Manual for Design and Detailings of Reinforced Concrete to Code of ...

Manual for Design andDetailings ofReinforced Concrete toCode of Practice forStructural Use of Concrete2004Housing DepartmentMay 2008(Version 2.3)

AcknowledgementWe would like to express our greatest gratitude to Professor A.K.H. Kwanof The University of Hong Kong who has kindly and generously providedinvaluable advice and information during the course of our drafting ofthe Manual. His advice is most important for the accuracy andcompleteness of contents in the Manual.

ContentsPage1.0 Introduction 12.0 Some highlighted aspects in Basis of Design 33.0 Beams 104.0 Slabs 495.0 Columns 686.0 Column Beam Joints 937.0 Walls 1028.0 Corbels 1169.0 Cantilever Structures 12410.0 Transfer Structures 13211.0 Footings 13712.0 Pile Caps 14513.0 General R.C. Detailings 15614.0 Design against Robustness 16315.0 Shrinkage and Creep 16816.0 Summary of Aspects having significant Impacts on Current Practices 184References 194AppendicesAppendix A – Clause by Clause Comparison between “Code of Practice forStructural Use of Concrete 2004” and BS8110Appendix B – Assessment of Building AccelerationsAppendix C – Derivation of Basic Design Formulae of R.C. Beam sectionsagainst FlexureAppendix D – Underlying Theory and Design Principles for Plate Bending ElementAppendix E – Moment Coefficients for three side supported SlabsAppendix F – Derivation of Design Formulae for Rectangular Columns to RigorousStress Strain Curve of ConcreteAppendix G – Derivation of Design Formulae for Walls to Rigorous Stress StrainCurve of ConcreteAppendix H – Estimation of support stiffnesses of vertical support to transferstructuresAppendix I – Derivation of Formulae for Rigid Cap AnalysisAppendix J – Mathematical Simulation of Curves related to Shrinkage and CreepDetermination

Version 2.3 May 20081.0 Introduction1.1 Promulgation of the Revised CodeA revised concrete code titled “Code of Practice for Structural Use of Concrete2004” was formally promulgated by the Buildings Department of Hong Kongin late 2004 which serves to supersede the former concrete code titled “TheStructural Use of Concrete 1987”. The revised Code, referred to as “the Code”hereafter in this Manual will become mandatory by 15 December 2006, afterexpiry of the grace period in which both the revised and old codes can be used.1.2 Main features of the CodeAs in contrast with the former code which is based on “working stress” designconcept, the drafting of the Code is largely based on the British StandardBS8110 1997 adopting the limit state design approach. Nevertheless, thefollowing features of the Code in relation to design as different from BS8110are outlined :(a)(b)(c)(d)(e)Provisions of concrete strength up to grade 100 are included;Stress strain relationship of concrete is different from that of BS8110for various concrete grades as per previous tests on local concrete;Maximum design shear stresses of concrete ( v max) are raised;Provisions of r.c. detailings to enhance ductility are added, togetherwith the requirements of design in beam-column joints (Sections 9.9and 6.8 respectively);Criteria for dynamic analysis for tall building under wind loads areadded (Clause 7.3.2).As most of our colleagues are familiar with BS8110, a comparison tablehighlighting differences between BS8110 and the Code is enclosed inAppendix A which may be helpful to designers switching from BS8110 to theCode in the design practice.1.3 Outline of this ManualThis Practical Design Manual intends to outline practice of detailed design anddetailings of reinforced concrete work to the Code. Detailings of individual1

Version 2.3 May 2008types of members are included in the respective sections for the types, thoughSection 13 in the Manual includes certain aspects in detailings which arecommon to all types of members. Design examples, charts are included, withderivations of approaches and formulae as necessary. Aspects on analysis areonly discussed selectively in this Manual. In addition, as the Department hasdecided to adopt Section 9.9 of the Code which is in relation to provisions for“ductility” for columns and beams contributing in the lateral load resistingsystem in accordance with Cl. 9.1 of the Code, conflicts of this section withothers in the Code are resolved with the more stringent ones highlighted asrequirements in our structural design.As computer methods have been extensively used nowadays in analysis anddesign, the contents as related to the current popular analysis and designapproaches by computer methods are also discussed. The background theoryof the plate bending structure involving twisting moments, shear stresses, anddesign approach by the Wood Armer Equations which are extensively used bycomputer methods are also included in the Appendices in this Manual fordesign of slabs, flexible pile caps and footings.To make distinctions between the equations quoted from the Code and theequations derived in this Manual, the former will be prefixed by (Ceqn) andthe latter by (Eqn).Unless otherwise stated, the general provisions and dimensioning of steel barsare based on high yield bars with f = 460 N/mm 2 .y1.4 Revision as contained in Amendment No. 1 comprising major revisionsincluding (i) exclusion of members not contributing to lateral load resistingsystem from ductility requirements in Cl. 9.9; (ii) rectification of ε 0 in theconcrete stress strain curves; (iii) raising the threshold concrete grade forlimiting neutral axis depths to 0.5d from grade 40 to grade 45 for flexuralmembers; (iv) reducing the x values of the simplified stress block forconcrete above grade 45 are incorporated in this Manual.2

Version 2.3 May 20082.0 Some highlighted aspects in Basis of Design2.1 Ultimate and Serviceability Limit statesThe ultimate and serviceability limit states used in the Code carry the usualmeaning as in BS8110. However, the new Code has incorporated an extraserviceability requirement in checking human comfort by limiting accelerationdue to wind load on high-rise buildings (in Clause 7.3.2). No method ofanalysis has been recommended in the Code though such accelerations can beestimated by the wind tunnel laboratory if wind tunnel tests are conducted.Nevertheless, worked examples are enclosed in Appendix B, based onapproximation of the motion of the building as a simple harmonic motion andempirical approach in accordance with the Australian Wind Code AS/NZS1170.2:2002 on which the Hong Kong Wind Code has based in derivingdynamic effects of wind loads. The relevant part of the Australian Code isAppendix G of the Australian Code.2.2 Design LoadsThe Code has made reference to the “Code of Practice for Dead and ImposedLoads for Buildings” for determination of characteristic gravity loads fordesign. However, this Load Code has not yet been formally promulgated andthe Amendment No. 1 has deleted such reference. At the meantime, the designloads should be therefore taken from HKB(C)R Clause 17. Nevertheless, thedesigner may need to check for the updated loads by fire engine for design ofnew buildings, as required by FSD.The Code has placed emphasize on design loads for robustness which aresimilar to the requirements in BS8110 Part 2. The requirements include designof the structure against a notional horizontal load equal to 1.5% of thecharacteristic dead weight at each floor level and vehicular impact loads(Clause 2.3.1.4). The small notional horizontal load can generally be coveredby wind loads required for design. Identification of key elements and designfor ultimate loads of 34 kPa, together with examination of disproportionatecollapse in accordance with Cl. 2.2.2.3 can be exempted if the buildings areprovided with ties determined by Cl. 6.4.1. The usual reinforcement provisionsas required by the Code for other purposes can generally cover the requiredties provisions.3

Version 2.3 May 2008Wind loads for design should be taken from Code of Practice on Wind Effectsin Hong Kong 2004.It should also be noted that there are differences between Table 2.1 of theCode that of BS8110 Part 1 in some of the partial load factors γ f . Thebeneficial partial load factor for earth and water load is 1. However, lowervalues should be used if the earth and water loads are known to beover-estimated.2.3 Materials – ConcreteTable 3.2 has tabulated a set of Young’s Moduli of concrete up to grade 100.The values are generally smaller than that in BS8110 by more than 10% andalso slightly different from the former 1987 Code. The stress strain curve ofconcrete as given in Figure 3.8 of the Code, whose initial tangent isdetermined by these Young’s Moduli values is therefore different from Figure2.1 of BS8110 Part 1. Furthermore, in order to achieve smooth (tangential)connection between the parabolic portion and straight portion of the stressstrain curve, the Code, by its Amendment No. 1, has shifted the ε0value to1.34(f / γ )Ecucminstead of staying atf cu−42.4× 10 which is the value inγmBS8110. The stress strain curves for grade 35 by the Code and BS8110 areplotted as an illustration in Figure 2.1.1816Comparison of stress strain profile between the Code andBS8110 for Grade 35The Code BS8110Stress (MPa)141210864200 0.2 0.4 0.6 0.8 1Distance ratio from neutral axisFigure 2.1 - Stress Strain Curves of Grade 35 by the Code andBS81104

Version 2.3 May 2008From Figure 2.1 it can be seen that stress strain curve by BS8110 envelops thatof the Code, indicating that design based on the Code will be slightly lesseconomical. Design formulae for beams and columns based on these stressstrain curves by BS8110, strictly speaking, become inapplicable. A fullderivation of design formulae and charts for beams, columns and walls aregiven in Sections 3, 5 and 7, together with Appendices C, F and G of thisManual.Table 4.2 of the Code tabulated nominal covers to reinforcements underdifferent exposure conditions. However, reference should also be made to the“Code of Practice for Fire Resisting Construction 1996”.To cater for the “rigorous concrete stress strain relation” as indicated in Figure2.1 for design purpose, a “simplified stress approach” by assuming arectangular stress block of length 0.9 times the neutral axis depth has beenwidely adopted, as similar to BS8110. However, the Amendment No. 1 of theCode has restricted the 0.9 factor to concrete grades not exceeding 45. For 45< f cu ≤ 70 and 70 < f cu , the factors are further reduced to 0.8 and 0.72respectively as shown in Figure 2.20.0035 for f cu ≤ 600.0035 – 0.0006(f cu – 60) 1/2 for f cu > 600.67f cu /γ m0.9x for f cu ≤ 45;0.8x for 45 < f cu ≤ 70;0.72x for 70 < f custrainstressFigure 2.2 – Simplified stress block for ultimate reinforced concrete design2.4 Ductility Requirements (for beams and columns contributing to lateral loadresisting system)As discussed in para. 1.3, an important feature of the Code is the incorporationof ductility requirements which directly affects r.c. detailings. By ductility werefer to the ability of a structure to undergo “plastic deformation”, which is5

Version 2.3 May 2008comparatively larger than the “elastic” one prior to failure. Such ability isdesirable in structures as it gives adequate warning to the user for repair orescape before failure. The underlying principles in r.c. detailings for ductilityrequirements are highlighted as follows :(i)Use of closer and stronger transverse reinforcements to achieve betterconcrete confinement which enhances both ductility and strength ofconcrete against compression, both in columns and beams;confinement by transversere-bars enhances concretestrength and ductility of theconcrete core within thetransverse re-barsaxial compressionFigure 2.3 – enhancement of ductility by transverse reinforcements(ii)Stronger anchorage of transverse reinforcements in concrete by meansof hooks with bent angles ≥ 135 o for ensuring better performance ofthe transverse reinforcements;(a) 180 o hook (b) 135 o hook (c) 90 o hookAnchorage of link in concrete : (a) better than (b); (b) better than (c)Figure 2.4 – Anchorage of links in concrete by hooks(In fact Cl. 9.9.1.2(b) of the Code has stated that links must beadequately anchored by means of 135 o or 180 o hooks and anchorage bymeans of 90 o hooks is not permitted for beams. Cl. 9.5.2.2, Cl. 9.5.2.3and 9.9.2.2(c) states that links for columns should have bent angle at6

Version 2.3 May 2008(iii)(iv)least 135 o in anchorage. Nevertheless, for walls, links used to restrainvertical bars in compression should have an included angle of not morethan 90 o by Cl. 9.6.4 which is identical to BS8110 and not a ductilityrequirement;More stringent requirements in restraining and containing longitudinalreinforcing bars in compression against buckling by closer andstronger transverse reinforcements with hooks of bent angles ≥ 135 o ;Longer bond and anchorage length of reinforcing bars in concrete toensure failure by yielding prior to bond slippage as the latter failure isbrittle;Ensure failure by yielding hereinstead of bond failure behindLonger and strongeranchoragebar in tensionFigure 2.5 – Longer bond and anchorage length of reinforcing bars(v)Restraining and/or avoiding radial forces by reinforcing bars onconcrete at where the bars change direction and concrete cover is thin;Radial force by bartending to cause concretespalling if concrete isrelatively thinRadial force by barinward on concretewhich is relatively thickFigure 2.6 – Bars bending inwards to avoid radial forces on thin concrete cover(vi)Limiting amounts of tension reinforcements in flexural members asover-provisions of tension reinforcements will lead to increase of7

Version 2.3 May 2008neutral axis and thus greater concrete strain and easier concrete failurewhich is brittle;ε cε cxxLesser amount of tensilesteel, smaller x, smaller ε cGreater amount of tensilesteel, greater x, greater ε cFigure 2.7 – Overprovision of tensile steel may lower ductility(vii)More stringent requirements on design using high strength concretesuch as (a) lowering ultimate concrete strain; (b) restricting percentageof moment re-distribution; and (c) restricting neutral axis depth ratiosto below 0.5 as higher grade concrete is more brittle.Often the ductility requirements specified in the Code are applied to locationswhere plastic hinges may be formed. The locations can be accuratelydetermined by a “push over analysis” by which a lateral load with step by stepincrements is added to the structure. Among the structural members met at ajoint, the location at which plastic hinge is first formed will be identified as thecritical section of plastic hinge formation. Nevertheless, the determination canbe approximated by judgment without going through such an analysis. In acolumn beam frame with relatively strong columns and weak beams, thecritical sections of plastic hinge formation should be in the beams at theirinterfaces with the columns. In case of a column connected into a thick pilecap, footing or transfer plate, the critical section with plastic hinge formationwill be in the columns at their interfaces with the cap, footing or transfer plateas illustrated in Figure 2.8.8

Version 2.3 May 2008Critical section withplastic hinge formationPile cap / footing /transfer structureStrong column / weak beamFigure 2.8 – locations of critical section with plastic hinge formation2.5 Design for robustnessThe requirements for design for robustness are identical to BS8110 and moredetailed discussions are given in Section 14.2.6 Definitions of structural elementsThe Code has included definitions of slab, beam, column and wall inaccordance with their dimensions in Clause 5.2.1.1, 5.4 and 5.5 which arerepeated as follows for ease of reference :(a)(b)(c)(d)(e)(f)Slab : the minimum panel dimension ≥ 5 times its thickness;Beam : for span ≥ 2 times the overall depth for simply supported spanand ≥ 2.5 times the overall depth for continuous span, classified asshallow beam, otherwise : deep beam;Column : vertical member with section depth not exceeding 4 times itswidth;Wall : vertical member with plan dimensions other than that of column.Shear Wall : wall contributing to the lateral stability of the structure.Transfer Structure : horizontal element which redistributes vertical loadswhere there is a discontinuity between the vertical structural elementsabove and below.This Manual is based on the above definitions in delineating structuralmembers for discussion.9

Version 2.3 May 20083.0 Beams3.1 Analysis (Cl. 5.2.5.1 & 5.2.5.2)Normally continuous beams are analyzed as sub-frames by assuming nosettlements at supports by walls, columns (or beams) and rotational stiffnessby supports provided by walls or columns as 4 EI / L (far end of column /wall fixed) or 3 EI / L (far end of column / wall pinned).Figure 3.1 – continuous beam analyzed as sub-frameIn analysis as sub-frame, Cl. 5.2.3.2 of the Code states that the followingloading arrangements will be adequate for seeking for the design moments :1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.4G K +1.6Q KFigure 3.2a – To search for maximum support reactions1.4G K +1.6Q K 1.0G K 1.4G K +1.6Q K 1.0G K 1.4G K +1.6Q K 1.0G KFigure 3.2b – To search for maximum sagging moment in spans with1.4G K +1.6Q K1.0G K 1.0G K 1.4G K +1.6Q K 1.4G K +1.6Q K 1.0G K 1.0G KFigure 3.2c – To search for maximum hogging moment at supportadjacent to spans with 1.4G K +1.6Q K10

Version 2.3 May 2008However, most of the commercial softwares can actually analyze individualload cases, each of which is having live load on a single span and the effectson itself and others are analyzed. The design value of shears and moments atany location will be the summation of the values of the same sign created bythe individual cases. Thus the most critical loads are arrived at easily.With wind loads, the load cases to be considered will be 1.2(G K +Q K +W K ) and1.0G K +1.4W K on all spans.3.2 Moment Redistribution (Cl. 5.2.9 of the Code)Moment redistribution is allowed for concrete grade not exceeding 70 underconditions 1, 2 and 3 as stated in Cl. 5.2.9.1 of the Code. Nevertheless, itshould be noted that there would be further limitation of the neutral axis depthratio x / d if moment redistribution is employed as required by (Ceqn 6.4)and (Ceqn 6.5) of the Code which is identical to the provisions in BS8110. Therationale is discussed in Concrete Code Handbook 6.1.2.3.3 Highlighted aspects in Determination of Design Parameters of Shallow Beam(i)Effective span (Cl. 5.2.1.2(b) and Figure 5.3 of the Code)For simply supported beam, continuous beam and cantilever, theeffective span can be taken as the clear span plus the lesser of half of thestructural depth and half support width except that on bearing where thecentre of bearing should be used to assess effective span;(ii)Effective flange width of T- and L-beams (Cl. 5.2.1.2(a))Effective flange width of T- and L-beams are as illustrated in Figure 5.2.of the Code as reproduced as Figure 3.3 of this Manual:b eff,1b effb eff,2b 1 b 1 b w b 2 b 2b eff,1 =0.2×b 1 +0.1l pib eff,2 =0.2×b 2 +0.1l pib eff, =b w +b eff,1 +b eff,2Figure 3.3 – Effective flange Parameters11

Version 2.3 May 2008Effective width (b eff ) = width of beam (b w ) + ∑(0.2 times of half thecentre to centre width to the next beam (0.2b i ) + 0.1 times the span ofzero moment (0.1l pi ), with the sum of the latter not exceeding 0.2 timesthe span of zero moment and l pi taken as 0.7 times the effective span ofthe beam). An example for illustration as indicated in Figure 3.4 is asindicated :Worked Example 3.14002000 400 2000 400 2000 400Figure 3.4 – Example illustrating effective flange determinationThe effective spans are 5 m and they are continuous beams.The effective width of the T-beam is, by (Ceqn 5.1) of the Code :lpi= 0 .7 × 5000 = 3500 ;b eff , 1= b eff ,2= 0.2×1000 + 0.1×3500 = 550As b eff , 1= b eff ,2= 550 < 0.2×3500 = 700 , ∴b eff , 1= beff,2= 550 ;beff= 400 + 550×2 = 400 + 1100 = 1500So the effective width of the T-beam is 1500 mm.Similarly, the effective width of the L-beam at the end isb b = 400 + 550 950.w+ eff , 1=(iii) Support Moment Reduction (Cl. 5.2.1.2 of the Code)The Code allows design moment of beam (and slab) monolithic with itssupport providing rotational restraint to be that at support face if thesupport is rectangular and 0.2Ø if the support is circular with diameter Ø.But the design moment after reduction should not be less than 65% ofthe support moment. A worked example 3.2 as indicated by Figure 3.5for illustration is given below :Worked Example 3.212

Version 2.3 May 2008350 kNm atsupport250 kNm at 0.2 Ø intothe support face200 kNm atsupport facecentre line of beamcolumn elementsidealized as lineelements in analysis8000.2×800Bending MomentDiagramFigure 3.5 – Reduced moment to Support Face forsupport providing rotational restraintIn Figure 3.5, the bending moment at support face is 200 kNm which canbe the design moment of the beam if the support face is rectangular.However, as it is smaller than 0.65×350 = 227.5 kNm. 227.5 kNmshould be used for design.If the support is circular and the moment at 0.2Ø into the support and thebending moment at the section is 250 kNm, then 250 kNm will be thedesign moment as it is greater than 0.65×350 = 227.5 kNm.For beam (or slab) spanning continuously over a support considered notproviding rotational restraint (e.g. wall support), the Code allowsmoment reduction by support shear times one eighth of the support widthto the moment obtained by analysis. Figure 3.6 indicates a numericalWorked Example 3.3.Worked Example 3.3By Figure 3.6, the design support moment at the support under0.8consideration can be reduced to 250 − 200×= 230kNm.813

Version 2.3 May 2008250 kNm230 kNmF Ed,sup = 200 kN800Figure 3.6 – Reduction of support moment by support shear for supportconsidered not providing rotational restraint(iv) Slenderness Limit (Cl. 6.1.2.1 of the Code)The provision is identical to BS8110 as1. Simply supported or continuous beam :Clear distance between restraints ≤ 60b c or 250b 2 c /d if less; and2. Cantilever with lateral restraint only at support :Clear distance from cantilever to support ≤ 25b c or 100b 2 c /d if lesswhere b c is the breadth of the compression face of the beam and d isthe effective depth.Usually the slenderness limits need be checked for inverted beams orbare beam (without slab).(v)Span effective depth ratio (Cl. 7.3.4.2 of the Code)Table 7.3 under Cl. 7.3.4.2 tabulates basic span depth ratios for varioustypes of beam / slab which are deemed-to-satisfy requirements againstdeflection. The table has provisions for “slabs” and “end spans” whichare not specified in BS8110 Table 3.9. Nevertheless, calculation can becarried out to justify deflection limits not to exceed span / 250. Inaddition, the basic span depth ratios can be modified due to provision oftensile and compressive steels as given in Tables 7.4 and 7.5 of the Codewhich are identical to BS8110. Modification of the factor by 10/span for14

Version 2.3 May 2008span > 10 m except for cantilever as similar to BS8110 is also included.Flanged Beam One or two-waySupport condition Rectangular Beam b w /b < 0.3 spanning solidslabCantilever 7 5.5 7Simply supported 20 16 20Continuous 26 21 26End span 23 18.5 23 (2)Note :1. The values given have been chosen to be generally conservative and calculation mayfrequently show shallower sections are possible;2. The value of 23 is appropriate for two-way spanning slab if it is continuous over one long side;3. For two-way spanning slabs the check should be carried out on the basis of the shorter span.Table 3.1 – effective span / depth ratio(vi) Maximum spacing between bars in tension near surface, by Cl. 9.2.1.4 ofthe Code, should be such that the clear spacing between bar is limited by70000βbclear spacing ≤ ≤ 300 mm where βbis the ratio of momentfyredistribution. Or alternatively, clear spacing ≤47000 ≤ 300 mm. So the70000βb 70000×1simplest rule is = = 152 mm when using high yieldfy460bars and under no moment redistribution.f s(vii) Concrete covers to reinforcements (Cl. 4.2.4 and Cl. 4.3 of the Code)Cl. 4.2.4 of the Code indicates the nominal cover required in accordancewith Exposure conditions. However, we can, as far as our buildingstructures are concerned, roughly adopt condition 1 (Mild) for thestructures in the interior of our buildings (except for bathrooms andkitchens which should be condition 2), and to adopt condition 2 for theexternal structures. Nevertheless, the “Code of Practice for Fire ResistingConstruction 1996” should also be checked for different fire resistanceperiods (FRP). So, taking into account our current practice of usingconcrete not inferior than grade 30 and maximum aggregate sizes notexceeding 20 mm, we may generally adopt the provision in our DSEGManual (DSEDG-104 Table 1) with updating by the Code except forcompartment of 4 hours FRP. The recommended covers are summarizedin the following table :15

Version 2.3 May 2008DescriptionNominal Cover (mm)Internal30 (to all rebars)External40 (to all rebars)Simply supported (4 hours FRP)80 (to main rebars)Continuous (4 hours FRP)60 (to main rebars)Table 3.2 – Nominal Cover of Beams3.4 Sectional Design for Rectangular Beam against Bending3.4.1 Design in accordance with the Rigorous Stress Strain curve of ConcreteThe stress strain block of concrete as indicated in Figure 3.8 of the Code isdifferent from Figure 2.1 of BS8110. Furthermore, in order to achieve smoothconnection between the parabolic and the straight line portions, the ConcreteCode Handbook has recommended to shift the ε 0 to the right to a value of1.34 fcu, which has been adopted in Amendment No. 1. With the values ofγmEcYoung’s Moduli of concrete, Ec, as indicated in Table 3.2 of the Code, thestress strain block of concrete for various grades can be determined. The stressstrain curve of grade 35 is drawn as shown in Figure 3.7.Stress Strain Profile for Grade 35Stress (MPa)1816141210864200.3769 whereε 0 = 0.0013190 0.2 0.4 0.6 0.8 1Distance Ratio from Neutral axisFigure 3.7 – Stress strain block of grades 35Based on this rigorous concrete stress strain block, design formulae for beam16

Version 2.3 May 2008can be worked out as per the strain distribution profile of concrete and steel asindicated in Figure 3.8.d’εult= 0.0035xdneutral axisStress DiagramStrain DiagramFigure 3.8 – Stress Strain diagram for BeamThe solution for the neutral axis depth ratio dxfor singly reinforced beam isthe positive root of the following quadratic equation where εult= 0. 0035 forconcrete grades not exceeding 60 (Re Appendix C for detailed derivation) :0.67 fγmcu⎡⎢−⎢⎣12+13ε0εult−112⎛ ε0⎜⎝ εult⎞⎟⎠2⎤⎛⎥⎜⎥⎦⎝xd⎞ ⎟⎠20.67 f +γmcu⎛ 1 ε0⎞⎜1−3⎟⎝ εult ⎠xdM−bd2= 0(Eqn 3-1)With neutral axis depth ratio determined, the steel ratio can be determined byAstbd1 0.67 fcu⎛ 1 ε=f⎜ −010.87yγm ⎝ 3 εult⎞⎟⎠xd(Eqn 3-2)As dx is limited to 0.5 for singly reinforcing sections for grades up to 45under moment redistribution not greater than 10% (Clause 6.1.2.4 of the Code),Mby (Eqn 3-1), will be limited to K ' values as in2bdf cuK ' = 0.154 for grade 30; K ' = 0. 152 for grade 35;K ' = 0.151 for grade 40; K ' = 0. 150 for grade 45which are all smaller than 0.156 under the simplified stress block.However, for grades exceeding 45 and below 70 where neutral axis depth ratiois limited to 0.4 for singly reinforced sections under moment redistribution not17

Version 2.3 May 2008greater than 10% (Clause 6.1.2.4 of the Code), again by (Eqn 3-1)bdM2f cuwill be limited toK ' = 0.125 for grade 50; K ' = 0. 123 for grade 60;K ' = 0.121 for grade 70.which are instead above 0.120 under the simplified stress block as AmendmentNo. 1 has reduce the x / d factor to 0.8. Re discussion is in Appendix C.It should be noted that the x / d ratio will be further limited if momentredistribution exceeds 10% by (Ceqn 6.4) and (Ceqn 6.5) of the Code (withrevision by Amendment No. 1) asx≤ ( βb− 0.4)for fcu≤ 45; anddx≤ β − 0.5 for 45 f ≤ 70dwhere( )b< cuβbus the ratio of the moment after and before moment redistribution.MWhen exceeds the limited value for single reinforcement,2bd f cucompression reinforcements at d ' from the surface of the compression sideshould be added. The compression reinforcements will take up the differencebetween the applied moment and2K' bd f and the compressioncureinforcement ratio is⎛ M⎟ ⎞⎜− K'f2cuA ⎝ bd fsccu ⎠=bd ⎛ d'⎞0.87 fy ⎜1− ⎟⎝ d ⎠(Eqn 3-3)And the same amount of reinforcement will be added to the tensilereinforcement :⎛ M⎟ ⎞⎜− K'f2cuA⎛ ⎞ ⎝ bd fst 1 0.67 f⎠=cu 1 ε⎜ −0cu1⎟η+bd 0.87 fyγm ⎝ 3 εult ⎠ ⎛ d'⎞0.87 fy ⎜1− ⎟⎝ d ⎠(Eqn 3-4)where η is the limit of neutral axis depth ratio which is 0.5 for f ≤ 45, 0.4for 45 f ≤ 70 and 0.33 for 70 f ≤ 100 where moment redistribution< cudoes not exceed 10%.< cucuIt follows that more compressive reinforcements will be required for grade 50than 45 due to the limitation of neutral axis depth ratio, as illustrated by thefollowing Chart 3-1 in which compression reinforcement decreases from grade18

Version 2.3 May 2008M30 to 40 for the same2 , but increases at grade 45 due to the change of thebdlimit of neutral axis depth ratio from 0.5 to 0.4 with moment redistribution notexceeding 10%. The same phenomenon applies to tensile steel also. Withmoment redistribution exceeding 10%, the same trend will also take place.Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.1Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bdGrade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Grade 50 Ast/bd Grade 50 Asc/bd141210M/bd 2864200 0.5 1 1.5 2 2.5 3 3.5 4Reinforcement ratios A/bd (%)Chart 3-1 – Reinforcement Ratios of Doubly Reinforced Beams for Grade 30to 50 with Moment Redistribution limited to 10% or belowzAs similar to BS8110, there is an upper limit of “lever arm ratio” which is dthe depth of the centroid of the compressive force of concrete to the effectivedepth of the beam section of not exceeding 0.95. Thus for calculated values ofzd≥ 0.95approach,xor ≤ 0. 111 in accordance with the simplified stress blockdAstM=bd 0.87 f ( 0.95d)bdyDesign Charts for grades 30 to 50 comprising tensile steel and compressionsteel ratiosA stA and sc are enclosed at the end of Appendix C.bd bd3.4.2 Design in accordance with the Simplified Stress BlockThe design will be simpler and sometimes more economical if the simplified19

Version 2.3 May 2008rectangular stress block as given by Figure 6.1 of the Code is adopted. Thedesign formula becomes :MFor singly reinforced sections where K = ≤ K'where K ' = 0. 1562fcubdfor grades 45 and below and K ' = 0. 120 for 45 < f cu≤ 70; K '= 0. 094 for70 < f cu≤ 100.zdxdK= 0.5+ 0.25 − ≤ 0.95;0.9⎛ z ⎞ 1 ⎛K ⎞= ⎜1− ⎟ = ⎜0.50.25 ⎟0.45− −⎝ d ⎠0.9⎝⎠10.45;AstM= (Eqn 3-5)0.87 f zyFor doubly reinforced sectionszdA= 0.5 +scK'0.25 −0.92( K − K') fcubd0.87 f ( d − d')MK = > K',2f bdcuxd⎛ z ⎞ = ⎜1 − ⎟⎝ d ⎠K'f10.452cu= Ast= + Ascy0.87 fyzbd(Eqn 3-6)3.4.3 Ductility Requirement on amounts of compression reinforcementIn accordance with Cl. 9.9.1.1(a) of the Code, at any section of a beam(participating in lateral load resisting system) within a “critical zone” thecompression reinforcement should not be less than one-half of the tensionreinforcement at the same section. A “critical zone” is understood to be a zonewhere a plastic hinge is likely to be formed and thus generally include sectionsnear supports or at mid-span. The adoption of the clause will likely result inproviding more compression reinforcements in beams (critical zones).3.4.4 Worked Examples for Determination of steel reinforcements in RectangularBeam with Moment Redistribution < 10%Unless otherwise demonstrated in the following worked examples, therequirement in Cl. 9.9.1.1(a) of the Code as discussed in para. 3.4.3 byrequiring compression reinforcements be at least one half of the tensionreinforcement is not included in the calculation of required reinforcements.20

Version 2.3 May 2008Worked Example 3.4Section : 500 (h) × 400 (w), fcu= 35 MPacover = 40 mm (to main reinforcement)(i) M = 1286 kNm;d = 500 − 40 −16= 4441.34 fcu 1.34×35ε0ε0= == 0.0013192 = 0. 3769γ E 1.5×23700mc6M1 286×10== 0.104 < 0.152 , so singly reinforced22fcubd35×400×444xSolving the neutral axis depth ratio by (Eqn 3-1) d20.67 f ⎡ 1 1 1 ⎤cuε0⎛ ε0⎞⎢−+ − ⎥ = −60.38⎢ 2 3 12⎜⎟ ;γmεult⎣⎝ εult ⎠ ⎥⎦60.67f cu⎛ 1 ε0⎞M 286×10⎜1− = 13.6693⎟ ; − == −3.6272 2γm ⎝ εult ⎠bd 400×444xd−13.699+=13.69922×ε ult− 4×( − 60.38) × ( − 3.627)( − 60.38)= 0.307 ≤ 0.5Ast1 0.67 f ⎛ 1 0⎞ 1=cuε x1 = × 13.699×0.307 = 0.01050.87⎜ −3⎟bd fyγm ⎝ εult ⎠ d 0.87×460⇒ A = 1865 mm 2 Use 2T32 + 1T25st(ii) M = 2486 kNm;d = 500 − 40 − 20 = 4401.34 fcu 1.34×35ε0ε0= == 0.0013192 = 0. 3769γ E 1.5×23700mc6M2 486×10== 0.179 > 0.152 , so doubly reinforced22fcubd35×400×440d'50d ' = 40 + 10 = 50 = = 0. 114 (assume T20 bars)d 440⎛ M2 ⎟ ⎞⎜− K fcuA ⎝ bd fsccu ⎠ ( 0.179 − 0.152)× 35By (Eqn 3-3) === 0.267 %bd ⎛ d'⎞ 0.87×460×( 1−0.114)0.87 fy ⎜1− ⎟⎝ d ⎠A = 0 .00267×400×440 = 469 mm 2 Use 2T20scε ult21

Version 2.3 May 2008⎛ M⎟ ⎞⎜− K f2cuA⎛ ⎞ ⎝ bd fst 1 0.67 f⎠By (Eqn 3-4) =cu 1 ε⎜ −0cu1⎟η+bd 0.87 fyγm ⎝ 3 εult ⎠ ⎛ d'⎞0.87 fy ⎜1− ⎟⎝ d ⎠A 1st = 13.699×0.5 + 0.00267 = 1.978 %bd 0.87×460A = 0 .01978×400×440 = 3481mm 2 Use 3T40stWorked Example 3.5(i) and (ii) of Worked Example 3.4 are re-done in accordance with Figure 6.1of the Code (the simplified stress) block by (Eqn 3-5) and (Eqn 3-6)6zK286×10(i) = 0.5+ 0.25 − = 0.5 + 0.25 −= 0. 8672d0.935×400×444 × 0.96AstM286×10=== 0.010452 2bd bd × 0.87 fy( z / d ) 400×444 × 0.87×460×0.867⇒ A = 1856 mm 2 Use 2T32 + 1T25st6M 486×10(ii) K = == 0.179 > 0. 156 , so doubly reinforcing2 2fcubd35×400×440zsection required, = 1 − 0.5×0.9×0.5 = 0. 775d22( K − K' ) fcubd( 0.179 − 0.156)× 35×400×440Asc === 399 mm0.87 fy( d − d') 0.87×460×( 440 − 50)2 >0.2% in accordance with Table 9.1 of the Code, Use 2T1622K' fcubd0.156×35×400×440Ast= + Asc=+ 399 = 3498 mm 20.87 fyz 0.87×460×0.775×440Use 3T40(Note : If the beam is contributing in lateral load resisting system and thesection is within “critical zone”, compressive reinforcements has to be atleast half of that of tension reinforcements Asc= 3498 / 2 = 1749mm 2 byCl. 9.9.1.1(a) in the Code (D). So use 2T25 + 1T32.)Results of comparison of results from Worked Examples 3.4 and 3.5 (with theomission of the requirement in Cl. 9.9.1.1(a) that compressive reinforcementsbe at least half of that of tension reinforcements) are summarized in Table 3.3,indicating differences between the “Rigorous Stress” and “Simplified Stress”Approach :22

Version 2.3 May 2008SinglyReinforcedDoublyReinforcedAst(mm 2 ) Asc(mm 2 ) Ast(mm 2 ) Total(mm 2 )Based on Rigorous 1865 469 3481 3950Stress ApproachBased on Simplified 1856 399 3498 3897stress ApproachTable 3.3 – Summary of Results for comparison of Rigorous stress andsimplified stress Approaches.Results by the two approaches are very close. The approach based on thesimplified stress block are slightly more economical.3.4.5 Worked Example 3.6 for Rectangular Beam with Moment Redistribution >10%If the Worked Example 3.4 (ii) has undergone a moment redistribution of 20%> 10%, i.e. β = 0. 8 , by (Ceqn 6.4) of the Code, the neutral axis depth isxdbxβb,dlimited to ≤ ( − 0 .4) ⇒ ≤ 0.8 − 0.4 = 0. 4zand the lever arm ratio becomes = 1 − 0.4×0.9×0.5 = 0. 82 .dSo theMK= value become 0 .5 + 0.25 − = 0.82 ⇒ K = 0. 1320.9K2bd f cu2( K − K ) fcubd0.87 f ( d − d')( 0.176 − 0.132)'2× 35×400×440Asc === 764 mm 2 > 0.2 %0.87×460×yas required by Table 9.1 of the Code;( 440 − 50)22K' fcubd0.132×35×400×440Ast= + Asc=+ 764 = 3242 mm 20.87 f z 0.87×460×0.82×440ySo total amount of reinforcement is greater.3.5 Sectional Design of Flanged Beam against Bending3.5.1 Slab structure adjacent to the beam, if in flexural compression, can be used toact as part of compression zone of the beam, thus effectively widen thestructural width of the beam. The use of flanged beam will be particularlyuseful in eliminating the use of compressive reinforcements, as in addition to23

Version 2.3 May 2008reducing tensile steel due to increase of lever arm. The principle of sectionaldesign of flanged beam follows that rectangular beam with an additionalflange width ofb − b as illustrated in Figure 3.9.effwb eff0.67γmf cuh f0.9xxdb wFigure 3.9 – Analysis of a T or L beam sectionDesign formulae based on the simplified stress block are derived in AppendixC which are summarized as follows :(i)(ii)Singly reinforcing section where η × neutral axis depth is insideflange depth by checking where η = 0. 9 for fcu≤ 45 ; η = 0. 8 for45 < f cu≤ 70; η = 0. 72 for 70 < f cu≤100.xK hfMη = 1−1−≤ where K =2d 0.225 dfcubeffd(Eqn 3-7)If so, carry out design as if it is a rectangular beam of width beff.Singly reinforcing section where η × neutral axis depth is outsidex hfflange depth, i.e. η >d dandMb d0.67 f=γ⎛ b⎜⎝ b⎞ hf−1⎟⎠ d⎛ ⎜1−⎝hf ⎞⎟ + 0.67 fd ⎠ γm⎛ x ⎞⎟⎜⎛ η x⎜η1 −⎝ d ⎠⎝dcu effcu2w m w212⎞⎟⎠x be solved by the quadratic equation :d2 20.67 f ⎛ ⎞ 0.67 M − Mcu η x fcuxf⎜ ⎟ − η + 02γ 2 ⎝ d ⎠ γ d b dmmw=(Eqn 3-8)wherebM0.67 f=fcu2wdγmhfd⎛ b⎜⎝ beffw⎞⎛ 1−1⎟⎜1 −⎠⎝2hfd⎞⎟⎠(Eqn 3-9)24

Version 2.3 May 2008AndAstb dw0.67 fcu=γ 0.87 fmy⎡⎛b⎢⎜⎣⎝beffw⎞ hf−1⎟⎠ dx ⎤+ η ⎥d ⎦(Eqn 3-10)(iii) Doubly reinforcing section :By following the procedure in (ii), if dxobtained by (Eqn 3-8)exceeds ϕ where ϕ = 0. 5 for fcu> 45 ; 0.4 for fcu> 70 and 0.33for f > 100 , then double reinforcements will be required withcurequiredAscandA stasA 1 ⎡ 0.67 ⎡⎛b ⎞ h ⎛ 1 h ⎞ ⎛ ⎞⎤⎤scM fcueff ff 1= ⎢ − ⎢⎜−1⎟⎜1−⎟ + ηϕ⎜1− ϕ ⎟⎥b d 0.87 fwy( ) ⎥ ⎥ 21−d'/ d⎟ ⎜⎢⎣b⎜wdγm ⎣⎝bw⎠ d ⎝ 2 d ⎠ ⎝ 2 ⎠⎦⎦A 0.67 f ⎡⎛beff⎞ hf⎤cu= ⎢⎥bwdmf⎜ −1+y ⎣ b⎟ ηϕγ 0.87 ⎝ w ⎠ d ⎦st+25Ascb dw(Eqn 3-11)(Eqn 3-12)3.5.2 Worked Examples for Flanged Beam, grade 35 ( η = 0. 9 )x hf(i) Worked Example 3.7 : Singly reinforced section where 0 .9 ≤d dConsider the previous example done for a rectangular beam 500 (h) ×400 (w), fcu= 35 MPa, under a moment 486 kNm, with a flangedsection of width = 1200 mm and depth = 150 mm :b = 400 , d = 500 − 40 − 20 = 440 , b = 1200 h = 150wFirst check ifK =fcuMbeffdx hf0 .9 ≤ based on beam width of 1200,d d6486×10=0.0598235×1200×4402=x ⎛K ⎞ 1By (Eqn 3-5), = ⎜0 .5 0.25 ⎟− − = 0. 1590.9;d ⎝⎠ 0.45x hf 150z x∴0 .9 = 0.143 < = = 0.341. = 1 − 0.45 = 0. 928 ; Thusd d 440 d d6AstM486×10=== 0.005632 2b d b d × 0.87 f z / d 1200×440 × 0.87×460×0.928effeffy( )bw400> 0.18% (minimum for = = 0.33 < 0. 4 in accordance withbeff1200Table 9.1 of the Code)∴ A st= 2974mm 2 . Use 2T40 + 1T25As in comparison with the previous example based on rectangularefff

Version 2.3 May 2008section, it can be seen that there is saving in tensile steel (2974 mm 2 vs3498 mm 2 ) and the compression reinforcements are eliminated.(ii)x hfWorked Example 3.8 – Singly reinforced section where η > , andd dη = 0.9 for grade 35.Beam Section : 1000 (h) × 600 (w), flange width = 2000 mm,flange depth = 150 mm fcu= 35 MPa under a moment 4000 kNmb = 600 , d = 1000 − 50 − 60 = 890 , b = 2000 h = 150wh f 150 beff2000 = = 0.169 ; = = 3. 333d 890bw600x hfFirst check if 0 .9 ≤ based on beam width of b w= b eff= 2000d d6M 4000×10K = == 0.07212 2fcubeffd 35×2000×890By (Eqn 3-7)x ⎛K ⎞ hf 1500 .9 = 2⎜0.50.25 ⎟− − = 0.176 > = = 0.169d0.9⎝⎠ d 890So 0.9 × neutral axis depth extends below flange.Mf 0.67 f hf ⎛ beff⎞⎛ 1 h ⎞cuf= ⎜ 1⎟⎜1⎟ ⇒ = 2675.652 −−2MfkNmbwdγmd ⎝ bw⎠⎝d ⎠xSolve by (Eqn 3-8) with = 0. 9defffη .2⎛ x ⎞x M − Mf0.1809fcu ⎜ ⎟ − 0.402 fcu+ = 02⎝ d ⎠d b d2w( 4000 − 2675.65)⎛ x ⎞x⇒ 0.1809×35⎜⎟ − 0.402×35 +2⎝ d ⎠d 600×890x⇒ = 0.2198 ; d× 106= 0 ;By (Eqn 3-10)A 1 0.67 ⎡⎛b ⎞ h⎤stfcueff f=⎢ 1 + 0.9×0.2198⎥= 0.023090.87⎜ −⎟bwdfyγm ⎣⎝bw⎠ d⎦= 12330 mm 2 , Use 10-T40 in 2 layersAst(iii)Worked Example 3.9 – Doubly reinforced sectionBeam Section : 1000 (h) × 600 (w), flange width = 1250 mm,flange depth = 150 mm f = 35 MPa under a moment 4000 kNmcub = 600 , d = 1000 − 50 − 60 = 890 , b = 1250 h = 150wefff26

Version 2.3 May 2008h f 150 beff1250 = = 0.169 ; = = 2. 083 ; η = 0. 9d 890bw600x hfFirst check if η ≤ based on beam width of beff= 1250d d6M 4000×10K = == 0.1152 2fcubeffd 35×1250×890By (Eqn 3-7)x 0.115 hf 1500 .9 = 1−1−= 0.302 > = = 0.169d 0.225 d 890So 0.9 × neutral axis depth extends below flange.Mf 0.67 f hf ⎛ beff⎞⎛ 1 h ⎞cuf= ⎜ 1⎟⎜1⎟ ⇒ = 1242.262 −−2MfkNmbwdγmd ⎝ bw⎠⎝d ⎠xSolve by (Eqn 3-8) with = 0. 9dη2⎛ x ⎞x M − Mf0.1809fcu ⎜ ⎟ − 0.402 fcu+ = 02⎝ d ⎠d b d2w( 4000 −1242.26)⎛ x ⎞x⇒ 0.1809f cu ⎜ ⎟ − 0.402×35 +2⎝ d ⎠d 600×890x= 0 .547 >dBy (Eqn 3-11)× 106= 00.5 . Double reinforcement required. d ' = 50 + 20 = 70A= 1 ⎡ 0.67 ⎡⎛b ⎞ h ⎛ 1 h ⎞ ⎛ ⎞⎤( ) ⎥ ⎥ ⎤scM fcueff ff 1⎢ − ⎢⎜ −1⎟⎜1−⎟ + ηϕ⎜1− ⎟⎥0.87 1−'/⎢⎣γ ⎣⎝⎠ ⎝ 2 ⎠ ⎝ 2ϕ2bwdfyd d bwdmbwd d⎠⎦⎦= 0 .001427 = 0.143 %A = 763 mm 2 > 0.4% on flange as per Table 9.1 of the Code which issc0 .004× 1250×150 = 750 mm 2 . Use 6T20By (Eqn 3-12)Ast0.67 fcu=b d γ 0.87 fwmy⎡⎛b⎢⎜⎣⎝beffw⎞ hf−1⎟⎠ d⎤ Asc+ ηϕ⎥+⎦ bwd= 0.02614⇒ A = 13958 mm 2 , Use 10-T40 + 2-T32 in 2 layers (2.65%)st3.6 Detailings of longitudinal steel for bendingThe followings should be observed in placing of longitudinal steel bars forbending. Re Cl. 9.2.1 and 9.9.1 of the Code. The requirements arising from“ductility” requirements are marked with “D” for beams contributing in lateralload resisting system:(i)Minimum tensile steel percentage : For rectangular beam, 0.13% inaccordance with Table 9.1 of the Code and 0.3% in accordance with Cl.27

Version 2.3 May 2008(ii)(iii)(iv)9.9.1 of the Code (D); except for beams subject to pure tension whichrequires 0.45% as in Table 9.1 of the Code;Maximum tension steel percentage : 2.5% (Cl. 9.9.1.1(a)) for beamscontributing in lateral load resisting system(D); and 4% for others (Cl.9.2.1.3 of the Code);Minimum compressive steel percentage : When compressive steel isrequired for ultimate design, Table 9.1 of the Code should be followedby providing 0.2% for rectangular beam and different percentages forothers. In addition, at any section of a beam within a critical zone (e.g.a potential plastic hinge zone as discussed in Section 2.4) thecompression reinforcement ≥ one-half of the tension reinforcement inthe same region (Cl. 9.9.1.1(a) of the Code) (D);For flanged beam, Figure 3.10 is used to illustrate the minimumpercentages of tension and compression steel required in various partsof flanged beams (Table 9.1 of the Code), but not less than 0.3% inaccordance with Cl. 9.9.1.1(a) of the Code (D);hfb effLongitudinal bars in flangeAst ≥ 0. 0026bwh(T-beam)Ast ≥ 0. 002bwh(L-beam)A ≥ 0. 004b hscefffhTransverse bars in flangeA ≥ 0 .0015h× 1000 persunit metre of flange lengthfb wLongitudinal bars in web:Ast ≥ 0. 0018bwhif b / < 0. 4w b effAst ≥ 0. 0013bwh if b / ≥ 0. 4Asc ≥ 0. 002bwhw b effFigure 3.10 – Minimum steel percentages in various parts of flanged beams(v)For beams contributing in lateral load resisting system, calculation ofanchorage lengths of longitudinal bars anchored into exterior columns,bars must be assumed to be fully stressed as a ductility requirementaccording to Cl 9.9.1.1(c) of the Code. That is, stresses in the steelshould bef yinstead of0 .87 fyin the assessment of anchoragelength. As such, the anchorage and lap lengths as indicated in Tables8.4 and 8.5 of the Code should be increased by 15% as per (Ceqn 8.4)28

of the Code in whichlbbuVersion 2.3 May 2008fyφ≥ which is a modification (by changing4 f0 .87 fyto fy) where fbu= β fcuand β is 0.5 for tension(vi)(vii)anchorage and 0.63 for compression anchorage for high yield bars inaccordance with Table 8.3 of the Code. Lap lengths can be taken asidentical to anchorage lengths (D);Full strength welded splices may be used in any location according toCl. 9.9.1.1(d) of the Code;For beams contributing in lateral load resisting system, no portion ofthe splice (laps and mechanical couplers) shall be located within thebeam/column joint region or within one effective depth of the memberfrom the critical section of a potential plastic hinge (discussed inSection 2.4) in a beam where stress reversals in lapped bars couldoccur (Cl. 9.9.1.1(d) of the Code). However, effects due to wind loadneed not be considered as creating stress reversal (D);potentialplastic hingesectionno lap /mechanicalcoupler zonestress reversal couldoccurddFigure 3.11 – Location of no lap / mechanical coupler zone in beamcontributing to load resisting system(viii) For beams contributing in lateral load resisting system, longitudinalbars shall not be lapped in a region where reversing stresses at theultimate state may exceed0 .6 fyin tension or compression unlesseach lapped bar is confined by links or ties in accordance with (Ceqn9.6) reproduced as follows (D) :Astr⋅ fy≥ φ (Eqn 3-13)48 fytAccording to the definitions in the Code, φ is the diameter of thelongitudinal bar; Atris the smaller of area of transversereinforcement within the spacing s crossing a plane of splitting29

Version 2.3 May 2008normal to the concrete surface containing extreme tension fibres, ortotal area of transverse reinforcements normal to the layer of barswithin a spacing s , divided by n (no. of longitudinal bars) in mm 2 ;s is the maximum spacing of transverse reinforcement within the laplength,fytis the characteristic yield strength of the transversereinforcement.As the “just adequate” design is by providing steel so that thereinforcing bars are at stress of0 .87 fy, overprovision of the sectionby 0.87/0.6 – 1 = 45% of reinforcing bars at the laps should fulfill therequirement for lapping in regions with reversing stress. Or else,transverse reinforcement by (Ceqn 9.6) will be required. Figure 3.12shows the occurrence of the plane of splitting at lapping.Potential split faces by the bar forcetransmitting lapping force by shearfriction(a)Figure 3.12 – splitting of concrete by shear friction in lapping of barsConsider the example (a) illustrated in Figure 3.12, transverse(b)reinforcement required will simply beAstrφ ⋅ f≥48 fyytφ=48if high yieldbars are used for both longitudinal and transverse reinforcements. If40φ = 40 (i.e. the longitudinal bars are T40), ≥ = 0. 833 . Thes 48A trtotal area of transverse reinforcement is∑ A tr= 4 Atras there are 430

∑A trVersion 2.3 May 2008no. of bars. So ≥ 0 .833×4 = 3. 333. Using T12 – 4 legs – 125 iss(ix)∑ A tradequate as provided is 3.619. It should be noted that case (b)sis generally not the controlling case.At laps in general, the sum of reinforcement sizes in a particular layershould not exceed 40% of the beam width as illustrated by a numericalexample in Figure 3.13 (Cl. 9.2.1.3 of the Code);bar diameterd = 40bar diameterd = 40Sum of reinforcementsizes = 40 × 8 = 320< 0.4 × 900 = 360.So O.K.beam width b = 900Figure 3.13 – Illustration of sum of reinforcement sizes at laps ≤ 0.4 ofbeam width(x)(xi)Minimum clear spacing of bars should be the greatest of bar diameter,20 mm and aggregate size + 5 mm (Cl. 8.2 of the Code);Maximum clear spacing between adjacent bars near tension face of abeam ≤ 70000β b /f y ≤ 300 mm where β b is the ratio of momentredistribution (ratio of moment after redistribution to moment beforeredistribution) or alternatively ≤ 47000/f s ≤ 300 mm wherefs2 fyAs,req 1= × . Based on the former with β b = 1 (no3Aβs,provb(xii)redistribution), the maximum clear spacing is 152 mm (Cl. 9.2.1.4 ofthe Code);Requirements for containment of compression steel bars is identical tothat of columns (Cl. 9.5.2.2 of the Code) : Every corner bar and eachalternate bar (and bundle) in an outer layer should be supported by alink passing around the bar and having an included angle ≤ 135 o . Linksshould be adequately anchored by means of hook through a bent angle≥ 135 o . No bar within a compression zone be more than 150 mm froma restrained bar (anchored by links of included angle ≥ 135 o ) asillustrated in Figure 3.14;31

Version 2.3 May 2008(xiii) No tension bars should be more than 150 mm from a vertical leg whichis also illustrated in Figure 3.14 (Cl. 9.2.2 of the Code);Links bent throughangle ≥ 135 o foranchorage in concreteSpacing of tensionbar ≤150 from avertical legAlternate bar in an outer layerrestrained by link of includedangle ≤135 ocompression zone≤135 o≤ 150≤ 150≤ 150≤ 150restrained longitudinalcompression bars beanchored by links ofincluded angle ≤ 135 o≤ 150≤ 150bar in compression ≤ 150from a restrained barFigure 3.14 – Anchorage of longitudinal bar in beam section(xiv)At an intermediate support of a continuous member, at least 30% of thecalculated mid-span bottom reinforcement should be continuous overthe support as illustrated in Figure 3.15 (Cl. 9.2.1.8 of the Code);Calculatedmid-span steelarea A s2≥ and≥0.3As10.3As2Calculatedmid-span steelarea A s1Figure 3.15 – At least 30% of the calculated mid-span bottom bars becontinuous over intermediate support(xv)In monolithic construction, simple supports should be designed for15% of the maximum moment in span as illustrated in Figure 3.16 (Cl.9.2.1.5 of the Code);32

Version 2.3 May 2008section designedfor 0.15 M maxmaximum bendingmoment M maxSimple support bybeam or wallBending moment diagramFigure 3.16 – Simple support be designed for 15% of the maximum spanmoment(xvi)For flanged beam over intermediate supports, the total tensionreinforcements may be spread over the effective width of the flangewith at least 50% inside the web as shown in Figure 3.17 reproducedfrom Figure 9.1 of the Code;b effat most 50% ofreinforcementsoutside the webat least 50% ofreinforcementsinside the webbFigure 3.17 – distribution of tension rebars of flanged beam over support(xvii) For beam with depths > 750 mm, provision of sides bars of size (inmm) ≥sbb/ fywhere bs is the side bar spacing (in mm) and bis the lesser of the beam breadth (in mm) under consideration and 500mm.f is in N/mm 2 . In addition, it is required that s ≤ 250 mm andybside bars be distributed over two-thirds of the beam’s overall depthmeasured from its tension face. Figure 3.18 illustrate a numericalexample (Cl. 9.2.1.2 of the Code);33

Version 2.3 May 20081500T161000b is the lesser of 600 and 500, sob = 500s b chosen to be 200 mm ≤ 250mm,So size of side bar iss b / fby= 200 × 500 / 460= 14.74Use T16.The side bars be distributed over2× 1500 = 1000 from bottom3which is the tension side.tension side600Figure 3.18 – Example of determination of side bars(xviii) When longitudinal bars of beams contributing to lateral load resistingsystem are anchored in cores of exterior columns or beam studs, theanchorage for tension shall be deemed to commence at the lesser of 1/2of the relevant depth of the column or 8 times the bar diameter asindicated in Figure 3.19. In addition, notwithstanding the adequacy ofthe anchorage of a beam bar in a column core or a beam stud, no barshall be terminated without a vertical 90 o standard book or equivalentanchorage device as near as practically possible to the far side of thecolumn core, or the end of the beam stud where appropriate, and notcloser than 3/4 of the relevant depth of the column to the face of entry.Top beam bars shall be bent down and bottom bars must be bent up asindicated in Figure 3.19. (Cl. 9.9.1.1(c) of the Code) (D);Danchoragecommences at thissection generally.Notpermitted≥ 0.5Dor 8Ø≥ 500mm or hXh≥ 0.75Danchorage can commence at this sectionif the plastic hinge (discussed in Section2.4) of the beam is beyond XFigure 3.19 – Anchorage of reinforcing bars at support for beams contributing tolateral load resisting system34

Version 2.3 May 2008(xix)Beam should have a minimum support width by beam, wall, column asshown in Figure 3.20 as per Cl. 8.4.8 of the Code;≥2(4Ø+c) if Ø ≤ 20≥2(5Ø+c) if Ø > 20centre line ofsupportc3Ø if Ø ≤ 20;4Ø if Ø > 20bar ofdiameter Ø≥0Figure 3.20 – Support width requirement(xx)Curtailment of tension reinforcements except at end supports should bein accordance with Figure 3.21 (Cl. 9.2.1.6 of the Code).Bar of diameter ØdSection beyondwhich the bar is nolonger required≥12Ø and d at least;if the bar is inside tensionzone, ≥ 1.0 bond lengthFigure 3.21 – curtailment of reinforcement barsWorked Example 3.10Worked example 3.10 is used to illustrate the arrangement of longitudinal barsand the anchorages on thin support for the corridor slab beam of a typicalhousing block which functions as coupling beam between the shear walls onboth sides. Plan, section and dimensions are shown in Figure 3.22(a). Concretegrade is 35.35

Version 2.3 May 2008300Slab beam1400Plan200SectionFigure 3.22(a) – Layout of the slab beam in Worked Example 3.10The designed moment is mainly due to wind loads which is 352 kNm,resulting in required longitudinal steel area of 3910 mm 2 (each at top andbottom). The 200 mm thick wall can accommodate at most T16 bars as2 ( 4× 16 + 25) = 178 < 200 as per 3.6(xix). So use 20T16 ( Astprovided is4020 mm 2 . Centre to centre bar spacing is ( 1400 − 25×2 −16) /19 = 70 mm.For anchorage on support, lap length should be 34 × 16 = 544 mm. The factor34 is taken from Table 8.4 of the Code which is used in assessing anchoragelength. Anchorage details of the longitudinal bars at support are shown inFigure 3.22(b);20T16T16cross bar544T10 – 10 legs – 200 c/c256420011Anchorage commences atcentre line of wall as200/2=100

3.7.1 Checking of Shear Stress and provision of shear reinforcementsVersion 2.3 May 2008Checking of shear in beam is based on the averaged shear stress calculatedfrom (Ceqn 6.19)Vv =bvdwhere v is the average shear stress, V is the ultimate shear, d is theeffective depth of the beam and bvis beam width. b vshould be taken as theaveraged width of the beam below flange in case of flanged beam)If v is greater than the values of vc, termed “design concrete shear stress” inTable 6.3 of the Code which is determined by the formula11133cus ⎛ 400 ⎞ 4 1c= ⎜ ⎟25 ⎜ ⎜ ⎟bvd⎟listed in Table 6.3 of the Code with⎝ d ⎠ γm⎛ f ⎞ ⎛100A⎞v 0.79 ⎜ ⎟⎝ ⎠ ⎝ ⎠the following limitations :(i) γm= 1. 25;100As(ii) should not be taken as greater than 3;b dv14⎛ 400 ⎞(iii) ⎜ ⎟⎠ should not be taken as less than 0.67 for member without shear⎝ dreinforcements and should not be taken as less than 1 for members withminimum links. For others, calculate according to the expression;Asvbv( v − vc) bvvrThen shear links of = ≥ should be provided (Tablesv0.87 fyv0.87 fyv6.2 of the Code) where vr= 0. 4 for fcu≤ 40 MPa and 0.4(f cu /40) 2/3 for> 40 , or alternatively, less than half of the shear resistance can be taken upfcud − d'by bent up bars by 0.5V≥ Vb= Asb( 0.87 fyv)( cosα + sin α cot β ) as persb(Ceqn 6.20) and Cl. 6.1.2.5(e) of the Code and the rest by vertical links.Maximum shear stress not to exceedv = 0. 8 f or 7 MPa, whichever istucuthe lesser by Cl. 6.1.2.5(a).3.7.2 Minimum shear reinforcements (Table 6.2 of the Code)Ifv < 0. 5v , no shear reinforcement is required;cIf .5v < v < ( v + v )0 , minimum shear links ofccrAssv v r= along thevb v0.87 fyv37

