project.doc

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1.1 GENERAL
Building construction is the engineering dealing with the construction of building
such as residential houses. In simple building can be defined as an enclosed space by
walls with roof, food, cloth and the basic needs of human beings. In the ancient times
humans lived in caves, over trees or under trees, to protect themselves from wild animals,
rain, sun, etc. as the time passed, humans being started living in huts made of timber
branches. The shelters of those have been developed nowadays into beautiful houses.
Rich people live in sophisticated condition houses.
Buildings are the important indicator of social progress of the county. Every
human has a desire to own a comfortable home. On an average generally one spends his
two-third life times in the houses.. These are the few reasons which are responsible that
the person do utmost effort and spend hard earned saving in owning houses.
Nowadays the building house is major work of the social progress of the county.
Daily new techniques are being developed for the construction of houses economically,
quickly and fulfilling the requirements of the community engineers and architects do the
design work, planning and layout, etc, of the buildings. Draughts man is responsible for
doing the drawing works of building as for the direction of engineers and architects. The
draughtsman must know his job and should be able to follow the instruction of the
engineer and should be able to draw the required drawing of the building, site plans and
layout plans etc, as for the requirements.
A building frame consists of number of bays and storey. A multi-storey, multi-
paneled frame is a complicated statically intermediate structure. A design of R.C building
of G+3 storey frame work is taken up. The building in plan (44.5 x 19.5) consists of
columns built monolithically forming a network. The size of building is 44.5 x 19.5 m.
The numbers of columns are 85. It is a residential complex.

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The design is made using software on structural analysis design (staad-pro). The
building is subjected to both the vertical loads as well as horizontal loads. The vertical
load consists of dead load of structural components such as beams, columns, slabs etc and
live loads. The horizontal load consists of the seismic forces thus building is designed for
dead load, live load and seismic load as per IS 1893. The building is designed as three
dimensional vertical frame and analyzed for the maximum and minimum bending
moments and shear forces by trial and error methods as per IS 456-2000. The help is
taken by software available in institute and the computations of loads, moments and shear
forces are obtained from this software.
Structural design is an art and science of designing the structure with economy,
elegance, safe, serviceable and durability.
The process of structural design involves the following data:
1. Structural planning
2. Estimation of loads
3. Analysis of structural elements
4. Design of structural elements
The principal elements of an R.C.C. building frame are as follows:
1. Slab to cover to the entire area.
2. Beams to support slabs and walls.
3. Columns to support beams
4. Footings to distribute the column loads over larger area of soil.
1.2. ANALYSIS
Structural analysis involves the determination of internal forces like axial forces,
bending moments, shear forces etc., in the component members for which these members
are to be designed under the action of given external loads.
The different approaches to structural analysis are
1. Elastic analysis based on elastic theory
2. Limit state analysis based on plastic theory

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1.3. DESIGN
Reinforced concrete structural elements can be designed by using any one of the
following design philosophies.
1. Working stress method (WSM)
2. Ultimate load method (USM)
3. Limit state method (LSM)
Working stress method used over the decades is now practically outdated and is not
used at all in many of the advanced countries of the world because of its inherent
drawbacks. The latest I.S.456-2000 code gives emphasis on the limit state method which
is the modified form of the Ultimate load method. It is judicious amalgamation of the
WSM and the USM removing all the inherent draw backs of these methods but
maintaining all their good points. The LSM proves to have an edge over the WSM from
viewpoint of economy.

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2.1 GENERAL
The present project work is a multistoried residential apartment planned to be
constructed at Tirupati. So as to cater the needs of the fast growing population.
It is a four storied building (G+3) provided with a dog-legged stair-case and a lift
for easy ascending and descending from one floor to another. Each floor consists of eight
flats.
Each flat of the apartment is facilitated with the following:
1. Two bed rooms with attached toilets.
2. Drawing room
3. Dining
4. Kitchen cum utility.
2.2 DATA:
Type of structure : Multistoried residential building.
Building plan : As shown in plan at end.
Floor to floor height : 3.2m
Height of plinth : 600 mm above ground level.
Depth of foundation : 1.8m below ground level.
Bearing capacity of soil : 350 kN/Sq.m
Thickness of walls : External -230mm.
Internal-115mm.
Materials : Concrete: M-25
Steel: Fe-415
Design basis : Limit state method based on I.S.456-2000.

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3.1. General
Method of analysis of statistically indeterminate portal frames:
1. Method of flexibility coefficients.
2. Slope displacements methods(iterative methods)
3. Moment distribution method
4. Kani’s method
5. Cantilever method (Approximate Methods )
6. Portal method
7. Matrix method
3.1.1. Method of flexibility coefficients:
The method of analysis comprises reducing the hyper static structure to a
determinate structure form by removing the redundant support (or) introducing
adequate cuts (or) hinges.
Limitations:
 It is not applicable for degree of redundancy greater than 3
3.1.2. Slope displacement equations:
It is advantageous when kinematic indeterminacy is less than static
indeterminacy. This procedure was first formulated by axle bender in 1914 based
on the applications of compatibility and equilibrium conditions.
The method derives its name from the fact that support slopes and
displacements are explicitly comported. Set up simultaneous equations are formed
to obtain the solution of these parameters and the joint moments in each element
or computed from these values.
Limitations:
 A solution of simultaneous equations makes methods tedious for manual
computations. This method is not recommended for frames larger than too
bays and two storeys.

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3.1.3. Moment Distribution methods:
These methods involve distributing the known fixed and moments of the
structural member to adjacent members at the joints in order satisfy the conditions
of compatibility.
3.1.4. Kani’s method:
This method over comes some of the disadvantages of hardy cross method.
Kani’s approach is similar to H.C.M to that extent it also involves repeated
distribution of moments at successive joints in frames and continues beams.
However there is a major difference in distribution process of two methods.
H.C.M distributes only the total joint moment at any stage of iteration. The most
significant feature of kani’s method is that process of iteration is self corrective.
Any error at any stage of iterations corrected in subsequent steps consequently
skipping a few steps error at any stage of iteration is corrected in subsequent
consequently skipping a few steps of iterations either by over sight of by intention
does not lead to error in final end moments.
Advantages:
 It is used for side way of frames.
Limitations:
 The rotation of columns of any storey should be function of a single
rotation value of same storey.
 The beams of storey should not undergo rotation when the column
undergoes translation. That is the column should be parallel.
 Frames with intermediate hinges cannot be analyzed.
3.1.5. Cantilever Method (Approximate method):
Approximate analysis of hyper static structure provides a simple means of
obtaining a quick solution for preliminary design. It makes some simplifying
assumptions regarding Structural behavior so to obtain a rapid solution to complex
structures.
The usual process comprises reducing the given indeterminate configuration to
determine structural system by introducing adequate no of hinges. it is possible to

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sketch the deflected profile of the structure for the given loading and hence point of
inflection can be located.
Since each point of inflection corresponds to the location of zero moment in the
structures. The inflection points can be visualized as hinges for the purpose of
analysis. The solution of structures is simple once the inflection points are located.
The analysis carried out separately for both vertical and horizontal load cases.
3.1.5.1. Horizontal cases:
The behavior of a structure subjected to horizontal forces depends upon its heights
to width ratio. It is necessary to differentiate between low rise and high rise frames in
this case.
Low rise structures : Height < width
It is characterized predominately by shear deformation.
High rise buildings : Height > width
It is dominated by bending action
3.1.6. Matrix analysis of frames:
The individual elements of frames are oriented in different directions unlike those
of continues beams, so their analysis is more complex .stiffness method is more useful
because of its adaptability to computer programming. stiffness method is used when
degree of redundancy is greater than degree of freedom.

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4.1. GENERAL
This project is mostly based on software and it is essential to know the details
about these software’s.
List of software’s used
 Staad pro(v8i)
 Auto cad
4.2. STAAD
Staad is powerful design software licensed by Bentley .Staad stands for structural
analysis and design.
Any object which is stable under a given loading can be considered as structure.
So first we find the outline of the structure, where as analysis is the estimation of what are
the type of loads that acts on the beam and calculation of shear force and bending moment
comes under analysis stage. Design phase is designing the type of materials and its
dimensions to resist the load. This we do after the analysis.
To calculate SFD and BMD of a complex loading beam it takes about an hour. So
when it comes into the building with several members it will take a week. Staad pro is a
very powerful tool which does this job in just an hour. Staad is a best alternative for high
rise buildings.
Now a day’s most of the high rise buildings are designed by staad which makes a
compulsion for a civil engineer to know about this software.
This software can be used to carry RCC, steel, bridge, truss etc according to
various country codes.

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4.3 Alternatives for staad:
Struts, E-tabs, SAP, Tekla which give details very clearly regarding reinforcement
and manual calculations.
4.4 Staad Editor:
Staad has very great advantage to other softwares i.e., staad editor. Staad editor is
the programming. For the structure we created and loads we applied, all details are
presented in programming format in staad editor. This program can be used to analyze
another structure also by just making some modifications, but this require some
programming skills. So load cases created for a structure can be used for another structure
using staad editor.
4.4.1. Limitations of Staad pro:
 Huge output data
 Even analysis of a small beam creates large output.
4.5. AutoCAD
AutoCAD is powerful software licensed by auto desk. The word auto came from
auto Desk Company and cad stands for computer aided design. AutoCAD is used for
drawing different layouts, details, plans, elevations, sections and different sections can be
shown in auto cad.
It is very useful software for civil, mechanical and also electrical engineer.
The importance of this software makes every engineer a compulsion to learn this
software’s.
We used AutoCAD for drawing the plan, elevation of a residential building. We
also used AutoCAD to show the reinforcement details and design details of a stair case.
AutoCAD is a very easy software to learn and much user friendly for anyone to
handle and can be learn quickly
Learning of certain commands is required to draw in AutoCAD.

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5.1. General
Planning of a building is a systematic assembling or grouping of a building and
arrangement of its component parts in a systematic manner and proper order. While
planning a building it is necessary to follow certain laws and basic principles, which are
given below. The work of the structural Engineer starts with planning of structural
members in the given architectural plan. It commences with deciding positions of
columns followed by positioning of beams and spanning of slabs.
5.2. FRONT OPEN SPACE:
Every building facing a street shall have a front yard forming an integral part of
the site and should have a minimum width of 3 m and in case of two or more sides facing
street, an average width of 3m, and in no case less than 1.8m
Each site shall have a minimum frontage of 6m on any street for the building up to
a height of 10m.
No construction work of a building shall be undertaken within 7.5m from the
center line of any street as determined by the authority for the individual road/street width
taken into account for the traffic flow.
5.3. REAR OPEN SPACE:
Every building shall have a back yard forming an integral part of site. It should
have an average width of 3m and in no case less than 1.8m. Except in the case of back to
back site, the width of the rear yard shall be 3m throughout.

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5.4. SIDE OPEN SPACE:
Every semi-detached and detached building shall have a permanent open air space
forming an integral part of the site and should have a minimum side width of 3m.
5.5. DISTANES FROM ELECTRICAL LINES:
No verandah balcony shall be allowed to be erected or re-erected or any additions
or alterations made to a building with in the distance quoted below in accordance with the
current Indian Electrical Rules and its Amendments form time to time, between the
building and any overhead electrical supply line.
TABLE 5.1 DISTANCES FROM ELECTRICAL LINES
Vertically Horizontally
1 Low and medium voltage line
and service line
2.4m 1.22m
2 High voltage lines up to and
including 33KV
3.66m 1.83m
3 Extra high voltage greater
than 33 KW
3.66+3m for every
additional 33KW
1.83+0.3m for every
additional 33KW
5.6. AREA OF OPENING:
For light and ventilation, a clear window opening in the wall abutting to air space
either through an open verandah or gallery should not be less than one tenth of the floor
area of the room for dry hot climate and one sixth for wet hot climate. The total area of
door and window opening should not be less than one seventh of the floor area.
The building where doors and windows need to be closed for sake of privacy or
security, total area may be either by windows or doors. It will be possible in the case of
living room, dining hall as such rooms abut on an open verandah or gallery.

