混凝土结构理论 英文版重点

－PAGE1－《混凝土结构理论》课程重点内容总结-版本2015.12.30华南理工大学混凝土结构课程组版本2015年12月30日星期三(初稿，此处指明的是重点，需根据新规范逐步调整，最终版会根据大家的整理重新上传）(如此文件与课本有冲突，以课本说法为准）第一章绪论建筑结构名词:结构，构件，构造分类:按材料混凝土、砌体、木、钢、组合按受力砖混、单层厂房、框架、框剪、剪力墙、筒体、巨型、悬挂、大跨度。钢筋混凝土结构优点:耐久性好、耐火性好、可模性好、整体性好、就地取材、节约钢材缺点:自重大、工期长、抗裂性差混凝土结构的发展参见课本[chapter1,2罗晗吴志浩]Chapter1Introduction1.Thegeneralconceptofconcretestructures(1)Classification.①Plainconcretestructure:Withoutsteelbarsorreinforcement.②Reinforcementconcretestructure(R.C.S)—It’sakindofstructurethatreinforcementconcreteismainload-bearingskeleton.(ReinforcedConcreteisakindofmaterialsofconstructionthatconcreteandsteelarecombinedincertainmethodforconcertedaction.)③Prestressedconcretestructure—It’sakindofstructurethatreinforcementconcreteequipswithpre-stressbyallocatingpre-stressedsteelbarsmethodandtensilemethodetc.④Steelreinforcedconcretestructure(Encasedconcretestructure)⑤Concretefilledtubestructure⑥Compositestructure(Hybridstructure)(2)Theroleandrequirementsofreinforcement.①Role:(a)ImprovingtheloadcapacityofRCmembers.(Beamcanworkhealthilywithcracks.)(b)ImprovingthedeflectioncapacityofRCmembers.(Beambecomeductile,haveobviousfailureomenfromcrackingtocollapse.)②Requirement:(a)Reinforcementandconcretecanworktogether—Thereissufficientlystrongbondbetweenreinforcementandconcrete.(Reason:i.Thethermalexpansioncoefficientofthebarsandtheconcreteisnearlythesame.ii.Theconcretecovercanpreventthesteelbarsinsidefrombucklingundercompressionanrusting.)(b)Theallocationandnumberofreinforcementshouldbedeterminedbycalculationanddetailingrequirements.(3)Mainadvantagesanddisadvantagesofconcretestructure①Advantages(a)Locallyavailableresources.(b)Durabilityandfireresistance.(c)Thevariousformscanbedeposited.(d)highstiffness.②Disadvantages(a)Largedeadmass:density25kN/m3(b)Easilycracked:inherentweakness.(c)Longcuringperiod.2.Thedevelopmentandapplicationofconcretestructure(1)Development.①Firststage:1850tothelatenineties(a)ApplicationofR.Ccomponent.(b)Capacityiscalculation:i.elasticanalysismethodii.allowablestressdesignprincipal.②Secondstage:theearlytwentiestothesecondworldwar(a)Spacestructureandpre-stressedconcrete,thinshellsstructure.(b)Capacitycalculation:i.plasticanalysismethod,ii.failurestatedesignprinciple③Thirdstage:thesecondworldwartonow(a)industrializationconstructionmethod.(b)limitstatedesignprinciplebasedonprobabilitytheory.(c)Materials:highstrengthconcreteandreinforcement.(d)Structure:high-riseandoff-shorestructure.(2)Theapplicationandprospectsincivilengineeringapplication:①Industrialandcivilbuildings.②Transportfacilities.③Waterconservancyandhydro-powerconstructions.④Foundationworks.etc.Prospects:①Lunarsurfaceengineering.②Largespanbuilding.③High-risebuilding.etc.第二章材料的力学性能混凝土 1．基本概念:抗压好，抗拉差； 2．组成:水泥凝胶+砂+石；3．强度:3.1单轴强度a．立方体抗压强度150mm×150mm×150mm的试件，在20±3℃的温度下，和90％以上的湿度中，按标准试验方法测得。C15，20，25，30，35，40，45，50，55，60，65，70，75，80，其中C50及C50以上是高强混凝土。一般结构中，要求C20以上，预应力中需要C30或者C40注意点:套箍效应b．