civil-Net zero residential building.docx

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PROJECT REPORT on

PLANNING AND DESIGN OF NET ZERO ENERGY RESIDENTIAL BUILDING Submitted in partial fulfillment for the award of the degree of

BACHELOR OF TECHNOLOGY in

CIVIL ENGINEERING by

KARTHIK V SASIDHAR K.V NEERAJ PORWAL ABHINAV N

(1010910090) (1010910092) (1010910118) (1010910119)

Under the guidance of

Mrs. VASANTHI.P Assistant Professor (O.G)

DEPARTMENT OF CIVIL ENGINEERING FACULTY OF ENGINEERING AND TECHNOLOGY

1


SRM UNIVERSITY (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 SASIDHAR K.V NEERAJ PORWAL ABHINAV N

(1010910090) (1010910092) (1010910118) (1010910119)

Under the guidance of

Mrs. VASANTHI.P Assistant Professor (O.G)

2


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),

REDDY.K.V(1010910092),

NEERAJ

PORWAL

SASIDHAR (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 Assitant Professor (O.G) Department of Civil Engineering Engineering SRM University Kattankulathur- 603203

Dr. R. ANNADURAI Professor & Head Department of Civil SRM University Kattankulathur- 603203

INTERNAL EXAMINER EXAMINER

EXTERNAL

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 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.

5


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

1

TITLE

PAGE

ABSTRACT

iv

ACKNOWLEDGEMENT

v

LIST OF TABLES

ix

LIST OF FIGURES

x

ABBREVATIONS

xi

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

6


1.12 2

FUTURE SCOPE

5

INTRODUCTION

6

2.1

GENERAL

6

2.2

LITERATURE REVIEW

7

2.3

DEVELOPMENT CONTROL RULES FOR CHENNAI

2.4

METROPOLITAN AREA, 2004

8

2.3.1

8

Primary Residential Use Zone

CONFORMATION TO NATIONAL BUILDING CODE OF INDIA 2.4.1 2.4.2

3

4

9

Fire Safety, Detection And Extinguishing System

10

Security Deposits

10

OBJECTIVE AND SCOPE

11

3.1

OBJECTIVE

11

3.2

SCOPE

12

3.3

MATERIALS AND METHODOLOGY

12

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

16

Design of Hall 4.2.2

Design of Bedroom

20 4.2.3

Design of Bedroom

23

4.2.4

Design of Bathroom

27

7


4.3

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

51

Design of Stair Case

DESIGN OF SOLAR PANEL AND ITS COMPONENTS 4.3.1

54

54

Solar power system components 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

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

4.3.4

4.7

ESTIMATION

65

4.7.1

65

Abstract Estimate of Conventional Building 4.7.2 Abstract Estimate of NZERB

67 4.7.3 5

Rate Analysis

70

CONCLUSION

72 8


5.1

CONCLUSION

72

5.2

FUTURE SCOPE

72

REFERENCES

73

LIST OF TABLES TABLE PAGE

TITLE 1.1

Codes Used 3

1.2

Earlier Course Work Used

4 9


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

10


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

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

11


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

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

12


PWD

-

Public works department

NBC

-

National Building Code

KKNP

-

Kudankulam Nuclear Power Plant

W.h

-

Watt hour

A.h

-

Ampere hour

CHAPTER 1

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.

13


iii. 1.2

Comparing the net zero energy building with conventional building.

NECESSITY The basic necessities of such a building are: i.

As the country is developing day by day the consumption of power is also

ii.

very high. Now if we are going for NZERB building we can save energy locally

iii.

which mean to save energy in global level. The use of this technology used in residential buildings has shown huge

iv.

amount savings in the electricity bill. The proper design and alignment of the building can make the building

v.

cheaper than that of the conventional type of buildings. Usage of hollow bricks and avoidance of columns and beams will result

vi.

1.3

in lowering of temperature inside the building To achieve sustainability.

SCOPE i. ii. iii. iv.

Functional planning of G+1 Residential building Design of load bearing structure using hollow bricks Design of solar panels Comparison of room temperature between NZERB and conventional

v.

building Comparison of energy consumption between NZERB and conventional building.

1.4

METHODOLOGY This entire project is a planning and design in nature and the methodology followed in this project is listed as below. 14


1.5

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

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

1.6

i.

Design of slabs

ii.

Design of footings

iii.

Design of wall using Hollow bricks

iv.

Design of solar panels

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. 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

15


1.8

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 Construction

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 CE 0104- Computer aided building drawing CE0102- Elements of building science and Architecture

Context Computer aided building drawing layout and planning Byelaws, Setbacks, Open space, Floor area ratio R.C.C Design

CE0209- Building technology CE0303-Structural Design II

16


CE0304-Structural Design III

1.9

R.C.C Design

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. It also involves interaction with software people to learn about the

ii.

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. ii. iii.

1.11

Auto CAD MS EXCEL MS WORD

CONCLUSION The two types of buildings are analyzed with respect to cost, time, availability of skilled labour and ease in construction. ELECTRICIT COST NORMAL CONVENTIONA L BUILDING

Low

Y

It requires an active source

17

AVAILABILITY OF RESOURSES

Easily Available


NZERB

1.12

High

Produced on its own

Difficult

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

INTRODUCTION 2.1 GENERAL

18


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).

2.2

LITERATURE REVIEW

Anna Joanna, Aalborg University, Department of Civil Engineering, 19


According to ANNA, “With energy conservation arrangements, such as highinsulated 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. ii.

Solar panels are designed to harness. Solar energy in buildings include systems that capture heat (such as Solar water

iii.

heating systems and passive heating). 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 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.