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Every room shall have ventilator of at least 0.3sq.m in area near the top of each of
the rooms and these ventilators preferably be placed opposite to each other for through
ventilation.
Generally, the aggregate area of ventilation is provided at the rate of 0.1sq.m. For
each 10cu.m of space of such rooms. Where no ventilators are provided, the windows
should extend practically to ceiling and preferably for the windows, two sheets of shutters
are provided one in the upper half and another in lower half. Half lower shutters are
closed for the sake of privacy while the upper half will remain in open condition as to
facilitate ventilation.
5.7. SUNSHADES OVER WINDOWS AND VENTILATORS:
Projection of sunshades over windows or ventilators when permitted by
the authority shall fulfill the following conditions.
1. Sunshades shall not be permitted over the road, over any drain or over any portion
outside the boundaries of the site below the height of 2.8m from the road level.
2. Sunshades provided above the height of 2.8m from ground level shall be
permitted to project up to a maximum width of 60cm.
3. No sunshades shall be permitted on roads less than 9m in width or on roads
having no footpath.
5.8. KITCHEN:
(a) Height:
The height of a kitchen measured form the surface of the floor to the
lowest point of the ceiling shall not be less than 2.75m.
(b) Size:
The area of the kitchen shall not be less than 5.5 sq.m and with a minimum
width of 1.5m. If there is a separate store it will reduce to 4.5sq.m.

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5.9. BATH ROOMS AND WATERS CLOSETS:
(a) Height:
The height of a bathroom or water closets shall not be less than 2.2m.
(b) Size:
The size of the bathroom shall not be less than 1.5mx1.2m or 1.84sq.m
The minimum floor area of water closets shall be 1.1 sq.m If the
bathrooms and water closets are combined, the floor area should not be
less than 2.8 sq.m with a minimum width of 1.2m.
5.10. FIRE PROTECTION:
This deals with safety from fire and emergencies.
TYPE 1: All structural components shall be of 4 hours fire resistance.
For type 1 to 3 constructions, a doorway opening in a separating wall of a floor
shall be limited to 5sq.m in area with a maximum height of 2.75m and a maximum width
of 2.1m.
The separating walls which are provided for fire resistance such as shutter etc.,
shall have fire resistance of not less than 4 hours.
Building erected in fire zone shall confirm to the construction of type 1, 2 or 3.

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Fig. 5.1 Stilt Floor (Parking) Plan

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Fig. 5.2 Typical Floor Plan

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6.1. General:
A structure can be defined as a body which can resist the applied loads without
appreciable deformations.
Civil engineering structures are created to serve some specific functions like
human habitation, transportation, bridges, storage etc. in a safe and economical way. A
structure is an assemblage of individual elements like pinned elements (truss elements),
beam element, column, shear wall slab cable or arch. Structural engineering is concerned
with the planning, designing and thee construction of structures.
Structure analysis involves the determination of the forces and displacements of
the structures or components of a structure. Design process involves the selection and
detailing of the components that make up the structural system.
The main object of reinforced concrete design is to achieve a structure that will
result in a safe economical solution.
The objective of the design is
Foundation design
Column design
Beam design
Slab design
These all are designed under limit state method
6.2. Limit state method:
The object of design based on the limit state concept is to achieve an acceptability
that a structure will not become unserviceable in its life time for the use for which it is
intended. i.e., it will not reach a limit state. In this limit state method all relevant states
must be considered in design to ensure a degree of safety and serviceability.
6.3. Limit state:
The acceptable limit for the safety and serviceability requirements before failure
occurs is called a limit state.

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6.4. Limit state of collapse:
This corresponds to the maximum load carrying capacity.
Violation of collapse limit state implies failures in the source that a clearly defined
limit state of structural usefulness has been exceeded. However it does not mean complete
collapse.
This limit state corresponds to:
Flexural
Compression
Shear
Torsion
6.5. Limit state of serviceability:
This state corresponds to development of excessive deformation and is used for
checking member in which magnitude of deformations may limit the rise of the structure
of its components.
Deflection
Cracking
Vibration
6.6. Objectives of present project:
1. To design a (G+3) residential building manually and check out the same design in
Staad.pro software and compare the results.
2. To design the same building in staad. Pro software with the effects of seismic
loads and compares the results without the effects of seismic loads and find out
the percentage variation of steel.
3. To detail the drawings of all the structural members of the building in
AUTOCADD according to SP: 34

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7.1. General
The various types of loads acting on the structure, which need consideration in
building design, are as follows:
1. Dead load
2. Live load
3. Seismic load
7.1.1. Dead loads:
Dead loads are the loads which do not vary in magnitude and in position. The
dead load of a structure is not known before it is designed. After designing, the
assumed dead load is compared with actual dead load. If the difference is significant,
the assumed dead load is revised and the structure is redesigned. The deal load
calculation should also include the superimposed loads that are permanently attached
to the structure.
The dead load includes:
1. Self weight of members
2. Weight of finishes
3. Weight of partitions, walls etc.
The unit weight of different materials taken from IS: 875-1987 is as follows:
Reinforced concrete : 25kN/m3
Brick masonry : 20kN/m3
Hollow brick masonry : 19kN/m3

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7.1.2. Live loads:
Live loads are the loads which vary in magnitude and in position. Live loads
on roofs and floors are taken according to N.B.C and are given below.
For Residential buildings:
1. All rooms and Kitchens : 2KN/m2
2. Toilets and Bath rooms : 2KN/m2
3. Corridors, Passages, Stair cases,
including fire escapes and store rooms : 3KN/m²
4.
Balconies : 3KN/m2
7.1.3. Seismic loads:
Buildings and portions shall be designed and constructed, to resist the
effects of design lateral forces as specified in IS 1893(Part 1): 2002.The design
lateral force shall first be computed for the building as a whole. This design lateral
force shall then be distributed to the various floor levels. The overall design
seismic force thus obtained at each floor level shall be distributed to individual
lateral load resisting elements depending on the floor diaphragm action.
The total design lateral force or design seismic base shear (VB) along any
principal direction shall be determined by the following expression.
W
A
V h
B 
7.2. Design loads for residential buildings:
7.2.1. General
Loads are primary consideration in any building design because they
define how a structure must resist providing a reasonable performance (i.e., safety
and serviceability) through out the structure’s useful life. The anticipated loads are
influenced by a building’s intended use (occupancy and function), configuration
(size and shape) and location (climate and site conditions).Ultimately, the type
and magnitude of design loads affect critical decisions such as material collection,
construction details and architectural configuration.
Thus, to optimize the value (i.e., performance versus economy) of the
finished product, it is essential to apply design loads realistically.

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Since building codes tend to vary in their treatment of design loads the
designer should, as a matter of due diligence, identify variances from both local
accepted practice and the applicable code relative to design loads as presented in
this guide, even though the variances may be considered technically sound.
Complete design of a home typically requires the evaluation of several different
types of materials. Some material specifications use the allowable stress design
(ASD) approach while others use load and resistance factor design (LRFD).
7.2.2. Dead Loads
Dead loads consist of the permanent construction material loads
comprising the roof, floor, wall, and foundation systems, including claddings,
finishes and fixed equipment. Dead load is the total load of all of the components
of the building that generally do not change over time, such as the steel columns,
concrete floors, bricks, roofing material etc.
In staad pro assignment of dead load is automatically done by giving the
property of the member.
In load case we have option called self weight which automatically
calculates weights using the properties of material i.e., density and after
assignment of dead load the skeletal structure looks red in color as shown in the
figure.
Fig 7.1 Dead Load on the skeletal structure

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Fig 7.2 Dead Load on the Frame
7.2.3. Live Loads:
Live loads are produced by the use and occupancy of a building. Loads
include those from human occupants, furnishings, no fixed equipment, storage,
and construction and maintenance activities. As required to adequately define the
loading condition, loads are presented in terms of uniform area loads,
concentrated loads, and uniform line loads.. Concentrated loads should be applied
to a small area or surface consistent with the application and should be located or
directed to give the maximum load effect possible in endues conditions.
In staad we assign live load in terms of U.D.L .we has to create a load case
for live load and select all the beams to carry such load. After the assignment of
the live load the structure appears as shown below.
For our structure live load is taken as 25 N/mm for design.
Live loads are calculated as per IS 875 part 2

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Fig 7.3. Live Load on the Frame
7.2.4. Floor load:
Floor load is calculated based on the load on the slabs. Assignment of floor
load is done by creating a load case for floor load. After the assignment of floor
load our structure looks as shown in the below figure.
The intensity of the floor load taken is -3.85 kN/m². -ve sign indicates that
floor load is acting downwards.
Fig 7.4. Diagram of floor load

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7.2.5. Load combinations:
All the load cases are tested by taking load factors and analyzed as per IS
456 for all the load combinations and results are taken and maximum load
combination is selected for the design.
Fig 7.5. Diagram of combination load case

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8.1 Slab design:
Slab is plate elements forming floor and roofs of buildings carrying distributed
loads primarily by flexure.
8.1.1. One way slab:
One way slab are those in which the length is more than twice the breadth,
it can be simply supported beam or continuous beam.
8.1.2. Two way slab:
When slabs are supported on four sides, two ways spanning action occurs.
Such slabs are simply supported on any or continuous on all sides. The deflections
and bending moments are considerably reduced as compared to those in one way
slab.
8.1.3. Checks:
There is no need to check serviceability conditions, because design
satisfying the span for depth ratio.
 Simply supported slab
 Continuous beam

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Fig.8.1. Diagrams of slab deflection in one way and two way slabs
Following figures shows the load distributions in two slabs.
Fig. 8.2. Diagram of load distribution of one way slab

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Fig. 8.3. Diagram of load distribution of two way slab
Slabs are designed for deflection. Slabs are designed based on yield theory
This diagram shows the distribution of loads in two way slabs.
Fig.8.4. Distribution of loads in slabs.
In order to design a slab we have to create a plate by selecting a plate cursor.
Now select the members to form slab and use form slab button. Now give the thickness of
plate as 0.115 m. Now similar to the above designs give the parameters based on code and
assign, design slab command and select the plates and assign commands to it. After
analysis is carried out go to advanced slab design page and collect the reinforcement
details of the slab. Slabs are also designed as per IS: 456-2000.

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Fig.8.5. Typical floor diagram.

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Fig.8.6. Typical Slab diagram.

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Fig.8.7. Slab reinforcement details.

30
8.2. Design of slabs (Flat 1)
8.2.1. Two Adjacent edges discontinuous (S1)
Span: long span = Ly = 4.267 m
Short span = Lx = 3.556 m
Ly / Lx = 1.20 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Assume dia of bars 8 mm
The slab will be designed as two - way continuous slab
Trial depth:
In the case of two - way slab the shorter span is used for calculating L/d ratio for
deflection check. Since live load in this case 4 KN/m² which is greater than 3 KN/m² the
serviceability requirements for deflection should be worked out as per basic values of
(span / effective depth) ratio L/d basic for spans up to 10 m.
End condition Basic L/d ratio
Cantilever 7
Simply supported member 20
Fixed or continuous member 26
In this case of two - way slab the loads are distributed in both directions, the design
moments are small compared to one - way slab.
The percentage of steel required in general is less between 0.2% to 0 .35% for HYSD
bars.
Assume Pt = 0.25%
For Pt = 0.25%
Modification factor α1 = 1.60
For Fe 415 = 415 N/mm²
Now basic L/d = 26 for continuous slab
Required d = Lx (L/d xα1)
= 85.48 mm
Required D = 109.48 mm or 110 mm

31
Effective depth provided (d) = 86 mm
Effective depth for mid span steel in y direction = 78 mm
Loads
Consider 1 m width of slab
Dead load = 2.750 KN/m
Floor finish = 1.000 KN/m
Live load = 2.000 KN/m
Total = 5.750 KN/m
Wu = 8.625 KN/m
Design moments
The boundary conditions for slab are Two adjacent edges discontinuous
Case No 2 BMC
The design moment are worked out using the formula
Mu = α x Wu x Lx²
Short span
Negative moment at continuous edge = 6.43 KNm
Positive moment at mid span = 4.85 KNm
Long span
Table 8.1 Reinforcement details of Slab (S1)
Span
position
Mu d
(Ast)
reqd
Dia spacing
(Ast)
provided
spacing
KN-m mm mm² mm mm mm² mm
a) Along Short span
Support 6.43 86 219 8 230 219 230
Mid span 4.85 86 163 8 309 219 230
b) Along Long span
Support 5.13 78 192 8 262 219 230
Mid span 3.82 78 141 8 357 219 230
Spacing = 3d or 300 mm whichever is less

32
Check for deflection:
Pt (reqd) = 0.189% < 0.25% assumed Hence safe
However detailed check is carried out for verification
Pt (prov) = 0.25 %
fs = 179 N/mm²
for Pt = 0.25 % α1 = 1.60
d(reqd) = 85 mm < 86 mm Hence safe
Distribution steel using Fe 415 steel
Ast = 93.6 mm²
Using 8 mm bars spacing = 537 mm
Assume 230 mm c/c
= 219 mm² > 93.6 mm² Hence safe
Torsion steel
a) At the corner contained by discontinuous edges
Torsional steel required = 0.75 x Ast = 122 mm²
Provide 8 mm bars spacing = 412 mm
Assume 230 mm c/c
= 219 mm² > 122 mm² Hence safe
In both directions at right angles in each of the two meshes one at the top and other at
the bottom for a length of
Lx/5 = 711.2 mm
b) At the corner at which one edge is discontinuous and the other continuous
Torsional steel required = 3/8 x Ast = 61 mm²
Assume 400 mm c/c = 126 mm² > 61 mm² Hence safe
For a distance of Ly/5 = 853.44 mm in both directions
------

