design of single storey rcc framed building

CAPSTONE PROJECT REPORT (Project Term January-April, 2014)

(DESIGN OF SINGLE STOREY MUNICIPAL BUILDING) Department of Civil engineering Submitted by:Mysum Shabir Yavar Ahad Mohd. Tajamul Jaspreet Kaur Amol Kumar

Registration Number: 11007722 Registration Number: 11011945 Registration Number: 11012119 Registration Number: 11001586 Registration Number: 11008366 Project Group Number:Under the Guidance of Miss Damanpreet Kaur Discipline of Civil Lovely Professional University, Phagwara January to April, 2014 i|Page

Declaration:We hereby declare that the project work entitled “DESIGN OF SINGLE STOREY MUNICIPAL BUILDING” is an authentic record of our own work carried out as requirements of Capstone Project for the award of degree of B.Tech in Civil engineering from Lovely Professional University, Phagwara, under the guidance of Miss Damanpreet Kaur during January to April 2014.

Mysum Shabir Registration Number: 11007722 Yavar Ahad Registration Number: 11011945 Mohd. Tajamul Islam Registration Number: 11012119 Jaspreet Kaur Registration Number: 11001586 Amol Kumar Registration Number: 11008366

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CERTIFICATE This is to certify that the declaration statement made by this group of students is correct to the best of my knowledge and belief. The Capstone Project Proposal based on the technology / tool learnt is fit for the submission and partial fulfillment of the conditions for the award of B.Tech in Civil Engineering from Lovely Professional University, Phagwara.

Name: Damanpreet Kaur U.ID: 17418 Designation: Assistant Professor

Signature of Faculty Mentor

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Table of contents Abstract………………………………………………..v LIST OF FIGURES………………………………. ….vii LIST OF TABLES………………………………..........viii 1.0 Introduction……………………………………… 1-5 1.1 General Introduction…………………………1-3 1.2 Proposals……………………………………….4 1.3 Statement of project……………………………4 1.4 Elements of RCC Framed Building……………4-5 2.0 Basic codes for design……………………………….5-6 3.0 Aim of design………………………………………. 7 4.0 Method of design……………………………………..7 5.0 Requirement of reinforcement for structural member………………………. ……. 8-16 5.1 Beams…………………………………………….8-9 5.2 Slabs……………………………………………….10 5.3 Columns…………………………………………..10-11 5.4 Shear……………………………………………….11-14 5.5 Development length of bars………………………14-16 6.0 Structural design (Manual Method Of Design)………17- 76 6.1 Study of architectural drawing…………………….17 6.2 Finalization of structural configuration……………18 6.3 Design of slabs………………………………………..18-43 6.4 Design of beams………………………………………44-53 6.5 Design of columns……………………………………53-66 6.6 Design of foundation…………………………………67-76 7.0 Conclusion……………………………………………….77 8.0 References……………………………………………….78

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Abstract Structural design is the primary aspect of civil engineering. The very basis of construction of any building, residential house or dams, bridges, culverts, canals etc. is designing. Structural engineering has existed since humans first started to construct their own structures. The foremost basic in structural engineering is the design of simple basic components and members of a building viz., Slabs, Beams, Columns and Footings. In order to design them, it is important to first obtain the plan of the particular building that is, positioning of the particular rooms (Drawing room, bed room, kitchen toilet etc.) such that they serve their respective purpose and also suiting to the requirement and comfort of the inhabitants. Thereby depending on the suitability; plan layout of beams and the position of columns are fixed. Thereafter, the loads are calculated namely the dead loads, which depend on the unit weight of the materials used (concrete, brick) and the live loads, which according to the code IS:875-1987 is around 2kN/m2. Once the loads are obtained, the component takes the load first i.e. the slabs can be designed. Designing of slabs depends upon whether it is a one-way or a two-way slab, the end conditions and the loading. From the slabs, the loads are transferred to the beam. The loads coming from the slabs onto the beam may be trapezoidal or triangular. Depending on this, the beam may be designed. Thereafter, the loads (mainly shear) from the beams are taken by the columns. For designing columns, it is necessary to know the moments they are subjected to. For this purpose, frame analysis is done by Moment Distribution Method. After this, the designing of columns is taken up depending on end conditions, moments, eccentricity and if it is a short or slender column. Most of the columns designed in this mini project were considered to be axially loaded with biaxial bending. Finally, the footings are designed based on the loading from the column and also the soil bearing capacity value for that particular area. Most importantly, the sections must be checked for all the four components with regard to strength and serviceability. Overall, the concepts and procedures of designing the basic components of a single storey building are described. Apart from that, the planning of the building with regard to appropriate directions for the respective rooms, choosing position of beams and columns are also properly explained. The future of structure engineering mainly depends on better and more effective methods of designing the structures so that they serve better and are also economical.

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LIST OF FIGURES S.No.

FIGURE NAME

PAGE NO.

1

Load bearing masonry building

2

2

Framed Structural system

3

3

Elements of RCC framed building

5

4

Architectural plan of building

17

5

Load distribution in one way slab for slab S9

22

6

Two way slab load distribution and action for slab S3

27

7

Two slab load distribution for slab S1

32

8

Two way slab load distribution and action for slab S4

37

9

Cross section of beams

52,53

10

Layout of beams and columns

63

11

Sectional and cross sectional view of column (C1)

64

12

Cross sectional view of column (C2)

64

13

Sectional view of column (C2)

65

14

Cross sectional view of column (C3)

65

15

Sectional view of column (C3)

66

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List of tables TABLENO.

NAME OF TABLE

PAGE NO.

1

Values of partial safety factor for loads (IS 456:2000, TABLE 18 )

7

2

Maximum shear stress, τcmax , N/mm2 (Table 20 of IS 456:2000)

12

3

Design shear strength of concrete τc, N/mm2 (table

13

19 of IS 456:2000) 4

Development length for fully stressed deformed

15

bars shear reinforcement (stirrups) 5

Live loads on floors as per IS-875(Part-2)-1987

16

6

Live loads on roofs as per IS-875(Part-2)-1987

16

7

Beam design data

51

8

Beam design data

51

9

column design data

62

10

Foundation design data

76

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Acknowledgement:We have taken efforts in this project. However, it would not have been possible without the kind support and help of many individuals. We would like to extend my sincere thanks to all of them. We are highly indebted to Dept. of Civil Engineering for their guidance and constant supervision as well as for providing necessary information regarding the project & also for their support in completing the project. We would like to take this opportunity to express our gratitude towards all those people who have helped us in the successful completion of this capstone project, directly or indirectly. We would also like to express our sincere gratitude towards Miss Damanpreet Kaur for her guidance and help which she willingly provided at every step of the project.

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1.0 INTRODUCTION 1.1 General Introduction The procedure for analysis and design of a given building will depend on the type of building, its complexity, the number of stories etc. First the architectural drawings of the building are studied, structural system is finalized sizes of structural members are decided and brought to the knowledge of the concerned architect. The procedure f or structural design will involve some steps which will depend on the type of building and also its complexity and the time available for structural design. Often, the work is required to start soon, so the steps in design are to be arranged in such a way the foundation drawings can be taken up in hand within a reasonable period of time. Further, before starting the structural design, the following information of data is required: (i)

A set of architectural drawings;

(ii)

Soil Investigation report (SIR) of soil data in lieu thereof;

(iii)

Location of the place or city in or der to decide on wind and seismic loadings;

(iv)

Data for lifts, water tank capacities on top, special roof features or loadings, etc.

Choice of an appropriate structural system for a given building is vital for its economy and safety. There are two types of building systems: (a) Load Bearing Masonry Buildings. (b) Framed Buildings. (a) Load Bearing Masonry Buildings:- Small buildings like houses with small spans of beams, slabs generally constructed as load bearing brick walls with reinforced concrete slab beams. This system is suitable for building up to four or less stories.( as shown in fig. below). In such buildings crushing strength of bricks shall be 100 kg/cm2 minimum for four stories. This system is adequate for vertical loads it also serves to resist horizontal loads like wind & earthquake by box action. The design of Load Bearing Masonry Buildings are done as per IS: 1905 - 1980 (Indian Standards Code of Practice f or Structural Safety of Buildings: Masonry Walls(Second Revision) . 1|P a g e

Fig 1. Load bearing masonry building ( b) Framed Buildings: -In these types of buildings reinforced concrete frames are provided in both principal directions to resist vertical loads and the vertical loads are transmitted to vertical framing system i.e. columns and Foundations. This type of system is effective in resisting both vertical & horizontal loads. The brick walls are to be regarded as non load bearing filler walls only. This system is suitable for multi - storied building which is also effective in resisting horizontal loads due to earthquake. In this system the floor slabs, generally 100 - 150 mm thick with spans ranging from 3.0 m to 7.0 m. In certain earthquake prone areas, even single or double storey buildings are made framed structures for safety reasons. Also the single storey buildings of large storey heights (5.0m or more ) ,like electric substation etc. are made framed structure as brick walls of large heights are slender and load carrying capacity of such walls reduces due to slenderness.

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Fig 2. Framed Structural system

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1.2 proposals The municipal building housed in a single storey shall have following accommodation:Bed room = 4 No. 3.2m x 3.15 m Kitchen = 4 No. 3.25m x 2.40 m Bathroom = 4 No. 2.15m x 1.25 m Dressing room = 4 No. 1.25m x 0.95m Living room = 4 No. 3.050m x 4.25m Balcony = 4 No. 3.00m x 1.475m Lobby = 1 No. 4.4m x 2.2m

1.3 statement of the project Utility of the building------ Residential building. No.of storeys-----------------Single storey (G). Shape of the building-------- Rectangular. No. of staircases-------------- 1 No. Type of foundation----------isolated footing. Type of building------------RCC Framed Building.

1.4 Elements of RCC Framed Building:1. Slab: - The flat ceiling of a storey is called a 'Slab'. 2. Beams:- The peripheral horizontal member supporting the slab is called 'Beam'. 3. Plinth Beam:-The beam at ground level or plinth level is called 'Plinth Beam'. 4|P a g e

4. Column:-The vertical member supporting the beam is called 'Column'. 5. Foundation: - The system below ground transferring the entire load of the structure to the soil is called 'Foundation'.

• slab • Beams •Plinth beam •Columns •Foundation

Fig 3. Elements of RCC framed building 2.0 Basic Codes for Design :The design should be carried so as to conform to the following Indian code for reinforced concrete design, published by the Bureau of Indian Standards, New Delhi. Purpose of Codes:- National building codes have been formulated in different countries to lay down guidelines for the design and construction of structure. The codes have evolved from the collective wisdom of expert structural engineers, gained over the years. These codes are periodically revised to bring them in line with current research, and often, current trends. Firstly, they ensure adequate structural safety, by specifying certain essential minimum requirement for design.