20


(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

Side setback according to CMDA code is shown in Table 2.3. Table 2.3 Side Set Back Width of Plot Not more than 6m

Side Set Back 1.0m on one side

21


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. 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

22


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 3.1

OBJECTIVE i.

Design a building with Net zero energy concept.

23


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 1

Brick Material

NZERB

CONVENTIONAL

Hollow brick

Normal brick More than NZERB

2

Temperature

4 to 5 degree less compared To conventional building

3

Electricity

Produced on its own

4

Initial Cost

High

5

Solar Panels

250 w panels Provided in NZERB

It requires an active source Less compared to NZERB Not provided

Uses less energy

Uses more energy

6 3.2

Energy Efficient SCOPE i ii iii iv v

3.3

Functional planning of G+1 Residential building. Design of load bearing structure using hollow bricks. Design of solar panels. Comparison of room temperature between NZERB and conventional building. Comparison of energy consumption between NZERB and conventional building.

METHODOLOGY

24


This entire project is 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 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

25


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

26


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.

27


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)

28


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 L y 5.1 = Aspect ratio = = 1.56<2 L x 3.26 Hence Two Way Slab 2. Load Calculation Assuming Slab Thickness L x 3260 d= = =101.875 mm=100 mm 32 32 Assume 10 ∅ bar, Clear Cover 20mm 10 + 20=125 mm=130 mm 2 ∴ Actual Depth (d) = 130-5-20 = 105 mm D 130 Self Weight of a Slab= ×25= ×25 1000 1000 kN ¿ 3.25 2 m Assume Floor Finish = 40 mm

D=100+

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

kN m2

(Ref 8)

= 6.21

kN 2 m

Factored Load (Wu) = 1.5 × 6.21 = 9.315 Consider 1m width of slab 29

kN m2

kN 2 m


∴ Load per meter Length = 9.127

kN m

3. Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous Ly =1.56 (already found out ) Lx 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. 0.36× X u max 1−0.42× X u max f ck b d 2 Mu,lim= d d

[

]

(4.2) (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 30


= 0.138 × 20 × 1000 × 1052 =30.42 kN.m (Mu Limit) > (Mu Short) ∴ Hence its ok 4. Calculation of Steel Ast(+) Short = b d

f ck [1 - √ 1-4.598 2f y

R ] f 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 R=

fy = Yield stress of steel Mu 1 = 6.731 × 10 6 2 105 2 ×1000 bd

Ast(+) Short = 1000 × 105 × (

20 0.61 ¿ ) × 415 [1 - √ 1-4.598 × 2 20

= 184.27 mm2 Minimum Steel = 0.12% × D × B 0.12 Ast,min = ( )× 130 × 1000 = 156 mm2 100 Ast(+) Short
3d = 3 × 105 = 315 mm 300

Max Spacing = 300 mm ∴ d for long span bars

d = D – Clear Cover – = 130 – 20 –

10 2

d 2

= 0.6105

- ∅

- 10 = 95 mm

6. Calculation of Ast for Long Span

31


R f ck [1 - √ 1- 4.598 ] f ck 2f y Same as equation 4.3 Mu 2 R= d b 3.662 ×10 6 = × 952 1000 = 0.4057 20 0.4057 Ast(+)Long = 1000 × 95 × ( )× 415 [1 – √ 4.598 × ] 2 20 = 109.37 mm2 Ast(+)Long
4.2.2

Design of Bed Room

Using M20 Concrete Fe415 steel Live Load = 2

kN m2

1. Effective Span Lx = 3 m 32


Ly = 3.5 m

Ly 0.23 + 0.23=1.154< 2 Hence Two Way Slab = 3.5+ Lx 3 2. Load Calculation Lx 3230 d= = =100 mm Assuming Slab Thickness 32 32 Assume 10 ∅ bar, Clear Cover = 20mm 10 D=100+ + 20=125 mm 2 ∴ Actual Depth (d) = 125-5-20 = 100 mm D 125 kN ×25= ×25=3.125 Self Weight of a Slab = 1000 1000 m Assume 40 mm Floor Finish Aspect ratio =

∴ Weight of Floor Finish ∴ Imposed Load

= 6.08

= 0.04 X 24 = 0.96 kN m2

=2

kN m2

Total Load

kN 2 m

Factored Load (Wu)

= 1.5× 6.96 = 9.127

kN m2

Consider 1m width of slab ∴ Load per meter Length

= 9.127

kN m2

3. Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous Ly =1.154 ( already found out ) Lx Refer Table 26 Short Span αx = 0.043 Long Span αy = 0.035 [ Note that Lx only to be taken, where it is long span or short span only coefficient varies ].

33


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=

[

]

0.36× X u max 1−0.42× X u max f ck b d 2 d d (or)

Mu,lim = 0.138 fckbd2 Same as equation 4.2 = 0.138 × 20 × 1000 × 1002

Mu,lim

kN m2 (Mu Limit) > (Mu Short) ∴ Hence its ok 4. Calculation of Steel Mu,lim

= 27.6

Ast(+) Short = b d

f ck [1 - √ 1-4.598 2f y

R ] f ck

Same as equation 4.3 R=

Mu 1 = 4.1 × 10 6 × 1000 = 0.41 2 100 2 bd

Ast(+) Short = 1000× 100 ×

20 0.41 × 415 [1 - √ 1-4.598 × ] 2 20

= 116.37 mm2 Minimum Steel = 0.12% × D × B Ast,min =

0.12 × 125 × 1000 = 150 mm2 100

Ast(+) Short
34


i. 3d = 3 X 100 = 300 mm ii. 300 Max Spacing = 300 mm ∴ d for long span bars d d= D – Clear Cover – - ∅ 2 d= 125 – 20 – 10/2 - 10 d= 90 mm 6. Calculation of Ast for Long Span R f ck Ast(+)Long ¿ b d [1 - √ 1- 4.598 ] f ck 2f y Same as equation 4.3 R=