33
8.2.2. Four edges continuous interior panel (S2)
Ly / Lx = 1.64 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Assume Dia of bars 8 mm
Trial depth:
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
For Fe 415 corresponding to fs = 240 N/mm²
Required d = Lx (L/d x α1)
= 80.60 mm
Effective depth provided d = 86 mm

34
Loads
Total = 5.750 KN/m
Wu = 8.625 KN/m
Design moments:
The boundary conditions for slab are four edges continuous interior panel
Case No 1 BMC
Mu = α x Wu x Lx²
Short span
Long span
Span
position
Mu d (Ast)reqd Dia spacing (Ast)prov spacing
a) Along Short span
Support 5.51 86 186 8 271 219 230
Mid span 4.19 86 140 8 360 219 230
a) Along Long span
Support 3.10 78 114 8 442 219 230
Mid span 2.33 78 85 8 595 219 230
Spacing = 3d or 300mm whichever is less

35
Pt (reqd) = 0.16 % < 0.25% assumed Hence safe
Pt (prov) = 0.25 %
fs = 154 N/mm²
for Pt = 0.25 % α1 = 1.60
------

36
8.2.3. One long edge discontinuous (S3)
Ly / Lx = 1.54 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25% Modification factor α1 = 1.70
= 52.29 mm

37
Loads
Total = 5.500 KN/m
Wu = 8.250 KN/m
Design moments:
The boundary conditions for slab are one long edge discontinuous
case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
Span
position
Mu d (Ast)
reqd
Dia spacing (Ast)
provided
spacing
a) Along Short span
Support 3.02 76 114 8 442 219 230
Mid span 2.30 76 86 8 585 219 230
b) Along Long span
Support 2.30 68 97 8 520 219 230
Mid span 1.23 68 51 8 984 219 230

38
Pt (prov) = 0.29 %
fs = 95N/mm²
for Pt = 0.29 %
α1 = 1.51
Distribution steel:
Using Fe 415 steel
Ast = 81.6 mm²
230 mm c/c
Torsion steel:
230 mm c/c
the bottom for a length of Lx/5 = 462.28 mm
230 mm c/c

39
8.2.4. One shore edge Discontinuous (S4)
Ly / Lx = 1.68 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
Modification factor (α1)= 1.60
= 34.80 mm

40
Loads
Total = 5.500 KN/m
Wu = 8.250 KN/m
Design moments:
The boundary conditions for slab are one short edge discontinuous
Case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
span
position
Mu d (Ast)
reqd
dia spacing (Ast)
prov
spacing
a) Along Short span
Support 1.07 76 40 8 1269 219 230
Mid span 0.81 76 30 8 1685 219 230
b) Along Long span
Support 0.64 68 26 8 1913 219 230
Mid span 0.48 68 20 8 2533 219 230

41
Pt(prov) = 0.29%
fs = 33N/mm²
For Pt = 0.29%
α1 = 1.51
Ast = 81.6 mm²
using 8 mm bars spacing = 616 mm
230 mm c/c
Torsion steel:
Torsion steel required = 0.75xAst = 22 mm²
230 mm c/c
Torsion steel required = 3/8 x Ast = 11 mm²
230 mm c/c

42
8.2.5. Design of One – Way slab:
Balcony (S5)
Ly / Lx = 2.23 > 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Since length of the slab is more than twice the width, it is a one - way slab.
Load will be transferred to the supports along the shorter span. Consider a 100 cm
wide strip of the slab parallel to its shorter span.
Minimum depth of slab’d’ = L/αβγλδ
Let α = 20
β = 1
γ = 1
δ = 1
λ = 1
d = 72.39 mm
Let us adopt overall depth ‘D’ = 92.39 mm
Loads
Total = 4.310 KN/m
Wu = 6.465 KN/m
Assume steel consists of 8 mm bars with 20 mm clear cover
Effective depth = 92.39 - 20 - 4
= 68.39 mm bars with
Effective span of slab = 1.45 + 0.068
= 1.52 m

43
Factored Moment (Maximum at midspan)
Mu = Wu x l ² /8
(6.5 x 1.52 ²) / 8 = 1.86 kNm
Max. Shear Force = Wu x lc /2
= 6.465 x 1.45/2
= 4.68 kN
Depth of the slab is given by
BM = 0.138 fck bd²
d = 26 mm
Adopt effective depth d = 80 mm
and over all depth D = 100 mm
Area of Tension Steel is given by
M = 0.87 fyAt(d-((fyAt)/(fck b)))
1.86 x 10^6 = 0.87 x 415x At ((80 - 415At/20 x 1000))
5145.1 = 80 At - 0.02075At²
At = 65.42 mm² (from Calculator)
Use 8 mm bars @ 230 mm c/c giving total area
= 218.63 mm² > 65.42 mm² Hence O.K.
Bend alternate bars at L/7 from the face of support where moment reduces to less
than half of its maximum value. Temperature reinforcement equal to 0.15% of the gross
concrete area will be provided in the longitudinal direction.
= 0.15 x 1000 x 100 / 100
= 150 mm²
= 218.63 mm² > 150 mm² Hence O.K.

44
Check for shear:
Present Tension Steel = 100At / bd
Use 8 mm bars @ 230 mm c/c
= 100 x 218.63 / 1000 x 80
= 0.27 %
Shear Strength of concrete for 0.27 % steel
τc = 0.38 N/mm²
τc' = k x τc
for 100 mm thick slab k = 1.3
τc' = 1.3 x 0.378
τc' = 0.49 N/mm²
Nominal shear stress τv = Vu/bd
= 4.68 x 1000 /1000 x 80
= 0.06 < 0.49 Hence O.K.
Check for Development length
Moment of Resistance offered by 8 mm bars @ 230 mm c/c
= 0.87 x 415 x 218.6335 x (80 - ((415 x 218.6335 / (20 x 1000)))
= 5956899.1 Nm = 5.96 kNm
V = 4679.74 N
Let us assume anchorage length Lo
Ld ≤ 1.3 M1/V
47 Ф ≤ 1.3 x 5956899.1 / 4679.74
1655 mm
Ф < 35 mm Hence O.K.
The Code requires that bars must be carried in to the supports by at least Ld/3 = 92.39
mm

45
Check for Deflection:
Percent tension steel at mid span
= 100 x 218.63 / 1000 x 80
= 0.27 %
γ= 1.42 (fig 10.1, page no 230 in ak jain)
β = 1
δ = 1
λ = 1
Allowable L/d = 20 x 1.42 = 28.4
Actual L/d = 1516.19 / 80 = 19.0 < 28.4 Hence O.K.
The detail of reinforcement
Fig.8.8. Reinforcement details of One way slab.

46
8.3. Design of Slabs (Flat 2)
8.3.1. Two adjacent edges discontinuous (S1)
Ly / Lx = 1.20 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
Cantilever 7
moments are small compared to one - way slab
bars.
Assume Pt = 0.25%
For Pt = 0.25%
For Fe 415 = 415 N/mm²
= 85.48 mm
Effective depth provided (d) = 86 mm

47
Loads
Total = 5.750 KN/m
Wu = 8.625 KN/m
Design moments
The boundary conditions for slab are Two adjacent edges discontinuous
Case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
Span
position
Mu d
(Ast)
reqd
Dia spacing
(Ast)
provided
spacing
a) Along Short span
Support 6.43 86 219 8 230 219 230
Mid span 4.85 86 163 8 309 219 230
b) Along Long span
Support 5.13 78 192 8 262 219 230
Mid span 3.82 78 141 8 357 219 230

48
Pt (prov) = 0.25 %
fs = 179 N/mm²
for Pt = 0.25 % α1 = 1.60
Ast = 93.6 mm²
Assume 230 mm c/c
Torsion steel
Assume 230 mm c/c
the bottom for a length of
Lx/5 = 711.2 mm
Assume 400 mm c/c = 126 mm² > 61 mm² Hence safe

49
Ly / Lx = 1.64 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
servicebility requirements for deflection should be worked out as per basic values of
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
= 80.60mm

50
Loads
Total = 5.750 KN/m
Wu = 8.625 KN/m
Design moments:
Case No 1 BMC
Mu = α x Wu x Lx²
Short span
Long span
Span
position
Mu d (Ast)reqd Dia spacing (Ast)prov spacing
a) Along Short span
Support 5.51 86 186 8 271 219 230
Mid span 4.19 86 140 8 360 219 230
a) Along Long span
Support 3.10 78 114 8 442 219 230
Mid span 2.33 78 85 8 595 219 230

51
Pt(reqd) = 0.16% < 0.25% assumed Hence safe
Pt (prov) = 0.25 %
fs = 154 N/mm²
for Pt = 0.25 % α1 = 1.60
d (reqd) = 81 mm < 86 mmHence safe

52
Ly / Lx = 1.54 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25% Modification factor α1 = 1.70
= 52.29 mm

53
Loads
Total = 5.500 KN/m
Wu = 8.250 KN/m
Design moments
case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
Span
position
Mu d (Ast)
reqd
Dia spacing (Ast)
provided
spacing
a) Along Short span
Support 3.02 76 114 8 442 219 230
Mid span 2.30 76 86 8 585 219 230
b) Along Long span
Support 2.30 68 97 8 520 219 230
Mid span 1.23 68 51 8 984 219 230

54
Check for deflection
Pt (prov) = 0.29 %
fs = 95 N/mm²
for Pt = 0.29 %
α1 = 1.51
Distribution steel
using Fe 415 steel
Ast = 81.6 mm²
230 mm c/c
Torsion steel
230 mm c/c
230 mm c/c

55
8.3.4. One shore edge Discontinuous (S4)
Ly / Lx = 1.68 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
Cantilever 7
In this case of two - way slab the loads are distributed in both directions, the
design moments are small compared to one - way slab.
bars.
Assume Pt = 0.25%
For Pt = 0.25%
Modification factor (α1) = 1.60
= 34.80 mm

56
Loads
Total = 5.500 KN/m
Wu = 8.250 KN/m
Design moments
Case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
span
position
Mu d (Ast)
reqd
dia spacing (ast)
prov
spacing
a) Along Short span
Support 1.07 76 40 8 1269 219 230
Mid span 0.81 76 30 8 1685 219 230
b) Along Long span
Support 0.64 68 26 8 1913 219 230
Mid span 0.48 68 20 8 2533 219 230

57
Pt (prov) = 0.29%
fs = 33N/mm²
For Pt = 0.29%
α1 = 1.51
d (reqd) = 37 mm < 76 mm Hence safe
Ast = 81.6 mm²
230 mm c/c
Torsion steel:
230 mm c/c
Torsion steel required = 3/8 x Ast = 11 mm²
230 mm c/c

58
8.3.5. Design of one – way slab
Balcony (S5)
Ly / Lx = 2.23 > 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Load will be transferred to the supports along the shorter span. Consider a 100 cm wide
strip of the slab parallel to its shorter span.
Let α = 20
β = 1
γ = 1
δ = 1
λ = 1
d = 72.39 mm
Loads
Total = 4.310 KN/m
Wu = 6.465 KN/m

59
= 1.52 m
Factored Moment (Maximum at midspan)
Mu = Wu x l ² /8
(6.5 x 1.52 ²) / 8 = 1.86 kNm
= 6.465 x 1.45/2
= 4.68 kN
BM = 0.138 fck bd²
d = 26 mm
1.86 x 10^6 = 0.87 x 415x At ((80 - 415At/20 x 1000))
5145.1 = 80 At - 0.02075At²
At = 65.42mm² (from Calculator)
= 218.63 mm² > 65.42 mm² Hence O.K.
concrete area will be Provided in the longitudinal direction.
= 0.15 x 1000 x 100 / 100
= 150 mm²

60
= 218.63 mm² > 150 mm² Hence O.K.
Check for shear
= 100 x 218.63 / 1000 x 80
= 0.27 %
τc = 0.38 N/mm²
τc' = k x τc
τc' = 1.3 x 0.378
τc' = 0.49 N/mm²
= 4.68 x 1000 /1000 x 80
= 0.06 < 0.49 Hence O.K.
= 0.87 x 415 x 218.6335 x (80 - ((415 x 218.6335 / (20 x 1000)))
= 5956899.1 Nm = 5.96 kNm
V = 4679.74 N
Ld ≤ 1.3 M1/V
47 Ф ≤ 1.3 x 5956899.1 / 4679.74
1655 mm
mm

61
Percent tension steel at midspan
= 100 x 218.63 / 1000 x 80
= 0.27 %
β = 1
δ = 1
λ = 1
Allowable L/d = 20 x 1.42 = 28.4
Actual L/d = 1516.19 / 80 = 19.0 < 28.4
Hence O.K.

