civil engineering project

(1)

1

PROJECT REPORT

on

PLANNING AND DESIGN OF NET ZERO ENERGY

RESIDENTAL BUILDING

Submitted in partial fulfillment for the award of the degree

of

BACHELOR OF TECHNOLOGY

in

CIVIL ENGINEERING

by

KARTHIK V

(1010910090)

SASIDHAR K.V (1010910092)

NEERAJ PORWAL (1010910118)

ABHINAV N (1010910119)

Under the guidance of

Mrs. VASANTHI.P

Assistant Professor (O.G)

DEPARTMENT OF CIVIL ENGINEERING

FACULTY OF ENGINEERING AND TECHNOLOGY

SRM UNIVERSITY

(2)

2

(Under section 3 of UGC Act, 1956)

SRM Nagar, Kattankulathur- 603203

Kancheepuram District

MAY 2013

PROJECT REPORT

on

PLANNING AND DESIGN OF NET ZERO ENERGY

RESIDENTAL BUILDING

Submitted in partial fulfillment for the award of the degree

of

BACHELOR OF TECHNOLOGY

in

CIVIL ENGINEERING

by

KARTHIK V

(1010910090)

SASIDHAR K.V (1010910092)

NEERAJ PORWAL (1010910118)

ABHINAV N (1010910119)

Under the guidance of

Mrs. VASANTHI.P

Assistant Professor (O.G)

(3)

3

DEPARTMENT OF CIVIL ENGINEERING

FACULTY OF ENGINEERING AND TECHNOLOGY

SRM UNIVERSITY

(Under section 3 of UGC Act, 1956)

SRM Nagar, Kattankulathur- 603203

Kancheepuram District

MAY 2013

BONAFIDE CERTIFICATE

Certified that this project report titled “PLANNING AND DESIGN OF

NET ZERO ENERGY RESIDENTAL BUILDING” is the bonafide

work

of

KARTHIK.V(1010910090),

SASIDHAR

REDDY.K.V(1010910092),

NEERAJ

PORWAL

(1010910118),

ABHINAV. N (1010910119) who carried out the research under my

supervision. Certified further, that to the best of my knowledge the work

reported herein does not form part of any other project report or dissertation

on the basis of which a degree or award was conferred on an earlier

occasion or any other candidate.

(4)

4

Signature of the Guide Signature of the HOD

Mrs. VASANTHI .P Dr. R. ANNADURAI

Assitant Professor (O.G) Professor & Head Department of Civil Engineering Department of Civil

Engineering

SRM University SRM University

Kattankulathur- 603203 Kattankulathur- 603203

INTERNAL EXAMINER

EXTERNAL

EXAMINER

DATE:

ABSTRACT

The proposed Net zero residential building is located at

Urapakkam. The NZERB has G+1 floor. The total land surface covered by

the Net zero energy residential building is 99 square meters. A complete

design shall be done for the proposed NZERB using Indian standard

codes. There are three main phases in a construction project which are

planning, designing and estimation. The first stage in a project is planning,

in which preparation of layout of plot has to be done. To conclude the

project a detailed estimate of the residential building has also been prepared.

(5)

5

ACKNOWLEDGEMENT

The author wish to acknowledge my indebtedness to alma mater for congenial cooperation and granting me permission to accomplish a work on “PLANNING AND

DESIGN OF NET ZERO ENERGY RESIDENTIAL BUILDING”

The author is grateful and records his sincere thanks to Dr. T. P.

GANESAN

Pro Vice Chancellor (P&D) and Dr. C. MUTHAMIZHCHELVAN, Director, (E&T), SRM UNIVERSITY for providing all the necessary facilities for carrying out this work.

(6)

6

The author expresses his sincere thanks and Gratitude to HOD Dr. R.

ANNADURAI, Department of Civil Engineering, for his valuable suggestions and

advice in carrying out this thesis work.

The author expresses his sincere thanks to Department Coordinator/Civil

Dr.K.GUNASEKARAN, Professor, Department of Civil Engineering, for initiative and

motivation during the course of this work.

The author hereby acknowledges with deep sense of gratitude the valuable guidance given by the Guide Mrs.VASANTHI P, Assistant Professor, Department of Civil Engineering, for initiative and motivation during the course of this work.

The

author is extremely grateful to

the

valuble advices given by the class incharge Mr.K.PRASANNA, Assistant professor,Department of Civil Engineering, for constant support.

The author is grandly indebted to all the Faculty Members of Department of

Civil Engineering, for their valuable help rendered during the course of study.

Finally, the author expresses his hearty thanks to Friends for their kind help and encouragement throughout the course of this thesis work.

TABLE OF CONTENTS

CHAPTER TITLE PAGE

(7)

7 ACKNOWLEDGEMENT v LIST OF TABLES ix LIST OF FIGURES x ABBREVATIONS xi 1 OVERVIEW 1 1.1 OBJECTIVE 1 1.2 NECESSITY 1 1.3 SCOPE 2 1.4 METHODOLOGY 2

1.5 MAJOR DESIGN EXPERIENCE 2

1.6 REALISTIC DESIGN CONSTRAINTS 3

1.7 REFERENCE TO CODES AND STANDARS 3

1.8 APPLICATION OF EARLIER COURSE WORKS 4

1.9 MULTIDISCIPLINARY AND TEAM WORK 4

1.10 SOFTWARE USED 5 1.11 CONCLUSION 5 1.12 FUTURE SCOPE 5 2 INTRODUCTION 6 2.1 GENERAL 6 2.2 LITERATURE REVIEW 7

2.3 DEVELOPMENT CONTROL RULES FOR CHENNAI

METROPOLITAN AREA, 2004 8

2.3.1 Primary Residential Use Zone 8

2.4 CONFORMATION TO NATIONAL BUILDING

CODE OF INDIA 9

2.4.1 Fire Safety, Detection And Extinguishing

(8)

8

2.4.2 Security Deposits 10

3 OBJECTIVE AND SCOPE 11

3.1 OBJECTIVE 11

3.2 SCOPE 12

3.3 MATERIALS AND METHODOLOGY 12

4 RESULTS AND DISCUSSIONS 13

4.1 PLANNING 13

4.1.1 Selection of Site 13

4.1.2 Plot Layout 14

4.1.3 Plan of the Building 15

4.2 DESIGNS 16 4.2.1 Design of Hall 16 4.2.2 Design of Bedroom 20 4.2.3 Design of Bedroom 23 4.2.4 Design of Bathroom 27 4.2.5 Design of Portico 30 4.2.6 Design of Kitchen 33

4.2.7 Design of Dining Room 37

4.2.8 Design of Wall 40

4.2.9 Design of Footing 44

4.2.10 Design of Hollow Brick Wall 44

4.2.11 Design of Footing (Hollow Brick) 49

(9)

9

4.3 DESIGN OF SOLAR PANEL AND ITS COMPONENTS 54 4.3.1 Solar power system components

54 4.3.2 Working of Solar Panel 55

4.3.3 Description of Individual Solar Panel components 55

4.3.3.1 Solar Panels 55

4.3.3.2 Solar Regulator 55

4.3.3.3 Power Inverter 56

4.3.3.4 Solar Batteries 56

4.3.4 Designing of Solar Panel 57

4.4 RATE ANALYSIS OF SOLAR PANELS 59

4.5 INFRARED THERMOMETER 60

4.6 HOLLOW BRICK 62

4.6.1 Parameters of Hollow Brick 62

4.6.2 Advantages of Hollow Bricks 64

4.7 ESTIMATION 65

4.7.1 Abstract Estimate of Conventional Building 65 4.7.2 Abstract Estimate of NZERB

67 4.7.3 Rate Analysis 70 5 CONCLUSION 72 5.1 CONCLUSION 72 5.2 FUTURE SCOPE 72 REFERENCES 73

(10)