Mu bd 2

= 3.33 ×

10 6 1000 ×100 2

= 0.33 Ast(+)Long = 1000 × 100 ×

[

20 0.33 × 415 1 – 4.598 × 2 20

= 93.2 mm2 Ast(+)Long
35

]


Modified Basic Value = 20 × 1.7 = 34 Span 3230 = =32.3 d 100

32.3 < 34

∴ Hence its ok

4.2.3

Design of Bed Room


kN 2 m

1. Effective Span Lx = 3.85 m Ly = 3.95 m Ly 3.95 Aspect ratio = = = 1.027<2 Lx 3.85 Hence Two Way Slab 2. Load Calculation Assuming Slab Thickness L x 3850 = d ¿ = 120.31mm = 120 mm 32 32 Assume 10 ∅ bar, Clear Cover 20 mm 10 D = 120+ +20 = 145mm = 150 mm 2 ∴ Actual Depth (d) = 150-5-20 = 125 mm D 150 Self Weight of a Slab = × 25 = ×25 1000 1000 kN = 3.75 m2 Assume 40 mm Floor Finish ∴ Weight of Floor Finish = 0.04 × 24 = 0.96 Imposed Load = 2

kN m2

∴ Total Load

= 6.75

kN 2 m

36

kN m2


kN m2

Factored Load (Wu) = 1.5 × 6.75 = 10.125 Consider 1m width of slab Load per meter Length = 10.125

kN m

3. Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous Ly = 1.027 (already found out) Lx Refer Table 26 Short Span αx = 0.048 Long Span αy = 0.047 [ 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 = 0.048 × 10.125 × 3.852 = 7.203 kN.m Mu(+) Long = 0.047 × 10.125 × 3.852 = 7.063 kN.m Take the Highest Moment and check for adequacy of the section. 0.36× X u max 1−0.42× X u max f ck b d 2 Mu,lim= d d (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 R f ck Ast(+) Short = b d [1 - √ 1-4.598 ] f ck 2f y Same as equation 4.3 Mu 1 R= = 7.203 × 106 = 0.460 2 125 2 ×1000 bd 20 0.46 Ast(+)Short = 1000 × 125 × ( ) × 415 [1 - √ 1-4.598 × ] 2 20 = 164.08 mm2 Minimum Steel = 0.12% × D × B

[

]

37


0.12 )× 150 × 1000 100 = 180 mm2 Ast(+) Short
38


Span d

4.2.4

3850 = 30.8 125 30.8 < 35.6 ∴ Hence its ok =

Design of Bathroom


kN m2

1. Effective Span Lx = 2.38 m Ly = 4.28 m Ly 4.28 Aspect ratio = = = 1.798<2 Lx 2.38 Hence Two Way Slab 2. Load Calculation Assuming Slab Thickness L x 2380 = d= = 74.375 mm = 80 mm 32 32 Assume 10 ∅ bar, Clear Cover 20 mm 10 D=80+ +20=105mm=110 mm 2 ∴ Actual Depth d = 110-5-20 = 85 mm D 110 ×25= ×25 Self Weight of a Slab = 1000 1000 kN = 2.75 2 m Assume 40 mm Floor Finish ∴ Weight of Floor Finish = 0.04 × 24 = 0.96

Imposed Load = 2

kN m2

∴ Total Load = 5.75

kN 2 m

39

kN m2


Factored Load (Wu) = 1.5 × 5.75 = 8.625

kN m2

Consider 1m width of slab ∴ Load per meter Length = 8.625

kN m

3. Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous Ly Lx

= 1.798 (already found out)

Refer Table 26 Short Span αx = 0.085 Long Span αy = 0.047 [ 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 = 0.085 × 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. 0.36× X u max 1−0.42× X u max f ck b d 2 Mu,lim= d d (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 R f ck Ast(+) Short = b d [1 - √ 1-4.598 ] f ck 2f y Same as equation 4.3 Mu 106 R = = 4.1527 × = 0.574 2 (85 2× 1000) bd 20 0.57 Ast(+) Short = 1000 × 85 × ( ) × 415 [1 - √ 1-4.598 × ] 2 20

[

]

40


= 139.92 mm2 Minimum Steel = 0.12% × D × B 0.12 Ast,min= ( ) × 110 × 1000 = 132 mm2 100 Ast(+) Short
3d = 3 × 855 = 255 mm 300

Max Spacing = 255 mm ∴ d for long span bars d d= D – Clear Cover – - ∅ 2 10 d= 110 – 20 – - 10 2 d= 75 mm 6. Calculation of Ast for Long Span R f ck Ast(+)Long ¿ b d [1 - √ 1- 4.598 ] f ck 2f y Same as equation 4.3 Mu R= bd 2 2.29 ×10 6 = 1000 ×75 2 = 0.4082 20 0.4082 ¿ = 86.88 Ast(+) Long = 1000 × 75 × ( ) × 415 [1 – √ 4.598 × 2 20 mm2 Ast(+) Long
3d = 3 × 75 = 225 mm 300

MaxSpacing = 225 mm 8. Check for Deflection Short Span Lx = 2380 mm Ast(+) Short = 139.92 mm2 Basic Value = 20 139.92 Fs = 0.58 × 415 × = 240.2 139.92 +¿ ¿ (139.92× 100) Short Pt = (1000× 85) A st ¿ ¿¿

41

N mm2


= 0.1646% Modification Factor = 1.9 Modified Basic Value = 20 × 1.9 = 38 Span 2380 = = 28 d 85 28 < 38 ∴ Hence its ok