62
8.4. Design of slabs (Flat 3)
Span: long span = Ly = 5.49m
Short span = Lx = 3.51m
Ly / Lx = 1.57 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
(span / effective depth ) ratio L/d basic for spans up to 10 m.
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
= 84.26 mm

63
Loads
Total = 5.750 KN/m
Wu = 8.625 KN/m
Design moments
Case No 1 BMC
Mu = α x Wu x Lx²
Short span
Long span
span
position
mu d (ast)
reqd
dia spacing (ast)
prov
spacing
a) Along Short span
Support 5.81 86 197 8 256 219 230
Mid span 4.46 86 149 8 338 219 230
a) Along Long span
Support 3.39 78 125 8 404 219 230
Mid span 2.54 78 93 8 543 219 230

64
Pt(reqd) = 0.17% <0.25% assumed Hence safe
Pt(prov) = 0.25%
fs = 164N/mm²
for Pt = 0.25%
α1 = 1.59
d(reqd) = 85mm < 86mm Hence safe

65
Ly / Lx = 1.17 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
(span / effective depth ) ratio L/d basic for spans up to 10 m.
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
= 87.92 mm

66
Loads
Total = 5.875 KN/m
Wu = 8.813 KN/m
Design moments
Case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
span
position
Mu d (Ast)
reqd
dia spacing (Ast)
provided
spacing
a) Along Short span
Support 5.82 91 185 8 272 219 230
Mid span 4.36 91 137 8 367 219 230
b) Along Long span
Support 4.36 83 151 8 332 219 230
Mid span 3.30 83 113 8 443 219 230

67
Pt(reqd) = 0.151% < 0.25% Hence safe
Pt(prov) = 0.24%
fs = 151N/mm²
for Pt = 0.24%
α1 = 1.62
d(reqd) = 87mm < 91 mm Hence safe
Distribution steel
using Fe 415 steel
Ast = 99.6 mm²
230 mm c/c
Torsion steel
230 mm c/c
230 mm c/c = 219 mm² > 51 mm² Hence safe

68
Ly / Lx = 1.54 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth:
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
= 52.29 mm

69
Loads
Total = 5.500 KN/m
Wu = 8.250 KN/m
Design moments
case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
Span
position
Mu d (Ast)
reqd
Dia spacing (Ast)
provided
spacing
a) Along Short span
Support 3.02 76 114 8 442 219 230
Mid span 2.30 76 86 8 585 219 230
b) Along Long span
Support 2.30 68 97 8 520 219 230
Mid span 1.23 68 51 8 984 219 230

70
Pt (prov) = 0.29 %
fs = 95 N/mm²
for Pt = 0.29 %
α1 = 1.51
d (reqd) = 59 mm < 76 mm Hence safe
Distribution steel
using Fe 415 steel
Ast = 81.6 mm²
230 mm c/c
Torsion steel
230 mm c/c
230 mm c/c

71
8.4.4. Design of One – Way slab
Balcony (S4)
Ly / Lx = 2.23 > 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Let α = 20
β = 1
γ = 1
δ = 1
λ = 1
d = 72.39 mm
Loads
Total = 4.310 KN/m
Wu = 6.465 KN/m

72
= 1.52 m
Factored Moment (Maximum at mid span)
Mu = Wu x l ² /8
(6.5 x 1.52 ²) / 8 = 1.86 kNm
= 6.465 x 1.45/2
= 4.68 kN
BM = 0.138 fck bd²
d = 26 mm
1.86 x 10^6 = 0.87 x 415x At ((80 - 415At/20 x 1000))
5145.1 = 80 At - 0.02075At²
At = 65.42mm² (from Caluculator)
= 218.63 mm² > 65.42 mm² Hence O.K.
concrete area will be Provided in the longitudinal direction.
= 0.15 x 1000 x 100 / 100
= 150 mm²
= 218.63 mm² > 150 mm² Hence O.K.

73
Check for shear
= 100 x 218.63 / 1000 x 80
= 0.27 %
τc = 0.38 N/mm²
τc' = k x τc
τc' = 1.3 x 0.378
τc' = 0.49 N/mm²
= 4.68 x 1000 /1000 x 80
= 0.06 < 0.49 Hence O.K.
= 0.87 x 415 x 218.6335 x (80 - ((415 x 218.6335 / (20 x 1000)))
= 5956899.1 Nm = 5.96 kNm
V = 4679.74 N
Ld ≤ 1.3 M1/V
47 Ф ≤ 1.3 x 5956899.1 / 4679.74
1655 mm
mm

74
Percent tension steel at midspan
= 100 x 218.63 / 1000 x 80
= 0.27 %
β = 1
δ = 1
λ = 1
Allowable L/d = 20 x 1.42 = 28.4
Actual L/d = 1516.19 / 80 = 19.0 < 28.4 Hence O.K.

75
8.5. Design of Slabs (Flat 4)
8.5.1. Interior Panel (S1)
Span: long span = Ly = 5.49m
Short span = Lx = 3.51m
Ly / Lx = 1.57 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
= 84.26 mm

76
Loads
Total = 5.750 KN/m
Wu = 8.625 KN/m
Design moments
Case No 1 BMC
Mu = α x Wu x Lx²
Short span
Long span
span
position
mu d (ast)
reqd
dia spacing (ast)
prov
spacing
a) Along Short span
Support 5.81 86 197 8 256 219 230
Mid span 4.46 86 149 8 338 219 230
a) Along Long span
Support 3.39 78 125 8 404 219 230
Mid span 2.54 78 93 8 543 219 230

77
Pt(reqd) = 0.17% <0.25% assumed Hence safe
Pt(prov) = 0.25%
fs = 164N/mm²
for Pt = 0.25%
α1 = 1.59
d (reqd) = 85mm < 86mm Hence safe

78
8.5.2. One Long Edge Discontinuous (S2)
Ly / Lx = 1.11 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
= 84.26 mm

79
Loads
Total = 5.750 KN/m
Wu = 8.625 KN/m
Design moments:
Case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
span
position
Mu d (Ast)
reqd
dia spacing (Ast)
provided
spacing
a) Along Short span
Support 4.74 86 159 8 317 219 230
Mid span 3.55 86 118 8 427 219 230
a) Along Long span
Support 3.55 78 131 8 385 219 230
Mid span 2.97 78 109 8 463 219 230

80
Pt(reqd) = 0.137 % < 0.25% Hence safe
Pt(prov) = 0.25%
fs = 130N/mm²
for Pt = 0.25%
α1 = 1.60
d(reqd) = 84mm < 86 mm Hence safe
Ast = 93.6 mm²
230 mm c/c
Torsion steel:
230 mm c/c
= 219 mm² > 88mm² Hence safe
230 mm c/c

81
8.5.3. One Long Edge Discontinuous (S3)
Ly / Lx = 1.22 < 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Trial depth
Cantilever 7
bars.
Assume Pt = 0.25%
For Pt = 0.25%
= 58.62 mm

82
Loads
Total = 5.500 KN/m
Wu = 8.250 KN/m
Design moments
Case No 2 BMC
Mu = α x Wu x Lx²
Short span
Long span
span
position
Mu d
(Ast)
reqd
dia spacing
(Ast)
prov
spacing
a) Along Short span
Support 2.38 76 89 8 565 219 230
Mid span 1.79 76 67 8 755 219 230
b) Along Long span
Support 1.81 68 76 8 664 219 230
Mid span 1.37 68 57 8 883 219 230

83
Pt(reqd) = 0.09% < 0.25% Hence safe
Pt(prov) = 0.29%
fs = 73N/mm²
for Pt = 0.29%
α1 = 1.51
Distribution steel
Using Fe 415 steel
Ast = 81.6 mm²
230 mm c/c
Torsion steel
230 mm c/c
Torsional steel required =3/8xAst= 25 mm²
230 mm c/c =219 mm² > 25 mm²
Hence safe
for a distance of Ly/5 = 594.36 mm in both directions

84
8.5.4. Design of One – Way slab
Balcony (S4)
Ly / Lx = 3.16 > 2.00 m
fck = 20 N/mm²
fy = 415 N/mm²
Let α = 20
β = 1
γ = 1
δ = 1
λ = 1
d = 80.01 mm
Loads
Total = 4.500 KN/m
Wu = 6.750 KN/m
Effective depth = 100.01 – 20 - 4
= 1.68 m

85
Factored Moment (Maximum at mid span)
Mu = Wu x l ² /8
(6.8 x 1.68 ²) /8 = 2.37 kNm
= 6.750 x 1.60 / 2
= 5.40 kN
BM = 0.138 fck bd²
d = 29 mm
2.37 x10^6 = 0.87 x 415x At ((80-415At/ 20 x 1000))
6566.4=80At -0.02075At²
At =83.9mm² (from Calculator)
=218.63 mm² > 83.9 mm² Hence O.K.
concrete area will be provided in the longitudinal direction.
= 0.15 x 1000 x 100 /100
= 150 mm²
= 218.63 mm² > 150 mm² Hence O.K.

86
Check for shear
= 100 x 218.63 / 1000 x 80
= 0.27%
τc = 0.38 N/mm²
τc' = k x τc
τc' = 1.3 x 0.378
τc' = 0.49 N/mm²
= 5.40 x 1000 / 1000 x 80
= 0.07 < 0.49 Hence O.K.
The slab is safe in shear
= 0.87 x 415 x 218.63 x (80- ((415 x 218.63) / (20 x 1000)))
= 5956899.1 Nm = 5.96 kNm
V = 5400.98 N
Let us assume anchorage length Lo =
Ld ≤ 1.3 M1/V
47 Ф ≤ 1.3 x 5956899.1 / 5400.98
1434 mm
The Code requires that bars must be carried in to the supports by at least Ld/3 =100.01
mm
Percent tension steel at mid span

87
= 100 x 218.63 / 1000 x 80
= 0.27 %
γ = 1.42 (fig 10.1, page no 230 in ak jain)
β = 1
δ = 1
λ = 1
Allowable L/d =20 x 1.42 = 28.4
Actual L/d =1676.21 / 80 = 21.0 < 28.4 Hence O.K.

88
8.6. DESIGN OF STAIR CASE
Data: Stair case Room = 4496mm x 2438 mm
Floor to floor height (H) = 3.2 m = 3200mm
Live load for Residential buildings = 3 KN/m²
Functional Design:
Rise(R) = 150 mm
Tread (T) = 300 mm
T
T
R
Sec
2
2



= 1.118
No. of risers required = H/R = 21
No. of risers in each of the two flights = 11
No. of Treads per flight = 10
Going = 300x10 = 3000 mm
Width of landing at end = 4266 - 3000
= 1266 mm
Design of Flight-I:
Assume thickness of waist slab = 175mm
Dead Load = 25 x 0.175 x 1.118 = 4.89 kN/m²
Live Load = = 3.00 kN/m²
Floor Finish = = 1.00 kN/m²
Weight of steps = 24 x 150 / 2 = 1.80 kN/m²
Total Load = = 10.69 kN/m²
Factored Load (w) = = 1.5 x 10.69 = 16.04 kN/m²
Mu = wl²/8
= 16.04 x 4.50 x 4.50/8
= 40.5182 kNm
Span L = 3000 + 1266 + 230
= 4496 mm horizontally,

89
Mu/bd² = 40.518 / 1000 x 175x 175
= 1.32
pt = 0.4 (refer pillai and menon page no 839 to 842)
Modification factor α = 1.32 (refer IS 456-1978, PAGE NO: 57)
Basic L/d Ratio = 20 for simply supported slab.
= 20 x 1.32 = 26.4
L/d Ratio = 4496 / 26.4 = 170.3
Actual L/d Ratio = 4496 / 175 = 25.7 < 26.4 Hence o.k.
Ast = 0.4 x 1000 x 175 / 100
= 700 mm²
Using 10 mm dia
Spacing = 0.7853 x 10² x 1000/ 700
= 112.2 mm
Provide 10 mm dia @ 110 mm c/c
Distribution:
For Fe415, Pt = 0.12% of bd
Ast= 0.12 x 1000 x 175 / 100 = 210 mm
Using 8 mm dia
Spacing = 0.7853 x 8² x 1000 /210
= 239.3 mm
Provide 8mm dia @ 230 mm
Design of Flight-II
Assume thickness of waist slab = 175 mm
Mu = wl²/10
= 16.04 x 4.50 x 4.50/10
= 32.4145 kNm
Mu/bd² = 32.415 / 1000 x 175 x 175
= 1.0584
pt = 0.420 (refer pillai and menon)
Modification factor α = 1.36 (refer IS 456-1978, PAGE NO:57)
Basic L/d Ratio = 20 for simply supported slab.
= 20 x 1.36 = 27.2

