PROJECT REPORT onPLANNING AND DESIGN OF NET ZERO ENERGY RESIDENTAL BUILDING Submitted in partial fulfillment for the award of the degree of BACHELOR OF TECHNOLOGY in CIVIL ENGINEERING by KARTHIK V (1010910090) SASIDHAR K.V (1010910092) NEERAJ PORWAL (1010910118) ABHINAV N (1010910119) Under the guidance of Mrs. VASANTHI.P Assistant Professor (O.G) DEPARTMENT OF CIVIL ENGINEERING FACULTY OF ENGINEERING AND TECHNOLOGY SRM UNIVERSITY 1 (Under section 3 of UGC Act, 1956) SRM Nagar, Kattankulathur- 603203 Kancheepuram District MAY 2013 PROJECT REPORT on PLANNING AND DESIGN OF NET ZERO ENERGY RESIDENTAL BUILDING Submitted in partial fulfillment for the award of the degree of BACHELOR OF TECHNOLOGY in CIVIL ENGINEERING by KARTHIK V (1010910090) SASIDHAR K.V (1010910092) NEERAJ PORWAL (1010910118) ABHINAV N (1010910119) Under the guidance of Mrs. VASANTHI.P Assistant Professor (O.G) 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), NEERAJ PORWAL SASIDHAR (1010910118), REDDY.K.V(1010910092), 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 Mrs. VASANTHI .P Assitant Professor (O.G) Department of Civil Engineering Engineering SRM University Kattankulathur- 603203 Signature of the HOD Dr. R. ANNADURAI Professor & Head Department of Civil SRM University Kattankulathur- 603203 INTERNAL EXAMINER EXAMINER DATE: EXTERNAL 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 (E&T). C. MUTHAMIZHCHELVAN. Pro Vice Chancellor (P&D) and Dr. T. 5 . P. Director.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 GANESAN and records his sincere thanks to Dr. SRM UNIVERSITY for providing all the necessary facilities for carrying out this work. TABLE OF CONTENTS CHAPTER TITLE PAGE ABSTRACT 6 iv . ANNADURAI. The author is grandly indebted to all the Faculty Members of Department of Civil Engineering. the author expresses his hearty thanks to Friends for their kind help and encouragement throughout the course of this thesis work. for his valuable suggestions and advice in carrying out this thesis work. The author is extremely grateful to the valuble advices given by the class incharge Mr.GUNASEKARAN. Assistant professor.Department of Civil Engineering. Professor. Finally.VASANTHI P. for their valuable help rendered during the course of study.K. Department of Civil Engineering. for initiative and motivation during the course of this work.PRASANNA. The author expresses his sincere thanks to Department Coordinator/Civil Dr. for initiative and motivation during the course of this work.The author expresses his sincere thanks and Gratitude to HOD Dr. The author hereby acknowledges with deep sense of gratitude the valuable guidance given by the Guide Mrs. Assistant Professor.K. R. for constant support. Department of Civil Engineering. Department of Civil Engineering. ACKNOWLEDGEMENT v LIST OF TABLES LIST OF FIGURES ABBREVATIONS 1 OVERVIEW 1.3 1.1 1. Detection And Extinguishing System 10 9 7 .4 1.3.5 OBJECTIVE NECESSITY SCOPE METHODOLOGY MAJOR DESIGN EXPERIENCE ix x xi 1 1 1 2 2 2 1.2 1.9 1.8 1.1 2.11 1.12 2 REALISTIC DESIGN CONSTRAINTS REFERENCE TO CODES AND STANDARS APPLICATION OF EARLIER COURSE WORKS MULTIDISCIPLINARY AND TEAM WORK SOFTWARE USED CONCLUSION FUTURE SCOPE 3 3 4 4 5 5 5 6 6 7 INTRODUCTION 2.4 Primary Residential Use Zone 8 8 CONFORMATION TO NATIONAL BUILDING CODE OF INDIA 2.1 2.4.3 GENERAL LITERATURE REVIEW DEVELOPMENT CONTROL RULES FOR CHENNAI METROPOLITAN AREA.10 1.7 1. 2004 2.1 Fire Safety.6 1.2 2. 7 4.2 Selection of Site Plot Layout Plan of the Building DESIGNS 4.9 4.2.4 4.2.2.1.4.3 4.2 Security Deposits 10 3 OBJECTIVE AND SCOPE 3.1 3.11 4.2 Design of Bedroom 20 4.2.2.2.2 4.1 PLANNING 4.3 4.3 OBJECTIVE SCOPE MATERIALS AND METHODOLOGY 11 11 12 12 13 13 13 14 15 16 16 4 RESULTS AND DISCUSSIONS 4.8 4.2.6 Design of Kitchen 33 4.1.5 Design of Bedroom Design of Bathroom Design of Portico 23 27 30 4.10 4.1 Design of Hall 4.2.2.1 4.2.2.2.1.2.2 3.12 Design of Dining Room Design of Wall Design of Footing Design of Hollow Brick Wall Design of Footing (Hollow Brick) Design of Stair Case 37 40 44 44 49 51 8 . 2 CONCLUSION FUTURE SCOPE 72 72 72 73 REFERENCES 9 .2 Abstract Estimate of NZERB 65 67 4.3.1 Solar power system components 54 54 55 4.6.3.3.2 Working of Solar Panel Description of Individual Solar Panel components 4.2 4.3.1 Abstract Estimate of Conventional Building 4.5 4.3.6.3 Rate Analysis 70 5 CONCLUSION 5.2 4.3 DESIGN OF SOLAR PANEL AND ITS COMPONENTS 4.3.4.3.3.7.6 Designing of Solar Panel RATE ANALYSIS OF SOLAR PANELS INFRARED THERMOMETER HOLLOW BRICK 4.7 Advantages of Hollow Bricks 64 65 ESTIMATION 4.1 4.4 Solar Panels Solar Regulator Power Inverter Solar Batteries 55 55 55 56 56 57 59 60 62 62 4.7.3.3.4 4.3.1 5.7.4 4.3 4.3.3 4.1 Parameters of Hollow Brick 4. 1 3 1.LIST OF TABLES TABLE PAGE 1.2 4 TITLE Codes Used Earlier Course Work Used 10 . 2 9 2.1 48 4.3 9 4.2.4 49 4.7 67 4.3 49 4.2 48 4.5 57 4.8 70 4.1 8 2.9 71 Front Setback Rear Setback Side Setback Values of slenderness ratio Stress reduction factor for slenderness ratio Calculation of permissible stress Safe allowable load Calculations of loads Abstract Estimate of Conventional Building Abstract Estimate of NZERB Rate Analysis of Proposed Conventional Building Rate Analysis of Proposed NZERB 11 .6 65 4. 3 4.Values 13 14 15 51 54 60 63 12 .6 4.LIST OF FIGURES FIGURE TITLE PAGE 4.4 4.1 4.7 Plot Layout Ground Floor Plan First Floor Plan Footing Design Working of Solar Panels Infrared Thermometer U.2 4.5 4. c D b Mu.lim fck Mu Pt Pc Ast Asc - Effective depth Clear cover Total depth Width Ultimate limiting moment of resistance Characteristic compressive strength of concrete Ultimate moment Percentage of tension reinforcement Percentage of compression reinforcement Area of steel in tension zone Area of steel in compression zone 13 .ABBREVIATIONS deff c. NY.L L. Nq CMDA PWD NBC - Spacing of stirrups Yield stress of steel Total cross sectional area of stirrup legs Modification factor for tension reinforcement Modification factor for compression reinforcement Reduction factors for flanged beams Ultimate load Permissible shear stress Gross area of cross section Length in y direction Length in x direction Ultimate load Bending moment coefficient for short span Bending moment coefficient for long span Moment in short span direction Moment in long span direction Required depth Provided depth Maximum ultimate moment Area of steel required Area of minimum steel required Area of 1 bar Dead Load Live Load Angle of internal friction Bearing capacity factors Chennai metropolitian development authority Public works department National Building Code - - 14 .L Φ NC .max Ast( reqd) Ast (min) ast D.Sv fy Asv kt kc kf Pu τc Ag Ly Lx Wu αx αY Mx My dreq dprov Mu. h A.h - Kudankulam Nuclear Power Plant Watt hour Ampere hour CHAPTER 1 15 .KKNP W. To eliminate the necessity of active energy loads on the building. 