United Arab Emirates University College of Engineering Civil and Environmental Engineering Department Graduation Project II
1
Analysis and Design of a Multi-storey Reinforced Concrete Building Prepared
Sultan Saif Saeed Alneyadi Sultan Khamis AL-shamsi Hasher Khamis AL-azizi Rashed Hamad AL-Neyadi Abdulrahman Abdulla Jarrah
200203903 200101595 200106031 200204018 200210915 Adviser Dr. Usama Ebead
Second Semester 2007/2008
Outline 2
Objectives Summary General Approach Building Types Concrete Structural Elements
Slabs
Columns
Rectangular Columns Design of Rectangular Columns
Shear walls
Flat Slab Design of Flat Slab
Design of Shear Walls
Foundations
Pile Group Design of Pile Group
Economic Impact Environmental Impact Conclusion
Objectives 3
The Objectives of the Project are: Carrying out a complete analysis and design of the main structural elements of
a multi-storey building including slabs, columns, shear walls and foundations Getting familiar with structural softwares ( SAFE ,AutoCAD) Getting real life experience with engineering practices
Summary 4
Our graduation project is a residential building in Abu- Dhabi.
This building consists of 12 repeated floors.
General Approach 5
Obtaining an architectural design of a regular residential multi-
storey building.
Al-Suwaidy residential building in Abu Dhabi.
Establishing the structural system for the ground, and repeated
floors of the building. The design of column, wind resisting system, and type of
foundations will be determined taking into consideration the architectural drawings.
Types of building 6
Buildings are be divided into:
Apartment building Apartment
buildings are multi-story buildings where three or more residences are contained within one structure.
Office building The
primary purpose of an office building is to provide a workplace and working environment for administrative workers.
Residential buildings 7
Office buildings 8
Concrete Mixtures 9
Concrete is a durable material which is ideal for many jobs. The concrete mix should be workable. It is important that the desired qualities of the hardened concrete
are met. Economy is also an important factor.
Structural Elements 10
Any reinforced concrete structure consists of : Slabs Columns Shear walls Foundations
Flat Slab Structural System 11
Flat slab is a concrete slab which is reinforced in two directions
Advantages
Disadvantages
Types of Flat slab 12
Defining properties 13
Slab thickness = 23 cm Concrete compressive strength = 30 MPa Modules of elasticity of concrete = 200 GPa Yielding strength of steel = 420 MPa Combination of loads (1.4Dead Load + 1.6 Live Load)
ACI 318-02 14
ACI 318-02 contains the current code requirements for
concrete building design and construction. The design load combinations are the various
combinations of the prescribed load cases for which the structure needs to be checked. 1.2 DL + 1.6 LL
Flat Slab Analysis and Design Analyzing of flat slab mainly is done to find 1.
Shear forces.
2.
Bending moment.
3.
Deflected shape.
4.
Reactions at supports.
15
Results and Discussion Deflection
16
Results and Discussion Reactions at supports must be checked by a simple method.
17
Flat Slab Reinforcement
18
Columns It is a vertical structural member supporting axial
compressive loads, with or with-out moments. Support vertical loads from the floors and roof and transmit these loads to the foundation.
19
Types of column • Tied Columns Over 95% of all columns in building in non-seismic regions are tied columns • Spiral Columns Spiral columns are generally circular. It makes the column more ductile. Spiral column
Rectangular column
20
Steel Reinforcement in Columns 21
The limiting steel ratio ranges between 1 % to 8 %. The concrete strength is between 25 MPa to 45 Mpa. Reinforcing steel strength is between 400 MPa to 500 Mpa.
Design procedure 22
1. Calculate factored axial load Pu 2. Select reinforcement ratio 3. Concrete strength = 30 MPa, steel yield strength = 420 MPa 4. Calculate gross area 5. Calculate area of column reinforcement, As, and select rebar number and size.
Columns to be designed 23
Guidelines for Column Reinforcement 24
Long Reinforcement
Min. bar diameter Ø12 Min. concrete covers 40 mm Min. 4 bars in case of tied rectangular or circular Maximum distance between bars = 250 mm
Short Reinforcement ( Stirrups)
Least of:
Asp
(16)×diameter of long bars least dimension of column (48)×diameter of ties
S
dc
Column Design 25
As = 0.01Ac
8- # of bars =
Reinforcement of Columns 26
Shear walls 27
A shear wall is a wall that resists
lateral wind loads which acts parallel to the plane of the wall.