轴心抗压强度(较接近于实际构件）150mm×150mm×300mm 3.2双轴受力:要点(双向应力状态下混凝土的破坏包络图）结论:1）双向受拉强度与单向受拉接近；2）一向受拉，一向受压时，受压或受拉强度均降低；3）双向受压时抗压强度比单向受压强度高(在双向应力比值为0.5～0.7时可以达到最大）【要点】:需要明白这个双向受力图怎么画出来的 3.3三向受力:近似公式: 4．变形 4.1受力变形4.1.1短期一次加载(单轴）a）应力应变曲线:是一种非线性材料，弹塑性材料，有上升和下降段，应力图由抛物线段和直线段组成。0.002，0.0033；b）泊松比:μ=0.2；c）模量:混凝土的模量是一个变值，分原点模量(弹性模量）、切线，割线模量(割线模量或弹塑性模量）。4.1.2长期荷载(徐变）:概念见书，分线性徐变和非线性徐变。需要了解影响徐变的因素:内在因素；环境因素；应力因素。 4.1.3重复荷载:疲劳破坏:特征是裂缝小而变形大疲劳强度与重复作用时应力变化的幅度关系很大，变化幅度越小，疲劳强度越高 4.2体积变形(温度变形，收缩变形）非荷载变形 收缩:混凝土在空气中凝结硬化时，水分析出引起了混凝土的收缩。 影响因素:水泥，骨料，养护条件，制作方法，使用环境等。影响:引起混凝土开裂；使预应力构件产生预应力损失钢筋分类劲性钢筋(角钢，槽钢，工字形钢）柔性钢筋(普通钢筋，钢丝束）钢丝束+普通钢筋【包括光面钢筋HPB235+变形钢筋HRB335(II级），HRB400(III级），RRB400(余热处理III级）】要求记住普通钢筋的强度品种:【2.1碳素钢:低、中、高碳钢。随碳增加，强度大塑性小】；【低合金钢锰硅钒钛】强度屈服强度，极限强度变形(画变形图）0.0015，0.01，塑性指标:伸长率和冷弯性能加工性能(冷拉(抗拉强度提高、抗压强度降低），冷拔(同时提高抗拉强度和抗压强度））钢筋与混凝土的粘结粘结的重要意义:粘结是钢筋和混凝土这两种性质不同的材料能形成整体，共同工作的基础。粘结应力的产生:粘结应力与梁的剪力变化规律一致粘结的分类:a锚固粘结和延伸粘结b裂缝附近的局部粘结粘结的机理:【光面钢筋:胶接力，摩阻力】；【变形钢筋(主要是机械咬合力）:胶接力(化学，较小），摩阻力(混凝土收缩引起），咬合力(表面的凹凸）】影响粘结的因素:见书各种措施:锚固长度:；;还有第四章的las，及搭接长度[chapter1,2罗晗吴志浩]Chapter2thephysicalandmechanicalpropertiesofconcretestructurematerialsI.ReinforcementTheclassificationstructuralsteel(Angles,channels,I-sectionsteel)flexiblereinforcement(ordinarysteelbar,roundwires,strandedcable)VarietiesCarbonsteelLowcarbonsteel,medium-carbonsteel,highcarbonsteel.Thestrengthincreaseswhiletheplasticdecreasewiththeincreasingofcarbon.lowalloysteel:manganese,silicon,vanadium,titaniumStrength:fy—yieldstrength(designstrength)fsu—ultimatestrengthYieldratio:fy/fsu,fy/fsu≤0.8Forthehardsteelthathasnoobviousyieldpoint,theyieldstrengthisspecified.DeformationInChineseCode,thestress-straincurveissimplifiedtobilinearcurves.[0.0015，0.01]Plasticityindex:elongationandthecoldbendingperformanceColdworkingColdstretched(冷拉）--Timeeffectofcoldstretching(冷作硬化效应)Cold-stretchedsteelcannotbewelded.Tensilestrengthofcold-stretchedsteelisincreased.Thecompressivestrengthcannotbeincreased.Theelongationofthatisobviouslydecreased.Colddrawn(冷拔）:Bothtensilestrengthandcompressivestrengthofcold-drawnsteelisincreased.Thereinforcementrequirementsforconcretestructures①Strength②Ductility③Mechanism(weldability可焊性,processing施工性)④BondwithconcreteⅡ.ConcreteBasicconcept:Goodcompressivestrength,whilerelativeweaktensionstrengthComposition:cementgel+sand+gravelStrengthuniaxialstrengthCubiccompressivestrength:Cubicconcretespecimens(150mm×150mm×150mm)curedat20+-3°Candrelative90％humidityfor28days,usingstandardtestmethods.Thereare14concretegrades.(C15，20，25，30，35，40，45，50，55，60，65，70，75，80)，C50andabovearecalledashigh-strengthconcrete.C20oraboveisappliedincommonstructure,whileC30orC40isappliedinprestressedconcretestructure.(payattentiontoconfiningeffects.)Axialcompressivestrength(moresimilartotheactualstructuremember）150mm×150mm×300mmBiaxialstrength:StrengthofconcreteinbiaxialstressstateUnderbi-axialtension,thetensilestrengthisessentiallythesameasuniaxialstrength.Whenthespecimenisundertensioninonedirectionandcompressionintheotherdirection,boththetensileandcompressivestrengtharelowerthanthatinuniaxialloadsUnderbi-axialcompression,asshowninthefirstquadrant,themaximumcompressivestrengthoccurswhentheratiois2or0.5andthestrengthisabout27%higherthantheuniaxialcompressivestrength.