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邯郸市高新宾馆设计毕业设计计算书

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'邯郸市高新宾馆设计毕业设计计算书137 摘要本设计为河北省邯郸市高新宾馆设计。全楼主体共十层,建筑面积为10335.46㎡。设计分为两部分:第一部分是建筑部分,包括对建筑的周围的环境设计,建筑的平面设计,建筑立面的设计,建筑剖面设计,建筑构造做法及材料的选用。本建筑设计采用了内廊式的平面组合,使宾馆内各房间相对独立、安静,不被穿越但又有一定的联系,建筑布局因地制宜,灵活布置,功能分区明确、流线清晰、便于管理,满足使用要求,特别是消防和疏散要求。第二部分为结构设计,本设计的结构采用横向框架承重体系,选取一榀框架进行水平力和竖向力计算,并进行内力组合,以确定最不利内力,并进行截面设计。包括框架梁、板、柱、楼梯、雨篷及基础的截面设计等。墙体采用粉煤灰蒸压加气混凝土砌块砌筑,有利于减轻建筑物自重;其中对框架梁、板、柱的设计都采用了比较常用的弹性理论方案。本设计为7度抗震设防烈度设计,因此设计过程中还进行了地震作用下弹性位移验算和罕遇地震作用下的弹塑性变形验算。设计后达到了强度、刚度和稳定性方面的要求。关键词:建筑框架混凝土抗震设计137 AbstractThisdesignforHebeiProvinceHandanhighnewguesthousedesign.Entirebuildingmainbodyaltogetherten,floorspacefor10335.46㎡.Thedesigndividesintotwoparts:Thefirstpartconstructsthepart,includingtotheconstructionperipheryenvironmentdesign,theconstructionplanedesign,theconstructionsetsupthesurfacethedesign,constructionsectionplanedesign,constructionstructureprocedureandmaterialselection.Thisarchitecturaldesignhasusedintheporch-likeplanecombination,causesintheguesthousevariousroomsrelativeindependence,peaceful,butdoesnotpassthroughalsohascertainrelation,thearchitecturalcompositionactsascircumstancespermit,thenimblearrangement,thefunctiondistrictisclearabout,thestreamlineclearly,isadvantageousforthemanagement,satisfiestheoperationrequirements,speciallythefirepreventionanddispersestherequest.Thesecondpartisthestructuraldesign,thisdesignstructureusesthecrosswiseframeload-bearingsystem,selectsapinframetocarryonthehorizontalforceandtheverticalstrengthcomputation,andcarriesontheendogenicforcecombination,bydeterminedthemostdisadvantageousendogenicforce,andcarriesonthesectiondesign.Includingframebeam,board,column,staircase,rainawningandfoundationsectiondesignandsoon.Thewallusesthepulverizedcoalashtosteampressestheaeroconcreteartificialbrickmasonryandbuilding,isadvantageousinreducingthebuildingdeadweight;Inwhichtoframebeam,theboard,thecolumndesignhasallusedthequitecommonlyusedtheoryofelasticityplan.Thisdesignis7degreeearthquakeresistancefortificationintensitydesign,thereforeinthedesignprocesshasalsocarriedonundertheearthquakefunctiontheelasticdisplacementcheckingcalculationandmeetsundertheearthquakefunctionrarelytheelastoplasticitydistortioncheckingcalculation.Afterthedesignhasachievedtheintensity,therigidityandthestableaspectrequest.Keywords:Constructionframeconcreteearthquakeresistancedesign137 目录摘要IAbstractII第1章建筑设计11.1建筑的总平面设计11.2建筑平面设计11.2.1房间的平面设计11.2.2楼梯设计21.3建筑立面设计31.4建筑剖面设计3第2章结构设计说明42.1框架的承重方案42.2梁柱截面尺寸的确定42.3框架结构的计算简图52.4结构计算5第3章工程概况及结构布置73.1工程概况73.2结构的布置及计算简图7第4章重力荷载的计算94.1屋面及楼面的永久荷载标准值94.2屋面及楼面可变荷载的标准值104.3梁、柱、墙、门的重力荷载计算104.4重力荷载代表值12第5章侧移刚度的计算135.1横向框架侧移刚度的计算135.2柱的侧移刚度计算14137 第6章横向水平荷载作用下框架结构的内力和侧移计算166.1横向水平地震作用下框架结构的内力和侧移计算166.1.1横向自振周期的计算166.1.2水平地震作用及楼层地震剪力计算176.1.3水平地震作用下的位移验算196.1.4水平地震作用下框架内力计算206.2横向风荷载作用下框架结构内力和位移的计算246.2.1风荷载标准值246.2.2风荷载作用下的水平位移276.2.3风荷载作用下框架结构内力计算28第7章竖向荷载作用下框架结构的内力计算327.1计算单元327.2荷载计算337.3弯矩计算387.3.1恒荷载作用下的内力计算387.3.2活荷载作用下的内力计算397.4内力计算407.5梁端剪力及柱轴力计算40第8章横向框架的内力组合488.1结构抗震等级488.2框架梁的内力组合488.3框架柱内力组合55第9章截面设计669.1框架梁设计669.1.1梁的正截面受弯承载力计算669.1.2梁斜截面受剪承载力计算689.2框架柱设计719.2.1剪跨比和轴压比验算719.2.2正截面承载力计算729.2.3柱斜截面受剪承载力计算76137 9.3框架梁柱节点核芯区截面抗震验算79第10章罕遇地震作用下弹塑性变形验算8210.1罕遇地震作用下的楼层剪力8210.2楼层受剪承载力计算83第11章楼梯设计8711.1梯段板设计8711.2平台板设计8911.3平台梁设计90第12章雨篷的设计及承载力验算9212.1雨篷板的设计及承载力的验算9212.2雨篷梁设计及承载力计算9312.3雨蓬抗倾覆验算97第13章楼板的设计9813.1楼板类型的选择9813.2设计参数9813.3弯矩计算9913.4截面设计101第14章基础设计104结论109致谢110参考文献111附录一:112附录二:120137 第1章建筑设计1.1建筑的总平面设计(1)根据地形考虑,该地段地处邯郸市高新技术开发区内主入口的南侧,北临高新区内城市广场,西临邯郸市东环大街,其它两侧为办公用地,地理位置十分优越。考虑到已有建筑,所以正立面和左侧立面依城市主要街道而设,更加突出了此宾馆选址的优越。(2)为满足使用要求及当地规划部门的要求,在宾馆正面留有足够的空间设置相应的停车场地。(3)为创造一个良好舒适的环境,在用地范围内布置适当的绿化,以及布置景观小品设计和园林绿化设计,并适当考虑与城市绿化带的相互关系,创造一个和谐、舒适的环境。1.2建筑平面设计建筑平面设计是针对建筑的室内使用部分进行的,即有机地组合内部使用空间,使其更能满足使用者的要求,本设计是从平面设计入手,着眼于建筑空间的组合,结合高层宾馆的具体特点进行设计。本设计按照各基本单元空间的功能性质、使用顺序进行功能分析和功能分区。处理好各建筑空间的关系,合理组织好交通流线,使各种流线符合使用顺序并做到简洁明确、顺畅直接、不交叉迂回,避免相互干扰,并布置良好的朝向,满足采光和通风条件。1.2.1房间的平面设计137 根据设计任务书中对建筑总面积、层数及房间数量及使用面积的要求,对各个房间在每层平面中所占的比例初步确定每层及各房间的面积、形状与尺寸,根据功能分析、流线分析等进行平面组合设计。首先确定组合方式。且为了提供一个较舒适宽敞的的住宿的空间,使得内部的各种功能划分比较灵活,本设计采用了开间7.8m,进深为7.2m柱网,且采用蒸压粉煤灰加气混凝土砌块对大空间进行隔断,其材料较轻并有很好的隔音效果。基于以上各项要求本设计的平面为矩形。楼梯设计在大厅内体现了导向明确、疏散快捷方便的理念。1.2.2楼梯设计根据人流出行和疏散的要求,有大厅的显著位置设置有电梯和楼梯,电梯参数确定客梯为2400mm×2500mm,且设置了一部消防楼梯,依据《建筑设计防火规范》设封闭式楼梯间兼消防与疏散,乙级防火门向楼梯间开启,设前室33.61m2,单独电源,备有电话。(1)根据《旅馆建筑设计规范》中确定楼梯的踏步尺寸与楼梯段净宽:旅馆的楼梯踏步高度底层取175mm,宽度取300mm。其它层楼梯踏步高度取150mm,宽度取300mm;单跑踏步数均为12。(2)楼梯形式的选择应便于疏散迅速、安全,尽量减少交通面积并有利于房间平面布置,根据高层旅馆的平面的布置、形状与尺寸,确定楼梯形式为两跑楼梯。(3)确定楼梯开间进深尺寸,在满足各种功能要求的情况下将开间尺寸定为3900mm,则梯段宽度为1740mm;根据平台宽度大于等于梯段宽度的规定,平台宽度亦取1760mm。(4)楼梯踏步数:首层踏步数:第一跑梯段踏步数为2100/175=12;第二跑梯段踏步数是2100/175=12;标准层踏步数:第一跑踏步数为1800/150=12;第二跑梯段踏步数是1800/150=12;(5)梯段长度:首层梯段长度(12-1)×300=3300mm标准层梯段长度:(12-1)×300=3300mm(6)确定楼梯间的位置楼梯应布置均匀、位置明显、空间导向性强,有利于人员的出行的疏散。本设计的楼梯入口设置在了大厅内,并在另一侧设有楼楼,并满足防火间距的要求。137 1.3建筑立面设计高层旅馆建筑和其它公共建筑一样,不仅能给人们提供工作、活动空间,满足使用要求,而且能够反映出建筑的主题思想。在满足使用要求的同时,照顾到立面造型,本设计的的底层层高为4.2m,标准层的层高为3.6m。建筑体型设计是高层旅馆建筑设计中的重要环节。建筑体型是建筑空间组合的外在因素,它是内在诸因素的反映。建筑的内部空间与外部体型是建筑造型艺术处理问题中的矛盾双方,是互为依存不可分割的,往往完美和谐的建筑艺术形象,总是内部空间合乎逻辑的反映。立面设计是反映整个建筑的一个方面,是生动的、富有表现力的信息来源。通过立面门、窗及各种构配件的位置、大小、外形等变化,使建筑的外观与使用功能、经济技术的合理性达到统一,给人以简洁、明快、朴素、大方的感受。立面处理的好坏,将影响建筑设计的效果。1.4建筑剖面设计剖面设计的目的主要是确定内部空间的使用高度,以确保建筑空间的满足使用要求,考虑到旅馆是人流密集的公共建筑,要求有空调,消防等设备高度及主梁的高度,吊顶等。所以确定底层层高为4.2m,以上每层层高为3.6m,经初步估算,净高符合要求。137 第2章结构设计说明2.1框架的承重方案本设计采用框架结构承重体系,根据框架结构的承重体系的不同,可分为横向框架承重体系纵向框架承重体系和纵横向框架承重体系。根据本设计的结构布置特点,采用横向框架承重体系。2.2梁柱截面尺寸的确定楼盖和屋盖均采用现浇钢筋混凝土结构,楼板厚度选用100mm。梁截面高度按跨度的1/8~1/12进行估算,宽度取b=(1/2~1/3)h。柱的截面尺寸按公式进行估算。-框架柱轴压比限值。本设计为二级抗震,查《抗震规范》,取值为0.8,柱的截面尺寸由下式边柱:中柱:式中A-柱的负荷面积()n-柱上的层数q-重力荷载代表值()取第一层柱截面边柱:137 中柱:取截面为正方形,则边柱和中柱的截面尺寸为535mm和567mm。各层柱的截面尺寸变化范围在100~150之间。由此估算框架梁和柱的截面尺寸,并考虑其它因素影响,在设计中的取值见表2-1。表2-1梁截面尺寸(mm)混凝土等级横梁纵梁次梁C35ABCDBC350700300500C35300600300500表2-2柱的截面尺寸(mm)层次混凝土等级1层C408008002~7层C357007008~10层C356006002.3框架结构的计算简图计算简图用梁柱轴线表示,梁柱轴线取各自的形心线。