TRIBHUVAN UNIVERSITY
INSTITUTE OF ENGINEERING
HIMALAYA COLLEGE OF ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
FINAL YEAR PROJECT REPORT
ON
STRUCTURAL ANALYSIS AND DESIGN OF
COMMERCIAL COMPLEX
Submitted to the Department of Civil Engineering
In the partial fulfillment of requirements for the Bachelor’s Degree in Civil Engineering
(Course Code: CE 755)
SUPERVISOR
Er. Debendra Dev Khanal
SUBMITTED BY:
Aagat Pyakurel
(BCE/068/01)
Abhigya Bhandari (BCE/068/03)
Aman Dawadi
(BCE/068/05)
Dibas Kathayat
(BCE/068/22)
September, 2015
ACKNOWLEDGEMENT
We would like to express our humble and sincere gratitude to the Department of Civil
Engineering of Himalaya college of Engineering for providing us the opportunity to choose
“Structural Analysis and Design of Commercial Complex” as the project work which is enlisted
in the syllabus of 4th year II part as per the course designed by IOE, Tribhuwan University.
We would like to extend our heartfelt thanks and gratitude to the college for arrangement and
support for working in the academic project. We are particularly indebted to our project
supervisor Er.Debendra Dev Khanal (M.Sc. in Structure Engineering) for his constructive and
encouraging suggestion, regarding the project and leading us towards the goal of the project work.
We also would like to sincerely acknowledge our hearty gratitude to H.O.D. of civil engineering
Er. Kishor Bagale Thapa. We greatly value the love and financial support provided for the project
work by the college administration.
We would also like to acknowledge those persons who have indirectly helped to successfully
complete this project and apologise them to whom we have left to extend acknowledgement
unknowingly.
Last but by no means least; we also would have high sense of appreciation to our own project
group for a unit co-ordination among the group during the project work.
ABSTRACT
Kathmandu is an earthquake hot zone, topping the earthquake risk list (with Nepal ranked 11th
globally), with Kathmandu's geological structure cited as the main reason for it being so. The
entire part of Nepal falls in a high earthquake intensity belt, and National Society for Earthquake
Technology (NSET-Nepal) estimates that nearly 100,000 people would lose their lives in
Kathmandu if the quake with equal magnitude of 1990 BS (8.4 Richter scale earthquake that
killed 8519 people, destroyed 80893 building then) was repeated and nearly 60 per cent buildings
would be collapsed. On top of that, the unplanned urbanization and unregulated building by-laws
by the related government offices in Nepal has lead citizens to build unregulated and illegal
structures in the city with almost a one-upmanship of outdoing others.
The overwhelming interest of people in the rush to economic supremacy seeks and sucks the
designer for his profound knowledge in his design. The trend of supermarkets and commercial
malls in our municipal areas requires a well learned and adequately trained designers in a
handsome quantity in near future. It is a responsibility of a young engineer to combat any
circumstances that appear in his professional market. In light of all these, the project entitled
“Structural Analysis and Design of Commercial Complex” is the need of time and an
appropriate selection for the project work.
The report starts with the seismic considerations being summed up that needs to be considered
during the building design. Analysis works are done using SAP2000 v16 as analytical tool.
Meanwhile MS-Excel is used for general calculations and Programming. Designs are carried as a
special moment resisting framed building. The report owes a complete conformity with various
stipulations in Indian Standards. The entire design works is ruled upon by the Limit State Design
Philosophy; however capacity approaches are also adopted in required sections. Necessary
Ductile Detailing is drafted using AutoCAD2007. Finally the report winds up with concluding
thoughts at the end. Thus the report seems to be a valuable academic achievement from Civil
Engineering perspective.
PREFACE
A course entitled “Civil Engineering Project” is prescribed by the TU, Institute of Engineering
as a practicing of case study and helping tool to get familiar with the practical problems that every
professional has to face in their professional life.
This project is the practical use of theoretical knowledge that we acquire during the four years of
civil engineering course with application of knowledge we gained from our respectable teachers
and superiors.
We have chosen the project “EARTHQUAKE ANALYSIS AND DESIGN OF
COMMERCIAL BUILDING”. The course offered on 4th year namely “Design of Reinforced
Concrete Structure” is a strong base. This course really helped us while designing the structure
and provided the knowledge to design the structure in terms of safety, economy, stability and
efficiency.
During the project work, we got to know thoroughly that how to analyze and tackle the problems
and got the optimal result which will safe guard the lives of people and the structure itself in the
state of seismic disasters.
This project work also helped us to work with Team spirit and the coordination for the long-term
work and getting through the problems effectively.
In gist, it was a real enthusiasm and full supportive to work under the guidance of our project
supervisor Er. Debendra Dev Khanal who always guided us with valuable tips while tackling
the problem and gave in-depth knowledge of Structural Engineering. We believe that his valuable
guidance and support is profoundly appreciable and will always help us in our future professional
life.
LIST OF SYMBOLS
List of Symbols:
Symbols
Ac
Ah
Ag
Ast
Asc
Asv
bf
bw
D
d
Df
fck
fy
I
Ix.Iy
hi
k
leff
lx
ly
l
lo
Ld
Mu
Mulim
Mux
Muy
Muxl
Muyl
P
Pu
pc
pt
Qi
Sv
T
V
Descriptions
Area of concrete
Horizontal seismic coefficient
Gross area of section
Area of tension reinforcement
Area of compression reinforcement
Area of vertical stirrup
Effective width of flanged section
Breadth of web in T or L-section
Overall depth of the section
Effective depth of the section
Thickness of the flange ‘T or L-section’
Characteristics compressive strength of concrete
Characteristics yield strength of steel
Importance factor of the structure
Moment of inertia about X and Y axis respectively
Height of the ith floor base of frame
Performance factor depending on the structural framing
system and brittleness or ductility of the construction
Effective length of the element
Span of the slab in the shorter direction
Span of slab in the longer direction
Unsupported length or clear span of element
Distance between point of inflection
Development length of the bar
Factored moment, design moment for limit state design
Limiting moment of resistance
Factored moment about X-axis
Factored moment about Y-axis
Maximum uni-axial moment capacity of the section with
axial load, bending about X-axis.
Maximum uni-axial moment capacity of the section
section with axial load, bending about Y-axis
Axial load on the element
Factored axial load, Design axial load for limit state design
Percentage of compression reinforcement
Percentage of tension reinforcement
Base shear distributed in the ith floor
Spacing of stirrup
Estimated natural of fundamental time period of the building
Shear force
Vu
Vus
Vb
Wi
Xu
Xumax
αx,αy
Design shear force for limit sate design, Factored shear force
Strength of shear reinforcement in the limit state of design
Total base shear
Lump load on the ith floor
Depth of the neutral axis in limit state of collapse
Maximum depth of neutral axis in limit state of design
Bending moment coefficient for slab about X-axis and Y-axis
respectively
Coefficient depending upon the
1)span longer than 10m
2)soil foundation system
Coefficient depending upon bf/bw ratio
Coefficient depending upon percentage of compressive
reinforcement
Allowable shear stress in concrete
Allowable bond stress in concrete
Allowable maximum shear stress in concrete width shear reinforcement
Nominal shear stress
Diameter of bar
β
λ
δ
τc
τbd
τcmax
τv
Φ
ABBREVIATIONS
CM
CR
DL
EQ
IS
LL
RCC
Center of Mass
Center of Rigidity
Dead Load
Earthquake Load
Indian Standard
Live Load
Reinforced Cement Concrete
COPYRIGHT©
The author has agreed that the library, Department of Civil Engineering, Himalaya
College of Engineering may make this project freely available for inspection.
Moreover the author agreed that the permission for the extensive copying of this
project for scholarly purpose may by granted by the Supervisor who supervised the
project work recorded here in or, in his absence by Head of Department concerning
BCE program coordinator or principal of the Institute in which project work was
done. It is understood that the recognition will be given to the author of this project
and to the Department of Civil Engineering, Himalaya college of Engineering, in
any use of the material of project. Copying or publication or other use of the
material of this for financial gain without approval al of Department of Civil
Engineering, Himalaya college of Engineering and author's written permission is
prohibited.
Request for permission to copy or to make any use of the material in this in whole
or part should be addressed to:
Head of Department civil Engineering
Himalaya College of Engineering
Chysal-9, Lalitpur, Nepal.
TABLE OF CONTENTS
Acknowledgement
Abstract
Preface
Symbols and abbreviations
Copyright
CHAPTER 1: INTRODUCTION
1. Background……………………………………..........................................................................1
2. Theme of Project Work………………………………………………………………………....1
3. Objectives………………………………………………………………………………………2
4. Scope…………………………………………………………………………………………...2
5. Methodology…………………………………………………………………………………...2
6. Time Schedule………………………………………………………………………………….5
CHAPTER 2: ASPECTS OF SEISMIC PERFORMANCE AND BUILDING
DESCRIPTION ...............................................................................................................................4
2.1 Seismic Performance of building.............................................................................................4
2.2 Configuration Issues in building ..............................................................................................5
2.3 Vertical Configuration .............................................................................................................6
2.4 Structural Ductility ..................................................................................................................6
2.5 Structural Layout .....................................................................................................................7
2.6 Isolation of Structure ...............................................................................................................7
2.7 General Principles for the design .............................................................................................7
CHAPTER 3: DESIGN METHOD................................................................................................9
3.1 Understanding and Design Philosophy ....................................................................................9
3.1.1 Background .......................................................................................................................9
3.1.2 Design Philosophies ..........................................................................................................9
3.2 Compression Members ..........................................................................................................10
3.3 Earthquake Resistance Design of Structure ...........................................................................10
3.4 Assessment of loads ...............................................................................................................11
3.5 Study of the Building .............................................................................................................12
CHAPTER 4: PRELIMINARY ANALAYSIS AND DESIGN .................................................14
4.1 Introduction ............................................................................................................................14
4.2 Need of Preliminary Design ..................................................................................................14
4.3 Preliminary Design ................................................................................................................14
4.3.1 Slab .................................................................................................................................14
4.3.2 Beam ...............................................................................................................................15
4.3.3 Column ............................................................................................................................16
4.4 Centre of Mass……………………………………………………………………………..
4.5 Centre of Stiffness
4.6 Base Shear..............................................................................................................................17
4.6.1 Base Shear Calculation ...................................................................................................19
4.6.2 Distribution of Base Shear ..............................................................................................20
4.6.2 Lumped Mass Calculation ..............................................................................................19
CHAPTER 5: STRUCTURAL ANALAYSIS
5.1 Salient feature of SAP2000
5.2 Load combinations
5.3 SAP application procedure
5.4 Analysis Features
5.5 Inputs and Outputs
CHAPTER 6: STRUCTURAL DESIGN
6.1 Limit state method of Design
6.2 Slab
6.2.1 Flowchart of Slab Design
6.2.2 Design
6.3 Beam
6.3.1 Flowchart of Beam Design
6.3.2 Design Calculation
6.4 Column
6.4.1 Flow chart of Column Design
6.4.2 Design Calculation
6.5 Staircase
6.6 Foundation
6.6.1 Design of Mat Foundation
6.7 Design of Basement Wall
6.8 Design of Lift Shear Wall
CALCULATION
ARCHITECTURAL DRAWING
CHAPTER 7:
DETAILING OF STRUCTURAL ELEMENT
CHAPTER 8: CONCLUSION
CHAPTER 9: RECOMMENDATION
CHAPTER 10: BIBILOGRAPHY
ANNEX A
CHAPTER 1: INTRODUCTION
BACKGROUND
Structural analysis is the determination of the effects of loads on physical structure and
their components. Structures subject to this type of analysis include all that must withstand loads,
such as buildings, bridges, vehicles, machinery, furniture, attire, soil strata, prostheses and
biological tissue. Structural analysis incorporates the fields of applied mechanics, materials
science and applied mathematics to compute a structure’s deformations, internal forces, stresses,
support reactions, accelerations, and stability. The results of the analysis are used to verify a
structure’s fitness for use, often saving physical tests. Structural analysis is thus a key part of
the engineering design of structures.
The project selected by our group is a multi-storey commercial building located at
Teenkune, Kathmandu. According to IS 1893:2002, Kathmandu valley lies in Zone V, the
severest one. Hence the effect of earthquake is predominant than the wind load. So, the building is
analysed for earthquake as lateral load. The dynamic analysis method as stipulated in IS
1893:2002 is applied to analyze the building for earthquake. Especial reinforced concrete moment
resisting frame is consider as the main structural system of building we might have in our hands
in the future.
The project report has been prepared in complete conformity with various stipulation in Indian
standards, code of practice for plain and reinforced concrete IS 456:2000, Design Aids for
Reinforced Concrete to IS456:2000(SP-16), Criteria Earthquake Resistant Design Structure IS
1893:2000, Ductile Detailing Of Reinforced Concrete Structure Subjected To Seismic ForcesCode Of Practice IS 13920:1993, Handbook On Concrete Reinforcement And Detailing SP34.Use of these codes have emphasized on providing sufficient safety, economy, strength and
ductility besides satisfactory serviceability requirements of cracking and deflection in concrete
structures. These codes are based on principles of Limit State Of Design.
This project has been undertaken as a partial requirement for B.E. degree in Civil Engineering.
This project work contains structural analysis, design and detailing of building located in
Kathmandu District . All the theoretical knowledge on analysis and design acquired on the course
work are utilized with practical application. The main objective of the project is to acquaint in the
practical aspect of Civil Engineering. We, being the budding engineers of tomorrow, are
interested in such analysis and design of structures which will, we hope, help us in similar jobs
that we might have in our hands in the future.
2. Theme of Project Work
This group under the project work has undertaken the structural analysis and design of
multistoried commercial building. The main aim of the project work under the title is to acquire
knowledge and skill with an emphasis of practical application. Besides the utilization of analytical
methods and design approaches, exposure and application of various available codes of practices
is another aim of the work.
3. Objectives
The specific objectives of the project work are:
 Identification of structural arrangement of plan.
 Modeling of the building for structural analysis.
 Detail structural analysis using structural analysis program.
 Sectional design of structural components.
 Structural detailing of members and the system.
4. Scopes
 To achieve above objectives, the following scope or work is planned.
 Identification of the building and the requirement of the space.
 Determination of the structural system of the building to undertake the vertical
horizontal loads.
 Estimation of loads including those due to earthquake.
 Preliminary design for geometry of structural elements.
 Determination of fundamental time period by free vibration analysis.
 Calculation of base shear and vertical distribution of equivalent earthquake load.
 Identification of load cases and load combination cases.
 Finite element modeling of the building and input analysis.
 The structural analysis of the building by SAP2000 for different cases of loads.
 Review of analysis outputs for design of individual components.
 Detailing of individual members and preparation of drawings as a part of working
construction document.
and
5. Methodology
A) Study of the Architectural
Initially, the architectural drawing of the building was studied. Rooms within the building
were allocated to various purposes.
B) Preliminary Design
Estimation of various structural elements such as beam, column were designed and checks were
done with the help of deflection criteria and moment criteria. For the column, vertical axial
capacity was taken for the design and percentage of steel was checked.
C) Load Calculation
After the study of architectural drawing and preliminary design, load calculation was done. In
vertical direction, dead load was obtained by the size determined in preliminary design and live
load was determined by using code for design loads (IS 875 part II). In horizontal direction,
earthquake load was determined by calculating lumped mass at floor level for each frame and
then horizontal base shear was calculated by Seismic Coefficient Method (IS 1893:2002).
Earthquake load being pre-dominate between two lateral loads, its effect was only considered.
D) Modeling and Analysis:
The building is modeled as a space frame. SAP2000 is adopted as the basic tool for the execution
of analysis. SAP2000 program is based on Finite Element Method. Due to possible actions in the
building, the stresses, displacements and fundamental time periods are obtained using SAP2000
which is used for the design of the members: Lift wall, Staircase, Slabs are analyzed separately.
Initially, the characteristics of the materials used were defined such as concrete M20, M25 and
reinforcement- Fe500. Then, the load cases as well as their combination with load factors were
introduced. Structures were then analyzed for different load combinations and the final output
was determined in the form of Bending Moment, Shear Force, Torsion and Axial Force.
E) Design:
Design was done on the basis of limit state of design for collapse and serviceability. The sample
calculations of various structural elements were done with numerous checks and with the help of
MS-excel, the formulation was done for each and every structural member in the building.
The following materials are adopted for the design of the elements:
•
Concrete Grade: M25
•
Reinforcement Steel: Fe415
•
Limit state method is used for the design of RC elements. The design is based on
IS: 456-2000, SP-16, IS: 1893-2002.
F) Detailing:
Detailing was done by determining number, size, layout and location of reinforcement giving the
element dimensions and areas of steel required. Certain details such as lap and development
lengths, hook requirements, cut-off points etc. were covered by the code. The detailing is based
on SP-34 and IS: 13920-1993.
Time frame of our work completion:
Activities
Poush
1
1.Plan
2.Literature
Review
3.Plan
Modification
4.Load
Calculation
5.Analysis and
interpretation
of result
6.Detail design
7.Structural
drawing with
ductile drawing
2
Magh
3
4
1
2
Falgun
3
4
1
2
Chaitra
3
4
1
2
3
Baishak
4
1
2
3
Jestha
4
1
2
Asar
3
4
1
2
Shrawn
3
4
1
2
Bhadra
3
4
1
2
CHAPTER 2: ASPECTS OF SEISMIC PERFORMANCE AND BUILDING
DESCRIPTION
Nepal has witnessed several major disasters due to earthquakes in the
current decade. Earthquakes do not kill people but poorly designed or constructed
buildings do. These earthquakes have clearly brought out that we need to have a comprehensive
strategy for disaster mitigation which should include planning, design and construction of
arthquake resistant buildings through strict compliance of Codal provisions for earthquake
countermeasures. The general philosophy of earthquake resistant building design is that:
 Under minor but frequent
shaking,
the main members
of the building that carry
vertical and horizontal forces
should not be damaged;
however building parts that
do not carry load may
sustain repairable damage.
 Under moderate but occasional
shaking, The main members
may sustain repairable damage,
while the other parts of the building may be damaged such that they may even have to be
replaced after the earthquake.
 Under strong but rare shaking, the main members may sustain severe (even irreparable)
damage, but the building should not collapse.
2.1 Seismic Performance of building
Level of Performance of a building in an earthquake depends upon its overall configuration.
Generally it is common that an architect fixes the configuration i.e. shape size and geometry of a
building and the structural engineer adds the structural design. Contribution of building
configuration in seismic performance of building is rarely considered. It is a frequent mistake that
earthquake load consideration in structural design guarantees earthquake resistance of a
building regardless of the configuration. In this context the emphasis of Henry Degenkolb, a
prominent American structural engineer may be noteworthy. He stressed that: “If we have a poor
configuration to start with, all the engineer can do is to provide a band aid – improve a basically
poor solution as best he can. Conversely, if we start up with a good configuration and a
reasonable framing scheme, even a poor engineer can’t harm its ultimate performance too
much” The building configuration stated thereof that effect the seismic performance of building
are: Architectural shape and size. Type, size and location of structural elements.
2.2 Configuration Issues in Building
Plan of building

Size of Buildings: The horizontal movement of the floors during ground shaking is large in
tall buildings. In short but very long buildings, the damaging effects during earthquake
shaking are many. In buildings with large plan area, the horizontal seismic forces can be
excessive to be carried by columns and walls.

