Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Planning Analysis Designing and Estimation of Residential Building

19,139 views

Published on

Project is based on designing of a two storey residential building in chennai. It is 2BHK flats for two families.

Published in: Engineering
  • Unlock Her Legs(Official) $69 | Get 90% Off + 8 Special Bonus? ■■■ http://t.cn/AijLRbnO
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
  • Earn a 6-Figure Side-Income Online... Signup for the free training HERE ■■■ https://tinyurl.com/y3ylrovq
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
  • DOWNLOAD FULL eBOOK INTO AVAILABLE FORMAT ......................................................................................................................... ......................................................................................................................... 1.DOWNLOAD FULL. PDF eBook here { https://tinyurl.com/y3nhqquc } ......................................................................................................................... 1.DOWNLOAD FULL. EPUB eBook here { https://tinyurl.com/y3nhqquc } ......................................................................................................................... 1.DOWNLOAD FULL. doc eBook here { https://tinyurl.com/y3nhqquc } ......................................................................................................................... 1.DOWNLOAD FULL. PDF eBook here { https://tinyurl.com/y3nhqquc } ......................................................................................................................... 1.DOWNLOAD FULL. EPUB eBook here { https://tinyurl.com/y3nhqquc } ......................................................................................................................... 1.DOWNLOAD FULL. doc eBook here { https://tinyurl.com/y3nhqquc } ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... .............. Browse by Genre Available eBooks ......................................................................................................................... Art, Biography, Business, Chick Lit, Children's, Christian, Classics, Comics, Contemporary, CookeBOOK Crime, eeBOOK Fantasy, Fiction, Graphic Novels, Historical Fiction, History, Horror, Humor And Comedy, Manga, Memoir, Music, Mystery, Non Fiction, Paranormal, Philosophy, Poetry, Psychology, Religion, Romance, Science, Science Fiction, Self Help, Suspense, Spirituality, Sports, Thriller, Travel, Young Adult,
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
  • ◆◆◆ http://scamcb.com/ezpayjobs/pdf
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
  • The #1 Woodworking Resource With Over 16,000 Plans, Download 50 FREE Plans... ➤➤ http://tinyurl.com/yy9yh8fu
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here

Planning Analysis Designing and Estimation of Residential Building

  1. 1. 1 | P a g e PLANNING, ANALYSING, DESIGNING AND ESTIMATION OF RESIDENTIAL BUILDING MINI PROJECT REPORT SUBMITTED IN THE PARTIAL FULLFILLMENT OF THE REQUIREMENTS FOR THE AWARD OF DEGREE IN BACHELOR OF TECHNOLOGY IN CIVIL ENGINEERING (HABITAT DEVELOPMENT) BY MOHAMED PEER THAVOOD.R (13209016) IV YEAR RURAL TECHNOLOGY CENTRE THE GANDHIGRAM RURAL INSTITUTE-DEEMED UNIVERSITY (FULLY FUNDED BY MHRD) GANDHIGRAM-624 302 DINDIGUL DISTRICT, TAMIL NADU
  2. 2. 2 | P a g e CERTIFICATE This is to certify that the mini project work titled “PLANNING, DESIGNING & ESTIMATION OF TWO STOREY RESIDENTIAL BUILDING” is a bonafide work done by R.MOHAMED PEER THAVOOD under the guidance and supervision of Dr.K.MAHENDRAN, Director i/c in Rural Technology Centre of Gandhigram Rural Institute (Deemed University), Gandhigram Dr.K.Mahendran Guide Director i/c, RTC Dr.K.Mahendran Director i/c, RTC Place: Date: Submission for viva-voce held on: Internal examiner External examiner
  3. 3. 3 | P a g e DECLARATION CERTIFICATE I submit this mini project work titled “PLANNING,ANALYSIS DESIGN AND ESTIMATION OF TWO STOREY RESIDENTIAL BUILDING” to Rural Technology Centre, Gandhigram Rural Institute(DU), Gandhigram, Dindigul, in partial fulfilment of requirements for the award of degree of Bachelor of Technology in Civil Engineering(Habitat Development) and this is my original and independent work carried out. R. MOHAMED PEER THAVOOD (13209016)
  4. 4. 4 | P a g e
  5. 5. 5 | P a g e Content S.No TITLE Page 1 Introduction 8 2 Methodology 9 3 Specification 10 4 Chapter I (Planning) 18 5 Chapter II (Analysis) 24 6 Chapter III (Design) 29 7 Chapter IV (Estimation) 49 8 Chapter V (Management) 54 9 Chapter VI(Electrification and plumbing) 60
  6. 6. 6 | P a g e NOTATIONS a - Area b - Breadth of beam Ast - Area of reinforcement Mu - Ultimate Moment Vu - Ultimate shear force D - Overall depth of beam DL - Dead load D - Effective width of beam Et - Modulus of elasticity of concrete Es - Modulus of elasticity of steel e - Eccentricity fd - Design strength fck - Compressive Strength of Concrete fy - Characteristic strength of steel Hwe - Effective height of wall Ief - Effective moment of wall Igs - Moment of inertia of cracked section K - Stiffness of member K - Constant Ld - Development length LL - Live load Lw - Horizontal distance between centers of lateral restraint L - Length of column Lef - Effective span of beam Lct - Effective length about y axis
