Worked Examples for the Design of Concrete Buildings - ARUP

WORKED 1116 EXAMPLES OF CONCRETE oil BUILDING r0= FOR THE DESIGN S Based on BSI publication DD ENV 1992-1-1:1992. Eurocode 2: Design of concrete structures. Part 1. General rules and rules for buildings. ..: I This book of worked examples has been prepared by: British Cement Association Ove Arup & Partners E06 and S. B. Tietz & Partners The work was monitored by the principal authors: A. W. Beeby BSc, PhD, CEng, MICE, MIStructE, FACI Professor of Structural Design, Dept of Civil Engineering, University of Leeds (formerly Director of Design and Construction, British Cement Association), .L-- c O-0 c0) Partner, S. B. Tietz & Partners, Consulting Engineers, and R. Whittle MA(Cantab), CEng, MICE Associate Director, Ove Arup & Partners, and edited by: A. J. Threlfall BEng, DIC Consultant (formerly a Principal Engineer at the British Cement Association). This publication was jointly funded by the British Cement Association and the Department of the Environment to promote and assist the use of DD ENV 1992-1-1 Eurocode 2: Part 1. aa)) a.- R. S. Narayanan (p- BE(Hons), MSc, DIC, CEng, FIStructE foil The British Cement Association, BCA, is a research and information body dedicated to furthering the efficient and proper design and execution of concrete construction. Membership of BCNs Centre for Concrete Information is open to all involved in the construction process. BCA is funded by subscriptions from cement producers, through joint ventures, sales of publications, information and training courses, and the carrying out of research contracts. Full details are available from the Centre for Concrete Information, British Cement Association, Century House, Telford Avenue, Crowthorne, Berkshire RG11 6YS. Telephone (0344) 725700, Fax (0344) 727202. Qom. `c_ cio woo 2-0 .-" U)- ?l. Ove Arup industry. ((D & Partners is an international firm offering a wide range of design and specialist services for the construction N:3 S. B. Tietz & Partners offer consultancy services A catalogue and prices for BCA publications can be obtained from Publication Sales, Centre for Concrete Information, Qom) at the above address. 43.505 First published 1994 ISBN 0 7210 1446 1 pin ,.: in civil, structural and traffic engineering. c Published by British Cement Association Century House, Telford Avenue, Crowthorne, Berks RG11 6YS Telephone (0344) 762676 Fax (0344) 761214 From 15 April 1995 the STD Code will be (01344) °J2 Price group M © British Cement Association 1994 advice or information from the British Cement Association is intended for those who will evaluate the significance and limitations of its contents and take responsibility for its use and application. No liability (including that for negligence) for any loss resulting from such advice or information is accepted. Readers should note that all BCA publications are subject to revision from time to time and should therefore ensure that they are in possession of the latest version. 50" m-0 All (0m Lma Nod ..E odd 0.9 FOR THE DESIGN OF CONCRETE BUILD 1 N G S Based on BSI publication DD ENV 1992-1-1:1992. Eurocode 2: Design of concrete structures. Part 1. General rules and rules for buildings. Published by the British Cement Association in conjunction with: Ove Arup 13 & Partners Fitzroy Street London W1 P 6BQ Tel: 071-636 1531 S.B. Tietz & Partners 14 Clerkenwell Close Clerkenwell London ECiR OPQ Tel: 071-490 5050 The Department of the Environment 2 Marsham Street London SW1P 3EB Tel: 071-276 3000 July 1994 FOREWORD Eurocode 2: Design of concrete structures, Part 1: General rules and rules for buildings (EC2)(') sets out both the principles for the design of all types of concrete structure, and design rules for buildings. Rules for other types of structure and particular areas of technology, including precast concrete elements and structures, will be covered in other parts of EC2. x)00 Q(° .-. EC2 contains a considerable number of parameters for which only indicative values are given. The appropriate values for use in the UK are set out in the National Application Document (NAD)(') which has been drafted by BSI. The NAD also includes a number of amendments to the rules in EC2 where, in the draft for development stage of EC2, it was decided that the EC2 rules either did not apply, or were incomplete. Two such areas are the design for fire resistance and the provision of ties, where the NAD states that the rules in BS 8110(2) should -CO be applied. Attention is drawn to Approved Document A (Structure) related to the Building Regulations 1991(3), which states that Eurocode 2, including the National Application Document, is considered to provide appropriate guidance for the design of concrete buildings in the United Kingdom. Enquiries of a technical nature concerning these worked examples may be addressed to the authors directly, or through the BCA, or to the Building Research Establishment. (0^D 3>a) vii 'Fn -C-0 ca) CONTENTS 1 1.1 1.2 INTRODUCTION AND SYMBOLS 5 Introduction 5 Symbols ................. .................... ................ 8 8.1 8.2 SPECIAL DETAILS Corbels Nibs 8.3 180 ..................... 185 188 Simply supported ends ................... ...... ........ 2 2.1 COMPLETE DESIGN EXAMPLE 15 Introduction Basic details of structure, 8.4 9 Surface reinforcement 191 2.2 2.3 materials and loading 2.4 2.5 2.6 2.7 2.8 2.9 3 3.1 Floor slab Main beam Edge beam (interior span) Columns Foundation Shear wall Staircase .................. ................. ..... ................... ................. ................. .................. ................ ..... ......... PRESTRESSED CONCRETE Introduction Design data Serviceability limit state Ultimate limit state Mi n i 15 17 9.1 9.2 9.3 9.4 20 30 34 39 43 49 ............... ............... ...... .......... 193 193 195 204 9.5 9.6 mum and max mum areas of 207 reinforcement 3.2 3.3 3.4 3.5 4 4.1 4.2 5 5.1 5.2 L(7 (D. 5.3 5.4 5.5 5.6 5.7 CON- .............. ..... i BEAMS Introduction 53 Design methods for shear 53 Shear resistance with concentrated loads close to support 63 70 Design method for torsion Slenderness limits 81 SLABS Solid and ribbed slabs Flat slabs COLUMNS Introduction Capacity check of a section by strain compatibility Biaxial bending capacity of a section Braced slender column Slender column with biaxial bending Classification of structure Sway structures 6 6.1 WALLS Introduction Example ca) 10 10.1 10.2 Cracking (fl Reinforcement summary 207 SERVICEABILITY CHECKS BY CALCULATION Deflection ................. 208 .p° .................. 219 ........ ..... ........... ....... DEEP BEAMS Introduction 11.2 Example 11 11.1 12 ................. 82 109 12.1 12.2 12.3 ............... .......... .................. 132 12.4 132 12.5 137 141 LOAD COMBINATIONS Introduction Example 1 frame Example 2 continuous beam 1 Example 3 continuous beam 2 Example 4 WOE ............... .................. ............... .................. ....... ....... 222 222 236 237 COY) - - 240 243 ...... - tank ................... 245 13 .................. ..... ............ 143 147 151 13.1 13.2 13.3 13.4 13.5 6.2 ............... .................. ............. ..... 154 154 DESIGN OF BEAM AND COLUMN SECTIONS Concrete grades 246 Singly reinforced rectangular beam sections 246 Compression reinforcement ... 248 Flanged beams 249 Symmetrically reinforced rectangular columns 249 7 7.1 7.2 FOUNDATIONS Ground bearing footings Pilecap design 158 172 REFERENCES ........... .............. ............ ........ ................. -fl 256 :-I INTRODUCTION AND SYMBOLS 1.1 Introduction and symbols The main objective of this publication is to illustrate through worked examples how EC20) may be used in practice. It has been prepared for engineers who are generally familiar with design practice in the UK, particularly to BS 8110(2). The worked examples relate primarily to in-situ concrete building structures. The designs are in accordance with EC2: Part 1 as modified by the UK National Application Document'). Where necessary, the information given in EC2 has been supplemented by guidance taken from other documents. The core example, in Section 2, is a re-design of the in-situ concrete office block used in the BCA publication Designed and detailed (SS 8110: 1985), by Higgins & Rogers(). Other design aspects and forms of construction are fully explored by means of further examples in Sections 3 to 12. CAD °°6 Equations and charts for the design of beam and column sections, taken from the Concise Eurocode for the design of concrete buildings(5), are given in Section 13. Publications used in the preparation of this book, and from which further information may be obtained, are listed in the References. Unless otherwise stated, all references to BS 8110 refer to Part 1. wow >-o ((DD (Q, moo- W-0 -CD been adopted in the preparation of this book. Statements followed by OK' mark places where the calculated value is shown to be satisfactory, Green type is used to draw attention to key information such as the reinforcement to be provided. Two conventions have '....... The calculations are cross-referenced to the relevant clauses in EC2 and, where appropriate, to other documents; all references in the right-hand margins are to EC2 unless indicated otherwise. The symbols used throughout the publication are listed and defined below, and are generally those used in EC2 itself. -pp 1.2 Symbols A Ac Act Act,ext Area of cross-section Area of concrete cross-section Area of concrete within tensile zone Area of concrete tensile zone external to links Ak AP Area enclosed within centre-line of thin-walled section Area of prestressing tendons Area of tension or, in As columns, total longitudinal reinforcement E A's As,min Area of compression reinforcement Minimum area of tension or, in columns, total longitudinal reinforcement As, prov As,feq As,surf Area of tension reinforcement provided Area of tension reinforcement required Area of surface reinforcement Ast Area of transverse reinforcement within flange of beam Area of tension reinforcement effective at a section of additional longitudinal reinforcement E ivy CD- (OD ((DD J7- Coa c>3 L_i 0;Y Qom) tea) 0 E E __0 As/ Asw Asw,min or, for torsion, area Area of shear reinforcement or torsion links Minimum area of shear reinforcement Effective modulus of elasticity of concrete E.,eff INTRODUCTION AND SYMBOLS Er E ,, ES Secant modulus of elasticity of concrete at transfer Secant modulus of elasticity of concrete Modulus of elasticity of reinforcement or prestressing steel Force due to concrete in compression at ultimate limit state F FS Force in tension reinforcement or prestressing tendons at ultimate limit state FSd Design value of tie force in pilecap Design value of support reaction Tie force in corbel or due to accidental action Fsd,sup Ft F Gk Gkf Vertical force applied to corbel or, for sway classification of structures, sum of all vertical loads under service conditions Characteristic value of permanent action or dead load Characteristic dead floor load Characteristic dead roof load Overall depth of tank Horizontal force applied to corbel .., --E Gkr H Hc I It Second moment of area of cross-section Second moment Second moment Second moment E E E E E E of of of of of of area of uncracked concrete section area of cracked concrete section area of beam section area of concrete section Iu Ib Second moment I.., slab Second moment Second moment I v J Jtot Second moment of Second moment of area area area area of column section of slab section of section in x direction of section in y direction St Venant torsional stiffness of rectangular section St Venant torsional stiffness of total section K K, K2 Deflection-curvaturefactor dependent upon the shape of the bending moment diagram Reduction factor for calculation of second order eccentricity Coefficient taking account of decrease force CD. in curvature due to increasing axial M Mc Mc, MCX Bending moment Moment of force, FC, about tension reinforcement Moment causing cracking Moment of force, Nc, about x axis MCy Moment of force, N., about y axis First order moment M. MRd MRd,c Design moment of resistance Moment of force, NRd,c' about mid-depth of section M/Rd,c MRd,s Moment of force, N'Rd,c' about mid-depth of section Moment of force, NRd,s' about mid-depth of section 6 INTRODUCTION AND SYMBOLS MSd Design value of applied moment Design moment in x direction Msdx MSdy MShc Design moment in y direction First order moment at end (1) On. 1 Msd2 Msd,cs MSd,ms First order moment at end 2 Design moment in column strip Design moment in middle strip Mspan Moment in span M SUP Moment at support Mt",,. MX E Maximum moment transfer value Moment about x axis Moment about y axis Axial force M N y N NRd NRd,c Axial force due to concrete in compression Design resistance to axial force Design resistance to axial force due to concrete Design resistance to axial force due to concrete of hypothetical section of depth x > h N'Rd,c Design resistance to axial force due to reinforcement Design value of applied axial force Mean applied axial force Prestressing force or point load Average prestressing force along tendon profile Initial prestressing force at transfer Mean effective prestressing force at time t Final prestressing force after all losses Maximum initial prestressing force at active end of tendon Required prestressing force Final prestressing force at service Characteristic value of variable action or imposed load Characteristic value of imposed floor load Characteristic value of imposed roof load Reaction at support A Reaction at support B First moment of area of reinforcement about centroid of section First moment of area of reinforcement about centroid of uncracked section First moment of area of reinforcement about centroid of cracked section Design value of tensile force in longitudinal reinforcement Maximum torsional moment resisted by concrete struts Maximum torsional moment resisted by reinforcement INTRODUCTION AND SYMBOLS Tsd Tsd,e Tsd,tot _`L Design value of applied torsional moment Torsional moment applied to flange Total applied torsional moment Tsd VA W Torsional moment applied to web Shear force at support A VB Vod Vex, Shear force at support B Design shear resistance provided by concrete Shear force at exterior support Vnt VRdl VRd2 VRd3 Shear force at interior support Design shear resistance of member without shear reinforcement Maximum design shear force to avoid crushing of notional concrete struts Design shear resistance of member with shear reinforcement Design value of applied shear force Design shear force in x direction Design shear force in y direction COD COD COD Vsd Vsdx Vsdy Vsd,max Maximum design shear force Design shear resistance provided by shear reinforcement V,d Wb Section modulus at bottom fibre Wcp Wk Section modulus at centroid of tendons Characteristic value of wind load Section modulus at top fibre Wt a al all ac acs a, 0-. Distance or deflection or maximum drape of tendon profile Deflection based on uncracked section Deflection based on cracked section Distance of load from face of support (corbel) or from centre-line of hanger bars (nib) Deflection due to concrete shrinkage Distance from face of support to effective centre of bearing COD COD al ato, av ax 0-_ Horizontal displacement of the envelope line of tensile force Total deflection Distance between positions of zero and maximum bending COD Deflection at distance x along span aj,a2 Values of ai at ends of span Width of section or flange width or lateral cover in plane of lap b bav be betf bmin Average width of trapezoidal compression zone Width of effective moment transfer strip Effective width of flange Minimum width of support beam Width of rib Width of support br bsup COD INTRODUCTION AND SYMBOLS bt bW Mean width of section over the tension zone Minimum width of section over the effective depth Cover to longitudinal torsion reinforcement c ct,c2 Support widths at ends of beam Effective depth of section d d' day Depth to compression reinforcement Average effective depth for both directions db dcnt c"'- Depth to bar considered Distance of critical section for punching shear from centroid of column Effective depth of flange Effective depth for punching shear check in column head df dm dm. dmin 'PL Maximum effective depth for both directions Minimum effective depth for both directions Effective depth in x direction Effective depth in y direction dX dy d, d2 ea eay Effective depth to bars in layer 1 Effective depth to bars in layer 2 Additional eccentricity due to geometrical imperfections Additional eccentricity in the y direction Additional eccentricity in the z direction Q1) eaz ee eoy Equivalent eccentricity at critical section First order eccentricity in y direction First order eccentricities at ends of column Total eccentricity eot,eo2 etot ey eZ Eccentricity in y direction Eccentricity in z direction e2 e2y Second order eccentricity Second order eccentricity in y direction Second order eccentricity in z direction Stress in concrete at bottom fibre e2, fb fbd Design value of ultimate bond stress Design cylinder strength of concrete cd f t k Cube strength of concrete at transfer Characteristic cylinder strength of concrete ct,eff Effective tensile strength of concrete at time cracking is expected to occur f tm 1 fpd Mean value of axial tensile strength of concrete Characteristic cube strength of concrete Design tensile strength of prestressing steel fk Characteristic tensile strength of prestressing steel INTRODUCTION AND SYMBOLS Design value of ultimate bearing stress Stress in reinforcement Stress in concrete at top fibre Design