Symmetry: The building as a whole or its various blocks should be kept symmetrical about
both the axes. Asymmetry leads to torsion during earthquakes and is dangerous. Symmetry is
also desirable in the placing and
sizing of door and window openings,
as far as possible.

Regularity: Simple rectangular
shapes behave better in an
earthquake than shapes with
many projections. Torsional
effects of ground motion
are seen in long narrow
rectangular blocks. Therefore,
it is desirable to restrict
the length of a block to three
times its width. If longer
lengths are required two
separate blocks with sufficient
separation in between should be
provided.
Separation of Blocks: Separation of a large building into several blocks may be required so
as to obtain symmetry and regularity of each block. For preventing hammering or pounding
damage between blocks a physical separation is advisable.


Simplicity: Decoration invo1ving large cornices, vertical or horizontal cantilever projections
etc. are dangerous and undesirable from a seismic viewpoint. Simplicity is the best approach.
Where ornamentation is insisted upon, it must be reinforced with steel, which should
be properly embedded or tied into the main structure of the building.

Enclosed Area: A small building enclosure with properly interconnected walls acts like a
rigid box. Since the earthquake strength which long walls derive from transverse walls
increases as their length decreases. Therefore structurally it will be advisable to have
separately enclosed rooms rather than one long room.

Separate Buildings for Different Functions: In view of the difference in importance of
buildings, it may be economical to plan separate blocks for different functions so as to affect
economy in strengthening costs.
2.3 Vertical Configuration
The earthquake forces developed at different floor levels in a building need to be brought down
along the height to the ground by the shortest path. Any deviation or discontinuity in this load
transfer path results in poor performance of the building. In addition, all sections in load paths
should be detailed as ductile elements. Those parts of the load path that cannot be detailed as
ductile elements must be designed to resist their forces elastically. In other words, non-ductile
connections must be able to elastically resist forces greater than the maximum probable strength
of the elements framing into the connection. Setbacks, unequal building Element height,
discontinuous column, weak storey etc. are common examples of ill vertical configuration.
2.4 Structural Ductility
Ductility is the most desirable quality for good earthquake performance and can be
incorporated to some extent in brittle masonry constructions by introduction of steel reinforcing
bars at critical sections. Some materials are ductile, such as steel, wrought iron and wood while
others are not, such as cast iron, plain masonry, concrete, i.e. they break suddenly, without
warning. Brittle materials can be made ductile, usually by the addition of modest amounts of
ductile materials, Such as steel reinforcing in masonry and concrete constructions. For these
ductile materials to achieve a ductile effect in the overall behavior of the component, they must
be proportioned and placed so that they come in tension and are subjected to yielding. Thus, a
necessary requirement for good earthquake resistant design is to have sufficient ductile materials
at points of tensile stresses.
2.5 Structural Layout
When creating a frame building, structural member in regard to their stiffness are to be
uniformly distributed and these should be well framed up in both orthogonal directions with
nearly uniform spans. It is always advisable to provide stiffer elements such as walls or bracings
along the perimeter of the building rather than concentrating them in the center of the building,
whatever be the structural system. It results in enhanced torsional resistance of the building
giving it additional earthquake protection. It helps to maintain similar stiffness in both the
directions. An additional force Viz. Torsion emerges when the center of Gravity does not
coincides the center of stiffness.
2.6 Isolation of Structure
Beside traditional approach of resisting the Seismic forces, an alternative approach which is
presently emerging is to avoid these forces, by isolation of the structure from the ground
motions which actually impose the forces on the structure.
The concept of base isolation is explained through an example building resting on frictionless
rollers. When the ground shakes, the rollers freely roll, but the building above does not move.
Thus, no force is transferred to the building due to the shaking of the ground; simply, the building
does not experience the earthquake. Now, if the same building is rested on the flexible pads that
offer resistance against lateral movements (as shown in figure aside), then some effect of the
ground shaking will be transferred to the building above. The main feature of the base isolation
technology is that it introduces flexibility in the structure.
2.7 General Principles for the design
On starring at the overview of structural action, mechanism of damage and modes of failure of
buildings, we can come up with following considerations: 




Structures should not be brittle or collapse suddenly. Rather, they should be tough, able to
deflect or deform a considerable amount.
Resisting elements, such as bracing or shear walls, must be provided evenly throughout
the building, in both directions side-to-side, as well as top to bottom.
All elements, such as walls and the roof, should be tied together so as to act as an
integrated unit during earthquake shaking, transferring forces across connections and
preventing separation.
The building must be well connected to a good foundation and the earth. Wet, soft soils
Should be avoided, and the foundation must be well tied together, as well as tied to the
wall. Where soft soils cannot be avoided, special strengthening must be provided.


Care must be taken that all materials used are of good quality, and are protected from rain,
sun, insects and other weakening actions, so that their strength lasts.
Unreinforced earth and masonry have no reliable strength in tension, and are brittle
in compression. Generally, they must be suitably reinforced by steel or wood.
Adherence to above mentioned simple rules, a designer can give a structure that does not prevent
all damage in moderate or large earthquakes, but life threatening collapses can be prevented, and
damage limited to repairable proportions. These principles fall into several broad categories,
some of which are listed as under: 


Planning and layout of the building involving consideration of the location of rooms and
walls, openings such as doors and windows, the number of storeys, etc. At this stage, site
and foundation aspects should also be considered.
Layout and general design of the structural framing system with special attention to
furnishing lateral resistance
Consideration of highly loaded and critical sections with provision of reinforcement as
required.
CHAPTER 3: DESIGN METHOD
3.1 Understanding and Design Philosophy
3.1.1 Background
The aim of design is the achievement of an acceptable probability that structure being designed
will perform satisfactorily during their intended life. We are mainly dealing with seismic analysis
and structural design of RCC framed concrete structure. Structure and structural element shall
normally be designed by limit state method.
3.1.2 Design Philosophies
There are three philosophies for the design of reinforced concrete.
i.
ii.
iii.
Working stress method
Ultimate load method
Limit State method
Working stress method:
The working stress method of design is based on the behavior of structure at working load. The
stress distribution in concrete and steel at working load is assumed to be linear. Hence the design
is made by assuming the linear stress-strain relationship ensuring that the stresses in steel and
concrete do not exceed their permissible values at service load which is taken as the fixed
proportion of the ultimate or yield strength of the material.
Ultimate load method:
This is an aspect of limit design, which confines the structural usefulness up to the plastic strength
or ultimate load carrying capacity. This method is based on failure condition rather than working
load condition. In plastic design method, working load is multiplied by load factor and the cross
section of members is selected and design on the basis of collapse Strength.
Limit state method:
It is a judicious amalgamation of working stress method and ultimate load method, removing the
drawback of both of these methods but retaining their good points. In the method of design based
on the limit state concept, the structure shall be designed to withstand safely all loads liable to act
on it throughout its life; it shall also satisfied the serviceability requirements, such as limitation on
deflection and cracking should be based on characteristic value for materials strength and applied
load .The designed value are derived from characteristics value through, the use of partial factor
of safety for load and strength.
3.2 Compression Members
Limit state of collapse:
The limit state of collapse of structure could be assessed from rupture of one or more critical
section and from buckling due to elastic or plastic instability or overturning. The resistance
bending, shear, torsion and axial load at every section should not be less than the value at the
section produced by the most unfavorable combination of load using partial factor of safety.
Limit state of serviceability:
This state corresponds to the development of the excessive deformation and is use for checking
members in which magnitude of deformation may limit the use of the structure or its
component. This state may correspond to:
a) Deflection
b) Cracking
c) Vibration
3.3 Earthquake Resistance Design of Structure
Assumptions

Earthquake causes impulsive ground motions, which are complex and irregular in
character, changing in period and amplitude is lasting for a small duration. Therefore
resonance of the type as visualized under steady state sinusoidal excitations will not
occur as it would need time to build up such amplitudes.

Earthquake is not likely to occur simultaneously with wind or maximum flood or
maximum sea waves.

The value of elastic modulus of materials, wherever required, may be taken as for static
analysis unless a more definite value is available for use in such condition.
The criteria adopted by codes for fixing the level of the design seismic loading are generally as
follows:




Structure should be able to resist minor earthquake without damage.
Structure should be able to resist moderate earthquake without significant structural
damage, but with some non-structural damage and
Structure should be able to resist major earthquake without collapse but with some
structural as well as non- structural damage.
There are basically two methods to determine the earthquake force in the building:
o Seismic coefficient method or static method
o Dynamic method.
3.4 Assessment of Loads
a) Dead load
The correct assessment and calculation of dead loads is a most important step. This can be done
precisely if the architectural drawings are complete and include the roof, ceiling, floor, wall
finishes, parapet and railings, overhead water storage tanks placed on the roof, position thickness
and specifications of fixed partitions, etc
The correct thickness/size of structural member (i.e. slab, beams and columns) cannot be
ascertained before the structural analysis and design are finalized. Thus , some sizes need to be
assigned by experience and architectural considerations to begin with, checked and modified
during preliminary design, and finalized during analysis and checking. The dead load of each
member has been separately calculated as per IS 875 (part 1): 1987 for obtaining seismic weight
and compute design base shear and compare it with the actual base shear obtained from SAP2000.
b) Live load
These are to be chosen from codes as IS: 875(part 2) for carious occupancies where required.
These codes permit certain modifications in the load intensities where large contributory areas are
involved, or when the building consists of many stories. For economy in design, such reductions
should be utilized. Lateral and vertical loads on parapets and railings, and higher loads intensities
on entrance halls, stairs, must be considered. It will be useful to mark the design load classes or
intensities on small- scale.
c) Wind load
Wind pressure occurs on all exposed surfaces and acts normal to the surface. The intensities of the
pressure specified by the coefficient of wind pressure Cp in Is 875(part 3) depends on a large
number of parameter such as direction of wind relative to the axes of the building, its shape in
plan and elevation, and size of exposed individual elements. The terrain and building height, the
topography etc. IS 875(Part 3) is an attempt to codify the wind-tunnel perimental results by
considering the various parameters. It gives a very exhaustive treatment and thus may appear
complex for simple application to building of usual shape and size.
When adopting IS 875(Part 3) for Nepal, the wind velocity zoning of the country require will be
that applicable to 10m height, peak gust velocity averaged over a three second, and a return period
of 50 years.
d) Seismic load
Seismic weight is the total dead load plus appropriate amount of specified imposed load. While
computing the seismic load weight of each floor, the weight of columns and walls in any story
shall be equally distributed to the floors above and below the storey. The seismic weight of the
whole building is the sum of the seismic weights of all the floors. It has been calculated according
to IS: 1893(Part I) – 2002.
Seismic load or earthquake load on a building depends upon its geographical location, lateral
stiffness and mass, and is reversible. During an earthquake, the mass is imparted by the building
whereas the acceleration is imparted by the ground disturbance. In order to have minimum force,
the mass of the building should be as low as possible. The point of application of this inertial
force is the centre of gravity of the mass on each floor of the building.
This load on a structure is a function of the site dependent probable maximum earthquake
intensity or string ground- motion and the local soil, the stiffness, and its orientation in relation to
the incident seismic waves. For designing purpose, the resultant effects are usually represented by
the horizontal and vertical seismic coefficient αh, αv. Alternatively, a dynamics analysis of the
building is required under the action of the specified ground motion or design response spectra.
Since the probable maximum earthquake occurrences are not so frequent, designing building for
such earthquake so as to ensure that they remain elastic and damage- free is not considered
economically prudent. Instead, reliance is placed on kinetic energy dissipation in the structure
through plastic deformation of elements and joints. Thus, the philosophy of a seismic design is to
obtain no collapse of structure rather than a no damage of structure.
3.5 Study of the Building
Structural and Architectural Features of the Building
Type of the building
Commercial building
Location
Tinkune, Kathnmandu
Plinth Area
299.64 sq.m
Building Perimeter
99.01 m
Structural system
Moment Resisting Frame structure
Soil type
Medium Soil
Seismic zone
Zone V
Foundation
Mat Foundation
Number of storey
Basement + Ground floor + 5 Stories
Basement height
3 m
Ground storey height
3m
Floor to floor height
3m
Width of wall
230 mm
Types of loads
i. Dead Load
ii. Live Load as per IS 875 part II
iii. Earthquake induced load as per IS 1893
Analysis Tools
SAP2000 v16
Size of Beam
400 X 450 mm
Size of Column
550 X 550 mm
Depth of Slab
175 mm
Type of Staircase
Open well Staircase
Grade of Concrete
M25
Grade of Steel
Fe4150
No. of Beam from ground to
5th floor
24 nos.
No. of Beam in 6th floor
4 nos.
No. Column from basement to
5th floor
16 nos.
No. of Column in 6th floor
4 nos.
CHAPTER 4: PRELIMINARY ANALYSIS AND DESIGN
4.1 Introduction
This deals about preliminary analysis and design of structure. To begin with, the size of basic
structural members, unit weights, live loads, earthquake loads and wind loads are assessed. Based
on these dimensions and loads, the structure is analyzed and stresses on members are calculated.
From these calculated loads adequacy of the assumed sections is checked. In due course, a tradeoff is also identified for material type to be used in the structural members. For preliminary
purpose torsion analysis is not done.
4.2 Need of Preliminary Design
It is necessary to know the preliminary section of the structure for the detail analysis. As the
section should be given initially while doing analysis in almost all software, the need of
preliminary design is vital. Only dead loads and live loads are considered while doing preliminary
design. Preliminary design is carried out to estimate approximate size of the structural members
before analysis of structure.
4.3 Preliminary Design
4.3.1 Slab
Slab design are assumed to be monolithically casted with beams. Hence all slabs are designed as
continuous slabs as per end conditions defined in IS 456:2000.
Dimension of the biggest room=7.468m*4.953m
Shorter span=4.953m
Depth of slab,
Dmin =shorter span/(basic value*modification factor); IS 456:2000 Cl. 23.2.1
Where
; Modification factor=1.2
; Basic value=26; continuous beam
Provide depth, d=124.83mm <125mm , provide depth=125mm
So for 125mm secondary beam is not needed
4.3.2 BEAM
Main Beam
Maximum span of main beam in X- direction, L=4953mm
Depth of beam =L/18 =275.17mm
Clear cover =25.00mm
Overall depth, D=315.67mm
Provide overall depth, D=350mm
Provide width of beam, B=350/2=175mm
So B=200mm
4.3.3COLUMN
From Load Calculation Method:
Working load (P) = 2710.1475KN
Ultimate load (Pu) =1.5*P=4065.22KN
We know:
Pu=0.4Fck*Ac+0.67* Fy* Asc
Provide 2% steel
Size of column =514.44mm
Adopting greater size.
Provide sq column of 520mm*520mm
4.4 Calculation of Center of mass
Center of mass in x-dir. (Cmx) =
=
∑Mi
∑Mi∗Xi of slab+∑Mi∗Xi ofbeam+∑Mi∗Xi of column+∑Mi∗Xi of wall
∑Mi of slab+∑Mi of beam+∑Mi of column+∑Mi of wall
Center of mass in y-dir. (Cmy) =
=
∑Mi∗Xi
∑Mi∗Yi
∑Mi
∑Mi∗Yi of slab+∑Mi∗Yi ofbeam+∑Mi∗Yi of column+∑Mi∗Yi of wall
∑Mi of slab+∑Mi of beam+∑Mi of column+∑Mi of wall
Center of mass (Cmx,Cmy) = (7.207, 10.330)
4.5 Calculation of Center of stiffness
Center of stiffness
In x-dir.(Csx)=
In y-dir.(Csy)=
∑EIx∗Xi
∑EIx
∑EIy∗Yi
∑EIy
=
=
∑EIx∗Xi of column
∑EIx of column
∑EIy∗Yi of column
∑EIy of column
Center of stiffness (Csx, Csy) = (7.201, 10.573) m
Static eccentricity:
Static eccentricity in x-dir.(esx)= Cmx- Csx
= 7.207-7.201= 0.006m
Static eccentricity in y-dir.(esy)= Cmy- Csy = 10.573-10.33= 0.243m
Minimum eccentricity (emin) = 5% of smaller dimension of building.
=5/100*14.402 = 0.7201 m
4.6. Lateral load calculation (Seismic Load):
Seismic load:
Seismic weight is the total dead load plus approximate amount of specified superimposed
load. While computing the seismic weight of each floor, the weight of columns and walls in any
storey shall be equally distributed to the floors above and below the storey (equals to half the storey
height). The seismic weight of the whole building is the sum of the seismic weights of all the floors. It
has been calculated according to IS 1893: 2002 (Part I).
Seismic load or earthquake load on a building depends upon its geographical location, lateral
stiffness and mass, and is reversible. Its effect should be considered along both axes of building
taken at a time. A force is defined as the product of mass and acceleration. During an earthquake, the
mass is imparted by the building whereas the acceleration is imparted by the ground disturbance. In
order to have minimum force, the mass of the building should be as low as possible. The point of
application of this inertial force is the centre of mass of each floor of the building.
There are two methods to determine the earthquake force in a building:
1. Seismic coefficient method (static method)
2. Response spectrum method or modal analysis method of spectral acceleration method
(dynamic method)
The seismic coefficient method is generally applicable to buildings up to 40m in height and
those are more or less symmetrical in plan and elevation.
A building may be modeled as a series of 2D plane frames into orthogonal direction. Each
node will have 3 degrees of freedom: two translations and one rotation. Alternatively, a building may
be modeled as a 3D space frame. Each node will have 6 degrees of freedom: 3 translations and 3
rotations.
 Response Spectrum
The representation of the maximum response of idealized single degree of freedom system
having certain period of vibration and damping during earthquake is referred to as response
spectrum. The maximum response, i.e., maximum absolute acceleration, maximum velocity or
maximum relative displacement of the single degree of freedom system is plotted against the undamped natural period and for various damping values.
Calculation of FUNDAMNETAL natural period (Ta):
The approximate fundamental natural period of vibration (Ta) in seconds, of a Moment
resisting frame building panels may be estimated by the empirical expression:
Ta = 0.075h 0.75 (without Brick infill)
= 0.09h/√Ds (with Brick infill panels)
(IS 1893: 2002 Clause 7.6.1)
(IS 1893: 2002 Clause 7.6.2)
Where,
h = Height of building in m. This excludes the basement storey, where basement walls
are connected with the ground floor deck or fitted between the building columns but
it includes the basement storey, when they are not so connected.
Ds = Dimension of building in meter in a direction parallel to the applied earthquake
force
4.6.1. Calculation of base shear And Design lateral force:
Base shear (Vb) = Ah  W
(IS 1893: 2002 Clause 7.5.3)
Where, Ah = Design horizontal seismic coefficient
h = Height of building (m)
Ah 
Z I Sa
2R g
(IS 1893: 2002 Clause 6.4.2)
Where,
Z = zone factor as give
(IS 1893 (Part 1): 2002)
I = importance factor, depending upon the functional use of the
structure
R = response reduction factor
Sa/g = Average Response Acceleration
W = Seismic Weight of the building (KN) which include
a. Floor wise dead load consisting of weight of floor, beams, parapet, fixed
permanent equipment and half the walls and columns etc above and below.
b. Reduce live load on building (25 % of live load for live load ≤ 3 KN/m 2 and 50
% for live load > 3 KN/m2)
After calculating the base shear VB, the distribution of earthquake force on different
floor is determined as follows:
Wi hi
Qi =
 VB
2
 Wi hi
Where, Qi = horizontal force acting at any floor i
Wi = weight of ith storey assumed to be lumped at ith floor
hi = height of ith floor above base of frame
Calculation:
Storey height = 3 m
W = seismic weight of building = 25152.885 KN (from table of Lateral Load Calculation)
Z = Zone factor
= 0.36
(From Table 2, IS 1893: 2002 Clause 6.4.2)
For zone V (very severe)
I = Importance factor = 1.0
(Table 6, s1 No. 1( i ), IS 1893: 2002 Clause 6.4.2)
R = Response reduction factor = 5.0 for Zone V
Assuming the frame to be special moment resisting (table 7)
For medium soil sites
1  15T 0.00  T  0.10

