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Term Project for Seismic design
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THE FIRST UNIT
TERM PROJECT
ADVANCED REINGORCED CONCRETE
13CIVL10H
Submitted by:
ABDELRHMAN OKASHA
ABDULLAH SAAD
AMGAD ALAA
NOHA HANY
April, 2014
i
Contents LIST OF FIGURES ........................................................................................................................ ii
LIST OF TABLES ......................................................................................................................... iii
1.Project Summary .......................................................................................................................... 1
2.Assumptions ................................................................................................................................. 1
3.Axial Load Calculations .............................................................................................................. 1
4.Column Dimensions..................................................................................................................... 2
5.Columns and Shear walls Arrangement ....................................................................................... 3
6.Lateral Loads Calculations .......................................................................................................... 4
6.1Wind Load ............................................................................................................... 4
6.2 Seismic load .......................................................................................................... 11
7.Shear Walls Design .................................................................................................................... 18
7.1Shear Wall 1 .......................................................................................................... 18
7.2Shear Wall 2 .......................................................................................................... 22
References ..................................................................................................................................... 27
ii
LIST OF FIGURES
Figure 1 Plan View ......................................................................................................................... 3
Figure 2 Vertical cross section ........................................................................................................ 4
Figure 3 Elevation in X-direction ................................................................................................... 5
Figure 4 Bending moment diagram due to wind load in X-direction ............................................. 6
Figure 5 Shear force diagram due to wind load in X-direction ...................................................... 7
Figure 6 Elevation in Y-direction ................................................................................................... 8
Figure 7 Bending moment diagram due to wind load in Y-direction ............................................. 9
Figure 8 Shear force diagram due to wind load in Y-direction .................................................... 10
Figure 9 Bending Moment diagram due to seismic load along the building height. .................... 13
Figure 10 Shear Force diagram due to Seismic load along the building height. .......................... 14
Figure 11 The location of center of mass and center of rigidity ................................................... 15
Figure 12 Load- Capacity ratios for shear wall 1 ......................................................................... 21
Figure 13 Interaction diagram for shear wall 1 ............................................................................. 21
Figure 14 Load- Capacity ratios for shear wall 2 ......................................................................... 25
Figure 15 Interaction diagram for shear wall 1 ............................................................................. 25
Figure 16 Reinforced detailing of the shear walls. ....................................................................... 26
iii
LIST OF TABLES
Table 1 Moment due to wind load in X-directions ......................................................................... 6
Table 2 Shear force due to wind load in X-direction ...................................................................... 7
Table 3 Moment due to wind load in Y-direction........................................................................... 9
Table 4 Shear force due to wind load in Y-direction .................................................................... 10
Table 5 Building Description ........................................................................................................ 11
Table 6 Building Total Weight Calculations ................................................................................ 11
Table 7 Calculations of force/floor and accidental torsion due to seismic load ........................... 13
Table 8 inertia of the shear walls .................................................................................................. 15
Table 9 Total force on each shear wall due to seismic load ......................................................... 17
Table 10 Straining actions on each shear wall due to seismic load. ............................................. 17
1
1. Project Summary
The project objective is to locate the columns for a tall building, estimate the slab thickness
and locate the shear walls locations to resist the lateral forces due to earthquake and wind
loads. After locating of the shear walls it is required to calculate the lateral loads and
analyse the structure. Eventually, the design of lateral resisting elements should be
conducted then the reinforcement detailing should be presented.
Project
No. Soil Type
Building
Type
Building
Location Footprint Dimensions
4 Type A Commercial South Egypt 35X36
2. Assumptions
Fcu = 30 N/mm2
Fy = 400 N/mm2
(High grade steel).
Fy = 240 N/mm2 (horizontal steel).
L.L. = 5 KN/m2
(Commercial).
L.L. = 10 KN/m2 (Storage).
S.D.L. = 1.5 KN/m3
γ concrete = 25 KN/mm3
γ wall = 15 KN/mm3
Flat slab is used in all floors
Slab thickness =
=
= 220 mm
3. Axial Load Calculations
Dead load = 1.4 (F.C. + slab thickness* γ concrete) * No. of floors.
Dead load = 1.4 (1.5 + 25*0.22) *12 = 117.6 KN/m2.
Live load = 1.6 (L.L. comm.*No. of floors + L.L. storage * No. of floors).
Live load = 1.6 (5*11 + 10*1) = 104 KN/m2.