4.2.5

Design of Portico


kN m2

1. Effective Span Lx = 3.78 m Ly = 6.93 m L y 6.93 = Aspect ratio = = 1.83<2 L x 3.78 Hence Two Way Slab 2. Load Calculation Assuming Slab Thickness L x 3780 = d ¿ = 118.124 mm = 120 mm 32 32 Assume 10 ∅ bar, Clear Cover 20 mm +10 +20 = 150 mm D = 120 2 ∴ Actual Depth (d) = 150-5-20 = 125 mm D 150 ×25= Self Weight of a Slab = ×25 1000 1000 kN = 3.75 m2 Assume 40 mm Floor Finish ∴ Weight of Floor Finish = 0.04 × 24 = 0.96 Imposed Load = 2

kN m2

42

kN m2


kN m2

∴ Total Load = 6.75

kN m2

Factored Load (Wu) = 1.5 × 6.75 = 10.125 Consider 1m width of slab ∴ Load per meter Length = 10.125

kN m

3. Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous Ly =1.83 (already found out) Lx Refer Table 26 Short Span αx = 0.087 Long Span αy = 0.047 [ 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 = 0.087 × 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. 0.36× X u max 1−0.42× X u max f ck b d 2 Mu,lim= d d (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 R f ck Ast(+) Short = b d [1 - √ 1-4.598 ] f ck 2f y Same as equation 4.3 Mu 12.58× 106 R= = = 0.805 2 bd 125 2 ×1000

[

]

43


Ast(+) Short = 1000 × 125 ×

20 2 × 415

[1 - √ 1-4.598 ×

0.8 ] 20

= 184.27 mm2 Minimum Steel = 0.12% × D × B 0.12 Ast,min = ( )× 130 × 1000 100 = 156 mm2 Ast(+) Short
3d = 3 ×105 = 315 mm 300

Max Spacing = 300 mm ∴ d for long span bars d d= D – Clear Cover – - ∅ 2 d= 130 – 20 – 10/2 - 10 d= 95 mm 6. Calculation of Ast for Long Span R f ck Ast(+)Long ¿ b d [1 - √ 1- 4.598 ] f ck 2f y Same as equation 4.3 Mu R= 2 bd 3.662 ×10 6 = 952×1000 = 0.4057 20 0.4057 Ast(+)Long = 1000 × 95 × ( )× 415 [1 – √ 4.598 × ] 2 20 = 109.37 mm2 Ast(+)Long
44


+¿ ¿ Short Pt A st ¿ ¿¿

=

184.27 × 100 (1000 × 105 )

= 0.175% Modification Factor = 1.62 Modified Basic Value = 20 × 1.62 =32 Span 3260 = = 31.047 d 105 31.047 < 32.8 ∴ Hence its ok 4.2.6

Design of Kitchen


kN m2

1. Effective Span Lx = 2.23 m Ly = 3.73 m Ly 0.23 =3.73+ + 0.23=1.67<2 Aspect ratio ¿ Lx 2.23 Hence Two Way Slab 2. Load Calculation Assuming Slab Thickness L x 2230 = d= = 65 mm 32 32 Assume 10 ∅ bar, Clear Cover 20 mm 10 D = 65+ +20 = 90 mm 2 ∴ Actual Depth (d) = 90-5-20 = 65 mm D 90 ×25= ×25 Self Weight of a Slab = 1000 1000 kN = 2.25 m2 Assume Floor Finish = 40 mm Weight of Floor Finish = 0.04 × 24 = 0.96

45

kN m2


kN m2

∴ Imposed Load = 2 ∴ Total Load

= 5.25

kN m2

Factored Load (Wu) = 1.5 × 5.25 = 7.875

kN m2


kN m

Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 Two adjacent edges are discontinuous Ly = 1.67 (already found out) Lx Refer Table 26 Short Span αx = 0.06 Long Span αy = 0.035 [ 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 = 0.06 × 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. 0.36× X u max 1−0.42× X u max f ck b d 2 Mu,lim= d d (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 R f ck Ast(+) Short = b d [1 - √ 1-4.598 ] f ck 2f y Same as equation 4.3

[

]

46


Mu 10 6 = 2.735 × = 0.55 2 bd 100 0 × 602 20 0.55 Ast(+) Short = 1000 × 65 × × 415 [1 - √ 1-4.598 × ] 2 20 = 103.56 mm2 Minimum Steel = 0.12% × D × B 0.12 Ast,min ¿ × 90 × 1000 100 = 108 mm2 Ast(+) Short
i. ii.

3d = 3 × 65 = 195 mm 300

Max Spacing = 195 mm ∴ d for long span bars d d= D – Clear Cover – - ∅ 2 10 d= 90 – 20 – - 10 2 d= 55 mm 5. Calculation of Ast for Long Span R f ck Ast(+)Long ¿ b d [1 - √ 1- 4.598 ] f ck 2f y Same as equation 4.3 Mu 10 6 R= = 1.373 × 2 2 bd 1000 ×55 = 0.115 20 0.115 Ast(+) Long =1000 × 55 × × 415 [1 – 4.598 × ] 2 20 = 71.05 mm2 Ast(+) Long
47


Fs = 0.58 × 415 × +¿ ¿ Pt Short A st ¿ ¿¿ = 0.16%

=

103.56 103.56

= 240.7

N mm2

103.56 ×65 1000

Modification Factor

= 1.8

Modified Basic Value = 20 × 1.8 = 36 Span 2230 = =34.3 d 65 34.3 < 36 ∴ Hence its ok 2.4.7