Sa 
  2.5
0.10  T  0.55
g  1.36
0.55  T  4.00
T


The fundamental time period of the vibration
Ta = 0.075 h0.75 (assuming no brick infill faces)
(IS 1893: 2002 Clause 7.6.1)
h = 3  7 =21 m
Ta = 0.736sec
The fundamental time period of vibration, Ta=0.736 sec
Hence,
Sa
 1.36 / Ta  1.848
g
 Ah   ZIS a

0.36  1  1.364
= 0.0665
25
2 Rg
Now, seismic base shear
(IS 1893: 2000 Clause 6.4.2)
Vb    Ah   W
= .0665 x 25152.885 =1672.67 KN
Base Shear =1672.67 KN
4.6.2 LUMPED MASS CALCULATION:
LUMPED MASS CALCULATION
FLOOR
SLAB
BEAM
Ground 1310.77 554.480
COLUMN
WALL
MASS(KN)
363.000
915.787
3144.037
1st
1310.77 554.480
363.000
1831.330
4059.580
2nd
1310.77 554.480
363.000
1831.330
4059.580
3rd
1310.77 554.480
363.000
1831.330
4059.580
4th
1310.77 554.480
363.000
1831.330
4059.580
5th
1310.77 554.480
226.875
1071.396
3163.521
45.375
155.609
442.101
cover
146.9
94.217
4.6.3 BASE SHEAR DISTRIBUTION:
Base Shear (Vb) =1673.370 KN
Distribution of Base Shear:((Wi*hi2)/sum(Wi*hi2))*Vi
Level
7
6
5
4
3
2
1
Wi (KN)
442.100
3163.519
4435.623
4435.623
4435.623
4435.623
3804.774
hi (m)
21.000
18.000
15.000
12.000
9.000
6.000
3.000
Wi*hi2
194966.100
1024980.156
998015.175
638729.712
359285.463
159682.428
34242.966
3409902.000
Qi
95.677
502.997
489.764
313.449
176.315
78.362
16.804
Vi
95.677
598.675
1088.439
1401.888
1578.203
1656.566
1673.370
Chapter: 5
STRUCTURAL ANALYSIS
5.1 Salient feature of SAP2000
SAP2000 represents the most sophisticated and user-friendly release of SAP series of
computer programs. Creation and modification of the model, execution of the analysis, and
checking and optimization of the design are all done through this single interface. Graphical
displays of the results, including real-time display of time-history displacements are easily
produced.
The finite element library consists of different elements out of which the three
dimensional FRAME element was used in this analysis. The Frame element uses a general, threedimensional, beam-column formulation which includes the effects of biaxial bending, torsion,
axial deformation, and biaxial shear deformations.
Structures that can be modeled with this element include:
• Three-dimensional frames
• Three-dimensional trusses
• Planar frames
• Planar grillages
• Planar trusses
A Frame element is modeled as a straight line connecting two joints. Each element has its
own local coordinate system for defining section properties and loads, and for interpreting output.
Each Frame element may be loaded by self-weight, multiple concentrated loads, and multiple
distributed loads. End offsets are available to account for the finite size of beam and column
intersections. End releases are also available to model different fixity conditions at the ends of the
element. Element internal forces are produced at the ends of each element and at a user-specified
number of equally-spaced output stations along the length of the element.
Loading options allow for gravity, thermal and pre-stress conditions in addition to the
usual nodal loading with specified forces and or displacements. Dynamic loading can be in the
form of a base acceleration response spectrum, or varying loads and base accelerations.
1.1.
Load combinations
Different load cases and load combination cases as per IS 875 are considered to obtain most
critical element stresses in structure in the course of analysis. There are altogether six load cases
considered for the structural analysis and are mentioned as below:
i.
Dead load (DL)
ii.
Live load (LL)
iii.
Earthquake load in X-direction (EQx)
iv.
Earthquake load in Y-direction (EQy)
v.
Response spectrum in X-direction (RSx)
vi.
Response spectrum in Y-direction (RSy)
Following load combinations are adopted:
1. COMB1 = 1.5 DL
2. COMB2 = l.5 ( DL + LL)
3. COMB3 = 1.2 ( DL + LL + EQx)
4. COMB4 = 1.2 ( DL + LL – EQx)
5. COMB5 = l.2 ( DL +LL+ EQy)
6. COMB6 = 1.2 ( DL+LL – EQy)
7. COMB7 = 1.5 ( DL + EQx)
8. COMB8 = l.5 ( DL –EQx)
9. COMB9 = l.5 ( DL + EQy)
10. COMB10 = l.5 ( DL – EQy)
11. COMB11 = 0.9 DL + 1.5 EQx
12. COMB12 = 0.9 DL – 1.5 EQx
13. COMB13 = 0.9 DL + 1.5 EQy
14. COMB14 = 0.9 DL – 1.5 EQy
15. COMB15= 1.5 DL
16. COMB16= l.5 ( DL + LL)
17. COMB17= 1.2 ( DL + LL + RSx)
18. COMB18= l.2 ( DL + LL+ RSy)
19. COMB19= 1.5 ( DL + RSx)
20. COMB20= 1.5 ( DL + RSy)
21. COMB21= 0.9 DL + 1.5 RSx
22. COMB22=0.9 DL + 1.5 RSy
23. Envelope (Max. of all combinations)
5.2. SAP application procedure
 Construction of structural framework (Grid model) according to given drawing
 Assigning size, material, name etc. of the members used in the structure
 Loading all the vertical loads to every horizontal member according to their location,
whether it is point load, triangular, trapezoidal or uniformly distributed (rectangular load)
 Assigning calculated earthquake forces with their magnitude at particular location (Centre
of mass) to each floor level in both X and Y directions. For this slab is considered as rigid
by constructing diaphragm at different floor level
 Fix the structure to the ground
 Run the program
 Check each and every elements and verify if they safely pass through assigned loads
 Modification of geometry of elements, property of materials, reinforcement in case of
inadequacy of previous ones
 Export the required diagrams and data from SAP
Filtering of the data by making small programs
5.3. Analysis Features
The CSI analysis engine offers the following features;
1. Static and dynamic analysis
2. Linear and non-linear
3. Dynamic seismic analysis and static push over analysis
4. Vehicle live load analysis for bridge
5. Geometric non linearity, including P-delta and large- displacement effects
6. Staged (incremental) construction
7. Creep, shrink age and aging effects
8. Buckling analysis
9. Steady state and power-spectral-density analysis
10. Frame and shell structural elements, including beams, column truss, membrane
and plate behavior.
11. Two dimensional plane and axis symmetrical solid elements
12. Three dimensional solid elements
13. Non linear link and support elements
14. Frequency –dependent link and support properties
15. Multiple co-ordinates systems
16. Many types of constraints
17. Wide verity of loading options
The following general steps are required to analyze and design a structure using SAP200:
1. Create or modify a model that numerically defines the geometry, properties, loading, and
analysis parameters for the structure.
2. Perform an analysis of the model.
3. Review the results of the analysis.
4. Check and optimize the design of the structure.
Constraints:
Constraints are used to enforce certain types of rigid-body behavior, to connect together
different parts of the model, and to impose certain types of symmetry conditions. A constraint
consists of a set of two or more constrained joints. The displacements of each pair of joints in the
constraint are related by constraint equations. The types of behavior that can be forced by
constraints are:
1. Rigid body behavior, in which the constrained joints translate and rotate together as if
connected by links. The types of rigid behavior that can be modeled are:
a. Rigid Body: fully rigid for all displacements.
b. Rigid Diaphragm: rigid for membrane behavior in plane.
c. Rigid plate: rigid of plate bending in a plane.
d. Rigid rod: rigid for extension along an axis.
e. Rigid beam: rigid for beam bending on an axis.
2. Equal displacement behavior, in which the translations and rotations are equal at the
constrained joints.
3. Symmetry and anti – symmetry conditions:
The use of constraints reduces the number of equations in the system to be solved and
will usually result in increased computational efficiency. Most constraints types must
be defined with respect to some fixed co -ordinate system. The co ordinate system may
be the global co ordinate system or an alternate co ordinate system, or it may be
automatically determined from the location of the constrained joints. The local
constraint does not use a fixed co ordinate system, but references each joints using its
own local co ordinate system.
Body constraint:
A body constraint causes all of its constrained joints to move together as a three
dimensional rigid body. By default, all degree freedom at each connected joint
participates. However, you can select a sub set of the degrees of freedom to be
constrained.
This constraint can be used to:
1. Model rigid connections, such as where several beams and column or frame
together.
2. Connect together different parts of the structural model that were defined using
separate meshes.
3. Connect frame elements that are acting as eccentric stiffeners to shell elements.
5.4. Inputs and Outputs
The design of earthquake resistant structure should aim at providing appropriate dynamic
and structural characteristics so that acceptable response level results under the design earthquake.
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 misuse and fire.
For the purpose of seismic analysis of our building we used the structural analysis
program SAP2000. SAP2000 has a special option for modeling horizontal rigid floor diaphragm
system.
A floor diaphragm is modeled as a rigid horizontal plane parallel to global X-Y plane, so
that all points on any floor diaphragm cannot displace relative to each other in X-Y plane.
This type of modeling is very useful in the lateral dynamic analysis of building. The base shear
and response spectra are calculated as per code IS 1893(part1)2002 and response spectra is
implemented in SAP2000 for analysis and design
SOME OUTPUT DIAGRAM FROM SAP:
A) 3D MODEL
Fig: Axial Force Diagram
Fig: Shear Force Diagram
Fig:Deformed structure
Fig: Bending Moment Diagram
CHAPTER: 6
STRUCTURAL DESIGN
The structure should be designed in such a way that it fulfills the targeted requirement
throughout its life. The objective of structural design is to design such kind of building that
gives complete resonance with Safety (in terms of Strength, Stability and Structural
integrity), adequate Serviceability (in terms of deflection and crack) and economy.
It is necessary that reinforced concrete structure should satisfy the Serviceability limit
state, i.e. if a section is of sufficient Strength to support the design loads, there should not be
excessive deformation, deflection, cracking etc., which may affect its appearance. Safety
implies that the likelihood of (partial or total) collapse of the structure is acceptably low not
only under the normal expected loads (service load) but also under abnormal but probable
overloads (such as earthquake or extreme wind). The objective here is to minimize the
likelihood of progressive collapse.
But through the increment of design margins we can resist the problem regarding
structural failure but at the same time cost also increases with the increase in design margins
for Safety and Serviceability. So, considering overall economy the cost associated with
increased Safety and Serviceability should be weighed against the potential losses and the best
cost is selected.
Limit state method of design:
Limit state is the state of impending failure beyond which structure ceases to perform its
intended function satisfactorily in terms of either safety or serviceability. The limit state method of
design is based on the behavior or structure at different limit state insuring adequate Safety against
each limit state. The two principle limit states are the ultimate limit state and the Serviceability limit
state. The ultimate limit state is reached when the structure collapses. It requires that structure
must withstand the load for which it is designed with adequate factor of Safety without collapse.
The limit state of Serviceability corresponds to development of excessive deformation and is
used for checking members in which magnitude of deformation may limit the use of structure or its
components .The limit state corresponds to deflection, cracking and vibration. It requires that the
appearance, durability and of structure must not be effected by deflection and cracking.
i.
ii.
iii.
iv.
v.
Assumptions for flexure members ( IS 456:2000, clause 38.1)
Plane sections normal to the axis of the member remain plane after bending.
The maximum strain in concrete at the outer most compression fiber is 0.0035.
The relationship between the compressive stress distribution in concrete and the strain in
concrete may be assumed to be rectangle, trapezoid, parabola or any other shape which
results in prediction of strength in substantial agreement with the results of text. For
design purposes, the compressive strength of concrete in the structure shall be assumed to
be 0.67 times the characteristics strength .The partial Safety fact γm = 1.5 shall be applied
in addition to this .
The tensile strength of concrete is ignored.
The stresses in the reinforcement are derived from representative stress-strain curve for
the type of steel used. For the design purposes the partial Safety factor γ m = 1.15 shall be
applied.
vi.
The maximum strain in the tension reinforcement in the section at failure shall not be less
than:
i. 0.002+0.87 fy/Es
Where, fy = characteristics strength of steel Es =modulus of elasticity of steel
Limit state of collapse for compression
(IS 456:2000, clause 39.1)
Assumptions:
In addition to the assumptions given above from (a) to (e) for flexure, the following shall be
assumed:
a. The maximum compressive strain in concrete in axial compression is taken as 0.002
b. The maximum compressive strain at the highly compressed extreme fiber in concrete subjected
to axial compression and bending and when there is no tension on the section shall be 0.0035
minus 0.75 times the strain at the least compressed extreme fiber.
The limiting values of the depth of neutral axis for different grades of steel based on the
assumptions are as follows:
Fy
xu,max
250
0.53d
415
0.48d
500
0.46d
Materials adopted in our design:
Concrete M25
Reinforcement Fe 415
5.3. Slab
Slabs are plate elements forming floors and roofs of building and carrying distributed loads
primarily by flexure. Inclined slabs may be used as ramps for multistory car parks. Soffit of
staircases can be considered as inclined slabs. A slab may be supported by beams or walls or
continuous over one or more supports. One-way Slabs are those in which the length is more than
twice the breadth. A one way slab can be simply supported or continuous.When slabs are those
supported on four sides, two way spanning action occurs. Such slabs may be simply
supported or continuous on any or all sides. A two way slab may be considered to consist of
a series of interconnected beams.
Flow chart of slab design:
Determine factored
load w=1.5(DL+LL)
w D = 1.5DL
w L =1.5LL
Determine ratio l y /lx
No
if l y/lx <2
One way slab
Yes
Determine moment coefficient
IS code 456, Table 12
Two way slab
Determine type of panel
e.g. Two adjacent cont. edge
Determine moment
coefficients, IS code 456,
Table 26, e.g. long, short span,
edge, mid
Calculate moment at mid, edge
M=MD+ML
MD= α D wl x 2
ML= αL wl x 2
A s >A
Calculate M x =α x wl x 2
M y = αy wl x 2
Calculate area of steel A
M = 0.87 f Y A s (d - f Y A s
t
t
s
t fck
r
b)
/
Determine spacing of bars
S v = A ba /A gross ×1000
st min
t
=0.12%bD
S v < 300mm
or 3d
Calculation of slab:
Slab Panel S1
1. Design data
Clear span
Edge condition
Material
Concrete grade
:
:
:
:
two adjacent edge continuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4953 + 140mm or 4953 + 230/2 + 230/2
= 5093 mm or 5183 mm
Since 5093 < 5183,
Adopt lxe = 5093mm
lye = 7010+ 130 or 7010+ 230/2 + 230/2
= 7140 mm or 7240 mm
Since7140 < 7240
Adopt lye = 7140 mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 4 KN/m2 * 1 = 4 KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 9.49 KN/ m
Factored load (Wu) = 1.5 × 9.49 = 14.235 KN/ m
Moment coefficient from IS 456: 2000 Table 26
l ey
Long span to short span ratio,
= 1.40(Two way slab)
l ex
For negative moment: (x) = 0.071
(y) = 0.047
For positive moment: (x) = 0.053
(y) = 0.035
Moment Calculation:
At mid span
Mx = (x)×w×lx2 = 0.053 × 14.235 × 5.0932 = 19.57 KNm /m
My = (y)×w×lx2 = 0.035 × 14.235 × 5.0932 =12.92 KNm/m
At edge
Mx = (x)×w×lx2 = 0.071 × 14.235 × 5.0932 = 26.22KNm/m
My = (y)×w×lx2 = 0.047 × 14.235× 5.0932 = 17.35 KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
Mx
d
0.138  f ck  b
26.22 1000 1000
0.138  25 1000
= 87.18mm < 140mm O.K.
d
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
 b  d  f ck 