2
4. Column Dimensions
Interior Columns
D.L. = 117.6 * (7*6) = 4939.2 KN.
L.L. = 104 * (7*6) = 4368 KN.
Assume µst= 2 %
Ac =
=
= 645518.285 mm
2.
C1 (600*1100 mm).
Edge Columns
D.L. = 117.6 * (7*3) = 2469.6 KN.
L.L. = 104 * (7*3) = 2184 KN.
Assume µst= 2 %
Ac =
=
= 322759.14 mm
2.
C2 (600*550 mm).
Corner Columns
D.L. = 117.6 * (3.5*3) = 1234.8 KN.
L.L. = 104 * (3.5*3) = 1092 KN.
Assume µst= 2 %
Ac =
=
= 161379.57 mm
2.
C3 (300*550 mm).
3
5. Columns and Shear walls Arrangement
Figure 1 Plan View
4
6. Lateral Loads Calculations
6.1Wind Load
Pe = q Ce K
K = 1
q = 0.5 *10-3
(ρv2CtCs)
ρ = 1.25 Kg/m2
V = 36 m/sec (South Egypt)
Ct = 1 (topography coefficient)
Cs = 1 (<60 m)
q = 0.5 *10-3
(1.25*362*1*1) = 0.81 KN/m
2.
Vertical Cross Section
Pe (compression) = 0.81*0.8*1 = 0.648 KN/m2.
Pe (suction) = 0.81*0.5*1 = 0.405 KN/m2.
Figure 2 Vertical cross section
5
Wind Forces in X-direction (compression)
F1 = 0.648*2.75*35 = 62.37 KN.
F2 = 0.648*3.5*35 = 79.38 KN.
F3 = 0.648*3*35 = 68.04 KN.
F4 = 0.648*1.5*35 = 34.02 KN.
Wind Forces in X-direction (suction)
F5 = 0.405*2.75*35 = 38.2725 KN.
F6 = 0.405*3.5*35 = 49.6125 KN.
F7 = 0.405*3*35 = 42.525 KN.
F8 = 0.405*1.5*35 = 21.2625 KN.
Figure 3 Elevation in X-direction
6
Overturning Moment Calculations
Table 1 Moment due to wind load in X-directions
SLAB
NO.
HEIGHT
(m)
COMPRESSION
(KN)
SUCTION
(KN)
TOTAL
(KN) Moment (KN.m)
1 4 62.37 38.2725 100.6425 402.57
2 8 79.38 49.6125 128.9925 1031.94
3 11 68.04 42.525 110.565 1216.215
4 14 68.04 42.525 110.565 1547.91
5 17 68.04 42.525 110.565 1879.605
6 20 68.04 42.525 110.565 2211.3
7 23 68.04 42.525 110.565 2542.995
8 26 68.04 42.525 110.565 2874.69
9 29 68.04 42.525 110.565 3206.385
10 32 68.04 42.525 110.565 3538.08
11 35 68.04 42.525 110.565 3869.775
12 38 34.02 21.2625 55.2825 2100.735
Over turning moment at the base 26422.2
Figure 4 Bending moment diagram due to wind load in X-direction
0
5
10
15
20
25
30
35
40
0 5000 10000 15000 20000 25000 30000
Bu
ildin
g h
eig
ht
(m)
B.M.D (KN.m)
Bending Moment Along the building height
7
Shear Force Calculations
Table 2 Shear force due to wind load in X-direction
SLAB
NO.
HEIGHT
(m) COMPRESSION
(KN)
SUCTION
(KN)
TOTAL
(KN)
Force/Shear wall
(W2)
1 4 62.37 38.2725 100.6425 50.32125
2 8 79.38 49.6125 128.9925 64.49625
3 11 68.04 42.525 110.565 55.2825
4 14 68.04 42.525 110.565 55.2825
5 17 68.04 42.525 110.565 55.2825
6 20 68.04 42.525 110.565 55.2825
7 23 68.04 42.525 110.565 55.2825
8 26 68.04 42.525 110.565 55.2825
9 29 68.04 42.525 110.565 55.2825
10 32 68.04 42.525 110.565 55.2825
11 35 68.04 42.525 110.565 55.2825
12 38 34.02 21.2625 55.2825 27.64125
Total base shear 1280.0025
Figure 5 Shear force diagram due to wind load in X-direction
0
5
10
15
20
25
30
35
40
0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 1400.00
Bu
ildin
g h
eig
ht
(m)
Shear Force (KN)
Shear Force Along the building height
8
Wind Forces in Y-direction (compression)
F1 = 0.648*2.75*36 = 64.15 KN.
