Multistorey building

1
CHAPTER-1
INTRODUCTION
1.1 GENERAL
Due to the concentration and increase of population into urban cities there is
a need to accommodate the influx in the urban cities. However, due to rapid increase
of land cost, and limited availability of land the trend is to build high rise building.
The advantages of high rise buildings include but not limited to high ratio rentable
floor space per unit area of land. These high rise buildings are sky scrapers are built
not just for economy of space they are considered icons of a city’s economic power
and the city’s identity.
Various types of structural system have been used to facilitate the demand of
high rise structures. Thousands of high rise buildings are being built all over the world
with steel as well as reinforced concrete. Many of the high rise buildings are designed
with structural components consisting of various systems such as flat slab, flat plate
system, and shear wall core with or without perimeter beams. High rise buildings are
used for densely populated areas where mix uses high rise buildings including
commercial and residential uses because the systems have various following
advantages.
High rise buildings are characterized by their high susceptibility to lateral
drift under the effect of lateral loads such as wind and earthquake loads. Providing
shear walls and or knee bracings in the building system greatly helps in improving its
resistance behavior to lateral loads. It may be possible by engaging the perimeter
columns with the shear wall core which will increase the effective depth of structure
participating in lateral load resistance. Outrigger braced tall building structure usually
consists of a stiff central core, connected to the exterior columns by flexural stiff
cantilevers at the outrigger floors and floor members (slab and floor beam) at typical
floors.
Structure: It is a G+5 multistory building with a height of 5 floors 15 m. The
typical floor height is 3.00 m. Area of the building is 207.27 Sqm. The entire columns
size up to 5th floor is 0.45 x 0.55 m. The no. of columns is 20 per each floor. The
columns have been provided at required spacing. Perimeter columns are connected

2
with knee bracings to resist the lateral deflections. The shear walls with a thickness
of 0.15 m are used in resisting the lateral loads.
1.2 LOADING:
Loading on tall buildings is different from low-rise buildings in many ways such
as large accumulation of gravity loads on the floors from top to bottom, increased
significance of wind loading and greater importance of dynamic effects. Thus, multi-
storied structures need correct assessment of loads for safe and economical design.
Excepting dead loads, the assessment of loads cannot be done accurately. Live loads can
be anticipated approximately from a combination of experience and the previous field
observations. But, wind and earthquake loads are random in nature. It is difficult to predict
them exactly. These are estimated based on probabilistic approach. The following
discussion describes the influence of the common loads.
Gravity loads: Dead loads due the weight of every element within the structure and
live loads that are acting on the structure when in service constitute gravity loads. The
dead loads are calculated from the member sizes and estimated material densities.
Live loads prescribed by codes are empirical and conservative based on experience and
accepted practice.
Live loads: Live loads are those loads produced by the use and occupancy of the
building or other structure and do not include construction or environmental loads
such as wind load, snow load, rain load, earthquake load, flood load, or dead load.
Live loads on a roof are those produced (1) during maintenance by workers,
equipment, and materials; and (2) during the life of the structure by movable objects
such as planters and by people.
Lateral loads and resisting systems: The lateral loads from wind and earthquakes
are known as lateral loads and are resisted by a set of steel frames in orthogonal
directions or by reinforced concrete shear walls. Steel frames are broadly classified as
braced-frames and moment-resisting frames depending on the type of configuration
and beam-to-column connection provided.

3
CHAPTER-2
PLANNING OF MULTISTOREY BUILDING
2.1 BASIC AND ARCHITECTURAL PLANNING AND DESIGN
Building: Building is defined as any structure for whatsoever purpose and of
whatsoever materials constructed and every part there of whether used as human
habitation or not and includes Foundation, Plinth walls, Floors, Chimneys, Plumbing
and building services, fixed platforms, Verandah, balcony. Cornice(or projection), and
signs and outdoor display structures. Broadly speaking, building consist of three parts,
namely
(i) Foundation
(ii) Plinth and
(iii) Superstructure.
General principles of site selection: Site selection has an important bearing on
planning and designing of buildings. Generally, therefore an architect has either to
make a choice of suitable site or to plan his building structure to suit the available site.
Natural defects of a site will involve considerable expenditure on construction and
maintenance of the building.
 A site which comes within the limits of an area where the by-laws of the local
authority enforce restrictions regarding proportions of plots to built up, vacant
spaces to be left in front and sides, heights of buildings etc. should be preferred.
 The site should be situated on an elevated place and also leveled on with uniform
slopes from one end to the other so as to provide good and quick drainage of rain
water.
 The soil surface of the site should be good enough to provide economical
foundations for the intended building without causing any problem. Generally for
most satisfactory instructions, the site should have rock, sand or firm soil below
60 to 120cm. layer of light or even black cotton soil.
 The situation of the site should be such as to ensure unobstructed natural light and
air.

4
 The site should have a good land scape but away from quarries, kilns, factories
etc.
 A site should be abandoned under adverse circumstances such as (a) Un healthy,
noisy or crowded localities (b) Immediate neighborhood of rivers carrying heavy
floods, badly maintained drains (c) Reclaimed soil of water-logged areas, subject
to substance or settlement and (d) Industrial having smoke and obnoxious odours.
Fig2.1: Plan and detailing of building

5
2.1.1 ORIENTATION OF A-BUILDING
Orientation is defined as a method of setting or fixing the direction of the plan
of the building in such a way that it derives maximum benefit from elements of nature
such as a sun, wind and rain.
Faulty housing conditions cause poor health and spread of various types of
diseases. Resistance to disease may be increased by living in fresh air and exposing
the body to sunshine. Proper orientation of a house increases fresh air and sunshine in
the house and decrease possibility of direct infection.
FACTORS AFFECTING ORIENTATION:
The factors affecting the orientation of a building are:
 Climatic factors solar radiation and wind directions.
 Local topography
 Pleasure of a view
 Requirements of privacy
 Reduction of noise
Of the above, climatic factors are more important than all the other factors. For this it
is essential to know the following:
 Sun’s path and its relative position with respect to the locality.
 The direction of prevalent wind and Intensity and direction of rain.
HELPFUL POINTS REGARDING THE ORIENTATION:
 Long walls of building should face North and South, short wall’s should face East
and West.
 A verandah or balcony can be provided towards East and West to keep the rooms
cool.
 To protect building from sub and rain Chajjas are required for window facing the
east, west and South.

6
2.1.2 COMPONENT PARTS OF A BUILDING:
A building generally consists of the following parts.
 Foundations and basement.
 Superstructure.
 Doors and Windows.
 Lintels.
 Sun Shades.
 Flooring.
 Stair Case.
2.1.3 FUNCTIONS OF THE DIFFERENT COMPONENTS:
1. Foundations and basement: The foundation is that parts of the building normally
below the ground level. It receives the Load from the superstructure and uniformly
transmits it to the soil below it. The Strength and stability of a structure depend on the
foundation and hence to be carefully designed.
Basement is that part of the structure which separates the foundation and the
superstructure. The height of the basement decides the level of the flooring in the
building. The top of the basement B called the Plinth.
2. Super structure: This is the part of the wall above the basement. It carries the
superimposed loads from the floors and roof. It performs the following functions.
 It supports the floors and roof.
 It encloses the space or divides if.
3. Doors and Windows: Doors and Windows are provided in openings in a wall.
Doors enable people to move from in and out of rooms, or building windows provide
free circulation of air control entry of natural light in to the building.
4. Lintels: A Lintel is a horizontal member of RCC stone or Wood placed over an
opening to support the structure above it.
5. Sun shade: A sun shade is a horizontal member projecting from the face of the
wall placed above an opening to prevent sun’s rays or rain entering the room.
Generally there are made of RCC and cast monolithically with an RCC lintel.

