm 350≅ mm=
smax
3t 3 200 600mm=×=
500mm governs←
≤
s 350mm 500mm<=
t 200mm 210mm<=
Pf 1500kN=
Mf 5100kNm=
Ag 1000mm t× 1000mm 200× 200 10
3
mm
2
×= = =
Avmin 0.0015Ag=

13-36
Set
Determine the required bar spacing.
15M bars
[3.29]
Check whether the spacing is within the limits prescribed by the CSA A23.3 (Cl.14.1.8.4).
Since
Hence, use
Actual area:
• Determine the moment resistance .
i) Determine the total vertical reinforcement area along the wall length as
ii) Calculate the parameters , , and .
, [3.7]
[13.9]
[13.10]
0.0015 200 10
3
mm
2
×( )× 300mm
2
m⁄==
As Avmin 300mm
2
m⁄= =
Ab 200mm
2
, Table A.1=( )
s Ab
1000
As
------------ 200
1000
300
------------ 666mm= =≤
smax
3t 3 200 600mm=×=
500mm governs←
≤
s 500mm=
Av
Ab
s
------ 1000×
200
500
--------- 1000× 400mm
2
m⁄= = =
Mr( )
Avt( )
Avt As lw× 400mm
2
m⁄( ) 5m( ) 2000mm
2
= = =
ω α
c
lw
----
α1 0.8≅ β1 0.9≅
ω
φsfyAvt
φcfc′lwt
-------------------=
0.85 400MPa 2000mm
2
××
0.65 30MPa 5000mm 200mm×××
----------------------------------------------------------------------------------------- 0.035==
α
Pf
φcfc′lwt
-------------------
1500 10
3
N×
0.65 30MPa 5000mm 200mm×××
----------------------------------------------------------------------------------------- 0.077= = =

13-37
[13.8]
iii) Find the value.
[13.7]
Since
it follows that the moment resistance is not adequate.
However, let us take into account the effect of concentrated reinforcement as well.
Assume the minimum concentrated reinforcement according to A23.3 Cl.14.1.8.8, that is, 2-15M
bars at each end of the wall section.
2-15M bars:
(see the sketch above)
Moment resistance provided by the concentrated reinforcement can be determined as
The total wall moment resistance is
Since
it can be concluded that the total moment resistance provided by the combined distributed and
concentrated wall reinforcement is adequate.
c
lw
---- ω α+
2ω α1 β1×+
--------------------------------
0.035 0.077+
2 0.035 0.8 0.9×+×
-------------------------------------------------- 0.142= = =
Mr
Mr 0.5φsfyAvtlw 1
Pf
φsfyAvt
-----------------+ 1
c
lw
----–=
0.5 0.85 400MPa 2000mm
2
( )××× 5000mm( ) 1
1500 10
3
N×
0.85 400MPa 2000mm
2
××
---------------------------------------------------------------------+ ×=
1 0.142–( ) 4676kNm=×
Mr 4676kNm Mf 5100kNm=<=
As 2 200× 400mm
2
= =
x 4850mm=
Mr∆ T x× φsfyAs( )x= =
0.85 400MPa 400mm
2
××( ) 4850mm( ) 659.6kNm 660kNm≅==
Mr
total
Mr Mr∆+ 4676 660+ 5336kNm= = =
Mr
total
5336kNm Mf 5100kNm=>=

13-38
__________________________________________________________________________
1. Determine the design axial loads, bending moments, and shear forces.
Factored wind loads (NBC 2005 Table 4.1.3.2):
Factored gravity loads (NBC 2005 Table 4.1.3.2):
Develop the bending moment, shear force, and axial force diagrams.
The critical wall section is at the base of the wall, where
13.8.
H1f 1.4 H1× 1.4 400kN× 560kN= = =
H2f H1f 560kN= =
H3f 1.4H3 1.4 200kN× 280kN= = =
P1f 1.25P1 1.25 500× 625kN= = =
P2f P1f 625kN= =
P3f 1.25 600× 750kN= =
Mf 8820kNm= 8800kNm≅
Vf 1400kN=

13-39
Design shear force:
Wall length:
Wall thickness:
The effective shear depth is
Find (A23.3 Cl.11.3.6.3), assuming a minimum shear reinforcement
Find .
(A23.3 Eq.11.6) [6.12]
Since
it follows that the shear reinforcement is required in this case.
Find the required spacing of shear reinforcement
where
[6.9]
Use 10M bars in 2 layers (because the wall thickness is larger than 210 mm according to A23.3
Cl.14.1.8.3), hence
Find (A23.3 cl.11.3.6.3) as
The spacing of horizontal reinforcement can be determined from equation (6.9) as
Pf 2000kN=
Vf 1400kN=
lw 6000mm=
t 250mm=
dv( )
dv 0.8lw 0.8 6000mm× 4800mm= = =
β
β 0.18=
Vc
Vc φcλβ fc′tdv=
0.65 1.0 0.18 30MPa 250mm 4800mm×××× 769kN==
Vf 1400kN Vc 769kN=>=
Vs Vf Vc–≥ 1400 769– 631kN= =
Vs
φsAhfydv θcot
s
----------------------------------=
Ah 2 100× 200mm
2
= =
θ
θ 35°= θcot 1.43=
s
φsAhfydv θcot
s
----------------------------------=

