1Analysis and Design of a Multi-storey Reinforced Concrete

Building

United Arab Emirates University College of Engineering

Civil and Environmental Engineering DepartmentGraduation Project II

Second Semester 2007/2008

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

Adviser Dr. Usama Ebead

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Outline

Objectives Summary General Approach Building Types Concrete Structural Elements

Slabs Flat Slab Design of Flat Slab

Columns Rectangular Columns Design of Rectangular Columns

Shear walls Design of Shear Walls

Foundations Pile Group Design of Pile Group

Economic Impact Environmental Impact Conclusion

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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 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.

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Types of building

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.

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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 concrete

are met.Economy is also an important factor.

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Structural Elements

Any reinforced concrete structure consists of : Slabs Columns Shear walls Foundations

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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 GPaYielding 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 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

1515

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 Discussion

Deflection

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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 axial compressive loads, with or with-out moments.

Support vertical loads from the floors and roof and transmit these loads to the foundation.

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Types of column

Spiral columnSpiral column Rectangular Rectangular columncolumn

• 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 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.

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Columns to be designed

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Guidelines for Column Reinforcement

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:

(16)×diameter of long bars least dimension of column (48)×diameter of ties

dc

S

Asp

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Column Design

cs AA 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 resists lateral wind loads which acts parallel 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 =0.85 Windward Wall, Cp = +0.8Leeward 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/hKz = To be determined from the equationsKzt = 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

3333

Velocity Exposure Coefficient ( Kz)

3434

Design of the wind force

North south direction

3535

Shear wall axial reactions

3636

Calculating Velocity Pressure

145 km/h

0.85 11V

(km/hr)145

α 9.5Zg 274.32Kzt 1Kd 0.85I 1

G 0.85Cp

(windward)0.8

Cp (leeward) -0.5B (m) 26.18

LevelHeight

(z)

Tributary Height

(ht )Kz qz (kn/m2)

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

3737

Design of the wind pressure

G 0.85

Cp (windward) 0.8

Cp (leeward) -0.5

B (m) 26.18

qb = qz (at the top of the building)

LevelHeight(z) m

Tributary Height(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

3838

Computing total moment acting toward N-S Direction

M = total floor level *height (z)

3939

W-E Direction Computation

L= 26.18

B=

18.

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LevelHeight(z) m

Tributary Height(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

4040

Design of Shear Wall

East west direction

North south direction

4141

Interaction Diagram

4242

Shear Wall Reinforcement

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Foundations

Foundations are structural components used to support columns 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 which can’t resist the loads coming from our proposed building, so we have to choose pile foundation.

Pile cap

PilesWeak 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 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

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Concrete piles

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

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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 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.

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Pile Cap Reinforcement

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

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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 = 50 Diameter = 0.9 m For 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 carry only vertical load, and the equation used to determine the resistance of cap is

Where P 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

n

QP 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 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 Actual forces on each pile = = 1536.8 kN

niQ

iP

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Second type

Second typeThis 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

2r

rM

n

QP 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 = 14072.11561 Q = 12285.6 Assume 8 piles

KNPSoP

KNPSoP

r

rM

n

QP

PileofCapacity

PileofCapacity

75.2035,676.24

)26.4(*11561.14072

8

6.12285

75.2035,676.24

)909.1(*11561.14072

8

6.12285

2

2

2

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Economical impact

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.

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Environmental impact

Although the cement production is environmentally challenging, 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 project

We 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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