  7. 7. 7 | P a g e La - Clear span lx - Length of shorter side of slab ly - Length of longer side of slab lo - Distance between points of zero moment in abeam li - Span in the direction in which moment are determined c/c of supports l2 - Span transverse to l1 m - Bending moment M - Modular ratio P - Axial load on compression member qe - Calculated maximum bearing pressure qn - Calculated maximum bearing pressure of soil r - Radius s - Spacing of stirrups T - Wall thickness V - Shear force W - Total load WL - Wind load W - Distributed load per unit area Wd - Distributed dead load per unit area X - Depth of neutral axis Z - Lever arm
  8. 8. 8 | P a g e Introduction Shelter is one of the basic need for Habitation. In this project, I have completed the Planning, Analysis, design and Estimation of a Residential Building of (2BHK) two flats in Chennai, Kuratur (Near Avadi). The Project is completed with reference to the Indian Standard codes. In planning, I have used Autocadd, Revit, sweet Home 3D software for Plan, Elevation, Interior and Exterior design is with reference to National Building Code 2005 completed. Analysis of the structure is done in manual as well as using STAAD Pro Vi8 software. Designing of structural components are carried out using Indian Standard Code (Given in reference) in Limit state Method. Estimation is completed by using rates from Schedule of Rates (2015-2016) by Public Work Department. Management of construction is very much Important hence we need shelter with good Quality in appreciative cost and time. Here I have used Critical Path Method as well as PERT Analysis with Probability for Time and Cost Management (Time-cost Trade off).
  9. 9. 9 | P a g e Methodology The methodology followed for completing the mini Project was given below
  10. 10. 10 | P a g e SPECIFICATION 1 Slab: A flat piece of concrete, typically used as a walking surface, but may also serve as a load bearing device as in slab homes. It also act as a beam but thick is less and width is more when compare to beam. Design Consideration: Among the design function that need to be taken in consideration for construction of ground floor slab is  The provision of a uniform, level surface  Sufficient strength and stability  Exclusion of dampness from inside of building  Thermal insulation (max.0.45 W/square meter)  Resistance to fire R.C.C slabs are divided into two: 1. One way slab 2. Two way slab. One way slab: When the slab is supported on all the four edges , and when the ratio of the long span to short span is large (>2) bending takes place along one the span , such as slab is known as a one way slab.
  11. 11. 11 | P a g e Two way slab: When the slab is supported on all the four edges , and when the ratio of the long span to short span is small bending takes place along both the span , such as slab is known as a two way slab. 2 BEAM: Beam is the horizontal member of a structure, carrying transverse loads. Beam is rectangular in cross-section. Beam carry the Floor slab or the roof slab. Beam transfer all the loads including its self-weight to the columns or walls .R.C.C. Beam is subjected to bending moments and shear. Due to the vertical external load , bending compresses the top fibers of the beam and elongates the bottom fibers. The strength of R.C.C. beam depends on the composite action of concrete and steels. 4.2.1.2 TYPES OF BEAMS :  Simply Supported Beam.  Fixed Beam.  Cantilever Beam.  Continuous Beam.  Overhanging Beam. Simply supported beam : Simply supported beam supported freely at the two ends on walls or columns. In actual practice, no beam rests freely on the supports (walls or columns).
  12. 12. 12 | P a g e Fixed beam : Fixed beam both ends of the beam are rigidly fixed into the supports. Also, Main reinforcement bars and stirrups are provided. Cantilever beam : Cantilever beamis fixed in a wall or column at one end and the other end is free, it is called cantilever beam. It has tension zone in the top side and compression zone in the bottom side. Continuous beam : Continuous beam is supported on more than two supports . This beam is more economical for any span lengths. R.C.C. Beams :  Singly Reinforced Beam  Doubly Reinforced Beam 4.2.1.3 Types of Loading on Beams :  Concentrated Load.  Uniformly Distributed Loads.  Uniformly Varying Loads.  Arbitrary Loading. 3 LINTELS: Lintel is a horizontal structural member which is placed across any opening (window, door, almirah, wardrobes etc.) to support load of masonry coming over it.
  13. 13. 13 | P a g e 4.3.2.2 FUNCTIONS OF LINTEL: Lintel is a beam supporting the load of masonry above it which is transmitted to the adjacent wall portions over which it is laid. The bearing of lintel The bearing of lintel beam should be the minimum of the following: (1) 100 mm (2) Height of lintel (3) 1/10 th to 1/12 th of the span of the lintel (4) CLASSIFICATION OF LINTELS: Lintels are classified into the following types according to the materials used in their construction: (1)Timber lintels (2)Stone lintels (3)Brick lintels (4)Reinforced cement concrete (R.C.C) lintels (5)Steel lintels (1)Timber lintels: These are the oldest types of lintels. At present, they are not used expect in hilly regions where timber is cheaply available. (2)Stone lintels: These are commonly used in stone masonry. The least thickness of the stone lintel is a about 75 mm and as a thumb rule, thickness is taken as at least 1 mm per 10mm length of the span of opening. (3)Brick lintels: These lintels are not structurally strong, and they are used only when the opening is small (less than 1 m) and loads are light. The depth of brick lintel varies from 100 mm to 200 mm, depending upon the span.