yield strength of reinforcement Characteristic yield strength of reinforcement Design yield strength of longitudinal torsion reinforcement Design yield strength of shear reinforcement or torsion links Characteristic yield strength of shear reinforcement or torsion links Characteristic dead load per unit area Overall depth of section or liquid in tank Reduced value of h for separate check about minor axis with biaxial eccentricities ha of column section Active height of deep beam Overall depth of corbel at face of support Overall depth of flange hC hf hH Depth of column head Larger dimension of rectangular section hmax hmin htot Smaller dimension of rectangular section Total height of structure in metres i k kA kB kbottom ko Radius of gyration of section Coefficient or factor Restraint coefficient at end A Restraint coefficient at end B Restraint coefficient at bottom Minimum reinforcement coefficient associated with stress distribution Restraint coefficient at top ktop ki Crack spacing coefficient associated with bond characteristics k2 Crack spacing coefficient associated with strain distribution Length or span Length of tendon over which anchorage slip is taken up l if Ib Ib,min Basic anchorage length Minimum anchorage length Ib,net IC Required anchorage length Diameter of circular column Height of column between centres of restraints Effective span Effective span of slab Icon left leff,slab lH In Distance from column face to edge of column head Clear distance between faces of support 10 INTRODUCTION AND SYMBOLS to Distance between positions of zero bending or effective height of column or, for deep beams, clear distance between faces of support Length of compression flange between lateral supports lot is ls,min Required lap length or floor to ceiling height in metres Minimum lap length It Greater of distances in metres between centres of columns, frames or walls supporting any two adjacent floor spans in direction of tie under consideration Effective span in x direction Effective span in y direction COD lX ly 1,,12 Lengths between centres of supports or overall dimensions of rectangular column head c(0 u0) mSd Minimum design moment per unit width n Ultimate design load per unit area or number of tendons or number of sub-divisions p' q qk Average loss of prestressing force per unit length due to friction Equivalent load per unit length due to prestressing force profile ... Characteristic imposed load per unit area Radius of bend or radius of curvature Radius of curvature based on uncracked section Radius of curvature based on cracked section Radius of curvature due to concrete shrinkage Radius of curvature due to concrete shrinkage based on uncracked section Radius of curvature due to concrete shrinkage based on cracked section Total radius of curvature (DD r rI rH r, rcsu riot n"" r.I s sf Smax Srm Spacing of shear reinforcement or torsion links or horizontal length of tendon profile Spacing of transverse reinforcement within flange of beam Maximum spacing of shear reinforcement or torsion links Average final crack spacing Thickness of supporting element or wall of thin-walled section Minimum thickness of wall t train u uk VRdl Circumference of concrete section or critical section for punching shear Circumference of area Ak Design shear resistance per unit length shear reinforcement 0)8 of critical perimeter, for '-. r-. -f6 VRd3 VSd Design value of shear force per unit length of critical perimeter w wk Support width or quasi-permanent load per unit length Design crack width 0 CAD '23 slab without VRd2 Maximum design shear resistance per unit length of critical perimeter, for slab with shear reinforcement Design shear resistance per unit length shear reinforcement of critical perimeter, for slab with INTRODUCTION AND SYMBOLS wmin Minimum width of support xc Depth of concrete in compression at position of minor axis for column section with biaxial eccentricities y yt Drape of tendon at distance x along profile or column dimension direction Distance from centroid of uncracked section to extreme tension fibre Lever arm of internal forces 07Q x' Maximum depth of concrete in compression in direction of minor axis for column section with biaxial eccentricities 5'3 z ZCP an imperfections aWasy Moment coefficients in x and y directions Effectiveness coefficient for lap a, a Coefficient with several applications including shear resistance enhancement, effective height of column, St Venant torsional stiffness, punching shear magnification, design crack width ared Reduced value '-' CC)) yG,inf yG,sup Partial safety factor for permanent action, in calculating upper design value yP To 'YS Partial safety factor for actions associated with prestressing force Partial safety factor for variable action or imposed load Partial safety factor for steel material properties of reinforcement or prestressing tendons Ratio of redistributed moment to moment before redistribution Strain in concrete at bottom of section S Eb ECS Basic concrete shrinkage strain Final concrete shrinkage strain EeS- EP Minimum strain in tendons to achieve design tensile strength Strain in tendons corresponding to prestressing force Pm,t EPm o E 'C$ x Neutral axis depth or distance along span from face along tendon or column dimension in x direction of support or distance in y Distance from centroid of section to centroid of tendons a aI Reduction factor for concrete compressive stress or modular ratio or deformation parameter 0 Value of parameter based on uncracked section Value of parameter based on cracked section au as ae Effectiveness coefficient for anchorage Effective modular ratio Reduction coefficient for assumed inclination of structure due to of shear resistance enhancement coefficient a, a2 -y' Coefficient associated with bond characteristics Coefficient associated with duration of load Partial safety factor for concrete material properties yF TG Partial safety factor for actions Partial safety factor for permanent action or dead load Partial safety factor for permanent action, in calculating lower design value u°' INTRODUCTION AND SYMBOLS ES Strain in reinforcement es(t,t) esm EU Estimated concrete shrinkage strain Mean strain in reinforcement allowing for tension stiffening effect of concrete Ultimate compressive strain in concrete Initial yield strain in reinforcement Eyd Distribution coefficient n Moment coefficient 6 X Xcrit Xm Xmin Angle of rotation or angle between concrete struts and longitudinal axis Slenderness ratio Critical slenderness ratio Mean slenderness ratio of all columns in storey considered Slenderness ratio beyond which column is considered slender Coefficient of friction between tendon and duct or applied moment ratio Limiting value of applied moment ratio for singly reinforced section (OD µ µlim P Efficiency factor or assumed inclination of structure due to imperfections vred YU Reduced value of assumed inclination Longitudinal force coefficient ate) of structure P Tension reinforcement ratio or density of liquid P' PI Compression reinforcement ratio Longitudinal tension reinforcement ratio Longitudinal tension reinforcement ratios in x and y directions Effective reinforcement ratio (n. PIX'PIy Pr Pw Pw,min Shear reinforcement ratio Minimum shear reinforcement ratio P1,P2 UC9 b0) Principal and secondary reinforcement ratios in solid slabs Stress in concrete adjacent to tendons due to self-weight and any other permanent actions Average stress in concrete due to axial force Initial stress in concrete adjacent to tendons due to prestress app acpo upo stress in tendons immediately after stressing (pre-tensioning) or immediately after transfer (post-tensioning) Initial as asr 7' TRd Stress in tension reinforcement calculated on basis of cracked section Value of as under loading conditions causing first cracking Basic design shear strength Factor defining representative value of variable action ,yo Value of Value of for rare load combination for frequent loading 0t 02 W Value of 0 for quasi-permanent loading Mechanical ratio of tension reinforcement Mechanical ratio of compression reinforcement W' (On 00.. INTRODUCTION AND SYMBOLS wl.m Limiting value of w for singly reinforced section Total vertical force E. applied to frame at floor A,/ AFd OHM Anchorage slip or wedge set Variation of longitudinal force in section of flange over distance Equivalent horizontal force acting on frame at floor imperfections Moment of force cctea) c,) j a, j due to assumed o-° AMRd,c ANRd,c about mid-depth of section AMSd ANRd,c Reduction in design moment at support Design resistance to axial force due to concrete in area of hypothetical section lying outside actual section Average loss of prestressing force due to elastic deformation of concrete Loss of prestressing force at active end of tendon due to anchorage slip Loss of prestressing force due to creep, shrinkage and relaxation at time t Loss of prestressing force due to friction between tendon and duct at distance x from active end of tendon Variation of stress in tendon due to relaxation Bar size or duct diameter or creep coefficient ((DD -_p APC APst APt(t) APA(x) DTT 0o pr 0(t,to) Creep coefficient, defining creep between times t and to, related to elastic deformation at 28 days 0(-,td Final creep coefficient 1 2 COMPLETE DESIGN EXAMPLE 2.1 Introduction Design calculations for the main elements of a simple in-situ concrete office block are set out. The structure chosen is the same as that used in Higgins and Rogers' Designed and detailed (BS 8110: 1985)t4i. Calculations are, wherever possible, given in the same order as those in Higgins and Rogers enabling a direct comparison to be made between BS 8110(2) and EC20) designs. For the same reason, a concrete grade C32/40 is used. This is not a standard grade recognized by EC2 or ENV 200), which gives grade C35/45 in Table NA.1. Some interpolation of the tables in EC2 has, therefore, been necessary. ma` +-O (Nn 0)- 0)- The example was deliberately chosen to be simple and to cover a considerable range of member types. Comparison shows that, for this type of simple structure, there is very little difference between BS 8110 and EC2 in the complexity of calculation necessary or the results obtained. 2.2 Basic details of structure, materials and loading These are summarized in Table Table 2.1 2.1 'Or and Figure 2.1. Design information Intended use Laboratory and office block Fire resistance 1 hour for all elements .r. Loading (excluding self-weight of structure) Roof - - imposed (kN/m2) finishes (kN/m2) ("O CAD (CA E E NAD Table 5 E (OD 2.3.6 Shear Shear resistance VRd1 of the slab without shear reinforcement is given by + 40pl) bwd (OD 4.3.2.3 = TRdk (1.2 Eqn 4.18 where TRd = 0.35 N/mm2 Table 4.8 E k = 1.6 - d = 1.6 = 0.149 = 1.451 0. 0038 565 Pi 1000 x 149 Hence VRd1 = 102.3 kN/m > Vsd = 33.9 kN/m ...... ......... OK No shear reinforcement required Note: Since shear is rarely a problem for normally loaded solid slabs supported on beams, as the calculation has shown, it is not usually necessary to check in these instances. COD COMPLETE DESIGN EXAMPLE 2.3.7 Deflection Reinforcement ratio provided in span = 377 1000 x 149 = 0.0025 Using NAD Table 70) and interpolating between 48 for 0.15% and 35 for 0.5%, a basic span/effective depth ratio of 44 is given. By modifying according to the steel stress, the ratio becomes '2O _.a NAD Table 7 4.4.3.2(4) 44 (400 x 377) 460 x 342 = (OD 42.2 The actual spanleffective depth ratio is 5000 149 = 33.6 ........... OK Had EC2 Table 4.14 been used instead of NAD Table 7, the basic ratio before modification would have been 35, which would not have been OK. 2.3.8 Cracking For minimum area of reinforcement assume fot,ett 4.4.2.2 = 3 N/mm2 kC k Act = = = 0.4 0.8 0.5 x 175 x 1000 = 87500 mm2 E Hence As = = kckfct ettA ct/as Eq n 4.78 E 0.4 x 0.8 x 3 x 87500/460 x = E = 183 mm2/m x Area of reinforcement provided No further check is necessary as h Maximum bar spacing x k 377 mm2/m = 175 (1 - yk/805) = 0.429 Increase w' to (1 1 - 0.429) 0.546 0.075 = 0.0943 A' S = 0.0943 x 300 x 440 x 32/460 E = 866 mm2 E Use 4T25 (1960 mm2) top Use 2T25 (982 mm2) bottom COMPLETE DESIGN EXAMPLE 2.4.4.2 Near middle of 8 m span From bending moment envelope M = 325 kNm Effective flange width = 300 + 0.2 x 0.85 x 8000 = 1660 mm 2.5.2.2.1 E E Eqn 2.13 µ = 325 x 106 1660 x 4502 x 32 = 0.030 x/d = 0.068 (Section 13, Table 13.1) Neutral axis is in flange since x w = 31 < 175 mm = 0.035 (Section 13, Table 13.1) AS = 0.035 x 1660 x 450 x (Y) 32/460 = 1819 mm2 E Use 4T25 (1960 mm2) E 2.4.4.3 Left-hand end of 8 m span From bending moment envelope M S = = 126 kNm 0.7 and µ,.m = 0.0864 (Section 13, Table 13.2) 106 It _ 300 126 x x 4402 x 32 = 0.0678 < µlim Therefore no compression reinforcement is required. W = 0.084 (Section 13, Table 13.1) = 0.084 x 300 x 440 x 32/460 AS = 772 mm2 Using 2T25 bent-up bars, minimum diameter of mandrel 5.2.1.2 NAD = 130 (As,regIA,.prov) = 100 Table 8 Use 2T25 (982 mm2) with r = 50 24 COMPLETE DESIGN EXAMPLE 2.4.4.4 Right-hand end of 6 m span From bending moment envelope M = 76kNm M 76 x 106 A _ = 0.041 bd2 ck W 300 x 4402 x 32 = 0.049 (Section 13, Table 13.1) 450 mm2 = AS = Use 2T25 (982 mm2) with r E 4¢ minimum 2.4.4.5 Near middle of 6 m span From bending moment envelope M = 138 kNm Effective flange width = 138 300 + 0.2 x 106 x 0.85 x 6000 = 1320 mm E u = = 1320 0.0161 32 x 4502 x W 0.019 (Section 13, Table 13.1) As = 0.019 x 1320 x 450 x 32/460 = 785 mm2 Use 2T25 (982 mm 2) 2.4.4.6 Minimum reinforcement As kc kfct,eff A 07c ct /a s 4.4.2.2 Eq n 4.78 where kC = = = 0.4 k ct.eff 0.68 3 N/mm2 Act or S = 300 = x 325 mm2 E E CY) 460 N/mm2 Therefore AS 0.6b d t >_ 173 mm2 .................................... = 203 mm2 OK 0.0015 btd _o' 4 ....................... OK 5.4.2.1.1(1) fYk COMPLETE DESIGN EXAMPLE 2.4.5 Shear reinforcement 2.4.5.1 Minimum links Here, for comparison with BS 8110 design, grade 250 reinforcement will be 4.3.2 5.4.2.2 used. Interpolation from EC2 Table 5.5 gives Minimum pW = = 0.0022 0.0022 A3W Is If Vsd x 300 = 0.66 mm2/mm E VRd2 < = ( 5) VRd2 - refer to Section 2.4.5.3 for smax lesser of 300 mm or 0.8d = 300 mm = 0.75 mm2/mm) Eqn 5.17 Use R12 links @ 300 mm crs. (AsW/s 2.4.5.2 Capacity of section without shear reinforcement VRd, 0 4.3.2.3 = 7Rdk(1.2 + 40pl) bald + Assume 2T25 effective pi = 982/(300 x = 440) = 0.00743 k k¢¢ = = 1.6-d 0.35 1.6-0.44 = 1.16 TRd Table 4.8 VRdl = 300 x 440 x 0.35 x 1.16 x (1.2 + 40 x 0.00743) x 10-3 = 80.2 kN 2.4.5.3 Shear reinforcement by standard method Maximum capacity of section v .R+ 4.3.2.4.3 = 0.7 - ck/200 = 0.7 x 32/200 = 0.54 4 0.5 Eq n 4.21 VRd2 0.5 x 0.54 x (32/1.5) x 300 x 0.9 x 440 x 10-3 = 684 kN Eqn 4.25 4.3.2.2(10) Design shear force is shear at a distance d from the face of the support. This is 590 mm from the support centreline. A SW s 0.9 CC) Vsd x 80.2 0.87 x 440 x 250 = 0.0116 (Vsd - Eqn 4.23 80.2) Design of shear reinforcement is summarized in Table 2.3. COMPLETE DESIGN EXAMPLE Table 2.3 Location 8 Design of shear reinforcement Vsa A.Is s for 12 mm links Links 223 m span LH end RH end 203 1.42 248 1.95 159 116 R12 @ 150 R12 @ 100 m span LH end RH end 6 S23 202 128 1.41 160 min. max. R12 @ 150 R12 @ 300 R12 @ 300 Minimum 2.4.6 Deflection Reinforcement percentage at centre of 8 m span 4.4.3.2 = 100 x 1960/(450 x 1660) = 0.26% NAD Interpolating between 0.15 and 0.5%, basic span/effective depth ratio for end span = 40 0 -Ca Table 7 To modify for steel stress multiply by 400/460 E To To modify for T section multiply by 0.8 modify for span > 7 m multiply by 0 a`) Q0) Therefore permissible ratio E Actual ratio 2.4.7 Cracking For estimate of steel stress under quasi-permanent loads Ultimate load Assuming 02 Quasi-permanent load (nn Approx. steel stress at midspan = Approx. steel stress at supports allowing for 30% redistribution = 196/0.7 These are conservative figures since they do not allow for excess reinforcement over what is needed or for moment calculated at centreline of support rather than at face of support. Check limits on either bar size or spacing. 3C. From EC2 Table 4.11, 25 mm bars in spans are satisfactory at any spacing since steel stress < 200 N/mm2 OK °01 From EC2 Table 4.12, bar spacing at supports should be 25, hence column is slender in N-S direction The slenderness in the E-W direction will be found to be approximately the same. The structure is braced and non-sway (by inspection), hence the Model Column Method may be used with the column designed as an isolated column. Xorc = 25(2 - eo1/eo2) = 50 in both E-W and N-S directions 0-0 4.3.5.5.3 Eq n 4.62 Slenderness ratios in both directions are less than Xcr.t, hence it is only necessary to ensure that the column can withstand an end moment of at least NSdh/20 = 1801 x 0.3/20 = (Y) 27.0 kNm 4.3.5.5.3 Eqn 4.64 This exceeds the first order moments. Hence NSd = 1801 kN and MSd = 27.0 kNm NSd 0.62 bh ,k Msd bh 2f Assume _ 27.0 x 106 = 0.031 ck 3003 x 32 d' = 45 mm E COMPLETE DESIGN EXAMPLE Then d'lh AS yk = 45/300 = 0.15 = 0.16 (Section 13, Figure 13.2(c)) bh ck Hence AS = 1002 mm2 Use 4T20 (1260 mm2) Note: In the design by Higgins and Rogers, the slenderness ratio exceeds the equivalent of Xcrc but the design moment is still Nh120. EC2 requires less reinforcement due to the smaller design load and the assumption of a smaller a1) cover ratio. If the same cover ratio is used in the Higgins and Rogers design, 4T20 are sufficient in both cases. 