90
L/d Ratio = 4496 / 27.2 = 165.29
Actual L/d Ratio = 4496/175 =25.69 < 27.2 Hence o.k.
Ast = 0.42 x 1000 x 175 / 100 = 735 mm²
Using 10 mm dia
Spacing = 0.7853 x 10² x 1000/ 735
= 106.8 mm
Provide 10mm dia @ 100 mm c/c
Distribution
For Fe415, Pt = 0.12% of bd
Ast = 0.12 x 1000 x 175 / 100
= 210 mm
Using 8 mm dia
Spacing = 0.7853 x 8² x 1000 /210
= 239.3 mm
Provide 8 mm dia @230 mm c/c

91
Fig.8.12. Reinforcement details of Staircase (Flight I).

92
Fig.8.13. Reinforcement details of Staircase (Flight II).

93
8.7. BEAMS
Beams transfer load from slabs to columns .Beams are designed for bending.
In general we have two types of beam: single and double. Similar to columns
geometry, perimeters of the beams are assigned. Design beam command is assigned and
analysis is carried out, now reinforcement details are taken.
A reinforced concrete beam should be able to resist tensile, compressive and shear
stress induced in it by loads on the beam.
There are three types of reinforced concrete beams
 single reinforced beams
 double reinforced concrete
 flanged beams
8.7.1. Singly reinforced beams:
In singly reinforced simply supported beams steel bars are placed near the bottom
of the beam where they are more effective in resisting in the tensile bending stress. In
cantilever beams reinforcing bars are placed near the top of the beam, for the same.
8.7.2. Doubly reinforced concrete beams:
It is reinforced under compression and tension regions. The necessity of steel of
compression region arises due to two reasons. When depth of beam is restricted. The
strength availability singly reinforced beam is in adequate. At a support of continuous
beam where bending moment changes sign such as situation may also arise in design of a
beam circular in plan.
Figure shows the bottom and top reinforcement details at three different sections.
These calculations are interpreted manually.

94
8.8. Mid Landing Beam (Manual)
Beam Mark : Mid Landing Level
End Condition : Simply Supported at Both Ends
Room size : 4496mm x 2438 mm
Span L : 2.667 m = 2667 mm
Section:
b = 230 mm = 0.23 m
D = 380 mm = 0.38 m
Cover = 40 mm = 0.04 m
d = 340 mm = 0.34 m
Slab thickness = 175 mm = 0.175 m
fck = 20 N/mm²
fy = 415 N/mm²
Loads: Due to supported slabs
Left : Landing of stair flights - I & II
w1 = 10.69 kN/m² (from staircase design)
L =
1000
)
230
4496
( 
= 4.27 m
Right: Nil
Load from : Stair flights - I & II =
2
27
.
4
69
.
10 
= 22.80 kN/m
Grill 0.9144 m high x
2
667
.
2
= 1.22 kN/m
Self Weight, 25 x 0.23 x 0.21 = 1.18 kN/m
Total Working Load = 25.20 kN/m
Ultimate Load: 1.5 x 25.20 = 37.80 kN/m
a) Design Moment: wL²/8 = 37.80 x 2.667²/8
= 33.61 kN m

95
b) Max. Ultimate Moment of Resistance of Rectangular Section
Mu Max. = RuMax.bd²
RuMax = 2.76 N/mm² for M20, Fe415
= 2.76 x 230 x340²/10^6
= 73.38 kN m > 33.61 kN m
Main Steel:
Req. Ast = bd
bd
f
M
f
f
ck
u
y
ck









 2
6
.
4
1
1
5
.
0
= 340
230
340
230
20
10
61
.
33
6
.
4
1
1
415
20
5
.
0
2
6


















= 297.41 mm²
Provide Bottom st.: 3 no.s of 12 mm dia
= 3
12
4
2



= 339.12 mm²
Provide Top st.: 3 no.s of 12 mm dia
= 3
12
4
2



= 339.12 mm²
Check for width:
Req. b = n1 x Ø + (n1 + 1) x 25+2xØ/2 (n1 = No. of rods)
= )
2
12
2
(
)
25
)
1
3
(
)
12
3
(( 





= 148 mm < 230 mm
Check for Effective Cover:
For Mild Environment nominal cover = 20 mm
Assuming 6 mm stirrups
d' = Nominal cover + Dia of stirrups + Ø / 2
= 20 + 6 + 6 = 32 mm < 40 mm

96
Design for Shear:
Vu Max. : wL/2 = 37.80 x 2.667 /2 = 50.41 kN
Vuc : Tuc bd, Tuc depends upon pt% at support
Ast1: 3 bars of 12 mm dia
= 3
12
4
2



= 339.12 mm² (tension steel at support)
bd
A
p st
t
100
 =
340
230
12
.
339
100


= 0.43 %
For M20, Fe415:
0.25% = 0.36
0.5% = 0.48
0.43% =?
By interpolation: (IS: 456-2000, table: 19)
= )
25
.
0
43
.
0
(
25
.
0
50
.
0
36
.
0
48
.
0
36
.
0 









 = 0.45N/mm²
Tuc : 0.45 N/mm²(by interpolation)
Vuc: 0.45 x 230 x 340 / 1000 = 35.05 kN
Shear Resistance of Minimum Stirrups:
Vusv.min: 0.4bd (from page No:64 in Limit State Theory for R.C.
Members Book:4.4.8)
=0.4 x 230 x 340 /1000 = 31.28 kN
Vur.min: Vuc + Vusv.min
35.05 + 31.28 = 66.33 kN > 50.41 kN
Min. stirrups are sufficient
Provide 2 bars of 6 mm dia = 2
6
4
2



= 56.52 mm²
Pitch S =
b
a
f st
y
4
.
0
87
.
0 

fy =250 N/mm²
=
230
4
.
0
54
.
56
250
87
.
0



= 133.62 mm
Provide 6mm dia @ 130 mm c/c < 0.75 x 340 = 255 mm and < 300 mm

97
Check for Development Length:
Req. Ld = 47Ø = 47 x 12 = 564 mm < 0
1
3
.
1
L
V
M

M1 = Moment of Resistance of the beam at the section, assuming all bars stressed to
0.87 fy. When 50% bars are available at support, M1 can be approximately taken equal
to Mmax./2,
M1 = Mu = 








bd
f
A
f
d
A
f
ck
st
y
st
y 1
87
.
0
Ast1: 3 bars of 12 mm dia = 339.12 mm²
M1 = 6
10
340
230
20
12
.
339
415
1
340
12
.
339
415
87
.
0 














 = 37.9 kNm
V = Vu max.= 50.41 kN
Lo =
2
3
s
d b
L
 Where bs is width of support
Assume bs = 230 mm
Lo =
2
230
3
564
 = 73 mm
Available Ld = 73
41
.
50
1000
9
.
37
3
.
1



= 1050.38 mm > 564mm O.k.
fs =
prov
st
req
st
y
A
A
f
.
.
58
.
0
=
12
.
339
41
.
297
415
58
.
0 

= 211.09 N/mm² say 240 N/mm²
Pt prov =
bd
A ov
st Pr
.
100
=
340
230
12
.
339
100


= 0.43 %
For fs = 240 & Pt prov = 0.43 %
α = 1.3 (IS 456:2000, fig 4, page No: 38)
Basic L/d ratio = 20 for simply supported beams
Required d =
3
.
1
20
2667

= 102.58 mm << 340 mm
Load on Column = 37.80 x 2.667 /2 = 50.41 kN

98
Fig.8.14. Reinforcement details of Mid landing Beam (Manual)

99
8.9. Mid Landing Beam (using Staad without earthquake, Beam No:
519)
Fig.8.15. Reinforcement details of mid landing Beam (Using Staad without EQ)

100
Fig.8.16. Deflection of Mid landing Beam (Using Staad without EQ)

101
Fig.8.17. Shear & Bending values of mid landing Beam (Using Staad without EQ)

102
8.10. Mid Landing Beam (using Staad with earthquake, Beam No: 1673)
Fig.8.18. Reinforcement details of Mid landing Beam (Using Staad with EQ)

103
Fig.8.19. Deflection of Midlanding Beam (Using Staad with EQ)

104
Fig.8.20. Shear & Bending values of mid landing Beam (Using Staad with EQ)

105
8.11. Design of floor beam (Manual, Trapezoidal load on beam)
Given data:
End Condition : One end Simply Supported and other continuous
Span : Lx : 3.58 m = 3580 mm
Ly : 4.27 m = 4267 mm
Floor to Floor Height = 3200 mm = 3.2 m
Section:
b = 230 mm = 0.230 m
D = 500 mm = 0.500 m
Cover = 40 mm = 0.040 m
d = 460 mm = 0.460 m
Slab thickness Df = 115 mm = 0.115 m
fck = 20 N/mm²
fy = 415 N/mm²
Self Weight (ws) 25 x 0.230 x 0.39 = 2.21 kN/m
Wall load 19.5 x 0.230 x 3.20 = 14.35 kN/m
Loads: self weight + wall +slab (trapezoidal two way)
w = 6.13 kN/m² (from slab design)
wu = 1.50 x (weqb)
In the case of simply supported at one end and continuous at the other reduce the
load at simply supported end by 10% (i.e., take shear coefficient = 0.45) & increase the
same by 20% at the continuous end (i.e., take shear coefficient = 0.6) & 25% at
continuous end of two span continuous beam.
Shear Co-efficient = 0.45

106
For bending 








 2
3
1
1
2 
x
eqb
wL
w where β is Ly/Lx
19
.
1
58
.
3
27
.
4



= 













 
2
19
.
1
3
1
1
2
27
.
4
13
.
6
=10.00 kN m
= 1.50((ws + wall load +shear coefficient + weqb)
= 1.5 x (2.21+14.35+0.45+10) = 40.52 kN/m
For Shear = 










2
1
1
2
x
eqs
wL
w where β is Ly/Lx
= 













 
19
.
1
2
1
1
2
27
.
4
13
.
6
= 7.58 kN m
= 1.50 (ws + wall load + shear coefficient + weqs)
= 1.5 x (2.21+14.35+0.45+7.58) = 36.90 kN/m
Design Moment:
Mu = wL²/10 = 40.52 x 4.27 ²/10
= 73.79 kN m at support as well as mid span.
Main Steel:
a) At intermediate support:
Mu = wL²/10 = 40.52 x 4.27²/10
= 73.79 kN m
ku limit = 0.40
Mur limit = 0.36 fck kulimit (1-0.42kulimit)bd
= 16.62 kN m > 73.79 kN m
The Section is singly reinforced
Req. Ast = bd
bd
f
M
f
f
ck
u
y
ck









 2
6
.
4
1
1
5
.
0
= 460
230
460
230
20
10
79
.
73
6
.
4
1
1
415
20
5
.
0
2
6


















= 492 mm²

107
b) At Mid span Section:
Mu = wL²/10 = 40.52 x 4.27²/10
= 73.79 kN m
The beam is designed as L - Beam,
Lo = 0.8 x 4267 = 3413.76 mm
w
f
o
f b
D
L
b 

 3
12
= 230
)
115
3
(
12
76
.
3413


 = 859.5 mm
for xu=Df,(Mur1) = 0.36 fck bf Df (d-0.42Df)
= 0.36 x 20 x 859.5 x 115 x(460-(0.42 x 115))x 10^-6
= 293 kN m > 73.79 kN m
Req. Ast = bd
bd
f
M
f
f
ck
u
y
ck









 2
6
.
4
1
1
5
.
0
= 460
5
.
895
460
5
.
895
20
10
79
.
73
6
.
4
1
1
415
20
5
.
0
2
6


















= 455 mm²
c) Detailing:
Provide
Top (at mid span) 2 no.s of 12 mm dia = 226 mm²
Top (at continuous end) 3 no.s of 16 mm dia = 603 mm²
Bottom (at mid span) 3 no.s of 16 mm dia = 603 mm²
Bottom (at continuous end) 2 no.s of 16mm dia = 402 mm²
Ast At Mid span At continuous end
Required At top ------- 492
At Bottom 455 -------
Provided At top 226 603
At Bottom 603 402

108
Design for Shear:
a) At Continuous End:
Vu Max. 0.6 wusL = 0.60 x 36.90x 4.27
= 94.48 kN
Ast1 : 4 bars of 16 mm dia = 804 mm² (tension steel at support)
bd
A
p st
t
100
 =
460
230
804
100