对于与钢筋混凝土楼盖整体现浇的框架梁,可取楼板底面处作为梁轴线。底层的框架柱取自基础的顶面。由于各层柱的截面尺寸不同而导致形心线不重合时,取顶层柱的形心作为柱子的轴线。按此计算简图计算出的内力是计算简图轴线上的内力,由于此轴线不一定是各截面的形心线,所以在截面配筋计算时,应将计算简图轴线上的内力转化为柱截面形心处的内力,即计入上、下柱截面形心间的偏心距对弯矩的影响。2.4结构计算137 本设计进行了重力荷载代表值计算,水平地震作用的计算,风荷载的计算,水平荷载作用下框架结构的内力和位移计算,竖向荷载作用下框架的内力计算,框架梁和框架柱的内力组合。梁和柱的的截面设计及一些小构件的计算,包括楼梯、雨篷、楼板和基础的计算。并进行罕遇地震作用下的弹塑性变形验算。137 第3章工程概况及结构布置3.1工程概况建设地点:河北省邯郸市建筑类型:框架填充墙结构,楼盖及屋盖均采用现浇钢筋混凝土结构,楼板厚度取100mm。填充墙采用蒸压粉煤灰加气混凝土砌块。主体结构10层,底层层高m,其余各层为m。局部突出的塔楼为电梯机房和水箱间,层高为3.9m。门窗:门为木门,部分为钢制防火门,窗为铝合金窗。材料选用:框架柱底层选用C40混凝土,2~10层选用C35混凝土。框架梁在底层用C40混凝土,210层用C35混凝土。受力钢筋采用HRB335和HRB400级钢筋。箍筋采用HPB235级钢筋。钢筋的连接优先采用焊接。柱网:本建筑采用内廊式平面组合,柱距为7.8m,边跨跨度为7.2m,中间跨度为2.1m。自然条件:基本风压为,基本雪压为。地质条件:建筑场地为二类场地,设防烈度为7度,地震分组为第二组。基本地震加速度为,水平地震影响系数为=。3.2结构的布置及计算简图根据该建筑的使用功能和建筑设计的要求,进行了建筑的平面、立面及剖面设计。主体结构共10层,底层层高为4.2m,其余各层层高3.6m,主体高度为36.6m。137 填充墙外墙采用390厚的蒸压粉煤灰加气混凝土砌块砌筑,内填充墙采用190厚蒸压粉煤灰加气混凝土砌块砌筑,底层窗高2.5m,其余各层窗高2.0m。在计算简图中,2~10层的柱高为层高,即3.6m,底层柱高从基础顶面至一层板底,即5.8m。其计算简图见2-1图。图2-1结构计算简图137 第4章重力荷载的计算4.1屋面及楼面的永久荷载标准值屋面永久荷载标准值(上人)30厚细石混凝土保护层三毡四油防水层20厚水泥找平层150厚水泥蛭石保温层100厚钢筋混凝土结构层V型轻钢龙骨吊顶0.25KN/m2合计4.85KN/m2客房部分:3~9层楼面木块地面(加防腐油膏铺砌厚76mm)100厚钢筋混凝土结构层V型轻钢龙骨吊顶0.25KN/m2合计3.45KN/m2餐厅、办公、走道部分磨光大理石板20厚,水泥砂浆擦缝30厚1:3干硬性水泥砂浆,面上撒2厚素水泥水泥浆结构层一道1.16KN/m2100厚现浇钢筋混凝土结构层V型轻钢龙骨吊顶0.25KN/m2合计3.91KN/m2卫生间:137 陶瓷锦砖5厚,铺实拍平,干水泥擦缝1:3干硬水泥砂浆结合层30厚表面撒水泥粉聚氨脂防水层1.5厚1:3水泥砂浆找坡层,最薄处20厚1.70KN/m2100厚现浇钢筋混凝土结构层20厚石灰砂浆板底抹灰0.02×17=0.34KN/m2合计4.54KN/m24.2屋面及楼面可变荷载的标准值楼面活荷载标准值上人屋面均布活荷载标准值屋面雪荷载标准值(为屋面积雪荷载分布系数)梁柱的密度蒸压粉煤灰加气混凝土砌块4.3梁、柱、墙、门的重力荷载计算梁、柱重量按其尺寸及材料容重标准值确定。梁两侧面层重量也应计入梁自重内,墙体重量可根据厚度及材料容重标准值确定,其两侧的面层应计入墙体自重内。本设计的的墙体采用蒸压粉煤灰加气混凝土砌块填充,以砌块厚度和材料标准值计入墙体自重,门、窗按其材料、种类、规格,按《荷载规范》进行计算。计算结果见表4-1。表4-1梁、柱重力荷载代表值层次构件b/mh/mrKN/mgKN/mLi/mn/KN/KN137 1层柱0.80.8251.0516.85.8322204.164730.87纵梁0.350.7251.056.437.0281260.28横梁0.30.6251.054.736.416483.840.30.5251.053.942.1866.19次梁0.30.5251.053.947.113389.800.30.5251.053.946.014380.602层柱0.70.7251.0512.863.6321440.324009.57纵梁0.350.35251.056.437.1281278.28横梁0.30.6251.054.736.516491.280.30.5251.053.942.1866.19次梁0.30.5251.053.946.614364.060.30.5251.053.947.113368.803~7层柱0.70.7251.0512.863.61928461.9119685.26纵梁0.350.7251.056.437.1843848.80横梁0.30.6251.054.736.5962951.520.30.5251.053.942.140330.96次梁0.30.5251.053.944.5681205.610.30.5251.053.947.1701958.230.20.3251.051.582.2220762.308~10层柱0.60.6251.059.453.6963175.2110324.82纵梁0.350.7251.056.437.2843888.92横梁0.30.6251.054.736.6481496.880.30.5251.053.942.124198.58次梁0.30.5251.053.947.2421191.460.30.5251.053.947.112335.720.20.3251.051.582.21138.12注:表中为考虑梁柱粉刷层重力荷载的增大系数;g表示单位长度构件重力荷载;n为构件数量;梁长度取净长;柱长度取层高。137 4.4重力荷载代表值重力荷载代表值是集中于各楼层标高处的重力荷载为计算单元范围内各层楼面的重力荷载代表值及其上、下各半层的墙柱重量。图4-1重力荷载代表值计算Gi时,各可变荷载的组合系数查相应的规范采用,无论是否是上人层面,其屋面上可变荷载均取雪荷载。外墙为390mm厚蒸压粉煤灰加气混凝土砌块,内墙面为20mm厚抹灰,外墙面贴瓷砖(0.5KN/m2),则外墙单位面积重力荷载为内墙为190mm厚蒸压粉煤灰加气混凝土砌块,两侧均为20mm厚抹灰。则内墙的单位面积重力荷载为137 第5章侧移刚度的计算5.1横向框架侧移刚度的计算横向框架梁线刚度ib计算过程见表5-1;柱的线刚度ic计算过程见表5-2。C40:C30:表5-1横梁线刚度ib计算类别层次Ec/N/mm2b×h/mm2Io/mm4l/m边横梁13.25×104300×6005.4×10970002~103.15×104300×6005.4×1097200走道梁13.25×104300×5003.125×10921002~103.15×104300×5003.125×1092100续表5-1横梁线刚度ib计算类别层次EcIo/l/N.mm1.5EcIo/l/N.mm2EcIo/l/N.mm边横梁12.43×10103.65×10104.86×10102~102.36×10103.54×10104.72×1010走道梁14.84×10107.26×10109.68×10102~104.69×10107.04×10109.38×10108137 表5-2柱的线刚度ic计算层次hc/mmEc/N/mm2b×h/mm2Ic/mm4EcIc/hc/N.mm158003.25×104800×8003.4×101019.05×10102~736003.15×104700×7002.0×101017.5×10108~1036003.15×104600×6001.08×10109.45×1010注:ic柱的线刚度Ic=EcIc/h,Ic为柱的截面惯性矩,h为框架柱的计算高度。5.2柱的侧移刚度计算柱的侧移刚度按下式计算确定式中系数由下表确定表5-3柱侧移刚度修正系数位置边柱中柱一般层底层(固接)根据梁柱线刚度比的不同,柱可分为中框架中柱和边柱,边框架中柱和边柱以及电梯柱等。柱的侧移刚度计算结果见表5-4、5-5和5-6。表5-4中框架柱侧移刚度D值(N/mm)层次边柱(10根)中柱(10根)αcDi1αcDi28~100.4990.202175001.4920.42737363548625137 3~70.2690.119192820.8120.2884666765949020.2740.121193330.8200.2904699166324010.2520.334226970.5080.40227963506599表5-5边框架柱侧移刚度D值(N/mm)层次A-1,A-15,D-15B-1,B-15,C-15αcDi1αcDi28~100.3740.158138251.1190.359314131357133~70.2010.09143830.6010.2323726815495520.2120.095153930.6120.2343791715993110.1930.315214060.3820.37125143139646表5-6楼电梯间框架侧移刚度D值(N/mm)层次C-9,C-11C-1D-9,D-11,D-1αcDi1αcDi2αcDi38~100.7510.273238880.7400.271236250.3720.156136501123503~70.4040.168272220.4860.195315970.2130.0911458312979020.3420.145234950.5730.221356480.3020.1342106414583010.7610.456309870.4710.389265020.3840.37225143163905将上述各种不同情况下同层框架柱侧移刚度相加,即得各层层间侧移刚度,见表5-7。表5-7横向框架层间侧移刚度(N/mm)层次12345∑Di810150969000944235944235944235层次678910∑Di944235944235796687796687796687137 由上表可见,∑D1/∑D2=0.836>0.7,该框架为规则框架。137 第6章横向水平荷载作用下框架结构的内力和侧移计算6.1横向水平地震作用下框架结构的内力和侧移计算6.1.1横向自振周期的计算由公式将折算到主体结构的顶层,。表6-1结构顶点的假想位移计算层次Gi/KNVGi/KN∑Di/(N/mm)△ui/mmui/mm1010171.6410171.6479668712.8686.6910860.9021032.5479668726.4667.8810860.9031893.4479668740.0641.4711080.5042973.9494423545.5601.0612084.3155058.2594423558.3555.9512084.3167142.5694423571.1497.6412084.3179226.8794423583.9426.5312084.3191311.1894423596.0342.6211987.51103298.69969000106.6246.6110378.05113676.74810150140.0140.0按式计算周期,的量纲为,取=0.7,为计算结构基本自振周期用的结构顶点假想位移();137 为结构基本自振周期考虑非承重墙的影响的折减系数。6.1.2水平地震作用及楼层地震剪力计算本结构主体结构在40m以下,且质量和刚度沿高度分布比较均匀,变形以剪切变形为主。用底部剪力法计算水平地震作用。由公式计算总水平地震作用标准值,其中为相应于结构基本自振周期的水平地震影响系数值;为结构等效总重力荷载,多质点应取总重力荷载代表值的85%。=96481.8KN建筑场地为第二类,得,地震设防烈度为7度,查得,则=因为,所以应考虑顶部附加水平地震作用,顶部附加地震作用系数137 =0.08=0.0884各质点的水平地震作用按下式计算具体的计算过程见表6-2,各楼层地震剪力按进行计算。图6-1(a)水平地震作用分布图6-1(b)层间剪力分布表6-2各质点横向水平地震作用及楼层地震剪力计算层次Hi/mGi/KNGiHi/KN.mGiHi/∑GiHiFi/KNVi/KN42.11096.4346159.700.018587.8687.86137 1038.28907.30340258.860.136645.861194.43934.610860.90375787.140.150712.351906.63831.010860.90367687.900.147698.102604.73727.411080.50303605.700.123584.133188.86623.812084.31287606.580.115546.143735.00520.212084.31244103.060.099470.154205.07416.612084.31200599.550.080379.924585.07313.012084.31157096.030.063299.194884.2629.411987.51112682.590.045213.715097.9715.810378.0560192.690.024113.985211.95各质点水平地震作用及楼层剪力沿房屋层高分见图6-16.1.3水平地震作用下的位移验算水平地震作用下框架结构的间位移和顶点位移公式表中计算的各层的层间弹性位移角应满足,为多遇地震作用标准值产生的楼层内最大的层间弹性位移,h为计算楼层高度,为弹性层间位移角的限值。