DESIGN AND ANALYSIS OF G+3 RESIDENTIAL BUILDING BY S.MAHAMMAD FROM RAJIV GANDHI UNIVERSITY OF KNOWLEDGE TECHNOLOGIES

ANALYSIS AND DESIGN OF G+3
RESIDENTIAL BUILDING
A project report is submitted in partial fulfillment of the requirement for the
Award of the degree of
BACHELOR OF TECHNOLOGY
in
CIVIL ENGINEERING
Submitted by
SHAIK MAHAMMAD (R092251)
Under the esteemed guidance of
A.RAM BHUPAL REDDY
Lecturer in
Department of Civil Engineering
DEPARTMENT OF CIVIL ENGINEERING
RAJIV GANDHI UNIVERSITY OF KNOWLEDGE TECHNOLOGIES
APIIIT,R.K.VALLEY-516329(A.P.)
May 2015

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RAJIV GANDHI UNIVERSITY OF KNOWLEDGE
TECHNOLOGIES
DEPARTMENT OF CIVIL ENGINEERING
CERTIFICATE
This is to certify that the Project report entitled ANALYSIS AND DESIGN OF
G+3 RESIDENTIAL BUILDING ,submitted by SHAIK MAHAMMAD (R092251) to
Rajiv Gandhi University of Knowledge Technologies (RGUKT) is a record of
bona fide project work during the academic year 2014-15.
PROJECTGUIDE HEAD OF THE DEPARTMENT
A.RAM BHUPAL REDDY A.RAM BHUPAL REDDY
Lecturer in Civil Engineering Dept. Lecturer in Civil Engineering Dept.
Date:
Place:RGUKT-RKValley.

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APPROVAL SHEET
This B.Tech project entitled “ANALYSIS AND DESIGN OF G+3
RESIDENTIAL BUILDING ” done and prepared by
SHAIK MAHAMMAD (R092251) is hereby approved for submission at Civil
Engineering Department, Rajiv Gandhi University of Knowledge Technologies,
R.K Valley.
…………………………………………
(Signature ofthe guide)
A.RAM BHUPALREDDY
………………………………………….
Headof the Department
Date:
Place: R.K Valley.

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DECLARATION SHEET
I declare that this written submission represents my ideas in my own words and
where other ideas or words have been included; I have adequately cited and
referenced the original sources. I have conformed to the norms and guidelines
given in the Ethical Code of Conduct of the Institute. Whenever I have used
materials from other sources, I have given due credit to them by citing them in the
text of the report and giving their details in the references. Further, I have taken
permission from the copyright owners of the sources, whenever necessary.
…………………………….
SHAIK MAHAMMAD (R092251)
(Signature of the Student)
Date:
Place: RGUKT-RKValley.

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ACKNOWLEDGEMENT
We are extremely thankful to A.RAM BHUPAL REDDY, lecturer and Head of the
department of civil engineering for his valuable guidance and constant co-operation throughout
this work.
We are very much thankful to department of Civil Engineering for the Successful
Completion of my project.
It is my glowing feeling to place on record my best regards, deepest sense of gratitude to
all the professors for their encouragement and cooperation in carrying out the project.
Finally, yet importantly, I would like to express my heart full thanks to my beloved
parents for their blessings, my friends/classmates for their help and wishes for successful
completion of this project.
DATE: Signature of the Student

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ABSTRACT
Structural planning and design is an art and science of designing with economy and
elegance, serviceable and durable structure. The entire process of structural planning and
designing requires not only imagination and conceptual thinking but also sound knowledge of
science of structural engineering besides knowledge of practical aspects, such as relevant design
codes and byelaws backed up by example experience.
The process of design commence with planning of structural primarily to meet the
defined as he is not aware of various implications involved in the process of planning and design.
The functional requirements and aspects of aesthetics are locked into normally be the architect
while the aspect of the safety, serviceability, durability and economy of the structure are attended
by structural designer.
For this purpose a site is selected in which the building has three floors including a
ground floor, it consists of all the rooms required for a residential house like bedroom , toilet,
living/dining, kitchen and store room.
Staad Pro is a software tool to design structural design of any plan and also it can give
loads of that structure. We can mention about which material we are going to use and what is the
strength of that member, it all comes under this software.
AutoCAD is a software tool to design functional design of any plan. It involves outer
appearance of the plan.
In this project work, an attempt is made according to Building Bye laws and design of
residential building as per IS: 456-2000, SP-16 and SP-34 specifications.

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INDEX
CERTIFICATE…………………………………………………………………………II
APPROVAL SHEET…………………………………………………………………..III
DECLARATION SHEET……………………………………………………………..IV
ACKNOWLEDGEMENT……………………………………………………………...V
ABSTRACT ……………………………………………………………………………VI
NOMENCLATURE…………………………………………………………………...IX
LIST OF TABLES AND FIGURES …………………………………………………XI
CHAPTER 1 INTRODUCTION……………………………………………………...1
1.1 Introduction ………………………………………………………………………….1
1.2 General Theory ………………………………………………………………………2
1.3 Statement of the project ……………………………………………………………...5
CHAPTER 2 LITERATURE REVIEW ……………………………………………..6
2.1 Method of flexibility coefficients …………………………………………………....6
2.2 Slope displacement method…………………………………………………………..6
2.3 Kane’s method ……………………………………………………………………….7
2.4 Approximate method ………………………………………………………………...7
2.5 Matrix analysis of frames …………………………………………………………....8
2.6 Design of multi storied residential building ………………………………………....8
2.7 Limit state method ………………………………………………………………...…8
CHAPTER 3DESIGN OF SLABS…………………………………………………...10
3.1 One way slab ………………………………………………………………………..10
3.2 Two way slab …………………………………………………………………….....10
CHAPTER 4 DESIGN OF BEAMS ………………………………………….…...…58
4.1 Theory………………………………………………………………….…………….58
4.2 Singly reinforced beams …………………………………………….…………....…58
4.3 Doubly reinforced beams ………………………………………………………...…59

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4.4 Load Calculation …………………………………………………………………….59
CHAPTER 5 STAIRCASE DESIGN …………………………………………………69
5.1 Theory……………………………………………………………………………..…69
5.2 Column Reactions……………………………………………………………………70
5.3 Column Grouping……………………………………………………………..……..70
5.4 Design Calculations……………………………………………………….…………71
CHAPTER 6 DESIGN OF FOOTING………………………………….……………..73
6.1 Footing Reactions………………………………………………………..……….…..73
6.2 Design and Calculations……………………………………………...……………….74
CHAPTER 7STAIR CASE DESIGN …………………………….…………………..82
CONCLUSION ……………………………………………………………………….…85
REFERENCES……………………………………………………………………….…86

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NOMENCLATURE
A : Area
b : Breadth of the beam, or short dimension of the member
b : Effective width of the slab
bf : Effective width of the flange
bw : Breadth of web or rib
D : Overall depth of beam or slab or diameter of column
Df : Thickness of the flange
DL : Dead Load
d : Effective depth of beam or slab
d : Depth of compression reinforcement from the highly compressed face
EC : Modulus of elasticity of concrete
EL : Earthquake Load
ES : Modulus of elasticity of steel
e : Eccentricity
fck : Characteristic cube compressive strength of concrete
fy : Characteristic strength of steel
fα : Modulus of rupture of concrete (Flexural tensile strength)
fct : Splitting tensile strength of concrete
fd : Design strength
Hw : Unsupported height of wall
Hwe : Effective height of wall
Ief : Effective Moment of Inertia
Igr : Moment of Inertia of the gross section excluding reinforcement
Ir : Moment of Inertia of cracked section
K : Stiffness of member
k : Constant (or) Coefficient of factor
Ld : Development length
LL : Live load (or) imposed load
Lw : Horizontal distance between centers of lateral restrains
l : Length of column (or) beam between adequate lateral restrain
lef : Effective span of beam or slab or effective length of column
lex : Effective length about x-x axis

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ley : Effective length about y-y axis
ln : Clear span, face to face of supports
ln, : ln for shorter of the two spans at right angles
lx : Length of shorter side of slab
ly : Length of longer side of slab
lo : Distance between points of zero moments in a beam
ll : Span in the direction in which moments are determined
l2 : Span transfer to l1 , center to center of supports
l2, : l2 for the shorter of the continuous spans
M : Bending Moment
m : Modular ratio
n : Number of samples
P : Axial load on a compression member
qau : Calculated maximum bearing pressure of soil
r : Radius
s : Spacing of stirrups (or) standard deviation
T : Torsional Moment
t : Wall thickness
V : Shear Force
W : Total load
WL : Wind load
w : Distributed load per unit area
wd : Distributed dead load per unit area
wl : Distributed live (imposed) load per unit area
x : Depth of neutral axis
Z : Modulus of section
z : Lever arm
α, β : Angle (or) ratio
ϒr : Partial safety factor for load
ϒm : Partial safety factor for material
τc : Shear stress in concrete
τc max : Maximum shear stress in concrete with shear reinforcement
τv : Nominal shear stress
ɸ : Diameter of bar

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LIST OF FIGURES and TABLES
List of Figures:
Fig 1.1 Plan of residential building
Fig 4.1 Loading diagram for floor
Fig 4.2 Bending Moment Distribution
List of Tables:
Table 1.1 Minimum grade of concrete for different exposure conditions
Table 1.2 Unit weights of common building materials
Table 1.3 Live load onto floor
Table 3.1 Bending Moments Coefficients
Table 3.2 Bending Moments Coefficients
Table 5. 1 Column Reactions
Table 5.2 Column Grouping
Table 6. 1 Footing Reactions
In this present project we are following the Bylaws and design of residential buildings asper the IS
456:2000, SP16,SP34, IS 875part2.