F2 = 0.648*3.5*36 = 81.648 KN.
F3 = 0.648*3*36 = 69.98 KN.
F4 = 0.648*1.5*36 = 34.99 KN.
Wind Forces in Y-direction (suction)
F5 = 0.405*2.75*36 = 40.095 KN.
F6 = 0.405*3.5*36 = 51.03 KN.
F7 = 0.405*3*36 = 43.74 KN.
F8 = 0.405*1.5*36 = 21.87 KN.
Figure 6 Elevation in Y-direction
9
Overturning Moment Calculations
Table 3 Moment due to wind load in Y-direction
SLAB
NO.
HEIGHT
(m) COMPRESSION
(KN)
SUCTION
(KN)
TOTAL
(KN) Moment (KN.m)
1 4 64.15 40.095 104.245 416.98
2 8 81.648 51.03 132.678 1061.424
3 11 69.98 43.74 113.72 1250.92
4 14 69.98 43.74 113.72 1592.08
5 17 69.98 43.74 113.72 1933.24
6 20 69.98 43.74 113.72 2274.4
7 23 69.98 43.74 113.72 2615.56
8 26 69.98 43.74 113.72 2956.72
9 29 69.98 43.74 113.72 3297.88
10 32 69.98 43.74 113.72 3639.04
11 35 69.98 43.74 113.72 3980.2
12 38 34.99 21.87 56.86 2160.68
Over turning moment at the base 27179.124
Figure 7 Bending moment diagram due to wind load in Y-direction
0
5
10
15
20
25
30
35
40
0 5000 10000 15000 20000 25000 30000
Bu
ildin
g h
eig
ht
(m)
B.M.D (KN.m)
Bending Moment Along the building height
10
Shear Force Calculations
Table 4 Shear force due to wind load in Y-direction
SLAB
NO.
HEIGHT
(m) COMPRESSION
(KN)
SUCTION
(KN)
TOTAL
(KN) Force/Shear wall (W1)
1 4 64.15 40.095 104.245 52.1225
2 8 81.648 51.03 132.678 66.339
3 11 69.98 43.74 113.72 56.86
4 14 69.98 43.74 113.72 56.86
5 17 69.98 43.74 113.72 56.86
6 20 69.98 43.74 113.72 56.86
7 23 69.98 43.74 113.72 56.86
8 26 69.98 43.74 113.72 56.86
9 29 69.98 43.74 113.72 56.86
10 32 69.98 43.74 113.72 56.86
11 35 69.98 43.74 113.72 56.86
12 38 34.99 21.87 56.86 28.43
Shear at the base 1317.263
Figure 8 Shear force diagram due to wind load in Y-direction
0
5
10
15
20
25
30
35
40
0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 1400.00
Bu
ildin
g h
eig
ht
(m)
Shear Force (KN)
Shear Force Along the building height
11
6.2 Seismic load
Calculate the total building weight W= D.L+ α L.L
Table 5 Building Description
TYPE No./Floor DIEMENSIONS
Slab 1 35 36
Shear walls 4 0.4 4.5
C1 14 0.6 1.1
C2 22 0.6 0.55
C2 4 0.3 0.55
For commercial buildings α=0.5, while for storage floor α=1.
Building total weight = 129018 KN.
Table 6 Building Total Weight Calculations
Floor Height α Live
Load
Dead Load Floor
Weight Slab Shear
wall C1 C2 C3 S.D.L. TOTAL
1 4 0.5 5 277.2 28.8 36.96 29.04 2.64 1890 11256 11258.5
2 4 0.5 5 277.2 28.8 36.96 29.04 2.64 1890 11256 11258.5
3 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
4 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
5 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
6 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
7 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
8 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
9 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
10 3 0.5 5 277.2 21.6 27.72 21.78 1.98 1890 10647 10649.5
11 3 1 10 277.2 21.6 27.72 21.78 1.98 1890 10647 10657
12 3 0.5 2 277.2 21.6 27.72 21.78 1.98 1890 10647 10648
Total Weight (KN) 129018
12
T = Ct*H3/4
= 0.05*(38)3/4
= 0.76 sec.
Soil type A
γ = 1.2
S =1
Tb = 0.05
Tc = 0.25
Td = 1.2
Tc < T <Td
Sd(t) = ag* γ*S*(2.5/R)*(Tc/T) ξ
Sd(t) = 0.1g*1.2*1*(2.5/5)*(0.25/0.76)*1 = 0.0197g.
