Final project documentation pdf

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DEPARTMENT OF CIVIL ENGINEERING, GNIT
CHAPTER-1
INTRODUCTION
1.1. BUILDING PLANNING ACCORDING TO VASTHU
Principles of Planning
The basic objective of planning of buildings is to arrange all the units of a building on all floors
and at level according to their functional requirements making best use of the space available for a
building. The shape of such a plan is governed by several factors such as climatic conditions, site
location, accommodation requirements, local by-laws, surrounding environment, etc. These principles or
factors, which govern the theory of planning, are common to all buildings of all classes intended to be
used for residential purposes. These principles, enunciated below, are right but just factors to be
considered in planning.
1) Aspect 2) Prospect
3) Privacy 4) Grouping
5) Roominess 6) Furniture Requirement
7) Sanitation 8) Sanitation
9) Circulation 10) Elegance
11) Economy 12) Practical Considerations
Aspect: It means peculiarity of the arrangements of doors and windows in the external walls of a building
which allows the occupants to enjoy the natural gifts such as sunshine, breeze, scenery etc. Aspect is a
very important consideration in planning as it provides not only comfort and good environment but
hygiene too.
A room which receives light and air from a particular side is said to have aspect of that direction.
From this angle, the following aspects for different rooms are preferred:
a) For kitchen – East Aspect
b) For dining room – South Aspect

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c) For drawing, living room – South/South East Aspect
d) For bedroom – South West/West Aspect
e) For verandah –South West/West Aspect
f) For reading, store room, class room, and stairs— North Aspect
FIG:1.1 Sun Path Diagram
Prospect: Prospect in its proper sense is the impression that al house is likely to make on a person who
looks it from outside. It includes the attainment of pleasing appearance by the use of natural beauties,
disposition of doors and windows and concealment of some undesirable parts of a building.
Privacy: Privacy is one of the most important principles in planning of all types of buildings in general
and residential building. Privacy requires consideration in two ways:
1) Privacy of one room from another.
2) Privacy of all parts of a building from the neighboring buildings.
Privacy of one room from another is attained by carefully planning the building with respect to grouping,
disposition of door/windows, provision of small corridors and lobbies. Privacy of a building is easy
secured by carefully planning the entrance and steering it with trees and creepers.
Grouping: Grouping is arranging the layout in typical fashion so that all the rooms are placed in proper
correlation of their functions in due proximity with each other. The objective of grouping is to maintain
the sequence of their functions with least interference.
It should be noted that while grouping a residential building provides efficiency, comfort and health to the
occupants whereas the buildings other than residential provide economical service and proper co-relation.

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Roominess: Roominess refers to the effect produced by deriving the maximum benefit from the
minimum dimensions of a room. In other words, it is the accomplishment of economy of space at the
same time avoiding cramping of the plan. It is essential particularly in case of residential buildings where
large storage space is required, to make maximum use of every nook and corner of built-up area of the
building before making an addition to the plinth area.
Furniture Requirements: The functional requirement of a room or an apartment governs the furniture
requirements. This is an important consideration in planning of building other than residential such as
offices, commercial building, schools etc.
In case of buildings other than residential, they are generally planned, with due thought to the
furniture, equipment and other fixtures, to meet the needs of particular function required to be performed.
This can be done by assuming the sufficient sizes of furniture pieces and then studying the circulation and
space requirements around them.
Sanitation: It consists of providing ample light, ventilation, facilities for cleaning and sanitary
conveniences in the following manner:
a) Light: Light has dual significance. Firstly it illuminates and secondly from hygienic point of
view. Light in interior building can be provided by natural or artificial lightening. Glare in the
light distracts and disables the vision and hence source of glare may be concealed or avoided.
Uniform distribution of light is necessary particularly in schools, workshop etc. A room should
get sun-light as long as and as much as possible. Vertical windows are therefore preferred to horizontal
ones.
b) Ventilation: Ventilation is the supply of outside air by either positive ventilation or by infiltration into
the building. Good ventilation is an important factor conductive to the comfort in building. Poor
ventilation or lack of fresh air in a building always produces head-ache, sleepiness, inability to fix
attention etc. Ventilation may be natural or mechanical. In natural ventilation, the outside air is supplied
into the building through windows, ventilation other opening due to wind outside. In mechanical
ventilation, the outside air is supplied either by mechanical device such as a fan or by infiltration by
reduction of pressure inside due to the exhaust of air or by combination of positive ventilation and
exhaust of air.

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1.2. BUILDING BYELAWS
A building code is a document containing standardized requirement for the design & construction
of most types of building, Codes regulate building construction & building use in order to protect
the health, safety & welfare of the occupant. Codes express all aspects of construction including
structural integrity, fire resistance, safe exists, lighting, electrical, energy conservation, plumbing,
sanitary facilities, ventilation, seismic design & correct use of construction materials. N.B.C is a
single document in which like a network the information contained in various INDIAN
STANDARDS is woven into a pattern of continuity with interdependent requirements of sections
carefully analyzed & fitted into to make the whole document, a continuous one.
The purpose of all these building codes is to ensure public safety, health & welfare as affected by
building construction. This purpose includes:
 STRUCTURAL STRENGTH
 SANITARY EQUIPMEN
 LIGHT & VENTILATION
 FIRE SAFETY
1.2.1. NEED & IMPORTANCE:
The basis for modern bye-laws, requirements look mysterious always to layman & even to many
officials & most of them are based on natural scientific laws, known properties of building
materials & inherent hazards of users. The need for bye-laws becomes more important from the
point of view in INDIA. The construction activity both in public & private sector in terms of
building for industrial, commercial & residential & administrative user amount to 50% of outlay of
any 5 year plan. With this certain projects can be delayed over a period of time to get completed.
Eg. ST. MARK’S PIAZZA, hence it would lose their identity in terms of MASS & SPACE
RELATIONSHIP. If we talk about the cities, if we lose control over the construction activity then
the problems like Visual disorder, traffic becomes uncontrollable, uncomfortable living,
environmental problems etc. In the absence of suitable bye-laws & machinery to enforce them, the
poor people will be left at the mercy of well to do people.

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1.2.2. Architectural Need & SETBACKS:
It defines the areas of varying densities for use. For example ZONING ORDINANCES describes,
in. parking lot size Maximum building height Yard requirement Other structures on the property
IN this the major area of concern is the PARKING LAODING ZONE & PRIVATE STREETS.
Roads can be widened due to presence of setbacks. Constructs one line parallel to axis of road,
resulting in improvement of road. Results in better condition of air, light & ventilation of
building. Reduces danger of fire. Width of setback varies from 1mts to 1.50mts for congested
areas& 4.50mts to 6mts for new underdeveloped areas. The side setbacks shall be optional.
Where left it shall not be less than 2 meter’s or 1/6th of the height of the building whichever
is more.
The width of the rear set back, if left at any point of building, it shall not be less than 3
meter’s or 1/5th of the height of the building whichever is more.
Sr.No Plot Area (sq.
yds.)
Site coverage Minimum front
set back
Height
permissible
F.R.R.
1 Upto 100 sq.
yds.
80% 5’ 38'-6" 1:2.00
2 Above 100 to
150"
75% 6’ 38'-6" 1:1'.90
3 Above 150 to
200"
70% 7’ 38'-6" 1:1.75
4 Above 200 to
300"
65% 10’ 38'-6" 1:1.65
5 Above 300 to
500"
605 15’ 38'-6" 1:1.50

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CHAPTER -2
LITERATURE
Building construction is the engineering deals with the construction of building such as residential houses.
In a simple building can be define as an enclose space by walls with roof, food, cloth and the basic needs
of human beings. In the early ancient times humans lived in caves, over trees or under trees, to protect
themselves from wild animals, rain, sun, etc. as the times passed as humans being started living in huts
made of timber branches. The shelters of those old 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 desire to
own comfortable homes on an average generally one spends his two-third life times in the houses. The
security civic sense of the responsibility. These are the few reasons which are responsible that the person
do utmost effort and spend hard earned saving in owning houses.
Nowadays the house building 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. Draughtsman 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+2 storey frame work is taken
up. The building in plan consists of columns built monolithically forming a network. The number of
columns is 13. It is residential building.
The design is made manually, by using software on structural analysis design (staad-pro). The
building subjected to both the vertical loads because for low rise buildings say up to 4-5 storeys the wind
load is not critical because the moment of resistance provided by the continuity of floor system to column
connection and walls provided between columns are sufficient to accommodate the effects of these forces.
The vertical load consists of dead load of structural components such as beams, columns, slabs etc and
live loads thus building is designed for dead load, live load as per IS 875. The building is designed as two
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 and obtained from this software.
2.1 Early modern and the industrial age:
With the emerging knowledge in scientific fields and the rise of new materials and technology,
architecture engineering began to separate, and the architect began to concentrate on aesthetics and the
humanist aspects, often at the expense of technical aspects of building design.
Meanwhile, the industrial revolution laid open the door for mass production and consumption. Aesthetics
became a criterion for the middle class as ornamental products, once within the province of expensive
craftsmanship, became cheaper under machine production.

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Vernacular architecture became increasingly ornamental. House builders could use current architectural
design in their work by combining features found in pattern books and architectural journals.
2.1.1 Modern architecture:
The Bauhaus Dessau architecture department from 1925 by Walter Gropius.
The dissatisfaction with such a general situation at the turn of the 20th century gave rise to many new
lines of thought that served as precursors to modern architecture. Notable among these is detachers’
deskbound, formed in 1907 to produce better quality machine made objects. The rise of the profession of
industrial design is usually placed here. Following this lead, the Bauhaus school, founded in Weimar,
Germany in 1919, redefined the architectural bounds prior set throughout history viewing the creation of a
building as the ultimate synthesis—the apex—of art, craft and technology.
When modern architecture was first practiced, it was an avant-garde moment with moral, philosophical,
and aesthetic underpinning. Immediately after World War I, pioneering modernist architects sought to
develop a completely new style appropriate for a new post-war social and economic order, focused on
meeting the needs of the middle and working classes. They rejected the architectural practice of the
academic refinement of historical styles which served the rapidly declining aristocratic order.
2.2 Statement of project
Utility of building: Residential Building
No of stories: G+2
Shape of the plan: Irregular
No of staircases: 2
Type of construction: R.C.C framed structure
Types of walls: brick wall
Geometric details:
Ground floor: 3m
Floor to floor height : 3m.
Height of plinth: 0.6m
Depth of foundation: 500mm
Materials:
Concrete grade: M20
All steel grades: Fe415 grade
Bearing capacity of soil: 200

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2.3 Design of multi storied residential building:
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
1. Foundation design
2. Column design
3. Beam design
4. Slab design
These all are designed under limit state method
2.3.1 Limit state method:
The object of design based on the limit state concept is to achieve an acceptability that a structure will not
become unserviceable in its life time for the use for which it is intended. I.e. it will not reach a limit state.
In this limit state method all relevant states must be considered in design to ensure a degree of safety and
serviceability.
Limit state:
The acceptable limit for the safety and serviceability requirements before failure occurs is called a limit
state.
Limit state of collapse:
This is corresponds to the maximum load carrying capacity.
Violation of collapse limit state implies failures in the source that a clearly defined limit state of structural
usefulness has been exceeded. However it does not mean complete collapse.
This limit state corresponds to:
a) Flexural
b) Compression
c) Shear

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d) Torsion
Limit state of survivability:
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.
a) Deflection
b) Cracking
c) Vibration

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CHAPTER-3
SOFTWARES USED
This project is mostly based on manual design, software and it is essential to know the details
about these software’s.
List of software’s used
1. STAAD PRO (v8i)
2. STAAD PRO 5(v8i)
3. AUTO CADD
STAAD PRO, STAAD, AUTO CADD:
3.1. 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 find the
outline of the structure, whereas 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.

