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TRANSCRIPT
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A MINI PROJECT REPORT
Submitted by
GEENA GEORGE
RINCY SUNNY
SONIJA T.
TISNA THOMAS
I n partial ful f il lment for the award of the degree
of
BACHELOR OF ENGINEERING
IN
CIVIL ENGINEERING
MAHENDRA ENGINEERING COLLEGE NAMAKKAL
ANNA UNIVERSITY: CHENNAI 600025
NOVEMBER 2012
PLANNING, ANALYSING AND
DESIGNING OF A BANK
BUILDING
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PLANNING, ANALYSING
AND DESIGNING OF A BANK
BUILDING
A MINI PROJECT REPORT
Submitted by
GEENA GEORGE
RINCY SUNNY
SONIJA T.
TISNA THOMAS
(090104608012)
(090104608046)
(090104608051)
(090104608055)
I n partial ful fi llment for the award of the degree
of
BACHELOR OF ENGINEERING
IN
CIVIL ENGINEERING
MAHENDRA ENGINEERING COLLEGE, NAMAKKAL
ANNA UNIVERSITY: CHENNAI 600025
NOVEMBER 2012
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ANNA UNIVERSITY: CHENNAI 600025
BONAFIDE CERTIFICATE
Certified that this project report on PLANNING, ANALYSING AND
DESIGNING OF A BANK BUILDING is the bonafide work of
GEENA GEORGE
RINCY SUNNY
SONIJA T.
TISNA THOMAS
(090104608012)
(090104608046)
(090104608051)
(090104608055)
Who carried out the project work under my supervision
SIGNATURE
Dr.K.JAGADESAN
HEAD OF THE DEPARTMENT
Department of Civil Engineering,Mahendra Engineering College
SIGNATURE
Dr.S.SAMSON
SUPERVISOR
Professor, Dept. of Civil Engg.Mahendra Engineering College
Submitted for the practical examination held on--------------at----------------
Internal Examiner External Examine
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ACKNOWLEDGEMENT
The satisfaction and euphoria of successful completion of any task could
be incomplete with mentioning the people who made it possible, whose constant
guidance and encouragement crown our efforts with success.
We take this opportunity to express our sincere gratitude to
Shri.M.G.BHARATHUMAR, the chairman and Er.Ba.MAHENDRAN,Managing
Director, Mahendra Engineering College, Mahendrapuri, Namakkal for their
encouragement and granting permission to do this project.
We are also express our gratitude to Dr.S.SAMSON RAVINDRAN,
Principal, Mahendra Engineering college, for allowing us to use all facilities that are
available in the college to complete our project.
We express our sincere thanks to Dr.K.JAGADESAN, Head of the
Department for this extended co-operation, persistent encouragement and support.
We wish express our thanks and acknowledgement to our project guide
Dr.S.SAMSON for this guidance and timely help to complete the project in time.
We also thank all teaching and non teaching staff members of Civil
Engineering Department for their valuable suggestion and co-operation in
completing this project. Also we thank our friends for their support.
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TABLE OF CONTENTS
CHAPTER NO. TITLES PAGE NO.
ABSTRACT
i
LIST OF SYMBOLS ii
1. INTRODUCTION 1-4
1.1 GENERAL 1
1.2 PRACTICAL CONSIDERATIONS 1
1.3 PLANNING CONSIDERATIONS 1
1.4 SPECIFICATIONS 2
1.4.1 FOOTING 2
1.4.2 DAMP PROOF COARSE 2
1.4.3 PLINTH 2
1.4.4 FRAMES 2
1.4.5 SUPER STRUCTURE 3
1.4.6 ROOF 3
1.4.7 FLOORING 3
1.4.8 PLASTERING 3
1.4.9 DOORS AND WINDOWS 3
1.4.10 STAIRCASE 4
1.4.11 WHITE WASHING ,COLOUR
WASHING AND PAINTING4
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2. METHODOLOGY 5
3. ANALYSIS 6-12
3.1 INTRODUCTION 6
3.2 ANALYSIS 8
4. DESIGN 13-35
4.1 INTRODUCTION 13
4.1.1 DESIGN OF SLABS 14
4.2.DESIGN OF BEAMS 22
4.3 DESIGN OF COLUMNS 24
4.4 DESIGN OF FOOTING 30
4.5 DESIGN OF STAIRCASE 33
5. REINFORCEMENT DETAILS 36
6. PLANS 39
7. CONCLUSION 44
8. REFERENCE 46
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ABSTRACT
The primary objective of this project is to gain sufficient knowledge in
planning, analysis, and design of building.Our project deals with the plan and design of a Bank building. It is a
reinforced concrete framed structure consisting of G +2 with adequate facilities.