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CHAPTER 1
INTRODUCTION
1.1 INTRODUCTION
My project involves analysis and design of multi-storied [G + 3] using a very popular
designing software STAAD.Pro v8i. I have chosen STAAD Pro because of its following
Advantages:
Easy to use interface,
Conformation with the Indian Standard Codes,
Versatile nature of solving any type of problem,
Accuracy of the solution.
STAAD.Pro v8i features a state-of-the-art user interface, visualization tools, powerful analysis
and design engines with advanced finite element and dynamic analysis capabilities. From model
generation, analysis and design to visualization and result verification, STAAD.Pro v8i is the
professional’s choice for steel, concrete, timber, aluminum and cold-formed steel design of low
and high-rise buildings, culverts, petrochemical plants, tunnels, bridges, piles and much more.
STAAD.Pro v8i consists of the following:
The STAAD.Pro v8i Graphical User Interface: It is used to generate the model, which can then
be analyzed using the STAAD engine. After analysis and design is completed, the GUI can also
be used to view the results graphically.
The STAAD analysis and design engine: It is a general-purpose calculation engine for structural
analysis and integrated Steel, Concrete, Timber and Aluminum design.
To start with I have solved some sample problems using STAAD Pro and checked the accuracy
of the results with manual calculations. The results were to satisfaction and were accurate. In the
initial phase of my project I have done calculations regarding loadings on buildings and also
considered seismic and wind loads.Structural analysis comprises the set of physical laws and
mathematics required to study and predicts the behavior of structures. Structural analysis can be
viewed more abstractly as a method to drive the engineering design process or prove the
soundness of a design without a dependence on directly testing it.

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To perform an accurate analysis a structural engineer must determine such information as
structural loads, geometry, support conditions, and materials properties. The results of such an
analysis typically include support reactions, stresses and displacements. This information is then
compared to criteria that indicate the conditions of failure. Advanced structural analysis may
examine dynamic response, stability and non-linear behavior. The aim of design is the
achievement of an acceptable probability that structures being designed will perform
satisfactorily during their intended life. With an appropriate degree of safety, they should sustain
all the loads and deformations of normal construction and use and have adequate durability and
adequate resistance to the effects of seismic and wind. Structure and structural elements shall
normally be designed by Limit State Method. Account should be taken of accepted theories,
experiment and experience and the need to design for durability. Design, including design for
durability, construction and use in service should be considered as a whole. The realization of
design objectives requires compliance with clearly defined standards for materials, production,
workmanship and also maintenance and use of structure in service.
The design of the building is dependent upon the minimum requirements as prescribed in the
Indian Standard Codes. The minimum requirements pertaining to the structural safety of
buildings are being covered by way of laying down minimum design loads which have to be
assumed for dead loads, imposed loads, and other external loads, the structure would be required
to bear. Strict conformity to loading standards recommended in this code, it is hoped, will not
only ensure the structural safety of the buildings which are being designed.
1.2 GeneralTheory
Table1.1Minimum Grade of Concrete for different exposure conditions:
Exposure Minimum Grade of Concrete for RCC
Mild M20
Moderate M25
Severe M30
Very Severe M35
Extreme M40
Tensile Strength:
The tensile strength of concrete is very low and hence it is not taken in to account in the
design of reinforced concrete. But it is an important property which affects the extent and width
of cracks in the structure. According to IS 456-2000, the tensile strength of concrete can be
calculated from the compressive strength using the following relation
𝑓𝑐𝑟 = 0.7√𝑓𝑐𝑘 𝑁/𝑚𝑚2

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Wherefck is the characteristic cube compressive strength of concrete
Modulus of Elasticity:
Modulus of elasticity of concrete is an important property required for computation of
deflections of structural concrete members. In the absence of test data the modulus of elasticity
fck concrete is related to compressive strength by the following relation as per IS 456-2000
𝐸𝑐 = 5000√𝑓𝑐𝑘 𝑁/𝑚𝑚2
Where Ec is the short term static modules of elasticity in N/mm2
Unit weight:
The unit weight of concrete depends up on the type of aggregates and amount of voids.
The unit weight as specified by the IS 456-2000 for plain concrete and reinforced concrete are 24
KN/m3 and 25 KN/m3 respectively
Table 1.2: Unit weight of common Building Materials(From IS 875 part 1)
S.No Material Unit weight KN/m3
1 Plain concrete 24
2 Reinforced concrete 25
3 Brick masonry 20
4 Stone masonry 24
5 Wood 8
6 Steel 78.5
7 Floor finish 0.6-1.2
Table 1.3: Live loads on Floors(From IS 875 part 2)
S.No Type of Floors Minimum Live Load KN/m2
1 Floors in dwelling houses, tenements, hospital wards,
hostels and dormitories
2.0
2 Office floors other than entrance halls, floors of light 2.5-4.0 (2.5,when separate storage work
rooms facility is provided , other wise
4.0)
3 Floors of banking halls, Office entrance halls and
reading rooms
3.0
4 Shops, educational buildings, assembly buildings,
restaurants
4.0
5
Office floors for storage, assembly floor space without
fixed seating, public rooms in hotels, dance halls and
waiting halls
5.0
6 Ware houses. Workshops and factories
(a)light weight loads
(b)Medium weight loads
(c)Heavy weight loads
5.0
7.5
10.0

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7 Stairs, landing , Balconies and corridors for floors
mentioned in 1, but not liable to over crowding
stairs, landings and corridors for floors mentioned in
1, but liable to overcrowding and for all other floors
3.0
5.0
8 Flat slabs, sloped roofs
(a)Access provided
(b)Access not provided
1.5
0.75
PLAN OF RESIDENTIAL BUILDING:
Fig: 1.1 Plan of residential building

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1.3 STATEMENT OF THE PROJECT:
Utility of building: residential complex
No of stories: G+3
No.of staircase:1
Type of construction: R.C.C framed structure
Types of walls: brick wall
Ground floor: 3m
Floor to floor height: 3m.
Walls: 230 mm thick brick masonry walls for external and 115 mm internal wall
Materials:
Concrete grade: M20
All steel grades: HYSD bars of Fe415 grade
Bearing capacity of soil: 350 KN/M2

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CHAPTER 2
LITERATURE REVIEW
Method of analysis of statistically indeterminate portal frames:
1. Method of flexibility coefficients.
2. Slope displacement method (iterative method)
3. Kane’s method
4. Approximate method
5. Matrix method
2.1 Method of flexibility coefficients
The method of analysis is comprises reducing the hyper static structure to a determinate structure
form by removing the redundant support (or) introducing adequate cuts (or) hinges.
Limitations: It is not applicable for degree of redundancy>3
2.2 Slope displacement method
It is advantageous when kinematic indeterminacy <static indeterminacy. This procedure was first
formulated by axle bender in 1914 based on the applications of compatibility and equilibrium
conditions. The method derives its name from the fact that support slopes and displacements are
explicitly comported. Set up simultaneous equations is formed the solution of these parameters
and the joint moment in each element or computed from these values.
Limitations:
A solution of simultaneous equations makes methods tedious for manual computations. This
method is not recommended for frames larger than too bays and two storeys. This method
involves distributing the known fixed and moments of the structural member to adjacent
members at the joints in order satisfy the conditions of compatibility.
Limitations of hardy cross method:

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It presents some difficulties when applied to rigid frame especially when the frame is susceptible
to side sway. The method cannot be applied to structures with intermediate hinges.
2.3 Kane’s method
This method over comes some of the disadvantages of hardy cross method. Kane’s approach is
similar to H.C.M to that extent it also involves repeated distribution of moments at successive
joints in frames and continues beams. However there is a major difference in distribution process
of two methods. H.C.M distributes only the total joint moment at any stage of iteration. The most
significant feature of Kane’s method is that process of iteration is self-corrective. Any error at
any stage of iterations corrected in subsequent steps consequently skipping a few steps error at
any stage of iteration is corrected in subsequent consequently skipping a few steps of iterations
either by over sight of by intention does not lead to error in final end moments.
Advantages:
It is used for side way of frames.
Limitations:
The rotational of columns of any storey should be functioning single rotation value of same
storey. The beams of storey should not undergo rotation when the column undergoes translation.
That is the column should be parallel. Frames with intermediate hinges cannot be analysis.
2.4 Approximate method
Approximate analysis of hyper static structure provides a simple means of obtaining a quick
solution for preliminary design. It makes some simplifying assumptions regarding Structural
behavior so to obtain a rapid solution to complex structures. The usual process comprises
reducing the given indeterminate configuration to determine structural system by introducing
adequate no of hinges. It is possible to sketch the deflected profile of the structure for the given
loading and hence by locate the print inflection. Since each point of inflection corresponds to
the location of zero moment in the structures. The inflection points can be visualized as hinges
for the purpose of analysis. The solution of structures is sundered simple once the inflection
points are located. The loading cases are arising in multistoried frames namely horizontal and
vertical loading. The analysis carried out separately for these two cases.
Horizontal cases:

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The behavior of a structure subjected to horizontal forces depends upon its heights to width ratio
among their factor. It is necessary to differentiate between low rise and high rise frames in this
case.
Low rise structures: Height< width
It is characterized predominately by shear deformation. High-rise buildings
Height > width
It is dominated by bending action
2.5 Matrix analysis of frames
The individual elements of frames are oriented in different directions unlike those of continues
beams so their analysis is more complex .never the less the rudimentary flexibility and stiffness
methods are applied to frames stiffness method is more useful because its adaptability to
computer programming stiffness method is used when degree of redundancy is greater than
degree of freedom. However stiffness method is used degree of freedom is greater than degree of
redundancy especially for computers.
2.6 Design of multi storied residential building
2.6.1 General:
A structure can be defined as a body which can resist the applied loads without appreciable
deformations. Civil engineering structures are created to serve some specific functions like
human habitation, transportation, bridges, storage etc. in a safe and economical way. A structure
is an assemblage of individual elements like pinned elements (truss elements), beam element,
column, shear wall slab cable or arch. Structural engineering is concerned with the planning,
designing and the construction of structures. Structure analysis involves the determination of the
forces and displacements of the structures or components of a structure. Design process involves
the selection and detailing of the components that make up the structural system. The main
object of reinforced concrete design is to achieve a structure that will result in a safe economical
solution. The Design of each part may be designed separately as follows
1. Beam design
2. Column design
3. Slab design
4. Foundation design
These all are designed under limit state method

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2.7 Limit state method
The object of design based on the limit state concept is to achieve an acceptability that a
structure will not become unserviceable in its life time for the use for which it is intended. I.e. it
will not reach a limit state. In this limit state method all relevant states must be considered in
design to ensure a degree of safety and serviceability.
Limit state: The acceptable limit for the safety and serviceability requirements before failure
occurs is called a limit state.
Limit state of collapse:
This is corresponds to the maximum load carrying capacity. Violation of collapse limit state
implies failures in the source that a clearly defined limit state of structural usefulness has been
exceeded. However it does not mean complete collapse.
This limit state corresponds to:
a) Flexural
b) Compression
c) Shear
d) Torsion
Limit State of Serviceability.

