香港混凝土结构设计规范-5(最新版)

406ULTIMATELIMITSTATES6.1MEMBERSINFLEXURE6.1.1GeneralThefollowingclausesdealwiththedesignofmemberspredominantlyinflexurei.e.beamsandslabs.Thedesignofslabelementshasbeenseparatedintothevariousslabtypesnamelysolidslabs(clause6.1.3),ribbedslabs(clause6.1.4)andflatslabs(clause6.1.5).Thegeneralrequirementsfordesignformomentandsheararegiveninclause6.1.2forbeams.Whereadditionalrequirementsarerequiredforeachtypeofslab,thesearegivenintherelevantclauses.6.1.2Beams6.1.2.1General(a)DesignlimitationsThissub-clausedealswiththedesignofbeamsofnormalproportions.Deepbeams(seeclause5.2.1.1(a))arenotconsidered.Forthedesignofdeepbeams,referenceshouldbemadetospecialistliterature.(b)EffectivespanofbeamsTheeffectivespanofabeamshouldbetakenasstatedinclause5.2.1.2(b).(c)EffectivewidthofflangedbeamTheseshouldbeasspecifiedinclause5.2.1.2(a).(d)SlendernesslimitsforbeamsforlateralstabilityThecleardistancebetweenrestraintsshouldnotexceed:•forsimply-supportedorcontinuousbeams:60bcor250bc2/difless;or•forcantileverswithlateralrestraintonlyatsupport:25bcor100bc2/difless.6.1.2.2ContinuousbeamsContinuousbeamsmaybeanalysedinaccordancewithsection5ordesignedanddetailedtoresistthemomentsandshearforcesgivenbyclause6.1.2.3,asappropriate.6.1.2.3Uniformly-loadedcontinuousbeamswithapproximatelyequalspans:momentsandshearsTable6.1maybeusedtocalculatethedesignultimatebendingmomentsandshearforces,subjecttothefollowingprovisions:•characteristicimposedloadQkmaynotexceedcharacteristicdeadloadGk;•loadsshouldbesubstantiallyuniformlydistributedoverthreeormorespans;and•variationsinspanlengthshouldnotexceed15%oflongest.AtoutersupportNearmiddleofendspanAtfirstinteriorsupportAtmiddleofinteriorspansAtinteriorsupportsMoment00.09Fl-0.11Fl0.07Fl-0.08FlShear0.45F-0.6F-0.55FNote:1.Noredistributionofthemomentscalculatedfromthistableshouldbemade.Table6.1-Designultimatebendingmomentsandshearforces6.1.2.4Designresistancemomentofbeams(a)AnalysisofsectionsIntheanalysisofacross-sectiontodetermineitsultimatemomentofresistancethefollowingassumptionsshouldbemade:•thestraindistributionintheconcreteincompressionandthestrainsinthereinforcement,whetherintensionorcompression,arederivedfromtheassumptionthatplanesectionsremainplane;41•thestressesintheconcreteincompressionmaybederivedfromthestress-straincurveinfigure3.8withγm=1.5.Alternatively,thesimplifiedstressblockillustratedinfigure6.1maybeused;•thetensilestrengthoftheconcreteisignored;•thestressesinthereinforcementarederivedfromthestress-straincurveinfigure3.9withγm=1.15;and•whereasectionisdesignedtoresistonlyflexure,theleverarmshouldnotbeassumedtobegreaterthan0.95timestheeffectivedepth.Intheanalysisofacross-sectionofabeamthathastoresistasmallaxialthrust,theeffectofthedesignultimateaxialforcemaybeignoredifitdoesnotexceed0.1fcutimesthecross-sectionalarea.Figure6.1-Simplifiedstressblockforconcreteatultimatelimitstate(b)LimitationondepthofneutralaxisInordertoensurelargestrainsaredevelopedinthetensilereinforcement,thedepthoftheneutralaxisfromthecompressionfaceshouldbelimited.Whereredistributionofmomentsisnotcarriedoutordoesnotexceed10%,thedepthtotheneutralaxis,x,shouldbelimitedasfollows:x≤0.5d,forfcu≤40N/mm2;6.1x≤0.4d,for40fcu≤70N/mm2;or6.2x≤0.33d,for70fcu≤100N/mm2andnomomentredistribution.6.3Whereredistributionofmomentsexceeds10%,thedepthtotheneutralaxis,x,shouldbelimitedasfollows:x≤(βb-0.4)d,forfcu≤40N/mm2;6.4x≤(βb-0.5)d,for40fcu≤70N/mm2;or6.5where:βb=6.6(c)DesignformulaeforrectangularbeamsThefollowingequations,whicharebasedonthesimplifiedstressblockoffigure6.1,arealsoapplicabletoflangedbeamswheretheneutralaxislieswithintheflange.momentatthesectionafterredistributionmomentatthesectionbeforeredistribution42K=M/bd2fcu6.7Whereredistributionofmomentsisnotcarriedoutordoesnotexceed10%:K’=0.156forfcu≤40N/mm2,x≤0.5d;or6.80.132for40fcu≤70N/mm2,x≤0.4d;or0.113for70fcu≤100N/mm2,x≤0.33dandnomomentredistribution.Whereredistributionexceeds10%,K’=0.402(βb-0.4)-0.18(βb-0.4)2,forfcu≤40N/mm2;or6.90.402(βb-0.5)-0.18(βb-0.4)2,for40fcu≤70N/mm2.IfK≤K’,compressionreinforcementisnotrequiredand:−+=9.025.05.0Kdz6.10butnotgreaterthan0.95d.45.0/)(zdx−=6.11zfMAys87.0/=6.12lfKK’,compressionreinforcementisrequiredand:−+=9.0'25.05.0Kdz6.1345.0/)(zdx−=6.14)'(87.0)'('2ddfdbfKKA−−=yccus6.15'Azf.dbf'KAsy2ccus870+=6.16Ifd'/xexceeds0.43(forfy=460N/mm2),thecompressionstresswillbelessthan0.87fyandshouldbeobtainedfromfigure3.9.(d)DesignformulaeforflangedbeamswheretheneutralaxisfallsbelowtheflangeProvidedthatthedesignultimatemomentislessthanβffcubd2andthatnotmorethan10%ofredistributionhasbeencarriedout,therequiredareaoftensionsteelmaybecalculatedusingthefollowingequation:)5.0(87.0)9.0(1.0fyfwcushdfhxdbfMA−−+=6.17Equation6.17isonlyapplicablewhen:hf0.45dforfcu≤40N/mm2,x≤0.5d;orhf0.36dfor40fcu≤70N/mm2,x≤0.4d;orhf0.30dfor70fcu≤100N/mm2,x≤0.33d43Ifthedesignultimatemomentexceedsβffcubd2ormorethan10%redistributionhasbeencarriedout,thesectionmaybedesignedbydirectapplicationoftheassumptionsgiveninclause6.1.2.4(a).βfinthisexpressionisafactorgiveninequation6.18.