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  1. 1. DESIGN AIDSFORREINFORCED CONCRETETO IS : 456-l 978
  2. 2. As in the Original Standard, this Page is Intentionally Left Blank
  3. 3. DesignAidsForReinforcedConcreteto IS : 4564978BUREAU OF INDIAN STANDARDSBAHADUR SHAH ZAFAR MARC, NEW DLEHI 110 002
  4. 4. SP16:1980FIRST PUBLISHED SEPTEMBER 1980ELEVENTH REPRINT MARCH 1999(Incorporatinp Amendment No. I)0 BUREAU OF INDIAN STANDARDSUDC 624.0 12.45.04 (026)PRICE Rs.500.00I’KiNTED 1N INDIA ATVlB, PRESS PVT. LTD., 122 DSIDC SHEDS. OKHLA INDL!STRIAL ARtA. PfIASE-I. NEW DELHI 110(!20AND PI II3LISHED BYI<I!REAI OF INDIAN STANDARDS. NEW DELI11 II0002
  5. 5. FOREWORDUsers of various civil engineering codes have been feeling the need for explanatory hand-books and other compilations based on Indian Standards. The need has been further emphasizedin view of the publication of the National Building Code of India 1970 and its implementation.In 1972, the Department of Science and Technology set up an Expert Group on Housing andConstruction Technology under the Chairmanship of Maj-Gen Harkirat Singh. This Groupcarried out in-depth studies in various areas of civil engineering and constr,uction practices.During the preparation of the Fifth Five Year Plan in 1975, the Group was assigned the taskof producing a ,Science and Technology plan for research, development and extension workin the sector of housing and construction technology. One of the items of this plan was theproduction of design handbooks, explanatory handbooks and design aids based on the NationalBuilding Code and various Indian Standards and other activities in the promotion of NationalBuilding Code. The Expert Group gave high priority to this item and on the recommendationof the Department of Science and Technology the. Planning Commission approved the follow-ing two projects which were assigned to the Indian Standards Institution:a) Development programme on Code implementation for building and civil engineeringconstruction, andb) Typification for industrial buildings.A Special Committee for Implementation of Science and Technology Projects (SCIP)consisting of experts connected with different aspects (see page viii ) was set up in 1974 to advisethe IS1 Directorate General in identification and for guiding the development of the work underthe Chairmanship of Maj-Gen Harkirat Singh, Retired Engineer-in-Chief, Army Headquartersand formerly Adviser ( Construction) Planning Commission, Government of India. TheCommittee has so far identified subjects for several explanatory handbooks/compilationscovering appropriate Indian Standards/Codes/Specifications which include the following:Functional Requirements of BuildingsFunctional Requirements of Industrial BuildingsSummaries of Indian Standardsfor Building MaterialsBuilding Construction PracticesFoundation of BuildingsExplanatory Handbook on Earthquake Resistant Design and Construction (IS : 1893.Des& %?for Reinforced Concrete to IS : 456-1978Explanatory Handbook on Masonry CodeCommentary on Concrete Code ( IS : 456 )Concrete MixesConcrete ReinforcementForm WorkTimber EngineeringSteel Code ( IS : 800 )Loading CodeFire SafetyPrefabricationTall Buildings,Design of Industrial Steel StructuresInspection of Different Items of Building WorkBulk Storage Structures in SteelBulk Storage Structures in ConcreteLiquid Retaining Structures
  6. 6. .Construction Safety PracticesCommentaries on Finalized Building Bye-lawsConcrete Industrial StructuresOne of the explanatory handbooks identified is on IS : 456-1978 Code of practice forplain and reinforced concrete ( third revision). This explanatory handbook which is underpreparation would cover the basis/source of each clause; the interpretation of the clause andworked out examples to illustrate the application of the clauses. However, it was felt that somedesign aids would be of help in designing as a supplement to the explanatory handbook. Theobjective of these design aids is to reduce design time in the use of certain clauses in the Codefor the design of beams, slabs and columns in general building structures.For the preparation of the design aids a detailed examination of the following handbookswas made :4‘4cl4CP : 110 : Part 2 : 1972 Code of practice for the structural use of concrete : Part 2Design charts for singly reinforced beams, doubly reinforced beams and rectangularcolumns. British Standards Institution.AC1 Publication SP-17(73) Design Handbook in accordance with the strength designmethods of AC1 318-71, Volume 1 ( Second Edition). 1973. American ConcreteInstitute.Reynolds ( Charles E ) and Steadman ( James C ). Reinforced Concrete Designer’sHandbook. 1974. Ed. 8. Cement and Concrete Association, UK.Fintel ( Mark ), Ed. Handbook on Concrete Engineering. 1974. Published by VanNostrand Reinhold Company, New York.The charts and tables included in the design aids were selected after consultation withsome users of the Code in India.The design aids cover the following:a) Material Strength and Stress-Strain Relationships;b) Flexural Members ( Limit State Design);c) Compression Members ( Limit State Design );d) Shear and Torsion ( Limit State Design );e) Development Length and Anchorage ( Limit State Design );f) Working Stress Method;g) Deflection Calculation; andh) General Tables.The format of these design aids is as follows:a) Assumptions regarding material strength;b) Explanation of the basis of preparation of individual sets of design aids as relatedto the appropriate clauses in the Code; andc) Worked example illustrating the use of the design aids.Some important points to be noted in the use of the design aids are:4b)4d)4viThe design units are entirely in SI units as per the provisions of IS : 456-1978.It is assumed that the user is well acquainted with the provisions of IS : 456-1978before using these design aids.Notations as per IS : 456-1978 are maintained here as far as possible.Wherever the word ‘Code’ is used in this book, it refers to IS : 456-1978 Code ofpractice for plain and reinforced concrete ( third revision ).Both charts and tables are given for flexural members. The charts can be used con-veniently for preliminary design and for final design where greater accuracy is needed,tables may be used.
  7. 7. f) Design of columns is based on uniform distribution of steel on two faces or on fourfaces.g) Charts and tables for flexural members do not take into consideration crack controland are meant for strength calculations cnly. Detailing rules given in the Code shouldbe followed for crack control.h) If the steel being used in the design has a strength which is slightly different from theone used in the Charts and Tables, the Chart or Table for the nearest value may beused and area of reinforcement thus obtained modified in proportion to the ratio ofthe strength of steels.j) In most of the charts and tables, colour identification is given on the right/left-handcorner along with other salient values to indicate the type of steel; in other charts/tables salient values have been given.These design aids have been prepared on the basis of work done by Shri P. Padmanabhan,Officer on Special Duty, ISI. Shri B. R. Narayanappa, Assistant Director, IS1 was alsoassociated with the work. The draft Handbook was circulated for review to Central PublicWorks Department, New Delhi; Cement Research Institute of India, New Delhi; Metallurgicaland Engineering Consultants (India) Limited, Ranchi, Central Building Research Institute,Roorkee; Structural Engineering Research Centre, Madras; M/s C. R. Narayana Rao, Madras;and Shri K. K. Nambiar, Madras and the views received have been taken into considerationwhile finalizing the Design Aids.vii
  8. 8. .SPECIAL COMMIlTEE FOR IMPLEMENTATION OF SCIENCE ANDTECHNOLOGY PROJECTS (SCIP)MembersSEIR~A. K. BANERJEEPROF DINESH MOHANDR S. MAUDOALDR M. RAMAIAHSHRI T. K. SARANSHRI T. S. VEDAGIRIDR ‘H. C. VISVESVARAYASHRI D. AJITHA SIMHA(Member Secrewv)...VlllChairmanMAJ-GEN HARKIRAT SINGHW-51 Greater Kailash I, New Delhi 110048Metallurgical and Engineering Consultants (India) Limited,Ran&iCentral Building Research Institute, RoorkeeDepartment of Science and Technology, New DelhiStructural Engineering Research Centre, MadrasBureau of Public Enterprises, New DelhiCentral Public Works Department, New DelhiCement Research Institute of India, New DelhiIndian Standards Institution, New Delhi
  9. 9. CONTENTSPageLIST OF TABLES M THE EXPLANATORY -TEXT ... ... xLIST OF CHARTS ... ... xiLIST OF TABLES ... ... XivSYMBOLS ... ... xviiCONVERSK)N FACTORS ... ... xix1. MATERIAL STRENGTH AND STRESS-STRAIN RELATIONSHIPS 31.1 Grades of Concrete1.2 Types and Grades of Reinforcement1.3 Stress-strain Relationship for Concrete1.4 Stress-strain Relationship for Steel2. FLEXURAL MEMBERS2.12.22.32.3.12.3.22.42.5AssumptionsMaximum Depth of Neutral AxisRectangular SectionsUnder-Reinforced SectionsDoubly Reinforced SectionsT-SectionsControl of Deflection3. COMPRESSION MEMBERS3.13.23.2.13.2.2Axially Loaded Compression MembersCombined Axial Load and Uniaxial BendIngAssumptions3.2.33.33.4Stress Block Parameters when the Neutral iAxisLiesOutside the SectionConstruction of Interaction DiagramCompression Members Subject to BiaxialBendingSlender Compression Members4. SHEAR AND TORSION4.1 Design Shear Strength of Concrete4.2 Nominal Shear Stress4.3 Shear Reinforcement4.4 Torsion. .............................................*.............................. 3... 3... 4... 4... 9... 9... 9... 9... 10... 12... 14... 14... 99... 99... 99... 100... 101... 101... 104... 106... 175..* 175... 175... 175 .... 175ix
  10. 10. Page5.5.15.26.6.16.1.16.1.26.1.36.26.36.47.7.17.2DEVELOPMENT LENGTH AND ANCHORAGE ...Development Length of Bars ...Anchorage Value of Hooks and Bends ...WORKING STRESS DESIGN ...Flexural Members ...Balanced Section ...Under-Reinforced Section ...Doubly Reinforced Section ...Compression Members ...Shear and Torsion ...Development Length and Anchorage ...DEFLECTION CALCULATION ...Effective Moment of Inertia ...Shrinkage and Creepl)eflections ...... 183... 183... - 183... 189... 189... 189... 189... 190... 190... 1911.. 191... 213... 213... 213LIST OF TABLES IN THE EXPLANATORY TEXTTableABCDEFGHIJKLMXSalient Points on the Design Stress Strain Curve for Cold WorkedBars ... ...Values of F for Different Grades of Steel ... ...Limiting Moment of Resistance and Reinforcement Index for SinglyReinforced Rectangular Sections ... ...Limiting Moment of Resistance Factor Mu,ii,/bd’, N/mm2 for Singly.Reinforced Rectangular Sections ... ...Maximum Percentage of Tensile Reinforcement Pt,lim for SinglyReinforced Rectangular Sections ... ..,Stress in Compression Reinforcement, fX N/mma in DoublyReinforced Beams with Cold Worked Bars ... ...Multiplying Factors for Use with Charts 19 and 20 ... ...Stress Block Parameters When the Neutral Axis Lies Outside theSection ... ...Additional Eccentricity for Slender Compression Members ... .,.Maximum Shear Stress rc,max . . . . . .Moment of Resistance Factor M/bd’, N/mm” for BalancedRectangular Section ... ...Percentage of Tensile Reinforcement P1,b.i for Singly ReinforcedBalanced Section .*. ...Values of the Ratio A,/&, ... ...691010101313101106175I89189190
