civil engineering project

PROJECT REPORTon

PLANNING AND DESIGN OF NET ZERO ENERGY RESIDENTAL BUILDING

Submitted in partial fulfillment for the award of the degreeof

BACHELOR OF TECHNOLOGYin

CIVIL ENGINEERINGby

KARTHIK V (1010910090) SASIDHAR K.V (1010910092)

NEERAJ PORWAL (1010910118) ABHINAV N (1010910119)

Under the guidance of

Mrs. VASANTHI.PAssistant Professor (O.G)

DEPARTMENT OF CIVIL ENGINEERINGFACULTY OF ENGINEERING AND TECHNOLOGY

SRM UNIVERSITY

1

(Under section 3 of UGC Act, 1956)

SRM Nagar, Kattankulathur- 603203Kancheepuram District

MAY 2013PROJECT REPORT

on

PLANNING AND DESIGN OF NET ZERO ENERGYRESIDENTAL BUILDING

Submitted in partial fulfillment for the award of the degreeof

BACHELOR OF TECHNOLOGYin

CIVIL ENGINEERINGby

KARTHIK V (1010910090) SASIDHAR K.V (1010910092)

NEERAJ PORWAL (1010910118) ABHINAV N (1010910119)

Under the guidance of

Mrs. VASANTHI.PAssistant Professor (O.G)

2

DEPARTMENT OF CIVIL ENGINEERINGFACULTY OF ENGINEERING AND TECHNOLOGY

SRM UNIVERSITY(Under section 3 of UGC Act, 1956)

SRM Nagar, Kattankulathur- 603203Kancheepuram 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.

3

Signature of the Guide Signature of the HOD Mrs. VASANTHI .P Dr. R. ANNADURAIAssitant Professor (O.G) Professor & Head Department of Civil Engineering Department of Civil EngineeringSRM University SRM UniversityKattankulathur- 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.

4

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

5

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.

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

6

ABSTRACT iv

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

System 10

7

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

4.2.12 Design of Stair Case 51

4.3 DESIGN OF SOLAR PANEL AND ITS COMPONENTS 54

4.3.1 Solar power system components

8

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

9

LIST OF TABLES

TABLE TITLE

PAGE 1.1 Codes Used

3

1.2 Earlier Course Work Used

4

2.1 Front Setback

8

2.2 Rear Setback

9

10

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

11

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

4.7 U- Values 63

12

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

Asc - Area of steel in compression zone

Sv - Spacing of stirrups

fy - Yield stress of steel

Asv - Total cross sectional area of stirrup legs

kt - Modification factor for tension reinforcement

13

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

NBC - National Building Code

KKNP - Kudankulam Nuclear Power Plant

W.h - Watt hour

A.h - Ampere hour

14

CHAPTER 1

OVERVIEW

1.1 OBJECTIVE

15

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

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

16

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

1.6 REALISTIC DESIGN CONSTRAINTS

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.

17

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 -1,2Design loads for buildings and structures

(Dead load , Imposed load )

IS 456 :2000Design 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 Construction

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

18

Course Code and Name ContextCE 0104- Computer aided building drawing Computer aided building drawingCE0102- Elements of building science and

Architecturelayout and planning

CE0209- Building technologyByelaws, Setbacks, Open space, Floor area

ratioCE0303-Structural Design II R.C.C DesignCE0304-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

1.11 CONCLUSION

The two types of buildings are analyzed with respect to cost, time,

availability of skilled labour and ease in construction.

ELECTRICITY AVAILABILITY

OF

19

COST RESOURSES

NORMAL

CONVENTIONAL

BUILDING

LowIt requires an

active sourceEasily Available

NZERBHigh

Produced on its

ownDifficult

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.

CHAPTER 2

20

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.

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).

21

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).

2.3 DEVELOPMENT CONTROL RULES FOR CHENNAI METROPOLITAN AREA, 2004

2.3.1 Primary Residential Use Zone

22

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.

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

23

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.

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.

24

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.

CHAPTER 3

OBJECTIVE AND SCOPE

25

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

1Brick Material

Hollow brick Normal brick

2 Temperature4 to 5 degree less compared

To conventional buildingMore than NZERB

3 Electricity Produced on its ownIt requires an active

source

4 Initial Cost HighLess compared to

NZERB

5 Solar Panels 250 w panels Provided in NZERB Not provided

6 Energy Efficient Uses less energy Uses more energy

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.

26

http://en.wikipedia.org/wiki/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

v. Comparison of the NZERB building with other conventional building

CHAPTER 4

RESULTS AND DISCUSSIONS

4.1 PLANNING

27

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

Fig. 4.1 Key Plan

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

28

Fig.4.2 Ground Floor Plan

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.

29

The First Floor plan is shown in Figure 4.3

Fig.4.3 First Floor plan

4.2 ANALYSIS AND DESIGNS

SLAB DESIGN (Ref 6)

30

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 kN

m2 (Ref 7)

1. Effective Span

Lx = 3.26 m

Ly = 5.1 m

Aspect ratio = L yL x

= 5.13.26

= 1.56<2

Hence Two Way Slab

2. Load Calculation

Assuming Slab Thickness

d= L x32

=326032

=101.875 mm=100 mm

Assume 10∅ bar, Clear Cover 20mm

D=100+ 102

+20=125 mm=130 mm

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

Self Weight of aSlab= D1000

× 25= 1301000

× 25

¿3.25 kN

m2

Assume Floor Finish = 40 mm

∴Weight of Floor Finish = 0.04 × 24 = 0.96 kN

m2

Imposed Load = 2 kN

m2 (Ref 8)

31

∴Total Load = 6.21 kN

m2

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

m2

Consider 1m width of slab

∴Load per meter Length = 9.127 kNm

3. Finding Design Bending Moment

Refer Table 26, Page No.91 of IS456

Two adjacent edges are discontinuous

L yL x

=1.56 (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= 0.36 × X umax

d [ 1−0.42 × X umaxd ] f ck b d2

(4.2)

