Planning Analysis Designing and Estimation of Residential Building
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PLANNING, ANALYSING, DESIGNING AND ESTIMATION
OF RESIDENTIAL BUILDING
MINI PROJECT REPORT SUBMITTED IN THE PARTIAL FULLFILLMENT OF THE
REQUIREMENTS FOR THE AWARD OF DEGREE IN
BACHELOR OF TECHNOLOGY
IN
CIVIL ENGINEERING (HABITAT DEVELOPMENT)
BY
MOHAMED PEER THAVOOD.R (13209016)
IV YEAR
RURAL TECHNOLOGY CENTRE
THE GANDHIGRAM RURAL INSTITUTE-DEEMED UNIVERSITY
(FULLY FUNDED BY MHRD)
GANDHIGRAM-624 302
DINDIGUL DISTRICT, TAMIL NADU
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CERTIFICATE
This is to certify that the mini project work titled “PLANNING, DESIGNING & ESTIMATION
OF TWO STOREY RESIDENTIAL BUILDING” is a bonafide work done by
R.MOHAMED PEER THAVOOD under the guidance and supervision of
Dr.K.MAHENDRAN, Director i/c in Rural Technology Centre of Gandhigram Rural Institute
(Deemed University), Gandhigram
Dr.K.Mahendran Guide
Director i/c, RTC Dr.K.Mahendran
Director i/c, RTC
Place:
Date:
Submission for viva-voce held on:
Internal examiner External examiner
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DECLARATION CERTIFICATE
I submit this mini project work titled “PLANNING,ANALYSIS DESIGN AND ESTIMATION
OF TWO STOREY RESIDENTIAL BUILDING” to Rural Technology Centre, Gandhigram
Rural Institute(DU), Gandhigram, Dindigul, in partial fulfilment of requirements for the award of
degree of Bachelor of Technology in Civil Engineering(Habitat Development) and this is my
original and independent work carried out.
R. MOHAMED PEER THAVOOD (13209016)
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Content
S.No TITLE Page
1 Introduction 8
2 Methodology 9
3 Specification 10
4 Chapter I (Planning) 18
5 Chapter II (Analysis) 24
6 Chapter III (Design) 29
7 Chapter IV (Estimation) 49
8 Chapter V (Management) 54
9 Chapter VI(Electrification and plumbing) 60
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NOTATIONS
a - Area
b - Breadth of beam
Ast - Area of reinforcement
Mu - Ultimate Moment
Vu - Ultimate shear force
D - Overall depth of beam
DL - Dead load
D - Effective width of beam
Et - Modulus of elasticity of concrete
Es - Modulus of elasticity of steel
e - Eccentricity
fd - Design strength
fck - Compressive Strength of Concrete
fy - Characteristic strength of steel
Hwe - Effective height of wall
Ief - Effective moment of wall
Igs - Moment of inertia of cracked section
K - Stiffness of member
K - Constant
Ld - Development length
LL - Live load
Lw - Horizontal distance between centers of lateral restraint
L - Length of column
Lef - Effective span of beam
Lct - Effective length about y axis
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La - Clear span
lx - Length of shorter side of slab
ly - Length of longer side of slab
lo - Distance between points of zero moment in abeam
li - Span in the direction in which moment are determined c/c of supports
l2 - Span transverse to l1
m - Bending moment
M - Modular ratio
P - Axial load on compression member
qe - Calculated maximum bearing pressure
qn - Calculated maximum bearing pressure of soil
r - Radius
s - Spacing of stirrups
T - Wall thickness
V - Shear force
W - Total load
WL - Wind load
W - Distributed load per unit area
Wd - Distributed dead load per unit area
X - Depth of neutral axis
Z - Lever arm
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Introduction
Shelter is one of the basic need for Habitation. In this project, I have completed the
Planning, Analysis, design and Estimation of a Residential Building of (2BHK) two flats in
Chennai, Kuratur (Near Avadi). The Project is completed with reference to the Indian Standard
codes. In planning, I have used Autocadd, Revit, sweet Home 3D software for Plan, Elevation,
Interior and Exterior design is with reference to National Building Code 2005 completed. Analysis
of the structure is done in manual as well as using STAAD Pro Vi8 software. Designing of
structural components are carried out using Indian Standard Code (Given in reference) in Limit
state Method. Estimation is completed by using rates from Schedule of Rates (2015-2016) by
Public Work Department. Management of construction is very much Important hence we need
shelter with good Quality in appreciative cost and time. Here I have used Critical Path Method as
well as PERT Analysis with Probability for Time and Cost Management (Time-cost Trade off).