按上式计算的框架各层层间位移应满足规范要求,当不满足时,说明梁、柱截面尺寸偏小,应调整梁、柱截面尺寸或提高混凝土等级,并重新计算。表6-3横向水平地震作用下的位移计算137 层次Vi/KN∑Di/(N/mm)△ui/mmui/mmhi/mθe=ui/hi101194.437966870.9235.473600391391906.637966871.8034.553600200082604.737966872.7132.753600133373188.869442352.9230.053600124163735.009442353.5127.153600102854205.079442353.9623.65360090944585.079442354.3716.69360082434884.269442354.6815.32360076925097.979690004.7810.64360075315211.958101505.865.865800989由上表可见,最大层间弹性位移角发生在第二层,其值为1/753〈1/550,满足要求。6.1.4水平地震作用下框架内力计算在框架平面布置图中,以轴线横向框架为例,其余框架内力计算从略。框架柱端剪力及弯矩按式M其中取自表5-4,取自表5-7,层间剪力Vi取自表6-2,各层柱反弯点的高度比y按式确定,其中由规范查得,为上、下层梁线刚度变化的反弯点高度比的修正值;和为上、137 下层高度变化时反弯点高度比的修正值;本设计中底层柱需考虑修正值,第二层柱需考虑和其余柱无需修正,各层柱的柱端弯矩计算见表6-4、6-5。表6-4各层柱端弯矩及轴力计算表层次hi/mVi/KNDij/N/mm边柱Di2Vi2yMi2bMi2u103.61194.437966871750016.120.4990.2514.5143.5293.61906.637966871750031.760.4990.3540.1074.3283.62604.737966871750047.090.4990.4067.81101.7173.63188.869442351928255.700.2690.3876.21124.3263.63735.009442351928266.860.2690.4096.27144.4253.64205.079442351928276.460.2690.45123.86147.4343.64585.079442351928284.230.2690.45136.40166.7233.64884.269442351928290.320.2690.53172.30152.8023.65097.979690001933392.510.2740.78259.7473.2615.85211.9581015022697113.810.2550.92607.8960.98表6-5各层柱端弯矩及轴力计算表层次hi/mVi/KNDij/N/mm中柱Di2Vi2yMi2bMi2u103.61194.437966873736334.471.4920.4252.1271.9793.61906.637966873736367.811.4920.45109.85134.2683.62604.7379668737363100.541.4920.47170.11191.8373.63188.8694423546667134.820.810.45218.41266.9463.63735.0094423546667161.810.810.45262.13320.6053.64205.0794423546667184.050.810.50333.12333.1243.64585.0794423546667203.810.810.50366.86366.8633.64884.2694423546667218.600.810.50393.48393.48137 23.65097.9796900046991224.850.820.56445.16320.5515.85211.9581015027963163.910.5080.65617.90332.85梁端弯矩、剪力及轴力分别按下式计算:其中梁线刚度取自表5-1,具体计算结果见表6-6。137 图6-2水平地震作用下的弯矩图137 图6-3水平地震作用下的剪力及轴力图137 .表6-6梁端弯矩、剪力及轴力的计算层次边梁走道梁柱轴力MblMbrlVbMblMbrlVb边柱N中柱N1043.5224.097.29.3932.6232.622.131.06-9.39-21.67988.8362.397.220.9084.7384.732.180.69-30.29-81.468141.8189.877.232.20122.17122.172.1116.35-62.93-165.617192.13146.307.247.00198.26198.262.1188.82-109.49-307.436220.63180.437.255.69245.30245.302.1233.62-165.18-484.365243.70199.267.264.52270.72270.722.1257.83-229.70-678.874290.58234.227.272.90318.62318.622.1303.45-302.60-909.223289.20255.207.275.61346.81346.812.1330.30-378.21-1163.912245.56180.497.259.20244.26244.262.1232.17-437.41-1336.881320.72256.627.280.20348.26348.262.1331.67-517.61-1588.36注:(1)柱轴力中的负号表示拉力。当为左震时,左侧两根柱子为拉力;对应的右侧两根柱子为压力。(2)表中M单位为KN.m,V单外为KN,N单位为KN,l单位为m。水平地震作用下框架的弯矩图剪力图及轴力图见6-2图、6-3图。6.2横向风荷载作用下框架结构内力和位移的计算6.2.1风荷载标准值风荷载标准值按式计算。基本风压,类场地,查得;,,查得,,背面137 。则建筑结构在高度处的风振系数,按下式计算。仍取轴线框架,其负载宽度为7.8米,由式,得沿房屋高度的分布风荷载标准值,根据各楼层标高处的高度,查得,代入上式可得各楼层标高处的,见下表6-7,沿高度分布见图6-2。表6-7沿房屋高度分布风荷载标准值层次Hi/mHi/Hμzβzq1(z)/kN.mq2(z)/KN.m1038.21.001.5351.3995.3623.352934.60.9061.4801.3725.0643.163831.00.8121.4341.3514.8323.021727.40.7171.3801.3234.5512.843623.80.6231.2561.3124.1102.572520.20.5291.2531.2613.9422.461416.60.4341.1751.2213.5842.243313.00.3401.0841.1923.2212.01129.40.2461.001.1532.8711.79315.80.1521.001.0822.6961.684<<荷载规范>>规定对于高度大于30米且高度比大于1.5的房屋结构,应采用风振系数来考虑风压脉动的影响。本设计房屋高度H=38.2m>30m,H/B=2.06>1.5,因此该房屋框架应考虑风压脉动的影响。由上表可见,沿房屋高度在1.08~1.399范围内变化,即风压脉动影响较大,因此该房屋应考虑风压脉动的影响。137 框架结构分析时,应按静力等效原理将分布风荷载转化为节点集中荷载,如第8层的集中荷载F8的计算过程如下:+=27.90KN图6-4风荷载沿房屋高度的分布(单位:KN/m)137 图6-5等效节点集中风荷载(单位:KN)6.2.2风荷载作用下的水平位移由图所示的水平荷载,由公式计算层间剪力,然后依据表5-7求出轴线框架的层间侧移刚度,再按式137 计算各层的相对侧移刚度和绝对侧移刚度,计算结果见表6-8。表6-8风荷载作用下框架层间剪力及侧移刚度计算层次12345Fi/KN18.4516.7818.5020.5922.50Vi/KN219.60201.14184.36165.86145.27∑D/(N/mm)101320132648131898131898131898△ui/mm2.171.501.401.261.10Ui/mm2.173.675.076.337.43△ui/hi1/26731/24001/25711/28571/3272层次678910Fi/KN24.026.2127.9029.4218.45Vi/KN122.2798.7772.5644.6615.24∑D/(N/mm)131898131898109725109725109725△ui/mm0.930.750.660.400.14Ui/mm8.369.119.7710.1710.30△ui/hi1/38701/48001/54551/90001/25899由上表可见风荷载作用下框架的最大层间位移角为1/2400,远小于1/550,满足规范要求。6.2.3风荷载作用下框架结构内力计算风荷载作用下的框架结构内力计算过程与水平地震荷载作用相同。各层柱端弯矩及轴力计算见表6-9和6-10。梁端弯矩、剪力及轴力计算见表6-9、6-10、6-11。横向框架在风荷载作用下的弯矩、梁端剪力及柱轴力图见图6-6和6-7。表6-9各层柱端弯矩及轴力计算层次hiViDij边柱137 /m/KN/N/mmDi1Vi1yMi1bMi1u103.615.24109725175002.430.4990.252.186.5693.644.66109725175007.120.4990.358.9716.7083.672.561097251750011.570.4990.4016.7024.9973.698.771318981928214.410.2690.3819.6932.1463.6122.771318981928217.910.2690.4025.7938.6853.6145.271318981928221.230.2690.4534.3941.9243.6165.861318981928224.210.2690.4539.2247.9433.6184.361318981928226.720.2690.5350.9845.2123.6201.141326481933329.310.2740.7882.3023.2115.8219.601013202269749.190.2550.852205.3435.82表6-10各层柱端弯矩及轴力计算层次hi/mVi/KNDij/N/mm中柱Di2Vi2yMi2bMi2u103.615.24109725373635.191.4920.427.8510.8493.644.661097253736315.211.4920.4524.6430.1283.672.561097253736324.701.4920.4741.8047.1373.698.771318984666734.950.810.4556.6169.2063.6122.771318984666743.430.810.4570.3685.9953.6145.271318984666751.400.810.5092.5292.5243.6165.861318984666758.680.810.50105.62105.6233.6184.361318984666765.230.810.50117.41117.4123.6201.141326484699171.250.810.50128.25128.2513.6219.601013202796360.610.820.50228.49123.04表6-11梁端弯矩、剪力及柱轴力计算层次边梁走道梁轴力MblMbrlVbMblMbrlVb边柱中柱137 106.563.627.21.414.894.892.14.66-1.41-3.25918.8812.677.24.3817.3517.352.116.53-5.79-15.40833.9623.997.28.0432.6732.672.131.11-13.83-38.47748.8437.127.211.9450.5650.562.147.87-25.77-74.40658.3747.667.214.7364.8264.822.161.73-40.50-121.40567.7154.497.216.9774.1874.182.170.65-57.47-175.08482.3366.277.220.6490.1290.122.185.91-78.11-240.35384.4374.477.222.07101.33101.332.196.50-100.18-314.78274.1954.567.217.8874.2774.272.170.73-118.06-367.631118.12111.327.231.87151.37151.372.1144.16-149.93-479.92图6-6水平风荷载作用下的弯矩图137 图6-7水平风荷载作用下的剪力及轴力图137 第7章竖向荷载作用下框架结构的内力计算7.1计算单元取轴线横向框架进行计算,计算单元宽度取7.8m,由于房间内布置有次梁,故直接传给框架的楼面荷载如下图所示。