32

(or)

= 0.138fckb d2

Mu,lim is calculated by equation 4.2

Where,

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 = b df ck2 f y

[1 - √1-4.598 Rf ck

]

(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 = M u

bd2 = 6.731 ×10 61

105 2 ×1000 = 0.6105

Ast(+) Short = 1000 × 105 × (

202

) × 415 [1 - √1-4.598 ×0.6120

¿

= 184.27 mm2

Minimum Steel = 0.12% × D × B

33

Ast,min = (0.12100

)× 130 × 1000 = 156 mm2

Ast(+) Short <Ast,min

5. Check for maximum Spacing

i. 3d = 3 × 105 = 315 mm

ii. 300

Max Spacing = 300 mm

∴d for long span bars

d = D – Clear Cover – d2

- ∅

= 130 – 20 – 102

- 10 = 95 mm

6. Calculation of Ast for Long Span

Ast(+)

Long¿b df ck2 f y

[1 - √1- 4.598Rf ck

]

Same as equation 4.3

R = M ub

d2

= 3.662× 10 6

1000× 952

= 0.4057

Ast(+)

Long = 1000 × 95 × (202

)× 415 [1 – √4.598 ×0.4057

20]

= 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

34

Ast(+) Short = 116.37 mm2

Basic Value = 20

Fs = 0.58 × 415×184.27184.27

=240.2 N

mm2

Pt¿ A st ¿¿ =( 184.27 ×1001000 ×105

)

= 0.175%

Modification Factor = 1.62

Modified Basic Value = 20 × 1.62 = 32.8

Spand

= 3260105

= 31.047

31.047 < 32.8

∴ Hence its ok

4.2.2 Design of Bed Room

Using M20 Concrete

Fe415 steel

Live Load = 2 kN

m2

1. Effective Span

Lx = 3 m

Ly = 3.5 m

Aspect ratio =L yL x

= 3.5+ 0.233

+0.23=1.154<2Hence Two Way Slab

2. Load Calculation

Assuming Slab Thickness d= Lx32

=323032

=100 mm

Assume 10∅ bar, Clear Cover = 20mm

D=100+ 102

+20=125 mm

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

35

Self Weight of a Slab =D

1000×25= 125

1000×25=3.125

kNm

Assume 40 mm Floor Finish

∴Weight of Floor Finish = 0.04 X 24 = 0.96 kN

m2

∴Imposed Load = 2 kN

m2 Total Load =

6.08 kN

m2

Factored Load (Wu) = 1.5× 6.96 = 9.127 kN

m2


∴Load per meter Length = 9.127 kN

m2




L yL x

=1.154 ( 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


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

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


36


d [ 1−0.42 × X umaxd ] f ck b d2

(or)

Mu,lim = 0.138 fckbd2


Mu,lim = 0.138 × 20 × 1000 × 1002

Mu,lim = 27.6 kN

m2


∴ Hence its ok


Ast(+)


[1 - √1-4.598 Rf ck

]


R = M u

bd2 = 4.1 ×10 61

100 2× 1000 = 0.41

Ast(+) Short = 1000 ×100 ×

202

× 415 [1 - √1-4.598 ×0.4120

]

= 116.37 mm2


Ast,min = 0.12100

× 125 × 1000 = 150 mm2



i. 3d = 3 X 100 = 300 mm

ii. 300



37

d= D – Clear Cover – d2

- ∅

d= 125 – 20 – 10/2 - 10

d= 90 mm


Ast(+)

Long¿b df ck2 f y

[1 - √1- 4.598Rf ck

]


R = M u

bd2

= 3.33 ×10 6

1000× 1002

= 0.33

Ast(+)

Long = 1000 × 100 ×202

× 415[1– 4.598×0.3320 ]

= 93.2 mm2

Ast(+)

Long<Ast,min

7. Spacing for Steel

Ast¿π4

×102 = 78.5 mm2

Ast(+)

Short = 78.5

116.37× 1000 = 674.5 mm

Ast(+)

Long = 78.593.2

× 1000 = 842.27 mm




Basic Value = 20

Fs = 0.58 × 415 ×121.32121.32

=240.7N

mm 2

Pt ¿ A st ¿¿

38

= (1.22 X 10-3) ×100

= 0.122%


Modified Basic Value = 20 × 1.7 = 34

Span

d=3230

100=32.3 32.3 < 34

∴Hence its ok

4.2.3 Design of Bed Room

Using M20 Concrete

Fe415 steel

Live Load = 2 kN

m2

1. Effective Span

Lx = 3.85 m

Ly = 3.95 m


= 3.953.85

= 1.027<2

Hence Two Way Slab

2. Load Calculation


d ¿L x32

=385032

= 120.31mm = 120 mm

Assume 10∅ bar, Clear Cover 20 mm

D = 120+102

+20 = 145mm = 150 mm

∴ Actual Depth (d) = 150-5-20 = 125 mm

Self Weight of a Slab = D

1000× 25 =

1501000

×25

= 3.75 kN

m2

39



m2

Imposed Load = 2 kN

m2


m2


m2


Load per meter Length = 10.125 kNm




L yL x

= 1.027 (already found out)

Refer Table 26







Mu(+) Short = 0.048 × 10.125 × 3.852 = 7.203 kN.m

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



d [ 1−0.42 × X umaxd ] f ck b d2

(or)

40

= 0.138fckb d2


= 0.138 × 20 × 1000 × 1252

= 43.125 kN.m


∴ Hence its ok


Ast(+)


[1 - √1-4.598 Rf ck

]


R = M u

b d2 = 7.203 × 106 11252 ×1000

= 0.460

Ast(+)

Short = 1000 × 125 × (202

) × 415 [1 - √1-4.598 ×0.4620

]

= 164.08 mm2


Ast,min = (0.12100

)× 150 × 1000

= 180 mm2



i. 3d = 3 × 125 = 375 mm

ii. 300


∴ d for long span bars


- ∅

d= 150 – 20 – 102

- 10

d= 115 mm


41

Ast(+)