IS 456:2000 codes is the basic code for general construction in concrete
structures, hence all the structural members are designed using limit state method in
accordance with the IS 456:2000 code and design aids.
The planning of any building in India will be recognized by National Building
Code (NBC) , hence the building is planned in accordance with the National
Building Code of India.
The building includes the following:
1. ATM Centre2. Safety locker room3. Conference hall4. Documentary room5. Manager room6. Power control room7. Assistant managers room8. Counters9. Toilet
The bank building has proper ventilation, it is provided with sufficient doors,
windows. Water supply and electrification are also provided.
The ceiling height is provided as 3.2m, for assembly buildings as mentioned
Building Code (NBC).
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LIST OF SYMBOLS
SYMBOL DESCRIPTION
D Effective depth in mm
D Overall depth in mm
ly Longer span length
lx Shorter span length
W Factored load
x Bending moment coefficient along shorter direction
y Bending moment coefficient along longer direction
dreq Depth required
Mu Bending moment
Mulim Ultimate moment
Ast Area of steel in tension
Asc Area of steel in compression
fy Grade of steel
fck Grade of concrete
Xu Depth of neutral axis
B Breadth
Ast(req) Reinforcement required
Ast(pro) Reinforcement provided
P Percentage of steel
Diameter of bar used
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Peq Equivalent load
v Nominal shear stress
c Shear capacity of concrete
D.L Dead load
L.L Live load
Lx Effective length in shorter span
Ly Effective length in longer span
Sv Spacing of stirrups
V Shear force
Vu Design Shear force
Pu Ultimate axial load
Ag Gross sectional area of column
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CHAPTER-1
INTRODUCTION
1.1 GeneralThe main objective of our project is to know the various design aspects
like planning, analysis and design etc.We have planned to design a bank
building consisting of three floors (G+2).The planning is done as per the
requirements and regulations given by the National Building Code (NBC).
1.2 Practical considerationsBesides all the fundamentals of planning discussed, following practical
points should be additionally considered:
1) The elements of the building should be strong and capable to withstandthe likely adverse effects of natural agencies.
2)
Strength, stability, convenience and comfort of the occupants should bethe first consideration in planning.
3) Elevation should be simple but attractive. The number of doors andwindows provided should be less for a bank building.
4) The provisions of built in furniture at proper places are useful from thepoint of view of utility.
5) Since the plan is for a bank building, the locker rooms must be securedwith thicker walls than usual.
1.3Planning considerationsThe plan and detailing was drawn using Auto CAD. The proposed area
of the bank is 324sq.m. The shape of the building is rectangular in plan. The
building consists of ground floor, first floor and second floor.The parking
space is provided around the building. The floor height of the building is
3.2m.The height of the parapet wall is 1m.The staircase is provided with
enough safe.
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Area of each floor is given below
Ground floor = 108 sq.m
First floor = 108 sq.m
Second floor = 108 sq.m
Total area = 324 sq.m
1.4 Specifications1.4.1 Footing
Earth work excavation for foundation is proposed to a depth of
1.50m.below the ground level. For design, the safe bearing capacity of soil is
assumed as 200KN/.Isolated footings are provided with a concrete grade ofM20. The maximum axial load 2210KN as arrived from analysis result is
taken for the design of the footing.
1.4.2 Damp proof coarseThe damp proof coarse is to be provided around the plinth level using
C.M 1:3 with a thickness of 20mm. The column below the ground level are
also provided with damp proof coarse of C.M 1:3
1.4.3 Plinth
The plinth beam will be at a level of 0.5m above the ground level. M20
grade of concrete is used and Fe415 steel was used for plinth design.