[Note]:shouldneedtounderstandhowthisdiagramisobtainedTriaxialstressstate:DeformationLoadingdeformationShort-termmonotonicloading(uniaxial)Stress-straincurve:Concreteisanonlinearmaterial,plasticmaterials.Thecurvehaveascendinganddescendingbranches.InChineseCode,thiscurveissimplifiedtoapowerexponentsegmentandlinesegment.[KeyNumber]0.002,0.0033Poissonratio:μ=0.2Modulus::Therearethreemethodstoexpressthemodulusofconcrete.1)Initialmodulusofelasticity2)Secantmodulus3)Tangentmodulus.Long-termload(creep)Whenastressorloadisappliedtoaconcretespecimenandkeptconstant,thespecimenshowsanimmediatestrain(elasticstrain)followedbyafurtherdeformationwhichmaybecomeseveraltimestheoriginalimmediatestrainandisreferredtoasthecreepstrain.Orsimply,creepisincreasedinstrainwithtimeduetoasustainedload.Twokinds:Linearandnonlinearcreep.Causes(a)Plasticflowingofcementgel;(b)Internalmicro-fissuresdevelop.Thefactorsaffectingcreepconcreteformation:theamountofcement,watercementratio(水灰比)More→↑themixproportionbetter→↓concreteage(加载龄期)longer→↓curingcondition:temperature↑,humidity↓→↑stresscondition:larger→↑Repeatedloading:Fatiguefailure:Cracksaresmallwhiledeformationislarge.Themagnitudeoftherepeatedloadhaslargeeffectonthefatiguestrength.Thesmallertheamplitudeis,thehigherthefatiguestrengthis.Non-loadingdeformation(temperaturedeformation,shrinkagedeformation)Shrinkage:Theshrinkageofconcreteisthoughttobeduemainlytothelossofabsorbedwaterinthegel.Influencefactors:cement,aggregate,maintenancecondition,productionmethods,theenvironment,etc.a).Barsaresubjectedtocompressivestress,whileconcretegainstensilestress.b).tocauselossofprestressing.c).tocausesecondaryinternalforcesinthestaticallyindeterminatestructuresⅢ.Bondofconcreteandreinforcement(1)Thedefinitionandimportanceofbinder.①Definitionofbond:Thebondstressisshearstressthatthesurfaceofreinforcementandconcreteproduces,inordertoresistdifferentofdeformation.②Theimportantmeaningofthebond:Forreinforcedconcretetobehaveasintended,itisessentialthatbondforcesbedevelopedontheinterfacebetweenconcreteandsteel.Itisthebondforcewhichensuresthesetwokindsofdifferentmaterialstoformawholeandworktogether.(2)Theclassificationofbond ①anchoragebond ②localbondnearthecracks(3)Formationofbondaction①Adhesionaction—Chemicaladhesionactionbetweencementgelandbars’surface.②Frictionaleffect—Derivedfromgrippingeffect,whichisresultedfromthedryingshrinkageofthesurroundingconcrete.③Interlockingeffect:Resultingfromthebarsurfacedeformationorribsandconcrete.ForHPB300gradebars:①②;Fordeformedbars:①②③(5)Thefactorsaffectingbondstrengthτu5)Thediameterandsurfaceshapeofbars→τuofplainbarissmaller.Ifthediameterofbarislarger,therelativesuperficialarea(相对表面积)oftheribissmallerbecausetheparametersofribsarenotincreasedlinearlywithdiameter,sotheτuissmaller.Iftheepoxyresin(环氧树脂[eˈpɔksiˈrezɪn])isusedtopreventrustingofbars,τuissmallerbecausethesurfaceissmoother.