Minimum limit for Sd(t) = 0.2*ag*γ = 0.2*0.1*1.2 = 0.024g.
Sd(t) = 0.0197g < Sd(t) minimum limit = 0.024g.
Therefore, Sd(t) = 0.024g.
Fb = Sd(t)*λ*(W/g).
Fb = 0.024g*0.76*1*(W/g) = 0.01824*W.
The base shear due to seismic load, Fb = 0.01824 W.
Fb = 0.01824*129018= 2353.288 KN.
Distribute the base shear on the floor levels using the equation, Fi= Fb*
13
Table 7 Calculations of force/floor and accidental torsion due to seismic load
Floor Height Cum. Height Floor Weight Force/Floor Ty Tx
1 4.00 4.00 11258.5 38.62 69.51 67.58
2 4.00 8.00 11258.5 77.23 139.02 135.15
3 3.00 11.00 10649.5 100.45 180.81 175.79
4 3.00 14.00 10649.5 127.84 230.12 223.73
5 3.00 17.00 10649.5 155.24 279.43 271.67
6 3.00 20.00 10649.5 182.63 328.74 319.61
7 3.00 23.00 10649.5 210.03 378.05 367.55
8 3.00 26.00 10649.5 237.42 427.36 415.49
9 3.00 29.00 10649.5 264.82 476.67 463.43
10 3.00 32.00 10649.5 292.21 525.99 511.37
11 3.00 35.00 10657 319.83 575.70 559.71
12 3.00 38.00 10648 346.96 624.52 607.17
Total Weight 129018.00
Base Shear 2353.29
Figure 9 Bending Moment diagram due to seismic load along the building height.
0
5
10
15
20
25
30
35
40
0 10000 20000 30000 40000 50000 60000 70000
Bu
ildin
g h
eig
ht
(m)
B.M.D. (KN)
Bending Moment Along the building height
14
Figure 10 Shear Force diagram due to Seismic load along the building height.
0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500
Bu
ildin
g h
eig
ht
(m)
S.F.D (KN)
Shear Force Along the building height
15
Horizontal forces on each shear wall
Shear walls are symmetric in both directions X and Y axis. Therefore, the center of rigidity
and the center of mass lies on the same point and torsional moment does not exist.
Figure 11 The location of center of mass and center of rigidity
Moment of inertia = b*H3/12
Table 8 inertia of the shear walls
Wall No. Length (m) Thickness (m) Inertia (m4)
1 4.5 0.4 7.63
2 4.5 0.4 7.63
16
Torsional moment equals to Mt= Fb* (emin)
Torsional Moment Calculation for Floor 1:
Y-direction: Mt= 38.62*(0.05*36) = 69.516 KN.m.
X-direction: Mt= 38.62*(0.05*35) = 67.585 KN.m.
The torsional moment applies a shear force on the shear wall cross section. The shear force
will be calculated according to this equation:
Fx =
Fx for wall 1=
*7.63*12 = 1.59 KN.
Fx for wall 2=
*7.63*10.5 = 1.4 KN.
Fy =
Fy for wall 1 =
*7.63*12 = 1.64 KN.
Fy for wall 2=
*7.63*10.5 = 1.44 KN.
The shear force on the shear walls is calculated according to this equation,
Shear force/shear wall = shear force/floor *
Shear force/shear wall = 38.62 *
= 19.31 KN.
After obtaining the forces due to torsion need to get the resultant forces applied on shear
walls, Torsion forces + Horizontal Force from base shear
Therefore, total force on wall 1 = 1.64+19.31 = 20.95 KN,
& total force on wall 2 = 1.4+19.31 = 20.71 KN.
17
Table 9 Total force on each shear wall due to seismic load
Floor
#
Torsion Force Shear Force Total Force/Shear wall
(Force/Shear wall)
Y-dir.
(Force/Shear wall)
X-dir.
(Force/Shear wall)
Y-dir.
(Force/Shear wall)
X-dir.
(Force/Shear wall)
Y-dir.
(Force/Shear wall)
X-dir.