1.1 OBJECTIVE i.3 SCOPE i. ii. The use of this technology used in residential buildings has shown huge amount savings in the electricity bill. Now if we are going for NZERB building we can save energy locally which mean to save energy in global level. Comparing the net zero energy building with conventional building. iii. Functional planning of G+1 Residential building 16 . ii.OVERVIEW 1.2 NECESSITY The basic necessities of such a building are: i. iv. To achieve sustainability. Usage of hollow bricks and avoidance of columns and beams will result in lowering of temperature inside the building vi. Design a building with Net zero energy concept. iii. As the country is developing day by day the consumption of power is also very high. 1. The proper design and alignment of the building can make the building cheaper than that of the conventional type of buildings. v. 6 Design of slabs Design of footings Design of wall using Hollow bricks Design of solar panels REALISTIC DESIGN CONSTRAINTS 17 .5 MAJOR DESIGN EXPERIENCE Design experience in the following areas has been gained during the course of the project i. 1. iii. iv. ii. iii. i. iv. Selection of site where renewable energy is available Study the climate conditions of area Aligning the building to utilize maximum amount of renewable resources Planning and design of proposed NZERB building Comparison of the NZERB building with other conventional building 1. v. iii. ii.4 METHODOLOGY This entire project is an planning and design in nature and the methodology followed in this project is listed as below. Comparison of energy consumption between NZERB and conventional building.ii. Design of load bearing structure using hollow bricks Design of solar panels Comparison of room temperature between NZERB and conventional building v. iv. 1. Limit state design method and Fixing of dimensions are shown in Table 1. Economic: Building shall be designed such that the entire energy requirements are met by solar energy only due to shortage of conventional energy. Sustainability Constraints: The design shall be such that the requirement of cooling do not fluctuate throughout the year.i. iii. Limit state design method used for slab and footing Design of Hollow bricks Structural use of Unreinforced Masonry Handbook of Masonry design and Construction IS 875 :1987 -1.1 Reference to codes and standards Codes /Standards Context Design loads for buildings and structures (Dead load . ii. Design co-efficient. 1.2 IS 456 :2000 IS 2572-1963(R 1997) IS 1905 :1987 SP 20 :1991 18 . Imposed load ) Design co-efficient. Economic Constraint: The materials adopted for construction are economical compared to conventional materials.7 REFERENCE TO CODES AND STANDARDS The codes for design of buildings and structures.1 Table 1. Open space. Open space.Elements of building science and Architecture CE0209. Floor area ratio R.C.9 MULTIDISCIPLINARY COMPONENT AND TEAM WORK i. layout and planning and Byelaws. 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. Setbacks. 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.Computer aided building drawing CE0102. Auto CAD MS EXCEL MS WORD 19 .2 Table 1. ii.1. Setbacks. ii.C. Floor area ratio are shown in Table 1.C Design 1.8 APPLICATION OF EARLIER COURSE WORK The codes for Computer aided building drawing.2 Application of earlier course work Course Code and Name CE 0104. 1.C Design R.10 SOFTWARE USED i. iii.Building technology CE0303-Structural Design II CE0304-Structural Design III Context Computer aided building drawing layout and planning Byelaws. analyse and estimation of the parts of the structure. thus we can save a huge amount in electricity bill.12 FUTURE SCOPE OF THE PROJECT The building is designed as a NET ZERO ENERGY BUILDING which produces its own electricity. AVAILABILITY COST ELECTRICITY OF RESOURSES NORMAL CONVENTIONAL BUILDING High Low It requires an active source Produced on its own Easily Available Difficult NZERB 1.11 CONCLUSION The two types of buildings are analyzed with respect to cost. 20 . time.1. availability of skilled labour and ease in construction. Hence requirement has brought in new building technologies by utilizing the renewable energy resources. But the projects have got delayed. In housing aspects it is necessary to design the material adopted structurally in a proportion with reference standard codes. This is the major problem faced.CHAPTER 2 INTRODUCTION 2. Due to increase in consumption of electricity the Tamil Nadu electricity board is unable to fulfill the requirements of the public and industrial sectors . also abiding by the Government Rules.1 GENERAL Fast rate of urbanization and increase in the consumption of electricity has become a major problem in Tamil Nadu. which would generate 14. nuclear and other power projects. with the KKNP turning out to be a big challenge . 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. 21 . from thermal. Designing of building is the most essential work to be proposed in any projects. Most of these should have been completed by 2012. As a whole we have incorporated all the needs for a building to be built with efficient. Officials were banking on a number of projects.000 MW of power. eco-friendly and economic. Overall cost of the project should be economical so estimation of building is very important.In Tamil Nadu. “With energy conservation arrangements. such as highinsulated constructions. According to ANNA. Saitoh.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. Solar energy in buildings include systems that capture heat (such as Solar water heating systems and passive heating).” i. It converts solar energy into electrical energy. Finally this project will end up with the preparation of an estimation of the prepared plan (Ref 1). 2. Extra Energy supply for the electric installations in the house is taken from the municipal mains” (Ref 2).2 LITERATURE REVIEW Anna Joanna. ii. 22 . solar heating system. solar energy. the natural underground coldness and sky radiation cooling are utilized. (1988) (JAPAN) According to SAITOH. For this purpose. Solar panels are designed to harness. Aalborg University. its done with the help of photovoltic (PV) systems (Ref 3). iii. Department of Civil Engineering. “… 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. (d) Parks and playgrounds occupying an area not exceeding 2 hectares. 2004 Primary Residential Use Zone 2. mutton stall and milk kiosks.5m Rear setback according to Chennai Metro Development Authority (CMDA) code is shown in Table 2. buildings shall be permitted only for the following purposes and accessory uses.3. cycle repair shops and tailoring shops.1. newspapers. primary and high school.2.3 DEVELOPMENT CONTROL RULES FOR CHENNAI METROPOLITAN AREA. Front setback according to the CMDA code is shown in Table 2. (a) Professional consulting offices of the residents and incidental uses there to occupy a floor area not exceeding 40 square meters. 23 . Table 2.1 In this primary residential use zone. (e) Taxi stands and car parking.1 Front Set Back Abutting Road Width Front Set Back Above 30m Above 15m but less than 30m Above 10m but less than 15m Below 10m 6. (c) Nursery.2. cigarettes.0m 4.5m 3. tea stalls. (b) Petty shops dealing with daily essentials including retail sale of provisions. soft drinks.0m 1. 3.5m on one side 1. (ii) a stand-by electric generator of adequate capacity for running lift and water pump. fire protections. (iii) a room of not less than 6 meters by 4.5m 3.3 Side Set Back Width of Plot Side Set Back Not more than 6m More than 6m but not more than 9m More than 9m 1.4 meters for every 10 consumers or three floor whichever is less. The meter room shall be provided in the ground floor. building services.4 meters by 2. Table 2. 24 .4 CONFORMATION TO NATIONAL BUILDING CODE OF INDIA In so far as the determination of sufficiency of all aspects of structural designs. and (iv) at least one meter room of size 2.0m on one side 1. Every multi-storied development erected shall be provided with (i) Lifts as prescribed in National Building Code.5m on either side 2. standards and code of practices recommended in the National Building Code of India (Ref. shall be fully confirmed to any breach thereof shall be deemed to be a breach of the requirements under these rules.Table 2.5 meters in area with a minimum head room of 3 meters to accommodate electric transformer in the ground floor.0m 4.5m Side setback according to CMDA code is shown in Table 2. and a room to accommodate the generator. plumbing.2 Rear Set Back Depth of Plot Rear Set Back Up to 15m Between 15m to 30m Above 30m 1. construction practice and safety are concerned the specifications.4). 