10

LIST OF TABLES

TABLE TITLE PAGE 1.1 Codes Used

3

1.2 Earlier Course Work Used

(11)

11 2.1 Front Setback 8 2.2 Rear Setback 9 2.3 Side Setback 9

4.1 Values of slenderness ratio 48

4.2 Stress reduction factor for slenderness ratio 48

4.3 Calculation of permissible stress 49

4.4 Safe allowable load

49

4.5 Calculations of loads

57

4.6 Abstract Estimate of Conventional Building 65

4.7 Abstract Estimate of NZERB

67

4.8 Rate Analysis of Proposed Conventional Building 70

4.9 Rate Analysis of Proposed NZERB 71

(12)

12

LIST OF FIGURES

FIGURE TITLE PAGE

4.1 Plot Layout 13

4.2 Ground Floor Plan 14

4.3 First Floor Plan 15

4.4 Footing Design 51

4.5 Working of Solar Panels 54

4.6 Infrared Thermometer 60

(13)

13

ABBREVIATIONS

deff - Effective depth

c.c - Clear cover

D - Total depth

b - Width

Mu,lim - Ultimate limiting moment of resistance

fck - Characteristic compressive strength of concrete

Mu - Ultimate moment

Pt - Percentage of tension reinforcement

Pc - Percentage of compression reinforcement

Ast - Area of steel in tension zone

(14)

14

Sv - Spacing of stirrups

fy - Yield stress of steel

Asv - Total cross sectional area of stirrup legs

kt - Modification factor for tension reinforcement

kc - Modification factor for compression reinforcement

kf - Reduction factors for flanged beams

Pu - Ultimate load

τc - Permissible shear stress

Ag - Gross area of cross section

Ly - Length in y direction

Lx - Length in x direction

Wu - Ultimate load

αx - Bending moment coefficient for short span

αY - Bending moment coefficient for long span

Mx - Moment in short span direction

My - Moment in long span direction

dreq - Required depth

dprov - Provided depth

Mu,max - Maximum ultimate moment

Ast( reqd) - Area of steel required

Ast (min) - Area of minimum steel required

ast - Area of 1 bar

D.L - Dead Load

L.L - Live Load

Φ - Angle of internal friction

NC , NY, Nq - Bearing capacity factors

CMDA - Chennai metropolitian development authority PWD - Public works department

(15)

15

KKNP - Kudankulam Nuclear Power Plant W.h - Watt hour

A.h - Ampere hour

(16)

16

OVERVIEW

1.1 OBJECTIVE

i. Design a building with Net zero energy concept.

ii. To eliminate the necessity of active energy loads on the building. iii. Comparing the net zero energy building with conventional building.

1.2 NECESSITY

The basic necessities of such a building are:

i. As the country is developing day by day the consumption of power is also very high. ii. Now if we are going for NZERB building we can save energy locally

which mean to save energy in global level.

iii. The use of this technology used in residential buildings has shown huge amount savings in the electricity bill.

iv. The proper design and alignment of the building can make the building cheaper than that of the conventional type of buildings.

v. Usage of hollow bricks and avoidance of columns and beams will result in lowering of temperature inside the building

vi. To achieve sustainability.

1.3 SCOPE

(17)

17

ii. Design of load bearing structure using hollow bricks iii. Design of solar panels

iv. Comparison of room temperature between NZERB and conventional building

v. Comparison of energy consumption between NZERB and conventional building.

1.4 METHODOLOGY

This entire project is an planning and design in nature and the methodology followed in this project is listed as below.

i. Selection of site where renewable energy is available ii. Study the climate conditions of area

iii. Aligning the building to utilize maximum amount of renewable resources iv. Planning and design of proposed NZERB building

v. Comparison of the NZERB building with other conventional building

1.5 MAJOR DESIGN EXPERIENCE

Design experience in the following areas has been gained during the course of the project

i. Design of slabs ii. Design of footings

iii. Design of wall using Hollow bricks iv. Design of solar panels

(18)

18

i. Economic: Building shall be designed such that the entire energy requirements are met by solar energy only due to shortage of conventional energy.

ii. Sustainability Constraints: The design shall be such that the requirement of cooling do not fluctuate throughout the year.

iii. Economic Constraint: The materials adopted for construction are economical compared to conventional materials.

1.7 REFERENCE TO CODES AND STANDARDS

The codes for design of buildings and structures, Design co-efficient, Limit state design method and Fixing of dimensions are shown in Table 1.1

Table 1.1 Reference to codes and standards

Codes /Standards Context

IS 875 :1987 -1,2 Design loads for buildings and structures (Dead load , Imposed load )

IS 456 :2000 Design co-efficient, Limit state design method used for slab and footing IS 2572-1963(R 1997) Design of Hollow bricks

IS 1905 :1987 Structural use of Unreinforced Masonry SP 20 :1991 Handbook of Masonry design and

(19)

19

1.8 APPLICATION OF EARLIER COURSE WORK

The codes for Computer aided building drawing, layout and planning and Byelaws, Setbacks, Open space, Floor area ratio are shown in Table 1.2

Table 1.2 Application of earlier course work Course Code and Name Context

CE 0104- Computer aided building

drawing Computer aided building drawing CE0102- Elements of building science

and Architecture layout and planning

CE0209- Building technology Byelaws, Setbacks, Open space, Floor area ratio

CE0303-Structural Design II R.C.C Design CE0304-Structural Design III R.C.C Design

1.9 MULTIDISCIPLINARY COMPONENT AND TEAM WORK

i. This project involves in multidisciplinary team work and helps interacting with the builders who deal with the non conventional building

methods and use of waste and cost effective building materials.

ii. It also involves interaction with software people to learn about the function and operation of the software‟s used in this project for the design, analyse and estimation of the parts of the structure.

1.10 SOFTWARE USED

i. Auto CAD ii. MS EXCEL iii. MS WORD

(20)

20

1.11 CONCLUSION

The two types of buildings are analyzed with respect to cost, time, availability of skilled labour and ease in construction.

COST ELECTRICITY AVAILABILITY OF RESOURSES NORMAL CONVENTIONAL BUILDING Low It requires an active source Easily Available NZERB High Produced on its own Difficult

1.12 FUTURE SCOPE OF THE PROJECT

The building is designed as a NET ZERO ENERGY BUILDING which produces its own electricity, thus we can save a huge amount in electricity bill.

(21)

21

CHAPTER 2 INTRODUCTION

2.1 GENERAL

Fast rate of urbanization and increase in the consumption of electricity has become a major problem in Tamil Nadu. Due to increase in consumption of electricity the Tamil Nadu electricity board is unable to fulfill the requirements of the public and industrial sectors .In Tamil Nadu, This is the major problem faced. Officials were banking on a number of projects, which would generate 14,000 MW of power, from thermal, nuclear and other power projects. Most of these should have been completed by 2012. But the projects have got delayed, with the KKNP turning out to be a big challenge .Hence requirement has brought in new building technologies by utilizing the renewable energy resources.