Design of Dinning Room


kN m2

1. Effective Span Lx = 2.6 m Ly = 3.73 m Ly 0.23 =3.73+ + 0.23=1.43<2 Aspect ratio ¿ Lx 2.6 Hence Two Way Slab 2. Load Calculation Assuming Slab Thickness L x 2600 = =80 mm d= 32 32 Assume 10 ∅ bar, Clear Cover 20 mm 10 D ¿ 8 0+ +20=105 mm 2 ∴ Actual Depth (d) = 105-5-20 = 80 mm D 105 × 25= × 25 Self Weight of a Slab= 1000 1000 kN = 2.625 m2

48


Assume Floor Finish

= 40 mm

∴ Weight of Floor Finish = 0.04 × 24 = 0.96

Imposed Load ∴ Total Load

kN m2

kN m2

=2

= 5.62

kN m2

Factored Load (Wu) = 1.5 × 5.62 = 8.43

kN 2 m


kN m

3. Finding Design Bending Moment Refer Table 26, Page No.91 of IS456 One edge discontinuous Ly = 1.43 (already found out) Lx Refer Table 26 Short Span αx = 0.049 Long Span αy = 0.028 [ 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 = 0.049 × 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. 0.36× X u max 1−0.42× X u max f ck b d 2 Mu,lim= d d (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

[

]

49


4. Calculation of Steel f ck R Ast(+) Short = b d [1 - √ 1-4.598 ] 2f y f ck Same as equation 4.3 Mu 106 R= = 2.79 × = 0.435 2 bd 1000 × 802 20 0.435 Ast(+) Short = 1000 × 80 × × 415 [1 - √ 1-4.598 × ] 2 20 = 99.15 mm2 Minimum Steel = 0.12% × D × B 0.12 Ast,min= × 105 × 1000 100 = 126 mm2 Ast(+) Short
d= D – Clear Cover – d= 105 – 20 –

10 2

d 2

- ∅

- 10

d= 70 mm 6. Calculation of Ast for Long Span R f ck Ast(+)Long ¿ b d [1 - √ 1- 4.598 ] f 2f y ck Same as equation 4.3 Mu R= bd 2 10 6 = 1.595 × 2 1000 ×70 = 0.325 1 0.325 (+) 20 Ast Long =1000 × – 4.598 × ] 70 × × 415 ¿ 20 2 = 64.34 mm2 (+) Ast Long

π × 102 = 78.5 mm2 4 78.5 × 1000 = 791.7 mm Ast(+) Short = 99.15 78.5 Ast(+) Long = × 1000 = 1200.08 mm 64.34 8. Check for Deflection Short Span Lx = 2600 mm Ast(+) Short = 99.15 mm2 Basic Value = 20 99.15 N =240.7 Fs = 0. 58 ×415 × 99.15 mm 2 +¿ ¿ Short Pt A st ¿ ¿¿ 99.15 = 1000 ×70 = 0.14% Modification Factor = 1.8 Modified Basic Value = 20 × 1.8 = 36 Span 2600 = = 32.5 d 80 32.5 < 36 ∴ Hence its ok Ast =

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

=3m

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

=1m 51


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 h 5.8 =25.21 Slenderness Ratio ¿ = t 0.23 4. Shape modification factor: Crushing Strength of Modular Brick = 5

kN m2

H 75 = =0.652 w 115 ∴ 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 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

52

(4.4)


= 0.21

N m²

8. Safe Load Q= =

( f1c + 6te ) A 0.21×1000 × 300 1

= 63

kN m

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 = 91.05 m3

Total wall volume 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 =

23.81 ×100 67.25

= 35.4 % ∴ Thickness = 1 Brick thick wall (using nomograms)

11. For Hall :

53


Wu=

Wl x lx × {3- [( )] 6 ly

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 (3.03+0.23) Wu = 9.315 ×¿ } ¿ ¿ Wu = 13.11 × 103

kN m

For bed room: Wu=

Wl x lx × {3−[( )]² } 6 ly

= (9.127 ×

(3+0.23) × {3¿¿ 6

= 11.055 × 103

[

]

3+0.23 2 } 3.5+0.23

kN m

For dinning room: Wu= =

Wl x × {36

[( )] lx ly

2

}

(2.37+0.23) 2 (8.43 × ( 2.37+ 0.23 )) × {3- [ ]} (3.5+0.23) 6

= 9.81 × 103

kN m

Total : 13.11 + 11.055 + 9.81 = 33.98 63>33.98

kN m 54

× 103

kN m


Hence the design is ok

4.2.9

Design of Footing kN m

Load from Walls = 126.7

10% for the weight of the Building = 63+6.3 = 70 1. Area of Footing = Assume SBC

Load SBC

= 150 A =

kN m

70 150

kN m2 = 0.47 m2

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

55


The thickness of each step is given by modular brick = 200 mm ∴ The offset in the brick is also given as modular

4.2.10

= 100 mm

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 2 m

Step 2: Assumptions Height of the Plinth from ground

= 0.5 m

Height of the Plinth above Footing

=1m

Height of the Parapet Wall

=1m

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)

56


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 Effective height

= 0.75 × H = 3.15 m

Slenderness ratio

=

H t

=

3.15 0.2

= 15.75

Step 4: Shape modification factor Crushing Strength of Hollow Brick= 4.1 H W

=

20 20

N 2 mm

=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)

57


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 N mm2

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: H = 2.6+0.8 = 3.4 m Effective height = 0.75 × H = 2.55 m Slenderness ratio =

H t

=

2.55 0.2

= 12.75

Step 8: Shape modification factor Crushing Strength of Hollow Brick = 4.1 =

H W

=

20 20

=

1

Shape Modification Factor = Kp =1.2

Step 9: Stress Reduction Factor ks = 0.81

Step 10: Area reduction factor

58

N 2 mm

kN m


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 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