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CHAPTER 3
DESIGN OF SLABS
Slabs are plane structural members whose thickness is small as compared to its length
and breadth. Slabs are most frequently used as roof coverings and floors in various shapes such
as square, rectangular, circular, triangular etc, in building. Slabs supports mainly transverse loads
and transfers them to the supports by bending action in one or more directions. Beams or walls
are the common supports for the slabs.
Types of Slabs:
Depending up on the ratio of longer span to short span(ly/lx) the slabs are classified in to:
a. One way slab
b. Two way slab
3.1 One way slab
Slabs which are supported on all four edges and the ration of longer span to the shorter
span (ly/lx) are greater than 2 are called as one way slabs. One way slabs bends in one direction
.i.e. along the shorter span and hence span and hence it needs main reinforcement in one
direction only (along the shorter span) to resist one way bending However minimum
reinforcement known as distribution steel is provided along the longer span above the main
reinforcement to distribute the load uniformly and to resist temperature and shrinkage stresses.
3.2 Two way slab
When the slabs are supported on all the four edges and the ratio of longer span to the
shorter span (ly/lx) is less than or equal to 2, the slabs are likely to bend along the two spans and
such slabs are called as two way slabs. The load is transferred in both the direction to the four
supporting edges and hence main reinforcement has to be designed in both directions to resist
two way bending.
General Design Requirements for slabs as per IS 456:2000

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A. Effective Span: The effective span of a simply supported slab shall be taken as clear pan
plus effective depth of the slab or center to center distance between the supports whichever is
less.
The effective span of a cantilever slab shall be taken as its length to the face of the
support plus half the effective depth except where it forms the end of a continuous slab where the
length to the center of support shall be taken.
B. Limiting Stiffness: The stiffness of slabs is governed by the span to depth ratio. As per
Clause 23.2 of IS 456 for spans not exceeding 10m, the span to depth ration (Basic values)
should not exceed the limits given below
Cantilever – 7
Simply supported – 20
Continuous – 26
Depending upon the type of steel and percentage of steel, the above values have to be modified
as per Fig .4 of IS-456
For two way slabs, the shorter span be used for calculating the span to effective depth ration
C. Minimum Reinforcement: The reinforcement in either direction of span shall not be
less than 0.15% of gross cross sectional area if mild steel is use. However, this value is reduced
to 0.12% where high strength deformed bars or welded wire fabrics are used. (Clause 26.5.2.2 of
IS -456)
D. Maximum Reinforcement: the diameter of the bars shall not exceed one eighth of the
total thickness of slab (clause 26.5.2.2 of IS-456)
E. Spacing of Main Reinforcement: The spacing of main reinforcement in slabs shall not
be more than three times the effective depth of solid slab or 300 mm whichever is less (clause
26.3.3 of IS-456)

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Slab: A1-A2-B1-B2
Continuous Two way Slab (5.29X3.96)
Thickness of slab = 170 mm
Live load = 2KN/m2
Floor finish = 1 KN/m2
Self weight of slab = 25×0.170 = 4.25 KN/m2
Total load = 2+1+4.25= 7.25 KN/m
Factored load = 7.25×1.5= 10.875 KN/m
Dimension of the slab lx = 4.11m, ly = 5.44m
Ratio of longer span to shorter span (ly/lx) =5.44/4.11 = 1.32
Two adjacent edges are discontinuous:
Bending moment coefficients (Clauses D-1.1 and 24.41)
Table:3.1 Bending moment coefficients
Type of panel and Moments
Considered
Short Span Coefficients:
(αx)
Long Span Coefficients:
(αy )
Positive Moment at mid span
0.050 0.035
Negative Moment at continuous
edge
0.066 0.047
Moments along short span Mx and long span (My) are given by
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.050×10.875×4.112=9.185×106Nmm
My =αy Wly
2= 0.035×10.875×5.442= 11.26×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.066×10.875×4.112=12.12×106Nmm
My = αy Wly
2= 0.047×10.875×5.442=15.12×106Nmm

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Minimum Depth:
Mu = 0.138 fck b d2
15.12 x106 = 0.138x20x1000xd2
d=74 mm< 150 mm, Hence safe.
D=Overall depth=150+20=170 mm.
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
𝑓𝑦 𝐴 𝑠𝑡
𝑓𝑐𝑘 𝑏𝑑
)}
9.18×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.221Ast
2-54157.5Ast+9.18×106=0
Ast required = 173.61 mm2
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (78.5/173.61)×1000=452.16mm
2) 3d=3x150=450 mm
3)300mm
provide c/c spacing 300mm and 10 mm bars
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
12.124×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.221Ast
2-54157.5Ast+12.124×106=0
Ast required = 230 mm2
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (78.5/230)×1000=341 mm
2) 3d=3x150=450 mm
3)300mm

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Ast provided = (78.5/300) ×1000 = 261.62mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
11.26×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.221Ast
2-54157.5Ast+11.26×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
15.12×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.221Ast
2-54157.5Ast+15.126×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (113.04/290.55)×1000=389mm
2) 3d=3x150=450 mm
3)300mm
Ast provided = (113.04/300) ×1000 = 376.8 mm2

Page | 15
Torsion reinforcement:
Area of tension steel at each of the corner is = 75% of steel of max reinforcement
= 0.75 X 290.55= 217.91 mm2
Length of torsion steel = 1/5 X short span
= 1/5 X 3960 = 792mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing =(28.26/217.91)x1000=129.6 mm
Provide 6mm dia bars @ 120mm
Reinforcement in edge strips:
Ast = 0.12 X bD = 0.12 X 1000X 170 = 204 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/204)x1000=246.27 mm
2)5d= 5x150=750 mm
3)450 mm
Provide 8mm dia @ 230 c/c
Deflection check
𝑓𝑠 = 0.58 × 415 ×
𝐴 𝑠𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
𝐴 𝑠𝑡 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
= 0.58×415× (290.55/376.8)=185.60
% steel = (100×376.8)/ (150×1000 =0.25
Basic values of span to effective depth ratios for spans up to 10 m;
Continuous = 26

Page | 16
From the graph Modification factor = 1.9 (Note: Based on values of fs and %Steel)
Effective length to depth ratio = 1.9×26 = 49.4
Theoretical value of length to depth ratio = lx /d = 4.11×103/150 = 27.4< 49.4
Hence it is safe.
Slab: A1-A2-B1-B2(TERRACE)
Live load = 1.5KN/m2
Self weight of slab = 25×0.170 = 4.25 KN/m2
Total load = 1.5+1+4.25= 6.75 KN/m

Page | 17
Considered
(αx)
(αy )
0.050 0.035
edge
0.066 0.047
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.050×10.125×4.112=8.51×106Nmm
My =αy Wly
2= 0.035×10.125×5.442= 10.48×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.066×10.125×4.112=11.28×106Nmm
My = αy Wly
2= 0.047×10.125×5.442=14.08×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
14.08 x106 = 0.138x20x1000xd2
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
8.51×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.491Ast
2-54157.5Ast+8.51×106=0
use 10 mm dia bars.