  14. 14. 14 | P a g e (4) Reinforced cement concrete (R.C.C) lintels: These lintels are commonly used these days. They have high rigidity, fire resistance and are simple in construction. R.C.C lintels are available as precast units. Precast R.C.C lintels are preferred for small spans upto 2metres. Generally 1:2:4 concrete is used in the construction of these lintels. The depth of lintel and the amount of the reinforcement is governed by the intensity of load, the type of support and the span. For the cast-in-situ lintels which are quite common, form work is required for construction. (5)Steel lintels: These lintels are provided when the opening is large and where the super-imposed loads are heavy. Rolled steel joints are used either singly or in combination of two or three units.S COLUMN: Column is a vertical structural member. It transmits the load from ceiling/roof slab and beam, including its self-weight to the foundation. Columns may be subjected to a pure compressive load. R.C.C. columns are the most widely used now-a-days. 4.4.2 Types of Column  Long Column Or Slender  Short Column  Intermediate Column  Columns of square, rectangular and circular sections 4.4.3 Definition of Columns:  A column is defined as a compression member the effective length of which exceeds three times its lateral dimensions .Compression members whose lengths do not exceed three times their least lateral dimensions are classified as pedestals.
  15. 15. 15 | P a g e  R.C Columns concrete has a high compressive strength and a low tensile strength .Hence, theoretically concrete should need no reinforcement when it is subjected to compression. Reinforcements are provided in order to reduce the size of columns. Though a column is mainly a compression member, it liable to some moment due to eccentricity of loads or transfer loads or due to its slenderness. Such moments may occur in any direction and so it is necessary to provide reinforcement near all faces of the column. These reinforcements form the longitudinal steel. In order to maintain the position of the longitudinal reinforcement and also to prevent their buckling which may cause splitting of concrete, it is necessary to provide transverse reinforcement in the form of lateral ties or spirals at close pitch. The transverse reinforcement also assists in confining the concrete. 4.4.4 Classification of Columns:  A column may be classified on the basis of its shape, its slenderness ratio, the manner loading and the type of lateral reinforcement provided. A column may have a section which may be square, rectangle, circular or a desire polygon .Depending on the slenderness ratio, column may be a short or long column.  The slenderness ratio of column is the ratio of the effective length of the column to its least lateral dimension. A column whose slenderness ratio exceeds 12 is a long column. A column whose slenderness ratio does not exceeds the above limit is a short column. Based on the manner of loading, column may be classified into  Axially loaded columns  Column subjected to axial load and uniaxial bending.  Columns subjected to axial load and biaxial bending. 1 STAIR CASE: Stair cases are used in almost all buildings. A staircase consist of a number of steps arranged in a series, with landing at appropriate locations, for the purpose of giving access to different floors of a building. The width of a staircase may depend on the purpose for which it is
  16. 16. 16 | P a g e provided, and may generally vary between 1m for residential buildings to 2m for public buildings. A flight is the length of the staircase situated between two landings. The number of steps in a flight may vary between 3 to 12. The rise of a step and the tread should be proportioned that it gives comfortable access. Generally the sum of tread plus twice the rise is kept about 500mm and the product of the tread and the rise is kept about 40000 to 42000. In residential buildings, the rise may vary between 150mm to 180mm and tread between 200 to 250mm. in public buildings, rise is kept between 120 to 150mm and tread between 200 to 300mm. 4.5.2 Classification of stairs: (1) Straight stair (2) Quarter turn stair (3) Half turn stair (open newel type or open well stair) (4) Dog-legged stair (5) Open newel stair with quarter space landing (6) Geometrical stairs such as circular stair, spiral stair etc. Footings: All structures supported on earth of superstructure and substructure. The foundation can be defined as the substructure which interfaces the superstructure and the supporting ground. Its purpose is to transfer all loads from the superstructure to ground safely and to provide stable base to the superstructure. 4.6.2 Types of foundations: Foundations are classified as shallow and deep foundations. 1. Shallow foundations:  It has smaller depth limited to the width of footing. It spreads the load from superstructure on a larger area of soil so that stress intensity is reduced to a value which can be carried safely by soil. They are classified as isolated and combined footing and combined footings.
  17. 17. 17 | P a g e  An isolated footing supports one well or one column. A wall footing is a continuous strip, either flat or stepped which distributes the load of wall on the soil. An isolated footing supporting a single column is commonly used where the load on columns are small and are not closely spaced. They are squared, rectangular ,circular or of other shapes.  A combined footing supports two or more columns. They may classified as those which supports two columns and those which support more than two columns. combined footing supporting two columns may be used for columns on property lines. Such footings may be rectangular, trapezoidal or T shaped in plan or may consist of isolated footing connected with a narrow beam. 2. Deep foundations:  When the top layer of the soil is too weak to support the structure on the shallow foundation, the depth of foundation is increased till more suitable soil is found to support the structure. Such a foundation is termed as deep foundation because of its large depth. Different types of deep foundation are pile and well foundation. Bearing capacity of the soil:  Bearing capacity of soil is the maximum intensity of the load or pressure developed under the foundation without causing failure of soil and in settlement of superstructure supported on foundation.