1.4.4 Frames
All the R.C.C. structural components are designed using M20 grade
steel. Each member is designed separately for its loading condition
And its location as per the IS 456:2000 and SP 16 codes. The dimension of
slab, beam, column and footing are designed according to the IS 456:2000
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code. The column is designed as per the design principles given in SP-16 and
the axial load was taken from the analysis results.
1.4.5 Super Structure
The super structure is proposed in CM.1:6 using second class brick
work. Brick partition walls of 110mm thick are also proposed using the C.M
1:4 with a width of 300mm as a safety measure.
1.4.6 Roof
R.C.C Roof in M20 concrete is to be laid. A layer of weathering coarse
using brick jelly lime mortar is to be used. Considering the future expansion
of the structure, the roof slab is also designed as same as that of the floor
slabs.
1.4.7 FlooringIn each floor, all the rooms are to be provided with P.C.C. 1:5:10 as
flooring base. The floors of entrance, toilet floors, staircase and entire flat are
to be finished with granite tiles over the P.C.C. 1:2:4 flooring.
1.4.8 Plastering
All walls and structural members including the basement will be
plastered smooth with C.M. 1:5 externally and internally, using 12mm thick
plastering mortar.
1.4.9 Doors and windows
The main door will be of steel having a sliding shutter. The other doors
inside the bank are to be provided with aluminium panel. The windows are to
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be provided with steel and glazing is provided to supply a good light from
outside.
1.4.10 Staircase
The stair will be of M20 grade concrete and Fe415 steel with a rise of
150mm and tread of 300mm. The staircase is designed as spanning parallel to
landing slab referring to IS 456-2000.
1.4.11 White washing, Colour washing, Painting
All the inner walls are to be finished with a first coat of white cement
wash and then colouring as required. All the joiners and iron works are to be
finished with two coats of synthetic enamel paint. The toilet walls are to be
provided with mat finishing.
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CHAPTER2
METHODOLOGY
NBCPlanning
Drawing
Analysis
Designing
Collection of data
NBC
AUTOCAD
BENDING MOMENT
SHEAR FORCE
SLABS
BEAMS
COLUMNS
FOOTING
STAIRCASE
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CHAPTER-3
ANALYSIS
3.1 INTRODUCTION
Structural analysis is the application of solid mechanics to predict the
response (in terms of force and displacements) of a given structure (existing or
proposed) subjected to specified loads.
Based on degree of indeterminacy the structure will be classified as
i. Determinate structureii. Indeterminate structure
The determinate structure can be completely analyzed by using equilibrium
equation. I.e.M =0;V=0; & H=0.
Example: simply supported beam, cantilever beam, overhanging beam.
In the indeterminate structure, cant be complete analyzed by equilibrium
equations.
Example: Fixed beam, continuous beam, and propped cantilever beam.
Moment Area method:
This method is used for analyzing cantilever and fixed beam.
Theorem of three moment equation:
It is more suitable for continuous beam.
Moment distribution method:
It is the iterative technique.
Slope- deflection method:
When the beam has more than four spans then the calculation is difficult.
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Stiffness method:
Force and displacements play on important role in the structural
analysis. In this method the force is measured to produce a unit displacement.
Flexibility method:
It is the inverse of stiffness. It is defined as the measure of displacement
caused by the unit load. The moment distribution method for the analysis of
beam is adopted in this project.