Page | 18
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
11.28×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+11.28×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (78.5/214.81)×1000=365 mm
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.48×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+10.48×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars

Page | 19
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
14.08×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+14.08×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
providec/c spacing 300mm and 12 mm bars
Ast provided = (113.04/300) ×1000 = 376.8 mm2
= 0.75 X 270.06= 202.545 mm2
= 1/5 X 3960 = 792mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing =(28.26/202.54)x1000=139.52 mm
Ast = 0.12 X bD = 0.12 X 1000X 170 = 204 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/204)x1000=246.27 mm
2)5d= 5x150=750 mm
3)450 mm

Page | 20
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (270.06/376.8)=172.51
% steel = (100×376.8)/ (150×1000) =0.25
Continuous = 26
From the graph Modification factor = 2 (Note: Based on values of fs and %Steel)
Effective length to depth ratio = 2×26 = 52
Theoretical value of length to depth ratio = lx /d = 4.11×103/150 = 27.4< 52
Hence it is safe.
Slab: A2-A3-B2-B3
Live load = 2KN/m2
Self weight of slab = 25×0.160 = 4KN/m2
Total load = 2+1+4= 7 KN/m
Factored load = 7×1.5
= 10.5 KN/m
Dimension of the slab lx = 4.94m,ly = 5.43m
Considered
(αx)
(αy )
0.040 0.035

Page | 21
edge
0.063 0.047
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.040×10.5×4.94.2=10.24×106Nmm
My =αy Wly
2= 0.035×10.5×5.432= 10.84×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.063×10.5×4.942=16.14×106Nmm
My = αy Wly
2= 0.047×10.5×5.432=14.55×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
14.55x106 = 0.138x20x1000xd2
d=72.60 mm< 140 mm, Hence safe.
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.24×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.491Ast
2-50547Ast+10.24×106=0
Ast required = 209 mm2
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (78.5/209)×1000=375 mm
2) 3d=3x140=420 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2

Page | 22
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
16.14×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.49Ast
2-50547Ast+16.14×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (113.04/336.03)×1000=336.39 mm
2) 3d=3x140=420 mm
3)300mm
Ast provided = (113.04/300) ×1000 = 376.8 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.83×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.49Ast
2-50547Ast+10.83×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x140=420 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}

Page | 23
14.55×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.49Ast
2-50547Ast+14.55×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars
2) 3d=3x140=420 mm
3)300mm
Ast provided = (113.04/300) ×1000 = 376.8 mm2
= 0.75 X 336.03
= 250.02 mm2
= 1/5 X 4940 = 988 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing =(28.26/250.02)x1000=113.03 mm
Ast = 0.12 X bD = 0.12 X 1000X 160 = 192 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/192)x1000=261.6 mm
2)5d= 5x140=700 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (336.03/376.8)=214.6
% steel = (100×376.8)/ (140×1000)=0.269

Page | 24
Continuous = 26
Hence it is safe.
Slab: A2-A3-B2-B3(TERRACE)
Live load = 1.5 KN/m2
Total load = 1.5+1+4= 6.5 KN/m
Dimension of the slab lx = 4.94m ,ly = 5.43m
Considered
(αx)
(αy )
0.040 0.035
edge
0.063 0.047
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.040×9.75×4.94.2=9.51×106Nmm
My =αy Wly
2= 0.035×9.75×5.432= 10.06×106Nmm
Negative moment:

Page | 25
Mx = αx Wlx
2= 0.063×9.75×4.942=14.98×106Nmm
My = αy Wly
2= 0.047×9.75×5.432=13.51×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
14.98x106 = 0.138x20x1000xd2
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
9.51×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.491Ast
2-50547Ast+9.51×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x140=420 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
14.98×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.49Ast
2-50547Ast+14.98×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars

Page | 26
2) 3d=3x140=420 mm
3)300mm
Ast provided = (113.04/300) ×1000 = 376.8 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.06×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.49Ast
2-50547Ast+10.06×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x140=420 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
13.51×106 = 0.87×415 × 140 × Ast{1-(415Ast/20×1000×140)}
7.49Ast
2-50547Ast+13.51 ×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars

Page | 27
2) 3d=3x140=420 mm
3)300mm
Ast provided = (113.04/300) ×1000 = 376.8 mm2
= 0.75 X 310.65= 232.98 mm2
= 1/5 X 4940 = 988 mm
Use 8 mm bars
ast=πd2/4= 50.24 mm2
spacing =(50.24/232.98)x1000=215 mm
Ast = 0.12 X bD = 0.12 X 1000X 160 = 192 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/192)x1000=261.6 mm
2)5d= 5x140=700 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (310.65/376.8)=198.44
% steel = (100×376.8)/ (140×1000) =0.269
Continuous = 26
Hence it is safe.
Slab: B1-B2-C1-C2

Page | 28
Live load = 2KN/m2
Self weight of slab = 25×0.170 = 4.25KN/m2
Total load = 2+1+4= 7.25 KN/m
One long edge discontinuous:
Considered
(αx)
(αy )
0.0415 0.028
edge
0.0545 0.037
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.0415×10.875×4.112=8.26×106Nmm
My =αy Wly
2= 0.028×10.875×5.142= 8.044×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.0545×10.875×4.112=10.01×106Nmm
My = αy Wly
2= 0.037×10.875×5.142=10.63×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
10.63x106 = 0.138x20x1000xd2
Positive Mx :

Page | 29
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
8.26×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.491Ast
2-54157.5Ast+8.26×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.01×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+10.01×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
8.044×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}

Page | 30
7.49Ast
2-54157.5Ast+8.044×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.46mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.630×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+10.630×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.6 mm2
= 0.75 X 201.91
= 151.43 mm2
= 1/5 X 4111 = 822 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing = (28.26/151.43)x1000=186.62 mm

Page | 31
Ast = 0.12 X bD = 0.12 X 1000X 170 = 204 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/204)x1000=246.27 mm
2)5d= 5x150=750 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (201.91/261.6)=185.77
% steel = (100×261.62)/ (150×1000) =0.17
Continuous = 26
Hence it is safe.
Slab: B1-B2-C1-C2(TERRACE)
Self weight of slab = 25×0.170 = 4.25KN/m2
Total load = 1.5+1+4= 6.75 KN/m

Page | 32
Considered
(αx)
(αy )
0.0415 0.028
edge
0.0545 0.037
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.0415×10.125×4.112=7.08×106Nmm
My =αy Wly
2= 0.028×10.125×5.142= 7.48×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.0545×10.125×4.112=9.32×106Nmm
My = αy Wly
2= 0.037×10.125×5.142=9.89×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
9.89 x106 = 0.138x20x1000xd2
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
7.08×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.491Ast
2-54157.5Ast+7.08×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars

Page | 33
2) 3d=3x150=450 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.46mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
9.32×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+9.32×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
7.48×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+7.48×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm

Page | 34
Ast provided = (50.24/300) ×1000 = 167.46mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
9.89×106 = 0.87×415 × 150 × Ast{1-(415Ast/20×1000×150)}
7.49Ast
2-54157.5Ast+9.89×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.6 mm2
= 0.75 X 187.4= 140.55 mm2
= 1/5 X 4111 = 822 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing = (28.26/140.05)x1000=201.06 mm
Ast = 0.12 X bD = 0.12 X 1000X 170 = 204 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/204)x1000=246.27 mm
2)5d= 5x150=750 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×

Page | 35
= 0.58×415× (187.4/261.6)=172.44
% steel = (100×261.62)/ (150×1000) =0.17
Continuous = 26
Hence it is safe.
Slab: B2-B3-C2-C3
Live load = 2KN/m2
Factored load = 8×1.5= 12 KN/m
Considered
(αx)
(αy )
0.0295 0.028
edge
0.039 0.037
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:

Page | 36
Mx = αx Wlx
2= 0.0295×12×4.982=8.77×106Nmm
My =αy Wly
2= 0.028×12×5.172= 8.98×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.039×12×4.982=11.60×106Nmm
My = αy Wly
2= 0.037×12×5.172=11.86×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
11.86x106 = 0.138x20x1000xd2
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
8.77×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.491Ast
2-64989Ast+8.77×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x180=540 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.4mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
11.60×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.49Ast
2-64989Ast+11.60×106=0
use 10 mm dia bars.

Page | 37
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x180=540 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
8.98×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.49Ast
2-64989Ast+8.98×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x180=540 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.46mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
11.86×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.49Ast
2-64989Ast+11.86×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars

Page | 38
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.6 mm2
= 0.75 X 186.5= 139.85 mm2
= 1/5 X 4980 = 996 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing = (28.26/139.86)x1000=202 mm
Ast = 0.12 X bD = 0.12 X 1000X 200 = 240 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/240)x1000=209 mm
2)5d= 5x180=900 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (186.5/261.6)=171.5
% steel = (100×261.62)/ (180×1000)=0.14
Continuous = 26
Hence it is safe.
Slab: B2-B3-C2-C3(Terrace)

Page | 39
Dimension of the slab lx = 4.98 m ,ly = 5.17m
Considered
(αx)
(αy )
0.0295 0.028
edge
0.039 0.037
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.0295×11.25×4.982=8.23×106Nmm
My =αy Wly
2= 0.028×11.25×5.172= 8.41×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.039×11.25×4.982=10.88×106Nmm
My = αy Wly
2= 0.037×11.25×5.172=11.12×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
11.12 x106 = 0.138x20x1000xd2

Page | 40
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
8.23×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.491Ast
2-64989Ast+8.23×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x180=540 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.4mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.88×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.49Ast
2-64989Ast+10.88×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x180=540 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}

Page | 41
8.41×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.49Ast
2-64989Ast+8.41×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x180=540 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.46mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
11.12×106 = 0.87×415 × 180 × Ast{1-(415Ast/20×1000×180)}
7.49Ast
2-64989Ast+11.12×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x150=450 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.6 mm2
= 0.75 X 174.62= 130.965 mm2
= 1/5 X 4980 = 996 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing = (28.26/130.965)x1000=215 mm

Page | 42
Ast = 0.12 X bD = 0.12 X 1000X 200 = 240 mm2
Use 8 mm bars
ast=πd2/4= 50.24 mm2
1)spacing =(50.24/240)x1000=209 mm
2)5d= 5x180=900 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (130.69/261.6)=120.48
% steel = (100×261.62)/ (180×1000)=0.14
Continuous = 26
Hence it is safe.
Slab: C1-C2-D1-D2
Live load = 2KN/m2
Dimension of the slab lx = 3.45 m ,ly = 4.06 m
Two edges are discontinuous:

Page | 43
Considered
(αx)
(αy )
0.043 0.035
edge
0.057 0.047
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.043×9×3.452=4.60×106Nmm
My =αy Wly
2= 0.035×9×4.062= 5.19×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.057×9×3.452=6.10×106Nmm
My = αy Wly
2= 0.047×9×4.062=6.97×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
6.97 x106 = 0.138x20x1000xd2
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
4.60×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.491Ast
2-36105Ast+4.60×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars

Page | 44
2) 3d=3x100=300 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.4mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
6.10×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+6.10×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000 = (78.5/175.32)×1000=447.75 mm
2) 3d=3x100=300 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
5.19×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+5.19×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm

Page | 45
Ast provided = (50.24/300) ×1000 = 167.46mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
6.97×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+6.97×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (78.5/201.46)×1000= 389.65mm
2) 3d=3x100=300 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.6 mm2
= 0.75 X 201.46= 151.095 mm2
= 1/5 X 3450 = 690 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing = (28.26/151.09)x1000=187.03 mm
Ast = 0.12 X bD = 0.12 X 1000X 120 = 144 mm2
Use 6 mm bars
ast=πd2/4= 28.26 mm2
1)spacing =(28.26/144)x1000=196.25 mm
2)5d= 5x100=500 mm
3)450 mm
Provide 6mm dia@ 190mmc/c
Deflection check
𝑓𝑠 = 0.58 × 415 ×

Page | 46
= 0.58×415× (201.46/261.6)=185.36
% steel = (100×261.62)/ (100×1000) =0.261
Continuous = 32(lx<3.5)
Hence it is safe.
Slab: C1-C2-D1-D2(Terrace)
Live load = 1.5 KN/m2
Considered
(αx)
(αy )
0.043 0.035
edge
0.057 0.047
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:

Page | 47
Mx = αx Wlx
2= 0.043×8.25×3.452=4.22×106Nmm
My =αy Wly
2= 0.035×8.25×4.062= 4.75×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.057×8.25×3.452=5.59×106Nmm
My = αy Wly
2= 0.047×8.25×4.062=6.39×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
6.39x106 = 0.138x20x1000xd2
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
4.22×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.491Ast
2-36105Ast+4.22×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.4mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
5.59×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+5.59×106=0
use 10 mm dia bars.

Page | 48
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
4.75×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+4.75×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.46mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
6.39×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+6.39×106=0
Ast required = 184mm2
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars

Page | 49
1)spacing = (ast/Ast) ×1000 = (78.5/184)×1000= 426.63mm
2) 3d=3x100=300 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.6 mm2
= 0.75 X 184= 138 mm2
= 1/5 X 3450 = 690 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing = (28.26/138)x1000=204 mm
Provide 6mm dia bars @ 200 mm
Ast = 0.12 X bD = 0.12 X 1000X 120 = 144 mm2
Use 6 mm bars
ast=πd2/4= 28.26 mm2
1)spacing =(28.26/144)x1000=196.25 mm
2)5d= 5x100=500 mm
3)450 mm
Provide 6mm dia@ 190 mmc/c
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (184/261.6)=169.26
% steel = (100×261.62)/ (100×1000) =0.261
Hence it is safe.
Slab: C2-C3-D2-D3

Page | 50
Live load = 2KN/m2
Considered
(αx)
(αy )
0.053 0.035
edge
0.071 0.047
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.053×9×3.452=5.67×106Nmm
My =αy Wly
2= 0.035×9×4.902= 7.56×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.071×9×3.452=7.69×106Nmm
My = αy Wly
2= 0.047×9×4.902=10.15×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
10.15x106 = 0.138x20x1000xd2

Page | 51
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
5.67×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.491Ast
2-36105Ast+5.67×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000 = (50.24/162.52)×1000=309 mm
2) 3d=3x100=300 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.4mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
7.69×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+7.69×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261.62 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}

Page | 52
7.56×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+7.56×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm
Ast provided = (78.5/300) ×1000 = 261 mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
10.15×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+100.15×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000= (113.04/299.76)×1000= 377.10mm
2) 3d=3x100=300 mm
3)300mm
Ast provided = (113.04/300) ×1000 = 376.8 mm2
= 0.75 X 299.76= 224.82 mm2
= 1/5 X 3450 = 690 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing = (28.26/224.82)x1000=125 mm

Page | 53
Ast = 0.12 X bD = 0.12 X 1000X 120 = 144 mm2
Use 6 mm bars
ast=πd2/4= 28.26 mm2
1)spacing =(28.26/144)x1000=196.25 mm
2)5d= 5x100=500 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×
= 0.58×415× (299.26/376.8)=191.48
% steel = (100×376.8)/ (100×1000) =0.37
Hence it is safe.
Slab: C2-C3-D2-D3(Terrace)

Page | 54
Considered
(αx)
(αy )
0.053 0.035
edge
0.071 0.047
Mx = αy Wlx
2
My =αy Wly
2
Positive Moment:
Mx = αx Wlx
2= 0.053×8.25×3.452=5.20×106Nmm
My =αy Wly
2= 0.035×8.25×4.902= 6.93×106Nmm
Negative moment:
Mx = αx Wlx
2= 0.071×8.25×3.452=3.97×106Nmm
My = αy Wly
2= 0.047×8.25×4.902=9.30×106Nmm
Minimum Depth:
Mu = 0.138 fck b d2
9.30x106 = 0.138x20x1000xd2
d=58.04< 100 mm, Hence safe.
Positive Mx :
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
5.20×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.491Ast
2-36105Ast+5.20×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars

Page | 55
2) 3d=3x100=300 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.4mm2
Negative-Mx:
𝑀 𝑥 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
3.97×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+3.97×106=0
use 8 mm dia bars.
ast=πd2/4=50.24 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm
Ast provided = (50.24/300) ×1000 = 167.4 mm2
Positive My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
6.93×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+6.93×106=0
use 10 mm dia bars.
ast=πd2/4=78.5 mm2
Spacing of bars
2) 3d=3x100=300 mm
3)300mm

Page | 56
Ast provided = (78.5/300) ×1000 = 261 mm2
Negative My :
𝑀 𝑦 = 0.87 × 𝐴 𝑠𝑡 × 𝑓𝑦 × 𝑑 × {1 − (
)}
9.30×106 = 0.87×415 × 100 × Ast{1-(415Ast/20×1000×100)}
7.49Ast
2-36105Ast+9.30×106=0
use 12 mm dia bars.
ast=πd2/4=113.04 mm2
Spacing of bars
1)spacing = (ast/Ast) ×1000 = (113.04/273.04)×1000= 414 mm
2) 3d=3x100=300 mm
3)300mm
Ast provided = (113.04/300) ×1000 = 376.8 mm2
= 0.75 X 273.04= 204.78 mm2
= 1/5 X 3450 = 690 mm
Use 6 mm bars
ast=πd2/4= 28.26 mm2
spacing= (28.26/204.78)x1000=138 mm
Ast = 0.12 X bD = 0.12 X 1000X 120 = 144 mm2
Use 6 mm bars
ast=πd2/4= 28.26 mm2
1)spacing =(28.26/144)x1000=196.25 mm
2)5d= 5x100=500 mm
3)450 mm
Deflection check
𝑓𝑠 = 0.58 × 415 ×

Page | 57
= 0.58×415× (273.04/376.8)=174.41
% steel = (100×376.8)/ (100×1000) =0.37
Hence it is safe.

Page | 58
CHAPTER 4
DESIGN OF BEAMS
4.1. Theory
Concrete is fairly strong in compression but very weak in tension. Hence Plain concrete
cannot be used in situations where considerable tensile stresses develop. If flexural members like
beams and slabs are made of plain concrete their load carrying capacity is very low due to its low
tensile strength. Since steel is very strong in tension, steel bars are provided to resist tensile
stresses at a place where the maximum tensile stresses are developed
In case of simply supported beam, tensile stresses are induced in bottom layers because
of positive bending moment (sagging bending moment) and hence steel bars are provided near
the bottom of the beam. In cantilever beams steel bars are placed near the top of the beam to
resist the tensile stress developed in top layers due the negative bending moment (hogging
bending moment)
There are three types of reinforced concrete beams:
(A) Singly reinforced beams
(B) Doubly reinforced beams ,and
(C) Singly or doubly reinforced flanged beams.
4.2 Analysis of Singly Reinforced Sections:
If the reinforcing bars are provided only on tension side in the beam section, it is called as
singly reinforced beams.
Consider a simply supported beam subjected to bending under factored loads. Since plane
sections are assumed to remain plane before and after bending strain are proportional to
distance from the neutral axis. Above the neutral axis, the entire cross section is in
compression and below the neutral axis, the cross section is in tension. All the tensile stresses
are assumed to be resisted by the steel bars as the tensile strength of concrete is ignored. The
resultant tensile force, thus acts at the centroid of reinforcing bars.