Version 2.3 May 2008whole length of the beam be provided where v = 0. 4 for f ≤ 40 and2 / 3( / 40) 0.4fcufor f > 40 , but not greater than 80;curcu3.7.3 Enhanced shear strength close to support (Cl. 6.1.2.5(g))At sections of a beam at distancecan be increased by a factorv = 0. 8 or 7 MPa as illustrated by Figure 3.23.tuf cua va v≤ 2dfrom a support, the shear strength2 d , bounded by the absolute maximum ofda vsection under consideration, shear strength2denhanced to vcavFigure 3.23 – Shear enhancement near support3.7.4 Where load is applied to the bottom of a section, sufficient verticalreinforcement to carry the load should be provided in addition to anyreinforcements required to carry shear as per Cl. 6.1.2.5(j);Vertical rebars toresist beambottom loadswhich mayincrease requiredprovisions oflinksHanging load at beam bottomFigure 3.24 – Vertical rebars to resist hanging load at beam bottom(e.g. inverted beam)3.7.5 Worked Examples for Shears38

Version 2.3 May 2008(i)Worked Example 3.11 – shear design without shear enhancement inconcreteSection : b = 400 mm;100A d = 700 − 40 −16= 644 mm; st= 1. 5 ; fcu = 35 MPa;bdV = 700 kN;vc13⎛ fcu⎞= 0.79⎜⎟⎝ 25 ⎠14⎛100A⎞s⎜⎟⎝ bvd⎠1314⎛ 400 ⎞⎜ ⎟⎝ d ⎠⎛ 400 ⎞where ⎜ ⎟⎠⎝ d3700×10v = = 2.72 MPa,400×644Asvb v − vc400 2.72 − 0.81= =s 0.87 f 0.87×460v1γm= 0.81MPa;be kept as unity for d > 400 .( ) ( )yv= 1.91; Use T12 – 200 c/c d.s.(ii)Worked Example 3.12 – shear design with shear enhancement inconcrete.Re Figure 3.25 in which a section of 0.75 m from a support, as a heavypoint load is acting so that from the support face to the point load alongthe beam, the shear is more or less constant at 700 kN.a v = 750700 kNd = 644T32Figure 3.25 – Worked Example 3.11Section : b = 400 mm; cover to main reinforcement = 40 mm;100A d = 700 − 40 −16= 644 mm; st= 1. 5 ; fcu = 35 MPa; V = 700 kN;bd13⎛ fcu⎞vc= 0.79⎜⎟⎝ 25 ⎠3.11.⎛100A⎞s⎜⎟⎝ bvd⎠1314⎛ 400 ⎞⎜ ⎟⎝ d ⎠1γm= 0.81MPa as in Worked Example39

Version 2.3 May 20082d2×644Concrete shear strength enhanced to vc= × 0.81 = 1. 39 MPa

Version 2.3 May 2008(iii)(iv)At least 50% of the necessary shear reinforcements be in form of links(Cl. 6.1.2.5(e) of the Code);The maximum spacing of links in the direction of span of the beamshould be the least of the followings as illustrated by a numericalexample in Figure 3.27 (Cl. 6.1.2.5(d), 9.2.1.10, 9.5.2.1 (BS8110 Cl.3.12.7.1), 9.5.2.2, 9.9.1.2(a) of the Code) :(a) 0.75d;(b) the least lateral dimension of the beam (D);(c) 16 times the longitudinal bar diameter (D);(d) 12 times the smallest longitudinal bar diameter for containmentof compression reinforcements.width of web of beam= 4003T32 (tension bar)d = 6443T25 (compression bar)For beam contributing tolateral load resistingsystem, maximum linkspacing be the least of(a) 0.75d = 483;(b) b = 400;(c) 16 × 32 = 512;(d) 12 × 25 = 300So maximum link spacingshould be 300 mm.For beam not contributingto lateral load resistingsystem, (b) & (c) not countFigure 3.27 – Maximum spacing of shear links in the span direction of beam(v)At right angle of the span, the horizontal spacing of links should besuch that no longitudinal tension bar should be more than 150 mmfrom a vertical leg and ≤ d as per Cl. 6.1.2.5(d) of the Code and shownin Figure 3.28;≤ d≤ d≤150≤150Figure 3.28 – Maximum spacing of shear links at right angle to the spandirection of beam41

Version 2.3 May 2008(vi) Links or ties shall be arranged so that every corner and alternatelongitudinal bar that is required to function as compressionreinforcement shall be restrained by a leg as illustrated Figure 3.14;(vii) By Cl. 9.9.1.2(b) of the Code, links in beams contributing to lateralload resisting system should be adequately anchored by means of 135 oor 180 o hooks in accordance with Cl. 8.5 of the Code as shown inFigure 3.29 (D);(viii) Anchorage by means of 90 o hook is only permitted for tensile steel inbeams not contributing to lateral load resisting system;(ix) Links for containment of compression longitudinal bars in generalmust be anchored by hooks with bent angle ≥ 135 o in accordance withCl. 9.2.1.10 and 9.5.2.1 of the Code. Links with different angles ofhooks are shown in Figure 3.29. (Reference to Cl. 9.5.2.1 should beincluded in Cl. 9.2.1.10 as per Cl. 3.12.7.1 of BS8110)Link with 180 ohooksLink with 135 ohooksLink with 90 o hooks (notpermitted for containmentof compression bars ofbeam in general and alllinks in beams contributingto lateral load resistingsystem)Figure 3.29 – Links with hooks with different bent angles3.9 Design against Torsion3.9.1 By Cl. 6.3.1 of the Code, in normal slab-and-beam and framed construction,checking against torsion is usually not necessary. However, checking needs becarried out if the design relies entirely on the torsional resistance of a membersuch as that indicated in Figure 3.30.42

Version 2.3 May 2008Beam carryingunbalancedtorsion inducedby slab needs becheckedFigure 3.30 – Illustration for necessity of checking against torsion3.9.2 Calculation of torsional rigidity of a rectangular section for analysis (ingrillage system) is by (Ceqn 6.64) of the Code1 3C = βhminhmaxwhere β is to be read from Table 6.16 of the Code2reproduced as Table 3.4 of this Manual.h max /h min 1 1.5 2 3 5 >5β 0.14 0.20 0.23 0.26 0.29 0.33Table 3.4 – Values of coefficient β3.9.3 Calculation of torsional shear stressUpon arrival of the torsion on the rectangular section, the torsional shear stressis calculated by (Ceqn 6.65) of the Code2Tv t=2 ⎛ hmin⎞hmin⎜hmax− ⎟⎝ 3 ⎠and in case of a section such as T or L sections made of rectangles, the section3should be broken up into two or more rectangles such that the ∑h minhmaxismaximized and the total Torsional moment T be apportioned to eachrectangle in accordance with (Ceqn 6.66) of the Code as⎛3⎞⎜ hminhmaxT × ⎟ .⎜3⎟⎝ ∑ hminh max ⎠43

Version 2.3 May 2008If the torsional shear stress exceeds0 .067 f (but not more than 0.6MPa),cutorsional reinforcements will be required (Table 6.17 of the Code).Furthermore, the torsional shear stress should be added to the shear stressinduced by shear force to ensure that the absolute maximumv = 0. 8tufcuor 7MPa is not exceeded, though for small section where y 1(the largercentre-to-centre dimension of a rectangular link) < 550mm, v tuwill bedecreased by a factor y1/ 550 . Revision of section is required if the absolutemaximum is exceeded (Table 6.17 of the Code).3.9.4 Calculation of torsional reinforcementsTorsional reinforcement in forms of close rectangular links and longitudinalbars are to be calculated by (Ceqn 6.67) and (Ceqn 6.68) of the Code asAsvT= (Ceqn 6.67)sv0.8x1y1( 0.87 fyv)A is the area of the 2 legs of the link)(svAs( x y )Asvfyv 1= (Ceqn 6.68)s fv1 +yIt should be noted that there is no reduction by shear strength ( v c) of concrete.The derivation of the design formula (Ceqn 6.67) of the Code for closerectangular links is under the assumption of a shear rupture length of stirrupwidth + stirrup depth x1+ y1as shown in Figure 3.31. A spiral torsionalfailure face is along the heavy dotted line in the figure. It is also shown in thefigure that the torsional moment of resistance by the stirrups within theRegions X and Y are identical and is the total resistance is therefore0.87Asvfyx1y1sv. SoTAsv0.87 fyx1y1AsvT= ⇒ =. An additionals s 0.87 f x yvvy11factor of 0.8 is added and the equation becomes (Ceqn 6.67) by whichAsvT= . The derivation of the longitudinal bars is based on thes .8xy 0.87 fv01 1( )yuse of same quantity of longitudinal bars as that of stirrups with evendistribution along the inside of the stirrups. Nevertheless, the Code allowsmerging of the flexural steel with these longitudinal bars by using larger44

diameter of bars as will be illustrated in the Worked Example 3.14.Version 2.3 May 2008shear rupturespiral face45 oy 10 .5Asv0.87 f y0 .5Asv0.87 f yMoment provided by thisstirrup in region X is0.5Asv0.87 f y y1. Totalnos. of stirrup within X isx 1 / s v . So total momentis 0.5A sv0.87f y y1x1/ svx1Ty145 o 1y 1xRegion YRegion X0.5Asv0.87 f yx 10 .5Asv0.87 f yMoment provided by thisstirrup in region Y is0.5Asv0.87 f yx1. Totalnos. of stirrup within Y isy 1 / s v . So total momentis 0.5A sv0.87f yx1y1/ svFigure 3.31 – Derivation of Formulae for torsional reinforcements3.9.5 Worked Example 3.14 – Design for T-beam against torsionA total torsion of T = 200 kNm on a T-section as shown in Figure 3.32 withan average vertical shear stress on the web of 0.82 N/mm 2 . The section is alsounder bending requiring flexural steel area of 2865 mm 2 at bottom. Concretegrade is 35.15004001000450Option AOption BFigure 3.32 – Section of a T section resisting torsion for Worked Example 3.14100As2865×100For vertical shear, taking = = 0. 477b d 450×1334v45

Version 2.3 May 2008vcAsv13⎛ fcu⎞= 0.79⎜⎟⎝ 25 ⎠sv⎛100A⎞s⎜⎟⎝ bvd⎠1314⎛ 400 ⎞⎜ ⎟⎝ d ⎠( v − v ) 450( 0.82 − 0.55)b=0.87 fcyv=0.87×4601γm= 0.3= 0.55 , again taking14⎛ 400 ⎞⎜ ⎟⎠⎝ das unity.For torsion, Option A is made up of two rectangles of 525× 400 and onerectangle of 450× 1400 .33311( h ) = 2 × 525×400 + 1400×450 = 1.94775×∴ ∑h mm 4min max10optionAOption B is made up of one rectangle of 1500× 400 and one rectangle of450× 1000 .33311( h ) = 1500 × 400 + 1000×450 = 1.87125×∴ ∑ h mm 4min max10optionBAs Option A has a larger torsional stiffness, it is adopted for design.The torsional moment is apportioned to the three rectangles of Option A as :3525×400For the two 525× 400 rectangles T1 = 200×= 34. 50kNm;111.94775×10Torsional shear stress is62T12×34.5×10v t 1=== 1.101N/mm 22⎛hmin⎞ 2⎛400 ⎞hmin⎜hmax− ⎟ 400 ⎜525− ⎟⎝ 3 ⎠ ⎝ 3 ⎠> 0 .067 f = 0. 396 N/mm 2 (< 0.6 N/mm 2 )cuSo torsional shear reinforcement is requiredx = 400 − 40×2 − 6×2 308 ; y = 525 − 40×2 − 6×2 4331=1=AssvvT=0.8xy11( 0.87 f )634.5×10=0.8×308×433×0.87×4601=yv0.808Use T12 – 200 C.L. x 308 , y / 2 = 525/ 2 262. 5 ; use s = 200 ≤ 200 ;x 11 =1=≤ and ≤ / 2 as per Cl. 6.3.7 of the Code.y 1( x1+ y1) 0.808×460×( 308 + 525)AsvfyvAs=== 673 mm 2s f460Use 4T16vyv46

Version 2.3 May 200831400×450For the 1400× 450 rectangle T2 = 200×= 131kNm111.94775×1062T22×131×10v t 1=== 1.035N/mm 22⎛hmin⎞ 2⎛450 ⎞hmin⎜hmax− ⎟ 450 ⎜1400− ⎟⎝ 3 ⎠ ⎝ 3 ⎠The total shear stress is 1 .035 + 0.82 = 1. 855 N/mm 2 < v = 4. 73 MPatu>cuAs 1 .035 0.067 f = 0. 396N/mm 2 , torsional shear reinforcement isrequired.x = 450 − 40×2 − 6×2 358 mm; y = 1400 − 40×2 − 6×2 1308mm1=1=AssvvT=0.8xy11( 0.87 f )6131×10=0.8×358×1308×0.87×4602=yv0.87 mmAsvAdding that for vertical shear, total = 0 .87 + 0.3 = 1. 17svUse T12 – 175 C.L. x 358 , y / 2 = 1308/ 2 654 ; use s =175 ≤ 200;x 11 =1=≤ and ≤ / 2 as per Cl. 6.3.7 of the Code.y 1It should be noted that the torsional shear link should be closed links of shapeas indicated in Figure 9.3 of the Code.As=Asvfyvs( x1+ y1) 0.87×460×( 358 + 1308)vfy=460= 1449 mm 2 . Use 13T12Incorporating the bottom 3T12 into the required flexural steel, the bottom steelarea required is 2865 + 113.1×3 = 3205 mm 2 . So use 4T32 at bottom and10T12 at sides.vThe sectional details is shown in Figure 3.33.1500T12 – 200 C.L.4001000T12T16T324T32450Figure 3.33 – Arrangement of torsional reinforcements47

Version 2.3 May 2008It should be borne in mind that these torsional reinforcements are in additionto others required for flexure and shear etc.3.10 Placing of Torsional reinforcementsThe followings (in Cl. 6.3.7, Cl. 6.3.8 and Cl. 9.2.3 of the Code) should beobserved for the placing of shear reinforcements :(i) The torsional shear link should form the closed shape as in Figure 9.1of the Concrete Code Handbook reproduced as Figure 3.34. It shouldbe noted that the second configuration in the Figure is not included inFigure 9.3 of the Code though it should also be acceptable;Full lap lengthFull lap lengthFullanchoragelengthFigure 3.34 – Shape of Torsional shear links(ii)(iii)The value s vfor the closed link should not exceed the least of x 1,y1/ 2 or 200 mm as per Cl. 6.3.7 of the Code;In accordance with Cl. 9.2.3 of the Code, provision of the longitudinaltorsion reinforcement should comply the followings :(a) The bars distributed should be evenly round the inside perimeterof the links as illustrated in Figure 3.33;(b) Clear distance of the bars not to exceed 300 mm;(c) Additional longitudinal bars required at the level of the tension orcompression reinforcements may be provided by using largerbars than those required for bending alone, as illustrated inWorked Example 3.14 and Figure 3.33;(d) The longitudinal bars should extend a distance at least equal tothe largest dimension of the section beyond where it theoreticallyceases to be required.48

Version 2.3 May 20084.0 Slabs4.1 Types of SlabsSlabs can be classified as “one way slab”, “two way slab”, “flat slab”, “ribbedslab” with definition in Cl. 5.2.1.1 of the Code.4.1.1 One way slab is defined by the Code as one subjected predominantly to u.d.l.either(i) it possesses two free and parallel edge; or(ii) it is the central part of a rectangular slab supported on four edges witha ratio of the longer to the shorter span greater than 2.4.1.2 Two way slab is a rectangular one supported on four sides with length tobreadth ratio smaller than 2.4.1.3 Flat slab is a slab supported on columns without beams.4.1.4 Ribbed or Waffled Slab is a slab with topping or flange supported by closelyspaced ribs. The Code allows idealization of the ribbed slab or waffled slab asa single slab without treatment as discretized ribs and flanges in analysis in Cl.5.2.1.1(d) of the Code. If the stiffness of the ribbed or waffled slab is requiredfor input, the bending stiffness in the X and Y directions can be easily foundby summing the total bending stiffness of the composite ribs and flangestructure per unit width as illustrated in Figure 4.1. The twisting stiffness ismore difficult to assess. However, it should be acceptable to set the twistingstiffness to zero which will end up with pure bending in the X and Y directionsas the slab, with its ribs running in the X and Y directions are clearlypredominantly strong in bending in the two directions.Figure 4.1 illustrates the computation of “I” value of a waffle slab about theX-direction which is the total stiffnesses of the nos. of “flanged ribs” withinone metre. “I” value in the Y-directions can be worked out similarly.49

Version 2.3 May 2008200Y200X200750600750Dimensions ofa flanged rib750200100600Centroid of a flanged rib is at220.55×0.1 / 2 + 0.2×0.6 / 20.55×0.1+0.2×0.6=0.2214m from topI of a rib is13 3( 0.55×0.1 + 0.2×0.6 )12( 0.2214 0.05) 2( 0.3 0.2214) 2+ 0.55×0.1×−+ 0.2×0.6×−=0.006m 4Within one metre, there will be1000/750 = 1.333 nos. of ribs.So the I per metre width is0 .006× 1.333 = 0.008 m 4 /mFigure 4.1 – Illustration of calculation of I value about X-direction of a waffle slab4.2 Analysis of Slabs without the use of computer method4.2.1 One way slab can be analyzed as if it is a beam, either continuous or singlespan. As we aim at simple analysis for the slab, we tend to treat it as a singleelement without the necessity to consider the many loading cases forcontinuous spans, Cl. 6.1.3.2(c) of the Code allows the design against momentand shear arising from the single-load case of maximum design load on allspans provided that :(i) the area of each bay (defined in Figure 6.5 of the Code and reproducedin Figure 4.2) > 30 m 2 ;(ii) the ratio of the characteristic imposed load to characteristic dead load ≤1.25; and(iii) the characteristic imposed load ≤ 5 kN/m 2 excluding partitions.panelbayFigure 4.2 – Definition of panels and bays50

Version 2.3 May 20084.2.2 Two way rectangular slab is usually analyzed by treating it as if it is a singleslab in the absence of computer method. Bending moment coefficients forcalculation of bending moments are presented in Table 6.6 of the Code fordifferent support restraint conditions. Nevertheless, simplified formulae for thebending coefficients in case of rectangular simply supported two way slab areavailable in the Code (Ceqn 6.26 and 6.27) and reproduced as follows :mxsx2x= α nl andm2y= αsynlxwhere n is the u.d.l. xrespectively the shorter and longer spans and42( ly/ lx)( ly/ lx)αsx=; α81+l / l4 sy=.81+l / l4[ ( ) ]yx[ ( ) ]yxl and l yare4.2.3 Flat slabs, if of regular arrangement, can be analyzed as frames in thetransverse and longitudinal directions by such methods as moment distributionmethod as if they are separate frames. Analyzed moments and shears shouldbe apportioned to the “column strip” and “Middle strip” as per Table 6.10 ofthe Code. In addition, the bending moment and shear force coefficients for theone way slab can also be used as a simplified approach.4.2.4 More bending moment and shear force coefficients of rectangular slabs withvarious different support and loading conditions can be found from otherpublished handbooks, the most famous one being “Tables for the Analysis ofPlates, Slabs and Diaphragms based on the Elastic Theory”.4.3 Analysis of Slabs with the use of the computer methodAnalysis of slabs with the use of the computer method is mainly by the finiteelement method in which the slab is idealized as an assembly of discrete “platebending elements” joined at nodes. The support stiffnesses by the supportingwalls and columns are derived as similar to that for beams as “sub-frames”. Acomplete set of results including bending moments, twisting moment, shearforce per unit width (known as “stress” in finite element terminology) can beobtained after analysis for design purpose. The design against flexure is mostcommonly done by the Wood Armer Equations which calculate designmoments in two directions (conveniently in two perpendicular directions) andthey are adequate to cater for the complete set of bending and twistingmoments. The design based on node forces / moments should be avoided dueto its inadequacy to cater for twisting effects which will result in under-design.A discussion of the plate bending theory and the design approach by the Wood51

Version 2.3 May 2008Armer Equations is enclosed in Appendix D, together with the “stressapproach” for checking and designing against shear in the same appendix. Anexample of the mathematical modeling of a floor slab by the software SAFEand results of subsequent analysis is illustrated in Figure 4.3.Figure 4.3 – Modeling of an irregular floor slab as 2-D mathematical model,subsequent analytical results of bending moments and twisting moment, design ofreinforcements by the Wood Armer Equations.The finite element mesh of the mathematical model is often very fine. So it isa practice of “lumping” the design reinforcements of a number of nodes overcertain widths and evenly distributing the total reinforcements over the widths,52

Version 2.3 May 2008as is done by the popular software “SAFE”. However, care must be taken innot taking widths too wide for “lumping” as local effects may not be wellcaptured.4.4 Detailing for Solid SlabsGenerally considerations in relation to determination of “effective span”,“effective span depth ratio”, “moment redistribution”, “reduced designmoment to support”, “maximum and minimum steel percentages”, “concretecovers” as discussed in Section 3.3 for design of beam are applicable to designof slab. Nevertheless, the detailing considerations for slabs are listed asfollows with h as the structural depth of the slab (Re 9.3.1.1 of the Code) :(i)Minimum steel percentage (Cl. 9.3.1.1(a) of the Code):Main Reinforcing bars:0.24% for f = 250 MPa and 0.13% for f = 460 MPa;yDistribution bars in one way slab ≥ 20% of the main reinforcements(ii) Maximum reinforcements spacing (Cl. 9.3.1.1(b) of the Code):(a) In general areas without concentrated loads :the principal reinforcement, max. spacing ≤ 3h≤ 400 mm; andthe secondary reinforcement, max. spacing ≤ 3.5h ≤ 450 mm.(b) In areas with concentrated loads or areas of maximum moment:the principal reinforcement, max. spacing ≤ 2h ≤ 250 mm; andfor the secondary reinforcement, max. spacing ≤ 3h≤ 400 mm.(iii) In addition to (ii), if either :(a) h ≤ 250 mm (grade 250 steel);(b) h ≤ 200 mm (grade 460 steel); or(c) the percentage of required tension reinforcement is less than0.3%.no direct crack widths check by calculation is required. If none ofconditions in (a), (b) & (c) is satisfied, bar spacing to comply with Cl.9.2.1.4 of the Code as discussed in 3.3(vi) of this Manual if steelpercentage > 1%. Otherwise, increase the spacing by 1/percentage;(iv) Requirements pertaining to curtailment and anchoring of tensionreinforcements should be similar to that of beams;(v) Reinforcements at end supports (Cl. 9.3.1.3 of the Code)(a) At least 50% of the span reinforcements should be provided andy53

Version 2.3 May 2008(vi)well anchored on supports of simply supported slabs and endsupports of continuous slabs as illustrated in Figure 4.4;(b) If support shear stress v < 0. 5vc, the arrangement in Figure 4.4can be considered as effective anchorage.Minimum bottom reinforcements at internal supports : 40% of thecalculated mid-span bottom reinforcements as illustrated in Figure 4.4.(Cl. 9.3.1.4 of the Code)bat least 0 .5Asprovided at top ofend support, dia. Ø= the greater of 1/3b and30 mm if v < 0. 5v c ,otherwise 12ØAsat spanat least 0 .5Asanchored into endspan supportat least 0 .4Ascontinuous throughinternal supportFigure 4.4 – Anchorage of bottom reinforcements into supports(vii)Reinforcements at free edge should be as shown in Figure 4.5 (Cl.9.3.1.6 of the Code)h≥2hFigure 4.5 – Free edge reinforcements for Slabs(viii) Shear reinforcements not to be used in slabs < 200 mm. (Cl. 9.3.2 ofthe Code)4.5 Structural Design of SlabsThe structural design of slab against flexure is similar to that of beam. Thedetermination of reinforcements should be in accordance with Section 3.4 of54

Version 2.3 May 2008this Manual listing the options of either following the rigorous or simplified“stress strain” relationship of concrete. Design against shear for slabs underline supports (e.g. one-way or two-way) is also similar to that of beam.However for a flat slab, the checking should be based on punching shear inaccordance with the empirical method of the Code or based on shear stressesrevealed by the finite element method. They are demonstrated in the WorkedExamples in the following sub-Section 4.6 :4.6 Worked ExamplesWorked Example 4.1 – One Way SlabA one-way continuous slab with the following design data :(i) Live Load = 4.0 kN/m 2 ;(ii) Finishes Load = 1 kN/m 2 ;(iii) Concrete grade : 35 with cover 25 mm;(iv) Slab thickness : 200 mm;(v) Fire rating : 1 hour, mild exposure;(vi) Span : 4 m4m4m4m4mFigure 4.6 – Slab in Worked Example 4.1Sizing : Limiting Span depth ratio = 23 × 1.1 = 25. 3 (by Table 7.3 and Table7.4 of the Code, assuming modification by tensile reinforcement to be 1.1 asthe slab should be lightly reinforced). Assuming 10mm dia. bars under 25mmconcrete cover, effective depth is d = 200 − 25 − 5 = 170 . Span effectivedepth ratio is 4000 /170 = 23.5 < 25. 3. So OK.Loading : D.L. O.W. 0 .2 × 24 = 4.8kN/m 2Fin. 1.0 kN/m 2Total 5.8 kN/m 2L.L. 4.0 kN/m 2The factored load on a span is F = ( 1 .4 × 5.8 + 1.6 × 4.0) × 4 = 58. 08kN/m.55

Version 2.3 May 2008Based on coefficients of shear and bending moment in accordance with Table6.4 of the Code listed as follows :B.M. -0.04Fl 0.075 Fl -0.086Fl 0.063Fl -0.063Fl 0.063Fl -0.086Fl 0.086 Fl 0FlS.F. 0.46F 0.6F 0.6F 0.5F 0.5F 0.6F 0.6F 0.4FContinuous(e.g. overwall)simplysupported(a) End span support moment (continuous) = 0 .04 × 58.08×4 = 9. 29 kNm/m6M 9.29×10K = == 0.00922 2fcubd35×1000×170zK= 0 .5 + 0.25 − = 0.989 > 0.95d0.96AstM9.29 × 10=== 0.08%2bd bd × 0.87 f z / d 1000 × 170 × 0.87 × 460 × 0.95 0.95d0.96AstM19.98×10=== 0.18%2bd bd × 0.87 f z / d 1000×170 × 0.87×460×0.95>2y0.13%Ast= 0 .18 ÷ 100×1000×170 = 309 mm 2 Use T10 – 250( Astprovided = 314mm 2 )(c) First interior support moment = 0 .086 × 58.08×4 = 19. 98kNm/m, samereinforcement as that of end span reinforcement.(d) Interior span or support moment = 0 .063× 58.08×4 = 14. 64 kNm/m;6AstM14.64 × 10=== 0.133% > 0.13%2 2bd bd × 0.87 f z / d 1000 × 170 × 0.87 × 460 × 0.95(e)Figure 4.7 – Bending Moment and Shear Force coefficients for Continuous SlabyAst= 0 .133 ÷ 100×1000×170 = 227 mm 2 Use T10 – 300( Astprovided = 261mm 2 )End span span moment to continuous support56

Version 2.3 May 2008(f)= 0 .075× 58.08×4 = 17. 42kNm/m6AstM17.42 × 10=== 0.159% > 0.13%2 2bd bd × 0.87 f z / d 1000 × 170 × 0.87 × 460 × 0.95yAst= 0 .159 ÷ 100×1000×170 = 270 mm 2 . Use T10 – 250( Astprovided = 314mm 2 )Check ShearMaximum shear = 0 .6 × 58.08 = 34. 85 kN/m.By Table 6.3 of the Codevc1 1⎛100A3 400 4s ⎞ ⎛ ⎞ 1 ⎛ fcu= .79⎜⎟ ⎜ ⎟ ⎜⎝ bd ⎠ ⎝ d ⎠ γm ⎝ 250 ⎟⎠⎞1314131⎛ 400 ⎞ 1 ⎛ 35 ⎞v 0.79( 0.13) 3c= ⎜ ⎟ ⎜ ⎟ = 0. 44 N/mm 2 , based on minimum⎝ 170 ⎠ 1.25 ⎝ 25 ⎠34850steel 0.13%; v = = 0. 205 N/mm 2 < vc= 0. 44N/mm 2 .1000×170No shear reinforcement required.Worked Example 4.2 – Two Ways Slab (4 sides simply supported)A two-way continuous slab with the following design data :(i) Live Load = 4.0 kN/m 2 ;(ii) Finishes Load = 1 kN/m 2 ;(iii) Concrete grade : 35;(iv) Slab thickness : 200 mm(v) Fire rating : 1 hour, mild exposure, cover = 25mm;(vi) Span : Long way : 4 m, Short way, 3 mSizing : Limiting Span depth ratio = 20 (by Table 7.3). So effective depthtaken as d = 200 − 25 − 5 = 170 as 3000/170 = 17.65 < 20.Loading : D.L. O.W. 0 .2 × 24 = 4.8kN/m 2Fin. 1.0 kN/m 2Total 5.8 kN/m 2L.L. 4.0 kN/m 2The factored load is = ( 1 .4 × 5.8 + 1.6 × 4.0) = 14. 52F kN/m 2(Ceqn 6.26) and (Ceqn 6.27) of the Code are used to calculate the bending57

Version 2.3 May 2008moment coefficients along the short and long spans :4( ly/ lx)+ ( l / l )4[ ]αsx== 0.095 ;81yxαsy=81So the bending moment along the short span is2( ly/ lx)+ ( l / l )4[ ]yx= 0.053Mx= 0.095×14.52×32 =12.41kNm/m6M 12.41×10K = == 0.01232 2fcubd35×1000×170zK= 0 .5 + 0.25 − = 0.986 > 0.95d0.96AstM12.41×10=== 0.113%2bd bd × 0.87 f z / d 1000 × 170 × 0.87 × 460 × 0.95

Loading : D.L. O.W. 0 .2× 24 = 4.8kN/m 2Fin. 1.0 kN/m 2Sum 5.8 kN/m 2L.L. 4.0 kN/m 2The factored load is = ( 1 .4 × 5.8 + 1.6 × 4.0) = 14. 52F kN/m 2Version 2.3 May 2008From Table 1.38 of “Tables for the Analysis of Plates, Slabs and Diaphragmsbased on Elastic Theory” where γ = 4 / 5 = 0. 8, the sagging bending momentcoefficient for short way span is maximum at mid-span of the free edge which0.1104 (linear interpolation between γ = 0. 75 and γ = 1. 0 ). The coefficientsrelevant to this example are interpolated and listed in Appendix E.Mx= 0.1104×14.52×42 =25.65 kNm/m6M 25.65×10K = == 0.02542 2fcubd35×1000×170zK= 0 .5 + 0.25 − = 0.971 > 0.95d0.96AstM25.65×10=== 0.233%2bd bd × 0.87 f z / d 1000 × 170 × 0.87 × 460 × 0.95>2y0.13%Ast= 0 .233 ÷ 100×1000×170 = 397 mm 2 Use T10 – 175( Astprovided = 449mm 2 )At 2 m and 4 m from the free edge, the sagging moment reduces to0.0844× 14.52×42 = 19.608 kNm/m and 0.0415× 14.52×42 = 9. 64 kNm/mand Astrequired are reduces to 303 mm 2 and 149 mm 2 .Use T10 – 250 and T10 – 300 respectively.The maximum hogging moment (bending along long-way of the slab) is atmid-way along the supported edge of the short-way spanMy= 0.0729 × 14.52 × 52 =26.46 kNm/m6M 26.46×10K = == 0.02622 2fcubd35×1000×170zK= 0 .5 + 0.25 − = 0.97 > 0.95d0.96AstM26.46 × 10=== 0.241%2bd bd × 0.87 f z / d 1000 × 170 × 0.87 × 460 × 0.95>2y0.13%Ast= 0 .241÷100 × 1000×170 = 409 mm 2 Use T10 – 175( Astprovided = 449mm 2 )59

Version 2.3 May 2008The maximum sagging moment along the long-way direction is at 2 m fromthe free edge which isMy= 0.0188×14.52×52 =6.82 kNm/m. The moment is small.Use T10 – 300Back-check compliance of effective span ratio (Re Tables 7.3 and 7.4 of theCode) by considering only the short span which is simply supported,fs2 fyA=3Ast,reqst,prov1×βb2×460×397 1=× = 271N/mm 2 ;3×449 1The modification factor (Table 7.4) is( 477 − f s) ( 477 − 271)= 0.55 +⎛ M ⎞ 120( 0.9 + 0.0254×35)0.55 +120⎜0.9+⎝ bd2⎟⎠= 1.51Allowable effective span depth ratio is 1 .51× 20 = 30. 2 > 4000 = 23. 5 . O.K.170Finally the reinforcement arrangement on the slab is (Detailed curtailment, topsupport reinforcements at simple supports (0.5A s ) omitted for clarity.)T10 – 300 T11200T10 – 175T1T10 – 300B2800800T10 – 300 B1 T10 – 250 B1 T10 – 175B11200100020002000Figure 4.9 – Reinforcement Details for Worked Example 4.3Worked Example 4.4 – Flat Slab by Simplified Method (Cl. 6.1.5.2(g))60

Version 2.3 May 2008Flat slab arrangement on rectangular column grid of 7.5 m and 6 m as shownin Figure 4.10 with the following design data :(i) Finish Load = 1.5 kPa(ii) Live Load = 5.0 kPa.(iii) Column size = 550 × 550(iv) Column Drop size = 3000 × 3000 with d h = 200 mm(v) Fire rating : 1 hour, mild exposure, cover = 25 mm(vi) Concrete grade 35;As the number of panel is more than 3 and of equal span, the simplifiedmethod for determining moments in accordance with Cl. 6.1.5.2(g) of theCode is applicable and is adopted in the following analysis.Effective dimension of column headh max= lc+ 2( dh− 40)= 550 + 2( 200 − 40) = 870 mmEffective diameter of column head (Cl. 6.1.5.1(c))870 2 × 41hc= = 982 mm < × 6000 = 1500mmπ4l (Ceqn 6.37)l ho =30007500L hmax =870250d h =200l c =550Column head details75001500150015001500600060006000Figure 4.10 – Flat Slab Plan Layout for Worked Example 4.4In the simplified method, the flat slab is effectively divided into (i) “columnstrips” containing the columns and the strips of the linking slabs and of “stripwidths” equal to the widths of the column drops; and (ii) the “middle strips”61

Version 2.3 May 2008between the “column strips”. (Re Figure 6.9 of the Code.) They are designedas beams in flexural design with assumed apportionment of moments amongthe strips. However, for shear checking, punching shears along successive“critical” perimeters of column are carried out instead.Sizing : Based on the same limiting span depth ratio for one way and two wayslab which is 26 × 1.15 = 30 (by Table 7.3 and Table 7.4 of the Code,6000assuming modification by tensile reinforcement to be 1.15), d = = 200 .30As cover = 25 mm, assuming T12 bars, structural depth should at least be200 + 25 + 12 ÷ 2 = 231mm, so use structural depth of slab of 250 mm.Loading : D.L. O.W. 0 .25× 24 = 6.0kN/m 2Fin. 1.5 kN/m 2Total 7.5 kN/m 2L.L. 5.0 kN/m 2The factored load is = ( 1 .4 × 7.5 + 1.6 × 5.0) = 18. 5F kN/m 2Design against Flexure (Long Way)The bending moment and shear force coefficients in Table 6.4 will be used asper Cl. 6.1.5.2(g) of the Code. Total design ultimate load on the full width ofpanel between adjacent bay centre lines is F = 18 .5×7.5×6 = 832. 5 kN. Thusthe reduction to support moment for design, as allowed by Cl. 6.1.5.2(g) of theCode, is 0 .15Fh= 0.15×832.5×0.982 = 122. 63 kNm for internal support andc0 .15Fh= 0.15×832.5×0.982 / 2 = 61.32 kNm for outer support.cThe design moment at supports are :Total moment at outer support is 0 .04 × 832.5×7.5 = 249. 75 kNm, which canbe reduced to 249 .75 − 61.32 = 188. 43kNm;Total moment at first interior support is 0 .086 × 832.5×7.5 = 536. 96 kNm,which can be reduced to 536 .96 −122.63= 414. 33 kNmTotal moment at interior support is 0 .063× 832.5×7.5 = 393. 36 kNm, whichcan be reduced to 393 .36 −122.63= 270. 73 kNm62

Version 2.3 May 2008The flat slab is divided into column and mid strips in accordance with Figure6.9 of the Code which is reproduced as Figure 4.11 in this Manual.columnstripmiddle stripl y – l x /2columnstripl x /4l x /4columnstripmiddlestripl xl x /4l x /4columnstripl x /4 l x /4l yl x /4 l x /4l x is the shorterspan whilst l y isthe longer spanFlat slab without dropColumn strip = drop sizeIgnore drop if dimension < l x /3Dropmiddle strip– drop sizeColumn strip= drop sizeIgnore drop ifdimension

Version 2.3 May 200825% respectively as per Table 6.10 of the Code,Column Strip (75%) Mid Strip (25%)Total Mt Mt/width Total Mt Mt/widthOuter Support 141.32 47.11 47.11 15.701 st interior support 310.75 103.58 103.58 34.53Middle interior support 203.05 67.68 67.68 22.56The reinforcements – top steel are worked out as follows, (minimum of 0.13%in brackets) ( d = 450 − 25 − 6 = 419 over column support andd = 250 − 25 − 6 = 219 in other locations)Column Strip (75%) Mid Strip (25%)Area (mm 2 )/m Steel Area (mm 2 )/m SteelOuter Support 281 (548) T12 – 200 178 (548) T12 – 2001 st interior support 618 T12 – 150 392 (548) T12 – 200Middle interior support 404 (548) T12 – 200 256 (548) T12 – 200Sagging Moment :Total moment near middle of end span is 0 .075× 832.5×7.5 = 468. 28 kNmTotal moment near middle of interior span 0 .063× 832.5×7.5 = 393. 36 kNmThese moments are to be apportioned in the column and mid strips inaccordance with the percentages of 55% and 45% respectively as per Table6.10, i.e.Column Strip (55%) Mid Strip (45%)Total Mt Mt/width Total Mt Mt/widthMiddle of end span 257.55 85.85 210.73 70.24Middle of interior span 216.35 72.12 177.01 59.00The reinforcements – bottom steel are worked out as follows :Column Strip (55%) Mid Strip (45%)Area (mm 2 )/m Steel Area (mm 2 )/m SteelMiddle of end span 980 T12 – 100 801 T12 – 125Middle of interior span 823 T12 – 125 673 T12 – 150Design in the short way direction can be carried out similarly.Design against ShearDesign of shear should be in accordance with Cl. 6.1.5.6 of the Code which isagainst punching shear by column. For the internal column support, in the64

Version 2.3 May 2008absence of frame analysis, the shear for design will beV = 1. 15Vwhere V tefftis the design shear transferred to column calculated on the assumption of alladjacent panels being fully loaded by Cl. 6.1.5.6(b) of the Code.V = 7 .5×6×18.5 = 832.5 kN; V 1 .15V= 957. 38 kNteff=tCheck on column perimeter as per Cl. 6.1.5.6(d) of the Code :V3eff957.38×10≤ 0.8 fcuor 7= 1.04 ≤ 0.8 fcu= 4.73MPa; O.K.ud( 4 × 550)× 419Check on 1 st critical perimeter – 1.5dfrom column face, i.e.1 .5× 0.419 = 0.6285 . So side length of the perimeter is( 550 + 628.5×2) = 1807 mmLength of perimeter is 4 × 1807 = 7228mmShear force to be checked can be the maximum shear 957.38 kN afterdeduction of the loads within the critical perimeter which is957.38 −18.5×1.8072 = 896.97 kN3896.97×10Shear stress = = 0. 296 N/mm 2 . < vc= 0. 48N/mm 2 in accordance7228×4191/ 3with Table 6.3 ( 0.43 ( 35/ 25) 0. 48× = ). No shear reinforcement is required.No checking on further perimeter is required.Worked Example 4.5 – Design for shear reinforcement(Ceqn 6.44) and (Ceqn 6.45) of the Code gives formulae for reinforcementdesign for different ranges of values of v .( v − vc) udFor v ≤ 1. 6vc, ∑ Asvsinα ≥0.87 fFor1.6v < v ≤ 2. 0v,cc∑Asv5sinα ≥65y( 0.7v− v )0.87 fAs a demonstration, if v = 0. 7 N/mm 2 in the first critical perimeter which is vc= 0. 48N/mm 2 in the Example 4.4. By Table 6.8of the Code, as v v < 0. 4 , v = 0. 4− crvrud0.4×7228×419∑ Asvsin α ≥ == 3027 mm 2 . If vertical links is chosen0.87 fy0.87×460as shear reinforcement, α = 90 0 ⇒ sinα= 1 . So the 3027 mm 2 should bedistributed within the critical perimeter as shown in Figure 4.12.In distributing the shear links within the critical perimeter, there arerecommendations in Cl. 6.1.5.7(f) of the Code thatcyud

Version 2.3 May 2008(i)(ii)at least two rows of links should be used;the first perimeter should be located at approximately 0.5d from theface of the loaded area (i.e. the column in this case) and should containnot less than 40% of the calculated area of reinforcements.So the first row be determined at 200 mm from the column face with total rowlength 950 × 4 = 3800 . Using T10 – 225 spacing along the row, the total steel10 2 π × = mm 2 > 40% of 3027 mm 2 .area will be ( / 4) 3800/ 225 1326The second row be at further 300 mm (≤ 0.75d = 314) away where row lengthis 1550 × 4 = 6200 . Again using T10 – 225 spacing along the row, the total10 2 π × = mm 2 > 60% of 3027 mm 2 .steel area will be ( / 4) 6200/ 225 2164Total steel area is 1326 + 2164 = 3490 mm 2 for shear. The arrangement isillustrated in Figure 4.12.628.51 st row , usingT10 – 225 (or 17 nos. of T10)(area = 1335mm 2 )>40% of 3027 mm 2550628.5200 ≈ 0.5d300 ≤ 0.75d275 2003001 st criticalperimeter2 nd row , usingT10 – 225 (or 27 nos. of T10)(area = 2120mm 2 )Note : It should be notedthat the link spacings asarrived in this Exampleare for demonstrationpurpose. In actualpractice, they shouldmatch with thelongitudinal bar spacing.T10 – 225 link1 .5d=628.5Figure 4.12 – Shear links arrangement in Flat Slab for Worked Example 4.566

Version 2.3 May 2008Design for Shear when ultimate shear stress exceeds 1.6v cIt is stated in (Ceqn 6.43) in Cl. 6.1.5.7(e) that if∑Asv( 0.7v− v )1.6< v ≤ 2. 0v,5cudsinα ≥which effectively reduces the full inclusion of0.87 fyvcfor reduction to find the “residual shear to be taken up by steel” atv = 1. 6v cto zero inclusion at v = 2. 0vc.ccWorked Example 4.6 – when 1.6v c < v ≤ 2.0v cIn the previous Example 4.5, if the shear stress v = 0. 85 N/mm 2 which liesbetween 1 .6v = 1.6×0.48 = 0. 77 N/mm 2 and 2 .0v = 2×0.48 = 0. 96 N/mm 2c( 0.7v− v ) ud 5( 0.7×0.85 − 0.48)5c× 7228×419∑ Asvsin α ≥== 4351 mm 2 . If0.87 fy0.87×460arranged in two rows as in Figure 4.12, use T12 – 225 for both rows : the inner12 2 π × =row gives ( / 4) 3800/ 225 1908row gives ( / 4) 6200/ 225 3114cmm 2 > 40% of 4351 mm 2 ; the outer12 2 π × = mm 2 . The total area is 1908 + 3114 =5022mm 2 > 4351 mm 2 .Cl. 6.1.5.7(e) of the Code says, “When v > 2vcand a reinforcing system isprovided to increase the shear resistance, justification should be provided todemonstrate the validity of design.” If no sound justification, the structuralsizes need be revised.67

Version 2.3 May 20085.0 Columns5.1 Slenderness of ColumnsColumns are classified as short and slender columns in accordance with their“slenderness”. Short columns are those with ratios l ex/ h and l ey/ b < 15(braced) and 10 (unbraced) in accordance with Cl. 6.2.1.1(b) of the Codewhere lexand l eyare the “effective lengths” of the column about the majorand minor axes, b and h are the width and depth of the column.As defined in Cl. 6.2.1.1 of the Code, a column may be considered braced in agiven plane if lateral stability to the structure as a whole is provided by wallsor bracing or buttressing designed to resist all lateral forces in that plane. Itwould otherwise be considered as unbraced.The effective length is given by (Ceqn 6.46) of the Code asl e= β ⋅l 0where l 0is the clear height of the column between restraints andthe value β is given by Tables 6.11 and 6.12 of the Code which measures therestraints against rotation and lateral movements at the ends of the column.Generally slenderness limits for column : l / b 60 as per Cl. 6.2.1.1(f) of0≤100b2the Code. In addition, for cantilever column l0 = ≤ 60b.hWorked Example 5.1 : a braced column of clear height l0 = 8 m and sectionaldimensions b = 400 mm, h = 550 mm with its lower end connectedmonolithically to a thick cap and the upper end connected monolithically todeep transfer beams in the plane perpendicular to the major direction but beamof size 300(W) by 400(D) in the other direction.By Tables 6.11 and 6.12 of the CodeLower end condition in both directions : 1Upper end condition about the major axis : 1Upper end condition about the minor axis : 2For bending about the major axis : end conditions 1 – 1, βx= 0. 75 ,lex= 0 .75×8 = 6lex/ 550 = 10.91 < 15. ∴ a short column.For bending about the minor axis : end conditions 1 – 2, β = 0. 8 ,y68

ley= 0 .8×8 = 6.4l / 400 = 16 > 15 ∴ a slender column. l / 400 = 16 < 60 , O.K.eyeyVersion 2.3 May 2008For a slender column, an additional “deflection induced moment” Maddwillbe required to be incorporated in design, as in addition to the workingmoment.5.2 Design Moments and Axial Loads on Columns5.2.1 Determination of Design moments and Axial Loads by sub-frame AnalysisGenerally design moments, axial loads and shear forces on columns are thatobtained from structural analysis. In the absence of rigorous analysis, (i)design axial load may be obtained by the simple tributary area method withbeams considered to be simply supported on the column; and (ii) moment maybe obtained by simplified sub-frame analysis as illustrated in Figure 5.1 :K uK u1.4G k +1.6QK b1.0G k 1.4G k +1.6QK b2K b1K LM eKMuu =KL+ Ku+ 0.5KbM eKMLL =KL+ Ku+ 0.5KbK LM esKMuu =KL+ Ku+ 0.5Kb1 + 0. 5Kb2M esKMLL =KL+ Ku+ 0.5Kb1 + 0. 5Kb2Symbols:M e : Beam Fixed End Moment. K u : Upper Column StiffnessM es : Total out of balance Beam Fixed End Moment. K L : Upper Column StiffnessM u : Upper Column Design Moment K b1: Beam 1 StiffnessM L : Upper Column Design Moment K b2: Beam 2 StiffnessFigure 5.1 – Diagrammatic illustration of determination of column designmoments by Simplified Sub-frame AnalysisWorked Example 5.2 (Re Column C1 in Plan shown in Figure 5.2)Design Data :Slab thickness : 150 mm Finish Load : 1.5 kN/m 269

Version 2.3 May 2008Live Load : 5 kN/m 2Beam size : 550(D) × 400(W)Upper Column height : 3 mLower Column Height : 4 mColumn size : 400(W) × 600(L)Column Load from floors above D.L. 443 kN L.L. 129 kN5mB3B4C1B1B24m3m3mFigure 5.2 – Plan for illustration for determination of design axial load andmoment on column by the Simplified Sub-frame MethodDesign for Column C1 beneath the floorCheck for slenderness : As per Cl. 6.2.1.1(e) of the Code, the end conditions ofthe column about the major and minor axes are respectively 2 and 1 at theupper end and 1 at the lower end for both axes (fixed on pile cap). The clearheight between restraints is 4000 − 550 = 3450 . The effective heights of thecolumn about the major and minor axes are respectively 0 .8× 3.45 = 2. 76 mand 0 .75× 3.45 = 2. 59 m. So the slenderness ratios about the major and minor27602590axes are = 4.6 < 15 and = 6.475 < 15 . Thus the column is not600400slender in both directions.Loads :Slab: D.L. O.W. 0 .15× 24 = 3. 6 kN/m 2Fin. 1.5 kN/m 25.1 kN/m 2L.L. 5.0 kN/m 2Beam B1 D.L. O.W. 0 .4 × 0.55×24 × 4 = 21. 12 kN70

Version 2.3 May 2008End shear of B1 on C1 is D.L. 21 .12 ÷ 2 = 10. 56 kNBeam B3 D.L. O.W. 0 .4 ( 0.55 − 0.15) × 24 = 3. 84× kN/mSlab 5 .1× 3.5 = 17. 85 kN/m21.69 kN/mL.L. Slab 5 .0 × 3.5 = 17. 5 kN/mEnd shear of B3 on C1 D.L. 21 .69 × 5 ÷ 2 = 54. 23kNL.L. 17 .50 × 5 ÷ 2 = 43. 75 kNBeam B4 D.L. O.W. 0 .4 ( 0.55 − 0.15) × 24 = 3. 84× kN/mSlab 5 .1× 3 = 15. 3 kN/m19.14 kN/mL.L. Slab 5 .0 × 3 = 15. 0 kN/mEnd shear of B4 on B2 D.L. 19 .14 × 5 ÷ 2 = 47. 85 kNL.L. 15 .00 × 5 ÷ 2 = 37. 50 kNBeam B2 D.L. O.W. 0 .4 × 0.55×24 × 6 = 31. 68 kNB447.85 kN79.53 kNL.L. B437.50 kNEnd shear of B2 on C1 , D.L. 79 .53 ÷ 2 = 39. 77 kNL.L. 37 .5 ÷ 2 = 18. 75kNTotal D.L. on C1 O.W. 0 .4 × 0.6 × 24 × 4 = 23. 04kNB1 + B2 + B3 10 .56 + 39.77 + 54.23 = 104. 56 kNFloor above443.00 kNSum570.60 kNTotal L.L. on C1 B1 + B2 +B3 0 + 18.75 + 43.75 = 62. 5 kNFloor above129.00 kNSum191.50 kNSo the factored axial load on the lower column is1 .4 × 570.6 + 1.6 × 191.5 = 1105.24 kNFactored fixed end moment bending about the major axis (by Beam B3 alone):Me=1 ×212=( 1.4×21.69 + 1.6×17.5) × 5 121. 6kNm71

Version 2.3 May 2008Factored fixed end moment bending about the minor axis by Beam B2:⎛ 1 1 ⎞ ⎛ 1 ⎞Meb2 = 1.4×⎜ × 31.68 + × 47.85⎟ × 6 + 1.6×⎜ × 37.5⎟ × 6 = 95.24 kNm⎝128 ⎠ ⎝ 8 ⎠Factored fixed end moment bending about the minor axis by Beam B1:⎛ 1 ⎞Meb2 = 1.0×⎜ × 21.12⎟ × 4 = 7.04 kNm⎝12⎠So the unbalanced fixed moment bending about the minor axis is95 .24 − 7.04 = 88.2 kNmThe moment of inertia of the column section about the major and minor axes30.4×0.6are Icx= = 0. 0072 m 4 30.6×0.4, Icy= = 0. 0032 m 41212The stiffnesses of the upper and lower columns about the major axis are :4EIcx 4E× 0.0072Kux= == 0. 0096EL 3u4EIcx 4E× 0.0072KLx= == 0. 0072EL 4LThe stiffnesses of the upper and lower columns about the minor axis are :4EIcy 4E× 0.0032Kuy= == 0. 004267ELu34EIcy 4E× 0.0032KLy= == 0. 0032EL 4LThe moment of inertia of the beams B1, B2 and B3 are0.4×0.55123= 0.005546m 4The stiffness of the beams B1, B2 and B3 are respectively4E× 0.055464E× 0.05546= 0.005546E; = 0.003697E; and464E× 0.05546= 0.004437E5Distributed moment on the lower column about the major axis isMexKLx121.6 × 0.0072EMLx==K + K + 0.5K30.0096E+ 0.0072E+ 0.5×0.004437EuxLxb= 46.03 kNmDistributed moment on the lower column about the minor axis isMLy=Kuy+ KLyM+Key Ly0.5b1b2( K + K )72

Version 2.3 May 200888.2×0.0032E=0.004267E+ 0.0032E+ 0.5×( 0.005546 + 0.003697)E= 23.35 kNmSo the lower column should be checked for the factored axial load of1105.24kN, factored moment of 46.03 kNm about the major axis and factoredmoment of 23.35 kNm about the minor axis.5.2.2 Minimum EccentricityA column section should be designed for the minimum eccentricity equal tothe lesser of 20 mm and 0.05 times the overall dimension of the column in theplane of bending under consideration. Consider Worked Example in 5.2, theminimum eccentricity about the major axis is 20 mm as0 .05× 600 = 30 > 20 mm and that of the minor axis is 0 .05× 400 = 20 mm. Sothe minimum eccentric moments to be designed for about the major and minoraxes are both 1105 .24 × 0.02 = 22. 1kNm. As they are both less than thedesign moment of 46.03 kNm and 23.35 kNm, they can be ignored.5.2.3 Check for SlendernessIn addition to the factored load and moment as discussed in 5.2.1, it is requiredby Cl. 6.2.1.3 of the Code to design for an additional moment Maddif thecolumn is found to be slender by Cl. 6.2.1.1. The arrival of Maddis aneccentric moment created by the ultimate axial load N multiplied by apre-determined lateral deflection a uin the column as indicated by thefollowing equations of the Code.Madd= Na u(Ceqn 6.52)a = βaKh(Ceqn 6.48)ul e21 ⎛ ⎞βa= ⎜ ⎟2000 ⎝ b ⎠(Ceqn 6.51)Nuz− NK =Nuz− Nbal≤ 1 (conservatively taken as 1) (Ceqn 6.49)or by Nuz= 0 .45 fcuAnc+ 0. 87 fyAsc(Ceqn 6.50)Nbal= 0. 25 fcubdFinal design moment Mtwill therefore be the greatest of(1) M2, the greater initial end moment due to design ultimate load;73

Version 2.3 May 2008(2) Mi+ Maddwhere M i= 0.4M1+ 0.6M2≥ 0. 4M2(with M2aspositive and M1negative.)(3) M1+ Madd/ 2in which M1is the smaller initial end moment due to design ultimateload.(4) N × emin(discussed in Section 5.2.2 of this Manual)where the relationship between M 1, M2, Maddand the arrival of thecritical combination of design moments due to Maddare illustrated in Figure5.3 reproduced from Figure 6.16 of the Code.Endconditions ofcolumnInitial moment(from analysis)AdditionalmomentDesign momentenvelope+ M add =M 2+ =M iM addM i + M addLargermoment M 2M iM addM i + M addSmallermoment M 10.5M addFigure 5.3 – Braced slender columnsM1 + 0. 5M addIn addition to the above, the followings should be observed in thedetermination of Mtas the enveloping moment of the 4 cases described inthe previous paragraph (Re Cl. 6.2.1.3 of the Code) :(i) In case of biaxial bending (moment significant in two directions), Mt74

Version 2.3 May 2008should be applied separately for the major and minor directions withb in Table 6.13 of the Code be taken as h , the dimension of thecolumn in the plane considered for bending. Re Ceqn 6.48;(ii) In case of uniaxial bending about the major axis where l e/ h ≤ 20 andh < 3b , Mtshould be applied only in the major axis;(iii) In case of uniaxial bending about the major axis only where eitherl e/ h ≤ 20 or h < 3bis not satisfied, the column should be designedas biaxially bent, with zero Miin the minor axis;(iv) In case of uniaxial bending about the minor axis, Maddobviously beapplied only in the minor axis only.Worked Example 5.3 :A slender braced column of grade 35, cross sectionsb = 400 , h = 500l l = 8 m, N = 1500 kNex= ey(i)Moment due to ultimate load about the major axis only, the greater andsmaller bending moments due to ultimate load are respectivelyM2 x= 153kNm and M1 x= 96kNmAs l ex/ h = 16 ≤ 20 ; h = 500 < 3b= 1200So needs to check for additional bending in the major axis but withMaddbased on the minor axis.Take K = 1βa=12000l e⎛⎜⎝ b2⎞⎟⎠=120002⎛ 8000 ⎞⎜ ⎟⎝ 400 ⎠= 0.2= Kh = 0 .2×1×0.5 = 0.1au βaM addx= Nau= 1500 × 0.1 = 150M ix= 0.4M1+ 0.6M2= 0.4 − 96 + 0.6×153 = 53.4 < 0.4M2= 0.4×153 =( ) 61. 2The design moment about the major axis will be the greatest of :(1) M 1532 x=ix+ Maddx= 61 .2 + 150 =1 x+ Maddx/ 2 = 96 + 150 / 2 =×emin = 1500×0.02 =(2) M211. 2(3) M171(4) N 30 as e = 20 < 0.05×500 2575min=Mix+ Maddx= 211.So the greatest design moment is case (2) 2Thus the section needs only be checked for uniaxial bending withN = 1500 kN and M = 211. 2 kNm bending about the major axis.x

Version 2.3 May 2008(ii)Moments due to ultimate loads about the minor axis only, the greater andsmaller moments are identical in magnitudes to that in (i), but about theminor axis, repeating the procedure :M2 y= 143kNm and My791=kNmAs l ex/ h = 16 ≤ 20 ; h = 500 < 3b= 1200So needs to checked for additional bending in the major axis.Take K = 1l e21 ⎛ ⎞ 1 ⎛ 8000 ⎞= ⎜ ⎟ = ⎜ ⎟2000 ⎝ b ⎠ 2000 ⎝ 400 ⎠au = βaKh = 0 .2×1×0.4 = 0.08M addyNa = 1500 × 0.08 = 120βaM2= 0.2=uiy= 0.4M1 y+ 0.6M2 y= 0.4×2 y=( − 79) + 0.6×143 = 54.2 < 0.4M= 0.4×143 57. 2The design moment will be the greatest of :(1) M2 y= 143(2) Miy+ Maddy= 57 .2 + 120 = 177. 2(3) M1 y+ Maddy/ 2 = 79 + 120/ 2 = 139(4) N ×emin = 1500×0.02 = 30 as emin= 20 ≤ 0.05×400 = 20So the greatest design moment is case (2) Miy+ Maddy= 177. 2Thus the section need only be checked for uniaxial bending withN = 1500 kN and M = 177. 2 kNm bending about the minor axis.y(iii) Biaxial Bending, there are also moments of Mx153 kNm andM1 x= 96kNm; M2 y= 143kNm and M1 y= 79 kNm. By Cl. 6.1.2.3(f),Maddabout the major axis will be revised as follows :2=Bending about the major axis :βa=12000l e⎛⎜⎝ h2⎞⎟⎠=120002⎛ 8000 ⎞⎜ ⎟⎝ 500 ⎠= 0.128a= Kh = 0 .128×1×0.5 = 0.064uβaaddx= NauM= 1500 × 0.064 = 96 kNm.Thus items (2) and (3) in (i) are revised as(2) M M = 61 .2 + 96 = 157. 2ix+addx1 x+ Maddx/ 2 = 96 + 96 / 2 =(3) M144So the moment about major axis for design is 157.2 kNm76