2. 2. In building of such size. 25 .4. if not.2 Security Deposits The applicant shall deposit a sum at the rate of Rs.4. 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.1 Fire Safety. smoke. and for the satisfaction of the Director of Fire Services by obtaining a no objection certificate from him (Ref.100 per square meters of floor area as a refundable non-interest earning security and earnest deposit.4). 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. The deposit shall be refunded on completion of development as per the approved plan as certified by CMDA. 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. so that they may escape. fumes and also panic arising from these or similar other causes. it would be forfeited. To eliminate the necessity of active energy loads on the building. iii. 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.no 1 2 3 4 5 6 Brick Material Temperature Electricity Initial Cost Solar Panels Energy Efficient NZERB Hollow brick 4 to 5 degree less compared To conventional building Produced on its own High 250 w panels Provided in NZERB Uses less energy CONVENTIONAL Normal brick More than NZERB It requires an active source Less compared to NZERB Not provided Uses more energy 26 . The comparison of NZERB and conventional building is shown in Table 3. 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.1 Comparison of NZERB and Conventional Building Sl. Solar panels is one of the technologies used to achieve net-zero status. Comparing the net zero energy building with conventional building.1 Table 3.1 OBJECTIVE i. Design a building with Net zero energy concept. ii. usually as measured over the course of a year.CHAPTER 3 OBJECTIVE AND SCOPE 3. The hottest part of the year is late May to early June. Design of load bearing structure using hollow bricks. The weather is hot and humid for most of the year. The highest recorded temperature is 45 °C (113 °F) iii. Design of solar panels. v. maximum temperatures is around 35– 40 °C (95–104 °F). ii.3.3 METHODOLOGY This entire project is an planning and design in nature and the methodology followed in this project is listed as below. Comparison of room temperature between NZERB and conventional building. iii. Hence solar energy is available on the site which makes the site suitable to harness solar energy ii. South-facing windows harvest solar energy. Comparison of energy consumption between NZERB and conventional building. i. Functional planning of G+1 Residential building. Study the climate conditions of area The city lies on the thermal equator and is also on the coast. Aligning the building to utilize maximum amount of renewable resources Elongated east-west and oriented to astronomic south (Ref 5).2 SCOPE i. iv. Planning and design of proposed NZERB building Comparison of the NZERB building with other conventional building 27 . iv. v. 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. 3. which prevents extreme variation in seasonal temperature. 4. The key plan of the site is shown in Figure 4.1 Fig.CHAPTER 4 RESULTS AND DISCUSSIONS 4.1 Key Plan 28 .1 PLANNING The key plan of the residential building is drawn by considering the alignment of the building with respect to the CMDA. 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. two bedrooms. one dinning. The Ground Floor plan is shown in Figure 4.2 Fig. one kitchen.The ground floor of the building consist of one hall. The plan has been prepared using Auto CAD software.4.2 Ground Floor Plan 29 . 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.3 Fig.The first floor of the building consist of one hall. two bedrooms.4.3 First Floor plan 30 . one dinning. The First Floor plan is shown in Figure 4. Dinning. Bathroom.56<2 Hence Two Way Slab 2. Load Calculation Assuming Slab Thickness (Ref 7) Assume 10 bar.1 Design of Hall Using M 20 Concrete Fe 415 steel Live Load = 2 1.1 m Aspect ratio = = 1.2 ANALYSIS AND DESIGNS SLAB DESIGN (Ref 6) The analysis and designs of the slab for Hall. 4. Kitchen.26 m Ly = 5. Stair case.04 × 24 = 0.21 31 (Ref 8) .2. Portico are done with proper considerations as per IS 456:2000.4. Clear Cover 20mm Actual Depth (d) = 130-5-20 = 105 mm Assume Floor Finish = 40 mm Weight of Floor Finish = 0.96 Imposed Load = 2 Total Load = 6. Effective Span Lx = 3. Bedroom. 068 Long Span αy = 0.262 = 6.lim= (or) = 0.315 × 3.lim is calculated by equation 4.1 Where.315 Consider 1m width of slab Load per meter Length = 9.127 3.5 × 6.262 = 3. Mu = Moment in short span direction Wu= Ultimate load Lx = Length in x direction Mu(+) Short = 0.9) Mu(+) Long = 0.662 kN.Factored Load (Wu) = 1.2) 32 . Mu. Page No.2 Where.m Take the Highest Moment and check for adequacy of the section. Mu = Wu × Co-efficient × Lx2 (4.21 = 9.138fckb d2 Mu.037 × 9.731 kN.91 of IS456 Two adjacent edges are discontinuous (already found out ) Refer Table 26 Short Span αx = 0.068 × 9.315 × 3. * + (4. Finding Design Bending Moment Refer Table 26. where it is long span or short span only coefficient varies].1) Mu is calculated by equation 4.m (Ref.037 [Note that Lx only to be taken. 598 ] (4.138 × 20 × 1000 × 1052 =30. ii.min 5.1-4.3 Where. 3d = 3 × 105 = 315 mm 300 Max Spacing = 300 mm 33 .27 mm2 Minimum Steel = 0. Ast(+) Short = Area of steel required b = Width d = Effective depth fck = Characteristic compressive strength of concrete R= fy = Yield stress of steel = 6.12% × D × B Ast.Mu. Calculation of Steel Ast(+) Short = [1 .3) Ast(+) Short is calculated by equation 4.lim = Ultimate limiting moment of resistance fck = Characteristic compressive strength of concrete b = Width d = Effective depth = 0.min = ( )× 130 × 1000 = 156 mm2 Ast(+) Short <Ast.1-4.m (Mu Limit) > (Mu Short) Hence its ok 4.42 kN.6105 Ast(+) Short = 1000 × 105 × ( ) × 415 [1 . Check for maximum Spacing i.731 × = 0.598 = 184. Calculation of Ast for Long Span Ast(+)Long [1 .598 × = 109.2 =( ) Fs = 0.62 Modified Basic Value = 20 × 1.4057 Ast(+)Long = 1000 × 95 × ( )× 415 [1 – 4.047 34 .58 × 415× Pt ( ) = 0. 3d = 3 × 95 = 285 mm 300 ] d2 × 952 Spacing = 285 mm 8. Check for Deflection Short Span Lx = 3260 mm Ast(+) Short Basic Value = 116.8 = = 31.3 R= = = 0.37 mm2 Ast(+)Long<Ast. ii. Spacing for all Steel i.598 ] Same as equation 4.min 7.175% Modification Factor = 1.4.62 = 32.1.d for long span bars d = D – Clear Cover – = 130 – 20 – .10 = 95 mm 6.37 mm2 = 20 =240. 127 = 0.96 = 9. Load Calculation Assuming Slab Thickness Assume 10 bar.31.04 X 24 = 0.96 =2 = 6.5 m Aspect ratio = = Hence Two Way Slab 2.127 = 3.2 Design of Bed Room Using M20 Concrete Fe415 steel Live Load = 2 1.2. Effective Span Lx = 3 m Ly = 3.08 = 1. Finding Design Bending Moment 35 .8 Hence its ok 4.5× 6.047 < 32. Clear Cover = 20mm Actual Depth (d) = 125-5-20 = 100 mm Self Weight of a Slab Assume 40 mm Floor Finish Weight of Floor Finish Imposed Load Total Load Factored Load (Wu) Consider 1m width of slab Load per meter Length = 9. 13 × 3.33 kN.09 kN.m Take the Highest Moment and check for adequacy of the section.232 = 4.lim = 0. Calculation of Steel Ast(+) Short = [1 .lim = 0.035 [ Note that Lx only to be taken. Mu.043 × 9.13 × 3.91 of IS456 Two adjacent edges are discontinuous ( already found out ) Refer Table 26 Short Span Long Span αx = 0.2 Mu.6 (Mu Limit) > (Mu Short) Hence its ok 4.m Mu(+) Long= 0.138 × 20 × 1000 × 1002 = 27. Page No.1 Mu(+) Short = 0.3 36 . Mu = Wu× Co-efficient × Lx2 Same as equation 4.043 αy = 0.1-4.lim= * (or) Mu.598 ] + Same as equation 4.Refer Table 26.lim Mu.138 fckbd2 Same as equation 4.232 = 3.035 × 9. where it is long span or short span only coefficient varies ]. 