In housing aspects it is necessary to design the material adopted structurally in a proportion with reference standard codes. Designing of building is the most essential work to be proposed in any projects. Before starting the project it is necessary to prepare layout and plan in a plot as per the Government Rules and Regulation for getting an approval without any delay and to execute the project. Overall cost of the project should be economical so estimation of building is very important. As a whole we have incorporated all the needs for a building to be built with efficient, eco-friendly and economic, also abiding by the Government Rules.

(22)

22

This project envisages the preparation of a Residential layout by incorporating the Tamil Nadu Government rules and the preparation of a plan for a residential building in a plot by using software AutoCAD. Finally this project will end up with the preparation of an estimation of the prepared plan (Ref 1).

2.2 LITERATURE REVIEW

Anna Joanna,

Aalborg University, Department of Civil Engineering,

According to ANNA, “With energy conservation arrangements, such as high-insulated constructions, solar heating system. Extra Energy supply for the electric installations in the house is taken from the municipal mains” (Ref 2).

Saitoh, (1988) (JAPAN)

According to SAITOH, “… a multi-purpose natural energy autonomous house will meet almost all the energy demands for solar panel and cooling as well as

supply of hot water. For this purpose, solar energy, the natural underground coldness and sky radiation cooling are utilized.”

i. Solar panels are designed to harness.

ii. Solar energy in buildings include systems that capture heat (such as Solar water heating systems and passive heating).

iii. It converts solar energy into electrical energy, its done with the help of photovoltic (PV) systems (Ref 3).

(23)

23

2.3 DEVELOPMENT CONTROL RULES FOR CHENNAI METROPOLITAN AREA, 2004

2.3.1 Primary Residential Use Zone

In this primary residential use zone, buildings shall be permitted only for the following purposes and accessory uses.

(a) Professional consulting offices of the residents and incidental uses there to occupy a floor area not exceeding 40 square meters.

(b) Petty shops dealing with daily essentials including retail sale of provisions, soft drinks, cigarettes, newspapers, tea stalls, mutton stall and milk kiosks, cycle repair shops and tailoring shops.

(c) Nursery, primary and high school.

(d) Parks and playgrounds occupying an area not exceeding 2 hectares. (e) Taxi stands and car parking.

Front setback according to the CMDA code is shown in Table 2.1.

Table 2.1 Front Set Back

Abutting Road Width Front Set Back

Above 30m 6.0m

Above 15m but less than 30m 4.5m

Above 10m but less than 15m 3.0m

Below 10m 1.5m

Rear setback according to Chennai Metro Development Authority (CMDA) code is shown in Table 2.2.

(24)

24

Table 2.2 Rear Set Back

Depth of Plot Rear Set Back

Up to 15m 1.5m

Between 15m to 30m 3.0m

Above 30m 4.5m

Side setback according to CMDA code is shown in Table 2.3.

Table 2.3 Side Set Back

Width of Plot Side Set Back

Not more than 6m 1.0m on one side

More than 6m but not more than 9m 1.5m on one side

More than 9m 1.5m on either side

2.4 CONFORMATION TO NATIONAL BUILDING CODE OF INDIA

In so far as the determination of sufficiency of all aspects of structural designs,

building services, plumbing, fire protections, construction practice and safety are concerned the specifications, standards and code of practices recommended in the National Building Code of India (Ref.4), shall be fully confirmed to any breach thereof shall be deemed to be a breach of the requirements under these rules.

Every multi-storied development erected shall be provided with (i) Lifts as prescribed in National Building Code; (ii) a stand-by electric generator of adequate capacity for running lift and water pump, and a room to accommodate the generator; (iii) a room of not less than 6 meters by 4.5 meters in area with a minimum head room of 3 meters to accommodate electric transformer in the ground floor; and (iv) at least one meter room of size 2.4 meters by 2.4 meters for every 10 consumers or three floor whichever is less. The meter room shall be provided in the ground floor.

(25)

25

2.4.1 Fire Safety, Detection and Extinguishing Systems

All building in their design and construction shall be such as to contribute to and ensure individually and collectively and the safety of life from fire, smoke, fumes and also panic arising from these or similar other causes.

In building of such size, arrangement or occupancy than a fire may not itself provide adequate warning to occupants automatic fire detecting and alarming facilities shall be provided where necessary to warn occupants or the existence of fires, so that they may escape, or to facilitate the orderly conduct of fire exit drills. Fire protecting and extinguishing system shall conform to accepted standards and shall be installed in accordance with good practice a recommended in the National Building Code of India, and for the satisfaction of the Director of Fire Services by obtaining a no objection certificate from him (Ref.4).

2.4.2 Security Deposits

The applicant shall deposit a sum at the rate of Rs.100 per square meters of floor area as a refundable non-interest earning security and earnest deposit. The deposit shall be refunded on completion of development as per the approved plan as certified by CMDA, if not, it would be forfeited.

(26)

26

CHAPTER 3 OBJECTIVE AND SCOPE

3.1 OBJECTIVE

i. Design a building with Net zero energy concept.

Net-zero energy buildings start with energy-conscious design A zero-energy residential building is a building with zero net energy consumption A net-zero energy (NZE) building is one that relies on renewable sources to produce as much energy as it uses, usually as measured over the course of a year.

ii. To eliminate the necessity of active energy loads on the building.

Solar panels is one of the technologies used to achieve net-zero status. To eliminate the necessity of active energy loads solar techniques are used which include the use of photovoltaic panels and solar thermal collectors to harness the energy.

iii. Comparing the net zero energy building with conventional building.

The comparison of NZERB and conventional building is shown in Table 3.1

Table 3.1 Comparison of NZERB and Conventional Building

Sl.no NZERB CONVENTIONAL

1 Brick Material Hollow brick Normal brick

2 Temperature 4 to 5 degree less compared

To conventional building More than NZERB 3 Electricity Produced on its own It requires an active

source

4 Initial Cost High Less compared to

NZERB 5 Solar Panels 250 w panels Provided in NZERB Not provided

(27)

27

3.2 SCOPE

i. Functional planning of G+1 Residential building. ii. Design of load bearing structure using hollow bricks. iii. Design of solar panels.

iv. Comparison of room temperature between NZERB and conventional building.

v. Comparison of energy consumption between NZERB and conventional building.

3.3 METHODOLOGY

This entire project is an planning and design in nature and the methodology followed in this project is listed as below.

i. Selection of site where renewable energy is available

Urappakam has a tropical wet and dry climate. The weather is hot and humid for most of the year. The hottest part of the year is late May to early June. Hence solar energy is available on the site which makes the site suitable to harness solar energy

ii. Study the climate conditions of area

The city lies on the thermal equator and is also on the coast, which prevents extreme variation in seasonal temperature. The weather is hot and humid for most of the year. maximum temperatures is around 35– 40 °C (95–104 °F). The highest recorded temperature is 45 °C (113 °F)

iii. Aligning the building to utilize maximum amount of renewable resources

Elongated east-west and oriented to astronomic south (Ref 5). South-facing windows harvest solar energy.

iv. Planning and design of proposed NZERB building

(28)

28

CHAPTER 4 RESULTS AND DISCUSSIONS

4.1 PLANNING

The key plan of the residential building is drawn by considering the alignment of the building with respect to the CMDA.

The key plan of the site is shown in Figure 4.1

(29)

29

The ground floor of the building consist of one hall, two bedrooms, one dinning, one kitchen. The allocations of the rooms in the plan has been done with due consideration of sun diagram as per the requirement of zero energy building. The plan has been prepared using Auto CAD software.