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Secondly, they r ender the task of the designer relatively simple; often, the result of sophisticate analyses is made available in the form of a simple formula or chart. Thirdly, the codes ensure a measure of consistency among different designers. Finally, they have some legal validity in that they protect the structural designer from any liability due to structural failures that are caused by inadequate supervision and/or faulty material and construction. (i) IS 456 : 2000 – Plain and reinforced concrete – code of practice (fourth revision) (ii) Loading Standards These loads to be considered for structural design are specified in the following loading standards: IS 875 ( Part 1 - 3) : 1987 – Code of practice f or design loads (other than earthquake) f or buildings and structures (second revision ) Part 1: Dead loads Part 2: Imposed (live) loads Part 3: Wind loads IS 13920: 1993 – Ductile detailing of reinforced concrete structure subject to seismic forces. Design Handbooks The Bureau of Indian standards has also published the following handbooks, which serve as useful supplement to the 1978 version of the codes. Although the handbooks need to be updated to bring them in line with the recently revised ( 2000 version) of the Code, many of the provisions continue to be valid ( especially with regard to structural design provisions). SP 16 : 1980 – Design Aids (for Reinforced Concrete) to IS 456 : 2000

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3.0 AIM OF DESIGN:The aim of design is achievement of an acceptable probability that structures being designed shall, with an appropriate degree of safety – 

Perform satisfactorily during their intended life.



Sustain all loads and deformations of normal construction & use



Have adequate durability



Have adequate resistance to the effects of misuse and fire. 4.0 METHOD OF DESIGN:-

Structure and structural elements shall normally be designed by Limit State Method. DESIGN LOAD:- Design load is the load to be taken for use in appropriate method of design. It is Characteristic load with appropriate partial safety factors for limit state design.

Table 1:- values of partial safety factor for loads (IS 456:2000, TABLE 18 )

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5 REQUIREMENT OF REINFORCEMENT FOR STRUCTURAL MEMBER 5.1 Beams 5.1.1 Tension reinforcement (a) Minimum reinforcement:- The minimum area of tension reinforcement shall not be less than that given by the following:As/bd = 0.85/fy where As = minimum area of tension reinforcement. b = breadth of beam or the breadth of the web of T-beam. d = effective depth, and fy = characteristic strength of reinforcement in M/mm2 (b) Maximum reinforcement:- the maximum area of tension reinforcement shall not exceed 0.04bD. 5.1.2 Compression reinforcement The maximum area of compression reinforcement shall not exceed 0.04 bd. Compression reinforcement in beams shall be enclosed by stirrups for effective lateral restraint. 5.1.3 Maximum spacing of shear reinforcement Maximum spacing of shear reinforcement means long by axis of the member shall not exceed 0.75 d for vertical stirrups and d for inclined stirrups at 45‖ where d is the effective depth on the section under consideration. In no case shall be spacing exceed 300mm.

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5.1.4 Minimum shear reinforcement Minimum shear reinforcement in the form of stirrups shall be provided such that: Asv / bsv > 0.4/0.87 fy Where Asv = total cross-sectional area of stirrups legs effective in shear. Sv= stirrups spacing along the length of the member b = breadth of the beam or breadth of the web of flange beam, and fy = characteristic strength of the stirrups reinforcement in N/mm2which shall not taken greater than 415 N/mm2

5.1.5 Minimum Distance between Individual Bars (a) The horizontal distance between two parallel main reinforcing bars shall usually be not-less than the greatest of the following: (i) Dia of larger bar and (ii) 5 mm more than nominal maximum size of coarse aggregate. (b) When needle vibrators are used it may be reduced to 2/3rdof nominal maximum size of coarse aggregate,Sufficient space must be left between bars to enable vibrator to be immersed. (c) Where there are two or more rows of bars, bars shall be vertically in line and the minimum vertical distance between bars shall be 15 mm, 2/3rd of nominal maximum size of aggregate or the maximum size of bars, whichever is greater.

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5.2. Slabs:5.2.1 Minimum reinforcement:- The mild steel reinforcement in either direction in slabs shall not be less than 0.15 percent of the total cross-sectional area. However, this value can be reduced to 0.12 percent when high strength deformed bars or welded wire fabric are used. 5.2.2 Maximum diameter. The diameter of reinforcing bars shall not exceed one eight of the total thickness ofslab. 5.2.3 Maximum distance between bars - Slabs 1) The horizontal distance between parallel main reinforcement bars shall not be more than three times the effective depth of solid slab or 300 mm whichever is smaller. 2) The horizontal distance between parallel reinforcement bars provided againstshrinkage and temperature shall not be more than five times the effective depth of a solid slab or 300 mm whichever is smaller. 5.2.4 Torsion reinforcement - Slab Torsion reinforcement is to be provided at any corner where the slab is simply supported on both edges meeting at that corner. It shall consist of top and bottom reinforcement, each with layers of bars placed parallel to the sides of the slab and extending from the edges a minimum distance of one-fifth of the shorter span. The area of reinforcement in each of these four layers shall be three-quarters of the area required for the maximum mid-span moment in the slab. 5.3. Columns 5.3.1 Longitudinal Reinforcement a. The cross sectional area of longitudinal reinforcement shall be not less than 0.8% nor more than 6% of the gross sectional area of the column. Although it is recommended that the maximum area of steel should not exceed 4% to avoid practical difficulties in placing & compacting concrete.

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b. In any column that has a larger cross sectional area than that required to support the load, the minimum percentage steel must be based on the area of concrete resist the direct stress & not on the actual area. c. The bar should not be less than 12 mm in diameter so that it is sufficiently rigid to stand up straight in the column forms during fixing and concerting. d. The minimum member of longitudinal bars provided in a column shall be four in rectangular columns & six in circular columns. e. A reinforced concrete column having helical reinforcement must have at least six bars of longitudinal reinforcement with the helical reinforcement. These bars must be in contact with the helical reinforcement & equidistance around its inner circumference. f. Spacing of longitudinal should not exceed 300 mm along periphery of acolumn. g. In case of pedestals, in which the longitudinal reinforcement is not taken into account in strength calculations, nominal reinforcement should be not be less than 0.15% of cross sectional area. 5.3.2 Transverse Reinforcement a. The diameter of lateral ties should not be less than ¼ of the diameter of the largest longitudinal bar in no case should not be less than 6 mm. b. Spacing of lateral ties should not exceed least of the following:

Least lateral dimension of the column



16 times the smallest diameter of longitudinal bars to be tied.



300mm.

5.4 SHEAR 5.4.1 Nominal Shear Stress The nominal shear stress in beams of uniform depth shall be obtained by the following equation: 11 | P a g e

τv =Vu/ b.d where Vu = shear force due to design loads; b = breadth of the member, which for flanged section shall be taken as the breadth ofthe web, bw; and d = effective depth. With Shear Reinforcement Under no circumstances, even with shear reinforcement, shall the nominal shear stress in beams should not exceed given in Table 20 of IS 456:2000.  For solid slabs, the nominal shear stress shall not exceed half the appropriate values given in Table 20 of IS 456:2000.

CONCRETE

M15

M20

M25

M30

M35

M40 & above

2.5

2.8

3.1

3.5

3.7

4.0

GRADE

Τc max , N/mm2

TABLE 2:- MAXIMUM SHEAR STRESS, τcmax , N/mm2 (Table 20 of IS 456:2000)

5.4.2 Minimum Shear Reinforcement When τv, is less than τc given in Table 19 of IS 456:2000 , minimum shear reinforcement shall be provided in accordance with clause 26.5.1.6 of IS 456:2000.

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100 Ast / bd

Concrete grade M15

M20

M25

M30

M35

M40 & above

<0.15

0.28

0.28

0.29

0.29

0.29

0.30

0.25

0.35

0.36

0.36

0.37

0.37

0.38

0.50

0.46

0.48

0.49

0.50

0.50

0.51

0.75

0.54

0.56

0.57

0.59

0.59

0.60

1.00

0.60

0.62

0.64

0.66

0.67

0.68

1.25

0.64

0.67

0.70

0.71

0.73

0.74

1.50

0.68

0.72

0.74

0.76

0.78

0.79

1.75

0.71

0.75

0.78

0.80

0.82

0.84

2.00

0.71

0.79

0.82

0.84

0.86

0.88

2.25

0.71

0.81

0.85

0.88

0.90

0.92

2.50

0.71

0.82

0.88

0.91

0.95

0.95

2.75

0.71

0.82

0.90

0.94

0.96

0.98

3.00 & above

0.71

0.82

0.92

0.96

0.99

1.01

TABLE 3:- design shear strength of concrete τc, N/mm2 (table 19 of IS 456:2000) 5.4.3 Design of Shear Reinforcement When τv, is exceeds τc , given in Table 19, shear reinforcement shall be provided in any of the following forms: a) Vertical stirrups, b) Bent-up bars along with stirrups, and Where bent-up bars are provided, their contribution towards shear resistance shall not be more than half that of the total shear reinforcement. Shear reinforcement shall be provided to carry a shear equal to Vu/ τc b d the strength of shear reinforcement Vus shall be calculated as below: a) For Vertical Stirrups: Vus = 0.87 fy Asv d/ Sv 13 | P a g e

b) For inclined stirrups or a series of bars bent up at different cross –section: Vus = 0.87 fy Asv d (Sin ά + Cos ά) / Sv c) For single bar or single group of parallel bars, all bent up at the same cross sections: Vus = 0.87 fy Asv Sin ά Where Asv = total cross –sectional area of stirrups legs or bent-up bar within a distance Sv Sv = spacing of the stirrups or bent-up bars along the length of the member. τv = nominal shear stress, τc = design shear strength of the concrete, b = breadth of the member which for flanged beams, shall be taken as the breadth of the web bw. fy = characteristic strength of the stirrup or bent-up reinforcement which shall not be taken greater than 415 N/mm2ά = angle between the inclined stirrup or bent up bar and the axis of the member not less than 45o and d = effective depth 5.5 DEVELOPMENT LENGTH OF BARS 5.5.1 Development of Stress in Reinforcement The calculated tension or compression in any bar at any section shall be developed on each side of the section by an appropriate development length or end anchorage or by a combination thereof. Development length Ld is given by:Ld = Φσst /4τbd Φ

= nominal diameter of bar,

τbd = design bond stress 14 | P a g e

σst = stress in bar at the section considered at design load 

Design bond stress in limit state method for plain bars in tension is given in clause 26.2.1.1 in IS 456:2000



For deformed bars conforming to IS 1786 these values are to be increased by 60 %.

 For bars in compression, the values of bond stress for bars in tension is to be increased by 25 percent

 TABLE 4 :- DEVELOPMENT LENGTH FOR FULLY STRESSED DEFORMED BARS SHEAR REINFORCEMENT (STIRRUPS) Development length and anchorage requirement is satisfied, in case of stirrups and transverse ties, when Bar is bent – • Through an angle of at least 90 degrees (round a bar of at least its own dia) & is continued beyond for a length of at least 8 υ, or • Through an angle of 135 degrees & is continued beyond for a length of at least 6 υ or • Through an angle of 180 degrees and is continued beyond for a length of at least 4 υ

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Type of floor usage

Live load (KN/m2)

RESIDENTIAL

2.0

OFFICIAL – WITH SEPARATE STORAGE

2.5

– WITHOUT SEPARATE

4.0

STORAGE SHOPS,CLASS ROOMS,WAITINGS ROOMS, RESTAURANTS,WORK ROOMS,THEATRES ETC -WITHOUT FIXED SEATING

4.0

- WITH FIXED SEATING

5.0

FACTORIES & WAREHOUSES

5.0 - 10.0

STACK ROOM IN LIBRARIES ,BOOK

10.0

STORES GARRAGES –LIGHT VEHICLES

4.0

–HEAVY VEHICLES

7.5

STAIRS -NOT LIABLE TO OVER

4.0

CROWDING - LIABLE TO OVER CROWDING

5.0

Table 5 :- LIVE LOADS ON FLOORS AS PER IS-875(Part-2)-1987

Type of access to roof

Live load (KN/m2)

Roof with access

1.5

Roof without access

0.75

Table 6:- LIVE LOADS ON ROOFS AS PER IS-875(Part-2)-1987

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6.0 STRUCTURAL DESIGN (MANUAL METHOD OF DESIGN):6.1: STUDY OF ARCHITECTURAL DRAWINGS:- Before proceeding for structural design of any building it is ensure that approved working drawings are available in the office. All working drawings i.e. each floor plan, elevations, sections, are studied thoroughly & discrepancy if any brought to the notice of concern Architect for rectification/correction. The problems coming in finalization of structural configuration may also be intimated to concern Architect for rectification/correction if any.

