1

Design of 12- Storey Reinforced Concrete Office

Building in Amarah

STUDENT’S NAME

1- Ahmed Hatif Obaid

2- Mohammed Abass Kshaen

3- Ammar Abass Abd

A project report submitted in partial fulfilment of the

requirements for the award of the degree of

Bachelor of Civil Engineering

Civil Engineering Department

Engineering College

University of Misan

Iraq

2015-2016

2

DECLARATION

We hereby declare that this project report is based on our

original work except for citations and quotations which have

been duly acknowledged.

Signature :

Name : Ahmed Hatif Obaid

Date : / /

Signature :

Name : Mohammed Abass Kshaen

Date : / /

Signature :

Name : Ammar Abass Abd

Date : / /

3

APPROVAL FOR SUBMISSION

I certify that this project report

Design of 12- Storey Reinforced Concrete Office

Building in Amarah

was prepared by Ahmed Hatif Obaid , Mohammed Abass and Ammar

abass has met the required standard for submission in partial fulfilment

of the requirements for the award of Bachelor of Civil Engineering at

University of Misan.

Approved by,

Signature :

Supervisor : Ass.Prof.Dr. Abbas Oda Dawood

Date : / /

4

Dedicated

All praise to Allah, today we fold the days' tiredness and the errand

summing up between the cover of this humble work.

To the utmost knowledge lighthouse, to our greatest and most honored

prophet Mohamed - May peace and grace from Allah be upon him

To the Spring that never stops giving, to my mother who weaves my

happiness with strings from her merciful heart... to my mother.

To whom he strives to bless comfort and welfare and never stints what he

owns to push me in the success way who taught me to promote life stairs

wisely and patiently, to my dearest father

To whose love flows in my veins, and my heart always remembers them,

to my brothers and sisters .

5

ACKNOWLEDGEMENTS

I would like to thank everyone who had contributed to the successful

completion of this project. I would like to express my gratitude to my

research supervisor, Ass.Prof.Dr. Abbas Oda Dawood for his invaluable

advice, guidance and his enormous patience throughout the development

of the research.

In addition, I would also like to express my gratitude to my loving

parent and friends who had helped and given me encouragement......

6

Design of 12- storey reinforced concrete office

building in Amarah

ABSTRACT

The high rise building represent the optimum benefit from areas

especially in dense and expensive areas. Misan local government have

plans to build many new high rise building in Amarah to satisfy two

goals, the first one is benefit from area and the second to reflect the

progress in Amarah and gives an image for modern Amarah. The present

project focused on design high rise building in Amarah. The building is

located in a densely populated quarters of Amarah. It is a reinforced

concrete framed building with twelve storeys above the ground. The

building not included basements due to high level of underground water

in Amarah. The building is being assumed as an office building, therefore

it is evaluated for the Life Safety (LS) level of seismic performance,

meaning that its occupants should survive during earthquake and be able

the building to be safe. The lateral loading due to both wind and seismic

loadings was investigated. The wind forces are calculated based on

analytical method of ASCE7-10 and a wind speed of 42 m/sec was

adopted according to Iraqi codes for Misan Province. The seismic loads

was represented by equivalent static method adopted by both Iraqi and

ASCE7-10. Its found that seismic load more critical than wind loads. The

building was designed according to ACI –Code and requirements for

seismic forces according to ASCE7-10. According to Seismic Category

of Misan province SDC and for building height 64m the suitable

structural system according to ASCE7-10 is Dual system with special

frame resistance moment and special shear walls. The soil properties for

7

both bearing strength and seismic classifications are accomplished from

soil investigation report for the specified area. It's found that Amarah soil

is classified as Class D according to ASCE7-10 which based on standard

penetration tests. According to soil investigation the bored piles of

diameter 1.2m are used with capacity of 180 ton.

8

TABLE OF CONTENTS

Declaration 2

Approval For Submission 3

Dedicated 4

Acknowledgements 5

Abstract 6

Table Of Contents 8

Chapter one

1.1 Introduction 10

1.2 Structural systems for tall buildings 11

1.3 Tall building in Missan 18

1.4 Tall building codes 20

1.5 Difference between law rise buildings and tall buildings 20

Chapter two

2.1 Loads 22

2.1.1Dead Load 22

2.1.2 Live load 22

2.2 Wind load 26

2.3 Procedure for determined wind load 28

2.4 design wind load cases 34

2.5 Earthquake Load 38

2.6 Seismic Load Calculation 49

9

2.7 Loads on building 52

Chapter three

3.1 selection building system 53

3.2 selection of slab system 58

3.3 description of project 60

Chapter four

4.1 Methods of Slab design 66

4.2 Slab design 68

4.3 beams design 104

4.4 Shear Wall design 118

4.5 Columns design 125

4.6 foundation design 139

References 143

Appendices 144

10

Chapter one

1.1 Introduction

The high rise building represent the optimum benefit from areas

especially in dense and expensive areas, , including excellence in

execution and aesthetically also other considerations that all over the

world. some high rise buildings about the world See figure (1.1)

Figure (1.1 ): some high rise buildings about the world

11

Tall buildings are closely related to the growth of the city Their

developments are due to. [3]

- economic growth

- human ego to build higher

- the natural response to dense population

- scarcity of land in urban areas

- high land costs

- technological advancement

1.2 Structural systems for tall buildings

The following classification is proposed for the structural systems of tall

buildings for all the types namely, steel buildings, reinforced concrete

buildings, and composite buildings.[5]:

1. rigid frame systems

2. braced frame and shear-walled frame systems

3. outrigger systems

4. framed-tube systems

5. braced-tube systems

6. bundled-tube systems

1.2.1 Rigid frame systems

Rigid frame systems are utilized in both steel and reinforced concrete

construction. Rigid frame systems for resisting lateral and vertical loads

have long been accepted for the design of the buildings. Rigid framing,

namely moment framing, is based on the fact that beam-to-column

connections have enough rigidity to hold the nearly unchanged original

12

angles between intersecting components. Owing to the natural

monolithically behavior, hence the inherent stiffness of the joist, rigid

framing is ideally suitable for reinforced concrete buildings [5].

For a rigid frame the strength and stiffness are proportional to the

dimension of the beam and the column dimension, and inversely

proportional to the column spacing. ,Especially for the buildings

constructed in seismic zones, a special attention should be given to the

design and detailing of joints, since rigid frames are more ductile and less

vulnerable to severe earthquakes when compared to steel braced or shear-

walled structures . The 21-storey-highLever House (1952) (Fig. 1) in New

York (fig.1.2)

Fig. 1.2 lever house , new York ,USA 1952

13

1.2.2 Braced frame and shear-walled frame systems

Rigid frame systems are not efficient for buildings taller than 30 stories,

because lateral deflection due to the bending of columns causes the drift

to be too large . on the other hand, steel bracing or shear walls with or

without rigid frame (brace systems and shear wall systems), increases the

total rigidity of the building and the resulting system is named as braced

frame or shear-walled frame system. Namely, systems composed of steel

bracing or shear walls alone, or interacting with the rigid frames can be

accepted as an improvement of the rigid frame system. These systems are

stiffer when compared to the rigid frame system, and can be used for

buildings over 30 stories, but mostly applicable for buildings about 50

stories in height. .[5]

1.2.2.1 Braced frame systems

Braced frame systems are utilized in steel construction. This system is a

highly efficient and economical system for resisting horizontal loading,

and attempts to improve the effectiveness of a rigid frame by almost

eliminating the bending of columns and girders, by the help of additional

bracing. It behaves structurally like a vertical gravity loads, and diagonal

bracing components so that the total set of members forms a vertical

cantilever truss to resist the horizontal loading.