2.6.5 External column 2.6.5.1 Loading and moments at various levels These are summarized in Table 2.6. Table 2.6 Loading and moments for external column Beam loads (kN) Total E Column design loads (kN) 'L7 Column moments (kNm) Top Imposed 2 2 Dead 2 Bottom E 2 2 Load case 1 1 1 1 1 Roof Main Edge Self-weight 184 55 186 55 39 41 145 55 9 145 55 9 209 104 107 39 41 209 93 98 3rd floor Main Edge Self-weight 235 55 240 55 109 c°0 114 126 55 9 (f) 126 55 9 399 93 98 148 155 399 93 98 2nd floor Main Edge Self-weight 235 55 240 55 109 114 126 55 9 126 55 9 589 93 98 257 269 589 CIO 103 109 1st floor Main Edge Self-weight N.. 233 55 238 55 108 113 125 55 9 (f) 125 (t) 68 72 55 9 Foundations 365 382 778 778 (D00 COMPLETE DESIGN EXAMPLE 2.6.5.2 Design for column between first floor and foundation 675 x 106 x 0.5 ktop 675 x 106 3125 x 106 = 0.71 4000 3500 8000 kbottom = 00 Hence a = 0.85 Figure 4.27 Effective height = 0.85 x 4000 = 3400 mm E Slenderness ratio = l°/i = 3400 12 = 39.3 300 v° will be small so be less than 25 Hence xmin . = 25 X > 25, therefore column L() is slender in N-S direction E Calculate Xcrt e,,, eo2 bottom moment top moment _ 0 = 0 4.3.5.5.3 85 Hence Xcrit = 25 (2 + 0) = 50 Slenderness ratios in the E-W and N-S directions are both less than 50, hence it is only necessary to ensure that the end moment is at least NW20. The worst condition occurs with load case 2 at section just above the first floor, where MSd is greatest. Nsd = 589 + 0.8 x 269 = 804 kN 0)) Nh 20 804 x 0.3 20 = 12.0 kNm Design end moment = 109 > 12 kNm Hence NSd = 804 kN and MSd = 109 kNm COMPLETE DESIGN EXAMPLE 2.6.6 Reinforcement details Maximum spacing of links for internal column Generally Above and below floor 5.4.1.2.2(3) x 20 = 240 mm 0.67 x 240 = 160 mm 12 NAD Table 3 5.4.1.2.2(4) E Maximum spacing of links for external column Generally At lap and below floor 12 x 25 c>) = 300 mm 0.67 x 300 = 200 mm The reinforcement details are shown in Figure 2.11. LL. INTERNAL COLUMN F2 EXTERNAL COLUMN F1 Links 150 Vertical bars Section 1"1 Links R8-6 Vertical bars Section o O m 350 ^ @ 1st. 4 f 4l 4 4 4 225 -° YI _7 7 o N 200 1 4 OOE 300 2() °o 300 F 4T20-1 i 1 ° 18(0) o 1 n 1 ' o 300 o o 'r 4T25-4 O _ a N F W Edge beam @ E 0 b rn _ _ 4 5 m 1 1 0 0 @ 30 4 m Cover to links= .-- Fdn'o g N T Fdn. f s Cover to links 40 2-14R8 300 0 +f in ss c N W St arters, see Fig. 2.13 , Figure 2.11 Column reinforcement details 2.7 Foundation Design typical pad footing for internal column. 2.7.1 Cover Use 50 mm nominal cover against blinding 4.1.3.3(9) a nominal cover of not less than 40 mm against blinding. EC2 specifies a minimum cover greater than 40 mm. This implies a nominal cover greater than 45 mm, hence the choice of 50 mm. BS 8110 specifies te) C?. 2.7.2 Loading Taken from internal column design. Ultimate design loads: Dead = = = 1226 Imposed Total 575 1801 kN °-. E COMPLETE DESIGN EXAMPLE Hence service loads: Dead Imposed Total = = 908 383 1291 kN = The assumption is made that the base takes no moment. Also it is assumed that the dead weight of the base less the weight of soil displaced is 10 kN/m2 over the area of the base. 2.7.3 Size of base Since, at the time of publication, EC7: Geotechnical design(s) and EC2, Part 3: Concrete foundations0o) have not been finalized, the approach used here is based on current UK practice. Use 2.75 m x 2.75 m x 0.6 m deep pad Bearing pressure under service loads _ 1291 + 10 2.752 = 181 < 200 kN/m2 .................... OK = 1801 Design pressure at ultimate limit state = 238 kN/m2 E 2.752 2.7.4 Flexural reinforcement Moment at face of column = (0) Q)) 238 x 2.75 x 1.2252/2 = 491 kNm E Average effective depth = 600 CAD - 50 x - 25 = 525 mm Msd bd2fck 491 x 106 = 0.020 2750 x 5252 x 32 AS yk bd>ck = 0.023 (Section 13, Table 13.1) Hence AS = 0.023 x E E x 2750 x 525 x (Y) 32/460 = 2310 mm2 Use 9T20 @ 300 mm crs. each way (2830 mm2) E x 2.7.5 Shear 2.7.5.1 Shear across base Shear force may be calculated at a critical section distance d from the face of the column. Design shear (VSd) 4.3.2.2(10) = 238 x 2.75 x (2.75 2 0.3) _ 0.525 = 458 kN x 0 COMPLETE DESIGN EXAMPLE calculating VRd1, the influence of the reinforcement will be ignored since, if straight bars are used, they will not extend d + Ib,net beyond the critical section. In (1) 4.3.2.3(1) VRdl = 0.35 x 1.075 x 1.2 x 2750 x 525/1000 x x = 652 kN Eqn 4.18 VRd, > VSd, hence no requirement for shear reinforcement c 2.7.5.2 Punching shear The critical perimeter is shown in Figure 2.12. Design load on base = 1801 kN Length of critical perimeter u x = [ 4 x 300 + 7r (2 x 1.5 x 525) ] /1000 = 6.15 m Figure 2.12 Critical perimeter for punching a)) VRdl = 0.35 x 1.075 x 1.2 x 525 x 6.15 ((7 = 1458 kN 4.3.4.5.1 Area within perimeter = 2.98 m2 Design shear (VSd) Area of base = 7.56 m2 = 1090 kN 4.3.4.1(5) _ (7.56 - 2.98) x 238 VSd hence no requirement for shear reinforcement C VRdl, 2.7.6 Cracking Approximate steel stress under quasi-permanent loads 460 1.15 X (908 + 0.3 x 383) 1801 x 2310 = 186 N/mm2 2830 From EC2 Table 4.11 bar size should not exceed 25 > 20 mm used. Hence cracking ........................................... OK 4.4.2.3 Table 4.11 41 COMPLETE DESIGN EXAMPLE 2.7.7 Reinforcement details The reinforcement details are shown in Figure 2.13 and given in Table 2.7. c T I 9T202 2 3 1 -300 B2 O AA 2 2 DA 9T20- -300 1 81 PLAN 1 rCo I e-2 2R8-3-300 Fdn. I Cover =40 ACOVER A - B1 = 50, end =75 Figure 2.13 Base reinforcement details Table 2.7 Commentary on bar arrangement Bar marks 1 Notes Straight bars extend full width of base less end cover of 75 mm. Bars should extend an anchorage length beyond the column face _-0 E Anchorage length = 32 °-' x 20 = 640 mm E 4.1.3.3(9) 5.2.3.4.1 Actual extension = 1150 mm 2 Column starter bars wired to bottom mat Minimum projection above top of base is a compression lap + kicker = 32 x 20 11o 5.2.4.1.3 + 75 = 715 mm + E 3 Links are provided to stabilize and locate the starters during construction COMPLETE DESIGN EXAMPLE 2.8 Shear wall 2.8.1 Structure The structure is shown in Figure 2.14. 1 st floor _, 0.5 x wind load on building 4000 250 i 14300 T Figure 2.14 Shear wall structure 2.8.2 Loading at foundation level Dead load from first to third floors and roof = Self-weight = 0.5 (3 x 23.5 + 28.5) = 49.5 kN/m 0.175 x 24 x 15.5 = 65.1 kN/m CJ) x x Characteristic dead load = x 49.5 + 65.1 = 114.6 kN/m Characteristic imposed load from slabs = 2.5 (1.5 + 3 (J1 x 4) x 0.7 = 23.6 kN/m V: Wind loading is taken as 90% of value obtained from CP3: Ch Total wind load on building in Part 201). N-S direction = 0.9 x 449 = 404 kN Wind load on wall = 404/2 = 202 kN Moment in plane of wall = 202 x 8 = 1616 kNm Hence Maximum force per unit length due to wind moment M x 6 P - + 1616 x 6 = + 47.4 kN/m 14.22 2.8.3 Vertical design load intensities at ultimate limit state Dead load + imposed load = 1.35 x ^O, z 114.6 + 1.5 x 23.6 = 190.1 kN/m Dead load + wind load 1.35 x x 114.6 + 1.5 x 47.4 = 225.8 kN/m; or = 1.0x114.6-1.5x47.4 x 43.5kN/m z NAD 4(c) Eqn 2.8(a) E Eqn 2.8(a) + COMPLETE DESIGN EXAMPLE Dead load + wind load + imposed load = = 1.35x 114.6 + 1.35 x 23.6 ± 1.35 x 47.4 Eq n 2.8(b) 250.6 kN/m or 122.6 kN/m NAD 4(c) Therefore maximum design load = 250.6 kN/m From analysis of slab (not presented), maximum moment perpendicular to plane of wall = 11.65 kNm/m E E 2.8.4 Slenderness ratio 0.5 4 + 1 kA 1 5 3.5 = 2.05 Eqn 4.60 kB = 00 Hence a 10 = 0.94 01c0l Figure 4.27 a)) = = 0.94 x 4 = 3.76 m 12 _ 3.76x1000x 175 = 74.4 Hence wall is slender 2.8.5 Vertical reinforcement Higgins and Rogers design the shear wall as unreinforced. Plain concrete walls will be covered in EC2 Part 1A which, at the time of publication, has not yet been finalized. The wall will, therefore, be designed here as a reinforced wall. As will be seen, the result is the same. Eccentricity due to applied loads eo1 = = 0 11.65 eo2 x 1000/250.6 = 46.5 mm Hence ee = 0.6 x 46.5+0=27.9 mm Eqn 4.66 Accidental eccentricity ea _ 1 200 x 3760 2 = 9.4 mm Eq n 4.61 COMPLETE DESIGN EXAMPLE Second order eccentricity e z x 1 10 L!7 1.15 x 200000 0.9 x 122 x _ 3760z x 2 x 460 x x Kz Eqns 4.72 & 4.69 = 51.5K2 1 Assuming K2 = Design eccentricity = 27.9 + 9.4 + 51.5 = 88.8 mm Design ultimate load = 250.6 kN/m Design ultimate moment = 88.8 M bh2fok x 250.6/1000 = 22.3 kNm/m 0.023 N bhf,k Asfyk = 0.045 0.01 (Section 13; Figure 13.2(d)) bhfck Hence As E = 122 mm2/m or 61 mm2/m in each face Minimum area of reinforcement = 0.004 x 1000 x 175 = 700 mm2/m 5.4.7.2 This exceeds the calculated value. Hence the minimum governs. Use T12 @ 300 mm crs. in each face (754 mm2/m) E E E E E E 2.8.6 Shear Design horizontal shear = 1.5 x 202 = 303 kN Shear stress = 303 x 1000 (=D 14300 x 175 = 0.12 N/mm2 E .................. x OK Note: not calculated since it must be > 0.12bw d by quick inspection of VRdt is EC2 Eqn 4.18. 2.8.7 Horizontal reinforcement Minimum at 50% of vertical reinforcement provided As 5.4.7.3 min = 188 mm2/m (E F) Minimum for controlled cracking due to restraint of early thermal contraction 4.4.2.2 45 COMPLETE DESIGN EXAMPLE As = kckf 1.0 t.ettAct/QS Eqn 4.78 kC k ct,eft = = = 0.8 1.9 N/mm2 (assuming concrete strength to be equivalent to Table 3.1 C16/20 at time of cracking) = as 360 N/mm2 (assuming 10 mm bars) Table 4.11 AS = 1.0 x 0.8 x 1.9 x 175 x 1000/360 = 739 mm2/m Use T10 @ 200 mm crs. in each face (785 mm2/m) 2.8.8 Tie provisions at first floor According to the NAD, these should follow the rules in BS 8110. Ft NAD 65(g) BS 8110 3.12.3 = 36 kN 2.8.8.1 Peripheral tie = As 36x103_78mm2 460 Use 1T10 (78.5 mm2) 2.8.8.2 Internal tie force Force = 2.5 x 36 (4.7 + 4.0) x 7.5 14.3 _ 299 kN 5 Hence As 299 x 460 103 = 650 mm2 Use 5T10 in each face (785 mm2) Hence T10 @ 200 mm crs. horizontal reinforcement below slab is adequate. 2.8.8.3 Wall tie i-0 in wall 0.5 m above and Take the greater of (a) and (b) (a) Lesser of 2.OFt or lsF 2.5 = 72 or 48 kN COMPLETE DESIGN EXAMPLE (b) 3% of total vertical load = 0.03 x 190.1 = 5.7 kN Hence Tie force = 48 kN 48 x 103 A S - = 104 mm2 460 Therefore reinforcement in slab will suffice 2.8.9 Strip footing EC2, Part 3: Concrete foundations, at the time of publication, has not yet been drafted, hence current UK practice is adopted. (1) Maximum pressure due to characteristic dead, imposed and wind loads = 114.6 + 23.6 + 47.4/0.9 = 191 kN/m 212 kN/m2 For 900 mm wide strip, pressure = 191 = 0.9 Allow extra 10 kN/m2 for ground floor loads soil in foundations. This gives 222 kN/m2. z and weight 250 > 222 z + of concrete displacing OK Allowable pressure = 1.25 x 200 = kN/m2........ Use 900 mm wide strip Calculate reinforcement for flexure Moment AS = 250.6 x ion (0.9 8 0.175)2 = 16.5 kNm/m E E = 209 mm2/m E E = = Minimum area 0.0015bd 0.0015 5.4.2.1.1 x 1000 x 200 = 300 mm2/m E Use T12 @ 300 mm crs. (377 mm2/m) COMPLETE DESIGN EXAMPLE 2.8.10 Reinforcement details The reinforcement details are shown in Figure 2.15 and given in Table 2.8. IT- A I 1T10-7 II r 2x1T10-8 T 1T10 B bars links O N 200 o° P NN -200 350 ci 19F1) Y -N (48N2 -48F2) -200 links a 96T12-3-300 c O bars N B iJ i 1st. SFL wall tie 8 v ooz- M3 1 L N 19R8-4 19T10-5 0 Cr Cr P P G N (48 N2+48 F2) t 96T12-1-300 L la 19T10-5 -T- 19R8-4 - 1000 %13R10 -9 EW i Fdn. -71 1 i- 2%4T12-2-300 B2 cover ends = = 40 75 - 2 A-A grid 2 omitted for clarity) WALL ELEVATION B-B COVER to EAST N1 = outer bars - 40, F1= 20 Figure 2.15 Shear wall reinforcement details COMPLETE DESIGN EXAMPLE Table 2.8 Bar marks E 1 Commentary on bar arrangement Notes Wall starters match vertical reinforcement The projection of the horizontal legs beyond the face of the wall form the tension reinforcement in the footing This extension must be at least a tension anchorage length _ 12 x 1.15 460 4 x 3.2 x 209 = 208 377 mm .............. .... OK 5.2.2.2 5.2.2.3 5.2.3.4.1 The minimum projection above the top of the base is a compression lap + 75 mm kicker = 32 x 12 + 75 = 459 mm This is detailed at 525 mm 2 .................................. 2 OK Minimum longitudinal reinforcement provided Minimum horizontal reinforcement provided 4,5,6 5.4.7.3 4.4.2.2 7,8 u') Peripheral tie at floor ,r. BS 8110 3.12.3.5 9 Wall spacers to maintain location of reinforcement 2.9 Staircase 2.9.1 Idealization The idealization of the staircase is shown in Figure 2.16. T 3500 5060 Figure 2.16 Idealization of staircase Design as end span of a continuous beam. Calculations will be given for width. 1 m 2.9.2 Durability and fire resistance As for floor slab, Section 2.3, 20 mm nominal cover will be satisfactory. COMPLETE DESIGN EXAMPLE 2.9.3 Loading Average slab thickness on plan Self-weight Finishes = 250 mm 6.0 kN/m 0.5 = 0.25 x 24 = = Characteristic dead load Characteristic imposed load Design ultimate load = 6.5 kN/m 4.0 kN/m 1.35 = = x 6.5 + 1.5 x 4 = 14.78 kN/m 2.9.4 Analysis Using coefficients in the Concise Eurocode Moment at interior support Moment near mid-span Shear = = 0.11 x x 14.78 14.78 14.78 x 5.062 = 41.6 kNm 0.09 0.6 x x x 5.062 = 34.1 kNm Concise Eurocode Table A.1 = x 5.06 = 44.9 kN 2.9.5 Reinforcement for flexure Effective depth = 175 Interior support, M - 20 103 13.1 6 = 149 mm 41.6 x 10s bd2fCk From Section 13, Table AS yk x 1492 x 32 E = 0.059 bdf Hence AS (L) = 0.072 = 746 mm2/m E Use T12 @ 150 mm crs. (754 mm2/m) E E Span M bd2ck Asf ,k = 0.048 bd ck Hence = 0.058 E A5 = 601 mm2/m Use T12 @ 150 mm crs. (754 mm2/m) L() E E E E COMPLETE DESIGN EXAMPLE 2.9.6 Shear Reinforcement ratio = 754 1000 x 149 = 0.0051 Near support VRd, = 0.35 x (1.6 - 0.175) x (1.2 + 40 x 0.0051) x 149 = 104.3 kN 4.3.2.3 Eqn 4.18 VRd, > Vsd = 44.9 kN, hence no shear reinforcement required 2.9.7 Deflection Reinforcement ratio at mid-span = 0.51% Concrete is lightly stressed, hence basic span/effective depth ratio is 32. Since Yk Table 4.14 = 460, this should be modified to: = 34.9 4.4.3.2(4) (+'j 32 x 400/460 x 754/601 Actual span/effective depth ratio = 5060/149 = 34 < 34.9....... OK 2.9.8 Cracking As for floor slab in Section 2.3.8 Minimum area of reinforcement Thickness of waist E c = 183 mm2/m E 4.4.2.2 = 175 < 200 mm 4.4.2.3 (1) No further check is necessary. 2.9.9 Tie provisions E-W internal Total tie, the minimum area required = x 3 = 273 mm2 91 mm2/m (see Section 2.3.9) BS 8110 3.12.3.2 area for staircase = 91 Provide 2T12 tie bars each side of staircase in adjacent slab 51 COMPLETE DESIGN EXAMPLE 2.9.10 Reinforcement details The reinforcement details are shown in Figure 2.17. 0 a 5T10-I11 -300 ST10-16-300 i 2 750 o 150 3- 2T10-8 3 6T12-3*6T12-141-125 alternate 3Qrd 16 _ L 7 117-.1 { I 10T12-12-150 '-1OT12-13-150 7 Cover = 40 - 10T12-15-150 10 16 -9 6 117 15 LANDING 0 Cover I to outer bars = 20 A-A 2nd V Aa FLIGHT A 4 10-2-300 L 1OT12-1-150 Figure 2.17 Staircase reinforcement details 52 3BEAMS 3.1 Introduction This Section covers the design of beams for shear and torsion, and supplements the examples given in Section 2. The requirements for adequate safety against lateral buckling are also examined. N c O O 0 0 O O O C 3.2 Design methods for shear - C the Code. (1) Standard Variable Strut Inclination (VSI). C C O C (2) The standard method assumes a concrete strut angle of 450 (coto = 1) and that the direct shear in the concrete, V d, is to be taken into account. This contrasts with the VSI method which permits the designer to choose strut angles between the limits set in the NADO), as shown in Figure 3.1, but ignores the direct shear in the concrete. ((] p30 (D N 3.2.1 Introduction EC20) differs from BS 8110(2) because the truss assumption used in shear design is explicit. Leading on from this, two alternative methods are given in O U C O C C O O U c c U C U O C O O U U C O U c A further disadvantage of this method is that with increasing values of cote, i.e., reductions in the concrete strut angle, the forces in the tension reinforcement U O Z O C C C_ C C - O U Figure O 3.1 Limits of cotA (VSI method) Because the direct shear in the concrete is not taken into account in the VSI method, no savings in shear reinforcement can be achieved until the applied shear exceeds three times the concrete shear (VSd > 3V d). O O U o U O c C O U O O U C C C O O O U C O C c L U U U O E U O O O ° U C U C C o U C O C BEAMS increase significantly and may well outweigh any notional savings in shear reinforcement. These forces are, it should be noted, explicitly checked in EC2 but not in BS 8110. Given special circumstances the VSI method may be required but for most practical situations, the standard method will provide the most economic design. COO CD- E °OM 3.2.2 Example 1 - uniformly distributed loading The beam shown in Figures 3.2 and 3.3 is to be designed for shear. AO) Figure 3.2 Figure 3.3 Typical section The material strengths are E Ultimate load = 385 kN/m 6m Beam span and loading iii 400 400 example 1 1000 *FJ Z Asl = 6434mm 2 = (BT32) 50mm Cover to links - example 1 fck fvWk = 30 N/mm2 (concrete strength class C30/37) = 250 N/mm2 (characteristic yield strength of links) The beam will be checked for shear reinforcement at three locations using both the standard and VSI methods for comparison. These are (1) 4.3.2.4.3 4.3.2.4.4 4.3.2.2(10) 4.3.2.2(2) d from support Where Vsd = VRd, i.e., the point beyond which only minimum shear reinforcement is required ' 0 (2) (3) An intermediate point between 1 and 2. 