Fig 4. Architectural plan of building

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6.2: FINALIZATION OF STRUCTURAL CONFIGURATION. After receiving corrected working drawing from the architectural wing, the structural system is finalized. The structural arrangements of a building is so chosen as to make it efficient in resisting vertical as well as horizontal loads due to earthquake. The span of slabs co chosen that thickness of slab 100-150mm and slab panels, floor beams, and columns, are all marked and numbered on the architectural plans. Now the building is ready for structural design to start.

6.3: DESIGN OF SLABS

One way slab (S9) : (LOBBY- 2.2*4.1 m² c/c) Step 1: Design constants and limiting depth of neutral axis :For, Fy = 415 N/mm2

=> Xu,max/d = 0.479 and Ru =2.761

Fck = 20 N/mm2 Step 2: Computation of design B.M. and S.F. :

Assume,L/d = 20 for simply supported slab (from the point of deflection) Pt = 20% for under reinforced section ft (Modification factor) = 1.5 Taking c/c dis. L = 3.2m =>

L/d = 20×1.5

{ L=4100mm}

4100/d = 20×1.5 d = 136.66 mm.

140mm.

Providing 20mm. nominal cover and 8mm. Ø bars D = 140+20+4 = 164 

165mm.

Design load (Wu) :For, depth (D) = 165mm. = 0.165m. 18 | P a g e

DL (wt. of slab/m2) = vol. × density = (0.165m.×1m.×1m.)×(25 KN/m2) LL+FF (imposed loads) = 1.5 KN/m2 + 1.5 KN/m2

= 4.125 KN/m2 = 3 KN/m2

So, Total load = 4.125 + 3 = 7.125 KN/m2 Design load (Wu) = 1.5 × 7.125 = 10.687 KN/m2 

Design shear force (Vu) :Vu = Wu × l/2 = 10.687 × 3.2 / 2 = 17.09 KN



Design bending moment (Mu):Mu = Wu ×l2/8

= ( 10.687 × 3.22) / 8 = 13.68 KN-m

Step 3: Computation of effective depth :For, b(width) = 1m. =1000mm d = √Mu / (Ru×b) = √(13.68×106) / (2.761×1000) = 70.38mm dassumed = D-20-4 = 165-20-4 = 141mm

=> So.provide d = 141mm

Step 4:- Steel reinforcement :Main reinforcement : Astreq. = 0.5fck/fy {1-√1- 4.6Mu/ / fck×b×d2 }×bd = 0.5 × 20/415 {1-√1 - 4.6(13.68×106)/(20×1000×1412)} ×1000×141 = 280.42 Spacing (S) = (1000 × 50.3) / 300 = 167

300 mm2 160mm c/c , which is < 3d =423mm and 300mm

Astactual = (1000×50.3)/160 = 314.3mm2 Pt = 100Ast/bd = (100×314.3)/(1000×141) =0.22 = 20%

(Page no.-73)

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Note :- Bend alternate bars of main reinforcement up at a dis. of l/7 = 4100/7 = 585.7mm from the face of support. Distribution reinforcement:Astmini. = 0.12% × bD

= (0.12/100)×1000×165 =198mm2

Spacing (S) = (1000×50.3)/198 = 254mm.

,which is <5d (705) and 450mm

Step 5:- Check for limit state of serviceability for deflection :Pt = 20% = 0.2 (page no.-38 table)

fs = 0.58fy.Astreq./Astprovided = 0.58×415×300/314.3 See,table for fs=240 and Pt=0.2

=229.7

240

=> ft (modification factor) =1.68

From the deflection pt. of view, L/dreq. = 20×ft 4100/dreq. = 20×1.68 =122

=> dreq. =125mm

But,available dprovided = 141mm Hence ,safe. Step 6:- Check for shear :Nominal shear stress (ʈv) : (page no. 72) ʈv = VuD/ bd =(15.58×103)/(1000×141) =0.110 N/mm2

{ VuD = S.F. due to design loads = Vu – Wu(0.141) = 17.09 -(10.687×0.141) =15.58 KN}

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Design shear strength of concrete (ʈc.k): * Page no.72:- 40.2.11

=> k=

for D=165mm.

*Page no.73:table19

=> ʈc =0.28

for M20 ,Pt = 10% ( at supports =1/2×20%)

So, k.ʈc =

×0.28 =

Now, ʈv <<< k.ʈc 0.110<<<<< Hence section is safe.

Step7:-Check for development length :- (page no.42) Page no.44:- At simple supports, Ø of bars should be so restricted that following requirement is satisfied

:- (1.3Mu1/Vu )+ LO > Ld

{*Ast1 at supports of short span=½ (314.3mm2)

*Mu1 = 0.87fyAst1/(d - 0.416xu) = 0.87×415×157.15 / (141-0.416×7.88)

*Xu = 0.87 fy.Ast1/(0.36 fck.b) =7.88mm }

= 411.98 Nmm *Vu (as calculated in step2) = 17090 N at centre of support * Assume that width of support ls =300mm and a side cover(x‘) of 25mm Providing 90° bend ; Lo= ls/2 – x‘+3

=300/2 -25+3×8 =149mm

*Ld (Development length of bars: page no.42):Ld = Øσs / 4 ʈbd

{page no.43:- ʈbd for M20 =1.2 and this

21 | P a g e

= 60 % of 0.87fyØ / (4ʈbd)

value should be increased by 60%}

= 0.6 ×{0.87×415×8 / 4×1.2 } =376 mm

So,

(1.3Mu1/Vu) + LO > Ld (1.3×6.5×106/15301) + 149 >Ld =>

701 >376

Hence,satisfied.

Step 8: Load distribution in one way slab :

Fig 5 load distribution in one way slab for slab S9

22 | P a g e

Two way slab (S3) :- (KITCHEN : 2.5×3.4 and BATHROOM+W.C. :2.5×3.4)

Step 1: Design constants and limiting depth of neutral axis :For, Fy = 415 N/mm2

=> Xu,max/d = 0.479 and Ru =2.761

Fck = 20 N/mm2

Step 2: Computation of design load and B.M. :

(Page no. 35):Assume,L/d = 35 for simply supported slab L/d=35×0.8

(If HYSD bars are used then L/d must be multiplied with 0.8, but if mild steel is used then no need to multiply)

Pt = 20% for under reinforced section

(can be taken upto 30% for HYSD bars and 40-50% for mild steel and these values are generally taken byexperience)

ft (Modification factor) = 1.68 Now,L/d =35×0.8×1.68

(*page no.-380) (for slabs spanning in two directions,

=> 2500/d= 35×0.8×1.68

L=shorter of two spans(3.4×2.5) should

=>

be used for calculating L/d ratio)

d = 53.14

60mm.

Providing nominal cover of 20mm. and 8mmØ bars D= 60+20+4=84mm

100mm

23 | P a g e



Design load(Wu) : for D=100mm =0.1m DL(wt. of slab/m2) = vol.×density =(0.1m×1m×1m)×25KN/m2

=2.5KN/m2

LL+FF(imposed loads) =1.5+1.5 =3KN/m2 So,total load =2.5+3 = 5.5KN/m2 Design load(Wu) =1.5 ×5.5 =8.25KN/m2 

Design moment(Mu) :

Taking d=60mm: ly = 3.4+0.06 =3.46mm

{page no.91 table:27 :- for ly/lx =3.46/2.56= 1.4 ,

lx =2.5+0.06 =2.56mm

=> αx =0.099,αy =0.051 }

Mux = αx..Wu..lx2 = 0.099×8.25×2.562 =5.353KN-m Muy = αy.Wu.ly2 = 0.051×8.25×3.462 = 5.037 KN-m (Page no. 90) :For long span, width of middle strip =3/4 ly = ¾( 3.46) =2.6m width of edge strip =½(3.46-2.6) = 0.43m For short span ,width of middle strip =3/4 lx= ¾(2.56) =1.92m Width of edge strip = ½(2.56-1.92) =0.32m

Step 3: Computation of effective depth :

For, b(width) = 1m. =1000mm d = √Mu/(Ru×b) = √(5.35×106) / (2.761×1000) = 44.01mm



but,,from point of deflection,Dassumed =100mm with nominl cover=20mm and 8mmØ bars

24 | P a g e

So,dshort span =100-20-4 =76mm dlong span = 100-20-4-8 = 68mm

Step 4: Computaion of steel reinforcement for short span (lx =2.56m):

Astx =0.5fck/fy{1-√1-4.6Mux/fck.b.d2}.b.d. =0.5×20/415 {1-√1 – (4.6×5.35×106)/(20×1000×762)}×1000×76 = 206.9

207mm2

Providing 8mmØ bars,

Spacing( Sx) =1000×50.3/207 =242.99mm

200mmc/c

Hence,providing 8mmØ bars @200mmc/c for middle strip of short span = ¾(lx) =1.92m 

Edge strip of lx =0.32m Reinforcement in edge strip =1.2D = 1.2(100) =120mm2 Spacing(S) = 1000×50.3/120 = 419mm c/c

Note: Bend half the bars up at a distance of 0.15lx =0.15(2.56) = 0.384m =384mm from centre of support. Step 5 :Reinforcement for long span (ly=3.46m): 

Asty=0.5×20/415{1-√1- (4.6×5.037×106)/(20×1000×682)} =220mm2 Providing 8mmØ bars,

Spacing (Sy) =1000×50.3/220 =229mm

220mm c/c

Hence,providing 8mmØ bars@220mmc/c in middle strip of ¾(ly) = ¾(3.46) =2.6m 

Edge strip of ly =0.43m Reinforcement in edge strip =1.2D =120mm2

25 | P a g e

Spacing(S) =1000×50.3/120 =419mm c/c

Step 6: Torsional reinforcement at corners: Size of torsional mesh = lx/5 =2.56/5 =0.512m from the centre of support Area of torsional reinforcement =¾Astx =¾(207) =155.25mm2 Spacing (S) =1000×50.3/155.25 =323mm c/c

Step 7 :Check for shear and development length in short span(lx) :Shear force at long edges(Vu) =Wu×lx×r/(2+r) =8250×2.56×1.4/(2+1.4) =8696.4 N Page no.72 :- ʈv =Vu/b.dx =8696.4/(1000×76) =0.114 N/mm2 Page no.44:- At simply supports, Ø of bars should be so restricted that following requirement is satisfied

*Mu1 = 0.87fyAst1/(d - 0.416xu)

:- (1.3Mu1/Vu )+ LO > Ld

{*Ast1 at supports of short span=½(1000×50.3/200) =125.75mm2

= 0.87×415×125.75 / (100-0.416×6.3) =

Nmm

*Xu = 0.87 fy.Ast1/(0.36 fck.b) =6.3mm }

*Assuming that width of support ls =160mm and a side cover(x‘) of 20mm Provinding no hooks ; Lo = ls/2 – x‘ =160/2 -20 =60mm *Ld (Development length of bars: page no.42):Ld = Øσs / 4 ʈbd = 60 % of 0.87fyØ / (4ʈbd)

{page no.43:- ʈbd for M20 =1.2 and this value should be increased by 60%}

26 | P a g e

= 0.6 ×{0.87×415×8 / 4×1.2 } =376 mm

So,

(1.3Mu1/Vu) + LO > Ld (1.3×3.3×106/8696.6) + 60 >Ld =>

553 >376

Hence,satisfied.