1.2.2.2 Shear-walled frame systems

resist lateral wind and seismic loads acting on a building and transmitted

to them by the floor diaphragms. Shear walls are generally parts of the

elevator and service cores , and frames to create a stiffer and stronger

structure. These elements can have various shapes such as, circular,

curvilinear, oval, box-like, triangular, or rectilinear. This system

14

structurally behaves like a concrete building with shear walls resisting all

the lateral loads. (Fig.1.3) in New York is a is a good example of it.

1.2.3 Outrigger system

Outrigger systems are modified form of braced frame and shear-walled

frame systems, and utilized in steel and composite constructions . As an

innovative and efficient structural system, the outrigger system comprises

a central core, including either braced frames or shear walls, with

horizontal ‘‘outrigger’’ trusses or girders connecting the core to the

external columns. Furthermore, in most cases, the external columns are

interconnected by exterior belt girder. If the building is subjected to the

horizontal loading, the rotation of the core is prevented by the column-

restrained outriggers. Outrigger structures can be used for buildings with

Fig. 1.3 :Metropolitan Tower, New York, USA, 1987

15

over 100 stories. The 42-storey-high First Wisconsin Center with its steel

structure (1974) (fig.1.4) , the 101-storey-high Taipei 101 (fig.1.5).

.

1.2.4 Framed-tube systems

Framed-tube systems, are proper for steel, reinforced concrete and

composite construction, and represent a logical evolution of the

conventional frame structure , The primary characteristic of a tube is the

employment of closely spaced perimeter columns interconnected by deep

spandrels, so that the whole building works as a huge vertical cantilever

to resist overturning moments.

Fig.1.5: the 101-storey-high Taipei 101 Fig. 1.4 : The 42-storey-high First Wisconsin Center

16

It is an efficient system to provide lateral resistance with or without

interior columns. The efficiency of this system is derived from the great

number of rigid joints acting along the periphery, creating a large tube.

Exterior tube carries all the lateral loading. The gravity loading is shared

between the tube and the interior columns or walls, if they exist. Besides

its structural efficiency, framed-tube.[5]

buildings leave the interior floor plan relatively free of core bracing and

heavy columns, enhancing the net usable floor area thanks to the

perimeter framing system resisting the whole lateral load. There are two

popular versions used currently for this system for composite

construction: spandrels while the other utilizes structural steel spandrels

instead of concrete ones. The 110-storey-high World Trade Center Twin

Towers (1972) (fig.1.6).

Framed-tube systems can be categorized into three groups:

1. systems without interior columns, shear walls, or steel bracings .

2. systems with interior columns, shear walls, or steel bracings.

3. tube-in-tube systems.

Fig.1.6 The 110-storey-high World Trade

Center Twin Towers (1972)

17

1.2.5. Braced-tube systems

Braced-tube systems can be utilized in steel, reinforced concrete, and

composite construction. By adding multistory diagonal bracings to the

face of the tube, the rigidity and efficiency of the framed tube can be

improved, thus the obtained braced-tube system, also known as trussed

tube or exterior diagonal-tube system, could be utilized for greater

heights, and allows larger spacing between the columns. It offers an

excellent solution by utilizing a minimum number of diagonals on each

face of the tube intersecting at the same point as the corner columns. In

steel buildings, steel diagonals/trusses, are used, while in reinforced

concrete buildings, diagonals are created by filling the window openings

by reinforced concrete shear walls to achieve the same effect as a

diagonal bracing. New York’s 50-storey-high 780 Third Avenue Building

, (fig.1.7).

Fig.1.7 : New York’s 50-storey-high 780 Third Avenue Building

18

1.2.6 Bundled-tube systems

Bundled-tube systems are proper for steel, reinforced concrete, and

composite construction. A single framed tube does not have an adequate

structural efficiency, if the building dimensions increase in both height

and width. Namely, the wider the structure is in plan, the less effective is

the tube. In such cases, the bundled tube, also known as modular tube,

with larger spaced columns is preferred. This concept, being created by

the need for vertical modulation in a logical fashion, can be defined as a

cluster of tubes interconnected with common interior panels to generate a

perforated multicell tube.[5]

The above classification is the expansion of the following basic structural

systems: frame systems, braced or shear walled systems, and tube

systems. ‘Brace systems and shear wall systems’ which are the systems

composed of only braces or shear walls, are the subsets of ‘braced frame

and shear-walled frame systems’ Nowadays, reinforced concrete and

composite structures are in serious competition with the steel structures,

and by the advancements in concrete technology, such as manufacturing

ultra- high-strength concrete, except ‘outrigger systems’, all the structural

systems classified above can be applied in reinforced concrete.[5]

1.3. Tall building in Missan

In Iraq we note a few are implemented so buildings in Iraq, particularly in

the province of Maysan and can causes due to a lack of skilled local

workers, lack of engineering expertise or lack of machinery and

equipment needed for implementation and the lack of materials locally.

An example of the Turkish hotel carried out by a Turkish company and

Turkish also the hospital and the al-sader Hospital, all these buildings

constructed by foreign companies see (fig 1.8 , fig 1.9 , fig 1.10).

19

Fig1.8:Turkish hotel

Fig 1.9 : AL - Sader hospital

20

1.4. Tall building codes

ASCE/SEI 7-10

(2006 – 2009) IBC

ACI 318 – 05 /08

( 303لمسودة االولية لمدونة الزلزال العراقية )م.ب.ع ا

1.5 Difference Between Law Rise Buildings And Tall Buildings

A tall- building is defined as a building 35 meters or greater in height,

which is divided at regular intervals into occupiable levels. To be

considered a high-rise building, an edifice must be based on solid ground

and fabricated along its full height.[3]

The cut-off between tall and low buildings is 35 meters.

A low-rise building is defined as any occupiable building which is

divided at regular intervals into occupiable levels and which is lower than

Fig 1.10 : Turkish hospital

21

a high-rise, i.e., lower than 35 meters. To be considered a low-rise

building, an edifice must be based on solid ground and fabricated along

its full height and have at least one floor above the ground[3].see (fig

1.11) for law rise and tall building difference.

Fig 1.11 : Tall and law buildings

22

Chapter two

Loads On High Rise Buildings In Misan Province

2.1 Loads

Structural members must be designed to support specific loads. Loads are

those forces for which a given structure should be proportioned. In

general, loads may be classified as dead or live.

2.1.1Dead Load

Dead loads consist of the weight of all materials of construction

incorporated into the building including but not limited to walls, floors,

roofs, ceilings, stairways, built-in partitions, finishes, cladding and other

similarly incorporated architectural and structural items, and fixed service

equipment including the weight of cranes.[1]

2.1.2 Live load

Live loads are those loads produced by the use and occupancy of the

building or other structure and do not include construction or

environmental loads such as wind load, snow load, rain load, earthquake

load, flood load, or dead load. Live loads on a roof are those produced :

(1) during maintenance by workers, equipment, and materials.

(2) during the life of the structure by movable objects such

as planters and by people.

The ACI Code does not specify loads on structures; however,

occupancy loads on different types of buildings are prescribed by

IBC-2012 and the American National Standards Institute (ANSI)

23

[5]. Some typical values are shown in Table 2.1. Table 2.2 shows

the density of various materials.

The live loads used in the design of buildings and other structures

shall be the maximum loads expected by the intended use or

occupancy but shall in no case be less than the minimum uniformly

distributed unit loads required by Table (2-1).