(6)ThebasicdetailrequirementstoensurereliableadhesionIftheanchoragelengthislarger,thetotalbondstressislarger.→Inapull-outtest,iftherearenotstirrupsandlateralforces,whentheanchoragelengthislargeenoughtomakethebarreachingyield.Itiscalledthebasedanchoragelength——lab.→Mechanicalanchorageeffectoftheendsofbarsthroughdevelopmentlength,splicing(叠接),hooks(弯钩),andcrossbars.τuismuchlarger.(CH.P33.F.2-28)Thebasedanchoragelength——(inbook,page32)Theanchoragelength——(inbook,page3)Thelaplength——ll=ξlla(inbook,page107)Theanchoragelength——las(inbook,page102)第三章:受弯构件的正截面受弯承载力引言主要受弯构件:梁板；四种基本受力形式:压(拉）弯剪扭；受弯构件三种截面:单筋，双筋，T形配筋率:保护层厚度的定义和作用；受弯构件正截面三种破坏形式:少筋破坏、适筋破坏、超筋破坏；受弯构件正截面三个受力阶段:I(包括Ia）、II、III(包括IIIa）受力全过程受力阶段现象特点IM-f图呈直线弯矩小,应变小，钢筋:弹性阶段；混凝土:弹性阶段，受拉区略显塑性拉力受钢筋和混凝土共同承担Ia混凝土即将开裂，转折点1混凝土达到受拉应变ct，是严格不允许出现裂缝构件的设计依据，可作为受弯构件抗裂度的计算依据。IIM-f呈曲线，裂缝出现并开展钢筋:弹性阶段；混凝土:压区出现塑性，拉区逐渐退出工作拉力主要由钢筋承担，是使用阶段的裂缝开展与宽度计算的依据。IIa钢筋即将屈服，转折点2钢筋达到屈服应变σy；IIIM-f呈曲线，裂缝宽度发展较快钢筋:屈服阶段；混凝土:塑性、压区应力图形更加饱满中性轴一直上升IIIa混凝土压碎，破坏极限承载力的设计依据，混凝土达到极限压应变cu承载力计算原理3.1四个基本假定平截面假定；不考虑混凝土的抗拉强度混凝土受压的应力应变关系，为一个曲线段加一个直线段【需知道关键点，当混凝土强度小于C50的时候，这个曲线段是一个二次抛物线】纵向受拉钢筋的极限拉应变取为0.01，钢筋的应力应变关系，为两个直线段【需知道关键点】3.2适筋梁公式的推导过程:要求看书理解【难点】3.3力的平衡方程:见书注:其中的h0的取法:h0=截面高度－保护层厚度－0.5纵筋直径，一般梁可取h0=h-as=h-35(单排钢筋）；h0=h-50~60(双排钢筋）；板可取h0=h-as=h-20，一般考试会直接给出as。考试的时候，只要是受弯或者压弯计算，一般都要验算两个适用条件，配筋率和界限受压高度，没有做的会扣分。当为适筋梁时，有两种情况，【难点】1.Mu已知，求As，x——(截面设计）2.As已知，求Mu，x——(截面复核）3.4问题的扩展几个相关的问题:a．防止超筋破坏:控制x<=ξbh0，在适筋梁范围内，随配筋率减小，构件的变形能力增强，延性增大。b.防止少筋破坏:规定最小配筋率:min＝min{45ft/fy％,0.2％}c.关于简化公式:α，ξ，γ；一般考试会给出后两者与α的关系式。3.5双筋梁的计算:【难点】1概念:一般用钢筋受压不是非常经济，只在如下情况用:1）受压区太大，在截面和强度都受到限制的情况下用；2）有异号弯矩的作用。注意点:双筋梁和双排钢筋梁的概念是不同的2计算公式:见书P62。注意点:公式中的受压钢筋达到屈服应变，这一点要求受压区不能太小，要x>=2as’，其受压区高度仍应该小于界限高度(适筋范围）3计算方法:1．直接求解方程法:1）假设As’已经知道；2）假设x=；3）根据As+As’为最小作为优化的约束条件2.叠加法(用于已知受压钢筋面积的情况，需在理解的基础上掌握）注意点:(1）若求得的x<2as’，则表明受压钢筋不能屈服，需重新计算，(2）若用双筋梁计算出来的横截面积还比按单筋梁截面计算的横截面积大时，可以按照单筋梁的计算结果作为设计面积。适用条件:(1）x(2）x2as’一般来说:钢筋最小配筋率是自然满足的，所以不需要验算。3.6T型截面和异型截面的抗弯设计【难点】概念:受拉区混凝土有很大部分的混凝土不起作用，将这部分的混凝土去掉之后可以节省材料，于是产生T型梁，工字型梁，箱型梁等等。计算原理:五个基本假定仍然成立，力平衡方程仍然成立。注意点:分清楚受压区和受拉区。判别受压区的范围(在腹板区还是在翼缘区），判别方法:假设法或者界限点法以T型梁为例作分析:T型梁的分类:第一类，中和轴在翼缘内；第二类:中和轴在腹板内计算公式:第一类T型梁，计算公式同普通单筋截面梁，不过梁宽b变为bf。第二类T型梁，具体见书如何判别第一或第二类T型梁1）假定法；2）直接判断法适用条件:(1）x(2）拓展内容:假如是矩形梁挖孔怎么作？假如是梯形梁挖圆孔怎么办？【难点】【注意:梁中空心的开方孔或者圆孔的箱形梁书上没有讲，其原理和工字型梁一样】本章小结:附加，适筋梁的Mmax＝[chapter3王卓炜吴梓楠彭出]chapter3Flexuralmembers1.IntroductionTypicalmemberswithflexure:beams,slabsetc；Fourbasicformstobearloadings:compression(tension）,flexure,shear,torsion；Threetypicaltypesofmemberswithflexure:singlyreinforcedrectangularbeam，doublyreinforcedrectangularbeam，Tsectionbeamthesteelratio:thedefinitionofconcretecover:Concretecover,inreinforcedconcrete,istheleastdistancebetweenthesurfaceofembeddedreinforcementandtheoutersurfaceoftheconcrete；thefunctionsofconcretecover:(1)anti-rust(2)anti-fire(3)completebondbetweenconcreteandsteelbarsThreefailuremodesofnormalsectionofflexuralmember:scarce-reinforcement,under-reinforcement,over-reinforcement；Threestagesintestforflexuralbeam:I(includeIa）、II、III(includeIIIa）stageBehaviorcharacteristicIM-φislinearlowbendingmoment,smallstrain;reinforcementandconcretesharethebearingforce.reinforcement:elastic;concreteincompressivezone:elastic;concreteintensionzone:slightlyplasticityIaconcreteisnearlytobecracked,turningpoint1concretetensionalstrainreachesεct.