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To calculate S.F.D and B.M.D 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 R.C.C steel, bridge, truss etc. according to various country codes.
3.1.1. Alternatives for STAAD:
Struts, robot, sap, adds pro which gives details very clearly regarding reinforcement and manual
calculations. But these software’s are restricted to some designs only whereas STAAD can deal with
several types of structure.
3.1.2. STAAD Editor:
STAAD has very great advantage to other software’s i.e., STAAD editor. STAAD editor is the
programming
For the structure we created and loads we taken all details are presented in programming format in
STAAD editor. This program can be used to analyze other structures 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.
Limitations of Staad pro:
1. Huge output data
2. Even analysis of a small beam creates large output.
3. Unable to show plinth beams.
3.1.3. STAAD foundation:
STAAD foundation is a powerful tool used to calculate different types of foundations. It is also
licensed by Bentley software’s. All Bentley software’s cost about 10 lakhs and so all engineers can’t use
it due to heavy cost.
Analysis and design carried in Staad and post processing in staad gives the load at various
supports. These supports are to be imported into these software to calculate the footing details i.e.,
regarding the geometry and reinforcement details.
This software can deal different types of foundations
SHALLOW (D<B)
► 1.Isolated (Spread) Footing
► 2.Combined (Strip) Footing
► 3.Mat (Raft) Foundation

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DEEP (D>B)
► 1.Pile Cap
►2.Driller Pier
1. Isolated footing is spread footing which is common type of footing.
2. Combined Footing or Strap footing is generally laid when two columns are very near to each other.
3. Mat foundation is generally laid at places where soil has less soil bearing capacity.
4. Pile foundation is laid at places with very loose soils and where deep excavations are required.
So depending on the soil at type we have to decide the type of foundation required.
Also lot of input data is required regarding safety factors, soil, materials used should be given in
respective units.
After input data is give software design the details for each and every footing and gives the
details regarding
1. Geometry of footing
2. Reinforcement
3. Column layout
4. Graphs
5. Manual calculations
These details will be given in detail for each and every column.
Another advantage of foundations is even after the design; properties of the members can be updated if
required.
The following properties can be updated
► Column Position
► Column Shape
► Column Size
3.2. 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

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quickly
Learning of certain commands is required to draw in AutoCAD.
3.3. PLAN:
The auto cad plotting no.1 represents the plan of a g+2 building. The plan clearly shows that it is an
individual residential.
This is located at UPPARI GUDA nearby Ibrahimpatnam which is surrounded by RED Soil.
This Plan consists of two floors including various different rooms in different dimensions.
It is a g+2 proposed building, so for 2 floors we have 3 Master Bed Rooms, a children bed room, kitchen,
library, hall etc.
The plan shows the details of dimensions of each and every room and the type of room and orientation of
the different rooms like bed room, bathroom, kitchen, hall etc.. All the two floors have different room
arrangement.
The entire plan area is about 126.38 square yards; build up area is 66.8676 square yards. There is some
space left around the building for parking of cars, surge tank. The plan gives details of arrangement of
various furniture like sofa etc.
The plan also gives the details of location of stair cases in different blocks. We have 2 stair cases
throughout the plan.
In the top floor we have a small area which is left for vertical harvesting and for planting purpose it also
provides the good view from top to the lake which is seen from the top, those who want to enjoy the
morning & evening sunrise and sun set .
So these represent the plan of our building and detailed explanation of remaining parts like elevations and
designing is carried in the next sections.

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3.4. Elevation:
Auto CADD represents the proposed elevation of building. It shows the elevation of a g+2
building representing the front view which gives the overview of a building block.
The figure represents the site picture of our structure which is taken at the site .the building is
actually under constructions and all the analysis and design work is completed before the
beginning of the project.
Each floor consists of height 3m which is taken as per LOCAL MUNICIPAL rules for
residential buildings.
The building is not designed for increasing the number of floors in future.so the number of floors
is fixed for future also for this building due to unavailability of the permissions of respective
authorities.
Also special materials like fly ash and stone dust were also used in order to reduce the dead load
and increase life of the structure and also improve economy. But these materials were not
considered while designing in staad to reduce the complexity and necessary corrections are made
for considering the economy and safety of the structure as it is a very huge building.
The construction is going to complete in the month of June 2017 and ready for the occupancy.
This is regarding the plan and details of the site and next section deals with the design part of the
building under various loads for which the building is designed.

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CHAPTER-4
MANUAL DESIGN
4.1 MIX DESIGN:
M20
IS 10262:1982
= + t x s
= Target Average Compressive Strength
= Characteristic Compressive Strength at 28 days
t = a statistic, depending upon accepted law Proportion
s = Standard deviation ; V = Total Volume
= 20 + 1.65 x 4.6 = 27.59
= 0.53, from table – 2% of voids
Water percentage according to table – 4 for 20 mm Aggregate
For 20 mm size 186lts water
C = = 351 kg but according to Table 4 of R.C.C (IS 456:2000), c = = 372 kg
Sand percentage according to Table 4 for 20 mm Aggregate
20 mm – 186 lts water – 35% of sand + 2% = 37% of sand
0.6 – 0.5 = 0.10
= 2
For FINE AGGREGATES: V = [w + ( )] x
Assume 1 – 2 % of voids= 100%
1 – = 1 – 0.02 = 0.98
0.98 = [186 + ( )] x

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= 660 kg
For COARSE AGGEGATES: 0.98 = [186 + ( )] x
= 1128.4 kg
372: 660: 1128.4
1: 1.7 : 3
0.150 x 0.150 x 0.150 = 3.375 x
S.No 1 block (kg) 3 blocks (kg) 6 blocks (kg)
Cement 1.26 4.158 8.316
Fine Aggregate 2.23 7.37 14.74
Coarse Aggregate 3.8 12.54 25.08
Water 1 cube = 2.1 ltr ; 3 cube = 6.3 ltr ; 6 cube = 12.6 ltr

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4.2.Slab design:
SLABS
A Slab is a common structural element of modern buildings. It is a covering provided over the
four walls or beams of a room in order to enclose it. A slab may be a floor or a roof depending upon it
location in the building. Slabs carry uniformly distributed loads primarily by flexure. Inclined slabs may
be used as ramps for multi-storey car parks.
4.2.1 Classification based on resistance to Torsion:
1) Restrained Slabs: A slab that may have its few or all edges restrained is called a restrained slab. The
degree of restraints may vary depending upon whether it is continuous or over its supports or cast
monolithically with its supporting beams. A hogging or negative bending moment will develop in top
face of the slab at supported sides. In these slabs the corners are prevented from lifting and provision is
made for torsion.
2) Simply Supported Slabs: A slab which is supported at its edges tends to lift off its supports near the
corners when loaded. Such a slab is the only truly simply supported slab. These slabs do not have
adequate provision to resist torsion at corners is to prevent the corners from lifting.
4.2.2 Classification based on type of support:
1) One way slab:
One way slabs are those in which the length is more than twice the breadth. A continuous way
slab can be analyzed in a manner similar to that of a continuous beam. The general recommendation for
curtailment of bars given in clause 26.2.3 of IS: 456-2000 applies for slabs also. Shear Stresses in slabs
are generally not critical under normal loads but should be checked in accordance with the requirements
of clause 40.2.1.1 of IS: 456-2000.
2) Two way slab: Two way slabs are those in which the length to breadth ratio is less than two. When
slabs are supported on four sides two spanning occurs. The deflection and bending moments in a two way
slab are considerably reduced as compared to that of a one way slab. A two way slab may be considered
to consist of a series of interconnected beams of unit width which transfer the load to the respective
supports

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3) Circular slab:
Circular slabs are more commonly used in the design of circular water tanks containers with flat
bottom and circular raft foundations. The analysis of stresses in these slabs is generally based on elastic
theory. Under uniformly distributed loads, these slabs deflect in the form of saucer and develop radial and
circumferential stresses. Tensile stresses develop on the convex surface and compressive stresses develop
on the concave surface. Thus the reinforcement should be provided in radial and circumferential
directions near the convex surface.
4) Flat slab:
The term flat slab means a reinforced concrete slab supported directly on columns without beams.
There are large bending moments and shears near the junctions with columns. Therefore, there may need
to flare the column at its top end or thicken the slab over the column. A flat slab system requires lesser
head room; hence, it is very economical for larger rooms.
Following figures shows the load distributions in slabs.

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Slabs are designed for deflection. Slabs are designed based on yield theory
This diagram shows the distribution of loads in two slabs.
Distribution of loads in two slabs.
The following figure shows the monolithic connection between beam, column and
slab :-

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Since we designed all slabs as two ways we have many types of lengths which are as follows
DESIGN OFTWO WAY SLABS (1, 5, and 11):
1. Data :
Short span, = 2936.24 mm
Long span, = 3469.6 mm
Live load = 1.5
Floor finish = 1.5
= 20
= 415
= 1.18 , < 2 two way slab
2. Thickness of slab :
Assume effective depth d = = = 104.87 mm
Adopt effective depth d = 125 mm
Overall depth D = 150 mm
3. Effective span :
= 3.4696 + 0.125 = 3.5946 m
= 2.936 + 0.125 = 3.061 m
= = 1.17 = 1.2
4. Loads :
Self-weight of the slab = 0.150 x 25 = 3.75
Live load = 1.5
Floor finish = 1.5
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
5. Design Moments and Shear Forces :
The slab is supported on all the four sides. The corners are not held down. Hence moment
coefficients are obtained from Table – 27 of IS: 456.
= 0.084
= 0.059
= 0.084 x 10.125 x = 7.96 KN-m
= 0.059 x 10.125 x = 5.6 KN-m
= = 15.49 KN