Table ( 2 – 1 ) Typical Uniformly Distributed Design live Loads

24

Table ( 2 – 1 ) Typical Uniformly Distributed Design live Loads

25

Table ( 2 – 2 ) Density of Various Materials

26

2.2 Wind load

Buildings and their components are to be designed to withstand the code-

specified wind loads. Calculating wind loads is important in design of the

wind force-resisting system, including structural members, components

and cladding, against shear, sliding, overturning and uplift actions. See

fig 2.1

Fig 2.1 : effect of wind load on building

27

2.2.1Methods of determination wind load

The design wind loads for buildings and other structures shall be

determined according to one of the following procedures[ASCE7-10]:

(1) Method – Simplified procedure for low-rise simple diaphragm

buildings.

(2) Method – Analytical procedure for regular shaped building and

structures.

2.2.1.1 Method – Simplified Procedure

The simplified procedure is used for determining and applying wind

pressures in the design of simple diaphragm buildings with flat, gabled,

and hipped roofs and having a mean roof height not exceeding the least

horizontal dimension or 60 feet (18.3 m), whichever is less, and subject to

additional limitations.

2.2.1.2 Method – Analytical Procedure

Wind loads for buildings and structures that do not satisfy the conditions

for using the simplified procedure can be calculated using the analytical

procedure provided that it is a regular shaped building or structure, and it

does not have response characteristics making it subject to a cross-wind

loading, vortex shedding, instability due to galloping or flutter, or does

not have a site location that require special consideration. [ASCE7-10]

Method 1 can't use for determination of wind load due to building

height ( 64m ) and it use for building with law rise (< 18.3 m) so

we will be use method 2 for determine wind load.

28

2.3 Procedure for determined wind load

2.3.1. Determine velocity pressure

Velocity pressure, qz, evaluated at height z shall be calculated by the

following equation [ASCE7-10]:-

qz = 0.613 Kz Kzt Kd V 2

(N/m2); V in m/s

where :-

qz = velocity pressure

Kd = wind directionality factor

Kz = velocity pressure exposure coefficient

Kzt = topographic factor defined

V= basic wind speed

2.3.1.1. determine Kd

Kd find from Table (2-3).

Table (2-3)

29

2.3.1.2 find kzt

Kzt find from

Kzt = (1 + K1K2K3)2

Where:-

K1, K2, and K3 are given in Table(2-4) shown below. If site conditions and

locations of structures do not meet all the conditions specified in Table(2-

4). then Kzt = 1.0.

Table (2-4)

30

2.3.1.3 find Kz

Kz find from equation below :-

1- The velocity pressure exposure coefficient Kz may be determine from

the following formula or from the Table (2-5):-

For 15 ft. ≤ z ≤ zg for z < 15 ft.

Kz = 2.01(z/zg)2/

kz = 2.01(15/zg)2/

2- and zg in Table (2-6)

3- Linear interpolation for intermediate values of height z is acceptable.

Table (2-5) . velocity pressure exposure coefficients , Kz and Kh

31

2.3.1.4 find V

Basic wind speed From Iraqi code by using wind map in (figure 2.2)

below[ المسودة االولية لمدونة الزلزال العراقية )م.ب.ع

Table (2-6)

Fig 2.2 wind speed zoning in Iraq

32

2.3.2 Determine the design wind pressure (p) or design wind load (F)

The design wind load pressure is given by following equation :-

P = qz G Cp - qi (GCpi) (N/m2)

Where :-

q = qz for windward walls evaluated at height z above the ground z.

q = qh for leeward walls, side walls, and roofs, evaluated at height h.

qi = qh for windward walls, side walls, leeward walls, and roofs of enclosed buildings

and for negative internal pressure evaluation in partially enclosed buildings.

G = gust-effect factor.

Cp = external pressure coefficient from.

( GCpi) = internal pressure coefficient from .

2.3.2.1 Find G

The gust-effect factor for a rigid building or other structure is permitted to

be taken as 0.85.

2.3.2.2 find Cp

Cp find from Table (2-7).

Table (2-7)

33

2.3.2.3 Find GCpi

GCpi Find from Table (2-8)

Table (2-8)

34

2.4 design wind load cases

The design wind load cases as defined in the figure below and take case

that give max design load .

Table 2-9: design cases

35

Due to difficult and long for four loading cases calculations procedure so

in the present project an excel sheet is used to determine design wind load

for every floor for the input information shown in Table ( 2-10 ):-

DESIGN SUMMARY

Max building horizontal force normal to building length, L, face = 3423.18 KN

Max base moment at wind normal to building length, L, face = 130314.21 KN.m

Max building horizontal force normal to building length, B, face = 2606.528 KN

Max base moment at wind normal to building length, B, face = 96203.365 KN.m

Feet unit Metric unit Input Data

B B Exposure category (A,B, C or D)

1 1 Importance factor (0.87, 1.0 or 1.15) I

137.76 42 Basic wind speed (V)

1 1 Topographic factor (Sec.6.5.7.2) (Kzt )

204.672 62.4 Building height to roof (H)

3.28 1 Parapet height (HP )

169.25 51.6 Building length (L)

127.92 39 Building width (B)

Table 2-10

36

Wind Normal to face L

Lateral P (KN)

z (m) Windward Leeward

4.6 84.1 -91.314

6.1 30.491 -91.314

7.6 32.461 -91.314

9.1 34.427 -91.314

12.2 74.757 -91.314

15.2 79.672 -91.314

18.3 83.609 -91.314

21.3 87.545 -91.314

24.4 91.477 -91.314

27.4 94.431 -91.314

30.5 97.38 -91.314

36.6 204.594 -91.314

42.7 214.433 -91.314

48.8 222.302 -91.314

54.9 230.170 -91.314

61 236.073 -91.314

63.4 94.858 -91.314

1992.78 -1552.338

Base shear = 1992.78 + 1552.338 = 3545

Table 2-11

37

Wind Normal to face B

Lateral P (KN)

z (m) Windward Leeward

4.6 63.566 -59.917

6.1 23.045 -59.917

7.6 24.535 -59.917

9.1 26.021 -59.917

12.2 56.503 -59.917

15.2 60.217 -59.917

18.3 63.192 -59.917

21.3 66.168 -59.917

24.4 69.139 -59.917

27.4 71.372 -59.917

30.5 73.601 -59.917

36.6 154.634 -59.917

42.7 162.071 -59.917

48.8 168.020 -59.917

54.9 173.966 -59.917

61 178.428 -59.917

63.4 71.693 -59.917

1506.171 -1018.59

Base shear =1506.171+1018.59 = 2524.76 KN

Table 2-12

38

2.5 Earthquake Load

Every building and its portions, as a minimum, shall be designed and

constructed to resist the effects of earthquake ground. The seismic

loadings are determined according to ASCE and IBC provisions.

2.5.1. Procedure for Calculation of Seismic Design Category (SDC)

1- Determine seismic use group as described in (Table 2-13).[1]

Table (2-13)

39

2- Based on the location of the building, determine the mapped spectral

accelerations for short periods, Ss , S1 as shown below or from Table

(2-14)

Fig 2.3 :Iraq contour map of the value of the spectral acceleration of

seismic ground motion at a period of time Short (0.2) seconds (Ss)

according US DOD Unified Facilities Criteria 3-310-01(US DOD UFC)

2007

40

Fig 2.4 :Iraq contour map of the value of the spectral acceleration of

seismic ground motion at a period of time Short (1) seconds (S1) :US

DOD Unified Facilities Criteria 3-310-01 (US DOD UFC) 2007

41

City Ss S1

Baghdad 1.24 0.56

Al-Amarah 1.31 0.52

Basrah 0.98 0.39

Kut 0.97 0.39

AL-Mijer Al Kaber 1.13 0.45

Samara 0.29 0.12

Arbil 1.38 0.55

3- Use Table (2-15) to determine site class based on the soil profile

name and properties of soil.