εct≈200με,σs≈40MPaAbeamwithoutcracksaredesignedonthebasisofstageIa.IItherelationshipbetweenMandφisNon-linear,crackingisvisiblethestageaftertheconcretehasbeencrackedandthesteelbarstakeupalmostallthetensionforcebuthasnotyetyielded.Non-linearitybeginstoshowintherelationshipbetweenloadanddeflection.Stressdistributionintheconcretecompressivezonealsoshowsnonlinearity.Butitisfoundfromexperimentsthatplane-sectionsremainasplanes.Abeaminserviceisusuallyinthisstage.IIaSteelstrainreachestheyieldturningpoint2steelstrainreachestheyield,ForHPB335steel,εy≈1500με,σs=300MPaIIItherelationshipbetweenMandφiscurvilinear,thewidthofcracksincreaserapidly.thestiffnessofthebeamismuchfurtherweakened.Considerableshiftintheneutralaxispositionoccurs.Load-deflectionrelationshipbecomesmorenon-linearIIIaConcretecrushedandfailedextensivedevelopmentsofcracksandcrushingoftheconcreteinthecompressionzoneleadtotheultimatecollapseofthebeam.ConcretecompressivestrainreachesεcuIII.Principleofthebearingcapacitycalculation3.1Fourbasicassumptions:1)Planesectionremainsplaneafterbending.2)Thetensilestrengthinconcretecanbeneglected. 3)Strain-stresscurveofconcreteconsistsofastraightlineandacurvedline.(WhenconcreteislessthanC50,strain-stresscurveisaquadraticparabola.)4)Theultimatetensilestrainofreinforcementintensionmustbe0.01.Strain-stresscurveofsteelconsistsoftwostraightlines.3.2Theprocessofbearingcapacitycalculationformulaofunder-reinforced-beama.SelectingIIIaforstudy,inthiskeystatus,steelbarshaveyielded(maybejustyieldedoryieldedforatime),concretejustcrushed.b.AccordingtothePlanesectionassumption,thestressofsteelisdefinedasfy,thelimitstrainofconcreteisεcu,theedgestressofconcreteisfc.Supposethepositionoftheneutralaxis,thestrainfromtheneutralaxistocompressionzoneislinear,thenthecurveofstresscanbedrawn.c.Atbalancestatus,totaltensileforceinsteelbars=totalcompressivestressinconcrete;steelandconcreteformamoment=externalmoment.Ifthevalueofexternalmomentisgiven,theneutralaxiscanbesolvedaccordingtothesetwoequations.d.Simplifythecalculationprocess:usinganequivalentrectanglestressblocktoreplacetheactualstressdistribution.WhentheconcretegradeislessthanC50,heightoftheequivalentrectangleis0.8timesastheoriginandwidthoftheequivalentrectangleisequal.3.3Balanceequationα1fcbx-σsAs=0Mu=α1fcbx(h0-0.5x)=σsAs(h0-0.5x)Methodtogeth0:h0=h–c–0.5dc:thicknessofconcretecoverd:diameteroflongitudinalreinforcementForbeam:h0=h-as=h-35(single-rowsteel)h0=h-50~60(double-rowsteel)Forslab:h0=h-as=h-20Under-reinforcedbeams:1.Muisgiven,calculateAs,x(sectiondesign).2.Asisgiven,calculateMu,x(sectioncheck).3.4ExpansionSomerelevantquestionsa.Topreventover-reinforcedfailuremode,weneedtocontrolx<=ξbh0,forunder-reinforcedbeams,ifthereinforcementratiobecomessmaller,thedeformationabilityofthebeamwillincrease,andtheductileabilitywillincrease.b.Topreventscarce-reinforcedfailuremode,thesmallestreinforcementratioρmin＝min{45ft/fy%,0.2%}c.Aboutthesimplifiedequation:therelationamongα，ξ，γwillbegivenintest.3.5Calculationofdoublereinforcedbeam:1Concept:It’snoteconomictousereinforcementforcompression;reinforcementisusedforcompressiononlyinthefollowingcircumstances:1)Thecompressionzoneistoolarge,andthesizeofthecrosssectionandstrengthofconcretearelimited;2)Differentsignsofmomentactonthestructure.2Formulas:Thespecificcontentsareatpage62ofourChinesetextbook(page49ofEnglishtextbooks).Caution:inordertomakethereinforcementincompressionzonesyield,thecompressionzonecan’tbetoosmall,sox>=2as’,thecompressionzoneisstillsmallerthantheboundarycompressionzoneheight.3Calculationmethods:Solvetheequationdirectly.Superpositionmethod(itcanbeusedonlywhentheareaofthecompressionzoneisknown)Caution:(1)Aftercalculation,ifx<2as’,thenthereinforcementincompressionzonewon’tyield,weneedtorecalculate.Ifthecrosssectionareacalculatedbydoublereinforcedbeamislargerthantheareacalculatedbysinglereinforcedbeams,wecanusetheresultcalculatedbysinglereinforcedbeams.Applicability:(1）xξbh0(2）x2as’Normally,wedon’tneedtochecktheleastreinforcementratiobecauseitwillsatisfyautomatically.3.6FlexuraldesignofT-sectionandnoncircularsectionConcept:Theconcreteintensioncontributeslittletotheflexurebearingcapacity.Removepartofconcreteintensionwon’treducethebearingcapacityofRCbeam.Furthermore,itcansavematerialandreduceself-weight.[T-section,I-section,box-sectionbeam]Calculationassumption:Fourbasicassumptionsandtheforcebalanceequationisstilleffective.Note:Determinetheneutralaxisisinflangeorinweb.Determinedmethod:AssumptionmethodCriticalmethodT-sectionBeam:TypesofT-sectionbeam:NeutralaxisinflangeNeutralaxisinwebFormula:Neutralaxisinflange:α1fcbf’x-fyAs=0Mu=α1fcbf’x(h0-0.5x)=fyAs(h0-0.5x)Neutralaxisinweb:α1fcbx-α1fc(bf’–b)hf’-fyAs=0Mu=α1fcbx(h0-0.5x)+α1fc(bf’–b)hf’(h0-0.5hf’)DeterminetypeofT-sectionbeam:HypothesismethodDirectlymethod

Note:xξbh0Expand:HowtoworkouttheproblemwhenrectangularbeamorT-sectionbeamwithholes?Summary:Mmax＝(Under-reinforcedbeam)第四章:斜截面承载力计算概述(实例和较简单理论） 1．内力:要求掌握:应力迹线图;剪应力的分布:对矩形截面，中性轴位置处最大，上下截面处为0。2．裂缝的产生:裂缝特征及其产生的原因和位置:1）腹剪斜裂缝;2）弯剪斜裂缝3．钢筋配置:实际中首选箍筋，不用弯筋，箍筋最好的角度是什么？为什么不这么做？原因:1）由于裂缝方向的不定性；2）弯起钢筋传力过于集中；3）弯筋数量有限，无法形成骨架。4．斜截面承载力分类:斜截面承载力理论主要是斜截面抗剪，也有斜截面抗弯问题存在。受弯构件斜截面的受力特点和破坏形态。1.受力特点:受弯矩，剪力的共同作用，采用剪跨比考虑这种相互关系。引入剪跨比的概念:广义剪跨比:，狭义剪跨比剪跨比不同，导致破坏形态也不相同。2.破坏形态:具体见书，包括斜压破坏；斜拉破坏；剪压破坏。相比之下，剪压破坏是比较好的，箍筋可以得到充分发挥，必须利用它。但应注意，三种情况都属于脆性破坏，只是脆性程度不同。要避免斜压破坏和斜拉破坏。后面会提到，用限制截面尺寸来避免斜压破坏，用最小配箍率和最小箍筋直径和间距来限制斜拉破坏。引出一句话:强柱弱梁，强剪弱弯。形式特点破坏过程箍筋备注斜压破坏无腹筋梁:剪跨比较小(）时，有腹筋梁:剪跨比很小或者剪跨比较大，但腹筋配置过多同时腹板较薄时，先在梁腹出现数条由集中力指向支座的斜裂缝，其将砼分割为数个倾斜受压柱体，最后小柱体压坏，箍筋未屈服。有点类似超筋破坏。脆性破坏。