计算单元内的其余楼面则通过次梁和纵向框架梁以集中力的形式传给横向框架,作用于各节点上。由于纵向框架梁的中心线与柱的中心线不重合,因此框架节点还有集中力矩。图7-1横向框架计算单元137 7.2荷载计算因在AB、CD跨处标准客房内卫生间内孔洞及两侧的内梁格均较小,为简化计算可不考虑其分担板的荷载,而将其自身及其上的墙体的荷载传递到两端的梁,忽略洞口的存在。1恒载计算图7-2各层梁上作用的恒荷载对第10层,、代表横梁自重,为均布荷载形式。=4.73KN/m=3.94KN/m和分别为房间和走道板传给横梁的梯形荷载和三角形荷载,且由图示的几何关系所得P1、P2分别为边纵梁、中纵梁直接传给柱的恒荷载,它包括梁自重、楼板自重和女儿墙等重力荷载。计算如下137 在客房部分有截面300mm500mm的次梁,对卫生间的隔墙起支承作用,此梁对横梁产生一集中力,则集中力矩对8~9层,q包括梁自重和其上横梁自重,为均布荷载,其它计算方法同10层集中力矩为137 对于2~7层对于1层137 2活荷载计算活荷载作用下各层框架梁上的荷载分布图图7-3各层梁上作用的活荷载对于第10层同理,在屋面雪荷载作用下对于8~9层137 对于2~7层对于1层将计算结果汇总,见表7-1、7-2。表7-1横向框架恒载汇总表层次/KN/m/KN/m/KN/m/KN/m/KN/KN/KN/KN.m/KN.m104.733.9418.910.19165.9154.528.3420.719.3137 899.253.9413.468.2157.8183.628.3419.722.95279.253.9413.468.2157.8183.0728.3427.632.019.253.9413.468.2160.74187.9828.3436.242.3表7-2横向框架活载汇总表层次/KN/m/KN/m/KN/KN/KN.m/KN.m107.8(1.7)4.2(0.63)30.42(4.56)44.6(6.69)3.8(0.57)5.58(0.84)897.84.230.4244.63.85.58277.84.230.4244.65.327.817.84.230.4244.66.810注:表中括号内数值为对应于屋面雪荷载作用情况。7.3弯矩计算7.3.1恒荷载作用下的内力计算等效于均布荷载与梯形、三角形荷载的叠加。。对于第10层==92.23KN.m137 对于1~9层=98.1KN7.3.2活荷载作用下的内力计算对于1~10层,137 7.4内力计算梁端、柱端弯矩采用弯矩二次分配法计算。由于结构对称,故计算时取半边框架,梁端剪力可根据梁上竖向荷载引起的剪力与梁端弯矩引起的剪力相叠加而得。柱轴力可由梁端剪力和节点集中力叠加而得。计算柱轴力还要考虑柱的自重。7.5梁端剪力及柱轴力计算1恒载作用下如第10层,由荷载引起的剪力由集中力引起的剪力由弯矩引起的剪力值柱A轴力:N顶=165.9+71.58=237.48KN柱重:0.6×0.6×3.6×25=32.4KNN底=N顶+32.4=237.48+32.4=269.88KN柱B轴力:N顶=154.5+9.49+76.7=240.69KNN底=240.69+32.4=273.09KN137 其它层的计算方法同10层,计算结果汇总于表7-2和表7-4。2活荷载作用下以第10层为例,荷载引起的剪力AB跨:BC跨:柱A轴力:N顶=N底=30.42+20.48=50.9KN柱B轴力:N顶=N底=44.6+2.21+20.48=67.18KN其它层的计算方法同第10层,计算结果汇总于表7-3和表7-4。表7-2恒载作用下梁端剪力(KN)层次荷载引起剪力弯矩引起剪力总剪力AB跨BC跨AB跨BC跨AB跨BC跨1071.5876.709.49-2.53069.0579.239.49973.5778.638.44-1.99071.5880.628.44873.5778.638.44-2.20071.3780.608.44773.5778.638.44-1.97071.3680.588.44673.5778.638.44-1.95071.6280.588.44573.5778.638.44-1.96071.6180.598.44473.5778.638.44-1.96071.6180.598.44373.5778.638.44-1.96071.6180.598.44273.5778.638.44-1.96071.6180.598.44173.5778.638.44-1.98071.5980.618.44表7-3活荷载作用下梁端剪力(KN)层次荷载引起剪力弯矩引起剪力总剪力AB跨BC跨AB跨BC跨AB跨BC跨137 1020.482.21-0.436020.0420.922.21920.482.21-0.11020.3720.592.21820.482.21-0.18020.3020.662.21720.482.21-0.11020.3720.592.21620.482.21-0.05020.4320.532.21520.482.21-0.08020.4020.562.21420.482.21-0.08020.4020.562.21320.482.21-0.08020.4020.562.21220.482.21-0.076020.4020.552.21120.482.21-0.109020.3720.592.21表7-4恒荷载及活荷载各自作用下的柱轴力层次恒荷载作用下柱轴力活荷载作用下柱轴力A柱B柱A柱B柱N顶N底N顶N底N顶=N底N顶=N底10237.48269.88240.69273.0950.9067.189501.22533.62543.79576.19101.80134.368764.96797.36846.89879.29152.70201.5471028.701072.801149.481193.56203.60268.7261304.141348.241463.731507.83254.50335.9051579.581623.681778.001822.10305.40403.0841855.021899.122092.232136.33356.30470.2632130.462174.562406.502450.60407.20537.4422405.902450.002720.772764.87458.10604.6212684.262777.063039.953132.75559.90672.24137 图7-4恒载作用下的弯矩二次分配表137 续图7-4恒载作用下的弯矩二次分配表137 图7-5活载作用下的弯矩二次分配表137 续图7-5活载作用下的弯矩二次分配表137 图7-6恒荷载作用下的弯矩图图7-7活荷载作用下的弯矩图137 第8章横向框架的内力组合8.1结构抗震等级结构的抗震等级可根据结构类型、地震烈度、房屋高度等因素,查看相应资料可知,本工程的框架为地级抗震等级。8.2框架梁的内力组合框架梁的内力组合中本方案考虑了四种内力组合,即,,及。对于本工程,这种组合与考虑地震作用的组合相比一般较小,对结构设计不起控制作用,故不予考虑。各层梁的内力组合结果见表8-1。表中、中梁端弯矩M为经过调幅后的弯矩(调幅系数取0.8,因为考虑到钢筋混凝土结构具有塑性内力重分布的性质,在竖向荷载作用下可以适当降低梁新局面弯矩,以减少负弯矩钢筋的拥挤现象)。为梁端剪力增大系数,抗震等级为二级时取1.2。跨间最大弯矩值的计算以第一层AB跨梁为例,说明计算方法和过程。计算理论:根据梁端弯矩的组合值及梁上荷载设计值,由平衡条件确定。对支座负弯矩按相应的组合情情况进行计算。由梯形荷载转化为均布荷载,由下列公式求得137 在均布荷载和集中荷载作用下,如下图所示图8-1在均布和集中荷载作用下由上式求得的,设x为最大弯矩至支座A支的距离当x>4.7时,令求得x值。由当时,令求得x值。由137 同时,三角形分布荷载和均布荷载作用下,、x、的计算公式图8-2在三角形分布荷载和集中荷载作用下由下式求得:本例中,梁上荷载设计值左震137 当时,令11.05-29.29x=0x=0.38m右震当x>4.7时,令219.53-34.01-29.29x=0x=6.33m剪力计算:AB净跨左震=-5.4KN右震137 表8-1框架梁内力组合表层次截面内力SGKSQKSWKSEK1.2SGK+1.26(SQK+SWK)→←一层AM-88.87-20.88118.12320.7215.88-281.78V71.5920.3731.8780.2071.42151.73M-100.30-20.50111.32256.62-287.71-7.19V80.8620.5931.8780.20162.8382.52M-8.49-1.94151.37348.26178.09-203.36V8.442.21144.16331.67-168.73194.55跨间M143.60124.23V139.41139.41三层AM-90.10-21.2484.83289.20-28.45-241.71137 V71.6120.4022.0775.6183.83139.44M-101.40-21.7074.47255.20-242.85-55.19V80.5920.5622.0775.61150.4294.81M-7.88-1.82101.33346.81115.93-139.43V8.442.2196.50330.30-108.68134.50跨间M165.85112.99V86.3286.32续表8-1框架梁内力组合表层次截面内力RE[1.2(SGK+0.5SQK)+1.3SEK]1.35SGK+SQK1.2SGK+1.4SQKV→←一层AM223.33-402.10-140.85-135.88194.34V-4.59151.79117.02114.43M-350.15150.24-156.91-150.46V160.003.62129.41125.56M331.04-348.07-13.40-12.90319.81V-314.79331.9713.6013.22跨间M224.92157.89136.13131.35V332.64332.699.239.15三层AM191.34-372.60-142.84-137.80169.81V-0.09147.35117.07114.49M-349.85147.80-158.59-152.06V155.508.06129.36125.49M330.22-346.05-12.46-12.00248.54V-313.45330.6313.6013.22跨间M194.11147.01122.14118.07V330.46330.468.788.78137 续表8-1框架梁内力组合表层次截面内力SGKSQKSWKSEK1.2SGK+1.26(SQK+SWK)→←八层AM-86.60-20.3233.96141.81-86.73-172.31V71.3720.308.0432.20101.10121.35M-99.10-21.3023.9989.87-175.99-115.53V80.8320.668.0432.20133.16112.90M-13.50-2.6832.67122.1721.59-60.74V8.442.2131.11116.35-26.2952.11跨间M89.2684.79V54.0954.09十层AM-75.13-18.176.5643.52-104.78-121.32V69.0520.041.419.39106.33109.89M-89.90-20.683.6224.00-138.50-129.38V79.2320.921.419.39123.21119.66M-22.40-4.984.8932.62-26.99-38.32V9.492.214.6631.068.3020.20跨间M57.5355.52V5.325.32续表8-1框架梁内力组合表层次截面内力RE[1.2(SGK+0.5SQK)+1.3SEK]1.35SGK+SQK1.2SGK+1.4SQKV→←八层AM51.18-225.35-137.23-132.36124.73V41.97104.76116.65114.06M-186.39-11.15-115.08-148.74V113.4450.65129.78125.92137 M105.76-132.47-20.91-19.95142.26V104.85122.0313.6013.22跨间M101.7592.5380.3376.30V112.85112.859.689.68十层AM-33.36-118.23-119.60-115.5992.09V62.0080.32113.26110.92M-113.70-66.73-142.05-136.83V89.8871.57127.88124.36M9.39-54.21-35.22-33.8587.33V-20.7539.8215.0214.48跨间M60.6654.6460.5262.69V50.3450.3412.5012.50注:表中8.3框架柱内力组合取每层柱顶和柱底两个截面,每个截面上有M、N、V。按下式进行组合。137 式中:、、为由恒荷、楼面活载及风荷载在柱端截面产生的弯矩标准值;、NQK、NWK为由恒载、楼面活荷及风载标准值在柱端产生的轴力标准值;MGE、NGE、MEK、NEK为由重力荷载代表值及水平地震作用标准值在柱端截面产生的弯矩、轴力标准值。