Table (2-14)

Table (2-15)

42

4- Using Table (2-16) determine site coefficient Fa based on mapped

maximum considered earthquake spectral response accelerations at

short periods, SS. Also using Table (2-17)determine site coefficient Fv

based on mapped maximum considered earthquake spectral response

accelerations at (1-s) period, S1

Table (2-16)

Table (2-17)

43

5- Calculate the maximum considered earthquake spectral response

accelerations for short periods for specific soil class, SMS, using (Eq. 1)

Also calculate the maximum considered earthquake spectral response

accelerations for (1-s) period for specific soil class, SM1, using (Eq.

2).[1]

SMS = Fa SS ---------- (Eq.1)

SM1 = Fv S1 ---------- (Eq.2)

Where :-

SMS = mapped maximum considered earthquake spectral response

accelerations for short periods adjusted for site class effect.

SM1 = mapped maximum considered earthquake spectral response

accelerations for 1-s period adjusted for site class effect.

6- Using (Eq.3) determine design spectral response acceleration

coefficient for short periods, SDS, and using (Eq.4) determine spectral

response acceleration coefficient for 1-s period, SD1.[1]

SDS = 2/3 SMS ------- (Eq.3)

SD1 = 2/3 SM1 ------- (Eq.4)

7- Determine SDC according to Tables (2-18) and (2-19).[1]

Table (2-18)

44

2.5.2 Analysis procedures

During the earthquake motions, the structure is subjected to the

deformation that produces internal forces and stresses. Earthquake

engineering philosophy is to relate earthquake dynamic forces to the

equivalent static forces, and then using static analysis of the structure,

determine deformations, internal forces, and stresses in the structure. IBC

describes two analysis procedures to determine the equivalent static

forces that will simulate an earthquake action on the structure[1] . these

are:-

1- The equivalent lateral force procedure (used for SDC B, C, D, E,

and F).

2- The simplified analysis (used for SDC B, C, D, E, and F, and for

constructions limited to two stories in height and three stories in

height for light frame constructions).

In the present project the equivalent lateral force procedure is

used due to the building consists of 12 stories and simplified

procedure is not applicable.

Table (2-19)

45

2.5.2.1 Equivalent Lateral Force Procedure

This procedure describes how to calculate the seismic base shear and

lateral seismic forces.[1]

2.5.2.1.1 Seismic Base Shear Calculation

The total seismic force that acts at the base of the structure, called seismic

base shear, can be determined according to the following equation:

V = Cs W

Where :-

Cs = seismic response coefficient

W = effective weight of structure including total dead load and other

loads listed.

2.5.2.1.2 Seismic Response Coefficient Calculation

The seismic response coefficient, Cs, shall be determined from:

Where :-

SDS = design spectral response acceleration parameter for short period.

R = response modification factor given in Table (2-20).

IE = seismic importance factor determined from Table (2-13).

46

The value of Cs should not exceed

SD1:- design spectral response acceleration parameter at a period of 1.0 s,

as determined from (Eq.4)

T or Ta = fundamental period of structure(s) determined in (Eq.5)

or(Eq.6)

Table (2-20)

47

where hn is the height in feet above the base to the highest level of the

structure and the coefficients Ct and x are determined from Table (2-21)

For the concrete moment-resisting frame buildings that do not exceed 12

stories in height and have an average minimum story height of 10 Ft., the

approximate period of vibration, T, can be determined using the following

equation(6):

where N is the number of stories in the building above the base.

Eq.5

Table (2-21)

Eq.6

48

Also, CS should not be less than:

2.5.2.2 Lateral Seismic Force Calculation

Vertical distribution of the base shear force produces seismic lateral

forces, Fx, at any floor level. Seismic lateral forces act at the floor levels

because masses of the structure are concentrated at the floor levels. It is

known that the force is a product of mass and acceleration.

The lateral force that will be applied to level x of the structure, Fx, can be

determined from the following equation:

Where :-

Cvx = vertical distribution factor

k = distribution exponent related to building period

= 1 for building having a period of T ≤ 0.5 s

= 2 for building having a period of T ≥ 2.5 s

= 2, or linear interpolation between 1 and 2, for building having a

period of 0.5 s ≤ T ≤ 2.5 s

hi, hx = height from base to level i and x

wi, wx = portion of W assigned to level i or x

n = number of stories

49

V = total design lateral force or shear at the base

Overturning Moment

The lateral seismic force Fx produces overturning moments. Overturning

moment Mx should be calculated using the following equation:

2.6 Seismic Load Calculation

2.6.1 Determine SDC

1- risk category = III and Ie = 1.25

2- Ss = 1.31 , S1 = 0.52

3- site class = D ( from report of soil investigation)

4- Fa = 1 , Fv = 1.5

5- Sms = 1* 1.31 = 1.31 , Sm1 = 1.5 * 0.52 = 0.78

6- SDs = 2

3∗ 1.31 = 0.8733 , SD1 =

2

3∗ 0.78 = 0.52

7- SDC = D

50

2.6.2 Determine Seismic Base Shear

Cs = 0.87337/1.25

= 0.156

T = 0.0488 * (63.4) 0.75

= 1.0964 s

Cs,max = 0.52

1.0964∗(7

1.25) = 0.0846

Cs,min = 0.044 * 0.8733 * 1.25 = 0.048 > 0.01 o.k.

Cs > Cs,min o.k.

Cs > Cs,max not o.k. so use Cs = Cs,max = 0.0846

WT = 29622 * 11 + 11506 =337348 KN

V= 0.0846 * 337348 = 28539

Table (2-22) below show calculation of Fx and moment

Fig 2.5 : weight for each storey and loads

51

Base shear due to earthquake = 28486 KN

Base shear due to wind = 3545.12 KN

Use earthquake loads in design because it is more than from wind

loads

floor

level wi hi wi hi

k Cvx Fx

(KN)

Shear

force

(KN)

Overturning

moment Mx

(KN.M)

12 11506 60 897910 0.068 1940 1940 0

11 29622 55 2107212 0.16 4566 6506 9700

10 29622 50 1903961 0.1443 4118 10624 42230

9 29622 45 1702014 0.129 3681 14305 95350

8 29622 40 1501504 0.1138 3247 17552 166875

7 29622 35 1302601 0.0987 2816 20368 254635

6 29622 30 1105520 0.0838 2391 22759 356475

5 29622 25 910546 0.069 1969 24728 470270

4 29622 20 718076 0.0544 1522 26250 572401

3 29622 15 528701 0.04 1141 27391 725160

2 29622 10 343411 0.026 742 28133 862115

1 29622 5 164232 0.0124 353 28486 1002780

∑ 337348 13185688 28486 28486 1145210

Table 2-22

52

2.7 Loads on building

dead load :- include the following :-

i. self-weight = 0.2 * 24=4.8

ii. super imposed dead load (tiles and mortar ) = 0.03 *24 + 0.05 * 20

= 1.72 use 2 KN/m2

iii. portions ( 2.5 – 5 KN/m2 ) use 2.5

total dead load = 4.8 +2+2.5= 9.3 KN/m2

live load :- from Iraqi code for office building use 2.5 KN/m2

earthquake loads :- from calculations in section (2.6.2).

2.8 Load combinations

Basic Combinations. Structures, components, and foundations shall be

designed so that their design strength equals or exceeds the effects of the

factored loads in the following combinations:-

a) 1.4(D + F)

b) 1.2(D + F + T ) + 1.6(L + H) + 0.5(Lr or S or R)

c) 1.2D + 1.6(Lr or S or R) + (L or 0.8W)

d) 1.2D + 1.6W + L + 0.5(Lr or S or R)

e) 1.2D + 1.0E + L + 0.2S

Where :-D = dead load , E = earthquake load , L = live load , Lr = roof

live load S = snow load, W = wind load;

In the present project we will be used the item (e) from list of

combinations load due to loads in the present project are dead , live

and earthquake without snow .