Step 8 :-Distribution of loading in two way slab :- (page no. 41)

Fig 6 two way slab load distribution and action for slab S3

27 | P a g e

Two way slab (S1) :- (BEDROOOM: 3.15 ×3.4m2 c/c)


=> Xu,max/d = 0.479 and Ru =2.761

Fck = 20 N/mm2


Assume,L/d = 35 for simply supported slab L/d=35×0.8





(*page no.-380) (for slabs spanning in two directions,

=> 3150/d= 35×0.8×1.68


=>


d = 66.9

70mm.

Providing nominal cover of 20mm. and 8mmØ bars D= 70+20+4=94mm 

100mm

Design load(Wu) : 28 | P a g e

for D=100mm =0.1m DL(wt. of slab/m2) = vol.×density =(0.1m×1m×1m)×25KN/m2

=2.5KN/m2

LL+FF(imposed loads) =1.5+1.5 =3KN/m2 So,total load =2.5+3 = 5.5KN/m2 Design load(Wu) =1.5 ×5.5 =8.25KN/m2



Design moment(Mu) : Taking d=70mm

{ page no.91 table:27

long span:_ly = 3.4+0.07 =3.47mm

for ly/lx =3.46/3.26= 1.06

short span:-lx =3.15+0.07 =3.22mm

=> αx =0.074,αy =0.061 }

Mux = αx..Wu..lx2 = 0.074×8.25×3.222 =6.329KN-m Muy = αy.Wu.ly2 = 0.061×8.25×3.472 = 6.059 KN-m (Page no. 90) :For long span(ly), width of middle strip =3/4 ly = ¾( 3.47) =2.60m width of edge strip =½(3.47-2.6) = 0.435m For short span(lx) ,width of middle strip =3/4 lx= ¾(3.22) =2.415m width of edge strip = ½(3.22-2.415) =0.402m


For, b(width) = 1m. =1000mm d = √Mux/(Ru×b) = √(6.329×106) / (2.761×1000) = 47.87mm

29 | P a g e



but,,from point of deflection,Dassumed =100mm with nominal cover=20mm and 8mmØ bars So,dshort span(x) =100-20-4 =76mm dlong span(y) = 100-20-4-8 = 68mm


Astx =0.5fck/fy{1-√1-4.6Mux/fck.b.d2}.b.dx. =0.5×20/415 {1-√1 – (4.6×6.327×106)/(20×1000×762)}×1000×76 = 247.40

250mm2


Spacing( Sx) =1000×50.3/250 =201.2mm

200mmc/c



Note: Bend half the bars up at a distance of 0.15lx =0.15(3.22) = 0.483m =483mm from centre of support . Step 5 :Reinforcement for long span (ly=3.47m): 

Asty=0.5×20/415{1-√1- (4.6×6.059×106)/(20×1000×682)}×1000×68 =268.9mm2 Providing 8mmØ bars,

270mm2 Spacing (Sy) =1000×50.3/270 =186mm

180mm c/c

Hence,providing 8mmØ bars@1800mmc/c in middle strip of ¾(ly) = ¾(3.47) =2.6m 30 | P a g e



Edge strip of ly =0.435m Reinforcement in edge strip =1.2D =120mm2 Spacing(S) =1000×50.3/120 =419mm c/c

Note: Bend half the bars up at a distance of 0.15ly =0.15(3.47) = 0.52m =520mm from centre of support.


Step 7 :Check for shear and development length in short span(lx) :Shear force at long edges(Vu) =Wu×lx×r/(2+r) =8250×3.22×1.4 /(2+1.4) = 10938N Page no.72 :- ʈv =Vu/b.dx = 10938 /(1000×76) =0.413 N/mm2 Page no.44:- At long edges, Ø of bars should be so restricted that following requirement is satisfied

:- (1.3Mu1/Vu )+ LO > Ld

*Mu1 = 0.87fyAst1/(dx - 0.416xu)

{*Ast1 at supports of short span=½(1000×50.3/250) =100.6mm2

= 0.87×415×100.6/(76-0.416×5.04) = 491.47 Nmm

*Xu = 0.87 fy.Ast1/(0.36 fck.b) =5.04mm

}

*Assuming that width of support ls =160mm and a side cover(x‘) of 20mm 31 | P a g e

Provinding no hooks ; Lo= ls/2 – x‘ =160/2 -20 =60mm *Ld (Development length of bars: page no.42):Ld = Øσs / 4 ʈbd = 60 % of 0.87fyØ / (4ʈbd)


= 0.6 ×{0.87×415×8 / 4×1.2 } =376 mm

So,

(1.3Mu1/Vu) + LO > Ld (1.3×

/10938) + 60 >Ld

=>

>376

Hence,satisfied. Step 8 :-Distribution of loading in two way slab :- (page no. 41)

Fig 7 two slab load distribution for slab S1

32 | P a g e

Two way slab (S4) : (LIVING ROOM:3.2*4.4mm c/c) Step 1: Design constants and limiting depth of neutral axis :For, Fy = 415 N/mm2

=> Xu,max/d = 0.479 and Ru =2.761

Fck = 20 N/mm2


(Page no. 35):Assume,L/d = 35 for simply supported slab L/d=35×0.8





(*page no.-380 (for slabs spanning in two directions,

=> 2500/d= 35×0.8×1.68


=>


d = 53.14

60mm.

Providing nominal cover of 20mm. and 8mmØ bars D= 60+20+4=84mm 

100mm

Design load(Wu) :

33 | P a g e

for D=100mm =0.1m DL(wt. of slab/m2) = vol.×density =(0.1m×1m×1m)×25KN/m2

=2.5KN/m2

LL+FF(imposed loads) =1.5+1.5 =3KN/m2 So,total load =2.5+3 = 5.5KN/m2 Design load(Wu) =1.5 ×5.5 =8.25KN/m2



Design moment(Mu) :

Taking d=60mm: ly = 3.4+0.06 =3.46mm

{page no.91 table:27 :- for ly/lx =3.46/2.56= 1.4

lx =2.5+0.06 =2.56mm

=> αx =0.099,αy =0.051 }

Mux = αx..Wu..lx2 = 0.099×8.25×2.562 =5.353KN-m Muy = αy.Wu.ly2 = 0.051×8.25×3.462 = 5.037 KN-m (Page no. 90) :For long span(ly), width of middle strip =3/4 ly = ¾( 3.46) =2.6m width of edge strip =½(3.46-2.6) = 0.43m For short span(lx),width of middle strip =3/4 lx= ¾(2.56) =1.92m Width of edge strip = ½(2.56-1.92) =0.32m


For, b(width) = 1m. =1000mm d = √Mu/(Ru×b) = √(5.35×106) / (2.761×1000) = 44.01mm



but,,from point of deflection,Dassumed =100mm with nominl cover=20mm and 8mmØ bars 34 | P a g e

So,dshort span =100-20-4 =76mm dlong span = 100-20-4-8 = 68mm


Astx =0.5fck/fy{1-√1-4.6Mux/fck.b.d2}.b.d. =0.5×20/415 {1-√1 – (4.6×5.35×106)/(20×1000×762)}×1000×76 = 206.9

207mm2


Spacing( Sx) =1000×50.3/207 =242.99mm

200mmc/c



Note: Bend half the bars up at a distance of 0.15lx =0.15(2.56) = 0.384m =384mm from centre of support. Step 5 :Reinforcement for long span (ly=3.46m): 

Asty=0.5×20/415{1-√1- (4.6×5.037×106)/(20×1000×682)} =220mm2 Providing 8mmØ bars,

Spacing (Sy) =1000×50.3/220 =229mm

220mm c/c

Hence,providing 8mmØ bars@220mmc/c in middle strip of ¾(ly) = ¾(3.46) =2.6m 

Edge strip of ly =0.43m Reinforcement in edge strip =1.2D =120mm2

35 | P a g e

Spacing(S) =1000×50.3/120 =419mm c/c


Step 7 :Check for shear and development length in short span :Shear force at long edges(Vu) =Wu×lx×r/(2+r) =8250×2.56×1.4/(2+1.4) =8696.4KN Page no.72 :-

ʈv =Vu/b.dx =8696.4/(1000×76) =0.114 N/mm2

Page no.44:- At simply supports, Ø of bars should be so restricted that following requirement is satisfied

:- (1.3Mu1/Vu )+ LO > Ld

*Mu1 = 0.87fyAst1/(d - 0.416xu)

{*Ast1 at supports of short span=1000×50.3/400 =125.75mm2

= 0.87×415×125.75 / (100-0.416×6.3) = 466.23 Nmm

*Xu = 0.87 fy.Ast1/(0.36 fck.b) =6.3mm }

*Assuming that width of support ls =160mm and a side cover(x‘) of 20mm Provinding no hooks ; Lo= ls/2 – x‘ =160/2 -20 =60mm *Ld (Development length of bars: page no.42):Ld = Øσs / 4 ʈbd = 60 % of 0.87fyØ / (4ʈbd)


36 | P a g e

= 0.6 ×{0.87×415×8 / 4×1.2 } =376 mm

So,

(1.3Mu1/Vu) + LO > Ld (1.3×3.3×106/8696.6) + 60 >Ld =>

553 >376

Hence,satisfied.

Step 8 :-Distribution of loading in two way slab :- (page no. 41)

Fig 8 two way slab load distribution and action

37 | P a g e

One way slab (S9,Staircase): Step 1 :General arrangement of stairs:Height of each flight =3/2 =1.5m. No. of Rises required=1.5/0.15 =10 in each flight. No. of treads reqired in each flight = 10-1 =9 Space occupied by treads = 9×230 = 2070mm. Landing =1m Space for Passage =0.55m

Step 2 :Computation of design constants : For ,

fy = 415N/mm2

=> Xumax / d =0.479 and Ru = 2.761

Step 3: Computation of Loading and Bending moment : Let the bearing of landing slab in the wall to be 160mm Effective span = 2.07+1m+(0.16/2) =3.15m Let thickness of waist slab =200mm. Weight of slab w‘ on slope =200/1000×1×1×25000 =5000 N/m2 Dead wt. on horizontal area (W1) = (w‘√R2 + T2 ) / T = 5000√1502 + 2302 /230 =5969 N/mm2 Dead wt. of steps (W2) = R / (2×1000) ×25000 = 150 /2000 ×25000

= 1875 N/m2

38 | P a g e

Total dead wt/mm = 5969+1875

=7844 N

Weight of finishing =100 N (assumed) LL

=2500 N

Total (LL +finishing) =10444 N Wu =1.5×1044 =1566 N/m Mu =15666×3.152 /8

= 19.93×106 N-mm

Step 4 :Design of waist slab : d = √Mu / Ru×b

=√19.43×106 / 2.761×1000

=83.88mm

Adopt overall depth (D) of 150mm ,using 20mm nominal cover and 10mmØ bars d= 150-20-5 = 125mm