  18. 18. 18 | P a g e CHAPTER I (PLANNING)
  19. 19. 19 | P a g e
  20. 20. 20 | P a g e
  21. 21. 21 | P a g e MODULAR KITCHEN LIVING ROOM
  22. 22. 22 | P a g e MASTER BED ROOM PRAYER HALL
  23. 23. 23 | P a g e EXTERIOR DESIGN
  24. 24. 24 | P a g e Chapter II (Analysis) KANI’S METHOD
  25. 25. 25 | P a g e JOINT MEMBER K ∑K R.F BA 2.67 EI -0.2755 B BC 0.919EI 4.839 EI -0.0945 BD 1.25 EI -0.1290 DB 1.25 EI -0.1786 D DE 0.919 EI 3.499 EI -0.1313 DF 1.33 EI -0.1900 FD 1.33 EI -0.1858 F FG 0.919 EI 3.579 EI -0.1280 FH 1.33 EI -0.1858 H HF 1.33 EI 2.249 EI -0.2955 HI 0.919 EI -0.2043
  26. 26. 26 | P a g e
  27. 27. 27 | P a g e
  28. 28. 28 | P a g e ANALYSIS (SOFTWARE)
  29. 29. 29 | P a g e CHAPTER III (DESIGNING) 3. DESIGNING OF STRUCTUTAL COMPONENTS DESIGN OF SLAB DATA DIMENSION: 5.9m X 4.35m LIVE LOAD: 3 KN/m2 FLOOR FINISH: 1 KN/m2 TYPE OF SLAB LY/LX= 5.9/4.35 = 1.35 < 2 TWO WAY SLAB DEPTH SPAN/DEPTH = 25 DEPTH = 174mm ASSUME CLEAR COVER = 20mm d = 200 mm EFFECTIVE SPAN Clear Span + eff Depth = 4.1+0.2 = 4.3m Centre to Centre = 4.1 + 0.25 = 4.35m Effective Span = 4.3 m
  30. 30. 30 | P a g e LOAD CALCULATION LIVE LOAD = 3 KN/m2 FLOOR FINISH = 1 KN/m2 DEAD LOAD = 0.2 X 25 = 5 KN/m2 TOTAL LOAD = 9 KN/m2 FACTORED LOAD = 1.5 X 9 = 13.5 KN/m2 ULTIMATE MOMENT AND SHEAR FORCE Mx = ᵅx X Wu X Lx2 = 0.096 X 13.5 X 4.352 = 24.523 KN-m My= ᵅy X WU X Lx2 = 0.053 X 13.5 X 4.352 = 13.553 KN-m Vu = 0.5 X Wu X Lx = 0.5 X 13.5 X 4.35 = 29.362 KN LIMITING MOMENT MU lim = 0.138 X FCK X b X d2 =.> Mu lim = 110.4 KN-m Hence section is Under-reinforced REINFORCEMENT MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d ) 24.523 X 10^6 = 0.87 X 415 X Ast X 200 (1- 415 X Ast /20 X 1000 X 200) Ast = 814.90 mm2 Provide 12mm Dia bars @ 150 mm c/c
  31. 31. 31 | P a g e DISTRIBUTION Ast = 0.12% BD = 0.0012 X 1000 X 200 = 240 mm2 Provide 8mm Dia bars @ 250mm c/c CHECK SHEAR STRESS ƮV = VU/bd = 29.36 X 10^3/1000X200 = 0.1468 Pt = (100 X Ast)/bd = 0.4523 KƮc = 1.2 X 0.456= 0.547 N/mm2 KƮC > ƮV SHEAR STRESS IS WITH IN SAFE PERMISSIBLE LIMIT CHECK FOR DEFLECTION (L/d)BASIC = 20 Pt = 0.2015 => Kt = 1.2 (L/d)MAX = 20 X 1.2 = 24 (L/d)ACT = 4350/200 = 21.75 < 24 HENCE DEFLECTION IS UNDER CONTROL. TORSION REINFORCEMENT AT CORNER
  32. 32. 32 | P a g e Area of torsion Reinforcement = 0.75 X Ast = 611.175 mm2 Length over which Torsional Reinforcement provided = Short span/5 = 870mm Provide 8mm dia bars at 160 mm C/C for 670mm at the corners. Reinforcement Edge Strips = 0.12% Gross Area = 240 mm2 Provide 8 mm bars at 160 mm C/C