7
2.2 FOUNDATION:
A foundation is that part of structure which is indirect contact with the ground. It
transfers the load of the structure to the soil below so as to avoid over loading of the
soil beneath. It prevents the differential settlement by evenly loading the substrata. It
provides a level surface for building operations. It also increases stability of structure
by taking the structure deep into the ground. Foundations are generally built of bricks,
Stones, Concrete and Steel etc. The selection of material and type of foundation
depends upon the type of structure and the nature of underlying soil.
The selection of foundation type suitable for a particular site depends on the following
considerations.
 Nature of Sub Soil.
 Nature and extent of difficulties, e.g., presence of boulder, buried tree trunks
etc. likely to be meet with.
 Availability of expertise and equipment. Depending upon their nature and
depth, foundations have been categorized as follows.
i). Open foundations or shallow foundations.
ii). Deep foundations.
General procedure in Foundation Design:
The following steps should be followed in the design of foundations.
 A site investigation should be carried out to determine
o The Physical and Chemical properties of the soil beneath the site.
o To find the position of water table.
o To obtain information on all factors affecting the design of the foundations
and their behavior.
 To determine the magnitude and distribution of Loading from the Super Structure.
 To determine the total and differential settlements which can be tolerated by the
structure.
 To evolve the most suitable type of foundation and its depth below ground level.

8
 To determine preliminary values of the allowable bearing pressure appropriate to
the type of foundation.
 To calculate the pressure distribution beneath the foundation.
 To prepare cost estimates of alternate design.
 To select the material for foundation.
 To prepare structural designs.
 To prepare working designs.
Depth of Foundation:
The foundation of a building should be taken down to such depth, or be so constructed
as to safeguard the building against damage by swelling, shrinkage or freezing of the
subsoil. The foundation should be taken down to a depth where the bearing capacity
of the soil is adequate to support the foundation loading without failure of the soul in
the shear or excessive consolidation of the soil. As far as possible the foundations
should be kept above ground water level in order to avoid the most of pumping, and
possible instability of the soil due to seepage of water into the bottoms of the
excavation. A depth of 0.9 to 1 M. is regarded as a minimum at which some seasonal
movement will occur but is unlikely to be of a magnitude sufficient to cause damage
to the superstructure or ordinary building finishes.
The depth of foundation can also be obtained by plotting the lines of angles
45o and 60o
Let h1 = Depth of footing
h2 = Total depth of foundation
h3 = depth if concrete block
Then h = h1 + h2
To find the minimum depth of foundation of for loose soil, the following
Rankine’s formula can be used.
h = p/w [(1 - Sin Φ) / (1 + Sin Φ)]
Where,
h = Minimum depth of foundation in meters

9
w = Weight of Soil in Kg/m3
Φ = Angle of soil in Kg/m2
2.3 MASONRY:
Superstructure is that portion of the building which above the floor level. Its main
function is to enclose or divide space. It may have to provide support in certain
situations. Masonry is essentially a well material. Broadly, wall can be divided into
following two types.
 Load – bearing walls and
 Non - load bearing wall or certain wall or filler wall or panel.
The functional requirements of a wall depend upon its form of construction.
The popular form of construction used is
 Masonry walls, and
 Monolithic walls.
Masonry wall: The wall is built of individual blocks of materials such as stone, brick
concrete, hollow bricks, cellular concrete, laterite etc. usually in horizontal courses
cemented together with some form of mortar. The binding strength of mortar is
usually disregarded as far as the strength of the wall is concerned.
Monolithic Walls: These are the walls built of a material requiring some sort of
shuttering in the initial stages. The popular form of monolithic walls one traditional
earth walls and modern CC and RCC walls.
Masonry Classification: Broadly, Masonry can be classified into following
categories.
 Stone Masonry
 Brick Masonry
 Hollow block concrete masonry
 Reinforced Masonry and
 Composite Masonry.

10
Stone Masonry: It is the art of building the structures in stones. In some parts if the
country building stones are abundantly available in nature these stones when cut and
dressed to the proper shape provide an economical material for the constructing of
various parts of a building which are located in hilly areas.
Uses of stone Masonry:
Stone Masonry Construction is used in
 Building foundations, dams, monumental structures
 Building walls, piers, columns, pillars, light houses and architectural works.
 Arches, domes, lintel and beams
 Roofs, floors, paving jobs and
 Railway ballast, blackboards and electrical switch boards.
Brick Masonry: Brick Masonry is a unified mass obtained by systematic
arrangement of laying bricks and bounding together with mortar. Brick is a building
unit of hard inorganic clay material of a size which can be conveniently handled.
They can be easily arranged into various shapes for most of the structure, e.g.,
foundations, walls, columns, buttresses, retaining structures, window sills, jambs
corbels, copings ornamental brick work, circular brick work, fire places, flumes, tall
chimneys, cavity walls, thresholds, culverts, steps, floors, arches etc. The strength of
brick masonry works depends upon the quality of brick and type of mortar used.
Bonds in Brick Masonry: Bonding is a process of arranging bricks and mortars to tie
them together in a mass of brick work.
Wall thickness in brick work: The wall thickness depends upon
 The anticipated load to come on the wall
 The quality of wall material
 The overall height
 The spacing between buttresses and cross wall
The different loads which can act on walls are:
 Dead weight of the wall
 Dead weight due to floor or roof slab or beams etc.

11
 Wind force acting on the roof and wall
 Live load coming on the building
 Earth pressure in course of walls constructed below the ground level.
Let the total load coming on wall due to above mentioned force be P
If, A – Area of wall
P – Allowable Compressive strength of the brick
L – Length of wall
T – Thickness of wall
Then A – P/P
T x L – P/P
Or T – P/PL
For unit length, thickness – P/P
It has been found that loads acting on walls are eccentric
If e – eccentricity of the load
Max – pressure per unit area
Then – P/T [1 6e/T] and
Where Max. Compressive stress in Masonry
P/T [1 6e/T] and Minimum Stress – P/T [1 6e]
Partitions or Partition Walls: They are constructed as these walls enclosing areas
for rooms within a building either on ground floor or for upper floors. They vest either
on flooring concrete or on beams spanning between the main walls. They may be
taken either up to full floor height or up to 2.5m.
The following are the advantages of partition walls:
 They divide the whole area into number of rooms.
 They provide privacy to the in match from sight and sound.
 They are light in weight and cheaper in cost of construction.
 Being his in cross-section, they occupy less area of the floor.