The Ground Floor plan is shown in Figure 4.2

(30)

30

The first floor of the building consist of one hall, two bedrooms, one dinning, one kitchen. The allocations of the rooms in the plan has been done with due consideration of sun diagram as per the requirement of zero energy building. The plan has been prepared using Auto CAD software.

The First Floor plan is shown in Figure 4.3

(31)

31

4.2 ANALYSIS AND DESIGNS SLAB DESIGN (Ref 6)

The analysis and designs of the slab for Hall, Bedroom, Bathroom, Dinning, Kitchen, Stair case, Portico are done with proper considerations as per IS 456:2000.

4.2.1 Design of Hall

Using M 20 Concrete Fe 415 steel

Live Load = 2 (Ref 7)

1. Effective Span

Lx = 3.26 m

Ly = 5.1 m

Aspect ratio = = 1.56<2 Hence Two Way Slab

2. Load Calculation

Assuming Slab Thickness

Assume 10 bar, Clear Cover 20mm

Actual Depth (d) = 130-5-20 = 105 mm

Assume Floor Finish = 40 mm

Weight of Floor Finish = 0.04 × 24 = 0.96 Imposed Load = 2 (Ref 8)

(32)

32 Factored Load (Wu) = 1.5 × 6.21 = 9.315

Consider 1m width of slab

Load per meter Length = 9.127

3. Finding Design Bending Moment

Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous (already found out )

Refer Table 26

Short Span αx = 0.068

Long Span αy = 0.037

[Note that Lx only to be taken, where it is long span or short span only coefficient varies].

Mu = Wu × Co-efficient × Lx2 (4.1)

Mu is calculated by equation 4.1

Where,

Mu = Moment in short span direction

Wu= Ultimate load

Lx = Length in x direction

Mu(+) Short = 0.068 × 9.315 × 3.262 = 6.731 kN.m (Ref.9)

Mu(+) Long = 0.037 × 9.315 × 3.262 = 3.662 kN.m

Take the Highest Moment and check for adequacy of the section. Mu,lim=

* + _(4.2)

(or) = 0.138fckb d2

Mu,lim is calculated by equation 4.2

(33)

33

Mu,lim = Ultimate limiting moment of resistance

fck = Characteristic compressive strength of concrete

b = Width d =Effective depth = 0.138 × 20 × 1000 × 1052 =30.42 kN.m (Mu Limit) > (Mu Short) Hence its ok 4. Calculation of Steel Ast(+) Short = [1 - 1-4.598 ] (4.3) Ast(+) Short is calculated by equation 4.3

Where,

Ast(+) Short =Area of steel required

b = Width

d = Effective depth

fck = Characteristic compressive strength of concrete

fy = Yield stress of steel

R = = 6.731 × = 0.6105 Ast(+) Short = 1000 × 105 × ( _{) × 415 [1 - 1-4.598} = 184.27 mm2 Minimum Steel = 0.12% × D × B Ast,min = ( )× 130 × 1000 = 156 mm 2

Ast(+) Short <Ast,min

5. Check for maximum Spacing

i. 3d = 3 × 105 = 315 mm

ii. 300

(34)

34 d for long span bars

d = D – Clear Cover – - = 130 – 20 – - 10 = 95 mm

6. Calculation of Ast for Long Span Ast(+)Long [1 - 1- 4.598 ] Same as equation 4.3 R = d2 = × 952 = 0.4057 Ast(+)Long = 1000 × 95 × ( )× 415 [1 – 4.598 × ] = 109.37 mm2

Ast(+)Long<Ast,min

7. Spacing for all Steel

i. 3d = 3 × 95 = 285 mm ii. 300

Spacing = 285 mm

8. Check for Deflection

Short Span Lx = 3260 mm Ast(+) Short = 116.37 mm2 Basic Value = 20 Fs = 0.58 × 415× =240.2 Pt ( ) = ( ) = 0.175% Modification Factor = 1.62

Modified Basic Value = 20 × 1.62 = 32.8

(35)

35 31.047 < 32.8

Hence its ok

4.2.2 Design of Bed Room

Using M20 Concrete Fe415 steel Live Load = 2 1. Effective Span Lx = 3 m Ly = 3.5 m

Aspect ratio = = Hence Two Way Slab

Assuming Slab Thickness Assume 10 bar, Clear Cover = 20mm

Actual Depth (d) = 125-5-20 = 100 mm

Self Weight of a Slab = Assume 40 mm Floor Finish

Weight of Floor Finish = 0.04 X 24 = 0.96

Imposed Load = 2

Total Load = 6.08

Factored Load (Wu) = 1.5× 6.96 = 9.127

(36)

36 Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous ( already found out )

Refer Table 26

[ Note that Lx only to be taken, where it is long span or short span only coefficient varies ].

Mu = Wu× Co-efficient × Lx2

Same as equation 4.1

Mu(+) Short = 0.043 × 9.13 × 3.232 = 4.09 kN.m

Mu(+) Long= 0.035 × 9.13 × 3.232 = 3.33 kN.m

Take the Highest Moment and check for adequacy of the section. Mu,lim= * + (or) Mu,lim = 0.138 fckbd2 Same as equation 4.2 Mu,lim = 0.138 × 20 × 1000 × 1002 Mu,lim = 27.6 (Mu Limit) > (Mu Short) Hence its ok 4. Calculation of Steel Ast(+) Short = [1 - 1-4.598 ] Same as equation 4.3

(37)

37 R = = 4.1 × × 1000 = 0.41 Ast(+) Short = _{[1 - 1-4.598 ×} ] = 116.37 mm2 Minimum Steel = 0.12% × D × B Ast,min = × 125 × 1000 = 150 mm 2

i. 3d = 3 X 100 = 300 mm ii. 300

Max Spacing = 300 mm d for long span bars d= D – Clear Cover – - d= 125 – 20 – 10/2 - 10 d= 90 mm

6. Calculation of Ast for Long Span Ast(+)Long [1 - 1- 4.598 ] Same as equation 4.3 R = = 3.33 × = 0.33 Ast(+)Long = 1000 × 100 × * – + = 93.2 mm2

Ast(+)Long<Ast,min

7. Spacing for Steel

(38)

38 Ast(+)Short = × 1000 = 674.5 mm Ast(+)Long = × 1000 = 842.27 mm

Short Span Lx = 3230 mm Ast(+) Short = 116.37 mm2 Basic Value = 20 Fs = 0.58 × 415 × =240. Pt ( ) = (1.22 X 10-3) ×100 = 0.122% Modification Factor = 1.7

Modified Basic Value = 20 × 1.7 = 34

32.3 < 34 Hence its ok

4.2.3 Design of Bed Room

Using M20 Concrete Fe415 steel Live Load = 2 1. Effective Span Lx = 3.85 m Ly = 3.95 m Aspect ratio = = = 1.027<2

Hence Two Way Slab

(39)

39 d = 120.31mm = 120 mm Assume 10 bar, Clear Cover 20 mm D = 120+ +20 = 145mm = 150 mm Actual Depth (d) = 150-5-20 = 125 mm Self Weight of a Slab = × 25 = ×25 = 3.75 Assume 40 mm Floor Finish

Weight of Floor Finish = 0.04 × 24 = 0.96 Imposed Load = 2

Total Load = 6.75

Factored Load (Wu) = 1.5 × 6.75 = 10.125

Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous = 1.027 (already found out)