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Methodology
The methodology followed for completing the mini Project was given below
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SPECIFICATION
1 Slab:
A flat piece of concrete, typically used as a walking surface, but may also serve as a load
bearing device as in slab homes. It also act as a beam but thick is less and width is more when
compare to beam.
Design Consideration:
Among the design function that need to be taken in consideration for construction of ground
floor slab is
The provision of a uniform, level surface
Sufficient strength and stability
Exclusion of dampness from inside of building
Thermal insulation (max.0.45 W/square meter)
Resistance to fire
R.C.C slabs are divided into two:
1. One way slab
2. Two way slab.
One way slab:
When the slab is supported on all the four edges , and when the ratio of the long span to
short span is large (>2) bending takes place along one the span , such as slab is known as a one
way slab.
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Two way slab:
When the slab is supported on all the four edges , and when the ratio of the long span to
short span is small bending takes place along both the span , such as slab is known as a two way
slab.
2 BEAM:
Beam is the horizontal member of a structure, carrying transverse loads. Beam is
rectangular in cross-section. Beam carry the Floor slab or the roof slab. Beam transfer all the loads
including its self-weight to the columns or walls .R.C.C. Beam is subjected to bending moments
and shear. Due to the vertical external load , bending compresses the top fibers of the beam and
elongates the bottom fibers. The strength of R.C.C. beam depends on the composite action of
concrete and steels.
4.2.1.2 TYPES OF BEAMS :
Simply Supported Beam.
Fixed Beam.
Cantilever Beam.
Continuous Beam.
Overhanging Beam.
Simply supported beam :
Simply supported beam supported freely at the two ends on walls or columns. In actual
practice, no beam rests freely on the supports (walls or columns).
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Fixed beam :
Fixed beam both ends of the beam are rigidly fixed into the supports. Also, Main reinforcement
bars and stirrups are provided.
Cantilever beam :
Cantilever beamis fixed in a wall or column at one end and the other end is free, it is called
cantilever beam. It has tension zone in the top side and compression zone in the bottom side.
Continuous beam :
Continuous beam is supported on more than two supports . This beam is more economical for any
span lengths.
R.C.C. Beams :
Singly Reinforced Beam
Doubly Reinforced Beam
4.2.1.3 Types of Loading on Beams :
Concentrated Load.
Uniformly Distributed Loads.
Uniformly Varying Loads.
Arbitrary Loading.
3 LINTELS:
Lintel is a horizontal structural member which is placed across any opening (window, door,
almirah, wardrobes etc.) to support load of masonry coming over it.
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4.3.2.2 FUNCTIONS OF LINTEL:
Lintel is a beam supporting the load of masonry above it which is transmitted to the adjacent wall
portions over which it is laid. The bearing of lintel
The bearing of lintel beam should be the minimum of the following:
(1) 100 mm
(2) Height of lintel
(3) 1/10 th to 1/12 th of the span of the lintel
(4)
CLASSIFICATION OF LINTELS:
Lintels are classified into the following types according to the materials used in their construction:
(1)Timber lintels
(2)Stone lintels
(3)Brick lintels
(4)Reinforced cement concrete (R.C.C) lintels
(5)Steel lintels
(1)Timber lintels:
These are the oldest types of lintels. At present, they are not used expect in hilly regions where
timber is cheaply available.
(2)Stone lintels:
These are commonly used in stone masonry. The least thickness of the stone lintel is a
about 75 mm and as a thumb rule, thickness is taken as at least 1 mm per 10mm length of the span
of opening.
(3)Brick lintels:
These lintels are not structurally strong, and they are used only when the opening is small
(less than 1 m) and loads are light.
The depth of brick lintel varies from 100 mm to 200 mm, depending upon the span.
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(4) Reinforced cement concrete (R.C.C) lintels:
These lintels are commonly used these days. They have high rigidity, fire resistance and are
simple in construction.
R.C.C lintels are available as precast units. Precast R.C.C lintels are preferred for small
spans upto 2metres. Generally 1:2:4 concrete is used in the construction of these lintels. The depth
of lintel and the amount of the reinforcement is governed by the intensity of load, the type of
support and the span.