Version 2.3 May 2008Bending about the minor axis :M2 y= 143kNm and My791=kNm; same as (ii);Thus the ultimate design moment about the major axis is 157.2 kNm andthat about the minor axis is 177.2 kNm.Worked Example 5.4 :A slender braced column of grade 35, cross sectionb = 400 , h = 1200l l = 8 m, N = 1500 kN, M 153kNm and M 96kNmex= ey2 x=1 x=As 3 b = h , Cl. 6.2.1.3(e) should be used.Take K = 1βa=12000l e⎛⎜⎝ b2⎞⎟⎠=120002⎛ 8000 ⎞⎜ ⎟⎝ 400 ⎠= 0.2au = βaKb = 0 .2×1×0.4 = 0.08 m > 20 mmMaddy= Nau= 1500 × 0.08 = 120So the minor axis moment is 120 kNml e8000As = = 6. 67 , the column is not slender about the major axis.h 1200So the major axis moment is simply 153 kNm.5.3 Sectional DesignGenerally the sectional design of column utilizes both the strengths ofconcrete and steel in the column section in accordance with stress strainrelationship of concrete and steel in Figures 3.8 and 3.9 of the Coderespectively. Alternatively, the simplified stress block for concrete in Figure6.1 of the Code can also be used.5.3.1 Design for Axial Load only(Ceqn 6.55) of the Code can be used which is77

Version 2.3 May 2008N = 0 .4 f A + 0. 75Af . The equation is particularly useful for a columncucscywhich cannot be subject to significant moments in such case as the columnsupporting a rigid structure or very deep beams. There is a reduction ofapproximately 10% in the axial load carrying capacity as compared with thenormal value of0 .45 f A + 0. 87Af accounting for the eccentricity ofcucscy0 .05h .Furthermore, (Ceqn 6.56) readingN = 0 .35 f A + 0. 67Af which iscucscyapplicable to columns supporting an approximately symmetrical arrangementof beams where (i) beams are designed for u.d.l.; and (ii) the beam spans donot differ by more than 15% of the longer. The further reduction is to accountfor extra moment arising from asymmetrical loading.5.3.2 Design for Axial Load and Biaxial Bending :The general section design of a column is accounted for the axial loads andbiaxial bending moments acting on the section. Nevertheless, the Code hasreduced biaxial bending into uniaxial bending in design. The procedure fordetermination of the design moment, either'xM or M ' bending about theymajor or minor axes is as follows :Determine b ' and h ' as defined by thediagram. In case there are more than onerow of bars, b ' and h ' can be measured tothe centre of the group of bars.M xh'bh(i)CompareM x andh'M y.b'MIf xMyh'≥ , use Mx' = Mx+ β Myh' b'b'b 'M MIf x yb'< , use My' = My+ β Mxh' b'h'where β is to be determined from Table 5.1 which is reproducedM yfrom Table 6.14 of the Code under the pre-determinedN ;bhf cu78

( ) bhf cuVersion 2.3 May 2008N / 0 0.1 0.2 0.3 0.4 0.5 ≥ 0. 6β 1.00 0.88 0.77 0.65 0.53 0.42 0.30Table 5.1 – Values of the coefficients β(ii) The Mx' or My' will be used for design by treating the section aseither (a) resisting axial load N and moment Mx' bending aboutmajor axis; or (b) resisting axial load N and moment My' bendingabout minor axis as appropriate.h5.3.3 Concrete Stress Strain Curve and Design ChartsThe stress strain curve for column section design is in accordance with Figure3.8 of the Code. It should be noted that Amendment No. 1 has revised the1.34( fcu/ γm)Figure by shifting ε 0to. With this revision, the detailedEcdesign formulae and design charts have been formulated and enclosed inAppendix F. Apart from the derivation for the normal 4-bar column, thederivation in Appendix F has also included steel reinforcements uniformlydistributed along the side of the column idealized as continuum ofreinforcements with symbol A sh. The new inclusion has allowed moreaccurate determination of load carrying capacity of column with many barsalong the side as illustrated in Figure 5.4 which is particularly useful forcolumns of large cross sections. The user can still choose to lump the sidereinforcements into the 4 corner bars, with correction to the effective depth asin conventional design by setting A = 0 in the derived formulae.shidealized ascontinuum of steel“strip” with areaequivalent to the rowof barsFigure 5.4 – Idealization of steel reinforcements in large columnFigure 5.5 shows the difference between the 2 idealization. It can be seen thatthe continuum idealization is more economical generally except at the peakmoment portion where the 4 bar column idealization shows slight over-design.79

Version 2.3 May 2008N/bh N/mm 250454035302520Comparison of Load Carrying Capacities of Rectangular Shear Walls with Uniform VerticalReinforcements Idealized as 4 bar column with d/h = 0.75 and Continuum to the Structural Use ofConcrete 2004 Concrete Grade 350.4% steel as 4 bar column0.4% steel as continuum1% steel as 4 bar column1% steel as continuum4% steel as 4 bar column4% steel as continuum8% steel as 4 bar column8% steel as continuum1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5M/bh 2 N/mm 2Figure 5.5 – Comparison between Continuum and 4-bar column IdealizationWorked Example 5.5 :Consider a column of sectional size b = 400 mm, h = 600grade 35 and under an axial load and moments ofN = 4000 kN, M = 250 kNm, M = 150 kNm,xymm, concretecover to longitudinal reinforcements = 40 mmAssume a 4-bar column and T40 bars, h' = 600 − 40 − 20 = 540mm;b' = 400 − 40 − 20 = 340 mm;N 4000000== 0.476 ;f bh 35×400×600cuβ = 0.446 from Table 5.1 or Table 6.14 of the Code;Mh'My= 0.463 >b'= 0.441x;h'540∴M x' = Mx+ β My= 250 + 0.446×× 150 = 356 kNmb'3406NM 356×10d 540= 16.67 ; == 2. 47 ;2 2= = 0. 9bhbh 400×600h 600Use Chart F-9 in Appendix F as extracted in Figure 5.6, 1.8% steel isapproximated which amounts to 400 × 600 × 0.018 = 4320 mm 2 , or 6-T32(Steel provided is 4826mm 2 ) The section design is also shown in Figure 5.6,80

Version 2.3 May 2008with two additional T20 bars to avoid wide bar spacing.N/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 35, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15M/bh 2 N/mm 2T32Extra T20Figure 5.6 – Design Chart and Worked Re-bar details for Worked Example 5.5Worked Example 5.6 :Consider a column of sectional size b = 800 mm, h = 1000grade 40 and under an axial load and momentsN =14400 kN, M = 2000kNm, M = 1500 kNm,xconcrete cover to longitudinal reinforcement = 40 mm;Approximation as a 4-bar column and assume d / h = 0. 8b' = 0.8×800 = 640 mm; h' = 0.8×1000 = 800 mm;N 14400000== 0.45fcubh40×800×1000the Code;ymm, concrete; β = 0. 475 from Table 5.1 or Table 6.14 of81

Mh'2000 M= = 2.5 >800 b'1500= = 2.34640x y;h'800∴M x' = Mx+ β My= 2000 + 0.475×× 1500 = 2890.6 kNmb'640Version 2.3 May 20086NM 2890.6×10=18 ; == 3. 61;2 2bhbh 800×1000Use Chart F-12 in Appendix F as extracted in Figure 5.7, 3.0% steel isapproximated which amounts to 0 .031× 800×1000 = 24, 800 mm 2 , or 32-T32(Steel provided is 25,736mm 2 ) The arrangement of steel bars is also shown inFigure 5.7. It should be noted that alternate lapping may be required if thecolumn is contributing in lateral load resisting system as the steel percentageexceeds 2.6% as per discussion in 5.4(ii) of this Manual.N/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 40, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12M/bh 2 N/mm 2Centre of mass of the reinforcements in one halfof the section below the centre line is5×434 + 5×370 + 4×222 + 2×74= 31616741481486456So h ' = 316 + 500 = 816d= 0 .816 > 0.8hdSo the original use of = 0. 8his OKFigure 5.7 – Chart and Column Section for Worked Example 5.682

Version 2.3 May 2008The back-calculation in Figure 5.7 has shown that the hdratio is the steel bararrangement is 0.816 which is greater than the original assumed value of 0.8.So the use of the chart is conservative.5.3.4 Alternatively, the design of reinforcements can be based on formulae derivedin Appendix F. However, as the algebraic manipulations are very complicated(may involve solution of 4 th polynomial equations) and cases are many, theapproach is practical only by computer methods. Nevertheless, spread sheetshave been prepared and 2 samples are enclosed at the end of Appendix F.5.3.5 The approach by the previous British Code CP110 is based on interactionformula by which the moments of resistance in both directions under the axialloads are determined with the pre-determined reinforcements and the“interaction formula” is checked. The approach is illustrated in Figure 5.8.P /bdPbd⎛ M+⎜⎝ Mxux⎞⎟⎠α⎛ M+ ⎜⎝ Myuy⎞⎟⎠α≤ 12M x/bd2M y/ b dFigure 5.8 – Interaction formula for design of biaxial bending5.3.6 Direct sectional analysis to Biaxial Bending without the necessity ofconverting the biaxial bending problem into a uniaxial bending problem :Though the Code has provisions for converting the biaxial bending probleminto a uniaxial bending problem by(i) searching for the controlling bending axis; and(ii) aggravate the moments about the controlling bending axis as appropriateto account for the effects of bending in the non-controlling axis;a designer can actually solve the biaxial bending problem by locating theorientation and the neutral axis depth (which generally does not align with theresultant moment except for circular section) of the column section bybalancing axial load and the bending in two directions. Theoretically, by83

Version 2.3 May 2008balancing axial load and the 2 bending moments, 3 equations can be obtainedfor solution of the neutral axis orientation, neutral axis depth and the requiredreinforcement. However the solution process, which is often based on trial anderror approach, will be very tedious and not possible for irregular sectionwithout computer methods. Reinforcements generally need be pre-determined.Figure 5.9 illustrates the method of solution.Neutral axisThe total sectional resistance of the sectionunder the stress strain profile resists theapplied axial loads and momentsStrain profile on sectionstrain profileacross sectionconcretestressprofileFigure 5.9 – General Biaxial Bending on irregular section5.4 Detailing requirements for longitudinal bars in columns (generally by Cl. 9.5and Cl. 9.9.2.1(a) of the Code, the ductility requirements applicable to columnscontributing in lateral load resisting system are marked with “D”)(i) Minimum steel percentage based on gross area of a column is 0.8% (Cl.9.5.1 of the Code);(ii) Maximum steel based on gross area of a column is (a) 4% except at lapwhich can be increased to 5.2% (D) for columns contributing to lateralload resisting system (Cl. 9.9.2.1(a) of the Code); and (b) 6% withoutlaps and 10% at laps for other columns (Cl. 9.5.1 of the Code);(iii) Bar diameter ≥ 12 mm (Cl. 9.5.1 of the Code);(iv) The minimum number of bars should be 4 in rectangular columns and 6in circular columns. In columns having a polygonal cross-section, atleast one bar be placed at each corner (Cl. 9.5.1 of the Code);(v) In any row of longitudinal bars in columns contributing to lateral load84

Version 2.3 May 2008resisting system, the smallest bar diameter used shall not be less than 2/3of the largest bar diameter used (Cl. 9.9.2.1(a) of the Code). For example,T40 should not be used with T25 and below (D);(vi)(vii)At laps, the sum of reinforcement sizes in a particular layer should notexceed 40% of the breadth at that section (Cl. 9.5.1 of the Code). Therequirement is identical to that of beam as illustrated by Figure 3.13;For columns contributing to lateral load resisting system, where thelongitudinal bars pass through the beams at column beam joints, columnbars shall satisfyφ ≤ 3.2h 0.8 f cu/ f as per Ceqn 9.7 where h is theybeam depth. For grade 35 concrete and based on high yield bar, thelimiting bar diameter is simply φ ≤ 0.0368h, i.e. if beam depth is 600mm, φ ≤ 22. 1 implying maximum bar size is 20 mm. If the column isnot intended to form a plastic hinge, the bar diameter can be increased by25% (Cl. 9.9.2.1(a) of the Code) (D);(viii) For columns contributing to lateral load resisting system, where thelongitudinal bars terminate in a joint between columns and foundationmembers with possible formation of a plastic hinge in the column, theanchorage of the column bars into the joint region should commence at1/2 of the depth of the foundation member or 8 times the bar diameterfrom the face at which the bars enter the foundation member. Where aplastic hinge adjacent to the foundation face cannot be formed,anchorage can commence at the interface with the foundation (Cl.9.9.2.1(c) of the Code) as illustrated in Figure 5.10 (D);Anchorage cancommence at this point ifplastic hinge cannotoccur at the column faceAnchorage shouldgenerally commence atthis pointD≥0.5Dor 8øLfoundationelementThe bends can be eliminated ifL ≥ anchorage lengthFigure 5.10 – Longitudinal Bar anchorage in foundation for columns contributingto lateral load resisting system85

Version 2.3 May 2008(ix)For columns contributing to lateral load resisting system, where thelongitudinal bars anchor into beam (transfer beam or roof beam), inaddition to the requirement in (viii), the bars should not be terminated ina joint area without a horizontal 90 o standard hook or an equivalentdevice as near as practically possible to the far side of the beam and notcloser than 3/4 of the depth of the beam to the face of entry. Unless thecolumn is designed to resist only axial load, the direction of bend mustalways be towards the far face of the column (Cl. 9.9.2.1(c) of the Code)as illustrated in Figures 5.11 and 5.12 (D);Anchorage can commenceat this point if plastic hingecannot occur adjacent tobeam faceAnchorage shouldgenerally commence atthis pointBar can bend outwards ifcolumn designed foraxial load onlyD≥0.75D if baranchored inbeam≥0.5Dor 8øBeamelementFigure 5.11 – Longitudinal Bar anchorage in Beam (Transfer Beam) for columnscontributing to lateral load resisting systemBar can bend outwards ifcolumn designed foraxial load only≥0.75D if baranchored inbeamRoof Beamor transferbeam≥0.5Dor 8øDAnchorage can commenceat this point if plastic hingecannot occur adjacent tobeam faceAnchorage shouldgenerally commence atthis pointFigure 5.12 – Longitudinal Bar anchorage in Beam (Transfer Beam / Roof Beam)for columns contributing to lateral load resisting system86

Version 2.3 May 2008(x)For laps and mechanical couplers in a column contributing to lateral loadresisting system, the centre of the splice must be within the middlequarter of the storey height of the column unless it can be shown thatplastic hinges cannot develop in the column adjacent to the beam faces.As per discussion in Section 2.4, such lapping arrangement should befollowed in locations such as column joining at pile caps or thickstructures. Normal lapping at other floors can usually be followed unlessthere are very stiff beams, e.g. transfer beams (Cl. 9.9.2.1(d) of theCode). Examples are illustrated in Figure 5.13 (D);T32T20T401125Consider a column ofstorey height 3m,grade 40 and withT20, T32 and T40bars. Half lap lengths(tension) arerespectively1.4×32×20÷2=448;1.4×32×32÷2=716.8;1.4×32×40÷2=896;≤716.8≤448≤896750 30001125The middle quarter ofthe column is at a levelof 1125 from the lowerfloor as shown. If thecentre of lap is withinthe middle quarter, thebars need be lapped atlevel higher than thefloor level.Section where plastichinge may developFigure 5.13 – Centre of lapping be within middle quarter of floor height in Columncontributing to lateral load resisting system(xi)(xii)Full strength welded splices may be used in any location (Cl. 9.9.2.1(d)of the Code);As similar to limitation of lapping of bars in beams as described inSection 3.6(vii), longitudinal bars in columns contributing to lateral loadresisting system shall not be lapped in a region where reversing stressesat the ultimate limit state may exceed0 .6 fyin tension or compression87

Version 2.3 May 2008unless each lapped bar is confined by adequate links or ties satisfying(Ceqn 9.6), as explained in Section 3.6(vii) and illustrated by Figure 3.11.Summing up, lapping should be avoided from region with potentialplastic hinge and with reversing stresses (Cl. 9.9.2.1(a) of the Code) (D);(xiii) Minimum clear spacing of bars should be the greatest of (1)bar diameter;(2) 20 mm; and (3) aggregate size + 5 mm (Cl. 8.2 of the Code).5.5 Detailing Requirements for transverse reinforcements in columns include thegeneral requirements by Cl. 9.5.2 and the ductility requirements in Cl. 9.9.2.2 ofthe Code (marked with “D”) for columns contributing to lateral load resistingsystem. Items (i) to (iv) below are requirements for columns not within “criticalregions”. “Critical region” is defined in item (v) and Figure 5.15 for columnscontributing to lateral load resisting system:(i)(ii)(iii)(iv)Diameter of transverse reinforcements ≥ the greater of 6 mm and 1/4 oflongitudinal bar diameter (Cl. 9.5.2.1 of the Code);The spacing of transverse reinforcement shall not exceed 12 times thediameter of the smallest longitudinal bar (Cl. 9.5.2.1 of the Code);For rectangular or polygonal columns, every corner bar and eachalternate bar (or bundle) shall be laterally supported by a link passingaround the bar and having an included angle ≤ 135 o . No bar within acompression zone shall be further than 150 mm from a restrained bar.Links shall be adequately anchored by hooks through angles ≥ 135 o . SeeFigure 5.14 which is reproduced from Figure 9.5 of the Code (Cl. 9.5.2.2of the Code);For circular columns, loops or spiral reinforcement satisfying (i) to (ii)should be provided. Loops (circular links) should be anchored with amechanical connection or a welded lap by terminating each end with a135 o hook bent around a longitudinal bar after overlapping the other endof the loop. Spiral should be anchored either by welding to the previousturn or by terminating each end with a 135 o hook bent around alongitudinal bar and at not more than 25 mm from the previous turn.Loops and spirals should not be anchored by straight lapping, whichcauses spalling of the concrete cover (Cl. 9.5.2.2 of the Code). Thedetails are also illustrated in Figure 5.14;88

Version 2.3 May 2008≤135 o , longitudinalbar considered to berestrained≥135 o>135 o ≤150longitudinal bar restrainingnot consideredlinkbarrestrained sinceanchoragerequiredenclosing angle>135 orestrainingbar≥135 oFigure 5.14 – Column transverse reinforcements outside “Critical Regions”(v)Transverse reinforcements in “critical regions” within columns of limitedductile high strength concrete (contributing to lateral load resisting system)as defined in Figure 5.15 (Re Cl. 9.9.2.2 of the Code) shall have additionalrequirements as :(a) For rectangular or polygonal columns, each (not only alternate)longitudinal bar or bundle of bars shall be laterally supported by alink passing around the bar having an included angle of not morethan 135 o . As such, Figure 5.16 shows the longitudinal baranchorage requirements in “critical region” (Cl. 9.9.2.2(b) of theCode) (D);(b) Spacing ≤ 1/4 of the least lateral column dimension in case ofrectangular or polygonal column and 1/4 of the diameter in case ofa circular column and 6 times the diameter of the longitudinal barto be restrained (Cl. 9.9.2.2(b) of the Code) (D);89

Version 2.3 May 2008H, Critical regionswith enhancedtransversereinforcementsNormaltransversereinforcementH, Critical regionswith enhancedtransversereinforcementsxM maxh mHeight of “criticalregion” , H, dependson N/A g f cu ratio :(a) 0

Version 2.3 May 2008Worked Example 5.7 – for determination of “critical regions” within columns oflimited ductile and high strength concreteConsider a rectangular column of the following details :Cross section 500× 600 mm; height 3 m; grade 65; re-bars : T32Loads and moments are as follows :Axial Load 4875 kNM = 800 kNm (at top), M = 500 kNm (at bottom)xM = 450 kNm (at top) M = 300 kNm (at bottom)yxyNA gf cu34875×10== 0.25500×600×40∴ x = 0.75 for determination of critical regionsh m for bending about X and Y directions are determined as per Figure 5.17.h m =462800kNmh m =450450kNm800×0.75=600kNm1846450×0.75=338kNm1800500×0.75=375kNm1154300×0.75=225kNm1200500kNmh m =289300kNmh m =300Bending about X-XBending about Y-YFigure 5.17 – Determination of critical heights in Worked Example 5.7As the h m are all less than 1.5h = 1.5×600 = 900, so the critical regions shouldthen both be 1200 mm from top and bottom and the design of transversereinforcements is as indicated in Figure 5.18 :91

Version 2.3 May 2008T10 @ 125T10 @ 350T10 @ 1259001200900Transverse re-bars(i) Within critical region (for columns oflimited high strength concrete andcontributing to lateral load resistingsystem only) :Bar size 0.25×32 = 8 mm > 6 mmSpacing : the lesser of0.25×500 = 125 mm6×32 = 192 mmSo spacing is 125 mm(ii) Within normal region (regardless ofwhether the column is contributing tolateral load resisting system) :Bar size 0.25×32 = 8 mm > 6 mmSpacing 12×32 = 384 mmFigure 5.18 –Transverse Reinforcement arrangement to Worked Example 5.792

Version 2.3 May 20086.0 Column-Beam Joints6.1 GeneralThe design criteria of a column-beam joint comprise (i) performance not inferiorto the adjoining members at serviceability limit state; and (ii) sufficient strengthto resist the worst load combination at ultimate limit state. To be specific, theaim of design comprise (a) minimization of the risk of concrete cracking andspalling near the beam-column interface; and (b) checking provisions againstdiagonal crushing or splitting of the joint and where necessary, providingvertical and horizontal shear links within the joint and confinement to thelongitudinal reinforcements of the columns adjacent to the joint.6.2 The phenomenon of “diagonal splitting” of jointDiagonal crushing or splitting of column-beam joints is resulted from “shears”and unbalancing moment acting on the joints as illustrated in Figure 6.1(a) and6.1(b) which indicate typical loadings acting on the joint. Figure 6.1(a) shows ajoint with hogging moment on the right and sagging moment on the left, whichmay be due to a large applied horizontal shear from the right. In contrast, Figure6.1(b) shows a joint with hogging moment on both sides which is the normalbehaviour of a column beam joint under dominant gravity loads. However, itshould be noted that the hogging moments on both sides may not balance.ColumnShear V c1Potential failuresurface (tension)sagging momentin beamC BLV b2V b1T BRhogging momentin beamT BLC BRColumnShear V c2Figure 6.1(a) – Phenomenon of Diagonal Joint Splitting by moments of oppositesigns on both sides of joint93

Version 2.3 May 2008TBLhogging momentin beamV b2ColumnShear V c1Potential failuresurface (tension)V b1TBR> T BLTBRhogging momentin beamC BLC BRColumnShear V c2Figure 6.1(b) – Phenomenon of Diagonal Joint Splitting by moments of same signon both sides of jointIn both cases, the unbalanced forces due to unbalanced flexural stresses by theadjoining beams on both sides of the joint tend to “tear” the joint off with apotential tension failure surface, producing “diagonal splitting”. In co-existencewith the bending moments, there are shears in the columns which usually tend toact oppositely. The effects by such shears can help to reduce the effects of shearson the column joints created by bending. Reinforcements in form of links maytherefore be necessary if the concrete alone is considered inadequate to resist thediagonal splitting.6.3 Design procedures :(i) Work out the total nominal horizontal shear force across the joint VjhinX and Y directions generally.Vjhshould be worked out by consideringforces acting on the upper half of the joint as illustrated in Figures 6.2(a)and 6.2(b). Figure 6.2(a) follows the case of Figure 6.1(a) in which themoments in the beams on both sides of the joint are of different signs (i.e.one hogging and one sagging). There is thus a net “shear” ofVjh= TBL+ TBR−Vcacting on the joint whereBRfyAsRT = andC = T = f A are the pull and push forces by the beams in which AsRBLBLysLandAsLare the steel areas of the beams. This approach which originates94

Version 2.3 May 2008from the New Zealand Code NSZ 3103 requires T BRand T BLbeincreased by 25% under the load capacity concept in which the reinforcingbars in the beam will be assumed to have steel stress equal to 125% yieldstrength of steel if such assumption will lead to the most adverseconditions. Thus the following equation can be listed :VjhBLBRcy( AsL+ AsR) −Vc= T + T −V= 1 . 25 f(Eqn 6.1)Columnshear V cC BL = T BLT = 1. 25 forBRf y AsRyAsRT = 1. 25 forBLf y AsRyAsLsagging momentin beamColumnshear V c ’C BRhogging momentin beamFigure 6.2(a) – Calculation ofVjh, opposite sign beam moments on both sidesHowever, there is a comment that New Zealand is a country of severeseismic activity whilst in Hong Kong the dominant lateral load is windload. The 25% increase may therefore be dropped and (Eqn 6.1) can bere-written asVjhBLBRcy( AsL+ AsR) −Vc= T + T −V= f(Eqn 6.2)Furthermore, as V ccounteracts the effects of T BRand T BLand V cisgenerally small, Vccan be ignored in design. Nevertheless, the inclusionof V can help to reduce steel congestion in case of high shear.cSimilarly Figure 6.2(b) follows the case of Figure 6.1(b) which may be thecase of unbalancing moments due to gravity load without lateral loads oreven with the lateral loads, such loads are not high enough to reverse anyof the beam moments from hogging to sagging. By similar argument andformulation as for that of Figure 6.2(a), (Eqn 6.3) and (Eqn 6.4) can beformulated for Figure 6.2(b)MLVjh= TBR−TBL−Vc= 1 . 25 fyAsR− −Vc(Eqn 6.3)zL95

Version 2.3 May 2008VjhML= TBR−TBL−Vc= fyAsR− −Vc(Eqn 6.4)zLTBLMT BL =zLLz LColumnshear V czRT BR > T BLT = 1. 25 forBRf y AsRyAsRC BRhogging momentin beam M LC BLhogging momentin beamFigure 6.2(b) – Calculation ofVjh, same sign beam moments on both sidesEquations (Eqn 6.2) and (Eqn 6.4) will be used in this Manual.(ii) With the V jhdetermined, the nominal shear stress is determined byVjh(Ceqn 6.71) in the Code. vjh=bjhcwhere h is the overall depth of the column in the direction of shearcbj= b cor bjbw+ 0. 5hc= whichever is the smaller when bc≥ bw;bj= b wor bj= bc+ 0. 5hcwhichever is the smaller when bc< bw;where bcis the width of column and bwis the width of the beam.Cl. 6.8.1.2 of the Code specifies that “At column of two-way frames,where beams frame into joints from two directions, these forces need beconsidered in each direction independently.” Sovjhshould be calculatedindependently for both directions even if they exist simultaneously andboth be checked that they do not exceed 0 .25 f .cu(iii) Horizontal reinforcements based on Ceqn 6.72 readingAjh*Vjh=0.87 fyh⎛ CjN⎜0.5−⎝ Agf*cu⎞⎟⎠should be worked out in both directions and96

Version 2.3 May 2008be provided in the joint as horizontal links. In Ceqn 6.72,*Vjhshould be*the joint shear in the direction (X or Y) under consideration and N bethe minimum column axial load. If the numerical values arrived at ispositive, shear reinforcements of cross sectional areasA jhshould beprovided. It may be more convenient to use close links which can serve asconfinements to concrete and horizontal shear reinforcements in bothdirections. If the numerical values arrived by (Ceqn 6.72) becomesnegative, no horizontal shear reinforcements will be required;(iv) Similarly vertical reinforcements based on (Ceqn 6.73) readingAjv( h / h )*0.4b cV CjN= should be worked out in both directions and0.87 f* jh−yvbe provided in the joint as vertical links or column intermediate bars (notcorner bars). Again if the numerical values arrived by (Ceqn 6.73) isnegative, no vertical shear reinforcements will be required;(v)Notwithstanding the provisions arrived at in (iii) for the horizontalreinforcements, confinements in form of closed links within the jointshould be provided as per Cl. 6.8.1.7 of the Code as :(a) Not less than that in the column shaft as required by Cl. 9.5.2 of theCode, i.e. Section 5.5 (i) to (iv) of this Manual if the joint has a freeface in one of its four faces;(b) Reduced by half to that provisions required in (a) if the joint isconnected to beams in all its 4 faces;(c) Link spacing ≤ 10Ø (diameter of smallest column bar) and 200 mm.Longitudinalbar dia. ØTransverse reinforcementsø ≥1/4(maxØ) and 6mm;with spacing ≤ 10(minØ)and 200mmFigure 6.3 – Minimum transverse reinforcements in Column Beam Joint97

Version 2.3 May 20086.4 Worked Example 6.1:Consider the column beam joints with columns and beams adjoining asindicated in Figure 6.4 in the X-direction and Y-directions. Concrete grade is 40.All loads, shears and moments are all ultimate values. The design is as follows :X-directionN* = 6000 kNBeam moment 550 kNm(hogging)Beam size 700 × 500(effective depth 630)VcxBeam moment 300 kNm(sagging)= 300 kNColumn size 900 × 800Beam size 700 × 500(effective depth 630)800900Column size 900 × 800Y-directionN* = 6000 kNBeam size 700 × 500(effective depth 630)Beam moment 300 kNm(hogging)Beam moment 550 kNm(hogging)Vcy= 0 kNBeam size 700 × 500(effective depth 630)Column size 900 × 800900800Figure 6.4 – Design Example for Column Beam Joint(i) Check nominal shear stress :X-direction98

Version 2.3 May 2008The moments on the left and right beams are of opposite signs. So Figure6.2(a) is applicable. The top steel provided on the right beam is 3T32, asdesigned against the ultimate hogging moment of 550kNm withA = 2413 mm 2 whilst the bottom steel provided on the left beam issR4T20 with A = 1257 mm 2 , again as designed against the ultimatesagging moment of 300kNm.sLC BLT BRT BLC BR−T = 460×2413×103 BR= 1109.98 kN; CBR= TBR−T = 460×1257×103 BL= 578.22 kN; CBL= TBLSo the total shear isVjx= TBL+ TBR−Vcx= 1109 .98 + 578.22 − 300 = 1388.2 kNIn the X-direction h = 900cAs b 800 > b = 500 ,the effective joint width is the smaller ofc=wb = 800 and b 0 .5h= 500 + 0.5×900 = 950 , so b = 800cw+cSo, checking against Cl. 6.8.1.3 of the Code,3Vjx 1388.2×10vjx= == 1.93MPa < 0 .25 fcu= 10 MPab h 800×900jcY-directionThe moments on the left and right beams are of equal sign, both hogging.So Figure 6.2(b) is applicable. As the moment on the right beam ishigher, the potential plastic hinge will be formed on the right beam.Again the top steel provided in the right beam is 3T32 as designedagainst the ultimate hogging moment of 550 kNm.j99

Version 2.3 May 2008TBLTBRz LT BR > T BLC BRC BL−T = 460×2413×103 BR= 1109.98 kN; CBR= TBRTBLis to be determined by conventional beam design method for theultimate hogging moment of 300 kNmM=1.512 zMPa, = 0. 948, A2 sL=1254. 49 mm 2 ;bddT 0 .87 f A = 502.05 kNBL=y sLAs the column shear is zero, by (Eqn 6.3)V = 1109 .98 − 502.05 = 607.93kNjyIn the Y-direction h = 800 , b = 900cand b 0 .5h= 500 + 0.5×800 = 900 , so b = 900w+cSo, checking against Cl. 6.8.1.3 of the Code,3Vjy 607.93×10vjy= == 0.84 MPa < 0 .25 fcu= 10 MPab h 900×800jccj(ii)To calculate the horizontal joint reinforcement by Ceqn 6.72, reading**V ⎛ ⎞jhCjNA = ⎜ − ⎟jh0.50.87 fyh ⎝ Agfcu⎠Vjhwhere Cj=V + VjxjyX-directionVjxCjx=V + VAjhxjx*Vjhx=0.87 f= 1232 mm 2jyyh1388.2== 0.6951388.2 + 607.93*⎛⎞3CjxN⎜⎟1388.2×100.5 − =⎝ Agfcu ⎠ 0.87×460⎛⎜0.5−⎝0.695×6000000 ⎞⎟900×800×40 ⎠Y-directionVjyCjy=V + Vjxjy607.93== 0.3051388.2 + 607.93100

Ajhy*Vjhy=0.87 f= 663 mm 2yh⎛ C⎜0.5−⎝ AjygNfcu*⎞⎟607.93×10=⎠ 0.87×4603⎛⎜0.5−⎝Version 2.3 May 20080.305×6000000 ⎞⎟900×800×40 ⎠Use 6T12 close stirrups (Area provided = 1357 mm 2 ) which canadequately cover shear reinforcements in both directions(iii)To calculate the vertical joint reinforcement by (Ceqn 6.73 of the Code),0.4( hb/ hc) V* *jh− CjNreading Ajv=0.87 fyhX-directionAjvx0.4=**( hb/ hc) Vjh− CjN 0.4( 700/900)0.87 fyh=× 1388200 − 0.695×60000000.87×460= −9341. So no vertical shear reinforcement is required.Y-directionAjvy0.4=**( hb/ hc) Vjh− CjN 0.4( 700/800)0.87 fyh=× 607930 − 0.305×60000000.87×460= −4041. Again no vertical shear reinforcement is required.(iv)The provision of outermost closed stirrups in the column shaft is T12 atapproximately 120mm which is in excess of the required confinement aslisted in 6.3(v). So no additional confinement is requirement.Closed links 6T12(spacing = 120 < 200and 10Ø = 320)Figure 6.5 – Details of Column Beam Joint Detail for Column Beam Joint(Plan) Design – Other details omitted for clarityT32101

Version 2.3 May 20087.0 Walls7.1 Design Generally7.1.1 Similar to column by design to resist axial loads and moments.7.1.2 The design ultimate axial force may be calculated on the assumption that thebeams and slabs transmitting force to it are simply supported. (Re Cl. 6.2.2.2(a)and Cl. 6.2.2.3(a) of the Code).7.1.3 Minimum eccentricity for transverse moment design is the lesser of 20 mm orh / 20 , as similar to columns.7.2 Categorization of WallsWalls can be categorized into (i) slender walls; (ii) stocky walls; (iii)reinforced concrete walls; and (iv) plain walls.7.3 Slender Wall Section Design7.3.1 Determination of effective height le(of minor axis generally whichcontrols) –(i) in case of monolithic construction, same as that for column; and(ii) in case of simply supported construction, same as that for plain wall.7.3.2 Limits of slender ratio (Re Table 6.15 of the Code) –(i) 40 for braced wall with reinforcements < 1%;(ii) 45 for braced wall with reinforcements ≥ 1%;(iii) 30 for unbraced wall.7.3.3 Other than 7.3.1 and 7.3.2, reinforced concrete design is similar to that ofcolumns.7.4 Stocky Wall7.4.1 As similar to column, stocky walls are walls with slenderness ratio < 15 forbraced walls and slenderness ratio < 10 for unbraced walls;102

Version 2.3 May 20087.4.2 Stocky reinforced wall may be designed for axial load n wonly by (Ceqn6.59) of the Code provided that the walls support approximately symmetricalarrangement of slabs with uniformly distributed loads and the spans on eitherside do not differ by more than 15%;n ≤ 0 .35 f A + 0. 67 fwcucyAsc7.4.3 Other than 7.4.2 and the design for deflection induced moment Maddof stocky wall is similar to slender walls., design7.5 Reinforced Concrete Walls design is similar to that of columns withcategorization into slender walls and stocky walls.7.6 Plain Wall – Plain wall are walls the design of which is without considerationof the presence of the reinforcements.7.6.1 Effective height of unbraced plain wall, where l 0is the clear height of thewall between support, is determined by :(a) l e= 1.5l0when it is supporting a floor slab spanning at right angles to it;(b) l e= 2.0l0for other cases.Effective height ratio for braced plain wall is determined by(a) l e= 0.75l0when the two end supports restraint movements and rotations;(b) l e= 2.0l0when one end support restraint movements and rotations andthe other is free;(c) l e= l 0' when the two end supports restraint movements only;(b) l e= 2.5l0' when one end support restraint movements only and the otheris free; where l0' in (c) and (d) are heights between centres of supports.7.6.2 For detailed design criteria including check for concentrated load, shear, loadcarrying capacities etc, refer to Cl. 6.2.2.3 of the Code.7.7 Sectional DesignThe sectional design of wall section is similar to that of column by utilizingstress strain relationship of concrete and steel as indicated in Figure 3.8 and3.9 of the Code. Alternatively, the simplified stress block of concrete asindicated in Figure 6.1 can also be used. Nevertheless, the Code has additional103

Version 2.3 May 2008requirements in case both in-plane and transverse moments are “significant”and such requirements are not identical for stocky wall and slender wall.7.7.1 Wall with axial load and in-plane momentConventionally, walls with uniformly distributed reinforcements along itslength can be treated as if the steel bars on each side of the centroidal axis arelumped into two bars each carrying half of the steel areas as shown in Figure7.1 and design is carried out as if it is a 4 bar column. Nevertheless, it issuggested in this Manual that the reinforcements can be idealized as acontinuum (also as shown in Figure 7.1) which is considered as a morerealistic idealization. Derivation of the formulae for the design withreinforcements idealized as continuum is contained in Appendix G, togetherwith design charts also enclosed in the same Appendix.d = 0.75hShear Wall SectionCurrent idealizationbased on 4-barcolumn design chartProposed idealizationwith reinforcing bars ascontinuum with areasequal to the barsFigure 7.1 – Idealization of Reinforcing bars in shear wallWorked Example 7.1Consider a wall of thickness 300 mm, plan length 3000 mm and under an axialload P = 27000 kN and in-plane moment M = 4500kNm. Concrete grade is45. The problem is an uniaxial bending problem. Thenx104

Version 2.3 May 200836P 27000×10M 4500×10== 30 and == 1. 67 .2 2bh 300×3000bh 300×3000If based on the 4-bar column chart with d / h = 0. 75, p = 3. 8 %, requiringT32 – 140 (B.F.)N/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10M/bh 2 N/mm 2If use chart based on continuum of bars, the reinforcement ratio can be slightlyreduced to 3.7%.N/bh N/mm 255504540353025Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 450.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10M/bh 2 N/mm 2By superimposing the two design charts as in Figure 7.2, it can be seen that theidealization of steel re-bars as continuum is generally more conservative.105

Version 2.3 May 2008Comparison of Idealization as 4-bar columns and Continuum of Steel to Code of Practice forStructural Use of Concrete 2004, Concrete Grade 45N/bh N/mm 255504540353025200.4% steel - 4 bar column0.4% steel - wall2% steel - 4 bar column2% steel - wall5% steel - 4 bar column5% steel - wall8% steel - 4 bar column8% steel - wall1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10M/bh 2 N/mm 2Figure 7.2 – Comparison of design curve between idealization of steel bars as 4 barcolumn and continuum7.7.2 Wall with axial load and transverse momentThe design will also be similar to that of column with the two layers oflongitudinal bars represented by the bars in the 4-bar column charts as shownin Figure 7.3idealized asBars carry total steel totalarea of the row of steelFigure 7.3 – Sectional design for column with axial load and transverse moment7.7.3 Wall with significant in-plane and transverse moments106

Version 2.3 May 2008The Code has not defined the extent of being “significant”. Nevertheless, ifsignificant in-plane and transverse moments exist, the Code effectivelyrequires the wall section be examined at various points (for stocky wall) andunit lengths (for slender wall) along the length of the wall at the splitting up ofthe axial load and in-plane moment as demonstrated in Figure 7.4.PMwallBy elastic analysisLLMx3/12P ±2PL6M−L2≥ 0P 6M+L LLxORPL6M−L2< 02P⎛ L M3⎜−⎝ 2 P⎞⎟⎠⎛ L3⎜⎝ 2− PM⎞⎟⎠Figure 7.4 – conversion of axial load (kN) and in-plane moment (kNm) into linearvarying load (kN/m) along wall sectionLWorked Example 7.2Consider a grade 45 wall of thickness 300 mm, plan length 3000 mm and107

Version 2.3 May 2008under an axial load P = 27000 kN and in-plane moment M = 4500kNm andtransverse moment M = 300 kNm as shown in Figure 7.5. By elastic analysis,yxthe load intensities at the 4 points as resolution of P and27000 6×4500A : + = 12000 kN/m;23 327000 4500×0.5B : + = 10000 kN/m33 3 /1227000 4500×0.5C : − = 8000 kN/m;33 3 /1227000 6×4500D : − = 6000 kN/m23 3The varying load intensities are as indicated in Figure 7.5.Mxare :P = 27000 kNMx= 4500kNmBy elastic analysis8000kN/mwall10000kN/m6000kNm12000kN/mDCBA1000 10001000Figure 7.5 – Conversion of axial load (kN) and in-plane moment (kNm) into linearvarying load (kN/m) along wall section for Worked Example 7.2(i)If the wall is considered stocky, each of the points with loadintensities as determined shall be designed for the load intensities asderived from the elastic analysis and a transverse moment of300 ÷ 3 = 100 kNm/m by Clause 6.2.2.2(f)(iv) of the Code. Considerone metre length for each point, the 4 points shall be designed for thefollowing loads with section 1000 mm by 300 mm as tabulated in108

Version 2.3 May 2008Table 7.1, i.e. all the points are undergoing uniaxial bending and thesectional design are done in the same Table in accordance with thechart extracted from Appendix F:Point A B C DAxial Load 12000 10000 8000 6000In-plane Mt 0 0 0 0Transverse Mt 100 100 100 100N / bh40 33.33 26.67 20/ bhp (%) 5.9 4.1 2.4 0.62M 1.11 1.11 1.11 1.11Re-bars (BF) T40 – 140 T40 – 200 T32 – 225 T20 – 300Table 7.1 – Summary of Design for Worked Example 7.2 as a stocky WallN/bh N/mm 25550454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.8ABCD0.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5M/bh 2 N/mm 2The Code is not clear in the assignment of reinforcements at varioussegments of the section based on reinforcements worked out atvarious points. The assignment can be based on the tributary lengthprinciple, i.e. the reinforcement derived from A shall be extendedfrom A to mid-way between A and B; the reinforcement derived fromB be extended from mid-way between A and B to mid-way betweenB and C etc. As such, the average reinforcement ratio is 3.25%.Nevertheless, as a more conservative approach, the assignment ofreinforcement design between A and B should be based on A and that109

Version 2.3 May 2008of B and C be based on B etc. As such the reinforcement ratio of thewhole section will be increased to 4.13% and the reinforcement ratioat D is not used.(ii)If the wall is slender, by Cl. 6.2.2.2(g)(i) of the Code, the wall shouldbe divided into “unit lengths” with summing up of loads. Considerthe three units AB, BC and CD. The loads and in-plane momentssummed from the trapezoidal distribution of loads are as follows,with the assumption that the transverse moment of 300 kNm hasincorporated effects due to slenderness :For Unit Length AB :12000 + 10000Summed axial load = × 1 = 11000kN212000 −10000⎛ 2 1 ⎞Summed in-plane moment × 1×⎜ − ⎟× 1 = 167 kNm.2 ⎝ 3 2 ⎠The summed axial loads and moments on the unit lengths BC andCD are similarly determined and design is summarized in Table 7.2,with reference to the design chart extracted from Appendix F. In thecomputation ofM x/ h'and M y/b', h ' and b ' are taken as 750and 225 respectively.Unit Length AB BC CDAxial Load 11000 9000 7000In-plane Mt ( Mx) 167 167 167Transverse Mt ( M ) 100 100 100yM x/ h'0.227 0.227 0.227M y/b'0.444 0.444 0.444N / f bh0.272 0.222 0.172cuβ 0.684 0.744 0.801M ' = M + β b'/ h'M 134.2 137.2 140.1yy( )xN / bh36.67 30 23.332M y'/hb1.49 1.524 1.556p (%) 5.4 3.7 1.9Re-bars (BF) T40 – 155 T32 – 145 T25 – 175Table 7.2 – Design of Wall for Worked Example 7.2 as a slender wall110

Version 2.3 May 2008N/bh N/mm 25550454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.8ABBCCD0.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5M/bh 2 N/mm 2The average steel percentage is 3.67%.So the reinforcement worked out by Clause 6.2.2.2(g)(i) of the Codefor a slender wall is between the results of the two methods ofreinforcement ratios assignments as described in sub-section (i)based on Clause 6.2.2.2(f)(iv) of the Code.(iii) Summary of the reinforcements design of the three approachesstocky wall –tributarylengthT20 –30 BFT32 – 225 BFT40 –200 BFT40 – 140 BFstocky wall –conservativeapproachT32 – 225 BF T40 –200 BF T40 – 140 BFslenderwallT25 – 175 BF T32 – 145 BF T40 – 155 BFFigure 7.6 – Summary of reinforcement details of Worked Example 7.2(iv) The approach recommended in the Code appears to be reasonableand probably economical as higher reinforcement ratios will be inregion of high stresses. However, it should be noted that if momentarises from wind loads where the direction can reverse, design for the111

Version 2.3 May 2008reversed direction may result in almost same provisions ofreinforcements at the other end.As the division of segments or points as recommended by the Codefor design of wall with significant transverse and in-plane momentsis due to the inaccurate account by the biaxial bending formula usedfor design of column, more accurate analysis can be done by truebiaxial bending analysis as discussed in Section 5.3.5 and Figure 5.8of this Manual, so long the “plane remain plane” assumption is valid,though the design can only be conveniently done by computermethods. The sections with reinforcement ratios arrived at in (i) and(ii) have been checked against by the software ADSEC, the section in(ii) has yielded an applied moment / moment capacity ratio of 0.8showing there is room for slight economy. Nevertheless, the firstreinforcement ratio in (i) is inadequate as checked by ADSEC whilstthe second one yielded an over design with applied moment /moment capacity ratio up to 0.68.7.8 The following Worked Example 7.3 serves to demonstrate the determinationof design moment for a slender wall section, taking into account of additionalmoment due to slenderness.Worked Example 7.3Wall Section : thickness : 200 mm, plan length : 2000 mm;Wall Height : 3.6 m,Concrete grade : 35Connection conditions at both ends of the wall : connected monolithically withfloor structures shallower than the wall thickness.Check for slendernessGenerally only necessary about the minor axis.End conditions are 2 for both ends, β = 0. 85 (by Table 6.11 of the Code);le= 0 .85×3.6 = 3.06 mAxial Load : N = 7200 kN, M = 1800bottom , M = 25 kNm at top and 24 kNm at bottom.yxkNm at top and 1200 kNm at112

Version 2.3 May 2008Determination of final design momentsimilar to (i)Mtabout the major and minor axes isFor bending about the major axis, / h = 3060/ 2000 = 1.53 < 15Madd= 0 , Mxwill be the greatest of(1) M 1800 ;= 2i+ MaddM = 0 × − + × + = ;(2) .4 ( 1200) 0.6 1800 0 600< 0 .4 × 1800 = 720(3) M M / 2 = 1200 + 0 1200 ; and1+add=× emin = 7200 × 0.02 =(4) N 144 .So M = 1800 kNm for design.xl e, soFor bending about the minor axis, l e/ b = 3060/ 200 = 15.3 > 15,βa=12000l e⎛⎜⎝ b2⎞⎟⎠=120002⎛ 3060 ⎞⎜ ⎟⎝ 200 ⎠= 0.117a= Kh = 0 .117 × 1×0.2 = 0.0234uβaMadd= Nau= 7200 × 0.0234 = 168.48 kNm, Mywill be the greatest of(1) M 25 ;= 2i+ Madd(2) M = 0 .4×25 + 168.48 = 178. 48 ;as 0 .4×( − 24) + 0.6×25 = 5.4 < 0.4×25 = 10(3) M M / 2 = 24 + 168.48/ 2 108. 3; and1+add=(4) N × e = 7200 × 0.02 144 . So M =178. 48kNm for design.min=So the factored axial load and moments for design areN = 7200 kN; M = 1800 kNm; M =178. 48kNmxDesign can be performed in accordance with Cl. 6.2.2.2(g) of the Code asdemonstrated in Worked Examples 7.2 and by calculations with the formulaederived in Appendices F and G. However, the calculations are too tedious andcases to try are too many without the use of computer methods. Spread sheetshave been devised to solve the problem with a sample enclosed in Appendix G.yy7.9 Detailing RequirementsThere are no ductility requirements in the Code for walls. The detailingrequirements are summarized from Cl. 9.6 of the Code :113

Version 2.3 May 2008Vertical reinforcements for reinforced concrete walls :(i) Minimum steel percentage : 0.4%. When this reinforcement controlsthe design, half of the steel area be on each side;(ii) Maximum steel percentage : 4%;(iii) All vertical compression reinforcements should be enclosed by a linkas shown in Figure 7.7;(iv) Maximum distance between bars : the lesser of 3 times the wallthickness and 400 mm as shown in Figure 7.7.≤ 3h and 400 mmhFigure 7.7 – Vertical reinforcements for wallsHorizontal and transverse reinforcements for reinforced concrete walls(i)If the required vertical reinforcement does not exceed 2%, horizontalreinforcements be provided as follows and in accordance with Figure7.8 :(a) Minimum percentage is 0.25% for fy= 460 MPa and 0.3% forfy= 250 MPa;(b) bar diameter ≥ 6 mm and 1/4 of vertical bar size;(c) spacing ≤ 400 mm.114

Version 2.3 May 2008h(a) 0.25% for f y = 460 MPa and 0.3% for f y = 250 MPa;(b) bar diameter ≥ 6 mm and 1/4 of vertical bar size;(c) spacing in the vertical direction ≤ 400 mmFigure 7.8 – Horizontal reinforcements for walls with verticalreinforcement ≤ 2%(ii)If the required vertical reinforcement > 2%, links be provided asfollows as shown in Figure 7.9 :(a) to enclose every vertical compression longitudinal bar;(b) no bar be at a distance further than 200 mm from a restrained barat which a link passes round at included angle ≤ 90 o ;(c) minimum diameter : the greater of 6 mm and 1/4 of the largestcompression bar;(d) maximum spacing : twice the wall thickness in both thehorizontal and vertical directions. In addition, maximum spacingnot to exceed 16 times the vertical bar diameter in the verticaldirection.Links of includedangle ≤ 90 o torestrain verticalbars(a) Spacing in vertical direction ≤ 2hand 16 Ø;(b) bar diameter ≥ 6 mm or 1/4 Ø≤ 200≤ 200restrainedverticalbarsh≤ 2h≤ 200≤ 200Figure 7.9 – Anchorage by links on vertical reinforcements of more than 2%Plain wallsIf provided, minimum reinforcements : 0.25% for f = 460 MPa and 0.3%for f = 250 MPa in both directions generally.yy115

Version 2.3 May 20088.0 Corbels8.1 General – A corbel is a short cantilever projection supporting a load-bearingmember with dimensions as shown :a v< dAppliedLoadhd≥ 0.5hTop steel barFigure 8.1 – Dimension requirement for a Corbel8.2 Basis of Design (Cl. 6.5.2 of the Code)8.2.1 According to Cl. 6.5.2.1 of the Code, the basis of design method of a corbel isthat it behaves as a “Strut-and-Tie” model as illustrated in Figure 8.2. The strutaction (compressive) is carried out by concrete and the tensile force at top iscarried by the top steel.Steel strain to bedetermined by linearextrapolationTie action byreinforcing bara vAppliedLoad VuTneutral axisdStrut action byconcreteVuFc0 .9xxβBalancing force polygonConcrete ultimatestrain ε ult = 0. 0035Concrete stress block at corbelsupportFigure 8.2 – Strut-and-Tie Action of a Corbel116

Version 2.3 May 20088.2.2 Magnitude of resistance provided to the horizontal force should be not lessthan one half of the design vertical load, thus limiting the value of the angleβ in Figure 8.2 or in turn, that the value of a cannot be too small.v8.2.3 Strain compatibility be ensured.8.2.4 In addition to the strut-and tie model for the determination of the top steel bars,shear reinforcements should be provided in form of horizontal links in theupper two thirds of the effective depth of the corbel. The horizontal linksshould not be less than one half of the steel area of the top steel.8.2.5 Bearing pressure from the bearing pad on the corbel should be checked andproperly designed in accordance with “Code of Practice for Precast ConcreteConstruction 2003” Cl. 2.7.9. In short, the design ultimate bearing pressure toultimate loads should not exceed(i) 0 .4 f for dry bearing;(ii)(iii)cu0 .6 f for bedded bearing on concrete;cu0 .8 f for contact face of a steel bearing plate cast on the corbel withcueach of the bearing width and length not exceeding 40% of the width andlength of the corbel.The net bearing width is obtained byeffectivebearingultimatelength ×loadultimatebearingstressThe Precast Concrete Code 2003 (in Cl. 2.7.9.3 of the Precast Concrete Code)has specified that the effective bearing length of a bearing be the least of :(i) physical bearing length;(ii) one half of the physical bearing length plus 100 mm;(iii) 600 mm.8.3 Design Formulae for the upper steel tieThe capacity of concrete in providing lateral force as per Figure 8.2 is0 .45 fcu× b×0.9x= 0. 405 fcubxwhere b is the length of the corbel.The force in the compressive strut is therefore Fc = 0.405fcubxcos β .By the force polygon, F sin β = V ⇒ 0.405 f bx sin β cos β = Vcucuu117

Version 2.3 May 2008Asd − 0. 45xtan β = ;a vcos β =a( d − 0. x) 2a2 + 45vvsin β =Soa0.405 fvcu( d − 0.45x)+ ( d − 0. x) 22 45abxa( d − 0.45x)( d − 0.45x)Expanding and re-arranging= V⇒ V0.405 f=bxav( d − 0.45x)( d − 0. x) 2vcu2 2 u ua245v+v+22 2( 0.2025V+ 0.18225 f ba ) x − 0.9d( V + 0.45 f ba ) x + V ( a + d ) = 0PuttingucuuvcuuA = 0 .2025V+ 0. 18225 f ba ; B = −0 .9d( V + 0. 45 f ba )C = V2 2( a d )u v+− B − B2 − 4ACx = (Eqn 8-1)2AVuavBy the equilibrium of force, the top steel force is T = Vucot β =d − 0.45x(Eqn 8-2)The strain at the steel level is, by extrapolation of the strain diagram in Figured − x d − x8.2 is εs= ε ult= × 0. 0035(Eqn 8-3)x xvcuvuuvcuv8.4 Design Procedure :(i) Based on the design ultimate load and av, estimate the size of the corbeland check that the estimated dimensions comply with Figure 8.1;(ii) Check bearing pressures;(iii) Solve the neutral axis depth x by the equation (Eqn 8-1).(iv) By the assumption plane remains plane and that the linear strain at thebase of the corbel is the ultimate strain of concrete εult= 0. 0035 , workout the strain at the top steel level as εs;6(v) Obtain the steel stress as σs= Esεswhere Es= 200×10 kPa.However, the stress should be limited to 0 .87 fyeven εs≥ 0. 002 ;(vi) Obtain the force in the top steel bar T by (Eqn 8-2)(vii) Check that T ≥ 0. 5V ;(viii) Obtain the required steel area of the top steel barsuAstbyAstT=σs118

Version 2.3 May 2008Vu(ix) Check the shear stress by v = . If v > vc(after enhancement asbdAsvb( v − vc)applicable), provide shear reinforcements by = over thes 0.87 f2upper d where3is the link spacing.A svis the cross sectional area of each link and s vAsv(x) Check that the total shear area provided which is d is not less thansvd 1half of the top steel area, i.e. Asv× ≥ Asteven if v < vr.s 2vvy8.5 Detailing Requirements(i)By Cl. 6.5.2.2 of the Code, anchorage of the top reinforcing bar shouldeither(a) be welded to a transverse bar of equivalent strength or diameter.The bearing area of the load should stop short of the transverse barby a distance equal to the cover of the tie reinforcement as shownin Figure 8.3(a); or(b) bent back to form a closed loop. The bearing area of the loadshould not project beyond the straight portion of the bars formingthe tension reinforcements as shown in Figure 8.3(b).(ii)By Cl. 6.5.2.3 of the Code, shear reinforcements be provided in theupper two thirds of the effective depth and total area not less than half ofthe top bars as shown in Figure 8.3(a) and 8.3(b).119

Version 2.3 May 2008Top main bara vV u> cc, cover totransverse bard2d3cShear reinforcementstransverse bar welded tothe main tension bar ofequal diameter orstrengthAdditional bar for shearlink anchorageFigure 8.3(a) – Typical Detailing of a CorbelV uTop main bara v> 0cd2d3Shear reinforcementsFigure 8.3(b) – Typical Detailing of a Corbel8.6 Worked Example 8.1Design a corbel to support an ultimate load of 600 kN at a distance 200 mmfrom a wall support, i.e. V = 600 kN, a = 200 mm. The load is transmittedufrom a bearing pad of length 300 mm. Concrete grade is 40.v120

Version 2.3 May 2008a v= 200 < d = 450V u= 600h = 500d = 450Net bearingwidth250 ≥ 0.5hFigure 8.4 – Worked Example. 8.11. The dimensions of the corbel are detailed as shown which comply withthe requirement of Cl. 6.5.1 of the Code with length of the corbelb = 300 mm;2. Check bearing stress :Design ultimate bearing stress is 0 .8 fcu= 0.8×40 = 32 MPa3600×10Net bearing width is = 62. 5mm.300×32So use net bearing width of bearing pad 70 mm.3. With the following parameters :Vu= 600 kN; fcu= 40 MPa; b = 300 mm;av= 200 mm; d = 450 mmsubstituted into (Eqn 8-1)22 2( 0.2025V+ 0.18225 f ba ) x − 0.9d( V + 0.45 f ba ) x + V ( a + d ) = 0uSolving x = 276. 77 mm.4. The strain at steel level,εd − xεxs=ultcuv450 − 276.77=× 0.0035 = 0.00219 > 0.002276.775. The stress in the top steel is 0 .87 fyas εs> 0. 002;(if εs≤ 0. 002, fs= E s× εswhere Es= 200 GPa)6. The force in the top steel isucuvuv121

Version 2.3 May 2008TV a 600×200=u v== 368.71d − 0.45x450 − 0.45×276.77kN > 0 .5× 600 = 300 kN;3687107. Steel area required is = 921. 32mm 2 , provide 3T20 (0.7%);0.87×4601/ 38. = 0.556×( 40/ 25) = 0. 65v MPa without enhancement. Withc2denhancement, it becomes × 0.65 = 2. 925 MPa.9. Check shear stressa v600000 = 4.444 MPa > vc= 2. 925MPa.450×300Asvb v − vc300 4.444 − 2.925So shear reinforcement = =s 0.87 f 0.87×460v( ) ( )y= 1.14 mm;A = 1 .14×450 = 513 mm 2 . So use 3T12 closed links over the top 300svmm.10. Area of 3T12 closed link is 678 mm 2 > half of area of tensile top steel =0.5×3×314 = 471 mm 2 .The details of the Corbel is finally as shown in Figure 8.5.V u3T20a v450300T20 anchor barT12 closed linksFigure 8.5 – Detailing of Worked Example 8.18.7 Resistance to horizontal forcesCl. 9.8.4 requires additional reinforcement connected to the supported member122

