Storey multi

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Analysis and Design of a Multi-storeyReinforced Concrete Building

U nited A rab E mirates U niversity C ollege of E ngineering

C ivil and E nvironmental E ngineering D epartment G raduation P roject II

Second Semester 2007/2008

PreparedSultan Saif Saeed Alneyadi 200203903Sultan Khamis AL-shamsi 200101595Hasher Khamis AL-azizi 200106031Rashed Hamad AL-Neyadi 200204018 Abdulrahman Abdulla Jarrah 200210915

Adviser Dr. Usama Ebead



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Outline

ObjectivesSummaryGeneral ApproachBuilding TypesConcreteStructural Elements

SlabsFlat SlabDesign of Flat Slab

ColumnsRectangular ColumnsDesign of Rectangular Columns

Shear wallsDesign of Shear Walls

Foundations

Pile GroupDesign of Pile GroupEconomic ImpactEnviromental ImpactConclusion



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Objectives

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



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Summary

Our graduation project is a residential building in Abu- Dhabi.This building consists of 12 repeated floors.



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General Approach

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 repeatedfloors of the building.

The design of column, wind resisting system, and type offoundations will be determined taking into consideration thearchitectural drawings.



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Types of building

Buildings are be divided into: Apartment building

Apartment buildings are multi-story buildings where three or moreresidences are contained within one structure.

Office building The primary purpose of an office building is to provide a workplace andworking environment for administrative workers.



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Residential buildings



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Office buildings



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Concrete Mixtures

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 concreteare met.Economy is also an important factor.



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Structural Elements

Any reinforced concrete structure consists of :Slabs

ColumnsShear wallsFoundations



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Flat Slab Structural System

Flat slab is a concrete slab which is reinforced in two directions

Advantages

Disadvantages



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Types of Flat slab



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Defining properties

Slab thickness = 23 cmConcrete compressive strength = 30 MPaModules of elasticity of concrete = 200 GPa

Yielding strength of steel = 420 MPaCombination of loads (1.4Dead Load + 1.6 Live Load)



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ACI 318-02

ACI 318-02 contains the current code requirements forconcrete building design and construction.

The design load combinations are the variouscombinations of the prescribed load cases for which thestructure needs to be checked.

1.2 DL + 1.6 LL



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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.



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Results and DiscussionReactions at supports must be checked by a simple method.



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Flat Slab Reinforcement



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Columns

It is a vertical structural member supporting axialcompressive loads, with or with-out moments.Support vertical loads from the floors and roof andtransmit these loads to the foundation.



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Types of column

Spiral column Rectangularcolumn

• Tied ColumnsOver 95% of all columns in building in non-seismic regions are tied columns • Spiral ColumnsSpiral columns are generally circular. It makes the column more ductile.



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Steel Reinforcement in Columns

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.



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Design procedure

1. Calculate factored axial load Pu2. Select reinforcement ratio3. Concrete strength = 30 MPa, steel yield strength = 420 MPa

4. Calculate gross area5. Calculate area of column reinforcement, As, and select rebarnumber and size.



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Columns to be designed



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Guidelines for Column Reinforcement

Long ReinforcementMin. bar diameter Ø12Min. concrete covers 40 mmMin. 4 bars in case of tied rectangular or circular

Maximum distance between bars = 250 mmShort Reinforcement ( Stirrups)

Least of:(16) ×diameter of long bars

least dimension of column(48) ×diameter of ties

dc

S

Asp



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Column Design

c s A A 01.0 8- # of bars =



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Reinforcement of Columns



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Shear walls

A shear wall is a wall that resistslateral wind loads which actsparallel to the plane of the wall.



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Shear walls

Wind results in a pressure on the surface of the buildingPressure increases with height

Positive Pressure, acts towards the surface of the building

Negative Pressure, acts away from the surface of the building(suction)



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Wind pressure

q = Velocity pressure

(Wind speed, height and exposure condition)G = Gust factor that depends on the building stiffnessCp = External pressure coefficient



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Gust G Factor & External pressure Cp coefficient

for Stiff Structures take G =0.85Windward 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



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Velocity Pressure

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)



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Important factor32



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Velocity Exposure Coefficient ( K z )



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Design of the wind force

North south direction



35 35

Shear wall axial reactions



36 36

Calculating Velocity Pressure

145 km/h

0.851 1V(km/hr)

145

α 9.5Zg 274.32Kzt 1Kd 0.85I 1

G 0.85Cp

(windward)0.8

Cp (leeward) -0.5

B (m) 26.18

LevelHeight

(z)

TributaryHeight

(h t )Kz qz (kn/m 2)

12 43 1.75 1.36 1.15022511 39.5 3.5 1.34 1.12984910 36 3.5 1.31 1.1079949 32.5 3.5 1.28 1.0843918 29 3.5 1.25 1.0586887 25.5 3.5 1.22 1.0304066 22 3.5 1.18 0.9988735 18.5 3.5 1.14 0.9630924 15 3.5 1.09 0.9214953 11.5 3.5 1.03 0.8713642 8 3.5 0.95 0.8072701 4.5 4 0.85 0.715176



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Design of the wind pressure

G 0.85

Cp (windward) 0.8

Cp (leeward) -0.5

B (m) 26.18

q b = q z (at the top of the building)

Level Height(z) m

TributaryHeight(ht ) m

Kz qz (kn/m2)

Design Wind Pressure(KN/m^2) Design Wind Force (KN)

wind ward(qz G CP)

lee ward(qb G CP)

wind ward(qz G CP)(B)(ht )

lee ward(qb G

CP)(B)(ht )