53

Chapter three

Building System And Description

3.1 selection building system

The basic structural system that may be used to resist earthquake forces

are listed in table (3-1) .

Table (3-1).

54

Table (3-1)cont.

cont.cont.

55

Table (3-1) cont.

56

Table (3-1) cont.

57

Table (3-1) cont.

58

According to seismic design category the system that suitable for

building are :-

A- Moment-resisting frame systems

- Special reinforcement concrete moment frame

B- Dual systems with special moment frames capable of resisting

at least 25% of prescribed seismic forces

- Special reinforced concrete shear wall

C- Bearing wall system

- Special reinforcement masonry shear wall

Use Dual Systems With Special Moment Frames Capable Of resisting

At Least 25% Of Prescribed Seismic Forces and Special reinforced

concrete shear wall

3.2 selection of slab system

Flat Plates

Flat plates are most suitable for spans of 6 to 7.6 m. and live loads

between 2.87 and 4.8 KN/m2. The advantages of adopting flat plates

include low-cost formwork, exposed flat ceilings, and fast construction.

Flat plates have low shear capacity and relatively low stiffness, which

59

may cause noticeable deflection. Flat plates are widely used in buildings

either as reinforced or Prestressed concrete slabs.

Flat Slabs

Flat slabs are most suitable for spans of 6 to 9 m. and for live loads of

3.8 to 7.1 KN/m2. They need more formwork than flat plates, especially

for column capitals. In most cases, only drop panels without column

capitals are used.

Waffle Slabs

Waffle slabs are suitable for spans of 9 to 14.6 m. and live loads of 3.83

to 7.2 KN/m2. They carry heavier loads than flat plates and have

attractive exposed ceilings. Formwork, including the use of pans, is quite

expensive.

Slabs on Beams:

Slabs on beams are suitable for spans between 6 and 9 m. and live loads

of 2.87 to 5.75 KN/m2. The beams increase the stiffness of the slabs,

producing relatively low deflection. Additional formwork for the beams

is needed.

One-Way Slabs on Beams:

One-way slabs on beams are most suitable for spans of 3 to 6.1 m. and a

live load of 2.87 to 4.8 KN/m2. They can be used for larger spans with

relatively higher cost and higher slab deflection. Additional formwork for

the beams is needed.

60

One-Way Joist Floor System

A one-way joist floor system is most suitable for spans of 6 to 9 m. and

live loads of 3.8 to 5.75 KN/m2. Because of the deep ribs, the concrete

and steel quantities are relatively low, but expensive formwork is

expected. The exposed ceiling of the slabs may look attractive.

Use slabs on beams is suitable for building due to system of our

building is Dual Systems With Special Moment Frames Capable Of

resisting At Least 25% Of Prescribed Seismic Forces and Special

reinforced concrete shear wall and spans for our building is 7 to 9 m.

3.3 Description of project

The building as shown in Figure (a) , is a 12 story office building, without

basements, a ground floor and eleven upper floors. The building’s overall

dimensions are 39.85 m wide 52.4 m long, and it is approximately 64m

tall from ground level, all storeys height are 5.2 m , all beams

dimensions are 800 mm in depth and 400 mm in width The building

system consists of slabs with beams and shear walls .

61

The building’s architectural and structural drawings are shown in Figure

(b) Figure (c) respectively

Figure (a)

62

1 st

2 st

63

3,4,5,6,7,8,9 st

10 st

Figure (b)

64

Ground floor framing

plan

65

Figure (c)

66

Chapter four

Design Calculation

4.1 Methods of Slab design

There are two methods for design two way slab given by ACI code:-

- Direct Design Method DDM

- Equivalent Frame Method EFM

4.1.1 Direct Design Method DDM

Conditions of the DDM are :-

1- Minimum of three spans in each direction. Minimum of 3 x 3 = 9

panels.

2- Rectangular panels with aspect ratio between 0.5 and 2.0.

3- Successive spans in each direction must not differ by more than

one third of the largest span. 4- Column offset from basic rectangular grid must not exceed 10 % of

span in offset direction. 5- Gravity loading 6- Live load less or equal to twice the dead load 7- For slabs with beams, the relative beam stiffness must be such :

67

4.1.2 Equivalent Frame Method EFM

The equivalent frame method involves the representation of the

three-dimensional slab system by a series of two dimensional

frames that are then analyzed for loads acting in the plane of the

frames.

- equivalent frame method applicable to slab with or without

beams

- The equivalent frame method may be used for lateral loads

analysis.

4.1.3 Equivalent Frame analysis by computer

It is clear that the equivalent frame method ,as described in the ACI

Code and the ACI Code commentary, is oriented toward analysis using

the method of moment distribution, most offices make use of computers

and frame analysis is done using general purpose programs based on the

direct stiffness method. plane frame analysis programs can be used for

slab based on the concepts of the equivalent frame method , but frame

must be specially modeled. Variable moments of inertia along the axis of

slab –beams and columns require nodal points (continuous joints)

between sections where I is to be considered constant .

Alternately , a three – dimensional frame analysis may be used , in which

the torsional properties of the transverse supporting beams may be

included directly .

4.1.3.1 SPSlab program

spSlab analyzes beams, one-way slab systems (including standard and

wide module joist systems), and two-way slab systems (including waffle

and slab beams). With capacity to integrate up to 20 spans and two

68

cantilevers of multiple floor system types in each model, the capabilities

of spSlab are not limited to new designs. spSlab can perform strength

investigation for evaluation and/or modifications of existing building

slabs. This program is sure to save engineers time crosschecking designs

with applicable design code provisions.[2]

Save in both material and labor using the moment redistribution feature.

It allows up to 20% reduction of negative moments over supports

reducing reinforcement congestions in these areas.

4.2 Slab design

4.2.1 slab thickness

The thickness for each slab panel depends on the average beam relative

stiffness which is the average of the values for the four beams of the

panel

The minimum thickness is determined as follows:

m 0.2 Use minimum thickness Table below for flat plate (and slabs

without interior beams)

Direct Design Method DDM can't use because of conditions not

satisfy so we will use equivalent frame method

69

We will check slab thickness for larger span internal panel slab

because it critical case for slab thickness .

m = 7.385

hmin = 8000∗(0.8+

420

1400)

36+9∗1.29 = 184.8 we will use 200 mm

Table 4-1

70

We will be design slabs for ground floor only and the other floors

obtain by similarity .

Due to separation joint in building so design of floor into two part.

Note in the input data we need moments due to earthquake loads it

found from analysis by staadpro program and used max values for

safety for each frame .

Due to difficult hand calculations by equivalent frame method so use

computer program SPSlab for design slab .[2]

71

Design frame 4 (large part )

The figures (4.1-4.5) below show the analysis and design of the frame 4

large part

Fig 4.1

72

Fig 4.2

73

Fig 4.3

74

Fig 4.4

75

Fig 4.5

76

Design frame E (large part )

The same loads are distributed the results are shown in the figures(4.6 -

4.10) below.

Fig 4.6

77

Fig 4.7

78

Fig 4.8

79

Fig 4.9

80

Fig 4.10

81

Design frame I and A for two parts

The same loads are distributed the results are shown in the figures(4.11-

4.15) below .

Fig 4.11

82

Fig 4.12

83

Fig 4.13

84

Fig 4.14

85

Fig 4.15

86

Design frame B (large part and small part )

The same loads are distributed the results are shown in the figures(4.16 -

4.20) below

Fig 4.16

87

Fig 4.17

88

Fig 4.18

89

Fig 4.19

90

Fig 4.20

91

Design frame 4 for small part

The same loads are distributed the results are shown in the figures(4.21-

4.25) below.