  11. 11. chartNo. PWFLEXURE - Singly Reinforced Section123456789101112131415161718CL= 15 N/mm’, fy= 250 N/mm’Lk - 15 N/mm*, fu= 250 N/mm*fetr- 15 N/mm*, fr= 250 N/mmsfck= 15 N/mm*, fy= 415 N/mm*fEd- 15 N/mm*, fi - 415 N/mm*fsk- 15 N/mm*, f, - 415 N/mm*fck= 15 N/mm*, fi - 500 N/mm*fctr= 15 N/mm*, fy - 500N/mm*f&- 15 N/mm*, ’fv- 500N/mm*f&x 20 N/mm*, .fy= 250 N/u&f&- 20N/mm*, fy - 250N/mm%fek= 20N/mm’, f, = 2% N/mm’fh - 20 N/mm*, I;- 415 N/mm*fdrI 20 N/mm*, fv- 415 N/mm’fck- 20 N/mm*, fy- 415 N/m’f&- 20N/mm’, fy - 500 N/mm*fd - 20 N/mm’, fr - 500 N/mm*hk - 20 N/mm*, & = 500 N/mm*d- 5 to 30 cm ... 17d = 30 to 55 cm .*. 18d - 55 to 80 cm ... 19d= 5 to 30 cm ... 21d I 30 to 55 cm ... 22d-55 to 80 cm ... 23d== 5 to 30 cm ... 25d = 30 to 55 cm ... 26d-55 to 80 cm ... 27,d P 5 to 30 cm ... 29d - 30 to 55 cm ... 30d = 55 to 80 cm ... 31d- 5 to 30 cm ... 33d-30 to 55 cm ... 34d- 55 to 80 cm ... 35d= 5 to 30 cm ... 37d-30 to 55 cm ... 38d I 55 to 80 cm ... 39FLEXURE - Doubly Reinforced Section19 fr I 250 N/mm’, d-d’ - 20 to 50 cm20 fr I 250 N/mm*, d-d’ - 50 to 80 cm... ... 41... ... 42CONTROL OF DEFLECTION21 fr - 250 N/mm’ ... ... 4322 fr I 415 N/mm’ ... ,.. 4423 fi - 500N/mm’ ... ... 45LIST OF CHARTSAXIAL COMPRESSION24 h - 250 N/mm’ ... .“. 10925 ft - 415 N/mm’ ... ... 11026 A-5OON/mm’ ... ... 111xi
  12. 12. ChartNo.27 f, = 250 N/mm% d’/D = 0.0528 fi = 250 N/mms d’/D = 0.1029 h I 250 N/mm9 d’/D = 0.1530 fv LI 250 N/mm9 d’/D = 02031 fr = 415 N/mm9 d’/D = 0.0532 fu= 415 N/mm9 d’/D - 0.1033 fr PD415 N/mm9 d’/D P 0.1534 fy - 415 N/mm9 d’/D - 0.2035 fr - 508 N/mm9 d’/D = 0.0536 fr - 500 N/mm4 d’/D = 0.1037 fr = 500 N/mm9 d’/D = 0.1538 fy- 500 N/mm3 d’/D - 0.2039 J, I) 250 N/mm* d’/D - 0.0540 .fx I 250 N/mm” d’/D = 0.1041 fy= 250 N/mm9 d’/D = 0.1542 fy - 250 N/mm2 d’lD = 0.2043 fy II 415 N/mm9 d’/D = 0.0544 frP 415 N/mm9 d’/D c 0.1045 fr I 415 N/mm9 d’/D = 0.1546 fr - 415 N/mm9 d’/D = 0.2047 fy = 500 N/mm3 d’/D P 0.0548 f, - 500 N/mm9 d’/D = 0.1049 fv- 500 N/mm’ d’lD = 0.1550 fu= 500 N/mm9 d’/D = 0.20COMPRESSION WlTH BENDlNG - Circular Section51 fx - 250 N/mm9 d’/D = 0.0552 fv P 250 N/mm2 d’/D = 0.1053 fy= 250 N/mm’ d’/D = 0.1554 fr= 250 N/mm’ d’/D = 0.2055 fyP 415 N/mm9 d’/D = 0.0556 fr- 415 N/mm9 d’!D = 0.1057 fy = 415 N/mm’ d’/D = 0.1558 fy - 415 N/mm9 d’/D I= 0.2059 fi - 500 N/mm9 d’/D = 0.0560 fu- 500 N/mm” d’/D = 0.1061 h-500 N/mm* d’/D = 0.1562 fv-500N/mm* d’/D = 020636465Values of Puz for Compression Members ... ...Biaxial Bending in Compression Members ... ...Slender Compression Members - Multiplying Factor k for ...Additional MomentsPageCOMPRESSION WITH BENDING - Rectangular Section -Reinforcement Distributed Equally on Two Sides...............1... . .. . .. . .. . .. . .. . .... 112... 113... 114... 115... 116... 117... 118... 119... 120... 121... 122... 123COMPRESSION WITH BENDING - Rectangular Section -Reinforcement Distributed Equally on Four Sides... ... 124... ... 125... ... 126... ... 127... ... 128... ... 129... ... 130... ... 131... ... 132... ... 133... ... 134... ... 135........................................ 136... 137... 138... 139... 140... 141... 142i.. 143... 144... 145... 146... 147148149150xii
  13. 13. ClUWtNo.TENSION WITH BENDING - Rectangular Section -Reinfomment Distributed Equally on Two Sides66 h - 250 N/mm’67 fr - 250 N/mm’68 fr - 415 N/mm’69 & = 415 N/mm*70 h - 415 N/mm’71 /r - 415 N/mm’72 II-=5OON/llltII’73 h-5OON/IIUU’74 h-SOON/mm’75 &-so0 N/lIlOI’TENSION WITH BENDING - Rectangular Section - ReinforcementDistributed Equally on Four Sides7677787980818283a4858687888990‘Pl92f, - 250 N/mm’ d’/D- 0.05 and 010 ... ...fr - 250 N/mm* d’/D- 0.15 and 020 ... .,./r - 415 N/mm’ d’/D= 0.05 ... .../r I 415 N/mma d’/DP 0.10 ... ...fr - 415 N/mm’ al/D - 0.15 ... ...fr = 415 N/mm’ d’/D- 090 ... ...h-5OON/mm’ d’/D= 0.05 ... ...Jt-5OON/lIllII’ d’/D= 0.10 ... ...fr- 500N/mm* d’/D- 0.15 ... ...A-5OON/IUlll’ d*/D= 020 ... ...Axial Compiession (Working Stress Design) 0, - 130 N/mm* ...Axial Compression (Working Stress Design) am- 190 N/mm* ...Moment of Inertia of T-Beams ...Effective Moment of Inertia for Calculating Deflection ...Percentage, Area and Spacing of Bars in Slabs ...EffectiveLength of Columns - Frame Restrained .AgainstSway...Effective Length of Columns - Frame Without Restraiht to Swayd’/D = @l5 and 020 ... ...d’/D-0.05 and 010 ... ...d’/D- @OS ... ...d’/Da 0.10 ... ...d’/DP 0.15 ..., ..,d’/D - 020 ... ...d’/D- 0.05 ... ...d’/D = 010 ... ...d’/D- 0.15 ... ...d’/D- O-20 ... ...Page151152153154155156157158159160161162163164165166l6il68169170193194215216217218219...xul
  14. 14. LIST OF TABLESTableNo. PageFLEXURE - Reinforcement Percentage, pI for Singly Reinforced Sections1 fft - 15 N/mm’ ...... 472 f CL = 20 N/mm’ ...... 483 fCL- 25. N/mm* ...... 494 fd = 30 N/mm’ ...... 50FLEXURE - Moment of Resistance of Slabs, kN.m Per Metre Width5678910111213141516171819202122232425262728293031323334fd - 15 N/mm’ /r- 250 N/mm* Thickness - 10.0 cmfck- 15N/llIHl’fy- 250 N/mm* Thickness - 11.0 cmfdr- 15 N/mm* fr I 250 N/mm* Thickness = 120 cmfdr-15 N/mm’ fy- 250 N/mm* Thickness - 13.0 cmf&- 15 N/mm’. fy w 250 N/mm* Thickness = 14.0 cmId-15 N/mm’ f,- 250 N/mm’ Thickness - 15.0 cmfd - 15 N/mm* fy- 250 N/mm* Thickness - 175 cmfck- 15 N/mm’ fy- 250 N/mm* Thickness = 20.0 cmfd - 15 N/mm* fi - 250 N/mm* Thickness - 22.5 cmfck - ,15 N/mm* fy - 250 N/mm* Thickness = 25.0 cmfck- 15 N/mm* fy - 415 N/mm* Thickness - 10.0 cm2: f :: :rGI i- - 415415 N/mm*N/mm* ThicknessThickness -- 11.0120 cmcmfd - 15 N/mm* fy I 415 N/mm* Thickness .- 13.0 cmfd - 15 N/mm* fvI 415 N/mm* Thcikness - 140 cmfck- 15 N/mm* f, - 415 N/mm* Thickness - 15.0 cmfck= 15 N/mm* fv- 415 N/mm* Thickness - 17.5 cmfdt- 15 N/mm* fr - 415 N/mm* Thickness - 20.0 cmfsk- 15N/mm’fy E 415 N/mm* Thickness I 225 cmfa- 15N/m* fy = 415 N/mm* Thickness - 25.0 cmfclr-m N/mm*$, - 250 N/mm* Thickness - 10.0 cmfck - 20 N/mm’ f, - 250 N/mm* Thickness - 11.0 cmfclr-mN/=’ &I- 250 N/mm* Thickness - 12.0 cmfck-2ON/mm* h - 250 N/mm* Thickness - 13.0 cmf: z $ :=I 2 -I 250250 N/mm*N/mm* ThicknessThickness -- 15.014.0 cmcmfd - 20N/m+ h I 250 N/mm* Thickness - 17.5 cmfck- 20N/mm*fr- 250 N/mm* Thickness I 20.; cmf&- 20 N/mm’ fy- 250 N/mm* Thickness - 22.5 cmfck - 20 N/mm’ f, I 250 N/mm* Thickness - 25.0 cm,... 51... 51... 52... 52... 53... 53... 54... 55... 56... 57... 58... 58... 59... 59... 60... 61... 62... 63... 64... 65... 66... 66... 67... 67... 68... 68a*- 69. . . 70. . . 71. . . 72xiv
  15. 15. TableNO.35363738394041424344f ck - 20 N/mm2 h - 415 N/mm2 Thickness - 100 cmfti- 20 N/mm2 h I 415 N/mm2 Thickness - 110 cm1;- 20 N/mm2 h - 415 N/mm2 Thickness - 12.0 cm- 415 N/mm2 Thickness - 13.0 cm- 415 N/mm2 Thickness - 14.0 cmf,+- 20 N/mm2 f, - 415 N/mm2 Thickness - 15.0 cmfck - 20 N/mm2 fy - 415 N/mm2 Thickness - 175 cm- 415 N/mm22: 1: i/z: i - 415 N/mm2Thickness - 200 cmThickness - 225 cmfck- 20 N/mm2 fy - 415 N/mm2 Thickness - 25.0 cmFLEXURE - Reinforcement Percentages for DoublyReinforced Sections454647484950515253545556f& - 15 N/mm2 fr - 250 N/mm2 ...fr - 250 N/mm2 ...& - 250 N/mm2 ...f;k- 30N/mm2 fr P 250 N/mm2 ...fek- 15N/mm2 fy I 415 N/mm’ ...fyI 415 N/mm2 ...fr - 415 N/mm2 ...fck- 30 N/mm2 fu- 415 N/mm2 ...fdr- 15 N/mm2 fr- 500N/mm2 ...fck - 20 N/mm2 fYI: 500 N/mm2 ...fck= 25 N/mm2 fy- 500N/mm2 ...fck- 30 N/mm2 fr-5OON/r.llm2 .....................................................................7373747475767778798081828384858687888990919257585960616263FLEXURE - Limiting Moment of Resistance Factor, Mo,u,&,# /a, forSingly Reinforced T-beams N/mm*fv- 250 N/mm’ ... ...fy- 415 N/mm= ... ..*fy- 500N/mm’ ... ...Slender Compression Members - Values of P, ... ...Shear - Design Shear Strength of Concrete, rc, N/mm* ... ...Shear - Vertical Stirrups ... ...Shear - Bent-up Bars ... ...939495171178179179DEVELOPMENT LENGTH64 Plain Bars ... ... 18465 Deformed bars,fuP 415 N/mm* ... ... 18466 Deformed bars,fr - 500N/mm* ... ... La567 Anchorage Value of Hooks and Bends ... 2.. 18627071WORKING STRESS METHOD - FLEXURE - Moment ofResistance Factor, M/bd’, N/mm’ for Singly Reinforced Sectionsa* - 5-ON/mm* ... ...u,bc- 7.0 N/mm’ ... ...uca - 8.5 N/mm’ ... .*.ucbc- 10.0 N/mm* ... ...195::198xv
  16. 16. TableNo.7273747576777879uck- 5.0 N/mm1 aa - 140 N/mm8 ...acbs- 79 N/mms w I 140 N/mm’ ...oti - 8.5 N}mn’ an I 140 N/mm’ ...a&C- 10.0 N/mm* au - 140.N/J& ...ati I 5.0 N/m& u,, - 230 N/mm’ ...aa - 7.0 N/mm* au P 230 N/inn+ ...oca - 8.5 v/mm’ a,( - 230 N/mm’ ...oek - 10.0 N/mm* ust- 230 N/mm’ ...WORKING STRJZSSMETHOD-SHEAR...................... ...808182Permiklble Shear Stress in Concrete rc, N/mm* ... ...Vertical Stirrups ... ...Bent-up Bars ... ...WORKING STRESS METHOD - DEVELOPMENT LENGTH83 Plain Bars ...84 Deformed Bars- uat- 230 N/mm*, e - 190 N/mm’ ...85 Deformed Bars- u,, - 275 N/mm*, uc - 190 N/mm’ ...86 Moment of Inertia - Values of b@/l2 000 ...87 &Id - 095 ... ...88 d’ld - 090 ... ...89 d’jd - 015 ... ...90 d’ld - 020 ... ..I91 d’/d I 00592 d’/d - 0.1093 d’/d - 0.1594 d’jd = 02095 Areas of Given Numbers of Bars in cm* ... ...96 Areas of Bars at Given Spacings ... ...97 Fixed End Moments for Prismatic Beams ... ...98 Detlection Formulae for Prismatic Beams ... ...WORKING STRESS DESIGN - FLEXURE - RchforccmcntPercentages for Doubly Reinforced S&ionsMOMENT OF INERTIA OF CRACKED SECTION-Values of Ir/DEPTH OF NEUTRAL &US - Values of n/d by ElasticTheory1.. ... 225l .. ... 226... ... 227... ... 228221222223224229230a31232xvi
  17. 17. SYMBOLSAC4A*ACAI”Asoact6bbrbwb,DDidd’,d’d,ECEShaC.Yfck= Area of concreteI Gross area of section= Area of steel in a column or in asingly reinforced beam or slab- Area of compression steel= Area of stirrupsDCArea of additional tensilereinforcement= Deflection due to creep= Deflection due to shrinkage= Breadth of beam or shorterdimensions of a rectangularcolumn= Effective width of flange in aT-beam= Breadth of web in a T-beam= Centre-to-centre distance betweencorner bars in the direction ofwidthI Overall depth of beam or slab ordiameter of column or large1dimension in a rectangularcolumn or dimension of arectangular column in thedirection of bendingLI Thickness of flange in a T-beam- Effective depth of a beam or slab= distance of centroid of com-pression reinforcement fromthe extreme compression fibreof the concrete sectionG Centre to centre distance betweencomer bars in the direction ofdepth= Modulus of elasticity of concrete= Modulus of elasticity of steelP Eccentricity with respect to majoraxis (xx-axis)= Eccentricity with respect tominor axis (yy-axis)= Minimum eccentricity= Compressive stress in concrete atthe level of centroid ofcompression reinforcement= Charircteristic compressivestrength of concreteE Flexural tensik strength(modulus of rupture) ofconcrete= Stress in steel- Compressive stress in steelcorresponding to a strain of0402= Stress in the reinforcementnearest to the tension face of amember subjected to combinedaxial load and bending= Cytrteristic yield strength ofP Design yield strength of steel= Effective moment of inertiaP Moment of inertia of the grosssection about centroidal axis,neglecting reinforcement= Moment of inertia of crackedsection= Flexural stiffness of beam:= Fkxural stiffness of column= Constant or coefficient or factor= Development length of bar= Length of column or span ofbeam= Effective length of a column,bending about xx-axis= Effective length of a column,bending about yy-axis= Maximum moment under serviceloads- Cracking moment= Design moment for limit stateDesign (factored moment)M u3h-n - Limiting moment of resistance ofa singly reinforced rectangularbeamMu, e Design moment about xx-axisMUY a Design moment about &-axisM”l, = Maximum uniaxial momentcapacity of the section withaxial load, bending aboutxx-axisxvii