  33. 33. 33 | P a g e DESIGN OF BEAM (1): DATA DIMENSION: 5.9m LIVE LOAD: 15.963KN/m DEPTH SPAN/DEPTH = 12 DEPTH = 550mm ASSUME CLEAR COVER = 50mm d = 600 mm EFFECTIVE SPAN Clear Span + eff Depth = 5.7+0.5 = 6.2m Centre to Centre = 5.7 + 0.2= 5.9m Effective Span = 5.9 m LOAD CALCULATION LIVE LOAD = 15.963 KN/m DEAD LOAD = 0.3 X 0.6 X 25 = 4.5 KN/m TOTAL LOAD = 19.963 KN/m Wu = 1.5 X 19.963 = 29.9445 KN/m ULTIMATE MOMENT AND SHEAR FORCE
  34. 34. 34 | P a g e MU = 0.125 X WU X L2 = 0.125 x 29.94X5.92 = 130.27 KN-m VU = 0.5 X WU X L = 0.5 X 29.94X5.9 = 88.32KN LIMITING MOMENT MU lim = 0.138 X FCK X b X d2 = 298.08KN-m MU < MU lim Hence section is Under-reinforced REINFORCEMENT MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d ) 130.27 X 10^6 = 0.87 X 415 X Ast X 600 (1- 415 X Ast /20 X600 X 300) Ast = 650 mm2 Provide #2 of 16mm dia bars with #2 of 10mm dia hanger bars CHECK FOR SHEAR STRESS ƮV = VU/bd = 88.32X 10^3/600X300 = 0.4906 N/mm2 Pt = (100 X Ast)/bd = 0.446 KƮc = 1.2 X 0.3212 = 0.38544 N/mm2 KƮC < ƮV Shear reinforcement required Vus = Vu – (tcbd) = 88.32 X 10^3 – (0.454 X 300 X 600) = 6.6 KN Since shear is very less
  35. 35. 35 | P a g e Provide 8mm dia stirrups @ 300mm c/c CHECK FOR DEFLECTION (L/d)BASIC = 20 Pt = 0.446 => Kt = 1.4 (L/d)MAX = 20 X 1.4 X 1 X 1 = 29.2 (L/d)ACT = 4350/200 = 21.75 < 29.2 HENCE DEFLECTION IS UNDER CONTROL.
  36. 36. 36 | P a g e DESIGN OF BEAM (2) DATA DIMENSION: 4.35m LIVE LOAD: 15.963KN/m DEPTH SPAN/DEPTH = 15 DEPTH = 350mm ASSUME CLEAR COVER = 50mm d = 400 mm EFFECTIVE SPAN Clear Span + eff Depth = 4.1+0.35 = 4.45m Centre to Centre = 4.35m Effective Span = 4.35 m LOAD CALCULATION LIVE LOAD = 15.963 KN/m DEAD LOAD = 0.3 X 0.4 X 25 = 3 KN/m TOTAL LOAD = 18.963 KN/m Wu = 1.5 X 19.963 = 28.445 KN/m
  37. 37. 37 | P a g e ULTIMATE MOMENT AND SHEAR FORCE MU = 0.125 X WU X L2 = 0.125 x 29.94X5.92 = 67.26 KN-m VU = 0.5 X WU X L = 0.5 X 29.94X5.9 = 61.857KN LIMITING MOMENT MU lim = 0.138 X FCK X b X d2 = 132.48 KN-m MU < MU lim Hence section is Under-reinforced REINFORCEMENT MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d ) 67.26 X 10^6 = 0.87 X 415 X Ast X 400 (1- 415 X Ast /20 X400 X 300) Ast = 572.73 mm2 Provide #3 of 16mm dia bars with #2 of 10mm dia hanger bars CHECK FOR SHEAR STRESS ƮV = VU/bd = 61.851X 10^3/400X300 = 0.515 N/mm2 Pt = (100 X Ast)/bd = 0.502 Ʈc = 0.4806 N/mm2
  38. 38. 38 | P a g e ƮC < ƮV Shear reinforcement required Vus = Vu – (tcbd) = 61.851 – (0.454 X 300 X 400) X 10^3 = 6.6 KN Since shear is very less Provide 8mm dia stirrups @ 300mm c/c CHECK FOR DEFLECTION (L/d)BASIC = 20 Pt = 0.502 => Kt = 1.46 (L/d)MAX = 20 X 1.46 X 1 X 1 = 29.2 (L/d)ACT = 4350/200 = 21.75 < 29.2 HENCE DEFLECTION IS UNDER CONTROL.