598 × ] Ast(+) Short = = 116.3 R= = 3.10 d= 90 mm 6. Spacing for Steel Ast ×102 = 78.1 × × 1000 = 0. Calculation of Ast for Long Span Ast(+)Long [1 .37 mm2 Minimum Steel = 0.min = × 125 × 1000 = 150 mm2 Ast(+) Short <Ast.12% × D × B Ast.33 × = 0.598 ] Same as equation 4.41 [1 .R= = 4. 3d = 3 X 100 = 300 mm 300 Max Spacing = 300 mm d for long span bars d= D – Clear Cover – d= 125 – 20 – 10/2 .min 7.4.2 mm2 Ast(+)Long<Ast.33 Ast(+)Long = 1000 × 100 × = 93.1-4.5 mm2 * – + 37 .min 5. Check for maximum Spacing i.1. ii. 3 < 34 Hence its ok 4. Fs = 0.95 m Aspect ratio = = = 1.027<2 Hence Two Way Slab 2.Ast(+)Short Ast(+)Long = = × 1000 = 674.22 X 10-3) ×100 = 0.85 m Ly = 3.58 × 415 × Pt ( ) = (1. Effective Span Lx = 3.122% Modification Factor = 1. Check for Deflection Short Span Lx = 3230 mm Ast(+) Short Basic Value = 116.5 mm × 1000 = 842.7 Modified Basic Value = 20 × 1.27 mm 8. Load Calculation Assuming Slab Thickness 38 .7 = 34 32.37 mm2 = 20 =240.2.3 Design of Bed Room Using M20 Concrete Fe415 steel Live Load = 2 1. 125 × 3.m 39 .047 [ Note that Lx only to be taken.203 kN. where it is long span or short span only coefficient varies ].125 Consider 1m width of slab Load per meter Length = 10.5 × 6.d = 120.852 = 7. Page No. Mu = Wu× Co-efficient × Lx2 Same as equation 4.1 Mu(+) Short = 0.048 Long Span αy = 0.31mm = 120 mm Assume 10 bar.75 = 10.125 3.75 Assume 40 mm Floor Finish Weight of Floor Finish = 0.75 Factored Load (Wu) = 1.96 Imposed Load = 2 Total Load = 6. Clear Cover 20 mm D = 120+ +20 = 145mm = 150 mm Actual Depth (d) = 150-5-20 Self Weight of a Slab = = 125 mm ×25 × 25 = = 3.91 of IS456 Two adjacent edges are discontinuous = 1. Finding Design Bending Moment Refer Table 26.04 × 24 = 0.027 (already found out) Refer Table 26 Short Span αx = 0.048 × 10. Check for maximum Spacing i.3 R= = 7.min = ( )× 150 × 1000 = 180 mm2 Ast(+) Short <Ast. Calculation of Steel Ast(+) Short = [1 .lim= (or) = 0.2 = 0.Mu(+) Long = 0.852 = 7.138fckb d2 Same as equation 4.047 × 10.460 ] Ast(+)Short = 1000 × 125 × ( ) × 415 [1 .598 × = 164.125 × 3. 3d = 3 × 125 = 375 mm 300 Max Spacing = 300 mm d for long span bars d= D – Clear Cover – d= 150 – 20 – .m Take the Highest Moment and check for adequacy of the section. ii.138 × 20 × 1000 × 1252 = 43.063 kN.203 × 106 = 0.1-4.08 mm2 Minimum Steel = 0.1-4.125 kN.12% × D × B Ast. Mu.m (Mu Limit) > (Mu Short) Hence its ok 4.598 ] * + Same as equation 4.10 40 .min 5. 3 R= = = 0.6 Hence its ok 41 .8 Modified Basic Value = 20 × 1.8 30.598 ] Same as equation 4. Check for Deflection Short Span Lx = 3850 mm Ast(+)Short Basic Value = 181.58 × 415 × Pt ( ) = 0.553 Ast(+)Long = 1000 × 115 × ( ) × 415 [1 – 4.8 = 36 = = 30.d= 115 mm 6. 3d = 3 × 115 = 345 mm 300 = 300 mm ] Spacing 8.4.1. Spacing for all Steel i.11 mm2 Ast(+)Long<Ast. ii.157% Modification Factor = 1.598 × = 181.min 7.2 =( ) Fs = 0.11 mm2 = 20 =240. Calculation of Ast for Long Span Ast(+)Long [1 .8 < 35. 38 m Ly = 4.4 Design of Bathroom Using M20 Concrete Fe415 steel Live Load = 2 1.625 Consider 1m width of slab Load per meter Length = 8.04 × 24 = 0. Effective Span Lx = 2. Clear Cover 20 mm Actual Depth Self Weight of a Slab d = 110-5-20 = 85 mm = = 2.96 Imposed Load = 2 Total Load = 5.2.75 Assume 40 mm Floor Finish Weight of Floor Finish = 0.798<2 Hence Two Way Slab 2.75 = 8.375 mm = 80 mm Assume 10 bar. Load Calculation Assuming Slab Thickness d= = 74.75 Factored Load (Wu) = 1.5 × 5.4.625 42 .28 m Aspect ratio = = = 1. Mu = Wu× Co-efficient × Lx2 Same as equation 4. Mu.382 = 4.047 × 8.625 × 2.047 [ Note that Lx only to be taken. Page No.798 (already found out) Refer Table 26 Short Span αx = 0.598 ] * + Same as equation 4.lim= (or) = 0.625 × 2.3.138fckb d2 Same as equation 4.m Take the Highest Moment and check for adequacy of the section.127 kN.1-4. where it is long span or short span only coefficient varies ].085 × 8.91 of IS456 Two adjacent edges are discontinuous = 1.94 kN.1-4. Calculation of Steel Ast(+) Short = [1 .m Mu(+) Long = 0.574 ] Ast(+) Short = 1000 × 85 × ( ) × 415 [1 .1527 ×( ) = 0.138 × 20 × 1000 × 852 =19.598 × 43 . Finding Design Bending Moment Refer Table 26.m (Mu Limit) > (Mu Short) Hence its ok 4.3 R = = 4.382 = 2.085 Long Span αy = 0.29 kN.2 = 0.1 Mu(+) Short = 0. 3 R= = = 0. Calculation of Ast for Long Span Ast(+)Long [1 . Check for maximum Spacing i.min= ( ) × 110 × 1000 = 132 mm2 Ast(+) Short <Ast.4082 Ast(+) Long = 1000 × 75 × ( ) × 415 [1 – 4.88 mm2 MaxSpacing = 225 mm 8.1. 3d = 3 × 75 = 225 mm 300 = 86.4.min 5. Check for Deflection Short Span Lx = 2380 mm Ast(+) Short = 139. ii.= 139.min 7.598 ] .92 mm2 Minimum Steel = 0.598 × Ast(+) Long <Ast. Spacing for all Steel i. ii. 3d = 3 × 855 = 255 mm 300 Max Spacing = 255 mm d for long span bars d= D – Clear Cover – d= 110 – 20 – d= 75 mm 6.92 mm2 44 .10 Same as equation 4.12% × D × B Ast. 58 × 415 × Pt ( ) = ( = 0.9 Modified Basic Value = 20 × 1.5 Design of Portico Using M20 Concrete Fe415 steel Live Load = 2 1.Basic Value = 20 = 240.124 mm = 120 mm = 1.1646% Modification Factor = 1. Load Calculation Assuming Slab Thickness d = 118.9 = 38 = = 28 28 < 38 Hence its ok 4.2. Effective Span Lx = 3.83<2 Assume 10 bar.78 m Ly = 6.2 ) ( ) Fs = 0. Clear Cover 20 mm D = 120 = 150 mm = 150-5-20 = 125 mm ×25 Actual Depth (d) Self Weight of a Slab = 45 .93 m Aspect ratio = Hence Two Way Slab 2. 125 × 3. Finding Design Bending Moment Refer Table 26.125 3.2 46 * + .125 × 3.m Take the Highest Moment and check for adequacy of the section.75 Factored Load (Wu) = 1.125 Consider 1m width of slab Load per meter Length = 10.75 = 10. Mu = Wu× Co-efficient × Lx2 Same as equation 4.047 × 10. Mu.75 Assume 40 mm Floor Finish Weight of Floor Finish = 0.1 Mu(+) Short = 0.087 Long Span αy = 0.91 of IS456 Two adjacent edges are discontinuous (already found out) Refer Table 26 Short Span αx = 0.782 = 12.lim= (or) = 0.79 kN.5 × 6.m Mu(+) Long = 0.96 Imposed Load = 2 Total Load = 6. where it is long span or short span only coefficient varies ].782 = 6.04 × 24 = 0.= 3.087 × 10.58 kN.138fckb d2 Same as equation 4. Page No.047 [ Note that Lx only to be taken. min 5.27 mm2 Minimum Steel = 0.1-4.805 [1 . Check for maximum Spacing i.4.125 kN. 3d = 3 ×105 = 315 mm 300 Max Spacing = 300 mm d for long span bars d= D – Clear Cover – d= 130 – 20 – 10/2 . ii.1-4.m (Mu Limit) > (Mu Short) Hence its ok 4.3 R= = = 0.598 ] Same as equation 4.598 × ] Ast(+) Short = 1000 × 125 × = 184. Calculation of Ast for Long Span Ast(+)Long [1 .598 ] Same as equation 4.12% × D × B Ast.= 0. Calculation of Steel Ast(+) Short = [1 .10 d= 95 mm 6.3 R= 47 .1.min = ( )× 130 × 1000 = 156 mm2 Ast(+) Short <Ast.138 × 20 × 1000 × 1252 = 43. 37 mm2 = 20 = 240.62 Modified Basic Value = 20 × 1. ii.047 < 32.175% Modification Factor = 1.598 × = 109.4057 Ast(+)Long = 1000 × 95 × ( )× 415 [1 – 4.8 Hence its ok 4.= = 0.37 mm2 Ast(+)Long<Ast.6 Design of Kitchen Using M20 Concrete Fe415 steel Live Load = 2 1. Spacing for all Steel i.min 7. Effective Span 48 .62 =32 = = 31. 3d = 3 × 95 = 285 mm 300 ] Spacing = 285 mm 8.2 ( ) Fs = 0.58 × 415× Pt ( ) = = 0.047 31. Check for Deflection Short Span Lx = 3260 mm Ast(+) Short Basic Value = 116.2. 73 m Aspect ratio Hence Two Way Slab 2.875 Finding Design Bending Moment Refer Table 26.5 × 5.Lx = 2.96 Imposed Load = 2 Total Load = 5. Page No.91 of IS456 Two adjacent edges are discontinuous = 1.04 × 24 = 0.23 m Ly = 3.25 Assume Floor Finish = 40 mm Weight of Floor Finish = 0.875 Consider 1m width of slab Load per meter Length = 7. Load Calculation Assuming Slab Thickness d= = 65 mm Assume 10 bar. Clear Cover 20 mm D = 65+ +20 = 90 mm Actual Depth (d) = 90-5-20 = 65 mm Self Weight of a Slab = = 2.67 (already found out) Refer Table 26 Short Span αx = 0.25 Factored Load (Wu) = 1.06 49 .25 = 7. 