  18. 18. &I - Maximum uniaxial momentcapacity of the section withaxial load, bending aboutyy-axisMel - Equivalent bending momentMU, - Additional moment, MU- Mn,timin doubly reinforced beamsMu,timrr= Limiting moment of resistancemPpbP”PPCPCPtrSTT”VVSV&lVWxof a T-beam= Modular ratio= Axial load- Axial load corresponding to thecondition of maximumcompressive strain of 0903 5 inconcrete and OQO2 in theoutermost layer of tensionsteel in a compression member= Design axial load for limit statedesign (factored load)P Percentage of reinforcement- Percentage of compressionreinforcement, 100 A,,/bdlet Percentage of tension reinforce-ment, -l,OOAst/bd- Additional percentage of tensilereinforcement ’ doublyreinforced beams, ‘I”00A,t,/bd- Spacing of stirrups- Torsional moment due tofactored loads- Shear forceI Strength of shear reinforcement(working stress design)= Sbear force due to factored loads= Stren8hof shear reinforcementimit state design)= Dept;: neutral axis at serviceXl = Shorter dimension of the stirrup&I = Depth of neutral axis at thelimit state of collapseXu,mox = Maximum depth of neutral axisin limit state designYc = Distance from centroidal axisof gross section, neglectingreinforcement, to extreme fibrein tensionYl = Longer dimension of stirrupz = Lever arma P AngleYr - Partial safety factor for loadYm - Partial safety factor for materialstrengtht = Creep strain in concreteecbc - Permissible stress in concrete inbending compression6X = Permjssible stress in concrete indirect compression01 = Stress in steel bares 3: Permissible stress in steel incompression011 = Permissible stress in steel intensiones, I Permissible stress in shearreinforcement7Y P Nominal shear stress7bd P Design bond stressk - Shear stress in concrete‘5w - Equivalent shear stressQ,mu - Maximum shear stress in concretewith shear reinforcement8 i Creep coefficient9 - Diameter of bar...XVlll
  19. 19. CONVERSION FACTORSTo Convert intoConverselyMu&ply by Multiplyby(1)ILoads and Forces(2) (3) (4)-~NewtonKilonewtonMoments and TorquesNewton metreKilonewton metreStresseskilogram o-102 0 9.807Tonne 0.102 0 9.807kilogram metre o-102 0 9.807Tonne metre o-102 0 9.807Newton per mm*Newton per mm’kilogram per mm’kilogram per cm2o-102 0 9.80710.20 O-0981xix
  20. 20. As in the Original Standard, this Page is Intentionally Left Blank
  21. 21. 1. MATERIAL STRENGTHS ANDSTRESS-STRAIN RELATIONSHIPSI.1 GRADES OF CONCRETEThe following six grades of concrete canbe used for reinforced concrete work asspecified in Table 2 of the Code (IS : 4%1978*):M 15, M 20, M 25, M 30, M 35 and M 40.The number in the grade designation refersto the characteristic compressive strength,fti, of 15 cm cubes at 28 days, expressed inN/mmZ; the characteristic strength beingdefined as the strength below which notmore than 5 percent of the test results areexpected to fall.*Code d practice for plain and reinforced concrete( third revision ).1.1.1 Generally. Grades ;ti IS and M 20 areusedfor flexural members. Charts for flexuralmembers and tables for slabs are, therefore,given for these two grades ordy. However,tables for design of flexural members aregiven for Grades M 15, M 20, M 25 and M 30.1.1.2 The charts for compression membersare applicable to all grades of concrete.1.2 TYPES AND GRADES OFREINFORCEMENT BARSThe types of steel permitted for use as re-inforcement bars in 4.6 of the Code and theircharacteristic strengths (specified minimumyield stress or O-2 percent proof stress)are as follows:Type oj SteelMild steel (plain bars)Mild steel (hot-rolled deform-ed bars)Medium tensile steel (plainbars)Medium tensile steel (hot-rolled deformed bars)High yield strength steel (hot-rolled deformed bars)High yield strength steel(cold-twisted deformedbars)Hard-drawn steel wire fabricIndian StandardIS : 432 (Part I)-1966* 1IS : 1139-1966t rI-IS : 432 (Part I)-1966*>IIS : 1139-1966t1IS : 1139-1966tIS : 1786-1979$ 7IS : 1566-19674 andIS : 432 (Part II)-19661Yield Stress or O-2PercentProof Stress26 z$fm;rni,or bars up to24 kgf/mm* for bars over20 mm dia36 lkfe;i2’ bars up to.34.5 kgf/mm* for bars over20 mm’dia up to 40 mmiiia33 kgf/mm” for bars over40 mm dia42.5 kgf/mm2 for all sizes4 15N/mm2 for all bar sizes500 N/mm* for all bar sizes49 kgf/mm*Nom-S1 units have been used in IS: 1786-19793; in other Indian Standards. SI units will be adoptedin their next revisions.*Specification for mild steel and medium tensile steel bars and hard-drawn steel wire for concretereinforcement: Part I Mild steel and medium tensile steel bars (second revision).tSpecification for hot rolled mild steel, medium tensile steel and high yield strength steel deformedbars for concrete reinforcement (revised).$Specificatiod for cold-worked steel high strength &formed bars for concrete reinforcement (secondWlSiO#).&+eciiication for hard-drawn steel wire fabric for concrete reinforcement (#rsr revisfon).ijSpecification for mild steel and medium tensile steel bars and hard-drawn steel wire forconcrete reinforcement: Part II Hard drawn steel wire (secondrevision).MATERIAL STRENGTHS AND STRESS-STRAIN RELATIONSHIPS 3
  22. 22. Taking the above values into consideration,most of the charts and tables have beenprepared for three grades of steel havingcharacteristic strength& equal to 250 N/mm*,415 N/mm2 and 500 N/mm2.1.2.1 If the steel being used in a design hasa strength which is slightly diflerent from theabove values, the chart or table for the nearestvalue may be used and the area ofreinforce-ment thus obtained be modi$ed in proportionto the ratio of the strengths.1.2.2 Five values of fY (includinglthe valuefor hard-drawn steel wire fabric) have beenincluded in the tables for singly reinforcedsections.1.3 STRESS-STRAIN RELATIONSHIPFOR CONCRETEThe Code permits the use of any appro-priate curve for the relationship between thecompressive stress and strain distributionin concrete, subject to the condition that itresults in the prediction of strength in subs-tantial agreement with test results [37.2(c)of the Code]. An acceptable stress-straincurve (see Fig. 1) given in Fig. 20 of the Codewill form the basis for the design aids in thispublication. The compressive strength of con-crete in the structure is assumed to be O-67fd.With a value of l-5 for the partial safetyfactor ym for material strength (35.4.2.1 ofthe Code), the maximum compressive stressin concrete for design purpose is 0.446 fck(see Fig. I).1.4 STRESS-STRAIN RELATIONSHIPFOR STEELThe modulus of elasticity of steel, E,, istaken as 200 000 N/mm2 (4.6.2 of the Code).This value is applicable to all types ofreinforcing steels.The design yield stress (or 0.2 percent proofstress) of steel is equal to fr/ym. With a valueof l-15 for ym (3.5.4.2.2 of the Code), thedesign yield stress fvstress-strain relations tp for steel in tension1.becomes 0#87f,. Theand compression is assumed to be the same.For mild steel, the stress is proportionalto strain up to yield point and thereafter thestrain increases at constant stress (see Fig. 2).For cold-worked bars, the stress-strainrelationship given in Fig. 22 of the Code willI/ /.II a.002 0’001STRAINFIG. 1 DESIGN STRKSS-STRAINCURVE FORCONCRETE. 200000 N/mm’? --STRAINFIG.2 STRESS-STRAINCURVE FORMILD STEELbe adopted. According to this, the stressis proportional to strain up to a stress of0.8 fY. Thereafter, the stress-strain curve isdefined as given below:Stress hu#aslic~srrainO*SOfy Nil0.85 fr OQOOl0*9ofy 0.0% 30*9sf, o*ooo 70.975 fy 0~0010l-O& 0.002 0The stress-strain curve for design purposes isobtained by substituting fYe for fY in theabove. For two grades of cold-worked barswith 0.2 percent proof stress values of415 N/mms and 500 N/mm2 respectively,the values of total strains and design stressescorresponding to the points defined aboveare given in Table A (see page 6). The stress-strain curves for these two grades of cold-worked bars have been plotted in Fig. 3.4 DESIGN AIDS FOR REINFORCED CONCRETE
  23. 23. T<2.5004504003503002502001501005001so0m2soo/‘iv’UC/l0 0.001 o-002 0.003 o-004 0*005STRAINFIG. 3 STRESS-STRAINCURVESFOR COLD-WORKED STEELEMATERIAL SrRENGTHS AND STRESS-STRAINRELATIONSHIPS 5‘1.0‘mnl1-n
  24. 24. TABLE A SALIENT POINTS ON THE DESlGN STRESS-STRAIN CURVE GORCOLD-WORKED BARS( Chse 1.4 )STRESS LEVEL f, 0 415 N/mm’ fy= 500 N/mm8f--*> ,_-.-k bStrain Stress Strain Stress(1) (‘1 (3) (4) (5)N/mm* N/mm*0.80 fyd 090144 288.1, woo174 347.80.85 fyd 0031 63 306.7 0.001 95 369.60.90&l 0~00192 324.8 0.002 26 391.30’95 fyd 0032 4 I 342.8 0.002 77 413.00.975 fyd 0.002 76 351.8 0.003 12 423.9l’ofyd 0.003 80 360.9 MO4 17 434.8NOTE-- Linear interpolation may be done for intermediate values.6
  25. 25. As in the Original Standard, this Page is Intentionally Left Blank
  26. 26. 2. FLEXURAL MEMBERS2.2 ASSUMPTJONS 2.2 MAXIMUM DEPTH OF NEUTRAL. AXISThe basic assumptions in the design offlexur&lmembers for the limit state of col-lapse are fcivenbelow (see 37.2 of the Code):Assumptions (b) and (f’)govern the maximumdepth of neutral axis in flexural members.4 Plane sections normal to the axis ofthe member remain plane after bending.This means that the strain at any pointon the cross section is directly propor-tional to the distance from the neutralRXiS.W Ihe maximum strain in concrete atthe outermost compression fibre is0903 5.T& strain distribution across a membercorresponding to those limiting conditionsis shown in Fig, 4. The maximum depth ofneutralaxis x,,, - is obtaineddirectlyfromthe strain diagram by considering similartriangles.x0,,_ 0.003 5d (0.005 5 f 0.87f,/&)d The design stress-strain relationshipfor concrete is taken as indicated inFig. 1.The values of * for three grades ofreinforcing steel are given in Table B.d) The tensile strength of concrete is TABLE Bignored.VALUES OF F FORe), Tbt design stresses in reinforcementDIFFERENT GRADES OF STEELare derived from the strains using(Cfuu.re2.2)the stress-strain relationship given -in f,, N/mms 250 415 500Fig. 2 and 3.f) The strain in the tension reinforcement 0531 0.479 O-456is to be not less than72.3 RECTANGULAR SECTIONSThis assumption is intended to ensure The compressive stress block for concreteductile fail&e, that is, the tensile is represented by the design stress-strainreinforcementhas to undergo a certaindegree of inelastic deformation beforecurve as in Fig. 1. It is.seen from this stressthe concrete fails in compression.block (see Fig. 4) that the centroid of com-pressive force in a rectangularsection lies0*0035fXtu,m*a!zzx +0*002E*STRAINOIAGRAMFIQ. 4 SINOLY REINFQRCSDSECTIONO=87 f-,STRESSDIAGRAMFLEXURAL MRMM3R.S