4.3.2.4.3 3.2.2.1 Standard method The shear force diagram is shown in Figure 3.4. V Sd 1155 k N I I VRdl a I I l 1155 kN Figure 3.4 Shear force diagram -example 1 BEAMS The design shear resistance of the section, = = VRd1, is given by 4.3.2.3(1) Eq n 4.18 VRd1 T IT I TRdk (1.2 + 40pl) + 0.15acp ] bwd Rd 0.34 N/mm2 for 1.6 fok = 1 30 N/mm2 Table 4.8 k = - d 4 1 = As, P, 6434 400 x 900 0.018 I> 0.02 bwd (assuming BT32 throughout span) NSd = 0 AC VRd1 = 0.34 x 1 (1.2 + 40 x 0.018) x 400 x 900 = 235 kN 3.2.2.1.1 Position VSd 1 = at d from support 1155 VRd1' - 0.9 x 385 = 808.5 kN 4.3.2.4.3 VSd > shear reinforcement is required The shear resistance of a section with shear reinforcement is given by VRd3 Vcd + V = d Eq n 4.22 V V d = = VRd1 235 kN Eq n 4.23 d A sw s (0.9d)fywd where A sw area of shear reinforcement of spacing 250/1.15 For VRd3 70C shear reinforcement 217.4 N/mm2 E E = > ? Al VSd Vwd VSd - Vcd; or A (0.90) ywd Sw ? VSd - Vd Therefore 3.25 mm2/mm E (Nn Asw S - (808.5 - 235) x 103 = E 0.9 x 900 x 217.4 Try R12 links @ 140 mm crs. (4 legs), Asw/s E = 3.23 mm2/mm E E E 55 BEAMS Check crushing of compression struts VRd2 = (2 )Pf bw0.9d(1 + cots) Eqn 4.25 For vertical links, cot« 0.7 = ck 0 V - = 0.55 200 = 20 N/mm2 4 0.5 Eqn 4.21 30 cd 1.5 Therefore VRd2 = = (2)x0.55x20x400x0.9x900x1 1782 kN > Vsd, max = 1155 kN ................ x x OK 4.4.2.3 Check maximum spacing of links ASw 452 140 Pw sbwsin a x 400 103 0.0081 Eqn 4.79 Vsd - 3V d - (808.5 - 3 x 235) x = CY) pwbwd 0.0081 x 400 x 900 = 300 mm 35 N/mm2 Table 4.13 Maximum spacing for crack control Since (5) Smax VRd2 < VSd :5 (3) VRd2 5.4.2.2(7) Eqn 5.18 = 0.6d > 300 mm 140 mm spacing .......................................... Pw OK Check minimum value of Table 5.5 Concrete strength class C30/37 Steel class S250 By interpolation from EC2 Table 5.5 Pw,mm = 0.0022 < E 0.0081 proposed Use R12 links @ 140 mm crs. (4 legs) Note: Using the standard method, the increase in force in the tension reinforcement is best covered by using the shift rule. It will, however, be calculated in this example to provide a comparison with the values obtained in the subsequent examples using the VSI method. ((1 Force in tension reinforcement Td - z + ( 2) VSd(Cote Cota) G.) MS - '-h O-in 4.3.2.1P(6) 5.4.2.1.3 C.- Eqn 4.30 BEAMS Msd Vsd = = = Td 884 kNm, 808.5 kN 1, LC) z = 0.9d = 810 mm cotO Therefore 3.2.2.1.2 Position 2 cota 1091 = 0 for vertical links 3 4.3.2.4.3(5) = + 404 = = 1495 kN = = where Vsd VRdl = 235 kN From Figure 3.4 Vsd 1155 ((0 - a x 385 a 2.39 m from support VRd2 From Section 3.2.2.1.1, LC) = 235 kN > E Vsd, max ........... . ...... OK The amount of shear reinforcement provided should be greater than Pw,min Re-arranging EC2 Eqn 5.16 in terms of E For vertical links sins Hence A aw s Maximum longitudinal spacing Vsd VRd2 Since Vsd Smax 3.2.2.1.3 Position 3 This is a point intermediate between the section at dfrom support and the point at which shear reinforcement is no longer required. E aro) E Pwmin Table 5.5 A Pwbwsina S Asw Use R10 links @ 300 mm crs. (4 legs) E Vsd VRdl E = 0.0022 A 5w s gives = 1 = 0.0022 x 400 x 1 = 0.88 mm2/mm E E (smax) is given by EC2 Eqns 5.17-5.19. = = 235 kN 1782 kN from Section 3.2.2.1.1 (5) CC) VRd2' EC2 Eqn 5.17 applies U') = 0.8d > 300 mm 0.88 Eqn 5.17 264 mm2, 4R10 x 300 = = 314 mm2 E - at 1.65 m from support E = = 1155 - 1.65 x 385 = LC) 520 kN 235 kN BEAMS Since VSd > VRdt' shear reinforcement is required Re-arranging EC2 Eqn 4.23 ASw _ VSd - Vd _ (520 0.9 - 235) x 103 217.4 s 0.9df M,d x 900 x = 1.81 = 1.62 mm 2 /mm E Try R12 links @ 250 mm crs. (4 legs) mm2/mm E E Check maximum spacing of links A SW Pw 4.4.2.3 Eqn 4.79 sb w sina For vertical links sina = 1 Hence _ Pw 452 250 x 400 = 0.0045 Vgd - 3V d (520 - 3 x 235) x x 103 pwbwd 0.0045 x 400 x 900 _ -114 N/mm 2 E E x Maximum spacing for crack control = 300 mm E .................. OK Table 4.13 Since (5)VRd2 < = VSd (3)VRd2 5.4.2.2(7) Eqn 5.18 0.6d P 300 mm smax From Section 3.2.2.1.1 VRd2 > VSd,max ....................................... E OK Provide R12 links @ 250 mm crs (4 legs) To optimize link spacing, check the point at which shear reinforcement is satisfied by R12 @ 200 mm crs. (4 legs). ASw S 452 200 = 2.26 mm2/mm A SW wd = s (0.9d)fywd = 2.26 x 0.9 x 900 x 217.4 = 398 kN VRd3 Vd + Vwd Equating VRd3 = VSd and noting that + V ,d + = Vd = VRdl VSd = VRdl 235 + 398 = 633 kN BEAMS Distance of point from support = 1155 (Y) 385 633 = 1.36 m The proposed link arrangement is shown in Figure 3.5. R12-140 4 legs R12-200 4 R12-300 4 legs R12-200 4 legs R12-140 4 legs legs 1-36m -2.39 m + + 6.Om between centres of supports a 2-39m+ Figure 3.5 Link arrangement (standard method) Note: - example 1 the centre portion of the beam R10 links are required by calculations but R12 (') are shown to avoid the possible misplacement on site. Distance from the support (+) could be reduced to 1.70 m in this case. In r-1. 3.2.2.2 Variable strut inclination method This method allows the angle of the concrete compression strut to be varied at the designer's discretion within limits stated in the Code. can give some economy in shear reinforcement but will require the provision of additional tension reinforcement. In most cases the standard method will It suffice. This reduced shear reinforcement will only be obtained at high levels of design shear and is counter-balanced by increased tension reinforcement. This can be seen by a comparison of EC2 Eqns 4.22 and 4.23 in the standard method and EC2 Eqn 4.27 in the variable strut inclination method. The standard method gives VRd3 ((DD CLL. 4.3.2.4.4 O-N-0 >"O d = Vd A Sw + twd (0.9d)ywd Eqn 4.22 Eqn 4.23 V = Re-arranging gives Asw VRd3 - Vd s (0.9d)ywd The VSI method gives A sw VRd3 (0.90)fywd Cot() s Eqn 4.27 Re-arranging gives A sw S VRd3 (0.9d)ywd cot() BEAMS Note: the above equation the contribution of the concrete, resistance of the section is not taken into account. In V d, to the shear With cot6 = 1.5 which is the maximum value permitted in the NAD, reductions in shear reinforcement will only occur when VRd3 0 VRd3 - Vcd Or (0.9d) fyWd x 1.5 G 1.5(VRd3 (0.9d) f,N,d VRd3 - Vd) Vsd Putting If Vsd Vsd = VRd3 gives > 3V d > 3V reinforcement. If d, then the VSI method will allow a reduction in shear of of this inequality is not satisfied, use produce an uneconomic amount standard method should be used. the variable strut inclination method will shear reinforcement. In this case the VRd2 DLO For elements with vertical shear reinforcement, bWzv is given by cd Putting Vsd = VRd2 + VRd2 cote + tan6 and re-arranging gives 1 Eqn 4.26 Vsd bWzv cd cotO + tan6 Figure 3.1 shows cotO plotted against 1/(cot6 + tan6) together with the EC2 and NAD limits for cot6. Hence for a given Vsd, the limits for cotO can be found. Increasing the value of cotO will reduce the shear reinforcement required but increase the force in the tension reinforcement. In this example, cotO will be chosen to minimize the shear reinforcement. (OD -'O 3.2.2.2.1 Position From above V11 1 7-O z P E 1 - at d from support bWzvfcd cotO + tan6 = = bW 400 mm 0.9 x 900 = 810 mm = 0.7 - fck = 0.55 K 0.5 Eq n 4.21 200 BEAMS 30 cd = 20 N/mm2 1.5 Vsd = 808.5 kN Therefore 1 _ 808.5 x 103 cotO + tanO 400 x 810 x 0.55 x 20 x x From Figure 3.1, this lies under the curve. Therefore, cotO = 1.5 can be chosen which is the maximum value allowed under the NAD limits. !z? Coo x = 0.22 V Rd3 (A s VRd3 ) Z ywdCOte Eqn 4.27 Now equating Asw to Vsd and re-arranging 808.5 810 Vsd x 103 s zywdcotO x 217.4 x = 1.5 3.06 mm2/mm E Check Asw ywd _ 1.66 - < ('z)ofcd E = 5.5 ......................... = 3.01 mm2/mm Try R12 links @ 150 mm crs. (4 legs), Asw/s Check maximum spacing of links. SIN OK E 4.4.2.3 pw = A sw = 0.0075 Eqn 4.79 sbW sin« (808.5 V- 3V sd cd = - 3 x 235) x (NA) 103 = pw bwd 0.0075 x 400 x 900 38.3 N/mm2 Maximum spacing for crack control pw = = 300 mm E Table 4.13 OK = smax 0.0075 > pw.min 0.0022 ................... Table 5.5 5.4.2.2(7) Check Vsd = 808.5 kN bwzv cd 400 x 810 x 0.55 x 20 2.167 VRd2 = 1644 kN cotO + tanO + Since (5) VRd2 smax VSd (3) VRd2 Eqn 5.18 = 0.6d D 300 mm BEAMS Use R12 links @ 150 mm crs. (4 legs) Check additional force in tension reinforcement. Msd Td = z + Td ( 2) = VSd(cote - cot«) = 1091 + 606 = 1697 kN Eqn 4.30 c.. This compares with 1495 kN using the standard method. E Note: Although not permitted by the NAD, values of cote up to 2.5 are given in EC2. check on shear reinforcement using cotO = 2.5 is now given to illustrate the effect of increasing values of 0 on shear and tension reinforcement. A ASW + - Vsd 810 808.5 x 103 = S z ywd tote x 217.4 x 2.5 1.84 mm2/mm Try R12 @ 225 mm crs. (4 legs), A,Wls = 2.01 mm2/mm Check maximum spacing of links pw = 0.005 Vsd 3 °d pwbwd - = 57.5 N/mm2 Maximum spacing for crack control Smax = 250 mm E ................ OK OK Table 4.13 Eqn 5.18 = 0.6d > 300 mm ............................. Use R12 links @ 225 mm crs. (4 legs) Check additional force in tension reinforcement. + (2)Vsd (tote Td CEO Td = Msd z - cot«) = 1091 + 1011 = 2102 kN This compares with = 1495 kN using the standard method. 3.2.2.2.2 Position 2 - where (D3 Vsd = VRd1 Since only minimum shear reinforcement is required this case is identical to that shown in Section 3.2.2.1.2. E 3.2.2.2.3 Position 3 Vsd = at 1.65 m from support 520 kN VSd AiW S 520 x 810 103 zywdCoto x 217.4 x = = 1.5 1.96 mm2/mm E E Try R12 links @ 225 mm crs. (4 legs), ASW/s 2.01 mm2/mm 2 BEAMS From Section 3.2.2.2.1 spacing is satisfactory. Use R12 links @ 225 mm crs. (4 legs) As in Section 3.2.2.1.3, check the point at which the shear requirement is satisfied by R12 @ 200 mm crs. (4 legs). E SW = 452 = 2.26 mm2/mm s ( 200 Ssw J VRd3 zfyWd cot6 = 2.26 x 810 x 1155 217.4 x 1.5 = 597 kN Eq n 4.27 Distance from support = - 597 = 1.45 m 385 The proposed link arrangement is shown in Figure 3.6. R12-150 4 legs R12-200 4 legs 300_ 4 legs 1 R12-200 4 legs R12-1SO 4 legs 1.45m 2-39m 6.Om between centres of supports a 2. 39m Figure 3.6 Link arrangement (VSI method) - example 1 Comparing this with the arrangement in Figure 3.5 obtained using the standard method, it can be seen that less reinforcement is required near the support but this needs to be carried further along the beam. There is little overall saving in this case. 5 3.3 Shear resistance with concentrated loads close to support 3.3.1 Introduction Where concentrated loads are located within 2.5d of a support, the value TRd may be modified by a factor a when calculating VRd1. This enhancement only applies when the section is resisting concentrated loads and the standard method is used. For a uniformly distributed load, an unmodified value of VRdt should be used. -T 4.3.2.2(9) 3.3.2 Example 2 - concentrated loads only The beam shown in Figures 3.7 and 3.8 is to be designed for shear. BEAMS 800 k N 800 kN Ultimate loads 1.35rr 1.4 I 1-35m 6m Figure 3.7 Beam span and loading - example 2 A 900 1000 A sl = 4825mm Z (6T321 50mm Cover to links = Figure 3.8 Typical section - example 2 The materials strengths are fck fVwk = 30 N/mm2 (concrete strength grade, C30/37) 250 N/mm2 (characteristic yield strength of links) = the example, VRd, will be calculated at positions between the support and 2.5d away at intervals of 0.5d. This is done to illustrate the effect even though the critical section will normally be at the position of the concentrated load. In ((DD 3.3.2.1 Shear reinforcement The shear force diagram is shown in Figure 3.9. Figure 3.9 Shear force diagram - example 2 BEAMS The basic design shear resistance of the section, VRd, VRdi, is given by 4.3.2.3(1) = [TRd k (1.2 + 40pl) + 0.15 acp]b,,d = 30 N/mm2 Eqn 4.18 Table 4.8 a-0 TRd = 0.34 N/mm2 for Ck For concentrated loads within 2.5d of the face of the support, an enhancement of shear resistance is permitted. -r Rd may be multiplied by a factor a when T E determining a Taking values of x between 0.5d and 2.5d gives values of Table 3.1. Table 3.1 * No enhancement taken, see Figure 3.9 The equation for VRdl can be modified to give a range of values corresponding to the distance from the support. VRdl(X) k Pt cp Values of design shear resistance, Table 3.2 x (m) (r) E VRd,. = 2.5d/x with 1.0 Vsd = 1100 kN .......... OK Check maximum spacing of links By comparison with example 2, requirements are satisfied 4.4.2.3 ......... m OK 5.4.2.2(7) Use R12 links @ 125 mm crs. (4 legs) for 0 < x < 2.25 For the remainder of the beam beyond x = 2.5d (2.25 m) provide minimum reinforcement as example given in Section 3.2.2. 3.4 Design method for torsion 3.4.1 Introduction The edge beam shown in Figure 3.13 carries the ends of simply supported floor slabs seated on the lower flange. The beam is fully restrained at its ends. The example chosen is the same as that used in Allen's Reinforced concrete design to BS 8110: Simply explained(12). Analysis of the structure and the design of the section for flexure is not included. The section will be checked for shear, torsion and the combination of both. BEAMS Figure 3.13 Beam section 3.4.2 Design data Design torsional moment (Tsd) Design shear (Vsd) = = 120 kNm 355 kN Concrete strength grade is C30/37, Nominal cover to links is 35 mm. Ck = 30 N/mm2 Assuming 25 mm bars and 10 mm links .E- 4.1.3.3 NAD Table 6 = 1441.5 say 1440 mm d = 1500 - 35 - 10 - 25 2 Assume 0.25% tensile reinforcement for flexure 3.4.3 Shear resistance Shear will be taken as acting on the web of the section only. When combined shear and torsion effects are to be considered, shear is to be checked using the variable strut inclination method. The angle 9 of the equivalent concrete struts is to be the same for both torsion and shear design. The design shear resistance, VRdl VRd1, 4.3.3.2.2(4) 4.3.2.3(1) with zero axial load is given by = = = 7- Rdk(1.2 + 40pl)bwd ick Eqn 4.18 = TRd 0.34 N/mm2 for 1.6 30 N/mm2 E Table 4.8 k - d = 1.6 - 1.44 = 0.16 X1.0 OWED C') 3.1.2.4 Table 3.1 E E .0) BEAMS Assuming 0.25% tensile reinforcement, pi = 0.0025 > 0.02 VRdl = 0.34 x 1(1.2 + 40 x 0.0025) x 250 x 1440 x 10-3 = 159.1 kN < 355 kN Therefore shear reinforcement required. ;;z Use the variable strut inclination method. The maximum design shear force, to avoid web crushing is given by VRd2' bWzu VRd2 cd 0 4.3.2.4.4(2) (cot6 + tang) Eqn 4.26 Re-arranging gives VRd2 1 bwzu cd cot6 + tan6 = 355 kN Vsd bW z = = 250 mm 0.9d E = 0.9 x 1440 = 0.7 1296 mm u = 0.7 = °k = 200 30 1.5 - 200 30 = 0.55 -9 0.5 4.3.2.4.2(3) ca = °k -/C = 20 N/mm2 Therefore 355 x 103 250 x 1296 x 0.55 x 20 = 0.1 VSd bzu od W 1 cot6 + tan6 should be >_ 0.1 By reference to Figure 3.1, it will be seen that the value of cot6 may be taken anywhere between the limits of 0.67 to 1.5. CD NAD Table 3 4.3.2.4.4(1) To minimize link reinforcement, take cot6 = 1.5 Design shear resistance, VRd3' for shear reinforcement is given by E VRd3 IA- I ZYwdcote 4.3.2.4.4(2) Eqn 4.27 BEAMS Re-arranging gives A SW s VRd3 Zfywdcote VRd3 Putting ASW equal to Vsd Vsd s Zfywdcote Using high yield reinforcement ywd YwK = ys - 460 1.15 = 400 N/mm2 E E Therefore Sw = 0.9 355 x 103 = 1.5 S x 1440 x 400 x 0.46 mm2lmm ASWfywd = 0.46 x 400 = 0.74 < " 250 Ed = 0.55 x 20 = 5.5 N/mm2.. OK 2 bw s 2 4.3.2.4.4(2) Eqn 4.27 Before choosing the reinforcement, the effects of torsion will be considered and the results combined. The force in the longitudinal reinforcement, NIA Td ( Td, ignoring flexure, is given by 4.3.2.4.4(5) 2) Vsd(cotO = 1.5 0 For vertical links, cot« Td SIN Iota) Eq n 4.30 - 355 2 x = 266.3 kN Additional area of longitudinal reinforcement Td 266.3 x 103 400 = 666 mm 2 ywd This area of reinforcement must be combined with the tension reinforcement required for flexure together with the longitudinal reinforcement required for torsion. 3.4.4 Torsional resistance Torsional resistance is calculated on the basis of a thin-walled closed section. Solid sections are replaced by an idealized equivalent thin-walled section. Sections of complex shape are divided into sub-sections with each sub-section treated as an equivalent thin-walled section. The torsional resistance is taken °.' C,) (1) as the sum V'. of the torsional resistances of the sub-sections. X17 The torsional moment, carried by each sub-section according to elastic theory, may be found on the basis of the St Venant torsional stiffness. Division of the section into sub-sections should be so arranged as to maximize the calculated stiffness. 7 BEAMS For this example the section will be divided into the sub-sections shown in Figure 3.14. 310 200 I> 250 1500 200 x 310 I Figure 3.14 Dimensions of sub-sections 3.4.4.1 St Venant torsional stiffnesses ah3minhmax J BS 8110: Part 2 3.4.4.1.1 Top and bottom hmax flanges (gy 2.4.3 Eq n 1 = hmax h min . _ 310 310 mm, hmin = 200 mm = 1.55 200 = From which R 0.203 BS 8110: Therefore J 3.4.4.1.2 Web hmax = 0.203 x 2003 x 310 = 0.5 x 109 mm4 Part 2 2.4.3 Table 2.2 = 1500 mm, 1500 hmin = 250 mm E h max hmin 6 250 (3 From which = 0.33 Therefore BS 8110: Part 2 2.4.3 7.7 J = 0.33 x 2503 x 1500 = x 109 mm4 Table 2.2 BEAMS 3.4.4.1.3 Total stiffness `Jtot _ [(2 x 0.5) + 7.7] x 109 = 8.7x 109mm4 3.4.4.2 Thicknesses of equivalent thin-walled sections the actual wall thickness 4.3.3.1(6) t where u A = '4 u >5 = = outer circumference of the section total area within the outer circumference 3.4.4.2.1 Top and bottom flanges u = = (310 + 200)2 = 1020 mm 62 x 103 mm2 A Therefore t 310 x 200 = _ 62x103 1020 x = 61 mm 4.3.3.1(6) t may not be less than twice the cover, c, to the longitudinal bars. Hence, with 10 mm links tmin E E = 2(35 + 10) = 90 mm 3.4.4.2.2 Web u + = (1500 + 250)2 1500 = 3500 mm 375 x 103 mm2 A = x 250 = Therefore 375 x 103 t 3500 Values of t between the limits of A/u and 2c may be chosen provided that the design torsional moment, Tsd, does not exceed the torsional moment that can be resisted by the concrete compression struts. 3.4.4.3 Torsional moments Tsd,tot = 120 kN m This total moment is shared between the flanges and web in proportion to their torsional stiffness. Therefore TsdJI CA) = 107 mm > 2c .................. OK 120 x 0'5 = 8.7 6.9 kNm Tsd,w 120 x 8.7 106 kNm Tsd must satisfy the following two conditions Tsd _ 20 mm 20 mm 20 mm NAD Table 6 Nominal cover Use nominal cover NAD 6.4(a) = 25 mm E NAD in all Note: 20 mm nominal cover is sufficient to meet the NAD requirements 0 respects. Table 3 4.1.3.3(8) Check requirements for fire resistance to BS 8110: Part 2. 