Long¿b df ck2 f y

[1 - √1- 4.598Rf ck

]


R = M u

bd2 = 7.603× 10 6

1000× 1152

= 0.553

Ast(+)

Long = 1000 × 115 × (202

) × 415 [1 – √4.598 ×0.553

20]

= 181.11 mm2

Ast(+)

Long<Ast,min


i. 3d = 3 × 115 = 345 mm

ii. 300

Spacing = 300 mm



Ast(+)

Short = 181.11 mm2

Basic Value = 20

Fs = 0.58 × 415 ×181.11181.11

=240.2 N

mm 2

Pt¿ A st ¿¿ = 181.11×100(1000 ×125 )

= 0.157%



Spand

= 3850125

= 30.8

30.8 < 35.6

∴ Hence its ok

42

4.2.4 Design of Bathroom

Using M20 Concrete

Fe415 steel

Live Load = 2 kN

m2

1. Effective Span

Lx = 2.38 m

Ly = 4.28 m


= 4.282.38

= 1.798<2

Hence Two Way Slab

2. Load Calculation


d = L x32

=238032

= 74.375 mm = 80 mm


D=80+102

+20=105 mm=110 mm

∴ Actual Depth d = 110-5-20 = 85 mm


1000×25= 110

1000×25

= 2.75 kN

m2



m2

Imposed Load = 2 kN

m2


m2

43


m2






L yL x


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



d [ 1−0.42 × X umaxd ] f ck b d2

(or)

= 0.138fckb d2


= 0.138 × 20 × 1000 × 852

=19.94 kN.m


∴ Hence its ok

44


Ast(+)


[1 - √1-4.598 Rf ck

]


R = M u

b d2 = 4.1527 ×106

(85 2 ×1000) = 0.574

Ast(+) Short = 1000 × 85 × (

202

) × 415 [1 - √1-4.598 ×0.5720

]

= 139.92 mm2


Ast,min= (0.12100

) × 110 × 1000 = 132 mm2



i. 3d = 3 × 855 = 255 mm

ii. 300




- ∅

d= 110 – 20 – 102

- 10

d= 75 mm


Ast(+)

Long¿b df ck2 f y

[1 - √1- 4.598Rf ck

]


R = M u

bd2

= 2.29 × 1061000× 75 2

= 0.4082

45

Ast(+) Long = 1000 × 75 × (

202

) × 415 [1 – √4.598 ×0.4082

20¿= 86.88 mm2

Ast(+) Long <Ast,min


i. 3d = 3 × 75 = 225 mm

ii. 300

MaxSpacing = 225 mm




Basic Value = 20

Fs = 0.58 × 415 ×139.92139.92

= 240.2 N

mm 2

Pt¿ A st ¿¿= (139.92 ×100)(1000 ×85)

= 0.1646%



Spand

= 2380

85 = 28

28 < 38

∴ Hence its ok

4.2.5 Design of Portico

Using M20 Concrete

Fe415 steel

Live Load = 2 kN

m2

1. Effective Span

Lx = 3.78 m

Ly = 6.93 m

46


=6.933.78

= 1.83<2

Hence Two Way Slab

2. Load Calculation


d ¿L x32

=378032

= 118.124 mm = 120 mm


D = 120+10

2+20= 150 mm

∴ Actual Depth (d) = 150-5-20 = 125 mm


1000×25= 150

1000×25

= 3.75 kN

m2



m2

Imposed Load = 2 kN

m2


m2


m2






47

L yL x

=1.83 (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



d [ 1−0.42 × X umaxd ] f ck b d2

(or)

= 0.138fckb d2


= 0.138 × 20 × 1000 × 1252

= 43.125 kN.m


∴ Hence its ok


Ast(+)


[1 - √1-4.598 Rf ck

]


R = M u

bd2 = 12.58 ×106

1252 ×1000 = 0.805

Ast(+) Short = 1000 × 125 ×

202× 415

[1 - √1-4.598 ×0.820

]

= 184.27 mm2

48


Ast,min = (0.12100

)× 130 × 1000

= 156 mm2



i. 3d = 3 ×105 = 315 mm

ii. 300




- ∅

d= 130 – 20 – 10/2 - 10

d= 95 mm


Ast(+)

Long¿b df ck2 f y

[1 - √1- 4.598Rf ck

]


R = M u

b d2

= 3.662× 10 6952× 1000

= 0.4057

Ast(+)

Long = 1000 × 95 × (202

)× 415 [1 – √4.598 ×0.4057

20]

= 109.37 mm2

Ast(+)

Long<Ast,min


i. 3d = 3 × 95 = 285 mm

ii. 300

Spacing = 285 mm

49




Basic Value = 20

Fs = 0.58 × 415×184.27184.27

= 240.2 N

mm2

Pt¿ A st ¿¿ = 184.27 ×100(1000× 105 )

= 0.175%


Modified Basic Value = 20 × 1.62

=32

Spand

= 3260105

= 31.047

31.047 < 32.8

∴ Hence its ok

4.2.6 Design of Kitchen

Using M20 Concrete

Fe415 steel

Live Load = 2 kN

m2

1. Effective Span

Lx = 2.23 m

Ly = 3.73 m

Aspect ratio ¿L yL x

=3.73+ 0.232.23

+0.23=1.67<2

Hence Two Way Slab

2. Load Calculation


50

d = L x32

=223032

= 65 mm


D = 65+102

+20 = 90 mm

∴ Actual Depth (d) = 90-5-20 = 65 mm


1000×25= 90

1000×25

= 2.25 kN

m2


Weight of Floor Finish = 0.04 × 24 = 0.96 kN

m2

∴Imposed Load = 2 kN

m2


m2


m2



Finding Design Bending Moment



L yL x


Refer Table 26



51





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

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



d [ 1−0.42 × X umaxd ] f ck b d2

(or)

Mu,lim = 0.138 fckb d2


= 0.138 × 20 × 1000 × 652

=11.66 kN.m


∴ Hence its ok


Ast(+)