Version 2.3 May 2008to transmit external horizontal force exerted to the corbel in its entirety.However, it should be on the conservative side if strain compatibility is alsoconsidered in designing the corbel to resist also this horizontal force N as inaddition to the vertical load V u. This is in consistency with the Coderequirement. The force polygon as modified from Figure 8.2 will becomescSteel strain to bedetermined by linearextrapolationTie action byreinforcing bara vAppliedVuAppliedNcTConcrete ultimatestrain ε ult = 0. 00350 .9xneutral axisxdConcrete stress block at corbelsupportβStrut action byconcreteV uNcRFcBalancing force polygonFigure 8.6 – Strut-and-Tie Action of a Corbel with inclusion of horizontal forceFrom Figure 8.6 and formulae derivation in Section 8.3 of this Manual, it canbe seen that the determination of the neutral axis depth x and subsequentlythe strain profile of the root of the corbel is independent of Nc. Thus the steps(i) to (v) in Section 8.4 of this Manual can be followed in calculation of x ,εsand σ sas ifN cdoes not exist.VuavHowever, the tension in the top bar will be T = Nc+(Eqn 8-4)d − 0.45xTAnd the steel area of the top bar can be worked out as Ast=σs8.8 Worked Example 8.2If an additional horizontal force of 200kN is exerted on the corbel in Example8.1, tending to pull away from the root of the corbel, the total tensile force tobe resisted by the top bars will be T = 368 .71+200 = 568. 71kN and the top3568.71×10bar area required is = 1421. 06mm 2 , as the strain at the steel level0.87×460has exceed 0.002. The top bar has to be increased from 3T25.123

Version 2.3 May 20089.0 Cantilever Structures9.1 Cl. 1.4 of the Code defines “Cantilever Projection” as “a structural element thatcantilevers from the main structure, for example, canopies, balconies, baywindows, air conditioning platforms.” In addition, PNAP 173 which refers tocantilevered reinforced concrete structures in general indicates more clearlydesign and construction criteria to be complied with.9.2 Design ConsiderationsDesign considerations for a cantilevered structure from the Code (Table 7.3, Cl.9.4 etc. of the Code) and PNAP 173 are summarized as follows :Slabs and Beams in General(i) The span to overall depth of cantilever beams or slabs should not begreater than 7;(ii) For cantilever span more than 1000 mm, a beam-and-slab type ofarrangement should be used instead of pure slab cantilever wherepracticable (PNAP173 App. A 1(a));(iii) The minimum percentage of top tension longitudinal reinforcement basedon the gross cross-sectional concrete area should be 0.25% for allreinforcement grades generally (PNAP173 App. A 6(c)). However, if thecantilever structure is a flanged beam where the flange is in tension, theminimum steel percentage is 0.26% for T-section and 0.2% for L-sectionbut based on the gross area of the rectangular portion of width of the webtimes the structural depth as per Table 9.1 of the Code. The more stringentrequirement shall prevail;(iv) Diameter of the longitudinal reinforcement ≥ 10 mm as illustrated inFigure 9.1 (PNAP173 App. A 6(c));(v) The centre-to-centre spacing of the top tension longitudinal bars ≤ 150mmas illustrated in Figure 9.1 (PNAP173 App. A 6(c));(vi) For cantilevered structure exposed to weathering, cover to allreinforcement ≥ 40 mm (PNAP173 App. A 8(a));(vii) Anchorage of tension reinforcement shall be based on steel stress of0 .87 fyand (a) full anchorage length should be provided with location ofcommencement in accordance with Cl. 9.4.3 of the Code as illustrated in124

Version 2.3 May 2008Figures 9.1 and 9.2; and (b) minimum anchorage length of 45 times thelongitudinal bar diameter in accordance with PNAP 173 App. A 6(d). Thedifferent commencement points of anchorage lengths as indicated byPNAP 173 Appendices B and C are not adopted in this Manual. However,requirements for the lengths of curtailment of tension reinforcement barsPNAP173 and Cl. 9.2.1.6 of the Code in relation to curtailment of tensionreinforcements are amalgamated. They are shown in Figures 9.1 and 9.2.T.L.bar dia. Ø ≥ 10 mm≥0.5dor 0.5Lcover to all reinforcements≥ 40 if the beam is subjectto weathering≤150≤150dSupport providingrotational restraintLK(slab similar)T.L.bar dia. Ø ≥ 10 mmcover to all reinforcements≥ 40 if the beam is subjectto weatheringSupport not providingrotational restraintKdT.L. should be the greatest of(i) To point of zero moment +the greater of anchorage lengthand d; (ii) 1.5K; (iii) 45Ø; (iv)0.3×next span length (for slabonly)LFigure 9.1 – Anchorage and maximum longitudinal bar spacing in Cantilevers asrequired by the Code and PNAP 173Beam in particular(viii) The overall depth at support should be at least 300 mm as shown in Figure9.2;(ix) For cantilever beam connected with continuous beams, requirements forcurtailment of longitudinal bars into the next continuous span are similarto slab except that half of the bars can be curtailed at 0.75K + L/2 as125

Version 2.3 May 2008shown in Figure 9.2;Half of bars becurtailed at 0.75Kcover to all reinforcements ≥ 40 ifthe beam is subject to weatheringd≥300LKFigure 9.2 – Particular requirements for cantilever beams as required by the Codeand PNAP 173Slab in particular(x) Minimum overall slab thickness (PNAP173 App. A 6(a)):(a) 100 mm for span ≤ 500mm;(b) 125 mm for 500 mm < span ≤ 750mm;(c) 150 mm for span > 750 mm;(xi) Reinforcements be high yield bars in both faces and in both directions(PNAP173 App.A 6(c));(xii) Particular attentions to loads as shown in Figure 9.3 should be given :Care be taken notto ignore loadsfrom this parapetCare be taken (1) not toignore loads from thisarea; (2) change ofdirection of main barsFigure 9.3 – Loads on cantilever slab (PNAP173 App.A 6(e))(xiii) For a cantilever slab with a drop at the supporting end, top reinforcementbars ≤ 16 mm in diameter should be used in order that an effective andproper anchorage into the supporting beam and internal slab can be126

developed as illustrated in Figure 9.4. (PNAP173 App. A 6(d))Version 2.3 May 2008bar dia. ≤ 16 mmFigure 9.4 – Cantilever slab with drop at supporting end(xiv) Cantilevered slabs exposed to weathering should satisfy :(1) maximum crack width at the tension face ≤ 0.1 mm underserviceability check OR stress of deformed high yield steel bar ≤ 100MPa when checking the flexural tension under working loadcondition (PNAP173 App. A 8(a));(2) Cover to all reinforcement at the exposed surface ≥ 40 mm.(PNAP173 App. A 8(a)).9.3 Worked Example 9.1R.C. design of a cantilevered slab as shown in Figure 9.5 is subject toweathering. Concrete grade is 35.1001000(150) 900200 wall supportPlanElevationFigure 9.5 – Cantilever slab in Worked Example 9.1Loading D.L. O.W. 0 .15× 24 = 3. 6 kN/m 2Fin 2.0 kN/m 25.6 kN/m 2Para. 0 .1× 1.0×24 = 2. 4 kN/m127

L.L. 1.5 kN/m 2Effective span is taken to be 900 + 0.5×150 = 975Version 2.3 May 2008Moment =( 1 .4×5.6 + 1.6×1.5) × 0.975×0.975 / 2 + 1.4×2.4×0. 925= 7.975 kNm/mDesign for ultimate state,d = 150 − 40 − 5 = 105Mbd7.975×10=1000×1056=2 20.72367.975×10Ast== 200 mm 2 /m.0.87×460×0.95×105Use T10 – 150 (Area provided is 523 mm 2 /m)If the slab is subject to weathering, check the service stress by equationin item (2) in Table 7.4 of the Code readingfs2 fyA=3Ast,reqst,prov×1βbNote : β = 1 as no moment redistribution in cantilever.bIff is to be limited to 100 N/mm 2 , A 200 mm 2 /msst, req=AfA2y st,req 1 2×460×200 1st, prov= × =× =3 fsβb3×100 1613 mm 2Use T10 – 100 (area provided is 785 mm 2 /m or 0.52%)Alternatively, crack width is checked by (Ceqn 7.1) and (Ceqn 7.2)To calculate crack width, it is first necessary to assess the neutral axisdepth x by the elastic theory in accordance with the cracked sectionof Figure 7.1 of the Code on the basis of a cracked section.ε =cf / Eccf cxhdf s / E sstrainfsstressFigure 9.6 – Stress/strain relation of a cracked R.C. section128

Version 2.3 May 2008Ecis the long term value which, by Cl. 7.2.3 of the Code is taken ashalf of the instantaneous value which is 23 .7 ÷ 2 = 11. 85 GPaEs= 200 kN/mm 2Consider equilibrium of the section in Figure 9.6.12f bx = fc1 2sAst⇒1Ecεcbx= Es2εc( d − x)⇒ Ecbx+ EsAstx − EsdAst= 0(Eqn 9.1)2Consider 1 m width of the section in Worked Example 9.1, b = 1000(Eqn 9.1) becomes1 2× 11.85×1000x+ 200×785x− 200×105×785 = 02Solving x = 41. 14 mmTaking moment about the centroid of the triangular concrete stressblock (the moment should be the unfactored moment which is5.817kNm/m as it is a checking on serviceability limit state), the steeltensile stress can be worked out as6⎛ x ⎞M 5.817×10M = fsAst⎜d− ⎟ ⇒ fs= =(Eqn 9.2)⎝ 3 ⎠ ⎛ x ⎞ ⎛ 41.14 ⎞Ast⎜d− ⎟ 785⎜105− ⎟⎝ 3 ⎠ ⎝ 3 ⎠= 81.17 N/mm 2xAstSo the strain of the steel is81.170.8 fyεs= = 0.000406 < = 0.0184 .3200×10ESo checking of crack width by (Ceqn 7.1) is applicable.sAt the extreme fibre of the concrete at the tension side, the strain isε( h − x)( d − x)150 − 41.14= 0.000406×105 − 41.141= ε s=0.000692By (CEqn 7.2), to include the stiffening effect of cracked concrete,bt( h − x)( a'−x)1000( 150 − 41.14)( 150 − 41.14)ε m= ε1−= 0.000692 −33EA d − x3×200×10 × 785×105 − 41.14= 0.000298ss( )( )The expected shrinkage strain, in accordance with Cl. 3.1.8 of theCode isε = c K K K K K wherecssLcejs129

c = 3.0 ;Version 2.3 May 2008s−6KL= 275×10 for normal air according to Figure 3.6 of the Code;K = 1.17 according to Figure 3.3 of the Code for cement content 434ckg/m 3 and water cement ratio 0.47 for grade 35;K = 0.91 according to Figure 3.7 of the Code for h = 150 ;eK =1 according to Figure 3.5 at time at infinity.jeK s11= =1+ραe1+0.0052×20011.85So the expected shrinkage strain is= 0.919according to (Ceqn 3.3)−6ε = 3.0×275×10 × 1.17×0.91×1.0×0.919 = 0.000807 > 0.0006 .csThus it is subjected to “abnormally high shrinkage” according to theCode and half of the expected strain be added to ε .∴εm= 0 .000298 + 0.000807×0.5 = 0.000702mThe cracked width should be the greatest at the concrete surfacemid-way between steel bars as illustrated in Figure 9.7;a cr=502+ 402= 6440100Figure 9.7 – Illustration ofacrin Worked Example 9.1By (CEqn 7.1) the cracked width is3a3×64×0.000702ω =crε m== 0.0935 mm ≤ 0.1 mm as⎛ acr− cmin⎞ ⎛ 64 − 40 ⎞1+2⎜⎟ 1+2⎜⎟⎝ h − x ⎠ ⎝150− 41.14 ⎠required by PNAP 173. So O.K.As PNAP 173 requires either checking of working stress below 100MPa or crack width ≤ 0.1 mm, it should be adequate if any one of theconditions is satisfied. Apparently it would be simpler to check only130

Version 2.3 May 2008the former.Summing up, reinforcement details is as shown :T10 – 100Adequate anchoragelength as determined byFigure 9.1 and 9.2T10 – 300Figure 9.8 – Reinforcement Details for Worked Example 9.19.4 R.C. DetailingApart from the requirements stipulated in the preceding sections, reference canalso be made to the drawings attached at the Appendices B and C of PNAP 173,especially for the locations of anchorage length commencement. However, itshould be noted that not all sketches in PNAP 173 indicate locations ofanchorage length commence from mid-support widths.131

Version 2.3 May 200810.0 Transfer Structures10.1 According to Cl. 5.5 of the Code, transfer structures are horizontal elementswhich redistribute vertical loads where there is a discontinuity between thevertical structural elements above and below.10.2 In the analysis of transfer structures, consideration should be given to thefollowings as per Cl. 5.5 of the Code :(i) Construction and pouring sequence – the effects of constructionsequence can be important in design of transfer structures due to thecomparatively large stiffness of the transfer structure and sequential builtup of stiffness of structures above the transfer structure as illustrated inFigure 10.1;(ii) Temporary and permanent loading conditions – especially importantwhen it is planned to cast the transfer structures in two shifts and use thelower shift to support the upper shift as temporary conditions, thuscreating locked-in stresses;(iii) Varying axial shortening of elements supporting the transfer structures –which leads to redistribution of loads. The phenomenon is more seriousas the transfer structure usually possesses large flexural stiffness incomparison with the supporting structural members, behaving somewhatbetween (a) flexible floor structures on hard columns; and (b) rigidstructures (like rigid cap) on flexible columns;(iv) Local effects of shear walls on transfer structures – shear walls willstiffen up transfer structures considerably and the effects should be takeninto account in more accurate assessment of structural behaviour;(v) Deflection of the transfer structures – will lead to redistribution of loadsof the superstructure. Care should be taken if the structural model abovethe transfer structure is analyzed separately with the assumption that thesupports offered by the transfer structures are rigid. Re-examination ofthe load redistribution should be carried out if the deflections of thetransfer structures are found to be significant;(vi) Lateral shear forces on the transfer structures – though the shear is lateral,it will nevertheless create out-of-plane loads in the transfer structureswhich needs be taken into account;(vii) Sidesway of the transfer structures under lateral loads and unbalancedgravity loads should also be taken into account. The effects should beconsidered if the transfer structure is analyzed as a 2-D model.132

Version 2.3 May 2008Stage (1) :Transfer Structure (T.S.)just hardenedStage (2) :Wet concrete of 1/F justpouredStage (3) :1/F hardened and 2/F wetconcrete just pouredG/FStress/force in T.S. : {F T } dueto own weight of T.S.Stiffness : the T.S onlyG/F1/FStress/force in T.S. : {F T } +{F 1 }, {F 1 } being forceinduced in transferstructure due to weight of1/F structure.Stiffness : the T.S. only.G/F2/F1/FStress/force in T.S. being dueto {F T } + {F 1 } + {F 2 }, {F 2 }being force induced intransfer structure due toweight of 2/F structure.Stiffness : the T.S. + 1/F.Stage (4) :2/F hardened and 3/F wetconcrete just poured3/F2/FStage (5) :3/F hardened and 4/F wetconcrete just poured4/F3/F2/FStage (6) and onwardsStructure above transferstructure continues to bebuilt. Final force induced onT.S. becomes {F n } + {F n-1 }+ {F n-2 } + ........... + {F 2 } +{F 1 } + {F T }.1/F 1/FG/FG/FStress/force in T.S. : {F T } +{F 1 } + {F 2 } + {F 3 }, {F 3 }being force induced in T.S.due to weight of 3/Fstructure.Stiffness : T.S. + 1/F + 2/FStress/force in T.S. : {F T }+ {F 1 } + {F 2 } + {F 3 } +{F 4 }, {F 4 } being forceinduced in T.S. due toweight of 4/F structure.Stiffness : T.S. + 1/F + 2/F+ 3/FFigure 10.1 – Diagrammatic illustration of the Effects of Construction Sequence ofloads induced on transfer structure133

Version 2.3 May 200810.3 Mathematical modeling of transfer structures as 2-D model (by SAFE) :The general comments in mathematical modeling of transfer structures as 2-Dmodel to be analyzed by computer methods are listed :(i) The 2-D model can only be analyzed against out-of-plane loads, i.e.vertical loads and out-of-plane moments. Lateral loads have to beanalyzed separately;(ii) It is a basic requirement that the transfer structure must be adequatelystiff so that detrimental effects due to settlements of the columns andwalls being supported on the transfer structure are tolerable. In view ofthe relatively large spans by comparing with pile cap, such settlementsshould be checked. Effects of construction sequence may be taken intoaccount in checking;(iii) The vertical settlement support stiffness should take the length of thecolumn/wall support down to a level of adequate restraint against furthersettlement such as pile cap level. Reference can be made to Appendix Hdiscussing the method of “Compounding” of vertical stiffness and theunderlying assumption;(iv) Care should be taken in assigning support stiffness to the transferstructures. It should be noted that the conventional use of either 4 EI / Lor 3 EI / L have taken the basic assumption of no lateral movements atthe transfer structure level. Correction to allow for sidesway effects isnecessary, especially under unbalanced applied moments such as windmoment. Fuller discussion and means to assess such effects are discussedin Appendix H;(v) Walls which are constructed monolithically with the supporting transferstructures may help to stiffen up the transfer structures considerably.However, care should be taken to incorporate such stiffening effect in themathematical modeling of the transfer structures which is usually doneby adding a stiff beam in the mathematical model. It is not advisable totake the full height of the wall in the estimation of the stiffening effect ifit is of many storeys as the stiffness can only be gradually built up in thestorey by storey construction so that the full stiffness can only beeffected in supporting the upper floors. Four or five storeys of walls maybe used for multi-storey buildings. Furthermore, loads induced in thesestiffening structures (the stiff beams) have to be properly catered forwhich should be resisted by the wall forming the stiff beams;134

Version 2.3 May 200810.4 Modeling of the transfer structure as a 3-dimensional mathematical model caneliminate most of the shortcomings of 2-dimensional analysis discussed insection 10.3, including the effects of construction sequence if the software hasprovisions for such effects. However, as most of these softwares may not havethe sub-routines for detailed design, the designer may need to “transport” the3-D model into the 2-D model for detailed design. For such “transportation”,two approaches can be adopted :(i)Transport the structure with the calculated displacements by the 3-D software(after omission of the in-plane displacements) into the 2-D software forre-analysis and design. Only the displacements of the nodes with external loads(applied loads and reactions) should be transported. A 2-D structure will bere-formulated in the 2-D software for re-analysis by which the structure isre-analyzed by forced displacements (the transported displacements) withrecovery of the external loads (out-of-plane components only) and subsequentlyrecovery of the internal forces in the structure. Theoretically results of the twomodels should be identical if the finite element meshing and the shape functionsadopted in the 2 models are identical. However, as the finite element meshing ofthe 2-D model is usually finer than that of the 3-D one, there are differencesincurred between the 2 models, as indicated by the differences in recovery ofnodal forces in the 2-D model. The designer should check consistencies inreactions acting on the 2 models. If large differences occur, especially whenlesser loads are revealed in the 2-D model, the designer should review hisapproach;External nodalforce is {F 3D }External nodalforce is {F 2D } ≠{F 3D } afterre-analysis3-D model (usually coarser meshing)with displacements at nodes withexternal loads marked with2-D model (usually finer meshing) withnodal forces recovered by forceddisplacement analysis at nodes markedwithFigure 10.2 – 3-D model to 2-D with transportation of nodal displacements135

Version 2.3 May 2008(ii)Transport the out-of-plane components of the external loads (applied loads andreactions) acting on the 3-D model to the 2-D model for further analysis. Thistype of transportation is simpler and more reliable as full recovery of loadsacting on the structure is ensured. However, in the re-analysis of the 2-Dstructure, a fixed support has to be added on any point of the structure foranalysis as without which the structure will be unstable. Nevertheless, no effectsdue to this support will be incurred by this support because the support reactionsshould be zero as the transported loads from the 3-D model are in equilibrium.The out-of-planecomponents of all loadsacting on the structureincluding reactions betransported3-D model with external loads obtainedby analysis2-D model with out-of-plane components ofexternal forces transported from 3-D model andre-analyzed with a fixed support at any pointFigure 10.3 – 3-D model to 2-D with transportation of nodal forces10.5 Structural Sectional Design and r.c. detailingThe structural sectional design and r.c. detailing of a transfer structure membershould be in accordance with the structural element it simulates, i.e. it should bedesigned and detailed as a beam if simulated as a beam and be designed anddetailed as a plate structure if simulated as a plate structure. Though not socommon in Hong Kong, if simulation as a “strut-and-tie” model is employed,the sectional design and r.c. detailing should accordingly be based on the tie andstrut forces so analyzed.The commonest structural simulation of a transfer plate structure is as anassembly of plate bending elements analyzed by the finite element method. Assuch, the analytical results comprising bending, twisting moments andout-of-plane shears should be designed for. Reference to Appendix D can bemade for the principles and design approach of the plate bending elements.136

Version 2.3 May 200811.0 Footings11.1 Analysis and Design of Footing based on the assumption of rigid footingCl. 6.7.1 of the Code allows a footing be analyzed as a “rigid footing”provided it is of sufficient rigidity with uniform or linearly varying pressuresbeneath. As suggested by the Code, the critical section for design is at columnor wall face as marked in Figure 11.1, though in case of circular columns, thecritical section may need be shifted into 0.2 times the diameter of the column,as in consistency with Cl. 5.2.1.2(b) of the Code.critical sections for designFooting under pure axial loadcreating uniform pressure beneathFooting under eccentric load creatinglinearly varying pressure beneathFigure 11.1 – Assumed Reaction Pressure on Rigid FootingAs it is a usual practice of treating the rigid footing as a beam in the analysisof its internal forces, Cl. 6.7.2.2 of the Code requires concentration of steelbars in areas with high stress concentrations as illustrated in Figure 11.2.area with 2/3 of therequiredreinforcementsarea with 2/3 of therequiredreinforcements1.5dc1.5d1.5dc1.5dl c is the greaterof l c1 and l c2d is theeffective depthl c2 2l c1 PlanFigure 11.2 – Distribution of reinforcing bars when l c > (3c/4 + 9d/4)137

Version 2.3 May 2008Cl. 6.7.2.4 of the Code requires checking of shear be based on (i) sectionthrough the whole width of the footing (as a beam); and (ii) local punchingshear check as if it is a flat slab. (Re Worked Example 4.5 in Section 4).11.2 Worked Example 11.1Consider a raft footing under two column loads as shown in Figure 11.3.Design data are as follows :Column Loads (for each): Axial Load: D.L. 800 kN L.L. 200 kNMoment D.L. 100kNm L.L. 20 kNmOverburden soil : 1.5 m deepFooting dimensions : plan dimensions as shown, structural depth 400 mm,cover = 75 mm; Concrete grade of footing : grade 35400 4001000D.L. 100 kNmL.L. 20kNm foreach column1250 250040012501000PlanFigure 11.3 – Footing layout for Worked Example 11.1(i) Loading Summary :D.L. Column: 2 ×800 = 1600 kN;O.W. 5 .0× 2.0×0.4×24 = 96 kNOverburden Soil 5 .0× 2×1.5×20 = 300 kNTotal1996 kNMoment (bending upwards as shown in Figure 11.3)2 × 100 = 200 kNmL.L. Column 2 × 200 = 400 kN.Moment (bending upwards as shown in Figure 11.3)2 × 20 = 40 kNmFactored load : Axial load 1 .4× 1996 + 1.6×400 = 3434. 4 kNMoment 1 .4× 200 + 1.6×40 = 344 kNm138

Version 2.3 May 2008(ii) The pressure beneath the footing is first worked out as :At the upper end :3434.4 6×344+ = 343.44 + 103.2 = 446. 64 kN/m 225×2 5×2At the lower end :3434.4 6×344− = 343.44 −103.2= 240. 24 kN/m 225×2 5×2Critical section3434.4 344×0.2+ = 343.44 + 20.64 = 364. 08 kN/m 235×2 5×2 /12The pressures are indicated in Figure 11.3(a)400Section forcritical design400446.64 kN/m 2400364.08kN/m 21250 25001250240.24 kN/m 2(iv) At the critical section for design as marked in Figure 11.3(a), the totalshear is due to the upward ground pressure minus the weight of thefooting and overburden soil ( 1 .4( 0.4× 24 + 1.5×20) = 55. 44 kN/m 2 )⎛ 446.64 + 364.08 ⎞which is ⎜⎟× 0.8×5 − 55.44×0.8×5 = 1399. 68kN⎝ 2 ⎠The total bending moment is0.822⎛ 446.64 − 364.08 ⎞⎜⎟×⎝ 2 ⎠2( 364.08 − 55.44) × × 5 +0.8 × × 5= 581.89 kNmFigure 11.3(a) – Bearing Pressure for Worked Example 10.123(v) Design for bending : Moment per m width is :581.89= 116.38 kNm/m;5d = 400 − 75 −8= 317 mm, assume T16 bars6M 116.38×10K = == 1.158 ,bd2 21000×317By the formulae in Section 3 for Rigorous Stress Approach,p0= 0.306 %; Ast= 969 mm 2 /mAs = 1250 > 3c/ 4 + 9d/ 4 = 3×400 / 4 + 9×317 / 4 = 1013 , two thirdsl c139

Version 2.3 May 2008of the reinforcements have to be distributed within a zone of c + 1 .5×2dfrom the centre and on both sides of the column, i.e. a total width of400 + 1.5×317×2 = 1.351m about the centre line of the columns.Total flexural reinforcements over the entire width is969 × 5 = 4845 mm 2 , 2/3 of which in 1 .351× 2 = 2. 702 m.So 4845 × 2 / 3 / 2.702 = 1195 mm 2 /m within the critical zone.So provide T16 – 150.Other than the critical zone, reinforcements per metre width is4845 / 3 / 5 − 2.702 = mm 2 /m. Provide T16 – 275.( ) 703Design for Strip Shear : Total shear along the critical section is1399.86 kN, thus shear stress is31399.68×10v == 0.883 N/mm 25000×31711143400 1 353⎛ ⎞ ⎛ ⎞> vc= 0.79×0.306 ⎜ ⎟ × ⎜ ⎟ = 0. 505 N/mm 2 as per⎝ 317 ⎠ 1.25 ⎝ 25 ⎠Table 6.3 of the Code.So shear reinforcement required isAsvb( v − vc) 5000( 0.883 − 0.505)5000×0.4= =< = 4.998 mm 2 /mmsv0.87 fyv0.87×460 0.87×460Within the two-thirds (of total width 2.675 m) with heavier shearreinforcement :24 .998× ÷ 2.702 = 1.233mm 2 /m. Use T10 – 175 s.w. and – 300 l.w.3In the rest of the footing,14 .998× ÷ 2.298 = 0.724 mm/m. Use T10 – 300 BWs.3(vi) Check punching shear along perimeter of columnFactored load by a column is 1 .4× 800 + 1.6×200 = 1440 kN. By Cl.6.1.5.6(d), along the column perimeter,Veffud31440×10= 2.84 < 0.84×400×317= fcu= 4.7 MPa. O.K.Locate the next critical perimeter for punching shear checking as shownin Figure 11.3(b) which is at 1.5d from the column face.Weight of overburden soil and weight of footing is221.351 × 55.44 −1.4×0.4 × 1.5×20 = 94.47 kNUpthrust by ground pressure is 3434 .4× 1.3512 627. 035 2= kN×140

Version 2.3 May 2008Net load along the critical perimeter is1440 + 94.47 − 627.03 = 907.44 kNCritical perimeter forpunching shear checking4004001000400+1.5d×2=1351400400+1.5d×2=135112501000PlanFigure 11.3(b) – checking punching shear for Worked Example 10.1By (Ceqn 6.40)Veff⎛ 1.5 ⎞⎜M ⎛ 1.5×172 ⎞1 +tV ⎟t= 907.44⎜1+⎟ = 1098.41kN⎝ Vtx ⎠ ⎝ 907.44×1.351⎠=sp31098.41×10Punching shear stress is v == 0. 641N/mm 21351×4×317As v < 1 .6vc= 0. 808 , use (Ceqn 6.44) in determining punching shearreinforcement,( v − vc) ud ( 0.641−0.489)× 1351×4×317 0.4×1351×4×317=

Version 2.3 May 2008So the provision by the strip shear obtained in (v) which is greater isadopted as per Cl. 6.7.2.4 of the Code which requires the more “severe”provision for checking of strip and punching shears.(vii) Checking of bending and shear in the direction parallel to the line joiningthe columns can be carried out similarly. However, it should be notedthat there is a net “torsion” acting on any section perpendicular to theline joining the two columns due to linearly varying ground pressure. Tobe on the conservative side, shear arising due to this torsion should bechecked and designed accordingly as a beam as necessary. Nevertheless,one can raise a comment that the design has to some extent be duplicatedas checking of bending has been carried out in the perpendiculardirection. Furthermore, for full torsion to be developed for design inaccordance with (Ceqn 6.65) to (Ceqn 6.68) of the Code, the “beam”should have a free length of beam stirrup width + depth to develop thetorsion (as illustrated in Figure 3.31 in Section 3) which is generally notpossible for footing of considerable width. As unlike vertical shear whereenhancement can be adopted with “shear span” less than 2 d or 1 .5d,no similar strength enhancement is allowed in Code, though by the samephenomenon there should be some shear strength enhancement. So fulldesign for bending in both ways together with torsion will likely result inover-design.(viii) The flexural and shear reinforcements provisions for the directionperpendicular to the line joining the columns is13451345400400T16 – 275(B1)400T16 – 150(B1)PlanShear links T10 –175 s.w. 300 l.w.Shear links T10 –175 s.w. 300 l.w.Shear links T10 –300 BWs. inother areasFigure 11.3(c) – Reinforcement Details for Worked Example 11.1 (in thedirection perpendicular to the line joining the two columns only)142

Version 2.3 May 200811.3 Flexible Footing Analysis and DesignAs contrast to the footing analyzed under the rigid footing assumption, theanalysis of footing under the assumption of its being a flexible structure willtake the stiffness of the structure and the supporting ground into account bywhich the deformations of the structure itself will be analyzed. Thedeformations will affect the distribution of the internal forces of the structureand the reactions which are generally significantly different from that by rigidfooting analysis. Though it is comparatively easy to model the cap structure, itis difficult to model the surface supports provided by the ground because :(i) the stiffness of the ground with respect to the hardness of the subgradeand geometry of the footing are difficult to assess;(ii) the supports are interacting with one another instead of beingindependent “Winkler springs” supports. However, we are currentlylacking computer softwares to solve the problem. Use of constant“Winkler springs” thus becomes a common approach.As the out-of-plane deformations and forces are most important in footinganalysis and design, flexible footings are often modeled as plate bendingelements analyzed by the finite element method as will be discussed in 11.4 inmore details.11.4 Analysis and Design by Computer MethodThe followings are highlighted for design of footing modeled as 2-D model(idealized as assembly of plate bending elements) on surface supports:(i)(ii)The analytical results comprise bending, twisting moments andout-of-plane shears for consideration in design;As local “stresses” within the footing are revealed in details, the rulesgoverning distribution of reinforcements in footing analyzed as a beamneed not be applied. The design at any location in the footing can bebased on the calculated stresses directly. However, if “peak stresses”(high stresses dropping off rapidly within short distance) occur at certainlocations as illustrated in Figure 11.4 which are often results of finiteelement analysis at points with heavy loads or point supports, it would bereasonable to “spread” the stresses over certain width for design.143

Version 2.3 May 2008Nevertheless, care must be taken not to adopt widths too wide for“spreading” as local effects may not be well captured.peak stresswidth overwhich thepeak stress isdesigned forFigure 11.4 – Spreading of peak stress over certain width for design(iii) The design against flexure should be done by the “Wood ArmerEquations” listed in Appendix D, together with discussion of itsunderlying principles. As the finite element mesh of the mathematicalmodel is often very fine, it is a practice of “lumping” the designreinforcements of a number of nodes over certain widths and evenlydistributing the total reinforcements over the widths, as is done by thepopular software “SAFE”. Again, care must be taken in not takingwidths too wide for “lumping” as local effects may not be well captured.The design of reinforcements by SAFE is illustrated on the right portionof Figure 11.4;(iv) The principle together with a worked example for design against shear isincluded in Appendix D, as illustrated in Figure D5a to D5c. It should benoted that as the finite element analysis give detailed distribution ofshear stresses on the structure, it is not necessary to carry out sheardistribution into column and mid-strips as is done for flat slab underempirical analysis in accordance with the Code. The checking of shearand design of shear reinforcements can be based directly on the shearstresses revealed by the finite element analysis.144

Version 2.3 May 200812.0 Pile Caps12.1 Rigid Cap analysisCl. 6.7.3 of the Code allows a pile cap be analyzed and designed as a “rigidcap” by which the cap is considered as a perfectly rigid structure so that thesupporting piles deform in a co-planar manner at their junctions with the cap.As the deformations of the piles are governed, the reactions by the piles canbe found with their assigned (or assumed) stiffnesses. If it is assumed that thepiles are identical (in stiffnesses), the reactions of the piles follow a linearlyvarying pattern. Appendix I contains derivation of formulae for solution ofpile loads under rigid cap assumption.pile loads,magnitude followslinear profileunder assumptionof equal pilestiffnessFigure 12.1 – Pile load profile under rigid cap assumptionUpon solution of the pile loads, the internal forces of the pile cap structurecan be obtained with the applied loads and reactions acting on it as a freebody. The conventional assumption is to consider the cap as a beam structurespanning in two directions and perform analysis and design separately. It isalso a requirement under certain circumstances that some net torsions actingon the cap structure (being idealized as a beam) need be checked. As thedesigner can only obtain a total moment and shear force in any section of fullcap width, there may be under-design against heavy local effects in areashaving heavy point loads or pile reactions. The Code (Cl. 6.7.3.3) thereforeimposes a condition that shear enhancement of concrete due to closeness ofapplied load and support cannot be applied.Cl. 6.7.3.5 of the Code requires checking of torsion based on rigid body145

theory which is similar to discussion in Section 11.2 (vii).Version 2.3 May 200812.2 Worked Example 12.1 (Rigid Cap Design)The small cap as shown in Figure 12.2 is analyzed by the rigid capassumption and will then undergo conventional design as a beam spanning intwo directions.Design data : Pile cap plan dimensions : as shownPile cap structural depth : 2 mPile diameter : 2 mConcrete grade of Cap : 35Cover to main reinforcements : 75 mmColumn dimension : 2 m squareFactored Load from the central column :P = 50000 kN= 2000 kNm (along X-axis)MxMy= 1000 kNm (along Y-axis)15003000P1 P2 P3400Y30001500P4 P5 P64001500 40004000 1500Xcritical sectionsfor shearcheckingFigure 12.2 – Pile cap layout of Worked Example 12.1(i) Factored Loads from the Column :P = 50000 kN= 2000 kNm (along positive X-axis)MxMy= 1000 kNm (along positive Y-axis)O.W. of Cap 11 × 9×2×24 = 4752 kNWeight of overburden soil 11 × 9×1.5×20 = 2970 kNFactored load due to O.W. of Cap and soil is146

( 4752 + 2970) 10811Version 2.3 May 20081 .4× = kNSo total axial load is 50000 + 10811 = 60811kN(ii)Analysis of pile loads – assume all piles are identical(Reference to Appendix I for general analysis formulae)I of pile group = 6×32 = 54x2I of pile group = 4×4 + 2×0 = 64yPile Loads on P1 :60811 2000×4 1000×3− + = 10065. 72 kN6 64 54P2:60811 2000×0 1000×3− + = 10190. 72 kN6 64 54P3:60811 2000×4 1000×3+ + = 10315. 72 kN6 64 54P4:60811 2000×4 1000×3− − = 9954. 61kN6 64 54P5:60811 2000×0 1000×3− − = 10079. 61kN6 64 54P6:60811 2000×4 1000×3+ − = 10204. 61kN6 64 54(iii) Design for bending along the X-directionThe most critical section is at the centre line of the capMoment created by Piles P3 and P6 is( 10315 .72 + 10204.61) × 4 = 82081. 32kNmCounter moment by O.W. of cap and soil is10811 ÷ 2×2.75 = 14865.13 kNmThe net moment acting on the section is82081 .32 −14865.13= 67216.19 kNmd = 2000 − 75 − 60 = 1865 (assume 2 layers of T40); b = 90006M 67216.19×10z== 2.147 ; = 0. 926 p = 0.58 %2 2bd 9000×1865dA = 97210 mm 2 , provide T40 – 200 (2 layers, B1 and B3)st(iv) Design for shear in the X-directionBy Cl. 6.7.3.2 of the Code, the critical section for shear checking is at20% of the diameter of the pile inside the face of the pile as shown inFigure 12.2.Total shear at the critical section is :147

Version 2.3 May 2008Upward shear by P3 and P6 is 10315 .72 + 10204.61 = 20520. 33kNDownward shear by cap’s O.W. and soil is2.110811 × = 2063.92 kN11Net shear on the critical section is 20520 .33 − 2063.92 = 18456. 41kN318456.41×10v == 1.10N/mm 2 > vc= 0. 58N/mm 2 by Table 6.3 of9000×1865the Code.No shear enhancement in concrete strength can be effected as per Cl.6.7.3.3 of the Code because no shear distribution across section can beconsidered.Shear reinforcements in form of links per metre width isAsvb( v − vc) 1000( 1.10 − 0.58)= == 1.299s 0.87 f 0.87×460vyvUse T12 links – 200 in X-direction and 400 in Y-direction by whichAsvprovided is 1.41.sv(v)Design for bending along the Y-directionThe most critical section is at the centre line of the capMoment created by Piles P1, P2 and P310065 .72 + 10190.72 + 10315.72 × 3 = 30572.16×3 = 91716. kNm( ) 48Counter moment by O.W. of cap and soil is10811 ÷ 2×2.25 = 12162.38kNmThe net moment acting on the section is91716 .48 −12162.38= 79554.11kNmd = 2000 − 75 − 60 − 40 = 1825 ; (assume 2 layers of T40) b = 110006M 76554.11×10z== 2.09; = 0. 929 p = 0.55%2 2bd 11000×1825dA =110459 mm 2 , provide T40 – 200 (2 layers, B2 and B4)st(vi) Checking for shear in the Y-directionBy Cl. 6.7.3.2 of the Code, the critical section for shear checking is at20% of the diameter of the pile inside the face of the pile as shown inFigure 12.2Total shear at the critical section is :Upward shear by P1, P2 and P3 is 30572.16 kN148

Version 2.3 May 2008Downward shear by cap’s O.W. and soil is2.110811 × = 2522.57 kN9Net shear on the critical section is 30572 .16 − 2522.57 = 28049. 59 kN328049.59×10v == 1.397 N/mm 2 > vc= 0. 579 N/mm 2 by Table 6.311000×1825of the Code.Similar to checking of shear checking in X-direction, no shearenhancement of concrete strength can be effected.Shear reinforcements in form of links per metre width isAsvb( v − vc) 1000( 1.397 − 0.579)= == 2.044s 0.87 f 0.87×460vAsAssvvyvin Y-direction is greater than that in X-direction, so adopt thisfor shear reinforcement provision.Use T12 links – 200 BWs by whichAssvvprovided is 2.82.(vii) Punching shear :Punching shear check for the column and the heaviest loaded piles attheir perimeters in accordance with Cl. 6.1.5.6 of the Code :31.25×50000×10Column : = 4. 28 MPa < 0 .8 fcu= 4. 73 MPa.4×2000×182531.25×10315×10Pile P3 : = 1. 12 MPa < 0 .8 fcu= 4. 73 MPa.2000π× 1825Not necessary to check punching shear at the next critical perimetersas the piles and column overlap with each other to very appreciableextents;(viii) Checking for torsion : There are unbalanced torsions in any full widthsections at X-Y directions due to differences in the pile reactions.However, as discussed in sub-section 11.2(vii) of this Manual forfooting, it may not be necessary to design the torsion as for that forbeams. Anyhow, the net torsion is this example is small, being361 .11× 3 = 1083.33 kNm (361.11kN is the difference in pile loadsbetween P3 and P4), creating torsional shear stress in the order of149

v t=h2min2T⎛ h⎜hmax−⎝ 3minVersion 2.3 May 200862×1083.33×10== 0.065 N/mm 2 . So the⎞2⎛2000 ⎞⎟ 2000 ⎜9000− ⎟⎠ ⎝ 3 ⎠torsional shear effects should be negligible;(ix) Finally reinforcement details are as shown in Figure 12.3,1500300030001500P1 P2 P3T40-200 (T2, B2,B4)T40-200 (T1,B1,B3)P4 P5 P6YXShear linksT12 – 200 BWs onthe whole cap1500 40004000 150012.3 Strut-and-Tie ModelFigure 12.3 – Reinforcement Design of Worked Example 12.1Cl. 6.7.3.1 of the Code allows pile cap be designed by the truss analogy, ormore commonly known as “Strut-and-Tie Model” (S&T Model) in which aconcrete structure is divided into a series of struts and ties which arebeam-like members along which the stress are anticipated to follow. In aS&T model, a strut is a compression member whose strength is provided byconcrete compression and a tie is a tension member whose strength isprovided by added reinforcements. In the analysis of a S&T model, thefollowing basic requirements must be met (Re ACI Code 2002):(i) Equilibrium must be achieved;(ii) The strength of a strut or a tie member must exceed the stress inducedon it;(iii) Strut members cannot cross each other while a tie member can crossanother tie member;(iv) The smallest angle between a tie and a strut joined at a node shouldexceed 25 o .150

Version 2.3 May 2008The Code has specified the following requirements in Cl. 6.7.3.1 :(i) Truss be of triangular shape;(ii) Nodes be at centre of loads and reinforcements;(iii) For widely spaced piles (pile spacing exceeding 3 times the pilediameter), only the reinforcements within 1.5 times the pile diameterfrom the centre of pile can be considered to constitute a tensionmember of the truss.12.4 Worked Example 12.2 (Strut-and-Tie Model)Consider the pile cap supporting a column factored load of 6000kNsupported by two piles with a column of size 1m by 1 m. The dimension ofthe cap is as shown in Figure 12.4, with the width of cap equal to 1.5 m.6000kN2500Elevation1000dia.3000 30001000dia.1500PlanFigure 12.4 – Pile Cap Layout of Worked Example 12.2(i)Determine the dimension of the strut-and-tie modelAssume two layers of steel at the bottom of the cap, the centroid of thetwo layers is at 75 + 40 + 20 = 135 mm from the base of the cap. So the151

Version 2.3 May 2008effective width of the tension tie is 135 × 2 = 270 mm. The dimensionsand arrangement of the ties and struts are drawn in Figure 12.5.(ii)A simple force polygon is drawn and the compression in the strut can besimply worked out as (C is the compression of the strut) :2Csin38.250= 6000 ⇒ C = 4845.8 kN;And the tension in the bottom tie is T = C cos38.25 0 = 3805. 49 kN.1000top strutwidth =619.09mm6000kNconcretestrut22302500270Bottom strutwidth =831.13mm1000dia.bottom tie, strengthbe provided by steel3000 3000Elevation1000dia.6000 kN3000 kNC = 4845.8kNC = 4845.8kN2230+38.25 o T = 3805.49kN38.25 o270÷2=23653000 kN30003000Figure 12.5 – Analysis of strut and tie forces in Worked Example 12.2(iii) To provide the bottom tension of 3805.49 kN, the reinforcement steel333805.49×10 3805.49×10required is == 9509 mm 2 . Use 8–T40;0.87 0.87×460(iv) Check stresses in the struts :f y152

Version 2.3 May 2008Bottom section of the strut, the strut width at bottom is001000sin38.25 + 270cos38.25 = 831.13mmAs the bottom part is in tension, there is a reduction of compressivestrength of concrete to 1 .8 f = 1.8 35 = 10. 08 MPa as suggested bycuOAP, which is an implied value of the ultimate concrete shear strength of0 .8 as stated in the Code and BS8110.f cuAs a conservative approach, assuming a circular section at the base ofthe strut since the pile is circular, the stress at the base of the strut is4845.8×102831 π / 43= 8.93 MPa < 10.08MPaFor the top section of the strut, the sectional width is2× 500sin38.250 = 619.09 mmAs the sectional length of the column is 1 m, it is conservative to assumea sectional area of 1000mm × 619.09 mm.The compressive stress of the strut at top section is34845.8×10= 7.83 MPa < 0 .45 fcu= 15. 75MPa1000×619.09(v)The reinforcement details are indicated in Figure 12.6. Side bars areomitted for clarity.10004T25 T1T16 s.s. – 20025004T40 B1 & 4T40 B21000dia.3000 3000Elevation1000dia.Figure 12.6 – Reinforcement Details of Worked Example 12.2153

Version 2.3 May 200812.5 Flexible Cap AnalysisA pile cap can be analyzed by treating it as a flexible structure, i.e., as incontrast to the rigid cap assumption in which the cap is a perfectly rigid bodyundergoing rigid body movement only with no deformation upon theapplication of loads, the flexible pile cap structure will deform and thedeformations will affect the distribution of internal forces of the structure andthe reactions. Analysis of the flexible cap structure will require input of thestiffness of the structure which is comparatively easy. However, as similar tothat of footing, the support stiffness of the pile cap which is mainly offeredby the supporting pile is often difficult, especially for the friction pile whichwill interact significantly with each other through the embedding soil. Effectsby soil restraints on the piles can be considered as less significant inend-bearing piles such large diameter bored piles.Similar to the flexible footing, as the out-of-plane loads and deformation aremost important in pile cap structures, most of the flexible cap structures aremodeled as plate structures and analyzed by the finite element method.12.6 Analysis and Design by Computer MethodAnalysis and design by computer method for pile cap are similar to Section11.3 for footing. Nevertheless, as analysis by computer methods can oftenaccount for load distribution within the pile cap structure, Cl. 6.7.3.3 of theCode has specified the followings which are particularly applicable for pilecap design :(i)shear strength enhancement of concrete may be applied to a width of3 φ for circular pile, or pile width plus 2 × least dimension of pile asshown in Figure 12.7 as shear distribution across section has generallybeen considered in flexible cap analysis;154

Version 2.3 May 2008φφφArea where shearenhancement mayapplyBBBa va vava vFigure 12.7 – Effective width for shear enhancement in pile cap around a pile(ii) averaging of shear force shall not be based on a width > the effectivedepth on either side of the centre of a pile, or as limited by the actualdimension of the cap.d : effectivedepth of capXφdWidth over whichthe shear force canbe averaged in thecap for designX≤ dpeak shear atpile centreShear force diagram along X-XFigure 12.8 – Width in cap over which shear force at pile can beaveraged for DesignIllustration in Figure 12.8 can be a guideline for determination of“effective widths” adopted in averaging “peak stresses” as will oftenbe encountered in finite element analysis for pile cap structure modeledas an assembly of plate bending elements under point loads and pointsupports, as in the same manner as that for footing discussed in 11.4(ii)of this Manual.155

Version 2.3 May 200813.0 General Detailings13.1 In this section, the provisions of detailing requirements are general onesapplicable to all types of structural members. They are mainly taken fromSection 8 of the Code. Requirements marked with (D) are ductility ones forbeams and columns contributing in lateral load resisting system.13.2 Minimum spacing of reinforcements (Cl. 8.2 of the Code) – clear distance(horizontal and vertical) is the greatest of(i) maximum bar diameter;(ii) maximum aggregate size (h agg ) + 5 mm;(iii) 20 mm.13.3 Permissible bent radii of bars. The purpose of requiring minimum bend radiifor bars are(i) avoid damage of bar;(ii) avoid overstress by bearing on concrete in the bend.Table 8.2 of the Code requires the minimum bend radii to be 3 φ forφ ≤ 20 mm and 4 φ for φ > 20mm (for both mild steel and high yield bar)and can be adopted without causing concrete failures if any of the conditionsshown in Figure 13.1 is satisfied as per Cl. 8.3 of the Code.TL2≥8∅ or D/2Bar ofdiameter ∅Bar ofdiameter ∅4∅TL14∅Point beyond which barassumed not be stressedat ultimate limit statecondition (a)condition (b)TL1 ≥ requiredanchorage length forbeam contributing tolateral load resistingsystem;TL2 ≥ requiredanchorage length forbeam notcontributing tolateral load resistingcross bar of diameter≥ ∅ inside the bendBar ofdiameter ∅condition (c)Figure 13.1 – Conditions by which concrete failure be avoided by bend of bars156

Version 2.3 May 2008If the none of the conditions in Figure 13.1 is fulfilled, (Ceqn 8.1) of the Code,reproduced as (Eqn 13.1) in this Manual should be checked to ensure thatbearing pressure inside the bend is not excessive.bearing stress =Fbt2 fcu≤rφ⎛ φ ⎞⎜1+ 2⎟⎝ ab⎠(Eqn 13.1)In (Eqn 13.1), Fbtis the tensile force in the bar at the start of the bend; rthe internal bend radius of the bar; φ is the bar diameter, a bis centre tocentre distance between bars perpendicular to the plane of the bend and in casethe bars are adjacent to the face of the member, ab= φ + cover.Take an example of a layer of T40 bars of centre to centre separation of 100mm and internal bend radii of 160mm in grade 35 concrete.F = 0 .87×460×1257 = 503051NbtF bt 503051 2 f= =cu 2×3578.6 > == 38. 89rφ 160×40 ⎛ φ ⎞ ⎛ 40 ⎞1 2 ⎜1+2×⎜ +⎟⎟⎝ a ⎠ ⎝ 100b⎠So (Ceqn 8.1) is not fulfilled. Practically a cross bar should be added as inFigure 13.1(c) as conditions in Figure 13.1(a) and 13.1(b) can unlikely besatisfied.13.4 Anchorage of longitudinal reinforcements(i)(ii)Anchorage is derived from ultimate anchorage bond stress with concreteassessed by the (Ceqn 8.3) of the Code.fbu= β f cuwhere for high yield bars β = 0. 5 for tension andβ = 0.65 for compression. For example, fbu= 0 .5 35 = 2. 96 MPa forgrade 35. For a bar of diameter φ , the total force up to 0 .87 fy⎛φ 2 πis⎟ ⎞0.87 f⎜y. The required bond length L will then be related by⎝ 4 ⎠2⎛φπ ⎞0.87 fyφ0.87 f⎜ = β πφ ⇒ = = 33.8φ≈ 34φ4⎟yfcuL Lwhich agrees⎝ ⎠4βfcuwith Table 8.5 of the Code;Notwithstanding provision in (i), it has been stated in 9.9.1.1(c) of theCode which contains ductility requirements for longitudinal bars ofbeams (contributing in lateral load resisting system) anchoring into157

Version 2.3 May 2008exterior column requiring anchorage length to be increased by 15% asdiscussed in Section 3.6 (v); (D)(iii) With the minimum support width requirements as stated in Cl. 8.4.8 ofthe Code, bends of bars in end supports of slabs or beams will startbeyond the centre line of supports offered by beams, columns and walls.By the same clause the requirement can be considered as not confiningto simply supported beam as stated in Cl. 9.2.1.7 of the Code asillustrated in Figure 13.2.≥2(4Ø+c) if Ø ≤ 20≥2(5Ø+c) if Ø > 20c3Ø if Ø ≤ 20;4Ø if Ø > 20≥0centre line of supportFigure 13.2 – Support width requirement13.5 Anchorage of links – Figure 8.2 of the Code displays bend of links of bendangles from 90 o to 180 o . However, it should be noted that the Code requiresanchorage links in beams and columns contributing in lateral load resistingsystem to have bent angles not less than 135 o as ductility requirements (D);13.6 Laps arrangement – Cl. 8.7.2 of the Code requires laps be “normally”staggered with the followings requirement for 100% lapping in one singlelayer:(i) Sum of reinforcement sizes in a particular layer must not exceed 40% ofthe breadth of the section at that level, otherwise the laps must bestaggered;(ii) Laps be arranged symmetrically;(iii) Details of requirements in bar lapping are indicated in Figure 8.4 of theCode reproduced in Figure 13.3 for ease of reference;158

Version 2.3 May 20080≤50mm≤4øa : distance betweenadjacent lapsIf clear distancebetween 2 lappingbars > 4ø or 50mm by x, l 0should beincreased by x≥20mm≥2øFigure 13.3 – Lapping arrangement for tension laps(iv) When Figure 13.3 is complied with, the permissible percentage of lappedbars in tension may be 100% (but still required to be staggered, i.e. notin the same section)where the bars are all in one layer. When the bars arein several layers, the percentage should be reduced to 50%;(v) Compression and secondary reinforcements can be lapped in one section.The Clause effectively requires tension laps to be staggered with arrangementas shown in Figure 13.3 which is applicable in to the flexural steel bars inbeams, slabs, footings, pile caps etc. Fortunately, the Code allows compressionand secondary bars be lapped in one section, i.e. without the necessity ofstaggered laps. As such staggered laps can be eliminated in most of thelocations in columns and walls.13.7 Lap Lengths (Cl. 8.7.3 of the Code)The followings should be noted for tension lap lengths:(i) Absolute minimum lap length is the greater of 15 φ and 300 mm;(ii) Tension lap length should be at least equal to the design tensionanchorage length and be based on the diameter of the smaller bar;(iii) Lap length be increased by a factor of 1.4 or 2.0 as indicated in Figure13.4 which is reproduced from Figure 8.5 of the Code.159

Version 2.3 May 2008Top barsBottom bars< 2φ< 2φ< 75or 6 φ< 75or 6 φ< 2φ≥ 75and 6 φ≥ 75and 6φ≥ 75 ≥ 75and 6 φ and 6 φ > 2φFactor 1.4 1.4 1.0 1.4≥ 2φ≥ 75and 6 φFactor 2 2 1.4 1.0 1.4< 2φ< 2φNoteCondition 1:Lap at top as castand cover < 2φ;Condition 2 :Lap at corner andcover < 2φCondition 3 :Clear distancebetween adjacentlaps < 75 or 6 φAny one of the 3conditions, factoris 1.4.Condition 1 + 2 orconditions 1 + 3 :factor is 2.0Figure 13.4 – Factors for tension lapping barsThe compression lap length should be at least 25% greater than the designcompression anchorage length as listed in Table 8.4 of the Code.13.8 Transverse reinforcement in the tension lap zone (Cl. 8.7.4 of the Code)For lapped longitudinal bars in tension, the transverse reinforcement is used toresist transverse tension forces. 3 cases be considered as :(i) No additional transverse reinforcement is required (existing transversereinforcement for other purpose can be regarded as sufficient to resistthe transverse tension forces) when the longitudinal bar diameterφ < 20 mm or percentage of lapping in any section < 25%;(ii) When φ ≥ 20 mm, the transverse reinforcement should have area∑ A st≥ A swhere sA is the area of one spiced bar and be placedbetween the longitudinal bar and the concrete surface as shown inFigure 13.5;160

Version 2.3 May 2008≤ 4φ≤ 4φFor T40 bars in tension lap in concretegrade 35 with spacing 200 mm (≤ 10ø= 400mm) and lap lengthl 0 = 1.4×standard lap = 1920 mm.Transverse reinforcement arearequired is A = 1257 mm 2 .∑ stl 0One bar be outside lap ifthe lap is in compressionUse 12T12, spacing is 175 mm alongthe lapped length.Transverse bars betweenlongitudinal bars andconcrete surfaceFigure 13.5 – Transverse reinforcement for lapped splices –not greater than 50% ofreinforcement is lapped at one section and φ ≥ 20 mm(iii)If more than 50% of the reinforcement is lapped at one point and thedistance between adjacent laps ≤ 10φ, the transverse reinforcementshould be formed by links or U bars anchored into the body of thesection. The transverse reinforcement should be positioned at the outersections of the lap as shown in Figure 13.6;It should be noted that effectively condition (ii) requires area of transversereinforcement identical to that of (iii), except that the bars need not beconcentrated at the ends of the laps and the transverse reinforcements be inform of links or U bars.13.9 Transverse reinforcement in the permanent compression lap zoneThe requirement will be identical to that of tension lap except for an additionalrequirement that one bar of the transverse reinforcement should be placedoutside each of the lap length and within 4 φ of the ends of the lap length alsoshown in Figure 13.5 and 13.6.161

Version 2.3 May 2008∑ A st / 2l 0 /3∑ A st / 2l 0 /3≤150mmFor T40 bars in tension lap inconcrete grade 35 with spacing 200mm with lap lengthl 0 = 1.4×standard lap = 1920 mm.l 0 /3=640mm≤ 4φl 0≤ 4φTransverse reinforcement arearequired is A = 1257 mm 2 .∑ stFor tension lap, on eachl 0 /3=640mm, 1257/2 = 629mm 2 isrequired. So use 6T12, areaprovided is 678mm 2 over1950/3=640 mm zone, i.e. spacingis 128mm < 150mm.One bar be outside lap if thelap is in compressionTransverse bars inform of U bars orlinksFor compression lap, also use 7T10,(628mm 2 =1257/2) with 6T12within 640mm (equal spacing =128mm) and the 7 th T10 at 160mm(=4∅) from the end of lap.Lapping longitudinal barsFigure 13.6 – Transverse reinforcement for lapped splices – more than 50% islapped at one section and clear distance between adjacent laps ≤ 10∅162

Version 2.3 May 200814.0 Design against Robustness14.1 The Code defines the requirement for robustness in Clause 2.1.4 as “astructure should be designed and constructed so that it is inherently robust andnot unreasonably susceptible to the effects of accidents or misuse, anddisproportionate collapse.” By disproportionate collapse, we refer to thesituation in which damage to small areas of a structure or failure of singleelements may lead to collapse of large parts of the structure.14.2 Design requirements comprise :(i)(ii)building layouts checked to avoid inherent weakness;capable to resist notional loads simultaneously at floor levels and roof asshown in Figure 14.1. (Re Cl. 2.3.1.4(a) of the Code which also requiresthat applied ultimate wind loads should be greater than these notionalvalues);0.015W RW R , characteristicdead weight between0.015W N+10.015W N0.015W 1}}}}roof and next midfloor heightsW N+1 , characteristicdead weight betweenmid floor heightsW N , characteristic deadweight between midfloor heightsW 1 , characteristic deadweight between midfloor heightsFigure 14.1 – Illustration of notional loads for robustness design(iii) provides effective horizontal ties (in form of reinforcements embeddedin concrete) (a) around the periphery; (b) internally; (c) to externalcolumns and walls; and (d) vertical ties as per Cl. 6.4.1 of the Code, thefailure of which will lead to requirement of checking key elements inaccordance with Cl. 2.2.2.3 of the Code.14.3 Principles in Design of ties (Cl. 6.4.1.2 and Cl 6.4.1.3 of the Code)163

Version 2.3 May 2008(i) The reinforcements are assumed to be acting at fyinstead of0 .87 fy;(ii) To resist only the tying forces specified, not any others;(iii) Reinforcements provided for other purpose can also act as ties;(iv) Laps and anchorage of bars as ties similar to other reinforcements;(v) Independent sections of a building divided by expansion joints haveappropriate tying system.14.4 Design of ties(i) Internal Ties be provided evenly distributed in two directions in slabs –design force is illustrated in Figure 14.2 with example. The tiereinforcements can be grouped and provided in beam or wall.l ry1l ry2l rx1T.A.L.l rx1 =6m, l rx2 =4m,l rx ≥ l rx1 and l rx2 ;l rx =6m;l ry1 =l rx1 =6ml ry ≥ l ry1 and l ry2l ry =6mB1l rx2Design force (kN/m)Gk+ Q lk ry≥Ftand 1 .0Ft7.5 5where Ft≤ 60 and 20 + 4n0and n 0 is the no. of storeys40 storeys building with averagefloor dead load as 9 kPa and averagefloor live load as 3 kPaInternal ties in Y-directionn 0 = 40G = 9 kPa; Q = 3 kPalkry= 6 mF ≤ 60 and 20 + 4n0 = 180t∴ F t= 60kGk+ QklrFt= 115.27.5 51 .0Ft = 60Design tensile force in ties is115.2 kN/mRequired A for internal ties isFf yst115.2 × 10=4603= 250 mm 2 /mUse T10 – 300(can likely be met by DB or rebarsprovided for strength purpose)If tying bars grouped in beam, totalrebars in the middle beam B1 maybe 250 × 5 = 1250 mm 2 . So likelycan be met by the longitudinal bars.Figure 14.2 – Derivation of internal tie reinforcement bars in slabs (evenly distributed)164