斜拉破坏无腹筋梁在剪跨比较大(），有腹筋梁在(）同时腹筋很少时斜裂缝一旦出现，很快形成一条主斜裂缝而破坏。箍筋立即屈服，有点类似少筋破坏危险性非常大。脆性破坏。剪压破坏无腹筋梁在剪跨比适当(），有腹筋梁:剪跨比不论，但是截面尺寸适当，且腹筋配置适当时，荷载增加，首先在受拉边缘出现一些竖向裂缝，在沿竖向延伸一小段之后就斜向延伸成斜裂缝，数条裂缝中，最终有一条发展为破坏裂缝，称为临界斜裂缝，随荷载增大，斜裂缝不断向上延伸，【而后与临界斜裂缝相交的箍筋达到屈服】，剪压区混凝土在剪力和压力的共同作用下达到极限强度破坏，箍筋先于混凝土破坏达到屈服变形能力较差，仍为脆性破坏影响斜截面受剪承载力的主要因素。(叙述各个因素）混凝土强度；2）配箍率和箍筋强度3）剪跨比:斜压破坏的强度是最高，(由于抗压能力比抗拉强），4）纵筋配筋率:纵筋起的作用是销栓作用，但是影响比较小。5）加载方式:直接加载和间接加载(附图），间接加载导致受剪承载力的降低。6）截面形式:翼缘的存在对于有腹筋梁适当提高，但不是很明显，一般可以忽略不计。梁斜截面承载力的计算公式。受剪机理:具体见书。1）带拉杆的梳形拱模型:适用于无腹筋梁；2）拱形桁架模型，适合于有腹筋梁；3）桁架模型:适合于有腹筋梁计算方法对于无腹筋梁:在均布荷载作用下，在集中荷载作用下；注意点:1）Vc和Vs不是独立的，他们共同作用；2）未考虑纵筋的销栓作用3）需要考虑部分箍筋可能达不到屈服强度4）理解剪跨比λ的取值方式，剪跨比只有在集中力作用明显的情况下才考虑计算公式:有无弯起钢筋时:见书。对于公式的一些解释钢筋的作用是有限的，当截面过小而剪力过大的时候，会发生斜压破坏，在薄腹梁中更容易产生这个问题，因此限制更加严格，因此必须限制截面。对于厚腹()梁:；对于薄腹()梁:注意截面的腹板高度在不同截面形式中的范围。箍筋的最小配箍率集中力占主要作用的情况下，的取值要求限制在1.5～3之间。厚板的问题，，，当h0<800时取1。连续梁的问题，比较复杂。斜截面承载力的计算方法和步骤。(计算和复核）要点:截面的选取，计算步骤见书。需特别注意的，必须掌握书课后4.7及4.8这种弯剪综合的题目。纵向钢筋的弯起、截断和锚固。斜截面受弯问题:为什么a0.5h0，需要知道原理，什么是斜截面的受弯问题钢筋弯起的问题名词:抵抗弯矩图，充分利用点，不需要点。一些原则:1）抵抗弯矩图需包住整个外力弯矩图。2）弯筋和轴线的交点应该在该钢筋的不需要点之外，弯起点必须在充分利用点的0.5h0之外。钢筋截断的问题名词:锚固长度；支座锚固长度；抗震锚固长度；搭接长度；抗震搭接长度具体作法:L1L2≥1.2la≥20d≥1.2la+h0≥Max{20d，h0}截断点处于负弯矩区≥1.2la+1.7h0≥Max{20d，1.3h0}构造要求:见书[Chapter4刘子彦杨凯越颜博]Chapter4UltimateBearingCapacityofDiagonalSectionforFlexuralMembersI.Introduction(samplesandbasictheories)Internalforce:theprincipalstresstrajectoriesdiagraminthebeam;thedistributionoftheshearstressinthesection:forrectanglesection,themaximumstresslocatesattheneutralaxis,andzerostressatbothtopandbottomofthesection.Theoccurrenceofcracks:thecharacteristics,reasonsandlocationsofcracks:1)Web-shearcracks;2)Flexural-shearcracksClassificationofthereinforcement:webreinforcementincludesstirrups(transverseties)andbent-upbars,andskeletonofsteelincludeslongitudinalreinforcement,webreinforcementandsomeotherdetailingbars.Reinforcedconcretebeamswithoutwebreinforcementstillhavelongitudinalreinforcement.Distributionofthereinforcement:Thereasonsthatstirrupswouldbethefirstchoiceswhilebent-upbarsarenotusedingeneral,arethat1)theuncertaintyofthedirectionsofthecracks;2)theundueconcentrationofforcetransmittedbybent-upbars;3)thefinitenumberofthebent-upbarsfailingtoformskeletonofsteelinabeam.Theclassificationofthediagonalsectioncapacity:1)shearingresistanceinthediagonalsection;2)bendingresistanceinthediagonalsection.II.TheStressFeaturesandFailureModesofDiagonalBendingSection1.Stressfeature:Giventhecombinedactionsofthemomentandshearforce,weintroducetheconceptofshearspanratio:Generalizedshearspanratio:,calculatedshearspanratio:Differentshearspanratiowouldresultindiversefailuremodes.2.FailuremodePleaserefertothetextbook(ForEnglishP67,ChineseP79)orPPT.FailuremodesIncludesshearcompressionfailure,diagonalcompressionfailureanddiagonalsplittingfailure,andbestofwhichisshearcompressionfailureasitcanmakefulladvantageofthestirrups.Thoughthebrittlenessisdifferent,allthreeshearfailuresbelongtobrittlefailure,whichmeansdiagonalcompressionfailureanddiagonalsplittingfailureshouldbeavoided.Toassuretheshearcompressionfailure,thelimitationonsectionalsizeisintroducedtopreventdiagonalcompressionfailure,andminimumstirrupratio,diameterandspacingtoavoiddiagonalsplittingfailure.Aconceptispresented:strongcolumnandweakbeam,andstrongshearandweakbending.ModeCharacteristicsofreinforcedconcretebeamswithoutwebreinforcementCharacteristicsofreinforcedconcretebeamswithwebreinforcementFailureprocessStirrupNotesDiagonalcompressionfailureShearspanratiocanbeeitherquitesmallorbig.Webreinforcementisundueoritisthin-webbeam(TorIsectionbeam)Severaldiagonalcracksfromactingpointoftheconcentratedforcetothesupport,separatetheconcreteintodiagonalcompressioncylinders.Finally,thetinycylindersarecrushedunderthepressureThestirrupsdonotyield,similartoover-reinforcedbeamBrittlefailure.DiagonalsplittingfailureandsimultaneouslytheamountofwebreinforcementislowOncethediagonalcrackoccurs,anditwouldextendrapidlyupwardandsplitthememberintotwo.Thestirrupsyieldimmediatelyafterthecrack,similartoscare-reinforcedbeamVerydangerous.BrittlefailureShearcompressionfailureThewebreinforcementisproperlyset，orandthewebreinforcementisnotlessFirstly,verticalcracksoccurintensilezone,andformacriticaldiagonalcracklater.Withconcreteshearcompressivezonegettingsmaller,theconcreteiscrushedunderincreasingnormalstressandshearstress.ThestirrupsyieldbeforetheconcreteiscrushedTheabilityoftransformationisbad.Brittlefailure3.FactorsAffectingtheStrengthofDiagonalSectionStrengthofConcrete;StirrupsRatioandtheStrengthofStirrupShearSpanRatioλ:theshearresistanceofdiagonalcompressionfailureisthehighestduetothestrongercompressionresistanceofconcrete;LongitudinalSteelRatio:Affecttheextendingofdiagonalcracksinthecompressionzoneowingtothedowelactionforce,butindirectly.LoadingWay:Indirectloading;Directloading,whichresultsinthedeclineoftheshearcapacitySectionForm:Flangeslightlyincreasethestrength,butcanbeignored.4.StrengthCalculationMethodsofMembersUnderShearMechanismCombarchmodewithtierod:usedforbeamswithoutwebreinforcement.Arch-trussmode:usedforbeamswithwebreinforcement.Trussmode:usedforbeamswithwebreinforcement.CalculationFormembersunderuniformloads,forbeamsunderbothconcentratedloadsanddistributedloads,whentheshearforceproducedbyconcentratedloadsatthesupportisequaltoormorethan75%ofthetotalforce.:Vc,Vsisnotindependent;rather,theyinfluenceeachother.Thedowelactionsuppliedbylongitudinalbarsisnottakenintoconsideration.It’snotingthatpartofthestirrupsmaynotreachtheiryieldstrength.Asλ<1.5,letλ=1.5;asλ>3,letλ=3.Andλisonlytakenwhenconcentratedloadsdominate.,Thefunctionofreinforcementisnotunlimited.Diagonalcompressionfailurewillhappenwhenthesectionistoosmallwhiletheshearforcetobetoolarge,especiallyinsmall-depth-of-webbeams.Andthusaminimumsizeonthedimensionsofsection(upperboundofshearresistance)isset.Thishelptorestrictthewidthofthediagonalcracksandtopreventfromdiagonalcompressionfailure.For,;for,.Noticewhath0standsforindifferentsituation.Lowerboundofshearstrengthiscontr