232 = 1. Calculation of Steel Ast(+) Short = [1 .66 kN.35 kN.035 × 7.1-4.88 × 2.735 × = 0. Mu = Wu× Co-efficient × Lx2 Same as equation 4. where it is long span or short span only coefficient varies ].min × 90 × 1000 = 108 mm2 Ast(+) Short <Ast.035 [ Note that Lx only to be taken.m (Mu Limit) > (Mu Short) Hence its ok 3.56 mm2 Minimum Steel = 0.2 = 0.138 fckb d2 Same as equation 4.m Mu(+) Long = 0.1-4.598 ] * + Same as equation 4.55 ] Ast(+) Short = 1000 × 65 × × 415 [1 .3 R= = 2.lim = 0.lim= (or) Mu.232 = 2. Check for maximum Spacing 50 .m Take the Highest Moment and check for adequacy of the section.min 4.598 × = 103.06 × 7.12% × D × B Ast. Mu.138 × 20 × 1000 × 652 =11.Long Span αy = 0.88 × 2.1 Mu(+) Short = 0.373 kN. Calculation of Ast for Long Span Ast(+)Long [1 .05 mm2 Ast(+) Long <Ast. Spacing for all Steel Ast = ] × 102 = 78.1. ii.16% 51 .i.10 Same as equation 4. 3d = 3 × 65 = 195 mm 300 Max Spacing = 195 mm d for long span bars d= D – Clear Cover – d= 90 – 20 – d= 55 mm 5.58 × 415 × Pt ( ) = = 0.5 mm2 × 1000 = 758.4.598 ] .56 mm2 = 20 = 240.373 × = 0.3 R= = 1.01 mm × 1000 = 1104.7 Fs = 0.min 6.115 Ast(+) Long =1000 × 55 × × 415 [1 – 4.598 × = 71. Check for Deflection Short Span Lx = 2230 mm Ast(+) Short Basic Value = 103.8 mm Ast(+) Short = Ast(+) Long = 7. 96 Imposed Load =2 52 .04 × 24 = 0.625 Assume Floor Finish = 40 mm Weight of Floor Finish = 0. Clear Cover 20 mm D Actual Depth (d) = 105-5-20 = 80 mm Self Weight of = 2.4. Load Calculation Assuming Slab Thickness d= Assume 10 bar. Effective Span Lx = 2.Modification Factor = 1.6 m Ly = 3.7 Design of Dinning Room Using M20 Concrete Fe415 steel Live Load = 2 1.8 Modified Basic Value = 20 × 1.73 m Aspect ratio Hence Two Way Slab 2.8 = 36 34.3 < 36 Hence its ok 2. m Take the Highest Moment and check for adequacy of the section.028 × 8.049 × 8.2 Mu.595 kN.049 Long Span αy = 0. Page No.91 of IS456 One edge discontinuous = 1.lim= (or) Mu. where it is long span or short span only coefficient varies ].66 kN. Mu.Total Load = 5. Mu = Wu× Co-efficient × Lx2 Same as equation 4.43 3.62 = 2.43 (already found out) Refer Table 26 Short Span αx = 0.62 = 8.62 Factored Load (Wu) = 1.028 [ Note that Lx only to be taken.43 Consider 1m width of slab Load per meter Length = 8.m (Mu Limit) > (Mu Short) Hence its ok 4.43× 2. Calculation of Steel * + 53 .lim = 0.m Mu(+) Long = 0.79 kN.lim = 0.5 × 5. Finding Design Bending Moment Refer Table 26.62 = 1.138 × 20 × 1000 × 802 Mu.1 Mu(+) Short = 0.lim =17.138fckb d2 Same as equation 4.43 × 2. 15 mm2 ] Minimum Steel = 0.325 Ast(+) Long =1000 × = 64.435 × 415 [1 .1-4. Calculation of Ast for Long Span Ast(+)Long [1 .598 ] .min= × 105 × 1000 = 126 mm2 Ast(+) Short <Ast.10 Same as equation 4.598 ] Same as equation 4.min 5.79 × = 0.598 Ast(+) Short = 1000 × 80 = 99.34 mm2 – 4.4.598 × ] 54 . Check for maximum Spacing i. 3d = 3 × 80 = 240 mm 300 Max Spacing = 240 mm d for long span bars d= D – Clear Cover – d= 105 – 20 – d= 70 mm 6.12% × D × B Ast.595 × = 0.Ast(+) Short = [1 .3 R= = 2.1-4. ii.1.3 R= = 1. 14% Modification Factor = 1.5 mm2 Ast(+) Short = Ast(+) Long = 1000 = 791.7 mm × 1000 = 1200.Calculation of Loads Maximum short span Width of corridor Height of the storey Live load 2. Spacing for Steel Ast = × 102 = 78.min 7.15 mm2 = 20 Modified Basic Value = 20 × 1.08 mm 8.Ast(+) Long <Ast.5 32.8 ( ) = 99.8 = 36 = 32.2.60 m = 1.50 m =3m =2 55 . Pt = = 0. Check for Deflection Short Span Lx = 2600 mm Ast(+) Short Basic Value Fs = 0.8 Design of Wall Design of a wall 1.5 < 36 Hence its ok 4. Assumptions = 3. 5 m Height of the Plinth above Footing = 1 m Height of the Parapet Wall Thickness of Roof Slab Brick Size =1m =110 mm = 230 × 115 × 75 3.3 = 1.797 m Slenderness Ratio 4. Permissble Stress Fc = Ks × Ka × Kp× Basic compressible stress Fc is calculated by equation 4.1.5X0.2 m2 Ka = 1 ( From clause 5.5 A = 0. Ks= Stress reduction factor Ka= Area reduction factor Kp= Shape modification factor (4.73 = 5. Slenderness Ratio and Stress Factor Ground Floor + First Floor H = 3+0.15 A > 0. Shape modification factor: Crushing Strength of Modular Brick = 5 Shape Modification Factor = Kp = 1 ( From table 10 of IS: 1905-1987) 5.73 m Effective Height (h) = 0.75 × 7.Height of the Plinth from ground = 0.4 Where.7 + 1.5+3+0.115+1 = 7.4) 56 .7 + 1. Area reduction factor: Area Reduction Factor Ka = 0.75H = 0.46 ( from table 9) 7.115+0.2) 6.4. Stress Reduction Factor: ks = 0. 81 = 67.4 % Thickness = 1 Brick thick wall (using nomograms) 11.22 m3 Total wall volume 10.5×2×3) = 29.5) = 91.098+1.89+1.746+1.25 m3 % Opening = = 35.5 57 . Safe Load Q=( = = 63 9.098 = 14.31×2) + (8.4884+1.87×3) + (4.05 m3 ) Wu is calculated by equation 4.( ) 2} (4.81 m3 Total wall volume – Total Deductions = 91.Fc = 0.44 × 0.4884+1. For Hall : Wu= × {3.93×2))×3 = 40.042 m3 Inner Deductions = 1.87×3) + (3.05 -23.089+2.4884+ 1.22 + 21 = 50.746 = 9.89+2.496+1.089+1.4884+1. Wall Area Outer wall = Total Perimeter x 3(floor height) = ((11.768 m3 Total Deduction =23.48 × 1 × 1 = 8.83 m3 Inner wall = (4.4884+1. Deductions: Outer Deductions = 1.226+1. [ ( ( ) 2 ) ]} = 9.Where.055 + 9.3 = 70 58 .98 Hence the design is ok 4.127 × × {3.81 = 33.2.98 63>33.* + } = 11.11 + 11.9 Design of Footing Load from Walls = 126. Wu= Factored load W=Load from the slab Lx=Short span Ly=Long span Wu =( ( )) * } Wu = 13.055 103 For dinning room: Wu= = ( × {3.*( )+2} ( )) × {3.81 103 103 Total : 13.7 10% for the weight of the Building = 63+6.11 103 For bed room: Wu= * ( ( ) } )) = (9. C beyond the brick work should not be more than ½ of the thickness of P.C thickness Provide = 300 mm The Projection of P.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 Step 2: Assumptions 59 .C.C.47 m2 Consider 1m Length room Breadth of the Footing Required = 2.47 Actual work of Brick work = 760 – 300 = 460 mm Brick work projection beyond the wall 1.C= 760 mm It is customary to provide 150 to 300 mm P.C Projection = = 150 mm = 0. Area of Footing = Assume SBC = 150 A = = 0.C.2.C. Minimum Width = (2w+300)mm = (2x230+300) = 760 mm Provide Width of P.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. 88 m (continuous on one end & discontinuous on other end) Wall E = 2.907 m (continuous on one end & discontinuous on other end) Wall I = 3.23×0.9 =2.9 =1.2×0.82 × 0.15 × 0.9 =2.9=3.438 m (continuous on one end & discontinuous on other end) Wall K = 3.9=3.52 m (continuous on one end & discontinuous on other end) Wall G = 3.5 m =1m =1m = 0.584 m (continuous on both ends & supported by cross wall) Wall C = 3.Height of the Plinth from ground Height of the Plinth above Footing Height of the Parapet Wall Thickness of Roof Slab Hollow Brick Size = 0.7 m (discontinuous on both ends and braced by cross wall) Wall H = 3.20 ×0.57×0.120 m = 0.40 × 0.8=2. 20 m EFFECTIVE LENGTH OF WALL (From Table 5 of IS 1905-1987) Wall A = 3.33 m (continuous on one end & discontinuous on other end) Wall D = 3.935 m (continuous on one end & discontinuous on other end) Wall M = 5.438 m (continuous on one end & discontinuous on other end) Wall B = 3.8=2.438 m (continuous on one end & discontinuous on other end) Wall J = 3.8×0.23×0.82×0.9 =3.9 =2.7×0.056 m (continuous on both ends & supported by cross wall) Wall F = 2.07 m (continuous on one end & discontinuous on other end) 60 .72 × 0.976 m (continuous on both ends & supported by cross wall) Wall L = 2.8=2. 2 m Effective height Slenderness ratio = 0.75 × H = 3.1.2 × 1× 0.74( From table 9) Step 6: Area reduction factor Gross area = 200 × 1000 = 200000 mm2 A = 0.2 m2 Ka= 1( From clause 5.6+1 = 4.2) Step 7: Permissible stress Fc= 0.6+0.44 = 0.6+0.39 Safe allowable load per meter length is q = 0.8 = 3.4.15 m = = = 15.2( From table 10 of IS: 1905-1987) Step 5: Stress Reduction Factor ks = 0.75 Step 4: Shape modification factor Crushing Strength of Hollow Brick= 4.Step 3: Slenderness ratio and stress factor Ground floor: H = 2.39 × 2 × 105 = 78 Step 7: Slenderness ratio and stress factor First floor: H = 2.1 = =1 Shape Modification Factor = Kp =1.4 m 61 .74 × 1. 