59


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 Table 4.3 Calculation of permissible stress Wall type A B C D E F G H I J K L M

Fc=ks×kp×ka× basic compressive stress Permissible stress- ground Permissible stress- first floor(N/mm2) floor(N/mm2) 0.528×0.74=0.390 0.528×0.81=0.427 0.528×0.81=0.427 0.528×0.81=0.427 0.528×0.74=0.390 0.528×0.81=0.427 0.528×0.75=0.396 0.528×0.81=0.427 0.528×0.89=0.469 0.528×0.89=0.469 0.528×0.83=0.4382 0.528×0.83=0.4382 0.528×0.74=0.390 0.528×0.81=0.427 0.528×0.75=0.396 0.528×0.81=0.427 0.528×0.75=0.396 0.528×0.81=0.427 0.528×0.74=0.390 0.528×0.81=0.427 0.528×0.75=0.396 0.528×0.81=0.427 0.528×0.88=0.4646 0.528×0.88=0.4646 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 A B C D E

q = fc×2×105 kN/m(ground floor) 78 85 78 79.2 93 60

q = fc×2×105 kN/m(first floor) 85.4 85.4 85.4 85.4 93.8


F G H I J K L M

4.2.11

87.6 78 79.2 79.2 78 79.2 92 78

87.6 85.4 85.4 85.4 85.4 85.4 92 85.4

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

Load from Wall = 78

kN m

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

kN m

kN m

Factored load= 1.25 × 100=125 1. Area of Footing = Assume SBC = 150

Load SBC

kN m =

125 50

= 0.833 m2

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

61


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

Fig.4.4 Footing Design 4.2.12

Design of Stair Case:

Length

=4m

Live load

=2

kN m2

62


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 D × 1000

Self weight =

√ R2 +T 2

× 2 = 5.83

T

kN m2

(4.6) Self weight is calculated by equation 4.6 Where, D =Diameter R =Rise T =Thread Step 2: Calculation of load on waist slab 1. Assume 40 mm Floor finish Floor Finish = 2. Weight of steps=

40 1000 1 2

× 24 = 1

kN m2

×R×T×

1 T

kN m2

3. Live load

=3

4. Self weight

= 5.83

Wu = 11.075

kN m2

kN m2

Wu=1.5 × 11.075 =17.55

kN m2

63

× 25 = 1.875

kN m2


Step3: Calculation of Mu Mu

=

Mu,lim =

17.55 ×16 8

= 35 kN.m

[

]

0.36× X u max 1−0.42× X u max f ck b d 2 d d

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

Step 4: Calculation of Ast 6

R=

35 ×10 1000 ×120 × 120

Ast =

1000 ×120 × 20 2 × 415

Pt =

Ast bd

= 970.6 mm2

970.6 π ×20 2 4

Number of Bars = Ast actual = 4 ×

= 2.43

π 4

× 202

× 1000 =

= 4 bars = 1256 mm2

1256 1000 ×120

× 1000 = 1.04%

Step 5: Check for deflection Basic value = 20 Fs = 0.58fy

Ast reqiired Ast provided

= 0.58 × 415 ×

64

970 1256

= 185.89

N 2 mm


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

drequired =

=

4000 1.2 ×32

= 104.166 mm

dactual = 120 mm Step 6: Providing distribution steel Astmin=

0.12bd × 100

0.12×1000 × 150 100

Spacing of 8mm diameter = Main steel

1000 480

×

= 480 mm2 π 4

× 82 = 270 mm

= 4No.s 20 bars

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

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.

65


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.

4.3.2

Working of solar panels

66


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.

67


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

68


Table 4.5 Calculation Of Loads (Ref 12) PARTICULA RS HALL

BED ROOM 1 BED ROOM 2 KITCHEN

DINING ROOM

TOILET 1 TOILET 2

UNIT S

USAG E IN HRS

VOLTA GE W

CONSUM PTION

INVER TORS

4

5

20

400

80

2 1

5 5

50 80

500 400

100 80

CFL

2

3

15

90

30

FAN

1

10

50

500

50

CFL

2

3

15

90

30

FAN OVEN CFL EXHAUS T Mixer AUTOFRIDGE CFL FAN CFL HEATER CFL

1 1 3

10 1 4

50 900 15

500 900 180

50 900 45

1

4

50

200

50

1

1

450

450

450

1

18

150

2700

195

3 1 1 1 1

4 3 1 1 2

15 50 15 150 15

180 150 15 150 30

45 50 15 150 15

1

1

750

750

750

1

2

90

180

90

8365

3175

ITEMS CFL (Ref 13) FAN T.V

WATER PUMP WASHING MACHINE

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.

69


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 =

1673 250

=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

404 50 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. ii. iii. iv.

7x2x 250W solar panels 4 x 20A solar regulators x 105A.H deep cycle batteries( 3 in series) 1 x 3500W modified sine wave power inverter 70


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

71


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

II.

emitted by the object being measured. 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

III.

refer Figure.4.6 By knowing the amount of infrared energy emitted by the object and

IV.

its emissivity, the object's temperature can often be determined. Infrared thermometers are a subset of devices known as "thermal radiation thermometers".

72


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 temperature. This configuration facilitates temperature measurement from a distance without contact with the object

VI.