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6. Minimum Depth Required :
7.96 x = 0.138 x 20 x 1000 x
d = 53.70 mm < 125 mm, provided depth
Hence provided depth is adequate
7. Reinforcement : Along x-direction
7.96 x = 0.87 x 415 x x 125 (1- )
= 185 =215
Using 10 mm diameters bar, spacing of bars
S = =
No of bars = = 2.7 = 3
Maximum spacing is (i) 3d = 3 x 125 = 375 mm
(ii) 300 mm whichever is less
Hence provide 3 no 10 mm bars at 300 mm c/c
Along y-direction:
These bars will be placed above the bars in x-direction.
Hence, d = 125 – 10 = 115 mm
5.6x = 0.87 x 415 x x 115 (1- )
= 138 = 180
S = = x 1000 = 436.33 mm
No of bars = = 2.29 = 3
8. Reinforcement in Edge Strip :
= x 1000 x 150 = 180
S = = x 1000 = 279.25 mm = 290 mm
No of bars = = 2.29 = 3.58 = 4

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Hence provide 4 no 8 mm bars at 290 mm c/c in edge strips in both directions
9. Check for deflection :For simply supported slabs basic value of
Of steel at mid span
x 100 = 0.24 %
= 0.58 x 415 = 240
From figure of IS: 456, modification factor = 1.6
Maximum permitted ratio = 1.6 x 20 = 32
Provided = = 24.488 < 32
Hence deflection control is safe.
10. Check for Shear : (Normally slabs will be safe in shear and hence this check may be omitted)
Normal shear stress = = 0.123
Percentage of main steel at support (50% bars curtailed)
x 100 = 0.144 %
Referring to the Table – 19 of IS: 456, Shear strength of concrete for beam is
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
Maximum shear stress in concrete from Table – 19 of IS: 456
= 2.8
<
Hence slab is safe in shear
10mm-
300mm
c-c
2.9

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DESIGN OF TWO WAY SLABS (2, 6, and 12):
1. Data :
Short span, = 2.209 m
Live load = 1.5
Floor finish = 1.5
= 20
= 415
3. Effective span :
= 2.936 + 0.125 = 3.061 m
= 2.209 + 0.125 = 2.334 m
= = 1.31 = 1.3
4. Loads :
Live load = 1.5
Floor finish = 1.5
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
= 0.093
= 0.055
= 0.093 x 10.125 x = 5.13 KN-m
= 0.055 x 10.125 x = 3.03 KN-m
= = 11.81 KN
5.13 x = 0.138 x 20 x 1000 x

24
5.13 x = 0.87 x 415 x x 125 (1- )
= 115 = 135
S = =
No of bars = = 1.71 = 2
Along y-direction:
Hence, d = 125 – 10 = 115 mm
3.03 x = 0.87 x 415 x x 115 (1- )
= 73 = 120
S = = x 1000 = 654.49 mm
No of bars = = 1.52 = 2
= x 150 x 1000 = 180
S = = x 1000 = 279 mm
No of bars = = 3

25
x 100 = 0.22 %
= 0.58 x 415 = 240
Provided = = 18.67 < 32
x 100 = 0.096 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8
<

26
DESIGN OF TWO WAY SLABS (3,7and 13):
1. Data :
Live load = 1.5
Floor finish = 1.5
= 20
= 415
3. Effective span :
= 3.774 + 0.125 =3.899 m
= 2.2098 + 0.125 = 2.3348 m
= = 1.66
4. Loads :
Live load = 1.5
Floor finish = 1.5
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
= 0.104 + (0.113 - 0.104) x = 0.1076
= 0.046 - (0.046 – 0.0.037) x = 0.0424
= 0.1076 x 10.125 x = 9.42 KN-m
= 0.0424 x 10.125 x = 2.34 KN-m
= =11.82 KN
9.42 x = 0.138 x 20 x 1000 x

27
9.42 x = 0.87 x 415 x x 125 (1- )
= 216.50 =240
S = =
No of bars = = 3.05 = 3
Along y-direction:
Hence, d = 125 – 10 = 115 mm
2.34 x = 0.87 x 415 x x 115 (1- )
= 56.94 = 120
S = = x 1000 = 654 mm
No of bars = = 1.65 = 2
= x 1000 x 150 = 180
S = = x 1000 = 279.25 mm = 290 mm
No of bars = = 2.29 = 3.58 = 4

28
x 100 = 0.24 %
= 0.58 x 415 = 240
Provided = = 18.67 < 32
x 100 = 0.096 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8
<

29
DESIGN OF TWO WAY SLABS (4, 10 and 16):
1. Data :
Live load = 1.5
Floor finish = 1.5
= 20
= 415
3. Effective span :
= 3.7742 + 0.125 = 3.8992 m
= 3.4692 + 0.125 = 3.5942 m
= = 1.08 = 1.1
4. Loads :
Live load = 1.5
Floor finish = 1.5
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
= 0.074
= 0.061
= 0.074 x 10.125 x = 9.679 KN-m
= 0.061 x 10.125 x = 7.978 KN-m
= = 18.195 KN
9.679 x = 0.138 x 20 x 1000 x

30
9.679 x = 0.87 x 415 x x 125 (1- )
= 222.69 =235
S = =
No of bars = = 2.9 = 3
Along y-direction:
Hence, d = 125 – 10 = 115 mm
7.978x = 0.87 x 415 x x 115 (1- )
= 199.3 = 210
S = = x 1000 = 373.9 mm
No of bars = = 2.67 = 3
= x 1000 x 150 = 180
S = = x 1000 = 279.25 mm = 290 mm
No of bars = = 2.29 = 3.58 = 4

31
x 100 = 0.24 %
= 0.58 x 415 = 240
Provided = = 28.39 < 32
x 100 = 0.168 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8
<

32
DESIGN OF TWO WAY SLAB (8):
1. Data :
Long span, = 3075 mm
Live load = 1.5
Floor finish = 1.5
= 20
= 415
3. Effective span :
= 3.075 + 0.125 =3.2 m
= 2.2098 + 0.125 = 2.3348 m
= = 1.37
4. Loads :
Live load = 1.5
Floor finish = 1.5
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
= 0.093 + (0.099 - 0.093) x = 0.0954
= 0.055 - (0.055 – 0.051) x = 0.0566
= 0.0954 x 10.125 x = 5.26 KN-m
= 0.0566 x 10.125 x = 3.03 KN-m
= =11.82 KN

33
5.26 x = 0.138 x 20 x 1000 x
5.26 x = 0.87 x 415 x x 125 (1- )
= 118.89 =150
S = =
No of bars = = 1.90 = 2
Along y-direction:
Hence, d = 125 – 10 = 115 mm
3.03 x = 0.87 x 415 x x 115 (1- )
= 73.96 = 120
S = = x 1000 = 654 mm
No of bars = = 1.65 = 2
= x 1000 x 150 = 180
S = = x 1000 = 279.25 mm = 290 mm
No of bars = = 2.29 = 3.58 = 4

34
x 100 = 0.24 %
= 0.58 x 415 = 240
Provided = = 18.67 < 32
x 100 = 0.096 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8
<

35
DESIGN OF TWO WAY SLABS (9, 15):
1. Data :
Live load = 1.5
Floor finish = 1.5
= 20
= 415
3. Effective span :
= 3.7742 + 0.125 = 3.8992 m
= 3.1242 + 0.125 = 3.2492 m
= = 1.19 = 1.2
4. Loads :
Live load = 1.5
Floor finish = 1.5
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
= 0.084
= 0.059
= 0.084 x 10.125 x = 8.98 KN-m
= 0.059 x 10.125 x = 6.31 KN-m
= = 16.44 KN
8.98 x = 0.138 x 20 x 1000 x

36
8.98 x = 0.87 x 415 x x 125 (1- )
= 206.02 =235
S = =
No of bars = = 2.9 = 3
Along y-direction:
Hence, d = 125 – 10 = 115 mm
6.31 x = 0.87 x 415 x x 115 (1- )
= 156.38 = 210
S = = x 1000 = 373.9 mm
No of bars = = 2.67 = 3
= x 1000 x 150 = 180
S = = x 1000 = 279.25 mm = 290 mm
No of bars = = 2.29 = 3.58 = 4

37
x 100 = 0.24 %
= 0.58 x 415 = 240
Provided = = 25.99 < 32
x 100 = 0.168 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8
<
Hence slab is safe in shear.

38
DESIGN OF TWO WAY SLAB (14):
1. Data :
Long span, = 3075 mm
Live load = 2
Floor finish = 1.5
= 20
= 415
3. Effective span :
= 3.075 + 0.125 =3.2 m
= 2.2098 + 0.125 = 2.3348 m
= = 1.37
4. Loads :
Live load = 2
Floor finish = 1.5
Total load = 7.25
Factored load = 1.5 x 6.75 = 10.875
cofficientsare obtained from Table – 27 of IS: 456.
= 0.093 + (0.099 - 0.093) x = 0.0954
= 0.055 - (0.055 – 0.051) x = 0.0566
= 0.0954 x 10.875 x = 5.65 KN-m
= 0.0566 x 10.875 x = 3.35 KN-m
= =12.7 KN
5.65 x = 0.138 x 20 x 1000 x

39
5.65 x = 0.87 x 415 x x 125 (1- )
= 127.90 =150
S = =
No of bars = = 1.90 = 2
Along y-direction:
Hence, d = 125 – 10 = 115 mm
3.35 x = 0.87 x 415 x x 115 (1- )
= 81.89 = 120
S = = x 1000 = 654 mm
No of bars = = 1.65 = 2
= x 1000 x 150 = 180
S = = x 1000 = 279.25 mm = 290 mm
No of bars = = 2.29 = 3.58 = 4

40
x 100 = 0.24 %
= 0.58 x 415 = 240
Provided = = 18.67 < 32
x 100 = 0.096 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8
<

41
DESIGN OF CANTILEVER SLAB FOR STAIRCASE:
1. Data :
Live load = 3
Floor finish = 1.5
= 20
= 415
The depth may be reduced to D = 100 mm gradually at the free end
3. Loads :
Live load = 3
Floor finish = 1.5
Total load = 8.25
Factored load = 1.5 x 8.25 = 12.375
= = 4.46 KN-m
= 12.375 x0.941 = 11.64 KN
4.46 x = 0.138 x 20 x 1000 x
d =40.198 mm < 125 mm, provided depth
6. Reinforcement :
4.46 x = 0.87 x 415 x x 125 (1- )
= 100.49 =150
S = =
No of bars = = 1.90 = 2

42
7. Distribution Reinforcement:
= x 1000 x 150 = 180
S = = x 1000 =436.33 mm = 430 mm
No of bars = = 2.29 = 3
Hence provide 3 no 10 mm bars at 430 mm c/c in transverse directions
7. Anchorage Length at Support:
The anchorage length required at the support is given by
= = 376 mm
x 100 = 0.24 %
= 0.58 x 415 = 240
Maximum permitted ratio = 1.6 x 7 = 11.2
Provided = = 6.53 < 11.2
x 100 = 0.144 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8 , < safe..

