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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
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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 : / /
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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 .
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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......
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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).
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Fig1.8:Turkish hotel
Fig 1.9 : AL - Sader hospital
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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
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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
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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)
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[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
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Table ( 2 – 1 ) Typical Uniformly Distributed Design live Loads
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Table ( 2 – 2 ) Density of Various Materials
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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
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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.
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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)
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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)
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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
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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
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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)
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2.3.2.3 Find GCpi
GCpi Find from Table (2-8)
Table (2-8)
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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
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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
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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
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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
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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)
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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
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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)
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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)
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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)
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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