Step 5 : Computation of reinforcement :Ast = (0.5×20 /415) ×1000×125×{1-√1-(4.6×19.43×106 / 2.761×1000×1252)} = 1390 mm2 No. of bars for 1.05 width =1.05×1390/ 78.54 =19 bars Spacing , S = 1050 /19 =55mm c/c Asd = 0.12 ×150×1000 /100 =180 mm2 Spacing of 8mmØ bars = 1000×50.3 /180 =279 mm c/c

39 | P a g e

CANTILEVER SLAB ,S2 : (BATHROOM 1.475 overhang )


=> Xu,max/d = 0.479 and Ru =2.761

Fck = 20 N/mm2 Step 2: Computation of design B.M. and S.F. :

Assume,L/d = 7 Pt = 20%

for cantilever slab (from the point of deflection) for under reinforced section

ft (Modification factor) = 1.68 Taking c/c dis. L = 1.535m => L/d = 7×1.68

{Leffective =1.475+(.12/2) =1.535 mm}

1535/d = 7×1.68 d = 130 mm. Provide D=146mm at fixed end and 110mm at free end Averagw thickness = (146+110)/2 =128 mm 

Design load (Wu) :For, depth (D) = 128mm. = 0.128m. DL (wt. of slab/m2) = vol. × density = (0.128m.×1m.×1m.)×(25 KN/m2) LL+FF (imposed loads) = 1.5 KN/m2 + 1.5 KN/m2

= 3.2 KN/m2 = 3 KN/m2

So, Total load = 3.2 + 3 = 6.2 KN/m2 Design load (Wu) = 1.5 × 6.2 = 9.3 KN/m2 40 | P a g e



Design shear force (Vu) :Vu = Wu × l = 9.3×1.475 = 13.717 KN



{for cantilever Vu =Wu×l}

Design bending moment (Mu):Mu = Wu ×l2/2

= ( 13.71×1.4752) /2 = 10.116 KN-m

{for cantilever Mu =Wu×l2/2}

Step 3: Computation of effective depth :For, b(width) = 1m. =1000mm d = √Mu / (Ru×b) = √(10.116×106) / (2.761×1000) = 60.53mm providing nominal cover of 15mm and using 8mmØ bars dassumed = D-15-4 = 146-15-4 = 127mm

{D=146mm}

Reducing D =100mm at free end Step 4:- Steel reinforcement :Main reinforcement : Astreq. = 0.5fck/fy {1-√1- 4.6Mu/ / fck×b×d2 }×bd = 0.5 × 20/415 {1-√1 - 4.6(10.116×106)/(20×1000×1272)} ×1000×127 239 mm2

= 239 Spacing (S) = (1000 × 50.3) / 300 = 167

160mm c/c

,

which is < 3d and 300mm

Distribution reinforcement:Astmini. = 0.12% × bD

= (0.12/100)×1000×146 = 175.2mm2

Spacing (S) = (1000×50.3)/175.2 = 287 Pt =100Ast /bd = 100×186/1000×127

270mm c/c

, which is <5d and 450mm

=0.146%

Also, provide distribution reinforcement in the form of 8mmØ bars @300mm c/c

41 | P a g e

Step 6:- Check for shear :Nominal shear stress (ʈv) : (page no. 72) ʈv = VuD/ bd =(13.71×103)/(1000×127) =0.108 N/mm2

Design shear strength of concrete (ʈc.k): * Page no.72:- 40.2.11

=> k= 1.3

*Page no.73:table19

=> ʈc =0.28

for D=146mm. for M20 ,Pt = 10%

So, k.ʈc = 1.3×0.28 = 0.364 N/mm2

Now, ʈv <<< k.ʈc 0.110<<<<<0.364 Hence section is safe.

Step7:-Check for development length :- (page no.42) Page no.44:- At simple supports, Ø of bars should be so restricted that following requirement is satisfied

:- (1.3Mu1/Vu )+ LO > Ld

*Mu1 = 0.87fyAst1/(d - 0.416xu) = 0.87×415×157.15 / (141-0.416×7.88)

{*Ast1 at supports of short span=½ (314.3mm2) *Xu = 0.87 fy.Ast1/(0.36 fck.b) =7.88mm }

42 | P a g e

=

Nmm

*Vu (as calculated in step2) = 17090 N at centre of support * Assume that width of support ls =300mm and a side cover(x‘) of 25mm Providing 90° bend ; Lo= ls/2 – x‘+3

=300/2 -25+3×8 =149mm

*Ld (Development length of bars: page no.42):Ld = Øσs / 4 ʈbd = 60 % of 0.87fyØ / (4ʈbd)


= 0.6 ×{0.87×415×8 / 4×1.2 } =376 mm

So,

(1.3Mu1/Vu) + LO > Ld (1.3×6.5×106/15301) + 149 >Ld =>

701 >376

Hence,satisfied.

43 | P a g e

6.4:- DESIGN OF BEAMS DESIGN OF BEAM OF BED ROOM ALONG X-AXIS (beam-1) Span of beam (L) = 3.4 m Width of beam = 230 mm Take Depth of beam= 350 mm Concrete mix=M20 Characteristic strength= 415N/mm2 COMPUTATION OF DESIGN BENDING MOMENT:Slab load = (1/2(3.4+.25) x1.575x6.25)/3.4 = 5.28 KN/m Self-weight of beam= 0.23x0.35x1x25 = 2.125 KN/m Total design load (WD) = 1.5x (5.28+2.0125) = 11.11 KN/m Factored resisting moment (MR) = WDxL2/8= 11.11x3.42/8 = 16.05 KN-m COMPUTATION OF EFFECTIVE DEPTH (d):Mu=.36 x Xu, max/d [1 - .42 x Xu,max/d] fck bd2 Xu, max/d = 700/ [1100 + .87fy] = .479 . .. Mu= 2.761 bd2 And from above equation d=√Mu/2.761 bd2 = √16.05x106/2.761x230 = 159mm Provide 25mm clear cover and using 20mm dia. bars with 8mm stirrups Overall beam Depth (D) = 159+25+8+20/2 = 202 mm However keep beam Depth= 350mm Therefore, effective depth of beam (d) = 350-25-8-20/2 = 307 mm 44 | P a g e

STEEL REINFORCEMENT:Ast = 0.5 fck/fy [1 - √1-(4.6 Mu/fck bd2)] bd= 151.62mm2 Astmin = 0.85bd/fy = 144.62 mm2 Astmax = 0.04bD = 3220 mm2 No. of 20mm dia. Bars = 151.62/π/4x122 = 1.34≈2bars of 12 mm dia. Horizontal Spacing should be lesser of greatest of the following:a) Diameter of the bar b) 5mm more than the nominal size of aggregate Hence provide horizontal spacing of 12 mm between the bars

[IS 456:2000 Clause 26.3.2 (a) ]

Actual Ast = 2x(π/4x122) = 226.2 mm2 Moment of resistance at mid span is :MR = 16.705 KN-m (evaluated above). Mu,lim = 2.761 bd2 = 58.85 KN-m Since Mu < Mu,lim. The design is ok. SHEAR REINFORCEMENT:The critical section for shear is at a distance of d=(0.307 m) from face of support. Note that distance of theoretical centre of support from face = d/2=0.307/2=0.16 Therefore, VuD = Wu L/2 – Wu (d/2 + d)=(11.11x3.4)/2 – 11.11x(0.307+0.16) = 13.70 KN τv = VuD/bd= 13.70x103/(230x307) = 0.194N/mm2 100Ast/bd (at the support) = 100x226.2/(230x307) = 0.32% From table 19 of IS 456:2000 τc = 0.39N/mm2 Since τv < τc , min. shear reinforcement is given according to clause 26.5.1.6 of IS456:2000. 45 | P a g e

As per IS 456:2000, nominal shear reinforcement is given by the expression:Asv/b.sv > 0.4/0.87 fy

or

sv = 2.175 Asv . fy/b

Using 2 legged 8mm dia. Stirrups, Asv = 2 x π/4 (8)2 = 100.5 mm2 Sv= 2.175x100.65x415/230 = 335.66 mm Max. spacing is least of:0.75 d= 0.75x307 = 230.25 mm

Or (i)

300 mm

Hence provide 8 mm dia. 2-legged stirrups @225 mm c/c throughout the length of the beam. Provide 2-10 mm dia. holding bars at the top. CHECK FOR DEVELOPMENT LENGTH:Ld ≤ 1.3 M1/V + Lo Xu = 0.87 fy Ast1/0.36 fck b = 0.87x415x226.2/0.36x20x230 = 49.31mm ≈ 50mm M1 = 0.87 fy Ast1 (d – 0.42 Xu) = 0.87x415x226.2 (307 – 0.42x50) = 23.35x106 KN-m Vu = Wu L /2 = 11.11x3.4 /2 = 18.88 KN Ld = 0.87 fy x diameter of bar / 4τb

(clause 26.2.1 of IS456:2000)

Ld = 0.87x415x12/4x1.2x1.6 = 564.14 mm ≈ 565 mm Lo = sum of anchorage beyond centerline of support and anchorage hook value. If no hook is provided, then Lo = Ls/2 – x‘ Where Ls = wall thickness x‘ = side cover which is assumed to be 40 mm

46 | P a g e

Therefore, Lo = 200/2 -40 = 60 mm Ld/3 = 565/3 = 188.33 mm ≈ 189 mm Available distance beyond face of support = Lo + Ls/2 = 60+ 200/2 = 160 mm < Ld/3. Hence hook is required. According to codal provision, Anchorage Value of standard U-type hook shall be equal to 16 times the diameter of bar (clause 26.2.2.1, IS456:2000). Hence, Lo = 16x diameter of bar = 16x12 = 192 mm Therefore, 1.3 M1/V + Lo = 1.3 x (23.35x106/18.88x103) +192 = 1799.78 mm > Ld . Hence O.K.

DESIGN OF LOBBYBEAM (beam-10,11) Span of beam (L) = 2.2 m Width of beam = 230 mm Take Depth of beam= 350 mm Concrete mix=M20 Characteristic strength= 415N/mm2 COMPUTATION OF DESIGN BENDING MOMENT:staircase load = 10.44/2.2 = 4.74 KN/m Self-weight of beam= 0.23x0.35x1x25 = 2.0125 KN/m Total design load (WD) = 1.5x (4.74+2.0125) = 10.13 KN/m Factored resisting moment (MR) = WDxL2/8= 10.13x2.22/8 = 6.13 KN-m

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COMPUTATION OF EFFECTIVE DEPTH (d):Mu=.36 x Xu, max/d [1 - .42 x Xu,max/d] fck bd2 Xu, max/d = 700/ [1100 + .87fy] = .479 . .. Mu= 2.761 bd2 And from above equation d=√Mu/2.761 bd2 = √6.13x106/2.761x230 = 98.25mm ≈ 93mm Provide 25mm clear cover and using 20mm dia. bars with 8mm stirrups Overall beam Depth (D) = 93+25+8+16/2 = 134 mm However keep beam Depth= 350mm Therefore, effective depth of beam (d) = 350-25-8-16/2 = 309 mm

STEEL REINFORCEMENT:Ast = 0.5 fck/fy [1 - √1-(4.6 Mu/fck bd2)] bd= 55.88mm2 No. of 16mm dia. Bars = 55.88/π/4x102 = 0.71≈2bars of 10mm dia. Horizontal Spacing should be lesser of greatest of the following:a) Diameter of the bar b) 5mm more than the nominal size of aggregate Hence provide horizontal spacing of 12 mm between the bars [IS 456:2000 Clause 26.3.2 (a) ] Actual Ast = 2x(π/4x102) = 157 mm2 Moment of resistance at mid span is :MR = 6.13 KN-m (evaluated above). Mu,lim = 2.761 bd2 = 60.63 KN-m

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Since Mu < Mu,lim. The design is ok. SHEAR REINFORCEMENT:The critical section for shear is at a distance of d=(0.309 m) from face of support. Note that distance of theoretical centre of support from face = d/2=0.309/2=0.16 Therefore, VuD = Wu L/2 – Wu (d/2 + d)=(10.13x2.2)/2 – 10.13x(0.309+0.16) = 6.40 KN τv = VuD/bd= 6.40x103/(230x309) = 0.09N/mm2 100Ast/bd (at the support) = 100x157/(230x309) = 0.22% From table 19 of IS 456:2000 τc = 0.336N/mm2 Since τv < τc , min. shear reinforcement is given according to clause 26.5.1.6 of IS456:2000. As per IS 456:2000, nominal shear reinforcement is given by the expression:Asv/b.sv > 0.4/0.87 fy

or

sv = 2.175 Asv . fy/b

Using 2 legged 8mm dia. Stirrups, Asv = 2 x π/4 (8)2 = 100.5 mm2 Sv= 2.175x100.65x415/230 = 335.66 mm Max. spacing is least of:(i)

0.75 d= 0.75x307 = 230.25 mm Or

(ii)

300 mm

Hence provide 8 mm dia. 2-legged stirrups @230 mm c/c throughout the length of the beam. Provide 2-10 mm dia. holding bars at the top.