4.1.3.1.2 Materials Type 2 deformed reinforcement, yk 0 NAD 6.1(a) = 460 N/mm2 E _ vk _ 460 = 1.15 vd 400 N/mm2 l's 2.2.3.2P(1) Table 2.3 C25/30 concrete with 20 mm maximum aggregate size 4.1.3.1.3 Analysis model Span 6 m >_ >_ 4 4 x slab depth x 0.275 = = = 600 175 10 x slab depth = 2.75 m E Hence the ribbed slab may not be treated as a solid slab in the analysis under the terms of this clause unless intermediate transverse ribs are incorporated. This is not always desirable. 2.5.2.1(5) The model adopted in this example uses gross concrete section properties of the T shape in sagging regions and a rectangular section, based on the rib width, in the hogging region. EC2 Figure 2.3 has been used initially to define the extent of the hogging. This method can clearly be refined. SLABS 4.1.3.1.4 Effective span Jeff 2.5.2.2.2 In + al + a2 Eqn 2.15 Assume 300 mm wide supporting beams ln -t6 = 5700 mm = = a. taken a. Nip + a1 at edge beam as (2) t = = 150 mm 150 mm E Figure 2.4(a) Figure 2.4(b) a2 at central leff beam _ (2) t = 6000 mm For ratio of adjacent to spans between = 0.85 x 6000 1 and 1.5 = 5100 mm = 0.851, E 2.5.2.2.1(4) Figure 2.3 4.1.3.1.5 Effective width of flanges 2.5.2.2.1 L() ((DD Effective flange width is assumed constant across the span for continuous beams in 2.5.2.2.1(2) buildings. ` 0.02 Giving VRdl = 21.6 kN/rib < Vsd Therefore shear reinforcement must be provided. Use the standard design method for shear: VRd3 4.3.2.2(3) 4.3.2.2(7) > 4.3.2.4.3 VSd VRd3 Vd + VWd Eqn 4.22 SLABS where ('7 Vd Therefore = VRdl = 21.6 kN/rib 4.3.2.4.3(1) V A SW d x 0.9df,wd >_ 29.8 - 21.6 = 8.2 kN/rib Eqn 4.23 s 5.4.2.2(7) Check maximum longitudinal spacing of links VRd2 ( 2 ) v cdbw x 0.9d = 0 (1 + COta) For vertical stirrups, cot« P = 0.7 - ck = + Eqn 4.25 VRd2 = VRd2 0.5 16.7 (O < VSd < ( 3) VRd2 Therefore smax = 0.6d = 139 > 300 mm z Eqn 5.18 Table 5.5 Try mild steel links at 125 mm crs. SIN 0.575 >_ 0.5 200 Eqn 4.21 x 0.575 x x 125 x 0.9 x 232 x 10-3=125kN Pw,min = 0.0022 A 0.00226W s E = 35 mm2 E = 57 mm2) Use R6 links @ 125 mm crs. (Asw f ywd = 250 1.15 = 217 N/mm2 V d = 57 125 x 0.9 x 232 x 217 = 103 20.7 > 8.2 kN/rib ... OK Link spacing may be increased where VSd 300 = 185 mm Eqn 5.17 Use R6 links @ 175 mm crs. apart from region within 0.6 m of interior support V,d = 14.7 > 3.4 kN/rib ............................. OK SLABS ,It 4.1.3.1.9 Shear between web and flanges AFd VSd 4.3.2.5 Eqn 4.33 a a V _ (2) SIN to = 2550 mm Figure 4.14 Maximum longitudinal force in the flanges FC _ = = «fcd(0.8x)b x d F 0.075 at mid-span 14.2 x 0.8 x 0.075 x 232 x600 _ 103 122kN Force to one side of web AFd = 122 x 600 - 195 2 x 600 = 41.2 kN Therefore NIA Vsd _ 41.2 = 16.2 kN/m 2.55 0.2 cdhf = 0.2 x 16.7 + s, VRd2 = x 100 = 334 kN/m > VSd .. OK Eqn 4.36 Eq n 4.34 Eqn 4.37 V_ With Asf 2.5To,,h, = = 0 2.5 x 0.3 VRd3 x 100 = 75 kN/m > VSd ........... OK Eqn 4.35 A No shear reinforcement required 4.1.3.1.10 Topping reinforcement No special guidance is given in EC2 regarding the design of the flange spanning between ribs. The Handbook to BS 8110(13) gives the following guidance. .CJ)) car 361.5 Thickness of topping used to contribute to structural strength Although a nominal reinforcement of 0.12% is suggested in the topping (3.662), it is not insisted upon, and the topping is therefore expected to transfer load to the adjacent ribs without the assistance of reinforcement. The mode of transfer involves arching action and this is the reason for the insistence that the depth be at least one-tenth of the clear distance between the ribs. Minimum flange depths are the same in EC2 and BS 8110 and the above is therefore equally applicable. Provide minimum reinforcement transversely and where top bars in rib, which have been spread over width of flange, are curtailed. .v) 2.5.2.1(5) E_0 SLABS Asf 4 < 0.6btdflfyk hf g 0.0015btdf Eqn 5.14 df = 100 mm Therefore, conservatively Asf 4 150 mm2/m Use T8 @ 200 mm crs. (251 mm2/m) or consider fabric 7°t 4.1.3.1.11 Deflection 6000 Actual span/effective depth ratio 232 4.4.3.2 25.9 403 Mid-span reinforcement ratio, p = 600 x 232 = 0.0029 Therefore section is lightly stressed. 4.4.3.2.(5) NAD Basic span/effective depth ratio (interpolating for p) = 39.2 400 Table 7 x 403 1.19 Modification factor for steel stress = 460 x 295 Since flange width > 3 x rib width, a 0.8 modification factor is required. Since span > 7 m, no further modification is required. Permitted span/effective depth ratio = 39.2 x 1.19 x 0.8 = 37.3 > 25.9 ................................. OK 4.1.3.1.12 Cracking of 0.3 mm For exposure class 1, crack width has no influence on durability and the limit could be relaxed. However, the limit of 0.3 mm is adopted for this 4.4.2.1(6) example. Satisfy the requirements for control of cracking without calculation. Check section 4.4.2.3(2) 4.4.2.2(3) at mid-span: Minimum reinforcement, A. 0 Note: -O_ = kckct,effAct/as Eq n 4.78 can be conservatively taken as the area below the neutral axis for the plain concrete section, ignoring the tension reinforcement, as shown in Figure 4.10. Act 00^D SLABS 92 Neutral axis 1100 175 i 35 I 125 35 Figure 4.10 Tensile zone of plain concrete section Depth to neutral axis Act QS = 92 mm = = 160 x 175 + 600 (100 - 92) = 32800 mm2 100%f ,k = 460 N/mm2 E 4.4.2.2(3) fct,eff III = = = recommended value 3 N/mm2 0.4 for normal bending 0.8 kc k AS = 0.4 x 0.8 x 3 x 32800/460 = 69 mm2 < E E As,prov ..... OK Eqn 4.78 Table 4.11 Check limit on bar size. Quasi-permanent loads = Gk + 0.30 k = 6.1 kN/m2 4.4.2.3(3) 2.3.4 E Ratio of quasi-permanent/ultimate loads =136.7 = 0.45 Eqn 2.9(c) NAD Table Estimate of steel stress 0.45 x LC) 1 As,req x fyd = A S,Prav 0.45 x 295 x 400 403 (Y) = 132 N/mm2 Maximum bar size For cracks = 32 > 16 mm provided .................. OK Table 4.11 caused dominantly by loading, crack widths generally will not be 4.4.2.3(2) excessive. 4.1.3.1.13 Detailing Minimum clear distance between bars = 0 r, 20 mm Nominal clear distance in rib 5.2.1(3) = 49 mm ...................... OK 5 SLABS Bond and anchorage lengths: For h > 250 mm bottom reinforcement is in good bond conditions. Top reinforcement is in poor bond conditions. ((DD 5.2.2 5.2.2.1 Figure 5.1(c) Therefore, ultimate bond stresses are Bottom reinforcement, Top reinforcement, fbd fbd = 0.7 2.7 N/mm2 = x 2.7 E 5.2.2.2(2) Table 5.3 = 1.89 N/mm2 5.2.2.2(2) Basic anchorage length, lb = Ofyd 4fbd 5.2.2.3 Eqn 5.3 c4) For top reinforcement, 1b = x 400 4 x 1.89 = 530 For bottom reinforcement, lb 0 x 400 4 x 2.7 370 Anchorage of bottom reinforcement at end support. Treat as a solid slab and retain not less than half of the mid-span reinforcement. Use 2T12 L bars bottom at end support 5.4.2.1.4 5.4.3.2.2(1) 0 0 Anchorage force for this reinforcement with zero design axial load 0 a, 0 F where VSd = VSd X 0 5.4.2.1.4(2) d Eqn 5.15 = 21 kN/rib 5.4.2.1.3(1) For vertical shear reinforcement calculated by the standard method ar = z(1 - Iota)/2 4 0 a = 90° and z is taken as 0.9d 5.4.3.2.1(1) E Although this ribbed slab falls outside the solid slab classification requirements for analysis, treat as a solid slab for detailing and take al = d. c,5 Therefore Fs As,req 0 = 21 21 kN/rib - x 400 103 = 53 mm2 < As Prov OK Required anchorage length for bottom reinforcement at support: as 1bAs req 1b,net 19 5.2.3.4 5.2.3.4.1(1) Eqn 5.4 1b,min A s, Prov = = as 0.7 for curved bars in tension 0.31b lb,mn = 11.10 .9 100 or 100 mm Eqn 5.5 106 SLABS In calculations of should be taken Ib,net' As,req 4 As,spanA = 101 mm2 NAD 6.5(c) 5.4.2.1.4(3) A s,prov lb,net = 0.7 226 mm2 = x 37 x 12 x 101 = 139 mm 226 > lb,min ....... OK Eqn 5.4 Minimum transverse reinforcement (for indirect support): Ast E E 5.2.3.3 (`") = As14 = 226/4 = 57 mm2 Use 1T8 bar as transverse reinforcement Minimum top reinforcement at end support: MsUp E 5.4.2.1.2(1) = = (4) 26.7 = 6.7 kNm/rib M 0.040 Therefore nominal reinforcement is sufficient. Use 2T12 L bars top as link hangers The reinforcement details are shown in Figure 4.11. 2T12 per rib Figure 5.12 L TO R6 - 175 links L 2T12 2T16 per rib per rib 100 622 Is Lb, net b/3 Figure 4.11 Detail at edge support ls, Provide full lap length, is 1b,neta1 for bottom bars: 5.2.4.1.3 -K ls,min For 100% of bars lapped and b > 20, «1 = 1.4 Hence with « a = 1.0 and A s,req = A s,prov lb,net = = lb = 370 = 37 X 12 = 444 mm twin 0.3 «acxllb = 187 mm g 150 or 200 mm m m Eqn 5.7 NAD Table 3 Figure 5.6 Eq n 5.4 Eqn 5.8 SLABS Therefore is = 444 x 1.4 = 622 mm C"6 > ls,min ................ OK Transverse reinforcement at lapped splices should be provided as for a beam section. Since 0 < 16 mm, nominal shear links provide adequate transverse reinforcement. 5.2.4.1.2(1) 5.4.2.1.5 L() Anchorage of bottom reinforcement at interior support. Treat as a solid slab and continue 50% of mid-span bars into support. (OD OD) 5.4.3.2.2(1) Figure 5.13(b) The reinforcement details are shown in Figure 4.12. 4T12 per rib R6- 125 links L 2T16 '- per rib M6 per rib I b, net 10 0 = 160 Figure 4.12 ((D Detail at interior support cages because of the intersection of the bottom reinforcement with the supporting beam cage. It is suggested that providing suitably lapped continuity bars through the support should obviate the need to continue the main steel into the support. This detailing prohibits the easy use of prefabricated rib The arrangement of the reinforcement within the section including the anchorage of the links is shown in Figure 4.13. a(0 --I Figure 5.10 Figure 4.13 Arrangement of reinforcement 108 --+ -Z- 5.2.5 NAD Tables 3&8 5.4.2.1.2(2) SLABS 4.2 Flat slabs 4.2.1 Flat slabs in braced frames The same frame is used in each of the following examples, but column heads are introduced in the second case. 4.2.1.1 Design example of a flat slab without column heads Design the slab shown in Figure 4.14 to support an additional dead load of 1.0 kN/m2 and an imposed load of 5.0 kN/m2. Figure 4.14 Plan of structure The area shown is part of a larger structure which is laterally restrained in two orthogonal directions by core walls. (CD (n- The slab is 225 mm thick. All columns are 300 mm square and along grid 5 there is an edge beam 450 mm deep x 300 mm wide. 4.2.1.1.1 Durability For a dry environment, exposure class is 0 Table 4.1 ENV 206 Table NA.1 1. Minimum concrete strength grade is C25/30. Since a more humid environment is likely to exist at the edges of the slab, increase concrete strength grade to C30/37. 7-C For cement content and w/c ratio, refer to ENV 206 Table 3. v3) SLABS Nominal cover to reinforcement = 20 mm NAD Table 6 Nominal cover to all bars .9 bar size .9 NAD 6.4(a) = 20 mm . nominal aggregate size 20 mm OK 4.1.3.3(5) Use nominal cover = 4.2.1.1.2 Materials Type 2 deformed reinforcement, fvk = 460 N/mm2 NAD 6.3(a) C30/37 concrete with 20 mm maximum aggregate size 4.2.1.1.3 Load cases It is sufficient to consider the following load cases 2.5.1.2 (a) Alternate spans loaded with yGGk + yoQk and yGGk on other spans. (b) Any two adjacent spans carrying yGGk + yoQk and all other spans carrying Gk yGGk. = = 0.225 x 24 + 1.0 CJ) = 6.4 kN/m2 yGGk 1.35 x 6.4 = = 8.7 kN/m2 Table 2.2 yGGk + yoQk 8.7 + 1.5 x 5.0 = 16.2 kN/m2 Eqn 2.8(a) 4.2.1.1.4 Analysis Analyses are carried out using idealizations of both the geometry and the behaviour of the structure. The idealization selected shall be appropriate to the problem being considered. (On 2.5.1.1.P(3) and P(4) = No guidance is given in EC2 on the selection of analysis models for flat slabs, or on the division of panels into middle and column strips and the distribution of analysis moments between these strips. This is left to the assessment of individual engineers. The requirements set down in BS 8110 for the above points are taken as a means of complying with EC2 Clause 2.5.1.1P(3). (D3 cam -00 00, om' EC2 allows analysis of beams and slabs as continuous over pinned supports. It then permits a reduction in the support moment given by AMsa The analysis in this example includes framing into columns. Thus the reduction AMSd is not taken. Consider two frames from Figure 4.14 as typical: (i) cmn >-O (C) 2.5.3.3(3) 2.5.3.3(4) Fsasup bsup /8 (ii) Grid 3/A-D subframe Grid B. 110 SLABS Analysis results for the frames described above are given in Figure 4.15. The results for each frame are practically identical as the analysis for Grid B has an increased loaded width (5.2 m), since this is the first internal support for frames in the orthogonal direction. Member stiffnesses have been based on a plain concrete section O:3 Column moments and reactions are given in Table 4.1. 4250 5200 5200 (Dw in this analysis. Islab 2 3500 3500 /7777 /777 /7777 ANALYSIS MODEL -198 -199 -204 123 BENDING MOMENT ENVELOPE (kNm) Figures 4.15 Analysis of frame SLABS Table 4.1 Column moments and reactions Max reaction . Support °°_ (kN) E column moments (kNm) 37.9 Max. E column moments (kNm) 37.9 End 1st interior 156.4 444.7 6.8 21.4 4.2.1.1.5 Flexural design - Panel A-13/1-2 BS 8110 3.7.4.2 EC2 does not specifically address the problem of edge column moment transfer and the provisions of BS 8110 are adopted here. Mt,max - 0.15bed 2fCU Column A12 moment transfer Assuming 20 mm cover and 20 mm bars in the top d, d2 O.0 NAD 4.1.3.3(5) = = = = = 225 195 - 20 - 10 = 195 mm - 20 = 175 mm E be 300 + 300 (say) = 600 mm E 37 N/mm2 0.15 E fCU Mt,max + x 600 x This ought to be compared with an analysis for a loading of 1 AGk + 1.60k, which would give approximately 5% higher edge moments than the EC2 analysis results above. Mt,max c)' x E 1752 x 37 x 10-6 = 102 kNm = 102 > 1.05 x 37.9 = 39.8 kNm ................ E CA) OK Design reinforcement to sustain edge moment on 600 mm width. E x E Using ry = 1.5, a = 0.85 and -ys = 1.15 Table 2.3 Referring to Section 13, Table 13.1: Msd 37.9 x 106 bd2fck AS Yk 600 x 1752 x 30 CA) = 0.069 = 0.085 bdtck A S - 0.085 x 600 x 175 x 30 460 = 582 mm2 E = 970 mmz/m d = 0.163 CA) X < 0.45 (zero redistribution) ................. OK 2.5.3.4.2(5) SLABS Use T16 @ 150 mm crs. (1340 mm2/m) top at edge column Place over width = 900 mm (see Figure 4.16) Note: This approach gives more reinforcement than is necessary. Figure 4.16 Edge column moment transfer Check above moment against minimum value required for punching shear. msd 4.3.4.5.3 Eq n 4.59 ? Vsd For moments about axis parallel to slab 77 edge Table 4.9 _ ± 0.125 per m = 156.4 kN Vsd Therefore msd _ ± 0.125 x 156.4 = 37.9 = ± 19.6 kNm/m Edge moment = 63.2 > 19.6 kNm/m 0.6 ............... OK Design for msd above in region outside edge column moment transfer zone. msd bd2f ck _ 19.6 x 106 = 0.021 1000 x 1752 x 30 Minimum steel sufficient = 0t d .6b f 4 0.0015btd = 263 mm2/m E E 5.4.2.1.1 vk = 0.0015 x 1000 x 175 Use T12 at 300 mm crs. (373 mm2/m) top and bottom (minimum) Maximum spacing = 3h t> 500 = 500 > 300 mm ....... OK NAD Table 3 5.4.3.2.1(4) (Y) SLABS Column A/1 moment transfer Assume the design forces for the frame on grid for grid 3 in proportion to their loaded widths. Load ratio = (4.25/2) 1 are directly related to those = 0.41 5.2 The ratio of the edge column distribution factors for the frames is 2.0. Msd = 37.9x0.41 x2.0 = 31.1kNm Using design approach as for column A/2: be = 300 + 320 (say) = 450 mm Mt max = 0.15 x 450 x 1752 x 37 x 10-6 = 76 kNm > 1.05 x 31.1 = 32.7 kNm ..................... OK Design reinforcement to sustain edge moment on 450 mm width Msd _ 31.1 x 106 bd2lck AS rk 450 x 1752 x 30 = 0.075 = 0.093 bdlck AS _ 0.093 x 450 x 460 175 x 30 = 478 mm2 = 1062 mm2/m Use T16 @ 150 mm crs. (1340 mm2/m) top for a width of 600 mm Check above moment against minimum value required for punching shear. msd >_ 4.3.4.5.3 "f Msd where n = ± 0.5 per m for corner columns 0 0 0 _ Edge moment (O" + 0.5 x (0.41 x 156.4) say = = 69.1 ± 32.1 kNm/m x Eq n 4.59 Table 4.9 = 31.1/0.45 kNm/m > 32.1 .......... OK In region of slab critical for punching shear: Msd bd2fck 32.1 x G.) 106 (A) = 0.035 1000 x 1752 x 30 ASfyk bdfck As = 0.042 _ 0.042 x 1000 x 175 x 30 = 460 480 mm2/m E SLABS Use T16 @ 300 mm crs. (670 mm2/m) top and bottom outside 600 mm wide moment transfer zone and over area determined in punching calculation LOLL The division of panels into column and middle strips is shown in Figure 4.17. BS 8110 Figure 3.12 Although BS 8110 indicates a 2.36 m wide column strip at column 132, a 2.6 m width has been used in the following calculations. This is considered reasonable as a loaded width of 5.2 m has been taken in the analysis for grid B and grid 2. E 1 A 8 C 4-25m 5.2 I1 06m I I 4-25m I =I-I I 11.06ml --I I I 1 I I I I I rt ----TI I I I I I 1 I 1.06m I 1-301M I I I I 5-2m I I I I I I I I I -I--,--T 1.06m T -#-----II II 3 Figure 4.17 Assumed strip widths (arrangement symmetrical about diagonal A/1- C/3) E E Column 812 support moments Analysis moment = = 198 kNm in both directions 0 Column strip b Msd,cs Msd.cs 0.75 x 198 = 2600 E 149 kNm BS 8110 Table 3.20 = 1300 x 2 = mm = 0.062 E 149 x 106 bd2tck 2600 x 1752 x 30 11 SLABS Asyk bdyk A s = 0.076 0.076 = x 2600 x 175 x 30 460 = 2255 mm2 E E Use 13T16 (2613 mm2) top in column strip. Provide 9T16 @ 150 mm crs. in central 1.3 m and 2T16 @ 300 mm crs. on either side BS 8110 3.7.3.1 Check whether minimum moment required for punching shear has been met. With rt E 4.3.4.5.3 C`7 = -0.125 _ Table 4.9 = Msd " Vsd -0.125 x 444.7 = -55.6 kNm CJ) Eq n 4.59 This is to be carried over a width of 0.31. Since Vsd includes for a loaded width of 5.2 m, it is assumed that the larger panel width may be used. 0.31 = 0.3x5.2 = 1.56m By inspection reinforcement (9T16 