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3.2ANALYSIS
FIXED END MOMENTS
Type 1
Span AB
MFAB = KNmWl
12.3612
2)2.3(34.42
12
2
MFBA = KNmWl
12.3612
2)2.3(34.42
12
2
MFBC = MFCB
Distribution factor (DF)
@Joint B
BCBA
BA
AB
KK
KDF
= 5.03125.03125.0
3125.0
BABC
BC
BC
KK
KDF
= 5.03125.03125.0
3125.0
Distribution factor (DF) is same as each member
42.34KN/m
3.2m
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Table 3.1 Moment distribution method
Span
AB =
KNmWl
23.988
4.595.26
8
22
Span CE = KNmWl
92.288
37.25
8
22
Type 2
Span AB
MFAB =KNm
Wl
33.4512
2)2.3(13.53
12
2
MFBA = KNmWl
33.4512
2)2.3(13.53
12
2
MFBC = MFBC = MFCD
MFBA = MFCB = MFDC
Joint A B
C
Mem
ber
AB BA BC CB
D.F 0.5 0.5
I.M-
36.12
36.12 -
36.12
36.12
B 0 0 0 0
Net
B.M
-
36.12
36.12 -
36.12
36.12
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Distribution factor (DF)
@Joint B
BCBA
BA
AB
KK
KDF
= 5.03125.03125.0
3125.0
BABC
BC
BC
KK
KDF
= 5.03125.03125.0
3125.0
Table 3.2 Moment distribution method
Span AB = KNmWl
23.988
4.595.26
8
22
Span CE = KNmWl
92.288
37.25
8
22
Joint A B
C
Mem
ber
AB BA BC CB
D.F 0.5 0.5
I.M-
45.32
45.32 -
45.32
45.32
B 0 0 0 0
Net
B.M
-
45.32
45.32 -
45.32
45.32
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Type 3
Span AB
MFAB = KNmWl
50.4612
2)2.3(49.54
12
2
MFBA = KNmWl
50.4612
2)2.3(49.54
12
2
MFBC = MFBC = MFCD
MFBA = MFCB = MFDC
Distribution factor (DF)
@Joint B
BCBA
BA
AB
KK
KDF
= 5.03125.03125.0
3125.0
BABC
BC
BC
KK
KDF
= 5.03125.03125.0
3125.0
Table 3.3 Moment distribution method
Joint A B
C
Mem
ber
AB BA BC CB
D.F 0.5 0.5
I.M-
46.50
46.50 -
46.50
46.50
Net
B.M
-
46.50
46.50 -
46.50
46.50
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Type 4
SPAN AB
MFAB = KNmWl
50.4612
2)2.3(97.46
12
2
MFBA = KNmWl
50.4612
2)2.3(49.54
12
2
MFBC = MFBC = MFCD
MFBA = MFCB = MFDC
Distribution factor (DF)
@Joint B
BCBA
BA
AB
KK
KDF
= 5.03125.03125.0
3125.0
BABC
BC
BC
KK
KDF
= 5.03125.03125.0
3125.0
Table 3.4 Moment distribution method
Joint A B
C
Mem
ber
AB BA BC CB
D.F 0.5 0.5
I.M-
40.08
40.08 -
40.08
40.08
B 0 0 0 0
NetB.M
-40.08
40.08 -40.08
40.08
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CHAPTER-4
DESIGN
Introduction
Proper nomenclature of floors and storeys and also unified and improved
methods of designating the structural members eliminate the possible confusion and
led to less efforts and saving in time in the preparation of design calculation and
drawings.
There are two main methods to design the structural members, they are
working stress method and limit state method. Here, we adopt the limit state method
for designing all the structural members involved, in our project. The structures are
designed to its elastic limit in the working stress method, whereas in the limit state
method of design, the structural members are designed up to its plastic limits.
Both the methods are having the safety value. But, the most economical
method is the limits state method, which is adopted in every constructional design
nowadays. Hence we planned to go for the limit state method of design. For our
project work we took only for important structural members to design they are slab,
beam, column and footing. The slab is designed by assuming it as simply supported
with four edges discontinuous, for easier design calculation.
The beam is designed by knowing its span and its location (inner and outer).
The beam has to carry the self-weight of slab and live load of 4KN on its self-weight
also. The live load on each beam will be calculated separately by considering the
load transmission diagram. In some beams where the wall is constructed above it,the self-weight of wall has to be added.The column and footing design are made by
knowing the maximum axial load on each column. The column alone is designed by
following the SP16 codes.