[1 - √1-4.598 Rf ck

]


R = M u

bd2 = 2.735 ×10 6

100 0× 602 = 0.55

Ast(+) Short = 1000 × 65 ×

202

× 415 [1 - √1-4.598 ×0.5520

]

= 103.56 mm2


Ast,min¿0.12100

× 90 × 1000

= 108 mm2


52


i. 3d = 3 × 65 = 195 mm

ii. 300




- ∅

d= 90 – 20 – 102

- 10

d= 55 mm


Ast(+)

Long¿b df ck2 f y

[1 - √1- 4.598Rf ck

]


R = M u

bd2 = 1.373 ×10 6

1000× 552

= 0.115

Ast(+) Long =1000 × 55 ×

202

× 415 [1 – 4.598 ×0.115

20]

= 71.05 mm2



Ast =π4

× 102 = 78.5 mm2

Ast(+) Short =

78.5103.56

× 1000 = 758.01 mm

Ast(+) Long =

78.571.05

× 1000 = 1104.8 mm




53

Basic Value = 20

Fs = 0.58 × 415 ×103.56103.56

= 240.7 N

mm2

Pt¿ A st ¿¿ = 103.561000

× 65

= 0.16%



Span

d=2230

65=34.3

34.3 < 36

∴ Hence its ok

2.4.7 Design of Dinning Room

Using M20 Concrete

Fe415 steel

Live Load = 2 kN

m2

1. Effective Span

Lx = 2.6 m

Ly = 3.73 m

Aspect ratio ¿L yL x

=3.73+ 0.232.6

+0.23=1.43<2

Hence Two Way Slab

2. Load Calculation


d = L x32

=260032

=80 mm


D ¿8 0+ 102

+20=105 mm

54

∴ Actual Depth (d) = 105-5-20 = 80 mm

Self Weight of a Slab= D1000

×25= 1051000

×25

= 2.625 kN

m2



m2

Imposed Load = 2 kN

m2


m2


m2





One edge discontinuous

L yL x


Refer Table 26







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

55

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



d [ 1−0.42 × X umaxd ] f ck b d2

(or)

Mu,lim = 0.138fckb d2


Mu,lim = 0.138 × 20 × 1000 × 802

Mu,lim =17.66 kN.m


∴ Hence its ok


Ast(+) Short = b d

f ck2 f y

[1 - √1-4.598 R

f ck]


R = M u

bd2 = 2.79 ×106

1000× 802 = 0.435

Ast(+) Short = 1000 × 80 ×

202

× 415 [1 - √1-4.598×0.435

20]

= 99.15 mm2


Ast,min= 0.12100

× 105 × 1000

= 126 mm2



i. 3d = 3 × 80 = 240 mm

ii. 300



56


- ∅

d= 105 – 20 – 102

- 10

d= 70 mm


Ast(+)

Long¿b df ck2 f y

[1 - √1- 4.598Rf ck

]


R = M u

bd2

= 1.595 ×10 6

1000× 702

= 0.325

Ast(+) Long =1000 ×70 ×

202

× 415¿ – 4.598 ×0.325

20]

= 64.34 mm2


7. Spacing for Steel

Ast =π4

× 102 = 78.5 mm2

Ast(+) Short =

78.599.15

× 1000 = 791.7 mm

Ast(+) Long =

78.564.34

× 1000 = 1200.08 mm




Basic Value = 20

Fs = 0.58 × 415×99.1599.15

=240.7N

mm 2

Pt¿ A st ¿¿

57

= 99.15

1000× 70

= 0.14%



Spand

=260080

= 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 kN

m2

2. Assumptions

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 ¿ht= 5.8

0.23=25.21

4. Shape modification factor:

58

Crushing Strength of Modular Brick = 5 kN

m2

Hw

= 75115

=0.652

∴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

Kp= Shape modification factor

Fc = 0.44 × 0.48 × 1 × 1

=0.21Nm ²

8. Safe Load

Q = ( f c1

+6 et ) A

=0.21× 1000 ×300

1

= 63 kNm

9. Wall Area

Outer wall = Total Perimeter x 3(floor height)

59

= ((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 = 23.8167.25

×100

= 35.4 %

∴Thickness = 1 Brick thick wall (using nomograms)

11. For Hall :

Wu= Wl x

6× {3- [(

l xl y

)]2} (4.5)

Wu is calculated by equation 4.5

Where,

Wu= Factored load

W=Load from the slab

Lx=Short span

Ly=Long span

Wu =¿}

Wu = 13.11× 103 kNm

For bed room:

60

Wu= Wl x

6× {3−[( l x

l y)] ²}

= (9.127 ×(3+0.23)¿ ¿6 × {3- [ 3+0.23

3.5+0.23 ]2}

= 11.055× 103 kNm

For dinning room:

Wu= Wl x

6× {3- [( lx

ly )]2}

= (8.43 × (2.37+0.23 ))

6× {3- [

(2.37+0.23)(3.5+0.23)

]2}

= 9.81 × 103 kNm

Total : 13.11 + 11.055 + 9.81 = 33.98 × 103 kNm

63>33.98 kNm

Hence the design is ok

4.2.9 Design of Footing

Load from Walls = 126.7 kNm

10% for the weight of the Building = 63+6.3 = 70 kNm

1. Area of Footing =LoadSBC

Assume SBC = 150kN

m2

A = 70

150 = 0.47 m2

61

Consider 1m Length room

Breadth of the Footing Required = Area

L = 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 = 300

2 = 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 kN

m2

Step 2: Assumptions

Height of the Plinth from ground = 0.5 m

Height of the Plinth above Footing = 1 m

Height of the Parapet Wall = 1 m

62

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)

Wall M = 5.07 m (continuous on one end & discontinuous on other end)

Step 3: Slenderness ratio and stress factor

Ground floor:

H = 2.6+0.6+1= 4.2 m

63

Effective height = 0.75 × H = 3.15 m

Slenderness ratio = Ht

= 3.150.2

= 15.75

Step 4: Shape modification factor

Crushing Strength of Hollow Brick= 4.1 N

mm2

HW

= 2020

= 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 N

mm2

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

Step 7: Slenderness ratio and stress factor

First floor:

H = 2.6+0.8 = 3.4 m

Effective height = 0.75 × H = 2.55 m

64

Slenderness ratio = Ht

= 2.550.2

= 12.75

Step 8: Shape modification factor

Crushing Strength of Hollow Brick = 4.1N

mm2

= HW

= 2020

= 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 N

mm2

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)

Brickwork Ground floor First Floor

65

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

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

Table 4.3 Calculation of permissible stress

Fc=ks×kp×ka× basic compressive stress

Wall typePermissible stress- ground

floor(N/mm2)Permissible stress- first

floor(N/mm2)A 0.528×0.74=0.390 0.528×0.81=0.427B 0.528×0.81=0.427 0.528×0.81=0.427C 0.528×0.74=0.390 0.528×0.81=0.427

66

D 0.528×0.75=0.396 0.528×0.81=0.427E 0.528×0.89=0.469 0.528×0.89=0.469F 0.528×0.83=0.4382 0.528×0.83=0.4382G 0.528×0.74=0.390 0.528×0.81=0.427H 0.528×0.75=0.396 0.528×0.81=0.427I 0.528×0.75=0.396 0.528×0.81=0.427J 0.528×0.74=0.390 0.528×0.81=0.427K 0.528×0.75=0.396 0.528×0.81=0.427L 0.528×0.88=0.4646 0.528×0.88=0.4646M 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×2×105 kN/m(ground floor) q = fc×2×105 kN/m(first floor)A 78 85.4B 85 85.4C 78 85.4D 79.2 85.4E 93 93.8F 87.6 87.6G 78 85.4H 79.2 85.4I 79.2 85.4J 78 85.4K 79.2 85.4L 92 92M 78 85.4

4.2.11 Design of Footing for Hollow Brick wall: (Ref 11)

Load from Wall = 78 kNm

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 kNm

67

=100 kNm

Factored load= 1.25 × 100=125 kNm

1. Area of Footing = LoadSBC

= 12550

= 0.833 m2

Assume SBC = 150 kN

m2

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

= 300

2 = 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

These depth has to be Provided by means of series steps

The thickness of each step is given by hollow brick = 200 mm

The footing design is shown in the Figure 4.4

68

Fig.4.4 Footing Design

4.2.12 Design of Stair Case:

Length = 4 m

Live load = 2 kN

m2

Rise = 150 mm

Thread = 250 mm

Using M20 Concrete and Fe415

Step1: Calculation of self weight

Assume waist slab thickness = 4000

20 = 200 mm

D = 200 mm

Self weight = D

1000× √R2+T 2

T × 2 = 5.83

kN

m2 (4.6)

Self weight is calculated by equation 4.6

Where,

D =Diameter

69

R =Rise

T =Thread

Step 2: Calculation of load on waist slab

1. Assume 40 mm Floor finish

Floor Finish = 40

1000 × 24 = 1

kN

m2

2. Weight of steps= 12

× R × T × 1T

× 25 = 1.875 kN

m2

3. Live load = 3 kN

m2

4. Self weight = 5.83 kN

m2

Wu = 11.075 kN

m2

Wu=1.5 × 11.075 =17.55 kN

m2

Step3: Calculation of Mu

Mu = 17.55× 16

8 = 35 kN.m

Mu,lim = 0.36 × X umax

d [ 1−0.42 × X umaxd ] f ck b d2

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

70

Step 4: Calculation of Ast

R = 35 ×106

1000× 120 ×120 = 2.43

Ast = 1000× 120 ×20

2× 415 = 970.6 mm2

Number of Bars = 970.6

π4

× 202 = 4 bars

Ast actual = 4 × π4

× 202 = 1256 mm2

Pt = Astbd

× 1000 = 1256

1000× 120 × 1000 = 1.04%

Step 5: Check for deflection

Basic value = 20

Fs = 0.58fy Ast reqiired

Ast provided = 0.58 × 415 ×

9701256

= 185.89 N

mm2

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

drequired = Ast reqiired

Ast provided =

40001.2× 32

= 104.166 mm

dactual = 120 mm

Step 6: Providing distribution steel

Astmin= 0.12bd

100×

0.12× 1000 ×150100

= 480 mm2

Spacing of 8mm diameter = 1000480

× π4

× 82 = 270 mm

Main steel = 4No.s 20 bars

Distribution = 8 mm dia bars @ 270 mm c/c

71

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

72

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.

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

73

http://www.dc-ac-power-inverter.co.za/

http://www.solarpanel.co.za/solar-regulator.htm

http://www.solarpanel.co.za/solar-batteries.htm

http://www.solarpanel.co.za/photovoltaic-panel-retailers2.htm

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.

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

74

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

The calculation of loads for the Solar Panels are given below in Table 4.5

Table 4.5 Calculation Of Loads (Ref 12)

PARTICULARS ITEMS UNITSUSAGE IN HRS

VOLTAGE W

CONSUMPTION

INVERTORS

HALLCFL (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

EXHAUST 1 4 50 200 50

Mixer 1 1 450 450 450

DINING ROOMAUTO-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

75

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 = 1673250

=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

40450 A

= 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

iii. x 105A.H deep cycle batteries( 3 in series)

iv. 1 x 3500W modified sine wave power inverter

4.4 RATE ANALYSIS

Solar panels =Rs.32/W

Regulator = Rs 1800

Batteries = Rs 8000/series

Inverter = Rs 4800

Total Cost

Solar panels =14x250x32=Rs 112000

Regulator = Rs 1800

Batteries = Rs 8000x3=24000=Rs 24000

Inverter = Rs 4800

Total=112000+1800+24000+4800= Rs. 142600/-

The total cost of the solar panel is Rs. One lakh forty two thousand six hundred for

our residential building .In these solar panel cost is based on the solar panels, regulator,

batteries and inverter. The output of solar panel can be expected to vary by 0.25% for every 5

degrees variation in temperature.