由于柱是偏心受力构件且一般采用对称配筋,故应从上述组合中求出下列最不利内力:(1)、及相应的N。(2)、Nmax及相应的M。表8-2横向框架A柱的弯矩和轴力组合层次截面内力SGKSQKSwkSEK1.35SGk+SQk1.2SGk+1.4SQk10柱顶M72.3118.906.5643.52116.52113.26N237.4850.901.419.39371.49356.59柱底M-51.19-12.762.1814.51-81.87-79.39N269.8850.901.419.39415.24395.629柱顶M40.729.5316.7074.3264.5062.21N501.22101.805.7930.29778.47743.92柱底M-47.28-11.358.9740.10-75.18-72.69N533.62101.805.7930.29822.19782.948柱顶M41.3210.0024.99101.7165.8063.69N764.96152.7013.8362.491185.391132.35柱底M-35.14-8.6516.7067.81-56.09-54.39N797.36152.7013.8362.491229.141170.277柱M49.2315.8232.14124.3282.2881.29137 顶N1028.70203.6025.77109.491592.351519.67柱底M-45.26-12.2819.6976.21-73.38-71.56N1072.80203.6025.77109.491651.881572.336柱顶M40.399.3038.68144.4263.8361.59N1304.40254.5040.50165.182015.441921.62柱底M-42.51-10.6125.7096.27-67.99-65.92N1348.24254.5040.50165.182074.621974.535柱顶M42.5110.6141.92147.4367.9965.94N1579.58305.4057.47229.702437.832323.51柱底M-42.51-10.6134.39123.86-67.99-65.92N1623.68305.4057.47229.722497.402375.614柱顶M42.5110.6147.94166.7267.9965.95N1855.02356.3078.11302.602860.582725.68柱底M-42.51-10.6139.22136.40-67.99-65.96N1899.12356.3078.11302.602920.112778.733柱顶M42.5110.6145.21152.8067.9965.98N2130.46407.20100.18377.213283.323126.76柱底M-43.63-10.6650.98172.30-69.56-67.30N2174.56407.20100.18377.213342.863179.622柱顶M41.0810.5523.2173.2666.0164.30N2405.90458.10118.06437.413706.703528.34柱底M-45.78-11.7450.98259.74-73.54-71.46N2450.00458.10118.06437.413766.803581.211柱顶M29.067.5735.8260.9846.8045.52N2684.26559.90149.93517.614183.654005.60柱底M-14.53-3.79205.34607.89-23.41-22.76N2777.06559.90149.93517.614308.934116.77137 续表8-2横向框架A柱的弯矩和轴力组合层次截面内力1.2SGk+1.26(SQk+Swk)RE[1.2(SGk+0.5SQk)+1.3SEk]|Mmax|NNminMNmaxM→←→←10柱顶M102.34118.8231.32119.00119.0031.32116.52N347.27350.92228.70246.75246.75228.70371.49柱底M-74.71-80.36-37.76-86.29-86.29-37.76-81.87N386.79389.82256.70274.93274.93256.70415.249柱顶M39.8281.90-31.58123.51123.51-31.5864.50N722.65737.52467.42526.72526.72467.42778.47柱底M-59.25-82.32-8.63-86.82-86.82-8.63-75.18N761.73775.93496.65556.84556.84496.65822.198柱顶M30.7593.77-61.34150.25150.25-61.3465.80N1093.901128.26742.70872.65872.65742.701185.39柱底M-32.15-74.1232.65-108.49-108.4932.65-56.09N1131.851166.76773.84904.87904.87773.841229.147柱顶M38.71119.38-74.47184.49184.49-74.4782.28N1458.601523.62971.601199.271199.27971.601592.35柱底M-44.96-94.6329.97-128.66-128.6629.97-73.38N1511.341576.741013.841241.211241.211013.841651.886柱顶M13.19108.96-106.52193.61193.61-106.5263.83N1835.941936.931202.761546.521546.521202.762015.44柱底M-32.94-96.8254.24-146.31-146.3154.24-67.99N1887.601989.361245.381588.641588.641245.382074.625柱顶M11.59117.27-107.31196.42196.42-107.3167.99N2208.742352.561424.651902.651902.651424.652437.83柱底M-21.28-107.63-82.89-175.76-175.76-82.89-67.99N2260.922405.621467.471944.781944.781467.472497.40137 4柱顶M4.03124.71-127.25219.36219.36-127.2567.99N2516.832773.481637.382266.272266.271637.382860.58柱底M-14.99-113.8095.90-188.90-188.9095.90-67.99N2629.762826.721679.762308.692308.691679.762920.113柱顶M7.44121.30-113.33168.72168.72-113.3367.99N2943.883165.741847.882634.572634.571847.883283.32柱底M-1.65-129.98132.47-226.82-226.82132.47-69.56N2996.783248.891890.702676.742676.741890.703342.862柱顶M33.4291.88-31.72120.72120.72-31.7266.01N3315.963612.632075.602984.632984.632075.603706.70柱底M-5.59-133.80220.52-319.68-319.68220.52-73.54N3368.993665.522117.263026.753026.752117.263766.801柱顶M-0.7089.55-31.8394.9594.95-31.8346.80N3738.844115.642307.663384.863384.862307.664183.65柱底M236.58-280.97616.73647.36647.36616.73-23.41N3849.564226.712396.533473.733473.732396.534308.93注:表中M以左侧受拉为正,单位为KN.m,N以受压为正,单位为KN。表8-2横向框架A柱柱端组合弯矩设计值的调整层次10987截面柱顶柱底柱顶柱底柱顶柱底柱顶柱底REMc____171.38123.28209.65146.75REN____872.65904.871199.271241.21层次654截面柱顶柱底柱顶柱底柱顶柱底REMc168.16220.83199.17225.75236.19249.11REN1546.521588.641902.651944.782266.272308.69137 层次321截面柱顶柱底柱顶柱底柱顶柱底REMc210.90258.32137.44346.5699.54809.20REN2634.572767.702984.633026.753384.863473.73注:Mc=Mb表8-3横向框架A柱剪力组合层次SGKSQKSwkSEK1.35SGk+SQk1.2SGk+1.4SQk10-34.56-8.792.4316.12-55.45-53.789-24.44-5.807.1331.78-38.79-37.458-21.44-5.1811.5847.08-33.85-32.747-26.25-7.8014.3655.69-43.24-42.426-23.03-5.5317.8966.70-36.62-35.385-23.62-5.8921.2775.35-37.78-36.594-23.62-5.8924.1984.19-37.78-36.593-23.93-5.9126.6990.31-38.22-36.992-24.13-6.1929.3092.49-38.77-37.621-7.52-1.9641.5715.32-12.11-11.77续表8-3横向框架A柱剪力组合层次1.2SGk+1.26(SQk+Swk)RE[1.2(SGk+0.5SQk)+1.3SEk]见下式(1)→←→←10-49.49-55.61-19.34-50.9770.399-27.65-45.626.38-55.5972.118-17.42-46.5726.09-71.84101.037-23.23-59.4228.97-86.86122.19137 6-12.06-57.1544.61-94.13133.375-9.09-62.4452.86-103.87145.694-5.29-66.2462.06-113.06166.393-7.56-69.7967.93-119.73160.8820.16-72.4770.05-122.33166.94140.80-63.87111.77-128.09191.31注:(1)表8-4横向框架B柱弯矩和轴力组合层次截面内力SGKSQKSwkSEK1.35SGk+SQk1.2SGk+1.4SQk10柱顶M-65.52-13.9610.8471.97-102.41-98.17N240.6967.183.2521.63392.11382.88柱底M48.109.927.8552.1274.8671.59N273.0967.183.2521.63435.85421.769柱顶M-40.20-8.2530.12134.26-62.79-59.85N543.79134.3615.4081.46868.47840.72柱底M44.489.1824.63109.8569.2266.28N576.19134.3615.4081.46912.22880.948柱顶M-39.52-8.1247.13191.83-61.4758.80N846.89201.5438.47165.611345.631298.65柱底M33.626.8141.80170.1152.2049.93N879.29201.5438.47165.611389.341337.677柱顶M-48.99-9.8669.20266.94-76.25-72.62N1149.48268.7274.40307.431821.721755.69柱底M44.939.0256.61218.4169.6866.56N1193.56268.7274.40307.431880.851808.716柱M-40.86-8.2285.99320.60-63.38-60.59137 