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4.1 DESIGN OF SLAB
4.1.1 Slab 1: Two adjacent edges discontinuous
Data
Dimension of slab = 3 m 3 m fck = 20 N/m2
Support width = 230 mm fy = 415 N/m2
Live load = 4 KN/m2
Floor finish = 1 KN/m2
Depth of slab
Minimum depth = Span / B.V M.F
B.V = 26 (For continuous slab)
M.F = 1.4
Minimum depth d = 3000/(26 1.2)
= 96.15 mm 100 mm
Assume effective cover = 25 mm, Using 10 mm diameter bars
Effective depth = d = 100 mm
Overall depth = D = 100 + 25 + (10/2)
= 130 mm
d = 100 mm D = 130 mm
Effective span
The least value of:
(1)(Clear span + effective depth) = (3 + 0.1) = 3.1 m(2)(Centre to centre of supports = (3 + 0.23) = 3.23 m
Hence L = 3.1 m
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Loads
Self weight of slab = (0.12 25) = 3 KN/m2
Floor finish = 1 KN/m2
Live load = 4 KN/m2
Total service load = w=8 KN/m2
Ultimate load = wu= 1.5 8 = 12 KN/m2
Ultimate moments and shear forces
Ly / Lx = 3 / 3 = 1< 2
So it is a two way continuous slab
The coefficients for the positive and negative moments are taken from IS 456
2000
+x = 0.035
_ x = 0.047
+ y= 0.035
_ y= 0.047
Mux = x wuxx2
Muy = ywuLy2
Vux= 0.5 W lx
(1)B.M along span ( +ve) = 0.035 12 3.12Mux = 4.036 KN.m
(2)B.M along span ( -ve) = 0.047 12 3.12Muy = 5.420 KN
(3)Shear force Vux = 0.5 12 3.1= 18.6KN
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Check for depth
Mmax = 0.138fckb d2
d2 = 5.420 10
6/ (.138 20 10
3)
d = 44.31mm < 100mm
Reinforcements (Short and long span)
Mu = .87fy Astd [1(fy Ast / b dfck)]
Ast = 155.10mm2
Adopt 10mm diameter bars at 400 mm centers (Ast=185.71mm2)
Edge strip
Minimum area of steel = 0.12% Ag
= .12 /(100 1000 130)
= 156mm2
Provide 10mm diameter bars
ast = /4 d2
=78.5mm2
Spacing = ast/ Ast 1000
78.5/156 1000 = 509 mm
So adopt 10 mm bar @ 500mm c/c
Middle strip [+ve moment]
Mu = .87 fy Astd [1(fy Ast / b dfck)]
Ast = 201.44mm2
Provide 10mm diameter bars
ast = /4 d
2
ast = 78.5mm
2
Spacing = ast/ Ast 1000
S = 300mm
Provide 10 mm diameter bars @ 300mm c/c
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Torsional reinforcement
Single torsional reinforcement =l/5 ly/5
Area of torsional steel = 3/4 maximum mid span
= 3/4 201.44
= 151.08mm2
Provide 8mm bars
Spacing = 50.26/151.08 1000
= 332 mm c/c or 300 mm c/c.
Check for deflection
(Shorter span/depth)provided = 3000/130 = 23.07
(Shorter span/depth)permissible = B.V M.F
Fs = 0.58 415 (156/102)
Fs = 368.12
% Ast = 100 Ast/ bd
= (100 156) / (100 130)
= 1.2%
M.F = 1.2
(Shorter span/depth) = 26 1.2
= 31.2
Hence safe in deflection
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4.1.2 Slab 2: One short edge discontinues
Data:
Clear dimension of slab =1.2Step 1: Check the ratio
lx
ly
= 22.561.2
3
Hence it is one way slab
Step 2 Effective dept
d =
26
span
=26
3000
= 115 mmD = d + 25 mm
= 130+ 25+10/2 mm
= 160 mm
Step 3: Loads
Self wt of slab = 0.16 Finishes =1KN/Total dead load =5 KN/
Live load = 4 KN/The effective span = 3.2 mm
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Step 4: ultimate Moment and SF
Mu(-ve) =1.5( )
=1.5( ) = 12.75 KN.m
Mu(+ve) = ( )
=
(
) (
)= 11.025 KN.m
Vu = (1.5)(5+4) = 24.3 KN
Step 5 : Limiting Moment of Resistance
Mulimit = 0.138fckb.d2
= 0.138 20 1000 1302 10-6
= 46.64 KN.m
Mu< Mulimit, Section is under reinforced.
Step 7: Main reinforcement
Mu = .87 fy Astd [1(fy Ast / b dfck)]
A = 325 mm2
Spacing
a) Spacing = (ast/ Ast )1000
Use 10mm bar
Spacing = 1000325
210/4
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= 230mm
Step 6 : distribution Reinforcement
Ast =
= 192 mm2
Provide 10 mm bar at 300 mm c/c
Step 7: Check for shear stress
v =db.
uV
=1301000
31092.25
= 0.199N/mm2
Pt =db.