12
2.4 DOORS WINDOWS AND LINTELS:
Doors: A door is frame work of wood or steel secured in and opening left in a wall
for the purpose of free movement of people and material in and out of a room. A door
is provided with Shutters. If two Shutters are provided, it is called double Shuttered
door and it one Shutter is provided, it is called single Shuttered door.
Parts of doors: A few Common types if doors are
 Paneled doors
 Ledged and battened doors
 Ledges battened and braced doors.
A few special types of doors are
 Flush doors
 Revolving doors
 Collapsible doors
 Rolling Shutters
 Sliding doors.
WINDOWS: A Window is a frame work of wood or steel secured in an opening in a
wall for the purpose of controlling light and ventilation in a room.
Types of windows:
 Casement Window
 Bay Window
 Clerestory Window
Parts of windows:
A window consists
 Frames
 Shutters

13
LINTELS:
A lintel is a horizontal structural member which is fixed over the openings,
doors, windows recesses etc. to support the structure over opening. Lintels are usually
rectangular on shape and they afford facilities for fixing the door and window frames
wherever they are used.
Types of lintels: On the basis of material used in construction the lintel are classified
into the following types.
 Wooden Lintels
 Stone Lintels
 Brick Lintels
 Steel Lintels
 Reinforced concrete lintels
 Reinforced brick lintels (i.e. R.B. Lintels)
Depth of lintel: For ordinary loads 15cm depth for spans up to 1.2m and add another
2.5cm for every additional 40mm span.
2.5 ROOFS AND FLOORS:
Definition: A roof is the upper most part of a building whose function is to provide a
covering to keep out rain, snow, wind, etc.
Classification of roofs: Roofs are classified as follows. They are
(i) Pitched roofs. (ii) Flat roofs.
Pitched roofs: A pitched roof is a sloping roof. It is suitable for places where there is
heavy snow fall or rain fall.
Flat roofs: Common types of flat roofs are as follows
(i) Maura’s terrace roof. (ii) R.C.C roof.
R.C.C. roof: R.C.C Roofs are widely used in modern construction. For spans up to
2m for ordinary loads, a simple R.C.C slab is adequate. For greater spans a simple
R.C.C. beam and slab construction would be necessary. For constructing R.C.C.
floors or slabs the following procedure is adopted.

14
 Erection of forms.
 Typing and placing reinforcing grills.
 Batching, mixing, placing and compacting.
 Stripping of forms.
FLOORS:
Requirements of a good flooring: A good floor should floor should possess the
following features. They are as follows
 It should have sufficient resistance to fire wear and tear, temperature changes and
chemical reactions.
 It should have a sufficient resistance against dampness in buildings.
 It should have a pleating appearance.
 It should not cause noise.
 It should have a smooth and even surface.
 It should be cheap and economical to construct.
Ground and upper floors: Floors are the horizontal elements of a building structure
which divide the building in to different levels for the purpose of creating more
accommodations with in a restricted space one above the other and provided supports
for the occupants, furniture and equipment of a building.
A floor consists of the following two components.
 A Sub-floor.
 Floor coverings.

15
Floors suitable for different situations:
Sl.no. Type of flooring Requirements
Types of flooring
suggested
1
Laboratories,
hospitals, toilets
Should be non-absorbent,
capable of being cleaned
easily stain proof, to acid,
dust free, noise free
Polished stone
cuddapah slab
ceramic tiles mosaic
flooring
2
Public buildings such
as schools, offices.
Should be resistant to wear
and tear i.e. hard dust free
non-slippery pleasing in
appearance.
Flag stone flooring
mosaic cuddapah
slabs cement
plastering.
Sub-floor: The purpose of this component is to impart strength and stability to
support floor covering and all other super imposed loads.
Floor covering: This is the covering over the sub-floor and is meant to provide a
hard, clean, smooth, impervious, durable and attractive surface to the floor.
Generally the following materials are used for grant floor construction. They
are as follows.
 Bricks.
 Stones.
 Wooden blocks and
 Concrete.
The upper floor in addition to having a good wearing surface should be stronger to
sustain heavier loads and should provide adequate sound insulation in buildings.
Floor and Roof: The top most covering constructed over an enclosed space of a
building is called roof. The intermediate slab provided to divide the vertical space in a
building to provide a storey is called a floor. They carry the super imposed loads due
to men and materials and transfer them to the super structure. An additional function
of the roof is to provide a cover the super structure to prevent rain, sun, wind, etc.,
from entering the building.

16
2.5.1 FLOORING:
Floor is the surface over which people perform different activities. It is laid so
that it is flush with the top of basement. It is laid over the earth or sand filling in the
space within basement walls.
2.6 STAIRS AND STAIRCASES:
Definition: Stairs are a series of steps arranged to connect different floors of a
building an enclosure which contains the stair way is called stair case.
Stair case can be constructed with timber, stones, bricks, steel etc… but R.C.C. has
almost replaced all other materials in our country and is widely used.
Location of staircase: The following points should be observed in locating stairs in a
building.
 A stair case should be located so that it is easily accessible from the different
rooms of the building.
 In the case of public buildings it should be located near the entrance.
Types of staircases: The common types of staircases are as follows.
 Straight types of stairs.
 Quarter turn stairs.
 Half turn stairs.
o Dog legged stairs,
o Open well stairs.
 Three quarter turn stairs.
 Spiral stairs.
R.C.C. stairs: All types of stairs can be constructed with R.C.C. This stairs are
designed mainly in two ways they are as follows.
 Stairs spanning horizontally.
 Stairs spanning longitudinally.

17
Requirements and standards of staircases:
Minimum width:
Residential buildings =1.0m
Assembly buildings like auditoria =1.5m
Residential hotel buildings =1.5m
Theaters, cinema halls =1.5m
Educational buildings =1.5m
Intuitional buildings = 2.0m
Minimum treads: The maximum height of riser should be 190mm for residential
buildings and 150mm for other buildings. The number of steps in a flight should be
limited to 12.
To have easy climbing a descending the following relationships between going and
rise are adopted. They are as follows.
 Going in mm 2 x rises in mm – 600.
 Going in mm rise in mm – 4000 to 4100 (approximately)
Hand rails: The hand rails should have a minimum height of 0.9m from the centre of
the tread.
Landing: The width of landing should be the same as the width of stair.
2.7 PROTECTIVE AND DECORATIVE FINISHES:
Plastering:
Plastering is a thin coat of mortar applied on the surfaces of walls and ceilings,
plastering covers the uneven surface, scales and hides joints of walls and same times
used for decorative purposes. External plastering and other finishes applied for the
purpose of protection and decoration are rendering.
Objectives of plastering: The main objects of plastering are as follows.
 To protect the exposed surfaces from atmospheric influence.
 To cover decorative workmanship and interior quality materials.
 To improve the appearance of the structure.