顶N1463.73335.90121.40484.362342.362286.93柱底M42.468.5470.36262.1365.8662.91N1507.83335.90121.40484.362402.212335.335柱顶M-42.46-8.5492.52333.12-65.86-62.91N1778.00403.08175.08678.672880.932797.95柱底M42.468.5492.52333.1265.8662.91N1822.10403.08175.08678.672943.572870.224柱顶M-42.46-8.54105.62366.86-65.86-62.91N2092.23470.26240.35909.223406.713369.48柱底M42.668.54105.62366.8666.1363.15N2136.33470.26240.35909.223468.663421.963柱顶M-42.46-8.54117.41393.48-65.8662.91N2406.50537.44314.781163.913965.713904.22柱底M43.648.76117.41393.4867.6762.88N2450.60537.44314.781163.914025.823969.142柱顶M-40.65-8.18128.25320.55-63.05-60.23N2720.77604.62367.631336.884518.664445.19柱底M43.828.75128.25445.1667.9164.83N2764.87604.62367.631336.984580.684464.311柱顶M-28.61-5.72123.04332.85-44.34-42.34N3039.95672.64479.921588.365236.644985.63柱底M14.302.86228.49617.9022.1721.62N3132.75672.24479.921588.365290.735157.63续表8-4横向框架B柱弯矩和轴力组合层次截面内力1.2SGk+1.4(SQk+wk)RE[1.2(SGk+0.5SQk)+1.3SEk]|Mmax|NNminMNmaxM→←→←137 10柱顶M-108.86-82.68-135.264.76-135.26-135.26-102.41N369.76377.62225.76267.94225.76225.76392.11柱底M80.8360.3698.45-2.9598.4598.4574.86N408.31416.52254.92297.10254.92254.92435.859柱顶M-96.41-20.88-170.7490.95-170.74-170.74-62.79N802.69841.31470.58629.17470.58470.58868.47柱底M95.9633.91151.22-62.89151.22151.2269.22N841.32880.29499.74658.33499.74499.74912.228柱顶M-117.211.69-241.31157.64-241.31-241.31-61.47N1222.661318.77738.021081.49738.02738.021345.63柱底M101.60-3.75212.34-141.26212.34212.3452.20N1260.611357.67769.121112.59769.12769.121389.347柱顶M-158.3615.85-328.40224.88-328.40-328.40-76.25N1624.711811.96913.211551.77913.21913.211821.72柱底M136.67-5.97274.18-179.26274.18274.1869.68N1676.961864.83955.151594.08955.15955.151880.856柱顶M-167.6048.85-376.60290.25-376.60-376.60-63.38N2027.352332.921062.502070.291062.501062.502342.36柱底M150.37-26.98317.34-227.62317.34317.3465.86N2079.562385.841104.872112.631104.871104.872402.215柱顶M-178.3854.80-301.46301.46-301.46-301.46-65.86N2420.722862.461196.282604.441196.281196.282880.93柱底M178.33-54.85301.46-301.46301.46301.4665.86N2473.612915.731238.612646.771238.611238.612943.574柱顶M-194.6971.27-396.99336.61-396.99-396.99-65.86N2800.753406.521290.333178.201290.331290.333406.71柱底M194.97-70.96426.52-366.42426.52426.5266.13N2853.643459.611332.673220.541332.671332.673468.663M-209.3785.79-454.62364.90-454.62-454.62-65.86137 柱顶N3068.933961.931359.633777.091359.631359.633965.71柱底M210.88-84.23455.86-363.66455.86455.8667.67N3221.294015.761401.963819.131401.961401.964025.822柱顶M-175.28101.23-376.32290.42-376.32-376.32-63.05N3565.604490.381513.864290.451513.861513.864518.66柱底M223.61-96.70509.17-416.64509.17509.1767.91N3616.524543.521556.204332.791556.201556.204580.681柱顶M-196.52113.45-376.22315.79-376.22-376.22-44.34N3890.735100.541591.574890.871591.571591.575236.64柱底M308.69-267.30657.67-627.56657.67657.6722.17N4001.345211.621680.474979.761680.471680.475290.73表8-5横向框架B柱柱端组合弯矩设计值的调整层次10987截面柱顶柱底柱顶柱底柱顶柱底柱顶柱底REMc____249.04220.59341.18278.20REN____738.02769.12913.21955.15层次654截面柱顶柱底柱顶柱底柱顶柱底REMc382.12366.26352.31347.94424.45463.95REN1062.501140.871196.281238.611290.331332.67层次321截面柱顶柱底柱顶柱底柱顶柱底REMc438.87458.41378.88515.18373.83822.04REN1359.631401.961513.861556.201591.571680.47注:Mc=Mb137 表8-6横向框架B柱剪力组合层次SGKSQKSwkSEK1.35SGk+SQk1.2SGk+1.4SQk1031.566.335.1834.3948.9446.73923.524.8415.1767.7836.5935.00820.324.1524.69100.5031.5830.19726.095.2434.89134.4440.4638.64623.144.6643.39161.8335.9034.29523.594.7451.38185.0036.5734.94423.644.7458.61203.7836.6535.00323.924.8165.00218.8937.1035.44223.464.7070.56164.1136.3734.7017.391.4860.60163.9011.4610.94续表8-6横向框架B柱剪力组合层次1.2SGk+1.26(SQk+Swk)RE[1.2(SGk+0.5SQk)+1.3SEk]见下式(1)→←→←1052.3739.3264.78-2.2880.13953.4415.2189.43-42.74110.39860.72-1.49126.02-83.02136.87781.87-6.05167.38-112.26180.51688.31-22.57192.75-143.85218.10599.02-30.46217.32-167.48204.184108.19-39.51236.90-186.96258.903116.67-47.14252.92-202.37254.742122.98-54.83195.45-145.89245.23187.09-65.62178.26-162.65226.84137 注:(1)137 第9章截面设计9.1框架梁设计9.1.1梁的正截面受弯承载力计算从表中分别选出AB跨间截面及支支座截面最不利内力,并将支座中心处的弯矩换算成支座边缘控制截面的弯矩进行配筋。跨间弯矩取控制截面,即支座边缘处的正弯矩。可求得相应的剪力值为则支支座边缘处的当梁下部受拉时,按T形截面设计,当梁上部受拉时,137 按矩形截面设计。抗震设计中,对于楼面现浇的框架结构,梁支座负弯矩按矩形截面计算纵筋数量。跨中正弯矩按T形截面计算纵筋数量,跨中截面的计算弯矩,应取该跨的跨间最大正弯矩或支座弯矩与1/2简支梁弯矩之中的较大者,依据上述理论,得:考虑跨间最大弯矩处,按T形截面设计,翼缘计算宽度按跨度考虑取=7.2/3=2.4m=2400mm,按梁间距考虑=300+3600=3900mm,按翼缘厚度考虑时,,此时的,此种情况不起控制作用,故取。梁内纵向钢筋选用HRB400级钢筋,(),,下部跨间截面按单筋T形截面进行计算,由属于第一类T形截面。实配钢筋为4Ф20(,满足要求。将下部跨间截面的4Ф20钢筋伸入支座,作为支座负弯矩作用下的受压钢筋(1256)再计算相应的受拉钢筋,即支座A上部137 说明富裕,且达不到屈服,可近似取实配钢筋4Ф25(,支座的上部实配4Ф25(,,满足抗震构造要求。9.1.2梁斜截面受剪承载力计算以一层AB跨为例说明计算方法时,属厚腹梁。对于AB跨:截面满足要求验算是否需要计算配置箍筋:可知,按构造要求配筋。梁端加密区箍筋取4肢φ8@100137 级钢筋(),则加密区长度取0.85m,非加密区箍筋取4肢φ8@150,箍筋设置满足要求。BC跨:若梁端箍筋加密区取4肢φ10@100,则其承载力为由于非加密区长度较小,故全跨可按加密区配置。表9-1框架梁纵向钢筋计算表层次截面M/KN.mεAs//mm2As/mm2实配钢筋As/mm2As//Asρ/%10支座A-98.69<07637654Ф18(840)0.850.56Bl-90.20<07636994Ф18(840)0.850.56AB跨间59.140.016213Ф18(763)0.47支座Br-52.794524093Ф18(763)0.760.58BC跨间51.270.023624Ф12(763)0.448支座A-176.67<0101710184Ф20(1256)0.80.74Bl-156.19<010179824Ф20(1256)0.80.74AB跨间99.210.017544Ф18(1017)0.60支座Br-97.86<08047644Ф18(1017)0.790.73BC跨间90.220.0357714Ф16(804)0.583支A-308.12<0125616154Ф25(1964)0.641.16137 座Bl-283.07<0125614844Ф25(1964)0.641.16AB跨间189.250.01410654Ф20(1256)0.74支座Br-167.91<0100511034Ф20(1256)0.80.90BC跨间143.330.05610025Ф16(1005)0.721支座A-315.38<0125616524Ф25(1964)0.641.16Bl-293.68<0125615394Ф25(1964)0.641.16AB跨间219.460.01510794Ф20(1256)0.70支座Br-274.10<0152017714Ф20(1964)0.861.40BC跨间153.940.03611724Ф22(1520)1.09表9-2框架梁箍筋数量计算表层次截面REV/KN0.25βcfcbho/KN公式见下式(1)梁端加密区实配钢筋(Asv/s)10A、Bl92.09707.66>REV-0.13<0双肢φ8@100(1.01)Br87.33582.41>REV-0.04<0双肢φ8@100(1.01)8A、Bl124.73707.66>REV0.08双肢φ8@100(1.01)Br142.26582.41>REV0.42双肢φ8@100(1.01)3A、Bl169.81707.66>REV0.39双肢φ8@100(1.01)Br248.54582.41>REV1.23三肢φ10@100(2.36)1A、Bl194.34809.36>REV0.24四肢φ8@100(2.01)Br319.81666.11>REV0.57四肢φ10@100(3.14)续表9-2框架梁箍筋数量计算表层次截面REV/KN0.25βcfcbho/KN公式见下式(1)非加密区实配钢筋(ρsv%)10A、Bl92.09707.66>REV-0.13<0双肢φ8@150(0.223)Br87.33582.41>REV-0.04<0双肢φ8@100(0.340)8A、Bl124.73707.66>REV0.08双肢φ8@150(0.340)Br142.26582.41>REV0.42双肢φ8@100(0.340)137 3A、Bl169.81707.66>REV0.39双肢φ8@150(0.223)Br248.54582.41>REV1.23三肢φ8@100(0.590)1A、Bl194.34809.36>REV0.24四肢φ8@150(0.223)Br319.81666.11>REV0.57四肢φ10@100(0.520)注:(1)9.2框架柱设计9.2.1剪跨比和轴压比验算根据<<抗震规范>>,对于二级抗震和剪跨比大于2,轴压比小于0.8,下表给出了剪跨比和轴压比的计算结果。