For the cast-in-situ lintels which are quite common, form work is required for construction.
(5)Steel lintels:
These lintels are provided when the opening is large and where the super-imposed loads are
heavy. Rolled steel joints are used either singly or in combination of two or three units.S
COLUMN:
Column is a vertical structural member. It transmits the load from ceiling/roof slab and beam, including
its self-weight to the foundation. Columns may be subjected to a pure compressive load. R.C.C. columns
are the most widely used now-a-days.
4.4.2 Types of Column
Long Column Or Slender
Short Column
Intermediate Column
Columns of square, rectangular and circular sections
4.4.3 Definition of Columns:
A column is defined as a compression member the effective length of which exceeds three
times its lateral dimensions .Compression members whose lengths do not exceed three times their least
lateral dimensions are classified as pedestals.
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R.C Columns concrete has a high compressive strength and a low tensile strength .Hence,
theoretically concrete should need no reinforcement when it is subjected to compression.
Reinforcements are provided in order to reduce the size of columns. Though a column is mainly a
compression member, it liable to some moment due to eccentricity of loads or transfer loads or due to
its slenderness. Such moments may occur in any direction and so it is necessary to provide reinforcement
near all faces of the column. These reinforcements form the longitudinal steel. In order to maintain the
position of the longitudinal reinforcement and also to prevent their buckling which may cause splitting
of concrete, it is necessary to provide transverse reinforcement in the form of lateral ties or spirals at
close pitch. The transverse reinforcement also assists in confining the concrete.
4.4.4 Classification of Columns:
A column may be classified on the basis of its shape, its slenderness ratio, the manner
loading and the type of lateral reinforcement provided. A column may have a section which may be
square, rectangle, circular or a desire polygon .Depending on the slenderness ratio, column may be a
short or long column.
The slenderness ratio of column is the ratio of the effective length of the column to its least
lateral dimension. A column whose slenderness ratio exceeds 12 is a long column. A column whose
slenderness ratio does not exceeds the above limit is a short column.
Based on the manner of loading, column may be classified into
Axially loaded columns
Column subjected to axial load and uniaxial bending.
Columns subjected to axial load and biaxial bending.
1 STAIR CASE:
Stair cases are used in almost all buildings. A staircase consist of a number of steps
arranged in a series, with landing at appropriate locations, for the purpose of giving access to
different floors of a building. The width of a staircase may depend on the purpose for which it is
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provided, and may generally vary between 1m for residential buildings to 2m for public buildings.
A flight is the length of the staircase situated between two landings. The number of steps in a flight
may vary between 3 to 12. The rise of a step and the tread should be proportioned that it gives
comfortable access. Generally the sum of tread plus twice the rise is kept about 500mm and the
product of the tread and the rise is kept about 40000 to 42000. In residential buildings, the rise
may vary between 150mm to 180mm and tread between 200 to 250mm. in public buildings, rise
is kept between 120 to 150mm and tread between 200 to 300mm.
4.5.2 Classification of stairs:
(1) Straight stair
(2) Quarter turn stair
(3) Half turn stair (open newel type or open well stair)
(4) Dog-legged stair
(5) Open newel stair with quarter space landing
(6) Geometrical stairs such as circular stair, spiral stair etc.
Footings:
All structures supported on earth of superstructure and substructure. The foundation can be defined
as the substructure which interfaces the superstructure and the supporting ground. Its purpose is to
transfer all loads from the superstructure to ground safely and to provide stable base to the
superstructure.
4.6.2 Types of foundations:
Foundations are classified as shallow and deep foundations.
1. Shallow foundations:
It has smaller depth limited to the width of footing. It spreads the load from superstructure
on a larger area of soil so that stress intensity is reduced to a value which can be carried
safely by soil. They are classified as isolated and combined footing and combined footings.
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An isolated footing supports one well or one column. A wall footing is a continuous strip,
either flat or stepped which distributes the load of wall on the soil. An isolated footing
supporting a single column is commonly used where the load on columns are small and are
not closely spaced. They are squared, rectangular ,circular or of other shapes.
A combined footing supports two or more columns. They may classified as those which
supports two columns and those which support more than two columns. combined footing
supporting two columns may be used for columns on property lines. Such footings may be
rectangular, trapezoidal or T shaped in plan or may consist of isolated footing connected
with a narrow beam.