18
Types of plasters: Plastering can be applied with the following mortars.
 Lime mortar.
 Cement mortar.
 Combination mortar or cement-lime mortar.
The plaster can be one or two coats of thickness 12mm or 20mm respectively.
Pointing: The finishing and protection of mortar joints of walls with cement mortar
or lime mortar is known as pointing. Generally pointing is done with cement mortar
(1:3) or (1:4).
Objectives of pointing: The main objects of point are:
 To protect the joints from the disintegrating effects of wetness.
 To serve as an alternative to plastering.
 To cover weak mortar used.
 To enhance the natural beauty of the construction materials.
White washing and colour washing: The internal and external walls are treated with
one, two or three coats of white wash made lime and water. If a pigment is added then
it is called colour wash.
The objects of white washings and colour washings are as follows:
 To present a pleasing appearance.
 To provide better distribution of light in the rooms.
 To serve as a disinfectant.
Painting: Wooden and steel members and sometimes walls are painted for the
following reasons.
 To protect the surface from weathering effects of the atmosphere.
 To prevent decay of timber and rusting of steel.
 To provide a pleasing appearance.

19
2.8 ASPECT (VASTU)
‘Aspect or Vastu’ is the basic need of all of us. Everyone wants a home built for
oneself equipped with all the amenities for a comfortable living. But many people
experience difficulties after having a house constructed against the principles of Vastu
due to ignorance. Thus, owning a house built in accordance to Vastu principles is
desirous for everyone and some general rules are presented below:
 The southern and Western side compound wall should be thicker and higher than
the Northern and Eastern compound wall.
 The building portion of Southern and Western side should be higher than Northern
and East sides. Building should be leaving more space in East and North than
West and South.
 The gate should be in North-East in East for East facing house. The main door of
the house should be in North-East in East and if possible another door must be
placed in the Northern side for East facing home.
 The well and underground water storage (sump) must be dug in the North-East in
East or in the North-East in North.
 The generator, meter board and kitchen must be placed in South-East. Kitchen can
also be constructed in the North-West, West-South.
 The overhead tank should be in the South-West.
 The master bedroom is in the South-West. Balconies are built in the East & North
sides.
 The high trees should be grown in the Southern-Western portion of the compound.
 Bath-room, Latrine should be in the South-East, North-West, South or West.
 The roof and floor slope must be from South-North or from West to East.
 The stair-case in clock wise direction must be constructed in Southern and
Western sides.
 The gate should be in the North-East in North for a North facing home.
 The Septic tank must be dug in Ucchasthana of either in the North or East.
 The water closet (w.c) should never be placed in North-East.

20
 For South facing home gate should be in South-East in South. The main door of
the home in south facing must be placed in South-East in South. If possible the
second door must be placed in the North-East in North.
 For West facing home gate should be in West. The main door of the home must be
placed in North-West in West. The second gate must be placed in the Eastern
Ucchasthana for West facing home.

21
CHAPTER-3
DESIGN OF SLABS
SLABS:
Slabs are plate elements forming floors and roofs of buildings. Inclined
supported by beams or by walls and may be used as the flange of T/L beam. Further a
slab may be simply supported or continuous over one or more supports and is
classified according to the method support.
The provisions of IS: 456-2000.Clause no.23.2 for beams applied to slab also.
For slabs spanning in two directions, the shorter of the two spans should be used for
calculating the span to effective depth ratios. For two way slab of small span (up to
3.5m) with M.S. bars, the span to overall depth ratios given below may generally be
assumed to satisfy vertical deflection limits for loading class up to 3000 N/m2.
Simply supported slabs: 35
Continuous slabs: 40
For Fe415 steel the various given above should be multiplied by 0.8
a) One way slabs spanning in one direction
b) Two way slabs spanning in two directions.
c) Circular slabs.
d) Flat slabs resting directly in columns with no beams.
e) Grid floors and ribbed slabs.
Slabs are designed by using the same theories of bending and shear as used for beams.
The following methods are available for analysis.
A) Elastic analysis
B) Semi empirical co-efficient.
C) Yield line the theory
If the cross sectional area of the three basic structural elements, beam, slab and
column are related to the amount of the steel provided. It will be said that the percent
steel is usually maximum in a column than in a beam and a slab can beam made as
follows.
1) Slabs are analyzed and designed as having a unit width that is one meter wide
strips.

22
2) Compression reinforcement is used only in exceptional cases in a slab
3) Shear stresses are usually very low and reinforcement never provided in
slabs. It is preferred that to increase the depths. Of slabs and hence to decrease
the shear stresses.
4) Temperature reinforcement is invariably provided right angles to the main
longitudinal reinforcement in a slab.
5) Slabs are usually much thinner than beams.
TWO WAY SLABS:
When slabs are supported on four sides, two way spanning action occurs. Such
slabs be simply supported or continuous on any or all sides. The deflection and
bending moments in two-way slabs are considerably reduced as compared to those in
one-way slabs.
Thus a thinner slab can carry the same load when supported on four slabs,
where the length is greater than twice the breadth, two–way action effectively reduced
to one-way action the direction short span although the end beams to carry some slab
load.

23
SLAB TYPE 1:
Assume: fck = 20 N/mm2 fy = 415 N/mm2
Let Short span = Lx = 3.20 Longer span = Ly = 5.07
Ly/Lx = 5.07/3.20 = 1.584 <2
Two way slab
1. Thickness of slab:
Assume D=130mm
d=130-20=110
2. Loads on slab:
Dead load = 1 x 0.13 x 25 = 3.25 KN/m2
Live load = 2.0 KN/m2
Floor finish = 1.00 KN/m2
____________________
Total load = 6.25 KN/m2
Factored Load = 1.5 x 6.25 = 9.375 KN/m2
3. Design moments and shear forces:
Mux = αxwLx
2
Muy = αywLy
2
Where αx, αy are the bending coefficients for two way slabs for the one long edge and
one short edges are discontinued and negative.
For Ly/Lx = 5.28/3.31 = 1.595
(One long adjacent edge is discontinuous, case no. 3 in table no. 26 in IS 456:2000)
αx(-ve) = 0.067+(0.077-0.067) x 0.095/25 = 0.067
αx(+ve) = 0.051+(0.059-0.051) x 0.095/25 = 0.051

24
αy(-ve) = 0.037
αy(+ve) = 0.028
Mux(-ve)= αx(-ve)wLx
2
= 0.067 x 9.375 x 3.312
= 6.88 KN-m
Mux(+ve)= αx(+ve)wLx
2
= 0.051 x 9.375 x 3.312
= 5.23 KN-m
Muy(-ve)= αy(-ve)wLx
2
= 0.037 x 9.375 x 3.312
= 3.8 KN-m
Muy(+ve)= αy(+ve)wLx
2
= 0.028 x 9.375 x 3.312
= 2.87 KN-m
Vu = wu lx /2
= (9.375 x 3.31)/2
= 15.51 KN
4. Min depth required:
Depth required resisting the bending moment
Mu = 0.138 x fck x b x d2
6.88 x 106 = 0.138 x 20 x 1000 x d2
d = 49.92 <110
Hence provided depth is adequate