Ast100
= 1301000
314100
= 0.241
Permissible shear stress
Kc = 1.27 0.35
= 0.44 N/mm2
Kc > v
Hence the shear stresses are within safe permissible limits
Step 8: Check for deflection
maxdL = kfkckt
basicdL
Kt = 1.5
Kc = 1
Kf = 1 (Refer IS 4562000)
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max/dL = =
=34.5
actualdL / =130
3000
= 23 < 34
maxdL >
actualdL
Hence design is safe.
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4.2DESIGN OF BEAM
Dimension:
Shear force = 135 KN
Moment Mu = 67 KN.m
Beam size = 0.3 0.4m
Limiting moment of resistance
Mu limit = 0.138fckbd2
Mu limit = [0.138203004002] 10-6
= 132.48 KN.m
Since Mu
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Using 8 mm diameter 2 legged stirrups spacing
Sv = [0.87 fy Asv d/Vus]
= [(0.87 415 400 2 50)/(34.2 103)]
= 221 mm
Sv> 0.75 d = 0.75 450
= 337.5mm
Adopt spacing of 200mm near support, gradually increasing to 300mm towards
center of span.
Check For Deflection:
pt = [100Ast/bd]
= [(100 920)/(300 400)]
= 0.76
Neglecting bar in compression side
[L/d]ma = [L/d]basic Kf
= [2.6 1.2 1.2 0.92]
= 34.44
[L/d]actual = [3600/400]
= 9 < 34.4
Hence deflection control is satisfied.
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4.3 DESIGN OF COLUMN
4.3.1 COLUMN TYPE - 1
Data
Size of column = 450mm 450mm
Concrete mix = M20
Characteristic strength of reinforcement = 415 N/mm2
Factored load = 1660 KN
Factored moment = 60.12 KN.m
Assuming 25mm bars with 40mm cover,
d = 40 + 12.5
= 52.5mm
= 5.25cm
d/D = 5.25/45
= 0.12
By using SP16,
Chart for d/D = 0.15 will be used.
Pu /(fckbD) = (1660 103) / (20 45 45 102)
= 0.409
Mu/ (fck Bd2) = (60.12 106) / (20 45 45 45
103)
= 0.033
a) Reinforcement on two sides.
Refer Chart 33 of SP16,
P/fck = .02
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Percentage of reinforcement,
P = .01 20
= 0.4
As = PbD / 100
= 0.2 45 45 / 100
= 8.1cm2
b) Reinforcement on four sides.
Refer Chart 45 of SP16,
P/fck = 0.02
P = 0.02 20
= 0.4
As = 0.4 45 45 /100
= 8.1cm2
Provide 6 bars of 20mm diameter
Lateral Ties
Diameter of lateral ties
1 /4 of main bars = 1/4 x 20 = 5mm. Min diameter of bar = 6mm
Follow 6# bars.
Pitch of lateral ties
16n times main bars = 16 x 20 = 320 mm. Lateral dimension = 450mm 300mm
Provide 25mm diameter bars at 300mm c/c.
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4.3.2 COLUMN TYPE2
Data
Column = 450mm 450mm
Axial load = 1660KN
Unsupported length = 3.0mm
Concrete mix = M20
Characteristic strength of reinforcement = 415 N/mm2
Slenderness Ratio
(L/D) = (3000/450) = 6.7 < 12
Therefore column is short column.
Minimum Eccentricity
emin = ( )emin = ( )= 21.0 > 20mmAlso .05 D = .05 450 = 22.5 > emin
Factored Ultimate Load
Pu
= 1660KN
Longitudinal reinforcement
Pu = [.4fckAg+ (.6fy- .4fck )Asc]
3000 103= [(.4 20 450 450) +{ (.67 415 ) - (.4 20)} A sc ]
Asc = 5110.16mm2
The area of reinforcement provided is greater than the minimum
steel reinforcement of 0.08 percent = (.008 450 450) = 1620mm2.
Provide 6 bars of 25mm diameter
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Lateral Ties
Diameter of lateral ties
1 /4 of main bars = 1/4 x 25 = 6.25mm. Min diameter of bar = 6mm
Hence provide 8mm diameter ties.