Refer Table 26

(40)

40

Mu(+) Long = 0.047 × 10.125 × 3.852 = 7.063 kN.m

Take the Highest Moment and check for adequacy of the section. Mu,lim= _* ₊ (or) = 0.138fckb d2 Same as equation 4.2 = 0.138 × 20 × 1000 × 1252 = 43.125 kN.m (Mu Limit) > (Mu Short) Hence its ok 4. Calculation of Steel Ast(+) Short = [1 - 1-4.598 ] Same as equation 4.3 R = = 7.203 × 106 = 0.460 Ast(+)Short = 1000 × 125 × ( ) × 415 [1 - 1-4.598 × ] = 164.08 mm2 Minimum Steel = 0.12% × D × B Ast,min = ( )× 150 × 1000 = 180 mm2

i. 3d = 3 × 125 = 375 mm ii. 300

Max Spacing = 300 mm d for long span bars d= D – Clear Cover – - d= 150 – 20 – - 10

(41)

41 d= 115 mm

6. Calculation of Ast for Long Span Ast(+)Long [1 - 1- 4.598 ] Same as equation 4.3 R = = = 0.553 Ast(+)Long = 1000 × 115 × ( ) × 415 [1 – 4.598 × ] = 181.11 mm2

Ast(+)Long<Ast,min

i. 3d = 3 × 115 = 345 mm ii. 300

Spacing = 300 mm

Short Span Lx = 3850 mm Ast(+)Short = 181.11 mm2 Basic Value = 20 Fs = 0.58 × 415 × =240.2 Pt ( ) = ( ) = 0.157% Modification Factor = 1.8

= = 30.8 30.8 < 35.6 Hence its ok

(42)

42 4.2.4 Design of Bathroom Using M20 Concrete Fe415 steel Live Load = 2 1. Effective Span Lx = 2.38 m Ly = 4.28 m Aspect ratio = = = 1.798<2 Hence Two Way Slab

d = = 74.375 mm = 80 mm Assume 10 bar, Clear Cover 20 mm

Actual Depth d = 110-5-20 = 85 mm

Self Weight of a Slab = = 2.75

Assume 40 mm Floor Finish

Total Load = 5.75

Factored Load (Wu) = 1.5 × 5.75 = 8.625

(43)

43

Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous

= 1.798 (already found out)

Refer Table 26

Mu(+) Short = 0.085 × 8.625 × 2.382 = 4.127 kN.m

Mu(+) Long = 0.047 × 8.625 × 2.382 = 2.29 kN.m

Take the Highest Moment and check for adequacy of the section. Mu,lim= _* ₊ (or) = 0.138fckb d2 Same as equation 4.2 = 0.138 × 20 × 1000 × 852 =19.94 kN.m (Mu Limit) > (Mu Short) Hence its ok 4. Calculation of Steel Ast(+) Short = [1 - 1-4.598 ] Same as equation 4.3 R = = 4.1527 ×_{( )} = 0.574 Ast(+) Short = 1000 × 85 × ( ) × 415 [1 - 1-4.598 × ]

(44)

44 = 139.92 mm2 Minimum Steel = 0.12% × D × B Ast,min= ( ) × 110 × 1000 = 132 mm 2

i. 3d = 3 × 855 = 255 mm

ii. 300

Max Spacing = 255 mm d for long span bars d= D – Clear Cover – - d= 110 – 20 – - 10 d= 75 mm

6. Calculation of Ast for Long Span Ast(+)Long [1 - 1- 4.598 ] Same as equation 4.3 R = = = 0.4082 Ast(+) Long = 1000 × 75 × ( _{) × 415 [1 – 4.598 ×} = 86.88 mm 2

Ast(+) Long <Ast,min

i. 3d = 3 × 75 = 225 mm

ii. 300

MaxSpacing = 225 mm

Short Span Lx = 2380 mm

(45)

45 Basic Value = 20 Fs = 0.58 × 415 × = 240.2 Pt ( ) = ( ) ( ) = 0.1646% Modification Factor = 1.9

= = 28 28 < 38 Hence its ok 4.2.5 Design of Portico Using M20 Concrete Fe415 steel Live Load = 2 1. Effective Span Lx = 3.78 m Ly = 6.93 m Aspect ratio = = 1.83<2

Hence Two Way Slab

d = 118.124 mm = 120 mm Assume 10 bar, Clear Cover 20 mm D = 120 = 150 mm

Actual Depth (d) = 150-5-20 = 125 mm Self Weight of a Slab = ×25

(46)

46 = 3.75 Assume 40 mm Floor Finish

Total Load = 6.75

Factored Load (Wu) = 1.5 × 6.75 = 10.125

Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous (already found out)

Refer Table 26

Mu(+) Short = 0.087 × 10.125 × 3.782 = 12.58 kN.m

Mu(+) Long = 0.047 × 10.125 × 3.782 = 6.79 kN.m

Take the Highest Moment and check for adequacy of the section. Mu,lim= * + (or) = 0.138fckb d2 Same as equation 4.2

(47)

47 = 0.138 × 20 × 1000 × 1252 = 43.125 kN.m (Mu Limit) > (Mu Short) Hence its ok 4. Calculation of Steel Ast(+) Short = [1 - 1-4.598 ] Same as equation 4.3 R = = = 0.805 Ast(+) Short = 1000 × 125 × [1 - 1-4.598 × ] = 184.27 mm2 Minimum Steel = 0.12% × D × B Ast,min = ( )× 130 × 1000 = 156 mm2

i. 3d = 3 ×105 = 315 mm

ii. 300

Max Spacing = 300 mm d for long span bars d= D – Clear Cover – - d= 130 – 20 – 10/2 - 10 d= 95 mm

6. Calculation of Ast for Long Span Ast(+)Long

[1 - 1- 4.598 ] Same as equation 4.3

(48)

48 = = 0.4057 Ast(+)Long = 1000 × 95 × ( _{)× 415 [1 – 4.598 ×} ] = 109.37 mm2

Ast(+)Long<Ast,min

i. 3d = 3 × 95 = 285 mm ii. 300

Spacing = 285 mm

Short Span Lx = 3260 mm Ast(+) Short = 116.37 mm2 Basic Value = 20 Fs = 0.58 × 415× = 240.2 Pt ( ) = ( ) = 0.175% Modification Factor = 1.62 Modified Basic Value = 20 × 1.62 =32 = = 31.047 31.047 < 32.8 Hence its ok 4.2.6 Design of Kitchen Using M20 Concrete Fe415 steel Live Load = 2 1. Effective Span

(49)

49 Lx = 2.23 m

Ly = 3.73 m

Aspect ratio Hence Two Way Slab

Assuming Slab Thickness d = = 65 mm

Assume 10 bar, Clear Cover 20 mm D = 65+ +20 = 90 mm

Actual Depth (d) = 90-5-20 = 65 mm Self Weight of a Slab =

= 2.25 Assume Floor Finish = 40 mm

Total Load = 5.25

Factored Load (Wu) = 1.5 × 5.25 = 7.875

Load per meter Length = 7.875 Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous = 1.67 (already found out)

Refer Table 26

(50)