1 = = = 1 Shape Modification Factor = Kp =1.81 Step 10: Area reduction factor Gross area = 200 × 1000 = 200000 mm2 A = 0.Effective height = 0.75 × H = 2.1 Values of slenderness ratio (Ref 10) Brickwork Ground floor First Floor 62 .81 × 1.55 m Slenderness ratio = = = 12.427 × 2 × 105 = 85 The values of slenderness ratio for effective length and height of the building is given in Table 4.44 = 0.75 Step 8: Shape modification factor Crushing Strength of Hollow Brick = 4.427 Safe allowable load per meter length is q = 0.1 Table 4.2 Step 9: Stress Reduction Factor ks = 0.2 × 1 × 0.2 m2 Ka= 1 Step 11: Permissible stress Fc= 0. 81 E 0.15 3.6 12.55 2.15 3.55 2.4 10.75 12.81 L 0.55 2.28 12.75 0.75 10.3 Table 4.89 0.75 12.55 L 3.15 3.976 1.75 14.44 3.81 D 0.28 12.74 0.92 15.88 2.7 2.55 2.75 12.056 2.976 1.90 3.44 2.15 3.74 0.33 2.H A B C D E F G H I J K L M 3.7 2.R 15.15 L 3.44 2.55 2.75 14.75 9.89 F 0.44 3.88 0.81 I 0.81 K 0.5 15.75 14.6 15.3 Calculation of permissible stress 63 .75 0.75 0.58 3.81 J 0.675 12.33 2.88 9.81 B 0.55 2.15 3.74 0.81 C 0.75 0.2 Table 4.15 3.55 2.55 2.15 3.935 5.58 3.44 2.15 3.07 S.75 12.15 3.75 H 2.88 The calculation of permissible stress of the building is given in the Table 4.75 The values of stress reduction factor for slenderness ratio of the building is given in Table 4.75 15.52 3.75 12.75 12.75 12.55 2.056 2.15 3.55 2.07 S.75 12.81 0.83 G 0.74 0.52 3.88 2.44 2.2 Stress reduction factor for slenderness ratio Wall type Ground floor First floor A 0.83 0.R 12.81 H 0.55 2.7 15.9 3.15 3.55 2.15 3.935 5. 74=0.6 78 79.528×0.4 92 85.469 0.528×0.528×0.81=0.74=0.427 0.88=0.390 0.88=0.6 85.81=0.528×0.528×0.427 0.81=0.4 85.528×0.81=0.2 79.396 0.528×0.83=0.528×0.4382 0.528×0.4 85.89=0.4 85.2 93 87.75=0.8 87.528×0.75=0.528×0.427 0.528×0.74=0.81=0.first floor(N/mm2) floor(N/mm2) 0.4 Table 4.75=0.4 Safe allowable load Wall type A B C D E F G H I J K L M q = fc×2×105 kN/m(ground floor) 78 85 78 79.427 0.4 85.4 85.81=0.81=0.427 0.528×0.89=0.4 85.75=0.528×0.396 0.390 0.83=0.74=0.75=0.427 0.528×0.528×0.4646 0.528×0.2 92 78 q = fc×2×105 kN/m(first floor) 85.528×0.ground Permissible stress.427 0.390 0.427 The values of safe allowable load for the building is given in Table 4.81=0.4 4.528×0.427 0.427 0.528×0.528×0.528×0.528×0.2 78 79.4646 0.390 0.396 0.427 0.81=0.Wall type A B C D E F G H I J K L M Fc=ks×kp×ka× basic compressive stress Permissible stress.469 0.81=0.4382 0.81=0.2.390 0.528×0.528×0.4 93.396 0.528×0.11 Design of Footing for Hollow Brick wall: (Ref 11) 64 .4 85. C.125 + 4.Load from Wall = 78 Load from wall (critical wall M) +10% for the weight of the Building + weight of slab (hall.C.25 × 100=125 1.833 m2 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 65 .C thickness Provide = 300 mm The Projection of P.8 + 4.C = = 150 mm = = 0.5635 + 1 =100.C beyond the brick work should not be more than ½ of the thickness of P.236 =100 Factored load= 1.C.6575 + 4.C = 700 mm It is customary to provide 150 to 300 mm P. Area of Footing = Assume SBC = 150 Consider 1m Length room 2. Minimum Width = (2w+300) mm = (2 x 200+300) = 700 mm Provide Width of P. bed room & dining) + floor finish = 78 + 7.C. 83 (4.4.4 Footing Design 4.2. 66 .4 Fig.The thickness of each step is given by hollow brick = 200 mm The footing design is shown in the Figure 4.12 Length Live load Rise Thread Design of Stair Case: =4m =2 = 150 mm = 250 mm Using M20 Concrete and Fe415 Step1: Calculation of self weight Assume waist slab thickness = D = 200 mm Self weight = × × 2 = 5.6) = 200 mm Self weight is calculated by equation 4.6 Where. 55 × 24 = 1 ×R×T× =3 = 5.D =Diameter R =Rise T =Thread Step 2: Calculation of load on waist slab 1.5 × 11. Live load 4.875 Step3: Calculation of Mu Mu Mu.lim = = = 35 kN.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 67 .m * + 35 × 106 = 0. Weight of steps= 3.075 Wu=1.36 × 0.075 =17. Self weight Wu = 11.42 × 0.83 × 25 = 1. Assume 40 mm Floor finish Floor Finish = 2.48(1-0. R= Ast = Number of Bars = Ast actual = 4 × Pt = × 1000 = = 2.166 mm Step 6: Providing distribution steel Astmin= × = 480 mm2 × × 82 = 270 mm Spacing of 8mm diameter = Main steel = 4No.58 × 415 × = 185.43 = 970.04% .89 For Pt =1.58fy = 0.6 mm2 = 4 bars × 202 = 1256 mm2 × 1000 = 1.04% Step 5: Check for deflection Basic value = 20 Fs = 0.2 (Fig 4 of IS 456-2000) drequired = dactual = 120 mm = = 104.s 20 bars Distribution = 8 mm dia bars @ 270 mm c/c 68 . Modification factor = 1. 5 Fig.3. AC can be directly used for the appliances.3 4.4.4. 69 . A basic solar powered system is shown in Figure 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.1 DESIGN OF SOLAR PANEL AND ITS COMPONENTS Solar Power System Components Brief revision of the major components found in a basic solar power system. AC appliances need a power inverter to convert the DC electricity into 220 Volt AC power. 70 . This means that an 80W solar panel based on the average figure of 6 sun hours per day. Overcharging causes gassing and loss of electrolyte resulting in damage to the batteries.3. This prevents the battery from permanent damage and reduced life expectancy.3. which will switch off the supply to the load if the battery voltage falls below the cut-off voltage. As an example. DC appliances can then be powered directly from the battery. 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). is to regulate the current from the solar panels to prevent the batteries from overcharging. in Tamil Nadu. Different geographical locations receive different quantities of average peak sun hours per day. the amount of current flowing to the battery.3 Description of individual solar power components 4. 4. or charge controllers as they are also called. ii.3. 4.3. i. Most solar regulators also include a Low Voltage Disconnect feature. or decrease.4.1 Solar Panels Solar panels are classified according to their rated power output in Watts. Panels are rated at a nominal temperature of 25 degrees Celcius.2 Solar Regulator The purpose of solar regulators.25% for every 5 degrees variation in temperature. Solar panel output is affected by the cell operating temperature. iii.3.3. would produce a yearly average of around 480W.2 Working of solar panels The solar panel converts sunlight into DC power or electricity to charge the battery.H per day. the annual average is around 6am sun hours per day. The output of a solar panel can be expected to vary by 0. A Solar regulator is used to sense when the batteries are fully charged and to stop. Solar regulators are rated by the amount of current they are able to receive from the solar panel or panels. 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.3. This would equate to 1. Discharging beyond this level will significantly reduce the life of the batteries.4. This rating also includes a discharge rate. 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. Calculation of Loads The calculation of loads for the Solar Panels are given below in Table 4.2A per hour for 100 hours. Power inverters are generally rated by the amount of AC power they can supply continuously. This rating specifies the amount of current in Amps that the battery can supply over the specified number of hours.3. usually at 20 hours. 