Pitch of lateral ties
16n times main bars = 16 x 25 = 400 mm. Lateral dimension = 450mm 300mm
Provide 25mm diameter bars at 300mm c/c.
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4.3.3 COLUMN TYPE3
Data
Size of column = 450mm 450mm
Concrete mix = M20
Characteristic strength of reinforcement = 415 N/mm2
Factored load = 1660 KN
Factored moment MUX = 46.50 KN.m
MUY = 36.12KN.m
Moments due to minimum eccentricity are less than the values given above.
Reinforcement is distributed equally on four sides.
As a first trial assume the reinforcement percentage, P = 1.2
P/fck = 1.2/20 = 0.06
Uniaxial moment capacity of the section about XXaxis:
d/D = 5.25/45 = 0.117
Chart for d/D = 0.1 will be used.
Pu/fckbD = (1660x103) / (20 450 405)
= 0.409
Referring to chart 44
Mu /fckbD2
= 0.09
Mux1 = 0.09 20 45 452 103/106
= 164.02KN.m2
Uniaxial moment capacity of the section about YY-axis:
d/D = 5.25/45 = 0.117
Chart for d/D =0.15 will be used.
Referring to chart 45
Mu/fckbD2 = 0.083
Muy1 = 0.083 20 45 45
2 10
3/10
6
= 151.26KN.m
2
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Calculation of Puz:
Referring to chart 63 corresponding to
P = 1.2
fy = 415
fck = 20
Puz /Ag = 10.3N/mm2
Puz = 10.3 45 45 102/10
3
= 2085.7KN
Pu / Puz = 1660/2085.7 = 0.79
Mux/ Mux1 = 46.50/164.02 = 0.28
Muy/ Muy1 = 36.12/151.26 = 0.23
Referring to chart 64, the permissible value of Mux/ Mux1
Corresponding to the above Muy/ Muy1 and Pu / Puz is equal to 0.61
Hence the section is ok.
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4.4 DESIGN OF FOOTING
Data
Factored axial load Pu = 1660KN
Size of Column = 450 450 mm
Safe Bearing Capacity = 200 KN/m2
Assume self weight of footing 10% of load
Self weight of footing = 166 KN
Area of footing required = Total Load / SBC
= 1826/(2001.5)
=6.08m2
Side of square footing = 2.46m
Assume breadth of square footing = 3m
Provide a square footing of 3m3m = 9m2
Net upward design pressure W =1660/33
= 184.4
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Over all depth = 480+10+20+40
= 550mm
Tension Reinforcement
Mmax = 449.74
106
Mu = 0.87fyAstd[1-(Astfy/bdfy)]449.74106= 0.87 415 480[1-(Ast 415/20 480 3000)]
Ast = 2700mm2
Minimum Astto be provided = (0.15/100) 3000 550
= 2475 < 2700mm2
Provide 20 mm diameter bars,
Number of bars = 2700/314.15
= 8.59 10nos
Provide 10 numbers of 20mm bars (Actual Ast= 3141mm2)
Check for one way shear
The critical section for traverse shear (section yy) is at distance of 480mm (effective
depth).
Length of critical section = 1.2750.48= 0.795m or 795mm
Vu = 184.430.795
Vu = 439.79 KN
Nominal shear stress
v = Vu/bd
= 439.79/(3000550)= 0.26 N/mm2
From the table -19 of IS456-2000
100Ast/bd = 0.24 for M20.
v = 0.34
c > v , Hence safe.
Checking for punching shearThe critical section for shear is at distance of ( deff)
ie , 240mm around the face of the column.
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Side of the section = 450 + (2240)
=930mm
Punching shear across section zz
Vz = 184.4( 330.940.94 )
= 1496.99 KN
Nominal shear stress across section zz
v = (1496.66103)/(4940480)
v =0.82N/mm2
As per clause 31.6.31 of IS456:2000
Permissible shear stress in concrete = kc
For square column,
k = 1.0, c = 0.33
kc = 0.33
Check for Safe bearing capacity
Column of load = 1660 KN
Self weight of footing =330.5525
=123.75 KN
Total = 1783.75 KN
Pressure = 1783.75/(33)
=198.3< 300 (200 1.5)
Hence safe.