25
5. Reinforcement:
Along x-direction (short span)
Mux = 0.87 fy Ast d[1 - (Fy Ast/ Fck b d)]
6.88 x 106 = 0.87x 415x Ast x 110[1 - (415 x Ast/20x1000 x110)]
173.23 = Ast (1-Ast/5301.204)
Ast
2 – 5301.204Ast + 5301.204 x 173.23=0
Ast = 179.34 mm2
Using 8mm dia bars
6. Spacing of bars:
S = [ ast/ Ast]x 1000
= [(π/4) x 82/179.34] x 1000
= 280 mm
Max spacing:
(i) 3d = 3x110 = 330 mm
(ii). 300 mm
Whichever is less.
Hence provide 8 mm dia bars @ 280mm c/c spacing
Along y-direction (longer span)
These bars will be placed above the bars in x-direction
Hence d= 110-10 = 100mm
Muy = 0.87 fy Ast d2[1 -(Fy Ast/fckbd) ]
3.82 x106 = 0.87 x 415 x Ast x 100[1 - (415 x Ast/20x1000x100)]
105.24 = Ast[1 -( Ast/4819.277)
Ast = 107.41 mm2

26
Using 8 mm dia bars
Spacing of bars:
S = (ast/Ast)x 1000
= [(π/4)x82/107.41] x 1000
S = 467.97 mm
Max spacing:
i. 3d = 3x100 = 168 mm
ii. 300mm
Whichever is less
Hence provide 8mm dia. bars at 300 mm c/c spacing
7. Reinforcement in Edge Strip:
Ast = 0.12% of gross area
= (0.12/100 x 1000 x 130) = 156mm2
Using 8mm bars, Spacing
S =[(π/4)x82/156] x 1000 =322
Maximum spacing is
i. 5d = 5 x 110= 550
ii. 450mm whichever is less
Hence, provide 8mm bars at 320mm c/c in edge strips in both directions
8. Check for deflection:
For simply supported slab basic value of l/d ratio = 20
Modification factor for tension steel F1
% of steel = 0.12
fs = 0.58 fy
= 0.58 x 415 = 240.7 N/mm2
MF = 1.6
Max permitted l/d ratio = 1.6 x 20 = 32
l/d provided = 3310/110 = 30.09<32
Hence deflection control is safe.

27
SLAB TYPE 2:
Let fck = 20 N/mm2 fy = 415 N/mm2
Short span = Lx = 4.64 Longer span = Ly = 5.07
Ly/Lx = 5.07/4.64 = 1.09 <2
Two way slab
Assume D=130mm
d=130-20=110
2. Loads on slab:
Dead load = 1 x 0.13 x 25 = 3.25 KN/m2
Live load = 2.0 KN/m2
____________________
Factored Load = 1.5 x 6.25 = 9.375 KN/m2
3. Design moments and shear forces:
Mux = αxwLx
2
Muy = αywLy
2
Where αx, αy are the bending coefficients for two way slabs for the one side
discontinuous and three sides continuous.
For Ly/Lx = 5.18/4.75 = 1.09
(One long edge is discontinuous as per table no. 26 in IS 456:2000)
αx(-ve) = 0.037+(0.044-0.037) x 0.09/10 = 0.037
αx(+ve) = 0.028+(0.033-0.028) x 0.09/10 = 0.028
αy(-ve) = 0.037
αy(+ve) = 0.028

28
Mux(-ve) = αxwLx
2
= 0.037 x 9.375 x 4.642
= 7.826 KN-m
Mux(+ve) = αxwLx
2
= 0.028 x 9.375 x 3.312
= 5.922 KN-m
Muy(-ve) = αywLx
2
= 7.826 KN-m
Muy(+ve) = αywLx
2
= 5.922 KN-m
Vu = wulx /2
= (9.375 x 4.75)/2
= 22.265 KN
4. Min depth required:
Mu = 0.138 x fck x b x d2
7.826 x 106 = 0.138 x 20 x 1000 x d2
D = 48.38 <110
Hence provided depth is adequate
5. Reinforcement:
Along x-direction (short span)
Mux = 0.87 fy Ast d[1 - (Fy Ast/ Fck b d)]
7.826 x 106 = 0.87x 415x Ast x 110[1 - (415 x Ast/20x1000 x110)]
197.05 = Ast (1-Ast/5301.204)
Ast
2 – 5301.204Ast + 5301.204 x 197.05=0
Ast = 204.97 mm2
Using 10mm dia bars

29
Spacing of bars:
S = [ ast/ Ast]x 1000
= [(π/4)x102/204.97] x 1000
= 383.177 mm
Max spacing:
(i) 3d = 3x110 = 330 mm
(ii) 300 mm
Whichever is less.
Hence provide 10mm dia bars @ 300mm c/c spacing
Along y-direction (longer span)
These bars will be placed above the bars in x-direction
Hence d= 110-10 = 100mm
Muy = 0.87 fy Ast d2[1 -(Fy Ast/fckbd) ]
7.826 x 106 = 0.87 x 415 x Ast x 100[1 - (415 x Ast/20x1000x100)]
216.75 = Ast[1 -( Ast/4819.277)]
Ast = 226.42 mm2
Using 8 mm dia bars
Spacing of bars:
S = (ast/Ast)x 1000
=[(π/4)x82/226.42] x 1000
S = 346.57 mm
Max spacing:
ii. 3d = 3x100 = 168 mm
ii. 300mm
Whichever is less
Hence provide 8mm dia bars at 300 mm c/c spacing

30
6. Reinforcement in Edge Strip:
= (0.12/100 x 1000 x 130) = 156mm2
Using 8mm bars, Spacing
S =[(π/4)x82/156] x 1000 =322
Maximum spacing is
i. 5d = 5 x 110= 550
ii. 450mm whichever is less
Hence, provide 8mm bars at 320mm c/c in edge strips in both directions
For simply supported slab basic value of l/d ratio = 20
Modification factor for tension steel F1
% of steel = 0.12
fs = 0.58 fy
= 0.58 x 415 = 240.7 N/mm2
MF = 1.6
Max permitted l/d ratio = 1.6 x 20 = 32
l/d provided = 3310/110 = 30.09<32
Hence deflection control is safe.

31
CHAPTER-4
DESIGN OF BEAMS
BEAMS:
A reinforcement concrete beam should be able to resist tensile, Compressive
and shear stresses as induced in it by the loads on the beam. Concrete is fairly strong
in compression but very weak in tension. The plain concrete is overcome by the
provision of reinforcement in tension, thus the tensile weakness of concrete is
overcome by the provision of reinforcement in tension zone to make reinforced
concrete beam.
Mulim of the given section is calculated and is compared with the maximum
bending moment of the section. If Mulim is greater than Mu The section is designed as
singly reinforced section. If Mlim is less than Mu the section is designed as a doubly
reinforced section. Mu/bd2 is calculated and percentage of steel is required in tension
and compressions corresponding to grade of steel are obtained from S 16-1980.
Reinforcement required for bending and shear in beams in calculated in accordance
with the provision laid down in clauses 26.5.40.1 and 40.3 of IS456:2000
There are three types of reinforced concrete beams.
a) Singly reinforced beams.
b) Doubly reinforced beams
c) Flanged beams
SINGLY REINFORCED BEAMS:
In singly reinforced simply supported beams reinforcing steel bars are placed
near the bottom of the beak where they are most effective in resisting the tensile
bending stresses. In singly reinforced cantilever beams steel is placed near the top of
the beam for the same reason.
DOUBLY REINFORCED BEAMS:
A doubly reinforced beam is reinforced in both compression and tension
regions. The section of the beam may be rectangle or T/L in shape. The necessity of
using steel in the compression regions arises due to two reasons.
1) When depth of beam is restricted, the strength available from a singly
reinforced beam is inadequate.
2) As the supports of the continuous beams where the bending moment changes
sign. Such a situation may also arise in the design of a ring beam.

