1PLANNING, ANALYSING(2).docx

8/14/2019 1PLANNING, ANALYSING(2).docx

1/561

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


2/562

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


NOVEMBER 2012


3/563


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


4/564

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.


5/565

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


6/566

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


7/567

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).


8/568

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


9/569

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


10/5610

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.


11/5611

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


12/5612

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


13/5613

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.


14/5614

CHAPTER2

METHODOLOGY

NBCPlanning

Drawing

Analysis

Designing

Collection of data

NBC

AUTOCAD

BENDING MOMENT

SHEAR FORCE

SLABS

BEAMS

COLUMNS

FOOTING

STAIRCASE


15/5615

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.


16/5616

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.


17/5617

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


18/5618

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


19/5619


@Joint B

BCBA

BA

AB

KK

KDF

= 5.03125.03125.0

3125.0

BABC

BC

BC

KK

KDF

= 5.03125.03125.0

3125.0


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


20/5620

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


@Joint B

BCBA

BA

AB

KK

KDF

= 5.03125.03125.0

3125.0

BABC

BC

BC

KK

KDF

= 5.03125.03125.0

3125.0


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


21/5621

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


@Joint B

BCBA

BA

AB

KK

KDF

= 5.03125.03125.0

3125.0

BABC

BC

BC

KK

KDF

= 5.03125.03125.0

3125.0


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


22/5622

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.


23/5623

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


24/5624

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


25/5625

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


26/5626

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


27/5627

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


28/5628

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


29/5629

= 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)


30/5630

max/dL = =

=34.5

actualdL / =130

3000

= 23 < 34

maxdL >

actualdL

Hence design is safe.


31/5631

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


32/5632

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.


33/5633

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


34/5634

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.


35/5635

4.3.2 COLUMN TYPE2

Data

Column = 450mm 450mm

Axial load = 1660KN

Unsupported length = 3.0mm

Concrete mix = M20


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


36/5636

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.


37/5637

4.3.3 COLUMN TYPE3

Data

Size of column = 450mm 450mm

Concrete mix = M20


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


38/5638

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.


39/5639

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


40/5640

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.


41/5641

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.


42/5642

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


43/5643

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)


44/5644

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.


45/5645


46/5646

5.3 BEAM- REINFORCEMENT DETAILS


47/5647

5.4 COLUMNREINFORCEMENT DETAILS


48/5648


49/5649

150


50/5650


51/56


52/5652


53/5653


54/5654


55/5655

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.


56/56

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.

1PLANNING, ANALYSING(2).docx

Documents

Transcript of 1PLANNING, ANALYSING(2).docx