  27. 27. at a distance or U-416 xu (wnlcn nas oecnrounded off to 0.42 xu in the code) from theextreme compression fibre; and the total forceof compression is 0.36 fck bxu. The lever arm,that is, the distance between the centroidof compressive force and centroid of tensileforce is equal to (d - 0.416 x,). Hence theupper limit for the moment of resistance of asingly reinforced rectangular section is givenby the following equation:Mu,lim = O-36& bxu,,,x(d - 0.416 ~u,mu)Substituting for xu,- from Table B andtransposing fdr bd2, we get the values oftie limiting moment of resistance factors forsingly reinforced rectangular beams andslabs. These values are given in Table C.The tensile reinforcement percentage, pt,limcorresponding to the limiting moment ofresistance is obtained by equating the forcesof tension and compression.Substituting for xu,mPxfrom Table B, we getthe values of Pt,lim fYj& as given in Table C.TABLE C LIMITING MOMENT OFRESISTANCE AND REINFORCEMENT INDEXFOR SINGLY REl;~&FOR~N~ RECTANGULAR(Clause 2.3)j& N/mm* 250 415 500M*,lhl-- -Lk bd’0.149 W138 0.133Plrllrnfy/ ck21.97 19.82 18.87The values of the limiting moment of resis-tance factor Mu/bd2 for different grades ofconcrete and steel are given in Table D. Thecorresponding percentages of reinforcementsare given in Table E. These are the maximumpermissible percentages for singly reinforcedsections.TABLE D LIMITING MOMENT OFRESISTANCE FAVOR Mu,,im/bd’, N/mm’ FORSINGLY REINFC);&yE$sECTANGULAR(Clause 2.3)/CL,N/mm’fy, N/-YrK------ 50015 2.24Is:3.452.003: 2.983.73 2.663.3330 4.47 414 3.99TABLE E MAXIMUM PERCENTAGE OFTENSILE REINFORCEMENT pt,lim FORSINGLY REINFStRmTNSRE!aANGW(c%u.w 2.3)fdr, /y, Nhm’N/mm* r b250 415 u)o15 1.3241.76220;g “0%2% l.43 YE.2.3.1 Under-ReinforcedSectionUnder-reinforced section means a singlyreitiorced section with reinforcement per-centage not exceeding the appropriate valuegiven in Table E. For such sections, thedepth of neutral axis xu will be smaller thanx”,,,,~. The strain in steel at the limit stateof collapse will, therefore, be more than0.87 fy- + 0902 and, the design stress inE.steel will be 0937fy. The depth of neutralaxis is obtained by equating the forces oftension and compression.‘G (0.87 fr) - 0.36 fdrb xuThe moment of resistance of the section isequal to the prdduct of the tensile forceand the lever arm.Mu = pG (@87f,) (d - 0,416 xu)=O*87fy &( )(l- 0.4165)bd2Substituting foi $ we get_ _x 1C- 1.005 &$]bda2.3.Z.Z Charts 1 to 28 have been preparedby assigning different values to Mu/b andplotting d versuspt. The moment values inthe charts are in units of kN.m per metr$width. Charts are given for three grades ofsteel and, two grades of concrete, namelyM 15 and M 20, which are most commonlyused for flexural members. Tables 1 to 4cover a wider range, that ‘is, five values offy and four grades of concrete up to M 30.In these tables, the values of percentage ofreinforcement pt have been tabulated againstMu/bd2.10 DESIGN AIDS FOR WNFORCED CONCRETE
  28. 28. 2.3.2.2 The moment of resistance of slabs,with bars of different diameters and spacingsare given in Tables 5 to 44. Tables are givenfor concrete grades M 15 and M 20, withtwo grades of steel. Ten different thicknessesranging from 10 cm to 25 cm, are included.These tables take into account 25.5.2.2of the Code, that is, the maximum bardiameterdoesnot exceedone-eighth the thick-ness of the slab. Clear cover for reinforce-ment has been taken as 15 mm or the bardiameter, whichever is greater [see 25.4.1(d)of the Code]. Jn these tables, the zeros atthe top right hand comer indicate the regionwhere the reinforcement percentage wouldexceed pt,lim; and the zeros at the lowerleft hand comer indicate the region wherethe reinforcement is less than the minimumaccording to 25.5.2.1 of the Code.Example 1 Singly Reinforced BeamDetermine the main tension reinforcementrequired for a rectangular beam sectionwith the following data:Sixeof beam 3ox6OcmConcrete mix M 15Characteristicstrength 415N/mm’of reinforcement*Factored moment 170kN.m*Assuming 25 mm dia bars with 25 mmclear cover,Effectivedepth I 60 - 2.5 -2;- 5625 cmFrom Table D, for fr P 415 N/mm’ andfcrc- 15 N/mm*MWliUJM’ p 2.07 N/mm:v$g$ x (1000)’e; 2.07 x 101kN/m*:. &am - 2.07 x 1O’W30I 2-07 x 10’ x fa xI 1965 kN.m$%ua] moment. of. 170 kN.m is less *thanThe sectton 1stherefore to bedestgnedasu’~mm’singlyreinforced (unde&einforced)rectangular section.fVfM’HODOF RBFQIRINGTOFU3XURECHARTFor referring to Chart, we need the value ofmoment per metre width.Mu/b-g = 567kN.m per metrewidth.*The term ‘factoredmoment’means the momentdue to characteristic loads multiplied by the appro-priate value of p&rtialsafety factor yf.Retbrring to C/r& 6, corresponding toh&,/b - 567kN.m and d = 5625 cm,Percentageof steelpt - lOOAsM = 0.60.6 bd. .* A,= -jijiy0.6~30~5625 __O1 ,,*100For referring to Tables, we need the valueMuofwM” 170x IO’bd’ - -3m6.25 x 56.25 x IO’I 1.79 N/mm’From Table 1,Percentage of reinforcement,pt = 0.594* As-. .0.594 x 30 x 56.25 _ ,omo2,,*100Example 2 SlabDetermine the main reinforcement re-quired for a slab with the following data:Factored moment 9.60 kN.mE%etreDepth of slab 10 cmConcrete mix M 15Characteristic strength a) 415 N/mm2of reinforcement b) 250 N/mm*h&l-HODOFREPERRINGTOTABLESFOR SLABSReferring to Table 15 (for fy - 415 N/mmz),directly we get the following reinforcementfor a moment of resistance of 9.6 kN.mper metre width:8 mm dia at 13cm spacingor 10mm dia at 20 cm spacingReinforcement given in the tables is basedon a cover of 15 mm or bar diameter which-ever is greater.MFXHOD OF RFNRRJNG TO FLBXURB CHARTAssume 10 mm dia bars with 15 mm cover,d - 10- 1.5 - 9 =8cma) For fy= 415 N/mm’From Table D, Mu,tidb# = 2.07 N/mm*:. J%lirn - 2.07 x lOa x z x (A)’= 13.25kN.m ’ _’Actual bending moment of 960 kN.m is lessthan the limiting bending moment.FLExuRALmMBERs 11
  29. 29. Referring to Chart 4, reinforcement per-centage, pt 6 0.475Referring to Chart 90, provide8 mm dia at 13 cm spacingor 10 mm dia at 20 cm spacing.Alternately,A, = O-475 x 100 x &J = 3.8 cm* permetre width.From Table %, we get the same reinforce-ment as before.b) Forf, = 250 N/mm*From Table D, Mu&bd” = 2.24 N/mm2Mu&m = 2.24 x 10’ x 1 x(h)= 14.336 kN.m ‘---’Actual bending moment of 9.6 kN.m is lessthan the limiting bending moment.Referring to Chart 2, reinforcement per-centage, pt = 0.78Referring to Churf PO, provide 10 mm diaat 13 cm spacing.2.3.2 Doubly Reinforced Sections - Doublyreinforced sections are generally adoptedwhen the dimensions of the beam have beenpredetermined from other considerationsand the design moment exceeds the momentof resistance of a singly reinforced section.The additional moment of resistance neededis obtained by providing compression re-inforcement and additional tensile reinforce-ment. The moment of resistance of a doublyreinforced section is thus the sum of thelimiting moment of resistance Mu,lim of asingly reinforced .section and the additionalmoment of resistance Mu,. Given the valuesof Mu which is greater than M”,lim, the valueof Mu, can be calculated.Mu, = Mu - MuslimThe lever arm for the additional moment ofresistance is equal to the distance betweencentroids of tension reinforcement and com-pression reinforcement, that is (d-d’) whered’ is the distance from the extreme compres-sion fibre to the centroid of compressionreinforcement. Therefore, considering themoment of resistance due to the additionaltensile reinforcement and the compressionreinforcement we get the following:Mu, - Asts (0*87f,) (d - a,)also, Mu, =&Us-fQC)(d-J’)whereA1t2is the area of additional tensile rein-forcement,AK is the area of compression reinforce-ment,I= is the stress in compression reinforce-ment, andfee is the compressive stress in concrete atthe level of the centroid of compres-sion reinforcement.Since the additional tensile force is balancedby the additional compressive force,A, (l;c - fee)= At, (0*87&jAny two of the above three equations maybe used for finding Alt, and A,. The totaltensile reinforcement Ast is given by,Ast = Pblim mbd$ Asc,It will be noticed that we need the values offrc and J& before we can calculate Al.The approach, given here is meant for designof sections and not for analysing a givensection. The depth of neutral axis is, therefore,taken as equal to x,,,-. As shown in Fig. 5,strain at the level of the compression reinforce-ment will be equal to O-003 5(d’1- -XU,UWZ>12STRAIN OlAGRkMFIG. 5 DOUBLY REINKIRCED SECI-IONDESIGNAIDS FOR REINFORCED CONCRIXE
  30. 30. For values of d’/d up to 0.2, feeisequal to0446 fck; and for mild steel reinforcementfz would be equal to the design yield stressof 0.87 fY. When the reinforcement is cold-worked bars, the design stress in compressionreinforcement fw for different values ofd’/d up to 0.2 will be as given in Table F.TABLE F STRESS IN COMPRESSIONREINFORCEMENT ftc, N/mm* IN DOUBLYREINFORCED BEAMS WITH COLD-WORKED BARS(Clause 2.3 2)fY9N/mm’415500d’ld-A ,0.0s 0.10 O-15 0.20355 353 342 329424 412 395 3702.3.2.2 Astzhas been plotted against (d -d’)for different values of MU, in Charts 19 and20. These charts have been prepared forfs = 217.5 N/mm2 and it is directly appli-cable. for mild steel reinforcement with yieldstress of 250 N/mm*. Values of Aat? for othergrades of steel and also the values of A, canbe obtained by multiplying the value readfrom the chart by the factors given in Table G.The multiplying factors for A=, given inthis Table, are based on a value of fee corres-ponding to concrete grade M20, but it canbe used for all grades of concrete with littleerror.TABLE G MULTIPLYING FACTORS FORUSE WITH CHARTS 19 AND 20‘Clause 2.3.2.1)fN&PFACTOR FACTOR FOR A, FOR d’jdFORAc--at* 0.05 0.10 0.15 0.2250 1.00 1.04 1.04 1.04 1.04415 0.60 0.63 0.63 0.65 0.68500 0.50 0.52 0.54 0.56 0.602.3.2.2 The expression for the moment ofresistance of a doubly reinforced section mayalso be written in the following manner:Mu = Mu,lim + %(0*87fy) (d-d’)Mu Mu,limbj2 = bd”___ + -&(0*87f,)( I- ;>whereptz is the additional percentage of tensilereinforcement.Pt = phlim + pt2PC =P”[-L-]The values of pt and pc for four values ofd’jd up to 0.2 have been tabulated againstMU/bd2 in Tables 45 to 56. Tables are givenfor three grades of steel and four gradesof concrete.Example 3 Doubly Reinforced BeamDetermine the main reinforcements re-quired for a rectangular beam section withthe following data:Size of beam 30 x 6OcmConcrete mix M 15Characteristic strength of 415 N/mm2reinforcementFactored moment 320 kN.mAssuming 25 mm dia bars with 25 mmclear cover,d = fj0 - 2.5 - 225 = 56*25cmFrom Table D, for fy = 415 N/mm2 andfck = 15 N/mm2Mu,linJbd”=2.07 N/mm2 = 2.07 x IO2kN/m”.*. Mu,lim-2.07 x 103bd230 56.25 56.25-2.07 x 10”x loo x Ts- x -100-= 196.5 kN.mActual moment of 320 kN.m is greaterthan Mu,lim*. . The section is to be designed as a doublyreinforced section.Reinforcement from TablesMu 320$$ = O-562 5)2 x 103~~~~~ N/mm2d’/d c 2.5 + 1.25 i o,075625 >Next higher value of d’/d = 0.1 will be usedfor referring to Tables.Referring to Table 49 corresponding toMU/bd2 = 3.37 and $ = 0.1,Pt = 1.117,pc = 0.418.. . At - 18.85 cm2, A, = 7.05 cm2REINFORCEMENTFROM CHARTS(d-d’) = (56.25 - 3.75) - 52.5 cmMu2 - (320 - 196.5) = 123.5 kN.mChart is given only for fy = 250 N/mm2;therefore use Chart 20 and modificationfactors according to Table G.Referring to Chart 20,Art2 (for fY = 250 N/mm2) = 10.7 cm2FLEXURAL MEMBERS 13