32
FLANGED BEAMS:
In most of R.C. Structures, concrete slabs and beams are cast monolithic. Thus
beams form a part of the floor system together with the slab. In bending slab forming
the part of the beam at mid span would be in compression for a width of the rib, thus
increasing the moment of resistance for a given rib width. At continuous supports, the
position is reversed. The slab is in tension and part of the rib is in compression. Since
concrete is assumed to have cracked in tension, this beam is equivalent to a
rectangular section at the supports.
DESIGN OF BEAM 1:
Span of the Beam L= 5.07m
1. Depth of the Beam:
Selecting the depth in range of l/12 to l/15 Based on stiffness
d = 5070/12 =422
Adopt d = 450mm
D = 500mm
Let take B = 300mm
2. Loads per meter length of Beam:
Dead Load = .3x.5x25 = 3.75 KN/m
a. Load from Slab
𝑤𝑙𝑥
6
[3 − (
𝑙𝑥
𝑙𝑦
)2
]
=5.875 x (3.20)/6 [3-(3.20/5.07)2) = 8.151 KN/m
b. Load from Slab
=5.875 x (4.64)/6 [3-(4.64/5.07)2) = 9.824 KN/m
Load from wall =0.23 x2.7 x19 = 11.8 KN/m
____________
Total Load = 33.525 KN/m
____________
Factored Load 1.5 x 33.525 = 50.28 KN/m
Shear Force wul/2=(50.28 x 5.07/2) = 127.46 KN

33
Bending Moment
𝑤𝑙2
8
=(50.28×5.072/8) = 161.55 KN-m
3. Min. Depth Required:
Mu = Mu lim.
= 0.138 x fck x b x d2
161.55 x 106 = 0.138 x 20 x 300 x d2
d = 441.71 mm<450mm
Hence, provided depth is adequate
Mu lim. = 0.138x20x300x4502
= 167.67 x106
Mu < Mu lim
Hence it is a singly reinforced beam
4. Reinforcement:
Mu = 0.87 x fy x Ast x d[1-(fy xAst/fckbd)]
167.67 x 106 = 0.87 x 415 x Ast x 450[1-(415/20 x 300 x 450]
994 = Ast[1-( Ast /6506.02)]
Ast
2 - 6506.02 + (6506.02 x 994) =0
Ast= 1224.8mm2
Providing 16mm dia bars
[1146.98/ (π/4 x 162) = 5.70 = 6 no.of bars
By providing 16mm dia 4 bars
4 x π/4 x 162 = 804.24 mm2
4 x π/4 x 122 = 452.39 mm2
Hence, Ast Provided = 1256.63mm2
5. Check for shear:
% of steel = 100 x Ast/bd
= 100 x 1256.63/300x500
= 0.9%
As per IS456:2000 code book
τc = 0.6 N/mm2

34
and τv = Vu/(b x d)
= 127460/(300 x 450)
= 0.94 N/mm2
Since τv > τc , shear reinforcement has to be designed
Vus = Vu - 𝜏 𝑐bd
= 127460 – 0.6 x 300 x 450
= 46460N
6. Shear Resistance of Bent up Bars:
Shear Resistance of Bent up bars
Vusb =0.87fyAsbsinα
=0.87 x 415 x 2 x(π/4 x 162) x sin450
=102662N
7. Design of vertical Stirrups:
Shear resistance to be provided by the vertical stirrups= Vus/2= 46460/2= 23230N
Spacing of 2 legged 6mm stirrups
Asv =2 x (π/4 x62) =56.5mm2
Spacing from minimum shear reinforcement consideration as per IS 456
(Asv/b x Sv) = 0.4/0.87 x fy
Sv = (0.87 x fy x Asv /0.4 x b)
Sv =(0.87 x 415 x 56.5/0.4 x300)
= 170mm
Maximum allowed spacing
=0.75d = 0.75 x450=337.5
Or 300mm whichever is less.
Hence Provide 2-legged 6mm stirrups @ 170mm c/c throughout the span of the
beam
= 100 x 1256.63/300x450
= 0.93

35
Modification factor Kt = 1, Kc = 1 and Kf =1
(L/d)max = (L/d)basic x Kt x Kc x Kf = 20x1x1x1=20
(L/d)actual = 5070/450 = 11.2 < (L/d)max, Hence safe.
Fig4.1: Cross Section detailing of a beam

36
BEAM TYPE 2
Span of the Beam L= 5.07m
1. Depth of the Beam:
Selecting the depth in range of l/12 to l/15 Based on stiffness
d = 4640/12 =386
Adopt d = 400mm
D = 450mm
Let take B = 300mm
2. Loads per meter length of Beam:
Dead Load = .3x.45x25 = 3.375 KN/m
Load from Slab
2x
𝑤𝑙𝑥
3
= 2x (5.875 x 4.64/3) = 18.173 KN/m
Load from wall =0.23 x2.7 x19 = 11.8 KN/m
____________
Total Load = 33.348 KN/m
____________
Factored Load 1.5 x 33.348 = 50.02 KN/m
Shear Force wul/2=(50.02 x 4.64/2) = 116 KN
Bending Moment
𝑤𝑙2
8
=(50.02×4.642/8) = 134.61 KN-m
3. Min. Depth Required:
Mu = Mu lim.
= 0.138 x fck x b x d2
134.61 x 106 = 0.138 x 20 x 300 x d2
d = 403mm>400mm
Hence, provided depth is not adequate
Let take 420mm effective depth then

37
Mu lim. = 0.138x20x300x4002
= 146.05 x106
Mu < Mu lim
Hence it is a singly reinforced beam
4. Reinforcement:
Mu = 0.87 x fy x Ast x d[1-(fy xAst/fckbd)]
134.61 x 106 = 0.87 x 415 x Ast x 420[1-(415/20 x 300 x 420]
887.688 = Ast[1-( Ast /6072.28)]
Ast
2 – 6072.28 + (6072.28 x 887.688) =0
Ast Required= 1079.64mm2
Providing 16mm dia bars
[1079.64/(π/4 x 162) = 5.37 = 6 no.of bars
6 x π/4 x 162 = 1206.371 mm2
Hence, Ast Provided = 1206.371mm2
5. Check for shear:
= 100 x 1206.371/300x450
= 0.89%
As per IS456:2000 code book
τc = 0.6 N/mm2
and τv = Vu/(b x d)
= 116020/(300 x 420)
= 0.92 N/mm2
Since τv > τc , shear reinforcement has to be designed
Vus = Vu - 𝜏 𝑐bd
= 116020 – 0.6 x 300 x 420
= 40420N

38
6. Shear Resistance of Bent up Bars:
Shear Resistance of Bent up bars
Vusb =0.87fyAsbsinα
=0.87 x 415 x 2 x(π/4 x 162) x sin450
=102662N
7. Design of vertical Stirrups:
Shear resistance to be provided by the vertical stirrups= Vus/2= 40420/2= 20210N
Spacing of 2 legged 6mm stirrups
Asv =2 x (π/4 x62) =56.5mm2
Spacing from minimum shear reinforcement consideration as per IS 456
(Asv/b x Sv) = 0.4/0.87 x fy
Sv = (0.87 x fy x Asv /0.4 x b)
Sv =(0.87 x 415 x 56.5/0.4 x300)
= 170mm
Maximum allowed spacing
=0.75d = 0.75 x450=337.5
Or 300mm whichever is less.
Hence Provide 2-legged 6mm stirrups @ 170mm c/c throughout the span of the
beam
= 100 x 1206.371/300x420
= 0.95
Modification factor Kt = 1, Kc = 1 and Kf =1
(L/d)max = (L/d)basic x Kt x Kc x Kf = 20x1x1x1=20
(L/d)actual = 4640/420 = 11.04 < (L/d)max, Hence safe.