W1 W2 W1 W2 W1 W2 W1 W2 W1 W2 W1 W2
1 1.64 1.44 1.59 1.40 19.31 0.00 0.00 19.31 20.95 1.44 1.59 20.70
2 3.28 2.87 3.19 2.79 38.62 0.00 0.00 38.62 41.90 2.87 3.19 41.41
3 4.27 3.73 4.15 3.63 50.22 0.00 0.00 50.22 54.49 3.73 4.15 53.85
4 5.43 4.75 5.28 4.62 63.92 0.00 0.00 63.92 69.35 4.75 5.28 68.54
5 6.59 5.77 6.41 5.61 77.62 0.00 0.00 77.62 84.21 5.77 6.41 83.23
6 7.76 6.79 7.54 6.60 91.32 0.00 0.00 91.32 99.07 6.79 7.54 97.92
7 8.92 7.81 8.67 7.59 105.01 0.00 0.00 105.01 113.94 7.81 8.67 112.60
8 10.09 8.82 9.81 8.58 118.71 0.00 0.00 118.71 128.80 8.82 9.81 127.29
9 11.25 9.84 10.94 9.57 132.41 0.00 0.00 132.41 143.66 9.84 10.94 141.98
10 12.41 10.86 12.07 10.56 146.11 0.00 0.00 146.11 158.52 10.86 12.07 156.67
11 13.59 11.89 13.21 11.56 159.92 0.00 0.00 159.92 173.50 11.89 13.21 171.47
12 14.74 12.90 14.33 12.54 173.48 0.00 0.00 173.48 188.22 12.90 14.33 186.02
Table 10 Straining actions on each shear wall due to seismic load.
Floor Height Cum.
Height
Total Force Bending Moment Shear Force
W1 W2 W1 W2 W1 W2
1 4.00 4.00 20.95 20.70 33835.22 33439.65 1276.61 1261.68
2 4.00 8.00 41.90 41.41 33751.43 33356.84 1255.66 1240.98
3 3.00 11.00 54.49 53.85 33416.26 33025.59 1213.76 1199.57
4 3.00 14.00 69.35 68.54 32816.86 32433.19 1159.27 1145.72
5 3.00 17.00 84.21 83.23 31845.92 31473.61 1089.92 1077.18
6 3.00 20.00 99.07 97.92 30414.29 30058.71 1005.71 993.95
7 3.00 23.00 113.94 112.60 28432.80 28100.38 906.63 896.03
8 3.00 26.00 128.80 127.29 25812.27 25510.49 792.69 783.43
9 3.00 29.00 143.66 141.98 22463.54 22200.91 663.90 656.14
10 3.00 32.00 158.52 156.67 18297.44 18083.52 520.24 514.16
11 3.00 35.00 173.50 171.47 13224.81 13070.19 361.72 357.49
12 3.00 38.00 188.22 186.02 7152.20 7068.58 188.22 186.02
18
7. Shear Walls Design
7.1 Shear Wall 1
Obtain the straining actions on the wall form the above analysis.
D.L. = (117.6*(6*5.5) + (1.4*0.4*4.5*38*25)) = 6274.8 KN.
L.L. = 104*(6*5.5) = 3432 KN.
Normal Force = 6274.8+3432 = 9706.8 KN.
Q = 1317.26 KN.
M = 33835.22 KN.m.
a) Section Assumption:
Assume that the concrete section of wall 1 = 4500*400 mm
b) Vertical Reinforcement Assumption:
According to the code limitation,
Asv(min) = 0.4% Ac
The maximum limit for the vertical reinforcement is 4% Ac.
Therefore, assume that Asv = 2% Ac = (2/100)*(4500*400) = 36000 mm.
Then, design and check on the assumed dimensions and reinforcement.
19
c) Moment of Resistance Calculations:
Mr = 0.5 Asv L*(
)* (1+
)*(1-
).
L = 4500 mm.
t = 400 mm.
P = 9706.8 KN.
C/L=
=
= 0.436.
W=
(
)
=
(
)
= 0.347.
a =
=
= 0.269.
Mr = 0.5 (36000)* 4500*(
)* (1+
)*(1-0.436) = 28199.71118 KN.m.
Use diameter 28 uniformly distributed along the shear wall
Since, M applied > M resistance
Then, obtain additional Area steel concentrated on the boundaries.
Madd = Mapp-Mr = 33835.22-28199.7 = 5635.48 KN.m.
Area Steel additional =
= 3951.714724 mm2.
Then take As additional 6 Ф 28for both sides
Finally, recheck the area steel vertical compared to area concrete,
Area steel total = 36000+3951.71 = 39952.71 mm2.
Reinforcement ratio =
= 0.022
Therefore, reinforcement ratio = 2.2 % AC < Area steel max = 4% AC.