Page | 59
4.3 Doubly Reinforced Beams
Beams which are reinforced in both compression and tension sides are called as doubly
reinforced beam. These beams are generally provided when the dimensions of the beam are
restricted and it is required to resist moment higher than the limiting moment of resistance of
a singly reinforced section. The additional moment of resistance required can be obtained by
providing compression reinforcement and additional tension reinforcement.
Situations Under which Doubly Reinforced Beams are used:
(1) When the depth of the beam is restricted due to architectural or any construction
problems
(2)At the supports of a continuous beam where bending moment changes its sign
(3) In precast members (during handling bending moment changes its sign)
(4) In bracing members of a frame due to changes in the direction of wind loads
(5) To improve the ductility of the beams in earth quack regions
(6) To reduce long term deflection or to increase stiffness of the beam.
4.4 DESIGN CALCULATIONS
Load distribution:
Slab Area calculation:
Trapezoidal: [(ly-lx) + ly]/2 × lx/2
Triangular: ½ ×lx× lx/2
Load calculation on external Beams:
Beam -1 (AB&IJ):
Live load from slab = 2 kN/m2
Dead load from slab = 0.20 × 25 × 1 = 5 kN/m2
Floor Finishes = 1 kN/m2
Total load transferred from slab = 8 kN/m2
Slab area concentrated on beam = [(ly-lx) + ly]/2 × lx/2
= ½[ (4.19 – 5.52)+ 4.19]× 5.52 /2 = 3.94 m2
Total slab load = 8 × Area × 1

Page | 60
= 8 × 3.94 × 1 = 31.52kN
Dead load of Beam = 0.23 × 0.45× 5.52 × 25 = 14.28kN
Wall load = 5.52 × 20 × 3 × 0.23 = 76 .17kN
Load/ meter on Beam = 31.52 + 14.28 + 76.17 = 121.97/5.52 = 22.096kN/m
Factored load = 1.5 × 22.096 = 33.14 kN/m
Beam -2 (BC&JK):
= ½[ (4.19 – 5.22)+ 4.19]× 5.22 /2 = 4.12 m2
= 8 × 4.12 × 1 = 32.99kN
Wall load = 5.22 × 20 × 3 × 0.23 = 72.036 kN
Load/ meter on Beam = 32.99 + 13.50 + 72.036 = 118.52/5.22
=22.706kN/m
Beam -3 (CD&KL):
Slab area concentrated on beam = ½ × lx × lx/2
= ½ ×3.58×3.58/2 = 3.20 m2
= 8 × 3.20 × 1 = 25.63kN

Page | 61
Wall load = 3.58 × 20 × 3 × 0.23 = 49.404kN
Load/ meter on Beam = 25.63 + 9.26 + 49.404 =84.29/3.58 = 23.544kN/m
Beam:4(AE):
Live load from slab = 2 KN/m2
Dead load from slab = (0.20×25×1) = 5 KN/m2
Floor finishers = 1 KN/m2
Total load transferred from slab = 8KN/m2
= ½ ×4.19×4.19/2 = 4.38 m2
Total slab load = 8×Area ×1
= 8×4.38×1 =35.04KN
Dead load of beam = 0.23×0.45×4.19×25=10.84KN
Wall load = 4.19×20×3×0.23 =57.82 KN
Load /meter on beam = (35.04+10.84+57.82)/4.19 =24.74 KN/m
Factored load = 1.5×24.74 =37.12 KN/m
Beam:5(EI):
Live load from slab = 2 KN/m2
= ½ ×5.03×5.03/2 = 6.32 m2
Total slab load = 8×6.32 ×1
= 8×6.32×1 =50.60KN
Wall load = 5.03×20×3×0.23 =69.414 KN

Page | 62
Beam: 6 ( DH):
= ½[ (4.19 – 3.58)+ 4.19]× 3.58 /2 = 4.29m2
= 8×4.29×1 =34.36 KN
Wall load = 4.19×20×3×0.23 =57.82KN
Load/meter = (34.36+10.84+57.82)/4.19= 24.58KN/m
Factored load = 1.5×24.58 =36.88KN/m
BEAM:7(HL)
= ½[ (5.03 – 3.58)+ 5.03]× 3.58 /2 = 5.79m2
= 8×5.79×1 =46.39 KN
Wall load = 5.03×20×3×0.23 =69.41KN
Load/meter = (46.39+13.01+69.41)/5.03= 25.60KN/m
Beam -1 (AB&IJ):(TERRRACE)
Live load from slab = 1.5 kN/m2
Total load transferred from slab = 7.5 kN/m2

Page | 63
= ½[ (4.19 – 5.52)+ 4.19]× 5.52 /2 = 3.94 m2
= 7.5 × 3.94 × 1 = 29.55kN
Wall load = 5.52 × 20 × 3 × 0.23 = 76 .17kN
Load/ meter on Beam = 29.55 + 14.28 + 76.17 = 120/5.52 = 21.73kN/m
Beam -2 (BC&JK):
= ½[ (4.19 – 5.22)+ 4.19]× 5.22 /2 = 4.12 m2
Total slab load = 7.5× Area × 1
= 7.5 × 4.12 × 1 = 30.9kN
Wall load = 5.22 × 20 × 3 × 0.23 = 72.036 kN
Load/ meter on Beam = 30.9 + 13.50 + 72.036 = 116.43/5.22 =22.30kN/m
Beam -3 (CD&KL):
= ½ ×3.58×3.58/2 = 3.20 m2
Total slab load = 7.5 × Area × 1
= 7.5 × 3.20 × 1 = 24kN

Page | 64
Wall load = 3.58 × 20 × 3 × 0.23 = 49.404kN
Load/ meter on Beam = 24 + 9.26 + 49.404 =82.66/3.58 = 23.090kN/m
Beam:4(AE):
Live load from slab = 1.5 KN/m2
Total load transferred from slab = 7.5KN/m2
= ½ ×4.19×4.19/2 = 4.38 m2
Total slab load = 7.5×Area ×1
= 7.5×4.38×1 =32.85KN
Wall load = 4.19×20×3×0.23 =57.82 KN
Beam:5(EI):
Live load from slab = 1.5 KN/m2
= ½ ×5.03×5.03/2 = 6.32 m2
Total slab load = 7.5×6.32 ×1
= 7.5×6.32×1 =47.4KN
Wall load = 5.03×20×3×0.23 =69.414 KN

Page | 65
Beam: 6 ( DH):
= ½[ (4.19 – 3.58)+ 4.19]× 3.58 /2 = 4.29m2
= 7.5×4.29×1 =32.175 KN
Wall load = 4.19×20×3×0.23 =57.82KN
Load/meter = (32.175+10.84+57.82)/4.19= 24.06KN/m
BEAM:7(HL)
= ½[ (5.03 – 3.58)+ 5.03]× 3.58 /2 = 5.79m2
= 7.5×5.79×1 =43.42 KN
Wall load = 5.03×20×3×0.23 =69.41KN
Load/meter = (43.42+13.01+69.41)/5.03= 25.01KN/m
Beam Design:
(1) Continuous Beam (AK, DR)
MAB = -wl2/12 = -33.14 × 5.522/12 = -84.14 kN-m
MBA = wl2/12 = 33.14 × 5.522/12 = 84.14 kN-m
MBC = -wl2/12 = -34.059 × 5.222/12 = -77.33 kN-m

Page | 66
MCB = wl2/12 = 34.059 × 5.222/12 = 77.33 kN-m
MCD = -wl2/12 = -35.31 × 3.582/12 = -37.71 kN-m
MDC = wl2/12 = 35.31 × 3.582/12 = 37.7 1kN-m
Fig 4.1: Loading diagram for floor
Table 4.1: Distribution factors
Joints Member Relative stiffness Total Distribution
factor
B BA I/5.52 I/5.52 + I/5.22
= 0.37 I
1/0.5I × I/4.7
= 0.48
BC I/5.22 1/0.5I × I/3.45
= 0.51
C CB I/5.22 I/5.22+ I/3.58
=0.47I
1/0.47I × I/5.22
=0.40
CD I/3.58 1/0.47I × I/3.58
=0.59

Page | 67
Moment distribution method:
Fig 4.2: Moment distribution calculation
Net span moment:
Net span moment = 84.14 kN-m
(1) Span moments from table, d of SP16 for Fe415 and M20 :
Mu, lim/bd2 = 2.76
Mu, lim = 2.76 × (230 × 4502) = 128.54 kN-m
Mu = 84.14 < Mu, lim
Hence, design as a singly reinforced section.
Mu/bd2 = 84.14X106/(230 × 4502) = 1.80
(from, table of reinforcement % in singly reinforced beam)
Pt = 0.565

Page | 68
Ast = pt× bd/100 = 0.565 × 230 × 450/100 = 584mm2 ≈ 590 mm2
Provided steel = 678.24 > 590mm2.
Provide 6 – 12mm bars.
Check for maximum Ast:
Ast/bd >0.85/fy
678.25/(230 × 450) = 6.55 × 10-3> 0.85/415 = 2.04 × 10-3
Hence, safe.
Check for deflection:
For continuous l/d = 26
fs = 0.58 × fy × (Ast, req/Ast, provided)
= 0.58 × 415 × (590/678.24) = 209.38N/mm2
pt = 100 × 678.24/230× 450 = 0.65
Modification factor = 1.3
Allowable l/d = 26 × 1.3 = 33.8
Actual l/d ratio = 14.32/0.45 = 31.82< 33.8
Hence, safe.
Support moments:
Mu/bd2 = 25.93 × 106/230 × 4502 = 0.55< 1.6
Hence, designed as a singly reinforced section.
pt = 0.143
Ast = ptbd/100 = 0.143 × 230 × 450/100 = 148.00 mm2
Provided steel = 226.08 > 148.00 mm2
Provide 2 – 12 mm bars,
Maximum area of steel (Ast) = 0.04 bD
= 0.04 ×230 × 450 = 4140 mm2
Minimum area of steel (Ast) > 0.85/fy × bd = 276.5
223.7 > 226.08. Hence, safe.