39
CHAPTER-5
DESIGN OF COLUMNS
COLUMNS:
The load on columns is calculated by trapezium method. This method is used
when the loads on the beams coming from the slabs and walls are known prior ot
column design. The load on column at each floor level is given by,
Pu = Half of the loads coming from the beams which are rest on that column
on each floor + Pself(factored)
Pself = Self weight of the column at the floor level under consideration
DESIGN OF COLUMNS:
Assume: fck = 20 N/mm2 fy = 415 N/mm2
Load from Beam B3 =
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑
2
=50.28 x 5.07/2 = 127.46KN
Load from Beam B4 =
2
=50.28 x 5.07/2 = 127.46KN
Load from Beam B24 =
2
= 50.02 x 4.64/ 2 = 116.04KN
Load from Beam B25=
2
=41.55 x 3.20/2 = 66.48 KN
___________
Total Factored load = 437.44KN
Load transferred to the column from 5 floors is estimated to be = 2624.64 KN
Height of column = 3m
Assume 1% of steel,
Asc = 1% of Ag = 0.01 Ag, Ag = Ac + Asc
Ac = Ag - Asc = Ag – 0.01 Ag = 0.99 Ag
For axially loaded short column
pu = 0.4 fck Ac + 0.67 fy Asc
2624.64 x 103 = 0.4 x 20 x 0.99 Ag + 0.67 x 415 x 0.01 Ag
2624.64 x 103 = 7.92 Ag + 2.78 Ag

40
Ag = 245293.45mm2
Since the column is rectangular and take one side of the column is 450mm, dimension
of other side is = 245293.45/450
Adopt = 450 x 550= 247500 mm2 rectangle column
Asc = 0.01 x Ag, = 0.01 x 247500 = 2475 mm2
Ac = Ag – Asc = 247500 – 2475 = 245025 mm2
Provide 20mm dia, no. of bars
= 2475/(π/4 x 202) = 7.87=8 no. of bars
Asc provided= (8 x π/4 x 202) = 2513.27mm2
Lateral ties diameter of lateral ties should not be less than
Φ/4 = ¼ x 20 = 5mm
Hence adopt 6mm diameters bars
Pitch of the ties shall be minimum of
 Least lateral dimensions of column = 450 mm
 16 times the dia of longitudinal bar = 16 x 20 = 320 mm
 300 mm
Hence provide 6 mm lateral ties at 300 mm c/c.

41
CHAPTER-5
DESIGN OF FOOTINGS
FOOTINGS
Footing or Foundation is defined as the part of substructure, which transmits the load
from superstructure to surrounding soil stratum safely.
Foundations are classified in to two types.
1. Shallow foundation 2. Deep foundation
DESIGN OF FOOTING:
Axial Load = 2624.64KN
Size of the columns = 450×550mm
SBC of soil = 350 KN/m2
fck = 20 N/mm2
fy = 415 N/mm2
1. Size of the Footing:
Load from the column P = 1750KN
Self weight of footing = 10% of column load load = 1750/100= 175KN
Total load on soil = 1925KN
Area of the footing = (Toal load/SBC of soil) = 1925/350 = 5.5m2
Provide 2.2m×2.5m footing
2. Upward Soil Pressure:
Factored Load Pu = 1.5×1750 = 2624.26 KN
Soil Pressure at ultimate load
qu = (Pu/area of footing)
qu = (2426.64/2.2×2.5) = 441 KN/m2 = 0.441 N/mm2

42
3. Depth of Footing from Bending Moment Consideration:
Bending moment along longet direction
MuL = qu
𝐵(𝐿−𝑎)2
8
=0.441 × 2200
(2500 −550 )2
8
= 461.14×106 N-mm
Bending moment along shorter direction
MuB = qu
𝐿(𝐵−𝑏)2
8
= 0.441 × 2500
(2200 −450 )2
8
= 422.05×106 N-mm
For sloped footing, equivalent breadth,
be = b+(B-b)/8
= 450 + (2200-450)/8 = 668.75mm
MuL = 0.138fckbd2
461.14×106 = 0.138×20×668.75×d2
d = 499.83mm
Provide 750mm effective depth and 800mm overall depth. Increased depth taken due
to shear consideration.
4. Reinforcement:
Mu = 0.87 x fy x Ast x d[1-(fy xAst/fckbed)]
461.14 x 106 = 0.87 x 415 x Ast x 750[1-(415/20 x 668.75 x 750]
1702.9 = Ast[1-( Ast /24171.7)]
Ast
2 – 24171.7 + (24171.7 x 1702.9) =0
Using 16mm dia bars, spacing of bars
S = ast × (B/Ast) = (π/4×162)×(2200/1843.5)
= 239.9mm
Hence, provide 16mm bars at 220mm c/c in longer direction

43
5. Reinforcement along shorter direction:
MuB = 0.87 x fy x Ast x d[1-(fy xAst/fckbd)]
d = 550-16 = 534mm
422.05 x 106 = 0.87 x 415 x Ast x 534[1-(415/20 x 2500 x 534]
2189.04 = Ast[1-( Ast /64337.34)]
Ast
2 – 64337.34 + (64337.34 x 2189.05) =0
The above reinforcement is to be provided in a width of 2500mm. Area of steel to be
provided in the column width of 2200mm is,
Ast1 = (2×Ast/β+1) =
2
(
2500
2200
+1)
×2269.06 = 2124.22mm2
S = ast × (B/Ast) = (π/4×162)×(2200/2124.22) = 208.23mm
Hence, provide 16mm bars at 200mm c/c in the central band of width 2000mm along
shorter direction. The remaining reinforcement (2269.06-2124.22= 144.84mm2) has
to be provided in the outer band.
S = ast × (B/Ast) = (π/4×122)×(500/144) = 392.7mm
Hence, provide 12mm bars at 250mm c/c in the outer band of width 500 mm along
shorter direction.
6. Check for One Way Shear:
Factored shear force
Vu = Soil pressure from the shaded area
= 0.441×2200[
(2500 −550)
2
− 750]
= 218295N

44
At the critical section:
Width of footing at top = a+2d = 550+(2×750) = 2050mm
Average width A1 = (2050+2500)/2 = 2275mm
Depth of footing d1 = 275+(750-275)×(1000-750)/100
= 393.75
τv = Vu/A1d1 = 218295/2275×393.75= 0.215 N/mm2
it is less than minimum strength of M20 concrete = 0.28N/mm2 (Table -19 of IS:
456)
7. Check for Two Way Shear:
The critical section is at a distance of
𝑑
2
from the face of the column.
Perimeter of the critical section = 4(b+d) = 4(450+750) = 4800 mm
Depth of footing at critical section d2 = 225+(750-225)×(1000-375)/1000
= 553.125mm
Two way shear Vu2 = qu x area of shaded portion
= 0.441((2200 x 2500)-(1300 x 1200))
= 1737.54 x 103N
Two way shear stress =
𝑉𝑢2
𝐴
=
1737.54×1000
4800×553 .125
= 0.654 N/mm2
Permissible punching stress τp = 0.25 √ 𝑓𝑐𝑘
= 0.25 × √20 = 1.12N/mm2 > 0.654N/mm2
Hence, it is safe with respect to two way shear.