Shear walls 28
Wind results in a pressure on the surface of the building Pressure increases with height
Positive Pressure, acts towards the surface of the building Negative Pressure, acts away from the surface of the building (suction)
Wind pressure 29
q = Velocity pressure (Wind speed, height and exposure condition) G = Gust factor that depends on the building stiffness Cp = External pressure coefficient
Gust G Factor & External pressure Cp coefficient 30
for Stiff Structures take
G =0.85
Windward Wall, Cp = +0.8 Leeward Wall, Cp = varies between -0.2 & -0.5 Depending
on the L/B Ratio L/B = 18.84 m /26.18 m = 0.719 < 1 then , Cp = -0.5
Velocity Pressure 31
V = 160 km/h Kz = To be determined from the equations Kzt = 1 (level terrain adjacent to the building – not on hill) Kd = 0.85 (rectangular building) I = 1 (use group II)
Important factor 32
Velocity Exposure Coefficient ( Kz) 33
Design of the wind force 34
North south direction
Shear wall axial reactions 35
Calculating Velocity Pressure 36 V (km/hr)
145
α Zg
9.5 274.32
Kzt Kd I
1 0.85 1
Level
Height (z)
12 11 10 9 8 7 6 5 4 3 2 1
43 39.5 36 32.5 29 25.5 22 18.5 15 11.5 8 4.5
Tributary Height (ht ) 1.75 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 4
1 0.85 G Cp (windward) Cp (leeward)
0.85 0.8
B (m)
26.18
1
-0.5
Kz
qz (kn/m2)
1.36 1.34 1.31 1.28 1.25 1.22 1.18 1.14 1.09 1.03 0.95 0.85
1.150225 1.129849 1.107994 1.084391 1.058688 1.030406 0.998873 0.963092 0.921495 0.871364 0.807270 0.715176
145 km/h
Design of the wind pressure 37 G
0.85
Cp (windward)
0.8
Cp (leeward)
-0.5
B (m)
26.18
Level
qb = qz (at the top of the building)
Height (z) m
Tributary Height (ht ) m
Kz
qz (kn/m2)
Design Wind Pressure(KN/m^2) Design Wind Force (KN) wind ward lee ward wind ward lee ward Total (qz G CP) (qb G CP) (qz G CP)(B)(ht ) (qb G CP)(B)(ht (floor level) )
12
43
1.75
1.36
1.150225
0.782153
-0.488846
35.834345
-22.396465
58.230810
2503.924826
11
39.5
3.5
1.34
1.129849
0.768297
-0.488846
70.399094
-44.792931
115.192025
4550.084972
10
36
3.5
1.31
1.107994
0.753436
-0.488846
69.037332
-44.792931
113.830262
4097.889443
9
32.5
3.5
1.28
1.084391
0.737386
-0.488846
67.566683
-44.792931
112.359614
3651.687445
8
29
3.5
1.25
1.058688
0.719908
-0.488846
65.965161
-44.792931
110.758092
3211.984664
7
25.5
3.5
1.22
1.030406
0.700676
-0.488846
64.202965
-44.792931
108.995896
2779.395349
6
22
3.5
1.18
0.998873
0.679233
-0.488846
62.238149
-44.792931
107.031079
2354.683748
5
18.5
3.5
1.14
0.963092
0.654903
-0.488846
60.008720
-44.792931
104.801650
1938.830531
4
15
3.5
1.09
0.921495
0.626617
-0.488846
57.416871
-44.792931
102.209802
1533.147032
3
11.5
3.5
1.03
0.871364
0.592527
-0.488846
54.293292
-44.792931
99.086222
1139.491559
2
8
3.5
0.95
0.807270
0.548944
-0.488846
50.299721
-44.792931
95.092651
760.7412106
1
4.5
4
0.85
0.715176
0.486320
-0.488846
50.927427
-51.191921
102.119348
459.5370657
sum
Moment (KN.m)
1229.707452 28981.39785
Computing total moment acting toward N-S Direction 38
M = total floor level *height (z)
W-E Direction Computation B= 18.84
39
L= 26.18 Level
Height (z) m
Tributary Height (ht ) m
Kz
qz (kn/m2)
Design Wind Pressure(KN/m^2) Design Wind Force (KN) wind ward lee ward wind ward lee ward Total (qz G CP) (qb G CP) (qz G CP)(B)(ht ) (qb G CP)(B)(ht ) (floor level)
12 11
43 39.5
1.75 3.5
1.36 1.34
1.150225 1.129849
0.7821531 0.7682974
-0.48885 -0.48885
25.7875879 50.6615328
-16.1172424 -32.2344849
41.9048304 82.8960177
1801.907705 3274.392699
10 9 8 7 6 5 4 3 2 1
36 32.5 29 25.5 22 18.5 15 11.5 8 4.5
3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 4
1.31 1.28 1.25 1.22 1.18 1.14 1.09 1.03 0.95 0.85
1.107994 1.084391 1.058688 1.030406 0.998873 0.963092 0.921495 0.871364 0.807270 0.715176
0.7534359 0.7373860 0.7199079 0.7006763 0.6792333 0.6549025 0.6266165 0.5925275 0.5489438 0.4863200
-0.48885 -0.48885 -0.48885 -0.48885 -0.48885 -0.48885 -0.48885 -0.48885 -0.48885 -0.48885
49.6815633 48.6232356 47.4707271 46.2025923 44.7886449 43.1842734 41.3190931 39.0712612 36.1973543 36.6490728
-32.2344849 -32.2344849 -32.2344849 -32.2344849 -32.2344849 -32.2344849 -32.2344849 -32.2344849 -32.2344849 -36.8394113 sum
81.9160482 80.8577205 79.7052120 78.4370772 77.0231298 75.4187583 73.5535780 71.3057461 68.4318392 73.4884841 884.9384415
2948.977735 2627.875916 2311.451149 2000.145469 1694.508855 1395.247028 1103.30367 820.0160796 547.4547138 330.6981787 20855.9791983
Moment (KN.m)
Design of Shear Wall 40
East west direction
North south direction
Interaction Diagram 41
Shear Wall Reinforcement 42
Foundations 43
Foundations are structural components used to support
columns and transfer loads to the underlying Soil. Foundations Deep
Shallow Isolated footing
Combined Strap wall footing footing footing
Raft footing Caissons
Piles
Pile foundation 44
Our building is rested on a weak soil formation which
can’t resist the loads coming from our proposed building, so we have to choose pile foundation.