2. Deep foundations:
When the top layer of the soil is too weak to support the structure on the shallow
foundation, the depth of foundation is increased till more suitable soil is found to support
the structure. Such a foundation is termed as deep foundation because of its large depth.
Different types of deep foundation are pile and well foundation.
Bearing capacity of the soil:
Bearing capacity of soil is the maximum intensity of the load or pressure developed under
the foundation without causing failure of soil and in settlement of superstructure supported
on foundation.
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Chapter II (Analysis)
KANI’S METHOD
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JOINT MEMBER K ∑K R.F
BA 2.67 EI -0.2755
B BC 0.919EI 4.839 EI -0.0945
BD 1.25 EI -0.1290
DB 1.25 EI -0.1786
D DE 0.919 EI 3.499 EI -0.1313
DF 1.33 EI -0.1900
FD 1.33 EI -0.1858
F FG 0.919 EI 3.579 EI -0.1280
FH 1.33 EI -0.1858
H HF 1.33 EI 2.249 EI -0.2955
HI 0.919 EI -0.2043
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CHAPTER III (DESIGNING)
3. DESIGNING OF STRUCTUTAL COMPONENTS
DESIGN OF SLAB
DATA
DIMENSION: 5.9m X 4.35m
LIVE LOAD: 3 KN/m2
FLOOR FINISH: 1 KN/m2
TYPE OF SLAB
LY/LX= 5.9/4.35 = 1.35 < 2
TWO WAY SLAB
DEPTH
SPAN/DEPTH = 25
DEPTH = 174mm
ASSUME CLEAR COVER = 20mm
d = 200 mm
EFFECTIVE SPAN
Clear Span + eff Depth = 4.1+0.2 = 4.3m
Centre to Centre = 4.1 + 0.25 = 4.35m
Effective Span = 4.3 m
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LOAD CALCULATION
LIVE LOAD = 3 KN/m2
FLOOR FINISH = 1 KN/m2
DEAD LOAD = 0.2 X 25 = 5 KN/m2
TOTAL LOAD = 9 KN/m2
FACTORED LOAD = 1.5 X 9 = 13.5 KN/m2
ULTIMATE MOMENT AND SHEAR FORCE
Mx = ᵅx X Wu X Lx2
= 0.096 X 13.5 X 4.352
= 24.523 KN-m
My= ᵅy X WU X Lx2
= 0.053 X 13.5 X 4.352
= 13.553 KN-m
Vu = 0.5 X Wu X Lx = 0.5 X 13.5 X 4.35 = 29.362 KN
LIMITING MOMENT
MU lim = 0.138 X FCK X b X d2
=.> Mu lim = 110.4 KN-m
Hence section is Under-reinforced
REINFORCEMENT
MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d )
24.523 X 10^6
= 0.87 X 415 X Ast X 200 (1- 415 X Ast /20 X 1000 X 200)
Ast = 814.90 mm2
Provide 12mm Dia bars @ 150 mm c/c
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DISTRIBUTION
Ast = 0.12% BD = 0.0012 X 1000 X 200 = 240 mm2
Provide 8mm Dia bars @ 250mm c/c
CHECK SHEAR STRESS
ƮV = VU/bd = 29.36 X 10^3/1000X200 = 0.1468
Pt = (100 X Ast)/bd = 0.4523
KƮc = 1.2 X 0.456= 0.547 N/mm2
KƮC > ƮV
SHEAR STRESS IS WITH IN SAFE PERMISSIBLE LIMIT
CHECK FOR DEFLECTION
(L/d)BASIC = 20
Pt = 0.2015 => Kt = 1.2
(L/d)MAX = 20 X 1.2 = 24
(L/d)ACT = 4350/200 = 21.75 < 24
HENCE DEFLECTION IS UNDER CONTROL.
TORSION REINFORCEMENT AT CORNER
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Area of torsion Reinforcement = 0.75 X Ast = 611.175 mm2
Length over which Torsional Reinforcement provided = Short span/5 = 870mm
Provide 8mm dia bars at 160 mm C/C for 670mm at the corners.