Astx  415 
19.57  106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 406.78mm2
B) Calculation of reinforcement in long (Y) direction:

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



A st y  415 

12.92  106  0.87  415  A st y  115  1 
 1000  115  25 


Solving quadratic equation
(Ast) y = 285.69 mm2
Similarly, as above,
At Edge
(Ast)x = 555.29 mm2
(Ast)y = 388.97 mm2
Minimum Reinforcement
0.12
=
 1000×175 = 210 mm2 <(Ast) x , (Ast)y (OK)
100
(IS 456: 2000 clause 26.5.2.1)
1000
1000
= π ×102/4×
Ast
285.69
= 274.94 mm <3d = 3×130 = 390 mm or 300 mm (Smaller)
Therefore take Spacing as 250 mm.
Provide 10mm bar @ 250 mm c/c spacing giving total Area = 314.16 mm2.
Spacing required
= π ×102/4×
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 150mm c/c, Area = 523.6mm2.
Reinforcement in Y- direction 10 mm bars @ 150mm c/c, Area= 523.6 mm2.
b) At Edge:
Reinforcement in X And Y- direction 10 mm bars @ 130mm c/c spacing giving total
Area = 604.15 mm2
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   14.235  5.093 = 36.23KN
2
36.23  10
V
v  u 
= 0.259 N/mm2
b  d 1000  140
3
Shear strength of concrete is given by,
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
604.15
= 0.43%
1000  140
Design shear strength for 0.43% steel and M25 concrete from IS 456: 2000 Table19,
c = 0.454 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K x c = 1.25 x 0.454 = 0.568N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Moment of resistance offered by 10mm bars @ 130mm c/c

A st  f y 

M1  0.87  f y  A st  d  1 
 b  f ck  d 

604.15  415 
M1  0.87  415  302.1*140 * 1 

 1000  140  25  2 
= 14.72  10 6 N-mm
M1
+Lo
V
 σs
Also development length L d =
4 τ bd
Development length, Ld = 1.3
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000 page 44)
(IS 456: 2000 Clause 26.2.1.1)
τbd = 1.4 N/mm2
σ s = 0.87 x 415 MPa
14.72  106
 σs
= 1.3
+80
36.23  103
4 τ bd
47.011 x ϕ = 608.18mm
 = 12.94 mm > 10 mm
Check for Deflection
L
= 
d
A
Pt  100  st
bd
(IS 456: 2000, Clause 26.2.3.3)
(IS 456: 2000, Clause 26.2.3.3 c)
OK.
(IS 456: 2000, Clause 23.2.1)
604.15
= 0.43%
1000  140
A Required
f s  0.58  f y  st
A st Provided
555.29
f s  0.58  415 
604.15
= 221.23 KN
Pt  100 
Value of Coefficients from Code IS 456: 2000, Clause 24.1
 = 26
(For continuous slab, IS: 456-2000, Clause 24.1(a))
 =1.5
(IS 456: 2000, Fig 4)
=1, =1, =1
(IS 456: 2000, Clause 23.2.1(b, c, d))
L
We have,
= 
d min
Lx
dmin =
αβγλδ
5093
dmin =
= 130.59 < 175mm
Hence safe in Deflection.
26  1  1.5  1  1
For Temperature bar (for Ast min):
1000
1000
Spacing required = π ×82/4×
= π ×82/4×
=335 or 450 mm (smaller)
A st
150
(IS 456: 2000, Clause 26.3.3.b.2)
Provide 8mm bar @ 300 mm c/c spacing giving total Area = 167.551 mm2/m
Slab Panel S2
1. Design data
Clear span
Edge condition
Material
Concrete grade
:
:
:
:
one short edge discontinuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4496 + 140mm or 4496 + 230/2 + 230/2
= 4636 mm or 4726 mm
Since 4636 < 4726,
Adopt lxe = 4636mm
lye = 7010+ 130 or 7010+ 230/2 + 230/2
= 7140 mm or 7240 mm
Since7140 < 7240
Adopt lye = 7140 mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 3 KN/m2 * 1 = 3 KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 8.49 KN/ m
Factored load (Wu) = 1.5 × 8.49 = 12.735 KN/ m
Moment coefficient from IS 456: 2000 Table 26
l ey
Long span to short span ratio,
= 1.54(Two way slab)
l ex
For negative moment: (x) = 0.0581
(y) = 0.037
For positive moment: (x) = 0.0446
(y) = 0.028
Moment Calculation:
At mid span
Mx = (x)×w×lx2 = 0.0446× 12.735 × 4.6362 = 12.21 KNm /m
My = (y)×w×lx2 = 0.028 × 12.735 ×4.6362 =7.66 KNm/m
At edge
Mx = (x)×w×lx2 = 0.0581× 12.735 × 4.6362 = 15.90KNm/m
My = (y)×w×lx2 = 0.037× 12.735 × 4.6362 = 10.13KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
Mx
d
0.138  f ck  b
15.90 1000 1000
0.138  25 1000
= 67.88mm < 140mm O.K.
d
Minimum Reinforcement
0.12
 1000×175 = 210 mm2 .
=
100
(IS 456: 2000 clause 26.5.2.1)
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
 b  d  f ck 
For ‘x’ +ve

Astx  415 
12.21  106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 248.90 mm2
For -ve

Astx  415 
15.90 KNm/m  0.87  415  Astx  140  1 

 1000  140  25 
Solving this eqn,
(Ast)x =327.26mm2
B) Calculation of reinforcement in long (Y) direction:
For +ve

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



A st y  415 

7.66  106  0.87  415  A st y  130  1 
 1000  130  25 


Solving quadratic equation
(Ast) y = 166.75 mm2
Adopt +ve (Ast) y = 210 mm2
For –ve,

Ast y  415 

10.13  106  0.87  415  Ast y  115  1 
 1000  130  20 


Solving
(Ast) y =222.12mm2
Spacing required:
For x
+ve
Spacing = π ×102/4×
1000
248.9
=315.55mm
Adopt spacing 250mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=261.67mm
250
spacing
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
327.26
=240mm
Adopt spacing 200mm,
Actual area= π ×102/4×
For y
1000
1000
= π ×102/4×
=392.7mm
spacing
200
+ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
210
= 373.81 mm > 3d = 3×140 = 420 mm or 300 mm (Smaller)
Therefore take Spacing as 300 mm.
1000
1000
Actual area= π ×102/4×
= π ×102/4×
=261.67mm
300
spacing
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
222.12
=353.6mm
Adopt spacing 300mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=261.67mm
300
spacing
Provide 10 mm bar @ 300 mm c/c spacing giving total Area = 261.67mm
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 250mm c/c, Area = 261.67mm2.
Reinforcement in Y- direction 10 mm bars @ 300mm c/c, Area= 261.67mm2.
b) At Edge:
Reinforcement in X And Y- direction 10 mm bars @ 200mm c/c spacing giving total
Area = 392.7mm2 and 300mm c/c area=261.67mm2 respectively.
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   12.735  4.636 = 29.52KN
2
29.52  10
V
v  u 
= 0. 211N/mm2
1000  140
bd
3
Shear strength of concrete is given by,
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
392.7
1000  140
= 0.281%
Design shear strength for 0.281 % steel and M25 concrete from IS 456: 2000 Table19,
c = 0.376 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K×c = 1.25×0.376 = 0.47 N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Vu = 29.52 KN
Moment of resistance offered by 10 mm bars @ 300 mm c/c

A st  f y 

M1  0.87  f y  A st  d ' 
b

f
ck 


196.35  415 
M1  0.87  415  196.35140 

1000  25 

= 9.69  10 6 N-mm
Ld = 1.3
M1
+Lo
V
(IS 456; 2000, Clause 26.2.3.3)
Development length L d =  σs
4 τ bd
Assuming Lo = 4  +4  =8 
So L0=8*10=80mm
τbd = 1.6*1.2 N/mm2
σs = 0.87 x 415 MPa
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000, page 44)
(IS 456: 2000, clause 26.2.1.1)
M1
+Lo
V
9.69  106
 σs
= 1.3
+80
29.52  103
4 τ bd
Now, Ld =1.3
47.011 x ϕ = 506.72
 = 10.77 mm > 10 mm
(IS 456: 2000, Clause 26.2.3.3 c)
OK.
Check for Deflection
L
= 
d
Pt  100 
Ast
bd
 100 
(From IS 456: 2000, Clause 23.2.1)
392.7
1000  140
= 0.281%
A st Required
A st Provided
327.26
f s  0.58  415 
392.7
= 200.6 KN
f s  0.58  f y 
Value of Coefficients from Code
 = 26
 =1.75
=1, =1, =1
L
We have,
= 
d min
4636
L/d=
140
IS 456: 2000, Clause 24.1
(For continuous slab, IS: 456 2000 Clause 24.1(a))
(IS 456: 2000, Fig 4)
(IS 456: 2000, Clause 23.2.1(b, c, d))
=33.11
=26*1*1.75*1=45.5mm
Provide 10mm bar @ 300 mm c/c spacing giving total Area =261.67mm2/mm
Slab Panel S3
1. Design data
Clear span
Edge condition
Material
Concrete grade
:
:
:
:
Two Adjacent edge discontinuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4953 + 140mm or 4953 + 230/2 + 230/2
= 5093 mm or 5183 mm
Since 5093 < 5183,
Adopt lxe = 5093mm
lye = 7010+ 130 or 7010+ 230/2 + 230/2
= 7140 mm or 7240 mm
Since7140 < 7240
Adopt lye = 7140 mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 3 KN/m2 * 1 = 3 KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 8.49 KN/ m
Factored load (Wu) = 1.5 × 8.49 = 12.735 KN/ m
Moment coefficient from IS 456: 2000 Table 26
l ey
Long span to short span ratio,
= 1.40(Two way slab)
l ex
For negative moment: (x) = 0.071
(y) = 0.047
For positive moment: (x) = 0.053
(y) = 0.035
Moment Calculation:
At mid span
Mx = (x)×w×lx2 =0.053 × 12.735 × 5.0932 = 17.51 KNm /m
My = (y)×w×lx2 = 0.035 × 12.735 ×5.0932 =11.56 KNm/m
At edge
Mx = (x)×w×lx2 = 0.071× 12.735 × 5.0932 = 23.45KNm/m
My = (y)×w×lx2 = 0.047× 12.735 × 5.0932 = 15.53KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
d
Mx
0.138  f ck  b
23.45  1000  1000
0.138  25  1000
=82.44mm < 140mm O.K.
d
Minimum Reinforcement
0.12
=
 1000×175 = 210 mm2 .
100
(IS 456: 2000 clause 26.5.2.1)
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
b

d

f
ck


For ‘x’ +ve

Astx  415 
17.51 106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 361.94 mm2
For -ve

Astx  415 
23.45 *106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this eqn,
(Ast)x =492.71mm2
B) Calculation of reinforcement in long (Y) direction:
For +ve

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



Ast y  415 

11.56  106  0.87  415  Ast y  130  1 
 1000  130  25 


Solving quadratic equation
(Ast) y = 254.57 mm2
Adopt +ve (Ast) y = 210 mm2
For –ve,

A st y  415 

15.53  106  0.87  415  A st y  115  1 
 1000  130  20 


Solving
(Ast) y =346.17mm2
Spacing required:
For x
+ve
Spacing = π ×102/4×
1000
361.94
=217mm
Adopt spacing 150mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
492.71
=159.40mm
Adopt spacing 150mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
For y
+ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
254.57
= 308 mm < 3d = 3×130 = 390 mm or 300 mm (Smaller)
Therefore take Spacing as 300 mm.
1000
1000
Actual area= π ×102/4×
= π ×102/4×
=261.67 mm2
300
spacing
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
346.17
=226.88mm
Adopt spacing 200mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
Provide 10 mm bar @ 300 mm c/c spacing giving total Area = 261.67mm
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 150mm c/c, Area = 523.6mm2.
Reinforcement in Y- direction 10 mm bars @ 300mm c/c, Area= 261.67mm2.
b) At Edge:
Reinforcement in X And Y- direction 10 mm bars @ 150mm c/c spacing giving total
Area = 523.6mm2 and 200mm c/c area=392.7mm2 respectively.
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   12.735  5.093 = 32.43KN
2
32.43  103
Vu

v 
Shear strength of concrete is given by,
= 0. 2316N/mm2
b  d 1000  140
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
523.6
= 0.374%
1000  140
Design shear strength for 0.374% steel and M25 concrete from IS 456: 2000 Table19,
c = 0.425 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K×c = 1.25×0.425 = 0.531 N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Vu = 32.43 KN
Moment of resistance offered by 10 mm bars @ 300 mm c/c

A st  f y 

M1  0.87  f y  A st  d ' 
b

f
ck



261.8  415 
M1  0.87  415  261.8140 

1000  25 

= 12.82  10 6 N-mm
Ld = 1.3
M1
+Lo
V
(IS 456; 2000, Clause 26.2.3.3)
Development length L d =  σs
4 τ bd
Assuming Lo = 4  +4  =8 
So L0=8*10=80mm
τbd = 1.6*1.2 N/mm2
σs = 0.87 x 415 MPa
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000, page 44)
(IS 456: 2000, clause 26.2.1.1)
M1
+Lo
V
12.82  106
 σs
= 1.3
+80
32.43  103
4 τ bd
Now, Ld =1.3
(IS 456: 2000, Clause 26.2.3.3 c)
47.011 x ϕ = 593.90
 = 12.63 mm > 10 mm
OK.
Check for Deflection
L
= 
d
Pt  100 
Ast
bd
 100 
(From IS 456: 2000, Clause 23.2.1)
523.6
= 0.374%
1000  140
A st Required
A st Provided
492.71
f s  0.58  415 
523.6
2
= 226.5 N/mm
f s  0.58  f y 
Value of Coefficients from Code
 = 26
IS 456: 2000, Clause 24.1
(For continuous slab, IS: 456 2000 Clause 24.1(a))
 =1.62
=1, =1, =1
L
We have,
= 
d min
5093
L/d=
140
(IS 456: 2000, Fig 4)
(IS 456: 2000, Clause 23.2.1(b, c, d))
=36.38
=26*1*1.62*1=42.12mm
L/d< Hence safe in deflection.
Slab Panel S4
1. Design data
Clear span
Edge condition
Material
Concrete grade
:
:
:
:
Two Adjacent edge discontinuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4953 + 140mm or 4953 + 230/2 + 230/2
= 5093 mm or 5183 mm
Since 5093 < 5183,
Adopt lxe = 5093mm
lye = 7468+ 130 or 7468+ 230/2 + 230/2
= 7598 mm or 7698 mm
Since7598 < 7698
Adopt lye = 7598 mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 2.5 KN/m2 * 1 = 2.5KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 7.99 KN/ m
Factored load (Wu) = 1.5 × 7.99 = 11.985 KN/ m
Moment coefficient from IS 456: 2000 Table 26
l ey
Long span to short span ratio,
= 1.49(Two way slab)
l ex
For negative moment: (x) = 0.0666
(y) = 0.037
For positive moment: (x) = 0.0506
(y) = 0.028
Moment Calculation:
At mid span
Mx = (x)×w×lx2 =0.0506× 11.985 × 5.0932 = 15.73 KNm /m
My = (y)×w×lx2 = 0.028 × 11.985 ×5.0932 =8.704 KNm/m
At edge
Mx = (x)×w×lx2 = 0.0666× 11.985 × 5.0932 = 20.704KNm/m
My = (y)×w×lx2 = 0.037× 11.985 × 5.0932 = 11.502KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
Mx
d
0.138  f ck  b
d
20.70 1000 1000
0.138  25 1000
=77.47mm < 140mm O.K.
Minimum Reinforcement
(IS 456: 2000 clause 26.5.2.1)
=
0.12
 1000×175 = 210 mm2 .
100
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
 b  d  f ck 
For ‘x’ +ve

Astx  415 
15.73  106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 323.61 mm2
For -ve

Astx  415 
20.704 *106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this eqn,
(Ast)x =431.7mm2
B) Calculation of reinforcement in long (Y) direction:
For +ve

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



Ast y  415 

8.704  106  0.87  415  A st y  130  1 
 1000  130  25 


Solving quadratic equation
(Ast) y = 190.05 mm2
Adopt +ve (Ast) y = 210 mm2
For –ve,

Ast y  415 

11.502  106  0.87  415  Ast y  115  1 
 1000  130  20 


Solving
(Ast) y =253.24mm2
Spacing required:
For x
+ve
Spacing = π ×102/4×
1000
323.61
=242.70mm
Adopt spacing 200mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
431.7
=181.93mm
Adopt spacing 150mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
For y
+ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
210
= 374 mm < 3d = 3×130 = 390 mm or 300 mm (Smaller)
Therefore take Spacing as 300 mm.
1000
1000
Actual area= π ×102/4×
= π ×102/4×
=261.67 mm2
300
spacing
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
253.54
=309.77mm
Adopt spacing 300mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=261.8 mm2
spacing
300
Provide 10 mm bar @ 300 mm c/c spacing giving total Area = 261.67mm
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 200mm c/c, Area = 392.7mm2.
Reinforcement in Y- direction 10 mm bars @ 300mm c/c, Area= 261.67mm2.
b) At Edge:
Reinforcement in X And Y- direction 10 mm bars @ 150mm c/c spacing giving total
Area = 523.6mm2 and 300mm c/c area=261.67mm2 respectively.
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   11.985  5.093 = 30.52KN
2
30.52  103
Vu