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4.5 DESIGN OF STAIRCASE
DATA:
Spanning parallel to landing slab (stair-1)
Type: dog-legged
Number of steps in the flight = 11 per flight
Thread T = 300 mm
Rise R = 150 mm
Width of landing beam = 300 mm
Width of landing slab = 1850 mm
M-20 grade concrete (fck= 20 N/mm2)
Fe-415 hysd bars (fy= 415 N/mm2)
Solution:
Design of first flight
Effective span = (11 x 300) + 300/2 + 1850/2
= 4375 mm = 4.375 m
Thickness of waist slab = (span/20) =4375 /20)
= 218.75 mm
overall depth (D) = 218.75 mm
Assume using 25mm cover and 10mm diameter bars
Effective depth (d) = 188.75 mm
LOADS:
Dead loads of slab on slope (ws) = (0.188x1x25)= 4.717KN/m
Dead load of slab on horizontal span (w)
=[ws(R2+ T
2)]/T
= [4.717(1502+ 3002)]/300
= 5.28 KN/m
Dead load of one step = (0.5x0.15x0.3x25)
= 0.56 KN/m
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Load of step per meter length =[(0.56x1000)/300]=1.875KN/m
Finishes etc, = 0.6 KN/m
Total dead load = ( 4.72+1.86+0.6)
= 7.18 KN/m
Service live load = 5 KN/m2
Total service load = (7.18+5) = 12.18 KN/m
Factored load (wu) = (1.5x12.18) = 18.27 KN/m
BENDING MOMENTS:
Maximum bending moment at centre of span ( (Mu) = 0.125wuL2
= 0.125x18.27x4.32
= 42.235 KNm
CHECK FOR DEPTH OF WAIST SLAB
(d) = [Mu/(0.138fckb)]
= [(42.23x106)/(0.138x20x103)]
= 123.69 mm < 188.7mm Hence safe.
MAIN REINFORCEMENTS
Mu = (0.87fyAstd)[1-(Astfy/bdfck)]
42.23x106
= (0.87x415Astx188.75)[1-(415Ast/103x188.75x20)
Ast = 668 mm2
Provide 12 mm diameter bars at 150 mm centres (Ast= 678.6 mm2) as main
reinforcement.
DISTRIBUTION REINFORCEMENT
Distribution reinforcement = 0.12 per cent of cross section= (0.0012x1000x188.75)
= 226.5 mm2
Provide 8mm dia bars at 220 mm centres (Ast= 251.3mm2)
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Design of second flight
The design of the second flight loading above the landing slab will be the
same as that of the flight below the landing slab and that design is shown above.
There is only one change the landing beam and landing slab will be reversed.
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5.3 BEAM- REINFORCEMENT DETAILS
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5.4 COLUMNREINFORCEMENT DETAILS
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150
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CHAPTER -7
CONCLUSION
In this project, PLANNING DESIGNING AND ANALYSING OF A
BANK BUILDING. We all the members of our team has learned to plan a building
with referring to National Building Code of India -2005.
This bank building project has made us to learn Drawing and drafting the
building plans using Auto cad software.
In this bank building project we learnt to create the models by giving nodes
and property to the structural elements using analysis and also we learnt to the same
structure with corresponding loads as givenIS 875 part 1&2 using analysis.
This project is very useful in making us learn the design by referring to the IS
456:2000 for each slab and beam. SP: 16 codes alone are used for easier design of
columns yet we learned to design the columns.
The important thing that we done was referring to a lot of books for designing
and we are very much satisfied with exposing to field of design.
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CHAPTER -8
REFERENCES
1. Arul Manickam A.P (2004) Structural Engineering Pratheeba publishers.2. Is 456: 2000 Plain and Reinforced concrete code of practice (Fifth
revision).
3. SP 16 Design for reinforced concrete to 456 19784. Krishna raju (2002) Design of reinforced concrete structures5. Murugan.M (2007) Structural Analysis Samuthira Publications.6. Varghese P.C (2002) Limit state Design of Reinforced concrete second
Edition.
7. Vaidyanthan. R. Perumal. P (2005) Structural Analysis Volume I LaxmiPublications.
8.NBCNational building of India , Bureau of Indian Standards ,New Delhi.