  31. 31. usia Jl¶odibrion factors given infor BY= 415 N/nuns,I& - 10.7 x 0.60 r! 6-42cm*,& I 10.7 x 0.63 = 674cm’Referring to ruble E,Table Gpt,nm - 072* Ast,u,n -0.72 .x5625 x 30. . ,oo - 1215cm’A*: E 12.15+ 642 = 18.57cm’These values of At and AE are comparableto the values obtained from the table.2.4 T-SECTIONSThe moment of resistanceof a T-beam canbe considered as the sum of the moment ofresistance of the concrete in the web of widthb, and the contribution due to ,flanges ofwidth br.The maximum moment of resistance is ob-tained when the depth of neutral axis is x,,,~.When the thickness of flange is small,that is, lessthan about 0.2 d, the stress in theflange will be uniform or nearly uniform(see Fig. 6) and the centroid of the compres-sive force in the flange can be taken at Df/2from the extreme compression fibre. There-fore, the following expression is obtained forthe limiting moment of resistance of T-beamswith small values of Dfjd.x(br-bw)h( d-$)whereMll,llltniiveb30.36 fd bwxu,,,,.x(d-O.416 x0,,,,&.The equation givenin E-2.2 of the code is thesame as above, with the numericals roundedoff to two decimals. When the flange thick-,ness is greater than about 0.2 d, the aboveexpression is not corre4ztbecause the stressdistribution in the flanp would not be uni-form. The expression Bven in E-22.1 of theCode is an approximation which makes allo-wance for the variation of stress in the flange.This expression is obtained by substitutin#JYfor &in the equation of E-2.2 of the CO&yf beingequal to (0.15 X,,m&+ 065 or)but not greater than Dr. With this m&&a-tion,Mudin~~T9 Mu,lirn,web f 0446 f&Mr-WY+ - f )Dividing both sides by&kbwP,x(& l)$(l+$)wherexu;= + 0.65 !$but .f < $Using the above expression, the ~2:of the moment of resistanceMu,lim,T~ck b,# for different values of h/b*and &/d have been worked out and given inTables 57 to 59 for three grades of steel.2.5 CONTROL OF DEFLECTION2.5.2 The deffection of beams and slabswould generally be,within permissible limitsif the ratio of span to effectivedepth of themember does not exceed the values obtainedin accordance with 22.2.1 of the Code. Thefollowing basic values of span to effectivedepth are given:En!;;;EortedCantilever20“40.0.047 f”-* 0.0020.87 f,-_b E,STRAIN DIAGRAM STRESS DIAORAMho. 6 T-SECTION14 DBSIGN AIDS FOR REINPORCED CDNCRETE
  32. 32. Further modifying factors are given inorder to account for the effects of grade andpercentage of tension reinforcement andpercentage of compression reinforcement.2.5.2 In normal designs where the reinforce-ment provided is equal to that required fromstrength considerations, the basic values ofspan to effective depth can be multiplied bythe appropriate values of the modifyingfactors and given in a form suitable for directreference. Such charts have been preparedas explained below :4b)The basic span to effective depth ratiofor simply supported members is multi-plied by the modifying factor for ten-sion reinforcement (Fig. 3 of the Code)and plotted as the base curve in thechart. A separate chart is drawn foreach grade of steel. In the chart, spanto effective depth ratio is plotted onthe vertical axis and the tensilereinforcement percentage is dotted onthe horizontal axis.When the tensile reinforcement ex-.ceeds ~I,II,,, the section will be doublyreinforced. The percentage of compres-sion reinforcement is proportional tothe additional tensile reinforcement@t - PM,,) as explained in 2.3.2.However, the value of Pt,lim and pcwill depend on the grade of concretealso. Therefore, the values of span toeffective depth ratio according to basecurve is modified as follows for eachgrade of concrete:1)2)3)For values of pt greater thanthe appropriate value of pt,lim,the value of (pt - pt,lim) is cal-culated and then the percentage ofcompression reinforcement p= re-quired is calculated. Thus, thevalue of pc corresponding to a valueof pt is obtained. (For this purposed’/d has been assumed as 0.10 butthe chart, thus obtained can gene-rally be used for all values of d’/din the normal range, without signi-ficant error in the value of maximumspan to effective depth ratio.)The value of span to effective depthratio of the base curve is multipliedby the modifying factor for com-pression reinforcement from Fig. 4of the Code.The value obtained above is plottedon the same Chart in which the basecurve was drawn earlier. Hencethe span to effective depth ratio fordoubly reinforced section is plottedagainst the tensile reinforcementpercentage pt without specificallyindicating the value of pc on theChart.25.3 The values read from these Chartsare directly applicable for simply supportedmembers of rectangular cross section forspans up to 10 m. For simply supported orcontinuous spans larger than 10 m, the valuesshould be further multiplied by the factor(lo/span in me&es). For continuous spansor cantilevers, the values read from the chartsare to be modified in proportion to the basicvalues of span to effective depth ratio. Thetn.l~G$ing factors for this purpose are as..conned; spans&In the case of cantilevers which are longerthan 10 m the Code recommends that thedeflections should be calculated in order toensure that they do. not exceed permissiblelimits.2.54 For flanged beams, the Code recom-mends that the values of span to effectivedepth ratios may be determined as for rectan-Eoeons, subject to the followmg modi-..4b)The reinforcement percentage shouldbe,bcz&zm the area brd while referrmgThe value of span to effective depthratio obtained as explained earliershould be reduced by multiplying by thefollowing factors:b&v Factor>:::3For intermediate values, linear interpola-tion may be done.Nom --The above method for flanged beamsalay sometimes give anomalous mwlts. If the fhgesarcignored and the beam is considered as a rectangularsection, the value of span to effective depth ratio thusobtained (percenYof rciaforcemcnt being basedon the area l&) s ould always be oa the safe side.2.5.5 In the case of tw way slabs supportedon all four sides, the sPorter span should beconsidered for the purpose of calculating thespan to effective depth ratio (see Note 1 below23.2 of the Code).2.5.6 In the case of flat slabs the longer spanshould be considered (30.2.2 of the Code).When drop panels conforming to 30.2.2 ofthe Code are not provided, the values of spanto effective depth ratio obtained from theCharts should be multiplied by 0.9.Example 4 Control of DeflectionCheck whether the depth of the memberin the following cases is adequate for control-ling deflection :a) Beam of Example 1, as a simply suppor-ted beam over a span of 7.5 mFLBXURAL MEMBERS 15
  33. 33. b)Cla>Beam of Example 3, as a cantilever beamover a span of 4.0 mSlab .of Example 2, as a continuousslab spanning in two directions theshorter and longer spans being, 2.5 mand 3.5 m respectively. The momentgiven in Example 2 corresponds toshorter spa’n.Actual ratio ofSpanEflective depth= (56.;5;,oo) = 13.33Percentage of tension reinforcementrequired,pt = 0.6Referring to Char1 22, value of MaxSpan( >Tcorresponding to Pt = 0.6, is 22.2.Actual ratio of span to effective depth is lessthan the allowable value. Hence the depthprovided is adequate for controlling deflec-tion.b) Actual ratio ofSpanEtfective depth‘(d&J = 7.11Percentage of tensile reinforcement,pr = 1.117Referring to Churl 22,Max value of %!a? = 21.0( 1ClFor cantilevers, values read from theChart are to be multiplied by 0.35.:. Max value of 1I/d for ) =21.0x0*35=7.35cantilever J* The section is satisfactory for control. .of deflection.c) Actual ratio ofSpanEffective depth2.5=-= 31.250.08(for slabs spanning in two directions,the shorter of the two is to be con-sidered)(i) Forfv = 415 N/mm2pt = 0,475Referring to Chart 22,Max Span = 23.6(-> dFor continuous slabs the factorobtained from the Chart should bemultiplied by 1.3.:. Max “7 for continuous slab= 23.6 x 1.3 F 30.68Actual ratio of span to effective depth isslightly greater than the allowable. Thereforethe section may be slightly modified or actualdeflection calculations may be made to as-certain whether it is within permissible limits.(ii) F0r.j; = 250 N/mm2pt = 0.78Referring to Chart 21,Max Span = 31.3(-1 d:. For continuous slab,Max %% = 31.3 x 1.3d= 40.69Actual ratio of span to effective depth isless than the allowable value. Hence thesection provided is adequate for controllingdeflection.16 DESIGN AIDS FOR REINFORCED CONCRETE
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  41. 41. TABLE 1 FLEXURE - REINFORCEMENT PERCENTAGE, pc FOR SINGLYREINFORCED SECTIONSN/mm= bE040O-45O-501.10l-12l-14l-16l-181.201.22::zl-281.30l-321.34I::0.141 Oa5 0.074O-166 @lOO OQ868’E. 8’::;. o&EQ240 O-144 O-125O-276O-302O-329O-356O-383O-2650.290EO-3680.1590.175i-EOh0.1380.1510.164O-1780’1910.410O-4210.433IZO-394 O-237 Q-205O-405 0.244 0.211O-415 O-250 O-216O-426 O-257 o-222O-437 0’263 O-2278::;;0489O-500O-5120448 O-270O-458 0.2760469 02830.480 O-2890.491 0.2%0.523O-5350.5460.558O-570O-5020.5130.5248:::;0.303EJO-323O-3290.262o-267O-273O-279O-285O-5810393@6058’:;;.0.55803700.5810.592O-604EiO-350O-357O-364O-291O-297O-303O-309O-3150.641 O-615 0.371 0.3210.653 0.627 0.378 0.327O-665 O-639 0.385 0.333O-678 O-650 0.392 O-339O-690 O-662 O-399 O-3458E I:GO-727 O-698O-740 O-7100.752 O-7220.765O-778x:z0.816O-734o-7470.7590.771O-7840442X:t:YO-4650.472O-382O-3890.395x:z250 415 4800993l-0071*0211935l-049l-136l-151l-1661.181l-197l-21212281,2431.2591.275i’.O-503O-510O-398i;:0.423O-518O-5268:Z0’550O-4480.455O-4610468O-475O-430O-436O-4438:::0.558 0482 o-463O-566 0489 O-4690’574 0.496 O-476O-582 0’503 0483O-590 o-510 0.490O-5170.525O-5320.5390.546O-497o”:?i0.5180.52505540.5610.569O-576O-584O-5320.5390.685O-693O-7030.712O-592fY240250415480500 :7ck25NOTE-Blanks indicate inadmissible reinforcement percentage (see Table E).FLeXURAL MEMBERS
  42. 42. 240250415480fck20--TABLE 2 FLEXURE -REINFORCEMENT PERCENTAGE, pt FOR SINGLYREINFORCED ‘SECTIONSl-05 0.5381.10 O-5661’15OO’E:‘z *. O-650EO-678O-707E 0.736O-765l-50 0.7951.55 O-825;:g 8$;:.l-70 O-916l-75 0947l-80%!I-85 *E1’0411.073200 lm63gl-119E:::zl-159210 l-172f’::l-185. 1*199f’:;. :‘E.220 l-239“0%O-188O-213O-2370.517O-5430.5700.597@624O-651O-6790.7070.7350’763O-073Efl0.1110’123izi!FEJ.O-131O-143XEO-181O-201::E0.2420.255O-193FfgO-245::a:;O-297O-3110’3250.2588:;:;0.29803120.339O-354EL!.0397X:E ;:gjO-521 0.500o-537 O-5150.553 o-5310.5370’543O-550O-556O-562f& - zO.N/m1.253:‘26:1%LOZl-323E:‘:z.:‘z.l-423l-438l-452l-203l-216:z1.256l-338l-352196613801’39414671~482l-497:::;;1.542 1481l-558 1’495::z :::1’:l-604 l-5401.782:‘E.l-8331.632l-647l-663_g;;.l-711:‘%.l-760O-627O-633I::0.661pj.O-690o-6970.704o-711o-719O-726O-734O-741O-7480.7560.764x’%.O-6150.6210.628O-635xzz0.6550.662Num -Blanks lndk$teinadmissiblereinforcementpcmntagc (seeTableJ%48 DESIGNAIDSFORREINFORCEDCGNCRRI’E