Fig 4.21

92

Fig 4.22

93

Fig 4.23

94

Fig 4.24

95

Fig 4.25

96

Design Frame 7 (large part)

Fig 4.26

97

Fig 4.27

98

Fig 4.28

99

Fig 4.29

100

Fig 4.30

101

Fig 4.31

102

We will be design the slab for max moments and then other slabs

panels obtained by similarity

So we find reinforcement by hand calculation's as shown below

Max moments occurred at frame E

+Ve = 190.2 KN.m

-Ve = 251 KN.m

For max +ve = 190.2

Using db = 12 mm d = 200 – 20 -12/6 =178 mm

b = 3500 mm

fc' =25 MPa

fy = 420 MPa

= 0.9

Max spacing = 300 mm

As = 2966 mm2

As min1= 0.0018 b h = 0.0018* 3500 * 200 = 1260 mm2

As min 2 = Ab b/Smax = 113.04 * 3500 /300 = 1318.8 mm2

As > As min2

Number of bars = 2966 / 113.04 = 26 bars

103

Spacing between bars = 3500 /41 =130 mm

so use ∅ 12 @130mm c/c at bottom for all panels of slabs in two

directions

For max –ve = 251 KN.m use ∅16mm

d = 172 mm

As =4142.3 mm2

As > Asmin2 ….. O.K.

Number of bars = 4142.3/201 = 20.6 = 21 bar

Spacing between bars = 3500/21 = 166 use 160 mm

so use ∅16 @160 mm c/c at top for all panels of slabs in two

directions

Bar no location reinforced

A bottom ∅ 12 @130 c/c

B top ∅16 @160 c/c

200 mm thickness

104

4.3 beams design

We will choose four beams in design for ground level only due

to max axial force from earthquake loads effect on the ground

level the beams that chosen are the critical case due to loaded

max moment and shear force , for other beams in above storeys

are obtained by similarity , the beam plan design are shown in

the figures 4.32 , 4.33 and fig 4.34 below .

Fig 4.32

105

Fig 4.33

106

Notes

The red circles refers to critical beams that we will be

design .

Beam dimensions are (800*400 mm)

Moments for dead and live loads we will calculate by hand

according to the max moment coefficient for continuous

beam

Moments for earthquake effect we found from staadpro

program

Fig 4.34

107

4.3.1 requirement for flexural members for special moments

frames ACI- Code 318M -08

21.5.1.1 from ACI- Code 318M -08 :- Factored axial compressive force

on the member, Pu, shall not exceed Agfc′ /10.

21.5.1.2 from ACI- Code 318M -08 :- Clear span for member, ln, shall

not be less than four times its effective depth.

21.5.1.3 from ACI- Code 318M -08 :- Width of member, bw, shall not

be less than the smaller of 0.3h and 250 mm.

For our project all above requirement are satisfy

Requirement for longitudinal reinforcement (21.5.2 from ACI- Code

318M -08 )

for top as well as for bottom reinforcement, the amount of reinforcement

shall not be less than that given by

but not less than 1.4b wd/fy, and the reinforcement ratio, ρ, shall not

exceed 0.025. At least two bars shall be provided continuously at both top

and bottom.

108

Requirement for Transverse reinforcement (21.5.3 from ACI- Code

318M -08 )

4.3.2 Design calculations

All dimension for beams are (800*400 mm ) fc' = 25 MPa , fy =

420 MPa .

Dead load on slabs = 9.3 KN/m2

Live load on slabs = 2.5 KN/m2

Coefficient for max positive moment = 𝟏

𝟏𝟒

Coefficient for max negative moment = 𝟏

𝟏𝟎

109

Design B1

Wd on beam = (9.3* 7/3 )* (3-0.82/2)* 2 = 51 KN/m

WL on beam = (2.5 * 7/3) * (3-0.82/2) * 2 = 14 KN/m

Calculate As for max positive moment

MD = 51* 82/14 = 233 KN.M

ML = 14* 82 / 14 = 64 KN.M

ME = 571 KN.M

Mu = 1.2 * 233 + 64 + 571 = 915 KN.M

b = min {L/4 , 16 hf +bw , Lc1+Lc2/2 + bw }

b = 1750 mm

Muf = 0.9 * 0.85 * fc' * b * hf (d - hf/2 )

Assume two layer d= h – 95 = 800 -95 = 705 mm

Muf = 4050 KN . m

Muf > Mu …… design as rectangular

𝜌 max = 0.75 𝜌𝑏 = 0.75 ∗ 0.85 ∗ 𝛽1 ∗ (𝑓𝑐′

𝑓𝑦) ∗ (

600

600 + 𝑓𝑦)

𝛽1 = 0.85 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑐′ = 25 < 30 𝑀𝑃𝑎

𝜌 max = 0.0189

𝜌𝑚𝑖𝑛 = 1.4* bw /fy*b

𝜌 min = 7.6 * 10 -4

𝜌 = 1 −√1 −

2.622 𝑀𝑢𝑓𝑐′𝑏 𝑑2

1.18 ∗𝑓𝑦𝑓𝑐′

𝜌 = 0.01416 𝜌 min < 𝜌 < 𝜌 max 𝑂. 𝑘

110

As = 𝜌𝑏𝑑 = 0.01416 ∗ 400 ∗ 705 = 3994 𝑚𝑚2

Use ∅ 32 𝐴𝑏 = 804𝑚𝑚2

Number of bars = 3994/ 804 = 4.9 use 5 bars bottom

For negative moment

MD = 51∗ 82

10= 327 , ML =

14∗ 82

10 = 90 , ME = 571 ,Mu = 1054

𝜌= 0.0168 ……O.K

As = 4737 Use ∅ 32 𝐴𝑏 = 804𝑚𝑚2

Number of bars = 6 top

Check requirement for longitudinal reinforcement

As min 1 = 840 mm2

As min 2 = 1.4 * 400 *705 /420 = 940 mm2

As max = 0.025 * b * d = 0.025 * 400 * 705 =7050 mm2

As > As min …. O.K

As < AS max …. O.K

Design for shear

VD = 51∗8

2= 204 𝐾𝑁

VL = 14∗8

2= 56 KN

VE = 130 KN

Vu = 1.2 * 204 + 56 +130 = 431 KN

Vud = 355 KN

Vud / ∅ = 355/0.75 = 474 KN 4000

Vc = 0.17 √𝑓𝑐′ bw d = 242 KN

3Vc = 726 KN

431

710.5

Vud

111

Use ∅12 𝑠𝑡𝑢𝑖𝑟𝑟𝑢𝑝𝑠

Av = 2 * 113 =227

Vc < Vud / ∅ < 3Vc

S = min{d/2, 600 , 𝐴𝑣 𝑓𝑦

0.34 𝑏𝑤 ,

𝐴𝑣𝑓𝑦 𝑑𝑉𝑢

∅−𝑉𝑐

}

S = min{450 , 600 , 701 , 292}

Use 290

Check the requirement for transvers reinforcement

All satisfy except that of smallest spacing

Smallest spacing :-

1- d/4 = 700/4 = 175

2- 8 * 32 =256

3- 24 *12 =288

4- 200

Use 175

So we will use stirrups ∅12 @175 The section of B1 design are shown below

B1

112

Design B2

WD = (9.3 * 7/3) * 2 = 44 KN/m

WL = (2.5 * 7/3) * 2 = 12 KN/m

For positive moment

MD = 44*72 /14 = 154 KN.m

ML = 12*72 /14 = 42 KN.m

ME = 314 KN.m

Mu = 541 KN.m

Muf > Mu ….. rectangular

𝜌 max = 0.0189

𝜌 min = 7.6 * 10 -4

𝜌 = 7.8 * 10-3

…….. O.K

As = 2200 mm2

Use ∅ 25 Ab = 491 mm2

Number of bars = 5 ∅25 𝑏𝑜𝑡𝑡𝑜𝑚

For negative moment

MD = 44* 72 / 10 = 216 KN.m ,ML = 59 KN.m , ME = 345 KN.m

Mu = 663 KN.m

Muf > Mu …. Rectangular

𝜌 = 9.76 * 10-3

….. O.K.