Version 2.3 May 2008(ii)Peripheral ties – Continuous tie capable resisting 1.0F t , located within1.2 m of the edge of the building or within perimeter wall;perimeter wall1.2mPeripheral area of 1.2 m widewith ties to resist 1.0F t , i.e.the lesser of 60kN/m and( 20 + 4n0) kN/m; or withinperimeter wallFigure 14.3 – Location and determination of Peripheral ties(iii) External columns and wall to have ties capable of developing forces asindicated in Figure 14.4;Ties to take uptensile force beingthe greater of(i) 2.0F t or (l s /2.5)F tif less; and (ii) 3%of design ultimateload of column orwall where l s is floorto ceiling heightCorner column in 40 storeybuilding, tie design force >(i) 2.0F t =2×60=120kN;(l s /2.5)F t=(3/2.5)×60=72kN;Smaller is 72kN(ii) 400×600, grade 40 with8T32 steel, designultimate load is(400×600×0.45×40+0.87×460×6434)×10 -3=6895kN; 3% is 207kNSo design tie force is 207kN.Provide F/f y =207×10 3 /460 = 517mm 2 ; canlikely be provided by beamsteel anchoring into thecolumn.Figure 14.4 – Ties to external column and wall with example(iv) Vertical ties provided to wall and column should be continuous and becapable of carrying exceptional load. Use γ f ×[dead load + 1/3 imposedload + 1/3 wind load] of one floor to determine the design load for thevertical ties where γ f = 1.05.165

Version 2.3 May 20087m8m8mC16mDesign load for thevertical tie of the interiorcolumn C1 is(i) Dead load from onestorey 800kN;(ii) One third ofimposed load fromone storey 1/3×240= 80 kN;(iii) One third of windload = 1/3×90=30kNDesigned tensile load is1.05(800+80+30)= 956kN.Requiring 956000/460= 2078mm 2Figure 14.5 – Design of Vertical Ties in Columns and Walls14.5 Design of “Key Elements”By Cl. 2.2.2.3 of the Code, when for some reasons it is not possible tointroduce ties, key elements (usually columns or walls), the failure of whichwill cause disproportionate collapse should be identified. If layout cannot berevised to avoid them, design these elements and the supporting buildingcomponents to an ultimate load of 34 kN/m 2 , from any direction, to which nopartial safety factor shall be applied. The Code has not defined the extent of“disproportionate collapse” for the element to be qualified as a “key element”.However, reference can be made to the “Code of Practice for the StructuralUse of Steel 2005” Cl. 2.3.4.3 by which an element will be considered a keyelement if the removal of it will cause collapse of 15% of the floor area or70m 2 , whichever is the greater. The design is illustrated in Figure 14.6.166

Version 2.3 May 20087m8m8mB1C16m2mFigure 14.5 – Design of Key ElementsExamplesC1 is identified as the keyelement as the collapse ofwhich will lead todisproportionate collapseof area around it (morethan 15% collapse of thefloor).The tributary area is7.5×7=52.5m 2 ;The design load is34×52.5=1785kN.Similarly, beam B1 is alsorequired as the removal ofwhich will cause more than15% collapse of the floor.So B1 needs be designedfor a u.d.l. of 34kPa on thelinking slabs.14.6 Nevertheless, it should be noted that requirements in the Code for robustnessdesign often poses no additional requirements in monolithic reinforcedconcrete design in comparison with the criteria listed in 14.2 :(i) normally no inherent weakness in the structure for a reasonablestructural layout;(ii) ultimate wind loads normally applied to the structure according to thelocal Wind Code can usually cover the notional loads (1.5%characteristic dead weight) specified in 14.2(ii);(iii) requirements for various types of ties can normally be met by thereinforcements provided for other purposes. Nevertheless, continuity ofthe ties should be checked.167

Version 2.3 May 200815.0 Shrinkage and Creep15.1 ShrinkageShrinkage is the shortening movement of concrete as it dries after hardening.If the movement is restrained, stress and/or cracking will be created.(Ceqn 3.5) gives estimate of drying shrinkage of plain concrete underun-restrained conditions. Together with the incorporation of the“reinforcement coefficient” K , the equation can be written asssLcejssε = c K K K K K(Eqn 15-1)where c = 3. 0 and other coefficients can be found by Figures 3.3, 3.5, 3.6sand 3.7 and (Ceqn 3.4) of the Code, depending on atmospheric humidity,dimensions, compositions of the concrete, time and reinforcement content. Itshould be noted thatKjis a time dependent coefficient.The equation and the figures giving values of the various coefficients areadopted from BS5400 which in turn are quoted from CEB-FIP InternationalRecommendation for the Design and Construction, 1970 (CEB 1970) (MC-70).It should, however, be noted that the coefficient c is extra to CEB-FIP. Thevalue accounts for the comparatively higher shrinkage value (3 times as high)found in Hong Kong.sShrinkage is always in contraction.15.2 CreepCreep is the prolonged deformation of the structure under sustained stress.(Ceqn 3.2) and (Ceqn 3.3) give estimate of the creep strain :stressCreep strain = ×φc(Eqn 15-2)E28whereφ = K K K K K K(Eqn 15-3)cLmcejsAgain (Eqn 15-3) has incorporated the reinforcement coefficientKs.Thus creep strain depends on the stress in the concrete and various coefficientsrelated to parameters similar to that of shrinkage (which can be read from168

Version 2.3 May 2008Figures 3.1 to 3.5 of the Code) withKmandKjdependent on time. Asstress and strain are inter-dependent, it will be shown that assessment of strainwill require successive time staging in some cases.Creep creates deformation in the direction of the stress. In case shrinkagewhich results in tensile stress under restrained condition such as a floorstructure under lateral restraints, the creep strain will serve to relax the stressdue to shrinkage. Both stress and strain due to shrinkage and creep vary withtime, as can be shown in the analyses that follow.15.3 The determination of the time dependent coefficients KmandKjas listedin (Ceqn 3.3) and (Ceqn 3.5) will be tedious in calculation of stress and strainof a structure in a specified time step which may involve reading the figuresmany times. Curves in Figures 3.2 and 3.5 are therefore simulated bypolynomial equations as shown in Appendix J to facilitate determination of thecoefficients by spreadsheets.15.4 Worked Example 15.1A grade 35 square column of size 800 × 800 in a 4 storey building withreinforcement ratio 2% is under an axial stress from the floors as follows :Floor Height (m) Time of stress creation from floor (days) Stress (MPa)G 4 28 3.51 st 3 56 2.12 nd 3 84 2.13 rd 3 120 3.5Strain and shortening of the G/F column due to shrinkage and creep at 360days are determined as follows :ShrinkageThe coefficients for determination of the free shrinkage strain are as follows :K = 0.000275 for normal air from Figure 3.6;LBased on empirical formulae, for grade 35:Water / Cement ratio = − 0 .0054 f + 0.662 = −0.0054×35 + 0.662 = 0. 473cu169

Cement content = 3 .6 f + 308 = 3.6×35 + 308 = 434 kg/m 3From Figure 3.3 K =1. 17 ;ccuVersion 2.3 May 2008For the 800 × 800 column, the effective thicknesshe, defined as the ratio ofthe area of the section A, to the semi-perimeter, u/2 (defined in Cl. 3.1.7 of the800×800Code) is = 400 mm. So from Figure 3.7, Ke= 0. 55 ;800×4/ 2From Figure 3.5, time at 360 days K = 0. 51;1200K s= = 0.856 ; where ρ = 0. 02 (2% steel) and αe = = 8. 441+ραe23.7So the shrinkage strain under perfectly free condition is :jεs= csKLKcKeKjKs= 3.0×0.000275×1.17×0.55×0.51×0.856 = 231.76×10−6CreepFor estimation of creep strain, Creep strainstressεc= ×φcE28whereφ = KcLKmKcKeKjKsE28 = 23.7 GPa for grade 35 concrete.All coefficients are same as that for shrinkage except K = 2. 3 (Figure 3.1),K = 1.0 (Figure 3.2 – loaded at 28 days) and K = 0. 72 (Figure 3.4)meLLoadfromFloorConcreteage at timeof load(Day)K mTime sinceLoading(Day)K jφ cStress byFloor(MPa)ε c(×10 -6 )1/F 28 1 332 0.489 0.811 3.5 119.732/F 56 0.85 304 0.467 0.658 2.1 58.303/F 84 0.761 276 0.443 0.560 2.1 49.61Roof 120 0.706 240 0.412 0.482 3.5 71.15∑ ε c = 298.80So the creep strain at 360 days isεc= 298.80×10−6Elastic strainThe elastic strain is simplySo the total strain isσεe= = E3.5+ 2.1+2.1+3.5−623700= 472.57×10170

Version 2.3 May 2008−6−6−6−6ε = εs+ εc+ εe= 231.76×10 + 298.80×10 + 472.57×10 = 1003.13×10 .Total shortening of the column at G/F is at 360 days isε × H = 1003.13×10−6× 4000 = 4.01mm.15.5 Estimation of shrinkage and creep effect on restrained floor structureIt is well known that shrinkage and creep effects of long concrete floorstructures can be significant. The following derivations aim at providing adesign approach to account for such effects based on recommendations by theCode.Consider a floor structure spanning on vertical members of lateral supportstiffness Ksup1and Ksup 2as shown in Figure 15.1.Let the lateral deflections at supports 1 and 2 be δ 1and δ 2. At any timewhen the floor structure has an internal stress σ , a free shrinkage strain εs,creep strain εc, elastic strain εe, internal force, by displacementcompatibility, the followings can be formulated :σεc=cE φ σσAσA, εe= , Ksup1δ 1= Ksup 2δ2= F = σA⇒ δ1= ; δ2=EK K( εs− εc− εe) L = δ 1+ δ2sup1sup 2⎛ σ σ ⎞⇒ ⎜εs− φc− ⎟L= δ1+ δ2⎝ E E ⎠=σAKsup1σA+Ksup 2⇒ σ =1+φ +cAELEεs⎛⎜1⎝ Ksup11+Ksup 2⎞⎟⎠(Eqn 15-4)If the floor structure undergoes no net lateral deflection at a point P at L efrom support 2, it can be visualized as if the floor structure is divided into 2floor structures both fixed at P and undergoes deflection δ1at the left portionand δ2at the right portion. By constant strain (implying linearly varyingdisplacement) in the floor structure :L eδ L= L = δ + δ K12sup 2⎛⎜1⎝ Ksup1+K21sup 2⎞⎟⎠(Eqn 15-5)Substituting (Eqn 15-5) into (Eqn 15-4)171

sup 2sup 2Version 2.3 May 2008EεsEεsσ ==(Eqn 15-6)AEKb1+φc+ 1+φc+L KKeAEwhere Kb= , the equivalent axial stiffness of the floor.LeSo, as an alternative to using (Eqn 15-4), we may use (Eqn 15-5) to find outL and (Eqn 15-6) to calculate internal stress of the floor structure.eFloor structure of cross sectional area A and axial stiffness K bSupporting membersproviding lateralrestraints of stiffnessK sup1K sup1PPLK bSupporting memberproviding lateralrestraints of stiffnessK sup2idealized asK sup2HL e∆δL e=LKsup 2⎛⎜1⎝ Ksup11+Ksup 2⎞⎟⎠K bK sup∆ε sFigure 15.1 – Idealization of floor structure for shrinkage and creep estimationIn the determination of stress due to shrinkage and creep, the main difficultylies in the determination of φ which is time dependent. Stress in concrete hasctherefore to be determined in successive time steps and with numerical methodas demonstrated in Figure 15.2 for calculation of the creep strains. Instead ofL e172

Version 2.3 May 2008being treated as continuously increasing, the stress is split up into variousdiscrete values, each of which commences at pre-determined station of times.Fine divisions of time steps will create good simulation of the actualperformance.∆σ 2∆ σ 1∆σ 1t 12t 1(a) at t = t1– constant stress at ∆ σ 1from t 1 /2 to t 1 .t 12t 1 + t 22t 2(b) at t = t2– constant stress at ∆ σ 1from t 1 /2 to t 2 + constant stress ∆ σ 2from (t 1 + t 2 )/2 to t 2∆σ n∆ σ 3∆ σ 2∆σ 3∆σ 2∆ σ 1∆σ 1t 12t 1 + t 22t 2 + t 32t 3(c) at t = t3– constant stress at ∆ σ 1from t 1 /2 to t 3 + ∆ σ from (t21 + t 2 )/2 to t 3+ ∆ σ from (t32 + t 3 )/2 to t 3t 12t 1 + t 22t 2 + t 32t n− 1 + tnt n2(d) at t = tn– similarly adding up effectsof all stress incrementsFigure 15.2 – Estimation of Creep Strains by Numerical MethodConsider the floor structure shrinks for εs1at the time interval from timet = 0 to t = t , 1φc= KLKmKcKeKjKsshould be determined at concrete aget 1(for determination of2determination of⎛ t1⎞φc ⎜t1− ⎟⎝ 2 ⎠15-2(a). So by (Eqn 15-6)t1Km) and with the time since loading 2K j) which is the φ cvalue forto t 1(fort1t1− and denoted by2. The timing for determination of φ cis illustrated in Figure173

Version 2.3 May 2008Eεs1∆σ 1=(Eqn 15-7)Kb⎛ t1⎞1++ φc⎜t1− ⎟K ⎝ 2 ⎠sup 2At time t2after shrinkage commencement when the shrinkage strain is εs2,the creep strain ε c2can be regarded as made up of two time steps withstresses ∆ σ1and ∆ σ2(increment of concrete stress between time t 1and∆σ1 ⎛ t1⎞ ∆σ2 ⎛ t1+ t2⎞t2) as εc2= φc⎜t2− ⎟ + φc⎜t2− ⎟ as illustrated in FigureE ⎝ 2 ⎠ E ⎝ 2 ⎠15-2(b). So, similar to the above, we can listεs2⎡∆σ1 ⎛− ⎢ φc⎜t⎣ E ⎝2t1−2⎞⎟ +⎠∆σ2E⎛ t1+ tφc⎜t2−⎝ 22⎞⎤∆σ1+ ∆σ2⎟⎥− =⎠⎦E( ∆σ+ ∆σ)1KsupL2A⎡ ⎛ t1⎞ EA ⎤ ⎡ ⎛ t1+ t2⎞ EA ⎤⇒ ∆σ 1⎢φc ⎜t2− ⎟ + 1+⎥ + ∆σ2 ⎢φc⎜t2− ⎟ + 1+⎥ = Eε⎢⎣⎝ 2 ⎠ Ksup 2Le⎥⎦⎢⎣⎝ 2 ⎠ Ksup 2Le⎥⎦s2(Eqn 15-8)⎡ ⎛ t⎤1 ⎞ KbEεs2− ∆σ1⎢φc ⎜t2− ⎟ + 1+⎥⎢⎣⎝ 2 ⎠ Ksup 2 ⎥⇒ ∆σ =⎦2(Eqn 15-9)⎡ ⎛ t + ⎞ ⎤1t2Kb⎢φc ⎜t2− ⎟ + 1+⎥⎢⎣⎝ 2 ⎠ Ksup 2 ⎥⎦∆ σ 2can be determined with pre-determination of ∆ σ1by (Eqn 15-7)Similarly for time t3with 3 time steps where∆σ1 ⎛ t1⎞ ∆σ2 ⎛ t1+ t2⎞ ∆σ3 ⎛ t2+ t3⎞εc3= φc⎜t3− ⎟ + φc⎜t3− ⎟ + φc⎜t3− ⎟E ⎝ 2 ⎠ E ⎝ 2 ⎠ E ⎝ 2 ⎠⎡ ⎛∴∆σ⎢φc ⎜t⎢⎣⎝t⎤ ⎡1 ⎞ Kb⎛ t1+ t2⎞ K− ⎟ + 1+⎥ + ∆σ2 ⎢φc⎜t3− ⎟ + +2 ⎠ Ksup 2 ⎥⎦⎢⎣⎝ 2 ⎠ K1 31bsup 2⎤⎥⎥⎦⎡ ⎛ t2+ t3⎞ K ⎤b+ ∆σ 3 ⎢φc⎜t3− ⎟ + 1+⎥ = Eεs3(Eqn 15-10)⎢⎣⎝ 2 ⎠ Ksup 2 ⎥⎦Eεs∆σ=33⎡ ⎛− ∆σ1⎢φc ⎜t⎢⎣⎝3t⎤ ⎡1 ⎞ Kb⎛ t1+ t− ⎟ + 1+⎥ − ∆σ2 ⎢φc ⎜t3−2 ⎠ Ksup 2 ⎥⎦⎢⎣⎝ 2⎡ ⎛ t + ⎞ ⎤2t3Kb⎢φc ⎜t3− ⎟ + 1+⎥⎢⎣⎝ 2 ⎠ Ksup 2 ⎥⎦So for any time tnafter shrinkage commencement2⎞ K⎟ + 1+⎠ Kbsup 2(Eqn 15-11)⎤⎥⎥⎦174

Version 2.3 May 2008⎡ ⎛∆σ⎢φc ⎜t⎢⎣⎝t⎤ ⎡1 ⎞ Kb⎛ t1+ t2⎞ K− ⎟ + 1+⎥ + ∆σ2 ⎢φc ⎜tn− ⎟ + +2 ⎠ Ksup 2 ⎥⎦⎢⎣⎝ 2 ⎠ K1 n1⎡ ⎛ t⎤⎡2+ t3⎞ Kb⎛ tn−2+ tn1 ⎞ K∆σ3 ⎢φc ⎜tn− ⎟ + 1+⎥ + ..... + ∆σn−1⎢φc ⎜tn− ⎟ + 1+⎢⎣⎝ 2 ⎠ Ksup 2 ⎥⎦⎢⎣⎝ 2 ⎠ K+−bsup 2⎤⎥⎥⎦bsup 2⎤⎥⎥⎦⎡⎛ t + tn−1 nb+ ∆σ n ⎢φc ⎜tn− ⎟ + 1+⎥ =⎝ 2 ⎠ Ksup 2⎢⎣⎞K⎤⎥⎦Eεsn(Eqn 15-12)∆σn=Eεsn⎡ ⎛− ∆σ1⎢φc ⎜t⎢⎣⎝nt⎤⎡1 ⎞ Kb⎛ tn−1+ t− ⎟ + 1+⎥ −.....− ∆σn−1⎢φc ⎜tn−2 ⎠ Ksup 2 ⎥⎦⎢⎣⎝ 2⎡ ⎛ t + ⎞ ⎤n−1tnKb⎢φc ⎜tn− ⎟ + 1+⎥⎢⎣⎝ 2 ⎠ Ksup 2 ⎥⎦n−2(Eqn 15-13)⎞ K⎟ + 1+⎠ Kbsup 2⎤⎥⎥⎦Thus the solution for ∆ σncan be obtained by successive solution of (Eqn15-7), (Eqn 15-9), (Eqn 15-11) and (Eqn 15-13) or alternatively, in a morecompact form by a system of linear simultaneous equations of (Eqn 15-7),(Eqn 15-8), (Eqn 15-10) and (Eqn 15-12). The final stress up to t nis∆ σ + ∆σ+ ∆σ+ ...... + ∆1 2 3σ n.15.6 Worked Example 15.2A wide 200 mm slab of grade 35 is supported by 350×600 beams at spacing of3000mm under restraints at both ends as shown in Figure 15.3. The span of theslab beam structure between restraints is 10 m. The longitudinal steel ratio is0.5%. The free shrinkage strain and the stress developed due to shrinkage at360 days after casting are to be assessed.KK3EI= H3×23.7×10=33× 0.086sup 1=33EI= H3×23.7×10=33× 0.126sup 2=3210667 kN/m316000 kN/mL e=LKsup 2⎛⎜1⎝ Ksup11+Ksup 2⎞⎟ = 4 m⎠Area of a portion between centre line of two adjacent beams is175

A = 350 × 400 + 3000×200 = 740000 mm 2 for 3 m width;Half Perimeter of the portion in contact with the atmosphere is( 3000×2 + 400×2) ÷ 2 = 3400740000So the effective thickness is he= = 218 mm.3400Version 2.3 May 20082003mAll beams are 600(d)×350(w)3m3mCross sectionFloor structureSupporting membersproviding lateralrestraints of stiffnessI value = 0.08m 4 /mwidthSupporting membersproviding lateralrestraints of stiffnessI value = 0.12m 4 /mwidth3mL = 10mFigure 15.3 – Floor structure of Worked Example 15.2Determination of the coefficients for free shrinkage strainε = c K K K K K ,ssLcejsKL= 0.000275 for normal air from Figure 3.6 of the CodeBased on empirical formulae :Water / Cement ratio = − 0 .0054 fcu+ 0.662 = −0.0054×35 + 0.662 = 0. 473Cement content = 3 .6 f + 308 = 3.6×35 + 308 = 434 kg/m 3cuKc=1.17 from Figure 3.3 of the Code;For h = 218 mm thick slab, from Figure 3.7 of the Code, K = 0. 768 ;eKjis time dependent and is to be read from Figure 3.5 of the Code;e176

Version 2.3 May 2008K s1=1+ραe= 0.96from (Ceqn 3.4).Determination of the coefficients for creep strainstressCreep strain = ×φcwhere φc= KLKmKcKeKjKs,E28K = 2.3 for normal air from Figure 3.1 of the Code;LK is time dependent and is to be read from Figure 3.2 of the Code;mKc=1.17 from Figure 3.3 of the Code, same as ShrinkageFor h = 218 mm thick slab, from Figure 3.4 of the Code, K = 0. 831eKjis time dependent and is to be read from Figure 3.5 of the Code.eKs= 0.96 , same as Shrinkage.Stiffness per metre width:6EA 23.7×10 × 0.2467Kb= == 1461500 kN/mL4Ke3EI= H3×23.7×10=33× 0.126sup 2=3316000 kN/m;∴ KK bsup 2= 4.625The time history to 360 days can be divided into various time points, i.e.t = 3days,71t = 2days, t3 = 14 days ….. up to tn= 360 days in accordancewith Figure 15.2. Equations in accordance with the above can be formulatednumerically and stress at 360 days is calculated to be 1726.34 kN/m 2 . As ademonstration, the stress increment in the first two intervals are presented :At t 3, for h = 218 , for shrinkage K = 0. 0637 (Figure 3.5)1 =eεs1 = csKLKcKeKjKs= 3.0×0.000275×1.17×0.768×0.0637×0.96 = 45.22×10K = 1.743 at t 3 and K = 0. 0546 for time interval from 1.5 to 3 days.m1 =⎛ t1⎞ ⎛ t1⎞ ⎛ t1⎞φc ⎜t1− ⎟ = KLKm⎜⎟KcKeKj ⎜t1− ⎟Ks⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠= 2 .3×1.743×1.17×0.831×0.0546×0.96 = 0.2046By (Eqn 15-6)j6−6Eεs123.7×10 × 45.22×10∆σ 1=== 183.84kN/m 2 ;EA ⎛ t1⎞ 1+4.625 + 0.20461++ φc⎜t1− ⎟L K ⎝ 2 ⎠esup 2j−6177

Version 2.3 May 2008At t 7 , for h = 218 , for shrinkage K = 0. 0896 (Figure 3.5)2 =eε2c K K K K K = 3.0×0.000275×1.17×0.768×0.0896×0.96 = 63.85×10s=s L c e j sFor time step 1,K = 1.743 at t 3 and K = 0. 0818 for time interval from 1.5 to 7 days.m1 =⎛ t1⎞ ⎛ t1⎞ ⎛ t1⎞φc ⎜t2− ⎟ = KLKm⎜⎟KcKeKj ⎜t2− ⎟Ks⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠= 2 .3×1.743×1.17×0.831×0.0818×0.96 = 0.306 ;For time step 2,t1 + t2Km=1.4667 at t = = 5 and Kj= 0. 0572 from2t + t23 + 7t =1 = 52 2= to t = 7⎛ t1+ t2⎞ ⎛ t1+ t2⎞ ⎛ t1+ t2⎞∴φc ⎜t2− ⎟ = KLKm⎜⎟KcKeKj ⎜t2− ⎟K⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠= 2 .3×1.4667 × 1.17×0.831×0.0572×0.96 = 0.1800By (Eqn 15-9)⎡ ⎛Eεs2− ∆σ1⎢φc ⎜t⎢⎣⎝∆σ2=⎡ ⎛ t1+ t⎢φc⎜t2−⎢⎣⎝ 223.7×10=6× 63.8522jt1⎞ EA− ⎟ + 1+2 ⎠ LeK⎞ EA ⎤⎟ + 1+⎥⎠ LeKsup 2 ⎥⎦sup 2−6× 10 −183.84×( 0.306 + 1+4.625)( 0.1800 + 1+4.625)j⎤⎥⎥⎦s= 72.85kN/m 2 ;−6So the total stress at t = 27 is∆σ + ∆σ= 183.84 + 72.85 256. 69 kN/m 21 2=The process can be similarly repeated to calculate stress increments at latertimes. As the shrinkage rises rapidly in the beginning and slows down at latertimes, the time stations should be more frequent when t is small and be lessfrequent when t is high.The stress finally arrived at 360 days is 1726.34 kN/m 2 . The exercise stops at360 days because Figure 3.2 of the Code indicates the values of the coefficientK up to 360 days only. The stress induced in the structure is plotted inmFigure 15.4.178

Version 2.3 May 2008Stress due to shrinkage and creep on the slab structure underElastic Restraint of Worked Example 15.2200018001600Stress (kN/m 2 )14001200100080060040020000 50 100 150 200 250 300 350Days after CastFigure 15.4 – Increase of internal stress in concrete due to shrinkage and creep ofWorked Example 15.2By assuming Km= 0. 5 beyond 360 days, the exercise is repeated for variousspan lengths up to 80 m and finally at perfect restraint where the span is set atinfinity. The stress curves are plotted as indicated.Stress due to shrinkage and creep on the slab structure under ElasticRestraint of various spansL=10m L=20m L=40m L=60m L=80m L=infinity60005000Stress (kN/m 2 )400030002000100000 100 200 300 400 500 600 700 800 900 1000Days after CastFigure 15.5 – Increase of internal stress in concrete due to shrinkage and creep ofWorked Example 15.2 for various span lengths179

Version 2.3 May 200815.7 The followings are discussed as revealed from Figure 15.5 :(i) The stress induced increases with time and the increase becomes lesssignificant as time goes by;(ii) The magnitude of the stress increases with decrease of the K b/ Ksupratio. As Kbdecreases with increases of the floor length, longer floorlength will lead to higher stress. So particular attention in relation toshrinkage and creep should be paid to long floor structures;(iii) The particular case of perfect restraint is when K b/ Ksup = 0 , i.e. Ksup(iv)becomes infinity where the floor structure stress becomes maximum;Thicker floor structures are less prone to shrinkage and creep as thecoefficientsKjandK edecrease with increase of thickness;(v)Stronger lateral restraints will also induce higher shrinkage and creepstresses. The strong lateral restraints are often in form of core walls orshear walls whilst the columns are comparatively weak in lateralrestraints. For rough analysis, the columns can be ignored. Figure 15.6demonstrates the determination of floor span length for the assessmentof shrinkage and creep effects.Columns beignored due tosmall lateralrestraintsSpan length for shrinkage and creepStiffcorewallFigure 15.6 – Determination of floor span length for shrinkage and creep15.8 For single span floor structures, if only the stress at age near to the final onesuch as the 360 days age is to be estimated, the design parameters for aparticular concrete grade (ignoring reinforcements) can be reduced to180

Version 2.3 May 2008comprising only (i) effective thickness of the floor structure; and (ii) relativestiffness of the axial stiffness of the floor structure to the lateral stiffness of“support 2”, i.e. K bK sup 2. Charts as contained in Figure 15.7 for grades 30,35, 40 and 45 concrete are produced which can be for general use.Variation of stress of grade 30 concrete floor due to Shrinkage & Creep witheffective thickness and floor / end restraint ratios at 360 daysConcrete Stress at 360 days (kN/m 2 )55005000450040003500300025002000150010005000Effective thickness = 100mm Effective thickness = 200mmEffective thickness = 400mm Effective thickness = 800mm0 1 2 3 4 5 6 7 8 9 10Floor / End Restraint Ratio K b / K sup2Variation of stress of grade 35 concrete floor due to Shrinkage & Creep witheffective thickness and floor / end restraint ratios at 360 daysConcrete Stress at 360 days (kN/m 2 )55005000450040003500300025002000150010005000Effective thickness = 100mm Effective thickness = 200mmEffective thickness = 400mm Effective thickness = 800mm0 1 2 3 4 5 6 7 8 9 10Floor / End Restraint Ratio K b / K sup2181

Version 2.3 May 2008Variation of stress of grade 40 concrete floor due to Shrinkage & Creep witheffective thickness and floor / end restraint ratios at 360 days55005000Effective thickness = 100mmEffective thickness = 400mmEffective thickness = 200mmEffective thickness = 800mmConcrete Stress at 360 days (kN/m 2 )4500400035003000250020001500100050000 1 2 3 4 5 6 7 8 9 10Floor / End Restraint Ratio K b / K sup2Variation of stress of grade 45 concrete floor due to Shrinkage & Creep witheffective thickness and floor / end restraint ratios at 360 days55005000Effective thickness = 100mmEffective thickness = 400mmEffective thickness = 200mmEffective thickness = 800mmConcrete Stress at 360 days (kN/m 2 )4500400035003000250020001500100050000 1 2 3 4 5 6 7 8 9 10Floor / End Restraint Ratio K b / K sup2Figure 15.7 – Variation of 360 days stress due to shrinkage and creep of structuralfloor with effective thickness and span / support stiffness ratios15.9 The induced stress in the concrete structure can be resisted by the tensile182

Version 2.3 May 2008strength of concrete under no cracking conditions. Or if the tensile stress isexcessive, it should be resisted by reinforcements with cracks limited tovarious widths according to exposure conditions.Worked Example 15.3Consider a grade 35 floor structure of unit width under restraints at ends of thefollowing design parameters :Stress induced is 3MPa;Thickness h = 160 mm;Longitudinal reinforcement content : T10@100 (B.F.) ρ = 0. 982 %;The floor structure is now checked for pure tension created due to shrinkageand creep alone :Crack width is checked in accordance with Cl. 3.2.2 and Appendix B ofBS8007:1987 with limiting crack width of 0.2mm;Strain for coaxial tension :σ 2bth 1 ⎛ 2.5 2×1000×160 ⎞εm= ε1 − ε2= − = ⎜ −⎟ = 0.000934 ,Esρ3EsAs200000 ⎝ 0.00982 3×1571 ⎠0.8 fy< = 0.00184 ;Es( ε 1is the strain due to steel only without consideration of the tensile strengthof the concrete and ε 2represents the stiffening effect by the crackedconcrete.)Cover to reinforcement is c 25 mm;So the greatest valuecrmin =a (distance from the point under consideration to thenearest reinforcement) that will lead to greatest crack width is5022+ 25 = 55.9 mm;By equation 4 of Appendix B of BS8007, the crack width isω 3 a ε = 3×55.9×0.000934 = 0.157 mm < 0.2mm;=cr mThe crack width is acceptable for all exposure conditions as required by Table7.1 of the Code.183

Version 2.3 May 200816.0 Summary of Aspects having significant impacts on current Practice16.1 GeneralThough some of the new practices in the Code as different from BS8110 havesignificant impacts on our current design, detailing and construction practices,these practices are however generally good ones resulting in better design andworkmanship. The improvement in design lies mainly in enhancing ductilityof the structure which should be regarded as another “limit state” equally asimportant as the “ultimate” and “serviceability” limit states. This section tendsto summarize all these new practices and discuss the various impacts so as toalert the practitioners in switching from BS8110 to the Code.The aspects with the most significant impacts by the Code on our currentdesign are obviously the incorporation of the ductility requirements in Cl. 9.9of the Code for beams and columns contributing in lateral load resistingsystem, and the design of beam column beam joints in Cl. 6.8. Others includechecking building accelerations in Cl. 7.3.2. Nevertheless, minor ones suchmore stringent requirements in locations and provisions of transversereinforcements in lapping of longitudinal bars should also be noted. Inaddition, there are relaxations in design requirements such as raising theabsolute ultimate design shear stress (v tu ) to 7N/mm 2 and giving clearguidelines in choosing design moments at or near column faces in Cl. 5.2.1.2.These aspects are highlighted and briefly discussed in this Section. The effectsof different concrete stress strain curve as indicated in Figure 3.8 of the Codefrom that of BS8110 are, however, found to be insignificant on the calculationof longitudinal bars required in beams and columns.16.2 Ductility RequirementsThe followings are highlighted :(i)Bending and lapping of reinforcement bars(a)Though the Code includes BS8666 : 2000 in its list of acceptablestandards for the specifications of bending and dimensioning ofreinforcing bars, Table 8.2 of the Code has, however, indicated simplerules for the minimum internal bend radii of bar diameter as 3Ø for Ø ≤184

Version 2.3 May 200820 mm and 4Ø for Ø > 20 mm where Ø is the diameter of the reinforcingbar. The minimum internal bend radii are all greater than that required byBS8666 : 2000 which ranges from 2Ø to 3.5Ø. Furthermore, as unlikethe British Standards (including the former BS4466), no distinction ismade for mild steel and high tensile steel bar in the Code. The morestringent requirement in minimum bend radii creates greater difficultiesin r.c. detailings;(b)Cl. 9.9.1.2 and Cl. 9.9.2.2(c) of the Code under the heading of ductilityrequirements for “Beams” and “Columns” (contributing in lateral loadresisting system) state that “Links should be adequately anchored bymeans of 135 o or 180 o hooks in accordance with Cl. 8.5. (Presumably180 o bent hooks can be accepted as better anchorage is achieved.) So theall links in such beams and columns contributing in lateral load resistingsystem should be anchored by links with bent angle ≥ 135 o as indicatedin Figure 16.1.Link with 180 o bent hooksLink with 135 o bent hooksFigure 16.1 – Links with hooks for beams and columns contributing to lateral loadresisting system and for containment of beam compression reinforcementsFor structural elements other than beams and columns contributing inlateral load resisting system, the 90 o anchorage hooks can still be usedexcept for containment of compression reinforcements in beams whichshould follow Figure 16.1 (Re Cl. 9.2.1.10 and Cl. 9.5.2.2 of the Code.)Figure 16.2 – 90 o bent links : used in structural elements other than beams / columnscontributing in lateral load resisting system and except compression bar containment185

Version 2.3 May 2008(c)Cl. 9.9.1.1(c) of the Code under the heading of ductility requirement foranchorage of longitudinal bars in beams (contributing in lateral loadresisting system) into exterior column states that “For the calculation ofanchorage length the bars must be assumed to be fully stressed”. Thecalculation of anchorage length of bars should therefore be based onfyinstead of0 .87 fyas discussed in Sections 8.4.4 and 8.4.5 of thisManual, resulting in some 15% longer in anchorage and lap lengths ascompared with Table 8.4 of the Code. Thus longer anchorage length isrequired for longitudinal bars in beams contributing in lateral loadresisting system anchoring into exterior column;(d)Cl. 8.7.2 and Figure 8.4 of the Code have effectively required all tensionlaps to be staggered which are generally applicable in the flexural steelbars in beams, slabs, pile caps, footings etc. as per discussion in Section13.6 of this Manual. The practice is not as convenient as the practicecurrently adopted by generally lapping in one single section.Nevertheless, if staggered lapping is not adopted, lapping will likely begreater than 50% and clear distance between adjacent laps will likely be≤ 10φ, transverse reinforcement by links or U bars will be required by Cl.8.7.4.1 of the Code which may even be more difficult to satisfy.Fortunately, the requirements for staggered lapping (in Cl. 8.7.2) do notcover distribution bars and compression bars. So most of thelongitudinal bars in columns and walls can be exempted;(e)Cl. 8.7.4 of the Code requires additional transverse reinforcementsgenerally in lap zones of longitudinal bars which is not required byBS8110. Arrangement and form of transverse reinforcements (straightbars or U-bars or links) required are in accordance with the longitudinalbar diameter φ , spacing of adjacent laps and percentage of lapping atone point. Take an example : when T40 bars of transverse spacing ≤ 400mm ( 10 φ ) are lapped at one section, total area of transversereinforcements equal to 1 longitudinal bar which is 1257 mm 2 should bespaced along the lapped length of some 2000 mm. The transversereinforcement is therefore T12 – 125 mm spacing (providing 2261mm 2 )in form of U-bars or links at ends of the laps as demonstrated in Figure13.6. Apparently these transverse reinforcements should be in addition tothe transverse reinforcements already provided for other purposes unless186

Version 2.3 May 2008φ < 20 mm or percentage of lapping at a section < 25%. As it is difficultto perform lapping with percentage < 25% in any one section, such extratransverse reinforcement will normally be required for φ ≥ 20 mm.Nevertheless, with lapping ≤ 50% at one section, at least U-bars or linkscan be eliminated.(ii) Beam(a)Limitation of neutral axis depthsNeutral axis depths have been reduced from 0.5 to 0.4 for concretegrades 45N/mm 2 < f cu ≤ 70 N/mm 2 and further reduced to 0.33 if f cu > 70N/mm 2 as per Cl. 6.1.2.4(b) of the Code under Amendment No. 1. Theeffects should be insignificant as it is uncommon to design flexuralmembers with grade higher than 45.(b)Reduction of moment arm factors for high grades concrete in sectionaldesign of beam by the Simplified Stress Block from 0.9 to 0.8 and 0.72.(c)Steel PercentagesThe maximum and minimum tension steel percentages are respectively2.5% and 0.3% in Cl. 9.9.1.1(a) of the Code for beams contributing tolateral load resisting system. The lower maximum tension steelpercentages may force the designer to use larger structural sections forthe beams contributing in lateral load resisting system.In addition, Cl. 9.9.1.1(a) also imposes that “At any section of a beamwithin a critical zone (the Handbook gives an example of that “plastichinge zone” is a critical zone), the compression reinforcement should notbe less than one-half of the tension reinforcement at the same section.”The “critical zone” should likely include mid-spans and/or internalsupports in continuous beam. As plastic hinges will likely be extensivelyin existence in normal floor beams as per the discussion in Section 2.4,the requirement is expected to be applicable in many locations in beamscontributing in lateral load resisting system. The adoption of this clausewill obviously increase amounts of longitudinal bars significantly forthese beams.187

Version 2.3 May 2008(d)End Support AnchorageThe Code has clarified support anchorage requirements of reinforcementbars of beams as summarized in the following Figure 16.3 whichamalgamates contents in Figures 3.19, 3.20 and 13.1 of this Manual :cD ≥ 2(4Ø+c) if Ø ≤ 20≥2(5Ø+c) if Ø > 20≥0anchoragecommences at thissection generally.Longitudinalbar of dia. Øsupportcentreline≥0.5Dor 8Ø≥ 500mm or hXhcross bar ofdia. ≥ Ø≥0≥0≥0.75Danchorage can commence atthis section if the plastic hingeof the beam is beyond XBeam contributing to lateral load resisting systemD ≥ 2(4Ø+c) if Ø ≤ 20≥2(5Ø+c) if Ø > 20anchoragecommences at thissection generally.Longitudinalbar of dia. Øc≥0supportcentrelinehcross bar ofdia. ≥ Ø≥0Beam not contributing to lateral load resisting systemFigure 16.3 – Summary of longitudinal bar anchorage details at end support(1) Cl 8.4.8 clarifies the support widths to beams in form of beams,188

Version 2.3 May 2008columns and walls as in excess of 2(4Ø+c) if Ø ≤ 20 and 2(5Ø+c) ifØ > 20 where Ø is the diameter of the longitudinal bar and c is theconcrete cover to the bar. The clause has effectively required bendof bars to commence beyond the centre-line of support which is anexisting requirement in BS8110 stated for simply supported end (Cl.3.12.9.4 of the BS). The clause has extended the requirement tocover beam at supports restrained against rotation. Support widthsmay then require to be increased or bar size be reduced to satisfythe requirement, giving constraints in design;(2) Cl. 9.9.1.1(c) of the Code requires anchorage of longitudinal bar ofbeam contributing to lateral load resisting system to commence atthe centre line of support or 8 times the longitudinal bar diameterwhichever is the smaller unless the plastic hinge at the beam is atthe lesser of 500mm or a beam structural depth from the supportface of the beam. Effectively the requirement covers all suchbeams designed to be having rotational restraints at the exteriorcolumns or walls unless it can be shown that critical section ofplastic hinge is beyond X as shown in Figure 16.3. By therequirement, anchorage length needs be increased by the lesser ofthe half of the support width and or 8 times the longitudinal bardiameter for most of the end span beams contributing to lateralload resisting system and anchored into exterior column which isoften relatively stiffer than the beam as per discussion in Section2.4;(3) A method of adding a cross bar so as to avoid checking of internalstress on concrete created by the bend of the longitudinal bar (by(Ceqn 8.1) of the Code) has been added in Cl. 8.3 of the Code,even if checks on the bar indicates that anchorage of thelongitudinal bar is still required at 4Ø beyond the bend. The methodis not found in BS8110. The method is quite helpful as the designercan avoid using large bends of bars to reduce bearing stress inconcrete which may otherwise result in non-compliance with Cl.8.4.8 of the Code;(4) Cl. 9.9.1.1 (c) states clearly that top beam bars be bent downwardsand bottom beam bars bent upwards, again applicable to beamscontributing in lateral load resisting system. The requirement maycreate difficulties to the conventional construction work as, apartfrom aggravating steel bar congestion problems in the column189

Version 2.3 May 2008beam joint, the top bars may be required to be fixed prior tocolumn concreting if they have to be bent into the column shaft toachieve adequate anchorage. The practice is obviously notconvenient in the current construction sequence for buildings.A Solution to anchorage problem may be adding an “elongation” of thestructural beam, if possible, beyond the end column as shown in Figure 16.4.Longitudinalbar of dia. ØhelongationFigure 16.4 – “Elongation” for anchorage of longitudinal bars beyond end supports(iii)Column(a)Steel PercentagesCl. 9.9.2.1(a) of the Code has required the maximum longitudinalreinforcements to be 4% of the gross sectional area for columnscontributing to lateral load resisting system which are more stringentthan columns not contributing to lateral load resisting system (6% to10% in accordance with 9.5.1 of the Code). In addition, the clause alsoclarifies that the maximum longitudinal bar percentage at laps is 5.2%which effectively reduces the maximum steel percentage to 2.6% if theconventional lapping at single level (not staggered lap) is adopted inconstruction for columns contributing to lateral load resisting system.(b)Anchorage and lapping of longitudinal bars in supporting beams orfoundations190

Version 2.3 May 2008Figures 5.9, 5.10 and 5.11 of this Manual illustrate anchorage oflongitudinal bars of columns in supporting beam or foundations asrequired by Cl. 9.9.2.1(c) of the Code where the columns contribute inlateral load resisting system and plastic hinges will occur in the column.Generally anchorage lengths will be increased as anchoragecommences inside the beam and foundation element instead of columnfoundation interfaces for such columns. Furthermore, bars in columnsanchored into intersecting beams must be terminated with 90 o standardhooks (or equivalent anchorage device) and have to be bent inwardsunless the column is designed only for axial loads. All these lead tolonger anchorage lengths and stability problem in reinforcing barserection.(c)Splicing of longitudinal barsTo reduce weakening of the column in reinforcement splicing (lappingand mechanical coupling) in “critical zones” (potential plastic hingeformation zones), Cl. 9.9.2.1(d) of the Code requires the longitudinalsplicing locations of columns contributing in lateral load resistingsystem should, as far as possible, be away from these “critical zones”which are near the mid-storey heights as illustrated in Figure 5.9. Forsuch columns, the current practice of lapping at floor levels in buildingconstruction requires review.(d)Transverse ReinforcementsCl. 9.9.2.2 of the Code which is applicable to column contributing tolateral load resisting system defines “critical regions” along a columnshaft which are near the ends of the column resisting high bendingmoments and specifies more stringent transverse reinforcementrequirements in the same clause than the normal region nearmid-heights of the column. In this clause, the definition of “criticalregions” relies on axial stress in the column and has made no referenceto any potential plastic hinge formation zone. As it is not our usualpractice of specifying different transverse reinforcements along thecolumn shaft and the lengths of the “critical regions” are often morethan half of the column shaft (dictated also by the requirement of oneto two times the greater lateral dimension of the column), it seems191

Version 2.3 May 2008sensible to adopt the more stringent requirements along the wholecolumn shaft. The more stringent requirements of transversereinforcements comprise closer spacing and that every longitudinal bar(instead of alternate bar) must be anchored by a link. Whilst themaximum spacing in the normal region is 12Ø where Ø is thelongitudinal bar diameter and that in the critical region is the smaller of6Ø and 1/4 of the least lateral dimension in case of rectangular orpolygonal column and 1/4 of the diameter in case of circular column,the quantities of transverse reinforcements can be doubled.(iv)Column Beam JointsThe requirement of providing checking and design in column beamjoints as discussed in Section 6 constitutes a significant impact on thecurrent design and construction as, apart from increase of constructioncost due to increase of steel contents, the requirement aggravate theproblem of steel congestions in these joints. Enlargement of thecolumn head as indicated in Figure 16.6 may be required in case theshear stress computed by Ceqn 6.71 is excessive or the requiredreinforcements are too congested. In addition, it should also be notedthat even no shear reinforcement is required as per checking of shear inthe joints in accordance with Cl. 6.8 of the Code, transversereinforcements in accordance with Cl. 9.5.2 which are installed in thecolumn shaft outside “critical regions” shall also be installed within thecolumn beam joints as shown in Figure 6.3 of this Manual.Enlarged columnheadFigure 16.6 – Column head enlargement for column beam joint192

Version 2.3 May 200816.3 Building AccelerationsCl. 7.3.2 of the Code specifies that “where a dynamic analysis is undertaken,the maximum peak acceleration should be assessed for wind speeds based on a1-in-10 year return period of 10 minutes duration with the limits of 0.15m/sec 2for residential buildings and 0.25m/sec 2 for office or hotel. The term “dynamicanalysis” is not defined in the Code. However, if it is agreed that computationof wind loads in accordance with Wind Code 2004 Appendix F (titled“Dynamic Analysis”) is a dynamic analysis, the requirement will be applicableto all buildings defined as ones with “significant resonant dynamic response”in Clause 3.3 of the Wind Code, i.e. (i) taller than 100 m; and (ii) aspect ratio >5 unless it can be demonstrated that the fundamental natural frequency > 1 Hz.Thus most of the high-rise buildings are included.Empirical approaches for assessment of building accelerations are described inAppendix B. The second approach which is taken from the Australian Codeshould be compatible to the Hong Kong Wind Code as it is based on theAustralian Code that the Hong Kong Wind Code determines approaches ofdynamic analysis in its Appendix F. Furthermore, it can be shown in the chartattached in the Appendix that building acceleration generally increases withbuilding heights and thus pose another compliance criterion. Fortunately, theaccelerations approximated are not approaching the limiting criterion as perthe exercise on a square plan shaped building. However, the effects should bemore significant for buildings with large plan length to breadth ratios.193

Version 2.3 May 2008ReferencesThis Manual has made reference to the following documents :1. Code of Practice for the Structural Use of Concrete 20042. Concrete Code Handbook by HKIE3. Hong Kong Building (Construction) Regulations4. The Structural Use of Concrete 19875. Code of Practice on Wind Effects in Hong Kong 20046. Code of Practice for the Structural Use of Steel 20057. The Code of Practice for Dead and Imposed Loads for Buildings (Draft)8. BS8110 Parts 1, 2 and 39. BS5400 Part 410. Eurocode 211. Code of Practice for Precast Concrete Construction 200312. New Zealand Standard NZS 3101:Part 2:199513. ACI Code ACI 318-0514. Code of Practice for Fire Resisting Construction 199615. BS8666 : 200016. BS4466 : 198917. PNAP 17318. Practical design of reinforced and prestressed concrete structures – basedon CEP-FIP model code MC7819. CEB-FIP Model Code 199020. Standard Method of Detailing Structural Concrete – A Manual for bestpractice – The Institution of Structural Engineers21. The “Structural Design Manual” – Highways Department of HKSAR22. Reinforced and Prestressed Concrete 3 rd edition – Kong & Evans23. Reinforced Concrete Design to BS110 – Simply Explained A.H. Allen24. Design of Structural Concrete to BS8110 – J.H. Bungey25. Handbook to British Standard BS8110:1985 Structural Use ofConcrete – R.E. Rowe and others – Palladian Publication Ltd.26. Ove Arup and Partners. Notes on Structures: 17 April 198927. Australian/New Zealand Standard – Structural Design Actions Part 2 :Wind Actions AS/NZS 1170.2.200228. Examples for the Design of Structural Concrete with Strut-and-TieModels – American Concrete Institute29. Tables for the Analysis of Plates, Slabs and Diaphragms based on the194

Version 2.3 May 2008Elastic Theory – Macdonald and Evans Ltd.30. Some Problems in Analysis and Design of Thin/Thick Plate, HKIETransactions – Cheng & Law, Vol. 12 No. 1 2004195

Appendix AClause by Clause Comparisonbetween “Code of Practice forStructural Use of Concrete 2004”and BS8110

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contents1.1 – Scope The clause has explicitly stated that the Code applies Pt. 1 1.1 – The clause only explicitly excludes bridge structuresonly to normal weight concrete, with the exclusion of Scope and structural concrete of high alumina cement.(i) no fines, aerated, lightweight aggregate concreteetc; (ii) bridge and associated structures, precastconcrete (under the separate code for precastconcrete); and (iii) particular aspects of special typesof structures such as membranes, shells.RemarkThe exclusion of CoPConc2004should also be applied toBS8110. In addition, BS8110does not apply to high strengthconcrete.2.1.5 –Designworking lifeThe clause states that the Code assumes a designworking life of 50 years. Where design working lifediffers from 50 years, the recommendations should bemodified.– Nil No similar statement inBS8110.2.2.3.3 –Response towind loadsThe clause refers to clause 7.3.2 for the usual limitsof H/500 to lateral deflection at top of the buildingand accelerations of 1-in-10 year return period of 10minutes duration of 0.15m/sec 2 for residential and0.25m/sec 2 for office. However, there is norequirement on the inter-storey drift, though the draftsteel code has a requirement of storey height/400.Pt. 1 2.2.3.3 –Response towind loadsReference to specialist literature is required. Inaddition Pt. 2 3.2.2.2 stipulates a limit on inter-storeydrift of Storey height/500 for avoidance of damage tonon-structural elements.CoPConc2004 is more specific.However, method fordetermination of theacceleration is not given in theCode and in the HKWC-2004.2.3.2.1 –Loads forultimate limitstateTable 2.1 is generally reproduced from Table 2.1 ofBS8110 except that the partial factor for load due toearth and water is 1.0 for the beneficial case.Pt. 12.4.3.1.2 –Partial factorsfor earthpressuresIt is stated in the clause that when applying the loadfactor, no distinction should be made betweenadverse and beneficial loads.1.0 in CoPConc2004 may notbe adequately conservative asthere may even beover-estimation in thedetermination of the unfactoredsoil load. It is even a practice toset the load to zero in beneficialcase. ICU has raised thiscomment during the commentstage when the draft Code hasexactly the content of BS8110.2.3.2.3 & The clauses explicitly state that these effects need – No similar clauses in BS8110 The clause in CoPConcrete2004A-1

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contents2.3.2.4 – only be considered when they are significant forDifferential ULS. In most other cases they need not be consideredsettlement of provided ductility and rotational capacity of thefoundations, structure sufficient.creep,shrinkage,temperatureeffectsRemarkaffirms engineers to ignoreconsideration of these effects innormal cases which are theusual practices.2.4.3.2 –Values of γ mfor ULSTable 2.2 gives γ m for ULS for concrete and re-bars.γ m for re-bars is 1.15, implying strength of re-bars fordesign remain as 0.87f y .Pt. 1 2.4.4.1 –Values of γ mfor ULSTable 2.2 gives γ m for ULS for concrete and re-bars.γ m for re-bars is 1.05, implying strength of re-bars fordesign remain as 0.95f y .BD has been insisting on theuse of 0.87f y even if BS8110was used before thepromulgation of the newconcrete code3.1.3 –StrengthgradesTable 3.1 states concrete strength grades from 20MPa up to 100 MPa which is the range covered bythe Code.– BS8110 has not explicitly stated the concrete gradescovered by the BS, However, concrete gradescovered by the design charts in Part 3 of the Coderange from grade 25 to 50 whilst other provisionssuch as v c (Pt. 1 Table 3.8), lap lengths (Pt. 1 Table3.27) are up to grade 40.The coverage of CoPConc2004is wider.3.1.4 –DeformationofConcreteIt is stated in the 1 st paragraph of the clause that forULS, creep and shrinkage are minor, and no specificcalculation are required.Pt. 1 2.4.3.3 –creep,shrinkage andtemperatureeffectsThe clause states that “For the ULS, these effects willusually be minor and no specific calculations will benecessary.BS 8110 has includedtemperature effects be a minorone that can be ignored incalculation.3.1.5 – ElasticdeformationTable 3.2 stipulates short term static Young’sModulus of concrete of various grades based on theformula 3.46√f cu +3.21 in MPa derived from localresearch.Pt. 1 Figure2.1The determination of short term static Young’sModulus of concrete is given by the slope gradient inthe figure which is 5.5√(f cu /γ m ).Values in the CoPConc2004should be used as it is based onlocal research and concrete Evalues are affected byconstituents which are of localmaterials. Nevertheless, itA-2

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. ContentsRemarkshould be noted that E values inthe new Code are slightlyhigher than the previous ones in“The Structural Use ofConcrete – 1987” (Table 2.1).3.1.7 &3.1.8 – Creepand shrinkageThough it is stated in 3.1.4 that for ULS creep andshrinkage are minor and require no specificcalculations, these clauses contain detailed formulaeand charts for prediction of creep and shrinkagestrain. The approach is identical to the “StructuralDesign Manual” issued by Highways Department andthe charts are extracted from BS5400:Pt 4:1990. Asidentical to the previous version of “StructuralDesign Manual” (SDM), the c s value is 4.0 to suitlocal crushed granite. However, it is noted thatrecently the SDM has reduced the factor to 3.0 andthe Code has adopted the change in Amendment No.1.– No account has been given for creep and shrinkage,as stated in 2.4.3.3.As stated in the CoPConc2004Cl. 2.3.2.4, effects need only beconsidered if they aresignificant.3.1.9 –ThermalExpansionThe linear coefficient of thermal expansion given inthe Code for normal weight concrete is 10×10 -6 / o Cwhilst that stated in the “Structural Design Manual”issued by Highways Department is 9×10 -6 / o C inClause 2.4.4.– No account has been given for temperature, as statedin 2.4.3.3.The linear coefficient ofthermal expansion given byboth CoPConc2004 is slightlyhigher than SDM and both areindependent of concrete grades.3.1.10 –Stress-strainrelationshipfor designThe short term design stress-strain curve of concrete(Fig. 3.8) follows closely the traditional one inBS8110, though there are differences in the values ofthe Young’s Moduli. Furthermore, the ultimate strainis limited to below 0.0035 for concrete gradeexceeding C60. Nevertheless, the “plastic strain”,strain beyond which stress is constant remainsidentical as BS8110;Pt. 1 2.5.3 –Analysis ofsection forULSPt. 1 Fig. 2.1 shows stress-strain relation of normalweight concrete.(i) The Young’s Moduli ofconcrete stipulated inCoPConc2004 should beadopted as they are basedon local data and cover upto grade 100 concrete,together with the decreaseof ultimate strain for gradeA-3

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. ContentsRemarkabove C60, to account forthe brittleness of highstrength concrete;(ii) “Smooth” connectionbetween the paraboliccurve and the straight lineportion of the stress-straincurve cannot be effected ifε 0 = 2.4×10 -4 √f cu /γ m is keptand the Young’s moduli inTable 3.2 of CoPConc2004are used. For smoothconnection, ε 0 should berevised 2σ ult /E c or1.34f cu /γ m E c whereσ ult =0.67f cu /γ m .Nevertheless, ε 0 has beenrectified to 1.34f cu /γ m E c inAmendment No. 1.3.2.7 –Weldability(of re-bars)The clause states that re-bars can be welded providedthe types of steel have the required weldingproperties given in acceptable standards and underApproval and inspection by competent person.Further provisions for welding are given in 10.4.6.Pt. 1 3.12.8.167.6There are provisions for lapping re-bars by weldingin Pt. 1 3.12.8.16, and general welding requirementsin Pt. 1 7.6. Nevertheless, another BS7123 titled“Metal arc welding of steel for concretereinforcement” requires the re-bars be in compliancewith BS4449 or BS4482.Steel complying CS2 is likelyweldable as CS2 does not differsignificantly from BS4449.Section 4 –Durabilityand fireresistanceThe requirements are general. The followings arehighlighted :-(i) In 4.1.1, it is stated that requirements are basedon design working life of 50 years;(ii) In Table 4.1 under 4.2.3.2, exposure conditions1 to 5 are classified with headings similar to thatPt. 1 2.2.4 –Durability;Pt. 1 2.2.6 –Fireresistance; Pt.1 3.1.5.2 –The requirements are general. However, thefollowings are highlighted :(i) 3.1.5.2 states that when cement content > 400kg/m 3 and section thicker than 600 mm,measures for temperature control should beimplemented;The approaches of the twocodes are quite different.However, CoPConc2004 ismore related to local practice.A-4

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsin Pt. 1 3.3.4 of BS8110. However, detailed Design for (ii) A full description for fire resistance control isdescriptions are different except the last one Durabilitygiven in Pt. 2 Section 4, outlining methods of“abrasive”;Pt. 1 3.3 –determining fire resistance of structural(iii) In 4.2.7.3, control of AAR has beenConcreteelements and with reference to BS476 Pt. 8;incorporated from the previous PNAP 180; cover to(iv) Concrete covers are given in Table 4.2 for rebarsvarious concrete grades and exposurePt. 2 Section 4conditions;(v) By 4.3, the user has to refer to the current firecode for additional requirements against fireresistance.Remark5.1.3.2 –Load casesandcombinationsfor beams andslabs3 simplified load cases for Dead + Live loads arerecommended (with DL always present): (i) all spansloaded with LL; (ii) alternate spans loaded with LL;(iii) adjacent spans loaded with LL. The 2 nd case is toseek for max. sagging moment and the 3 rd case islikely to seek for max. hogging moment. But truemax hogging moment should also include alternatespans from the support being loaded.Pt. 1 3.2.1.2.22 simplified load cases are considered sufficient fordesign of spans and beams (with DL always present):(i) all spans loaded with LL; (ii) alternate spansloaded with LL;CoPConc2004 more reasonable,though not truly adequate.Nevertheless, the currentsoftwares mostly can accountfor the load case to search formax. hogging moment atsupport.5.2 – Analysisof StructureThe clause contains definition of beam, slab(included ribbed, waffle, one-way or two-ways),column in accordance with their geometries. Thefollowing differences with BS8110 are highlighted :(i)Conditions in relations to rib spacings andflange depths for analysis of a ribbed or waffleslab as an integral structural unit are given in5.2.1.1(d);(ii) By definition, the effective flange widths in Tand L-beams included in 5.2.1.2 (a) are slightlygreater than that in BS8110 by 0.1×clear ribspacing unless for narrow flange widthcontrolled by rib spacing. It is also stated that3.2 – Analysisof Structures3.4 – BeamsThe followings are highlighted :(i) Generally provisions are applicable to normalstrength concrete;(ii) Consideration for high-rise buildings (secondorder effects) and structures such as shear walls,transfer structures are not given;(iii) The effective span of a simply supported beamis the clear span plus the lesser of half effectivedepth and half of support width whilst that ofcontinuous beam and cantilever are spanbetween supports (presumably mid-supportwidth) except at end span in case of continuousbeam and a cantilever forming the end of aThe following remarks aremade :(i) CoPConc2004 is moresuitable for use in HongKong as it cover highstrengths concrete, highrise buildings, shears,transfer structures.(ii) Extended use of effectiveflange in beam instructural analysis andmoment reduced tosupport shear are explicitlyA-5