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

VII. VIII.

monitoring functions. A few examples provided to this article include: Detecting clouds for remote telescope operation Checking mechanical equipment or electrical circuit breaker boxes or

IX. X.

outlets for hot spots Checking heater or oven temperature, for calibration and control purposes Detecting hot spots / performing diagnostics in electrical circuit board

XI. XII.

manufacturing Checking for hot spots in fire fighting situations Monitoring materials in process of heating and cooling, for research and

XIII.

development or manufacturing quality control situations 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

XIV. XV.

the temperature of reflective (shiny) and non-reflective surfaces. The most common infrared thermometers is the: 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).

73


4.6

CHARACTERISTICS OF HOLLOW BRICKS 4.6.1 Parameters of Hollow Brick Used In Net Zero Energy Residential

Building I. II. III. IV. V.

LENGTH : 400 mm WIDTH : 200 mm HEIGHT: 200 mm WEIGHT: 11.1 kg DENSITY: 694 kg/m³

VI.

COMPRESSIVE STRENGTH : 4.1

VII. VIII. IX. X.

N 2 mm

WATER ABSORPTION : 15% U-VALUE : 1.1 W/m² SOUND INSULATION : 46 DB FIRE RESISTANCE 240 min

Available Sizes I. II. III. IV. V. VI.

400 X 200 X 200 mm 400 X 150 X 200 mm 400 X 100 X 200 mm 200 X 200 X 200 mm 200X 150 X 200 mm 200 X 100 X 200 mm

Hollow Brick Bigger Size I. II. III. IV.

Hollow brick is same in size as that of concrete blocks 1 Hollow brick = 9 Clay Bricks Less mortar joints, hence less plumb & alignment Faster construction

Light Weight I. II. III. IV.

Ease of handling, Transportation Saves labour Less dead load, Savings in Structural Cost (Steel & Concrete) by 10 to 15%

74


Thermal Insulation I. II. III. IV.

Savings on mortar Low ‘U’ Values – 1.0 W/m² Better Thermal Insulation = less energy loss through walls 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

75


i.

Highly Durable: The good concrete compacted by high pressure

ii.

and vibration gives substantial strength to the brick. Low Maintenance, Colour and brilliance of masonry withstands

iii. iv.

outdoor elements. Fire Resistant 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

v.

house cool in summer and warm in winter. Environment Friendly, fly ash used as one of the raw materials.

Constructional Advantages i.

No additional formwork or any special construction machinery is

ii. iii.

required for reinforcing the hollow brick masonry. Only skilled labour is required for this type of construction. It is a faster and easier construction system, when compared to the

iv. v. vi.

other conventional construction systems. Savings on RCC-Frame structure (steel/concrete). Faster construction and ease for handling at site. Hollow brick consist of four elements earth, water, fire, air which

vii. viii. ix. x.

makes easy work of construction. Less wastage through half bricks. Excellent thermal insulation. Reduction in energy consumption. 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

76


S.No

Description

Nos

77

Lengt h (m)

Breadt h (m)

Dept h (m)

Quantit y (m3)


1

2

3

3

4

5

Excavation Exterior wall Interior wall

1 1

39.48 22.51

0.76 0.76

0.53 0.53

15.9 9.06 24.96

P.C.C Exterior wall Interior wall

1 1

39.48 22.51

0.76 0.76

0.3 0.3

9 5.13 14.13

1 1

39.48 22.51

0.46 0.46

0.115 0.115

2.08 1.19

1 1

39.48 22.51

0.31 0.31

0.115 0.115

1.4 0.8

1 1

39.48 22.51

0.23 0.23

7.5 7.5

68.1 38.82 112.39

Deductions Window W Window W1 Window W2 Door D Door D1 Door D2 Spacing S1 Spacing S2

1×2 1×5 1×2 1×2 1×3 1 1 1

1.22 1.22 0.9 0.84 0.75 0.9 1.2 0.9

0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.23

0.9 1.22 1.21 2.1 2 2.1 2.08 2.1

0.51 1.71 0.5 0.81 1.035 1.035 0.57 0.43 6.6

Earth Filling Hall Water closet Bed Room1 Bed Room2 Dinning Kitchen

1 1 1 1 1 1

3.03 3.62 3.62 3.5 3.5 3.5

4.87 2.15 3.72 3 2.37 2.6

0.5 0.5 0.5 0.5 0.5 0.5

7.378 3.89 6.73 5.25 4.147 4.55 31.649

Flooring Concrete Hall Water closet

1 1

3.03 3.62

4.87 2.15

0.1 0.1

1.48 0.778

Brick work 1st Footing Exterior wall Interior wall 2nd Footing Exterior wall Interior wall Wall Exterior wall Interior wall

78


Bed Room1 Bed Room2 Dinning Kitchen 6

7

8 9

1 1 1 1

3.62 3.5 3.5 3.5

3.72 3 2.37 2.6

0.1 0.1 0.1 0.1

1.346 1.05 0.829 0.91 5.473

1×2 1×3 1 1 1×2 1×2 1×5 1×5 1×2 1×2

1.4 1.05 1.2 1.2 1.52 1.52 1.52 1.52 1.2 1.2

0.23 0.23 0.23 0.45 0.23 0.45 0.23 0.45 0.23 0.45

0.15 0.15 0.15 0.075 0.15 0.075 0.15 0.075 0.15 0.075

0.078 0.108 0.04 0.04 0.104 0.103 0.2622 0.26 0.083 0.081 1.1592

Roof Slab Hall, Water Closet, Bed Room1 Bed Room2,Dinning,Kitchen

1 1

6.65 3.5

4.87 7.97

0.1 0.1

3.24 2.79 6.8496

Plastering Exterior wall Interior wall

1 1

39.48 22.51

-

7.5 7.5

296.1 168.8 464.92

Deductions Window W1 Window W2 Window W3 Door D Door D1 Door D2 Spacing S Spacing S1