As = 2752 mm2

Use ∅ 25 Ab = 491 mm2

Number of bars = 6 Use ∅ 25 top

113

Design for shear

VD = 154 KN

VL = 42 KN

VE = 94 KN

Vu = 321 KN

Vud = 256 KN Vud / ∅ =341 KN 3500

Vc =242

3Vc = 726

Vc < Vud / ∅ < 3Vc

S = 350 mm

Check requirements for longitudinal reinforcement

As > As min ….. O.K

AS < As max …… . O.K

Check requirements for transvers reinforcement

Should be used smallest spacing = 175 mm

So use ∅12 𝑠𝑡𝑖𝑟𝑟𝑢𝑝𝑠@ 175 𝑚𝑚

The section of B2 design are shown below

710.5

321 Vud

B2

114

Design B3

WD = 44 KN/m

WL = 12 KN/m

For positive moment

MD = 154 KN.M , ML = 42 KN.M , ME = 501 KN. M

Mu = 728 KN.m

Muf > Mu …… rectangular

𝜌 max = 0.0189

𝜌 min = 7.6 * 10 -4

𝜌= 0.01

As = 3061

No.of bars = 7 ∅25 bot

For negative moment

MD = 216 , ML = 59 , ME = 467 , Mu = 785

Muf > Mu …. Rectangular

𝜌= 0.0118 ….. O.K

As = 3337 mm2

No . of bars = 7∅ 25 top

Design for shear

VD = 154 KN , VL = 42 KN , VE = 138 KN , Vu = 365 , Vud = 291

Vud / ∅ =388 KN

Vc = 242 KN

3VC = 726 KN

Vc < Vud / ∅ < 3Vc

115

S = 355 mm

All requirements are satisfy except that for transverse spacing

We should use spacing = 175 mm

So use ∅12 @ 175 mm

The section of B3 design are shown below

Design B4

WD = (9.3 * 4/3 ) * 2 = 25 KN/m

WL = (2.5 *4 /3 ) * 2 = 6.5 KN.m

For positive moment

MD = 25 * 42 /14 = 29 KN.M

ML = 6.5 * 42 /14 = 8 KN.M , ME = 127 KN.M , Mu = 170 KN.m

𝜌= 2.34 * 10 -3

….. O.K

As = 660 mm2

For negative moment

MD = 25 * 42 / 10 = 40 , ML = 6.5 * 4

2 /10 = 11 , ME = 107

116

Mu = 166 KN.M

𝜌= 2.6 * 10 -3

….. O.K ….. As = 637

Design shear

VD = 50 ,VL = 13 ,VE = 58 , Vu = 58 , Vud = 85

Vud / ∅ = 114 KN

Vc/2 = 121 KN

Vud / ∅ < Vc/2

No need stirrups

Check requirements for longitudinal reinforcement

As < As min ….. so use As min for top and bottom

Use ∅ 16 Ab = 201 mm2

No . of bars = 940/201 = 4.67 use 5 ∅ 16 at top and bottom

Check requirements for transvers reinforcement

According to 21.5.3.4 ACI –Code -08 it should use spacing

= 710.5/2 = 355mm

So use ∅12 @ 350 mm

The section of B4 design are shown below

117

B1

B2

B3

B4

Beam details According to ACI-Code

Table 4-2 : beams design section

118

4.4 Shear Wall design

A wall system is a structural system that provides support for all gravity

loads and all lateral loads applied to the structure. A structural wall

system is much stiffer than a frame system and its performance during an

earthquake is better than the performance of the frame system.

A structural wall should be properly designed to sustain all loads acting

on it. Boundary elements of structural walls are the areas around the

structural wall edges, as shown in Fig below, that are strengthen by the

longitudinal and transverse reinforcement. Boundary elements increase

the rigidity and strength of wall panels. The web reinforcement is

anchored into the boundary elements.

119

Design of shear wall

Factored load = 1.2D + L + E

For moment ( all result below from staadpro program )

MD = 291 KN.m

ML = 79 KN.m

ME = 27290 KN.m

MU = 27718.2 KN.m

For shear force

VD = 194 KN

VL = 52 KN

VE = 1693 KN

VU = 1977.8 KN

For axial force

PUD = 2874 KN

PUL = 772 KN

PUE = 5398 KN

PU = 9618.8 KN

250 mm

6700 mm

Check if we need to boundary element

120

I

MC

A

Pf

lbbhI

c

wwg

1212

33

Ig = 6.2658*1012

mm4

Fc = 20.56 Mpa > 0.2 Fc' = 5 Mpa o.k.

00

0 we need to boundary element

Acv = 250 * 6700 = 1675000 mm2

= 1675000* √25 * 1 = 8375KN > Vu

- Space between bars must be 450 mm (ACI code).

Minimum reinforcement in both direction

For horizontal direction

Acv = 250 * 64000 = 16 * 106 mm

2

As = 0.0025 * Acv = 40000 mm2

Use Ø12

Number of bars = 40000

𝐴𝑏= 353

Space = 180 mm

For vertical direction

As = 0.0015 * 1675000 = 2512.5 mm2

From ACI code minimum reinforcement for vertical and horizontal of

gross area with 420 Mpa , ρl = 0.0015 , ρt = 0.0025.

Use Ø12@180 mm

121

Number of bar = 23

Space = 291.3 mm

Check if we need two layer (curtains) of reinforcement

𝐀𝐜𝐯 √𝐟𝐜′

𝟔=

𝟐𝟓𝟎∗𝟔𝟕𝟎𝟎∗ √𝟐𝟓

𝟔= 1395.83 < Vu

Check shear requirement

𝒉𝒘

𝒍𝒘= 𝟗. 𝟓𝟓 αc = 0.17

Vn = Acv (αc √𝒇𝒄′ + ρt fy )

ØVc = 2386.875 KN > Vu o.k.

Use Ø12@290 mm

We need two layer of reinforcement in both direction

122

Design for flexural

d = 0.8 lw

d = 5360 mm

m = 𝑓𝑦

0.85 𝑓𝑐′

m = 19.76

R = 𝑀𝑢

Ø 𝑏 𝑑2 Ø = 0.9

R = 27718.2∗106

0.9∗250∗ 53602

Ρ = 1

𝑚 [ 1 − √1 −

2𝑚𝑅

𝑓𝑦

Ρ = 0.0115

ρmin = 1.4

𝑓𝑦

6.7 m

123

ρmax = 0.75 ρb

ρmax = 0.01897

ρmin < ρ < ρmax o.k.