13-40
Check whether the spacing is within the CSA A23.3 prescribed limits (Cl.14.1.8.4).
Since
Use
Horizontal reinforcement: 10M@500 (2 layers)
In this design, two layers of reinforcement are required according to A23.3 Cl.14.1.8.3 (wall
thickness larger than 210 mm).
3. Design the wall for flexure and axial load.
Design axial and flexural load effects:
• Determine the minimum vertical reinforcement (A23.3 Cl.14.1.8.5).
Assume
Determine the required bar spacing.
10M bars
Assume 2 layers of reinforcement, so
[3.29]
Check whether the bar spacing is within the limits prescribed by CSA A23.3 (Cl.14.1.8.4).
0.85 200mm
2
( )× 400MPa( ) 4800mm( )1.43
631 10
3
N×
------------------------------------------------------------------------------------------------------------ 740mm= =
smax
3t 3 250 750mm=×=
500mm governs←
≤
s 500mm=
Pf 2000kN=
Mf 8800kNm=
Ag 1000mm t× 1000mm 250mm× 250 10
3
mm
2
×= = =
Avmin 0.0015Ag=
0.0015 250 10
3
mm
2
×( )× 375mm
2
m⁄==
As Avmin 375mm
2
m⁄= =
Area 100mm
2
, Table A.1=( )
Ab 2 100× 200mm
2
= =
s Ab
1000
As
------------ 200
1000
375
------------× 533mm= =≤
smax
3t 3 250 750mm=×=
500mm governs←
≤

13-41
Since
use
Vertical reinforcement: 10M@500 (2 layers)
• Determine the moment resistance .
i) Determine the total vertical reinforcement area along the wall length as
ii) Calculate the parameters , , and .
[3.7]
[13.9]
[13.10]
[13.8]
iii) Find the value.
[13.7]
Since
it follows that the moment resistance is not adequate (deficiency on the order of 15%).
s 533mm 500mm>=
s 500mm=
Mr( )
Avt( )
As
Ab
s
------ 1000×
200mm
2
500mm
--------------------- 1000× 400mm
2
m⁄= = =
Avt As lw× 400mm
2
m⁄( ) 6m( ) 2400mm
2
= = =
ω α
c
lw
----
α1 0.8≅ β1 0.9≅
ω
φsfyAvt
φcfc′lwt
-------------------=
0.85 400MPa 2400mm
2
××
0.65 30MPa 6000mm 250mm×××
----------------------------------------------------------------------------------------- 0.028==
α
Pf
φcfc′lwt
-------------------
2000 10
3
N×
0.65 30MPa 6000mm 250mm×××
----------------------------------------------------------------------------------------- 0.068= = =
c
lw
----
ω α+
2ω α1 β1×+
--------------------------------
0.028 0.068+
2 0.028 0.8 0.9×+×
-------------------------------------------------- 0.12= = =
Mr
Mr 0.5φsfyAvtlw 1
Pf
φsfyAvt
-----------------+ 1
c
lw
----–=
0.5 0.85 400MPa 2400mm
2
( )××× 6000mm( ) 1
2000 10
3
N×
0.85 400MPa 2400mm
2
××
---------------------------------------------------------------------+=
1 0.12–( ) 7434kNm=×
Mr 7434kNm Mf 8800kNm=<=

13-42
Let us increase the amount of vertical reinforcement by decreasing the bar spacing to
.
Use 2 layers of 10M@300 vertical rebars:
Recalculate , , and as follows
(same as before)
[13.8]
Since
the moment resistance is still not adequate.
However, let us take into the account the effect of concentrated wall reinforcement as well.
Assume the minimum concentrated reinforcement according to A23.3 Cl.14.1.8.8, that is, 2-15M
bars at each end of the wall section.
2-15M bars:
Moment resistance provided by the concentrated reinforcement can be determined as
s 300mm=
Av Ab
1000
s
------------ 200mm
21000
300
------------ 666mm
2
m⁄= ==
Avt As lw× 666mm
2
m⁄( ) 6m( )× 3996mm
2
4000mm
2
≅= = =
ω α
c
lw
----
ω
4000
2400
------------ 0.028× 0.047= =
α 0.068=
c
lw
----
ω α+
2ω α1 β1×+
--------------------------------
0.047 0.068+
2 0.047 0.8 0.9×+×
-------------------------------------------------- 0.141= = =
Mr 0.5 0.85 400MPa 4000mm
2
( )××× 6000mm( ) 1
2000 10
3
N×
0.85 400MPa 4000mm
2
××
---------------------------------------------------------------------+=
1 0.14–( ) 8669kNm=×
Mr 8669kNm= Mf 8800kNm=<
As 2 200× 400mm
2
= =
x 5850mm=
Mr T x× φsfyAs( )x 0.85 400MPa 400mm
2
××( ) 5850mm( )== =∆
796kNm 800kNm≅=

13-43
The total wall moment resistance is
Since
it can be concluded that the total moment resistance provided by the distributed and concentrated
wall reinforcement is adequate.
Mr
total
Mr Mr∆+ 8669 800+ 9469kNm= = =
Mr
total
9469kNm Mf 8800kNm=>=

Brzev ch13 ism

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