43
DESIGN OF CANTILEVER SALB FOR 1’(WEST):
1. Data :
Live load = 2
Floor finish = 1
= 20
= 415
3. Loads :
Live load = 2
Floor finish = 1
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
= = 0.47 KN-m
= 10.125 x0.3048 = 3.08 KN
0.47 x = 0.138 x 20 x 1000 x
d =13.04 mm < 125 mm, provided depth
6. Reinforcement :
0.47 x = 0.87 x 415 x x 125 (1- )
= 10.43 =25
S = =
No of bars = = 1.90 = 2

44
7. Distribution Reinforcement :
= x 1000 x 150 = 180
S = = x 1000 =436.33 mm = 430 mm
No of bars = = 2.29 = 3
= = 376 mm
x 100 = 0.24 %
= 0.58 x 415 = 240
Provided = = 7.528 < 32
x 100 = 0.144 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8 < hence safe

45
DESIGN OF CANTILEVER SLAB FOR 2’(EXCEPT WEST):
1. Data :
Live load = 2
Floor finish = 1
= 20
= 415
3. Loads :
Live load = 2
Floor finish = 1
Total load = 6.75
Factored load = 1.5 x 6.75 = 10.125
= = 1.88 KN-m
= 10.125 x0.6096 = 6.17 KN
1.88 x = 0.138 x 20 x 1000 x
d =26 mm < 125 mm, provided depth
6. Reinforcement :
1.88 x = 0.87 x 415 x x 125 (1- )
= 41.94 =50
S = =
No of bars = = 1.90 = 2

46
= x 1000 x 150 = 180
S = = x 1000 =436.33 mm = 430 mm
No of bars = = 2.29 = 3
= = 376 mm
x 100 = 0.24 %
= 0.58 x 415 = 240
Provided = = 7.528 < 32
x 100 = 0.144 %
= 0.28
For solid slabs = 1.3 x 0.28 = 0.364
(k =1.3 as D < 150 mm)
= 2.8
<

47
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 and perimeters of
the beams are assigned. Design beam command is assigned and analysis is carried out, now reinforcement
details are taken.
4.3 Beam design:
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
1.) Single reinforced beams
2.) Double reinforced concrete
3.) Flanged beams
4.3.1 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. I cantilever beams reinforcing bars placed near
the top of the beam, for the same reason as in the case of simply supported beam.
4.3.2 Doubly reinforced concrete beams:
It is reinforced under compression 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.

48
DESIGN OF L- BEAMS (1, 5,13,19,30 and 36):
Effective span l = 3.296 3.3 m
Depth of flange = 150 mm
Breadth of web = 230mm
=20
= 415
1. Depth of the beam:
Selecting the depth in range of to based on stiffness
d = = 0.275 m = 0.300 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Self-weight of the beam = 0.45 x 0.23 x 1x 25 = 2.5875
Live Load of the Beam = = 16.09+10.493
Total load = = 29.1705
Factored load = 1.5 x 29.1705 = 43.755
Factored bending moment
= = 59.56 KN-m
3. Effective width of Flange :
=
= 0.955 m , limited to 3.3 m. Hence, = 0.96

49
4. Assuming is with in the flange :
Equating compressive force in concrete to tensile force in steel
= = 0.052
5. Reinforcement :
59.56 x = 0.87 x 415 x x (400 – 0.42(0.052 x ))
= 400.38 = 400
= 0.052 x 400 = 20.8 mm
= = 188.4 < provided
Maximum area of tension steel
= 0.04 x 230 x 450 = 4140 > provided
Provide 4 – 12 mm Bars.

50
DESIGN OF L BEAMS (2, 6,14,20,31 and 37):
=20
= 415
d = =0.180 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Live Load of the Beam = = 3.497
Total load = = 6.08
Factored load = 1.5 x 6.08 = 9.12
= =5.52 KN-m
=
= 0.863 m , limited to 2.5m. Hence, = 0.863 m

51
= = 0.058
5. Reinforcement :
5.52 x = 0.87 x 415 x x (400 – 0.42(0.058 x ))
= 38.31 = 50 =
0.058 x 50 = 2.9 mm
= = 188.4, provide has 190
= 0.04 x 230 x 450 = 4140 > provided

52
DESIGN OF L BEAMS (7, 9,21,23,38 and 40):
=20
= 415
d = =0.233 m =0.250 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Total load = = 6.76
Factored load = 1.5 x 6.76 = 8.64
= = 8.47 KN-m
=
= 0.913 m , limited to 2.8 hence, = 0.913 m

53
= = 0.055
5. Reinforcement :
8.47 x = 0.87 x 415 x x (400 – 0.42(0.055 x ))
= 58.84 = 100
= 0.055 x 100 = 5.5 mm
= 0.04 x 230 x 450 = 4140 > provided
Provide 4 – 12mm Bars.

54
DESIGN OF L-BEAMS (10, 12,24,26,41 and 43):
Effective span l = 3.83
=20
= 415
d = =0.319 m =0.320 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Factored load = 1.5 x 15.0975 = 22.64
= = 41.51 KN-m
=
= 0.999 m , limited to 3.83 m.Hence, = 1 m

55
= = 0.0501
5. Reinforcement :
41.51 x = 0.87 x 415 x x (400 – 0.42(0.0501 x ))
=291.90 =300
= 0.0501 x 300 = 15.03 mm
= = 188.4 < provide
= 0.04 x 230 x 450 = 4140 > provided

56
DESIGN OF L BEAMS (27, 29, 44 and 46):
=20
= 415
d = =0.258 m =0.260 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Factored load = 1.5 x 6.7585 = 10.14
= = 12.18 KN-m
= = 0.938 m , limited to 3.1 m
Hence, = 0.938 m

57
= = 0.053
5. Reinforcement :
12.18 x = 0.87 x 415 x x (400 – 0.42(0.053 x ))
= 84.73 = 100
= 0.053 x 100 = 5.3 mm
= = 188.4 provide 190
= 0.04 x 230 x 450 = 4140 > provided

58
DESIGN OF T-BEAMS (3, 15, 17, 32 and 34):
Effective span l = 3.296m =3.3 m
Breadth of web = 230 mm
= 20
= 415
d = = 0.275 m = 0.300 m
Adopt d = 400 mm
. D = 450 mm
2. Loads:
Self –weight of Beam = 0.45 x 0.23 x 1 x25 = 2.5875
Live Load of beam = 10.493 + 4.84 = 15.333
Total load = 17.92
Factored load ( = 17.92 x 1.5 = 26.88
= = 36.5904 KN- m
= + 0.230+ 6(0.150) = 1.680 m
= 1680 mm, limited to 3.3 m
Hence, = 1.68 m
4. Assuming is within the flange :
= = 0.029

59
5. Reinforcement:
36.59 x = 0.87 x 415 x x (400 – 0.42(0.029x ))
= 255.34 = 300
=0.029 X 300 = 8.7 mm
= 0.04 x 230 x 450 = 4140 > provided
Provide 4- 12mm Bars

60
DESIGN OF T-BEAMS (4, 16, 18, 33 and 35):
Effective span l = 2.1539 m =2.2 m
= 20
= 415
d = = 0.183 m
Adopt d = 400 mm
D = 450 mm
2. Loads :
Live Load of beam = 4.846 + 3.497 = 8.343
Total load = 10.9305
Factored load ( = 10.9305 x 1.5 = 16.395
= = 9.92 KN- m
= + 0.230+ 6(0.150) = 1.497 m
Hence, = 1.497 m

61
4. Assuming is within the flange:
= = 0.033
5. Reinforcement:
9.92 x = 0.87 x 415 x x (400 – 0.42(0.033x ))
= 68.85 = 100
= 0.033 x 100 = 3.3 mm
= = 188.4 , provide 190
= 0.04 x 230 x 450 = 4140 > provided

62
DESIGN OF T BEAM (8, 22 and 39):
Effective span l = 2.7635 m =2.8 m
= 20
= 415
d = = 0.233 m = 0.250 m
Adopt d = 400 mm
D = 450 mm
2. Loads:
Live Load of beam = 4.169 + 8.17 = 12.339
Factored load ( = 14.9265 x 1.5 = 22.38
= = 21.93 KN- m
3. Effective width of Flange:
= + 0.230+ 6(0.150) = 1.597 m
Hence, = 1.597 m

63
= = 0.031
5. Reinforcement:
21.93 x = 0.87 x 415 x x (400 – 0.42(0.031x ))
= 152.60 = 200
=0.031 X 200 = 6.2 mm
= 0.04 x 230 x 450 = 4140 > provided

64
DESIGN OF T-BEAM (11, 25 and 42):
Effective span l = 3.83 m
= 20
= 415
d = = 0.319 m = 0.320 m
Adopt d = 400 mm
D = 450 mm
2 Loads :
Live Load of beam = 8.171 + 12.51 = 20.681
Factored load ( = 23.2685 x 1.5 = 34.90
= = 63.99 KN- m
= + 0.230+ 6(0.150) = 1.768 m
Hence, = 1.768 m
= = 0.028

65
5. Reinforcement:
63.99x = 0.87 x 415 x x (400 – 0.42(0.028x ))
= 450
=0.028 X 450 = 12.6 mm
= 0.04 x 230 x 450 = 4140 > provided

66
DESIGN OF T-BEAM (28 and 45):
Effective span l = 3.068m =3.1 m
= 20
= 415
d = = 0.258 m = 0.260 m
Adopt d = 400 mm
D = 450 mm
2.Loads :
Live Load of beam = 8.165 + 4.171 = 12.336
Factored load ( = 14.9235 x 1.5 = 22.385
= = 26.88 KN- m
= + 0.230+ 6(0.150) = 1.646 m
Hence, = 1.646 m
= = 0.0305

67
5. Reinforcement:
26.88 x = 0.87 x 415 x x (400 – 0.42(0.0305x ))
= 187.24 = 200
=0.0305 X 200 = 6.09 mm
= 0.04 x 230 x 450 = 4140 > provided

68
DESIGN OF PLINTH BEAMS FOR 9” WALL (1, 7, 15 and 17):
=20
= 415
d = =0.275 m = 0.280 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Live Load of the Beam = 25 x 0.23 x 3 = 17.25
Factored load = 1.5 x 19.837 = 29.755
= = 40.5 KN-m
3. Reinforcement :
40.50 x = 0.87 x 415 x x 400 x (1- )
=298.26 = 300

69
= 0.04 x 230 x 450 = 4140 > provided
DESIGN OF PLINTH BEAMS FOR 9”( 2 and 8):
Effective span l = 2.156 m 2.16 m
=20
= 415
d = = 0.18 m = 0.180 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Factored load = 1.5 x 19.837 = 29.755
= = 17.35 KN-m
3. Reinforcement :
17.35 x = 0.87 x 415 x x 400 x (1- )
= 123 = 150

70
= = 188.4 provide as 190
= 0.04 x 230 x 450 = 4140 > provided
DESIGN OF PLINTH BEAMS FOR 9” WALLS (9 and 11):
=20
= 415
d = =0.233 m = 0.250 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Factored load = 1.5 x 19.837 = 29.755
= = 29.1599 KN-m
3. Reinforcement :
29.1599 x = 0.87 x 415 x x 400 x (1- )