Shear strength of concrete is given by,
= 0. 218N/mm2
v 
b  d 1000  140
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
523.6
1000  140
= 0.374%
Design shear strength for 0.374% steel and M25 concrete from IS 456: 2000 Table19,
c = 0.425 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K×c = 1.25×0.425 = 0.531 N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Vu = 30.52 KN
Moment of resistance offered by 10 mm bars @ 300 mm c/c

A st  f y 

M1  0.87  f y  A st  d ' 
b  f ck 


261.8  415 
M1  0.87  415  261.8140 

1000  25 

= 12.82  10 6 N-mm
Ld = 1.3
M1
+Lo
V
(IS 456; 2000, Clause 26.2.3.3)
Development length L d =  σs
4 τ bd
Assuming Lo = 4  +4  =8 
So L0=8*10=80mm
τbd = 1.6*1.2 N/mm2
σs = 0.87 x 415 MPa
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000, page 44)
(IS 456: 2000, clause 26.2.1.1)
M1
+Lo
V
12.82  106
 σs
= 1.3
+80
30.52  103
4 τ bd
Now, Ld =1.3
(IS 456: 2000, Clause 26.2.3.3 c)
47.011 x ϕ = 626.15
 = 13.12 mm > 10 mm
OK.
Check for Deflection
L
= 
d
Pt  100 
Ast
bd
 100 
(From IS 456: 2000, Clause 23.2.1)
523.6
= 0.374%
1000  140
A st Required
A st Provided
431.7
f s  0.58  415 
523.6
= 198.45 N/mm2
f s  0.58  f y 
Value of Coefficients from Code
 = 26
 =1.7
=1, =1, =1
L
We have,
= 
d min
5093
L/d=
140
=36.38
=26*1*1.7*1=44.2mm
IS 456: 2000, Clause 24.1
(For continuous slab, IS: 456 2000 Clause 24.1(a))
(IS 456: 2000, Fig 4)
(IS 456: 2000, Clause 23.2.1(b, c, d))
L/d< Hence safe in deflection.
Slab Panel S6
1. Design data
Clear span
Edge condition
Material
Concrete grade
:
:
:
:
One long edge discontinuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4953 + 140mm or 4953 + 230/2 + 230/2
= 5093 mm or 5183 mm
Since 5093 < 5183,
Adopt lxe = 5093mm
lye = 7468+ 130 or 7468+ 230/2 + 230/2
= 7598 mm or 7698 mm
Since7598 < 7698
Adopt lye = 7598 mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 3 KN/m2 * 1 = 3 KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 8.49 KN/ m
Factored load (Wu) = 1.5 × 8.49 = 12.735 KN/ m
Moment coefficient from IS 456: 2000 Table 26
l ey
Long span to short span ratio,
= 1.49(Two way slab)
l ex
For negative moment: (x) = 0.0666
(y) = 0.037
For positive moment: (x) = 0.0506
(y) = 0.028
Moment Calculation:
At mid span
Mx = (x)×w×lx2 =0.0506× 12.735 × 5.0932 = 16.715 KNm /m
My = (y)×w×lx2 = 0.028 × 12.735 ×5.0932 =9.249 KNm/m
At edge
Mx = (x)×w×lx2 = 0.0666× 12.735 × 5.0932 = 22 KNm/m
My = (y)×w×lx2 = 0.037× 12.735 × 5.0932 = 12.22 KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
Mx
d
0.138  f ck  b
d
22 1000 1000
0.138  25 1000
=79.85 mm < 140mm O.K.
Minimum Reinforcement
0.12
 1000×175 = 210 mm2 .
=
100
(IS 456: 2000 clause 26.5.2.1)
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
b

d

f
ck


For ‘x’ +ve

Astx  415 
16.715  106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 344.78 mm2
For -ve

Astx  415 
22 *106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this eqn,
(Ast)x =460.37mm2
B) Calculation of reinforcement in long (Y) direction:
For +ve

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



Ast y  415 

9.249  106  0.87  415  Ast y  130  1 
 1000  130  25 


Solving quadratic equation
(Ast) y = 202.28mm2
Adopt +ve (Ast) y = 210 mm2
For –ve,

Ast y  415 

12.22  106  0.87  415  Ast y  130  1 
 1000  130  20 


Solving
(Ast) y =269.68mm2
Spacing required:
For x
+ve
Spacing = π ×102/4×
1000
344.78
=227.79mm
Adopt spacing 200mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
460.37
=170.6mm
Adopt spacing 150mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
For y
+ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
210
= 374 mm < 3d = 3×130 = 390 mm or 300 mm (Smaller)
Therefore take Spacing as 300 mm.
1000
1000
Actual area= π ×102/4×
= π ×102/4×
=261.67 mm2
300
spacing
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
269.68
=291.23mm
Adopt spacing 200mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
Provide 10 mm bar @ 200 mm c/c spacing giving total Area = 392.7 mm2
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 200mm c/c, Area = 392.7mm2.
Reinforcement in Y- direction 10 mm bars @ 300mm c/c, Area= 261.67mm2.
b) At Edge:
Reinforcement in X - direction 10 mm bars @ 150mm c/c, Area = 523.6mm2
Reinforcement in Y- direction 10 mm bars @ 200mm c/c, Area= 392.7mm2.
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   12.735  5.093 = 32.43KN
2
32.43  103
Vu

Shear strength of concrete is given by,
= 0. 232N/mm2
v 
b  d 1000  140
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
523.6
1000  140
= 0.374%
Design shear strength for 0.374% steel and M25 concrete from IS 456: 2000 Table19,
c = 0.425 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K×c = 1.25×0.425 = 0.531 N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Vu = 32.43 KN
Moment of resistance offered by 10 mm bars @ 300 mm c/c

A st  f y 

M1  0.87  f y  A st  d ' 
b

f
ck 


261.8  415 
M1  0.87  415  261.8140 

1000  25 

= 12.82  10 6 N-mm
Ld = 1.3
M1
+Lo
V
Development length L d =  σs
4 τ bd
Assuming Lo = 4  +4  =8 
So L0=8*10=80mm
τbd = 1.6*1.2 N/mm2
σs = 0.87 x 415 MPa
Now, Ld =1.3
M1
+Lo
V
(IS 456; 2000, Clause 26.2.3.3)
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000, page 44)
(IS 456: 2000, clause 26.2.1.1)
12.82  106
 σs
= 1.3
+80
32.43  103
4 τ bd
(IS 456: 2000, Clause 26.2.3.3 c)
47.011 x ϕ = 593.91
 = 12.63 mm > 10 mm
OK.
Check for Deflection
L
= 
d
Pt  100 
Ast
bd
(From IS 456: 2000, Clause 23.2.1)
 100 
523.6
= 0.374%
1000  140
A st Required
A st Provided
460.36
f s  0.58  415 
523.6
= 211.63 N/mm2
f s  0.58  f y 
Value of Coefficients from Code
 = 26
 =1.68
=1, =1, =1
L
We have,
= 
d min
5093
L/d=
140
IS 456: 2000, Clause 24.1
(For continuous slab, IS: 456 2000 Clause 24.1(a))
(IS 456: 2000, Fig 4)
(IS 456: 2000, Clause 23.2.1(b, c, d))
=36.38
=26*1*1.7*1=43.68mm
L/d< Hence safe in deflection.
Slab Panel S7
1. Design data
Clear span
:
Edge condition
Material
Concrete grade
:
:
:
One long edge discontinuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4953 + 140mm or 4953 + 230/2 + 230/2
= 5093 mm or 5183 mm
Since 5093 < 5183,
Adopt lxe = 5093mm
lye = 6325+ 130 or 6325+ 230/2 + 230/2
= 6455 mm or 6555 mm
Since 6455 < 6555
Adopt lye = 6455mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 2.5 KN/m2 * 1 = 2.5 KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 7.99 KN/ m
Factored load (Wu) = 1.5 × 7.99 = 11.985 KN/ m
Moment coefficient from IS 456: 2000 Table 26
l ey
Long span to short span ratio,
= 1.27(Two way slab)
l ex
For negative moment: (x) = 0.06335
(y) = 0.047
For positive moment: (x) = 0.04768
(y) = 0.035
Moment Calculation:
At mid span
Mx = (x)×w×lx2 =0.04768× 11.985 × 5.0932 = 14.823 KNm /m
My = (y)×w×lx2 = 0.035× 11.985 ×5.0932 =10.881 KNm/m
At edge
Mx = (x)×w×lx2 0.06335× 11.985 × 5.0932 =19.694 KNm/m
My = (y)×w×lx2 = 0.047× 11.985 × 5.0932 = 14.611 KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
Mx
d
0.138  f ck  b
19.694  1000  1000
0.138  25  1000
=75.55 mm < 140mm O.K.
d
Minimum Reinforcement
0.12
 1000×175 = 210 mm2 .
=
100
(IS 456: 2000 clause 26.5.2.1)
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
b

d

f
ck 

For ‘x’ +ve

Astx  415 
14.823  106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 304.27 mm2
For -ve

Astx  415 
19.694 *106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this eqn,
(Ast)x =409.5mm2
B) Calculation of reinforcement in long (Y) direction:
For +ve

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



Ast y  415 

10.881 106  0.87  415  Ast y  130  1 
 1000  130  25 


Solving quadratic equation
(Ast) y = 221.06mm2
For –ve,

Ast y  415 

14.611 106  0.87  415  Ast y  130  1 
 1000  130  20 


Solving
(Ast) y =299.71mm2
Spacing required:
For x
+ve
Spacing = π ×102/4×
1000
304.27
=258.12mm
Adopt spacing 200mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
409.5
=191.79mm
Adopt spacing 150mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
For y
+ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
221.06
= 355.28 mm < 3d = 3×130 = 390 mm or 300 mm (Smaller)
Therefore take Spacing as 300 mm.
1000
1000
Actual area= π ×102/4×
= π ×102/4×
=261.67 mm2
300
spacing
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
299.71
=262.05 mm
Adopt spacing 200mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
Provide 10 mm bar @ 200 mm c/c spacing giving total Area = 392.7 mm2
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 200mm c/c, Area = 392.7mm2.
Reinforcement in Y- direction 10 mm bars @ 300mm c/c, Area= 261.67mm2.
b) At Edge:
Reinforcement in X - direction 10 mm bars @ 150mm c/c, Area = 523.6mm2
Reinforcement in Y- direction 10 mm bars @ 200mm c/c, Area= 392.7mm2.
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   11.985  5.093 = 30.52KN
2
30.52  103
Vu

v 
Shear strength of concrete is given by,
= 0. 218N/mm2
b  d 1000  140
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
523.6
1000  140
= 0.374%
Design shear strength for 0.374% steel and M25 concrete from IS 456: 2000 Table19,
c = 0.425 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K×c = 1.25×0.425 = 0.531 N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Vu = 30.52 KN
Moment of resistance offered by 10 mm bars @ 300 mm c/c

A st  f y 

M1  0.87  f y  A st  d ' 
b

f
ck 


261.8  415 
M1  0.87  415  261.8140 

1000  25 

= 12.82  10 6 N-mm
Ld = 1.3
M1
+Lo
V
(IS 456; 2000, Clause 26.2.3.3)
Development length L d =  σs
4 τ bd
Assuming Lo = 4  +4  =8 
So L0=8*10=80mm
τbd = 1.6*1.2 N/mm2
σs = 0.87 x 415 MPa
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000, page 44)
(IS 456: 2000, clause 26.2.1.1)
M1
+Lo
V
12.82  106
 σs
= 1.3
+80
30.52  103
4 τ bd
Now, Ld =1.3
47.011 x ϕ = 626.153
 = 13.31 mm > 10 mm
Check for Deflection
L
= 
d
(IS 456: 2000, Clause 26.2.3.3 c)
OK.
(From IS 456: 2000, Clause 23.2.1)
Pt  100 
Ast
bd
 100 
523.6
= 0.374%
1000  140
A st Required
A st Provided
409.5
f s  0.58  415 
523.6
= 188.4 N/mm2
f s  0.58  f y 
Value of Coefficients from Code
 = 26
 =1.8
=1, =1, =1
L
We have,
= 
d min
5093
L/d=
140
IS 456: 2000, Clause 24.1
(For continuous slab, IS: 456 2000 Clause 24.1(a))
(IS 456: 2000, Fig 4)
(IS 456: 2000, Clause 23.2.1(b, c, d))
=36.38
=26*1*1.8*1=46.8mm
L/d< Hence safe in deflection.
Slab Panel S8
1. Design data
Clear span
Edge condition
Material
Concrete grade
:
:
:
:
One long edge discontinuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4496 + 140mm or 4496 + 230/2 + 230/2
= 4636 mm or 4726 mm
Since 4636< 4726,
Adopt lxe = 4636mm
lye = 6325+ 130 or 6325+ 230/2 + 230/2
= 6455 mm or 6555 mm
Since 6455 < 6555
Adopt lye = 6455mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 4 KN/m2 * 1 =4 KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 9.49 KN/ m
Factored load (Wu) = 1.5 × 9.49 = 14.235 KN/ m
Moment coefficient from IS 456: 2000 Table 26
l ey
Long span to short span ratio,
= 1.4(Two way slab)
l ex
For negative moment: (x) = 0.055
(y) = 0.037
For positive moment: (x) = 0.041
(y) = 0.028
Moment Calculation:
At mid span
Mx = (x)×w×lx2 =0.041× 14.235 × 4.636 2 = 12.544 KNm /m
My = (y)×w×lx2 = 0.028× 14.235 × 4.636 2 =8.566 KNm/m
At edge
Mx = (x)×w×lx2 =0.055× 14.235 × 4636 2 =16.827 KNm/m
My = (y)×w×lx2 = 0.037× 14.235 × 4636 2 = 11.32 KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
d
Mx
0.138  f ck  b
16.827  1000  1000
0.138  25  1000
=69.84 mm < 140mm O.K.
d
Minimum Reinforcement
0.12
=
 1000×175 = 210 mm2 .
100
(IS 456: 2000 clause 26.5.2.1)
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
b

d

f
ck


For ‘x’ +ve

Astx  415 
12.54  106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 255.93 mm2
For -ve

Astx  415 
16.827 *106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this eqn,
(Ast)x =347.19mm2
B) Calculation of reinforcement in long (Y) direction:
For +ve

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



Ast y  415 

8.556  106  0.87  415  A st y  130  1 
 1000  130  25 


Solving quadratic equation
(Ast) y = 186.96mm2
Adopt area 210 mm2
For –ve,

Ast y  415 

11.32  106  0.87  415  Ast y  130  1 
 1000  130  20 


Solving
(Ast) y =249.10 mm2
Spacing required:
For x
+ve
Spacing = π ×102/4×
1000
255.93
=306.88mm
Adopt spacing 200mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
347.18
=226.22mm
Adopt spacing 150mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
For y
+ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
210
= 374 mm < 3d = 3×130 = 390 mm or 300 mm (Smaller)
Therefore take Spacing as 300 mm.
1000
1000
Actual area= π ×102/4×
= π ×102/4×
=261.67 mm2
300
spacing
-ve
spacing= π ×102/4×
Adopt spacing 250mm
1000
1000
= π ×102/4×
Ast
249.1
=315.3 mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=314.16 mm2
spacing
250
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 200mm c/c, Area = 392.7mm2.
Reinforcement in Y- direction 10 mm bars @ 300mm c/c, Area= 261.67mm2.
b) At Edge:
Reinforcement in X - direction 10 mm bars @ 150mm c/c, Area = 523.6mm2
Reinforcement in Y- direction 10 mm bars @ 250mm c/c, Area= 314.16mm2.
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   14.235  4.636 = 33KN
2
33  10
V
v  u 
= 0. 236N/mm2
1000

140
bd
3
Shear strength of concrete is given by,
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
523.6
1000  140
= 0.374%
Design shear strength for 0.374% steel and M25 concrete from IS 456: 2000 Table19,
c = 0.425 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K×c = 1.25×0.425 = 0.531 N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Vu = 30.52 KN
Moment of resistance offered by 10 mm bars @ 300 mm c/c

A st  f y
M1  0.87  f y  A st  d ' 
b  f ck

= 12.82  10 6 N-mm




261.8  415 
M1  0.87  415  261.8140 

1000  25 

Ld = 1.3
M1
+Lo
V
(IS 456; 2000, Clause 26.2.3.3)
Development length L d =  σs
4 τ bd
Assuming Lo = 4  +4  =8 
So L0=8*10=80mm
τbd = 1.6*1.2 N/mm2
σs = 0.87 x 415 MPa
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000, page 44)
(IS 456: 2000, clause 26.2.1.1)
M1
+Lo
V
12.82  106
 σs
= 1.3
+80
33  103
4 τ bd
Now, Ld =1.3
(IS 456: 2000, Clause 26.2.3.3 c)
47.011 x ϕ = 585.03
 = 12.44 mm > 10 mm
OK.
Check for Deflection
L
= 
d
Pt  100 
Ast
bd
 100 
(From IS 456: 2000, Clause 23.2.1)
523.6
= 0.374%
1000  140
A st Required
A st Provided
347.19
f s  0.58  415 
523.6
2
= 160 N/mm
f s  0.58  f y 
Value of Coefficients from Code
 = 26
 =2
=1, =1, =1
L
We have,
= 
d min
4636
L/d=
140
=33.11
=26*1*2*1=52mm
IS 456: 2000, Clause 24.1
(For continuous slab, IS: 456 2000 Clause 24.1(a))
(IS 456: 2000, Fig 4)
(IS 456: 2000, Clause 23.2.1(b, c, d))
L/d< Hence safe in deflection.
Slab Panel S9
1. Design data
Clear span
Edge condition
Material
Concrete grade
:
:
:
:
Two Adjacent edge discontinuous
Fe 415 grade steel
M25
2. Relevant codes
IS 456: 2000 and IS 875: 1987 (part1&2)
3. Allowable stresses
fy = 415 N/mm2
fck = 25 N/mm2
4. Assumed slab depth and local calculation
The slab depth, assumed to calculate the self weight, is taken as 175mm as per preliminary
design.
Taking effective cover as 30 mm and assuming reinforcement bar of 10 mm diameter
Effective Depth:
dx = 175 – 30 – 10/2 = 140 mm
dy = 140 – 10/2 – 10/2 = 130 mm
Effective Span:
lxe = 4953 + 140mm or 4953 + 230/2 + 230/2
= 5093 mm or 5183 mm
Since 5093< 5183,
Adopt lxe = 5093mm
lye = 6325+ 130 or 6325+ 230/2 + 230/2
= 6455 mm or 6555 mm
Since 6455 < 6555
Adopt lye = 6455mm
Calculation of loads on slab:
Self weight
= 25 * 0.175 * 1 = 4.375KN/m
Live load
= 4 KN/m2 *1 =4 KN/m
Floor finish and partition = 1.115 KN/m2 * 1 = 1.115 KN/m
Total load = 9.49 KN/ m
Factored load (Wu) = 1.5 × 9.49 = 14.235 KN/ m
Moment coefficient from IS 456: 2000 Table 26
Long span to short span ratio,
l ey
l ex
= 1.27(Two way slab)
For negative moment: (x) = 0.0635
(y) = 0.047
For positive moment: (x) = 0.0478
(y) = 0.035
Moment Calculation:
At mid span
Mx = (x)×w×lx2 =0.0478× 14.235 × 5.093 2 = 17.65 KNm /m
My = (y)×w×lx2 = 0.035× 14.235 × 5.093 2 =12.92 KNm/m
At edge
Mx = (x)×w×lx2 =0.0635× 14.235 × 5.093 2 =23.45 KNm/m
My = (y)×w×lx2 = 0.047× 14.235 × 5.093 2 = 17.35 KNm/m
As moment is critical at support, checking depth taking maximum moment at support,
Mx
d
0.138  f ck  b
23.45  1000  1000
0.138  25  1000
=82.44 mm < 140mm O.K.
d
Minimum Reinforcement
0.12
 1000×175 = 210 mm2 .
=
100
(IS 456: 2000 clause 26.5.2.1)
Reinforcement Area Calculation:
At Middle Strip (For positive moment)
A) Calculation of reinforcement in short(X) direction:

A st x  f y 

M ux  0.87  f y  A st x  dx  1 
 b  d  f ck 
For ‘x’ +ve

Astx  415 
17.65  106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this Quadratic equation,
(Ast) x = 364.97 mm2
For -ve

Astx  415 
23.45 *106  0.87  415  Astx  140  1 

 1000  140  25 
Solving this eqn,
(Ast)x =492.71mm2
B) Calculation of reinforcement in long (Y) direction:
For +ve

A st y  f y 

M uy  0.87  f y  A st y  dy  1 
 b  dy  f ck 



Ast y  415 

12.92  106  0.87  415  Ast y  130  1 
 1000  130  25 


Solving quadratic equation
(Ast) y = 285.69mm2
For –ve,

Ast y  415 

17.35  106  0.87  415  A st y  130  1 
 1000  130  20 


Solving
(Ast) y =388.97 mm2
Spacing required:
For x
+ve
Spacing = π ×102/4×
1000
364.97
=215.2mm
Adopt spacing 200mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=392.7 mm2
spacing
200
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
492.71
=159.40mm
Adopt spacing 150mm,
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
For y
+ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
285.69
= 274.91 mm
Therefore take Spacing as 200 mm.
1000
1000
Actual area= π ×102/4×
= π ×102/4×
=392.7 mm2
300
spacing
-ve
spacing= π ×102/4×
1000
1000
= π ×102/4×
Ast
388.97
=201.92 mm
Adopt spacing 150mm
Actual area= π ×102/4×
1000
1000
= π ×102/4×
=523.6 mm2
spacing
150
Hence Finally Adopted
a) At mid Span:
Reinforcement in X - direction 10 mm bars @ 200mm c/c, Area = 392.7mm2.
Reinforcement in Y- direction 10 mm bars @ 200mm c/c, Area= 392.7 mm2.
b) At Edge:
Reinforcement in X - direction 10 mm bars @ 150mm c/c, Area = 523.6mm2
Reinforcement in Y- direction 10 mm bars @ 150mm c/c, Area= 523.6mm2.
Check for shear (Along Short Span)
1
Shear force, V   w  l x
2
1
V   14.235  5.093 = 36.25KN
2
36.25  103
Vu

v 
Shear strength of concrete is given by,
= 0. 259N/mm2
b  d 1000  140
Percentage of tension steel,
Pt  100 
Ast
bd
 100 
523.6
1000  140
= 0.374%
Design shear strength for 0.374% steel and M25 concrete from IS 456: 2000 Table19,
c = 0.425 N/mm2
The value of K from IS 456: 2000, Clause 40.2.1
For slab overall depth of 175 mm, K (modification factor) = 1.25
K×c = 1.25×0.425 = 0.531 N/mm2 > v O.K. (Hence safe in shear).
Check for Development length at short edge:
Vu = 36.25 KN
Moment of resistance offered by 10 mm bars @ 300 mm c/c

A st  f y 

M1  0.87  f y  A st  d ' 
b  f ck 


261.8  415 
M1  0.87  415  261.8140 

1000  25 

= 12.82  10 6 N-mm
Ld = 1.3
M1
+Lo
V
(IS 456; 2000, Clause 26.2.3.3)
Development length L d =  σs
4 τ bd
Assuming Lo = 4  +4  =8 
So L0=8*10=80mm
τbd = 1.6*1.2 N/mm2
σs = 0.87 x 415 MPa
(IS 456: 2000, Clause 26.2.1)
(IS 456: 2000, page 44)
(IS 456: 2000, clause 26.2.1.1)
M1
+Lo
V
12.82  106
 σs
= 1.3
+80
36.25  103
4 τ bd
Now, Ld =1.3
47.011 x ϕ = 539.82
 = 11.48 mm > 10 mm
Check for Deflection
L
= 
d
(IS 456: 2000, Clause 26.2.3.3 c)
OK.
(From IS 456: 2000, Clause 23.2.1)
Pt  100 
Ast
bd
 100 
523.6
= 0.374%
1000  140
A st Required
A st Provided
492.71
f s  0.58  415 
523.6
= 226.5 N/mm2
f s  0.58  f y 
Value of Coefficients from Code
 = 26
 =1.7
=1, =1, =1
L
We have,
= 
d min
5093
L/d=
=36.38
140
IS 456: 2000, Clause 24.1
(For continuous slab, IS: 456 2000 Clause 24.1(a))
(IS 456: 2000, Fig 4)
(IS 456: 2000, Clause 23.2.1(b, c, d))
=26*1*1.7*1=44.2mm
L/d< Hence safe in deflection.
6.2. Beam:
Beam is horizontal structural flexural member which carries the load transferred from the
slab and ultimately transfers to the column. They are usually designed for the induced bending
moment due to combination of dead load, live load, partition load etc.
The design of beam requires the determination of steel for the section fixed from the safe
design from SAP analysis. The design of section may result as a singly or doubly reinforced
section which may be ascertained by comparing the design moment (Mu) with the moment of
resistance for balanced section (Mlim) and the section is usually designed as under reinforced
section.
6.2.1 FLOWCHART FOR DESIGN OF BEAM:
Take moment of each beam (Mu)
Calculate Muli
Mulim = 0.133fckbd2
Mlim=0.133fckbd2
No
If Mu< Mulim
Yes
Ast >Ast min= 0.12%
0f bD
Under reinforced section
Calculate Ast2 by
Ast2=M/(0.87*fy*(d-d’))
Calculate Ast from
Mu= 0.87fy Ast(d-0.42xu)
Calculate Asc by Asc=M/(fsc*(dd’))
Calculate numbers of bars
= Ast/Abar
Calculate Ast1 from Mulim by
Ast1= Mulim / (0.87*fy*(d-0.42*xulim))
Ast >Ast min= 0.12%
of bD
Calculate numbers of bars
= Ast/Abar
Fig: Flowchart of Beam Design
SECOND FLOOR
FOR BEAM ID 82
Ref.
Calculations
Output
Characteristic strength of concrete(fck)=25MPA
Grade of steel(Fy)=415MPA
Dimension of beam;
Breadth=350mm
Depth=450mm
IS 456
C.L.
26.5
Eff.cover(d')=30mm
d=450-30=420mm
Check for member size=b/d=350/450=0.778>0.3
From
13920
Minimum area of tensile reinforcement=0.85*b*d/fy
=301.08mm2
Maximum area of tensile reinforcement=0.04bD
=6300mm2
Minimum area of tension reinforcement=0.24
=455.421 mm
√𝐟𝐜𝐤
𝐟𝐲
*b*D
2
Check for depth=L/D=7468/450=16.6>4(OK)
IS 456
Limiting moment of resistence for single reinforced rectangular
section=0.138fckbd2
=213KN-M
At L=0m;
Torsion(Tu)=20.7501KnM
Maximum Moment=28.5717Knm
Minimum Moment=-213.9095Knm
Max shear=-167.968Kn
Moment due to Torsion;
1+𝐷/𝑏
or, (Mt)=Tu * 1.7
1+450/350
=20.7501* 1.7
=27.899 kn-m
Total moment(+ve)(Md+)=28.5717+27.899
=56.4707KnM
Total moment(-ve)(Md-)=-213.9095-27.899
=241.8085KnM
For +ve Moment;
Md+<Mu then;
Beam is design as singly reinforced section.
For single reinforced beam;
𝐴𝑠𝑡
or, Md =0.87fyAst(d-fy*𝑏∗𝐹𝑐𝑘 )
𝐴𝑠𝑡
From
13920
56.4707*106=0.87*415*Ast(420-415*420∗25)
or, Ast=386.45mm2<455.421mm2(Not ok)
So take Ast=455.421 mm2
Provide 2 nos.of 20mm dia bars.Then area of steel provided=628.318 mm2
For compression bar. area of compressive bars required=0.5*Ast
=314.158 mm2
Provide 2 nos. of 20mm bars.
For –ve Moment;
Md-<Mu then;
Beam is design as doubly reinforced section.
Area of tension steel corresponding to limiting moment;
Or,Ast1=Mu,lim/(0.87fy*(d-0.46xu,lim)
Or, Ast1=213/(0.87*415*(420-0.46*201.6)
or, Ast1=1802.66mm2
Remaining moment=Mu-Mu,lim
=241.8085-213
=28.8085Kn-m
Ast2=M/(0.87*fy*(d-d’))
=(28.8085*106)/(0.87*415*(420-30))
=190 mm2
IS 456
Ast=Ast1+Ast2
Or ,Ast =1802.66+190
=1992.66mm2
Provide 7-20mm dia bars in tension
Ast provided=2199.11mm2
Asc=M/(fsc*(d-d’))
For fsc; Esc=0.0035(1-d’/xu,lim)
Or,Esc=0.00297
For Esc=0.00297,fsc=353.505N/mm2
Asc=(28.8085*106)/(353.505*(420-30))
=208.96 mm2
Provide 2-20 mm dia bars in compression
Asc provided=402.12mm2
At L=3.734m;
Torsion(Tu)=-2.4953KnM
Maximum Moment=38.2491Knm
Minimum Moment=22.6128Knm
Max shear=Kn
Moment due to Torsion;
1+𝐷/𝑏
or,(Mt)=Tu * 1.7
1+450/350
=-2.4953* 1.7
=-3.84Kn-m
Total moment(+ve)(Md+)=38.2491+3.84
=42.089KnM
For +ve Moment;
Md+<Mu then;
Beam is design as single reinforced section.
For single reinforced beam;
𝐴𝑠𝑡
or, Md =0.87fyAst(d-fy*𝑏∗𝐹𝑐𝑘 )
𝐴𝑠𝑡
42.089*106=0.87*415*Ast(420-415*350∗25)
or, Ast=286.848mm2<455.421mm2(not ok)
hence,Ast=455.421mm2.
Provide 2 nos.of 20mm dia bars.
For compression bar;
Provide 2 nos. of 20mm dia bars.
At L=7.468m;
Torsion(Tu)=-21.0243KnM
Maximum Moment=30.9546Knm
Minimum Moment=-210.342Knm
Max shear=166.422Kn
Moment due to Torsion;
1+𝐷/𝑏
or,(Mt)=Tu * 1.7
1+450/350
=-21.0243* 1.7
=-2.42Kn-m
Total moment(+ve)(Md+)=30.9546+2.42
=33.3746KnM
Total moment(-ve)(Md-)=-210.342-2.42
=212.762KnM
For +ve Moment;
Md+<Mu then;
Beam is design as single reinforced section.
For single reinforced beam;
𝐴𝑠𝑡
or, Md =0.87fyAst(d-fy*𝑏∗𝐹𝑐𝑘 )
𝐴𝑠𝑡
33.3746*106=0.87*415*Ast(420-415*350∗25)
or, Ast=225.85mm2<455.421mm2(not ok)
so Ast=455.421mm2
Provide 2 nos.of 20mm dia bars.
For compression bar;
Provide 2 nos. of 20mm dia bars.
For –ve Moment;
Md-<Mu then;
Beam is design as single reinforced section.
For single reinforced beam;
𝐴𝑠𝑡
or, Md =0.87fyAst(d-fy*𝑏∗𝐹𝑐𝑘 )
𝐴𝑠𝑡
212.762*106=0.87*415*Ast(420-415*350∗25)
or, Ast=1748.177mm2>455.421mm2(ok)
Provide 6 nos.of 20mm dia bars.
Ast provided=1884.96mm2
For compression bar;
Provide 3 nos. of 20mm bars.
Summary
L=0m
L=3.734m
L=7.468m
IS 456
table 20
IS 456
table 19
Bar dia.(d)
Tens. bar no.(n) Bar dia. Com.bar.
20mm
20mm
20mm
7
2
6
20
20
20
2
2
3
Shear & Torsion Reinforcement:
Equivalent shear:
Ve=V+1.6(TU /b)
=167.968+1.6(20.75/0.35)
=262.825Kn
Equivalent shear stress:
tve=ve/bd=262.825*103/350*420=1.788N/mm2
Maximum shear stress for M25,tc max=3.1N/mm2
Area of longitudinal steel =2199.11mm2
% of steel=Ast/bd*100%=1.396>0.12%
<4%
Shear strength of concrete (tc)=0.7234N/mm2
since, tve> tc & tve< tc max (ok)
or,Asv =TU * Sv/b1d1(0.87fy) +VuSv/2.5d1(0.87fy)
Assume dia. of stirrups as 8mm.
From
or, d1=450-(30*2)-(8+8)=374mm
SP16
table62 or, b1=350-2(30+8)=274mm
then,area of two legs of the stirrups should satisfy the following:
or,Asv(0.87fy)/Sv =20.75*106/(274*374 ) + 167.968 *103/(2.5*374)
=382.13N/mm
=3.82Kn/cm
From
Area
of
all
legs
of
the stirrups should satisfy condition that Asv/Sv should not
13920
be less than (tve- tc)*b(0.87fy)
C.L.
or, Asv(0.87fy)/Sv=(tve- tc)*b
6.3.5
=(1.788-0.7234)*350
=372.61 Nmm(Ok)
Provide 8mm ø two legged vertical stirrups at 100mm spacing.
Spacing of stirrups shall not not exceed X1,(X1+Y1)/4 and 300mm.
where X1 &Y1 are the short & long dimensions of the stirrups.
or,X1=350-2(30-4)=298mm
orY1=450-2(30-4)=398mm
or,(X1+Y1)/4 =174mm
Hence,provide 8mmø two legged stirrups at 100mm spacing.
The spacing of hopps over a length of 2d=840mm at either end of beam
From IS shall not exceed
a)d/4 =420/4 =105mm
b)8ø=160mm
c)and not less than 100mm
hence,provide hoops at 150mm spacing at a distance of 2d from each end
with first hoop at a distance of 50mm from the joint face.
Also,Vertical hoops spacing shall not exceed d/2=210mm
Hence,provide vertical hoops at 150mm spacing over a length of
2d=840mm
After, the distance of 2d=840mm.Provide minimum spacing of vertical
stirrups of ;
From IS or, X=Asv *0.87*fy/0.4*b
=259.26mm
456 C.L
26.2.1
Hence,provide spacing at 200mm after the length of 2d from the end
support.
456 C.L
26.5.1.6
Check for development length;
Design bond stress tbd for tension for M25=2.24N/mm2
Design bond stress tbd for compression for M25=2.8N/mm2
Development length(Ld)=ø(0.87fy)/4 tbd
= 0.87*415*20/4*2.24
=805.92mm
At joint the bottom and the top bars of the beam shall be provided with
anchorage length,beyond the innerface of column equal to the development
length in tension plus 10 times the bar diameter minus the allowance for the
90◦ bend.
i.e.anchorage length=Ld+10ø=1005.92mm
Check for deflection:
L=7.468m
(L/d)< αβɣ∆λ;
α=26
λ =from IS 456:2000(clause 23.2.1 e)
fs=0.58*fy*(Ast required/Ast provided)
=0.58*415*(1992.66/2199.11)
=218.1
At 1.39% & 220.24;
λ =1.01
β=1
∆=1
ɣ=1
then;
(7468/420) < 26*1*1*1*1.01
17.78<26.26 (OK)
Special confinement:
Special confinement in beam is provided at the L/6 from the face of support.
LAP SPLICE:
The longitudinal bars shall be splice.
▪Not more than 50% of the bars shall be splices at one section.
Lap splice shall not be spliced within;
▪ Joint
▪Distance (2*d'eff) from face of joint
▪Quarter length of member where flexural yeiding may occur.
6.3. Column:
Columns are the vertical members that are subjected to axial loads and moment acting from
two directions (Bi-axially). All columns are subjected to some moment which may be due to
accidental eccentricity or due to end restraint imposed by monolithically placed beams or slabs. The
strength of column depends upon the strength of the material, shape and size of the cross section,
length and the degree of positional and directional restraint at its ends. The column section may be
rectangular, square or circular shaped depending upon the architectural or structural requirements.
A column may be classified as follows based on types of loading:
a. Axially loaded column
b. A column subjected to axial load and Uniaxial bending and
c. A column subjected to axial load and biaxial bending
The design of column section for given axial load and biaxial moments can be made by pre
assigning the section and then checking adequacy. The design of column depends upon the
eccentricity of loading and the moment acting in different directions. The minimum eccentricity
specified by the IS 456: 2000 Clause 39.2 is:
e min 
Where,
Lo
D