In NZERB, decrease in temperature for using of hollow bricks and solar panels

produces the electricity. When compared to conventional building, the intial cost is high but

in future the electricity cost is reduced.

4.5 INFRARED THERMOMETER

ii

Fig.4.6 Infrared Thermometer

The instrument Infrared Thermometer is shown in Figure 4.6

I. An infrared thermometer is a thermometer which infers temperature from a

portion of the thermal radiation sometimes called blackbody radiation emitted

by the object being measured.

II. They are sometimes called laser thermometers if a laser is used to help aim the

thermometer, or non-contact thermometers or temperature guns, to describe

the device's ability to measure temperature from a distance refer Figure.4.6

III. By knowing the amount of infrared energy emitted by the object and

its emissivity, the object's temperature can often be determined.

IV. Infrared thermometers are a subset of devices known as "thermal radiation

thermometers".

V. The most basic design consists of a lens to focus the infrared thermal radiation

on to a detector, which converts the radiant power to an electrical signal that

can be displayed in units of temperature after being compensated for ambient

iii

temperature. This configuration facilitates temperature measurement from

a distance without contact with the object to be measured.

VI. Infrared thermometers can be used to serve a wide variety of temperature

monitoring functions. A few examples provided to this article include:

VII. Detecting clouds for remote telescope operation

VIII. Checking mechanical equipment or electrical circuit breaker boxes or outlets

for hot spots

IX. Checking heater or oven temperature, for calibration and control purposes

X. Detecting hot spots / performing diagnostics in electrical circuit board

manufacturing

XI. Checking for hot spots in fire fighting situations

XII. Monitoring materials in process of heating and cooling, for research and

development or manufacturing quality control situations

XIII. The distance-to-spot ratio (D:S) is the ratio of the distance to the object and

the diameter of the temperature measurement area. For instance if the D:S

ratio is 12:1, measurement of an object 12 inches (30 cm) away will average

the temperature over a 1-inch-diameter (25 mm) area. The sensor may have an

adjustable emissivity setting, which can be set to measure the temperature of

reflective (shiny) and non-reflective surfaces.

XIV. The most common infrared thermometers is the:

XV. Spot Infrared Thermometer or Infrared Pyrometer, which measures the

temperature at a spot on a surface (actually a relatively small area determined

by the D:S ratio).

iv

4.6 CHARACTERISTICS OF HOLLOW BRICKS

4.6.1 Parameters of Hollow Brick Used In Net Zero Energy Residential Building

I. LENGTH : 400 mm

II. WIDTH : 200 mm

III. HEIGHT: 200 mm

IV. WEIGHT: 11.1 kg

V. DENSITY: 694 kg/m³

VI. COMPRESSIVE STRENGTH : 4.1 N

mm2

VII. WATER ABSORPTION : 15%

VIII. U-VALUE : 1.1 W/m²

IX. SOUND INSULATION : 46 DB

X. FIRE RESISTANCE 240 min

Available Sizes

I. 400 X 200 X 200 mm

II. 400 X 150 X 200 mm

III. 400 X 100 X 200 mm

IV. 200 X 200 X 200 mm

V. 200X 150 X 200 mm

VI. 200 X 100 X 200 mm

Hollow Brick

Bigger Size

I. Hollow brick is same in size as that of concrete blocks

II. 1 Hollow brick = 9 Clay Bricks

III. Less mortar joints, hence less plumb & alignment

IV. Faster construction

Light Weight

I. Ease of handling, Transportation

II. Saves labour

III. Less dead load, Savings in Structural Cost

v

IV. (Steel & Concrete) by 10 to 15%

Thermal Insulation

I. Savings on mortar

II. Low ‘U’ Values – 1.0 W/m²

III. Better Thermal Insulation = less energy loss through walls

IV. Savings on Energy consumption ,Comfortable inside temperature

U-value determines thermal Insulation.Lesser the Value higher the Insulation and vice

versa. U-values are mentioned in Figure 4.7

Fig.4.7 U-VALUES

4.6.2 Advantages Of Hollow Bricks

i. Highly Durable: The good concrete compacted by high pressure and

vibration gives substantial strength to the brick.

vi

ii. Low Maintenance, Colour and brilliance of masonry withstands

outdoor elements.

iii. Fire Resistant

iv. Provide thermal and sound insulation: The air in hollow of the brick,

does not allow outside heat or cold in the house. So it keeps house cool

in summer and warm in winter.

v. Environment Friendly, fly ash used as one of the raw materials.

Constructional Advantages

i. No additional formwork or any special construction machinery is

required for reinforcing the hollow brick masonry.

ii. Only skilled labour is required for this type of construction.

iii. It is a faster and easier construction system, when compared to the

other conventional construction systems.

iv. Savings on RCC-Frame structure (steel/concrete).

v. Faster construction and ease for handling at site.

vi. Hollow brick consist of four elements earth, water, fire, air which

makes easy work of construction.

vii. Less wastage through half bricks.

viii. Excellent thermal insulation.

ix. Reduction in energy consumption.

x. World class and best walling materials.