71
=210.82 = 250
= 0.04 x 230 x 450 = 4140 > provided
DESIGN OF PLINTH BEAMS FOR 9” WALLS (12 and 14):
=20
= 415
d = =0.319 m = 0.320 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Factored load = 1.5 x 19.837 = 29.755
= = 54.56 KN-m
3. Reinforcement :

72
54.56 x = 0.87 x 415 x x 400 x (1- )
=411 = 420
= 0.04 x 230 x 450 = 4140 > provided
DESIGN OF PLINTH BEAM FOR 4.6” WALL (4 and 6)
Effective span l = 2.156 m 2.16 m
=20
= 415
d = = 0.18 m = 0.180 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Live Load of the Beam = 25 x 0.12 x 3 = 9
Factored load = 1.5 x 11.5875 = 17.38
= = 10.13 KN-m

73
3. Reinforcement :
10.13 x = 0.87 x 415 x x 400 x (1- )
= 71.15 = 100
= = 188.4 provide as 190
= 0.04 x 230 x 450 = 4140 > provided
DESIGN OF PLINTH BEAMS FOR 4.6” WALLS (3, 5 and 16):
=20
= 415
d = =0.275 m = 0.280 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads
Factored load = 1.5 x 11.5875 = 17.38

74
= = 23.65 KN-m
3. Reinforcement :
23.65 x = 0.87 x 415 x x 400 x (1- )
=169.51 = 200
= 0.04 x 230 x 450 = 4140 > provided
DESIGN OF RECTANGULAR BEAMS FOR PLINTH WALL FOR 4.6” (10):
=20
= 415
d = =0.233 m = 0.250 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :
Factored load = 1.5 x 11.5875 = 17.38

75
= = 17.03 KN-m
3. Reinforcement :
17.03 x = 0.87 x 415 x x 400 x (1- )
= 120 = 150
= = 188.4 provide 190
= 0.04 x 230 x 450 = 4140 > provided
Provide 4 – 12 mm Bars
DESIGN OF RECTANGULAR BEAMS FOR PLINTH WALL FOR 4.6” (13):
=20
= 415
d = =0.319 m = 0.320 m
Adopt d = 400 mm
D = 400+50 = 450 mm
2. Loads :

76
Factored load = 1.5 x 11.5875 = 17.38
= = 31.86 KN-m
3. Reinforcement :
31.86 x = 0.87 x 415 x x 400 x (1- )
= 231.33 = 250
= 0.04 x 230 x 450 = 4140 > provided
DESIGN OF STAIRCASE BEAM:
Effective span l = 2.098 m 2.1m
=20
= 415
d = = 0.175 m = 0.180 m
Adopt d = 400 mm
D = 400+50 = 450 mm

77
2. Loads :
Live Load of the Beam = 25 x 0.12 x 1.5 = 4.5 + 4.36 = 8.86
Factored load = 1.5 x 11.4475 = 17.17
= = 9.465 KN-m
3. Reinforcement :
9.465 x = 0.87 x 415 x x 400 x (1- )
= 66.42 = 100
= = 188.4 provide as 190
= 0.04 x 230 x 450 = 4140 > provided

78
COLUMNS
A column in structural engineering is a vertical element that transmits the weight of the structure
above it to other structural elements below through compression. For the purpose of wind or earthquake
engineering, columns may be designed to resist lateral forces. A column or strut is a compression
member, which is used primarily to support axial compressive loads and with a height of at least three
times its least lateral dimensions.
A reinforced concrete column is said to be axially loaded when the line of the resultant thrust of
loads supported by column coincides with the line of center of gravity of the column in longitudinal
direction. Depending on the architectural requirements and the loads to be supported, reinforced concrete
columns may be cast in various shapes i.e., square, rectangle, hexagonal, octagonal and circular. Column
of L or T shaped are also sometimes.
The longitudinal bars in column help to bear the load in the combination with concrete. The
longitudinal bars are held in position by displacement of transverse reinforcement, lateral binders. The
binders prevent displacement of longitudinal bars during concreting operations and also check the
tendency of their buckling under loads.
Columns are designed with Fe-415 and M20 having a reinforcement on all four sides of column. The
longitudinal reinforcement should not be less than 0.8% and not more than 4% of the cross-sectional area
of the column.
1. The bars shall not be less than 12mm diameter.
2. There shall be minimum of 4 bars in rectangular column and 6 bars in circular column.
3. Spacing of longitudinal bars along periphery of column shall not exceed 300mm.This is
requirement of cracking.
4. If a column has larger cross-sectional area than that required to support the load, this minimum
area of bars shall be based on the concrete area required to resist the direct stress and not upon the
actual area.
4.4.1. Minimum eccentricity:
All columns shall be designed for minimum eccentricity equal to ( )
( ), subjected to minimum of 20mm.

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4.4.2. 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 value of the effective length for normal usage assuming
idealized end conditions shown in the appendix-D of IS 456-2000.
4.4.3. Cover:
The longitudinal reinforcing bar in a column shall have concrete cover, not less than 40mm, not less
than the diameter of bar. In the case of columns with minimum dimensions of 200mm or under whose
reinforcing bars do not exceed 12mm, a cover of 40mm may be used.
4.4.4. Short and Slender Column:
A compression member may be considered as short when the slenderness ratio and are less
than 12.All the columns in this building are designed as short columns.
4.4.5. Unsupported length:
The unsupported length, of a compression member shall be taken as the clear distance between end
restraints.
A column may be classified as follows based on the type of loading:
1. Axially loaded column.
2. A column subjected to axial load and uniaxial bending.
3. A column subjected to an axial load and biaxial bending.
4.4.6. Axially loaded column:
All compression members are to be designed for a minimum eccentricity of load in two principal
directions. In practice a truly axially loaded column is rare, if not – existent. Therefore every column
should be designed for a minimum eccentricity. Clause 22.4 in IS code specifies the following minimum
eccentricity, emin for the design of column.
Emin= , subjected to a minimum of 20mm where L is the unsupported length of the column and b is
the lateral dimensions of the column in the direction under the consideration.

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4.4.7. Axial load and uniaxial bending:
A member subjected to axial force and uniaxial bending on the basis of:
1. The maximum compressive strength in concrete in axial compression is taken as 0.002.
2. The maximum compressive strength at the highly compressed extreme fiber in concrete subjected
to axial compression and bending and where there is no tension shall be 0.0035-0.75 times the
strain at least compressed fiber.
Design charts for combined axial compression and bending are given in the form of interaction
diagrams in which curves of vs. are plotted for different values of where p is the percentage
of reinforcement.
4.4.8. Axial load and biaxial bending:
The resistance of members subjected to axial force and biaxial bending shall be obtained on the basis
of assumptions given in Clause 38.1 and Clause 38.2 with neutral axis so chosen as to satisfy the
equilibrium of load and moment about the two axes.
Alternatively such members may be designed by the following equation:
( ) ( )
Mux and Muy - moment about Y-axis due to design loads.
Mux1 and Muy1- maximum uniaxial moment capacity for an axial load of Pu bending about X-axis and Y-
axis respectively.
is related to
Puz=
For the values of = 0.2-0.8, the values of αn vary linearly from 1-2.For value of less than 0.2, αn= 1,
for values of greater than 0.8, αn= 2.

81
4.4.9. Guidelines for fixing the position and orientation of columns in the plan:
1. Column should generally be located at or near corners and intersection of walls.
2. If the site restrictions make it obligatory to locate column footings within the property line the column
may be shifted inside along a cross wall to accommodate footings within the property line.
3. While fixing the orientations of columns care should be taken that it does not change architectural
elevation. This can be achieved by keeping the column orientations and side restrictions as proposed in
plans by Architect.
4. As far as possible, column should be positioned so that the continuous frames from one end to other of
building in both x and y directions are available.
5. Columns should not obstruct the positions of doors and windows.
6. As far as possible, column projection outside the wall should be avoided.
7. When the locations of 2 columns are near to each other then as far as possible only 1 column should be
provided.
8. The column should not be closer than 2m c/c to avoid combined footings. Generally the maximum
distance between two columns is 8m c/c.
9. Columns should normally be provided around staircase and lift wells.

82
AXIAL COLUMNS ( )
1. Data :
Unsupported length L = 3 m
Effectively held in position at both ends,
Restrained against rotation at one end
Effective length l = 0.8L = 0.8 x 3 = 2.4 m
Column size 300 mm x 450 mm
= 20
= 415
2. Slenderness ratio :
Slenderness ratio = = 5.33 < 12
Hence, it may be deigned as short column.
3. Minimum eccentricity :
, subjected to a minimum of 20 mm
= = 19.8 < 20
Hence, = 20 mm
= = 0.044 0.05
Hence, it may be treated as axially loaded column.
Main Reinforcement:
Factored load 300 KN
Size of the column = 300 mm x 450 mm
Gross area = 300 x 450 = 135000
Area of concrete = 135000 -
Since the column is axially loaded short column, its load carrying capacity is given by
x = 0.4 x 20 x (135000 - ) + 0.67 x 415
300 x = 1080000 – 8 + 278.1
780000 = 270.1
=2888.35 = 2890

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Minimum reinforcement = 0.8 % of gross area = 1080 <
Maximum reinforcement = 6 % of gross area = 8100 >
Provide 6 bars of 22 mm diameter,
2 bars of 20 mm diameter, provided = 2890
Lateral ties:
Diameter of lateral ties should not be less than
(a) = = 5 mm
(b) 6 mm
Hence, adopt 6 mm diameter bars
Pitch of the ties shall be minimum of
(a) Least lateral dimension of column = 300 mm
(b) 16 times of diameter of longitudinal bar = 16 x 20 = 320 mm
(c) 48 times of diameter of tiles = 48 x 6 = 288 mm
(d) 300 mm
Provide 6 mm lateral ties at 280 mm c/c

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BI-AXIAL COLUMNS (C1, C2, C3, C6, C9, C10, C12):
1. Data:
b = 450 mm = 20
D = 450 mm = 415
= 75 kN.m = 50 mm
= 60 kN.m ( /D) = 0.10
2. Equivalent:
The reinforcement in section is designed for the axial compressive load and the equivalent
moment is given by the relation,
= 1.15√
1.15 √ = 110 kN.m
3. Non Dimensional Parameters:
[ ] = [ ] = 0.49
[ ] = [ ] = 0.06
4. Reinforcements:
Refer chart-44 of SP: 16 (equal reinforcement all faces) with ( /D) = 0.10 and read out the value
of (p/ ) = 0.06
p = (20 x 0.06) = 1.2
= ( ) = ( ) = 2430
Provide 8 bars of 20 mm diameter ( = 2512 ) with 3 bars in each face.
P = ( ) = 1.24 and the ratio ( ) = ( ) = 0.062
Refer Chart-44 (SP: 16) and read out the value of the ratio [ /( )] corresponding to
the value of ratio [ ] = 0.49 and (p/ ) = 0.062
[ ] = 0.06
= (0.06 x 20 x 450 x ) = 109 kN.m
Due to symmetry, = 109 kN.m
= [0.45 . ]
[(450 x 450) - 2512] + (0.75 x 4156 x 2512)
(2581 x )N
2581 kN
( ) = ( ) = 0.77
Read out the coefficient an = 1.95
5. Check for safety under Biaxial Loading:
+ 1
+ = 0.79 1
Hence, the section is safe under specified loading.