1×2 1×5 1×2 1×2 1×3 1 1 1

1.22 1.22 0.9 0.84 0.75 0.9 1.2 0.9

-

0.9 1.22 1.21 2.1 2 2.1 2.08 2.1

-

-

-

-

2.196 7.442 2.178 3.528 4.5 1.89 2.496 1.89 26.12 438.8 438.8

R.C.C Lintel & Sun shades Door D Door D1 Door D2 Sun Shade Window W Sun Shade Window W1 Sun Shade Window W2 Sun Shade

White Washing Colour Washing

79


The abstract estimate of NZERB is given in Table 4.7 Table 4.7 Abstract Estimate of NZERB

80


S.No 1

2

3

3

4

5

NoS

Lengt h (m)

Breadt h (m)

Dept h (m)

Quantit y (m3)

Excavation Exterior wall Interior wall

1 1

39.24 22.51

0.7 0.7

0.5 0.5

13.734 7.87 21.604

P.C.C Exterior wall Interior wall

1 1

39.24 22.51

0.7 0.7

0.3 0.3

8.24 4.72 12.96

1 1

39.24 22.51

0.4 0.4

0.2 0.2

3.13 1.8

1 1

39.24 22.51

0.25 0.25

0.2 0.2

1.96 1.12

Description

Brick work 1st Footing Exterior wall Interior wall 2nd Footing Exterior wall Interior wall Wall Exterior wall Interior wall Deductions Window W Window W1 Window W2 Door D Door D1 Door D2 Spacing S1 Spacing S2

1 1

39.24 22.51

0.2 0.2

7.5 7.5

58.86 33.76

1×2 1×5 1×2 1×2 1×3 1 1 1

1.22 1.22 0.9 0.84 0.75 0.9 1.2 0.9

0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2

0.9 1.22 1.21 2.1 2 2.1 2.08 2.1

0.4392 1.4884 0.432 0.705 0.9 0.378 0.504 0.378 95.02

Earth Filling Hall Water closet Bed Room1 Bed Room2 Dinning Kitchen

1 1 1 1 1 1

3.03 3.62 3.62 3.5 3.5 3.5

4.87 2.15 3.72 3 2.37 2.6

0.5 0.5 0.5 0.5 0.5 0.5

7.378 3.89 6.73 5.25 4.147 4.55 31.649

Flooring Concrete

81


Hall Water closet Bed Room1 Bed Room2 Dinning Kitchen 6

7

8

1 1 1 1 1 1

3.03 3.62 3.62 3.5 3.5 3.5

4.87 2.15 3.72 3 2.37 2.6

0.1 0.1 0.1 0.1 0.1 0.1

1.48 0.778 1.346 1.05 0.829 0.91 5.473

1×2 1×3 1 1 1×2 1×2 1×5 1×5 1×2 1×2

1.4 1.05 1.2 1.2 1.52 1.52 1.52 1.52 1.2 1.2

0.23 0.23 0.23 0.45 0.23 0.45 0.23 0.45 0.23 0.45

0.15 0.15 0.15 0.075 0.15 0.075 0.15 0.075 0.15 0.075

0.0684 0.0945 0.036 0.0405 0.07512 0.1026 0.228 0.0256 0.072 0.081 0.8237

Roof Slab Hall, Water Closet, Bed Room1 Bed Room2,Dinning,Kitchen

1 1

6.65 3.5

4.87 7.97

0.1 0.1

3.24 2.79 6.8496

Plastering Exterior wall Interior wall

1 1

39.24 22.51

-

7.5 7.5

294.3 168.82 463.12

Deductions Window W1 Window W2 Window W3 Door D Door D1 Door D2 Spacing S Spacing S1

1×2 1×5 1×2 1×2 1×3 1 1 1

1.22 1.22 0.9 0.84 0.75 0.9 1.2 0.9

-

0.9 1.22 1.21 2.1 2 2.1 2.08 2.1

-

-

-

-

2.196 7.442 2.178 3.528 4.5 1.89 2.496 1.89 26.12 437

R.C.C Lintel & Sun shades Door D Door D1 Door D2 Sun Shade Window W Sun Shade Window W1 Sun Shade Window W2 Sun Shade

White Washing

82


9

Colour Washing

-

83

-

-

-

437


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.N O

QTY in m³

DESCRIPTION OF WORK

QTY in cft

RATE

PE R

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

Flooring Work PCC 1:4:8

5.50

194.23

90.00

Cft

17480.00

6

R.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

15518.6 0

30.00

Sft

465559.00

439.00

15518.6 0

3.00

Sft

46555.80

Colour washing

439.00

15518.6 0

5.00

Sft

77593.00

Steel

501kg

60.00

kg

30060.00

5

8 9 10

White washing

TOTAL

84

1176181.30


85


The rate analysis proposed for NZERB building is given in the Table 4.9 Table 4.9 Proposed NZERB Building S.N O

DESCRIPTION OF WORK

QTY in m³

QTY in cft

RATE

PE R

AMOUNT

1

Earth Work Excavation

21.60

763.56

9.50

Cft

7253.82

2

Sand 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

Flooring Work PCC 1:4:8

5.50

194.23

90.00

Cft

17480.50

6

R.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

White washing

437.00

15448.00

3.00

Sft

46344.00

Colour washing

437.00

15448.00

5.00

Sft

77240.00

Steel

501.00

60.00

kg

30060.00

5

8 9 10 11

Solar Panel System

142600.00 TOTA L

86

1399046.82


CHAPTER 5

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

87


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

REFERENCES 1. 2. 3. 4.

http://en.wikipedia.org/wiki/Zero-energy_building http://energy.gov/energysaver/articles/annajohanna http://zeb.buildinggreen.com/saitoh 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

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