Reinforcement Edge Strips = 0.12% Gross Area = 240 mm2
Provide 8 mm bars at 160 mm C/C
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DESIGN OF BEAM (1):
DATA
DIMENSION: 5.9m
LIVE LOAD: 15.963KN/m
DEPTH
SPAN/DEPTH = 12
DEPTH = 550mm
ASSUME CLEAR COVER = 50mm
d = 600 mm
EFFECTIVE SPAN
Clear Span + eff Depth = 5.7+0.5 = 6.2m
Centre to Centre = 5.7 + 0.2= 5.9m
Effective Span = 5.9 m
LOAD CALCULATION
LIVE LOAD = 15.963 KN/m
DEAD LOAD = 0.3 X 0.6 X 25 = 4.5 KN/m
TOTAL LOAD = 19.963 KN/m
Wu = 1.5 X 19.963 = 29.9445 KN/m
ULTIMATE MOMENT AND SHEAR FORCE
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MU = 0.125 X WU X L2
= 0.125 x 29.94X5.92
= 130.27 KN-m
VU = 0.5 X WU X L = 0.5 X 29.94X5.9 = 88.32KN
LIMITING MOMENT
MU lim = 0.138 X FCK X b X d2
= 298.08KN-m
MU < MU lim
Hence section is Under-reinforced
REINFORCEMENT
MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d )
130.27 X 10^6
= 0.87 X 415 X Ast X 600 (1- 415 X Ast /20 X600 X 300)
Ast = 650 mm2
Provide #2 of 16mm dia bars with #2 of 10mm dia hanger bars
CHECK FOR SHEAR STRESS
ƮV = VU/bd = 88.32X 10^3/600X300 = 0.4906 N/mm2
Pt = (100 X Ast)/bd = 0.446
KƮc = 1.2 X 0.3212 = 0.38544 N/mm2
KƮC < ƮV
Shear reinforcement required
Vus = Vu – (tcbd)
= 88.32 X 10^3 – (0.454 X 300 X 600) = 6.6 KN
Since shear is very less
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Provide 8mm dia stirrups @ 300mm c/c
CHECK FOR DEFLECTION
(L/d)BASIC = 20
Pt = 0.446 => Kt = 1.4
(L/d)MAX = 20 X 1.4 X 1 X 1 = 29.2
(L/d)ACT = 4350/200 = 21.75 < 29.2
HENCE DEFLECTION IS UNDER CONTROL.
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DESIGN OF BEAM (2)
DATA
DIMENSION: 4.35m
LIVE LOAD: 15.963KN/m
DEPTH
SPAN/DEPTH = 15
DEPTH = 350mm
ASSUME CLEAR COVER = 50mm
d = 400 mm
EFFECTIVE SPAN
Clear Span + eff Depth = 4.1+0.35 = 4.45m
Centre to Centre = 4.35m
Effective Span = 4.35 m
LOAD CALCULATION
LIVE LOAD = 15.963 KN/m
DEAD LOAD = 0.3 X 0.4 X 25 = 3 KN/m
TOTAL LOAD = 18.963 KN/m
Wu = 1.5 X 19.963 = 28.445 KN/m
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ULTIMATE MOMENT AND SHEAR FORCE
MU = 0.125 X WU X L2
= 0.125 x 29.94X5.92
= 67.26 KN-m
VU = 0.5 X WU X L = 0.5 X 29.94X5.9 = 61.857KN
LIMITING MOMENT
MU lim = 0.138 X FCK X b X d2
= 132.48 KN-m
MU < MU lim
Hence section is Under-reinforced
REINFORCEMENT
MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d )
67.26 X 10^6
= 0.87 X 415 X Ast X 400 (1- 415 X Ast /20 X400 X 300)
Ast = 572.73 mm2
Provide #3 of 16mm dia bars with #2 of 10mm dia hanger bars
CHECK FOR SHEAR STRESS
ƮV = VU/bd = 61.851X 10^3/400X300 = 0.515 N/mm2
Pt = (100 X Ast)/bd = 0.502
Ʈc = 0.4806 N/mm2
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ƮC < ƮV
Shear reinforcement required
Vus = Vu – (tcbd)
= 61.851 – (0.454 X 300 X 400) X 10^3 = 6.6 KN
Since shear is very less
Provide 8mm dia stirrups @ 300mm c/c
CHECK FOR DEFLECTION
(L/d)BASIC = 20
Pt = 0.502 => Kt = 1.46
(L/d)MAX = 20 X 1.46 X 1 X 1 = 29.2
(L/d)ACT = 4350/200 = 21.75 < 29.2
HENCE DEFLECTION IS UNDER CONTROL.
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DESIGN OF COLUMN
DATA
DIMENSION: 0.3m X 0.3m
FACTORED LOAD: 926.508 KN
Fck = 20 N/mm2
Fy = 415 N/mm2
SLENDERNESS RATIO:
L/D= 3.2/0.3 = 10.66
Column is designed as slender column
MINIMUM ECCENTRICITY
Emin = (L/500 + D/30) = 6.4 + 10 = 16.4mm < 20mm
Also 0.05d = 0.05 X 300 = 15mm < 20mm
ULTIMATE MOMENT AND SHEAR FORCE
MU = 0.125 X WU X L2
= 0.125 x 13.5 x 4.1 2 = 28.36 KN-m
VU = 0.5 X WU X L = 0.5 X 13.5 X 4.1 = 27.67 KN
LIMITING MOMENT
MU lim = 0.138 X FCK X b X d2
= 110.4 KN-m
MU < MU lim
Hence section is Under-reinforced
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REINFORCEMENT
Pu = [0.4 X Fck X Ag + (0.67 Fy – 0.4 Fck)Asc]
911.508 X 10^3
= [0.4 X 20 X 300X 300 + (0.67 X 415- 0.4 X 20) Asc
Asc = 720 mm2
Provide #4 of 16mm Dia bars with two bars faced
LATERAL TIES:
1) DIA = φ </ {0.25 x DIA min = 4mm } => Provide 8mm Dia bars
2) Spacing
Least Dimension = 300mm
16 X Dia max = 256mm
Generally = 300mm
Provide 8 mm dia laterial ties @ 250 mm c/c
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DESIGN OF FOUNDATION
DATA
LIVE LOAD = 911.508 KN
DEAD LOAD = 25 X 0.3 X 0.3 X 3.2 = 7.2 KN
TOTAL LOAD = 918.708 KN
PERMISSIBLE STRESS = 500 KN/m2
ALLOWABLE BEARING CAPACITY = 200 KN/m2
BASE AREA
A = LOAD/ qa = 918.708/200 = 4.6 m2
Design a Square footing (since column is a square)
Length = 2.2m
Af = 2.2 X 2.2 = 4.84 m2
BENDING MOMENT
BM = [qa X B X (B-d)2
]/8 = 188.43 KN-m
MU = 1.5 X 188.43 = 282.645 KN-m
D = [MU/ 0.138 X Fck X b]^1/2
D= 0.21m = 210 mm = 300 mm (say)
REINFORCEMENT
MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d )
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282.645 X 10^6
= 0.87 X 415 X Ast X 300 (1- 415 X Ast /20 X2200 X 300)
Ast = 2868.05 mm2
Provide #15 of 16mm dia bars @ 190 mm c/c in both direction.
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DESIGN OF LINTEL BEAM
DEPTH
SPAN/DEPTH = 20
DEPTH = 70mm
ASSUME DEPTH D = 150mm
BREATH OF LINTEL = THICKNESS OF WALL = 250mm
DIMENSION = 0.25m X 0.15m
EFFECTIVE SPAN
Clear Span + eff Depth = 1.2 + 0.2 = 1.4m
Centre to Centre = 1.2 + 0.15 = 1.35m
Effective Span = 1.35 m
LOAD CALCULATION
WEIGHT OF WALL OVER LNTEL = 0.9 X 0.25 X 19.5= 4.3875 KN/m
DEAD LOAD = 0.15 X 0.25 X 25 = 0.9375 KN/m
TOTAL LOAD = 5.325KN/m
FACTORED LOAD (Wu) = 1.5 X 5.325 = 7.9875 KN/m
ULTIMATE MOMENT AND SHEAR FORCE
MU = 0.125 X WU X L2
= 0.125 x 7.9875X5.92
= 67.26 KN-m
LIMITING MOMENT
MU lim = 0.138 X FCK X b X d2
= 15.525 KN-m
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MU < MU lim
Hence section is Under-reinforced
REINFORCEMENT
MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d )
1.956 X 10^6
= 0.87 X 415 X Ast X 150 (1- 415 X Ast /20 X250 X 150)
Ast = 36.86 mm2
But Ast minimum = 0.87 X b X d / Fy = 78.61 mm2
Provide #2 of 10mm dia bars with #2 of 8mm dia hanger bars
Provide stirrups of 8mm dia @ 200 mm c/c.