but not less than 20mm
500 30
L o = unsupported length of column
D = lateral dimension in plane of bending
If emin is less than 0.05D, then column is designed as axially loaded column .If the eccentricity
exceeds 0.05D, then column is designed for both moment and axial load.
Select Maximum Mu=
|M2| + |M3|
Mux = |M2|
Muy = |M3|
6.3.1.Flow chart of column design :
Take corresponding axial
load (Pu)
Calculate minimum
eccentricity ex and ey
Calculate moment due to minimum
eccentricity by Muxe = Pu × ey and
Muye = Pu × ex
C
Take,
Mux = Max. of Mux and Muxe
Calculate Pu/Puz
Muy = Max. of Muy and Muye
Design as biaxial bending
Determine αn from table
from Pu/Puz and αn
Assume d’ and find ratio
d’/D
If (Mux/Mux1) +αn + (Mux/Mux1) αn >1
Assume suitable Asc and find
p = Asc/(B×D)
Yes
Calulate the ratios
Pu/(fck×BD) and p/fck
Determine Muxl, Muyl using
appropriate chart from SP-16 with
ratios p/fck, d’/D and Pu/ (fck×BD)
C
IncreaseAsc (steel
reinforcement) and find p.
No
The assumed reinforcement is OK.
6.4 Foundation:
Foundations are structural elements that transfer loads from the buildings or individuals columns to
the earth. Foundations must be designed to prevent excessive settlement or rotation, to minimize
differential settlement. Foundations are classified as:
a.
b.
c.
d.
a.
b.
c.
d.
e.
Isolated footing
Combined footing
Raft or mat foundation
Pile foundation
The type of foundations to be used in a given situation depends on a number of factors:
Soil strata
Bearing capacity of soil
Type of structure
Type of loads
Permissible differential settlement and
Economy
f.
CALCULATION OF BEARING PRESSURE OF SOIL
col
ID.
size
(m)
Approx. Size of Provide
Axial
BCS
Area
footing
size
load (KN) (KN/m2)
(m)
(m)
(m)
c1
c2
c3
c4
c5
c6
c7
c8
c9
c10
c11
c12
c13
c14
c15
c16
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
0.55
1118.447
1108.856
1085.503
1237.274
1093.477
1957.679
1816.765
1271.094
1000.985
1820.820
1834.211
1246.235
516.278
952.367
1014.653
688.245
150
150
150
150
150
150
150
150
150
150
150
150
150
150
150
150
8.202
8.132
7.960
9.073
8.019
14.356
13.323
9.321
7.341
13.353
13.451
9.139
3.786
6.984
7.441
5.047
2.864
2.852
2.821
3.012
2.832
3.789
3.650
3.053
2.709
3.654
3.668
3.023
1.946
2.643
2.728
2.247
2.9
2.9
2.9
3.1
2.9
3.8
3.7
3.1
2.8
3.7
3.7
3.1
2
2.7
2.8
2.3
Factored
Axial
load(KN)
Net soil
pressure
(KN/m2)
1677.67
1663.284
1628.254
1855.911
1640.216
2936.519
2725.148
1906.641
1501.477
2731.23
2751.317
1869.353
774.417
1428.55
1521.98
1032.367
199.485
197.775
193.609
193.123
195.032
203.360
199.061
198.402
191.515
199.505
200.973
194.522
193.604
195.960
194.130
195.154
Actual
Area of
footing
(m2)
8.41
8.41
8.41
9.61
8.41
14.44
13.69
9.61
7.84
13.69
13.69
9.61
4
7.29
7.84
5.29
150.24
PLINTH AREA OF BUILDING=14.402*20.803=299.6m^2
Since, area occupied by isolated footing is greater than 50% of plan area.
So, we use mat Foundation.
Since, area occupied by isolated footing is greater than 50% of plan area.
So, we use Mat Foundation.
DESIGN OF MAT FOUNDATION
Calculation of Corner stresses of mat foundation
Location of Geometric Centroid
(From bottom corner of grid 1-1)
x
y
7.201 m
10.402 m
Since, area occupied by isolated footing is greater than 50% of plan area.
So, we use Mat Foundation.
DESIGN OF MAT FOUNDATION
Safe Bearing Capacity of Soil (SBC) = 150 KN/m2
For Load Combination (DL+LL)
Summation of Forces (Σpi) = 19762.889KN
Column
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
C13
C14
C15
C16
reaction(P)
x
1118.447
1108.856
1085.503
1237.274
1093.477
1957.679
1816.765
1271.094
1000.985
1820.820
1834.211
1246.235
516.278
952.367
1014.653
688.245
19762.889
Y
0
4.953
9.449
14.402
0
4.953
9.449
14.402
0
4.953
9.449
14.402
0
4.953
9.449
14.402
C.G. of Column Load = (7.451,9.84)
ex = 7.451-7.201 = 0.25 m
ey = 10.402-9.84 = 0.562 m
Now,
0
0
0
0
7.01
7.01
7.01
7.01
14.478
14.478
14.478
14.478
20.803
20.803
20.803
20.803
p*x
p*y
0
5492.164
10256.91
17819.22
0
9696.386
17166.62
18306.3
0
9018.521
17331.46
17948.28
0
4717.072
9587.459
9912.1
147252.5
0
0
0
0
7665.276
13723.33
12735.52
8910.369
14492.26
26361.83
26555.71
18043
10740.13
19812.08
21107.83
14317.55
194464.9
b=14.902 m and d=21.303 m
Moment of inertia (assuming 0.50 m projection from corner column)
Ix = (b*d^3)/12 = 12006.454mm
Iy = (d*b^3)/12 = 5875.984 mm4
A = b*d = 317.478 m2
Factored load(P)=29644.334 KN
Mx = P*ey = 16660.116KNm
My = P*ex = 7411.08KNm
P/A =29644.334/317.478 =93.374 KN/m2
Now,
Stress (σ)=(P/A) ± (Mx/Ix)* y ± (My/Iy)* x
Column
C1
C2
C3
C4
C13
C14
C15
C16
C5
C9
C8
C12
X(m)
Y(m)
STRESS(KN/m^2)
-7.201
-10.42
117.11
-2.248
-10.42
110.24
2.248
7.201
-7.201
-2.248
2.248
7.201
-7.201
-7.201
7.201
7.201
-10.42
-10.42
10.383
10.383
10.383
10.383
-3.41
4.058
-3.41
4.058
110.24
117.11
117.06
110.19
110.19
117.06
108.27
109.08
108.27
109.08
Note: Here the maximum upward soil pressure (117.11KN/m2) is less than safe bearing capacity
(150 KN/m2) of foundation soil so it is not necessary to increase the strength of the foundation
soil by using geotechnical soil stabilizing process like certain depth of granular material packing,
grouting, etc.
Design of Mat Foundation
Concrete Grade = M25
Steel Grade = Fe415
Reference S.N
Calculations
Output
1
Known Data:
q=
Upward Soil Pressure, q = 117.01KN/m2
117.01
Max Span Length, L = 7.468 m
KN/m2
Maximum soil pressure in Ydirection ( For
Strip 4-4)
Moment Calculation:
Maximum Bending Moment, M = q L2 / 10
Ms=652.
Ms = 117.01x 7.468 2 / 10
57KNm/
= 652.57KNm/m
m
2
Known Data:
Upward Soil Pressure, q = 117.01 KN/m2
q=
Max Span Length, L = 4.953 m
117.01
Maximum soil pressure in X direction ( For
Strip A-A)
Moment Calculation:
Maximum Bending Moment, M = q L2 / 10
Ms = 117.01x 4.9532 / 10
= 287.05KNm/m
3
Size of
column is
550 * 550
mm
IS 456 :
2000, Cl.
31.6.3.1
Known Data:
Upward Soil Pressure, q = 108.27 KN/m2
Max Span Length, L = 4.953 m
Maximum soil pressure in X direction ( For
Strip C-C)
Moment Calculation:
Maximum Bending Moment, M = q L2 / 10
Ms = 125.64 x 4.9532 / 10
= 265.61KNm/m
Depth from two way shear consideration:
Considering intermediate column:
τv’=Vu/b0d
Vu= 1.5*1957.679=2936.52KN
d/2
d/2
d/2
KN/m2
Ms =
287.05KN
m/m
q = 108.27
KN/m2
Ms =
265.61
KNm/m
IS 456 :
2000, Cl.
31.6.2.1
d/2
b0=2(d+550)+2(d+550)
=4d+2100
550X
τv’=0.25*√𝑓𝑐𝑘
550
2
τv’=0.25 *250.5 =1.25N/mm
Ks =1.5>1 so Ks=1
then, τv’ = 1.25*1 = 1.25
1.12=(1957.679*1.5*1000)/(d*(2100+4d))
Upon Solving we get d=547.57mm
IS 456 :
2000, Cl.
31.6.3.1
d=547.5
72 mm
Considering corner column(C1):
τv’=Vu/b0d
Vu=1.5*1118.47KN
d/2=275
550
IS 456 :
2000, Cl.
31.6.2.1
b0=2(d/2+775)
=d+1550
’
τv =0.25*√𝑓𝑐𝑘
τv’=0.25 *250.5 =1.25N/mm2
Ks =1.5>1 so Ks=1
then,τv’ = 1.25*1 = 1.25
1.25=(1118.47*1.5*1000)/(d*(1550+d))
Upon Solving we get d=618.84 mm
Provide d=650mm
IS 456 :
2000 Cl.
31.6.3.1
d=618.84
mm
Considering edge column(C2):
τv’=Vu/b0d
Vu=1663.284KN
d/2
d/2
d/2
550
IS 456 :
2000, Annex
G Cl. 38.1 G
1.1 b
4
500
b0=[(d+550)+2(d/2+775)]
=(2100+2d)
’
τv =0.25*√𝑓𝑐𝑘
τv’=0.25 *250.5 =1.25N/mm2
Ks =1.5>1 so Ks=1
then,τv’ = 1.25*1 = 1.25
1.25=(1663.284*1000)/(d*(2100+2d))
Upon Solving we get d=445.02mm
On
comparing
d
=618.20m
m
adopt
d=650mm
IS 456 :
2000, Cl.
26.5.2.1
Taking highest of 3 depths.
Keep overall D=690mm giving 40mm effective
cover.
IS 456 :
2000, Cl.
26.2.1
D=
690mm
Calculation of Steel reinforcement:
Moment=0.87σyAt(d-σy*At/(σck*b))
652.57*106=0.87*415*At(650415*At/(25*1000))
Upon solving this we get:
At= 3012.4mm2
Minimum Steel=0.12*690*1000/100=828mm2
Provide, 28mm φ reinforcement bars.
Spacing = (1000/3012.4)*∏*282/4
= 204.41mm
Provide 28 mm φ rod @170mm c/c
For all strips along both directions in top and
bottom.
Development length
= (0.87*415*28)/(4*1.6*1.4)
= 1128.28mm
Min
Ast=828
mm2
Ast
provided=
3622.07
mm2
Ld=
1128.28m
m
Summary of Design of the Mat Foundation
Concrete Grade: M25
Safe Bearing Capacity: 150 KN/m2
Total Depth of Mat: 690 mm
Clear Cover: 26 mm
Effective cover: 40mm
Strips
All strips
Bottom Reinforcement
Diameter
Spacing
28mm
170mm
Steel Grade: Fe415
Top Reinforcement
Diameter
Spacing
28mm
170mm
6.5 Design of Staircase:
In this case landing slab A is spanning longitudinally along sec. 11 of Fig.9.20.21. Landing slab B
is common to spans of sec. 11 and sec. 22, crossing at right angles. Distribution of loads on
landing slab B shall be made 50 per cent in each direction (cl. 33.2 of IS 456). The effective span
for sec. 11 shall be from the centre line of edge beam to centre line of brick wall, while the
effective span for sec. 22 shall be from the centre line of landing slab B to centre line of landing
slab C (cl. 33.1b of IS 456).
A. (A) Design of landing slab A and going.
step 1: Effective span and depth of slab The effective span = 115 + 1220 + 1140 + 610 =
3085mm. The
depth of waist slab is assumed as 3085/20 = 154.25 mm, say 175mm. The effective depth = 175208/2 = 151mm. The landing slab is also assumed to have a total depth of 175 mm and effective
depth
of 151 mm.
Step 2: Calculation of loads (Fig.9.20.22)
(i) Loads on going (on projected plan area)
(a) Self weight of waist slab = 25*(0.175)*(322.06/285) = 4.944 kN/m
(b) Self weight of steps = 25*(0.5)*(0.15) = 1.875 kN/m
(c) Finish loads = 1.115 kN/m
(d) Live loads = 5.0 kN/m
Total = 12.934 kN/m
2
2
2
2
So, the factored loads = 1.5(12.934) = 19.401 kN/m
(ii) Landing slab A
(a) Self weight of slab = 25(0.175) = 4.375 kN/m
(b) Finish loads = 1.115 kN/m
(c) Live loads = 5.00 kN/m
Total = 10.49 kN/m
2
2
2
2
2
2
Factored loads = 1.5(10.49) = 15.735 kN/m
2
(iii) Landing slab B = 50 per cent of loads of landing slab A = 7.868 kN/m
The total loads of (i), (ii) and (iii) are shown in Fig.9.22.
Total loads (i) going = 19.401*(1.14)(1.22) = 26.983 kN
Total loads (ii) landing slab A = 15.735*(1.335)*(1.22) = 25.63 kN
Total loads (iii) landing slab B = 7.868*(0.61)*(1.22) = 5.855 kN
Total loads = 58.468 kN
The loads are shown in Fig. 9.20.22.
2
Step 3: Bending moment and shear force (width = 1.22 m)
V = {25.63*(3.085-0.6675) + 26.983(3.085 – 2.055) + 5.855(0.305)}/3.085
P
= 29.672 kN
VJ = 58.468 – 29.672 = 28.796 kN
The distance x where the shear force is zero is obtained from:
29.672 – 25.63 – 19.401*(1.22)*(x – 1.335) = 0
or x = 1.506 m
Maximum bending moment at x = 1.506 m (width = 1.22 m)
= 29.672(1.506) – 25.63*(1.506-0.668) – (19.401)(1.22)(0.171)(0.171)(0.5) = 22.862 kNm
Maximum shear force = 29.672 kN
Step 4: Checking of depth
½
3
From the maximum moment d = {22.862(10 )/1.22*(2.76)} = 85.5 mm < 175 mm.
Hence o.k.
2
From the maximum shear force, vτ = 29672/1220(175) = 0.139 N/mm . For the depth of
slab as 175 mm, k = 1.25(cl. 40.2.1.1 of IS 456) and cτ = 1.25(0.29) = 0.3625 N/mm
2
2
(Table 19 of IS 456). maxcτ = 3.1 N/mm (Table 20 of IS 456). Since, vτ < cτ < maxcτ, the
depth of slab as 175 mm is safe
Step 5: Determination of areas of steel reinforcement
Mu=0.87*fy*Ast*(d-fy*Ast/bd*fck)
Or, Ast=361.89mm2
2
Provide 8 mm diameter @ 120 mm c/c (= 418.88 mm ).
Distribution steel: The distribution steel is provided for both the slab as in waist slab. The
amount is =0.12*1220*175/100=256.2mm^2
Provide 8mm@180mm c/c (279.25mm^2)
Step 6: Checking of development length
Ld =
0.87fy∗∅
4τbd
=
0.87∗415∗∅
4∗2.24
= 40.3∅
For the slabs M1 for 8 mm diameter @ 120 mm c/c = (418.88)(22.862)/361.89 = 26.46
kNm.Shear force = 29.672 kN. Hence, 40.3 φ ≤ 1.3(26.46)/29.672 ≤ 1159.27 mm or the
diameter of main bar φ 28.76 mm. Hence, 8 mm diameter is o.k. The reinforcing bars are
shown in Fig.9.20.23.
(B) Design of landing slabs B and C and going (sec. 22 of Fig.9.20.21)
Step 1: Effective span and depth of slab
The effective span from the centre line of landing slab B to the centre line of landing slab
C = 610 + 2280 + 610 = 3500 mm. The depths of waist slab and landing slabs are
maintained as 175 mm like those of sec. 11.
Step 2: Calculation of loads (Fig.9.20.24)
(i) Loads on going (Step 2(i) of A) = 19.401 kN/m
2
(ii) Loads on landing slab B (Step 2(iii)) = 7.868 kN/m
2
(iii) Loads on landing slab C (Step 2(iii)) = 7.868 kN/m
Total factored loads are:
(i) Going = 19.401(2.28)(1.22) = 53.965 kN
(ii) Landing slab B = 7.868(0.61)(1.22) = 5.855 kN
(iii) Landing slab C = 7.868(0.61)(1.12) = 5.855 kN
Total = 65.675 kN
The loads are shown in Fig.9.20.24.
2
Step 3: Bending moment and shear force (width = 1.22 m, Fig.9.20.24)
The total load is 65.675 kN and symmetrically placed to give VG = VH = 32.8375 kN. The
maximum bending moment at x = 1.75 m (centre line of the span 3.5 m = 32.8375(1.75) –
5.855(1.75 – 0.305) – 19.401(1.22)(1.14)(1.14)(0.5) = 33.625 kNm. Maximum shear force
= 32.8375 kN.
Step 4: Determination of areas of steel reinforcement
Mu=0.87*fy*Ast*(d-fy*Ast/bd*fck)
Or, Ast=532.30mm2
2
Provide 8 mm diameter @ 90 mm c/c (= 558.505 mm ).
Step 5: Checking of development length
For the slab reinforcement 8 mm dia. @ 90 mm c/c, M1 = (558.505)(33.625)/532.3 = 35.28
kNm, V = 32.8375 kN. So, the diameter of main bar φ= {(1.3)(35.28)/(32.8375)}/40.3,
i.e., ≤34.65 mm. Hence, 8 mm diameter bars are o.k. Distribution steel shall remain the
same as in sec. 11, i.e., 8 mm diameter @ 90 mm c/c.
6.6. Design of basement wall
Introduction
Basement wall is constructed to retain the earth and to prevent moisture from seeping into
the building. Since the basement wall is supported by the mat foundation, the stability is ensured
and the design of the basement wall is limited to the safe design of vertical stem.
Basement walls are exterior walls of underground structures (tunnels and other earth
sheltered buildings), or retaining walls must resist lateral earth pressure as well as additional
pressure due to other type of loading. Basement walls carry lateral earth pressure generally as
vertical slabs supported by floor framing at the basement level and upper floor level. The axial
forces in the floor structures are , in turn, either resisted by shear walls or balanced by the lateral
earth pressure coming from the opposite side of the building.
Although basement walls act as vertical slabs supported by the horizontal floor framing ,
keep in mind that during the early construction stage when the upper floor has not yet been built
the wall may have to be designed as a cantilever.
Design of vertical stem
The basement wall is designed as the cantilever wall with the fixity provided by the mat
foundation.
Soil
Pressure
Due to Surcharge
Basement Wall
(Rear Face)
(Front Face)
Mat Footing
9.5 KN/m
23 KN/m
KN/m2
Fig: Basement Wall
Design of Basement Wall
Design Constants:
Concrete Grade = M25
Steel Grade = Fe415
Clear height between the floor (h) =3.0m
unit weight of soil, γ = 17 KN/m3
Angle of internal friction of the soil, ө = 300
surcharge produced due to vehicular movement is
Ws = 10 KN/m2
Safe bearing capacity of soil , qs = 150 KN/m2
Moment calculation
Ka 
1  sin  1  sin 30