  39. 39. 39 | P a g e DESIGN OF COLUMN DATA DIMENSION: 0.3m X 0.3m FACTORED LOAD: 926.508 KN Fck = 20 N/mm2 Fy = 415 N/mm2 SLENDERNESS RATIO: L/D= 3.2/0.3 = 10.66 Column is designed as slender column MINIMUM ECCENTRICITY Emin = (L/500 + D/30) = 6.4 + 10 = 16.4mm < 20mm Also 0.05d = 0.05 X 300 = 15mm < 20mm ULTIMATE MOMENT AND SHEAR FORCE MU = 0.125 X WU X L2 = 0.125 x 13.5 x 4.1 2 = 28.36 KN-m VU = 0.5 X WU X L = 0.5 X 13.5 X 4.1 = 27.67 KN LIMITING MOMENT MU lim = 0.138 X FCK X b X d2 = 110.4 KN-m MU < MU lim Hence section is Under-reinforced
  40. 40. 40 | P a g e REINFORCEMENT Pu = [0.4 X Fck X Ag + (0.67 Fy – 0.4 Fck)Asc] 911.508 X 10^3 = [0.4 X 20 X 300X 300 + (0.67 X 415- 0.4 X 20) Asc Asc = 720 mm2 Provide #4 of 16mm Dia bars with two bars faced LATERAL TIES: 1) DIA = φ </ {0.25 x DIA min = 4mm } => Provide 8mm Dia bars 2) Spacing Least Dimension = 300mm 16 X Dia max = 256mm Generally = 300mm Provide 8 mm dia laterial ties @ 250 mm c/c
  41. 41. 41 | P a g e DESIGN OF FOUNDATION DATA LIVE LOAD = 911.508 KN DEAD LOAD = 25 X 0.3 X 0.3 X 3.2 = 7.2 KN TOTAL LOAD = 918.708 KN PERMISSIBLE STRESS = 500 KN/m2 ALLOWABLE BEARING CAPACITY = 200 KN/m2 BASE AREA A = LOAD/ qa = 918.708/200 = 4.6 m2 Design a Square footing (since column is a square) Length = 2.2m Af = 2.2 X 2.2 = 4.84 m2 BENDING MOMENT BM = [qa X B X (B-d)2 ]/8 = 188.43 KN-m MU = 1.5 X 188.43 = 282.645 KN-m D = [MU/ 0.138 X Fck X b]^1/2 D= 0.21m = 210 mm = 300 mm (say) REINFORCEMENT MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d )
  42. 42. 42 | P a g e 282.645 X 10^6 = 0.87 X 415 X Ast X 300 (1- 415 X Ast /20 X2200 X 300) Ast = 2868.05 mm2 Provide #15 of 16mm dia bars @ 190 mm c/c in both direction.
  43. 43. 43 | P a g e DESIGN OF LINTEL BEAM DEPTH SPAN/DEPTH = 20 DEPTH = 70mm ASSUME DEPTH D = 150mm BREATH OF LINTEL = THICKNESS OF WALL = 250mm DIMENSION = 0.25m X 0.15m EFFECTIVE SPAN Clear Span + eff Depth = 1.2 + 0.2 = 1.4m Centre to Centre = 1.2 + 0.15 = 1.35m Effective Span = 1.35 m LOAD CALCULATION WEIGHT OF WALL OVER LNTEL = 0.9 X 0.25 X 19.5= 4.3875 KN/m DEAD LOAD = 0.15 X 0.25 X 25 = 0.9375 KN/m TOTAL LOAD = 5.325KN/m FACTORED LOAD (Wu) = 1.5 X 5.325 = 7.9875 KN/m ULTIMATE MOMENT AND SHEAR FORCE MU = 0.125 X WU X L2 = 0.125 x 7.9875X5.92 = 67.26 KN-m LIMITING MOMENT MU lim = 0.138 X FCK X b X d2 = 15.525 KN-m
  44. 44. 44 | P a g e MU < MU lim Hence section is Under-reinforced REINFORCEMENT MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d ) 1.956 X 10^6 = 0.87 X 415 X Ast X 150 (1- 415 X Ast /20 X250 X 150) Ast = 36.86 mm2 But Ast minimum = 0.87 X b X d / Fy = 78.61 mm2 Provide #2 of 10mm dia bars with #2 of 8mm dia hanger bars Provide stirrups of 8mm dia @ 200 mm c/c.
  45. 45. 45 | P a g e DESIGN OF STAIRCASE DATA Staircase space = 2.2m X 3.2m Live Load = 4 KN/m2 (As per IS 875 (part-2) 1987) DIMENSION Riser = 160mm; Tread = 300mm Width of stair = 2200/2 -100 = 1000mm Φ = (300 2 + 160 2 /250)^1/2 = 1.36 No of Riser = 3200/160 = 20 No of Riser per flight = 10 No of Tread = 10 – 1 = 9 Going = 300 X 9 = 2.7m Width of landing = 3.2 – 2.7 = 0.5 => 0.5 X 2 = 1m DESIGN OF FLIGHT Type = One way slab Span = 3.57m Depth (L/d) = 20 (simply supported slab) => d = 140mm Pt = 0.4 % ᵅ1= 1.32 for Fs = 240 N/mm2 D = 140 + 20 = 160 mm
  46. 46. 46 | P a g e LOAD CALCULATION Live Load = 4 KN/m2 Selfweight = 25 X D X φ = 25 X 0.16 X 1.187 = 4.748 KN/m2 Weight of steps = 25 X 0.16/2 = 2 KN/m2 Floor Finish = 1 KN/m2 Total Load = 11.748 KN/m2 Factored Load = 1.5 X 11.748 = 17.622 KN/m2 ULTIMATE MOMENT Mu max = Wul2/8 = 17.622 X 3.57 2/8 = 28.07 KN-m Limiting Moment MU lim = 0.138 X FCK X b X d2 = 54.096 KN-m Mu < Mu lim Hence Under reinforced section REINFORCEMENT MU = 0.87 X Fy X Ast X d (1- Fy X Ast/Fck X b X d ) Ast = 670 mm2 Provide 10mm dia bars @ 130mm c/c Distribution Reinforcement Ast min = 0.12% BD = 0.0012 X 1000 X160 = 180mm2 Provide 8mm dia bars @ 250mm c/c