  43. 43. TABLE 3 FLEXURE - REINFORCEMENT PERCENTAGE, pt FOR SINGLYREINFORCED SECTIONs/ck = 25 N/mms0.300.35g0.146O-171O-195:zO-271O-2960?321O-347O-3730.80O-858Clfl00.399t-I.4250.451O-477O-504l-051.101.15::zO-5308%O-611O-638l-30l-35:z1.500.6660.6930.7210.749O-7771.80 0.949l-83 09791.90 l-0091.95 l-0382-00 1,0682-052102.15$12”::zl-160l-1911.222l-25413283l-317l-350l-382250 415 480 500.0.140 0.0840.164 O-0990188 0.113O-211 Of270.236 O-142O-070:%zO-106O-118O-260!%0.333O-358O-1560.1710.186O-201O-2160.1300.142O-15401670.1790231X%0.276O-2910191O-2040.2160229O-2420.5096535O-561z-;;:.0.307O-3220.33803530.3690.2550.2670.2800.2930.3060.639X:%@7190.746o-7738:::0.856O-883O-385O-401w4170.433O-4490.3330.3478%O-3880.320O-3330.346O-3590373O-466O-4820.4990.515O-5320403x::::8:%O-387x%Y0.428O-442O-911O-9400968YZ.O-549O-566O-583O-601O-618O-415O-489fj:$0.4560.4700.484O-498o-5131.055: :%l-1431.173O-635O-6530.6710.689O-7078%O-5800.5960.611O-5270542O-557O-5720.587l-204l-234l-265: :%0.725O-7430.7620781o-799O-627O-643ZRO-691O-6020.617O-6320.6480.663Mlw2, fu,N/mm2N/mms 7’L--Y250 415 5001.415 1.3581448 l-3901482 l-4221.515 l-455l-549 1487%2902-953.00l-5841.6181.653l-689l-7243.053.103.15::::l-7601.7971.834l-871190933%:::t3-381.947 l-869l-962 l-884l-978 l-8991993 l-9142QO9 l-9292.025 l-944Z:% :‘zc.2.072 l-9892-088 2.0052.1042.120z:::2.1703.603.62::z3.682.1862.2032.219;:22:;3.70 2.2703-72 2.2873.74 2.304NOTE- Blanksindicate inadmissiblereinforcement percentage(see Table E).l-520:::zl-621l-655::%l-760l-7961.8322.0992.1152.1312.1472.1632.1792196O-8180.837::::zO-896::;:zO-956o-9770.9971.0181.039l-061l-0821.104::::;:::z1.162O-679:z:O-727O-7448%O-794O-8110.8280.845O-8630.8800.898O-9160.9350942fY2402504154805007ck25m3xuRAL MEMBERS 49
  44. 44. Y240250415480500fck30TABLE 4 FLEXURE - REINFORCEMENT PERCENTAGE, pi FOR SINGLYREINFORCED SECTIONS0.1408’::;.@2110235038004050.4290.4540.4790.52505520.578x’z.~~i?i0.71207390.766@631x:::0.7090’7358’;1;:oi49x:zi0.7620.788xz::0.8680.932 0.8950961 0.9220.989 0.9501.018 09771946 1.0051.075:x.1.1631.1921’173::i!f1.2601289250 500MUW,A. N/mm2N/mm2 r240*-7250 415 480 5000.0700.0820.0930.1050.117255 1.374 1.319fZ ::zi ::zi270 1467 1.408275 1.498 1.438c8”:tz3.003.053.103.1533:g0.2520.265Ct.2770.2900.3033.30 1.8593.35 1.893:z zi3.50 19980316xz0.3550.3683.553.603.653.703.75@381 3.800.394 3.85x:z it:z0.434 4.00oo:z0.47504880502fd = 30 N/mm21.5301.5621.5941.6261.659~~~1.5301.5611.5921.691 1.6241.725 1.6561.758 1.6871.791 1.7201.825 1.7521.785 1.075 0.9301.818 1.095 0.9471.851 1.115 0.964I.884 1.135 0.9811.918 1.156 0.9992z21052.1422.1781.952 l-1761.986 1.1972021 1.2182056 1.2392091 1.26022152253229123292367f:iz i::z2485 23862.525 2.4242566 2463NOTE- Blanksindicateinadmissiblereinforcementpyceniage(seeTableE).0.794 @6870.812 0.7020.830 0.7180.848 0.7330.866 0.749is;0.79708130.8290.978zz31.0361.0550.846 0.8120.862 0.8280.879 0.8440.896 0.8600.913 0.876KG;1.0531.0711.0891.2811.3031.3251.3471.3691.108:‘:z.1.1641.1841.3911.4140.7340.7500.7650.7810.7960976zf:1.0281.046KS1.0991.118so DESIGN AIDS FOR REmFORCJiDCONCREIB
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  47. 47. 3. COMPRESSION MEMBERS3.1 A$MU;,OADED COMPRESSION lower sectioqs would eliminate tho need forany calculation. This is particularly usefulAll compression members arc to be designedas an aid for deciding the sizes of columnsat the preliminary design stage of multi-for a minimum eccentricity of load in twooribcioal directions. Clause 24.4 of the Codesforeyed buildings.Ispecifiks the following minimum eccentri-city, eminfor the design of columns: Example 5 Axially Loaded Column1i-D subject to a minimum ofemin=ggg 3oDetermine the cross section and thereinforcement required for an axially loaded2 cm. columc with the following data:whereFactored loadConcrete gradeCharacteristic strength ofreinforcement3000kNM20415 N/mm’1 is the unsupported length of the column(see 24.1.3 of the Code for definition ofunsupported length), andD is the lateral dimension of the columnin the direction under consideration.Unyoyior;d length of 3.0 mAfter determining the eccentricity, the sectionshould be designed for combined axial loadand bending (see 3.2). However, as a simplifi-cation, when the value of the mininiumeccentricity calculated as above is less than orequal to 0*05D, 38.3 of the Code permitsthe design of short axially loaded compressionmembers by the following equation:P,=@4f,k AC-i-0.67fY ArThe cross-sectional dimensions required willdepend on the percentage of reinforcement.Assuming 1.0 percent reinforcement andreferring to Chart 25,Required cross-sectional area of column,A, - 2 700 cm*Provide a section of 60 x 45 cm.Area of reinforcement, A, - 1.0 x m~$,j2where 1: 27 cm8PUis the axial load (ultimate),A, is the area of concrete, andAsc is the afea of reinforcement.The above equation can be written asWe have to check whether the minimumeccentricity to be considered is within 0.05times the lateral dimensions of the column.In the direction of longer dimension,I D&in ---500 +30P” = 0.4 f& PA,A, - +$) t- 0.67fy loowhereAs is the gross area of cross section, andp is the percentage of reinforcement.Dividing both sides by A,,PU = @4&( 1 - j&) +“‘67fy $jCharts 24 to 26 can be used for designingshort columns in accordance with the aboveequations. In the lower section of thesecharts, P./A, has been plotted againstreinforcement percentage p for differentgrades of concrete. If the cross section ofthe column is known, PU/Al can be calculatedand the reinforcement percentage read fromthe chart. In the upper section of the charts,PU/Asis plotted against PUfor various valuesof AS. The combined use of the upper and3*0x 102 60=500+ jg P 0.6 j-2.0 - 2.6 cmor, e&D = 26160 = O-043In the direction of the shorter dimension,3.0 x 102emrn=500+4530= 0.6 + 1.5= 2.1 cmor, e,i,/b = 2*1/45 = @047The minimum eccentricity ratio is less than@05 in both directions. Hence the design ofthe section by the simplified method of 38.3of the Code is valid.3.2 COMBINED AXIAL LOAD ANDUNIAXIAL BENDINGAs already mentioned in 3.1, all com-pression members should be designed forCOMPRESSION MEMBERS 99
  48. 48. minimum eccentricity of load. It shouldalwaysbe ensuredthat the scotionis designedfor a moment whichis not lessthan that dueto the prescribedtinimum eccentricity.3.2.1 Amanptio&Assumptiom (a), (c),(d) and (e) for flexural members (see 2.1)are also applicable to members subjcotedto combined axial load and bending. ‘Theassumption (b) that the maximum strainin concrete at the outermost eom ressionifibre is 04N35 is also applicablew en theneutralaxis k withinthe seotionand in thelimitingcase when the neutralaxis lies alongone edge of the section; in the latter oascthe strain varies from 0@035 at the highlycompressed edge to zero at the opposiked~. For purely axial compression, thestrain is assumed to be uniformly equal00002 acxossthe seotion[see38.l(a) of theCode]. The strairidistributionlines for these~’ oases intersecteaeh other at a depth of~ffom the highly compressed edge. Thispoint is assumedto act as a fulcrum for thestrain distribution line when the neutralaxis lies outsidethe motion(see Fig. 7). Thisleads to the assumption that the strain atthe highly compresseded~ is 00035 minus0?5 times the strainat the leastcompressededge [see 38.Z(b) of the Cole],“-i t- - ---1-”I: 1q ‘* qIq*qqiqqcbq:00I I!* qtqJ: IHIWilmYq 00 EOOEC6MPRE SSE IICENTRO13AL AXIS+’+ ikh ROW OF REINFORCEMENTSTRAIN DIAGRAMS0035Neutral axiswlthln the scctlon-30/7-1-— -----Neutral axisoutside the sect ionFIG. 7 Cmramm Am- LOAD AND UNIAXIAL BENDINGNo DESIGNAIDSIK)RREINFORCEDCONCR81E
  49. 49. 3.2.2 Stress Block Parameters Wh&n the Area of stress blockNeutralAx& Lies O&sidethe Section- Whenthe neutral axis lies outside the section, g 4the shape of the stress block will be as.indi-- 0446f,D-5( >,-Dcated in Fig. 8. The stress is uniformly0446fd for a distance of Ly from the highly= 04461&D +gDcompressed edge because the strain is more - 0446fdr Dthan 0402 and thereafter the stress diagram[l-&&J]is parabolic.The centroid of the stress block will befound by taking moments about the highlycompressed edge.Moment about the highly compressed edgeDpO1446fckD i( 1-$ gDtiThe position of the centroid is obtained bydividing the moment by the area. For diier-ent values of k, the area of stress block andSTRAIN DIAORAM the position of its centroid are given inTable H.O-446 1,BTRESS OIAORAWFIG. 8 STRBSSBLOCK WHEN THE NEUTRALAm h¶ oUT?3IDE THE SECTIONLet x0- kD and let g be the ditferencebetween the strxs at the highly compressededgo and the stress at the least compressededge. Considering the geometric propertiesof a parabola,-o+Mf& &( 11TABLEH STRESSBLOCKPARAhUTTERSWHENTHE NETmmtA&mA?NLIES OUTSIDE(Clause 3.2.2)Nom-Values of stress block parametershavebeentabulatedforvaluesof k upto4’00forinfom-tion only. For constructionof interactiond@camsit b merally adaquatato considervaluesof k up toabout 1.2.33.3 Constructionof InteractionDiagram-Design chartsfor combined axialcompressionand bending are given in the form of inter-action diagmms in which curyes for PJbDfdversus MdbD* fb are plotted for differentvalues of p/f&, where p is the reinforcementpercentage.COMPRESSlONMEMBERS 101
  50. 50. 3.2.3.2 For the case of purely axial com-pression, the points plotted on the y-axisof the charts are obtained as follows:P,= 0446f,rbd + ‘g (A - 0.446 fek)wherefr is the compressive stress in steel corres-ponding to a strain of 0.002.The second term within parenthesis repre-sents the deduction for the concrete replacedby the reinforcement bars. This term isusually neglected for convenience. However,a9 a better approximation, a constant valuecorresponding to concrete grade M20 hasbeen used in the present work so that theerror is negligibly small over ;he range ofconcrete mixes normally used. An accurateconsideration of this term will necessitatethe preparation of separate Charts for eachgrade of concrete, which is not consideredworthwhile.3.2.3.2 When bending moments are alsoacting in addition to axial load, the pointsfor plotting the Charts are obtained byassuming different positions of neutral axis.For each position of neutral axis, the straindistribution across the section and thestress block parameters are determined asexplained earlier. The stresses in the rein-forcement are also calculated from theknown strains. Thereafter the resultant axialforce and the moment about the centroidof the section are calculated as follows:a) When the neutral axis lies outside thesectionliwhereCl -Pi -i-1coefficient for theblock to be taken(see 3.2.2);area of stressfrom Table HAdbxwhere A,i is the area of rein-forcement in the ith row;fii - stress in the ith row of reinformfci -n -102ment, compression being positiveand tension being negative;stress in concrete at the level ofthe ith row of reinforcement; andnumber of rows of reinforcement.The above expression can be written asnTaking moment of the forces about thecentroid of the section,+ x g (.Ai - fci).Yiwherei- 1C,D is the distance of the centroid of theconcrete stress block, measured fromthe highly compressed edge; andYi is the distance from the centroid of thesection to the ith row of reinforce-ment; positive towards the highlycompressed edge and negative to-wards the least compressed edge.Dividing both sides of the equation byfck bD”,c, (O-5-Cdn+X*(&i -&i)(s)i- 1b) When the neutral axis lies within thesectionIn this case, the stress block parametersare simpler and they can be directly incorpora-ted into the expressions which are otherwisesame as for the earlier case. Thus we get thefollowing r;xpressions:=@36k+ c &j&&di-1nwherek-Depth of neutral axisDDESIGN AIDS FOR REINFORCED CONCRETE