4. To maximize battery life. 50 % capacity remaining.e. i.4 Designing of Solar Panel Power rating of each appliance that will be drawing power from the system. There are three waveforms produced by modern solid state power inverters. Manufacturers generally also provide 5 second and ½ hour surge figures.3 Power Inverter The power inverter is the main component of any independent power system which requires AC power.g.3.5 71 . The power inverter will convert the DC power stored in the batteries and into Ac power to run conventional appliances. power from the grid. True Sine wave inverters provide AC power that is virtually identical to.H at the 100 hour rate can supply a total of 120A.H over a period of 100 hours. a battery rated at 120A. As an example. these are very rare. unlike conventional car batteries which are designed to provide a large amount of current for a short amount of time. as many appliances will not operate on a square wave. The simplest. Today.3. 100 hours) and recharged hundreds or thousands of times. Deep cycle batteries are rated in Ampere Hours (Ah). a square wave power inverter. and often cleaner than.3. deep cycle batteries should not be discharged beyond 50% of their capacity. used to be all that was available. 4. Table 4.5 Calculation Of Loads (Ref 12) PARTICULA RS HALL ITEMS CFL (Ref 13) FAN T.V CFL FAN CFL FAN OVEN CFL EXHAUS T Mixer AUTOFRIDGE CFL FAN CFL HEATER CFL UNITS 4 2 1 2 1 2 1 1 3 1 1 1 3 1 1 1 1 1 1 USAGE IN HRS 5 5 5 3 10 3 10 1 4 4 1 18 4 3 1 1 2 1 2 VOLTA GE W 20 50 80 15 50 15 50 900 15 50 450 150 15 50 15 150 15 750 90 CONSUM PTION 400 500 400 90 500 90 500 900 180 200 450 2700 180 150 15 150 30 750 180 8365 INVER TORS 80 100 80 30 50 30 50 900 45 50 450 195 45 50 15 150 15 750 90 3175 BED ROOM 1 BED ROOM 2 KITCHEN DINING ROOM TOILET 1 TOILET 2 WATER PUMP WASHING MACHINE 72 . 7x2x 250W solar panels 4 x 20A solar regulators .2. The regulators are put in series 14 x 4.Ah 105Ah batteries. thus 14x 4. Number of Batteries 250W panels produce 4. should be discharged to no more than 50%.2 So 14 solar panels would need 4 x 20 A solar regulators . thus we divide total amps by 105A x 50% = 50A. this would mean 3 in series of 3 batteries.8Amps.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.8 A = 67.8Amps. One 250W panel produces around 4. Size of Regulators Let‟s say we had 20A regulators at our disposal. A modified sine wave inverter with a 3500W continuous power rating will therefore be your obvious choice in this specific solar system design. (8365×1. ii.2)\6=1673 WATTS 250 Watt Solar Panel Total watt/ 250 watt solar panel = =7 PANELS = 7 x 250 W panels. 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.h = 8. For ease of possible 24V or 48V configuration. Complete the solar power system Well we have the following: i.8A=67.08 x 105Ah batteries.2A x 6 Hrs = 403. regulator. x 105A.H deep cycle batteries( 3 in series) 1 x 3500W modified sine wave power inverter 4.32/W Regulator Batteries Inverter Total Cost Solar panels =14x250x32=Rs 112000 Regulator = Rs 1800 Batteries = Rs 8000x3=24000=Rs 24000 Inverter = Rs 4800 = Rs 1800 = Rs 8000/series = Rs 4800 Total=112000+1800+24000+4800= Rs. 142600/- The total cost of the solar panel is Rs. decrease in temperature for using of hollow bricks and solar panels produces the electricity.In these solar panel cost is based on the solar panels. The output of solar panel can be expected to vary by 0. When compared to conventional building. the intial cost is high but in future the electricity cost is reduced. In NZERB.4 RATE ANALYSIS Solar panels =Rs.25% for every 5 degrees variation in temperature. batteries and inverter. 4. One lakh forty two thousand six hundred for our residential building . iv.iii.5 INFRARED THERMOMETER ii . to describe the device's ability to measure temperature from a distance refer Figure. 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. They are sometimes called laser thermometers if a laser is used to help aim the thermometer. By knowing the amount of infrared energy emitted by the object and its emissivity. II. Infrared thermometers are a subset of devices known as "thermal radiation thermometers". the object's temperature can often be determined. which converts the radiant power to an electrical signal that can be displayed in units of temperature after being compensated for ambient iii . or non-contact thermometers or temperature guns. IV.6 III.6 Infrared Thermometer The instrument Infrared Thermometer is shown in Figure 4. V. The most basic design consists of a lens to focus the infrared thermal radiation on to a detector.6 I.Fig.4.4. Checking for hot spots in fire fighting situations Monitoring materials in process of heating and cooling. The distance-to-spot ratio (D:S) is the ratio of the distance to the object and the diameter of the temperature measurement area. X. for calibration and control purposes Detecting hot spots / performing diagnostics in electrical circuit board manufacturing XI. This configuration facilitates temperature measurement from a distance without contact with the object to be measured. for research and development or manufacturing quality control situations XIII. VI. measurement of an object 12 inches (30 cm) away will average the temperature over a 1-inch-diameter (25 mm) area. Detecting clouds for remote telescope operation Checking mechanical equipment or electrical circuit breaker boxes or outlets for hot spots IX. iv . which measures the temperature at a spot on a surface (actually a relatively small area determined by the D:S ratio). XII. XV. Infrared thermometers can be used to serve a wide variety of temperature monitoring functions.temperature. VIII. Checking heater or oven temperature. A few examples provided to this article include: VII. For instance if the D:S ratio is 12:1. The sensor may have an adjustable emissivity setting. which can be set to measure the temperature of reflective (shiny) and non-reflective surfaces. The most common infrared thermometers is the: Spot Infrared Thermometer or Infrared Pyrometer. XIV. VII. X.6. IV. III. Hollow brick is same in size as that of concrete blocks 1 Hollow brick = 9 Clay Bricks Less mortar joints.6 CHARACTERISTICS OF HOLLOW BRICKS 4. VI. III.1 kg DENSITY: 694 kg/m³ COMPRESSIVE STRENGTH : 4. V. II.1 WATER ABSORPTION : 15% U-VALUE : 1. IX. V. II. III. hence less plumb & alignment Faster construction Light Weight I. III. IV.4. II. Transportation Saves labour Less dead load. IV. Savings in Structural Cost (Steel & Concrete) by 10 to 15% v . Ease of handling. LENGTH : 400 mm WIDTH : 200 mm HEIGHT: 200 mm WEIGHT: 11.1 Parameters of Hollow Brick Used In Net Zero Energy Residential Building I. II. IV. VIII. 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. VI.1 W/m² SOUND INSULATION : 46 DB FIRE RESISTANCE 240 min Available Sizes I. Thermal Insulation I. vi . ii.2 Advantages Of Hollow Bricks i.7 Fig. Savings on mortar Low „U‟ Values – 1. II. III.6. Colour and brilliance of masonry withstands outdoor elements.0 W/m² Better Thermal Insulation = less energy loss through walls Savings on Energy consumption .4. Highly Durable: The good concrete compacted by high pressure and vibration gives substantial strength to the brick. IV.Lesser the Value higher the Insulation and vice versa. Low Maintenance.Comfortable inside temperature U-value determines thermal Insulation. U-values are mentioned in Figure 4.7 U-VALUES 4. when compared to the other conventional construction systems.7 ESTIMATION 4.48 22.6 The abstract estimate of conventional building is given in Table 4.7. air which makes easy work of construction. water.6 Abstract estimate of conventional building S. v. World class and best walling materials. Less wastage through half bricks. It is a faster and easier construction system. ix. viii.53 0.No 1 Description Excavation Exterior wall Interior wall vii 1 1 39. Constructional Advantages i. iv.76 0.76 0. 