Pile cap
Weak soil
Bearing stratum
Piles
Pile foundation 45
Piles are structural members that are made of steel,
concrete or timber.
Function of piles 46
As with other types of foundation, the purpose of a pile
foundation is: To transmit a foundation load to a solid ground To resist vertical, lateral and uplift load
Piles can be
Timber Concrete Steel Composite
Concrete piles 47
General facts Usual length: 10m-20m Usual load: 300kN-3000kN
Advantages Corrosion resistance Can be easily combined with a concrete superstructure
Disadvantages Difficult to achieve proper cutoff Difficult to transport
Pile foundation 48
Piles can be divided in to two major categories: 1.
End Bearing Piles If the soil-boring records presence of bedrock at the site within a reasonable depth, piles can be extended to the rock surface
2.
Friction Piles When no layer of rock is present depth at a site, point bearing piles become very long and uneconomical. In this type of subsoil, piles are driven through the softer material to specified depths.
Pile Cap Reinforcement 49
Pile caps carrying very heavy point loads tend to produce high
tensile stresses at the pile cap. Reinforcement is thus designed to provide:
Resistance to tensile bending forces in the bottom of the cap Resistance to vertical shear
Design of the pile cap 50
bearing capacity of one pile:
Rs = α ⋅ Cu ⋅ As .L Length of pile penetration L = 18 meters Adhesion factor of soil (clay) α = 0.8 Untrained shear strength Cu = 50 Diameter = 0.9 m For piles with diameter 0.9 m Rs = 2035.75 KN
First type 51
This section shows how pile caps are designed to carry
only vertical load, and the equation used to determine the resistance of cap is
Qi Pi = n Where P
is the strength of the pile cap per one pile
Q n
is the total force acting on the pile cap is the number of piles used to support the pile cap
Columns layout & Reactions ( Vertical Load ) 52
Column
Reaction Total Reaction
kN
kN
1
129.63
1555.56
2
246.85
2962.2
8
382.66
4591.92
10
393.38
4720.56
21
458.35
5500.2
23
400.85
4810.2
24
627.74
7532.88
25
384.14
4609.68
30
158.3
1899.6
32
355.26
4263.12
Design of pile cap (Vertical Load only) 53 Pile Cap 2
Reaction = 4610.4 kN Pile diameter = 0.9 m Capacity for one pile = 0.8 * 50 * 18 * π * 0.9 = 2035.75 KN Need 3 piles Length between piles = (2*0.3) + (3*0.9) + (2*0.9)*2 =6.9 m Width = 1.5 meters Qi P = Actual forces on each pile = i = 1536.8 kN n
Second type 54
Second type This section shows how pile caps are designed to carry
vertical load and lateral loads ( Bending Moment), and the equation used to determine the resistance of cap is
Qi M i ∗ r Pi = ± 2 n ∑r
Shear walls layout & reactions 55
wall
M (KN.m)
N (KN)
W1
14072.12
12285.6
W2
366.048
3596.76
W3
366.048
3026.88
W4
5719.5
3605.04
W5
30.65295
4128
W6
301.6143
1899.6
W10
10141.2
32.80882
W11
2402.52
32.80882
W13
20978.4
6700.246
W14
3297.6
6700.246
W15
2040
262.4706
W16
5470.2
262.4706
W17
7262.76
7903.641
W18
8571.48
7086.706
Design of pile cap (Vertical Load & moment) 56
Shear wall # (1): M = 14072.11561 Q = 12285.6 Assume 8 piles
P=
Q Mr ± n ∑ r2
12285.6 14072.11561* (1.909) ± So, 〈 PCapacityof Pile = 2035.75 KN 8 24.6762 12285.6 14072.11561* (4.26) P= ± So, 〈 PCapacityof Pile = 2035.75 KN 8 24.6762 P=
Economical impact 57
Reinforced concrete is proven to be a very economical
solution in the UAE. the most affordable solution for multistory building such as the one we are making the analysis and design for.
Environmental impact 58
Although the cement production is environmentally
challenging, the final product of a reinforced concrete building is environmentally friendly.
Gantt Chart 59
Conclusion 60
We have applied our gained knowledge during our graduation
project We are able to use structural software ( SAFE ) We have practiced real life engineering practices This GP enables us to go into the market with an excellent background regarding design of RC At this point, we would like to thank all instructors, engineers, and Al Ain Consultant Office for their grateful effort.
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