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsthe effective flange can be adopted in structuralcontinuous beam, the effective span is the clearanalysis;span plus mid-support width should be used.(iii) Clearer definition of effective spans of beams isalso included in 5.2.1.2(a) with illustration bydiagrams, together with reduction of spanmoments due to support width in 5.2.1.2(b).In principle, the effective span of a simplysupported beam, continuous beam or cantileveris the clear span plus the minimum of halfeffective depth and half support width exceptthat on bearing where the span should be up tothe centre of the bearing.(iv) Furthermore reduction in support moments dueto support width is allowed which is notmentioned in BS8110;(v) The definition of effective flange for T- andL-beams are slightly different from BS8110though the upper bound values are identical tothat of BS8110. So more stringent.(vi) Moment redistribution is restricted to gradebelow C70 as per 5.2.9.1;(vii) Condition 2 in 5.2.9.1 relation to checking ofneutral axis depths of beam in adopting momentre-distribution is different from Cl. 3.2.2.1 b) ofBS8110. More stringent limit on neutral axis isadded for concrete grade higher than C40 as per6.1.2.4(b);(viii) Provision for second order effects with axialloads in 5.3 is added. The provisions are quitegeneral except the statement “second ordereffects can be ignored if they are less than10%”;(ix) Provisions for shears walls and transferstructures are added in 5.4 and 5.5 though theRemarkstated in CoPConc2004.(iii) Method of analysis in bothcodes are old-fashionedones that can be performedby hand calculations. Useof computer methods(extensively adoptedcurrently in Hong Kong)are not mentioned.A-6

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsprovisions are too general.Remark6.1 –Members inFlexureThe followings differences with BS8110 areidentified :(i) Ultimate strains are reduced below 0.0035 forconcrete grade exceeding C60 as per Fig. 6.1;(ii) In Figure 6.1, the factor 0.9, concrete stressblock depth ratio is revised to 0.8 for 45

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsreinforcements required is reduced to that canprovide shear resistance of 0.4(f cu /40) 2/3 MPafor grade over 40 whilst that for concrete at andbelow grade 40 remains to that can provide 0.4MPa;(vii) Partial strength factor for steel remains as 0.87f yin accordance with BS8110:1985 instead of0.95f y as in accordance with BS8110:1997, forboth flexure and shear;(viii) Ultimate shear strength of concrete increased to7.0 MPa as compared 5.0 MPa in BS8110. Theother limitation of 0.8√f cu is identical toBS8110;(ix) Table 6.3 is identical to Table 3.8 of BS8110except, the last note under Notes 2 for the effectof effective depth (d) on v c where shearreinforcement is required. Whilst BS8110effectively gives 1 for the factor (400/d) 1/4 for d≥ 400mm, Table 6.3 requires the factor todecrease beyond d=400mm. However assubsequent to discussion with experts, it isadvisable to keep the factor to 1 for d ≥ 400mmas there are no tests to justify the decrease;(x) By 6.1.3.5, the minimum shear reinforcement tocater for shear strength of 0.4 MPa is forconcrete grade below C40 (requirement byBS8110). Above grade C40, the required shearstrength is increased by factor (f cu /40) 2/3 as perTable 6.2;(xi) By 6.1.4.2, ribbed slabs (6.1.4.2) can bedesigned as two-ways spanning as similar to flatslab if they have equal structural properties intwo mutually perpendicular directions. BS8110does not explicitly require equal structuralRemarkflat slab, it should bequalified that the slab hasequal structural propertiesin two mutuallyperpendicular directions.A-8

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsproperties in mutually perpendicular directionsin adopting design method as flat slab (BS8110,3.6.2).Remark6.2 –Membersaxially loadedwith orwithoutflexureThe followings are highlighted:(i) The CoP contains no design charts for columndesign with moments. Strictly speaking, thedesign charts in BS8110 are not applicable (a)the Young’s Moduli of concrete are different;and (b) the ultimate strain for concrete grade >C60 are reduced;(ii) Due to the use of lower partial strength factor ofsteel (γ m = 1.15), the factors on steel inequations for strengths of column sections(6.55, 6.56) are lower than BS8110 (Eqn 28,29);(iii) The provisions for more accurate assessment ofeffective column height in accordance withBS8110 Pt. 2 2.5 are not incorporated in theCode.Pt. 1 3.8 –ColumnsPt 2 2.5 –Effectivecolumn heightThe followings are highlighted :(i) In the design aspects, the provisions are limitedto grade 40;(ii) A more tedious approach for determination ofeffective height of column by considerationstiffness of the connecting beams and columnsis outlined in Pt. 2 2.5.The followings are highlighted :(i) The limitation of neutralaxis depth as for beam isclearly not applicable tocolumns. However, itshould be noted thatmembers with axial loadscreating axial stress < 0.1f cumay be regarded as beam indesign. Re 6.1.2.4 ofCoPConc2004.6.3 – TorsionandCombinedEffectsThe followings are highlighted :Table 6.17 in relation to limitation of torsional shearstresses contains specific values for grades 25 to 80.For values below grade 40, they are identical toBS8110. Above grade 40, the Code provides valuesfor v tmin and v t for different grades up to 80, beyondwhich the values remain constant whilst BS8110 setthe values to that of grade 40 for grades above 40;Pt. 2 2.4The followings are highlighted(i) Table 2.3 contains specific values for v tmin andv t up to grade 40. The values remain constantfrom grade 40 thereon;(ii)CoPConc2004 contains morespecific values for ultimateconcrete shear stress.6.4 – Designfor robustnessThe followings are highlighted :(i) In 6.4.2 in relation to design of “bridgingelements” which is identical to BS8110 Pt.22.6.3, the words “where required in buildings ofPt 2 2.6.3 The followings are highlighted :(i) Pt.2 2.6.3 in relation to design of “bridgingelements” applies to buildings of five or morestoreys;CoPConcrete 2004 is morestringent.A-9

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsfive or more storeys” have been deleted. So theCode is more stringent as consideration to lossof elements is required for all buildings;Remark6.6 –StaircaseThe followings are highlighted :(i) 6.6.1 is in relation to design of staircase. Thereis no stipulation that the staircase may includelanding;Pt 1 3.10.1 The followings are highlighted :(i) A note in 3.10.1 in relation to design ofstaircase has stipulated that a staircase alsoinclude a section of landing spanning in thesame direction and continuous with flight;It is more reasonable to assumestaircase should includelanding, as in CoPConc2004.6.7 –FoundationsThe following differences with BS8110 arehighlighted :(i) In 6.7.1.1, the assumptions of uniform reactionor linearly varying reaction of footing and pilecap are based on use of rigid footings or pilecaps. So a pre-requisite for the use of theseassumptions is stated at the end of the clausewhich reads “if a base or pile cap is consideredbe of sufficient rigidity.”;(ii) In 6.7.3.1, a statement has been inserted that apile cap may be designed as rigid or flexible,depending on the structural configuration. Nosimilar provision is found in BS8110;(iii) In 6.7.3.3, 2 nd dot, it is stated that “where theshear distribution across section has not beenconsidered, shear enhancement shall not beapplied.” No similar provision is found inBS8110;(iv) In 6.7.3.3 3 rd dot, shear enhancement in pile capcan only be applied where due considerationhas been given to shear distribution acrosssection. No similar provision is found inBS8110;(v) In 6.7.3.3 4 th dot, determination of effectivePt 1 3.11 The followings are highlighted :(i) In Pt 1 3.11.2.1, the assumption of uniform orlinearly varying assumption is without thepre-requisite that the footing or pile cap issufficiently rigid. This is not good enough assignificant errors may arise if the footing orpile cap is flexible;(ii) In Pt 1 3.11.4.1, there is no mention of pile caprigidity which affects design;(iii) In Pt 1 3.11.4.3, there is no mention that shearenhancement shall not be applied to whereshear distribution across section has not beenconsidered;(iv) In Pt 1 3.11.4.4 b), there is no mention thatshear enhancement in pile cap shall be appliedto under the condition that shear distribution(v)across section has not been duly considered;There is no explicit stipulation on the limit ofeffective width for checking shear;(vi) No explicit statement that shear reinforcementis required if v

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentswidth of section for resisting shear is included.No similar provision is found in BS8110;(vi) In 6.7.3.3 5 th dot, it is explicitly stated that noshear reinforcement is required if v

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. ContentsEquation 14 of BS8110 Pt. 2 3.8.4.2 where thedifference in temperatures is divided into 2parts and a factor of 0.8 is employed in theestimation of thermal strain. Also it is allowedin the clause to take reinforcements intoconsideration;(v) In 7.3, pre-camber is limited to span/250 tocompensate excessive deflection. The limit isnot given in BS8110 Pt. 2 3.2.1. Also,deflection limit after construction for avoidanceof damage to structure is limited to span/500,whilst BS8110 Pt. 2 3.2.1.2 specifies span/500or span/350 in accordance with brittleness offinishes for avoidance of damage tonon-structural elements. In addition, 20 mm asan absolute value is also imposed in BS8110;(vi) In 7.3.2, limits (0.15 and 0.25 m/s 2 ) onaccelerations (10 years return period on 10 min.duration) are given as avoidance of “excessiveresponse” to wind loads whilst no numericalvalues are given in BS8110 Pt. 2 3.2.2.1.Furthermore, deflection limit due to wind loadis given as H/500 whilst BS8110 Pt. 2 3.2.2.2indicates limit of h/500 as inter-storey drift foravoidance of damage to non-structuralmembers;(vii) In 7.3.3, excessive vibration should be avoidedas similar in BS8110 Pt. 2 3.2.3. However,there is extra requirement in 7.3.3 that dynamicanalysis be carried out in case structuralfrequency less than 6 Hz for structural floor andfootbridge less than 5 Hz;(viii) Table 7.3 under 7.3.4.2 in relation todeem-to-satisfy requirement for basicRemarkA-12

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsspan/depth ratios of beam and slab contains,requirements for “two-way slabs” and “endspans” are included, as in comparison withTable 3.9 of BS8110 Pt 1;(ix) Table 7.4 in relation to modification factor toeffective span depth ratio by tensilereinforcement is identical to Table 3.10 ofBS8110 except that the row with service stress307 (f y = 460) has replaced that of 333 (f y =500);(x) 7.3.4.6 is identical to BS8110 Pt. 1 3.4.6.7except that the last sentence in BS8110 isdeleted;(xi) The provision of deflection calculation in 7.3.5is identical to BS8110 Pt 2 Cl. 3.7;(xii) Equation 7.7 in 7.3.6 is not identical toequation 9 in BS8110 Pt. 2 in the derivation ofshrinkage curvature;RemarkSection 8 –Reinf’trequirementsThe followings are highlighted :(i) In 8.1.1, it is declared that the rules given in theSection for re-bars detailings do not apply toseismic, machine vibration, fatigue etc andepoxy, zinc coated bars etc. No similarexclusion is found in BS8110;(ii) In 8.1.2, it is stated that bar scheduling shouldbe in accordance with acceptable standardswhilst BS8110 Pt. 1 3.12.4.2 requires standardbe in accordance with BS4466;(iii) In 8.2, the minimum spacing of bars should bethe lesser of bar diameter, h agg +5 mm and 20mm. BS8110 Pt. 1 3.12.11.1 does not include20 mm and bar diameter is only required whenbar size > h agg +5 mm;Pt. 1 Section 3Pt. 1 4.10 inrelation toanchorage oftendons inprestressedconcreteThe followings are highlighted :(i) The provisions are general;(ii) Consideration for ductility is not adequate.The followings are highlighted :(i) CoPConc2004 requires barscheduling to acceptablestandards. However,provisions in 8.3, 9.2.3 etchave requirements for bendof bars;(ii) ICU has commented thatrequirement in BS8110 Pt.1 3.12.9.4(a) should extendto all support conditions.8.4.8 of the Code seems toincorporate the comment;(iii) ICU has also suggested theuse of torsional linksA-13

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contents(iv) In 8.2, it is stated that the distance betweenhorizontal layer of bars should be sufficientwithout quantification whilst BS8110 Pt. 13.12.11.1 requires minimum be 2h agg /3;(v) 8.3 is essentially identical to BS8110 Pt. 13.12.8.22, 24, 25, except that(a) an additional provision of installing a crossbar inside a bend can eliminate checking ofbearing stresses of the bend in concrete;(b) a single Table 8.2 (in relation to minimumbend radii of re-bars) to replace Table 3 ofBS4466 to which BS8110 is makingreference. As such, no distinction is madebetween mild steel bars and HY bars –both adopting minimum radii of HY bars.Provisions to the newer BS – BS8666where the minimum bend radii aregenerally smaller is not adopted;(vi) 8.4 is essentially identical to BS8110 Pt. 13.12.8 except(a) It is mentioned in 8.4.1 that whenmechanical device is used, theireffectiveness has to be proven. No similarprovision is found in BS8110;(b) Type 1 bars are not included in the Code;(c) 8.4.6 is added with Figure 8.1 forillustration of bend anchorage;(d) 8.4.8 in relation to minimum supportwidths requires any bend inside support bebeyond the centre line of the support. Thiseffectively extend the requirement ofBS8110 Pt. 1 3.12.9.4(a) to supportconditions other than simply support;(vii) 8.5 in relation to anchorage of links and shearRemarksimilar to that in ACI code(135 o hood) which is ofshape other than shapecode 77 of BS4466.A-14

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsreinforcements contains more stringentrequirement for the length of the link beyondbends of bars than BS8110 Pt. 1 3.12.8.6. – (1)the greater of 4φ or 50 mm for 180 o bend in theCode but only 4φ in BS8110; (2) the greater of10φ or 70 mm for 90 o bend in the Code butonly 8φ in BS8110. Provisions for 135 o bendand welded bars are also added;(viii) 8.6 contains requirements for anchorage bywelded bars which is not found in BS8110;(ix) Except for the requirement that the sum ofre-bar sizes at lapping < 40% of the breadth ofthe section, 8.7 contains more requirements interms of “staggering laps”, minimumlongitudinal and transverse distances betweenadjacent lapping of bars which are not found inBS8110. The requirements are alsoschematically indicated in Fig. 8.4. Effectivelythe clause requires tension laps be alwaysstaggered. Fortunately compression laps andsecondary rebar lapping can be in one section;(x) 8.7.4.1 contains different and more detailedrequirements for transverse reinforcements inlapped zone than BS8110 Pt. 2 3.12.8.12;(xi) 8.8 and 8.9 in relation to large diameter bars(>40φ) and bundle bars which are not found inBS8110;(xii) The provision in BS8110 Pt. 1 3.12.8.16 forbutt joints of re-bars is not found in the Code.RemarkSection 9 –Detailing ofMembers andThe followings are highlighted :(i) Table 9.1 under Cl. 9.2 tabulates minimum steelpercentage equal to Table 3.25 of BS8110 forPt. 1 Section 3and 5.2.7The followings are highlighted :(i) The analysis procedures are largelyold-fashioned relying on old theories ofThe followings are highlighted :(i) The stress reduction indesign of cantileveredA-15

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsparticularbeams;rules (ii) In 9.2.1.4 in relation to maximum distance ofbars in tension as similar to BS8110 Pt. 13.12.11.2, the stipulation in BS8110 thatdemonstration of crack width < 0.3 mm can beexperiences;accepted is omitted;(ii)(iii) In 9.2.1.5, a requirement of 15% span momentbe used to design beam support even undersimply supported assumption is not found inBS8110. Furthermore, it is also stated in theclause that total tension re-bars of a flangedbeam over intermediate supports can be spreadover the effective width of the flange providedthat half of the steel within the web width.There is also no such provision in BS8110;(iv) 9.2.1.8 requiring 30% of the calculatedmid-span re-bars be continuous over supportappears to be adopted from Fig. 3.24 a) ofBS8110. However, the circumstances by whichthe Figure is applicable as listed in 3.12.10.2 ofthe BS is not quoted;(v)9.2.1.9 requires top steel of cantilever to extendbeyond the point of contraflexure of thesupporting span whilst Fig. 3.24 c) requires atleast half of the top steel to extend beyond halfspan of the cantilever or 45φ;(vi) In 9.2.2, maximum spacing of bent-up bars isstipulated whilst no such requirement is foundin BS8110;(vii) Torsional links has to be closed links (shapecode 77 of BS4466) as required by BS8110 Pt.2 2.4.8. However, 9.2.3 of the Code provides analternative of using closed links of 135 o bend;(viii) In 9.3.1.1 (b) in relation to maximum spacingJohansen, Hillerborg. Detailings to cater forbehaviours not well understood or quantifiedare thus provided, though the determination ofwhich are largely empirical or from pastThough ductility is not a design aid explicitlystated in the BS, the BS does requires 135 obend of links in anchoring compression bars incolumns and beams (Pt. 1 3.12.7.2).Remarkprojecting structures inPNAP173 is notincorporated is likelybecause the PNAP is basedon working stress designmethod. So there should besome other approaches andthis is not mentioned in theCoPConc2004;(ii) Ductility is moreemphasized inCoPConc2004 9.9 whichlargely stem from seismicdesign.A-16

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsof re-bars in slab is more detailed than BS8110Pt. 1 3.12.11.2.7 and appears more reasonable.The provisions are, in fact, more stringent :(a) for principal re-bars, 3h ≤ 400 mm whilstBS8110 is 3h ≤ 750 mm;(b) for secondary re-bars 3.5h ≤ 450 mmwhilst no provision in BS8110;(c) more stringent requirements are added forslabs with concentrated loads or areas ofmaximum moments whilst no similarrequirements are found in BS8110;(ix) The first para. in 9.3.1.3 requires half of thearea of the calculated span re-bars be providedat the simply supported slabs and end supportof continuous slabs. The requirement isidentical to BS8110 Pt. 1 3.12.10.3.2. However,the provision in BS8110 is under the conditionlisted in 3.12.10.3.1 that the slabs are designedpredominantly to carry u.d.l. and in case ofcontinuous slabs, approximately equal span.These conditions are not mentioned in theCode;(x) In 9.3.1.3, there is also a provision that if theultimate shear stress < 0.5v c at support, straightlength of bar beyond effective anchorage for1/3 of support width or 30 mm (whichever isthe greater) is considered effective anchorage.No similar provision is found in BS8110;(xi) 9.3.1.6 requiring closed loops of longitudinalrebars at free edge of slab is not found inBS8110;(xii) 9.3.2 is in relation to shear in slabs whichshould be identical to that for beams. Howeverit is stated that shears should be avoided inRemarkA-17

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsslabs < 200 mm thick;(xiii) 9.4 in relation to cantilevered projectingstructures has incorporated requirements forminimum thickness, minimum steel areas,maximum bar spacing, anchorage length fromPNAP173. However, other requirements suchas design with reduced stresses are not found;(xiv) 9.5.2.3 contains more stringent requirements oflinks in circular columns than that in BS8110Pt. 1 3.12.7.3 as the termination of links shouldbe at 135 o hook;(xv) There is an extra requirement in the maximumspacing of traverse reinforcement in wall in9.6.3 which is 400 mm, as in comparison withBS8110 Pt. 1 3.12.7.4;(xvi) 9.6.5 in relation to reinforcement provisions toplain walls include BS8110 Pt. 1 3.9.4.19 to 22.However, 3.9.4.23 in relation to plain wallswith more than 1/10 in tension to resist flexureis not included. Anyhow, this is not importantas the wall should not be regarded as plainwall;(xvii) In 9.7.1 and 9.7.2 in relation to pile caps andfootings, it is stipulated that the minimum steelpercentage should refer to Table 9.1 thoughTable 9.1 is under the heading “Beam”. So theminimum steel percentage of 0.13% for HYbars should be observed. There is no explicitprovision in BS8110 for minimum steel in pilecaps and footings;(xviii) 9.7.3 in relation to tie beams has includeda requirement that the tie beam should bedesigned for a load of 10 kN/m if the action ofcompaction machinery can cause effects to theRemarkA-18

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentstie beams. This is not found in BS8110;(xix) 9.8 in relation to design of corbels carries thesame content as 6.5.2.2 to 6.5.2.4 and isidentical to BS8110 Pt. 1 5.2.7;(xx) 9.9 contains requirements more stringent thanBS8110 in detailing with the aim to enhanceductility. The followings are highlighted :(a) 9.9.1.1(a) requires steel percentage inbeam > 0.3% and percentage of tensilesteel < 2.5%;(b) 9.9.1.1(c) requires anchorage of beam barinto exterior column to commence beyondcentre line of column or 8φ instead ofcolumn face unless the moment plastichinge can be formed at 500 mm or halfbeam depth from column face;(c) 9.9.1.1(d) imposes restriction in locationsof laps and mechanical couplers – (i) notwithin column/beam joints, (ii) not within1 m from potential plastic hinge locations;(iii) reversing stresses exceeding 0.6f yunless with specified confinement by links.In addition, bars be terminated by a 90 obend or equivalent to the far face ofcolumn;(d) 9.9.1.2(e) requires distribution andcurtailment of flexural re-bars be attainedin critical sections (potential plastic hingeregions);(e) 9.9.1.2(a) states the link spacing in beam

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentsanchored by means of 135 o or 180 o hooks.Anchorage by 90 o hooks or welded crossbars not permitted;(g) 9.9.2.1(a) states min. (0.8%) and max.steel% (4% with increase to 5.2% at lap)incolumn;(h) 9.9.2.1(a) requires the smallest dia. of anybars in a row > 2/3 of the largest bar;(i) 9.9.2.1(a) limits max. dia. of column re-barthrough beam by (eqn 9.7) dependent onbeam depth, with increase by 25% if notforming plastic hinge;(j) 9.9.2.1(b) requires spacing of links tolongitudinal bars not be spaced further than1/4 of the adjacent column dimension or200 mm;(k) 9.9.2.1(c) requires anchorage of columnbar into exterior beam or foundation tocommence beyond centre line of beam orfoundation or 8φ instead of interface unlessthe moment plastic hinge can be formed at500 mm or half beam depth from columnface;(l) 9.9.2.1(d) states restrictions in locations oflaps;(m) 9.9.2.2 describes the establishment of“critical regions” in columns where thereare extra requirements on links – (i) linkspacing in column < the lesser of 6φ andleast 1/4 of column lateral dimension; (ii)each longitudinal bar be laterally supportedby a link passing around the bar andhaving an included angle < 135 o . (Regionsother than “critical regions” fallow 9.5.2)RemarkA-20

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. ContentsRemarkSection 10 The followings are highlighted :(i) 10.2 lists figures for construction toleranceswhilst BS8110 refers most of the requirementsto other BS;(ii) 10.3.4 in relation to sampling, testing andcompliance criteria of concrete. They areextracted from HKB(C)R but withincorporation of 100 mm test cubes. Suchprovision is not found in BS8110;(iii) The sub-clause on “Concreting in coldweather” in BS8110 is not incorporated. 10.3.7on “Concreting in hot weather” is modifiedfrom BS8110 Pt. 1 6.2.5 (reference to BSdeleted);(iv) Table 10.4 is similar to BS8110 Pt. 1 Table 6.1.However the parameter t (temperature) isdeleted and the categorization of cement isOPC and “others” instead of the few types of(v)cement in BS8110;10.3.8.1 contains general requirements for“Formwork and falsework” similar (but notidentical) to BS8110 Pt. 1 6.2.6.1;(vi) 10.3.8.2 lists criteria for striking of formworkidentical to that in BS8110 Pt. 1 6.2.6.3.1. Inaddition, provisions for using longer or shorterstriking time for PFA concrete and climbingformwork are included;(vii) Minimum striking time in 10.3.8.2 are inaccordance with HKB(C)R Table 10 (with theaddition of props to cantilever requiring 10days striking time) instead of BS8110 Pt. 1Table 6.2. Furthermore, BS8110 Pt. 1 Table 6.2gives temperature dependent striking timePt. 1 2.3,Section 6,Section 7,Section 8The followings are highlighted :(i) Pt. 1 2.3 lists general requirements forinspection of construction;(ii) References to other BS are often stated inSection 6 and 7;(iii) Provisions of works in extreme temperaturesare given which are deleted in CoPConc2004.The followings are highlighted :(i) The first part of Section 10of CoPConc2004 mainlystems from HKB(C)R,CS1, CS2 whilst the secondpart incorporatesworkmanship requirementslisted in BS8110 Pt. 1Section 6;(ii)(iii)A-21

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentswhilst the striking times in CoPConc2004 arenot temperature dependent;(viii) The contents of 10.3.9 in relation to surfacefinish are extracted from BS8110 Pt. 1 6.2.7.However, the general requirements aredifferently written and the “classes of finish”have been deleted;(ix) 10.3.10 and 10.3.11 in relation to joints areidentical to BS8110 Pt. 1 6.2.9 and 6.2.10though the wordings are different, except thelast sentence of 6.2.9 last para. in relation totreating vertical joint as movement joint;(x) 10.4.1 contains general requirements on re-barsto standards CS2 and other acceptablestandards whilst BS8110 Pt. 1 7.1 requiresconformance to other BS;(xi) 10.4.2 in relation to cutting and bending ofre-bars is identical to BS8110 Pt. 1 7.2 except(a) conformance is not restricted to BS but toacceptable standards; and (b) the requirementof pre-heating re-bars at temperatures below5 o C is deleted;(xii) 10.4.3 is effectively identical to BS8110 Pt. 17.3 except that the requirement for spacerblocks be of concrete of small aggregates ofequal strength to the parental concrete isreplaced by spacer blocks to acceptablestandards;(xiii) 10.4.6 is effectively identical to BS8110 Pt. 17.6 except (a) conformance to BS changed toacceptable standards; (b) detailed descriptionsof the types of welding omitted; and (c)requirement to avoid welding in re-bar bendsomitted;RemarkA-22

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contents(xiv) 10.5.1 is identical to BS8110 Pt. 1 8.1 exceptconformance to BS is changed to acceptablestandards;(xv) 10.5.5.3 in relation to tensioning apparatus ofprestressing tendons is effectively identical toBS8110 Pt. 1 8.7.3 except that CoPConc2004has an additional requirements that apparatusbe calibrated within 6 months;(xvi) 10.5.5.4 in relation to pre-tensioning ofdeflected tendons, compressive and tensilestresses should be ensured not to exceedpermissible limits during transfer ofprestressing force to the concrete with therelease of holding-up and down forces. BS8110Pt. 1 8.7.4.3 has omitted the compressiveforces;(xvii) 10.5.5.5(b) requires anchorage of posttensioning wires to conform to acceptablestandards whilst BS8110 Pt. 1 8.7.5.2 requirescompliance with BS4447(xviii) 10.5.5.5(d) in relation to tensioningprocedures which is identical to BS8110 Pt. 18.7.5.4, the requirement of not carrying outtensioning below 0 o C is omitted. Further, theparagraph in BS8110 stipulating that when fullforce cannot be developed in an element due tobreakage, slip during the case when a large no.of tendons is being stressed is omitted inCoPConc2004;(xix) 10.5.7 contains detailed provisions for groutingof prestressed tendons whilst BS8110 Pt. 1 8.9requires compliance to BS EN 445, 446.RemarkSection 11 – This section outlines measures and procedures in – No similar provisions in BS8110. The control in CoPConc2004A-23

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. ContentsQuality general quality assurance and control, with referenceAssurance to local practice. The followings are highlighted :and Control (i) Control are on design, construction andcompleted products;(ii) Control can be by independent organization;(iii) Concrete must be from supplier certified underthe Quality Scheme for the Production andSupply of Concrete (QSPSC);(iv) Control on construction includes surveillancemeasures.Remarkare summaries of local goodpractice.Section 12 –PrestressedConcreteThis section is basically identical to Section 4 ofBS8110 Pt. 1. The followings are highlighted :(i) 12.1.5 in relation to durability and fireresistance makes reference to previousrecommendations in Sections 4 and 10 whilstBS8110 makes reference also to Part 2 ofBS8110;(ii) 12.2.3.1 in relation to redistribution of momentsis restricted to concrete grade C70, as inconsistency with reinforced concrete. BS8110Pt. 1 4.2.3.1 does not have this limitation. Butthe BS covers grades up to 40;(iii) The first loading arrangement in 12.3.3 forcontinuous beam is not found BS8110 Pt. 14.3.3. The loading arrangement is in consistencywith 5.1.3.2 for reinforced concrete beams.Though not truly adequate (per similarargument as above), CoPConc2004 is moreconclusive than BS8110;(iv) 12.3.8.2 gives ultimate concrete stress 7.0 MPa,as similar to r.c. works;(v) 12.8.2.2 in relation to 1000 h relaxation valuewhich is identical to BS8110 Pt. 1 4.8.2.2, “UK”Pt. 1 Section 4 The provisions are general. CoPConc2004 follows quiteclosely the provisions inBS8110 except for minorchanges.A-24

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix AHK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985)Clause No. Contents Clause No. Contentshas been deleted in description ofmanufacturer’s appropriate certificate;(vi) 12.8.4 and 12.8.5 in relation to shrinkage andcreep of concrete make reference to 3.1.8 and3.1.7 whilst BS8110 Pt 1. 4.8.4 and 4.8.5 listUK data;(vii) 12.10 makes reference to 8.10.2.2 fortransmission lengths in pre-stressed memberswhich is titled “transfer of prestress” which isidentical to BS8110 Pt. 1 4.10.1 except that the2 nd paragraph of the BS in relation to thedifficulty of determination of transmissionlength has been deleted;(viii) 12.12.3.1(a) is identical to BS8110 Pt. 14.12.3.1.1 except that not only protectionagainst corrosion is added. In 12.12.3.1(c),reference for protection against fire is notidentical to BS8110;RemarkSection 13 –Load Tests ofStructures orparts ofstructuresThis section contains testing of structures duringconstruction stage under circumstances such assub-standard works are suspected and visible defectsare identified.– No similar provisions in BS8110.A-25

Appendix BAssessment of BuildingAccelerations

Appendix BAssessment of “along wind” acceleration of Buildings (at top residential floor)Underlying principles :Two Approaches are outlined in this Appendix :(i)The first one is based on the assumption that the building will undergo simpleharmonic motion under wind loads. Thus the equation of governing simple2harmonic motion which is & x= −ω x where & x& is the acceleration, x is thedisplacement of the motion, ω is the circular frequency of the building equalto 2 πf( f is the natural frequency of the building) can be used. However,generally only the “dynamic resonant component” of the motion is needed forcalculating the acceleration. So if the G factor which is equal to22gfSE1 + 2IhgvB + in Appendix F of the Wind Code 2004 is used toζarrive at a total displacement which can be considered to be made of up ofthree components : (a) the static part which is 1 in the equation; (b) thedynamic background component which is2v2 I g B ; and (c) the dynamichresonant component2gfSE2 Ih, it is the last component that should beζ2multiplied to ω to arrive at the acceleration causing discomfort. So it isonly necessary to calculate the displacement due to the dynamic resonantcomponent by multiplying the total displacement by the factor2 ⎛2g⎞fSE ⎜2 gfSE2 I+ I g B +⎟h1 2ζ ⎜h v. Alternatively, the same result can beζ ⎟⎝⎠obtained by multiplying the factor2gfSE2 Ihto the static wind pressure,ζi.e. Table 2 of the Hong Kong Wind Code 2004. The circular frequency, ω ofthe building can either be obtained by detailed dynamic analysis or by someempirical formula such as 460/h.(ii)The second approach is that listed in Australian Wind Code AS/NSZB-1

Appendix B31170.2:2002 Appendix G2. It is based on the simple formula a = Mˆ2 bm hwhere m0is the average mass per unit height of the building, h is the0average roof height of the building above groundMˆ bis the resonantcomponent of peak base bending moment. By the “resonant component”, theapproach is also based on the same principle by using only the dynamicresonant component in deriving acceleration as the factorSEg Risζmultiplied to the overturning moment for assessment of acceleration. Theparameters comprising m0and h are used for assessment of the dynamicproperties of the building. In addition, there is a denominator of 1+ 2g vIhinthe expression forMˆ bin the Australian Code as different from Hong KongWind Code, the reason being that the Australian Code is based onVdes, θwhich is 3 second gust whilst Hong Kong Code is based on hourly mean windspeed. So this factor should be ignored when using Hong Kong Code which isbased on hourly mean speed.Furthermore, two aspects should also be noted :(i)The Concrete Code requires the wind load for assessment of acceleration to be1-in-10 year return period of 10 minutes duration whilst the wind load arrivedfor structural design in the Hong Kong Wind Code is based on 1-in-50 yearreturn period of hourly duration. For conversion, the formula listed inAppendix B of the Wind Code can be used (as confirmed by some experts thatthe formula can be used for downward conversion from 1-in-50 year to1-in-10 year return periods). The 10 minutes mean speed can also be taken asidentical to that of hourly mean speed (also confirmed by the experts.) Or1.27alternatively, as a conservative approach, the factor 1−0.62Iln( t / 3600)can be applied where I is the turbulence intensitytaken at top of the building and t = 600 sec;I⎛ h ⎞= 0.087⎜⎟⎝ 500 ⎠−0.11(ii)The damping ratio recommended in the Wind Code which is 2% is forB-2

Appendix Bultimate design. A lower ratio may need to be considered for serviceabilitycheck including acceleration. Nevertheless, a 10-year return period at dampingratio 2% should be accepted which is the general practice by the Americans.The worked examples follow are therefore based on damping ratio of 2%,though the readers can easily work out the same for damping ratio of 1%under the same principle.The procedures for estimation of acceleration are demonstrated by 3 worked examplesthat follow :Worked Example B-1For the 40-storey building shown in Figure B-1 which has been analyzed by ETABS,the acceleration of the top residential floor in the for wind in X-direction is to becomputed.YXFigure B-1 – 40 storeys building for Worked Example B-1Data : Building height h = 121. 05 m;Building plan width and depth are b = d = 43 m;Lowest building natural frequencies for the respective motion can beobtained with reference to the modal participating mass ratios as revealed bydynamic analysis in ETABS or other softwares:na1 = 0.297 Hz for rotation about Z axis (torsional)B-3

na2 = 0.3605 Hz for translation along Y-directionna1 = 0.3892 Hz for translation along X-directionFor wind in X-direction :−0.11−0.11⎛ h ⎞⎛121.05⎞I h= 0.1055⎜⎟ = 0.1055⎜⎟ = 0.1021; gv= 3. 7⎝ 90 ⎠⎝ 90 ⎠g 2 ln(3600n) = 2ln(3600×0.3892) = 3.8066 ;f=a0.25⎛ h ⎞ ⎛121.05⎞L h= 1000⎜⎟ = 1000⎜⎟ = 1865.35 ;⎝10⎠ ⎝ 10 ⎠11B === 0.6989 ;2 22236h+ 64b36×121.05 + 64×431+1+1865.35L h0.110.110.25Appendix B⎛ h ⎞ ⎛121.05⎞Vh= Vg⎜ ⎟ = 59.5⎜⎟ = 50.905m/sec;⎝ 500 ⎠ ⎝ 500 ⎠naLh0.3892×1865.35N = == 14.262 ;Vh50.90511S === 0.1019⎡ 3.5n⎤ 4 ⎤ ⎡ 3.5×0.3892×121.05⎤⎡ 4×0.3892×43⎤⎢1⎥⎢⎡ +ahn1 +ab⎥ ⎢1+150.905 ⎥⎢+⎣⎦⎣50.905 ⎥⎣ Vh⎦⎣Vh⎦⎦0.47N0.47×14.262E = == 0.0793;2 62 5 / 62 + N5 /2 + 14.262G = 1+2I=1.8155;( ) ( )h222gfSE2 3.8066 × 0.1019×0.0793gvB + = 1+2×.1021 3.7 × 0.6989 +ζ0.0222gfSE3.8066 × 0.1019×0.0793Gres = 2Ih= 2×.1021= 0.494 ;ζ0.02∴Procedures :G res/ G = 0.272Deflection (translation and rotation) of the centre of the top floor calculatedin accordance with Appendix G of HKWC2004 isX-direction : 0.069mY-direction : 0.00061mZ-direction : 0.000154radFor this symmetrical layout, the Y-deflection and Z-rotation are small andcan be ignored.(i)Conversion from 50 years return period to 10 years return period is by thefactor listed in Appendix B of HKWC2004. The factor isB-4

Appendix B( )2⎛ 5 + ln R ⎞⎜ ⎟⎝ 5 + ln50 ⎠⎛ 5 + ln10 ⎞= ⎜ ⎟⎝ 5 + ln50 ⎠2= 0.6714(ii)Conversion from hourly mean wind speed to 10 minutes mean wind speed isby the factor1.27( t /3600) = 1−0.62×0.1021 ln( 600/3600) = 1. 0611.271−0.62Iln(iii) So the displacements converted to contain only the dynamic resonantcomponent and to 10 years return period, 10 minutes wind speed can beobtained by multiplying the deflections obtained in accordance with AppendixG of HKWC2004 by the aggregate factor of 0 .272×0.6714×1.061 = 0. 1938(iv) The X-deflections for calculation of accelerations is therefore(v)0 .1938× 0.069 = 0.0134 m;The acceleration of the centre of the block in X-direction is therefore22( 2 n ) × 0.0134 = ( 2π× 0.3892) × 0.0134 0. 801π a 3=m/sec 2 as the fundamentalfrequency for X-translation is na3 = 0. 3892 Hz listed in the data.Worked Example B-2The acceleration of the block in Worked Example B-1 is redone by the AustralianCode AS/NSZ 1170.2:2002 Appendix G2 :Total dead load is 539693 kN and total live load is 160810 kNUsing full dead load and 40% live load for mass computation :Mass per unit height is( 539693 + 160810×0.4) ÷ 9.8 33m0 =× 10 = 509.165×10 kg/m121.05Overturning moment at 50 years return period is 1114040 kNm when wind is blocingin the X-direction. When the moment is converted to contain only the dynamicresonant component and to 10 years return period, 10 minutes wind speed, it becomesM ˆ = 0.272×0.6714×1.061×1114040 = 215857 kNm, the factors are quoted frombWorked Example 1.So the acceleration in the X-direction is33 ˆ 3×215857×10a = M= 0.0872 b=m/sec 2 .32m h 506.165×10 × 121.050which is greater than that in Worked Example B-1.B-5

Appendix BWorked Example B-3Another worked example for finding acceleration in Y-direction is demonstrated for abuilding shown in Figure B-2 where torsional effect is significant. The buildingsuffers significant torsion where the displacement and acceleration of Point A (atdistance 30m in the X-direction and 1.6m in the Y-direction from the centre of rigidityof the building) is most severe. The provision in the Australian Code should be quitelimited in this case and therefore not used.C, centre of rigidity1.6mA30mFigure B-2 – Layout of Building where torsional effect is significantThe first 3 fundamental frequencies are listed as follows. They can be read fromdynamic analysis of the building by ETABS with reference to the modal participatingmass ratios or other softwares. The dynamic resonant component factor2Ig2 h fSE /ζ are also calculated for the respective direction of motion whilst thedynamic magnification factor G for wind in Y-direction is calculated to be 1.8227.DirectionFundamentalPeriods (sec)Frequencyf(Hz)= 0.Circular2gfSEfrequency Gres= 2IhGω = 2πf (Hz) ζY-direction 2.6598 na376 ω = 2. 13623 0.512 0.2809Z-rotation 1.8712 na= 0. 5344 ω = 3. 23578 0.3185 0.1747X-direction 1.5652 n = 0. 6389 ω3= 4. 014 0.3036 0.1665aG resTable B-1 – Fundamental Frequencies of Worked Example B-2B-6

Appendix BThe displacements of centre of rigidity of the building at the top residential floor asper analysis in accordance with Appendix G of the Wind Code 2004 after applicationof the dynamic magnification factor, G is as follows in Table B-2. The correctedvalues after discount for (i) G resonant/ G ; (ii) 10 minutes duration (factor 1.06); and (iii)10 years return period (0.6714) are also listed.Displacement at Centre of Rigiditybefore adjustment (read from after adjustment for (i), (ii), (iii)ETABS output)X-displacement 0.0262m 0.00311mY-displacement 0.119m 0.0238mZ-rotation 0.00196rad 0.000244radTable B-2 – Displacement of Worked Example B-2The acceleration of the building at its centre of rigidity in X-direction, Y-direction andThe acceleration of the respective directions are :23∆ x=X-direction : ω = 4.0142 × 0.00311 0. 0501m/sec 2 ;21∆ y=Y-direction : ω = 2.36232 × 0.0238 0. 1328 m/sec 2 ;22θ z=Z-direction : ω = 3.35782 × 0.000244 0. 00275 rad/sec 2 ;The linear acceleration at point A will be the vector sum of that in the X andY-directions, each of which in turn comprises linear component equal to that in thecentre of rigidity and a component being magnified by the torsional effect.Linear acceleration due to Z-rotation acceleration is 0 .00275× 30 = 0. 0825m/sec 2 .Total acceleration in Y-direction is taken as the square root of sum of squares of thedirect linear Y acceleration and that induced by rotation. The reason why theacceleration is not taken as algebraic sum of both is because they do not occur at thesame frequency. So the total acceleration in Y-direction is0.132822+ 0.0825 = 0.1563m/sec 2 .Similarly, linear acceleration in the X-direction due to Z-rotation acceleration is0 .00275× 1.6 = 0.0044 m/sec 2 ;22Total acceleration in X-direction is 0.0501 + 0.0044 = 0. 0503 m/sec 2 .The vector sum of the acceleration of point A is thereforeB-7

Appendix B0.156322+ 0.0503 = 0.1642 > 0.15 m/sec 2 as required in Cl. 7.3.2 of the Code.So provisions should be made to reduce the acceleration.Thus it can be seen that, though the deflection complies with the limit of H/500, theacceleration exceeds the limit of 0.15m/sec 2 . However, compliance with the formermay be adequate as per Code requirements.Limitations of the two approachesIt should be borne in mind that the approaches described above are simplified ones.As the approaches are very much based on the assumed natural frequency of thebuilding or arrival of such value by empirical method (the Australian Code), it followsthat they should be used in care when the dynamic behaviour of the building iscomplicated such as having significant cross wind effects or coupling of buildingmodes is significant.B-8

Appendix CDerivation of Basic DesignFormulae of R.C. Beam sectionsagainst Flexure

Derivation of Basic Design Formulae of R.C. Beam sections againstBendingAppendix CThe stress strain relationship of a R.C. beam section is illustrated in Figure C-1.d’εult= 0.0035xdneutral axisStress DiagramStrain DiagramFigure C-1 – Stress Strain diagram for BeamIn Figure C-1 above, the symbols for the neutral axis depth, effective depth, cover tocompressive reinforcements are x, d, and d’, as used in BS8110 and the Code.To derive the contribution of force and moment by the concrete stress block, assumethe parabolic portion of the concrete stress block be represented by the equation2σ = A ε + Bε(where A and B are constants) (Eqn C-1)dσSo = 2 Aε+ Bdε(Eqn C-2)dσAs = Ec⇒ B = Ecwhere E cis the tangential Young’s Modulus ofdεε =0concrete listed in Table 3.2 of the Code.AlsodσdεB E= 0 ⇒ 2Aε0+ B = 0 ⇒ A = − = − cε = ε2ε02ε0 0(Eqn C-3)Asf cuσ = 0. 67 when ε = ε0γfmEE0.67 fcu c ccu ccu∴ 0.67− = − ⇒ = ⇒ ε0=(Eqn (C-4)2γ ε ε γmε020 mε02 Ecγ0m(accords with 3.14 of the Concrete Code Handbook)EC-11.34 f

Appendix CE cA = − where2ε 0ε0 =1.34 fE γcmcuSo the equation of the parabola isConsider the linear strain distributionσE c 2= − ε + E cε2ε0for ε ≤ ε0xε 0/ ε ultεult= 0.0035ε = ε 0uhxFigure C-2 – Strain diagram across concrete sectionAt distance u from the neutral axis,uε = ε ultxSo stress at u from the neutral axis up toεx 0 is εult22c 2 Ec⎛ u ⎞ ⎛ u ⎞ Ecεult 2 Ecεultε + Ecε= − ⎜εult ⎟ + Ec⎜εult ⎟ = − u u202ε0 ⎝ x ⎠ ⎝ x ⎠ 2ε0xxEσ = −+2ε(Eqn C-5)Based on (Eqn C-5), the stress strain profiles can be determined. A plot for grade 35 isincluded for illustration :Stress Strain Profile for Grade 35Stress (MPa)1816141210864200.3769 whereε 0 = 0.0013190 0.2 0.4 0.6 0.8 1Distance Ratio from Neutral axisFigure C-3 – Stress strain profile of grades 35C-2

Sectional Design of rectangular Section to rigorous stress strain profileAppendix CMaking use of the properties of parabola in Figure C-4 offered by the parabolicsection as Fc1given bybArea =2ab3centre of massa3a8Figure C-4 – Geometrical Properties of Parabolaf fF b2 ε0cu1.34εcucx0.6701= = bx(Eqn C-6)3 εultγm3γmεultand the moment exerted by Fc1about centre line of the whole sectionM⎡h⎛ ⎞⎢ −⎜ −0 3 ε0x 1⎟ − x⎣2⎝ εult ⎠ 8 εult⎤⎥ = F⎦⎡h⎛ 5 ε ⎞⎤⎢ − x⎜1−⎟⎥⎣2⎝ 8 ε ⎠⎦ε =0c1Fc1c1ult(Eqn C-7)The force by the straight portion isF0.67 f=γ⎛ ε0⎜ x − x⎝ εult⎞ 0.67 f⎟b=⎠ γmbx ⎛ ε⎜ −0⎝ εultcucuc2 1m⎞⎟⎠(Eqn C-8)The moment offered by the constant part about the centre line of the whole section isMc2⎡h⎛ ε ⎞ ⎤= ⎢ − ⎜ −0 xF⎟c21 ⎥⎣2⎝ ε ⎠ 2 ult ⎦(Eqn C-9)The compressive force by concrete as stipulated in (Eqn C-6) and (Eqn C-8) isFc1.34ε ⎛ ⎞⎛ = + =0fcu0.67 f +cubxε⎜ −00.67 f⎟ =cubxε⎜c cbx−0F1F2133γmεultγm ⎝ εult ⎠ 3γm ⎝ εult⎞⎟⎠For singly reinforcing sections, moment by concrete about the level of the tensile steelis, by (Eqn C-7) and (C-9)C-3

Appendix C⎡ ⎛ 5 ε ⎞⎤⎡ ⎛ ⎞ ⎤= + = ⎢ −⎜ −0ε⎟⎥+ ⎢ − ⎜ −0 xM M⎟c1Mc2Fc1d x 1 Fc2d 1 ⎥⎣ ⎝ 8 ε ⎠⎦⎣ ⎝ ⎠ 2ultεult⎦1.34ε⎡ ⎛ ⎞⎤⎛ ⎞⎡⎛ ⎞ ⎤=0fcu5 ε⎢ −⎜ −00.67 f +cubxε⎥⎜ −0ε⎟⎟⎢− ⎜ −0xbx d x 11 d 1⎟ ⎥3γmεult⎢⎣⎝ 8 εult⎠⎥⎦γm ⎝ εult ⎠⎢⎣⎝ εult⎠ 2⎥⎦M 1.34ε ⎡ ⎛ ⎞⎤⎛ ⎞⎡⎛ ⎞ ⎤⇒ =0fcu x x 5 ε⎢ −⎜ −00.67 f⎥ +cu x ε⎟⎜ −0 1 ε x⎟⎢−⎜ −01 11 1 1⎟2 ⎥bd 3γmεultd⎣ d ⎝ 8 εult ⎠⎦ γmd ⎝ εult ⎠⎣ 2 ⎝ εult ⎠ d⎦2M 0.67 f ⎪⎧⎛ ⎞ ⎡⎛ ⎞ ⎤⎪⎫⇒ =cu x 1 ε⎨⎜ −0 1 1 ε0 1 ε0 x1⎟ + ⎢−+ −⎜⎟ ⎥ ⎬bd2 γmd ⎪⎩ ⎝ 3 εult ⎠ ⎢ 2 3 εult⎣12 ⎝ εult ⎠ ⎥ d⎦ ⎪⎭2 20.67 f ⎡ 1 1 1 ⎤cuε0⎛ ε0⎞ ⎛ x ⎞ 0.67 f ⎛ 1 0⎞+cuε x M⇒ ⎢−+ − ⎥⎜⎟ 1 − = 022 3 12⎜⎟⎜ −3⎟γm ⎢ εult⎣⎝ εult ⎠ ⎥⎦⎝ d ⎠ γm ⎝ εult ⎠ d bd(Eqn C-10)which is a quadratic equation in dx .As dxis limited to 0.5 for singly reinforcing sections for grades up to 45 undermoment distribution not greater than 10% (Clause 6.1.2.4 of the Code), by (Eqn C-10),Mwill be limited to K ' values listed as2bdf cuK ' = 0.154 for grade 30K ' = 0.152 for grade 35K ' = 0.151 for grade 40K ' = 0.150 for grade 45which are all smaller than 0.156 under the simplified stress block.xHowever, for 45 f ≤ 70 where is limited to 0.4 for singly reinforcingd< cusections under moment distribution not greater than 10% (Clause 6.1.2.4 of the Code),Magain by (Eqn 3-1) will be limited to2bdK ' = 0.125 for grade 50K ' = 0.123 for grade 60K ' = 0.121 for grade 70f cuwhich are instead greater than 0.120 under the simplified stress block. This is becauseat concrete grade > 45, the Code has limited the rectangular stress block to 0.8 timesof the neutral axis depth.C-4

Appendix CxWith the analyzed by (Eqn C-9), the forces in concrete d1.34ε fcufcubx⎛ ⎞ Fcfcu⎛ xFcFcFcbxm ultmultbd⎟ ⎞= + =00.67 ε+⎜ −00.67 1 ε⎟ ⇒ =⎜ −01 2113γε γ ⎝ ε ⎠ γm ⎝ 3 ε ult ⎠ dcan be calculated which will be equal to the required force to be provided by steel,thusAst0.67 fcu⎛ 1 ε ⎞ x Astfcu⎛ xfybdmultd bd f⎟ ⎞= ⎜ −01 0.67 1 ε⎟ ⇒ =⎜ −00 .8711γ ⎝ 3 ε ⎠ 0.87yγm ⎝ 3 ε ult ⎠ d(Eqn C-11)MWhen exceeds the limited value for single reinforcement. Compression2bd f cureinforcements at d ' from the surface of the compression side should be added. Thecompression reinforcements will take up the difference between the applied moment2and K 'bd0.87 fyAscbd⎞ ⎛⎜⎛ d'1 − ⎟ = ⎜⎝ d ⎠⎝⎛ M⎟ ⎞⎜− K'f2cuM ⎞ A ⎝ bd fsccu⎟⎠− K'⇒ =(Eqn C-12)2bd fcu ⎠ bd ⎛ d'⎞0.87 fy ⎜1− ⎟⎝ d ⎠And the same amount of steel will be added to the tensile steel.Astbd⎛ M⎟ ⎞⎜− K'f2cu1 0.67 f ⎛ ⎞ ⎝ bd f ⎠=cu 1 ε⎜ −0cu1⎟η+0.87 fyγm ⎝ 3 εult ⎠ ⎛ d'⎞0.87 fy ⎜1− ⎟⎝ d ⎠(Eqn C-13)where η is the limit of dxratio which is 0.5 for grade 45 and below and 0.4 forgrades up to and including 70.Furthermore, there is a limitation of lever arm ratio not to exceed 0.95 which requires2 20.67 f ⎡ 1 1 1 ⎤cuε0⎛ ε0⎞ ⎛ x ⎞ 0.67 f ⎛ 1 012 3 123 ⎟ ⎞+cuε x⎢−+ −⎜⎟ ⎥⎜⎟ ⎜ −γm ⎢ εult⎣⎝ εult ⎠ ⎥⎦⎝ d ⎠ γm ⎝ εult ⎠ d≤ 0.950.67 fcu⎛ 1 ε0⎞ x⎜1−3⎟γn ⎝ εult ⎠ dC-5

Appendix C⇒xd≥⎛ 1 ε ⎞⎜ −00.05 1⎟⎝ 3 εult ⎠⎡11 ε0 1 ⎛ ε0⎢ − +⎜⎢23 εult12⎣⎝ εult⎞⎟⎠2⎤⎥⎥⎦(Eqn C-14)Thus the lower limits for the neutral axis depth ratios are 0.112, 0.113, 0.114 and0.115 for grades 30, 35, 40, 45 respectively. Thus for small moments acting on beamx Mwith not fulfilling (Eqn C-14), A = std 0 .87 f × 0.95d(Eqn C-15)yAs illustration for comparison between the rigorous and simplified stress blockMapproaches, plots of2 against steel percentages for grade 35 is plotted asbdComparison of Reinforcement Ratios for grade 35 according to the Rigorousand Simplfied Stress Block (d'/d = 0.1)M/bd 2Ast/bd - Rigorous Stress ApproachAsc/bd - Rigorous Stress ApproachAst/bd - Simplified Stress ApproachAsc/bd - Simplified Stress Approach141312111098765432100 0.5 1 1.5 2 2.5 3 3.5 4Reinforcement ratios A/bd (%)It can be seen that the differences are very small, maximum error is 1%.However, for high grade concrete where the K ' values are significantly reduced inthe rigorous stress block approach (mainly due to the switching of upper limits of theneutral axis depth ratios from 0.5 to 0.4 and 0.33 for high grade concrete), thedifferences are much more significant for doubly reinforced sections, as can be seenfrom the Design Charts enclosed in this Appendix that compressive steel ratiosincreased when concrete grade switches from grade 45 to 50 as the neutral axis depthratio changes from 0.5 to 0.4.C-6

Appendix CDetermination of reinforcements for Flanged Beam Section – T- or L-SectionsFor simplicity, only the simplified stress block in accordance with Figure 6.1 of theCode is adopted in the following derivation. The symbol η is used to denote theratio of the length of the simplified stress block to the neutral axis depth. Thusη = 0.9 for f ≤ 45 ; η = 0. 8 for 45 f ≤ 70; η = 0. 72 for 70 f ≤100.cu< cu< cuThe exercise is first carried out by treating the width of the beam asbeffand analyzethe beam as if it is a rectangular section. If η of neutral axis depth is within thex hfdepth of the flange, i.e. η ≤ , the reinforcement so arrived is adequate for thed dx hfsection. The requirement for η ≤ isd dxK hfη = 1−1−≤(Eqn C-15)d 0.225 d1The lever arm z = d − ηx2z 1 x 1 ⎛ K ⎞K⇒ = 1−η = 1−⎜11 ⎟ = 0.5 + 0.25 −d 2 d 2− −0.225(Eqn C-16)⎝⎠0.9If, however,C-5.xηdhf>d, the section has to be reconsidered with reference to Figurebeff0.67γmf cuh fηxxdbwFigure C-5 – Analysis of a T or L beam sectionFor singly reinforced sections, taking moment about the level of the reinforcing steel,C-7

Appendix C0.67f⎛ h ⎞cuf 0.67 fcu ⎛ η ⎞M = ( b − ) ⎜ −⎟effbwhfd + bw( ηx) ⎜d− x⎟ γm⎝ 2 ⎠ γm ⎝ 2 ⎠M 0.67 f ⎛ b ⎞ h ⎛ h ⎞cu eff f 1 f 0.67 f ⎛ x ⎞ ⎞⇒ =⎜ −⎟⎜⎟ + ⎜ ⎟⎜⎛ x−cu η1 1η 1 − ⎟ (Eqn C-17)2bwdγm ⎝ bw⎠ d ⎝ 2 d ⎠ γm ⎝ d ⎠⎝2 d ⎠Mf 0.67 f h ⎛ b ⎞⎛ hPutting⎟ ⎞cu f eff 1 f=⎜ −1⎟⎜1−(Eqn C-18)2bwdγmd ⎝ bw⎠⎝2 d ⎠The equation is in fact the contribution of the moment of resistance of the section bythe flange, (Eqn C-17) becomes0.67 fγmcu2η ⎛⎜2 ⎝xd2⎞⎟⎠0.67 f−γmcux M − Mη +2d b dwf= 0(Eqn C-19)xwhich is a quadratic equation for solution of where Mfcan be predetermineddby (Eqn C-18). Providedxd≤ ϕwhere ϕ = 0. 5 for f > 45 ; 0.4 for f > 70and 0.33 for fcu> 100 , single reinforcement be provided by the following equationwhich is derived by balancing the steel force and the concrete force.cucu0 .87fyAst0.67 f=γmcuA⎡⎛b ⎞ h ⎤st0.67 fcueff f x[( b − b ) h + b ηx] ⇒ = ⎜ −1⎟+ η ⎥⎦effwfwbwdγ 0.87 fmy⎢⎜⎣⎝bw⎟⎠ dd(Eqn C-20)xIf = ϕ , the maximum moment of resistance by concrete is reached which is (bydtaking moment about the tensile steel level)0.67 f ⎡ ⎛ h ⎞⎤cuf ⎛ ⎞⎢( ) + 1M⎜ ⎟c max= Mf max+ Mb max= beff− bwhfd − bwηϕd⎜d− ϕd⎟⎥ γm ⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎦M 0.67 ⎡⎛b ⎞ h ⎛ 1 h ⎞ ⎛ ⎞⎤cfcueff ff 1⇒ K' = = ⎢⎜ −1⎟⎜1−⎟ + ηϕ⎜1− ϕ2⎟⎥(Eqn C-21)bwdγm ⎣⎝bw⎠ d ⎝ 2 d ⎠ ⎝ 2 ⎠⎦and tensile steel required will be, by (Eqn C-20)A⎡⎤st,bal 0.67 f ⎛ b ⎞ hcu eff f= ⎢⎜ −1⎟ + ηϕ⎥(Eqn C-22)bwdγm0.87 fy ⎣⎝bw⎠ d ⎦If the applied moment exceedscompressive steel0 .87fyAsc( d − d' ) = M − McMc, the “excess moment” will be taken upAscwith cover to reinforcement'c .C-8

M − M⎡⎤c1 M Mc=⎢ −2 ⎥( 1−d'/ d ) 0.87 f 0.87 f ( 1−d'/ d ) b d b d ⎦Appendix CAsc⇒ =22bwdbwdyy ⎣ w w1 ⎡ M 0.67 f ⎡⎛b ⎞ h ⎛ 1 h ⎞ ⎛ ⎞⎤( ) ⎥ ⎥ ⎤cu eff ff 1= ⎢ − ⎢⎜ −1⎟⎜1−⎟ + ηϕ⎜1− ϕ2⎟⎥0.87 fy1−d'/d ⎢⎣bwdγm ⎣⎝bw⎠ d ⎝ 2 d ⎠ ⎝ 2 ⎠⎦⎦(Eqn C-23)The total tensile steel will beA Ast st,bal Asc= +b d b d b dwwwThe followings are stated for fcu≤ 45 where η = 0. 9 and ϕ = 0. 5 which is mostcommonly used in flexural members:For η × neutral axis depth below flange, (Eqn C-19) can be written as :2⎛ x ⎞x M − Mf0.1809fcu ⎜ ⎟ − 0.402 fcu+ = 0(Eqn C-24)2⎝ d ⎠d bwdwhere 0 .9x ≥ h and Mf 0.67 f h ⎛ b ⎞⎛ hf⎟ ⎞cu f eff 1 f=⎜ −1⎟⎜12−bwdγmd ⎝ bw⎠⎝2 d ⎠A⎡⎛b ⎞ h ⎤st0.67 fcueff f x= ⎢⎜ −1⎟ + 0. 9 ⎥(Eqn C-24)bwdγm0.87 fy ⎣⎝bw⎠ d d ⎦For double reinforcements whereϕ = 0.5 into (Eqn C-23)x > 0. 5dby (Eqn C-24), substituting η = 0. 9 andA= 1 ⎡ 0.67 ⎡⎛b ⎞ h ⎛ 1 h ⎞ ⎤( ) ⎥ ⎥ ⎤scM fcueff ff⎢ − ⎢⎜ −1⎟⎜1−⎟ + 0. ⎥0.87 1−'/33752bwdfyd d ⎢⎣bwdγm ⎣⎝bw⎠ d ⎝ 2 d ⎠ ⎦⎦(Eqn C-25)By (Eqn C-22),A A A f ⎡ beffhf⎤st st,bal sc0.67 ⎛ ⎞cuAsc= + = ⎢⎥ +bwdbwdbwdmf⎜ −1+y ⎣ b⎟ 0. 45(Eqn C-26)γ 0.87 ⎝ w ⎠ d ⎦ bwdC-9

Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.05Grade 30 As t/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bdGrade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Grade 50 Ast/bd Grade 50 Asc/bd141210M/bd 2864200 0.5 1 1.5 2 2.5 3 3.5 4Reinforcement ratios A/bd (%)Chart C-1

Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.1Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bdGrade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Grade 50 Ast/bd Grade 50 Asc/bd141210M/bd 2864200 0.5 1 1.5 2 2.5 3 3.5 4Reinforcement ratios A/bd (%)Chart C-2

Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.15Grade 30 As t/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bdGrade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Grade 50 Ast/bd Grade 50 Asc/bd141210M/bd 2864200 0.5 1 1.5 2 2.5 3 3.5 4Reinforcement ratios A/bd (%)Chart C-3

Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.05Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bdGrade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Grade 50 Ast/bd Grade 50 Asc/bd141210M/bd 2864200 0.5 1 1.5 2 2.5 3 3.5 4Reinforcement ratios A/bd (%)Chart C-4

Appendix DUnderlying Theory and DesignPrinciples for Plate BendingElement

Underlying Theory and Design Principles for Plate Bending ElementAppendix DBy the finite element method, a plate bending structure is idealized as an assembly ofdiscrete elements joined at nodes. Through the analysis, “node forces” at each node ofan element, each of which comprises two bending moments and a shear force can beobtained, the summation of which will balance the applied load at the node. FiguresD-1a and D-1b illustrates the phenomena.1F Z atNode 1M X atNode 12M X atNode 2F Z atNode 4F Z atNode 23M Y atNode 2F Z atNode 3M X atNode 2M Y atNode 3Note :ZYXM X , M Y , F Z representsrespectively the bendingmoments about X, Y axesand the force in the Z axisat the nodes of the platebending elementFigure D-1a – Diagrammatic illustration of the Node Forces at the four Nodes of aPlate Bending Element 1234.F A , external loadapplied at thecommon nodeF 1F 4F 2F 3For equilibrium, F 1 ,F 2 , F 3 and F 4 whichare node verticalshear of 4 elementsat the common nodewill sum up tobalance theexternally appliedforce F A such thatF 1 + F 2 + F 3 + F 4 =F A . Balancing ofmoments is similar.Figure D-1b – Diagrammatic illustration of balancing of Node shear forces at acommon node to 2 or more adjoining elements. The four elements joined at thecommon node are displaced diagrammatically for clarity.D-1

Appendix DThe finite element method goes further to analyze the “stresses” within the discreteelements. It should be noted that “stresss” is a terminology of the finite elementmethod which refer to bending moments, twisting moments and shear forces per unitwidth in plate bending element. They represent the actual internal forces within theplate structure in accordance with the plate bending theory. R.H. Woods (1968) hasdeveloped the famous Wood-Armer Equations to convert the bending moments andtwisting moments (both are moments per unit width) at any point to “designmoments” in two directions for structural design purpose.Outline of the plate bending theoryApart from bending moment in two mutually perpendicular directions as well knownby engineers, a twisting moment can be proved to be in existence by the plate bendingtheory. The bending and twisting moments constitutes a “moment field” whichrepresents the actual structural behaviour of a plate bending structure. The existenceof the twisting moment and its nature are discussed in the followings. Consider atriangular element in a plate bending structure with two of its sides aligning with theglobal X and Y directions as shown in Figure D-2 where moments MXand MY(both in kNm per m width) are acting respectively about X and Y. A moment MBwill generally be acting on the hypotenuse making an angle of θ with the X-axis asshown to achieve equilibrium. However, as the resultant of MXand MYdoes notnecessarily align with MB, so there will generally be a moment acting in theperpendicular direction of MBto achieve equilibrium which is denoted as MT. Thevector direction of MTis normal to the face of the hypotenuse. So instead of“bending” the element like MX, MYand MBwhich produces flexural stresses, it“twists” the element and produce shear stress in the in-plane direction. The shearstress will follow a triangular pattern as shown in Figure D-2 for stress-straincompatibility. MTis therefore termed the “twisting moment”. Furthermore, in orderto achieve rotational equilibrium about an axis out of plane, the shear stress will haveto be “complementary”. As the hypotenuse can be in any directions of the platestructure, it follows that at any point in the plate bending structure, there willgenerally be two bending moments, say MXand MYin two mutually perpendiculardirections coupled with a complementary twisting moment MXYas indicated inFigure 11a. The phenomenon is in exact analogy to the in-plane stress problem wheregenerally two direct stresses coupled with a shear stress exist and these componentsvary with directions. The equations relating MB, MTwith MX, MY, MXYandθ derived from equilibrium conditions are stated as follows:D-2

Appendix DMMBT1=21=212( M + M ) + ( M − M )X( M − M ) sin 2θ− M sin 2θXYYXXYYcos 2θ+ MXYsin 2θ(Eqn D-1)In addition, if θ is so varied that MTvanishes when θ = φ , then the element willbe having pure bending in the direction. The moments will be termed the “principalmoments” and denoted as M1, M2, again in exact analogy with the in-plane stressproblem having principal stresses at orientations where shear stresses are zero. Theangle φ can be worked out by1 −2Mtan1XYφ =(Eqn D-2)2 ( M − M )XYPlate StructureM TYM YθM BComplementary shear stress patternXM XFigure D-2 – Derivation and nature of the “Twisting Moment”D-3

Appendix DM XYYXM YM XM YM XYM XM 1M 1M XYM 2M 2φM XYFigure D-3a – General co-existenceof bending moments and twistingmoment in a plate bending structureFigure D-3b – Principal moment in aplate bending structureAgain, as similar to the in-plane stress problem, one may view that the plate bendingstructure is actually having principal moments “bending” in the principal directionswhich are free of “twisting”. Theoretically, it will be adequate if the designer designsfor these principal moments in the principal directions which generally vary frompoint to point. However, practically this is not achievable for reinforced concretestructures as we cannot vary the directions of the reinforcing steels from point to pointand from load case to load case.The “stress” approach for design against flexure would therefore involve formulae forproviding reinforcing steels in two directions (mostly in orthogonal directions)adequate to resist the “moment field” comprising the bending moments and twistingmoments. The most popular one is the “Wood Armer” Equations by Woods (1968),the derivation of which is based on the “normal yield criterion” which requires theprovided reinforcing steels at any point to be adequate to resist the normal momentwhich is the bending moment MBin any directions as calculated from (Eqn D-1).The effects of the twisting moments have been taken into account in the formulae.The Wood Armer Equations are listed as follows.For bottom steel reinforcement provisions:GenerallyM = M + M∗X∗∗If M < 0 , then M = 0 andXXXXY∗; MY= MY+ MXYM∗Y= MYM+M2XYX;D-4

Appendix D∗∗If M < 0 , then M = 0 andYYM∗X= MX+MM2XYYFor top steel reinforcement provisions:GenerallyM∗X= MX− M∗∗If M > 0 , then M = 0 andX∗∗If M > 0 , then M = 0 andYXYXY∗; MY= MY− MXYMM∗Y∗X= MY= MXM−M2XYXM−M2XYY;(Eqn D-3)The equations have been incorporated in the New Zealand Standard NZS 3101:Part2:1995 as solution approach for a general moment field.The “stress” approach is therefore based on the actual structural behaviour of the platebending structure which is considered as a direct and realistic approach. The approachis particularly suitable for structures analyzed by the finite element method whichproduces a complete set of components of the internal forces of the plate bendingstructures including twisting moments, Q . Design has to cater for all thesecomponents to ensure structural adequacy.maxDesign against shearAs an alternative to checking or designing punching shear in slab in accordance with6.1.5.7 of the Code by which the punching shear load created by column (or pile inpile cap) is effectively averaged over a perimeter, more accurate design or checkingcan be carried out which is based on finite element analysis by which an accurateshear stress distribution in the slab structure can be obtained. The finite elementanalysis outputs the “shear stresses” (shear force per unit width) in accordance withthe general plate bending theory at the “X-face” and “Y-face” of an element which arerespectivelyQXZ∂MY∂MXY= − + and∂x∂yQYZ∂MX∂MXY= − , as∂y∂ydiagrammatically illustrated in Figure D-4. It can be easily shown that the maximumshear after “compounding” these two components will occur in a plane at anorientation−1⎛QXZ⎟ ⎞θ = tan⎜ on plan and the value of the maximum shear is⎝ QYZ⎠D-5

Appendix Dmax22YZQ = Q XZ+ Q as per the illustration in the same Figure. Thus one can viewthat both QXZand Q YZare components of the actual shears in a pre-set global axissystem. The actual shear stress is Qmax, the action of which tends to produce shearfailure at the angle θ on plan as shown in Figure D-3. So the designer needs tocheck or design for Qmaxat the spot. There is no necessity to design for QXZandQYZseparately.Plate structureOut-of plane shear Q XZ , Q YZ and Q θYZXQ XZQ YZθQθPlanθQXZQYZThe triangular element formed fromthe rectangular element by cuttinginto half. By varying the angle θ,the maximum shear on thehypotenuse is obtained.A rectangular element extractedfrom a plate structure showingshear per unit width in the X andY directionsQmaxθXDerivationFor vertical equilibrium, Q θ can be expressed asQ θ × 1 = Q XZ sinθ+ QYZcosθFor Q θ to be maximum setdQθQ= 0 ⇒ QXZXZ cosθ− QYZsinθ= 0 ⇒ tanθ=dθQYZ22QXZQ⇒ QYZθ = Qmax=+2 2 2 2QXZ+ QYZQXZ+ QYZ=2 2Q XZ + Q YZPotential shear failure at theθ = tan −1Q /orientation ( )XZ Q YZQ maxFigure D-4 – Diagrammatic illustration of shear “stresses” in the X and Y facesof an element in Plate bending structure, potential shear failure and Derivation ofthe magnitude and orientation of the design shear stressFollowing the usual practice of designing against shear in accordance with the Code,if the Q does not exceed allowable shear strength of concrete based on v c (themaxD-6

Appendix Ddesign concrete shear stress with enhancement as appropriate) no shearreinforcements will be required. Otherwise, reinforcements will be required to caterfor the difference.The “stress” approach for shear design based on Qmaxcan best be carried out bygraphical method, so as to avoid handling the large quantity of data obtainable in thefinite element analysis. An illustration of the method for a raft footing is indicated inFigure D-5 as :(i) an enveloped shear stress (shear force per unit width) contour map of a structuredue to applied loads is first plotted as shown in Figure D-5(a);(ii) the concrete shear strength contour of the structure which is a contour mapindicating the shear strength of the concrete structure after enhancement of thedesign concrete shear stress (v c ) to closeness of supports in accordance withestablished code requirements (say BS8110) is plotted as shown in FigureD-5(b);(iii) locations where the stresses exceed the enhanced strengths be reinforced byshear links as appropriate in accordance with established code requirements asshown in Figure D-5(c).Figure D-5a – Stress contour of enveloped shear “stresses” of a raft footing due toapplied loadD-7

Appendix DFigure D-5b – Strength contour of the raft footing with enhancementFigure D-5c – Arrangement of shear reinforcementsD-8

Appendix EMoment Coefficients for threeside supported Slabs

Bending Coefficients in the plate of the indicated support conditions and lengthbreadth ratio are interpolated from Table 1.38 of “Tables for the Analysis ofPlates, Slabs and Diaphragms based on Elastic Theory”ba= 20.0933 0.1063 0.11040.0842 0.0956 0.09930.0721 0.0814 0.0844bSimply supported0.0516 0.0574 0.0594Simply supported0.0196 0.0207 0.0211Fixed supportaCoefficients for bending along X direction (+ve : sagging; –ve : hogging)Moments at various points isc . f × a2. × q where q is the u.d.l.E-1

a= 20 0 00.0116 0.0132 0.01380.0158 0.0180 0.0188bSimply supported0.0136 0.0155 0.0161Simply supported-0.0020 -0.0029 -0.0032-0.0607 -0.0698 -0.0729Fixed supportaCoefficients for bending along Y direction (+ve : sagging; –ve : hogging)Moments at various points isc . f q×b2. × where q is the u.d.l.E-2

Appendix FDerivation of Design Formulaefor Rectangular Columns toRigorous Stress Strain Curve ofConcrete

Appendix FDerivation of Design Formulae for Rectangular Columns to RigorousStress Strain Curve of Concrete(I) Computing stress / force contribution of concrete stress blockAssuming the parabolic portion of the concrete stress block as indicated in Fig. 3.8 ofHKCoP2004 be represented by the equation2σ = A ε + Bε(where A and B are constants) (Eqn F-1)dσSo = 2 Aε+ Bdε(Eqn F-2)dσAs = Ec⇒ B = Ecwhere E cis the tangential Young’s Modulus ofdεε =0concrete listed in Table 3.2 of the Code.AlsodσdεB E= 0 ⇒ 2Aε0+ B = 0 ⇒ A = − = − cε = ε2ε02ε0 0(Eqn F-3)Asf cuσ = 0. 67 when ε = ε0γfmEE0.67 f1.34 fcu c ccu ccu∴ 0.67− = − ⇒ = ⇒ ε0=(Eqn (F-4)2γ ε ε γmε020 mε02 Ecγ0m(accords with 3.14 of the Concrete Code Handbook)E cfcuA = − where ε2ε = 1.3400EcγmESo the equation of the parabola isConsider the linear strain distributionσE c 2= − ε + E cε2ε0for ε ≤ ε0xε 0/ ε ultεult= 0.0035ε = ε 0uxhFigure F-1 – Strain diagram across concrete sectionF-1

At distance u from the neutral axis,uε = ε ultxAppendix FSo stress at u from the neutral axis up toεx 0 is εult22c 2 Ec⎛ u ⎞ ⎛ u ⎞ Ecεult 2 Ecεultε + Ecε= − ⎜εult ⎟ + Ec⎜εult ⎟ = − u u202ε0 ⎝ x ⎠ ⎝ x ⎠ 2ε0xxEσ = −+2εBased on (Eqn F-5), the stress strain profiles for grade 35 within the concretecompression section are plotted in Figure F2 for illustration.(Eqn F-5)Stress Strain Profile for Grade 35Stress (MPa)1816141210864200.3769 whereε 0 = 0.0013190 0.2 0.4 0.6 0.8 1Distance Ratio from Neutral axisFigure F-2 – Stress strain profile of grades 35By the properties of parabola as shown in Figure F-3, we can formulate total forceoffered by the parabolic section as Fc1given bybArea =2ab3centre of massa3a8Figure F-3 – Geometrical Properties of ParabolaF-2

Appendix Ff fF b2 ε0cu1.34εcucx0.6701= = bx(Eqn F-6)3 εultγm3γmεultand the moment exerted by Fc1about centre line of the whole sectionM⎡h⎛ ⎞⎢ −⎜ −0 3 ε0x 1⎟ − x⎣2⎝ εult ⎠ 8 εult⎤⎥ = F⎦⎡h⎛ 5 ε ⎞⎤⎢ − x⎜1−⎟⎥⎣2⎝ 8 ε ⎠⎦ε =0c1Fc1c1ult(Eqn F-7)The force by the straight portion isF0.67 f=γ⎛ ε0⎜ x − x⎝ εult⎞ 0.67 f⎟b=⎠ γmbx ⎛ ε⎜ −0⎝ εultcucuc2 1m⎞⎟⎠(Eqn F-8)The moment offered by the constant part about the centre line of the whole section isMc2⎡h⎛ ε ⎞ ⎤= ⎢ − ⎜ −0 xF⎟c21 ⎥⎣2⎝ ε ⎠ 2 ult ⎦(Eqn F-9)Thus if full section of concrete in compression exists in the column sectionF F F f ⎛ ⎞cx= c1+c20.67cu 1 ε=bh bh bh⎜ −01⎟γm ⎝ 3 ε(Eqn F-10)ult ⎠ hM⎡⎤⎡⎤c11.34ε ⎛ ⎞⎛ ⎞ ⎛⎞⎥ = ⎜ ⎟⎢− ⎜ ⎟ ⎜ −⎟⎥⎣ ⎠⎛ ⎞=0fcu h 5 ε⎢ −⎜ −0 1 1.34ε0fcu x 1 x 5 ε0bx x 1⎟122bh 3γmεult⎣2⎝ 8 εult ⎠⎦ bh 3γmεult ⎝ h ⎠ 2 ⎝ h ⎝ 8 εult ⎠⎦M⎡⎤⎡⎤c20.67 f ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎥ = ⎜ ⎟ ⎜ −⎟⎢− ⎜ −⎟ ⎥⎠⎛ ⎞= cubxε⎜ −0 h ε⎟⎢− ⎜ −0 x 1 0.67 fcu x ε0ε0 x11⎟1 1 122bh γ ⎝ ⎠⎣ ⎝ ⎠⎦ bh⎝ h ⎝ ⎠⎣ ⎝ ⎠ hmεult2 εult2 2γmεultεult⎦M⎡⎞⎤⎞⎡⎛ ⎞ ⎤cMc1+ Mc21.34ε 0fcu⎛ x ⎞ 1 ⎛ x⎛= = ⎜ ⎟⎢− ⎜⎜⎟⎥+ ⎜ ⎟ ⎜ −⎟⎢− ⎜ −⎟ ⎥⎠⎛ ⎞⎟ −⎣ ⎠⎛ ⎞ 5 ε00.67 fcux ε0ε0 x11 1 12 2bh bh 3γmεult ⎝ h ⎠ 2 ⎝ h ⎝ 8 εult ⎠⎦2γm ⎝ h ⎝ εult ⎠⎣⎝ εult ⎠ h⎦20.67 f ⎪⎧ ⎡⎛ ⎞ ⎤⎪⎫cu ⎛ x ⎞ 1 1 ε0 1 1 ε0 1 ε0 x= ⎜ ⎟⎨− + ⎢−+ −⎜⎟ ⎥ ⎬(Eqn F-11)γm ⎝ h ⎠ 2 6 εult ⎪⎩⎢ 2 3 εult12⎣⎝ εult ⎠ ⎥ h⎦ ⎪⎭(II)Derivation of Basic Design Formulae of R.C. column sectionsCases 1 to 7 with different stress / strain profile of concrete and steel across thexcolumn section due to the differences in the neutral axis depth ratios, , are hinvestigated. The section is reinforced by continuous reinforcements Ashalong itslength h idealized as continuum and reinforcements at its end faces A with coverd ' .sbPursuant to the derivation of the stress strain relationship of concrete and steel, theF-3

Appendix Fstress strain diagram of concrete and steel for Cases 1 to 7 are as follows, under thedefinition of symbols as :b :h :x :Asb:d ' :Ash:width of the columnlength of the columnneutral axis depth of the columntotal steel area at the end faces of the columnconcrete cover to the centre of the end face steeltotal steel area along the length of the columnCase 1 (a) – where (i) x/h < 7/3(d’/h) for d’/h ≤ 3/14; and (ii) x/h < 7/11(1 – d’/h) ford’/h > 3/14Pursuant to the derivation of the stress strain relationship of concrete and steel, thestress strain diagram of concrete and steel for Case 1(a) is as indicated in FigureF-1(a) :It should be noted that F sc1 is in elastic whilst F st1 is in plastic range as d’/h < 3/14Steel compressive force in the portion steel elastic zone by Asbis⎛ x − d'⎞7 ⎛ d'⎞Fsc1= ⎜ ⎟× 0.87 fy × 0.5Asb= ⎜1− ⎟× 0.87 fy× 0. 5Asb(Eqn F-12)⎝ 4x/ 7 ⎠4 ⎝ x ⎠Steel compressive force in the portion steel plastic zone by AshisAsh⎛ 3x⎞⎛ 3 x ⎞Fsc2 = 0.87 fy× ⎜ ⎟ = 0.87 fy× Ash⎜ ⎟(Eqn F-13)h ⎝ 7 ⎠⎝ 7 h ⎠Steel compressive force in the portion steel elastic zone by AshisAsh⎛ 4x⎞ 1⎛ 2 x ⎞Fsc3 = 0.87 fy× ⎜ ⎟× = 0.87 fy× Ash⎜ ⎟(Eqn F-14)h ⎝ 7 ⎠ 2⎝ 7 h ⎠Steel tensile force in the portion steel plastic zone by AsbisFst1 = 0.87 fy× 0. 5Asb(Eqn F-15)Steel tensile force in the portion steel plastic zone by AshisAsh⎛ 11x⎞⎛ 11 x ⎞Fst2= 0.87 fy× ⎜h− ⎟ = 0.87 fy × Ash⎜1− ⎟(Eqn F-16)h ⎝ 7 ⎠⎝ 7 h ⎠Steel tensile force in the portion steel elastic zone by A isAsh⎛ 4x⎞ 1⎛ 2 x ⎞Fst3 = 0.87 fy× ⎜ ⎟ × = 0.87 fy× Ash⎜⎟(Eqn F-17)h ⎝ 7 ⎠ 2⎝ 7 h ⎠To balance the external loadNuc1+ Fc2+ Fsc1+ Fsc2+ Fsc3− Fst1− Fst2− Fst3⇒ Fc1+ Fc2+ Fsc1+ Fsc2− Fst1− Fst2= NuF = NF-4ush

⇒Appendix FNu 0.67f ⎛cuε ⎞0 x ⎛ x ⎞ Ash⎛ 3 7 d'h ⎞ A=sb3 0.87 fy2 10. 87 fybh 3⎜ −⎟ + ⎜ − ⎟ + ⎜ − ⎟γm ⎝ εult ⎠ h ⎝ h ⎠ bh ⎝ 8 8 h x ⎠ bh(Eqn F-18)d'xε 0/ ε ultε s = 0.002d 'ε = ε 0ε ult= 0.00354x/7 4x/7 3x/7uxhStrain diagram across whole sectionF c1F c20.67γmf cuConcrete stress BlockF sc1h–11x/7 4x/7F sc2F sc3F st3F st2 4x/7 3x/70.87f yF st1Steel stress BlockFigure F-1(a) – Concrete and steel stress strain relation for Case 1(a)Re-arranging (F-18)2⎡0.67fcu⎛ 1 ε ⎞A ⎤sh ⎛ x ⎞ ⎡ Nu ⎛ AshAsb⎞⎤x⎢⎜ −031⎟ + 2×0.87 fy ⎥⎜⎟ − ⎢ + 0.87 fy ⎜ − ⎟⎥⎣ γm ⎝ 3 εult ⎠bh⎦⎝h ⎠ ⎣ bh ⎝ bh 8 bh ⎠⎦h7 Asbd'− 0.87 fy= 0(Eqn F-19)8 bh hF-5

Appendix FF-6(Eqn F-19) can be used for solve for h xTo balance the external loaduMustststscscscccMMMMMMMMM =+++++++ 32132121ustststscscscccMxxhhFxhhFdhFxxhFxhFdhFMM=⎥⎦⎤⎢⎣⎡⎟ −⎠⎞⎜⎝⎛−−+⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−−+⎟⎠⎞⎜⎝⎛−+⎥⎦⎤⎢⎣⎡−−⎟ +⎠⎞⎜⎝⎛−⎟ +⎠⎞⎜⎝⎛−++2147112711212'22147321432'232132121⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−++=⇔'2211321432'2132121 dhFxhFxhFdhFMMMstscscscccu⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛+22129141132hxFxFstst (Eqn F-20)where⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎟ ⎠⎛ −⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛ −−= ultultmcuultultmcuchxhxfbhxhbxfbhMεεεγεεεεγε 00200218512131.341851231.34(Eqn F-21)⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝− ⎛ −⎟⎟⎠⎞⎜⎜⎝⎟ ⎠⎛ −⎞⎜⎝⎛=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝− ⎛ −⎟⎟⎠⎞⎜⎜⎝⎛ −= hxhxfbhxhbxfbhMultultmcuultultmcucεεεεγεεεεγ002002211120.67121210.67(Eqn F-22)⎟⎠⎞⎜⎝⎛−×⎟×⎠⎞⎜⎝⎛ −=⎟⎠⎞⎜⎝⎛−×⎟×⎠⎞⎜⎝⎛ −= hdbhAfxhhddhbhAfxdbhMsbysbysc'210.50.87'147'20.50.87'147221(Eqn F-23)⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛×=hxhxbhAfbhxhhxAfbhMshyshysc14321730.871143273.870 222(Eqn F-24)⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛×=hxhxbhAfbhxhhxAfbhMshyshysc211321720.8712113272.870 223(Eqn F-25)⎟⎠⎞⎜⎝⎛−×=⎟⎠⎞⎜⎝⎛−×=hdbhAfdhbhAfbhMsbysbyst'210.50.87'20.50.87221(Eqn F-26)⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛ −×=⎟ ⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛ −×= hxhxbhAfxhxbhAfbhMshyshyst141171110.8714117111.870 222(Eqn F-27)⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛×=⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛×=212129720.8712212972.870 223hxhxbhAfbhhxhxAfbhMshyshyst(Eqn F-28)Summing up⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−+−+−⎟⎠⎞⎜⎝⎛=+hxhxfbhMMultultultmcucc2000221121312161210.67εεεεεεγ(Eqn F-29)⎟⎠⎞⎜⎝⎛−×⎟ +⎠⎞⎜⎝⎛−×⎟×⎠⎞⎜⎝⎛ −=+hdbhAfhdbhAfxhhdbhMMsbysbystsc'210.50.87'210.50.87'147211

Appendix FAsb⎛ 1 d'⎞⎛117 d'h ⎞= 0 .87 fy ⎜ − ⎟⎜− ⎟bh ⎝ 2 h ⎠⎝8 8 h x ⎠(Eqn F-30)Msc2+ Mst 2Ash⎛ 3 x ⎞⎛1 3 x ⎞ A ⎛ ⎞⎛x ⎞= ⎜ ⎟⎜− ⎟ + ×sh 11 x 110.87 fy0.87 fy ⎜1− ⎟⎜⎟2bhbh ⎝ 7 h ⎠⎝2 14 h ⎠ bh ⎝ 7 h ⎠⎝14h ⎠2A ⎡⎤sh ⎛ x ⎞ 65 ⎛ x ⎞= 0.87 fy ⎢⎜⎟ − ⎜ ⎟ ⎥(Eqn F-31)bh ⎢⎣⎝ h ⎠ 49 ⎝ h ⎠ ⎥⎦Msc3+ Mst3Ash⎛ 2 x ⎞⎛1 13 x ⎞ Ash⎛ 2 x ⎞⎛29 x 1 ⎞= 0.87 f ⎜ ⎟⎜− ⎟ + 0.87 × ⎜ ⎟⎜− ⎟2yfybhbh ⎝ 7 h ⎠⎝2 21 h ⎠ bh ⎝ 7 h ⎠⎝21 h 2 ⎠2Ash32 ⎛ x ⎞= 0.87fy ⎜ ⎟ (Eqn F-32)bh 147 ⎝ h ⎠2M ⎪⎧⎡⎤⎪⎫sAsb⎛ 1 d'⎞⎛117 d'h ⎞ Ash⎛ x ⎞ 163 ⎛ x ⎞Total = 0.87 f ⎨ ⎜ − ⎟⎜− ⎟ + ⎢⎜⎟ − ⎜ ⎟2y⎥⎬(Eqn F-33)bh ⎪⎩bh ⎝ 2 h ⎠⎝8 8 h x ⎠ bh ⎢⎣⎝ h ⎠ 147 ⎝ h ⎠ ⎥⎦⎪⎭2M⎪⎧ ⎡⎤⎪⎫u0.67 fcu ⎛ x ⎞ 1 1 ε0 1 1 ε0 1 ⎛ ε0⎞ x∴ = ⎜ ⎟⎨− + ⎢−+ −⎜⎟ ⎥2⎬bh γm ⎝ h ⎠ 2 6 εult ⎪⎩⎢ 2 3 εult12⎣⎝ εult ⎠ ⎥ h⎦ ⎪⎭2⎪⎧A⎡⎤⎪⎫sb ⎛ 1 d'⎞⎛117 d'h ⎞ Ash⎛ x ⎞ 163 ⎛ x ⎞+ 0.87 fy ⎨ ⎜ − ⎟⎜− ⎟ + ⎢⎜⎟ − ⎜ ⎟ ⎥⎬(Eqn F-34)⎪⎩bh ⎝ 2 h ⎠⎝8 8 h x ⎠ bh ⎢⎣⎝ h ⎠ 147 ⎝ h ⎠ ⎥⎦⎪⎭Case 1 (b) – 7/11(1 – d’/h) ≤ x/h < 7/3(d’/h) where d’/h > 3/14Case 1(b) is similar to Case 1(a) except that both F sc1 and F st1 are in the elastic rangeas d’/h > 3/14.Re Figure F-1(b), the various components of stresses in concrete and in steel areidentical to that of Case 1(b) except that by Fst1, the stress of which ish − x − d'0.87 fy4x/ 7Fst1h − x − d'Asb7 ⎛ h d'h ⎞ AsbSo the = − 0.87 fy× 0.5 = − × 0.87 fy ⎜ −1−⎟bh 4x/ 7bh 8 ⎝ x h x ⎠ bhFst1Fsc1 7 ⎛ h ⎞ Asb+ = × 0.87 fy ⎜2− ⎟bh bh 8 ⎝ x ⎠ bhNu0.67fcu⎛ ε0⎞ x ⎛ x ⎞ Ash⎛ 7 7 h ⎞ A∴ =sb3 0.87 fy2 1⎟0.87fybh 3⎜ −⎟ + ⎜ − ⎟ + ⎜ −(Eqn F-35)γm ⎝ εult ⎠ h ⎝ h ⎠ bh ⎝ 4 8 x ⎠ bh2⎡0.67fcu⎛ 1 ε ⎞A ⎤sh ⎛ x ⎞ ⎡ ⎛ AsbAsh⎞ Nu⎤ x⇒ ⎢⎜ −071⎟ + 0.87 fy× 2 ⎥⎜⎟ + ⎢0.87fy ⎜ − ⎟ − ⎥⎣ γm ⎝ 3 εult ⎠bh⎦⎝h ⎠ ⎣ ⎝ 4 bh bh ⎠ bh ⎦ h7 A− 0 .87×sb= 0(Eqn F-36)8 bhF-7

Appendix FMsc 1+ Mst17 ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜⎛ d'h A= −sb 1 d'7 h d'h Asb1 d'1 ⎟× 0.87 fy ⎜ − ⎟ + 0.87 fy× ⎜ −1−⎟ ⎜ − ⎟2bh 8 ⎝ h x ⎠ bh ⎝ 2 h ⎠ 8 ⎝ x h x ⎠ bh ⎝ 2 h ⎠7 Asb⎛ 1 d'd ⎞ hfy ⎜ ⎟ ⎜ − ⎟bh h ⎠⎛ ⎞ '= 0 .87 × − 1 2(Eqn F-37)8 ⎝ 2 ⎝ h ⎠ x2M⎪⎧⎡⎤⎪⎫u0.67 fcu ⎛ x ⎞ ⎛ 1 1 ε0⎞ 1 ε0 1 1 ⎛ ε0⎞ ⎛ x ⎞∴ = ⎜ ⎟⎨⎜ −⎟ + ⎢ − −⎜⎟ ⎥⎜⎟2⎬bh γm ⎝ h ⎠⎝ ⎠⎪⎩ ⎝ 2 6 εult ⎠ ⎢3εult2 12⎣⎝ εult ⎠ ⎥ h⎦ ⎪⎭2⎪⎧7 A⎡⎤⎪⎫sb ⎛ 1 d'⎞ d'⎞ h A ⎛ ⎞ ⎛ ⎞⎨⎟ ⎜ −sh x 163 x+ 0.87 fy ⎜ − 1 2 ⎟ + ⎢⎜⎟ − ⎜ ⎟ ⎥⎬(Eqn F-38)⎪⎩8 bh ⎝ 2 h ⎠⎛⎝ h ⎠ x bh ⎢⎣⎝ h ⎠ 147 ⎝ h ⎠ ⎥⎦⎪⎭d'd 'ε = ε 0xε 0/ ε ultε s = 0.002ε ult= 0.00354x/7 4x/7 3x/7uxhStrain diagram across whole sectionF c1F c20.67γmf cuConcrete stress Blockh–11x/7 4x/7F sc1F st2F st3F sc3F sc24x/73x/70.87f yF st1Steel stress BlockFigure F-1(b) – Concrete and steel stress strain relation for Case 1(b)F-8

Case 2 – 7/3(d’/h) ≤ x/h < 7/11(1 – d’/h)Appendix FThere are two sub-cases to be considered in Case 2,i.e. Case 2(a) –d' ≥hFor Case 2(a), where314d' ≥hand Case 2(b) –314. However,doesn’t exist. For Case 2(b), wherezone as shown in Figure F-2.d'

Appendix FThe various components of stresses in concrete and steel are identical to that of Case2(a) except that of Asc1where⎛ 1 d'⎞Fsc1= 0.87 fy× 0. 5Asband Msc1= 0.87 fy× 0.5Asb⎜− ⎟⎝ 2 h ⎠It can be seen that Fst1and Fsc1are identical but opposite in direction, so cancel out.By formulation similar to the above,Nu0.67f ⎛cu 1 ε ⎞ x ⎛ x ⎞ A=sh+ fy ⎜ − ⎟bh⎜ −01⎟ 0.87 2 1(Eqn F-39)γm ⎝ 3 εult ⎠ h ⎝ h ⎠ bhNuAsh+ 0.87 fyx⇒ = bh bh(Eqn F-40)h ⎡0.67f ⎛ ⎞⎤cu1 ε⎢⎜ −0⎟ + Ash1 2×0.87 fy ⎥⎣ γm ⎝ 3 εult ⎠bh ⎦2M⎪⎧ ⎡⎛ ⎞ ⎤⎪⎫u0.67 fcu⎛ x ⎞ 1 1 ε0 1 1 ε0 1 ε0 x= ⎜ ⎟⎨− + ⎢−+ −⎜⎟ ⎥2⎬bh γm ⎝ h ⎠ 2 6 εult ⎪⎩⎢ 2 3 εult12⎣⎝ εult ⎠ ⎥ h⎦ ⎪⎭2⎪⎧A⎡⎤⎪⎫sb ⎛ 1 d'⎞ Ash⎛ x ⎞ 163 ⎛ x ⎞+ 0.87 fy ⎨ ⎜ − ⎟ + ⎢⎜⎟ − ⎜ ⎟ ⎥⎬(Eqn F-41)⎪⎩bh ⎝ 2 h ⎠ bh ⎢⎣⎝ h ⎠ 147 ⎝ h ⎠ ⎥⎦⎪⎭Case 3 – where 7/3(d’/h) ≤ x/h < 7/11 for d’/h > 3/14 and7/11(1 – d’/h) ≤ x/h < 7/11 for d’/h < 3/14The concrete / steel stress / strain diagram is worked out as indicated in Figure F-3 :The components of stresses are identical to Case 2 except that F st1 become elasticwhich is0.87 fy( h − x − d') ( h − x − d') 7⎛ h d'⎞ 7Fst1 =× 0.5Asb= 0.87 fyAsb= 0.87 fy ⎜ − −1⎟ Asb4x/ 7x 8⎝ x x ⎠ 8Fst1⎛ h d'h ⎞ 7 Asb⇒ = 0.87 fy ⎜ − −1⎟bh ⎝ x h x ⎠ 8 bhFsc1Fst1Asb⎛ h d'h ⎞ 7 AsbAsb⎛117 h∴ + = 0.87 fy× 0.5 − 0.87 fy ⎜ − −1⎟= 0.87 fy ⎜ − +bh bhbh ⎝ x h x ⎠ 8 bh bh ⎝ 8 8 x(Eqn F-42)Nu0.67fcu⎛ 1 ε ⎞ x Asb⎛ h d ⎞ ⎛ x ⎞ A∴ =shfy⎟ + fy ⎜ − ⎟bh⎜ −011 7 7 '1⎟ + 0.87 ⎜ − + 0.87 2 1γm ⎝ 3 εult ⎠ h bh ⎝ 8 8 x 8 x ⎠ ⎝ h ⎠ bh(Eqn F-43)Re-arranging (Eqn F-43)2⎡0.67f⎤cu⎛ 1 ε ⎞⎛ ⎞ ⎡ ⎛⎞ ⎤⎛⎞⎢⎜ −0AshxAsb11 AshNu x1⎟ + 2×0.87 fy ⎥⎜⎟ + ⎢0.87fy ⎜ − ⎟ − ⎥⎜⎟⎣ γm ⎝ 3 εult ⎠bh⎦⎝h ⎠ ⎣ ⎝ bh 8 bh ⎠ bh ⎦⎝h ⎠78d'⎞⎟x ⎠F-10

Appendix F7 Asb⎛ d'⎞+ 0 .87 fy ⎜ −1⎟= 0(Eqn F-44)8 bh ⎝ h ⎠d 'd 'ε = ε 0xε 0/ ε ultε s = 0.002ε ult= 0.0035xuhStrain diagram across concrete sectionF c14x/7F c20.67γmf cuConcrete stress BlockF sc1h–11x/7 4x/7F st2F st3F sc3F sc24x/73x/70.87f yF st1Steel stress BlockFigure F-3 – Concrete and steel stress strain relation for Case 3To balance the external moment Mu, all components are identical to Case 2 exceptthat by F st1 which isMst 17 ⎛ h d'⎞⎛h ⎞ 1 Asb7 ⎛ h d'⎞⎛1 d'⎞= 0.87fyAsb⎜ − −1⎟⎜− d'⎟ = 0.87 fy ⎜ − −1⎟⎜−22⎟bh8 ⎝ x x ⎠⎝2 ⎠ bh bh 8 ⎝ x x ⎠⎝2 h ⎠(Eqn F-45)F-11

Appendix F2M⎪⎧⎡⎤⎪⎫u0.67 fcu ⎛ x ⎞ ⎛ 1 1 ε0⎞ 1 ε0 1 1 ⎛ ε0⎞ ⎛ x ⎞∴ = ⎜ ⎟⎨⎜ −⎟ + ⎢ − −⎜⎟ ⎥⎜⎟2⎬bh γm ⎝ h ⎠⎝ ⎠⎪⎩ ⎝ 2 6 εult ⎠ ⎢3εult2 12⎣⎝ εult ⎠ ⎥ h⎦ ⎪⎭2⎪⎧⎛ 1 d'⎞⎛7 h 7 d'h 3 ⎞ A ⎡ ⎛ ⎞ ⎤ ⎪⎫sb ⎛ x ⎞ 163 x Ash+ 0 .87 fy ⎨⎜− ⎟⎜− − ⎟ + ⎢⎜⎟ − ⎜ ⎟ ⎥ ⎬ (Eqn F-46)⎪⎩ ⎝ 2 h ⎠⎝8 x 8 h x 8 ⎠ bh ⎢⎣⎝ h ⎠ 147 ⎝ h ⎠ ⎥⎦bh ⎪⎭ASo, by pre-determining the steel ratios for sbA and sh x, we can solve for bybh bhh(Eqn F-44) under the applied load Nu. The moment of resistance Mucan then beobtained by (Eqn F-46). The section is adequate if Muis greater than the appliedmoment.Case 4 – where x ≤ h < 11x/7, i.e. 7/11 ≤ x/h < 1The concrete / steel stress / strain diagram is worked out as in Figure 3-4. The stresscomponents are identical to Case 3 except that F st2 vanishes and F st3 reduces asindicated in Figure F-4 :Steel tensile force in the portion steel elastic zone byFA( h − x)Ashis2sh 1 h − x 77 ⎛ h xst3= 0.87 fy× ( h − x)× = 0.87 fy× Ash= 0.87 fy× ⎜ + − 2To balance the external loadh2 4x/ 7Nu8hx8 ⎝ x h(Eqn F-47)N⎛ ⎞⎛⎞⇒u0.67f =cubx1 ε⎜ −0Asb11 7 h 7 d'1⎟ + 0.87 fy ⎜ − + ⎟bh γm ⎝ 3 εult ⎠ bh ⎝ 8 8 x 8 x ⎠5 x 7 ⎛ h x ⎞+ 0 .87 fy× Ash− 0.87 fy× ⎜ + − 2⎟Ash7 h 8 ⎝ x h ⎠N⎛ ⎞⎛⎞⎛⎞⇒ u0.67f =cu 1 ε⎜ −0 x Asb11 7 h 7 d'Ash9 x 7 h 71⎟ + 0.87 fy ⎜ − + ⎟ + 0.87 fy ⎜−− + ⎟bh γ ⎝ 3 ⎠ h bh ⎝ 8 8 x 8 x ⎠ bh ⎝ 56 h 8 x 4mεult⎠(Eqn F-48)2⎡0.67f ⎛ ⎞⎤⎛⎞ ⎡ ⎛⎞ ⎤⎛⎞⇔cuε⎢⎜ −0 9 AshxAsb11 Ash7 Nu x3⎟ − × 0.87 fy ⎥⎜⎟ + ⎢0.87fy ⎜ + ⎟ − ⎥⎜⎟⎣ 3γm ⎝ εult ⎠ 56 bh ⎦⎝h ⎠ ⎣ ⎝ bh 8 bh 4 ⎠ bh ⎦⎝h ⎠7 ⎡ Asb⎛ d'⎞ Ash⎤+ 0 .87 fy1 = 08⎢ ⎜ − ⎟ − ⎥(Eqn F-49)⎣ bh ⎝ h ⎠ bh ⎦To balance the external load M about the centre of the column sectionMuMcMsc1+ Mst1Msc2Msc3Mst3= ++ + +2 222 2 2bh bh bh bh bh bhMuMcMsc1+ Mst1Fsc2⎛ h 3x⎞ Fsc⇒ = ++ ⎜ − ⎟ +2 222bh bh bh bh ⎝ 2 14 ⎠ bhu32⎛ h 3x4x⎞ Fst3⎡h⎛ h − x ⎞⎤⎜ − − ⎟ + ⎢ − ⎜ ⎟2⎥⎝ 2 7 21 ⎠ bh ⎣2⎝ 3 ⎠⎦(Eqn F-50)⎞⎟A⎠shF-12

M M MsTotal Moment = c+ , i.e.2 2 2bh bh bhM⎪⎧⎡u0.67 fcu ⎛ x ⎞ ⎛ 1 1 ε0⎞ 1 ε0 1 1 ⎛ ε0⎞= ⎜ ⎟⎨⎜ −⎟ + ⎢ − −⎜⎟2bh γm ⎝ h ⎠⎪⎩⎝ 2 6 εult ⎠ ⎢3εult2 12⎣⎝ εult ⎠⎪⎧⎛ 1 d'⎞⎡7⎛ h ⎞ 7 ⎛ d'⎞⎛h ⎞ 3⎤A ⎡sb 7 h+ 0 .87 fy ⎨⎜− ⎟⎢⎜ ⎟ − ⎜ ⎟⎜⎟ − ⎥ + ⎢ −⎪⎩ ⎝ 2 h ⎠⎣8⎝ x ⎠ 8 ⎝ h ⎠⎝x ⎠ 8⎦bh ⎢⎣48 x2⎤⎪⎫⎛ x ⎞⎥⎜⎟⎬⎥⎦⎝ h ⎠⎪⎭9112xh+Appendix F29 ⎛ x ⎞ ⎤ A ⎪⎫sh⎜ ⎟ ⎥ ⎬392 ⎝ h ⎠ ⎥⎦bh ⎪⎭(Eqn F-51)d 'd 'xε 0/ ε ultε s = 0.002ε ult= 0.0035ε 10.0035=( h − x)xε = ε 04x/7xuhStrain diagram across concrete sectionF c1 F c20.67γmf cuConcrete stress Blockh – xF st3F sc3F sc20.87f y4x/73x/7Steel stress BlockFigure 3-4 – Concrete and steel stress strain relation for Case 4F-13

Case 5 – where x>h>(1 – ε 0 /ε ult )x, i.e 1 ≤ x/h < 1/(1 – ε 0 /ε ult )Appendix FThe concrete / steel stress / strain diagram is worked out as follows. It should be notedthat the neutral axis depth ratio is greater than unity and hence becomes a hypotheticalconcept :d 'ε = ε 0xε 0/ ε ultd'ε s = 0.002ε ult= 0.0035ε 10.0035=( h − x)x4x/7uxhStrain diagram across concrete sectionF c1F c20.67γmf cuConcrete stress BlockF sc3F sc20.87f y4x/73x/7Steel stress BlockFigure F-5 – Concrete and steel stress strain relation for Case 5Concrete compressive stresses and forcesxε0 / εult2Ecεult 2 EcεFc1= ∫σbduwhere σ = − u +22εx xx−h0ultu(Eqn F-52)F-14

Appendix FF-15u duxbEduuxbEbduuxEuxEFultultultxhxultcxhxultcxhxultcultcc∫∫∫−−−+= −⎥⎦⎤⎢⎣⎡+−=εεεεεεεεεεεε//2202/2202100022⎟⎟⎠⎞⎜⎜⎝⎛−+⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−−⎟⎟⎠⎞⎜⎜⎝⎛−=⇒0233002220121612 εεεεεεεεεεultcultcultultcultultccEEhxEEbhF20202622⎟⎠⎞⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛−+xhExhEEultcultcultcεεεεε(Eqn F-53)hxfbhFbxfFultmcucultmcuc⎟ ⎟ ⎠⎞⎜⎜⎝⎛ −=⇔⎟⎟⎠⎞⎜⎜⎝⎛ −= εεγεεγ0202 10.6710.67(Eqn F-54)hxfxhExhEEEEhxEEbhFbhFbhFultmcuultcultcultcultcultcultultcultultcccc⎟⎟⎠⎞⎜⎜⎝⎛ −+⎟ ⎠⎞⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛−+⎟⎟⎠⎞⎜⎜⎝⎛−+⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−−⎟⎟⎠⎞⎜⎜⎝⎛−=+=εεγεεεεεεεεεεεεεεε02020202330022202110.6762221612⎟⎟⎠⎞⎜⎜⎝⎛−+⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛ −+⎟⎟⎠⎞⎜⎜⎝⎛−−⎟⎟⎠⎞⎜⎜⎝⎛−=0200330022202121612 εεεεεεεεεεεεεultcultcultcultultcultultccEEhxEEEbhF20202622⎟⎠⎞⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛−+xhExhEEultcultcultcεεεεε(Eqn F-55)Steel compressive force in the portion steel plastic zone bysbA isbhAfbhFAfFsbyscsbysc 50.0.870.50.8711 ×=⇔×= (Eqn F-56)( ) ( )bhAhdxhfbhAxhhdxhfbhAxdxhfbhFAxdxhfAxdhxfAxdhxfFsbysbysbyscsbysbysbysc87'110.8787'10.8787'10.8787'10.8787'0.870.57/4'0.871'1'⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛ −−=⎟⎠⎞⎜⎝⎛+−=⎟⎠⎞⎜⎝⎛+−=⇒⎟⎠⎞⎜⎝⎛+−=+−=×+−=bhAhdxhfbhAhdxhfbhAfbhFbhFsbysbysbyscsc⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛ −−=⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛ −−+×=+ '1878110.8787'110.870.50.871'1(Eqn F-57)Steel compressive force in the portion steel plastic zone byshA isbhAhxfbhFhxAfxhAfFshyscshyshysc⎟ ⎠ ⎞⎜⎝⎛=⇔⎟⎠⎞⎜⎝⎛×=⎟⎠⎞⎜⎝⎛×=730.87730.8773.870 22 (Eqn F-58)Steel compressive force in the portion steel elastic zone byshA is2173/ 7410.8773/ 740.873 ×⎟⎠⎞⎜⎝⎛−⎟×⎠⎞⎜⎝⎛−−⎟ +⎠⎞⎜⎝⎛−×−×=xhhAxhxfxhhAxhxfFshyshyscbhAhxxhfbhFshysc⎟⎠⎞⎜⎝⎛−−=⇒563387470.873 (Eqn F-59)

Appendix FF-16AsbhFbhFbhFbhFbhFbhNFFFFFNscscscsccuscscscsccu321'1321'1 ++++=⇔++++=hxbhAfEbhNshyultultultultcu⎥⎥⎦⎤⎢⎢⎣⎡−⎟⎟⎠⎞⎜⎜⎝⎛+−+−=⇒5690.876121216100220εεεεεεε⎟⎟⎠⎞⎜⎜⎝⎛++−+bhAfbhAfEEshysbyultcultc470.878110.872 0 2εεε202026870.871'870.8722⎟⎠⎞⎜⎝⎛−⎥⎦⎤⎢⎣⎡⎟ −⎠⎞⎜⎝⎛−+−+xhExhbhAfhdbhAfEEultcshysbyultcultcεεεεε(EqnF-60)Re-arranging (Eqn F-60)3002205690.8761212161⎟⎠⎞⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡−⎟⎟⎠⎞⎜⎜⎝⎛+−+−hxbhAfEshyultultultultcεεεεεεε202470.878110.872⎟⎠⎞⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−++−+hxbhNbhAfbhAfEEushysbyultcultcεεε06870.871'870.8722 0202=−⎥⎥⎦⎤⎢⎢⎣⎡⎟ −⎠⎞⎜⎝⎛−+−+εεεεεultcshysbyultcultcEhxbhAfhdbhAfEE(Eqn F-61)which is a cubic equation in h x .Summing the Moments as follows :Concrete compressive stresses and moments⎟⎠⎞⎜⎝⎛+−= ∫−uxhbduMultxhxc2/1ε 0 εσ where uxEuxEultcultcεεεσ += −22022duuxhbuxEuxEMultxhxultcultcc⎟ ⎠ ⎞⎜⎝⎛+−⎥⎥⎦⎤⎢⎢⎣⎡+−= ∫− 22/22021ε 0 εεεε⎥⎥⎦⎤⎢⎢⎣⎡⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛−+−+⎟⎠⎞⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−+⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−+⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎠⎞⎜⎝⎛−−⎥⎥⎦⎤⎢⎢⎣⎡⎥⎥⎦⎤⎢⎢⎣⎡+−+⎟⎠⎞⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−+⎥⎥⎦⎤⎢⎢⎣⎡−+⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎠⎞⎜⎝⎛−=224402330022330220214641413312131233131212121xhxhhxhxxhxhhxhxExhhxhxxhhxhxEbhMultultultcultultultccεεεεεεεεεεε⎥⎦⎤⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝− ⎛ −⎟⎟⎠⎞⎜⎜⎝⎟ ⎠⎛ −⎞⎜⎝⎛=⎥⎦⎤⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝− ⎛ −⎟⎟⎠⎞⎜⎜⎝⎛ −= hxhxfbhxhbxfbhMultultmcuultultmcucεεεεγεεεεγ002002211120.67121210.67⎟⎠⎞⎜⎝⎛−×=⎟⎠⎞⎜⎝⎛−×=hdbhAfdhbhAfbhMsbysbysc'210.50.87'20.50.87221bhAhdxhhdxhfhdhbhAxdxhfbhMsbysbysc⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+−= −⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+−= −'21'1870.871'287'10.8721'

7 ⎡ ⎛ d'⎞ h⎤⎛1 d'⎞ Asb= −0.87 fy ⎢1+ ⎜ −1⎟⎜ − ⎟8 h x⎥⎣ ⎝ ⎠ ⎦⎝2 h ⎠ bhMuMcM3Total Moment 1 cM2 scM1 scM M1' sc 2 sc= + + + + +2 2 2 2 2 2 2bh bh bh bh bh bh bh322M ⎛ 1 0 1 0 1 0 1 1 ⎞uε ε ε ε⎜ult⎟⎛x ⎞⇒ = E⎜ ⎟2 cεult− + − + −32bh24 6 4 6 24⎝ εεultε0 ⎠⎝hultεult⎠2⎡ 1 ε0 1 ε0 1 1 ε ⎤ult x ⎛ 1 1 εult⎞+ E ⎢−+ − + +2⎥ E12 4 4 12⎜ −012 12⎟cεultcεult⎣ ε εultε⎦ hult⎝ ε0 ⎠Asb⎛ 1 d'⎞⎡7d ⎞ h ⎤ ⎡ h xfy+ fy ⎢ − +bh h⎢ ⎜⎛ ' 3 7 9+ 0 .87 × ⎜ − ⎟ 1 − ⎟ −h x⎥ 0.87⎝ 2 ⎠⎣8⎝ ⎠ 8⎦⎢⎣48 x 112 hCase 6 – where (1 – ε 0 /ε ult )x>h>3x/7, i.e. 1/(1 – ε 0 /ε ult ) ≤ x/h < 7/3hx+ E ε9392cAppendix Fult1 εult ⎛ h ⎞⎜ ⎟24 ε ⎝ x ⎠2⎛ x ⎞ ⎤ Ash⎜ ⎟ ⎥⎝ h ⎠ ⎥⎦bh(Eqn F-62)Case 6 is similar to Case 5 except that F c1 vanishes. The concrete / steel stress / straindiagram is worked out as in Figure F-6 :Referring to (Eqn F-55) by replacing Fc1 + Fc2by0.67Nu0.67 fcu⎡117 h ⎛ d'⎞⎤Asb⎛ 7 7 h 9 x ⎞ A= +sh0.87 fy+ fy ⎜ − − ⎟bh⎢ − ⎜1− ⎟mx h⎥ 0.87γ⎣ 8 8 ⎝ ⎠⎦bh ⎝ 4 8 x 56 h ⎠ bh(Eqn F-63)29 A ⎡⎤sh ⎛ x ⎞ Nu0.67 fcu⎛11Asb7 Ash⎞ ⎛ x ⎞Re-arranging ⇔ 0.87 fy ⎜ ⎟ + ⎢ − − 0.87 fy ⎜ + ⎟⎥⎜⎟56 bh ⎝ h ⎠ ⎣ bh γm ⎝ 8 bh 4 bh ⎠⎦⎝h ⎠⎡7⎛ d'⎞ Asb7 Ash⎤− 0 .87 fy ⎢ ⎜ −1⎟− = 088⎥(Eqn F-64)⎣ ⎝ h ⎠ bh bh ⎦xwhich is a quadratic equation in which can be solvedhγmf cu02For Moment that can be provided by the section, similar to Case 5 except thatMc= 0 . SoMuAsbd ⎡ d ⎞ h ⎤ ⎡ h x ⎛ x ⎞ ⎤ Ashfyfy⎥bh bh h⎢ ⎜⎛ 2⎛ 1 ' ⎞ 7 ' 3 7 9 9= 0.87 × ⎜ − ⎟ 1 − ⎟ − + ⎢ − + ⎜ ⎟h x⎥ 0.872⎝ 2 ⎠⎣8⎝ ⎠ 8⎦⎢⎣48 x 112 h 392 ⎝ h ⎠ ⎥⎦bh(Eqn F-65)F-17

Appendix Fε s = 0.002d 'd'ε ult= 0.00354x/7xhStrain diagram across concrete sectionF c20.67γmf cuConcrete stress BlockF sc3F sc20.87f y4x/73x/7Steel stress BlockFigure 3-6 – Concrete and steel stress strain relation for Case 6Case 7 – where x/h ≥ 7/3In this case, the concrete and steel in the entire column section are under ultimatestress. The axial load will be simplyNu0.67 fcu ⎛ AsbAsh⎞= + 0.87 fy ⎜ + ⎟(Eqn F-66)bh γm ⎝ bd bd ⎠and the moment is zero.M u= 0 2bh(Eqn F-67)F-18

(III) Design formulae for 4-bar column sections for determination ofreinforcement ratiosAppendix FbhIt is the aim of the section of the Appendix to derive formulae for the determination ofA sb against applied axial load and moment under a pre-determined sectional size. InbhAthe following derivations, sh are set to zero. The process involves :bhA(i) For the 7 cases discussed in the foregoing, eliminate sb between equationsbhNobtained from balancing uM and uA2 by making sb subject of formulaebh bhbhNin the equation for balancing of u substitute into the equation for balancingbhMof ux. The equation obtained in a polynomial in2which can be solvedbhhby equations (if quadratic or cubic or even 4 th power) or by numerical methods.xSolution in will be valid if the value arrived at agree with thehpre-determined range of the respective case;(ii) Back substitute the accepted value of hx into the equation obtained bybalancingN uA to solve for sb.bhbhCase 1 (a) – where (i) x/h < 7/3(d’/h) for d’/h ≤ 3/14; and (ii) x/h < 7/11(1 – d’/h) ford’/h > 3/14Putting A sh = 0 ; bhN(Eqn F-18) ⇒ubh=.67 f0cu 1 0γm⎛ ε⎜1−⎝ 3 εult⎞⎟⎠xh⎛ 3+ ⎜ −⎝ 878d'h ⎞⎟0.87fh x ⎠yAsbbhF-19

Appendix FF-20⎟⎠⎞⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛ −−=⇒xhhdhxfbhNbhAfultmcuusby'87833110.670.870εεγ(Eqn F-68)Substituting into (Eqn F-34)⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−+−+−⎟⎠⎞⎜⎝⎛=⇒hxhxfbhMultultultmcuu20002121312161210.67εεεεεεγ⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛−+xhhdhdbhAfsby'87811'21.87032001213121830.67⎟⎠⎞⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−+−⇒hxfultultmcuεεεεγ22000 '967'43'162961210.67⎟⎠⎞⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛+−++−+hxhdhdhdfultultultmcuεεεεεεγhxhdfbhMhdbhNultmcuuu⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎠⎞⎜⎝⎛+⎟ −⎠⎞⎜⎝⎛−+ 131'870.6783'21811 022 εεγ0''2187'872 =⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−−+hdhdbhNhdbhMuu(Eqn F-69)Upon solving (Eqn F-69) for h x , back-substitution into (Eqn F-68) to calculate bhA sb .Case 1 (b) – 7/11(1 – d’/h) ≤ x/h < 7/3(d’/h) where d’/h > 3/14Putting 0=bhA sh ; (Eqn F-35) bhAfxhhxfbhNsbyultmcuu0.878747311.670 0⎟⎠⎞⎜⎝⎛−+⎟⎟⎠⎞⎜⎜⎝⎛ −=⇒εεγ⎟⎠⎞⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛ −−=⇒xhhxfbhNbhAfultmcuusby87473110.670.870εεγ(Eqn F-70)Substituting into (Eqn F-38), again putting 0=bhA sh and simplifying, 2200320096712716210.671213121470.67⎟⎠⎞⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛+−+⎟⎠⎞⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−+−⇒hxfhxfultultmcuultultmcuεεεεγεεεεγ0'212187'2'2161210.674722202 =⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛ −++⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−⎟⎟⎠⎞⎜⎜⎝⎛−+−bhNhdbhMhxhdhdfbhMuuultmcuuεεγ