= 0.76 %
for M20, Fe415:
0.75% = 0.56
1.00% = 0.62
0.76% =?
= )
75
.
0
76
.
0
(
75
.
0
00
.
1
56
.
0
62
.
0
56
.
0 









 = 0.56N/mm²
Tuc : 0.56 N/mm² (by interpolation) (IS:456-2000, table:19)
Vuc: 0.56 x 230 x 460/ 1000 = 59.25 kN
Vusv.min: 0.4bd (from page No: 64 in Limit State Theory for R.C.
Members Book: 4.4.8)
0.4 x 230 x 460/1000= 42.32 kN
59.25 + 42.32 = 101.57 kN > 94.48 kN
6
4
2



= 56.52 mm²
Pitch S =
b
a
f st
y
4
.
0
87
.
0 

fy =250 N/mm²
=
230
4
.
0
54
.
56
250
87
.
0



= 133.62 mm

109
Req. Ld = 47Ø = 47 x 16 = 752 mm < 0
1
3
.
1
L
V
M

0.87 fy. when 50% bars are available at support, M1 can be approximately taken equal
to Mmax./2,
M1 = Mu = 








bd
f
A
f
d
A
f
ck
st
y
st
y 1
87
.
0
Ast1 : 4 bars of 16 mm dia = 803.84 mm²
M1 = 6
10
460
230
20
84
.
803
415
1
460
84
.
803
415
87
.
0 














 =112.5 kNm
V = Vu max. = 94.48 kN
Lo =
2
3
s
d b
L
Assume bs = 230 mm
Lo =
2
230
3
752
 = 135.667 mm
Available Ld = 667
.
135
48
.
94
1000
5
.
112
3
.
1



= 1683.09 mm > 752 mm O.k.
fs =
prov
st
req
st
y
A
A
f
.
.
58
.
0
=
603
492
415
58
.
0 

= 193.39 N/mm² say 240 N/mm²
Pt prov =
bd
A ov
st Pr
.
100
=
460
230
603
100


= 0.57 %
For fs = 240 & Pt prov = 0.76 %
α1 = 1.09 (IS 456:2000, fig 4, page No: 38)
For the flanged section detailed check for deflection is carried out.

110
bw/bf = 230 / 859.5 = 0.27
α3 = 0.8 (IS 456:2000, fig 6, page No: 39)
d = L / ( Basinc L/d ratio x α1 x α3)
= 4267.20 / ( 20 x 1.09 x 0.8 )
= 244.68 mm << 460 mm
Load on Column = 94.48 kN

111
Fig.8.21. Reinforcement details of Beam (Manual, Trapezoidal load on beam)

112
8.12. Design of floor beam (Using Staad without EQ,
Trapezoidal load on beam, Beam No: 852)
Fig.8.22. Reinforcement details of Beam (Using Staad without EQ, Trapezoidal load on
beam)

113
Fig.8.23. Deflection details of Beam (Using Staad without EQ, Trapezoidal load on
beam)

114
Fig.8.24. Shear & Bending results of Beam (Using Staad without EQ, Trapezoidal load
on beam)

115
8.13. Design of floor beam (Using Staad with EQ, Trapezoidal
load on beam, Beam No: 852)
Fig.8.25. Reinforcement details of Beam (Using Staad with EQ, Trapezoidal load on
beam)

116
Fig.8.26. Deflection details of Beam (Using Staad with EQ, Trapezoidal load on beam)

117
Fig.8.27. Shear & Bending results of Beam (Using Staad with EQ, Trapezoidal load on
beam)

118
8.14. Design of floor beam (Manual, Triangular load on beam):
Given data:
End Condition : One end Simply Supported and other continuous
Span Lx : 3.581m = 3581 mm
Ly : 4.267m = 4267 mm
Floor to Floor Height = 3200 mm = 3.2 m
Section:
b = 230 mm = 0.230 m
D = 380 mm = 0.380 m
Cover = 40 mm = 0.040 m
d = 340 mm = 0.340 m
Slab thickness Df = 115 mm = 0.115 m
fck = 20 N/mm²
fy = 415 N/mm²
Self Weight (ws) = 25 x 0.230 x 0.27 = 1.52 kN/m
Wall load = 19.5 x 0.23 x 3.20 = 14.35 kN/m
Loads = self weight + wall + slab (tri two way)
w = 6.13 kN/m² (from slab design)
wu = 1.50x (weqb)
For Bending =
3
x
eqb
wL
w  =
3
581
.
3
13
.
6 
= 7.31 kN m
= 







3
5
.
1 x
s
wL
wallload
w
=  
31
.
7
35
.
14
52
.
1
5
.
1 
 = 34.78 kN/m
For Shear =
4
x
eqs
wL
w  =
4
581
.
3
13
.
6 
= 5.48 kN m
= 







4
5
.
1 x
s
wL
wallload
w
= 1.5 x (1.52 + 14.35 + 5.48) = 32.04 kN/m

119
a) Design Moment:
wL²/8 = 34.78 x 3.58²/8
= 55.77 kN m at mid span and at continuous support.
Mid span Section: L-Section
Lo = 0.7 x 3.58 = 2.51 m = 2510 mm
w
f
o
f b
D
L
b 

 6
12
= 230
)
115
6
(
12
2510


 = 1128.9 mm
for xu=Df,(Mur1) = 0.36 fck bf Df (d-0.42Df)
= 0.36 x 20 x 1128.9 x 115 x (340-(0.42 x 115))x 10^-6
= 273 kN m > 55.77 kN m
Main Steel:
Req. Ast = bd
bd
f
M
f
f
ck
u
y
ck









 2
6
.
4
1
1
5
.
0
= 340
9
.
1128
340
9
.
1128
20
10
77
.
55
6
.
4
1
1
415
20
5
.
0
2
6


















= 466 mm²
Provide Bottom st.: 2 no.s of 16 mm dia = 402 mm²
Provide Bottom st.: 1 no.s of 12 mm dia = 113 mm²
Total = 515 mm²
Support Reaction: Rect. Section
Murmax = RuMax.bd²
RuMax = 2.76 N/mm²
= 2.76 x 230 x340²/10^6
= 73.38 kN m > 55.77 kN m
Section is singly reinforced
Req. Ast = bd
bd
f
M
f
f
ck
u
y
ck









 2
6
.
4
1
1
5
.
0
= 340
230
340
230
20
10
77
.
55
6
.
4
1
1
415
20
5
.
0
2
6



















120
Req. Ast = 529 mm²
Provide 2 no.s of 16 mm dia= 402 mm²
Provide 2 no.s of 12 mm dia= 226 mm²
Total =628 mm²
Design for Shear:
a) at Continuous End: Ast = 628 mm²
Vu Max. : wu L/2 = 32.04 x 3.5814 /2
= 57.37 kN
Ast1 : 2 bars of 16 mm dia = 401.92 mm² (tension steel at support)
bd
A
p st
t
100
 =
340
230
92
.
401
100


= 0.51 %
for M20, Fe415:
0.50% = 0.48
0.75% = 0.56
0.51% =?
= )
50
.
0
51
.
0
(
50
.
0
75
.
0
48
.
0
56
.
0
48
.
0 









 = 0.48N/mm²
Tuc : 0.48 N/mm² (by interpolation) (IS:456-2000, table:19)
Vuc: 0.48 x 230 x 340 / 1000 = 37.54 kN
Vusv.min: 0.4bd (from page No: 64 in Limit State Theory for R.C.
Members Book: 4.4.8)
0.4 x 230 x 340 /1000 = 31.28 kN
37.54 + 31.28 = 68.82 kN > 57.37 kN

121
6
4
2



= 56.52 mm²
Pitch S =
b
a
f st
y
4
.
0
87
.
0 

fy =250 N/mm²
=
230
4
.
0
54
.
56
250
87
.
0



= 133.62 mm
Req. Ld = 47Ø = 47 x 16 = 752 mm < 0
1
3
.
1
L
V
M

0.87 fy. When 50% bars are available at support, M1 can be approximately taken equal
to Mmax. /2,
M1 = Mu = 








bd
f
A
f
d
A
f
ck
st
y
st
y 1
87
.
0
Ast1 : 2 bars of 16 mm dia = 401.92 mm²
M1 = 6
10
340
230
20
92
.
401
415
1
340
92
.
401
415
87
.
0 














 = 44.76 kNm
V = Vu max. = 57.37 kN
Lo =
2
3
s
d b
L
Assume bs = 230 mm
Lo =
2
230
3
752
 = 135.667 mm
Available Ld = 667
.
135
37
.
57
1000
76
.
44
3
.
1



=1149.92 mm > 752 mm O.k.

122
For the flanged section detailed check for deflection is carried out.
Pt prov =
bd
A ov
st Pr
.
100
=
340
230
628
100


= 0.80 %
fs =
prov
st
req
st
y
A
A
f
.
.
58
.
0
=
515
466
415
58
.
0 

= 217.79 N/mm² say 240 N/mm²
For fs = 240 & Pt prov = 0.80 %
α1 = 1.08 (IS 456:2000, fig 4, page No: 38)
bw/bf = 230 / 1128.915 = 0.20
α3 = 0.8 (IS 456:2000, fig 6, page No: 39)
d = L / (Basic L/d ratio x α1 x α3)
= 3581.4 / (20 x 1.08 x 0.8)
= 207.25 mm << 340 mm
Load on Column = 57.37kN

123
Fig.8.28. Reinforcement details of Beam (Manual, Triangular load on beam)

124
8.15. Design of floor beam (Using Staad without EQ, Triangular load on
beam, Beam No: 939)
Fig.8.29. Reinforcement details of Beam (Using Staad without EQ,
Triangular load on beam)

125
Fig.8.30. Deflection details of Beam (Using Staad without EQ, Triangular
load on beam)

126
Fig.8.31. Shear & Bending results of Beam (Using Staad without EQ,

127
8.16. Design of floor beam (Using Staad with EQ, Triangular load on
beam, Beam No: 939)
Fig.8.32. Reinforcement details of Beam (Using Staad with EQ, Triangular
load on beam)

128
Fig.8.33. Deflection details of Beam (Using Staad with EQ, Triangular load
on beam)

129
Fig.8.34. Shear & Bending results of Beam (Using Staad with EQ,

130
8.17. COLUMNS
A column or strut is a compression member, which is used primary to support
axial compressive loads and with a height of at least three its lateral dimension.
A reinforced concrete column is said to be subjected to axially loaded when line
of the resultant thrust of loads supported by column is coincident with the line of C.G 0f
the column in the longitudinal direction.
Depending upon the architectural requirements and loads to be supported, R.C
columns may be cast in various shapes i.e square, rectangle, and hexagonal, octagonal,
circular. Columns of L shaped or T shaped are also sometimes used in multistoried
buildings.
The longitudinal bars in columns help to bear the load in the combination with the
concrete. The longitudinal bars are held in position by transverse reinforcement, or lateral
binders.
The binders prevent displacement of longitudinal bars during concreting operation
and also check the tendency of their buckling towards under loads.
8.17.1 Positioning of columns:
Some of the guiding principles which help the positioning of the columns are as follows:-
 Columns should be preferably located at or near the corners of the building and at
the intersection of the wall, but for the columns on the property line as the
following requirements some area beyond the column, the column can be shifted
along a cross wall to provide the required area for the footing with in the property
line. Alternatively a combined or a strap footing may be provided.
 The spacing between the columns is governed by the spans of supported beams, as
the spanning of the column decides the span of the beam. As the span of the of the
beam increases, the depth of the beam, and hence the self weight of the beam and
the total.

131
8.17.2. Effective length:
The effective length of the column is defined as the length between the points of
contra flexure of the buckled column. The code has given certain values of the effective
length for normal usage assuming idealized and conditions shown in appendix D of IS -
456(table 24)
A column may be classified as follows based on the type of loading:
 Axially loaded column
 A column subjected to axial load and uni-axial bending
 A column subjected to axial load and biaxial bending.
8.17.2.1 Axially loaded columns:
All compression members are to be designed for a minimum eccentricity of load
into principal directions. In practice, a truly axially loaded column is rare, if not
nonexistent. Therefore, every column should be designed for a minimum eccentricity
.clause 22.4 of IS code
E min= (L/500) +(D/300) ,subjected to a minimum of 200 mm.
Where L is the unsupported length of the column (see 24.1.3 of the code for
definition unsupported length) and D is the lateral dimension of the column in the
direction under the consideration.
8.17.2.2. Axial load and uni-axial bending:
A member subjected to axial force and bending shall be designed on the basis of
 The maximum compressive strength in concrete in axial compression is
taken as 0.002
 The maximum compressive strength at the highly compressed extreme
fiber in concrete subjected to highly compression and when there is no
tension on the section shall be 0.0035-0.75 times the strain at least
compressed extreme fiber.
 Design charts for combined axial compression and bending are in the form
of Inter section diagram in which curves for Pu/fck bD verses Mu/fck bD2 are
plotted for different values of p/fck where p is reinforcement percentage.