  47. 47. 47 | P a g e CHECK FOR DEFLECTION Pt = 100 X 670 / 1000 X 140 = 0.47 % Fs = 232 N/mm2 ; Pt = 0.47 %; ᵅ1 = 1.31 D = 3170 /1.31 X 20 = 130 mm < 140mm Hence safe against Deflection
  48. 48. 48 | P a g e DESIGN OF SEPTIC TANK Flow of Sewage = 20 X140 /1000 = 2.8 m3/day Detention Period = 24 Hrs Tank Capacity = 2.8 X 24 /24 = 2.8 m3 Assume tank cleaned once in one years 1. Space required for storage of sludge = 0.0708 m3/capita = 20 X 0.0708 = 1.4 m3 = 1.5.m3 2. Space required for sludge digestion = 0.03 m3/ Capita = 20 X 0.03 = 0.6 m3 Add 25 % Extra = 25/100 X 2.8 = 0.7 m3 Total Tank Capacity = 2.8 + 1.5 + 0.6 + 0.7 = 5.6 m3 Let Depth of Liquid = 1m Plan Area = 5.6 /1 = 5.6 m2 L = 3 X B 3B X B = 5.6; B = 1.36 = 1.4 L = 4.2m Area of Tank = 4.2 X 1.4 = 5.88 m2 > 5.6m2 Assume free board = 0.3m Depth = 1 + 0.3 = 1.3m Size of Tank = 4.2m X 1.4m X 1.3m Liquid Capacity = 6.5 m3
  49. 49. 49 | P a g e CHAPTER IV (ESTIMATION)
  50. 50. 50 | P a g e ABSTRACT ESTIMATE S.NO DESCRIPTION REQUIRED QUANTITY RATE PER AMOUNT 1 Earth Work 58.08 300 Cubic metre 17,424 2 Sand Filling 2.904 220 Cubic metre 638.88 3 PCC 5.808 2116.3 Cubic metre 12,291.47 4 RCC 173.307 3437.54 Cubic metre 595,749.74 5 Brick Masonry 144.412 4054 Cubic metre 585446.248 6 Plastering 971.4 448.35 Square metre 435527.19 7 White washing 971.4 15.7 Square metre 15250.9.8 8 Distemper 971.4 15.7 Square metre 152509.8 9 Paint 971.4 30.0 Square metre 291420 10 Doors 24 6500 Numbers 156000 11 Windows 30 3000 Numbers 90000 Total 4979034.26 Contingencies 3% 149371.028 Work charged Establishment 2% 99580.68 Grand Total 52,27,985.9 /-
  51. 51. 51 | P a g e DETAILED ESTIMATE S.NO DESCRIPTION NO LENGTH BREATH DEPTH QUANTITY UNIT 1 Earthwork Foundation 12 2.2 2.2 1 58.08 (Cube.m) 2 Sandfill 12 2.2 2.2 0.05 2.904 (Cube.m) 3 Plan Cement Concrete (1:4:8) 12 2.2 2.2 0.1 5.808 (Cube.m) 4 Reinforced Cement Concrete (1:1.5:3) Foundation 12 2.2 2.2 0.3 17.424 (Cube.m) Column Ground Floor 12 0.3 0.3 3.2 4.212 (Cube.m) 1st & 2nd Floor 24 0.3 0.3 3 6.84 (Cube.m) Beam Type 1 6 5.625 0.3 0.6 6.015 Type 2 3 5.9 0.3 0.6 3.186 Type 3 8 4.35 0.3 0.4 4.176 X4 Total 53.748 (Cube.m) Slab Type I 4 5.625 4.35 0.2 19.575 Type II 2 5.9 4.35 0.2 10.266 X3 89.523 Type III 2 4 1.3 0.15 1.56 Total 91.083 (Cube.m) Grand Total 173.307 (Cube.m) 5 Brick Masonry Horizontal 2 15.6 0.25 3 23.4
  52. 52. 52 | P a g e 1 15.6 0.2 3 9.36 1 3 0.2 3 1.8 2 2 0.2 3 2.4 Vertical 2 11.7 0.2 3 14.04 2 11.7 0.25 3 17.55 1 4 0.2 3 2.4 2 2.4 0.2 3 2.88 Parapet 2 15.6 0.2 0.6 3.744 2 11.7 0.2 0.6 3.808 Total 155.212 (Cube.m) Deduction Door 12 1.2 2.1 0.25 7.56 Window 12 1.2 0.9 0.25 3.24 X2 Total 10.8 Total Actual 144.412 (Cubic.m) 6 Plastering (1:4) Living Room 1 29.6 3 88.8 (Sq.m) Bed Room 2 12.9 3 77.4 (Sq.m) Kitchen 1 16 3 48 (Sq.m) Store Room 1 8 3 24 (Sq.m) Toilet 2 9 3 54 (Sq.m) 1 8.4 3 25.2 (Sq.m) 1 7.6 3 22.8 (Sq.m) Outer 1 52.1 3 156.3 (Sq.m) X2 Total 993 (Sq.m) Deduction
  53. 53. 53 | P a g e 10.8 (Sq.m) Grand Total 971.4 (Sq.m) White Cement 971.4 (Sq.m) Primer 971.4 (Sq.m) Patti 971.4 (Sq.m) Primer 971.4 (Sq.m) Distemper 971.4 (Sq.m)
  54. 54. 54 | P a g e CHAPTER V (MANAGEMENT) Critical Path Method The Critical Path Method (CPM) is one of several related techniques for doing project planning. CPM is for projects that are made up of a number of individual "activities." If some of the activities require other activities to finish before they can start, then the project becomes a complex web of activities. Program Evaluation and Review Technique (PERT) PERT is a network analysis technique which is used to determine the time it will take to complete a complex process. Formula Expected Duration = (Optimistic + 4 X Likely + Pesimistic)/6 Standard Deviation= (Pesimistic – Optimistic)/6