in central 1.3 m) is sufficient. ..... OK Middle strip (using average panel width) Msd,ms _ 0.25 x 198 (4725 x 106 bd 2f x ck - 2600) x 1752 x 30 = 0.026 d Asyk = 0.059 < 0.45 ............................... OK 2.5.3.4.2(5) 0.031 bdick As b 0.031 x 1000 x 175 x 30 460 = 354 mm2/m E Use T16 @ 300 mm crs. (377 mm2/m) top in middle strip It is noted that EC2 Clause 2.5.3.3(5) would allow the use of the moment at the face of the support (subject to limits in EC2 Clause 2.5.3.4.2(7)), but this is considered more appropriate to beams or solid slabs and the peak moment over the support has been used in the above design. E E gym- _O) Span moments Q._0 be provided O-0 No special provisions are required in EC2. Hence the design basis of BS 8110 is adopted for the division of moments. The same pattern of reinforcement will in all panels. The column strip moments are given in Table 4.2 where Msd,cs = 0.55 Msd 11d = > 2035 mm .. OK length breadth Hence u Vsd = 1 2 ................................. OK = = 27r x 1.5 x 185 + 1200 = 2944 mm Figure 4.18 444.7 kN Note: No reduction in this value has been taken. The applied shear per unit length: Vsd R 4.3.4.3(4) v,, = u where Q internal column = 1.15 Eq n 4.50 Figure 4.21 444.7 x 103 x 1.15 Vsd = 174 N/mm 2944 4.3.4.5.1 `'/ 1000 x 185 Note: The amount of tensile reinforcement in two perpendicular directions -U) Assume plx + ply pl/ CC/) Shear resistance without links VRd1 = 7Rdk(1.2 + 40pi)d Eqn 4.56 Table 4.8 TRd = = 0.34 N/mm2 1.6 k pl - d = 1.6 - 0.185 = 1.415 >_ 1.0 = reinforcement ratio within zone 1.5d from column face (T16 (& 150 mm crs. top each way gives 1340 mm2/m) Pt = ), x ply > 0.015 0.0072 1340 > 0.5%. OK 4.3.4.1(9) = 2 (0.0072) > 0.005 ..................... SLABS Therefore VRdl VSd = = 0.34 x 1.415 x (1.2 + 40 x 0.0072) x 185 = 133 N/mm E 174 N/mm > VRdl Therefore shear reinforcement required such that Slab depth >_ VRd3 > VSd 4.3.4.3(3) 200 mm ..................................... = 2.0 x 1 OK 4.3.4.5.2(5) Check that applied shear does not exceed the maximum section capacity VRd2 - 2 .0 VRdl 33 = 266 > 174 N/mm ..... OK NAD Table 3 = 444.7 x 103 1200 x 185 4.3.4.5.2(1) Shea r stress aro und colu mn perimet er = 2.0 N/mm2 Therefore no shear reinforcement required 4.3.4.3(2) SLABS Column A12 (300 mm x 300 mm edge column) Critical perimeter located at 1.5d from face of column. U = = 900 + 2777r 156.4 kN = 1770 mm Vsd Applied shear per unit length, with a = 1.4 _ VSd Figure 4.21 = 124 N/mm u 1770 VRd1 = 133 N/mm (as for column B/2) VRd1 VSd Therefore no shear reinforcement required 4.2.1.1.7 Deflection Control by limiting span/effective depth ratio using NAD Table 7. For flat slabs the check should be carried out on the basis of the longer span. For span < 8.5 m, no amendment to basic span/effective depth ratio is required. Note 2 to NAD Table 7 states that modifications to the tabulated values for nominally reinforced concrete should not be carried out to take into account service stresses in the steel (refer to EC2: Clause 4.4.3.2(4)). However it is assumed that correction ought to be carried out for 0.15% Ri = 145 kN case for multi-bay frames unless the edge 4.3.5.3.5(2) columns carry large cladding loads. = l /i O where CD 128 0 SLABS Aol = 300 12 = 86.6 mm For a horizontally loaded flat slab frame determine the stiff nesses of the frame and thus the effective lengths of the columns using half the slab stiffness. Consider the centre column.from foundation to first floor. kA = EIcol/Icol ; Ealb/leff 3004 12 Ib _ = 4725 x 2253 2 x E lCol 3500 mm 5200 mm E 1.0 2.5.2.2.2 0 = ECM assumed constant Qom) 0.675 x 109 mm4 W°i 4.3.5.3.5(1) Eqn 4.60 = 2.24 x 109 mm4 12 leff = = a Therefore k A _ 2(0.675 x 109/3500) 2(2.24 x 10915200) = oo = 0.5 kB (pinned at foundation) R Assuming that EC2 Figure 4.27(b) is appropriate to determine to = alco = 2.15 x 3500 = 7500 mm Figure 4.27 Hence (fl x = 7500/86.6 = 87 A.3.2(3) For non-sway frames A 0.2 4.3.5.6.4(3) A reduced value of h, therefore, must be used in carrying out a check for bending in the y direction. The additional eccentricity in the z direction is 0.51/200 0 L!) = 20 mm E Hence ez + e az = 120 mm -vim (eYlb)l(e,lh) = 100 = < 0.2 .L. E E COLUMNS is assumed that the intention of EC2 Section 4.3.5.6.4(3) is that, using the reduced section, the applied load should just give zero stress at the least stressed face, i.e. as shown in Figure 5.9. It C/) Actual section Elastic stress distribution Reduced section h' Elastic stress distribution on reduced section Figure 5.9 It will Assumption for check in the y direction lie of be seen that the point of application of the load must the middle third of the reduced section. on the edge Hence h' = - eZ 3(200 - 120) 3(h12 eaz) = 240 mm 5.5.4 Check for bending in z direction 2 1.15 LC) This check uses the full section dimensions 1 x 460 x 10-6 K2 x r x 0.9 x 340 x 0.2 13.07K2 x 10-6 4.3.5.6.3 Eqn 4.72 COLUMNS Hence e2 = 0.1 x = 82 1 x 106 x 13.07K2 x 10-6 = (Y) 83.7K2 mm Eqn 4.69 (Since a > 35, K, etot in EC2 Eqn 4.69) = 100 + 20 + 83.7K2 mm 4.3.5.6.2 + As in the previous example, iterate using the design chart in Section 13 Figure 13.2(c) to find the appropriate value for K2 and hence ASyk/bh starting with K2 = 1. This procedure results in fck, E Eqn 4.65 K2 = = = 0.8 M1bh2 ck 0.194 0.417 NlbhfCk Hence AjA1bh f As ck= 0.55 0.55 = x 4002 x 30/460 = 5739 mm2 E E Use 12T25 (5890 mm2) 5.5.5 Check for bending in y direction The assumed section is shown in Figure 5.10. r.4 Figure 5.10 Reduced section for check in y direction e oy = 16 mm eay e2y = = eaz = e2z 20 mm 83.7K2 mm = Hence etot = 36 + 83.7K2 mm Nlbhf,k 2000 x 103 = 0.694 240 x 400 x 30 146 COLUMNS Mlbh`f°k = 0.694 x (36 + 83.7K2 ) 400 0.0625 + 0.145K 2 AS yklbh Using the same design chart as before, iterate to obtain This gives ok. K2 K2 and hence = 0.47 0.13 Mlbh2fCk = Hence AS vk/bhfCk= 0.57 0.57 x 240 A3 = x 400 x 30/460 = x 3569 mm2 OK This is less than required for z direction bending ................. An appropriate arrangement of reinforcement is shown in Figure 5.11. -1Z J =Z_. z Y Figure 5.11 Arrangement of reinforcement 5.6 Classification of structure 5.6.1 Introduction EC2 provides more detailed rules than BS 8110(2) for deciding whether or not a structure is braced or unbraced, or sway or non-sway. While it will normally be obvious by inspection how a structure should be classified (for example, with shear walls it will be braced and non-sway), there may be cases where direct calculation could give an advantage. The structure in the example following is chosen to illustrate the workings of EC2 in this area. It is entirely hypothetical and not necessarily practical. coo 5.6.2 Problem Establish an appropriate design strategy for the columns in the structure shown in Figures 5.12 and 5.13. The applied vertical loads in the lowest storey are set out in Table 5.1. ??w gin" (O] T7`- COLUMNS a c d x 300 x 300 c a O 11 C ') d 750 450 YLx 400 x 300 -10-O 0 C 300 x 300 Figure 5.12 General arrangement of columns Figure 5.13 Cross-section of structure Table 5.1 Column sizes and loads Column dimension (mm) 2nd moment of area (mm x 10-6) Service Ultimate (D3 Column type a b c load load (kN) y 300 300 750 x 400 400 450 Iv 900 900 15820 (7) 1x (kN) 1900 1600 2680 0 0 0 1600 5695 675 2100 0 0 0 2960 3300 Q') 0 4660 d 300 300 675 1200 1700 COLUMNS 750 x 450 columns forming the bracing elements To be considered as braced, the bracing elements must be sufficiently stiff to attract 90% of the horizontal load. Since all columns are the same length, this will be so if c,0 5.6.3 Check if structure can be considered as braced with the 5.6.3.1 y + Hence the four 750 x 450 columns can be treated as bracing elements carrying the total horizontal loads and columns type a, b and d can be designed as braced in the y direction 5.6.3.2 x direction 4 x 5695 EIbracing _ EItot x x 5.6.4 Check if structure can be considered as non-sway Classification of structures as sway or non-sway is covered 5.6.4.1 y in u1,-- 4.3.5.3.2(1) > 0.9 direction EIbracing 4 x 15820 6 x 900 + 4 x 15820 + = 2 0.904 x 675 direction Q)) For braced structures of four or more storeys, the frame can be classified as non-sway htot where htot Taking 0-0 6 x 1600 + 4 x 5695 + 2 x 675 = 0.68 Structure cannot be considered as braced in x direction EC2 Appendix 3. if F I Ecmlc _< 0.6 A3.2 Eq n A.3.2 c`') = height of frame in metres = 4 x 3.5 1 = 14 m F sum of all vertical loads taking yf = 4 x 1900 + 2 x 2100 + 4 x 3300 + 2 x 1200 = 27400 kN ECM I = sum of the stiff nesses of the bracing elements. 3.1.2.5.2 ECM as 32000 N/mm2 Table 3.2 ECMIc = = 4 x 15820 x 32000 x 109 106 Nmm2 E 2024960 x Nmm2 E E x COLUMNS Hence 27400 x 103 /mm = 0.000116/mm = 0.116/m 2024960 x 109 Note: Since the height of the building is stated to be in metres, it seems reasonable to assume that m units should be used for the other factors, though this is not stated in EC2. Hence htot = 14x0.116 = x o-(° 1.62 > 0.6 Therefore the bracing structure is a sway frame in the y direction 5.6.4.2 x direction CCD For frames without bracing elements, if X < greater of 25 or 15/ Ffor all elements carrying more than 70% of the mean axial force then the structure may be considered as non-sway. A.3.2(3) Mean axial force = sum of ultimate column loads no. of columns 4 x 2680 + 2 x 2960 + 4 x 4660 + 2 x 1700 x NSd,m = 700/oNsd,m 12 38680/12 = 3223 kN = 2256 kN Columns type d carry less than this and are therefore ignored. i6- Assume effective length of 400 x 300 columns is 0.8 x 3.5 = 2.8 m (i.e. value appropriate to a non-sway condition). X = 24.25 < 25 Therefore structure is non-sway 5.6.5 Discussion The results obtained in Sections 5.6.4.1 and 5.6.4.2 above are totally illogical as the structure has been shown to be a sway structure in the stiffer direction and non-sway in the less stiff direction. There are two possible areas where the drafting of EC2 is ambiguous and the wrong interpretation may have been made. m x (1) Eqn A.3.2 it is specifically stated that the height should be in metres. Nothing is stated about the units for C, FV and E.. Since the output from Eqn A.3.2 is non-dimensional, the statement of the units is unnecessary unless the units for F and Ec are different to that for htot. Should IC, FV and Ec be in N and mm units while htot is in m? If this were so, then the structure would be found to be 'braced' by a large margin. In 3`T I , 150 (1) ,..°E -CV c E COLUMNS (2) In A.3.2(3) it does not state whether X should be calculated assuming the columns to be sway or non-sway. In the calculation, the assumption was made that the X was a non-sway value. If a sway value had been adopted, the structure would have proved to be a sway frame by a considerable (Q. margin. Clearly, clarification is required It if A.3.2 is to be of any use at all. is possible to take this question slightly further and make some estimate at what the answer should have been. Considering the y direction, the ultimate curvature of the section of the 750 x 450 columns is 1 2x460xK2 0.2x106x1.15x0.9x700 E = r 6.35K2 2 x 10_6 Inspection of the design charts and levels of loading suggest K2 is likely to be about 0.6. Assuming an effective length under sway conditions of twice the actual height gives a deflection of: 0 (2 x X 3.5)2 x6.35x0.6 = 19 mm E E 10 This is an overestimate of the actual deflection. It corresponds to an eccentricity of 19/750 of the section depth or 2.5%. This must be negligible, hence, in the y direction, the structure must effectively be non-sway. 5.7 Sway structures 5.7.1 Introduction Although EC2 gives information on how to identify a sway structure, it does not give any simple approach to their design. However, Clause A.3.5.(2) states that "the simplified methods defined in 4.3.5 may be used instead of a refined analysis, provided that the safety level required is ensured". Clause A.3.5(3) amplifies this slightly, saying that ''simplified methods may be used which introduce ....... bending moments which take account of second order effects ...... provided the average slenderness ratio in each storey does not exceed 50 or 201 v, whichever is the greater' 5'w :(Q Cam. °-' -^. Z30 . EC2 Section 4.3.5 gives the 'model column' method which is developed only for non-sway cases, so it is left to the user to find a suitable method for sway frames on the basis of the Model Column Method. BS 8110 does this, so it is suggested that the provisions of 3.8.3.7 and 3.8.3.8 of BS 8110: Part 1 are adopted, but that the eccentricities are calculated using the equations in EC2. E 5.7.2 Problem Design columns type c in the structure considered in Section 5.6.2 assuming sway in the x direction. The column loads may be taken from Table 5.1. The design ultimate first order moments in the columns are as shown in Figure 0 5.14. _0) a has been assessed from EC2 Figure 4.27(b) as 1.6 for all columns. 151 (OD ((A OH- -LO O-0 COLUMNS Mo Column Mo type a b C kNm 90 110 176 d 25 Figure 5.14 First order moments 5.7.3 Average slenderness ratio The slenderness ratios are shown in Table 5.2. Table 5.2 Slenderness ratios No. Column type x a b C 4 2 4 2 48.5 48.5 43.1 d 64.7 49.4 Mean value (a) = Since Xm < 50, the simplified method may be used. 2K2 A.3.5(3) 1 x 460 0.0044K2 r Hence e2 200000 x 1.15 x 0.9d d 106 _ (1.6 x 10 3.5)2 x 0.0044K2 x x of _ 13800K2 mm of 100 14 200 2.5.1.3 Eqn 2.10 This may be multiplied by «n Where, with 12 columns «n Eqn 2.11 _ (1 + 1/12)/2 = 0.736 Hence V = = = 0.00368 ea etot 0.00368 x 1.6 x 3500/2 eo + ea = 10.3 mm Eqn 4.61 + 13800K2/d = eo + 10.3 + 13800K2/d mm Eqn 4.65 The total eccentricities are shown in Table 5.3. + 5 COLUMNS Table 5.3 Column type a Total eccentricities 3rD d (mm) 350 Mo (kNm) 80 eo (mm) 30 N eV `'' bhfck 0.744 (mm) 40.3 + 39K 2 b c 350 400 110 176 37 38 0.822 0.460 47.3 + 39K 2 48.3 + 35K 2 L() d 250 25 15 0.630 25.3 + 55K 2 As in the previous examples, the design charts can be used iteratively to establish K2 and hence e2. This process gives the values shown in Table 5.4. Table 5.4 Column type Lateral deflections K z (D3 e2 No. of (mm) columns 4 2 4 2 a b 0.39 0.41 15.2 16.0 c 0.50 0.45 Average deflection 17.5 24.8 = 17.7 mm ((') d columns will be assumed to deflect by the average value. The resulting designs are shown in Table 5.5 All ---1 E BS 8110 3.8.3.8 Table 5.5 Column type a b c Summary of designs e (mm) M bh2ck N bhfck 0.744 0.822 0.460 AsfYk A (mmz) bh ck 0.53 0.75 0.10 58.0 65.0 66.0 0.108 0.134 0.067 4148 5870 2201 d 43.0 0.090 0.630 0.38 2230 6.1 Introduction defined as a vertical load-bearing member with a horizontal length not less than four times its thickness. A wall is 2.5.2.1(6) The design of walls is carried out by considering vertical strips of the wall acting as columns. 0)c SOS = = 6.2 Example Design the lowest level of a 200 mm thick wall in an eight storey building supporting 250 mm thick solid slabs of 6.0 m spans on each side. The storey heights of each floor are 3.5 m, the height from foundation to the first floor being 4.5 m. The wall is fully restrained at foundation level. The building is a braced non-sway structure. 3-0(5. 6.2.1 Design data Design axial load (NSd) Design moment at first floor Design moment at foundation = = 700 kN/m 5 kNm/m = Concrete strength class is C30/37. (so 2.5 kNm/m nom 3.1.2.4 fCk 30 N/mm2 Table 3.1 6.2.2 Assessment of slenderness Consider a 1.0 m vertical strip of wall acting as an isolated column. The effective height of a column where 1COl to = 01col 4.3.5.3.5(1) actual height of the column between centres of restraint ko 0 is a factor depending upon the coefficients rigidity of restraint at the column ends. and kR relating to the kn = Assuming a constant modulus of elasticity for the concrete: E I o _ 1 x 12 0.23 = 6.67 x 10-4 x 10-3 m4 1 x 0.253 12 slab = = 1.3 m4 k A _ (6.67x10-4+6.67x10-4)I(2x1.3x 4.5 3.5 6 10-3) = 0.78 Base of wall is fully restrained. Therefore kB 4.3.5.3.5(1) = 0.40 which is the minimum value to be used for kA or kB. Figure 4.27(a) C)) E 4.3.5.3.5(1) slab/Ieff,slab Eqn 4.60 CA) col /lcol Q.- c,, WALLS Hence a = = 0.7 4500 mm = 0.7 x 4500 X = l 0li 3150 mm E The slenderness ratio = 4.3.5.3.5(2) where i = radius of gyration I _ 1000 x 2003 12 x 1000 TA Therefore X 57.7 mm E x 200 3150 57.7 = 54.6 Isolated columns are considered slender where or 15/ v. X exceeds the greater of 25 Nsd 4.3.5.3.5(2) '4ccd = Ac 700 M = 200 x E E 103 mm2 = 1000 x 200 = E °k yC cd Therefore vU = 30 1.5 = 20 Nlmm2 _ 700 x 103 200 x 103 x 20 = 0.175 Hence 15 u 15 = 35.9 0.175 Therefore the wall is slender 6.2.3 Design The wall may now be designed as an isolated column in accordance with EC2(1) Clause 4.3.5.6 and as illustrated in the example in Section 5. Although the column or wall has been classified as slender, second order effects need not be considered if the slenderness ratio X is less than the critical slenderness ratio Xcrlt. WALLS Xcrt = 25 (2 - eotleoz) 4.3.5.5.3(2) Eqn 4.62 of where eo1 and relating to the axial load. e of eoz are the first order eccentricities at the ends the member E = MSd, N Sd and e o2 = MSdz N Sd MSdt and MSdz are the first order applied moments. Therefore 25 (2 - MSdi/Msdz) where MShc ` MSdz These moments must be given their correct algebraic signs In in the equation. this example: 25 2 Xcrit - C)7 1-2.5) 50 L(7 = 62.5 > 54.6 The column or wall should therefore be designed for the following minimum conditions: 4.3.5.5.3(2) Design axial resistance (NRd) Design resistance moment (MRd) For this example MRd 6.2.4 Reinforcement The vertical reinforcement should not be less than 0.004Ao or greater than 0.04A C . Half of this reinforcement should be located at each face. 5.4.7.2(1) The maximum spacing for the vertical bars should not exceed twice the wall thickness or 300 mm. The area of horizontal reinforcement should be at least 50% of the vertical reinforcement. The bar size should not be less than one quarter of the vertical bar size and the spacing should not exceed 300 mm. The horizontal reinforcement should be placed between the vertical reinforcement and the wall face. ((DD te) CT0 E = Nsd Eqn 4.63 h Nsd = x 20 Eqn 4.64 = 700 x 0'2 20 = 7.0 > 5.0 kNm A 5 5.4.7.2(2) 5.4.7.2(3) 5.4.7.3 (1)-(3) WALLS Link reinforcement is required in walls where the design vertical reinforcement exceeds 0.02AC. 5.4.7.4(1) normal buildings it is unlikely that walls will be classified as slender. For practical considerations they will generally not be less than 175 mm thick and the vertical load intensity will normally be relatively low. Thus the limiting slenderness ratio given by 15/ vU will be high. _T_ In cases where the wall is slender, only slenderness about the minor axis need be considered. Even in this case it is likely that only the minimum conditions given in EC2 Clause 4.3.5.5.3(2) Eqns 4.63 and 4.64 will apply. In ((DD 15 can _--. 