132
8.17.2.3. Axial load and biaxial bending:
The resistance of a member subjected to axial force and biaxial bending shall be
obtained on the basis of assumptions given in 39.1 and 39.2 (IS 456:2000)
Alternatively such members may be designed by the following equation
(Mux/ Muy)αn +(Muy/ Muy1)αn<=1.0
Mux&Muy=moment about x and Y axis due to design loads
Mux1&Muy1=maximum uniaxial moment capacity for an axial load
of Pu bending about x and y axis respectively.
αn is related to Pu/puz
puz=0.45 x fck x Ac + 0.75 x fy x Asc
For values of pu/Puz=0.2 to 0.8, the values of αn vary linearly from 1.0 to 2.0 for
values less than 0.2, αn is values greater than 0.8, αn is 2.0
The main duty of column is to transfer the load to the soil safely. Columns are
designed for compression and moment. The cross section of the column generally
increases from one floor to another floor due to the addition of both live and dead load
from the top floors.
8.18. Column design
A column may be defined as an element used primary to support axial
compressive loads and with a height of a least three times its lateral dimension. The
strength of column depends upon the strength of materials, shape and size of cross
section, length and degree of proportional and dedicational restrains at its ends.
A column may be classify based on deferent criteria such as
 Shape of the section
 Slenderness ratio(A=L+D)
 Type of loading, land
 Pattern of lateral reinforcement.
The ratio of effective column length to least lateral dimension is defined as
slenderness ratio.

133
In our structure we have 3 types of columns.
1.) Column with beams on two sides
2.) Columns with beams on three sides
3.) Columns with beams on four sides
So we require three types of column sections. So create three types of column sections
and assign to the respective columns depending on the connection. But in these structure
we adopted same cross section throughout the structure with a rectangular cross section
.In foundations we generally do not have circular columns if circular column is given it
makes a circle by creating many lines to increase accuracy.
The column design is done by selecting the column and from geometry page assigns
the dimensions of the columns. Now analyze the column for loads to see the reactions and
total loads on the column by seeing the loads design column by giving appropriate
parameters like.
1. Minimum reinforcement, max, bar sizes, maximum and minimum spicing.
2. Select the appropriate design code and input design column command to the entire
column.
3. Now run analysis and select any column to collect the reinforcement details

134
8.19. Column Design (Manual)
Floor to Floor height = 3.20 m = 3200 mm
Height of Plinth above G.L. = 0.60 m = 600 mm
Depth of Foundation below G.L. = 1.83 m = 1829 mm
Depth of Footing = 0.23 m = 230 mm
Total height of column above top of footing = 5.40 m = 5399 mm
Depth of shallowest Beam = 0.42 m = 415 mm
Unsupported length of column L = 4.98 m = 4984 mm
fck = 20 N/mm²
fy = 415 N/mm²
Assuming effective length Leff = L since all columns are supported by beams in both the
directions and there are longitudinal and transverse external walls.
Effective length of the column Leff = 4.984 m = 4984 mm
Size of the Column = 230 x 450 or
= 0.23 x 0.45
Slenderness Ratio Leff / b = 4984 / 230
= 21.67 > 12 column is Slender
Allowance for slenderness = %
100
1
1










r
C
Where
b
L
C
eff
r
48
25
.
1 










230
48
4984
25
.
1
r
c = 0.798
Allowance for slenderness = 100
1
798
.
0
1







 = 25.31% say 26%
Factored self weight of column = 19.34 kN say 19.5 kN
Total Load on column = 49.32 + 94.48 + 57.37 = 201.17 kN
Self weight on column = 19.5 kN
Total axial Load on column (Pu) = 220.67 kN

135
Main Steel:
Assume Ast = 4 no.s of 12 mm dia
= 4
12
4
2



= 452.38mm²
Smaller End Moment Mu1 = 0 at the footing end since the footing is designed as rotation
free. i.e., for axial load only.
Larger End Moment Mu2 = 0 at the top of column because there is no unbalanced
moment from beams meeting at the column on opposite faces.
Initial Moment Mi = 0 due to external moments.
Minimum Eccentricity emin.y =
30
500
b
L
 =
30
230
500
4984

= 17.63 mm < 20mm
emin.y = 20 mm
Minimum Moment Mu.min.y = Pu x emin.y = 4.4035 kN m
Revised Initial Moment = Mi = Mu.min.y = 4.4035 kN m
The above steps calculate the design moment for the column. The steps that follow
calculate the additional moments due to the effect of slenderness.
Strength in axial compression is obtained by using following equations:
Pu = Puc + Pus
Where Puc =λ (0.4fck b D)
where λ = 0.9 for b = 230 mm
Puc = 745.2 kN
Where Pus = λ (0.67fy-0.4fck)Asc
Pus = 0.243045Asc kN
Similarly, values of Puz & Pub are required in calculation of additional moment in the
case of slender column.
Puz = 0.45fck Ac + 0.75fy Asc or
= 0.45fck Ag + (0.75fy - 0.45fck)Asc
= 931.5 + 0.3023Asc kN

136
Ignoring the contribution of steel to strength in axial compression, Pub can be obtained
using following equation assuming diameter of link equal to 6 mm
Pub = 0.36 fck b (7/11) d
Where d = 230 - (40 + 6 + Ø/2)
= 230 - (46 + Ø/2)
= 184 - Ø/2
Pub = 1.05x (184-Ø/2)
For 12 mm dia Pub = 187.58 kN
for 16 mm dia Pub = 185.47 kN
The values of Puc, Pu, Puz and Pub are calculated for different N-Ø. Combinations
using the above relations and the same are presented in below table
Table 8.15.Load carrying capacities Pu, Pub & Pub in kN of Axially Loaded
columns Grade of Concrete - M20 & Steel Grade - Fe 415
Section Puc
Number Diameter combinations of Main Reinforcement
No.s 4 6 8 4 6 8
b D kN dia (mm) 12 12 12 16 16 16
230 450 745.2 Asc ( mm²) 452 678 904 804 1206 1608
Pus (kN) 110 165 220 195 293 391
Pu (kN) 855 910 965 941 1038 1136
Puz (kN) 1068 1136 1205 1174 1296 1417
Pub (kN) 188 188 187.58 185 185 185
The percentage allowance for uniaxial bending and bi - axial bending is taken as 15% &
33% respectively. The allowance for slenderness has been worked out as 11%. Based on
this after adding allowance for Slenderness are given below
Table 8.16. Design Details of Column
Column
mark
section
floor
load kN
self
wt.
Total
load
p.a.f.
15% &
33%
Equ.
Axial
load
p.a.s.
11%
Design
Load
Steel
25 230 450 201.17 19.5 221 66.38 287 24.22 311 452

137
Design of lateral ties:
Though theoretically minimum diameter of lateral tie is 1/4 the diameter of main bar or
5mm whichever is less, in practice, minimum diameter adopted is 6 mm.
Provide 6 mm dia ties of grade Fe250 at spacing equal to least of the following
i) Least lateral dimension b = 230 mm
ii)
16 times of dia of main bar =
for 12 mm = 192 mm
for 16 mm = 256 mm
iii) Minimum Pitch 300 mm
i.e., s = 192 mm for main bar of 12 mm diameter &
s = 256 mm for main bar of 16 mm diameter

138
Table 8.17. Reinforcement Details of Column

139
8.20. Column Design (using Staad without EQ.)
Fig.8.35. Reinforcement details of Column (using staad without EQ)

140
Fig.8.36. Deflection details of Column (using staad without EQ)

141
Fig.8.37. Shear & Bending results of Column (using staad without EQ)

142
8.21. Column Design (using Staad with EQ.)
Fig.8.38. Reinforcement details of Column (using staad with EQ)

143
Fig.8.39. Deflection details of Column (using staad with EQ)

144
Fig.8.40. Shear & Bending results of Column (using staad with EQ)

145
8.22. Footings
Foundations are structural elements that transfer loads from the building or
individual column to the earth .If these loads are to be properly transmitted, foundations
must be designed to prevent excessive settlement or rotation, to minimize differential
settlement and to provide adequate safety against sliding and overturning.
 Footing shall be designed to sustain the applied loads, moments and forces and
the induced reactions and to assure that any settlements which may occur will
be as nearly uniform as possible and the safe bearing capacity of soil is not
exceeded.
 Thickness at the edge of the footing in reinforced and plain concrete footing at
the edge shall be not less than 150 mm for footing on the soil or less than
300mm above the tops of the pile for footing on piles.
8.22.1. Bearing capacity of soil:
The size of foundation depends on permissible bearing capacity of soil. The total
load per unit area under the footing must be less than the permissible bearing capacity of
soil to the excessive settlements.
8.22.2. Foundation design:
Foundations are structure elements that transfer loads from building or individual
column to earth this loads are to be properly transmitted foundations must be designed to
prevent excessive settlement are rotation to minimize differential settlements and to
provide adequate safety isolated footings for multi storey buildings. These may be square
rectangle are circular in plan that the choice of type of foundation to be used in a given
situation depends on a number of factors.
 Bearing capacity of soil
 Type of structure
 Type of loads
 Permissible differential settlements
 economy

146
A footing is the bottom most part of the structure and last member to transfer the load.
Fig 8.41 Elevation and Plan of Isolated Footing

147
8.22.3. Footing Design (Manual)
Load = 220.67 KN
fy = 415 N/mm²
fck = 20 N/mm²
SBC = 350 KN/m²
Depth of footing = 1.829 m
Column size = 230 mm x 450 mm
Size of Footing
Assuming the weight of the footing + backfill to constitute 10 percent of Pu and
assuming a load factor of 1.50
Base area required = 46
.
0
5
.
1
350
1
.
1
67
.
220



For economical proportions, the cantilever projections (for flexural design) should be
approximately equal in the two directions.
Area of Footing Provided:
L = B = 1.22 x 1.22 = 1.49 m² > 0.46 m² Hence OK
Design of pedestal:
D = 




 
2
tan
)
( 
b
L
D ≥
 













1
max
100
2
90
.
0
ck
f
q
b
L
D ≥ 575 mm when b = 230mm
Hence provide a concrete block of 550 mm
Further it is necessary to provide minimum reinforcement to provide for tie action and to
account for temperature and shrinkage effects.
Ast min. = 0.12%bd
=
100
550
1000
12
.
0 

= 660 mm²

148
Provide 12 mm dia bars 6No.s
= 6
12
4
2



= 678.6 mm² > 660 mm²
Spacing =
st
A
d 2
4

=
6
.
678
12
4
2


x 1000 = 166.66 mm
= 165 mm c/c
Thickness of footing based on shear
Factored net soil pressure qu = Load /Area
2
/
5
.
148
220
.
1
220
.
1
67
.
220
m
kN
qu 

 = 0.148 N/mm²
(a) One - way shear
The critical section is located d away from the column face (refer fig)
Vu1 = 0.148x 1220 x (385 -d) N
= (69611.9-181.00d) N
Assuming 36
.
0

C
 N/mm² for M20 concrete with nominal Pt = 0.25
Vuc = 0.36 x 1220 x d = 439.2d N
Vu1 ≤ Vuc
(69611.9 - 181.00 d) ≤ 439.2d) N
d = 112.24 mm
(b) Two - way shear
The critical section is located d/2 from the column periphery all around
Vu2 = 0.148 x (1220² - (230+ d) x (450+d))
Assuming d ≥ 115 mm
Vu2 = 191434 N
= 191.43 kN
Two-way shear resistance, limiting shear stress of concrete
 
ck
s
CZ f
k 25
.
0


Where
d
b
ks 
 5
.
0 = 01
.
1
450
230
5
.
0 

but limited to 1.00
  118
.
1
20
25
.
0
00
.
1 



cz


149
Vuc = (1.118 ×230 + d) +(450+d)) ×2d
d = 115 mm
Vuc = 215542N
= 215kN > 191.73kN Hence OK
Hence, one way shear governs the thickness. As a square footing is provided and the one
way shear requirement is equally applicable in both directions.
Assuming 75 mm clear cover and 10 mm dia bars.
D ≥ 115 + 75 + 5 = 195 mm
Provide D = 230 mm and consider the average effective depth
d = 150 mm while designing for flexure
Check maximum soil pressure:
Assuming unit weight of concrete and soil as 24 KN/m³ and 18 KN/m³ respectively at
the factored loads,
q max-gross = )
23
.
0
18
(
)
23
.
0
24
(
22
.
1
22
.
1
67
.
220