85
6. Reinforcements:
Provide 8 bars of 20 mm diameter as reinforcement and 8 mm lateral ties at 300 mm centers

86
4.5 STAIR CASE
A stair is a series of steps arranged in such a manner as to connect different floors of a building. Stairs are
designed to provide an easy and quick access to different floors. The stairs should be thoughtfully located,
carefully planned, tastefully designed, serving its purpose and at the same time being economical in
construction.
Stairs are provided in a building to afford a means of communication between various. Since they
have to perform the very important function, the slab over which the steps rest should be designed
properly to provide maximum comfort, easy and safety. The most important aspect in providing staircase
is its location. The location of staircase should be such as to provide an easy access so that in case of any
casualty, occupants should be placed in the centre or to the side of a building. The location depends upon
the positions of the rooms and the type of approach needed. In the commercial buildings, it should be
placed centrally so as to:
1. Provide easy access to all shops/offices.
2. Maintain privacy.
The inclined slab of a stair is known as height of stair while the straight portion other than the floor
level is known as landing. While going on flight, one travels vertically. The landing is provided mid-way
either to turn the position and lore to relax while going up. Vertical height to a stair is known as Rise and
available horizontal distance on stair is known as Tread.
1. Rise of the steps - 150mm to 180mm
2. Tread of the step - 200mm to 300mm
The width of the staircase – 1m in residential buildings to 2m in public buildings
DOGLEGGED STAIRCASE
1. Proportionating of stairs :
Dimensions of stair hall = 2.980 m x 3.6576 m
Height of the floor = 3 m
Height of one flight = = 1.5 m
Rise, R = 150 mm
Tread, T = 225 mm

87
Number of risers = = 10
Hence, number of treads = 10 – 1 = 9
Adopt width of stair = 0.9144 m
For 9 treads, the length required = 9 x 0.225 = 2.025 m
Width of landing = = 0.8163 m
2. Effective span :
As the stair slab is spanning longitudinally,
Effective span = Centre to Centre distance of walls
= 3.6576 + 0.23 = 3.8876 m
3. Thickness of Slab :
Assume effective depth
d = = = 194.38mm
Adopt d = 200 mm and D = 220 mm
4. Loads: Loads per meter horizontal which of stairs as follows .
Weight of waist slab = √
= √ = 6.61
Weight of steps = = R x
= x 25 = 1.875
Live load = 3
Floor finish = 1
Total load = 12.485
Factored load = 1.5 x 12.485 = 18.7275
5. Factored BM :Factored Bending moment
=

88
= 35.37 KN-m
35.37 x = 0.138 x 20 x 1000 x
7. Tension Reinforcement :
35.37 x = 0.87 x 415 x x 200 (1- )
= 517.61 = 550
S = =
No of bars = = 7
= x 1000 x 220 = 264
S = = x 1000 = 190.4 mm = 190 mm
No of bars = = 5.25 = 5

89
FOOTINGS
Footings are the structural members that transfer loads from the building or individual column to
the ground. If these loads are to be properly transmitted, foundations must be designed to prevent
excessive settlement or rotation and to minimize differential settlement and to provide adequate safety
against sliding and overturning.
Most foundations may be classified as follows:
1. Isolated footings under individual columns.
2. Strip footings and wall footings.
3. Combined footing supporting two or more columns.
4. Raft or mat foundations.
5. Pile foundations.
The size of the foundation depends on the permissible bearing capacity of the soil which in turn
depends on the type of sub strata. In general, the foundations are to resist vertical, horizontal loads and
moments.
A foundation is assumed to act as a rigid body which is in equilibrium under the action of applied
forces from the structure and the stress from the soil. It is further assumed that soil behaves elastically and
the stress and the strain distribution in soil immediately beneath the soil is linear. This permits the usage
of theory of bending to determine stress distribution in soil for the given axial load and moment.
4.6.1. Objectives of Footings:
1. To distribute the weight of the structure over a larger area so to avoid overloading of the soil
beneath.
2. To load the sub-structure evenly and thus prevent unequal settlement.
3. To provide a level surface for building operation.
4. To take the superstructure deep into the ground thus increasing its stability by preventing
overturning.
4.6.2 Assumptions of Footings:
1. The foundation is rigid so that the variation of pressure under the foundation is linear.
2. The distribution of pressure will be uniform if the centroid of the footing coincides with the
resultant of the applied loads.

90
DESIGN OF ISOLATED FOOTING:
ISOLATED FOOTING
Axial service load = 1000 KN
Size of the column = 300 mm x 300 mm
SBC of soil = 220
= 20
= 400
1. Size of the footing: Load from the column
P = 600 KN
Self-weight of footing = 10% of column load = = 60 KN
Total load on the soil = 660 KN
Area of the footing = = = 3
Size of the square footing
B = √ = 1.7 m
Adopt 2.4 m x 2.4 m square footing
2. Upward Soil Pressure:
Factored load = 1.5 x 600 =900 KN
Soil pressure at unlimited load
= 311 = 0.311
3. Depth of footing from Bending Moment Consideration:
The critical section for B.M will be at the face of the column
= = 130 x N-mm
d = √ = 170 mm
Provide 170 mm effective depth and 400 mm overall depth. Increased depth is taken due to shear
consideration

91
4. Reinforcement
Using 12 mm diameter bars, spacing of bars
= = 175 mm
Hence provide 12mm bars at 220 mm c/c in both directions
5. Check for One Way Shear:
The critical section for 1way shear is at a distance d from the face of the column.
Factored shear force
Soil pressure from the shaded area
=
=
= 0.311 x 1700 x 250 = 184450 N
= = 0.3616
It is less than the minimum strength of M20 concrete = 0.28 (table -19 of IS: 456)
Hence it is safe with respect to one way shear.
6. Check for Two Way Shear:
The critical section is at a distance of from the face of the column
Perimeter of the critical section = 4 (b + d) = 4(300+350) = 2600 mm
Two way shear x area of the shaded portion
= 0.311 x (1700 x 1700 – 650 x 650)
= 767.39 x N
Two way shear stress = = = 0.84
Permissible punching stress
0.25√ = 0.25√ = 1.12 > 0.46
Hence it is safe with respect to two way shear
7. Check for Development Length:
Length available beyond the column space
1.6 x 1.2 = 1.92
=
= = 564 mm
Length available beyond the column face =
= 700 mm > Hence, O.K

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CHAPTER-5
STADD ANALYSIS
5.1. STAAD.Pro
This chapter reviews about some of the fundamental concepts of structural design and
present them in a manner relevant to the design of light frame residential structures. The
concepts from the basis for understanding the design procedures and overall design approach
addressed in the remaining chapter of the guide. With this conceptual background, it is hoped
that the designer will gain a greater appreciation for creative and efficient design of home,
particularly the many assumptions that must be made.
The world is leading Structural Analysis and Design package for Structural Engineers.
• Starting the Program.
• Creating a New Structure.
• Creating Joints and Members.
• Switching On Node and Beam Labels.
• Specifying Member Properties.
• Specifying Material Constants.
• Specifying Member Offsets.
• Printing Member Information.
• Specifying Supports.
• Specifying Loads.
• Specifying the Analysis type.
• Specifying Post-Analysis Print Commands.
• Specifying Steel Design Parameters.
• Performing Analysis and Design.
5.1.1 Viewing the Output File:
Verifying results on screen both graphically and numerically.
World’s #1 Structural Analysis and Design software supporting Indian and major
International codes.
The choice of 0.2 million Structural Engineers worldwide, STAAD pro is guaranteed to
meet all your structural engineering needs. STAAD.pro features state of the art user interface,
visualization tools, powerful analysis and design engines with advanced finite element (FEM)

93
and dynamic analysis and design to visualization and result verification STAAD.pro is the
professional first choice.
STAAD.pro was developed by practicing engineers around the globe. It has evolved
over 20 years and meets the requirements of ISO 9001 certification. STADD.pro has building
codes for most countries including US, Britain, Canada, Australia, France, Germany, Spain,
Norway, Finland, Sweden, India, China, Euro Zone, Japan, Denmark and Holland.
5.1.2 Structural Analysis and Design:
STAAD-III, the world’s most powerful and popular structural analysis and design
software is in use across the globe since 1980. Now it is available in the form of STAAD.Pro
which consists of STAAD + STARDYNE +FEMkit + Visual Draw.
STAAD.Pro is comprehensive, general purpose software for integrated structural
analysis and design. STAAD.Pro may be utilized for analysing and designing practically all
types of structures - buildings, bridges, towers, transportation, industrial and utility structures.
STAAD.Pro implements the most modern technologies in today’s Computer-Aided-
Engineering.
5.2. LOADS CONSIDERED
5.2.1 DEAD LOADS:
All permanent constructions of the structure form the dead loads. The dead load
comprises of the weights of walls, partitions floor finishes, false ceilings, false floors and the
other permanent constructions in the buildings. The dead load loads may be calculated from the
dimensions of various members and their unit weights. the unit weights of plain concrete and
reinforced concrete made with sand and gravel or crushed natural stone aggregate may be taken
as 24 kN/m” and 25 kN/m” respectively.
5.2.2. IMPOSED LOADS:
Imposed load is produced by the intended use or occupancy of a building including the
weight of movable partitions, distributed and concentrated loads, load due to impact and
vibration and dust loads. Imposed loads do not include loads due to wind, seismic activity,
snow, and loads imposed due to temperature changes to which the structure will be subjected to,
creep and shrinkage of the structure, the differential settlements to which the structure may
undergo.

94
5.3. Working with STAAD.Pro
5.3.1 Input Generation:
The GUI (or user) communicates with the STAAD analysis engine through the STD
input file. That input file is a text file consisting of a series of commands which are executed
sequentially. The commands contain either instructions or data pertaining to analysis and/or
design. The STAAD input file can be created through a text editor or the GUI Modelling
facility. In general, any text editor may be utilized to edit/create the STD input file. The GUI
Modelling facility creates the input file through an interactive menu driven graphics oriented
procedure.
Generation of the structure:
The structure may be generated from the input file or mentioning the co-ordinates in the
GUI. The figure below shows the GUI generation method.
Fig 2.3 Generation of the structure through GUI.