7 FOUNDATIONS 7.1 Ground bearing footings 7.1.1 Pad footing Design a square pad footing for a 400 mm x 400 mm column carrying a service load of 1100 kN, 50% of this being imposed load with appropriate live load reduction. The allowable bearing pressure of the soil is 200 kN/m2. 7.1.1.1 Base size With 500 mm deep base, resultant bearing pressure Area of base required Use 2.5 m x 2.5 m x 0.5 m deep base 7.1.1.2 Durability ENV 206 Minimum concrete strength grade is C30/37. For cement content and w/c ratio refer to ENV 206 Table 3(6). Table NA.1 NAD Minimum cover to reinforcement is 30 mm. For concrete cast against blinding layer, minimum cover > 40 mm. Use 75 mm nominal cover bottom and sides E E Table 6 4.1.3.3(9) 7.1.1.3 Materials Type 2 deformed reinforcement with yk = 460 N/mm2 E z NAD 6.3(a) Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.1.4 Loading Ultimate column load 7.1.1.5 Flexural design = 1.35Gk + 1.50k = 1570 kN Eqn 2.8(a) Table 2.2 Critical section taken at face of column 2.5.3.3(5) Msd 8x2.5 CC) = 346 kN m Assuming 20 mm bars dave = 500 - 75 - 20 = 405 mm Figure 4.4 Using rectangular concrete stress diagram cd = Ck = 30 1.5 = 20 N/mm2 z 1570 (2.5 - 0.4)2 'Y' Eqn 4.4 Table 2.3 «cd = 0.85 x 20 = 17 N/mm2 Zoo For components in non-aggressive soil and/or water, exposure class is 2(a). LO[) = 200 - 0.5 x 24 = 188 kN/m2 = 1100 = 5.85 m2 188 Table 4.1 FOUNDATIONS For reinforcement fyd = vk = 460 1.15 = 400 N/mm2 -Y' 2.2.3.2P(1) Table 2.3 2.5.3.4.2(5) For the design of C30/37 concrete members without any redistribution of E moments, neutral axis depth factor x d .............................. 20 OK Bar crs. _ 2500 = 2(75) 6 - = 388 mm Maximum spacing 3h > 500 = 500 > 388 mm ....... OK NAD Table 3 5.4.3.2.1(4) 7T20 (EW) are sufficient for flexural design. Additional checks for punching and crack control require 9T20 (EM refer to Sections 7.1.1.7 and 7.1.1.8. - Use 9T20 (EW) 7.1.1.6 Shear Minimum shear reinforcement may be omitted in slabs having adequate provision for the transverse distribution of loads. Treating the pad as a slab, therefore, no shear reinforcement is required if Vsd _ kckct,eff Act/as Eq n 4.78 For Act it is considered conservative to use (h/2)b Us = 100% x fYk = 460 N/mm2 For cteff use minimum tensile strength suggested by EC2 kc - 3 N/mm2 = 0.4 for bending For k interpolate a value for h = 50 cm from values given k Therefore As,req As,prov = 0.5 + 0.3(80 3 50)/(80 - 30) = 0.68 = = 0.4 CC) x 0.68 x (V) x 250 x 2500/460 = 1109 mm2 2830 > 1109 mm2 .......................... OK 7.1.1.9 Reinforcement detailing Check that flexural reinforcement extends beyond critical section for bending for a distance >_ d + lb,net + lb 5.4.3.2.1(1) & 5.4.2.1.3 = 33.30 = 667 mm Assuming straight bar without end hook Ib, net 1.0x667x2179 lb,net = 514 mm Eqn 5.4 2830 = 919 mm of + = 405 + 514 2500 2 Actual distance = The reinforcement details are shown in Figure 7.1. + - 2 400 - 75 = 975 > 919 mm ..... OK FOUNDATIONS 500 75 cover I Figure 7.1 9T20 2500 - 300EW Detail of reinforcement in pad footing 7.1.2 Combined footing Design a combined footing supporting one exterior and one interior column. An exterior column, 600 mm x 450 mm, with service loads of 760 kN (dead) and 580 kN (imposed) and an interior column, 600 mm x 600 mm, with service loads of 1110 kN (dead) and 890 kN (imposed) are to be supported on a rectangular footing that cannot protrude beyond the outer face of the exterior column. The columns are spaced at 5.5 m centres and positioned as shown in Figure 7.2. (AD The allowable bearing pressure is 175 kN/m2, and because of site constraints, the depth of the footing is limited to 750 mm. _-j Figure 7.2 Plan of combined footing 7.1.2.1 Base size (J) Service loads = Gk +0 k Column A: 1340 kN and Column B: 2000 kN Distance to centroid of loads from LH end 0.3 + 2000 x 5.5 3340 = 3.593 m E 164 q.0 E c + FOUNDATIONS For uniform distribution of load under base Length of base = ((DD 2 x 3.593 say 7.2 m E With 750 mm deep base, resultant bearing pressure = 175 - 0.75 x 24 = = 157 kN/m2 Width of base required = 7.2 3340 x 157 2.96 say3.0m Use 7.2 m x 3.0 m x 0.75 m deep base 7.1.2.2 Durability For ground conditions other than non-aggressive soils, particular attention is needed to the provisions in ENV 206 and the National Foreword and Annex to that document for the country in which the concrete is required. In the UK it should be noted that the use of ISO 9690(15) and ENV 206 may not comply with the current British Standard, BS 8110: Part 1: 1985 Table 6.1(2) where (On (n. sulphates are present. Class 2(a) has been adopted for this design. Minimum concrete strength grade is C30/37. For cement content and w/c ratio refer to ENV 206 Table 3. c(° Table 4.1 ENV 206 Table NA.1 NAD Minimum cover to reinforcement is 30 mm. For concrete cast against blinding layer, minimum cover > 40 mm. However, of it is suggested that nominal cover > 40 mm the above clause. (ti Use 75 mm nominal cover bottom and sides and 35 mm top 7.1.2.3 Materials Type 2 deformed reinforcement with fYk Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.2.4 Loading Ultimate column loads Column A: 1896 kN and Column B: 2834 kN Distance to centroid of loads from LH end 0.3 + i.e. virtually at centre of 7.2 m long base = 4730 7.2 Assume uniform net pressure ((DD See Figures 7.3, 7.4 and diagrams respectively. E Table 6 4.1.3.3(9) is a sufficient interpretation E 2834 x 5.5 4730 E = 7.5 for loading, a)) E E = 460 N/mm2 NAD 6.3(a) 1.35Gk + 1.50k Eqn 2.8(a) Table 2.2 = 3.595 m = 657 kN/m = 219 kN/m2 shear force and bending moment FOUNDATIONS 2834 kN (/3160 kN/m 4723 kN/m f t t-" Fib 657 kN/m + 4.9 m i 6 1.1m .l Figure 7.3 Loading diagram Figure 7.4 Shear force diagram Figure 7.5 Bending moment diagram FOUNDATIONS 7.1.2.5 Flexural design 7.1.2.5.1 Longitudinal direction - top steel Mid-span MSd = = 2167 kNm d 750 - 35 - 20 - 32/2 = 679 say 675 mm E E Using the design tables for singly reinforced beams Mc14 E 2167 x 106 0.053 bd2fCk 3000 x 6752 x 30 = d ASfyk 0.123 < 0.45 limit with zero redistribution OK 2.5.3.4.2(5) = 0.064 bdf,k AS 0.064 x 3000 x 675 x 30 = 8452 mm2 = 2818 mm2/m E 460 E E Use 12T32 @ 250 mm crs. (3217 mm2/m) Continue bars to RH end of base to act as hangers for links. awe Particular attention is drawn to the clauses for bar sizes larger than 32 mm. These clauses are restrictive about laps and anchorages, such that designers may need to resort to groups of smaller bars instead. 5.2.6.3P(1) & P(2) Maximum spacing = 3h > 500 = 500 > 250 mm E ......... OK NAD Table 3 7.1.2.5.2 Longitudinal direction At column face MSd _Z) S - bottom steel 5.4.3.2.1(4) = 398 kNm 750 - 75 - 10 of = = 665 mm E Mlz 398 x (7N 106 0.010 bd2fck ASfyk 3000 x 6652 x 30 = 0.012 bdck A 0.012 x 3000 x 665 x 30 460 = 1561 mm2 = 520 mm2/m For minimum steel As,m.n = x 0.0015btd = 998 mm2/m E 5.4.2.1.1 Use 12T20 @ 250 mm crs. (1258 mm2/m) (.J X ......... E E E c E FOUNDATIONS 7.1.2.5.3 Transverse direction - bottom steel Msd = 1.5 2 0.45 x 2 219 = 178 kNm/m Minimum steel governs. Use T20 @ 250 mm crs. (1258 mm2/m) 7.1.2.6 Shear Critical shear section at distance d from face of column 4.3.2.2(10) Column Vsd B interior side = = = 1717 -T - 0.675 x 657 = 1273 kN VRdl TRd k(1.2 + 40pr)bwd 'Rd 0.34 N/mm2 1.6 k P, = - 4.3.2.3 Eqn 4.18 Table 4.8 C`7 d .9 1.0 = 1.0 = 0.00476 Ensure bars are continued sufficiently. VRdi = 957 kN VRdt VSd ((DD > Therefore shear reinforcement required. Shear capacity with links ((DD VRd3 = Vcd + V d = VRdl + Vwd 4.3.2.4.3 Eqn 4.22 Therefore (DD V d >- 1273 x 957 = 316 kN V A sw d 0.9dfywd Eqn 4.23 d = 675 mm C11 S fywd = 400 N/mm2, 316 x 103 E Asw >_ S 0.9 x 675 x 400 = 1.30 mm2/mm E Where shear reinforcement is required, the minimum amount is 100% of the E NAD EC2 Table 5.5 value. With = 460, = 0.0012 by interpolation Table 3 5.4.3.3(2) fyk Pwm n Table 5.5 168 FOUNDATIONS For links PW AIsb SW W Eqn 5.16 0.0012 x 3000 = 3.6 > 1.30 mm2/mm Therefore minimum links govern. Determine link spacing, using EC2 Eqn 5.17-19. (1) E VRd2 = = ofd bW(0.9 d)12 0.55 x 20 x 3000 x 0.9 x 675 x 10-3/2 = 10020 kN Eqn 4.25 Vsd/VRd2 = 1273/10020 = 0.13 < 0.2 Use EC2 Eqn 5.17 to determine link spacing. smax = 0.8d (Note: 300 mm limit in Eqn 5.17 does not apply to slabs) 5.4.3.3(4) P- 0.75d = 506 mm NAD 6.5(f) Transverse spacing of legs across section . 5.4.2.2(9) VRd, A SW s 12 x 78.5 = 250 3.77 > 3.6 mm2/mm E ........... OK Check diagonal crack control Vd Vsd 5.4.2.2(10) = = VRdl = 957 kN 1273 kN (max.) Vsd < 3V d No further check required. 4.4.2.3(5) VRdl Distances to where xa Vsd from face of columns A and B E 1502 - 957 = 657 = 0.830 m E xb = 1.157 m Check shear in areas where bottom steel is in tension and P, = 0.0015 (min. steel) (I1 E FOUNDATIONS VRdl = 0.34(1.2 + 0.06)3000 x 665 x 10-3 = 854 > 723 kN .. OK No links required at RH end of base 0 (C] In orthogonal direction, shear at d from column face 0 219(3.0-0.45-0.6x2) 2 E 148 kN/m From above VRdl _ 3.0 854 = 284 > 148 kN/m ................... OK No links required in orthogonal direction 7.1.2.7 Punching Length of one side of critical perimeter at 1.5d from face of column E 4.3.4.1P(4) & = 3 x 690 + 600 = 2670 mm = 4.3.4.2.2 This extends almost the full width of the base 3000 mm Hence it is sufficient just to check line shear as above and shear around perimeter of column face, where ,S- Vsdlud The shear stress at the column face perimeter with d = 675 mm is less than 4.9 N/mm2 in both cases (see Table 7.1) OK ill x _ 0.4 x 0.53 x 3 x 750 x 3000/(2 x 460) = 1555 mm2 12T32 gives As > 1555 mm2 E (Y) ................................. LH o') E OK E 7.1.2.9 Detailing Check bar achorage detail at ((D end. 5.4.2.1.4(2) The anchorage should be capable of resisting a tensile force F with al = VSdaild = d Vsd 5.4.3.2.1(1) Fs = U)- Vsd = column reaction = 1896 kN 5.2.2.1 & 5.2.2.2 The bond strength for poor conditions in the top of the pour = fbd 1b 0.7 x Table 5.3 value = 0.7 x 3 = 2.1 N/mm2 = (014)(fyd/fbd) = 47.60 = 1524 mm Eqn 5.3 Continuing all T32 bars to end A s. prov As,req = = 9650 mm2 Vsdl yd = 1896 x 1031400 4740 mm2 CA) FOUNDATIONS Hence required anchorage, (3 )lb,net at a direct support = Figure 5.12 (3) lb x 4740/9650 Anchorage up to face of column = The anchorage may be increased to the end of the bar. 500 mm > 0.31b .......... ..... OK OK 600 if - 75 = 525 mm lb,net' preferred, by providing a bend at 5.2.3.3 5.4.3.2.1 The requirement for transverse reinforcement along the anchorage length does not apply at a direct support. .--U Secondary reinforcement ratio for top steel p2 0-0 ? = Al >_ 0.2p1 = 0.2 x 0.00476 = 0.00095 d 750 - 35 - 10 = 705 mm As 670 mm2/m Use T16 @ 250 mm crs. (804 mm2/m) transversely in top Spacing 1090 mm2 E = 948 mm2 = 1090 mm2 ........................ OK 9.5.2 Maximum Maximum area of total tension and compression reinforcement 0.04AC = 0.04 x 635880 = 25435 > 2052 mm2 .... OK 9.6 Reinforcement summary 11 tendons throughout beam 0 2T16s top and bottom throughout beam. Additional 2T20s top at support 2 These areas are within maximum and minimum limits. 207 10 SERVICEABILITY CHECKS BY CALCULATION 10.1 Deflection Calculate the long term deflection of a 7.0 m span simply supported beam whose section is shown in Figure 10.1. The beam supports the interior floor spans of an office building. 5-0 C°. 1650 d' = 50 ffAss 450 2 r As 250 d = 390 Figure 10.1 Beam section Deflections will be calculated using the rigorous and simplified methods given in EC2(1), together with an alternative simplified method. The results will then be compared with the limiting span/effective depth ratios given in EC2. N(9 10.1.1 Design data Span Gk = :-' 7.0 m = = 19.7 kN/m Qk 19.5 kN/m A' AS ick S = = = 402 mm2 2410 mm2 30 N/mm2 (conc rete stre n gth class C30/37) CD 3.1.2.4 Table 3.1 10.1.2 Calculation method The requirements for the calculation of deflections are given in Section 4.4.3 and Appendix 4 of EC2. Two limiting conditions are assumed to exist for the deformation of concrete A4.3(1) sections (1) Uncracked Cracked. A4.3(2) (2) Members which are not expected to be loaded above the level which would cause the tensile strength of the concrete to be exceeded, anywhere in the member, will be considered to be uncracked. Members which are expected to crack will behave in a manner intermediate between the uncracked and fully cracked conditions. For members subjected dominantly to flexure, the Code gives a general equation for obtaining the intermediate value of any parameter between the limiting conditions SERVICEABILITY CHECKS BY CALCULATION a where = all + (1 - Dal A4.3(2) Eqn A.4.1 -(f) « is the parameter being considered «l and «n are the values of the parameter calculated for the uncracked and fully cracked conditions respectively is a distribution coeffient given by 1 - - 0 0102 ( a A4.3(2) )2 Eqn A.4.2 The effects of creep are catered for by the use of an effective modulus of elasticity for the concrete given by ECM Ec,eft 1+0 p"U A4.3.(2) Eqn A.4.3 Bond and deterioration of bond under sustained or repeated loading is taken account of by coefficients a1 and a2 in Eqn A.4.2 Curvatures due to shrinkage may be assessed from 1 E cs as e °(n A4.3(2) r cs I Eqn AAA Shrinkage curvatures should be calculated for the uncracked and fully cracked conditions and the final curvature assessed by use of Eqn A.4.1. In accordance with the Code, the rigorous method of assessing deflections is to calculate the curvatures at frequent sections along the member and calculate the deflections by numerical integration. The simplified approach, suggested by the Code, is to calculate the deflection assuming firstly the whole member to be uncracked and secondly the whole member to be cracked. Eqn A.4.1 is used to assess the final deflection. 10.1.3 Rigorous assessment The procedure (1) is, at frequent intervals along the member, to calculate 0 ago 0 E Moments Curvatures Deflections. (2) (3) Here, calculations will be carried out at the mid-span position only, to illustrate this procedure, with values at other positions along the span being tabulated. 0-0 _-' 10.1.3.1 Calculation of moments For buildings, duration. it will normally be satisfactory to consider the deflections under the quasi-permanent combination of loading, assuming this load to be of long A4.2(5) The quasi-permanent combination of loading is given, for one variable action, by Gk + w2Qk 2.3.4 P(2) Eqn 2.9(c) 09 SERVICEABILITY CHECKS BY CALCULATION 02 = 0.3 NAD Table = 19.7 + (0.3 CC] 1 Therefore Loading x 19.5) = 25.6 kN/m 25.6 x 72/8 Mid-span bending moment(" = = 156.8 kNm 10.1.3.2 Calculation of curvatures order to calculate the curvatures it is first necessary to calculate the properties of the uncracked and cracked sections and determine the moment at which In cracking will occur. 10.1.3.2.1 Flexural curvature CAD The effective modulus of elasticity (Ec,eff) 1 ECM + ¢ A4.3(2) Eqn A.4.3 + For concrete strength class C30/37, Ecm = 32 kN/mm2 3.1.2.5.2 Table 3.2 2A c U _ 2[(1650 x 150) + (250 x 300) ] 2(1650 + 300) = 165 mm For internal conditions and age at loading of 7 days 4) 3.1.2.5.5 Table 3.3 ('') = 3.1 Therefore Ec,eff = 1 32 + = 3.1 7.8 kN/mm2 Effective modular ratio ( ae) Modulus of elasticity of reinforcement Therefore a e = 200 7.8 + _ E5 Ec,eft (ES) = 200 kN/mm2 3.2.4.3(1) = 25.64 AS p bd 1650 x 390 A' PI C11 = _ 2410 = 3.75x10- 3 = S = 402 1650 x 390 bd - 6.25 x 10_4 For the uncracked section, the depth to the neutral axis is given by bh2/2 X = - (b - b) (h - h) (h hf + hf) + (ae - 1) (A'Sd' + Asd) 2 bhf + bN,(h - hf) + (ae - 1) (A's + A) = 165.2 mm E SERVICEABILITY CHECKS BY CALCULATION The second moment of the area of the uncracked section is given by bh 3 I' (h+hf_ X + b W(h 12 - h f )3 + bhf (x 122 12 - hil2)2 + bw(h - hf) + (ae-1) A'S(x-d')2 + (CYe-1) Ajd-x)2 = 7535 x 106 rnm4 2 For the cracked section the depth to the neutral axis is given by d = x - [Cep + (ae + [aeP + (ae = 134.6 mm E - 1)p'] 2 + 2 [«eP+(«e - 1)P' d /] = 0.345d The second moment of area of the cracked section is given by + cxep(1 'r) bd3 3 ( dl - I>( 1 3 2 ' 2 + (Cle Xl - 1)P' (d d In = 0.0556bd3 = 5448 x 106 mm4 The moment which will cause cracking of the section is given by M cr yt yt For concrete strength grade C30/37, Therefore The section is considered to be cracked, since Curvature of the uncracked section is given by 1 Curvature of the cracked section is given by 1 Having obtained the values for the two limiting conditions Eqn AAA is used to assess the intermediate value. Hence tip rl = h - x = 450 - 165.2 fctm = 284.8 mm 2.9 N/mm2 3.1.2.4 = E E Table 3.1 Mcr 2.9x7535x106x10-6 284.8 = 76.7 kNm Mcr < M = 156.8 kNm _ M Ec,effll _ 7.8 156.8 x 106 106 x 103 x 7535 x = 2.668 x 10-6 rad./mm _ M _ III 7.8 156.8 x 106 106 ru Ec,eff x 103 x 5448 x = 3.690 x 10-6 rad./mm E tip x i = ( ) " + (1 ' A4.3(2) Eqn A.4.1 SERVICEABILITY CHECKS BY CALCULATION _ , - a, a2 asr as )2 For high bond bars, a, = 1.0 For sustained loading, 62 Q = 0.5 s is the stress in the tension steel calculated on the basis of a cracked section Therefore aeM (d S - x) 25.64 x 156.8 x 106 (390 - 134.6) = 188.5 N/mm2 E III 5448 x 106 vsr is the stress in the tension steel calculated on the basis of a cracked section under the loading which will just cause cracking at the section considered. Therefore osr = aeMcr (d - X) III 25.64 x 76.7 x 106 390 5448 x Therefore ( 106 - 134.6 ) = 92.2 N/mm2 = Note: osr ors 1 - 0.5 ( 92.2 188.5 12 = 0.88 may be replaced by = M °r in the above calculation M III -ca 1 r [(0.88 x 3.69) + (1-0.88) x 2.668] x 10-6 = 3.567 x 10-6 rad./mm 10.1.3.2.2 Shrinkage curvature The shrinkage curvature is given by 1 S E cse of I A.4.3.2 res Eqn AAA where Ecs is the free shrinkage strain For internal conditions and 2AC/u ecs = 165 mm 3.1.2.5.5 = 0.60 x 10-3 cad Table 3.4 S is the first moment of area of the reinforcement about the centroid of the 0 section. 