Appendix F(Eqn F-71)x AUpon solving (Eqn F-71) for , back-substitution into (Eqn F-70) to calculate sb.h bhCase 2 – 7/3(d’/h) ≤ x/h < 7/11 – 7/11(d’/h) and d’/h > 3/14Using the equations summarized in Section 4 and settingbh= 0 in (Eqn F-39)x N ⎡ ⎛ ⎞⎤=u0.67 f ÷cu 1 ε⎢⎜ −01⎟⎥h bh⎣ γm ⎝ 3 εult ⎠⎦(Eqn F-72)xSubstituting h obtained in (Eqn F-72), substituting into (Eqn F-41) and calculateA shA sb asbhA ⎪⎧sbM= ⎨bh bh ⎪⎩u20.67 f−γmcu⎛ x ⎞⎡1⎞⎢ ⎜⎛ x 1 ε⎟ −0⎜ 1 ⎟ −⎝ h ⎠⎢2⎣⎝ h ⎠ 6 εult+13ε0εult⎛ x ⎞⎜ ⎟ −⎝ h ⎠112⎛ ε0⎜⎝ εult2⎞ ⎪⎫⎛ x ⎞⎤ ⎡⎟ ⎜ ⎟⎥⎬÷ ⎢0.87f⎠ ⎝ h ⎠⎥⎦⎪⎭⎣(Eqn F-73)y⎛ 1 d'⎞⎤⎜ − ⎟⎥ ⎝ 2 h ⎠ ⎦Case 3 – where 7/3(d’/h) ≤ x/h < 7/11 for d’/h > 3/14 and7/11(1 – d’/h) ≤ x/h < 7/11 for d’/h < 3/14and Case 4 – where 7/11 ≤ x/h < 1x NUsing the equation relating and uin (Eqn F-43) and setting A sh = 0 .h bhbhNu0.67f ⎛ ⎞⎛⎞= cuε⎜ −0 x Asb11 7 h 7 d'3⎟ + 0.87 fy ⎜ − + ⎟bh 3γ⎝ ⎠ h bh ⎝ x xmεult8 8 8 ⎠Nu0.67f ⎛ ⎞−cu 1 ε⎜ −0 x1⎟A bh γm ⎝ 3 ε ⎠ hsbult⇔ =(Eqn F-74)bh ⎛117 h 7 d'h ⎞0.87 fy ⎜ − + ⎟⎝ 8 8 x 8 h x ⎠A shSubstituting = 0 , (Eqn F-46) and simplifyingbh2 30.67 f 11 ⎡10 1 1 ⎤cuε ⎛ ε0⎞ ⎛ x ⎞⇔ ⎢ − − ⎥⎜⎟8 ⎢32 12⎜⎟γmεult⎣⎝ εult ⎠ ⎥⎦⎝ h ⎠2220.67 f ⎡217 0 13 ' 7 0 5 0 ' 7 0 ' ⎤cuε d ⎛ ε ⎞ ε d ⎛ ε ⎞ d ⎛ x ⎞+ ⎢ − − + + −⎥⎜⎟⎢1612 16 96⎜⎟12 96⎜⎟γmεulth⎣⎝ εult ⎠ εulth ⎝ εult ⎠ h ⎥⎦⎝ h ⎠⎧0.67f⎫cu 7 ⎛ d'⎞⎛1 ε0 d'1 d'ε0⎞ ⎛ ⎞⎛ ⎞⎨ ⎜ ⎟⎜ − − +⎟ − 3 1 d'Nu 11 Mu x+−11⎜ − ⎟ −2 ⎬⎜⎟⎩ γm8 ⎝ h ⎠⎝3 εulth 3 h εult ⎠ 8 ⎝ 2 h ⎠ bh 8 bh ⎭⎝h ⎠F-21

7 ⎛ 1 d'⎞⎛d'⎞ Nu 7 Mu ⎛ d'⎞− ⎜ − ⎟⎜−1⎟− 1 = 02⎜ − ⎟8 ⎝ 2 h ⎠⎝h ⎠ bh 8 bh ⎝ h ⎠⎛ x ⎞which is a cubic equation in ⎜ ⎟⎝ h ⎠⎛ x ⎞7 ⎛ d' ⎞Upon solving ⎜ ⎟ lying between ⎜1− ⎟⎝ h ⎠ 11⎝h ⎠back-substituting into (Eqn 5-7)Nu0.67f ⎛ ⎞−cuε⎜ −0 x3⎟A bh 3γm=⎝ ε ⎠ hsbultbh ⎛117 h 7 d'h ⎞0.87 fy ⎜ − + ⎟⎝ 8 8 x 8 h x ⎠Appendix F(Eqn F-75)Aand 1, sb can be obtained bybh(Eqn F-76)Case 5 – where 1 ≤ x/h < 1/(1 – ε 0 /ε ult )Referring to Case 5 of Section 3 and settingbd= 0 in (Eqn F-60)ASolving sb bybd22 32N ⎡uEcε⎛ultε ⎞ E ⎛ ⎞c ultEc⎛ ⎞⎤x ⎛ E ⎞c ult⎢ ⎜0ε ε⎟ ⎜0ε⎟ +0ε⎜ E⎟c ultbh⎜ −0ε=⎟−1−−11 ⎥ +ulthε −23⎢⎣2 ⎝ εult ⎠ 6ε0 ⎝ εult ⎠ 2 ⎝ ε ⎠⎥⎦⎝ 2ε0 ⎠22 2⎛ EcεultE ⎞cεult h Ecεult ⎛ h ⎞ ⎡ ⎛ d ⎞ h⎤A⎜⎟11 7 'sb+−− ⎜ ⎟ + 0.87 fyx x⎢ − ⎜ −1⎟h x⎥⎝ 2ε02 ⎠ 6ε0 ⎝ ⎠ ⎣ 8 8 ⎝ ⎠ ⎦ bh⇒ 0.87 fy2A ⎡ ⎛⎞ ⎤sbNu⎡ ⎛ ⎞ ⎤⎢ ⎜1 1 ε0 1 εult 1 ε0⎟x 11 7 d'h= − − − + + ⎥ ÷ ⎢ − ⎜ −1⎟⎥⎢2bh⎣bh⎝ 2 6 ε 6 ε ⎠ h⎥⎦ ⎣ ⎝ h ⎠ xult02 εult8 8 ⎦222 2⎡ ⎛ E ⎞ ⎛ E E ⎞ E ⎛ ⎞ ⎤ ⎡ ⎛ ⎞ ⎤+ ⎢−⎜cεult− ⎟ − ⎜cεult cεult− ⎟h cεult h 11 7 d'hE+ ⎜ ⎟ ⎥ ÷ ⎢ − ⎜ − ⎟ ⎥⎢cεult1⎣ ⎝ 2ε0 ⎠ ⎝ 2ε02 ⎠ x 6ε0 ⎝ x ⎠ ⎥⎦⎣ 8 8 ⎝ h ⎠ x⎦(Eqn F-77)Substituting into (Eqn F-62) and again setting = 0bd⎪⎧232M⎡⎛⎞ ⎤ ⎡⎛⎞⎤⎪⎫u 1 ⎛ 1 x ⎞ ε⎛ ⎞⎨ ⎢⎜0⎟x h 1 ε⎢⎜0⎟x x h= E ⎜ − ⎟ −1+ 2 − ⎥ + −1⎜ ⎟ + 3 − 3 + ⎥2 cεult⎬⎪⎩ ⎢2⎢3bh 2 ⎝ 2 h ⎠⎣⎝ε ⎠ h x⎥⎦3⎣⎝⎠⎝h ⎠ h xultεult⎥⎦⎪⎭2 ⎡3242E ⎛ ⎞⎛⎛⎞⎞ ⎛⎛⎞cεult 1 1 x⎢ ⎜ε⎜0⎟x h ⎛ h ⎞ ⎟ 1 ⎜ε⎜0⎟⎛x ⎞ x− ⎜ − ⎟ −1+ 3 − 3 + ⎜ ⎟ + −1⎜ ⎟ + 4 − 6 + 4⎢ ⎜3⎟ ⎜42ε03⎣⎝ 2 h ⎠⎝⎝ε ⎠ h x ⎝ x ⎠ ⎠4⎝⎝⎠⎝h ⎠ hultεult0.67 f ⎛ ⎞⎡⎛ ⎞ ⎤cu x⎛ ⎞⎡⎞ ⎤+ ⎜⎜⎟⎢⎜⎟ ⎥ + × ⎜ − ⎢ ⎜⎛ ⎟ − − −⎟ − ⎟ − ⎥⎠⎛ ⎞ ε0ε0 x Asb1 d'7 d'h 31 1 1 0.87 fy12γm ⎝ h ⎝ εult ⎠⎣ ⎝ εult ⎠ h⎦ bh ⎝ 2 h ⎠⎣8⎝ h ⎠ x 8 ⎦to solve for hx .A shA sh(Eqn F-78)hx⎛ h ⎞− ⎜ ⎟⎝ x ⎠2⎞⎤⎟⎥⎟⎠⎥⎦F-22

Appendix FBack-substituting into (Eqn 5-10) to solve forA sbbdCase 6 – where (1 – ε 0 /ε ult )x > h > 3x/7 i.e. 1/(1 – ε 0 /ε ult ) ≤ x/h < 7/3(1–d’/h)Referring to (Eqn F-63) of Case 6 of and settingbd= 0Nu0.67 fcu ⎡117 h ⎛ d'⎞⎤A= +sb0.87 fybh⎢ − ⎜1− ⎟γmx h⎥⎣ 8 8 ⎝ ⎠⎦bhAsb⎛ Nu0.67 fcu⎞ ⎡117 h ⎛ d'⎞⎤⇒ 0.87 f =⎜ −⎟y÷ ⎢ − ⎜1− ⎟bh⎥⎝ bh γm ⎠ ⎣ 8 8 x ⎝ h ⎠⎦Substituting into (Eqn F-64) of Case 6 and again settingbd= 0M Asb⎛ 1 d'⎞⎡7⎞ ⎤⎟⎢⎜d'h 3= 0.87 f × ⎜ − 1 − ⎟ −2ybh bh⎥⎝ 2 h ⎠⎣8⎝ h ⎠ x 8⎦7 ⎞⎡⎛⎞⎛⎞ ⎤⎜d'N −u0.67 fcu 1 d'M1 ⎟⎢⎜ −⎟⎜− ⎟ +2 ⎥x 8 ⎝ h ⎠⎣⎝bh γm ⎠⎝2 h ⎠ bh⇒ =⎦h ⎡11M ⎛ N 0.67 ⎞⎛1 ' ⎞ 3⎤ufcu d⎢ +⎜ −⎟⎜−2⎟ ⎥⎣ 8 bh ⎝ bh γm ⎠⎝2 h ⎠ 8⎦xWith hA determined, calculate sbbhA shA shbyNu0.67 fcu ⎡117 h ⎛ d'⎞⎤A= +sb0.87 fybh⎢ − ⎜1− ⎟γmx h⎥⎣ 8 8 ⎝ ⎠⎦bhNu0.67 fcu−Asbbh γm⇒ =bh ⎡117 h ⎛ d'⎞⎤0.87 fy ⎢ − ⎜1− ⎟⎥⎣ 8 8 x ⎝ h ⎠⎦(Eqn F-79)Case 7 – where x/h ≥ 7/3Nubh0.67 f=γM u= 0 2bhmcu+ 0.87 fyAsbbh⇒Asbbh⎛ Nu=⎜⎝ bh0.67 f−γmcu⎞⎟⎠10.87fy(Eqn F-80)(Eqn F-81)F-23

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 30, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9M/bh 2 N/mm 2Chart F - 1

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 30, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5M/bh 2 N/mm 2Chart F - 2

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 30, 4-bar column, d/h = 0.850.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13M/bh 2 N/mm 2Chart F - 3

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 30, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5M/bh 2 N/mm 2Chart F - 4

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 30, 4-bar column, d/h = 0.950.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5M/bh 2 N/mm 2Chart F - 5

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 35, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5M/bh 2 N/mm 2Chart F - 6

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 35, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12M/bh 2 N/mm 2Chart F - 7

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 35, 4-bar column, d/h = 0.850.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5M/bh 2 N/mm 2Chart F - 8

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 35, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15M/bh 2 N/mm 2Chart F - 9

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 35, 4-bar column, d/h = 0.950.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5M/bh 2 N/mm 2Chart F - 10

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 40, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5M/bh 2 N/mm 2Chart F - 11

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 40, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12M/bh 2 N/mm 2Chart F - 12

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 40, 4-bar column, d/h = 0.850.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5M/bh 2 N/mm 2Chart F - 13

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 40, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15M/bh 2 N/mm 2Chart F - 14

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 250454035302520Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 40, 4-bar column, d/h = 0.950.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17M/bh 2 N/mm 2Chart F - 15

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10M/bh 2 N/mm 2Chart F - 16

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5M/bh 2 N/mm 2Chart F - 17

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.850.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14M/bh 2 N/mm 2Chart F - 18

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5M/bh 2 N/mm 2Chart F - 19

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 45, 4-bar column, d/h = 0.950.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17M/bh 2 N/mm 2Chart F - 20

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 50, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10M/bh 2 N/mm 2Chart F - 21

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 50, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5M/bh 2 N/mm 2Chart F - 22

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 50, 4-bar column, d/h = 0.850.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14M/bh 2 N/mm 2Chart F - 23

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 50, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16M/bh 2 N/mm 2Chart F - 24

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 255504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 50, 4-bar column, d/h = 0.950.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5M/bh 2 N/mm 2Chart F - 25

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 55, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5M/bh 2 N/mm 2Chart F - 26

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 55, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13M/bh 2 N/mm 2Chart F - 27

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 55, 4-bar column, d/h = 0.850.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5M/bh 2 N/mm 2Chart F - 28

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 55, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16M/bh 2 N/mm 2Chart F - 29

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 55, 4-bar column, d/h = 0.950.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5M/bh 2 N/mm 2Chart F - 30

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 60, 4-bar column, d/h = 0.750.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5M/bh 2 N/mm 2Chart F - 31

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 60, 4-bar column, d/h = 0.80.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13M/bh 2 N/mm 2Chart F - 32

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 60, 4-bar column, d/h = 0.850.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5M/bh 2 N/mm 2Chart F - 33

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 60, 4-bar column, d/h = 0.90.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5M/bh 2 N/mm 2Chart F - 34

Appendix F – Summary of Design Chartsfor ColumnsN/bh N/mm 26055504540353025Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004Concrete Grade 60, 4-bar column, d/h = 0.950.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18M/bh 2 N/mm 2Chart F - 35

Rectangular Column R.C. Design to Code of Practice for Structural Use of Concrete 2004 - 4 bar columnProject :Column MarkFloorf cu = 35 N/mm 2 f y = 460 N/mm 2 E c = 23700 N/mm 2b = 400 h = 500 b' = 330.00 h' = 430.00 cover= 50 bar size = 40Basic Load CaseLoad Case No.Load CaseAxial Load P (kN)Moment M x (kNm)Moment M y (kNm)1 2 3 4 5 6D.L. L.L. Wx Wy W45 W1352304.7 582.1 -362.17 -545.1 56.92 82.0929.13 32.11 47.1 -75.12 98.1 8.93-31.33 16.09 2.15 44.2 76.99 35.21N Mx My N/bh M/bh 2 d/h / d/b x/h / y/h Steel Steel area(kN) (kNm) (kNm) (N/mm 2 ) (N/mm 2 ) (%) (mm 2 )Load Comb 1 1.4D+1.6L 4157.9 92.158 -18.118 Mx' = 99.411 20.79 0.9941 0.86 1.1701 1.9429 3885.7Load Comb 2 1.2(D+L+Wx) 3029.6 130.01 -15.708 Mx' = 140.12 15.148 1.4012 0.86 0.9702 0.8 1600Load Comb 3 1.2(D+L-Wx) 3898.8 16.968 -20.868 My' = 25.447 19.494 0.3181 0.825 1.3301 1.1905 2380.9Load Comb 4 1.2(D+L+Wy) 2810 -16.656 34.752 My' = 41.507 14.05 0.5188 0.825 1.0602 0.8 1600Load Comb 5 1.2(D+L-Wy) 4118.3 163.63 -71.328 Mx' = 192.82 20.591 1.9282 0.86 1.0405 2.4722 4944.4Load Comb 6 1.2(D+L+W45) 3532.5 191.21 74.1 Mx' = 231.22 17.662 2.3122 0.86 0.9448 2.0502 4100.4Load Comb 7 1.2(D+L-W45) 3395.9 -44.232 -110.68 My' = 125.49 16.979 1.5686 0.825 0.979 1.4098 2819.6Load Comb 8 1.2(D+L+W135) 3562.7 84.204 23.964 Mx' = 96.983 17.813 0.9698 0.86 1.0901 1.1681 2336.1Load Comb 9 1.2(D+L-W135) 3365.7 62.772 -60.54 My' = 81.79 16.828 1.0224 0.825 1.04 0.9815 1963Load Comb 10 1.4(D+Wx) 2719.5 106.72 -40.852 Mx' = 135.67 13.598 1.3567 0.86 0.9324 0.8 1600Load Comb 11 1.4(D-Wx) 3733.6 -25.158 -46.872 My' = 54.208 18.668 0.6776 0.825 1.15 1.226 2452.1Load Comb 12 1.4(D+Wy) 2463.4 -64.386 18.018 Mx' = 78.184 12.317 0.7818 0.86 0.9619 0.8 1600Load Comb 13 1.4(D-Wy) 3989.7 145.95 -105.74 Mx' = 192.25 19.949 1.9225 0.86 1.0306 2.2936 4587.2Load Comb 14 1.4(D+W45) 3306.3 178.12 63.924 Mx' = 215.64 16.531 2.1564 0.86 0.9327 1.663 3326.1Load Comb 15 1.4(D-W45) 3146.9 -96.558 -151.65 My' = 186.88 15.734 2.336 0.825 0.8772 1.6942 3388.3Load Comb 16 1.4(D+W135) 3341.5 53.284 5.432 Mx' = 56.433 16.708 0.5643 0.86 1.1401 0.8 1600Load Comb 17 1.4(D-W135) 3111.7 28.28 -93.156 My' = 103.6 15.558 1.2949 0.825 0.9803 0.8449 1689.7Load Comb 18 1.0D+1.4Wx 1797.7 95.07 -28.32 Mx' = 120.97 8.9883 1.2097 0.86 0.7146 0.8 1600Load Comb 19 1.0D-1.4Wx 2811.7 -36.81 -34.34 My' = 49.26 14.059 0.6158 0.825 1.0401 0.8 1600Load Comb 20 1.0D+1.4Wy 1541.6 -76.038 30.55 Mx' = 105.72 7.7078 1.0572 0.86 0.5779 0.8 1600Load Comb 21 1.0D-1.4Wy 3067.8 134.3 -93.21 Mx' = 193.56 15.339 1.9356 0.86 0.923 1.2156 2431.2Load Comb 22 1.0D+1.4W45 2384.4 166.47 76.456 Mx' = 226.37 11.922 2.2637 0.86 0.7836 0.8 1600Load Comb 23 1.0D-1.4W45 2225 -108.21 -139.12 My' = 191.32 11.125 2.3914 0.825 0.7278 0.8 1600Load Comb 24 1.0D+1.4W135 2419.6 41.632 17.964 Mx' = 55.564 12.098 0.5556 0.86 0.9858 0.8 1600Load Comb 25 1.0D-1.4W135 2189.8 16.628 -80.624 My' = 88.722 10.949 1.109 0.825 0.8718 0.8 1600Steel required = 2.4722 4944.4hbMxMy6000P versus Mx and My of the Column SectionP - Mx P - My Actual Loads Mx control Actual Loads My control50004000P (kN)30002000100000 100 200 300 400 500 600M (kNm)

Rectangular Column R.C. Design to Code of Practice for Structural Use of Concrete 2004Project :ColumnFloorf cu = 35 N/mm 2 f y = 460 N/mm 2 E c = 23700 N/mm 2b = 1500 h = 2000 b' = 1285.17 h' = 1684.31 cover= 50Steel provided : 15 Y 40 (Along each long sides h, excluding corner bars)12 Y 40 (Along each short sides b, excluding corner bars)4 Y 40 (Corner bars)Total Steel Area = 72885 mm 2 Steel Percentage = 2.43 % Max. Ultimate Load = 76069 kNArea of Steel per mm length for the long sides bars (including corner bars) = 21.36 mm 2 /mmArea of Steel along long sides (excluding corner bars) = 37699 mm 2Area of Steel per mm length for the short sides bars (including corner bars) = 23.46 mm 2 /mmArea of Steel along short sides (excluding corner bars) = 30159 mm 2Basic Load CaseLoad Case No.Load CaseAxial Load P (kN)Moment M x (kNm)Moment M y (kNm)1 2 3 4 5 6D.L. L.L. Wx Wy W45 W13537872 1101 -3628.1 -2611.1 -5692.3 8209.2-291.3 -37.11 470.81 -3700 -1750.3 4892.9-31.33 16.09 5.17 2700 2764 -3520.2P M x M yLoad Comb 1 1.4D+1.6L 54782 -467.2 -18.118 Mx' = 476.55 Mux = 16452 Section OKLoad Comb 2 1.2(D+L+Wx) 42413 170.88 -12.084 Mx' = 179.2 Mux = 23474 Section OKLoad Comb 3 1.2(D+L-Wx) 51121 -959.06 -24.492 Mx' = 973.01 Mux = 18743 Section OKLoad Comb 4 1.2(D+L+Wy) 43634 -4834.1 3221.7 Mx' = 6999.6 Mux = 22861 Section OKLoad Comb 5 1.2(D+L-Wy) 49900 4045.9 -3258.3 My' = 4639 Muy = 14871 Section OKLoad Comb 6 1.2(D+L+W45) 39936 -2494.5 3298.5 My' = 4352.2 Muy = 18945 Section OKLoad Comb 7 1.2(D+L-W45) 53598 1706.3 -3335.1 My' = 3865.6 Muy = 13130 Section OKLoad Comb 8 1.2(D+L+W135) 56618 5477.4 -4242.6 My' = 5801.2 Muy = 11590 Section OKLoad Comb 9 1.2(D+L-W135) 36916 -6265.5 4206 Mx' = 9507.3 Mux = 26076 Section OKLoad Comb 10 1.4(D+Wx) 47941 251.31 -36.624 Mx' = 273.77 Mux = 20575 Section OKLoad Comb 11 1.4(D-Wx) 58099 -1067 -51.1 Mx' = 1090.8 Mux = 14185 Section OKLoad Comb 12 1.4(D+Wy) 49365 -5587.8 3736.1 Mx' = 7805.2 Mux = 19771 Section OKLoad Comb 13 1.4(D-Wy) 56676 4772.2 -3823.9 My' = 5179.4 Muy = 11560 Section OKLoad Comb 14 1.4(D+W45) 45051 -2858.3 3825.7 My' = 4911.9 Muy = 16951 Section OKLoad Comb 15 1.4(D-W45) 60989 2042.6 -3913.4 My' = 4416.8 Muy = 9164 Section OKLoad Comb 16 1.4(D+W135) 64513 6442.2 -4972.2 My' = 6446.8 Muy = 7132 Section OKLoad Comb 17 1.4(D-W135) 41527 -7257.9 4884.4 Mx' = 10685 Mux = 23910 Section OKLoad Comb 18 1.0D+1.4Wx 32792 367.83 -24.092 Mx' = 387.89 Mux = 27896 Section OKLoad Comb 19 1.0D-1.4Wx 42951 -950.43 -38.568 Mx' = 976.72 Mux = 23206 Section OKLoad Comb 20 1.0D+1.4Wy 34216 -5471.3 3748.7 Mx' = 8512.2 Mux = 27278 Section OKLoad Comb 21 1.0D-1.4Wy 41527 4888.7 -3811.3 My' = 5808.5 Muy = 18343 Section OKLoad Comb 22 1.0D+1.4W45 29902 -2741.8 3838.3 My' = 5236.3 Muy = 22460 Section OKLoad Comb 23 1.0D-1.4W45 45841 2159.2 -3900.9 My' = 4707.8 Muy = 16626 Section OKLoad Comb 24 1.0D+1.4W135 49364 6558.7 -4959.6 Mx' = 9502.2 Mux = 19771 Section OKLoad Comb 25 1.0D-1.4W135 26379 -7141.3 4897 Mx' = 11689 Mux = 29921 Section OKP (kN)80000700006000050000400003000020000100000P versus Mx and My of the Column SectionP - Mx P - My Actual Loads Mx control Actual Loads My control0 5000 10000 15000 20000 25000 30000 35000M (kNm)

Appendix GDerivation of Design Formulaefor Walls to Rigorous StressStrain Curve of Concrete

Appendix GDerivation of Design Formulae for Shear Walls to Rigorous Stress StrainCurve of ConcretebhAs similar to the exercise in Appendix F for columns, the exercise in this Appendix isArepeated for walls by which sb are set to zero in the various cases 1 to 7, using thebhequations summarized in Appendix F.Cases 1 to 3 – where x/h ≤ 7/11By (Eqn F-18) or (Eqn F-32) or (Eqn F-36) of Appendix FNu0.67fcu⎛ ε ⎞ x ⎛ x ⎞ A=shfy ⎜ ⎟bh⎜ 3 −0⎟ + 0.87 2 − 13γm ⎝ εult ⎠ h ⎝ h ⎠ bhN 0 ⎛ ⎞u.67 f −cuε x⎜3−0⎟A bh 3γm=⎝ ε ⎠ hshultbh⎛ x ⎞0.87 fy ⎜2−1⎟⎝ h ⎠(Eqn G-1)Substituting into (Eqn F-31) or (Eqn F-34) or (Eqn F-38) and putting = 0bh⎪⎧⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪⎫⎡⎛⎞ ⎛ ⎞⎥ ⎥ ⎤⎨ ⎜⎛ 22Mu0.67 fcu ⎛ x ⎞ 1 x 1 ε −0 1 ε0 x 1 ε0 xAshx 163 x= ⎜ ⎟ 1 ⎟ − + ⎜ ⎟ −⎜⎟ ⎜ ⎟⎬+ 0.87 f ⎢⎜⎟ −2y⎜ ⎟bh γm ⎝ h ⎠ 2 ⎪⎩⎝ h ⎠ 6 εult3 εult ⎝ h ⎠ 12 ⎝ εult ⎠ ⎝ h ⎠ bh⎪⎭⎢⎣⎝ h ⎠ 147 ⎝ h ⎠ ⎦2 3220.67 f ⎪⎧ 16 131 0 1 0 ⎪⎫⎪⎧0.67 ⎡11 0 1 ⎤0 163⎪⎫cuε ⎛ ε ⎞ ⎛ x ⎞ fcuε ⎛ ε ⎞ Nu ⎛ x ⎞⇒ ⎨ + − ⎬⎜⎟ + ⎨ ⎢ − + ⎥ − ⎬⎜⎟147 441 6⎜⎟2 3 12⎜⎟γmεult⎝ ⎠147⎪⎩⎝ εult ⎠ h γm⎪⎭⎝ ⎠⎪⎩⎢ εult⎣⎝ εult ⎠ ⎥ bh h⎦ ⎪⎭⎡ N 2 0.67 1 1 ⎤uMufcu⎛ ε0⎞ x Mu+ ⎢ − −+ = 02 ⎜ − ⎥2⎣2 6⎟(Eqn G-2)bh bh γm ⎝ εult ⎠⎦h bhx AUpon solving , back substituting into (Eqn G-1) to calculate shh bhA sbCase 4 – where 7/11 < x/h ≤ 1G-1

Appendix GReferring to (Eqn F-39) of Appendix F and setting A sb = 0bh0.67fcu⎛ ε ⎞ x Ash⎛ x h ⎞ Nu⎜ −09 7 73⎟ + 0.87 fy ⎜−− + ⎟ =3γh bh h x bhm ⎝ εult ⎠⎝ 56 8 4 ⎠N 0 ⎛ ⎞u.67 f −cuε x⎜3−0⎟A bh 3γm⇒ =⎝ ε ⎠ hshultbh ⎛ 9 x 7 h 7 ⎞0.87 fy ⎜−− + ⎟⎝ 56 h 8 x 4 ⎠(Eqn G-3)A sbSubstituting into (Eqn F-41) with = 0bh⎪⎧⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎫⎨ ⎜⎛ 2Mu0.67 fcu⎛ x ⎞ 1 x 1 ε −0 1 ε0 x 1 ε0 x= ⎜ ⎟ 1 ⎟ − + ⎜ ⎟ −⎜⎟2⎜ ⎟ ⎬bh γm ⎝ h ⎠ 2 ⎪⎩⎝ h ⎠ 6 εult3 εult ⎝ h ⎠ 12 ⎝ εult ⎠ ⎝ h ⎠⎪ ⎭2⎡ 7 h 9 x 9 ⎛ x ⎞ ⎤ Ash+ 0 .87 fy ⎢ − + ⎜ ⎟ ⎥⎢⎣48 x 112 h 392 ⎝ h ⎠ ⎥⎦bh2M⎪⎧⎛⎞ ⎡⎛ ⎞ ⎤⎪⎫u ⎛ 7 9 x 7 h ⎞ 0.67 fcu⎛ x ⎞ 1 1 ε0⎛ ⎞ ⎛⎨+ 1 ε01 1 ε0x 7⇒ ⎜ − − ⎟ = ⎜ ⎟⎜ − ⎟ ⎢ − −⎜⎟ ⎥⎜⎟⎬⎜−2bh ⎝ 4 56 h 8 x ⎠ γm ⎝ h ⎠⎝ ⎠ ⎝⎪⎩ ⎝ 2 6 εult ⎠ ⎢3εult2 12⎣⎝ εult ⎠ ⎥ h 4⎦ ⎪⎭2⎡ 7 h 9 x 9 ⎛ x ⎞ ⎤⎡N ⎛ ⎞ ⎤u0.67 f⎢⎥⎢−cuε+ − + ⎜ ⎟⎜ −0 x3⎟ ⎥⎢⎣48 x 112 h 392 ⎝ h ⎠ ⎥⎦⎣ bh 3γm ⎝ εult ⎠ h⎦2 40.67 f ⎡ 45 9 0 3 ⎛ ⎞ ⎤cuε ε0 ⎛ x ⎞⇒ ⎢−+ − ⎥⎜⎟⎢ 784 196 224⎜⎟γmεult⎣⎝ εult ⎠ ⎥⎦⎝ h ⎠2⎡30.67 f ⎡7707 ⎛ ⎞ ⎤09⎤⎢cuε ε Nu⎥⎛ x ⎞+ ⎢ − + ⎥ − ⎜ ⎟⎢ 8 12 48⎜⎟γm392 ⎥⎝⎠⎣⎢ εult⎣⎝ εult ⎠ ⎥ bh h⎦ ⎦2⎡29 M 0.67 7 ⎡ 201 ⎛ ⎞ ⎤09⎤⎢ufcuε ε Nu⎛ x ⎞+ − + ⎢−1.5+ − ⎥ + ⎥2⎜ ⎟⎢ 568 3 12⎜⎟bh γm112 ⎥⎝⎠⎣⎢ εult⎣⎝ εult ⎠ ⎥ bh h⎦ ⎦7 M 0.67 7 ⎡ ⎤0 ⎡ 7 7 ⎤+uf3= 02 +cuε x M⎢ −uNu⎥ +24 3 12⎢−−8 48 ⎥(Eqn G-4)bh γm ⎣ εult ⎦ h ⎣ bh bh ⎦x AUpon solving , back-substitute into (Eqn G-3) to solve for shh bh956xh−7 h ⎞⎟8 x ⎠Case 5 – where 1 < x/h ≤ 1/(1 – ε 0 /ε ult )A sbReferring to (Eqn F-52) and setting = 0bd2N ⎡⎤u1 1 ε01 εult1 ε0x= Ecεult ⎢−− + +2⎥ + Ecεbh ⎣ 2 6 ε 6 ε02 εultult ⎦ hult⎛ 1 ε⎜ −ult1⎝ 2 ε0⎟ ⎞⎠G-2

Appendix GG-3bhAhxxhfxhExhEshyultcultultc⎟ ⎠ ⎞⎜⎝⎛−−+⎟⎠⎞⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛−+56987470.87621212020 εεεεε (Eqn G-5)Substituting forbhAfshy.870 into (Eqn F-57), again setting 0=bdA sb2002203302241614161241⎟⎠⎞⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−+−+−=hxEbhMultultultultultcεεεεεεεεε200002202411211211214141121⎟⎠⎞⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛−+⎥⎥⎦⎤⎢⎢⎣⎡+−+−+xhExhEhxEultultcultultcultultultultcεεεεεεεεεεεεεbhAhxhxxhfshy⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−+239291129487.870200220330256987472416141612415698747⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛−−⎟⎟⎠⎞⎜⎜⎝⎛−+−+−=⎟⎠⎞⎜⎝⎛−−⇒hxhxxhEhxxhbhMultultultultultcεεεεεεεεεxhhxxhEhxhxxhEultultcultultultultc⎟ ⎠ ⎞⎜⎝⎛−−⎟⎟⎠⎞⎜⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−−⎥⎥⎦⎤⎢⎢⎣⎡+−+−+569874712112156987471214141121000220εεεεεεεεεε⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−+⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛−−+220 392911294875698747241hxhxxhbhNxhhxxhEuultultcεεε⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−⎥⎥⎦⎤⎢⎢⎣⎡++−−−2002203929112948721616121hxhxxhhxEultultultultcεεεεεεε⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−⎟⎟⎠⎞⎜⎜⎝⎛−−⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−⎟⎟⎠⎞⎜⎜⎝⎛ −− 2020 39291129487212139291129487211hxhxxhxhEhxhxxhEultultcultultcεεεεεε⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+−⎟⎠⎞⎜⎝⎛+2202392911294876 hxhxxhxhEultc εε30022033031369196315684539294483⎟⎠⎞⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛+−+−⇒hxEultultultultultcεεεεεεεεε2002203303929470428929479167247967⎟⎠⎞⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡+⎟⎟⎠⎞⎜⎜⎝⎛−+−+−+hxbhNEuultultultultultcεεεεεεεεε⎟⎠⎞⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡−+⎟⎟⎠⎞⎜⎜⎝⎛+−+−+hxbhNbhMEuultultultultultc11295699408122958828932212471937200220330εεεεεεεεεxhbhNbhMEbhMEuultcultultultultc⎟ ⎟ ⎠⎞⎜⎜⎝⎛++−+⎥⎥⎦⎤⎢⎢⎣⎡−⎟⎟⎠⎞⎜⎜⎝⎛−+−+487871344125477056281215247727202200220εεεεεεεεε05767967302202=⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+xhExhEultcultcεεεε(Eqn G-6)which is in fact an equation of 6 th power in h x . h xis to be solved by numerical

Appendix Gxmethod. By back-substituting into (Eqn G-5) h2N ⎡⎤ ⎛⎟ ⎞u1 1 ε01 εult1 ε0x1 ε= ⎢−− + + ⎥ +⎜12−ultEcεultEcεultbh ⎣ 2 6 ε 6 ε02 εult ⎦ h ⎝ 2 εult0 ⎠2 2⎛ 1 εult 1 ⎞ h Ecεult ⎛ h ⎞ ⎛ 7 7 h 9 x ⎞ Ash+ Ecεult⎜ −⎟ − ⎜ ⎟ + 0.87 fy ⎜ − − ⎟⎝ 2 ε02 ⎠ x 6ε0 ⎝ x ⎠ ⎝ 4 8 x 56 h ⎠ bh2N ⎡⎤u 1 1 ε 0 1 ε ult 1 ε 0 x ⎛ 1 ε ⎞⎢⎥ −⎜ −ult− E − − + +⎟cεultEcεult 1 − Ecε2A bh ⎢⎣2 6 ε 6 ε ult ⎥⎦hshult 0 2 ε⎝ 2 ε 0 ⎠=bh⎛ 7 7 h 9 x ⎞0.87 f y ⎜ − − ⎟⎝ 4 8 x 56 h ⎠ult⎛ 1 ε ult⎜⎝ 2 ε 02ult1 ⎞ h Ecε−⎟ +2 ⎠ x 6ε0⎛ h ⎞⎜ ⎟⎝ x ⎠2Case 6 – where 1/(1 – ε 0 /ε ult ) < x/h ≤ 7/3Consider (Eqn F-58) and substituting A sb = 0bhNu0.67 fcu−Nu0.67fcu⎛ 7 7 h 9 x ⎞ AshAshbh γm= + 0.87 fy ⎜ − − ⎟ ⇒ =bh γ⎝ x h ⎠ bh bh ⎛ h xf ⎜ − − ⎟ ⎞m4 8 567 7 90.87y⎝ 4 8 x 56 h ⎠(Eqn G-7)Substituting into (Eqn F-60)2M⎡⎤uAsh7 h 9 x 9 ⎛ x ⎞= 0.87 f ⎢ − +2y⎜ ⎟ ⎥bh bh ⎢⎣48 x 112 h 392 ⎝ h ⎠ ⎥⎦329 ⎛ N 0.67 ⎞ ⎡ 9 9 ⎛ 0.67 ⎞⎤ufcu ⎛ x ⎞ MuNufcu⎛ x ⎞⇒⎜ ⎟ + ⎢ −2⎥⎜⎟392⎜ −⎟⎣56112⎜ −⎟⎝ bh γm ⎠⎝h ⎠ bh ⎝ bh γm ⎠⎦⎝h ⎠7 M ⎡ 7 ⎛ 0.67 ⎞ 7 ⎤u x NufcuMu− += 02 ⎢+2 ⎥4 ⎣48⎜ −⎟(Eqn G-8)bh h ⎝ bh γm ⎠ 8 bh ⎦xASolving the cubic equation for and substituting into (Eqn G-7) to solve for shh bhCase 7 – where x/h > 7/3By (Eqn F-61) and setting = 0bhNu0.67 fcuAshA ⎛shNu= + 0.87 fy⇒ =bhmbh bh⎜γ⎝ bhA sb 0.67 f ⎞cu 1−⎟γm0.87 fy⎠(Eqn G-9)M u= 0 2bhG-4

Appendix G – Summary of Design Charts for WallsN/bh N/mm 250454035302520Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 300.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9M/bh 2 N/mm 2Chart G - 1

Appendix G – Summary of Design Charts for WallsN/bh N/mm 250454035302520Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 350.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5M/bh 2 N/mm 2Chart G - 2

Appendix G – Summary of Design Charts for WallsN/bh N/mm 250454035302520Design Chart of Rectangular Shear Wall with Uniform Vertical reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 400.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel1510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5M/bh 2 N/mm 2Chart G - 3

Appendix G – Summary of Design Charts for WallsN/bh N/mm 255504540353025Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 450.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10M/bh 2 N/mm 2Chart G - 4

Appendix G – Summary of Design Charts for WallsN/bh N/mm 255504540353025Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 500.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10M/bh 2 N/mm 2Chart G - 5

Appendix G – Summary of Design Charts for WallsN/bh N/mm 26055504540353025Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 550.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5M/bh 2 N/mm 2Chart G - 6

Appendix G – Summary of Design Charts for WallsN/bh N/mm 26055504540353025Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practicefor Structural Use of Concrete 2004, Concrete Grade 600.4% steel1% steel2% steel3% steel4% steel5% steel6% steel7% steel8% steel201510500 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5M/bh 2 N/mm 2Chart G - 7

Shear Wall R.C. Design to Code of Practice for Structural Use of Concrete 2004 - Uniform ReinforcementsProject :Wall MarkFloorf cu = 35 N/mm 2 f y = 460 N/mm 2 E c = 23700 N/mm 2b = 200 h = 2000 b' = 165.00 h' = 1500.00 cover= 25 bar size = 20Basic Load Case My MxLoad Case No. 1 2 3 4 5 6Load Case D.L. L.L. Wx Wy W45 W135Axial Load P (kN) 3304.7 1582.1 -362.17 -245.1 56.92 82.09Moment M x (kNm) 29.13 32.11 2047.1 -1275.1 1098.1 888.93hMoment M y (kNm) -31.33 16.09 2.15 44.2 206.5 35.21bN Mx My N/bh M/bh 2 d/h / d/b x/h / y/b Steel Steel area(kN) (kNm) (kNm) (N/mm 2 ) (N/mm 2 ) (%) (mm 2 )Load Comb 1 1.4D+1.6L 7200 1500 100 Mx' = 1866.2 18 2.3328 - 0.8901 2.3359 9343.5Load Comb 2 1.2(D+L+Wx) 5429.6 2530 -15.708 Mx' = 2607.8 13.574 3.2597 - 0.7152 2.3059 9223.4Load Comb 3 1.2(D+L-Wx) 6298.8 -2383 -20.868 Mx' = 2473.2 15.747 3.0915 - 0.7772 2.5637 10255Load Comb 4 1.2(D+L+Wy) 5570 -1456.7 34.752 Mx' = 1624.9 13.925 2.0311 - 0.8365 1.0928 4371.1Load Comb 5 1.2(D+L-Wy) 6158.3 1603.6 -71.328 Mx' = 1918.9 15.396 2.3986 - 0.8314 1.7861 7144.5Load Comb 6 1.2(D+L+W45) 5932.5 1391.2 229.51 My' = 306.62 14.831 3.8328 0.825 0.7508 2.7599 11040Load Comb 7 1.2(D+L-W45) 5795.9 -1244.2 -266.09 My' = 336.52 14.49 4.2065 0.825 0.7255 3.0079 12032Load Comb 8 1.2(D+L+W135) 5962.7 1140.2 23.964 Mx' = 1249.5 14.907 1.5618 - 0.9185 0.9046 3618.4Load Comb 9 1.2(D+L-W135) 5765.7 -993.23 -60.54 Mx' = 1277.8 14.414 1.5972 - 0.9032 0.8122 3248.8Load Comb 10 1.4(D+Wx) 4119.5 2906.7 -40.852 Mx' = 3150.7 10.299 3.9384 - 0.6025 2.5143 10057Load Comb 11 1.4(D-Wx) 5133.6 -2825.2 -46.872 Mx' = 3068 12.834 3.835 - 0.6673 2.7996 11199Load Comb 12 1.4(D+Wy) 4283.4 -1744.4 18.018 Mx' = 1849.7 10.709 2.3121 - 0.6995 0.7417 2966.7Load Comb 13 1.4(D-Wy) 4969.7 1826 -105.74 Mx' = 2387.4 12.424 2.9842 - 0.7031 1.7913 7165.1Load Comb 14 1.4(D+W45) 4706.3 1578.1 245.24 My' = 350.54 11.766 4.3818 0.825 0.6532 2.626 10504Load Comb 15 1.4(D-W45) 4546.9 -1496.6 -332.96 My' = 435.07 11.367 5.4384 0.825 0.62 3.4305 13722Load Comb 16 1.4(D+W135) 4741.5 1285.3 5.432 Mx' = 1315.1 11.854 1.6439 - 0.8274 0.4 1600Load Comb 17 1.4(D-W135) 4511.7 -1203.7 -93.156 Mx' = 1731.6 11.279 2.1645 - 0.7373 0.6743 2697.1Load Comb 18 1.0D+1.4Wx 2797.7 2895.1 -28.32 Mx' = 3093.4 6.9942 3.8667 - 0.5051 2.1902 8760.9Load Comb 19 1.0D-1.4Wx 3811.7 -2836.8 -34.34 Mx' = 3050.1 9.5293 3.8126 - 0.5843 2.2839 9135.8Load Comb 20 1.0D+1.4Wy 2961.6 -1756 30.55 Mx' = 1966 7.4039 2.4576 - 0.5307 0.607 2428.1Load Comb 21 1.0D-1.4Wy 3647.8 1814.3 -93.21 Mx' = 2405.2 9.1196 3.0065 - 0.5941 1.3272 5308.9Load Comb 22 1.0D+1.4W45 3384.4 1566.5 257.77 My' = 381.82 8.461 4.7727 0.825 0.5596 2.3848 9539.2Load Comb 23 1.0D-1.4W45 3225 -1508.2 -320.43 My' = 442.13 8.0625 5.5266 0.825 0.5456 2.9149 11660Load Comb 24 1.0D+1.4W135 3419.6 1273.6 17.964 Mx' = 1390.7 8.5491 1.7384 - 0.6436 0.4 1600Load Comb 25 1.0D-1.4W135 3189.8 -1215.4 -80.624 Mx' = 1755.3 7.9744 2.1941 - 0.5706 0.4 1600Steel required = 3.4305 13722Plot of P (kN) versus M (kNm)14000P - Mx P - My Actual Loads Mx control Actual Loads My control1200010000P (kN)800060004000200000 500 1000 1500 2000 2500 3000 3500 4000 4500M (kNm)

Appendix HEstimation of Support Stiffnessesof vertical supports to TransferStructures

Appendix HEstimation of Support Stiffnesses of vertical supports to Transfer StructuresSimulation of Support Stiffness in Plate Bending StructureFor support stiffness, we are referring to the force or moment required to produce unitvertical movement or unit rotation at the top of the support which are denoted byKZ,K θ X, K θ Yfor settlement stiffness along the Z direction, and rotational stiffnesses aboutX and Y directions. These stiffnesses are independent parameters which can interact onlythrough the plate structure. Most softwares allow the user either to input numerical valuesor structural sizes and heights of the support (which are usually walls or columns) bywhich the softwares can calculate numerical values for the support stiffnesses as follows :(i) For the settlement stiffness KZ, the value is mostly simplyAE L where A isthe cross sectional of the support which is either a column or a wall, E is theYoung’s Modulus of the construction material and L is the free length of thecolumn / wall. The AE L simply measures the ‘elastic shortening’ of thecolumn / wall.Strictly speaking, the expression AE L is only correct if the column / wall isone storey high and restrained completely from settlement at the bottom. However,if the column / wall is more than one storey high, the settlement stiffnessestimation can be very complicated. It will not even be a constant value. Thesettlement of the support is, in fact, ‘interacting’ with that of others through thestructural frame linking them together by transferring the axial loads in thecolumn / wall to others through shears in the linking beams. Nevertheless, if thelinking beams (usually floor beams) in the structural frame are ‘flexible’, thetransfer of loads from one column / wall through the linking beams to the rest ofthe frame will generally be negligible. By ignoring such transfer, the settlementstiffness of a column / wall can be obtained by ‘compounding’ the settlementstiffness of the individual settlement stiffness at each floor asKZ=L1A E11L2+A E221L3+A E33Ln+ ......A Enn=∑1LiiA EiH-1

Appendix H(ii)For the rotational stiffness, most of the existing softwares calculate the numericalvalues either by4EILor3EI, depending on whether the far end of the supportingLcolumn / wall is assumed fixed or pinned (where I is the second moment of areaof the column / wall section). However, apart from the assumed fixity at the farend, the formulae4EILor3EILare also based on the assumption that both ends ofthe column / wall are restrained from lateral movement (sidesway). It is obviousthat the assumption will not be valid if the out-of-plane load or the structurallayout is unsymmetrical where the plate will have lateral movements. The errorsmay be significant if the structure is to simulate a transfer plate under wind loadwhich is in the form of an out-of-plane moment tending to overturn the structure.Nevertheless, the errors can be minimized by finding the force that will berequired to restrain the slab structure from sideswaying and applying a force ofthe same magnitude but of opposite direction to nullify this force. This magnitudeof this restraining force or nullifying force is the sum of the total shears producedin the supporting walls / columns due to the moments induced on the walls /columns from the plate analysis. However, the analysis of finding the effects onthe plate by the “nullifying force” has to be done on a plane frame or a spaceframe structure as the 2-D plate bending model cannot cater for lateral in-planeloads. This approach is adopted by some local engineers and the procedure foranalysis is illustrated in Figure H-1.Lateral force, S , toprevent sideswayh 1S1MU1S2MU 2S3MU 3h 2 h 3SS13S2Figure H-1 Diagrammatic illustration of the restraining shear or nullifying shearIn addition, the followings should be noted :H-2

Appendix HNote : 1. If the wall / column is prismatic and the lower end is restrained fromrotation, the moment at the lower end will be MLi= 0. 5MUi(carry-overfrom the top); if the lower end is assumed pinned, the moment at it will bezero;MUi+ MLi2. The shear on the wall / column will be Si= where MUiishiobtained from plate bending analysis and the total restraining shear isS S∑=iH-3

Appendix IDerivation of Formulae forRigid Cap Analysis

Appendix IDerivation of Formulae for Rigid Cap AnalysisUnderlying Principles of the Rigid Cap AnalysisThe “Rigid Cap Analysis” method utilizes the assumption of “Rigid Cap” in thesolution of pile loads beneath a single cap against out-of-plane loads, i.e. the cap is aperfectly rigid body which does not deform upon the application of loads. The capitself may settle, translate or rotate, but as a rigid body. The deflections of aconnecting pile will therefore be well defined by the movement of the cap and thelocation of the pile beneath the cap, taking into consideration of the connectionconditions of the piles. Consider a Pile i situated from a point O on the pile cap asshown in Figure I-1 with settlement stiffnessKiZYPile i+veMYyi+veMXOxiXFigure I-1 – Derivation of Pile Loads under Rigid CapAs the settlement of all piles beneath the Cap will lie in the same plane after theapplication of the out-of-plane load, the settlement of Pile i denoted by ∆iZcan be defined by bO+ b1 xi+ b2yiwhich is the equation for a plane in‘co-ordinate geometry’ where bO, b1and b2are constants.The upward reaction by Pile i is K b + b x + b y )iZ(O 1 i 2 iSumming all pile loads :∑ iZ O 1 i+K yBalancing the applied load K ( b + b x b y )P =2 i⇒ P = bO∑ KiZ+ b1∑ KiZxi+ b2∑iZ iI-1

Balancing the applied Moment X− K iZ( b O+ b x ib y i) y i⇒ MX= −bO∑KiZyi− b∑= ∑M1+2∑21KiZxiyi− b2KiZyiM ∑ K b + b x + b=2Balancing the applied Moment YiZ( O 1 i i) i2⇒ MY= bO∑ KiZxi+ b1∑ KiZxi+ b2∑∑KiZxiyiIt is possible to choose the centre O such that∑ KiZxi= ∑KiZyi= ∑KiZxiyi= 0 .So the three equations becomePP = b O ∑ KiZ⇒ bO=KMMXY∑22KiZyi= −b⇒∑21KiZxi= b⇒bb∑iZ− MX2=2KiZyiY1=2KiZxi∑The load on Pile i is thenP = ∑ KiZ( bO+ b1 xi+ b2yi)⎛⎜ P MYMX=iZ+ xi−⎜2⎝ ∑ KiZ ∑ KiZxi∑ KiZyPKiZMYKiZMXKiZ= + xi− y22 iK K x K yMK2i∑iZ∑∑iZiZii∑iZ= ∑ iZ i ∑iiZ⎞y ⎟i⎟⎠iiyxAppendix ITo effect K x K y = K x y = 0 , the location of O and theorientation of the axes X-X and Y-Y must then be the “principal axes” of thepile group.Conventionally, designers may like to use moments along defined axes instead ofmoments about defined axes. If we rename the axes and U-U and V-V after translationand rotation of the axes X-X and Y-Y such that the condition∑PK u K v = K u v = 0 can be satisfied, then the pile load becomeiZiPK= ∑ iZ i ∑MKiZ U iZY iZiZ= + ui+22KiZKiZuiKiZvi∑∑iZiiM∑KIf all piles are identical, i.e. all KiZare equal, then the formula is reducedP MUMVPiZ= + ui+ viwhere N is the number of piles.2 2N u v∑i∑iOr if we do not wish to rotate the axes to U and V , then only∑ iZxi∑ KiZyi=K = 0 and the moment balancing equations becomesviI-2

Appendix IMX= −bO∑KiZyi− b∑∑21KiZxiyi− b2KiZyi⇒ MX= −b∑∑21KiZxiyi− b2KiZyiandMY2= bO∑ KiZxi+ b1∑ KiZxi+ b2∑KiZxiyi2⇒ MY= b1∑ KiZxi+ b2SolvingP = bbb12O=−=−∑ KiZ⇒ bO=M∑∑PK222( K x y ) + ( K x K y )∑X− M∑iZiKiiZxiyi+ M∑KiZYiZiZxiy∑Ki ∑iZiy2i222( K x y ) + ( K x K y )∑YiZ∑iKiiZxiyi− M∑iZX∑Ki ∑iZiZx2iiZiiSo the pile load becomesPiZ=PKMKiZX iZ i i+∑ KiZ −iZ i i+− MYxy+ M222( K x y ) ( K x K y )∑∑∑YiZ∑ Ki ∑+( ) ( )iZ i−∑KiZ∑xiKyiiZ2xi+yi− M∑KiZX∑ K2i ∑x2iZxi2KiZyiIf all piles are identical, i.e. all K iZ are equal, then the formula is reducedPiZ=PNM+−X ∑∑ i22 2 i22 2( x y ) + ( x y ) − ( x y ) + ( x y )ixiyi+ M∑Y∑y∑i2i∑ix+iZKy2iiZ− M∑yYii.∑ixKiyiZixi− MFor a symmetrical layout where x = 0 , the equation is further reduced toy i i∑Xi∑∑x2iiyiPiZPM− MYX= +i+22N xiyi∑x∑yiI-3

Appendix JMathematical Simulation ofCurves related to Shrinkage andCreep Determination

Simulation of Curves for Shrinkage and Creep DeterminationAppendix JSimulation ofKjvaluesFigure 3.5 of the Code is expanded and intermediate lines are added for reading moreaccurate values. The intermediate values are scaled off from the expanded figure andlisted as follows (h e = 50 mm which is seldom used is ignored) :h e = 100 mm h e = 200 mm h e = 400 mm h e = 800 mmDays K j Days K j Days K j Days K j2 0.09 6 0.09 16.6 0.065 60 0.0653 0.108 7 0.095 20 0.08 70 0.0754 0.125 8 0.1 30 0.115 80 0.0845 0.145 9 0.105 40 0.145 90 0.0926 0.165 10 0.112 50 0.165 100 0.0997 0.185 11 0.12 60 0.185 200 0.178 0.2 12 0.13 70 0.2 300 0.229 0.213 13 0.138 80 0.22 400 0.26510 0.225 14 0.145 90 0.235 500 0.3120 0.33 20 0.18 100 0.25 600 0.3530 0.4 30 0.23 200 0.375 700 0.38640 0.45 40 0.275 300 0.46 800 0.4250 0.5 50 0.31 400 0.54 900 0.4560 0.543 60 0.345 500 0.6 1000 0.4870 0.57 70 0.37 600 0.64 2000 0.7380 0.6 80 0.4 700 0.67 3000 0.8390 0.625 90 0.425 800 0.7 4000 0.888100 0.645 100 0.445 900 0.72 5000 0.923200 0.775 200 0.61 1000 0.74 6000 0.95300 0.827 300 0.7 2000 0.87 7000 0.97400 0.865 400 0.75 3000 0.935 8000 0.98500 0.892 500 0.79 4000 0.97600 0.91 600 0.81 5000 0.99700 0.927 700 0.84800 0.937 800 0.855900 0.945 900 0.871000 0.955 1000 0.8831500 0.975 2000 0.955J-1

Appendix JCurves are plotted accordingly in Microsoft Excel as shown :Simulation of Kj ValuesEffective thickness = 100mmEffective thickness = 400mmEffective thickness = 200mmEffective thickness = 800mm10.90.80.70.6Kj0.50.40.30.20.101 10 100 1000 10000Time since loading, DaysThese curves are divided into parts and polynomial equations (x denote days) aresimulated by regression done by the Excel as follows :(i) Effectiveness thickness h = 100 mmefor 2 ≤ x ≤ 10K j = –1.5740740764×10 -6 x 6 + 7.1089743699×10 -5 x 5 – 1.2348646738×10 -3 x 4 +1.0396454943×10 -2 x 3 – 4.4218106746×10 -2 x 2 + 1.0785366750×10 -1 x –1.4422222154×10 -2 ;for 10 < x ≤ 100K j = –8.2638888726×10 -12 x 6 + 2.9424679436×10 -9 x 5 – 4.1646100361×10 -7 x 4 +2.9995170408×10 -5 x 3 – 1.1964688098×10 -3 x 2 + 3.0905446162×10 -2 x +9.3000049487×10 -3for 100 < x ≤ 1000K j = –9.9999999553×10 -18 x 6 + 3.7871794729×10 -14 x 5 – 5.7487179303×10 -11 x 4+ 4.4829720169×10 -8 x 3 – 1.9268813492×10 -5 x 2 + 4.6787198128×10 -3 x +3.3059999890×10 -1(ii) Effectiveness thickness h = 200 mmefor 1 ≤ x ≤ 10K j = –5.5555555584×10 -7 x 6 + 1.9230769236×10 -5 x 5 – 2.3632478631×10 -4 x 4 +1.1888111887×10 -3 x 3 – 1.8372455154×10 -3 x 2 + 5.1966197721×10 -3 x +5.0666667394×10 -2for 10 < x ≤ 100K j = –6.0905886799×10 -12 x 6 + 2.0287559012×10 -9 x 5 – 2.6706836340×10 -7 x 4 +J-2

Appendix J1.7840233064E×10 -5 x 3 – 6.6454331705×10 -4 x 2 + 1.7736234727×10 -2 x -1.3696178365×10 -2for 100 < x ≤ 1000K j = –4.1666665317×10 -19 x 6 + 4.6185897038×10 -15 x 5 – 1.2899038408×10 -11 x 4+ 1.6179152071×10 -8 x 3 – 1.0631842073×10 -5 x 2 + 3.8848713316×10 -3 x +1.4793333214×10 -1(iii) Effectiveness thickness h = 400 mmefor 1 ≤ x ≤ 16.6K j = 1.4187214466×10 -6 x 4 – 3.5464080361×10 -5 x 3 + 3.3384218737×10 -4 x 2 –2.2688256448×10 -5 x + 2.7836053347×10 -2for 16.6 < x ≤ 100K j = –1.5740740764×10 -6 x 6 + 7.1089743699×10 -5 x 5 – 1.2348646738×10 -3 x 4 +1.0396454943×10 -6 x 3 – 4.4218106746×10 -2 x 2 + 1.0785366750×10 -1 x –1.4422222154×10 -2for 100 < x ≤ 1000K j = –9.3749999678×10 -18 x 6 + 3.1193910157×10 -4 x 5 – 4.0436698591×10 -11 x 4+ 2.6279902314×10 -8 x 3 – 9.8112164735×10 -6 x 2 + 2.8475810022×10 -3 x +4.1166665811×10 -2for 1000 < x ≤ 5000K j = –8.3333333334×10 -16 x 4 + 1.4166666667×10 -11 x 3 – 9.6666666667×10 -8 x 2+ 3.3333333333×10 -4 x + 4.9000000000×10 -1(iv) Effectiveness thickness h = 800 mmefor 3 ≤ x ≤ 60K j = 9.5889348301×10 -12 x 5 – 1.5604725262×10 -8 x 4 + 1.8715280898×10 -6 x 3 –7.5635030550×10 -5 x 2 + 1.8805930655×10 -3 x + 1.4981311831×10 -2for 60 < x ≤ 100K j = –5.4210108624×10 -20 x 4 + 1.3010426070×10 -17 x 3 – 5.0000000012×10 -6 x 2+ 1.6500000000×10 -3 x – 1.6000000000×10 -2for 100 < x ≤ 1000K j = –3.9583333158×10 -18 x 6 + 1.4818910202×10 -14 x 5 – 2.1967147366×10 -11 x 4+ 1.6383442558×10 -8 x 3 – 6.5899851301×10 -6 x 2 + 1.8249511657×10 -3 x –3.1900000544×10 -2Simulation ofKmvaluesValues of Figure 3.2 of the Code for Ordinary Portland Cement are read, Excel chartis plotted and polynomial equations are simulated as :J-3

Appendix JSimulation of K m Values for Portland CementOrdinary Portland CementRapid Hardening Portland Cement21.81.61.41.2Km10.80.60.40.201 10 100 1000Age of Concrete at Time of Loading (Days)for 1 ≤ x ≤ 7K m = 8.3333333333×10 -3 x 2 – 1.3333333333×10 -1 x + 1.925for 7 < x ≤ 28K m = 7.3129251701×10 -4 x 2 – 4.4642857143×10 -2 x + 1.6766666667for 28 < x ≤ 90K m = 3.8967199783×10 -5 x 2 – 8.6303876389×10 -3 x + 1.2111005693for 90 < x ≤ 360K m = 2.3662551440×10 -6 x 2 – 1.9722222222×10 -3 x + 9.0833333333×10 -1J-4