50 Long Span αy = 0.035

Mu(+) Short = 0.06 × 7.88 × 2.232 = 2.35 kN.m

Mu(+) Long = 0.035 × 7.88 × 2.232 = 1.373 kN.m

Take the Highest Moment and check for adequacy of the section. Mu,lim= * + (or) Mu,lim = 0.138 fckb d2 Same as equation 4.2 = 0.138 × 20 × 1000 × 652 =11.66 kN.m (Mu Limit) > (Mu Short) Hence its ok 3. Calculation of Steel Ast(+) Short = [1 - 1-4.598 ] Same as equation 4.3 R = = 2.735 × = 0.55 Ast(+) Short = 1000 × 65 × × 415 [1 - 1-4.598 × ] = 103.56 mm2 Minimum Steel = 0.12% × D × B Ast,min × 90 × 1000 = 108 mm2

(51)

51

i. 3d = 3 × 65 = 195 mm

ii. 300

5. Calculation of Ast for Long Span Ast(+)Long [1 - 1- 4.598 ] Same as equation 4.3 R = = 1.373 × = 0.115 Ast(+) Long =1000 × 55 × × 415 [1 – 4.598 × ] = 71.05 mm2

Ast(+) Long <Ast,min

Ast = × 102 = 78.5 mm2 Ast(+) Short = × 1000 = 758.01 mm Ast(+) Long = × 1000 = 1104.8 mm

Short Span Lx = 2230 mm Ast(+) Short = 103.56 mm2 Basic Value = 20 Fs = 0.58 × 415 × = 240.7 Pt ( ) = = 0.16%

(52)

52 Modification Factor = 1.8

34.3 < 36 Hence its ok

2.4.7 Design of Dinning Room

Using M20 Concrete Fe415 steel Live Load = 2 1. Effective Span Lx = 2.6 m Ly = 3.73 m Aspect ratio Hence Two Way Slab

Assuming Slab Thickness d =

Assume 10 bar, Clear Cover 20 mm D

Actual Depth (d) = 105-5-20 = 80 mm

Self Weight of = 2.625

Assume Floor Finish = 40 mm

(53)

53 Total Load = 5.62

Factored Load (Wu) = 1.5 × 5.62 = 8.43

Refer Table 26, Page No.91 of IS456 One edge discontinuous

= 1.43 (already found out)

Refer Table 26

Mu(+) Short = 0.049 × 8.43× 2.62 = 2.79 kN.m

Mu(+) Long = 0.028 × 8.43 × 2.62 = 1.595 kN.m

Take the Highest Moment and check for adequacy of the section. Mu,lim= * + (or) Mu,lim = 0.138fckb d2 Same as equation 4.2 Mu,lim = 0.138 × 20 × 1000 × 802 Mu,lim =17.66 kN.m (Mu Limit) > (Mu Short) Hence its ok 4. Calculation of Steel

(54)

54 Ast(+) Short = [1 - 1-4.598 ] Same as equation 4.3 R = = 2.79 × = 0.435 Ast(+) Short = 1000 × 80 × 415 [1 - 1-4.598 ] = 99.15 mm2 Minimum Steel = 0.12% × D × B Ast,min= × 105 × 1000 = 126 mm2 Ast(+) Short <Ast,min

i. 3d = 3 × 80 = 240 mm ii. 300

6. Calculation of Ast for Long Span Ast(+)Long [1 - 1- 4.598 ] Same as equation 4.3 R = = 1.595 × = 0.325 Ast(+) Long =1000 × _{– 4.598} × ] = 64.34 mm2

(55)

55 Ast(+) Long <Ast,min

7. Spacing for Steel

Ast = × 102 = 78.5 mm2 Ast(+) Short = 1000 = 791.7 mm Ast(+) Long = × 1000 = 1200.08 mm

Short Span Lx = 2600 mm Ast(+) Short = 99.15 mm2 Basic Value = 20 Fs = 0. Pt ( ) = = 0.14% Modification Factor = 1.8

= 32.5 32.5 < 36 Hence its ok 4.2.8 Design of Wall Design of a wall 1.Calculation of Loads

Maximum short span = 3.60 m Width of corridor = 1.50 m Height of the storey = 3 m

Live load = 2

(56)

56 Height of the Plinth from ground = 0.5 m Height of the Plinth above Footing = 1 m Height of the Parapet Wall = 1 m Thickness of Roof Slab =110 mm

Brick Size = 230 × 115 × 75

3. Slenderness Ratio and Stress Factor

Ground Floor + First Floor

H = 3+0.115+0.5+3+0.115+1 = 7.73 m

Effective Height (h) = 0.75H = 0.75 × 7.73 = 5.797 m Slenderness Ratio

4. Shape modification factor:

Crushing Strength of Modular Brick = 5

Shape Modification Factor = Kp = 1 ( From table 10 of IS: 1905-1987)

5. Area reduction factor:

Area Reduction Factor

Ka = 0.7 + 1.5 A = 0.7 + 1.5X0.3 = 1.15

A > 0.2 m2

Ka = 1 ( From clause 5.4.1.2)

6. Stress Reduction Factor:

ks = 0.46 ( from table 9)

7. Permissble Stress

Fc = Ks × Ka × Kp× Basic compressible stress (4.4)

Fc is calculated by equation 4.4

Where,

Ks= Stress reduction factor

Ka= Area reduction factor

(57)

57 Fc = 0.44 × 0.48 × 1 × 1 = 8. Safe Load Q = ( ) = = 63 9. Wall Area

Outer wall = Total Perimeter x 3(floor height) = ((11.31×2) + (8.93×2))×3

= 40.83 m3

Inner wall = (4.87×3) + (4.87×3) + (3.5×2×3) = 29.22 + 21 = 50.22 m3

Total wall volume = 91.05 m3

10. Deductions: Outer Deductions = 1.098+1.089+2.226+1.089+1.4884+1.4884+1.4884+ 1.4884+1.4884+1.098 = 14.042 m3 Inner Deductions = 1.89+1.89+2.496+1.746+1.746 = 9.768 m3 Total Deduction =23.81 m3

Total wall volume – Total Deductions = 91.05 -23.81 = 67.25 m3 % Opening =

= 35.4 %

Thickness = 1 Brick thick wall (using nomograms)

11. For Hall : Wu= _{× {3- (} ) 2 } (4.5) Wu is calculated by equation 4.5

(58)

58 Where,

Wu= Factored load

W=Load from the slab Lx=Short span Ly=Long span Wu =( ( )) * } Wu = 13.11 103

For bed room: Wu=

* ( ) }

= (9.127 × ( )) × {3- * + } = 11.055 103

For dinning room: Wu= × {3- *( )+2 } = ( ( )) × {3- [( )_{( )}]2} = 9.81 103 Total : 13.11 + 11.055 + 9.81 = 33.98 103 63>33.98

Hence the design is ok

4.2.9 Design of Footing

Load from Walls = 126.7

(59)

59 1. Area of Footing =

Assume SBC = 150 A = = 0.47 m2 Consider 1m Length room

Breadth of the Footing Required = = 0.47 2. Minimum Width = (2w+300)mm

= (2x230+300) = 760 mm Provide Width of P.C.C= 760 mm

It is customary to provide 150 to 300 mm P.C.C thickness Provide = 300 mm

The Projection of P.C.C beyond the brick work should not be more than ½ of the thickness of P.C.C

Projection = = 150 mm

Actual work of Brick work = 760 – 300 = 460 mm Brick work projection beyond the wall

1.Depth of the Brick work = 115 × 2 = 230 mm

These depth has to be Provided by means of series steps The thickness of each step is given by modular brick = 200 mm The offset in the brick is also given as modular = 100 mm