As = ρ * b * d

As = 15410 mm2

Number of bar = 32

Space = 165 mm

Shear wall subjected to vertical load

Vc 1

6 √𝑓𝑐′ ℎ 𝑑 (equation 1)

Vc = 1116.6 KN

Vc = 0.27 √𝑓𝑐′ ℎ 𝑑 + 𝑃𝑢 𝑑

4 𝑙𝑤 (equation 2)

Vc = 3732.76 KN

Vc = h d

10 [ 0.5 √fc′ +

lw∗(√fc′+2Pu

lw∗hMu

Vu−

lw

2

] (equation 3 )

Vc = 1722.8 KN

Use smaller Vc = 1116.6 KN

ØVc = 949.11 KN < Vu

So shear reinforcement is required

S = 𝑨𝒗 𝒇𝒚 𝒅

𝑽𝒔

Vs = Vn – Vc

Use Ø25@165

mm

124

= 2386.875 – 949.11 = 1437.76 KN

Use Ø12

Space = 113∗420∗5360

1437.76∗ 103

Space = 175 mm

Use Ø12@175 mm

125

4.5 Columns design

Structural plans for columns design are shown below :-

126

C9

127

Design of C1(400*400)

Wu slab = 1.2 * 9.3 + 1.6 * 2.5 = 15 KN/m2

Axial load

From slab = 15* 3.5 * 3.5 = 184 KN

From beam

Beam stem = [(0.8 – 0.2 ) * 0.4 * 25 ] * 1.2 = 7.2 KN/m

(this for all beams )

Pstem = 7.2 * 3.5 +7.2 * 3.5 = 50 KN

From self-weight of column

Pcol =1.2 [0.4 * 5 * 0.4 * 25] = 24 KN

Pu = 184 + 50 + 24 = 258 KN

Wu B3 = WuB4 = 1.2 * 44 + 1.6 * 12 = 72 KN/m

Mux = Muy = 72∗(7−0.4)2

16= 221 𝐾𝑁. 𝑚

b/h = 400/400 = 1

Meq = ∅𝑀𝑛𝑥 = 221 + 221 (400

400) ∗ (

1−0.65

0.65) = 332 𝐾𝑁. 𝑚

Design as uniaxial with Mu = 332 , Pu = 258

𝛾 =400 − 2 ∗ 62.5

400= 0.7

𝐾𝑁 =258 ∗ 1000

0.7 ∗ 25 ∗ 16 ∗ 104= 0.092

𝑅𝑁 =332 ∗ 106

0.7 ∗ 25 ∗ 16 ∗ 104 ∗ 400= 0.3

From interaction diagram 𝜌 = 0.05

128

As = 0.05 * 400 * 400 =8000 mm2

Use 16 ∅25

Ties

S = min (16 * 25 , 48 * 12 , 400 ,400)

S= 400 use 300 mm

So use 3∅12 @300𝑚𝑚

C1

129

Design C2 (400*400)

Pu=338 KN , Mu = 221 KN.m , 𝛾 = 0.7 , Kn = 0.12 , Rn = 0.2

𝜌 = 0.03 , As = 4800 use 10∅25

Ties use 3∅12 @300𝑚𝑚

C2

130

Design C3 (400*400)

Pu=961 KN , Mu =0 , 𝛾 =0.7 , Kn = 0.343 , Rn = 0

Use 𝜌𝑚𝑖𝑛 = 0.01 , As = 1600 use 8∅16

Ties use 3∅12 @300𝑚𝑚

C3

131

Design C4(600*600)

Pu = 1728 KN , Mux =Muy = 221 KN.m , 𝛾 = 0.8 , Kn = 0.275 ,

Muq =332 KN.m , Rn = 0.087 , 𝜌𝑚𝑖𝑛 = 0.01 , As = 3600 mm2

Use 10∅25 , Ties use 3∅12 @300𝑚𝑚

C4

132

Design C5 (600*600)

Pu = 2548 KN , Mu = 221 KN , 𝛾 = 0.8 , Kn = 0.4 , Rn = 0.06 ,

𝜌𝑚𝑖𝑛 = 0.01 , As = 3600 , Use 10∅25 , Ties use 3∅12 @300𝑚𝑚

C5

133

Design C6 (600*600)

Pu = 6180 KN , Mu =0 , 𝛾 = 0.8 , Kn = 0.98 , Rn =0 , 𝜌 = 0.013

As = 4680 , , Use 10∅25 , Ties use 3∅12 @300𝑚𝑚

C6

134

Design C7 (800*800)

Pu = 3378 KN , Meq = 332 KN.m , , 𝛾 = 0.9 , Kn = 0.3 , Rn = 0.037

𝜌 = 0.01 , As = 6400 , Use 16∅25 , Ties use 3∅12 @300𝑚𝑚

C7

135

Design C8 (800*800)

Pu = 4641 KN , Mu = 221 KN.m , 𝛾 = 0.9 , Rn = 0.04 , Kn = 0.024

𝜌 = 0.01 , As = 6400 , Use 16∅25 , Ties use 3∅12 @300𝑚𝑚

C8

136

Design C9 (800*800)

Pu = 12306 KN , Mu = 0 , 𝛾 = 0.9 , Kn = 1.1 , Rn =0 , 𝜌 = 0.026

As= 16640 , Use 20∅32 , Ties use 3∅12 @300𝑚𝑚

C9

137

Columns Design section

C1

C2

C3

C4

C5

Table 4-3 columns section

138

C6

C7

C8

C9

Table (4-3 ) cont.

139

4.6 foundation design

The foundation types are dependent upon the size of the structures, the

anticipated loads and the allowable bearing pressure.

4.6.1 Pile Foundations

The use of precast, Franki or bored pile is considered appropriate for

supporting the relatively heavier structural loadings; when the applied

pressure exceeds that allowable bearing capacity of the supporting soil

and the estimated settlement greater than the allowable limits. For the

proposed eleventh stories building; pile foundation is the suitable type of

foundation to support the applied loads.[4]

In the present project we will use bored pile with bearing capacity 180

ton according to soil investigation report see table (4-4) below

Table 4-4

140

∑ Pu = 35 ∗ 12306 + 24 ∗ 4641 + 4 ∗ 3378= 555606 KN

this factored load it must use service load

= 555606

1.2 = 463005 KN = 46300.5 Ton

Minimum Number of piles used= ∑ 𝑃𝑢

𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑜𝑓 𝑝𝑖𝑙𝑒

∑ Pu = 46300.5 Ton

Min number =46300.5 /180 = 257 .225 use 285 for safety and

efficiency

We will use 285 bored pile with diameter = 120 cm ,

length 20m the distance between piles = 2.5 D = 3m

the distribution of piles shown below fig (4-35) below .

141

Fig (4-35)

142

Pile cap

Find depth of pile cap

Check wide beam shear

At distance d from face of column

Pu pile = 1800 KN

Vu = 2∗1800 = 3600 KN

∅𝑣𝑐 = 0.85 ∗ 0.17 ∗ √𝑓𝑐′ ∗ 𝑏𝑤 ∗ 𝑑

= 0.85∗0.17∗5∗6∗d∗1000

Vu =∅𝑣𝑐 → d= 0.83 m

Check punching shear

Vu = 4∗1800 14400 KN

∅𝑣𝑐 = 0.85 ∗ 0.33 ∗ √𝑓𝑐′ ∗ 𝑏𝑜 ∗ 𝑑 = 0.85∗ 0.33∗5∗ (3.2+4d) ∗d∗1000

Vu =∅𝑣𝑐 → 𝑑 = 1.25 𝑚

Use d= 1.25 +0.075+0.15 = 1.475 for practical use d=1.5m

Depth of cap = 1.5 m

The details of cap are shown in the figure (4-36) below :-

6m

Fig 4.36

143

References

[1] Structural Concrete Theory And Design Fifth Edition M. Nadim Hassoun

South Dakota State University Akthem Al-Manaseer

[2] Design Of Concrete Structures Fourteenth Edition Nilson

[3] Reinforced Concrete Design Of Tall Buildings Bungales

[4] Foundation Analysis and Design Fifth Edition Joseph E. Bowles, Re., S.E

[5] Building And Environment M. Halis Gunel, H. Emre Ilgin

[6] Website

144

APPENDIX A: staadpro program analysis

145

146

147

148

149

150

Appendix B

151

152

153

154