95
5.3.2 Material Constants:
The material constants are: modulus of elasticity (E); weight density (DEN); Poisson's
ratio (POISS); co-efficient of thermal expansion (ALPHA), Composite Damping Ratio, and beta
angle (BETA) or coordinates for any reference (REF) point. E value for members must be
provided or the analysis will not be performed. Weight density (DEN) is used only when self-
weight of the structure is to be taken into account. Poisson's ratio (POISS) is used to calculate
the shear modulus (commonly known as G) by the formula,
G = 0.5 x E/ (1 + POISS)
Supports:
Supports are specified as PINNED, FIXED, or FIXED with different releases (known as
FIXED BUT). A pinned support has restraints against all translational movement and none
against rotational movement. In other words, a pinned support will have reactions for all forces
but will resist no moments. A fixed support has restraints against all directions of movement.
Translational and rotational springs can also be specified. The springs are represented in terms
of their spring constants. A translational spring constant is defined as the force to displace a
support joint one length unit in the specified global direction. Similarly, a rotational spring
constant is defined as the force to rotate the support joint one degree around the specified global
direction.
5.4 Loads:
Loads in a structure can be specified as joint load, member load, temperature load and
fixed end member load. STAAD can also generate the self-weight of the structure and use it as
uniformly distributed member loads in analysis. Any fraction of this self-weight can also be
applied in any desired direction.
 Joint loads:

96
Joint loads, both forces and moments, may be applied to any free joint of a structure.
These loads act in the global coordinate system of the structure. Positive forces act in the
positive coordinate directions. Any number of loads may be applied on a single joint, in which
case the loads will be additive on that joint.
 Member load:
Three types of member loads may be applied directly to a member of a structure. These
loads are uniformly distributed loads, concentrated loads, and linearly varying loads (including
trapezoidal). Uniform loads act on the full or partial length of a member. Concentrated loads act
at any intermediate, specified point. Linearly varying loads act over the full length of a member.
Trapezoidal linearly varying loads act over the full or partial length of a member. Trapezoidal
loads are converted into a uniform load and several concentrated loads. Any number of loads
may be specified to act upon a member in any independent loading condition. Member loads
can be specified in the member coordinate system or the global coordinate system. Uniformly
distributed member loads provided in the global coordinate system may be specified to act along
the full or projected member length.
Fig: 2.4 Member load configuration.
 Area/floor load:
Many times a floor (bound by X-Z plane) is subjected to a uniformly distributed load. It
could require a lot of work to calculate the member load for individual members in that floor.
However, with the AREA or FLOOR LOAD command, the user can specify the area loads (unit

97
load per unit square area) for members. The program will calculate the tributary area for these
members and provide the proper member loads. The Area Load is used for one way
distributions and the Floor Load is used for two way distributions.
 Fixed end member load:
Load effects on a member may also be specified in terms of its fixed end loads. These
loads are given in terms of the member coordinate system and the directions are opposite to the
actual load on the member. Each end of a member can have six forces: axial; shear y; shear z;
torsion; moment y, and moment z.
5.4.1 Load Generator – Moving load:
Load generation is the process of taking a load causing unit such as wind pressure,
ground movement or a truck on a bridge, and converting it to a form such as member load or a
joint load which can be then be used in the analysis.
 Moving Load Generator:
This feature enables the user to generate moving loads on members of a structure.
Moving load system(s) consisting of concentrated loads at fixed specified distances in both
directions on a plane can be defined by the user. A user specified number of primary load cases
will be subsequently generated by the program and taken into consideration in analysis.
Section Types for Concrete Design:
The following types of cross sections for concrete members can be designed. For Beams
Prismatic (Rectangular & Square) & T-shape For Columns Prismatic (Rectangular, Square and
Circular)
5.5. Beam Design:
Beams are designed for flexure, shear and torsion. If required the effect of the axial force
may be taken into consideration. For all these forces, all active beam loadings are pre scanned to
identify the critical load cases at different sections of the beams. For design to be performed as

98
per IS: 13920 the width of the member shall not be less than 200mm. Also the member shall
preferably have a width-to depth ratio of more than 0.3.
 Design for Flexure:
Design procedure is same as that for IS 456. However while designing following criteria
are satisfied as per IS-13920:
1. The minimum grade of concrete shall preferably be M20.
2. Steel reinforcements of grade Fe415 or less only shall be used.
3. The minimum tension steel ratio on any face, at any section, is given by:
ρmin = 0.24√fck/fy.
The maximum steel ratio on any face, at any section, is given by ρmax = 0.025.
4. The positive steel ratio at a joint face must be at least equal to half the negative steel at that
face.

99
5. The steel provided at each of the top and bottom face, at any section, shall at least be equal to
one-fourth of the maximum negative moment steel provided at the face of either joint.
 Design for Shear:
The shear force to be resisted by vertical hoops is guided by the IS 13920:1993 revision.
Elastic sagging and hogging moments of resistance of the beam section at ends are considered
while calculating shear force. Plastic sagging and hogging moments of resistance can also be
considered for shear design if PLASTIC parameter is mentioned in the input file. Shear
reinforcement is calculated to resist both shear forces and torsional moments.

100
5.6. Column Design:
Columns are designed for axial forces and biaxial moments per IS 456:2000. Columns
are also designed for shear forces. All major criteria for selecting longitudinal and transverse
reinforcement as stipulated by IS: 456 have been taken care of in the column design of STAAD.
However following clauses have been satisfied to incorporate provisions of IS 13920:
1 The minimum grade of concrete shall preferably be M20
2. Steel reinforcements of grade Fe415 or less only shall be used.
3. The minimum dimension of column member shall not be less than 200 mm. For columns
having unsupported length exceeding 4m, the shortest dimension of column shall not be less
than 300 mm.
4. The ratio of the shortest cross-sectional dimension to the perpendicular dimension shall
preferably be not less than 0.
5. The spacing of hoops shall not exceed half the least lateral dimension of the column, except
where special confining reinforcement is provided.
6. Special confining reinforcement shall be provided over a length lo from each joint face,
towards mid span, and on either side of any section, where flexural yielding may occur. The
length lo shall not be less than a) larger lateral dimension of the member at the section where
yielding occurs, b) 1/6 of clear span of the member, and c) 450 mm.
7. The spacing of hoops used as special confining reinforcement shall not exceed ¼ of minimum
member dimension but need not be less than 75 mm nor more than 100.

101
5.7. Design Operations:
STAAD contains a broad set of facilities for designing structural members as individual
components of an analysed structure. The member design facilities provide the user with the
ability to carry out a number of different design operations. These facilities may design
problem. The operations to perform a design are:
 Specify the members and the load cases to be considered in the design.
 Specify whether to perform code checking or member selection.
 Specify design parameter values, if different from the default values.
 Specify whether to perform member selection by optimization.
These operations may be repeated by the user any number of times depending upon the
design requirements. Earthquake motion often induces force large enough to cause inelastic
deformations in the structure. If the structure is brittle, sudden failure could occur. But if the
structure is made to behave ductile, it will be able to sustain the earthquake effects better with
some deflection larger than the yield deflection by absorption of energy. Therefore ductility is
also required as an essential element for safety from sudden collapse during severe shocks.
STAAD has the capabilities of performing concrete design as per IS 13920. While designing it
satisfies all provisions of IS 456 – 2000 and IS 13920 for beams and columns.

102
5.8.STAIRCASE AND FOOTINGS:
After importing the reactions in the staad foundation the following input data is required
Regarding materials, Soil type, Type of foundation, safety factors.
� Type of foundation: ISOLATED.
� Unit weight of concrete: 25kn/m^3
� Minimum bar spacing: 50mm
� Maximum bar spacing: 500mm
� Strength of concrete: 30 N/mm^2
� Yield strength of steel: 415 n/mm^2
� Minimum bar size: 6mm
� Maximum bar size: 40mm
� Bottom clear cover: 50mm
� Unit weight of soil: 22 kn/m^3
� Soil bearing capacity: 300 kn/m^3
� Minimum length: 1000mm
� Minimum width: 1000mm
� Minimum thickness: 500mm
� Maximum length: 12000mm
� Maximum width: 12000mm
� Maximum thickness: 1500mm
� Plan dimension: 50mm
� Aspect ratio: 1
� Safety against friction, overturning, sliding: 0.5, 1.5, 1.5

103

104
RESULTS:

105
COMPARISION
S.NO MEMBER MANUAL DESIGN STAAD PRO.
1 BEAM
i) D.L
ii)L.L
iii) Dim.
iv) Length
v)
vi)
vii) R.C.C
1.5
2
0.23 x 0.45 m
3.1,3.3,3.83m
20
415
4-12mm-300mm c-c
1.0
1.5
0.3 x 0.23
2.7
25
415
3-10mm-270mmc-c(0-1.8m)
4-10mm-270mmc-c(1.8-2.7m)
2 COLUMNS
Axial column
Uni-axial
Bi-axial
5- 0.3 x 0.45m,6-22mm
-
7 – 0.45 x 0.45, 8-20mm
20
415
-
0.23 x 0.46
0.46 x 0.23
25
415
4 Staircase column 0.3 x 0.45 m 0.23 x 0.23
5 Cantilever slab
Dimension
Length
20
415
0.150 x 0.150 m
0.6083 m ,0.8163 m
25
415
0.23 x .23m
1.99 m
6 Slab edge strips 4-8mm – 290 mm c-c 7-8mm-140mm c-c
7 Reinforcement 6890 7833.413
8 Concrete 2402.615

106
CHAPTER-6
ESTIMATION & COSTING:
DETAILED ESTIMATIONS:-
GROUND FLOOR
S.No Description of work No’s Length
(m)
Breadth
(m)
Depth(m) Quantity
( )
1 Earthwork excavation
i) M.Bed
ii) Kitchen
iii) Pujamundir
&dining hall
iv) Living room
1
1
1
1
1
31.06
1.05
0.43
2.07
1.34
1.7
1.7
1.7
1.7
1.7
0.8
0.8
0.8
0.8
0.8
42.24
1.428
0.584
2.015
1.022
2 P.C.C BED 1 31.06 1.7 0.2 10.50
3 R.C.C (1:1.5:3) Using
HBG Metals for all
COLUMNS, FOOTINGS
and sloped portion column
up to G.L
13
13
13
1.7
0.23
1.7
0.3
0.23
0.2
0.25
7.514
3.763
0.1719
4 PLINTH BEAM 1 31.06 0.45 0.6 8.386
5 SUPER STRUCTURE
Brick work for all wall
I) M.Bed
II) Kitchen
III) Puja &Dining
hall
IV) Living room
DEDUCTIONS
I) M.Door
II) Door-1
III) Door-2
IV) Window
V) Window-1
VI) Ventilator
1
1
1
1
1
1
3
1
4
1
4
31.06
3.12
1.9
3.64
2.81
1.0
0.9
0.75
1.2
1.0
0.6
0.2
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
3
3
3
3
3
2.1
2.1
2.1
1.5
1.5
0.45
18.636
2.152
1.311
2.511
1.938
0.48
1.304
0.362
1.656
0.345
0.248
6 PLASTERING
IN-SIDE
OUTSIDE
1
1
22.153
22.153
-
-
0.12
0.2
2.65
4.43
7 SLAB 1 31.06 - 0.15 4.65
8 STAIR CASE 1 4.885 1.2 2.5 14.655
9 SAND FILLING 1 31.06 - 0.05 1.55
10 Cantilever slab
X-direction
Y-direction
2
2
10.349
5.676
0.609
0.609
0.15
0.15
1.8907
1.0370