4.2.10 Design of Hollow Brick Wall Step 1: Calculation of loads

Maximum short span = 3.6 m Height of the storey = 3 m Live load = 2

(60)

60 Height of the Plinth from ground = 0.5 m Height of the Plinth above Footing = 1 m Height of the Parapet Wall = 1 m Thickness of Roof Slab = 0.120 m

Hollow Brick Size = 0.40 × 0.20 ×0. 20 m

EFFECTIVE LENGTH OF WALL

(From Table 5 of IS 1905-1987)

Wall A = 3.82×0.9=3.438 m (continuous on one end & discontinuous on other end)

Wall B = 3.23×0.8=2.584 m (continuous on both ends & supported by cross wall)

Wall C = 3.7×0.9 =3.33 m (continuous on one end & discontinuous on other end) Wall D = 3.2×0.9 =2.88 m (continuous on one end & discontinuous on other end) Wall E = 2.57×0.8=2.056 m (continuous on both ends & supported by cross wall) Wall F = 2.8×0.9 =2.52 m (continuous on one end & discontinuous on other end) Wall G = 3.7 m (discontinuous on both ends and braced by cross wall)

Wall H = 3.23×0.9 =2.907 m (continuous on one end & discontinuous on other end)

Wall I = 3.82 × 0.9=3.438 m (continuous on one end & discontinuous on other end)

Wall J = 3.438 m (continuous on one end & discontinuous on other end) Wall K = 3.72 × 0.8=2.976 m (continuous on both ends & supported by cross wall)

Wall L = 2.15 × 0.9 =1.935 m (continuous on one end & discontinuous on other end)

(61)

61

Step 3: Slenderness ratio and stress factor

Ground floor:

H = 2.6+0.6+1 = 4.2 m

Effective height = 0.75 × H = 3.15 m Slenderness ratio = = = 15.75

Step 4: Shape modification factor

Crushing Strength of Hollow Brick= 4.1

= =1

Shape Modification Factor = Kp =1.2( From table 10 of IS: 1905-1987)

Step 5: Stress Reduction Factor

ks = 0.74( From table 9)

Step 6: Area reduction factor

Gross area = 200 × 1000 = 200000 mm2 A = 0.2 m2

Ka= 1( From clause 5.4.1.2)

Step 7: Permissible stress

Fc= 0.74 × 1.2 × 1× 0.44 = 0.39

Safe allowable load per meter length is q = 0.39 × 2 × 105 = 78

Step 7: Slenderness ratio and stress factor

First floor:

(62)

62 Effective height = 0.75 × H = 2.55 m Slenderness ratio = = = 12.75

Step 8: Shape modification factor

Crushing Strength of Hollow Brick = 4.1

= =

=

1

Shape Modification Factor = Kp =1.2

Step 9: Stress Reduction Factor

ks = 0.81

Step 10: Area reduction factor

Gross area = 200 × 1000 = 200000 mm2 A = 0.2 m2

Ka= 1

Step 11: Permissible stress

Fc= 0.81 × 1.2 × 1 × 0.44 = 0.427

Safe allowable load per meter length is q = 0.427 × 2 × 105 = 85

The values of slenderness ratio for effective length and height of the building is given in Table 4.1

Table 4.1 Values of slenderness ratio (Ref 10)

(63)

63 H L S.R H L S.R A 3.15 3.44 15.75 2.55 3.44 12.75 B 3.15 2.58 12.92 2.55 2.58 12.75 C 3.15 3.33 15.75 2.55 3.33 12.75 D 3.15 2.88 14.4 2.55 2.88 12.75 E 3.15 2.056 10.28 2.55 2.056 10.28 F 3.15 2.52 12.6 2.55 2.52 12.6 G 3.15 3.7 15.75 2.55 3.7 12.75 H 3.15 2.90 14.5 2.55 2.9 12.75 I 3.15 3.44 15.75 2.55 3.44 12.75 J 3.15 3.44 15.75 2.55 3.44 12.75 K 3.15 2.976 14.88 2.55 2.976 12.75 L 3.15 1.935 9.7 2.55 1.935 9.675 M 3.15 5.07 15.75 2.55 5.07 12.75

The values of stress reduction factor for slenderness ratio of the building is given in Table 4.2

Table 4.2 Stress reduction factor for slenderness ratio

Wall type Ground floor First floor

A 0.74 0.81 B 0.81 0.81 C 0.74 0.81 D 0.75 0.81 E 0.89 0.89 F 0.83 0.83 G 0.74 0.81 H 0.75 0.81 I 0.75 0.81 J 0.74 0.81 K 0.75 0.81 L 0.88 0.88

The calculation of permissible stress of the building is given in the Table 4.3

(64)

64

Fc=ks×kp×ka× basic compressive stress

Wall type Permissible stress- ground floor(N/mm2)

Permissible stress- first floor(N/mm2) A 0.528×0.74=0.390 0.528×0.81=0.427 B 0.528×0.81=0.427 0.528×0.81=0.427 C 0.528×0.74=0.390 0.528×0.81=0.427 D 0.528×0.75=0.396 0.528×0.81=0.427 E 0.528×0.89=0.469 0.528×0.89=0.469 F 0.528×0.83=0.4382 0.528×0.83=0.4382 G 0.528×0.74=0.390 0.528×0.81=0.427 H 0.528×0.75=0.396 0.528×0.81=0.427 I 0.528×0.75=0.396 0.528×0.81=0.427 J 0.528×0.74=0.390 0.528×0.81=0.427 K 0.528×0.75=0.396 0.528×0.81=0.427 L 0.528×0.88=0.4646 0.528×0.88=0.4646 M 0.528×0.75=0.390 0.528×0.81=0.427

The values of safe allowable load for the building is given in Table 4.4

Table 4.4 Safe allowable load Wall type q = fc×2×10 5 kN/m(ground floor) q = fc×2×105 kN/m(first floor) A 78 85.4 B 85 85.4 C 78 85.4 D 79.2 85.4 E 93 93.8 F 87.6 87.6 G 78 85.4 H 79.2 85.4 I 79.2 85.4 J 78 85.4 K 79.2 85.4 L 92 92 M 78 85.4

(65)

65 Load from Wall = 78

Load from wall (critical wall M) +10% for the weight of the Building + weight of slab (hall, bed room & dining) + floor finish

= 78 + 7.8 + 4.6575 + 4.125 + 4.5635 + 1 =100.236 =100 Factored load= 1.25 × 100=125 1. Area of Footing = = = 0.833 m2 Assume SBC = 150

Consider 1m Length room

2. Minimum Width = (2w+300) mm

= (2 x 200+300) = 700 mm Provide Width of P.C.C = 700 mm

It is customary to provide 150 to 300 mm P.C.C thickness Provide = 300 mm

The Projection of P.C.C beyond the brick work should not be more than ½ of the thickness of P.C.C

= = 150 mm

Actual work of Brick work = 700 – 300 = 400 mm Brick work projection beyond the wall

Depth of the Brick work = 200 x 2 = 400 mm

(66)

66

The thickness of each step is given by hollow brick = 200 mm The footing design is shown in the Figure 4.4

Fig.4.4 Footing Design 4.2.12 Design of Stair Case:

Length = 4 m Live load = 2

Rise = 150 mm

Thread = 250 mm Using M20 Concrete and Fe415

Step1: Calculation of self weight

Assume waist slab thickness = = 200 mm D = 200 mm

Self weight =

×

× 2 = 5.83 (4.6) Self weight is calculated by equation 4.6

(67)