  55. 55. 55 | P a g e S.NO DESCRIPTION NOTATION DURATION(DAYS) 1 Site Clearance A 2 2 Earthwork Excavation B 6 3 Shuttering C 12 4 Laying of foundation D 10 5 Earth filling cum foundation reinforcement E 15 6 Bar Bending Work F 10 7 Brick Work G 18 8 Roofing concrete with Lintel level beam laying with shuttering H 9 9 Ground floor work I 15 10 Laying Electrification and Plumbing J 15 11 Wood Work K 20 12 Plastering L 15 13 Sanitary Work M 10 14 Finishing Work N 15 15 White Washing O 12 16 Color Washing P 15
  56. 56. 56 | P a g e ACTIVTY OPTIMISTIC LIKELY PESIMISTIC EXPECTED ST.DEVIATION VAR. A 1 2 5 2.33 0.667 0.44 B 4 6 10 6.33 1 1 C 10 12 15 12.16 0.833 0.693 D 8 10 13 10.16 0.833 0.693 E 12 15 18 15 1 1 F 6 10 14 10 1.33 1.768 G 15 18 20 17.83 0.833 0.693 H 7 9 11 9 0.66 0.435 I 12 15 18 15 1 1 J 10 15 17 14.5 1.16 1.345 K 16 20 24 20 1.33 1.768 L 12 15 18 15 1 1 M 8 10 12 10 0.66 0.435 N 12 15 17 14.8 0.833 0.693 O 10 12 15 12.16 0.833 0.693 P 12 15 18 15 1 1
  57. 57. 57 | P a g e
  58. 58. 58 | P a g e Project Duration = 180.11 Days Standard Deviation of the project = 13.479 Days
  59. 59. 59 | P a g e PROBABILITY  Project to be completed after 190 days: Z = (190-180.11)/13.479 = 0.733 From Log table Probability = 1 – 0.7673 = 0.2327 Project to be completed after 190 days is 23.27%  Project to be completed before 160 days: Z = (160-180.110)/13.479 = -1.491 From Log table Probability = 0.0681 Project to be completed before 160 days is 6.81%  Project to be completed between 160 to 190 days Probability = 1-0.2327-0.0681 = 0.6992 Project to be between 160 days to 190 days is 69.92%  Number of days for completion of project with 90% probability From Log Table 1.28 + [(0.09-0.08) X (0.90-0.8997) / (6.9015-0.8997) = 1.281667 Days = 180.11+1.281667 X 13.429 = 197.32 Number of days for completion with 90% probability is 197.32 = 198 Days
  60. 60. 60 | P a g e CHAPTER VI (ELECTRIFICATION AND PLUMPING)
  61. 61. 61 | P a g e
  62. 62. 62 | P a g e REFERENCES Books used: 1. B.C. Punmia, “Limit state design of Reinforced concrete” Laxmi Publications (P) Ltd, New Delhi. 2. S.Ramamrutham (2010), “Design of Reinforced Concrete Structures” 3. Ashok kumar Jain,Arun kumar Jain “Soil Mechanics & Foundations” Laxmi Publications (P) Ltd, New Delhi. 4. Dr.K.R.Arorra, “Soil mechanics & Foundation Engineering” Standard publishers Distribution-New Delhi. 5. B.N.Dutta, “Estimation and Costing in Civil Engineering” USB Publishers Distributions (p)Ltd New Delhi. 6. KK.Chitkara “Construction Project Management-Planning” Tata McGraw-Hill publishing Limited, New Delhi.
  63. 63. 63 | P a g e IS CODES USED: 1. IS 456 : 2000 “Plain and Reinforced Concrete – code of practice” Fourth Revision – section 3 (clauses 19, 23, 26) – section 4(33) 2. IS 875 (Part 1, Part 2 & Part 3), “Code of practice for Design loads (other than earthquake) for buildings & structures” 3. SP 7 National Building Code (NBC) of India 1983 Group I – Part II, III, IV, V and X 4. SP 7 National Building Code of India (NBC)- 2000 5. SP 16 -1980 Design Aids for Reinforced Concrete – Section 2, 3 and 5 6. SP 34 - 1987 Hand book on Concrete Reinforced & Detailing. 7. IS 2911 (Part I/Sec2) – 1979 “Code of Practice for Design And Construction of Pile Foundations” 8. IS 1893 (Part 1) : 2002 “Criteria for Earthquake Resistant Design of structures” 9. IS 13920 : 1993 “Ductile Detailing Of Reinforced Concrete Structures Subjected to Seismic Forces - Code Of Practice”

×