4. ii.1 Abstract estimate of conventional building The quantities of the various materials in conventional building are calculated as shown in the Table 4. vi.96 Nos Length Breadth (m) (m) Depth (m) Quantity (m3) .53 15. iii.06 24. No additional formwork or any special construction machinery is required for reinforcing the hollow brick masonry. x. vii. Only skilled labour is required for this type of construction. So it keeps house cool in summer and warm in winter.51 0. fly ash used as one of the raw materials. Excellent thermal insulation. Fire Resistant Provide thermal and sound insulation: The air in hollow of the brick. v. iv.iii. fire. Hollow brick consist of four elements earth. Environment Friendly. Faster construction and ease for handling at site. does not allow outside heat or cold in the house. Savings on RCC-Frame structure (steel/concrete).9 9.6 Table 4. Reduction in energy consumption. 51 1.346 1.23 0.87 2.23 0.15 3.115 7.48 22.5 7.22 0.5 3.89 6.115 0.19 1.25 4.6 7.1 6 1×2 viii 1.5 3.81 1.473 3 1×2 1×5 1×2 1×2 1×3 1 1 1 1.37 2.5 0.51 39.43 6.87 2.5 4.23 0.5 5 1 1 1 1 1 1 3.1 4 Earth Filling Hall Water closet Bed Room1 Bed Room2 Dinning Kitchen Flooring Concrete Hall Water closet Bed Room1 Bed Room2 Dinning Kitchen R.48 0.1 2 2.48 22.03 3.5 3.08 2.05 0.115 0.23 0.23 0.C.37 2.035 0.23 0.5 0.03 3.5 2.21 2.C.5 0.13 14.72 3 2.2 0.46 0.5 0.62 3.1 38.23 0.1 0.8 68.4 0.1 2.3 9 5.4 0.22 1.62 3.48 22.91 5.6 0.31 0.15 3.C Exterior wall Interior wall 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.C Lintel & Sun shades Door D 1 1 1 1 1 1 3.2 P.778 1.39 0.62 3.5 4.9 0.5 0.13 3 1 1 1 1 1 1 39.51 39.6 0.378 3.23 0.55 31.23 0.73 5.1 0.5 0.75 0.115 0.84 0.76 0.62 3.035 1.23 0.1 0.31 0.147 4.22 1.5 3.9 0.9 1.72 3 2.15 0.82 112.46 0.1 0.9 1.23 0.3 0.51 0.08 1.48 22.649 1.57 0.1 0.078 .829 0.76 0.51 0.71 0. Bed Room1 Bed Room2.No 1 Description Excavation Exterior wall Interior wall P.23 0.23 0.104 0.9 0.45 0.Kitchen 7 Plastering Exterior wall Interior wall Deductions Window W1 Window W2 Window W3 Door D Door D1 Door D2 Spacing S Spacing S1 8 9 White Washing Colour Washing 1×3 1 1 1×2 1×2 1×5 1×5 1×2 1×2 1.9 - - 0.23 0.72 12.89 26.75 0.52 1.08 2.96 2 1 1 39.52 1.8 1 1 6.51 0.196 7.45 0.15 0.04 0.97 0.8 464.87 7.3 ix .04 0.Door D1 Door D2 Sun Shade Window W Sun Shade Window W1 Sun Shade Window W2 Sun Shade Roof Slab Hall.C.083 0.1 0.7 0.5 1×2 1×5 1×2 1×2 1×3 1 1 1 - 1.7 Depth (m) 0.2 1.52 1.15 0.22 1.1 2.604 8. Water Closet.8496 296.45 0.496 1.075 0.87 21.5 0.2 0.442 2.65 3.15 0.48 22.89 2.7 0.22 1.24 22.23 0.15 0.5 4.9 1.103 0.3 0.05 1.9 1.081 1.1592 3.51 0.52 1.1 1 1 39.1 2 2.23 0.2 1.12 438.24 22.1 - The abstract estimate of NZERB is given in Table 4.22 0.5 7.24 4.15 0.528 4.1 168.21 2.24 2.7 Abstract Estimate of NZERB S.075 0.84 0.2 0.5 1.51 - 7.92 2.2622 0.45 0.075 0.C Exterior wall Interior wall NoS 1 1 Length Breadth (m) (m) 39.79 6.075 0.734 7.2 1.7 Table 4.7 0.26 0.108 0.8 438.Dinning.5 Quantity (m3) 13.178 3. 3 3 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 Earth Filling Hall Water closet Bed Room1 Bed Room2 Dinning Kitchen Flooring Concrete Hall Water closet Bed Room1 Bed Room2 Dinning Kitchen R.C.C Lintel & Sun shades Door D Door D1 Door D2 Sun Shade Window W 1 1 1 1 1 1 1×2 1×5 1×2 1×2 1×3 1 1 1 39.24 22.51 39.24 22.51 39.24 22.51 1.22 1.22 0.9 0.84 0.75 0.9 1.2 0.9 0.4 0.4 0.25 0.25 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 7.5 7.5 0.9 1.22 1.21 2.1 2 2.1 2.08 2.1 3.13 1.8 1.96 1.12 58.86 33.76 0.4392 1.4884 0.432 0.705 0.9 0.378 0.504 0.378 95.02 7.378 3.89 6.73 5.25 4.147 4.55 31.649 1.48 0.778 1.346 1.05 0.829 0.91 5.473 4 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 5 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 6 1×2 1×3 1 1 1×2 x 1.4 1.05 1.2 1.2 1.52 0.23 0.23 0.23 0.45 0.23 0.15 0.15 0.15 0.075 0.15 0.0684 0.0945 0.036 0.0405 0.07512 Sun Shade Window W1 Sun Shade Window W2 Sun Shade Roof Slab Hall, Water Closet, Bed Room1 Bed Room2,Dinning,Kitchen 7 Plastering Exterior wall Interior wall Deductions Window W1 Window W2 Window W3 Door D Door D1 Door D2 Spacing S Spacing S1 8 9 White Washing Colour Washing 1×2 1×5 1×5 1×2 1×2 1.52 1.52 1.52 1.2 1.2 0.45 0.23 0.45 0.23 0.45 0.075 0.15 0.075 0.15 0.075 0.1026 0.228 0.0256 0.072 0.081 0.8237 3.24 2.79 6.8496 294.3 168.82 463.12 2.196 7.442 2.178 3.528 4.5 1.89 2.496 1.89 26.12 437 437 1 1 6.65 3.5 4.87 7.97 0.1 0.1 1 1 39.24 22.51 - 7.5 7.5 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 - 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 WORK QTY in m³ QTY in cft RATE PER AMOUNT 1 2 3 4 Earth Work Excavation Sand Filling with good river sand PCC 1:5:10, Brick Work in C.M. 1:5, using country brick For Basement level Flooring Work PCC 1:4:8 R.C.C (LINTEL,SUNSHADES & ROOF SLAB) Plastering in C.M 1:4, Inside and outside wall surface White washing Colour washing Steel 24.96 31.65 14.13 882.33 1118.82 499.51 9.50 35.00 90.00 Cft Cft Cft 8382.00 39158.00 44955.00 112.39 3973.00 90.00 Cft 357570.00 5 5.50 194.23 90.00 Cft 17480.00 6 7.19 253.91 350.00 Cft 88868.50 7 8 9 10 439.00 15518.60 439.00 15518.60 439.00 15518.60 501kg 30.00 3.00 5.00 60.00 TOTAL Sft Sft Sft kg 465559.00 46555.80 77593.00 30060.00 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 WORK QTY in m³ QTY in cft RATE PER AMOUNT 1 Earth Work Excavation Sand Filling with good river sand PCC 1:5:10, Brick Work in C.M. 1:5, using country brick For Basement level Flooring Work PCC 1:4:8 R.C.C (LINTEL,SUNSHADES & ROOF SLAB) Plastering in C.M 1:4, Inside and outside wall surface White washing Colour washing Steel Solar Panel System 21.60 763.56 9.50 Cft 7253.82 2 3 4 31.65 12.96 1118.80 458.10 35.00 90.00 Cft Cft 39158.00 41229.00 95.50 3376.00 125.00 Cft 422000.00 5 5.50 194.23 90.00 Cft 17480.50 6 6.84 241.55 350.00 Cft 84542.50 7 8 9 10 11 463.12 437.00 437.00 501.00 16371.29 15448.00 15448.00 30.00 3.00 5.00 60.00 Sft Sft Sft kg 491139.00 46344.00 77240.00 30060.00 142600.00 TOTAL 1399046.82 CHAPTER 5 xiii Thus we can conserve electricity locally and globally.The plan of the building was prepared by Auto-Cad software.The design for the building should be such that the requirement of temperature regulation does not fluctuate throughout the year. http://en.CONCLUSION 5. IS 456:2000 code book was used to design Slab and Footing. thus it can save a huge amount in electricity bill. http://zeb. REFERENCES 1. It would release zero carbon content that would help in controlling global warming).buildinggreen. http://energy.com/saitoh xiv . 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. Design of wall was done by using IS 1905:1987.gov/energysaver/articles/annajohanna 3.1 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 . Hence by using the renewable resources the impact on the active energy loads can be reduced. These kind of buildings are environmental friendly reducing the environmental hazards (eg.2 FUTURE SCOPE OF THE PROJECT The building designed as a NET ZERO ENERGY BUILDING produces its own electricity.wikipedia.org/wiki/Zero-energy_building 2. 5. 8. New Delhi. IS: 456 : 2000. Bureau if Indian Standards.solarpanel. New Delhi 7. SP 20 (S & T):1991 Handbook on masonry design and construction.co. 6. 5.(2010). IS: 1905 (1987). IS 2572:1963(R 1997) Code of practice for design of Hollow bricks 12.htm xv . National Building Code of India (NBC) and Chennai Metropolitan Development Authority (CMDA).za/solar-power-calculator.P. Dhanpat Rai publishing company limited. Building Construction . Arora and S. Fifth edition. S. 11. Code of Practice for Structural use of unreinforced masonry.Unit weights of building material and stored materials (Incorporating IS:1911-1967) 10. IS 875 : Part 1 : 1987 Code of practice for design loads (other than earthquake)for buildings and structures Part 1 Dead loads . IS 875 : Part 2 : 1987 Code of practice for design loads (other than earthquake) for buildings and structures: Part 2 Imposed loads 9.P Bindra . Indian Standard Code of practice for plain and reinforced concrete (Fourth Revision ).4. http://www.