Total(floor level)

Moment(KN.m)

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 1229.707452 28981.39785



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Computing total moment acting toward N-S Direction

M = total floor level *height (z)



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W-E Direction Computation

L= 26.18 B =

1 8

. 8 4

LevelHeight(z) m

TributaryHeight(ht ) m

Kz qz (kn/m2)Design Wind Pressure(KN/m^2) Design Wind Force (KN)

wind ward(qz G CP)

lee ward(qb G CP)

wind ward(qz G CP)(B)(ht )

lee ward(qb G CP)(B)(ht )

Total(floor level)

Moment(KN.m)

12 43 1.75 1.36 1.150225 0.7821531 -0.48885 25.7875879 -16.1172424 41.9048304 1801.907705

11 39.5 3.5 1.34 1.129849 0.7682974 -0.48885 50.6615328 -32.2344849 82.8960177 3274.392699

10 36 3.5 1.31 1.107994 0.7534359 -0.48885 49.6815633 -32.2344849 81.9160482 2948.977735

9 32.5 3.5 1.28 1.084391 0.7373860 -0.48885 48.6232356 -32.2344849 80.8577205 2627.875916

8 29 3.5 1.25 1.058688 0.7199079 -0.48885 47.4707271 -32.2344849 79.7052120 2311.451149

7 25.5 3.5 1.22 1.030406 0.7006763 -0.48885 46.2025923 -32.2344849 78.4370772 2000.145469

6 22 3.5 1.18 0.998873 0.6792333 -0.48885 44.7886449 -32.2344849 77.0231298 1694.508855

5 18.5 3.5 1.14 0.963092 0.6549025 -0.48885 43.1842734 -32.2344849 75.4187583 1395.247028

4 15 3.5 1.09 0.921495 0.6266165 -0.48885 41.3190931 -32.2344849 73.5535780 1103.30367

3 11.5 3.5 1.03 0.871364 0.5925275 -0.48885 39.0712612 -32.2344849 71.3057461 820.0160796

2 8 3.5 0.95 0.807270 0.5489438 -0.48885 36.1973543 -32.2344849 68.4318392 547.4547138

1 4.5 4 0.85 0.715176 0.4863200 -0.48885 36.6490728 -36.8394113 73.4884841 330.6981787

sum 884.9384415 20855.9791983



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Design of Shear Wall

East west direction

North south direction



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Interaction Diagram



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Shear Wall Reinforcement



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Foundations

Foundations are structural components used to supportcolumns and transfer loads to the underlying Soil.

Foundations

Isolated Combined Strap wall Raft

Shallow

footing footing footing footing footing

Caissons Piles

Deep



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Pile foundation

Our building is rested on a weak soil formation whichcan’t resist the loads coming from our proposed building,so we have to choose pile foundation.

Pile cap

Piles Weak soil

Bearing stratum



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Pile foundation

Piles are structural members that are made of steel,concrete or timber.



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Function of piles

As with other types of foundation, the purpose of a pilefoundation is:

To transmit a foundation load to a solid groundTo resist vertical, lateral and uplift load

Piles can beTimber

ConcreteSteelComposite



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Concrete piles

General factsUsual length: 10m-20mUsual load: 300kN-3000kN

AdvantagesCorrosion resistanceCan be easily combined with a concrete superstructure

DisadvantagesDifficult to achieve proper cutoffDifficult to transport



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Pile foundation

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 therock surface

2. Friction Piles

When no layer of rock is present depth at a site, point bearing piles become verylong and uneconomical. In this type of subsoil, piles are driven throughthe softer material to specified depths.



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Pile Cap Reinforcement

Pile caps carrying very heavy point loads tend to produce hightensile 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



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Design of the pile cap

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 = 50Diameter = 0.9 mFor piles with diameter 0.9 m

Rs = 2035.75 KN



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First type

This section shows how pile caps are designed to carryonly vertical load, and the equation used to determinethe resistance of cap is

WhereP is the strength of the pile cap per one pile

Q is the total force acting on the pile capn is the number of piles used to support the pile cap

nQ

P i

i



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Columns layout & Reactions ( Vertical Load )

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



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Design of pile cap (Vertical Load only)

Pile Cap 2Reaction = 4610.4 kNPile diameter = 0.9 mCapacity 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 mWidth = 1.5 metersActual forces on each pile = = 1536.8 kN

niQ

i P



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Second type

Second type This section shows how pile caps are designed to carryvertical load and lateral loads ( Bending Moment), and theequation used to determine the resistance of cap is

2

r

r M

n

Q P ii

i



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Shear walls layout & reactions

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



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Design of pile cap (Vertical Load & moment)

Shear wall # (1): M = 14072.11561Q = 12285.6Assume 8 piles

KN P So P

KN P So P

r

r M

n

Q P

Pileof Capacity

Pileof Capacity

75.2035,676.24

)26.4(*11561.140728

6.12285

75.2035,676.24

)909.1(*11561.140728

6.12285

2

2

2



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Economical impact

Reinforced concrete is proven to be a very economicalsolution in the UAE.the most affordable solution for multistory building suchas the one we are making the analysis and design for.



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Environmental impact

Although the cement production is environmentallychallenging, the final product of a reinforced concrete

building is environmentally friendly.



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Gantt Chart



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Conclusion

We have applied our gained knowledge during our graduation projectWe are able to use structural software ( SAFE )We have practiced real life engineering practicesThis GP enables us to go into the market with an excellent

background regarding design of RCAt this point, we would like to thank all instructors, engineers,and Al Ain Consultant Office for their grateful effort.



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