20
d) Axial Force Check:
Pu = 0.8*(0.35*Fcu*Ac*(1- (
)2))
As the K= 1.0 from the boundary condition of the shear wall (hinged- hinged) for typical
floors.
Pu = 0.8*(0.35*30*4500*400*(1- (
)2)) = 14289.43 KN.
The applied axial force = 1317.26 KN < Pult = 14289.43 KN.
Therefore, the wall is safe in axial.
e) Applied Shear Force Check:
qumax = 0.9α√
+µst*
= 0.9*0.25*√
+
*
= 1.201 N/ mm
2.
The applied shear stresses qu =
=
= 0.7318 N/mm
2.
Need to calculate shear strength of the shear wall, qcu = 0.24*√
= 1.07 N/mm
2.
qu < qcu
Therefore, use area steel horizontal minimum 5 Ф 8/m at spacing 200 mm.
21
Checking the capacity of shear wall 1 using interaction diagram
Figure 12 Load- Capacity ratios for shear wall 1
Figure 13 Interaction diagram for shear wall 1
22
7.2 Shear Wall 2
Obtain the straining actions on the wall form the above analysis.
D.L. = (117.6*(6*8) + (1.4*0.4*4.5*38*25)) = 8038.8 KN.
L.L. = 104*(6*8) = 4992 KN.
Normal Force = 8038.8+4992 = 13,030.8 KN.
Q = 1280 KN.
M = 33439.65 KN.m.
a) Section Assumption:
Assume that the concrete section of wall 2 = 4500*400 mm
b) Vertical Reinforcement Assumption:
According to the code limitation,
Asv(min) = 0.4% Ac
The maximum limit for the vertical reinforcement is 4% Ac.
Therefore, assume that Asv = 2% Ac = (2/100)*(4500*400) = 36000 mm.
Then, design and check on the assumed dimensions and reinforcement.
23
c) Moment of Resistance Calculations:
Mr = 0.5 Asv L*(
)* (1+
)*(1-
).
L = 4500 mm.
t = 400 mm.
P = 13030.8 KN.
C/L=
=
= 0.5.
W=
(
)
=
(
)
= 0.347.
a =
=
= 0.36.
Mr = 0.5 (36000)* 4500*(
)* (1+
)*(1-0.5) = 28666.73 KN.m.
Use diameter 28 uniformly distributed along the shear wall
Since, M applied > M resistance
Then, obtain additional Area steel concentrated on the boundaries.
Madd = Mapp-Mr = 33439.65-28666.73 = 4772.92 KN.m.
Area Steel additional =
= 3346.825915 mm2.
Then take As additional 6 Ф 28 for both sides
Finally, recheck the area steel vertical compared to area concrete,
Area steel total = 36000+3346.82 = 39346.83 mm2.
Reinforcement ratio =
= 0.022
Therefore, reinforcement ratio = 2.2 % AC < Area steel max = 4% AC.
24
d) Axial Force Check:
Pu = 0.8*(0.35*Fcu*Ac*(1- (
)2))
As the K= 1.0 from the boundary condition of the shear wall (hinged- hinged) for typical
floors.
Pu = 0.8*(0.35*30*4500*400*(1- (
)2)) = 14289.43 KN.
The applied axial force = 13030.8 KN < Pult = 14289.43 KN.
Therefore, the wall is safe in axial.
e) Applied Shear Force Check:
qumax = 0.9α√
+µst*
= 0.9*0.25*√
+
*
= 1.201 N/ mm
2.
The applied shear stresses qu =
=
= 0.71 N/mm
2.
Need to calculate shear strength of the shear wall, qcu = 0.24*√
= 1.07 N/mm
2.
qu < qcu
Therefore, use area steel horizontal minimum 5 Ф 8/m at spacing 200 mm.
25
Checking the capacity of shear wall 2 using interaction diagram
Figure 14 Load- Capacity ratios for shear wall 2
Figure 15 Interaction diagram for shear wall 1
26
Eventually, shear wall 1 and 2 have the same concrete section and steel reinforcement.
Figure 16 Reinforced detailing of the shear walls.
27
References
1. Egyptian code for loads, ECP 201-2012.
2. Egyptian code for structural detailing, ECP 203-2003.
3. Egyptian code of practice for designing concrete structures, ECP 203-2007.
4. MacGregor, J. G., Bartlett F. M. Reinforced concrete: mechanics and design.