67 D =Diameter

R =Rise T =Thread

Step 2: Calculation of load on waist slab

1. Assume 40 mm Floor finish Floor Finish = × 24 = 1 2. Weight of steps= × R × T × × 25 = 1.875 3. Live load = 3 4. Self weight = 5.83 Wu = 11.075 Wu=1.5 × 11.075 =17.55 Step3: Calculation of Mu Mu = = 35 kN.m Mu,lim = _* ₊ 35 × 106= 0.36 × 0.48(1-0.42 × 0.48) × 1000 × d2 × 20 d=113 mm

Assume Clear cover 20 mm Diameter of bar = 20 mm D= 113+20+10 = 143 mm D= 150mm (approximately) d= 150-20-10 = 120 mm

(68)

68 R = = 2.43 Ast = = 970.6 mm 2 Number of Bars = = 4 bars Ast actual = 4 × × 202 = 1256 mm2 Pt = × 1000 = × 1000 = 1.04%

Step 5: Check for deflection

Basic value = 20 Fs = 0.58fy = 0.58 × 415 × = 185.89

For Pt =1.04% , Modification factor = 1.2 (Fig 4 of IS 456-2000)

drequired = = = 104.166 mm dactual = 120 mm

Step 6: Providing distribution steel

Astmin= × = 480 mm 2 Spacing of 8mm diameter = × × 82 = 270 mm Main steel = 4No.s 20 bars

(69)

69

4.3 DESIGN OF SOLAR PANEL AND ITS COMPONENTS 4.3.1 Solar Power System Components

Brief revision of the major components found in a basic solar power system. A basic solar powered system is shown in Figure 4.5

Fig.4.5 Working of solar panels

The solar panel consists of solar regulator it is connected to DC storage battery and then DC is converted to AC by an inverter. AC can be directly used for the

appliances.

(70)

70

4.3.2 Working of solar panels

The solar panel converts sunlight into DC power or electricity to

charge the battery.

i. This DC electricity (charge) is controlled via a solar regulator which ensures the battery is charged properly and not damaged and that power is not lost/(discharged).

ii. DC appliances can then be powered directly from the battery.

iii. AC appliances need a power inverter to convert the DC electricity into 220 Volt AC power.

4.3.3 Description of individual solar power components

4.3.3.1 Solar Panels

Solar panels are classified according to their rated power output in Watts. Different geographical locations receive different quantities of average peak sun hours per day. As an example, in Tamil Nadu, the annual average is around 6am sun hours per day. This means that an 80W solar panel based on the average figure of 6 sun hours per day, would produce a yearly average of around 480W.H per day. Solar panel output is affected by the cell operating temperature. Panels are rated at a nominal temperature of 25 degrees Celcius. The output of a solar panel can be expected to vary by 0.25% for every 5 degrees variation in temperature.

4.3.3.2 Solar Regulator

The purpose of solar regulators, or charge controllers as they are also called, is to regulate the current from the solar panels to prevent the batteries from

overcharging. Overcharging causes gassing and loss of electrolyte resulting in damage to the batteries. A Solar regulator is used to sense when the batteries are fully charged and to stop, or decrease, the amount of current flowing to the battery. Most solar regulators also include a Low Voltage Disconnect feature, which will switch off the supply to the load if the battery voltage falls below the cut-off voltage. This prevents the battery from permanent damage and reduced life expectancy. Solar regulators are rated by the amount of current they are able to receive from the solar panel or panels.

(71)

71

4.3.3.3 Power Inverter

The power inverter is the main component of any independent power system which requires AC power. The power inverter will convert the DC power stored in the batteries and into Ac power to run conventional appliances. There are three waveforms produced by modern solid state power inverters. The simplest, a square wave power inverter, used to be all that was available. Today, these are very rare, as many appliances will not operate on a square wave. True Sine wave inverters provide AC power that is virtually identical to, and often cleaner than, power from the grid. Power inverters are generally rated by the amount of AC power they can supply continuously. Manufacturers generally also provide 5 second and ½ hour surge figures. The surge figures give an idea of how much power can be supplied by the inverter for 5 seconds and ½ an hour before the inverter‟s overload protection trips and cuts the power.

4.3.3.4 Solar Batteries

Deep cycle batteries are usually used in solar power systems and are designed to be discharged over a long period of time (e.g. 100 hours) and recharged hundreds or thousands of times, unlike conventional car batteries which are designed to provide a large amount of current for a short amount of time. To maximize battery life, deep cycle batteries should not be discharged beyond 50% of their capacity. i.e. 50 % capacity remaining. Discharging beyond this level will significantly reduce the life of the batteries. Deep cycle batteries are rated in Ampere Hours (Ah). This rating also includes a discharge rate, usually at 20 hours. This rating specifies the amount of current in Amps that the battery can supply over the specified number of hours. As an example, a battery rated at 120A.H at the 100 hour rate can supply a total of 120A.H over a period of 100 hours. This would equate to 1.2A per hour for 100 hours.

4.3.4 Designing of Solar Panel

Power rating of each appliance that will be drawing power from the system. Calculation of Loads

(72)

72

Table 4.5 Calculation Of Loads (Ref 12) PARTICULA RS ITEMS UNITS USAGE IN HRS VOLTA GE W CONSUM PTION INVER TORS HALL CFL (Ref 13) 4 5 20 400 80 FAN 2 5 50 500 100 T.V 1 5 80 400 80 BED ROOM 1 CFL 2 3 15 90 30 FAN 1 10 50 500 50 BED ROOM 2 CFL 2 3 15 90 30 FAN 1 10 50 500 50 KITCHEN OVEN 1 1 900 900 900 CFL 3 4 15 180 45 EXHAUS T 1 4 50 200 50 Mixer 1 1 450 450 450 DINING ROOM AUTO-FRIDGE 1 18 150 2700 195 CFL 3 4 15 180 45 FAN 1 3 50 150 50 TOILET 1 CFL 1 1 15 15 15 TOILET 2 HEATER 1 1 150 150 150 CFL 1 2 15 30 15 WATER PUMP 1 1 750 750 750 WASHING MACHINE 1 2 90 180 90 8365 3175

(73)

Power Invertor Sizing

Appliance total power draw = 3175 W

To provide a small buffer or margin your minimum size inverter choice should be around 3500W.

A modified sine wave inverter with a 3500W continuous power rating will therefore be your obvious choice in this specific solar system design.

Determining the Size And Number Of Solar Panels

Divide the total daily power requirement by the number of charge hours for that geographic region eg. (8365×1.2)\6=1673 WATTS

250 Watt Solar Panel

Total watt/ 250 watt solar panel =

=7 PANELS

= 7 x 250 W panels.

Number of Batteries

250W panels produce 4.8Amps, thus 14x 4.8 A = 67.2A x 6 Hrs = 403.2.Ah

105Ah batteries, should be discharged to no more than 50%, thus we divide total amps by 105A x 50% = 50A.h

= 8.08 x 105Ah batteries.

For ease of possible 24V or 48V configuration, this would mean 3 in series of 3 batteries.

Size of Regulators

Let‟s say we had 20A regulators at our disposal. One 250W panel produces around 4.8Amps. The regulators are put in series

14 x 4.8A=67.2

So 14 solar panels would need 4 x 20 A solar regulators .

Complete the solar power system Well we have the following:

i. 7x2x 250W solar panels ii. 4 x 20A solar regulators