其中剪跨比也可取,表中的,和都不应考虑承载力抗震调整系数。由表可见,各柱的剪跨比和轴压比均符合要求。表9-3柱的剪跨比和轴压比验算柱号层次b/mmho/mmfc/N/mm2Mc/KN.mVc/KNN/KNMc/VchoN/fcbhoA柱1060054016.7148.7587.90329.003.13>20.06<0.8860054016.7187.81126.291131.092.75>20.21<0.8370066016.7283.53201.103345.932.15>20.43<0.8180076019.1809.20237.854342.164.47>20.37<0.8B柱1060054016.7180.34100.16396.133.33>20.07<0.8860054016.7301.64171.091390.743.26>20.26<0.8370066016.7569.83318.434773.902.71>20.62<0.8180076019.1822.09283.556224.703.81>20.54<0.8137 9.2.2正截面承载力计算以第一层A柱为例进行计算。根据A轴内力组合表,将支座中心处弯矩换算至支座边缘,并与柱端组合弯矩的调整值比较后,选出最不利内力,进行配筋计算。A节点右梁端弯矩-536.13+178.580.8/2=-464.69KN.mA节点上、下柱端弯矩-399.6+70.05(0.6-0.1)=-364.58KN.m(上柱)118.69-139.710.1=104.72KN.m(下柱)在轴线处将其分配给上、下柱端一层柱底一层柱顶137 取20和偏心方向尺寸的1/30两者中的较大值,即800/30=26.67,故取=26.67。柱的计算长度按公式确定,式中分别柱的上端和下端节点处交会的各柱线刚度之和与交汇的各梁的线刚度之和的比值,即因为,故考虑偏心距增大系数。取取=1.39对称配筋:为大偏心受压情况。137 =1002mm2再按及相应的一组计算。73.45-38.770.1=69.66KN.m41.8-12.11(0.6-0.1)=40.75KN.m此组内力为非地震组合情况,且无水平荷载效应,故不必进行调整。取取取故考虑偏心距增大系数=1.89为小偏心受压。137 按上式计算时故可按构造配筋,且应满足,故选4Ф20()总配筋率,满足要求。表9-4其它层柱的配筋计算柱A柱层次13810截面800800700700600600600600组合一二一二一二一二M(KN.m)809.2023.41258.3269.56123.2856.0986.2981.87N(KN)3473.734308.932767.703342.86904.871229.14274.93371.49e0(mm)232.955.4393.3320.81136.2445.63313.86220.38ea(mm)26.6726.6723.3323.3320202020l0(m)11.035.8011.614.508.704.505.764.50ei(mm)259.6232.10116.6644.14156.2465.63343.86240.38l0/h13.797.2516.596.4314.507.509.607.50ξ11.01.01.01.01.01.01.01.0ξ21.01.00.981.01.01.01.01.0η1.391.892.081.441.541.341.111.10e(mm)720.90420.67552.65373.56500.61347.94641.68524.42137 ξ0.2990.3580.1610.049As=As’10021280147980<0720<0720实配单侧4Ф25(1964)4Ф20(1256)4Ф18(1017)2Ф16+2Ф18(1256)ρs0.97%>0.8%0.81%>0.8%0.91%>0.8%0.813%>0.8%偏心判断大小大小大小大小续表9-4其它层柱的配筋计算柱B柱层次13810截面800800700700600600600600组合一二一二一二一二M(KN.m)822.0422.17458.4167.67220.5952.2098.4578.46N(KN)1680.475290.731401.964025.82769.121389.34254.92435.85e0(mm)489.174.19326.9816.81286.8137.57386.20180.02ea(mm)26.6726.6723.3323.3320202020l0(m)7.985.806.274.505.384.504.324.50ei(mm)515.8430.86350.3140.14306.8157.57406.20200.02l0/h9.987.258.966.438.977.507.207.50ξ11.01.01.01.01.01.01.01.0ξ21.01.01.01.01.01.01.01.0η1.101.921.121.491.111.391.051.11e(mm)927.42419.25702.35369.81600.56340.02686.51482.02ξ0.1450.1820.1370.045As=As’14301280556980325720196720实配单侧4Ф25(1964)4Ф20(1256)4Ф18(1017)2Ф16+2Ф18(911)ρs0.97%>0.8%0.81%>0.8%0.91%>0.8%0.813%>0.8%偏心判断大小大小大小大小137 9.2.3柱斜截面受剪承载力计算仍以第一层柱为例进行计算,由前可知,上柱柱端弯矩设计值对二级抗震等级,柱底弯矩设计值则框架柱的剪力设计值满足要求取=3.0其中取较大的柱下端值,而且、不应考虑,故为将表中查得的值除以0.8,为表4-14查得值除以0.85。与相应轴力N=4231.08KN>0.3fcbh=0.319.18002/1000=3667.2KN取N=0.3fcbh=3667.2KN=故该层柱应按构造配置箍筋。柱端加密区的箍筋选用四肢φ10@100137 ,查表9-3一层柱的轴压比n=0.37,查表得最小配筋率特征值=0.087,则最小体积配箍率≥取φ10,,则,根据构造要求,取加密区箍筋为4φ@100,加密区位置及长度按规范要求确定。非加密区还应满足=200,故箍筋取4φ10@200。各层箍筋计算结果见表9-5。表9-5框架柱箍筋数量表柱号层次V/KN0.2bh0/KNN/KN0.3fcA/KNA柱1070.391122.44>V308.441803.608101.031122.44>V1090.811803.603160.881543.08>V3293.212454.901191.312322.56>V4231.083667.20B柱1080.131122.44>V282.201803.608136.871122.44>V922.531803.603254.741543.08>V1699.542454.901226.842322.56>V1989.463667.20续表9-5框架柱箍筋数量表柱号层次/s/mm(%)实配箍筋加密区非加密区A柱10<00.6364φ8@100(0.82)4φ8@150(0.56)8<00.6364φ8@100(0.73)4φ8@150(0.52)3<00.7164φ10@100(0.82)4φ10@150(0.62)1<00.7904φ10@100(0.92)4φ10@150(0.66)137 B柱10<00.6364φ8@100(0.82)4φ8@150(0.56)8<00.6364φ8@100(0.73)4φ8@150(0.52)3<00.6364φ10@100(0.82)4φ10@150(0.62)1<00.7274φ10@100(0.92)4φ10@150(0.66)9.3框架梁柱节点核芯区截面抗震验算以一层中节点为例,由节点两侧梁的受弯承载力计算节点核芯区的剪力设计值。因节点两侧梁不等高,计算时取两侧梁的平均高度,即本框架为二级抗震等级,按下式计算节点的剪力设计值,其中为柱的计算高度,取节点上下柱反弯点间的距离,即∑Mb=466.87+441.39=908.26KN.m(左震)剪力设计值因,,故取,1.5,则137 (满足要求)节点核芯区的受剪承载力计算,其中N取二层柱底轴力N=1945.25KN和0.5fcA=0.5×16.7×700×700=4091.5KN二者中的较小值,故取N=1945.25KN。设节点区配箍为4φ10@100,则=2668.42KN>Vj=1931.11KN故承载力满足要求。表9-6其它框架柱节点核芯区截面抗震验算层次13节点边节点中节点边节点中节点hb(mm)600550600550hb0(mm)555515555515Hc(m)3.273.763.313.60Mb(KN.m)536.13908.26496.80906.76Vj(KN)996.271931.11926.471910.14bj=bc(mm)800800700700hj(mm)800800700700公式(1)6471.523682.353682.352705.4配箍4φ10@1004φ10@1004φ10@1004φ10@100公式(2)2861.652668.422151.392012.39结论合格合格合格合格续表9-6其它框架柱节点核芯区截面抗震验算层次810137 节点边节点中节点边节点中节点hb(mm)600550600550hb0(mm)555515555515Hc(m)3.423.532.342.01Mb(KN·m)300.47389.53157.64161.25Vj(KN)565.53816.97255.07270.59bj=bc(mm)600600600600hj(mm)600600600600公式(1)2705.402705.402705.402705.40配箍4φ8@1004φ8@1004φ8@1004φ8@100公式(2)1417.121390.871382.851331.39结论合格合格合格合格注:(1))(2)137 第10章罕遇地震作用下弹塑性变形验算10.1罕遇地震作用下的楼层剪力仍以轴线的框架进行计算,查得罕遇地震的地震影响系数最大值,多遇地震作用下,罕遇地震与多遇地震影响系数之比为0.72/0.12=6,按照抗震设计规范,计算7度罕遇地震作用时,特征周期比多遇地震作用时不增加,罕遇地震时的楼层剪力为多遇地震作用下的楼层剪力乘以6,计算结果见表10-1,表中Vi表示轴线的框架的层间剪力。表10-1罕遇地震作用下的楼层剪力层次多遇地震作用下的楼层剪力/KN罕遇地震作用下的楼层剪力/KN101194.437166.58109725/796687=0.138988.9991906.6311439.78109725/796687=0.1381578.6982604.7315628.38109725/796687=0.1382156.7273188.8619133.16131898/944235=0.1392659.5163735.0022410.00131898/944235=0.1393114.9954205.1525230.90131898/944235=0.1393507.1044585.0727510.42131898/944235=0.1393823.9534884.3629306.16131898/944235=0.1394073.5625097.9730587.82132648/969000=0.1374190.5315211.9531271.70101320/810150=0.1253908.96137 10.2楼层受剪承载力计算1楼层实际截面承载力按框架梁、柱的实际配筋面积及材料强度标准值(柱还包括对应于重力荷载代表值的轴向力),分别由梁:柱:)(式中:、为纵向受拉钢筋的实配截面面积和强度标准值)。下面以框架底层为例进行计算说明,其中。AB跨梁:左端同理得右端BC跨梁:左(右)端A柱:)+137 其中B柱:其余各层的计算结果见图。2第i层柱的相对弹性线刚度k(i)式中为标准层柱的线刚度。此处以810层为标准层,设其相对线刚度为1,则2~7层,,1层的k值为3楼层受剪承载力计算采用简化的柱底塑性铰法计算。由式计算出各柱的受剪承载力,进而按式确定楼层受剪承载力。计算过程见表10-2。表10-2楼层受剪承载力计算(KN)楼层12345137 边柱328.71475.15384.86365.03341.28A、D343.75500.17409.87390.04366.29中柱374.54542.15438.63415.33389.73B、C397.23579.89470.78447.54421.94Vy(i)1444.232097.361704.141617.941549.24续表10-2楼层受剪承载力计算(KN)楼层678910边柱315.77288.47211.28187.18146.19A、D313.49313.49219.72195.63149.06中柱331.41331.41242.71215.29160.88B、C393.97363.62257.26229.84122.53Vy(i)1412.281296.77930.97827.94628.65137 图10-1罕遇地震下楼层正截面承载力4薄弱层弹塑性层间变形验算由式εy(i)=Vyi/Ve算出楼层的屈服强度系数,见表10-3。