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CHECK FOR DEVELOPMENT LENGTH:Ld ≤ 1.3 M1/V + Lo Xu = 0.87 fy Ast1/0.36 fck b = 0.87x415x157/0.36x20x230 = 34.22 mm ≈ 35mm M1 = 0.87 fy Ast1 (d – 0.42 Xu) = 0.87x415x157 (309 – 0.42x35) = 16.68x106 KN-m Vu = Wu L /2 = 10.13x2.2 /2 = 11.143 KN Ld = 0.87 fy x diameter of bar / 4τb

(clause 26.2.1 of IS456:2000)

Ld = 0.87x415x10/4x1.2x1.6 = 471 mm Lo = sum of anchorage beyond centerline of support and anchorage hook value. If no hook is provided, then Lo = Ls/2 – x‘ Where Ls = wall thickness x‘ = side cover which is assumed to be 40 mm Therefore, Lo = 200/2 -40 = 60 mm Ld/3 = 471/3 = 157 mm Available distance beyond face of support = Lo + Ls/2 = 60+ 200/2 = 160 mm < Ld/3. Hence hook is required. According to codal provision, Anchorage Value of standard U-type hook shall be equal to 16 times the diameter of bar(clause 26.2.2.1, IS456:2000). Hence, Lo = 16x diameter of bar = 16x10 = 160 mm Therefore, 1.3 M1/V + Lo = 1.3 x (16.68x106 /11.143x103) +160 = 2105.97 mm > Ld Hence O.K.

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 Other beams are similarly designed and the all values are tabulated below:S.NO

BEAM NO.

Span

Width

Depth

Design

Design moment

(m)

(mm)

(mm)

load(KN)

(KN-m)

1

Beam-1

3.40

230

350

11.11

16.05

2

Beam-2

3.15

250

400

10.39

12.88

3

Beam-3

5.00

250

400

14.06

47.52

4

Beam-4 &5

2.50

250

400

16.875

24.25

5

Beam-6

2.50

250

400

19.11

30.96

6

Beam-7

3.20

230

350

14.013

17.94

7

Beam-8 &12

4.40

250

400

24.36

58.96

8

Beam-9

3.20

250

400

19.95

25.54

9

Beam-10 & 11

2.20

230

350

10.13

6.13

10

Beam-13

3.65

230

350

11.56

19.25

Table 7:- beam design data BEAM NO.

Ast 2

Required(mm )

Dia of

No.of

Actual

Dia of shear

Spacing of

bars

bars(Ast/

Ast

reinforcement

shear

(mm2)

(mm)

reinforcement

(Φmm)

Φ

2

)

c/c (mm) Beam-1

151.62

12

2

226.2

8

225

Beam-2

102.41

10

2

157

8

260

Beam-3

407.45

10

6

471

8

260

Beam-4 &5

197.28

12

2

226.2

8

260

Beam-6

253.87

12

3

339.3

8

260

Beam-7

169.24

12

2

226.2

8

225

Beam-8 &12

520.68

10

5

565.5

8

260

Beam-9

208.33

12

2

226.2

8

260

Beam-10 &

55.88

10

2

157

8

225

183.67

12

2

226.2

8

225

11 Beam-13

Table 8 :- beam design data

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52 | P a g e

Fig 9. X-Section of beams

V:-DESIGN OF COLUMNS:There are usually three types of columns that are usually encountered while designing. These are as follows along with their nomenclature used . 1. Axially loaded columns (C1) 2. Columns with axial load and uniaxial bending (C2) 3. Columns with axial load and biaxial bending (C3) Design of column C1: Using fck=20N/mm2 ,fy=415N/mm2 Unsupported length= 3-0.4=2.6m Leff =1*2.6= 2.6m

(Cl:25.13) (Annex E)

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Load calculations: Load from beam 9=(19.95*3.20)/2=31.92KN Load from beam 8=(4.4*24.36)*2/2=107.184KN Load from beam4= (3.4*16.875)/2=28.6875KN Self weight= 1.5(25*B2*3*10-6) KN =112.5*10-6B2 KN (Assuming a square column) Total axial force due to D.L+L.L=(167.79+112.5*10-6B2 )KN 1)Size of Column: Assuming P=0.01 Ag= [p/{(0.4fck)(1-p/100)+(p/100)(0.67fy)} ] Ag = [(167.79*1000+0.1125Ag)/{(0.4*20)(1-1/100)+(1/100)(0.67*20)}] Ag =[(167790+0.1125Ag)/8.054] Ag= 21128.25 mm2 So B=D= 145.35 mm Let us provide a column size of 300*300 2) Slenderness ratio of a column Effective length (l)= 2.6m=2600mm l/D=2600/300=5.6<12

(so,it will behave as a short column)

3) Minimum Eccentricity emin =[(l/500)+(D/30)] emin= [(2600/500)+(300/30)]= 15.2 ( subject to a minimum of 20mm) 0.005D= 0.005*300= 15mm Since emin >0.005D, the column needs to be redesigned

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Lets provide a column section of 400x400 emin=[(2600/500)+(400/30)]=18.53~20mm 0.05(400)=20mm 4) Longitudinal Reinforcement:[cl 26.5.3.1] Asc =(1/100)(21142.10)=211.42mm2 Providing 12mm bars, so the no. of bars =211.42/(0.785x144)=1.87 However providing a minimum of 4 bars of HYSD grade (415),with a clear cover of 40mm So actual Asc provided =4x0.785x144=452.16mm2 5) Transverse reinforcement: [cl 26.5.3.2] Providing 8mm dia ties as 8>(12/4) Pitch of lateral ties= Least of lateral dimension=400 16(Dia of longitudinal rein.)=16*12=192 300 So provide 8mm ties@180mmc/c c/c distance between main bars=400-2 x 40-2x6=308mm<48(8).

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Design of column C2: Load calculations : Load from beam 13: (3.65x11.56)/2=21.097 KN --------------------6 : ( 0.5x3.4x19.11x2/3)=21.65 KN --------------------8 : (4.4x24.36/2) =53.592 KN ---------------------7: (3.2x14.013/2)=22.42KN --------------------5 : (3.4x16.875/2)(2/3)=19.125KN Total load (L.L+D.L)=137.884 KN Let us provide a column section of (350x400) Self weight=(25x0.35x0.4x3)x1.5=15.75KN So total axial load =153.634 KN~160KN Maximum Moment=30.96 KNm~50KNm Effective length of column= 2.6m

[cl 39.1 and cl 39.2]

Step1: check for slenderness of the column: [lex/D]= 2600/400= 6.5<12 [ley/B]= 2600/350 =7.42 <12 Therefore it is a short column in both axes of bending. Step2: Moment due to min eccentricity Eccentricity about X axis exmin=[(l/500)+(D/30)]=[(2600/500)+400/30]=18.53~20mm Moment due to exmin= (Pcu . exmin)= (160X20)/1000= 3.2 KNm

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Eccentricity about Y axis eymin=[(2600/500)+(350/30)]=16.86 ~20mm Moment due to eymin= (Pcu. eymin=)=(160x20)/1000= 3.2 KNm Let the cross sectional area of main reinforcement for the column be 1 percent and it is distributed equally on four sides. 16 mm diameter bars shall be used with 50mm effective cover Step 3: Calculation of various ratios Ratio of percentage of steel to characteristic strength of concrete (p/fck)= (1/20)=0.05 Ratio (Pcu/fckBD)=(160x1000/20x400x350)=0.053 Ratio of effective cover to overall depth (d‘/D)=(50/400)=0.125 Ratio of effective cover to width (d‘/B)=(50/350=0.142 From SP 16-1980 from charts 44 and 45 fy=415 N/mm2 and( Pcu/fck B.D)=0.053 p/fck=0.05 and reinforcement is distributed equally on all four sides (Mcu xlim/fck BD2)=0.1 and (Mcu ylim/fck B2D)=0.08 Bending strength of column about X axis Mcu xlim=112KNm Bending strength of column about Yaxis Mcu ylim=78.4KNm

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The column is subjected to biaxial bending (Viz-Given uniaxial bending about X-axis and moment due to eccentricity about y axis) Step 4. Check for interaction equation [cl 39.5] Column section shall be safe in case ,the following interaction equation is satisfied. [Mcu,x/Mcu,x1]αn+[Mcu,y/Mcu,y1]αn ≤1

(1)

Where Mcu,x1= Mcuxlim=maximum uniaxial moment capacity for an axial load Pcu,bending about x- axis Mcu,y1= Mcuylim=maximum uniaxial moment capacity for an axial load Pcu,bending about y- axis Ratio αn=(Pcu/Pcuz) Pcuz=(0.45fckAc+0.75fyAsc)= (0.45fckBD+(0.75fy-0.45fck)x P.B.D/100] Pcuz=[0.45x20x350x400/1000+(0.75x415-0.45x20)(1x350x400/100x1000)] Pcuz = 1683.1 [Pcu/Pcuz]=[160/1683.1]=0.095, since 0.095<0.2 Then αn=1

[cl 39.6]

[Mcu,x/Mcu,x1]αn+[Mcu,y/Mcu,y1]αn [50/112]1+[3.2/78.4]1=0.48<1 okay Therefore column section is safe with given size and assumed percentage of main reinfircement Step 5 . Crossectional area of steel reinforcement. From the above interaction equation calculations, it is seen that the column section is having the strength nearly equal to its value needed. Therefore, the assumed percentage of reinforcement is adopted. Hence the crossectional area of steel reinforcement:

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(1x350x400/100)=1400 mm2 Using 16mm diameter bars(HYSD) No. of bars= 1400/200.96=6.96 =~ 8 bars Provide 8 bars of 16mm diameter So, Ast provided = 8x200.96=107.68 Step 6: Transverse reinforcemernt The diameter of lateral ties should not be less than 5mm or (υ/4) of steel bars Using 8mm HYSD steel bars, Pitch = 1)Least dimension= 350 2)16υ=16x16=256 3)300 So, provide 8mm ties @250mmc/c Design of column C3: Load calculations: Load from beam 1=(11.11x3.4)/2= 18.887KN Load from beam 13= (3.65x11.56)/2=21.097 KN Let us provide a column section of 400x450 Self weight = (25x0.4x0.45x3)1.5= 20.25 KN Total factored axial load =Pu=60.227~ 70KN Mux=16.05 KNm ,Muy=19.25KNm

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lex=ley=2600mm lex/D=(2600/450)=5.77 <12 & ley/B=(2600/400)=6.5<12 so, it is a short column in both axes of bending.Take effective cover =60mm. Step 1: check for mimimum eccentricity ex= (Mux/Pu)=(16.05x1000/70)=229.285mm ey= (Muy/Pu)=(19.25x1000/70)=275mm exmin=[(lx/500)+(D/30)]= [(2600/500)+450/30]= 20.2>20mm eymin==[(ly/500)+(D/30)]=[(2600/500)+400/30]= 18.53~20mm Step 2: Selection of trial reinforcement ,taking a=1.15 Mu=a√(Mux2+Muy2)= 1.15√(16.152+19.252)=28.82KNm Since My>Mx, design the column for uniaxial bending with Mu=28.82 for bemding along y axis. d‘/B=[60/400]=0.15

[ chart no 45 of SP16]

[Pu/fckBd]=[(70x1000)/(20x400x450)]=0.019 [Mu/fckB2D]=[(28.82x106 ) /(20x4002x450)]= 0.020 We get[ P/fck]=0.01… P=0.01x20=0.2%= 100xAs/BD

However take p=1%, as p>0.8%

As=P.B.D/100=1x400x450/100=1800mm2 Provide 16mm [HYSD] bars No. of bars=1800/(0.785x256)=8.95 Provide 10 bars of 16mm dia, 60 | P a g e

Actual Ast= (10x0.785x256)=2009.6 mm2 P=(100xAs/B.D)= (100x2009.6/400x450)=1.116 [P/fck]=0.055 Step 3:Uniaxial moment capacity of the section about X-X d‘/D=(60/450)=0.133, Pu/fckBD=0.013 Hence from chart 45 of ISI hand book,For Pu/fckbD=0.019 and p/fck=0.055 we have We get [Mu/fckbd2]=0.075 Mu=[0.075x20x400x450x450]=`121.5KNm Step 4:Uniaxial moment capacity of the section about Y-Y d‘/b=[60/400]=0.15 Hence from chart 45 of ISI hand book,For Pu/fckbD=0.019 and p/fck=0.055 We‘ve =[Mu/fckb2d]=0.075 Mu =[0.075x20x4002x450]=108 KNm Step 5:Computation of Puz p=1.116 ,fy=415N/mm2 ;fck= 20N/mm2 d‘/D=(60/450)=0.133 check for the trial section For bending about x-axis αn=[Pcu/Pcuz] Pcuz=(0.45fckAc+0.75fyAsc)= (0.45fckBD+(0.75fy-0.45fck)x p.B.D/100] Pcuz=[(0.45x20x400x450/1000)+(0.75x415-0.45x20)(1.16x400x450/1000x100)]=2250.57KN αn=[Pcu/Pcuz]=[70/2250.57]=0.0311<0.2 61 | P a g e

Then αn=1

[cl 39.6]

[Mcu,x/Mcu,x1]αn+[Mcu,y/Mcu,y1]αn [16.05/121.05]1+[19.25/108]1=0.310 <1 ,so safe. Step 6: Design of transverse reinforcement Diameter of lateral ties >υ/4 Provide 8mm ties , Spacing of ties =min of 1) least lateral dimension=400mm 2) 16υ=16x16=256mm 3) 300mm So provide 8mm ties @250mmc/c

Column

Unsupported

Designati

Length (m)

on

No.

Size(L*B)

(L.L+D.L)m

Self

Total

Mmax

of

ax

Wt.

Axial

(KNm)

col

(KN)

(KN)

force(KN)

C1

3

2

400*400

167.79

12

179.79

N.A

C2

3

20

350*400

137.884

15.75

153.634

30.1=50

C3

3

8

400*450

39.984

20.25

60.234

Mx=16.05, My=19.25

Table 9:- column design data

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Fig10. layout of beams and columns

63 | P a g e

Fig11. Fig showing sectional view of column (C1)

Fig 12 cross sectional view of column (C2) 64 | P a g e

Fig 13 sectional view of column (C2)

Fig 14 cross sectional view of column (C3) 65 | P a g e

Fig 15 sectional view of column (C3)

66 | P a g e

6.6:- DESIGN OF FOOTING FOOTING FOR COLUMN (C1) Design footing for column C1 size 400*400, load on column is 181.79, and moment is 0KN-M. Soil bearing capacity is 200 KN /m2 Solution:1- Design constants For M20 and Fe 415 combination Xu max /d =0.479, Ru=2.761 2- Size of footing W=181.79, W‘=10% of W= 18.179, Total weight = 181.79 +18.179= 199.969 ~~ 200 KN Equating the maximum and minimum soil pressure to safebearing capacityof soil Total weight/B^2 + M /(B^3/6) = soil bearing capacity M=0, 200 = 199.969 B^2 B=1.1M So we provide the footing of size = (1.2*1.2) m^2 The maximum and soil pressure is given by po1 = W/B^2 +6M/B^3 po1= 200/(1.2)^2 = 138.88 KN here moment is zero therefore minimum and maximum soil pressure is equal Pressure intensity under the column is po = 0.5 (138.88) = 69.44 kN/m^2 3- Design of section for bending compression Intensity of net soil pressure below the column face is po‘ = 69.44+( (138.88-69.44)/0.6) therefore po‘ =92.586 KN/m^2 maximum B.M will occur at face AB of cantilever length = 0.5(B-b) = 0.5 (1.2-0.4) =0.4 m Total force under cantilever = (1.2*1)(92.586+138.79)/2 = 138.87KN Distance of its centroid from face AB = ( (92.58+(2*138.79))/(92.58+138.79))* 1/3 = 0.533m 67 | P a g e

Bending moment =138.87*0.533 = 73.96 KNm Alternatively M= (B(B-b)^2) /24 * (2po1 + po‘) M = 1.2 (1.2- 0.4)^2 *(2* 138.79 +92.58) M = 11.85 KNm Mu =1.5 * 0.1185 * 10^8 = 1.77 * 10^7 N-mm d = root (M/(Ru*b1)) d= root (1.77 * 10^7/ (2.761 * 400)) d= 126.84mm ~~~ 130mm However keep d= 150mm and D = 150 + 50 = 200mm , thus providing an effective cover of 50mm . Effective depth available for second layer = 150 -12 = 138mm, Using 12mm dia bars . also keep D= 100mm at the ends. So that available d at the end is equal to 100-50 = 50mm 4- Design for shear:The depth found above should be safe for shear. For two way shear (punching stress) . The critical plane lies at d/2 = 150/2 = 75mm from the column face for which width bo = b + d/2 = 400 + 75 = 475mm. The punching shear stress = ⌡v= Fu /(4*bo*do) Fu = 1.5(W-po*b^2) = 1.5 (181.79— 92.58 * 0.475^2) Fu = 241.35 KN do = d1 + (d- d1)/ ((B - b)/2) * (B-b-d)/2 = 50 + (150 - 50)/ ((1200 - 1500)/2) * (1200 – 400 – 150 ) / 2 do = 111.90 mm~~ 112mm ⌡v = (241.35 * 1000 ) / (4* 475* 112) Punching shear stress :⌡v = 1.029 N/mm^2 Permissible shear stress = ks * 0.25 root fck = 1 * 0.25 root 20 = 1.118 N / mm^2 Hence thickness found from point of view of bending compression is safe . For one way shear , the critical plane CD lies at distance d = 150 mm from column face. The cantilever length to the right of CD = 0.6 – 0.25 – 0.15 = 0.2 mm. Intensity of pressure po‘‘ at CD is given by po‘‘ = 138.79 + (138.79/ 1.2) * (1.2 – 0.20) po‘‘ = 254 KN/m^2 the section at CD will be trapezoidal ... the width at top , b = ((400 +1200 – 400) /1000) * 150 b = 520 mm effective depth d‘ = ((50 + 150 – 50 )/ 1000) * 181.79 d‘ = 69mm 68 | P a g e

for balanced section Xu max / d = 0.476 let Xu / d = 0.4 for under reinforced section so that Xu = 0.4 d‘ = 0.4 * 69= 27.6mm therefore width of section at NA is given by bn = 520 +9 (1200 – 520 )/ 19) * 27.6 = 1507mm bn =~ 1500mm Vu = 1.5 * 1.2 * 0.2 * 0.5 * (138.88 + 254) = 70.17 KN ⌡v = Vu‘/( bn * d‘) ⌡v = (70.71 * 1000) / (1500 * 69) = 0.638N/mm^2 Permissible shear stress at P = 0.3% (under reinforced section ) = 0.384 N/mm^2 with k =1 (IS code table 7.1 and 7.2) Hence thickness provided from point of view bending compression is adequate. 5- Design for bending tension for under reinforced section Ast = (0.5* fck) /fy )* (root (1-(( 4.6 *1.77*10^7) / (20 * 400*150^20))) Ast = 889.94 mm^2 ~~~ 890mm^2 :- therefore No. Of 8 mm dia bars = 890/ 50.24 = 17.86 ~~~ 18 bars 6- Check for development length Ld = 47* dia of bars = 47 * 8 = 376mm Available length , providing a side cover of 50 mm = 600-50 = 540 mm Hence safe ,:-

FOOTING FOR COLUMN (C2) Design footing for column C2 size 350*400, load on column is 153.634, and moment is 30kNm . Soil bearibg capacity is 200 kn /m2 Solution :1- Design constants For M20 and Fe 415 combination Xu max /d =0.479, Ru=2.761 2- Size of footing W=153.634KN , W‘=10% of W= 15.363 KN, Total weight = 168.99 KN ~~ 170 KN

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Equating the maximum and minimum soil pressure to safebearing capacityof soil Total weight/B^2 + M /(B^3/6) = soil bearing capacity M=0, 200 = 170 B^2 B=1.08M So we provide the footing of size = (1.2*1.2) m^2 The maximum and soil pressure is given by po1 = W/B^2 +6M/B^3 po1= 200/(1.2)^2 + (6*30)/1.2^3 = 243 KN- maximum po2 = W/B^2 - 6M/B^3 po1= 200/(1.2)^2 - (6*30)/1.2^3 = 34.65 KN - minimum Pressure intensity under the column is po = 0.5 (243 + 34.65) =138.825 kN/m^2 3- Design of section for bending compression Intensity of net soil pressure below the column face is po‘ = 138.825+( (243 - 138.825)/0.6) *0.2 therefore po‘ = 173.55 KN/m^2 maximum B.M will occur at face AB of cantilever length = 0.5(B-b) = 0.5 (1.2-0.35) =0.425 m Total force under cantilever = (1.2*1)( 173.55 +138.82)/2 = 187.425KN Distance of its centroid from face AB = ( (173.55 +(2*138.82))/(173.55 +138.79))* 1/3 = 0.481m Bending moment =187.425 *0. 481= 90.15 KNm Alternatively M= (B(B-b)^2) /24 * (2po1 + po‘) M = 1.2 (1.2- 0.35)^2 *(2* 234 +173.55) M = 556.22 KNm Mu =1.5 * 5.562 * 10^8 = 8.343 * 10^7 N-mm d = root (M/(Ru*b1)) d= root (8.343 * 10^7/ (2.761 * 400)) d= 126.84mm ~~~ 230mm However keep d= 250mm and D = 250 + 50 = 300mm , thus providing an effective cover of 50mm . Effective depth available for second layer = 350 -12 = 388mm, Using 12mm dia bars . also keep D= 100mm at the ends. So that available d at the end is equal to 100-50 = 50mm 70 | P a g e