Page | 69
CHAPTER 5
DESIGN OF COLUMN
5.1. Theory
A vertical member whose effective length is greater than 3 times its least lateral
dimension carrying compressive loads is called as column. Column transfer the loads from the
beams or slabs to the footings or foundations. The inclined member carrying compressive loads
as in case of frames and trusses is called as struts. Pedestal is a vertical compression member
whose effective length is less than 3 times its least lateral dimension. Generally the column may
be square, rectangular or circular in shape.
Necessity of Reinforcement in columns
Even though concrete is strong in compression , longitudinal steel bars are placed in the
column to reduce the size of the column or to increase the load carrying capacity and to resist
any tension that might develop due to bending of column due to horizontal loads, eccentric loads
or moments. To resist any tensile stresses likely to develop, the reinforcement should be placed
as near the surface as possible and should be evenly distributed ensuring the minimum
cover.Transverse reinforcement in the form of lateral ties or spiral reinforcement are provided to
resist longitudinal splitting of the column or splitting of concrete due to development of
transverse tension and to prevent buckling of longitudinal bars
Types of Columns
 Based on type of Reinforcement
(A) Tied column
(B) Spiral column
(C) Composite column
 Based on type of loading
(A) Axially loaded column
(B) Eccentrically loaded column(Uniaxial or Biaxial)
 Based on Slenderness ratio
(A) Short column

Page | 70
(B) Long column
In this G+3 apartment we are designing for the biaxial loaded column .Biaxial loaded column:
when the line of action of the resultant compressive force doesn’t coincide with the center of
gravity of the cross section of the column, it is called as eccentrically loaded column.
Eccentrically loaded columns have to be designed for combined axial force and bending
moments
5.2. Columnreactions
Table 5.1 column reactions.
Grid /
column
Load from
terrace(A)KN
Load from
Typical
floor(B) KN
Weight of the column
(C)KN
Total column reaction
(A+B+C)KN
A1 115.09 237.78x3 (0.23×0.4×13×20) =23.92 852.35
A2 186.16 344.08x3 (0.23×0.4×13×20) =23.92 1242.32
A3 99.105 198.99x3 (0.23×0.4×13×20) =23.92 719.99
B1 199.35 409.28x3 (0.23×0.5×13×20) =29.90 1457.09
B2 317.05 630.13x3 (0.23×0.5×13×20) =29.90 2237.34
B3 172.3 349.18x3 (0.23×0.4×13×20) =23.92 1243.76
C1 132.9 420.45x3 (0.23×0.5×13×20) =29.90 1424.15
C2 297.9 653.79x3 (0.23×0.5×13×20) =29.90 2289.17
C3 290.64 300.99x3 (0.23×0.4×13×20) =23.92 1217.53
D1 115.09 273.16x3 (0.23×0.4×13×20) =23.92 958.49
D2 263.67 287.40x3 (0.23×0.5×13×20) =29.90 1155.77
D3 86.32 185.00x3 (0.23×0.4×13×20) =23.92 665.24
5.3. Column Grouping
Table 5.2 Column Grouping
Column naming Load (KN) Mx My Column size
C1 900 12.30 17.39 230×400
C2 1300 17.75 25.12 230×400
C3 2400 32.78 54.40 230×500

Page | 71
5.4. DESIGN CALCULATIONS
1.C1=900KN
Size of the column = 230×400mm
Concrete mix = M 20
Characteristic strength of reinforcement = 415 N/mm2
emin,x = 3000/500 + 230/30 = 13.66 mm
emin,y = 3000/500 + 400/30 = 19.33 mm
Mux = Puxex KN-m =900x13.66=12.30 KN-m
Muy ==Puey KN-m=900x19.33=17.39 KN-m.
Assume the reinforcement Percentage = 1.5
P/fck = 0.075
Uniaxial moment capacity of the section about x-x axis :
(d’+dia)/D =( 40+12)/230 = 0.22
Pu/(fckbd) = 900×10^3/(20×230×400)=0.489
Referring chart 46
Mu/(fckbD2)= 0.06
Mux1 = 0.06×20×230×4002 = 44.16 KN-m
Uniaxial moment capacity of the section about y-y axis :
(d’+dia)/D = (40+12)/400 = 0.13
Pu/(fckbd )= 900×10^3/(20×230×400)=0.489
Referring chart 45
Mu/(fckbD2)= 0.07
Muy1 = 0.07×20×230×4002 = 51.52 KN-m
Calculating of Puz ;
P = 1.5 ,fy = 415 N/mm2 and fck= 20 N/mm2
Referring chart 63corresponding above values
Puz/Ag = 13 N/mm2
Puz = 13×230×400 = 1196KN
Pu/Puz= 0.752
Mux/Mux1 = 0.278

Page | 72
Muy/Muy1 = 0.33
Referring to chart 64 the permissible value of Mux/Mux1 Corresponding to the above values
Pu/Puzand Muy/Muy1 is equal to 0.93
0.93>0.27
Hence it is safe.
Ast =230X400X1.5/100 = 1380 mm2
Using 16 mm dia bars
No of bars = 1380/200.96=6.86= 7 bars
Provide 7, 16mm dia bars.
Provide lateral ties of 6 mm dia 230 mm c/c
2.C2=1300 KN
Procedure same as above, Provide 2.5 % of steel it is safe
Ast required =230X400X2.5/100 = 2300 mm2
Provide 18 mm dia bars 10 in number
Provide Lateral ties 6 mm @ 280 mm c/c
3.C3=2400 KN
Procedure same as above, Provide 3 % of steel it is safe
Astrequired =230X500X3/100 = 3450 mm2
Provide 20mm dia bars 10 in number
Provide Lateral ties of 6 mm dia 230 mm c/c

Page | 73
CHAPTER 6
DESIGN OF FOOTING
Foundation is an important part of the structure which transfers the load of the super
structure to the foundation soil. The foundation distributes the load over a larger area so that the
pressure on the soil does not exceed its allowable bearing capacity and restricts the settlement of
the structure with in the permissible limits. Foundation increases the stability of the structure.
Foundations may be shallow or deep foundation depending up on the load and type of foundation
soil. if the load to be supported is very high and soil is of low bearing capacity deep foundation
like pile foundation well foundation etc are provided if the soil with adequate bearing capacity is
available at reasonable depth, shallow foundations are provided.
6.1. Footing Reactions
Table 6.1 footing reactions:
Grid /
column
Load from
terrace(A)KN
Load from
Typical
floor(B) KN
Load from
Plinth(C)KN
Weight of the column
(D)KN
Total column
reaction
(A+B+C+D)KN
A1 115.091 237.78x3 10 (0.23×0.4×14.5×20) =26.68 865.11
A2 186.16 344.08x3 10 (0.23×0.4×14.5×20) =26.68 1255.08
A3 99.105 198.99x3 10 (0.23×0.4×14.5×20) =26.68 732.75
B1 199.35 409.28x3 10 (0.23×0.5×14.5×20) =33.35 1470.81
B2 317.05 630.13x3 10 (0.23×0.5×14.5×20) =33.35 2250.79
B3 172.33 349.18x3 10 (0.23×0.4×14.5×20) =26.68 1256.55
C1 132.90 420.45x3 10 (0.23×0.5×14.5×20) =33.35 1437.6
C2 297.90 653.79x3 10 (0.23×0.5×14.5×20) =33.35 2302.62
C3 290.64 300.99x3 10 (0.23×0.4×14.5×20) =26.68 1230.29
D1 115.09 273.16x3 10 (0.23×0.4×14.5×20) =26.68 971.25
D2 263.67 287.40x3 10 (0.23×0.5×14.5×20) =33.35 1169.22
D3 86.32 185.00x3 10 (0.23×0.4×14.5×20) =26.68 678.0

Page | 74
NOTE: UNIT WEIGHT OF THE COLUMN = thickness × width × unit weight of the rcc ×
column height
(Column height = 3m , plinth height = 1.5m parapet wall = 1m)
Total G +3 column height = 1+3+3+3+3+1.5=14.5m
6.2 DESIGN CALCULATIONS
F1 =900KN
Factored load =900KN
Size of column =230mm×400mm
S.B.C of Soil = 350 KN/m²
Un-factored =900/1.5=600 KN
1. Size of the footing:-
Load from the column = 600 KN
Self weight of footing =10% of the column loud=600/10 = 60
Total load of footing =600+60 = 660 KN
Area of the footing = Total load/SBC of soil
= 660/350 = 1.8m2
Size of the square footing
B = √1.8 = 1.37
Adopt 1.4m×1.4m square footing.
2. Upward soil pressure:-
Factor load = Pu= 900KN
Soil pressure at ultimate load
qu = Pu/area of footing
qu = 900/(1.4×1.4)= 459.18 KN/m2
qu = 0.459 N/mm2
3. Depth of Footing from Bending Moment consideration:
The critical section for B.M will be at the face of the column as shown in fig below
Mu = qu[
B(B−b)²
8
] = 0.459 [
1400 (1400 −230)²
8
]
Mu = 109.95×106 N-mm

Page | 75
Mu = 0.138fckBd2
109.95×106 = 0.138×20×1400×d2
d = 168.68 mm
Depth required to resist shear in footings is much higher than required for bending
Let us assume the effective depth as twice the depth required from bending consideration.
Provide effective depth = d =400mm
Overall depth D =450mm
4. Reinforcement:-
Mu = 0.87 fyAstd [1-
𝑓 𝑦 𝐴 𝑠𝑡
𝑓 𝑐𝑘 𝐵𝑑
]
109.95×106 = 0.87 × 415×Ast× 400 [1-
415 × 𝐴 𝑠𝑡
20 × 1400 × 400
]
109.95×106 = 144420 × Ast - 5.34Ast
2
Area of steel required ,Ast=784.05 mm2
Use 12mm diameter bars
ast=
𝜋
4
𝑑² = 113.04 mm²
Spacing: S =
𝑎 𝑠𝑡 × 𝐵
𝐴 𝑠𝑡
=
113.04× 1400
784.05
= 201.84 mm
Provide spacing = S = 200 mm
Provide Area of steel
Ast =
𝑎 𝑠𝑡 ×𝑩
𝑆
=
113.04×1400
200
Ast= 791.28 mm²
5. Check for one way shear:-
The critical section for one way shear is at a distance “d” from the face of the column as shown
in the fig
Factored shear force
Vu = Soil presser from the shaded area
Vu = qu B[
(B−b)
2
- d]
= 0.459×1400[
(1400−230)
2
- 400]
Vu = 118.8×103 N
Shear stress
τv=
𝑉𝑢
𝐵𝑑
=
118800
1400𝑋400
τv=0.21 N/mm²