  51. 51. An approximation is made for the valueOffci for M20, as in the case of 3.2.3.1. Forcircular sections the procedure is same asabove, except that the stress block para-meters given earlier are not applicable;hence the section is divided into strips andsummation is done for determining theforces and moments due to the stresses inconcrete.3.2.3.3 Chartsfor compression withbending -Charts for rectangular sections have beengiven for reinforcement on two sides (Charts27 to 38) and for reinforcement on foursides (Charts 39 to 50). The Charts for thelatter case have been prepared for a sectionwith 20 bars equally distributed on all sides,but they can be used without significant.error for any other number of bars (greaterthan 8) provided the bars are distributedequally on the four sides. The Charts forcircular section (Charts 51 to 62) have beenprepared for a section with 8 bars, but theycan generally be used for sections with anynumber of bars but not less than 6. Chartshave been given for three grades of steeland four values of d’/D for each case men-tioned above.The dotted lines in these charts indicatethe stress in the bars nearest to the tensionface of the member. The line for fs, I; 0indicates that the neutral axis lies along theoutermost row of reinforcement. For pointslying above this line on the Chart, all thebars in the section will be in compression.The line for fSt = fYd indicates that theoutermost tension reinforcement reaches thedesign yield strength. For points below thisline, the outermost tension reinforcementundergoes inelastic deformation while succes-sive inner rows may reach a stress of fyd.It should be noted that all these stress valuesare at the failure condition correspondingto the limit state of collapse and not at work-ing Ioads.3.2.3.4 Charts for tension with bending -These Charts are extensions of the Chartsfor compression with bending. Points forplotting these Charts are obtained by assum-ing low values of k in the expressions givenearlier. For the case of purely axial tension,Pu - g (O-87fy)hk (@87fy)Charts 66 to 75 are given for rectangularsections with reinforcement on two sidesand Charts 76 to 85 are for reinforcementon four sides. It should be noted that thesecharts are meant for strength calculationsonly; they do not take into account crackcontrol which may be important for tensionmembers.Example 6 Square Column with UniaxialBendingDetermine the reinforcement to be providedin a. square column subjected to uniaxialbending, with the following data:Size of column 45 x 45cmConcrete mix M 25Characteristic strength of 415 N/mm%reinforcementFactored load 2500kN(characteristic loadmultiplied by yr)Factored moment 200 kN.mArraugement ofreinforcement: (a) On two sides(b) On four sides(Assume moment due to minimum eccentri-city to be less than the actual moment).Assuming 25 mm bars with 40 mm cover,d = 40 + 12.5 OP52.5 mm z 5.25 cmd’/D = 5.25145- 0.12Charts for d’/D = 0.15 will be usedf& = 25 x245500x x45103x lo2_ = 0.494200 x 106_ -25x45~45~45~103 = 0.088a) Reinforcement on two sides,Referring to Chart 33,p/fck = 0.09Percentage of reinforcement,p = 0.09 x 25 - 2.25As = p bD/lOO = 2.25 x 45 x 45/100= 45.56 cm2b) Reinforcement on four sidesfrom Chart 45,p&k = 0.10p p. 0.10 x 25 = 2.5As = 2.5 x 45 x 45/100 = 50.63 cm”Example 7 Circular Column with UniaxialBendingDetermine the reinforcement to be pro-vided in a circular column with the followingdata:Diameter of column 50 cmGrade of concrete M20Characteristic strength 250 N/mm2 forof reinforcement bars up to20 mm+240 N/mm2 forbars over20mm#COMPRESSION MEMBERS 103
  52. 52. Factored load 16OOkNFactored moment 125 kN.mLateral reinforcement :(a) Hoop reinforcement(b) Helical reinforcement(Assume moment due to minimum eccentri-city to be less than the actual moment).Assuming 25 mm bars with 40 mm cover,d’ = 40 x 12.5 = 52.5 mm m 5.25 cmd’/D - 5.25150 = 0.105Charts for d’/D = 0.10 will be used.(a) Column with hoop reinforcement1600 x 10320 x 50 x 50 x ioa- o’32125 x 1020 x 50 x 50 x 50 x 103- 0.05Referring to Chart 52, for fy I 250 N/mm1p/fck = o-87= 0.87 x 20 = 1.74A: = pnD2/400= 1.74 x nx50x50/400=34*16cm2Forf, I 240 N/mm2,AS = 34.16 x 2501240 = 35.58 cm2(b) Column with Helical ReinforcementAccording to 38.4 of the Code, the $rengthof a compression member with hehcal re--inforcement is 1.05 times the strength of asimilar member with lateral ties. Therefore,the, given load and moment should be dividedby 1.05 before referring to the chart.Hence,From Chart 52, for fy = 250 N/mm2,p,$_k= 0.078p = 0.078 x 20 = 1.56As = 1.56 x x x 50 x 50/44X= 30.63 cm2For fy = 240 N/mm%,A, = 30.63 x 2501240= 31.91 cm2According to 38.4.1 of the Code the ratioof the volume of helical reinforcement to thevolume of the core shall not be less than0.36 (A,/Ac - 1) fck Ify where A, is thegross area of the section and Ac is the areaof the core measured to the outside diameterof the helix. Assuming 8 mm dia bars for thehelix,Core_diie;; = 50-2 (4-O - 0.8)AI/AC IP 5O’/43*6’ = 1.3150.36 (A,,& - 1)falJlrI egg; 0.315 x 201250Volume of helical reinforcementVolume of coreAarc .(42*8) 0.09 A,J,--_------=;(43*6%) Q, a,where, Ash is the area of the bar formingthe helix and sh is the pitch of the helix.In order to satisfy the coda1 requirement,0.09 Art&k > O*OO91For 8 mm dia bar, Ati = O-503cm2sh ( 0.09 x 0.503’ 0.0091‘__ < 4.97 cm3.3 COMPRESSION MEMBERS SUB-JECT TO BIAXIAL BENDINGExact design of members subject to axialload and biaxial bending is extremelylaborious. Therefore, the Code permits thedesign of such members by the followingequation :lhereM,,, M,, are the moments about x and yaxes respectively due to design loads,M MUYl“Xl, are the maximum uniaxialmoment capacities with an axial loadP,, bending about x and y axes res-pectively, andozn is an exponent whose value depends onPu/Puz (see table below) wherePuz = 0.45 fck A, + 0*75fy As:PUIPUZ ‘ango.2 1.0)0*8 2.0For intermediate values, linear interpo-lation may be done. Chart 63 can be usedfor evaluating Puz.For different values of Pu/Puz, the appro-priate value of azn has been taken and curvesfor the. equation(!$)“’ + (z)=” = 1.0 have beenplotted in Chart 64.104 DESIGN AIDS FOR REINFORCED CONCRETE .
  53. 53. ExampIe 8 Rectangular colrmu, with BiaxialBe?tdi?lgDeWmine the reinforcement to be pro-vided in a short column subjected to biaxialbending, with the following data:size of columnConcrete mix EPcmCharacteristic strengthof reinforcement415N/mm’Factored load, P,, 1600kNFactored moment acting 120kNparallel to the largerdimension, M,wFactored moment acting 90 kNparallel to the shorterdimension, Mu,Moments due to minimum eccentricityareless than the values given above.Reinforcement is distributed equally onfour sides.As a iirst trial assume the reinforcementpercentage, p= 1.2P&k - 1*2/l5 - 0.08Uniaxial moment capacity of the sectionabout xx-axis:Referring to Churn64, the permissible v&aMBaofns,,qrmsponding to the above v&seequal to 0.58.The actual value of 0.617 is only sli&lyhigher than the value read from the Chart.This can be made up by slight increase inreinforcement.A6 - 1-2x40x60 _2&8,.&10012bars of 18mm will give A.130.53 c&Reinforcement percentage provided,p _ 30.53 x 1006o x40 - 1.27With this percentage, the section may berechecked as follows:p/f&- l-27115= 0.084 7Referring to Chart 44,d’/D5.25- 6. - 0.087 5Chart for d’/D = 0.1 will be used.p&k bD =1600 x 10s15 x 40 x 60 x 10”- 0444Referring to Chart 44, lM&k bD= = 0.09:. MUX,- 0.09 x 15 x 40 x 60’ x loylo~- 194.4kN.mUniaxial moment capacity of the sectionabout yv_axis:5.25d’JD = 40- 0.131Chart for d’/D - 0.15 will be used.Referring to Chart 45,M&k bD’ - 0.083:. MuYl - 0.083 x 15 x 60 x40*x 10a/lO’I 119.52kN.mCalculation of P,,:Referring to Chart 63 corresponding top = 1.2,fu= 415 andfck= 15,10.3 x 40 x108/10SkN2 472kN60Xf$- 0,095Mw z 0.;9; &li x 40x 60’x 10*/10*. .Referring to Chart 45f+- 0.085M WI z W&354x&52 60 x 40’ x 10a/lO’Referring to Char; 63,PUZAl = 10.4N/mm2Puz - 10.4 x 60 x 40 x lO’/lO~- 2 496kNReferring to Chart 64,Corresponding to the above values ofMuy PUMuY, and z’the permissible value ofMUX- is 0.6.MUX,Hence the section is O.K.COMPRESSION, MEMBERS 105
  54. 54. 3.4 SLENDER COMPRESSIONMEMBERS&?.When the slenderness ratio D or # ofa compression member exceeds 12, it isconsidered to be a slender compressionmember (see 24.2.2 of the Code); In and i,being the effective lengths with respect tothe major axis and minor axis respectively.When a compression member is slender withrespect to the major axis, an additionalmoment Mu given by the following equation(modified as indicated later) should betaken into account in the design (see 38.7.1of the Code) :Similarly, if the column is slender about theminor axis an additional moment M.,, shouldbe considered.M = Pub &”ay ( 12000 bThe expressions for the additional momentscan be written in the form of eccentricitiesof load, as follows:Mu - P,euwhereTable 1 gives the values b or 3 fordifferent values of slenderness ratio.TABLE I ADDITIONAL ECCENTRICITY FORSLENDER COMPRESSION MEMBERS(Chuxe 3.4)In accordance with 38.7.1.1 of the Code,the additional moments may be reduced bythe multiplying factor k given below:whereP,, = 0.45 &k Ac + 0.75 fy A, whichmay be obtained from Chart 63, and Pb is theaxial load corresponding to the condition ofmaximum compressive strain of 0.003 5in concrete and tensile strain of O%Ml2inoutermost layer of tension steel.Though this modification is optional ac-cording to the Code, it should always betaken advantage of, since the value of kcould be substantially less than unity.The value of Pb will depend on arrangementof reinforcement and the cover ratio d’/D,in addition to the grades of concrete andsteel. The values of the coefficients requiredfor evaluating Pb for various cases are givenin Table 60. The values given in Table 60are based on the same assumptions as formembers with axial load and uniaxial bending.The expression for k can be written asfollows :Chart 65 can be used for finding the ratioof k after calculating the ratios P,/Pu, andpb/&z.Example 9 Slender Column (with biaxialbending)Determine the reinforcement required fora column which is restrained against sway,with the following data:Size of column 40 x 30 cmConcrete grade M 30Characteristic strength 415 N/mm1of reinforcementEffective length for 6-Ombending parallel tolarger dimension, Z,Effective length for 5.0 mbending parallel toshorter dimension, lyUnsupported length 70mFactored load 1500kNFactored.moment in the 40 kN.m at topd!““;f larger and 22.5 kN.mat bottomDESIGN AIDS FOR RRINFORCED CONCRKI-E