= 158 KN/m² < 350 kN/m²
Design of flexural reinforcement:
Maximum cantilever projection = 385 mm (from face of column)
Mu = 0.148 × 1219.2 ×385²=13.386 KNm
= 13386359 Nm = 13.38kNm
Mu/bd² = 2
150
1220
13386359

= 0.48 N/mm²
Pt/100 =









 2
6
.
4
1
1
5
.
0
bd
f
M
f
f
ck
m
y
ck
= 




 



20
48
.
0
6
.
4
1
1
415
20
5
.
0
= 0.001369
Pt = 0.1369
Ast min. = 0.0012BD
= 0.0012 x 1220 x 230
= 336.72 mm²

150
Pt min. =
Bd
Ast min
100
=
150
1220
72
.
336
100


= 0.1840 > 0.13919
However, this reinforcement is less than assumed for one way shear design
(‫ז‬c=0.36N/mm²)
For which Pt = 0.25 For M20 concrete
Ast req. = 0.25% x Bd
= 0.25% x 1220 x 150
= 457.5 mm²
Using 10 mm Ф bars, number of bars required
= Ast req /area of one bar
= 8
.
5
10
4
5
.
457
2



Say 6No.s
Corresponding spacing
)
1
(
)
)
2
(cov
(





n
er
B
s

)
1
6
(
)
10
)
2
75
(
1220
(





s
=212 mm is acceptable

151
Table 8.18. Reinforcement details of footing (manual)

152
Fig 8.42 Reinforcement details of Isolated Footing (Manual)

153
8.22.4. Footing Design (Using Staad without EQ)
Load = 507 KN
fy = 415 N/mm²
fck = 20 N/mm²
SBC = 350 KN/m²
Size of Footing
.
1
5
.
1
350
1
.
1
507



L = B = 1.52 x 1.52 = 2.32 m² > 1.06 m² Hence OK
Design of pedestal:
D = 




 
2
tan
)
( 
b
L
D ≥
 













1
max
100
2
90
.
0
ck
f
q
b
L
Ast min. = 0.12%bd
=
100
550
1000
12
.
0 

= 660 mm²

154
= 6
12
4
2



= 678.6 mm² > 660 mm²
Spacing =
st
A
d 2
4

=
6
.
678
12
4
2


x 1000 = 166.66 mm
= 165 mm c/c
2
/
219
520
.
1
520
.
1
507
m
kN
qu 

 = 0.219 N/mm²
(a) One - way shear
Vu1 = 0.219x 1520 x (540 -d) N
= (179553.15-332.68d) N
Assuming 36
.
0

C
Vuc = 0.36 x 1520 x d = 548.64d N
Vu1 ≤ Vuc
(179553.15 – 332.68 d) ≤ 548.64d) N
d = 203.73 mm
(b) Two - way shear
Vu2 = 0.219 x (1520² - (230+ d) x (450+d))
Vu2 = 439957 N
= 439.96 kN
 
ck
s
CZ f
k 25
.
0


Where
d
b
ks 
 5
.
0 = 01
.
1
450
230
5
.
0 

but limited to 1.00
  118
.
1
20
25
.
0
00
.
1 



cz

Vuc = (1.118 ×230 + d) + (450+d)) ×2d
d = 225 mm
Vuc = 568520N

155
= 568.52kN > 439.96kN Hence OK
D ≥ 225 + 75 + 5 = 305 mm
the factored loads,
q max-gross = )
33
.
0
18
(
)
33
.
0
24
(
52
.
1
52
.
1
507











= 232KN/m² < 350 kN/m²
Mu = 0.218 × 1524 ×540² = 48.45 KNm
= 48454383 Nm
Mu/bd² = 2
150
1220
13386359

= 0.52968 N/mm²
Pt/100 =









 2
6
.
4
1
1
5
.
0
bd
f
M
f
f
ck
m
y
ck
= 




 



20
53
.
0
6
.
4
1
1
415
20
5
.
0
= 0.0015148
Pt = 0.15148
Ast min. = 0.0012BD
= 0.0012 x 1520 x 230
= 594.36 mm²
Pt min. =
Bd
Ast min
100
=
245
1520
36
.
594
100


= 0.1592 > 0.15148

156
(‫ז‬c=0.36N/mm²)
= 0.25% x 1520 x 245
= 933 mm²
= 87
.
11
10
4
933
2



Say 12No.s
)
1
(
)
)
2
(cov
(





n
er
B
s

)
1
12
(
)
10
)
2
75
(
1520
(





s
=120 mm is acceptable

157
Table 8.18. Reinforcement details of footing (Using staad without EQ)

158
Fig 8.43 Reinforcement details of Isolated Footing (Using Staad without EQ)

159
8.22.5. Footing Design (Using Staad with EQ)
Load = 2800 kN
fy = 415 N/mm²
fck = 20 N/mm²
SBC = 350 KN/m²
Size of Footing
.
5
5
.
1
350
1
.
1
2800



m²
L = B = 3.05 x 3.05 = 9.29 m² > 5.87 m² Hence OK
Design of pedestal:
D = 




 
2
tan
)
( 
b
L
D ≥
 













1
max
100
2
90
.
0
ck
f
q
b
L
D ≥ 1450 mm when b = 525mm
D ≥ 1335 mm when b = 750mm
Ast min. = 0.12%bd
=
100
550
1000
12
.
0 

= 660 mm²
= 6
12
4
2



= 678.6 mm² > 660 mm²

160
Spacing =
st
A
d 2
4

=
6
.
678
12
4
2


x 1000 = 166.66 mm
= 165 mm c/c
2
/
4
.
301
05
.
3
05
.
3
2800
m
kN
qu 

 = 0.301 N/mm²
(a) One - way shear
Vu1 = 0.301x 3050 x (1466 -d) N
= (1346850.4-918.64d) N
Assuming 36
.
0

C
Vuc = 0.36 x 3048 x d = 1097.28d N
Vu1 ≤ Vuc
(1346850.4-918.64d) ≤1097.28d N
d = 668.11 mm
(b) Two - way shear
Vu2 = 0.301 x (3050² - (525+ d) x (750+d))
Vu2 = 2284624 N
= 2284.62 kN
 
ck
s
CZ f
k 25
.
0


Where
d
b
ks 
 5
.
0 = 01
.
1
450
230
5
.
0 

but limited to 1.00
  118
.
1
20
25
.
0
00
.
1 



cz

Vuc = (1.118 ×525 + d) + (750+d)) ×2d
d = 675 mm

161
Vuc = 3962033N
= 3962.03 kN > 2284.62 kN Hence OK
D ≥ 675 + 75 + 5 = 758 mm
the factored loads,
q max-gross = )
95
.
0
18
(
)
95
.
0
24
(
05
.
3
05
.
3
2800











= 341KN/m² < 350 kN/m²
Mu = 0.301 × 3050 ×1466²=987.338 KNm
= 987337542 Nm
Mu/bd² = 2
867
3050
987337542

= 0.43 N/mm²
Pt/100 =









 2
6
.
4
1
1
5
.
0
bd
f
M
f
f
ck
m
y
ck
= 




 



20
43
.
0
6
.
4
1
1
415
20
5
.
0
= 0.0012248
Pt = 0.122477
Ast min. = 0.0012BD
= 0.0012 x 3050 x 950
= 3474.72 mm²
Pt min. =
Bd
Ast min
100
=
867
3050
72
.
3474
100


= 0.1315 > 0.12248

162
(‫ז‬c=0.36N/mm²)
= 0.25% x 3050 x 867
= 6607 mm²
= 32
16
4
6607
2



Say 33No.s
)
1
(
)
)
2
(cov
(





n
er
B
s

)
1
33
(
)
16
)
2
75
(
3050
(





s
=90mm is acceptable

163
Table 8.19. Reinforcement details of footing (Using staad with EQ)

164
Fig 8.44 Reinforcement details of Isolated Footing (Using Staad with EQ)

165
C
C
CH
H
HA
A
AP
P
PT
T
TE
E
ER
R
R 9
9
9 R
R
RE
E
ES
S
SU
U
UL
L
LT
T
TS
S
S
9.1 GENERAL
The results of the present investigation are presented both in tabular and graphical
forms. In order to facilitate the analysis, interpretation of the results is carried out at each
phase of the experimental work. This interpretation of the results obtained is based on the
current knowledge available in the literature as well as on the nature of result obtained.
The significance of the result is assessed with reference to the standards specified by the
relevant IS codes.
Table 9.1 Comparison of Column Designs (Manually, using Staad without
EQ & using Staad with EQ)
Column
Mark
Manual
Using Staad without
EQ
Using Staad with EQ
Ast
(req)
Spacing
Ast
(Prov)
Spacing
Ast
(req)
Spacing
Ast
(Prov)
Spacing
Ast
(req)
Spacing
Ast
(Prov)
Spacing
mm² mm mm² mm mm² mm mm² mm mm² mm mm² mm
25 452 192 452 192 2907 190 3216 190 15120 100 15728 100
Provided Area of Steel (mm²)
452 = 4#12
3216 = 6#25 + 2 #20
15728 = 18#32 + 4#20

166
Table 9.2 Comparison of Beam Designs (Manual, using Staad without EQ
& using Staad with EQ)
Beam
Mark
Units
Mid landing beam
Trapezoidal load
on beam
Triangular load on
beam
Top Bottom Top Bottom Top Bottom
Manual Design
Ast (req) mm² 297 297 492 455 466 466
Spacing mm 130 130 130 130 130 130
Ast (Prov) mm² 339 339 515 515 515 515
Spacing mm 130 130 130 130 130 130
Using Staad without Earth Quake
Ast (req) mm² 495 308 1043 350 5223 4642
Spacing mm 120 120 150 150 130 130
Ast (Prov) mm² 515 339 1143 402 5227 4825
Spacing mm 120 120 150 150 130 130
Using Staad with Earth Quake
Ast (req) mm² 1778 1494 730 233 4431 4269
Spacing mm 100 100 150 150 130 130
Ast (Prov) mm² 1872 1583 804 339 4825 4825
Spacing mm 100 100 150 150 130 130
Provided Area of Steel (mm²)
339 = 3#12
402 = 2#16
515 = 2#16 + 1#12
804 = 4#16
1143 = 2#20 + 2#16 + 1#12
1583 = 2#25 + 3#16
1872 = 3#25 + 2#16
4825 = 6#32
5227 = 6#32 + 2#16

167
Table 9.3 Comparison of Footing Designs (Manually, using Staad without
EQ & using Staad with EQ)
Footing Details
Size Depth Reinforcement
L (mm) B (mm) mm
Manual 1220 1220 230
6no.s - T 10
Both ways
Using Staad
without EQ
1520 1520 325
12no.s - T 10
Both ways
Using Staad with
EQ
3050 3050 950
32no.s - T 16
Both ways

168
Fig. 9.1 Stilt Floor (Parking) Plan

169
Fig. 9.2 Typical Floor Plan

170
Fig 9.3 Skeletal structure

171
Fig 9.4 Shear Force Diagram of Frame (without EQ)

172
Fig 9.5 Bending Moment Diagram of Frame (without EQ)

173
Fig 9.6 Shear Force Diagram of Frame (With EQ)

174
Fig 9.7 Bending Moment Diagram of Frame (with EQ)

175
Fig 9.8 Dead Load on the Frame

176
Fig 9.9. Live Load on the Frame

177
Fig 9.10. Diagram of combination loads

178
Fig 9.11 Diagram of floor load

179
B
B
BI
I
IB
B
BL
L
LI
I
IO
O
OG
G
GR
R
RA
A
AP
P
PH
H
HY
Y
Y
I.S: 456-2000 code of practice for plain and reinforced concrete.
Design aids for reinforced concrete to SP 16:1980
Ductile detailing of Reinforced Concrete Structures Subjected to Seismic Forces – Code
of Practice IS 13920 – 1993
Criteria for Earthquake Resistant Design of Structures IS 1893(Part 1): 2002
Hand book on Concrete Reinforcement and Detailing SP 34: 1987
Illustrated design of reinforced concrete building Dr. S.R. Karve &
Dr. V.L. Shah
Limit state of design Dr. Ramachandra
Reinforced concrete structures Pillai & Menon
Reinforced concrete structures Syal. I.C
Goel. A.K.
Reinforced concrete structures P.C. Varghese
Reinforced concrete structures B.C. Punmia
Ashok. K. Jain
Limit state of design A.K. Jain
Limit state of design Dr. Ramamrutham
Theory of structures C.S. Reddy
Soil mechanics and foundation engineering K.R. Aroar

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