Page | 76
Percentage of steel
Pt =
𝐴 𝑠𝑡 ×100
𝐵𝑑
=
100×791.28
1400×400
Pt= 0.14 %
From table 19, IS 456:2000
τc>τv
So it is safe in one way shear
6. Check for two way shear :-
The critical section for two way shear is at a distance
𝑑
2
from the face of the column as shown in
the fig
Perimeter of the critical section (p) = 2[(230+300)+(400+300)]
= 2460 mm
Area of critical section (A) = p×d = 2460 × 400 = 768000.00mm²
Two way shear Vu2 = qu× area of shaded potion
Vu2 = 0.459[(1400×1400) – (630×800) =668.30 KN
Two way shear stress τv2 =
𝑉𝑢2
𝐴
=
668304
984000
τv2 = 0.679 N/mm²
Permissible punching shear
τp=0.25√ 𝑓𝑐𝑘 = 0.25√20
τp = 1.11 N/mm²
τv2<τp
Hence, it is safe with respect to two way shear.
F2=1300KN
Un-factored =1300/1.5=866.67 KN
Load from the column = 866.67 KN
Self weight of footing =10% of the column loud
=866.67/10 = 86.667
Total load of footing =866.67+86.667 = 953.33 KN

Page | 77
= 953.33/350 = 2.72m2
B = √2.72 = 1.65
qu= 1300/(1.7×1.7) = 449.8 KN/m2
qu = 0.449 N/mm2
Mu = qu[
B(B−b)²
8
] = 0.449 [
1700 (1700 −230)²
8
]
Mu = 206.17×106 N-mm
Mu = 0.138fckBd2
206.17×106 = 0.138×20×1700×d2
d = 209.6 mm
4. Reinforcement:-
]
206.17×106 = 0.87 × 415×Ast× 450 [1-
415 × 𝐴 𝑠𝑡
20 × 1700 × 450
]
Area of steel required ,Ast=1315 mm2
ast=
𝜋
4
𝑑² = 113.04 mm²
Spacing S =
𝐴 𝑠𝑡
=
113.04× 1700
1315
= 146.13 mm

Page | 78
Ast =
𝑆
=
113.04×1700
140
Ast= 1372.6 mm²
in the fig
Vu = qu B[
(B−b)
2
- d]
= 0.449×1700[
(1700−230)
2
- 450]
Vu = 217.92×103 N
Shear stress
τv=
𝑉𝑢
𝐵𝑑
=
217920
1700𝑋450
τv=0.284 N/mm²
Percentage of steel
Pt =
𝐴 𝑠𝑡 ×100
𝐵𝑑
=
100×1372 .6
1700×450
Pt= 0.17 %
τc>τv
𝑑
2
the fig
Perimeter of the critical section (p) = 2[(230+450)+(400+450)] = 3060 mm
Area of critical section (A) = p×d = 3060 × 450 =1377000.00mm²
Vu2 = 0.449 [(1700×1700) – (680×850)]
=1039.93 KN
𝑉𝑢2
𝐴
=
1039930
1377000
τv2 = 0.75 N/mm²

Page | 79
τp=0.25√ 𝑓𝑐𝑘 = 0.25√20
τp = 1.11 N/mm²
τv2<τp
F3 =2400KN
Un-factored =2400/1.5=1600 KN
Load from the column = 1600 KN
Self weight of footing =10% of the column loud=1600/10 = 160
Total load of footing =1600+160 = 1760 KN
= 1760/350 = 5.02m2
B = √5.02 = 2.24
qu = 2400/(2.25×2.25)= 474.07 KN/m2
qu = 0.474 N/mm2
Mu = qu[
B(B−b)²
8
] = 0.474 [
2250 (2250 −230)²
8
]
Mu = 543.96×106 N-mm
Mu = 0.138fckBd2
543.96×106 = 0.138×20×22500×d2

Page | 80
d = 295.96 mm
4. Reinforcement:-
]
543.96×106 = 0.87 × 415×Ast× 600 [1-
415 × 𝐴 𝑠𝑡
20 × 2250 × 600
]
Area of steel required, Ast=2615 mm2
ast=
𝜋
4
𝑑² = 113.04 mm²
Spacing S =
𝐴 𝑠𝑡
=
113.04× 2250
2615
= 97.26 mm
Ast =
𝑆
=
113.04×2250
90
Ast= 2826 mm²
in the fig
Vu = qu B[
(B−b)
2
- d]
= 0.47×2250[
(2250−230 )
2
- 600]
Vu = 437.26×103 N
Shear stress
τv=
𝑉𝑢
𝐵𝑑
=
437260
2250𝑋600
τv=0.32 N/mm²
Percentage of steel
Pt =
𝐴 𝑠𝑡 ×100
𝐵𝑑
=
100×2826
2250 ×600

Page | 81
Pt= 0.20 %
τc>τv
𝑑
2
the fig
Perimeter of the critical section (p) = 2[(230+600)+(500+600) = 3860 mm
Area of critical section (A) = p×d = 3860 × 600 = 2316000.00mm²
Vu2 = 0.474[(2250×2250) – (830×1100)]
=1966.86 KN
𝑉𝑢2
𝐴
=
1966860
2316000
τv2 = 0.849 N/mm²
τp=0.25√ 𝑓𝑐𝑘 = 0.25√20
τp = 1.11 N/mm²
τv2<τp

Page | 82
CHAPTER 7
DESIGN OF STAIRCASE
Stairs Provide access for the various floors of the building the stair consists of series of
steps with landings at appropriate intervals the stretch between the two landings is called flight.
The room or space where stairs are provided is called stair case.
The width of stair depends up on the type of building in which it is provided generally in
residential building the width of stair is kept as 1m and in case of public buildings it may be to
2m . to allow free flow of users the width of landings should be at least equal to the width of
stairs.
Each step has one tread (going) and one rise. Rise and tread are proportioned so as to
provide convenient and easy access. The rise may vary from 150mm to 200mm The tread is in
between 250 mm to 300mm As per IS : 456, the slope or pitch of the stairs should be in between
250to 400
Types of stair case
Depending up on the Geometry/shape
(A)Single Flight stair case
(B)Quarter Turn Stair case
(C) Doglegged Stair case
(D) Open well stair case
(E)Geometrical stair case
(F)Spiral stair case
Based on Structural Behavior (Support Conditions)
(1) Stairs Spanning Horizontally(with side supports)
(2) Stairs Spanning Longitudinally(with supports along sloping line)

Page | 83
1. DESIGN FACTORS:
Type of staircase = dog logged staircase.
Rise =150mm
Tread =300mm
Width of landing =300mm
Fck =20N/mm²
Fy =415N/mm²
To find total number of steps = (height of floor/rise)
= (3/0.15)
=20 nos
Number of steps in each flight=10 nos
2. EFFECTIVE SPAN:
Effective span = (9 x 300) + 300 = 3000mm = 3.0m
3. LOAD CALCULATION:
Thickness of waist slab= (3000/20) = 150 mm
Effective depth d = 130mm
Dead loads of slab on slope ws
Ws = (0.150x1x25)=3.75kN/m
Dead loads of slab on horizontal span ”w”
W = Ws(R²+T²)/T
W =3.75x(150²+300²)/300
W =4.91 kN/m
Dead loads of one step = (0.5x0.3x0.15x25)=0.56 kN/m
Finish =1 kN/m
Self weight of landing Portion (0.3x1x25) =7.5 kN/m
Total dead load = (4.19+0.56+7.5)= 13.25kN/m
Live load =3 kN/m
Total service load = 16.25 kN/m
Factor load =24.375 kN/m
Max B.M
Mu =0.125 mu xl2=0.125x24.375x32= 27.42 kN/m
CHECK FOR DEPTH OF WAIST SLAB:
d =√mu/0.138fckb
d =√27.42x106
/0.138x20x1000
=99.6mm < provide depth (130mm)
Hence safe

Page | 84
MAIN REINFORCEMENT:
Mu = (0.87 x fyAstd){1- (Astfy/bdfck)}
(27.42x106
) = (0.87x415xAstx130){1-(Astx415/1000x130x20)}
7. 49Ast2
-46936.15Ast +27.42x106
=0
Ast=652.03 mm2
Provide 16mm dia bars
Spacing = astx1000/Ast=200.96x1000/652.03
=308.20 mm say 300mm c/c
DISTRIBUTION REINFORCEMENT:
Distribution reinforcement = 0.12% of c/s
= (0.12/100) x (1000x150)
=180 mm2
Provide 8mm dia bars
Spacing = 50.26x1000/180
=279.20 mm say 250mmc/c
Provide 8mm dia of bars @ 250mm c/c

Page | 85
CONCLUSION
Construction of Residential building should be providing safety and it should be
economical. This project describes detailed planning and designing of the G+3 residential
building apartments. This project involves design of structures with various methods such as
conventional method and by software. Getting familiar with civil engineering softwares Auto
CAD and Staad Pro.

Page | 86
REFERENCES
1. M.R.Dheerendrababu, “Design of reinforced concrete structures”
2. AshokK.Jain,”Reinforced concrete (limit state design)”,6 th edition
3. Ramamrutham, “structural analysis”
4. S.S.Bhavikatti,”Structural analysis volume-1”
5. IS codes 456-2000,IS code 875 part-1,part-2 and part-3,sp 16and sp34.

DESIGN AND ANALYSIS OF G+3 RESIDENTIAL BUILDING BY S.MAHAMMAD FROM RAJIV GANDHI UNIVERSITY OF KNOWLEDGE TECHNOLOGIES

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DESIGN AND ANALYSIS OF G+3 RESIDENTIAL BUILDING BY S.MAHAMMAD FROM RAJIV GANDHI UNIVERSITY OF KNOWLEDGE TECHNOLOGIES