107
Net Quantity of Brick is 26.548 - 4.395 = 22.153
FIRST FLOOR
S.NO Description of work No’s Length(m) Breadth
(m)
Depth
(m)
Quantity( )
1 Super-structure
Outer wall
I) Main door
II) Toilet
III) Hall
IV) Children
bedroom
DEDUCTION
i) Main door
ii) Door -1
iii) Door-2
iv) Window
v) Window-1
vi) ventilator
1
1
1
1
1
1
1
2
3
2
2
31.01
3.12
1.9
5.44
2.89
1.0
0.9
0.75
1.2
1
0.6
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
3
3
3
3
3
2.1
2.1
2.1
1.5
1.5
0.45
21.43
2.152
1.311
3.75
1.99
0.483
0.434
0.724
1.242
0.69
0.124
2 Plastering
In-side
Out-side
1
1
26.94
26.94
-
-
0.12
0.2
3.23
5.38
3 SLAB 1 31.01 - 0.15 4.651
4 STAIR CASE 1 4.885 1.2 2.5 14.655
5 Cantilever slab
X- direction
Y -direction
2
2
10.349
5.676
0.609
0.609
0.15
0.15
1.8907
1.0379
Net Quantity of brick required = 30.63 – 3.697= 26.936 ( )
SECOND FLOOR
S.NO Description of work No’s Length
(m)
Breadth
(m)
Depth(m) Quantity( )
1 Brick work
i) M.BEDROOM
ii) Library
DEDUCTION
I) Main Door
II) Door
III) Window
IV) Window-1
V) Ventilator
1
1
1
1
1
2
1
1
24.58
2.89
5.44
1.0
0.9
1.2
1.0
0.6
0.23
0.23
0.23
0.23
0.23
0.23
0.23
0.23
3
3
3
2.1
2.1
1.5
1.5
0.45
16.96
1.994
3.753
0.483
0.434
0.828
0.345
0.062
2 PLASTERING
IN-SIDE
OUT-SIDE
1
1
20.555
20.555
-
-
0.12
0.2
2.46 ( )
4.11 ( )
3 SLAB 1 24.58 - 0.15 3.687( )

108
4 STAIR CASE 1 4.885 1.2 2.5 14.655
5 PARAPET WALL
X-direction
Y-direction
1
2
5.676
4.800
0.2
0.2
0.9
0.9
1.021
0.865
6 CANTILEVER SLAB
X- direction 1 5.6769 1 0.15 0.851
Net Quantity of bricks required = 22.707 - 2.152 = 20.555 ( )
CONSTRUCTION RATIO’S (1:6):
Volume consider for one Cu.m
1 volume of brick with mortar = 0.2 x 0.1 x 0.1= 0.002 Cu.m
No. of bricks for 1 Cu.m = = 500 No’s
Volume of 1 brick = 0.19 x 0.09 x 0.09 = 0.001539 Cu.m
Volume considered by 500 number of bricks = 500 x 0.001539 =0.7695 Cu.m
Volume of mortar = 1- 0.7695 = 0.2305 Cu.m (wet)
For dry mortar = 0.2305 x 1.33 = 0.30656 cu.m
Cement = 0.30656 x 1440 x = 63 kg.s
Sand = 0.306565 x x 35.3147 = 9.28 Cu. Feet
BRICKS (1:6)
Considered volume for 69.644 cu.m
No. of bricks for 69.644 cu.m = = 34822 no.s
Volume covered by 34822 bricks = 34822 x 0.001539 = 53.5910 Cu.m
Volume of Mortar = 69.644 - 53.5910 = 16.053 cu.m (wet)
For dry mortar = 16.053 x 1.33 = 21.350 cu.m
Cement required = 21.350 x x 1440 = 4392 Kgs
No. of bags = = 87.84 = 88 bags

109
Sand = 21.350 x x 35.3147 = 646.259 Cu.feet
PLASTERING RATIO (1:6) (1 cu.m):-
One side wall = length x breadth x thickness of plastering
= 10 x 3 x 0.02 = 0.6 cu.m
Dry density = 0.6+0.6 x 0.33 = 0.798 cu.m
Cement = 0.798 x = 0.114 cu.m
Density of cement = 1440 kg/cu.m
= 0.114 x 1440 = 164.16; for bags = 3.2832 for each cu.m
Sand = 0.114 x = 0.09771 cu.m
= 0.099771 x 35.3147 =3.450 cu.feet
PLASTERING OVER-ALL:
20mm & 12 mm thick plasters 1:6 for walls = = 0.22
Quantity of plastering for all floors = 22.26 sq.m
i) Cement = x 22.26 = 141.03 bags
ii) Sand = 0.22 x 6 x 22.26 = 29.38 cu.m
FOR R.C.C WORK (1:1.5:3):
= 0.28
Cement required for 1 cu.m of work is = 0.28 x 1 = = 8 bags for one cu.m
For 11.4489 cu.m of work
i) Cement = 11.4489 x 8 = 92.31 bags
ii) Sand required = 11.4489 x 0.28 x 1.5 = 4.8 cu.m
iii) Coarse aggregates (20mm) = 0.28 x 3 x11.448 = 9.61 cu.m
P.C.C BED (1:1.5:3):-
= 0.28

110
Cement quantity required for 10.5 cu.m of P.C.C is = 0.28 x 1 x 1440 x x 10.50 = 85 bags
Fine Aggregates = 0.28 x 1.5 x 10.50 = 4.41 =5 cu.m
Coarse aggregates = 0.28 x 3 x 10.5 = 8.82 = 10 cu.m
TOTAL MATERIALS REQUIREMENTS So far
i) Cement = 85 + 93 + 141 + 88 = 410 bags
ii) Sand = 5 + 4.8 + 29.38 +46.259 = 85.439 cu.m = 135 tons
iii) Coarse aggregates = 10 + 9.6 = 19.6 cu.m
iv) Total bricks required = for brick work = 34,822 No.
1. S.No Description
Qty or
No.'s
Rate
Amount
Rupees per
1 Brick Masonry
No.of bricks 34822 3600 1000 1,25,360.00
Cement Mortar(1:6) 0.2305cu.m
a) Cement (kg) 88 350 1 30,800.00
b) Sand 28.14 1500 1 ton 42,210.00
Labor:
Brick Layers 1.5 500 Each 750.00
Man Mazdoor 1 250 Each 250.00
Women Mazdoor 1 200 Each 200.00
Total amount for brickwork = 1,99,570.00
2 Plastering
First Coat-Cement Motar-1:4-16 mm thick b) Second Coat-Cement Motar-

111
1:3-4 mm thick with sponge finishing
a) First & second Coating
Materials: Cement Mortar 0.18 m3
Cement(kg) 141.03 350 1 7,051.50
Sand 45.2 1500 1 ton 63,750.00
Labor:
Mason 1.0 350 Each 350
Man Mazdoor 1 300 Each 300
Women Mazdoor 1 250 Each 250
Total= 71,701.5

112
4 R.C.C.( 1:1.5:3) M20
Materials:
Stone Aggregate (20mm gauge) 4.36 6400 t 27,904.00
Sand 3.12 1500 t 4,680.00
Cement(kg) 92.3 350 1 32,305.00
Steel ( bars, rings, binding wires ) 200kgs 50 kg 10,000.00
Labor:
Mason 1 500 Each 500.00
Man Mazdoor 2 300 Each 600.00
Women Mazdoor 2 200 Each 400.00
Add for Centering & Shuttering
Charges
5 sq.m 851.5
10
sq.m
425.75
Add for Fabrication Charges(
Cutting, bending & placing the
respective moulds)
200
kgs
5 kg 1,000.00
Add for Concrete Mixture - 400 day 400.00

113
Add for Vibrator - 2000 week 2000.00
Total amount for R.C.C. = 80,264.00
5 Plain Cement Concrete 1:3:6
Materials:
Cement (bags) 85 350 bag 29,750.00
Sand 7.69 1500 t 11,535.00
Coarse Aggregate 22.026 6400 t 1,40,966.40
Labor:
Mason 2 400 day 800.00
Man Mazdoor 2 450 day 900.00
Women Mazdoor 1 200 day 200.00
Total amount for PCC = 1,84,151.4

114
S.NO DESCRIPTION Quantity cost Total quantity Total cost
1 Reinforcement ton 41,000 6.5 tons 2,66,500.00
2 Paints Sq.m 650.00 75sq.m 48,750
3 Flooring Sq.m 4,000 210sq.m 8,40,000
4 Water tank Cu.m 1,000 5cu.m 5,000
5 Septic tank Cu.m 150 20cu.m 3,000
6 Bore-
well(electrical
expenses)
Feet’s 200 250 feets 1,00,000
7 Gate 1 - - 20,000
8 Miscellaneous
& furniture
works
Sq.m - - 4,50,000
9 Provision for water supply and sanitation arrangements @20%
10 Provision for electrification @10%
11
Provision for architectural appearance @10%
Total cost for project = Rs 22, 68,936.6 /-
Water supply & sanitation = Rs 4, 53,787.38 /-
Electrification charges = Rs 2, 26,893.66 /-
Architectural appearance = Rs 2, 26,893.66/-
TOTAL EXPENDITURE OF HOUSE = Rs31, 76,511.3/-

115
Conclusion
After obtaining STAAD Analysis and Design the results are seemed to be more accurate
when compared to manual design results. During the project work many obstacles were faced to
obtain error less STAAD output. The main drawback of STAAD is Designing slabs is tedious
job Designing using Software’s like Staad reduces lot of time in design work, Details of each
and every member can be obtained using staad pro. All the List of failed beams can be obtained
and also Better Section is given by the software. Accuracy is Improved by using software. In
Estimation the overall Cost of the project is determined and the Item rates are taken from
Standard Schedule of Rates.

116
Codes & references:
. Code of Practice (RCC) – IS-456-2000 & IS-875-1987 (part-1&2)
 SP-16 IS 456-1978
 MIX- DESIGN of M-20 was done according to IS 10262-1982.
References:
1 Theory of Structures by Ramamrutham for DESIGNS
2. ILLUSTRATED DESIGN OF RC BUILDINGS BY Dr. V.L SHAH & Dr. S.R. KARVE
3. REINFORCED CONCRETE DESIGN BY N. KRISHNA RAJU & R.N. PRANESH
4. REINFORCED CONCRETE STRUCTURES BY M.R. DHEERENDRA BABU

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