4- Design for shear:The depth found above should be safe for shear. For two way shear (punching stress) . The critical plane lies at d/2 = 250/2 = 125mm from the column face for which width bo = b + d/2 = 350 + 125 = 475mm. The punching shear stress = ⌡v= Fu /(4*bo*do) Fu = 1.5(W-po*b^2) = 1.5 (153.634 – 138.825 * 0.475^2) Fu = 183.46 KN do = d1 + (d- d1)/ ((B - b)/2) * (B-b-d)/2 = 50 + (250 - 50)/ ((1200 - 250)/2) * (1200 – 400 – 250 ) / 2 do = 176.31 mm~~ 176mm ⌡v = (183.46 * 1000 ) / (4* 475* 176) Punching shear stress :⌡v = 0.386 N/mm^2 Permissible shear stress = ks * 0.25 root fck = 1 * 0.25 root 20 = 1.118 N / mm^2 Hence thickness found from point of view of bending compression is safe . For one way shear , the critical plane CD lies at distance d = 150 mm from column face. The cantilever length to the right of CD = 0.6 – 0.25 – 0.15 = 0.2 mm. Intensity of pressure po‘‘ at CD is given by po‘‘ = 34.11 + ((243 – 34.11)/ 1.2) * (1.2 – 0.20) po‘‘ = 208.185 KN/m^2 the section at CD will be trapezoidal ... the width at top , b = ((400 +1200 – 400) /1000) * 150 b = 520 mm effective depth d‘ = 50 + ((250 – 50 )/ 1000) * 153.634 d‘ = 180.84mm~~ 180mm for balanced section Xu max / d = 0.476 let Xu / d = 0.4 for under reinforced section so that Xu = 0.4 d‘ = 0.4 * 180= 72mm therefore width of section at NA is given by bn = 520 +( (1200 – 520 )/ 19) * 72 = 3096mm bn =~ 3096mm Vu = 1.5 * 1.2 * 0.2 * 0.5 * (254 + 208.125) = 83.182 KN ⌡v = Vu‘/( bn * d‘) ⌡v = (83.183 * 1000) / (1500 * 180) = 0.308N/mm^2 Permissible shear stress at P = 0.3% (under reinforced section )

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= 0.384 N/mm^2 with k =1 (IS code table 7.1 and 7.2) Hence thickness provided from point of view bending compression is adequate. 5- Design for bending tension for under reinforced section Ast = (0.5* fck) /fy )* (root (1-(( 4.6 *1.77*10^7) / (20 * 400*150^20))) Ast = 889.94 mm^2 ~~~ 890mm^2 :- therefore No. Of 8 mm dia bars = 890/ 50.24 = 17.86 ~~~ 18 bars 6- Check for development length Ld = 47* dia of bars = 47 * 8 = 376mm Available length , providing a side cover of 50 mm = 600-50 = 540 mm Hence safe ,:-

FOOTING FOR COLUMN (C3) Design footing for column C3 size 400*450, load on column is 60.234, and moment is 16.02kNm . Soil bearibg capacity is 200 kn /m2 Solution :1- Design constants For M20 and Fe 415 combination Xu max /d =0.479, Ru=2.761 2- Size of footing W= 60.234KN , W‘=10% of W= 6.0234KN, Total weight = 66.25 KN ~~ 70 KN Equating the maximum and minimum soil pressure to safebearing capacityof soil Total weight/B^2 + M /(B^3/6) = soil bearing capacity M=0, 200 = 70 B^2 B=2.8M So we provide the footing of size = (2.8*2.8) m^2 The maximum and soil pressure is given by po1 = W/B^2 +6M/B^3 po1= 70/(2.8)^2 + (6*16.05)/2.8^3 =13.306 KN- maximum po2 = W/B^2 - 6M/B^3 po1= 70/(1.2)^2 - (6*30)/1.2^3 = 4.54KN - minimum Pressure intensity under the column is po = 0.5 (13.306 + 4.54) =8.923 kN/m^2 72 | P a g e

3- Design of section for bending compression Intensity of net soil pressure below the column face is po‘ = 8.923+( (13.306 – 8.923)/1.4) *0.25 therefore po‘ = 10.23 KN/m^2 maximum B.M will occur at face AB of cantilever length = 0.5(B-b) = 0.5 (2.8-0.45) =1.175 m Total force under cantilever = (2.8*1)( 10.23 +13.306)/2 = 32.950KN Distance of its centroid from face AB = ( (10.23 +(2*13.306))/(10.23 +32.950))* 1/3 = 0.521m Bending moment =32.950 *0.521 = 17.66 KNm Alternatively M= (B(B-b)^2) /24 * (2po1 + po‘) M = 2.8 (2.8- 0.45)^2 *(2* 13.306 +10.23) M = 569.577 KNm Mu =1.5 * 5.69 * 10^8 = 8.535 * 10^7 N-mm d = root (M/(Ru*b1)) d= root (8.535 * 10^7/ (2.761 * 450)) d= 262.094mm ~~~ 260mm However keep d= 260mm and D = 260 + 50 = 310mm , thus providing an effective cover of 50mm . Effective depth available for second layer = 310 -12 = 398mm, Using 12mm dia bars . also keep D= 150mm at the ends. So that available d at the end is equal to 150-50 = 100mm 4- Design for shear:The depth found above should be safe for shear. For two way shear (punching stress) . The critical plane lies at d/2 = 260/2 = 130mm from the column face for which width bo = b + d/2 = 450 + 130 = 320mm. The punching shear stress = ⌡v= Fu /(4*bo*do) Fu = 1.5(W-po*b^2) = 1.5 (60.234 – 8.923 * 0.320^2) Fu = 88.98 KN do = d1 + (d- d1)/ ((B - b)/2) * (B-b-d)/2 = 50 + (260 - 50)/ ((2800 - 260)/2) * (2800 – 450 – 260 ) / 2 do = 274 mm ⌡v = (88.98 * 1000 ) / (4* 320* 274) Punching shear stress :⌡v = 0.253 N/mm^2 Permissible shear stress = ks * 0.25 root fck = 1 * 0.25 root 20 73 | P a g e

= 1.118 N / mm^2 Hence thickness found from point of view of bending compression is safe . For one way shear , the critical plane CD lies at distance d = 150 mm from column face. The cantilever length to the right of CD = 1.4 – 0.25 – 0.45 = 0.7 mm. Intensity of pressure po‘‘ at CD is given by po‘‘ = 4.54 + ((13.30 – 4.54)/ 2.8) * (2.8 – 0.7) po‘‘ = 11.11 KN/m^2 the section at CD will be trapezoidal ... the width at top , b = 60.05 +((2800 – 450) /1000) *450 b = 1117mm effective depth d‘ = 100 + ((450 – 50 )/ 1000) * 60.05 d‘ = 124mm for balanced section Xu max / d = 0.476 let Xu / d = 0.4 for under reinforced section so that Xu = 0.4 d‘ = 0.4 * 124= 49.6mm therefore width of section at NA is given by bn = 1117 +( (2800 – 1117 )/ 24) * 49.6 = 4595mm bn =~ 3096mm Vu = 1.5 * 2.8 * 0.7 * 0.5 * ( 13.30+ 11.11) = 35.88 KN ⌡v = Vu‘/( bn * d‘) ⌡v = (35.88 * 1000) / (4595 * 124) = 0.0629N/mm^2 Permissible shear stress at P = 0.3% (under reinforced section ) = 0.384 N/mm^2 with k =1 (IS code table 7.1 and 7.2) Hence thickness provided from point of view bending compression is adequate. 5- Design for bending tension for under reinforced section Ast = (0.5* fck) /fy )* (root (1-(( 4.6 *8.535*10^7) / (20 * 450*260^2)))*450*260 Ast = 1140.22 mm^2 ~~~ 1140mm^2 :- therefore No. Of 12 mm dia bars =1140/ 113.4 = 10.08 ~~~ 10 bars 6- Check for development length Ld = 47* dia of bars = 47 * 10 = 470mm Available length , providing a side cover of 50 mm = 1400-250 =1150mm Hence safe ,:-

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Designation Of column

No. Of columns

Load on column

Depth of Size of footing footing

Diameter Of bars used 12mm

No.of bars used 18

C1

2

181.79 KN

200 mm

1.2m *1.2m

C2

20

153.634 KN

300 mm

1.2m * 1.2m

8mm

18

C3

8

60.234 KN

310 mm

2.8m* 2.8m

12mm

10

Table 10:- Foundation design data

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7. Conclusion: In an RCC framed structure, the load is transferred from a slab to the beams then to the columns and further to lower columns and finally to the foundation which in turn transfers it to the soil. The walls in such structures are constructed after the frame is ready and are not meant to carry any load. As against this, in a load bearing structure, the loads are directly transferred to the soil through the walls, which are capable of carrying them.  The floor area of a R.C.C framed structure building is 10 to 12 percent more than that of a load bearing walled building. Hence, there is actual economy in case of RCC framed structures especially where the cost of land is very high.  Also, in case of RCC framed structures, the inside planning of rooms, bathrooms, W.Cs etc. can be altered by changing the position of partition walls. Thus, there is greater flexibility in planning.  Monolithic construction is possible with R.C.C framed structures and they can resist vibrations, earthquakes and shocks more effectively than load bearing walled buildings.  Overall, the concepts and procedures of designing the basic components of a single storey building are described. Apart from that, the planning of the building with regard to appropriate directions for the respective rooms, choosing position of beams and columns are also properly explained.  The advancement of innovative and environmentally friendly building materials are also coming up. They can give a new direction to the structural engineering field as the availability of concrete and steel is not only decreasing but also they are harmful to the environment. Hence, ecofriendly materials which are economical and more effective methods of designing will decide the future of structure engineering.

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8. REFERENCES [1] www.theconstructor.org [2] www.wholebuilding design .org/structure.php [3] www.listdesign.co.uk/civil-engineering.html [4]

www.civilprojectsonline.com/building-construction/specification-for-

RCC-construction [5] www.scribd.com/basic-components-of-a-structure [6]

www.ghareexpert.com/articles/cement-concrete.1644/column-frame-

structure [7]B.C.Punmia, Ashok Kumar Jain, Arun Kumar Jain (2008) ‗Building Construction‘-10th edition: Laxmi publications [8] B.C.Punmia, Ashok Kumar Jain, Arun Kumar Jain (2008) ‗Limit State Design of Concrete Structures: Laxmi publications [9] IS codes (IS 456:2000, IS 875 (Part 1 - 3): 1987). [10] Design Handbooks - SP 16: 1980 – Design Aids (for Reinforced Concrete) to IS 456: 2000 [11] Dr.Ramachandra, Virendra Gehlot (2007) ―Limit State Design of Concrete Structures‖:- STANDARD PUBLISHERS DISTRIBUTORS [12] AutoCAD

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design of single storey rcc framed building

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