4.7 ESTIMATION

4.7.1 Abstract estimate of conventional building

The quantities of the various materials in conventional building are calculated as

shown in the Table 4.6

The abstract estimate of conventional building is given in Table 4.6

Table 4.6 Abstract estimate of conventional building

S.No Description NosLengt

h(m)

Breadth

(m)

Depth(m)

Quantity

(m3)

vii

1 Excavation

Exterior wall 1 39.48 0.76 0.53 15.9Interior wall 1 22.51 0.76 0.53 9.06 24.96

2 P.C.CExterior wall 1 39.48 0.76 0.3 9Interior wall 1 22.51 0.76 0.3 5.13 14.13

3 Brick work1st FootingExterior wall 1 39.48 0.46 0.115 2.08Interior wall 1 22.51 0.46 0.115 1.192nd FootingExterior wall 1 39.48 0.31 0.115 1.4Interior wall 1 22.51 0.31 0.115 0.8

3 WallExterior wall 1 39.48 0.23 7.5 68.1Interior wall 1 22.51 0.23 7.5 38.82 112.39DeductionsWindow W 1×2 1.22 0.23 0.9 0.51Window W1 1×5 1.22 0.23 1.22 1.71Window W2 1×2 0.9 0.23 1.21 0.5Door D 1×2 0.84 0.23 2.1 0.81Door D1 1×3 0.75 0.23 2 1.035Door D2 1 0.9 0.23 2.1 1.035Spacing S1 1 1.2 0.23 2.08 0.57Spacing S2 1 0.9 0.23 2.1 0.43 6.6

4 Earth FillingHall 1 3.03 4.87 0.5 7.378Water closet 1 3.62 2.15 0.5 3.89Bed Room1 1 3.62 3.72 0.5 6.73Bed Room2 1 3.5 3 0.5 5.25Dinning 1 3.5 2.37 0.5 4.147Kitchen 1 3.5 2.6 0.5 4.55 31.649

5 Flooring ConcreteHall 1 3.03 4.87 0.1 1.48Water closet 1 3.62 2.15 0.1 0.778Bed Room1 1 3.62 3.72 0.1 1.346Bed Room2 1 3.5 3 0.1 1.05Dinning 1 3.5 2.37 0.1 0.829Kitchen 1 3.5 2.6 0.1 0.91 5.473

6 R.C.C

viii

Lintel & Sun shadesDoor D 1×2 1.4 0.23 0.15 0.078Door D1 1×3 1.05 0.23 0.15 0.108Door D2 1 1.2 0.23 0.15 0.04Sun Shade 1 1.2 0.45 0.075 0.04Window W 1×2 1.52 0.23 0.15 0.104Sun Shade 1×2 1.52 0.45 0.075 0.103Window W1 1×5 1.52 0.23 0.15 0.2622Sun Shade 1×5 1.52 0.45 0.075 0.26Window W2 1×2 1.2 0.23 0.15 0.083Sun Shade 1×2 1.2 0.45 0.075 0.081 1.1592Roof SlabHall, Water Closet, Bed Room1 1 6.65 4.87 0.1 3.24Bed Room2,Dinning,Kitchen 1 3.5 7.97 0.1 2.79 6.8496

7 PlasteringExterior wall 1 39.48 - 7.5 296.1Interior wall 1 22.51 - 7.5 168.8 464.92DeductionsWindow W1 1×2 1.22 - 0.9 2.196Window W2 1×5 1.22 - 1.22 7.442Window W3 1×2 0.9 - 1.21 2.178Door D 1×2 0.84 - 2.1 3.528Door D1 1×3 0.75 - 2 4.5Door D2 1 0.9 - 2.1 1.89Spacing S 1 1.2 - 2.08 2.496Spacing S1 1 0.9 - 2.1 1.89 26.12

8 White Washing - - - - 438.89 Colour Washing - - - - 438.8

The abstract estimate of NZERB is given in Table 4.7

Table 4.7 Abstract Estimate of NZERB

S.No Description NoSLengt

h(m)

Breadth

(m)

Depth(m)

Quantity

(m3)1 Excavation Exterior wall 1 39.24 0.7 0.5 13.734 Interior wall 1 22.51 0.7 0.5 7.87 21.6042 P.C.C

ix

Exterior wall 1 39.24 0.7 0.3 8.24 Interior wall 1 22.51 0.7 0.3 4.72 12.963 Brick work 1st Footing Exterior wall 1 39.24 0.4 0.2 3.13 Interior wall 1 22.51 0.4 0.2 1.8 2nd Footing Exterior wall 1 39.24 0.25 0.2 1.96 Interior wall 1 22.51 0.25 0.2 1.123 Wall Exterior wall 1 39.24 0.2 7.5 58.86 Interior wall 1 22.51 0.2 7.5 33.76 Deductions Window W 1×2 1.22 0.2 0.9 0.4392 Window W1 1×5 1.22 0.2 1.22 1.4884 Window W2 1×2 0.9 0.2 1.21 0.432 Door D 1×2 0.84 0.2 2.1 0.705 Door D1 1×3 0.75 0.2 2 0.9 Door D2 1 0.9 0.2 2.1 0.378 Spacing S1 1 1.2 0.2 2.08 0.504 Spacing S2 1 0.9 0.2 2.1 0.378 95.024 Earth Filling Hall 1 3.03 4.87 0.5 7.378 Water closet 1 3.62 2.15 0.5 3.89 Bed Room1 1 3.62 3.72 0.5 6.73 Bed Room2 1 3.5 3 0.5 5.25 Dinning 1 3.5 2.37 0.5 4.147 Kitchen 1 3.5 2.6 0.5 4.55 31.6495 Flooring Concrete Hall 1 3.03 4.87 0.1 1.48 Water closet 1 3.62 2.15 0.1 0.778 Bed Room1 1 3.62 3.72 0.1 1.346 Bed Room2 1 3.5 3 0.1 1.05 Dinning 1 3.5 2.37 0.1 0.829 Kitchen 1 3.5 2.6 0.1 0.91 5.4736 R.C.C Lintel & Sun shades Door D 1×2 1.4 0.23 0.15 0.0684 Door D1 1×3 1.05 0.23 0.15 0.0945 Door D2 1 1.2 0.23 0.15 0.036 Sun Shade 1 1.2 0.45 0.075 0.0405 Window W 1×2 1.52 0.23 0.15 0.07512

x

Sun Shade 1×2 1.52 0.45 0.075 0.1026 Window W1 1×5 1.52 0.23 0.15 0.228 Sun Shade 1×5 1.52 0.45 0.075 0.0256 Window W2 1×2 1.2 0.23 0.15 0.072 Sun Shade 1×2 1.2 0.45 0.075 0.081 0.8237 Roof Slab