 0.333
1  sin  1  sin 30
Lateral load due to soil pressure, Pa = Ka x γ x h2/2
= 0.333x17x32/2
= 25.47 KN/m
Lateral Load due to surcharge load, Ps = Ka x Ws x h
= 0.333x10x3
= 10 KN/m
Characteristic Bending moment at the base of wall ,
Since weight of wall gives insignificant moment ,so this can be neglected in the design.
Mc = Pa x h/3 + Ps x h/2
= 25.47x3/3 + 10x3/2
= 40.47 KN-m
Design moment, M = 1.5*40.47=60.705KN-m
Approximate design of section
Let effective depth of wall = d
BM = 0.136 ƒckbd2
1.1(C)
IS 456:2000 ANNEX G
60.705x106 = 0.136x25x1000xd2
d = 133.62 mm
Let Clear cover is 30mm & bar is 20mm-Ф
Overall depth of wall , D =133.62+30+10
= 173.62mm
Take D = 200mm
So ,
d = 200 – 30- 10
= 160 mm
Calculation of Main Steel Reinforcement
Ast=
Ast=
bdf ck
2xf y

1  1  4.6M

f ck bd 2





IS 456:2000 ANNEX G 1.1(b)
1000 x160 x 25 
4.6 x60.705 x10 6
1 1

2 x 415
25 x1000 x160 2





Ast = 1635.94 mm2
Min.
Ast = 0.0012xbxD = 0.0012x1000x200
IS 456:2000 clause 26.5.2.1
= 240 mm2 < Ast
No. of bar =
1635.94
𝜋𝑥
202
4
= 5.21
Spacing of the vertical reinforcement bars=1000/5.21
=191.94mm
Max. Spacing = 3d = 3x160
= 480 mm or 300mm whichever is small.
Max. Dia. of bar = D/8 = 200/8
= 25 mm
Provide 20mm-Ф bar @180 mm c/c
So, Provided Ast =
𝜋𝑥202 1000
4
x 180
= 1745.33 mm2
% 0f steel, Pt =1745.33 x100/(1000x160)
= 1.09 %
Provide nominal vertical reinforcement 8mmФ@300mm
c/c at the front face.
Check for Shear
The critical section for shear strength is taken at a distance of ‘d ’ from the face of support .Thus ,
critical
section is at d = 0.16 m from the top of isolated foundation.
i.e. at (3- 0.16) = 2.84m below the top edge of wall.
Shear force at critical section is,
Vu = 1.5x(Ka x Ws x Z + Ka x γ x Z2/2)
= 1.5x(0.333x10x2.84 + 0.333x17x2.842/2)
= 48.48KN
Nominal shear stress ,  u 
Vu

bd
IS 456:2000 clause 40.1
= 44.17 x1000/(1000x160)
= 0.303 N/mm2
For, Pt=1.09%
Permissible shear stress ,
1.09−1
τc = 0.64+1.25−1 (0.7 − 0.64)=0.662 N/mm2
τc > τu , Hence safe.
IS456:2000,Table-19
Check for Deflection
Leff = 3+d = 3+0.16
= 3.16m
Allowable deflection = leff/250 = 3160/250
IS456:2000,clause 23.2(a)
= 12.64mm
Actual Deflection =
p s l 4 eff p a l 4 eff

8EI
30 EI
=
3160 4
 10 25.47 
 

1000 x 200^3
8
30 

x5000 25
12
=12.56<12.64
Which is less than allowable deflection, hence safe.
Calculation of Horizontal Reinforcement steel bar
Area of Hz. Reinforcement = 0.002Dh
= 0.002x200x3000 = 1200 mm2
As the temperature change occurs at front face of basement wall, 2/3 of horizontal reinforcement
is provided at front face and 1/3 of horizontal reinforcement is provided in inner face.
Front face Horizontal Reinforcement steel,
= 2/3x1200
= 800 mm2
Providing 12mm-Ф bar
No. of bar required, N =800 /113.1
= 7.07=8
Spacing = (h-clear cover at both sides- Ф)/(N-1)
= (3000-30-12)/(8-1)
= 422.57 mm
Provide 12mm-Ф bar @ 400 mm c/c
IS456:2000 clause.32.5.d
Inner face Horizontal Reinforcement steel,
= 1/3x1140 = 380 mm2
Providing 8mm-Ф bar
No. of bar required, N = 300/50.27
= 7.559≈ 8 nos.
Spacing = (h-clear cover at both sides- Ф)/(N-1)
= (2850-30-8)/(8-1)
= 401.71 mm
6.7DESIGN OF SHEAR LIFT WALL:
Table: Lateral Load Calculation
FLOOR
Baseme
nt
Ground
F1
F2
F3
F4
F5
F6
LUMP
MASS
Wi, (kN)
HEIGHT
hi, m
hi2
100.238
100.238
100.238
100.238
100.238
100.238
100.238
100.238
0
3
6
9
12
15
18
21
0
9
36
81
144
225
324
441
801.904
Vb=
0.0665*801.9
0
53.33KN
Wi*hi2
0.000
902.142
3608.568
8119.278
14434.272
22553.550
32477.112
44204.958
126299.88
0
Lateral
force
Qi (kN)
Moment
(kN-m)
0
0.38090714
1.52362857
3.42816429
6.09451429
9.52267857
13.7126571
18.66445
895.894
754.196
577.074
422.807
278.824
153.125
55.993
0.000
53.327
3137.913
Load Calculation for Lift Wall Design
25 Mpa
f=
Reference
Step
Calculation
1 Basement
Lift
wall
Characteristic load = 25(5.94*0.15*3)
=
66.825
kN
Factored load = 1.5 x 66.825=
2
IS1893(Par
t 1)
:2002
Cl.7.6.1
IS1893(Par
t 1)
:2002
100.237
kN
5
Load Calculation
Total Weight, ∑Wi = 6*100.2375
= 601.425 kN
Total Height, h = 6*3
21 m
Time Period, Ta = 0.075h^0.75
=
Table 2
Table 6
Table 7
For Ta = 0.655 sec
Sa/g =
1.67/Ta
2.550
=
Z=
0.36
I=
1
R=
5
Cl.6.4.2
Ah = (ZISa/2Rg)=
Cl.7.6.1
100.237
kN
5
Intermediate
Floor
Lift
wall
Characteristic load = 25(5.94*0.15*3)
=
66.825 kN
Factored load = 1.5 x 66.825 =
3
Result
0.0665
0.655 sec
Vb = Ah * ∑Wi = 0.0665 *
801.904 =
Cl.7.5.3
53.327 kN
Vb
=
53.327
Kn
Design of
Lift Wall
Reference
IS 456-2000
Cl.
32.2.4
Ste
Calculation
p
1 Known Data
Perimeter of lift wall=
Floor Height, H=
Assume wall
thickness, t =
Result
5.94 m
3 m
150 mm
2 Check for Slenderness ratio
Effective height, Heff= 0.75*H
=
=
IS 456-2000
Cl.
32.2.3
Slenderness ratio =
0.75* 3
2.25 m
Heff/t
=
=
IS 456-2000
Cl.
32.2.2
3 Minimum Ecentrcity
IS 456-2000
Cl.
32.2.5
4 Additional ecentrcity
ea = H2/(2500t) =
3^2/(2500*0.15)
=
15
15.000 < 30
e= emin=0.05t = 0.05*150
5
=
O.K.
7.5 mm
24.000 mm
Ultimate load carrying
capacity
Ultimate load carrying capacity per unit length of wall,
Puw= 0.3(t-1.2e-2ea)*fck
0.3(150-1.2*5-2*24)*25
= 549.000 N/mm
6 Calculation of Main Vertical
Pu
w=
549.00
0
N/mm
reinforcement
When lateral load is actiong in Ydirection
Assume clear cover =
15
Using rod of dia, Φ =
16
effective cover, d' =
23
Mu = 3137.914 =
Vu = 53.327/2 =
Pu = 601.425/3 =
SP-16
Chart
31
mm
mm
mm
3137.914 KN-m
26.664 KN
200.475 kN
d'/D =
0.0126 ≈0.05
Rectangular Section-Reinforcement Equally distributed on
both sides
Mu/fckbD2 =
(3137.914*106)/(25*150*1830^2) =
Pu/fckbD =
(200.475*1000)/(25*150*1830) =
so,
Pt/fck = 0.165
Pt =
4.125 %
11323.12
Asreq=
mm2
5
IS 456:2000
4.13%
= 0.12% * 100* 1830
=
Provide bar of Φ =
Area of single bar=
No. of bars, n =
Spacing of bar, Svy
=
Cl 32.5
b
Pt =
Minimum area of steel, Asmin = 0.12% of
bD
Cl 32.5
a
IS 456:2000
0.25
0
0.02
9
329.4 mm2
18 mm
254.469 mm2
44.497 ≈45
(1830/45)
=
41 mm
Maximum spacing = 3t or 450
mm
= 3*150 or 450 mm
= 450
or 450
Hence,
provide
45-18 mm Φ
bars @ 40
mm
mm
c/c.
Hence, provide 18 mm Φ bars @ 40 mm on both faces of
wall.
When lateral load is actiong in Xdirection
Mu = 1568.957 KN-m
Vu =
26.664 KN
Pu = 601.425*(3/2)
400.950 kN
=
SP-16
Chart
31
d'/D =
0.0126 ≈0.05
Rectangular Section-Reinforcement Equally distributed on
both sides
Mu/fckbD2 =
(1568.957*106)/(25*150*18302) =
Pu/fckbD = (400.95*1000)/(25*150*1830)
=
so,
Pt/fck = 0.14
Pt =
3.5 %
Asreq=
9607.5 mm2
IS 456:2000
3.50%
= 0.12% * 150* 1830
=
Provide bar of Φ =
Area of single bar=
No. of bars, n =
Spacing of bar, Svx
=
Cl 32.5
b
Pt =
Minimum area of steel, Asmin = 0.12% of
bD
Cl 32.5
a
IS 456:2000
0.12
5
0.05
8
329.4 mm2
18 mm
254.469 mm2
37.755
40
1830/40
46 mm
Maximum spacing = 3t or 450
mm
= 3*150 or 450 mm
= 450
or 450
mm
Hence,
provide
40-18 mm Φ
bars @ 46
mm c/c
Hence, provide 18 mm Φ bars @ 46 mm on both faces of
wall.
7
IS 456:2000
Calculation of Horizontal steel
reinforcement
Area of horizontal steel reinforcement = 0.2% of
bH
Cl 32.5
c
= 0.2% of 150*3000
900 mm2
=
Providing 12 mm dia rods,
No. of rods, n = 900/113.09 =
7.95826
≈8
3
Spacing of horizontal bars, S = 3000/8 =
IS 456:2000
375 mm
Maximum spacing = 3t or 450
mm
Cl 32.5
d
= 3*150 or 450 mm
= 450
or 450
mm
Hence, provide 12 mm Φ bars @ 375mm on both faces
of wall.
IS 456:2000
Cl
32.4.2
IS 456:2000
Cl 32.4.2.1
8 Check for Shear
When lateral load is actiong in Ydirection
Nominal Shear Stress,
τv = Vu/td
=(53.327*1000)/(150*0.8*1830)=
τalw = 0.17fck = 0.17*25
=
4.25 N/mm2
> τv
Hw/Lw=3000/1830
=
IS 456:2000
1.639
>1 (High
Wall)
0.24 N/mm
3 2
O.K.
Hence,
provide
12 mm Φ
bars
@ 350 mm
c/c.
Cl.
32.4.3
τcw should be lesser of
a.
b.
τcw = (3-(Hw/Lw))*K1*√fck = (3(1.639))*0.2*√25
=
1.361 N/mm2
τcw = K2*√fck*(((Hw/Lw)+1)/((Hw/Lw)1))
=
0.929 N/mm2
but,
should not be less than 0.15*√fck =
0.15*√f20=
τcw = 1.361 N/mm2
0.75
N/mm
2
> τv
Hence, safe in Shear.
When lateral load is actiong in Xdirection
Nominal Shear Stress,
τv = Vu/td
=(26.664*1000)/(152*0.8*1980)=
τalw = 0.17fck = 0.17*20
IS 456:2000
Cl 32.4.2.1
=
4.25 N/mm2
Hw/Lw=3000/1830
=
IS 456:2000
Cl.
32.4.3
1.639
> τv
0.12 N/mm
1 2
O.K
.
>1 (High
Wall)
τcw should be lesser
of
a.
b.
τcw = (3-(Hw/Lw))*K1*√fck = (3(1.639))*0.2*√25
=
1.361 N/mm2
τcw = K2*√fck*(((Hw/Lw)+1)/((Hw/Lw)1))
=
0.929 N/mm2
but,
should not be less than 0.15*√fck =
0.15*√25=
τcw = 1.361N/mm2
> τv
O.K.
0.75
N/mm
2
Hence, safe in Shear.
CHAPTER 8: CONCLUSION
After the completion of “Structural Analysis and Design of Multistoried Building”, we have
gained in-depth knowledge about the design of RCC Building.
The purpose of this project, though purely academic oriented, we have made every effort to make
it feasible for the real construction.
During our entire work, we were able to play with various codes for the Seismic design and
Analysis of Composite loads, moments, deflections, nature of impacts on each and every
member of the section through SAP Analysis.
Conclusion on Beam Design:
In case of design of beam sections, following conclusions can be extracted:
 For the analysis of beam, the envelope is taken as governing combinations.
 Negative moments is higher in support sides rather than in mid of the beam. So at
support sides we provided sufficient reinforcement.
 For tension reinforcement, curtailment was made at specified distance from edge of the
support as per IS 13920.
 For compression reinforcement, curtailment was made at mid part of beam.
 Spacing of stirrups are also designed as prescribed by Ductility Code i.e. IS
13920:1993.
Conclusion on Column Design:
 With increase in load with time, steel will attain yield strength before concrete attains its
full strength. The column will carry further load because steel will sustain yield stress
while concrete will carry additional load until it attains its full strength.
 The maximum axial load and moments acting along the length of the column was
considered for the design of the column section by Limit State Method. The design
required determination of area of longitudinal steel (load carrying capacity) and its
distribution and transverse steel (lateral support against buckling to every longitudinal
bar and confine concrete).
Conclusion on Slab Design:


Most of the slab panels were found to be two-way.
The entire slab panel are safe in Deflection and Shear check.
Conclusion on Staircase Design:
 Both geometrical design as well as structural designs was done by conservative
methods and not in SAP analysis.
CHAPTER 9: RECOMMENDATION
Design and analysis are two important tasks for the the project to be successful. Each part should
be done with great care and wisely to minimize the error.
Manual calculation is the initial job and it is with the reference to basic design principle and
various codes. But it is difficult for multi-storied building hence the use of computer aided design
and analysis is to be used. SAP 2000 version 16 and AutoCAD provide almost accuracy and time
saving in the analysis of the structure and other software like ETABS also can be used.
The design and analysis should be practically possible in construction materials and method.
Economic consideration should be considered for the analysis.
During our project, there were certain limitation and constraints which are enumerated here in
along with appropriate recommendations:




In the project only static analysis is done but dynamic analysis of the structure
should be done for more efficient result.
In the project the bearing capacity of the soil is assumed but for accurate analysis the
bearing capacity of soil should be measured in site.
Mostly IS codes are referred in the project but we should also refer NBC for more
accurate data.
Staircase modeling can be done for more precise analysis.
CHAPTER 10: BIBLIOGRAPHY
Menon, D. & Pillai S. U.: Reinforced Concrete Design,Tata Mcgraw Hill Education Pvt. Ltd.,
2012
Jain , A.K.: Reinforced Concrete Limit State Design, Nem Chand &Bros, 2012
Chopra, A.K.: Dynamics of Structure, Dorling Kindersley Publishing, Inc, 2014
Sinha, S.N.: Reinforced Concrete Design, Tata Mcgraw Hill Education Pvt. Ltd., 2011
Reynolds, C.E. & Steedmann, J.C.: Reinforced Concrete Designer’s Handbook
Agarwal, P. & Shrikhande, M.: Earthquake Resistant Design of Structure, Asoke k. Gosh, PHI
Learning Pvt. Ltd., 2013
Arora, K.R.: Soil Mechanics and Foundation Engineering, Standard publishers distributors, 2011
Shah, H.J. & Jain, S.K.: IITK-GSDMA Project on Building Codes
C.V.R. Murty: IITK-bmtpc Earthquake Tips, Learning Earthquake Design and
construction,National information canter of Earthquake engineering Indian Institute of
Technology Kanpur, 2005
IS 456:2000, Plain & Reinforced Concrete Code of Practice
IS 875:1987, Code of Practice for Design Loads Part 1: Dead Loads
IS 875:1987, Code of Practice for Design Loads Part 2: Imposed Loads
IS 1893(Part I):2002, Criteria for Earthquake Resistant Design of Structure
SP 16, Design Aids for Reinforced Concrete
SP 34(S & T):1987, Handbook on Concrete Reinforcement & Detailing
IS 13920:(1993), Ductile Detailing of Reinforced Concrete Structures