  55. 55. Factored moment in the 30 kN.m at topdirection of shorterdimension~tdJOcN.xnThe column is bent in double curvature.Reinforcement will be distributed equallyon four sides.Lx 6-o x 100‘- PD 40 = 15.0 > 12cy=I 5.0 x 100b 30= 16-7 > 12Therefore the column is slender aboutboth the axes.From Table I,For Lz P 15, eJD = 0.113layFor b = 167, e,/b = O-140Additional moments:M1x= Puex= 1 500 x0-1 13 x & -67.8kN.mMay = Pue, = 1 500 x0*14x &=63*0 kN.mThe above moments will have to be reducedin accordance with 38.7.1.1 of the Code;but multiplication factors can be evaluatedonly if the reinforcement is known.For first trial, assumep p: 3.0 (with reinforce-ment equally on all the four sides).&-=40x 30= 1200cm2From Chart 63, Puz/As = 22.5 N/mm2- Pu. . = 22.5 x 1200 x 102/10s =2 700 kNCalculation of Pb :Assuming 25 mm dia bars with 40 mm coverd’/D (about xx-axis) cs g = 0.13Chart or Table for d’/d P O-15 will beused.5.25d’/D (about yy-axis) = 3. = 0.17Chart or Table for d’/d = 0.20 will beused.From Table 60,Pb (about xx-axis) = (k, + k2 &rbD30.196 + 0.203 X ox 30 x 30 x 40 x 102/103= -779 kNPb (about yy-axis) i 0.184 -+O-028x 330x40x30x30x 10*/1os.pby - 672 kN* k, I p -;b; I ‘;ym- 17r. .= oG5kPuz - Pu 2700-1500.y = p “7.- pby - 2 700 - 672= o-592The additional moments calculated earlier,will now be multiplied by the above valuesof k.= 67.8 x O-625= 42.4 kN.m:: = 63.0 x 0.592 - 37.3 kN.mThe additional moments due to slendernesseffects should be added to the initial momentsafter modifying the initial moments asfollows (see Note 1 under 38.7.2 of the Code) :M,,=(O*6 x 40 - 0.4 x 22.5) = 15-OkN.mKY= (0.6 x 30 - 0.4 x 20) = 10.0 kN.mThe above actual moments should be com-pared with those calculated from minimumeccentricity consideration (see 24.4 of theCode) and greater value is to be taken as theinitial moment for adding the additionalmoments.&+$-7g 40ex = - + 3o= 2*73cm= 2.4 cmBoth e, and e, are greater than 20 cm.Moments due to minimum eccentricity:Mux = 1 500 x ‘g = 41.0 kN.m> 15.0 kN.mMuy - 1500x2’4100= 36.0 kN.m> 10.0 kN.m:. Total moments for which the columnis to be designed are:MUX- 41.0 + 42.4 = 83.4 kN.miu Uy= 36.0 + 37.3 = 73.3 kN.mThe section is to be checked for biaxialbetiding.Pul& bD =1500 x 10s30x 30 x40 x 102= 0.417COMPRESSION MEMBERS 107
  56. 56. iPlfck - .g = 0.10:. MUX1= 0.104 x 30 x 30 x 40 x 40 x103/10’= 149.8 kN.mRefep; to ihart 46 (d’/D P O-20),v cka = 0,0962:. M”Yl =0*096 x 30 x 40 x 30’~ 30 x103/106= 103.7 kN.mM”, 83.4- ic - e 0.56M”,, 149.8MUY 73.3-M Ic -103.7 = 0.71UYlPulPu, = - 15002700= 0.56Referring to Chciit 64, the maximum allow-able value of M,,/M,,, corresponding to theabove values of M,,/M,,, and PuIPuz is 0.58which is slightly higher than’ the actual valueof 0.56. The assumed reinforcement of 3.0percent is therefore satisfactory.A s = pbD/lOO - 3.0 x 30 x 40/100L=36.0 cm2108
  57. 57. Y250415500-fck15202530Chart 63 VALUES OF Puz for COMPRESSION MEMBERSti i i i i i i i i i i i i i i i i i i IWi148 ._ DESIGN AIDS FOR REINFORCED CONCRETE .a
  58. 58. -.. .Chart64 BIAXIALBENDINGIN COMPRESSION MEMBERS0 0.1 O-2 O-3 0.4 0.5 O-6 O-7 0.8 0.9 1.0%/LCOMPRFSSIONMEMBERS
  59. 59. a 0404O-30*2o-10Chart 65 SLENDER COMPRESSION MEMBERS -Multiplying Factor k for Additional MomentsPk+ur-puPUZ-Pe150 DESIGN AIDS FOR REINFORCED CONCRl3-b
  60. 60. TABLE 60 SLENDER COMPRESSION MEMBERS-VALUES OF AlktMgdW_:Valora of k,&IDr005 WlO9015 OQQ0219 om7 01% 01840172 @MO 0149 0.138V@mof4#IDfr r N,‘mrn~ O-05 0.10 &lS 020COMPRESSIONMEMBERS 171
  61. 61. As in the Original Standard, this Page is Intentionally Left Blank
  62. 62. As in the Original Standard, this Page is Intentionally Left Blank
  63. 63. Y”4. SHEAR AND TORSION4.f DESIGN SHEAR STRENGTH OFCONCRETE*The design shear strength of concrete isgiven in Table 13 of the Code. The valuesgiven in the Code are based on the followingequation:whereg =0.8 fck/6*89pl, but not less than 1.0,and Pt = 100 A&&.The value of ‘F~corresponding to pl varyingfrom 0.20 to 3.00 at intervals of 0.10 are givenin Table 61 for different grades of concrete.4.2 NOMINAL SHEAR STRESSThe nominal shear stress 7” is calculatedby the following equation:VU7” = -bdwhereV,,is the shear force.When rv exceeds 7c, shear reinforcementshould be provided for carrying a shearequal to Vu- Q bd. The shear stress rv shouldnot in any case exceed the values of T~,~,given in Table J. (If T”> T~,~~, the sectionis to be redesigned.)TABLE J MAXIMUM SHEAR STRESS w,muCON- GRADE Ml5 M20 M25 M30 M35 M40Q., mu, N/mm’ 25 2% 3-l 3.5 3-l 404.3 SHEAR REINFORCEMENTThe design shear strength of verticalstirrups is given by the following equation:v _ @87f,A,vd“I -svwhereFor a series of inclined stirrups, the valueof Vup/d for vertical stirrups should bemultiplied by (since i- coscc) where cc isthe angle between the inclined stirrups andthe axis of the member. The multiplyingfactor works out to 1.41 and 1.37 for 45”and 60” angles respectively.For a bent up bar,VuI= 0*87fY ASvsinceValues of V,,, for different sizes of bars,bent up at 45” and 60” to the axis of themember are given in Table 63 for two gradesof steel.4.4 TORSIONSeparate Charts or Tables are not givenfor torsion. The method of design for torsionis based on the calculation of an equivalentshear force and an equivalent bendingmoment. After determining these, some ofthe Charts and Tables for shear and flexurecan be used. The method of design fortorsion is illustrated in Example 11.Example 10 ShearDetermine the shear reinforcement (verticalstirrups) required for a beam section withthe following data:Ream size ‘30 x 60 cmDepth of beam acrnConcrete grade M 15Characteristic strength 250 N/mmaof stirrup reinforcementTensile reinforcement 0.8percentageFactored shear force, Vu 180 kNAssuming 25 mm dia bars with 25 mm cover,d = 60 -T - 2.5 = 56.25 cmShear stress, 7” =i g -30 ~8p,$o,“,,= l-07 N/mm*A,” is the total cross sectional area of From Table J for M15, 7c,max= 2.5 N/mm2the vertical legs of the stirrups, and T” is less than 7c,musv is the spacing (pitch) of the stirrups.From Table 61, for P1=0.8, ~~20.55 N/mm*The shear strength expressed as Vu/d are givenin Table 62 for different diameters and Shear capacity of concrete section = Q bdspacings of stirrups, for two grades of steel. = 0*55x 30 x 56.25 x 102/103=92*8 kNSHEAR AND TORSION 175
  64. 64. Shear to be carried by stirrups, VU,==V,-~bd= 180 - 92.8 = 87.2 kNV”, 87.2-EL-- =d 56251.55 kN/cmReferring to Table 62, for steelf, -250 N!mme.Provide 8 mm diameter two legged verticalstirrups at 14 cm spacing.Example II TorsionDetermine the reinforcements required fora rectangular beam section with the followingdata :Size of the beam 30 Y 6OcmConcrete grade M 15Characteristic strength 415 N/mm2of steelFactored shear force 95 kNFactored torsional 45 kN.mmomentFactored bending moment 11S kN.mAssuming 25 mm dia bars with 25 mm cover,d = 60 - 2.5 - ‘G - 56.25 cmEquivalent shear,Vc = V-I- 1*6(f ,45595-t 1*6x m = 95-l-240 = 335 kNEquivalent shear stress.V, 335 x 101%e = Fd = 3. x 56.25 >rlo2 = 1.99 N/mm*From Table J, for M 15, ‘F~,,,,.~= 2.5 N/mm”~~~is less than sc.,,,-; hence the section doesnot require revision.From Table 61, for an assumed value ofpt = 0.5,T. = 0.46 N/mm* c T”=.Hence longitudinal and transverse reinforcc-ments are to be designed Longitudinalreinforcement (see 40.4.2 of the Code):Equivalent bending moment,Me,.= M,C Mt *= 115-J-79.4= 194.4 kN.mM,Jbd2 =194.4x 10”30 x (56.292 x 103 = 2.05 N/mmaReferring to Table I, corresponding toMujbdz .= 2.05PI = 0.708A,, = O-708 x 30 x 5625/100 = 1l-95 cm*Provide 4 bars of 20 mm dia (A*= 12.56 cm*)on the flexural tensile face. As Mt is lessthan MU,we need not consider Me, accordingto 40.4.2.1 of the Code. Therefore, provideonly two bars of 12 mm dia on the compres-sion face, one bar being at each corner.As the depth of the beam is more than45 cm, side face reinforcement of 0.05 percenton each side is to be provided (see 25.5.1.7and 25.5.1.3 of the Code). Providing onebar at the middle of each side,Spacing of bar = 53.412 = 267 cmArea required for each0.05 x 30 x 26.7bar= .- ,oo= 040 cm*Provide one bar of 12 mm dia on each side.Transverse reinforcement (see 40.4.3 of theCode) :Area of two legs of the stirrup should satisfythe following:1Y&6cm-m-W---30 cm-b, = 23 cmkI76C1d, 953-4cm1cm,-FLEXURALTENSIONFACE*176 DESIGN AIDS FOR REINFORCED CONCREFE
  65. 65. Assuming diameter of stirrups as 10mmda = 60 - (2.5 + l-O)-(2*5+0%)-53.4 cmb1=30-2(25+1.0)=23cmAav(0*87&J 45 x 10’S” -23 x 53.4 x lOa+25 x9553.4x IO?x 10-366.4-l-71.2= 437.6N/mmP 438 kN/cmArea of all the legs of the stirrup shouldsatisfy the condition that A& should notFrom Table 61, for tensile reinforcementpercentage of @71, the value of o is O-53N/mm’- (I.99 - 0.53)30 x 10II 438N/mm -438 kN/cmNom-It is only a coincidence that the values ofAav(@87/rllSv cdcuhted by the hvo cqlm-tions 8rc the srmc.Referring T&e 62 (forfr - 415 N/mm’).Provide 10 mm + two legged stirrups at12.5 cm spacing.According to 25.5.2.7(u) of the Code, thespacing of stirrupa shall not exceed xl,(x, C yJ4 and 300 mm, where x1 andarc the short and long dimensions of1tLstirrup.xl - 30 - 2(2.5 - O-5)= 26 cmy,r60-2(25-05)=56cm(xl f y&/4- (26 i- 56)/4- 20.5cm10 mm + two legged stirru at 12.5 cmspacing willsatisfyall the cod$ requirements.SHEAR AND TORSION 177
  66. 66. fck152025303540-ptTABLE 61 SHEAR - DESIGN SHEAR STRENGTH OF CONCRETE, TC)N/mm2fck, N/mm*I - --.,IS 20 25 30 35 400.80%1.101.201.301’401.501*601.70:z2.002102202.30f %EE3.00032 0.33 0.33 0.33 0.34 0340.38 0.39 0.39 0.40 0.40 0.410.43 8:Z 0.45 045 046 0460.46 0.49 0.50 0.50 0.510.50 0.51 0.53 0.54 0.54 0.550.53 0.55 056 0.57 0.58 0590.55 0.57 0.59 iE 0.61 0620.57 8:Z 0.62O%O 0.64 0668:; 0.650.680.62 ~~~ 0.66 0.68 069 0.700.63 0.69 0.70 0.72 0.73065 0.68 0.71 0.72 0.74 0.750.67 0.70 072 0.74 076 0.77068 072 0.74 076 0.78 0.790.69 073 076 0.78 0.80 0.810.71 0.75 0.77 0.80 0.81 0.830.71 076 0.79 0.81 0.83 0.850.71 077 0.80 0.83 085 0860.71 0.79 0.82 084 0.86 0880.71 0.80 0.83 086 0.88 090071 0.81 0.84 087 0.89 0.910.71 0.82 0.86 088 0.91 0.930.71 082 0.87 092 0.940.71 0.82 0.88 ;I: 0.93 0.950.71 0.82 X:E 0.92 094 0970.71 0.82 0.93 0.96 0.98x::: @820.82 0910.92 0.940.95 0970.98 0.99l*oO0.71 0.82 0.92 0.96 0.99 1.01178 DESIGN AIDS FGR RHNFGRCBD CGNCmB
  67. 67. v250415STIRRUPSPmNO,em5,61537l-367-1.230TABLE 62 SHEAR -VERTICAL STIRRUPSValues of VW/dfor two legged stirrups, kN/cm.J, = 250 N/mm2DIAMEIZR,mm8 104373 683336443.124 6E2733 42712429 3.7962186 3,4161.988 3.1061.822 28471682 26281.5621.458 ::z::E213520101.215 1.8981.151 1.7981.093 1.7080.875 1.3670.729 1.1390625 0.9760.547 0.8540.486 0.759-712447241003.784Et3.0752894273325892460/x = 415 N/mm*DIAMETER,mm’ 6 8 1040833403t:;::22692042I.8561.7011.5711.4581.3611.2761.2011.1341.0751.0200.81706810’5830.5100.4547.259 11.342f :% 94528.1024537 70894.033 63023.630 5.6713.299 5.1563.025 47262792 43632593 40512420 3.7812269 3.5452135 3.336:8:83.151. 29851.815 2.836l-4521.210 :z1.037 1.6200907 1.4180.807 1.26074246806L%54455.1044804t:;:;:4083TABLE 63 SHEAR- BENT-UP BARSValues of Vu, for singal bar, kNBAR /i - 250 N/mm* fy = 415 N/mm2Dm, I , Imm a = 45’ a=60° Qi= 45”>a=60°:; 12081739 21.301479 20.0528.87 245635.36:B” 30.923914 47.9337.87 51.3364.97 %E20 48.32 5918 SO.21 98.235846 716075.49 924694.70 115.98123.69 151.4915654 191.7397.05_. __125.3215720205.3225986118.86:;;*z.251.47318-27NOTE- a is the angle between the bent-up bar and the axis of the member.8HEAR AND TOIlsION 179
  68. 68. As in the Original Standard, this Page is Intentionally Left Blank
  69. 69. As in the Original Standard, this Page is Intentionally Left Blank

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