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DESIGN OF STAIRCASE
DATA
Staircase space = 2.2m X 3.2m
Live Load = 4 KN/m2 (As per IS 875 (part-2) 1987)
DIMENSION
Riser = 160mm; Tread = 300mm
Width of stair = 2200/2 -100 = 1000mm
Φ = (300 2 + 160 2 /250)^1/2 = 1.36
No of Riser = 3200/160 = 20
No of Riser per flight = 10
No of Tread = 10 – 1 = 9
Going = 300 X 9 = 2.7m
Width of landing = 3.2 – 2.7 = 0.5 => 0.5 X 2 = 1m
DESIGN OF FLIGHT
Type = One way slab
Span = 3.57m
Depth
(L/d) = 20 (simply supported slab) => d = 140mm
Pt = 0.4 %
ᵅ1= 1.32 for Fs = 240 N/mm2
D = 140 + 20 = 160 mm
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LOAD CALCULATION
Live Load = 4 KN/m2
Selfweight = 25 X D X φ = 25 X 0.16 X 1.187 = 4.748 KN/m2
Weight of steps = 25 X 0.16/2 = 2 KN/m2
Floor Finish = 1 KN/m2
Total Load = 11.748 KN/m2
Factored Load = 1.5 X 11.748 = 17.622 KN/m2
ULTIMATE MOMENT
Mu max = Wul2/8 = 17.622 X 3.57 2/8 = 28.07 KN-m
Limiting Moment
MU lim = 0.138 X FCK X b X d2
= 54.096 KN-m
Mu < Mu lim
Hence Under reinforced section
REINFORCEMENT
MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d )
Ast = 670 mm2
Provide 10mm dia bars @ 130mm c/c
Distribution Reinforcement
Ast min = 0.12% BD = 0.0012 X 1000 X160 = 180mm2
Provide 8mm dia bars @ 250mm c/c
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CHECK FOR DEFLECTION
Pt = 100 X 670 / 1000 X 140 = 0.47 %
Fs = 232 N/mm2
; Pt = 0.47 %; ᵅ1 = 1.31
D = 3170 /1.31 X 20 = 130 mm < 140mm
Hence safe against Deflection
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DESIGN OF SEPTIC TANK
Flow of Sewage = 20 X140 /1000 = 2.8 m3/day
Detention Period = 24 Hrs
Tank Capacity = 2.8 X 24 /24 = 2.8 m3
Assume tank cleaned once in one years
1. Space required for storage of sludge = 0.0708 m3/capita = 20 X 0.0708 = 1.4 m3 = 1.5.m3
2. Space required for sludge digestion = 0.03 m3/ Capita = 20 X 0.03 = 0.6 m3
Add 25 % Extra = 25/100 X 2.8 = 0.7 m3
Total Tank Capacity = 2.8 + 1.5 + 0.6 + 0.7 = 5.6 m3
Let Depth of Liquid = 1m
Plan Area = 5.6 /1 = 5.6 m2
L = 3 X B
3B X B = 5.6; B = 1.36 = 1.4
L = 4.2m
Area of Tank = 4.2 X 1.4 = 5.88 m2 > 5.6m2
Assume free board = 0.3m
Depth = 1 + 0.3 = 1.3m
Size of Tank = 4.2m X 1.4m X 1.3m
Liquid Capacity = 6.5 m3
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10.8 (Sq.m)
Grand
Total 971.4 (Sq.m)
White Cement
971.4 (Sq.m)
Primer
971.4 (Sq.m)
Patti
971.4 (Sq.m)
Primer
971.4 (Sq.m)
Distemper
971.4 (Sq.m)
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CHAPTER V (MANAGEMENT)
Critical Path Method
The Critical Path Method (CPM) is one of several related techniques for doing project planning.
CPM is for projects that are made up of a number of individual "activities." If some of the activities
require other activities to finish before they can start, then the project becomes a complex web of
activities.
Program Evaluation and Review Technique (PERT)
PERT is a network analysis technique which is used to determine the time it will take to complete
a complex process.