45
Fig5.1: Section of the Footing
Fig5.2: Plan of the Footing

46
CHAPTER-6
DESIGN OF STAIRCASE
STAIR CASE:
The purpose of the staircase is to provide pedestrian access to different
levels within the building. The geometrical forms of staircase may be quite different
depending on the individual circumstances involved. The two main components of
staircase are stairs and landing slab. The stairs and landing slab can be arranged in
different forms together in different types of staircases.
1) Type of construction of the structure around the staircase i.e., load bearing brick
structure or reinforced concrete frame structure.
2) Availability of space.
Rise and tread are two terms associated with a stair. The term rise refers to
vertical height of a step and the term tread represents the horizontal dimension where
out foot placed.
GENERAL RULES:
 Between consecutive floors there should be an equal rise for every parallel
step. Similarly there should be equal tread.
 The sum of tread of single step, twice the rice should be in between 550mm an
700mm.
 The rice of the step should not be more than 200mm and tread should not be
less than 240mm
DESIGN PROCESS:
1. Proportioning of stair case:
Dimensions of stair hall = 3.1 × 3.2m
Height of the each floor = 3m
Height of each flight = 3/2 = 1.5m
Rise R = 150mm
Tread T = 240mm
Number of Rises = 1500/150 = 10 nos
Hence, No. of treads = 10-1 = 9
Adopt width of stair case = 1.5m

47
For 9 treads, Length required = 9 × 0.24 = 2.16m
Provide width of landing = 3200-2160 = 1040mm
Assume effective depth
d = span/25 = 3200/25 = 128mm
Hence provide d= 130mm and D = 160mm
3. Loads: Loads per meter horizontal width of stairs are as follows:
Weight of waist slab = D√1 + (𝑅/𝑇)2 × 25
= 0.16√1 + (0.15/0.24)2 × 25
= 4.716 KN/m2
Weight of steps =
1/2×𝑅𝑇
𝑇
×25 = R×25/2
= 0.15×(25/2) =1.875 KN/m2
Live Load = 3 KN/m2
Factored Load = 10.191 ×1.5 = 15.286 KN/m2
4. Factored Bending Moment:
Mu = wul2/8
= 15.286 × 3.22/8
= 19.56 KN-m
5. Minimum depth required:
Mu = 0.138 fck bd2
19.56 × 106 = 0.138×20×1000×d2
d = 84.18< 130mm
Hence, provided depth is adequate.
6. Tension Reinforcement:
Mu = 0.87 fy Ast d[1-(415 Ast/fck bd)]
19.56×106 = 0.87 x 415 x Ast x 130[1-(415/20 x 1000 x 130]
416.73 = Ast[1-( Ast /6265.06)]
Ast
2 – 6265.06 + (6265.06 x 416.73) =0

48
Using 12mm diameter bars, spacing of bars
S = (ast/Ast ×1000)
= (π/4×122/448.89)×1000
= 251.94mm
Hence, provide 12mm dia bars at 250mm c/c spacing
7. Distribution Reinforcement:
= 0.12 ×1000×160/100 = 192mm2
Using 8mm dia bars, spacing
S = (π/4×82×1000)/192
= 261.8mm
Hence, provide 8mm dia bars at 250mm c/c spacing.

49
CHAPTER 7
COST ESTIMATION
7.1 RATE ANALYSIS:
Cost estimation or Rate Analysis is a well-formulated prediction of the probable
construction cost of a specific building project. A cost estimate can be an important
management tool to library planners during the design phases of a project providing
information about the facility and the project budget. All projects begin with an idea
and end by filling a need. Most projects at conceptual design require changes to
present an acceptable workable solution. The conceptual cost estimate is becoming
more important to owners, architects, and builders. It is a tool for determining
required funding and to gauge the needs of a project. This tool continues to be refined
during the design stages of the project.
The cost estimate accounts for all items that will generally be included in the
general contractor’s bid. The cost estimate is prepared by breaking down the items of
work using a standard format and determining the cost of each item from experience
and a database of current construction cost information.
A cost estimate should not be confused with a project budget. A project budget
will include the total of the cost estimate, and will also include what are known as
“soft costs”. These soft costs will specifically be excluded from the cost estimate and
will typically include land acquisition, architectural and design fees, movable
furniture and equipment, building permits and fees, fire and all risk insurance. The
project budget will also include non-construction related costs such as fund raising
and moving costs.
7.2 TYPES OF COST ESTIMATE
Cost estimates fall into two groups: conceptual estimates and detailed
estimates. Each can be broadly defined as follows:
Conceptual Estimate
Conceptual estimating or parametric estimating is the process of establishing a
project’s cost, often before any graphical representation of a facility has been
developed.
Detailed Estimate
The detailed construction estimate is the product of a process whereby the cost of a
proposed Construction project is predicted. The estimate is prepared by breaking

50
down the items of work in an orderly and logical basis, determining the cost of each
item from experience, and summarizing the total.
The Estimation should be accompanied with:
 Report
 Specification
 Drawings (plans, elevation, sections)
 Design charts and calculations
 Standard schedule of rates
Factors to be considered while preparing Detailed Estimate:
i. Quantity and transportation of materials: for bigger project, the requirement of
materials is more. Such bulk volume of materials will be purchased and
transported definitely at cheaper rate.
ii. Location of site: The site of work is selected, such that it should reduce
damage or in transit during loading, unloading, stocking of materials.
iii. Local laboure charges: The skill, suitability and wages of local labours are
considered while preparing the detailed estimate
Fixing of Rate per Unit of an Item:
The rate per unit of an item includes the following:
i. Quantity of materials & cost: The requirement of materials is taken strictly in
accordance with standard data book (S.D.B). The cost of these includes first
cost, freight, insurance and transportation charges.
ii. Cost of labour: The exact number of labourers required for unit of work and
the multiplied by the wages/ day to get of labour for unit item work.
iii. Cost of equipment (T&P): Some works need special type of equipment, tools
and plant. In such case, an amount of 1 to 2% of estimated cost is provided.
iv. Overhead charges: To meet expenses of office rent, depreciation of equipment
salaries of staff postage, lighting an amount of 4% of estimate cost is
allocated.

Multistorey building

More Related Content

What's hot

Similar to Multistorey building

Multistorey building