1 is the second moment of area of the section. S and I should be calculated for both the uncracked and fully cracked conditions. Curvature of the uncracked section SI = AS(d - x) - A'S(x - d') = 495.5 x 103 mm3 SERVICEABILITY CHECKS BY CALCULATION 1 _ 0.60 resl x 10-3 x 25.64 x 495.5 x 7535 x 106 103 1.0 x 10-6 rad.lmm Curvature of the cracked section S 1 = AS(d - x) - A'S (x - L,, d') = 581.5 x 103 mm3 E _ 0.60x10-3x 25.64 x 581.5 x 103 resn 5448 x 106 1.64 x 10-brad./mm Therefore 1 =fix _ r CS 1 +(1 -0x (1 rCs« 1 rCS1 [(0.88 x 1.64) + - 0.88) x 1.0] x 10-6 = 1.563 x 10-6 rad./mm The total curvature at mid-span 1 1 + rtot 1 rcs = (3.567 + 1.563) x 10-6 = 5.130 x 10-brad./mm r The flexural, shrinkage and total curvatures at positions x11 along the span are given in Table 10.1. Table 10.1 Curvatures x 106 (rad./mm) 1 `I1 x/1 0 Moment (kNm) 0 0 (-) 1 1 1 1 r, T. r 0 0 r., 1.000 rw 1.000 0 0.1 56.4 100.4 0.960 1.708 2.363 0.708 0.960 2.171 1.000 1.960 0.2 1.453 1.531 3.624 0 0 0.3 131.7 2.241 2.561 (J) 3.100 3.542 3.690 0.830 0.870 0.880 2.954 3.414 3.567 4.485 4.971 0.4 0.5 0.6 0.7 0.8 L() 150.5 156.8 150.5 131.7 100.4 OW) 1.557 2.668 2.561 1.563 5.130 4.971 3.542 3.100 0.870 0.830 3.414 2.954 2.171 1.557 1.531 2.241 1.708 4.485 3.624 2.363 0.708 1.453 0.9 1.0 56.4 0 0.960 0 0 0 - 0.960 0 1.000 1.000 1.960 1.000 10.1.3.3 Calculation of deflections Having calculated the total curvatures, the deflections may be calculated by numerical integration using the trapezoidal rule. The uncorrected rotation at any point may be obtained by the first integral given by 1 eX = 6X_ + rX + r 2 x11 1 X_1 l xn 213 ()) SERVICEABILITY CHECKS BY CALCULATION Having calculated the uncorrected rotations, the uncorrected deflections may be obtained by the second integral given by (S= tea) ax = ax_, + 1 ex +2 ex-1 / n 1 +L+ -(3 where the subscript x denotes the values of the parameters at the fraction of the span being considered, and the subscript x-1 denotes the values of the parameters at the preceding fraction of the span. 3-m l n is the span number of span divisions considered. 0.11 is the Hence the uncorrected rotation at 8+( r°., l+ 2 1 r° 1 l n 10_6 = 0 + 1.96 + 1.0 2 I x 7000 10 ((D = 1.036 x 10_3 rad. and the uncorrected deflection at all 0 -9 2 0.11 n 10 _3 0 + (1.03 2 + 0) 0 x 7000 The uncorrected deflections may then be corrected to comply with the boundary conditions of zero deflection at both supports. This is done by subtracting from the uncorrected deflections the value of the uncorrected deflection at the right hand support multiplied by the fraction of the span at the point being considered. The values of the uncorrected rotations, uncorrected and corrected deflections at positions x/l along the span are given in Table 10.2. Table 10.2 x/! riot + = 0.363 mm 0.5 0.6 0.7 0.8 CJ) 3.624 22.356 Off) 54.998 71.331 88.711 70.969 79.840 88.711 -15.970 0.9 1.0 1.960 1.000 24.310 25.346 - 8.508 0 ((7 7-0 Deflections (mm) 1 x 106 1st integral 2nd integral Correction Deflection x 103 0 0.1 1.000 1.960 0 1.036 0 0.363 0 8.871 - 8.508 0.2 0.3 3.624 4.485 4.971 2.990 5.828 1.772 4.858 17.742 26.613 -15.970 -21.755 t() 0.4 9.138 12.673 10,096 17.730 35.484 44.356 53.227 62.098 -25.388 -26.626 -25.388 -21.755 5.130 4.971 16.208 27.838 40.342 4.485 19.518 (DD 0 SERVICEABILITY CHECKS BY CALCULATION Maximum deflection at mid-span atct = 26.6 mm E = span 263 span = < limit of 250 28 mm E E 10.1.4 Simplified approach The procedure for this approach is to (1) Calculate the maximum bending moment and the moment causing cracking Calculate the maximum deflections for the uncracked and fully cracked conditions, and use Eqn A.4.1 to assess the final maximum deflection. c (2) c From Section 10.1.3.2.1 the maximum bending moment M the moment causing cracking Mc = 76.7 kNm. The maximum deflection of the uncracked section due to flexure 5w14 E() = 156.8 kNm, and at w 1 384Ec,eff II = = 25.6 kN/m 7.0 m 7.8 kN/mm2 Ec,eff = = II 7535 x 106 mm4 Therefore a I The maximum deflection of the cracked section due to flexure 5 w14 an Therefore aII Final maximum deflection due to flexure .4III _ 5x25.6x74x 384 x 7.8 x 103 1012 = 106 x 7535 x 13.6 mm E 384Ec,ettlu = 5448 x 106 mm4 5x25.6x74x1012 384 x 7.8 x 103 E x 5448. x = 106 18.8 mm a = aII + (1 1 M a, r A4.3(2) Eqn A.4.1 = - 0102 )2 (_M_ SERVICEABILITY CHECKS BY CALCULATION = 0.5 (J) Therefore a It must be appreciated that the deflection calculated above is due to flexure due to shrinkage must also be assessed. The shrinkage curvature at mid-span from Section 10.1.3.2 only. The additional deflection rcs acs 1.563 C)) atot This figure is close to the rigorously assessed value of 26.6 mm. 10.1.5 Alternative An alternative simplified approach, which directly takes account of shrinkage, is given in BS 81100. The procedure here is to calculate the total curvature at one point, generally the point of maximum moment. Then, assuming the shape of the curvature diagram to be the same as the shape of the bending moment diagram, the deflection is given by a O-o = Kl 2 _ 1 n':7 where K is a factor For a simply supported beam with uniformly distributed load L'2 1 = 1.0 = 1 - 0.5 176.7 I2 = 156.8 0.88 = (0.88 x 18.8) + (1 - 0.88) x 13.6 = 18.2 mm = x 10-6 rad./mm l2 1.563 x 10-6 8 x 72 x 106 = 8 f BSI = 18.2 + 9.6 = 9.6 mm = a + acs = 27.8 mm E simplified approach BS 8110: Part 2 Section 3 rtot BS 8110: Part 2 3.7.2 (r) dependent upon the shape of the being moment diagram. Eqn 11 K Total = 0.104 curvature at mid-span, from Section 10.1.3.2 1 BS 8110: Part 2 3.7.2 = 5.130 x 10-6 rad./mm Table 3.1 riot Therefore maximum deflection at mid-span atot = 0.104 x 72 x 5.130 = 26.2 mm E Again this is close to the rigorously assessed value. SERVICEABILITY CHECKS BY CALCULATION 10.1.6 Comparison with span/effective depth ratio The procedure for limiting deflections by use of span/effective depth ratios is set out in EC2 Section 4.4.3. For the example considered A s, req P = 2392 mm2 100As,prov A s,prov 100 = 2410 mm2 2410 x = 0.37% bd 1650 x 390 Therefore the concrete is lightly stressed, p (L] 3.0 the basic span/effective depth ratio should be multiplied by a factor of 0.8 The span/effective depth ratios given in NAD Table 7 are based on a maximum service stress in the reinforcement in a cracked section of 250 N/mm2. The tabulated values should be multiplied by the factor of 2501as for other stress levels, where as is the service stress at the cracked section under the frequent load combination. As a conservative assumption the Code states that the factor may be taken as 250 as _ 400 yk(As, reg ) A s p rov , Therefore, for this example, allowable span/effective depth ratio I d I 28 x 0 8 . 400 460 x 2392/2410) ( l = 19 6 . d If (allowable) = 19.6 > d (actual) = 7000 390 = 180 the span/effective depth ratio is modified using the service stress in the reinforcement as calculated in Section 10.1.3.2.1 but adjusted for the frequent load combination aS = 188.5 x 31.4/25.6 = 231 N/mm2 d (allowable) = 28 x 0.8 x 2310 = 21.6 > 18.0 (actual) m00 (Nil .C. CAD ,,a^) SERVICEABILITY CHECKS BY CALCULATION It can be seen from this example that whilst the span/effective depth ratio based on the calculated steel service stress suggests that the deflection should be well within the prescribed limits, the deflection from the rigorous and simplified analysis proves to be much nearer to the limit of span/250. ,--. This is due to the contribution to the deflection from shrinkage, which in this example is approximately a third of the total deflection. 30°'= The values of shrinkage strain given in EC2 Table 3.4 relate to concrete having a plastic consistence of classes S2 and S3 as specified in ENV 206(6). For concrete of class S1 and class S4 the values given in the Table should be multiplied by 0.7 and 1.2 respectively. Table 4 of ENV 206 categorises the class in relation to slump as given in Table 10.3. C'7 000 E _.C 3.1.2.5.5(4) ENV 206 7.2.1 Table 10.3 Class S1 Slump classes Slump (mm) 10 °(O ENV 206 7.2.1 Table 4 - 40 90 S2 S3 S4 50 - 100 - 150 >_ 160 Thus for classes S2 and S3 the slump may vary between 50 mm and 150 mm. It is not logical that mixes with this variation of slump, and hence w/c ratio, should have a standard value of shrinkage strain. If the values in EC2 Table 3.4 are assumed to relate to the median slump for classes S2 and S3 of 100 mm, then the values for slumps of 40 mm to 100 mm should be multiplied by a factor between 0.7 and 1.0 and values for slumps of 100 mm to 160 mm should be multiplied by a factor between 1.0 and 1.2. v0) As most normal mixes will have a slump in the order of 50 mm the values of shrinkage strain for the example considered would be: 0.60x x 10-) x 0.75 = ADD 3(D 10 0.45 x 10-3 BS 8110: Part 2 7.4 Figure 7.2 This figure relates more closely to the value which would be given in BS 8110, for the same example, of 0.4 x 10-3. For the example considered, the calculated deflection due to shrinkage from the rigorous assessment would be E 9.1 x 0.75 = 6.8 mm and the total deflection from the rigorous assessment would be = atot 26.6 - 9.1 + 6.8 = 24.3 mm This is well within the limit of span _ 28 mm 250 218 SERVICEABILITY CHECKS BY CALCULATION 10.2 Cracking Check by calculation that the longitudinal reinforcement in the reinforced concrete wall section shown in Figure 10.2 is sufficient to control cracking due to restraint of intrinsic deformation resulting in pure tension. 0 r 0 T12- 125 T16- 200 M 200 a Figure 10.2 Wall section 10.2.1 Design data Concrete strength class is C30/37. Cover to reinforcement High bond bars with vk = = 35 mm 460 N/mm2 NAD Table 6 Exposure class 2(a) 10.2.2 Calculation method Requirements for the control of cracking are given in EC2 Section 4.4.2. Crack control is normally achieved by the application of simple detailing rules. The procedure for the calculation of crack widths is first to calculate the stress and hence the strain in the reinforcement, taking into account the bond properties of the bars and the duration of loading. Next, the average final crack spacing dependent on the type, size and disposition of the reinforcement and the form of strain distribution is established. -s; (CD O=-(D (/) The design crack width may then be obtained and compared with the limiting design crack width. In the absence of specific requirements, a limiting crack width of 0.3 mm will generally be satisfactory for reinforced concrete members in buildings with respect to both appearance and durability. (OD 4.4.2.1(6) 10.2.3 Check by calculation 10.2.3.1 Calculation Steel stress: a S of steel stress and strain kckct,e0Act 4.4.2.2(3) = AS (CD Eqn 4.78 where AS = = area of reinforcement within the tensile zone 905 x 2 = 1810 mm2/m E SERVICEABILITY CHECKS BY CALCULATION Act = = area of concrete within tensile zone 1000 x 200 = 200 x 103 mm2 of kc = a coefficient taking account 1.0 stress distribution = for pure tension k = = a coefficient allowing for the effect of non-uniform self-equilibrating stresses 0.8 for tensile stresses due to restraint of intrinsic deformations (h ct.eff h12 = CD) The effective tension area is generally the area of concrete surrounding the 100 mm E Therefore Ac0 = Pr III 1000 x 100 = 100 x 2 x 100 0 103 nlm2 1810 X 103 _ 0.009 50+(0.25x 0.8x0.8x 1.0x12) 0.009 Srm 263 mm 10.2.3.3 Calculation of crack width The design crack width wk = OSrmEsm 4.4.2.4 Eq n 4.80 where a a coefficient relating the average crack width to the design value 1.3 for restraint cracking in of 300 mm or less. members with a minimum dimension Therefore wk = 1.3 x 263 x 8.4 x 10-4 = 0.29 < 0.3 mm (limit) E 10.2.3.4 Concluding remark The Code suggests a minimum value of 3 N/mm2 be taken when the time of cracking cannot be confidently predicted as being less than 28 days. 003 (S-W the class specified. Consequently, unless strict site control is exercised, it would be prudent to adopt the apparently conservative figures given in EC2 Table 3.1. 9'C Cteff seem high, it is difficult at the design stage to assess accurately the as placed concrete strength because this often exceeds (fl Whilst the values given for 11z DD)) 11 DEEP BEAMS 11.1 Introduction The design of deep beams may be based on analyses applying: (a) linear elastic analysis; 2.5.1.1(5) (b) an equivalent truss consisting of concrete struts and arches with reinforcement, all preferably following the elastic field; (c) non-linear analysis. details of the analysis model and, therefore, much of the design are not given and it is left for the Engineer to satisfy the principal Code requirements. This can be achieved using CIRIA Guide 2, The design of deep beams in reinforced concrete("), which also provides recommendations on the detailed analysis and design. The Guide was written for use with the then current British Standard CP 11019). Here it has been assumed that a complete design to the CIRIA Guide would be carried out and then checks made to demonstrate compliance with the specific clauses for deep beams in EC2. highlight some of the differences between EC2 and design to the CIRIA Guide, the example in Appendix B of the Guide has been used. To A small number of EC2 clauses have been identified as relating specifically to deep beams: (a) 2.5.2.1(2) (b) 2.5.3.7.3 (c) 4.4.2.3(4) (d) 5.4.5 - definition of deep beams analysis modelling skin reinforcement reinforcement detailing 11.2 Example proposed arrangement of walls and columns is shown in Figure 11.1. Loading details are presented in Figure 11.2. It is intended to justify a design using the Simple Rules of Section 2 of the CIRIA Guide. A _G) The beam is a flat vertical plate and the thickness is small compared with other dimensions. 3((0 In EC20) o(° There are two loads which may be defined as concentrated and no indirect loads or supports. In EC2 depth. CIRIA Guide 2 classifies deep beams as 'Beams with span/depth ratios of less than 2 for single span beams or less than 2.5 for multi span beams', thus giving an extended range of elements to be designed as deep beams in comparison with EC2. 33°- (c1 2.5.3.7.3 >`N a beam is classified as a deep beam (On CIRIA Guide 2 CI.2.1.1(1) C1.2.1.1(4) C1.2.1.1(5) if the span is less than twice the 2.5.2.1(2) min 4'1 DEEP BEAMS 200 y 14.220 11 - 0 1800 900 20 h*- v I I I I 300 1 I I I I I I I 300 T j Iv 10.680 I 300 300 300 4 I-y TTT, A 250- 250 5000 1000 8 1 -1 1000 ELEVATION Figure 11.1 Structural arrangement DEEP BEAMS 57.6 kN/m dead load 475 kN dead load 475 kN V 14.220 10.680 +44.0 kN/m live load dead load 157.6 kN/m dead load+44.0 kN/m I live load 7 I 7.140 Load between faces of supports 3-in included bending moment _I 57.6 kN/m dead load+ 44.0 kN/m I live load 7 178.6 kN/m self weight + 57.6 kN/m dead load + 44.0 kN/m 3.600 live load 7_ Figure 11.2 Loading details 11.2.1 Durability For dry environment, exposure class is 1. Table 4.1 ENV 206 Table NA.1 Minimum concrete strength class is C25/30. The CIRIA Guide example uses ^^D fcu = 30 N/mm2 Use C25/30 for design to EC2 to keep examples broadly consistent. Minimum cement content and water cement ratio ENV 206 Minimum cover to reinforcement Assume nominal aggregate size Assume maximum bar size Nominal cover = = = >_ 15 mm Table 3 NAD 20 mm 20 mm 20 mm Table 6 NAD 6.4(a) 4 DEEP BEAMS Use 25 mm nominal cover Check requirements for fire resistance to BS 8110: Part 2(2). NAD 6.1(a) 11.2.2 Materials Type 2 deformed reinforcement with vk = 460 N/mm2 NAD 6.3(a) Concrete strength class C25/30, nominal aggregate size 20 mm 11.2.3 Effective dimensions of beam Effective span (0 CZ- = l0 + (c1/2 35 0.1835 0.1627 0.1418 0.1210 0.1001 0.1238 0.1099 0.272 0.85 0.80 0.75 0.70 20 25 30 Plastic design 0.328 0.288 0.232 0.192 )17 0.1292 0.1155 0.1013 0.0954 0.0803 0.0647 0.0485 (17 0.248 0.208 0.25 0.152 0.112 0.0864 0.1502 0.1293 0.1084 0.0792 0.0584 0.1020 0.1303 13.3 Compression reinforcement in Compression reinforcement is required amount can be calculated from A any section where it > µilm. The µlim 0.87(1 - d'Id) where W/ = mechanical ratio of compression steel s = x yk bd fk C DESIGN OF BEAM AND COLUMN SECTIONS d' A's = = depth from compression face to centroid of compression reinforcement area of compression reinforcement ((DD The area of tension reinforcement can now be obtained from Wlim + W' Equations above for w' and w are valid for d'/x