Formula
Expected Duration = (Optimistic + 4 X Likely + Pesimistic)/6
Standard Deviation= (Pesimistic – Optimistic)/6
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S.NO DESCRIPTION NOTATION DURATION(DAYS)
1 Site Clearance A 2
2 Earthwork Excavation B 6
3 Shuttering C 12
4 Laying of foundation D 10
5 Earth filling cum foundation
reinforcement
E 15
6 Bar Bending Work F 10
7 Brick Work G 18
8 Roofing concrete with Lintel level
beam laying with shuttering
H 9
9 Ground floor work I 15
10 Laying Electrification and Plumbing J 15
11 Wood Work K 20
12 Plastering L 15
13 Sanitary Work M 10
14 Finishing Work N 15
15 White Washing O 12
16 Color Washing P 15
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ACTIVTY OPTIMISTIC LIKELY PESIMISTIC EXPECTED ST.DEVIATION VAR.
A 1 2 5 2.33 0.667 0.44
B 4 6 10 6.33 1 1
C 10 12 15 12.16 0.833 0.693
D 8 10 13 10.16 0.833 0.693
E 12 15 18 15 1 1
F 6 10 14 10 1.33 1.768
G 15 18 20 17.83 0.833 0.693
H 7 9 11 9 0.66 0.435
I 12 15 18 15 1 1
J 10 15 17 14.5 1.16 1.345
K 16 20 24 20 1.33 1.768
L 12 15 18 15 1 1
M 8 10 12 10 0.66 0.435
N 12 15 17 14.8 0.833 0.693
O 10 12 15 12.16 0.833 0.693
P 12 15 18 15 1 1
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Project Duration = 180.11 Days
Standard Deviation of the project = 13.479 Days
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PROBABILITY
Project to be completed after 190 days:
Z = (190-180.11)/13.479 = 0.733
From Log table
Probability = 1 – 0.7673 = 0.2327
Project to be completed after 190 days is 23.27%
Project to be completed before 160 days:
Z = (160-180.110)/13.479 = -1.491
From Log table
Probability = 0.0681
Project to be completed before 160 days is 6.81%
Project to be completed between 160 to 190 days
Probability = 1-0.2327-0.0681 = 0.6992
Project to be between 160 days to 190 days is 69.92%
Number of days for completion of project with 90% probability
From Log Table
1.28 + [(0.09-0.08) X (0.90-0.8997) / (6.9015-0.8997) = 1.281667
Days = 180.11+1.281667 X 13.429 = 197.32
Number of days for completion with 90% probability is 197.32 = 198 Days
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CHAPTER VI (ELECTRIFICATION AND PLUMPING)
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REFERENCES
Books used:
1. B.C. Punmia, “Limit state design of Reinforced concrete”
Laxmi Publications (P) Ltd, New Delhi.
2. S.Ramamrutham (2010), “Design of Reinforced Concrete Structures”
3. Ashok kumar Jain,Arun kumar Jain “Soil Mechanics & Foundations”
Laxmi Publications (P) Ltd, New Delhi.
4. Dr.K.R.Arorra, “Soil mechanics & Foundation Engineering”
Standard publishers Distribution-New Delhi.
5. B.N.Dutta, “Estimation and Costing in Civil Engineering” USB
Publishers Distributions (p)Ltd New Delhi.
6. KK.Chitkara “Construction Project Management-Planning”
Tata McGraw-Hill publishing Limited, New Delhi.
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IS CODES USED:
1. IS 456 : 2000 “Plain and Reinforced Concrete – code of practice”
Fourth Revision – section 3 (clauses 19, 23, 26) – section 4(33)
2. IS 875 (Part 1, Part 2 & Part 3), “Code of practice for
Design loads (other than earthquake) for buildings & structures”
3. SP 7 National Building Code (NBC) of India 1983
Group I – Part II, III, IV, V and X
4. SP 7 National Building Code of India (NBC)- 2000
5. SP 16 -1980 Design Aids for Reinforced Concrete – Section 2, 3 and 5
6. SP 34 - 1987 Hand book on Concrete Reinforced & Detailing.
7. IS 2911 (Part I/Sec2) – 1979 “Code of Practice for Design And Construction of Pile
Foundations”
8. IS 1893 (Part 1) : 2002 “Criteria for Earthquake Resistant Design of structures”
9. IS 13920 : 1993 “Ductile Detailing Of Reinforced Concrete Structures Subjected to Seismic
Forces - Code Of Practice”