Hall, Water Closet, Bed Room1 1 6.65 4.87 0.1 3.24

Bed Room2,Dinning,Kitchen 1 3.5 7.97 0.1 2.79 6.84967 Plastering Exterior wall 1 39.24 - 7.5 294.3 Interior wall 1 22.51 - 7.5 168.82 463.12 Deductions Window W1 1×2 1.22 - 0.9 2.196 Window W2 1×5 1.22 - 1.22 7.442 Window W3 1×2 0.9 - 1.21 2.178 Door D 1×2 0.84 - 2.1 3.528 Door D1 1×3 0.75 - 2 4.5 Door D2 1 0.9 - 2.1 1.89 Spacing S 1 1.2 - 2.08 2.496 Spacing S1 1 0.9 - 2.1 1.89 26.128 White Washing - - - - 4379 Colour Washing - - - - 437

xi

4.7.3 Rate Analysis

The rate analysis for various description of work are calculated based on the PWD.

The rate analysis proposed for conventional building is given in the Table 4.8

Table 4.8 Rates Proposed Conventional Building

S.NO

DESCRIPTION OF WORKQTY in

m³QTY in

cftRATE

PER

AMOUNT

1 Earth Work Excavation 24.96 882.33 9.50 Cft 8382.00

2 Sand Filling with good river sand 31.65 1118.82 35.00 Cft 39158.00

3 PCC 1:5:10, 14.13 499.51 90.00 Cft 44955.00

4 Brick Work in C.M. 1:5, using country

brick For Basement level 112.39 3973.00 90.00 Cft 357570.00

5 Flooring Work

PCC 1:4:8 5.50 194.23 90.00 Cft 17480.00

6R.C.C (LINTEL,SUNSHADES & ROOF SLAB) 7.19 253.91 350.00 Cft 88868.50

7 Plastering in C.M 1:4, Inside and outside wall surface 439.00

15518.60 30.00 Sft 465559.00

8White washing 439.00

15518.60 3.00 Sft 46555.80

9Colour washing 439.00

15518.60 5.00 Sft 77593.00

10Steel 501kg 60.00 kg 30060.00

TOTA

L

1176181.30

xii

The rate analysis proposed for NZERB building is given in the Table 4.9

Table 4.9 Proposed NZERB Building

S.NO

DESCRIPTION OF WORKQTY in

m³QTY in cft RATE

PER

AMOUNT

1 Earth Work Excavation 21.60 763.56 9.50 Cft 7253.82

2Sand Filling with good river sand 31.65 1118.80 35.00 Cft 39158.00

3 PCC 1:5:10, 12.96 458.10 90.00 Cft 41229.00

4 Brick Work in C.M. 1:5, using

country brick For Basement level 95.50 3376.00 125.00 Cft 422000.00

5 Flooring Work

PCC 1:4:8 5.50 194.23 90.00 Cft 17480.50

6R.C.C (LINTEL,SUNSHADES & ROOF SLAB) 6.84 241.55 350.00 Cft 84542.50

7 Plastering in C.M 1:4, Inside and outside wall surface 463.12 16371.29 30.00 Sft 491139.00

8White washing 437.00 15448.00 3.00 Sft 46344.00

9Colour washing 437.00 15448.00 5.00 Sft 77240.00

10Steel 501.00 60.00 kg 30060.00

11Solar Panel System 142600.00

TOTA

L

1399046.82

CHAPTER 5

xiii

CONCLUSION

5.1 CONCLUSION

In this project we has completed the design of the Conventional building by using

modular bricks and Net Zero Energy Residential Building by using Hollow Brick .The plan

of the building was prepared by Auto-Cad software. IS 456:2000 code book was used to

design Slab and Footing. Design of wall was done by using IS 1905:1987.

The Comparison of the Conventional Building and NZERB was completed by using

the parameters such as the temperature by using instrument infrared thermometer which was

found to be 4oC less in NZERB compared to conventional building under same condition.

Hence by using the renewable resources the impact on the active energy loads can be

reduced, Thus we can conserve electricity locally and globally.

5.2 FUTURE SCOPE OF THE PROJECT

The building designed as a NET ZERO ENERGY BUILDING produces its own

electricity, thus it can save a huge amount in electricity bill. These kind of buildings are

environmental friendly reducing the environmental hazards (eg. It would release zero carbon

content that would help in controlling global warming).The design for the building should be

such that the requirement of temperature regulation does not fluctuate throughout the year.

REFERENCES

1. http://en.wikipedia.org/wiki/Zero-energy_building

2. http://energy.gov/energysaver/articles/annajohanna

3. http://zeb.buildinggreen.com/saitoh

xiv

http://zeb.buildinggreen.com/saitoh

http://energy.gov/energysaver/articles/annajohanna

http://en.wikipedia.org/wiki/Zero-energy_building

4. National Building Code of India (NBC) and Chennai Metropolitan

Development Authority (CMDA).

5. S.P. Arora and S.P Bindra .(2010), Building Construction , Fifth edition,

Dhanpat Rai publishing company limited, New Delhi.

6. IS: 456 : 2000, Indian Standard Code of practice for plain and reinforced

concrete (Fourth Revision ), Bureau if Indian Standards, New Delhi

7. IS: 1905 (1987), Code of Practice for Structural use of unreinforced masonry.

8. IS 875 : Part 2 : 1987 Code of practice for design loads (other than

earthquake) for buildings and structures: Part 2 Imposed loads

9. IS 875 : Part 1 : 1987 Code of practice for design loads (other than

earthquake)for buildings and structures Part 1 Dead loads - Unit weights of

building material and stored materials (Incorporating IS:1911-1967)

10. SP 20 (S & T):1991 Handbook on masonry design and construction.

11. IS 2572:1963(R 1997) Code of practice for design of Hollow bricks

12. http://www.solarpanel.co.za/solar-power-calculator.htm

xv

http://www.solarpanel.co.za/solar-power-calculator.htm

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