把这些值分别代入式底层顶层一般层表10-3楼层屈服强度系数楼层123456789100.3320.4930.4180.4230.4330.4530.4880.4320.5240.635以判断薄弱楼层的位置。经计算选出1,3,8层为薄弱层,其弹性层间位移计算结果见表10-4。由表可见,最大层间弹塑性位移角发生在第三层,其,故满足规范要求。表10-4弹塑性层间位移验算楼层Vei/KNDi/(N/mm)Δuei/mmηpΔupi/mmθp=Δupi/hi82156.7210972519.661.9137.559634073.5613189830.881.3340.148913908.9610132038.591.6161.7493137 137 第11章楼梯设计本楼梯标准层层高3.6m,底层层高4.2m,标准层踏步尺寸为150㎜×300㎜,采用C20混凝土,板采用HPB235钢筋,平台梁采用HRB335钢筋。楼梯上均布活荷载标准值qk=2.0kN/㎡。137 11.1梯段板设计取板厚h=100mm,板倾斜角tanα=175/300=0.58,cosа=0.865,取1m宽板带计算。1荷载计算恒荷载分项系数γG=1.2,活荷载分项系数γQ=1.4表11-1梯段板的荷载计算表恒荷载荷载种类荷载标准值(KN/m)水磨石面层(0.3+0.175)×0.65/0.3=1.03KN/m三角形踏步0.5×0.3×0.175×25/0.3=2.19混凝土斜板0.1×25/0.865=2.59板底抹灰0.02×17/0.865=0.39小计6.5活荷载2.0总荷载设计值P=1.2×6.5+1.4×2.0=10.6KN/m2截面设计板水平计算跨度Ln=3.6m。弯矩设计值M=1/10pln2=0.1×10.6×3.62=13.74KN.m板的有效高度h0=100-20=80mm查表得:=0.872选配φ12@110,As=1028mm2。137 分布筋每级踏步1根φ10筋。11.2平台板设计设平台板厚h=100mm,取1m宽板带计算。1荷载计算表11-2平台板的荷载计算恒荷载荷载种类荷载标准值(KN/m)水磨石面层0.65100厚混凝土板0.1×25=2.5板底抹灰0.02×17=0.34小计3.49活荷载2.0基本组合的总荷载设计值2截面设计平台板的计算跨度=1.76-0.2-0.2/2=1.56mm。弯矩设计值M==1/10×6.99×1.562=1.7KN.m板的有效高度h0=100-20=80mm=0.986则137 选配φ6/8@150的钢筋,AS=281mm2。11.3平台梁设计设平台梁的截面尺寸为200×3501荷载计算表11-3平台梁的荷载计算恒荷载荷载种类荷载标准值(KN/m)梁自重0.2×(0.35-0.07)×25=1.4梁侧粉刷0.2×(0.35-0.07)×25=0.19平台板传来3.49×1.76/2=3.07梯段板传来6.5×3.9/2=12.68小计17.34活荷载2.0×(3.9/2+1.76/2)=5.66KN总荷载设计值p=1.2×17.34+1.4×5.66=28.73kN/m2截面设计计算跨度:l0=1.05ln=1.05×(3.9-0.24)=3.84m弯矩设计值:M=1/8pl02=1/8×28.73×3.842=52.96KN.m剪力设计值:V=1/2pln2=1/2×28.73×3.84=55.16KN截面按倒L形计算,bf+5hf=200+5×100=700mm。梁的有效高度h0=350-35=315mm。经判别属于第一类T形截面as=52.96×106/1.0×9.6×700×3152=0.079查表得=0.959137 选配3Φ18的钢筋,As=763mm2。配置φ6@200的箍筋,则斜截面受剪承载力为:Vcs=0.7ftbh0+1.25fyvAsvh0/s=07×1.1×200×315+1.25×210×56.6×315/200=71910N>55160N满足要求137 第12章雨篷的设计及承载力验算12.1雨篷板的设计及承载力的验算1材料C10混凝土fc=19.1N/mm2ft=1.71N/mm2采用HRB335钢筋,fy=300N/mm22尺寸雨篷板悬挑长度=1.5m=1500mm雨篷板宽b=7.8m=7800mm根部板厚按/10计h=1500/10=150mm端部板厚hd=80mm雨篷梁尺寸b×h=300×450mm23荷载及内力计算20mm水泥沙浆面层1.2×0.02×20×1.0=0.48KN/m板自重(按平均厚度计)1.2×0.115×25=3.45KN/m板上厚20mm水泥沙浆1.2×0.02×17×1.0=0.408KN/m恒载设计值总计4.27KN/m137 施工或检修荷载1.4×1.0=1.4KN/m板的计算跨度=1.05=1.58m板固定端弯矩为4正截面受弯承载力计算雨篷板根部有效高度h=150-20=130mm。则查得可得选φ6@110(AS=257mm2),将钢筋置于板上部,并伸入支座内。分布钢筋采用φ6@200。12.2雨篷梁设计及承载力计算设梁的截面尺寸为300×450mm,ho=450-35=415mm。(两边伸入墙内30mm)1荷载计算梁上砌块自重1.2×3.9×2.49=11.65KN/m梁自重1.2×0.3×(0.45-0.07)×25=3.42KN/m梁侧粉刷1.2×0.03×(0.45-0.07)×2×17=0.47KN/m雨篷板传来均布荷载4.27×1.2=5.124KN/m总计20.66KN/m137 雨篷板传来的集中荷载1.4KN/m雨篷梁的计算跨度取雨篷板均布荷载对雨篷梁产生的力矩为雨篷板集中荷载对雨篷产生的力矩为扭矩(为为雨篷梁的净跨度)。2截面承载力计算雨篷梁为承受弯、剪、扭共同作用构件。(1)验算截面尺寸受扭塑性抵抗矩故截面尺寸满足要求。137 但要按计算配置钢筋。(2)确定是否考虑剪力的影响=0.175×1.71×15.75×106=4.71KN.m=0.351.71300415=74.51KN.m故进行剪扭承载力计算。(3)计算箍筋剪扭构件混凝土受扭承载力降低系数故取=1.0a.计算单侧受剪箍筋的数量则b.计算单侧受扭箍筋的数量137 则所需单肢箍筋总面积选φ10的HPB235级钢筋,其截面面积78.5mm2,所以箍筋间距为S=78.5/0.698=112.5mm,取S=110mm。(4)计算受扭纵筋(5)计算受弯纵筋查表得=0.805梁底受剪和受扭纵筋截面面积为选用4Φ28(As=2463mm2)。梁侧受剪和受扭纵筋截面面积为137 选用2Φ16(As=402mm2)。梁顶所需受扭纵筋截面面积为选用2Φ12(As=226mm2)。12.3雨蓬抗倾覆验算计算倾覆力矩的支点,取1倾覆力矩的计算雨篷板产生的力矩(恒载+集中荷载)=53.09KN.m2抗倾覆力矩抗倾覆不满足,但是由于是整体现浇框架,由于此建筑方案为框架结构,产生抗倾覆能力不满足要求的原因是因为框架填充墙自重太小所造成。要解决此问题,可将升高至底层框架顶,使雨蓬梁与底层框架梁相连,一起现浇。因为框架梁上有与雨蓬受力相反的楼板作用力,相互抵消,可以满足雨蓬满足使用要求。137 第13章楼板的设计13.1楼板类型的选择考虑民用建筑中楼板的最小板厚为80mm,考虑刚度要求,板厚在顶层和其它层均取100mm厚。楼板采用C30混凝土,板中钢筋采用HPB235级钢筋。13.2设计参数对于楼板,根据塑性理论时,在荷载作用下两个正交方向受力且都不可忽略。在本方案中,故属于双向板,设计时按塑性铰线法计算。1设计荷载对于1~9层楼面,活荷载恒荷载对于顶层屋面活荷载恒荷载2计算跨度137 内跨:(为轴线长,b为梁宽)边跨:13.3弯矩计算首先假定边缘板带跨中配筋率与中间板带相同,支座截面配筋率不随板带而变,取同一数值,跨中钢筋在离支座l1/4处间隔弯起。取(其中n为长短跨比值)取,然后利用下式进行连续运算:对于1-9层楼面,A区板格:l01=lc-250+50-b/2=3900-250+50-150=3550mml02=lc-250+50-b/2=7200-250+50-150=6850mm将上述各值代入公式137 =(和C的相等)(和B的相等)对其它区格板,亦按同理进行计算,详细计算过程从略,所得计算结果见表13-1。表13-1按塑性铰线法计算弯矩表(1~9层楼面)区格ABCDl01(m)3.553.601.801.80l02(m)6.856.853.553.60M15.96m1u5.95m1u3.10m1u3.15m1uM20.78m1u0.79m1u0.29m1u0.29m1uM1I-13.7m1u-13.7m1u-7.10m1u-7.20m1uM1II-13.7m1u-13.7m1u-7.10m1u-7.20m1uM2I-2.08m1u-2.11m1u-3.6m1u-3.6m1uM2II-2.08m1u-2.11m1u-3.6m1u-3.6m1um12.892.960.620.62m20.850.870.130.13m1I0-5.92-1.24-1.24137 M1II-5.780-1.24-1.24M2I0-1.740-0.26m2II-1.70-1.74-0.26-0.26同理,对顶层屋面计算弯矩见下图表13-2按塑性铰线法计算弯矩表(顶层屋面)区格ABCDl01(m)3.553.601.801.80l02(m)6.856.853.553.60M15.96m1u5.95m1u3.10m1u3.15m1uM20.78m1u0.79m1u0.29m1u0.29m1uM1I-13.7m1u-13.7m1u-7.10m1u-7.10m1uM1II-13.7m1u-13.7m1u-7.10m1u-7.20m1uM2I-2.08m1u-2.11m1u-3.6m1u-3.6m1uM2II-2.08m1u-2.11m1u-3.6m1u-3.6m1um13.493.580.750.75m21.021.050.160.16m1I0-7.16-1.50-1.50M1II-6.980-1.5-1.50M2I0-2.100-0.32m2II-2.04-2.10-0.32-0.3213.4截面设计受拉钢筋面积按公式计算,取0.9。对于四边都与梁整浇的板,中间跨截面及中间支支座处截面,其弯矩设计值减小20%。钢筋的配置:符合内力计算的假定,全板均匀布置。以第一层A区格板l1方向为例,截面有效高度h01=h-20=100-20=80mm。137 配筋φ8@200,实有As=50.3×1000/200=251.5mm2对于1~9层楼面,各区格板的截面计算与配筋见下表。表13-3按塑性塑性铰线法计算的截面计算与配筋表项目h0(mm)m(KN·m)As(mm2)配筋实有As跨中A区格L01方向812.89188.77φ8@200251.50L02方向730.8561.61φ6@30094.33B区格L01方向812.96193.35φ8@200251.50L02方向730.8763.06φ6@30094.33C区格L01方向810.6240.49φ6@30094.33L02方向730.139.42φ6@30094.33D区格L01方向810.4932.39φ6@30094.33L02方向730.117.54φ6@30094.33支座A-B81-1.74113.65φ6@200141.50A-C81-5.78377.56φ8@100503.00B-B81-1.74113.65φ6@200141.50B-D81-5.92386.71φ8@100503.00B-C81-5.92386.71φ8@100503.00C-D81-0.127.84φ6@30094.33同理,对顶层屋面,各区格板的截面计算与配筋见下表。表13-4按塑性塑性铰线法计算的截面计算与配筋表项目h0(mm)m(KN·m)As(mm2)配筋实有As跨中A区格L01方向813.49227.97φ8@200283.00L02方向731.0273.93φ6@30094.33B区格L01方向813.58233.85φ8@200283.00L02方向731.0576.10φ6@30094.33C区格L01方向810.7548.99φ6@30094.33137 L02方向730.1610.45φ6@30094.33D区格L01方向810.6039.18φ6@30094.33L02方向730.138.36φ6@30094.33支座A-B81-2.10137.17φ6@200141.50A-C81-6.98455.94φ8@100503.00B-B81-2.10131.17φ6@200141.50B-D81-2.20143.70φ8@100188.67B-C81-7.16467.70φ8@100503.00C-D81-0.3220.90φ6@30094.33137 第14章基础设计本设计采用桩基础,桩基础具有承载力高,稳定性好,沉降量小等特点,在计算中以B柱下桩基础为例进行计算。设计中拟采用预制桩基础,桩身采用方形截面,相应于荷载效应标准组合时柱底(标高为-1.05m)的荷载=3995.07KN,=470.41KN(作用于长边方向),=127.98KN,桩的方形截面尺寸取=400mm。取桩长为15m,由地质资料根据公式确定单桩的承载力特征值为,承台混凝土的强度等级取C25,配置HRB335级钢筋。对于C25的混凝土,HRB335级钢筋初选桩的根数为安全起见,并考虑采用矩形平台,初选6根桩。初选平台尺寸柱距纵向横向承台长边a=2(0.4+1.2)=3.2m承台矩边b=2(04+0.8)=2.4m暂取基础埋深为1.95m,承台高度为0.9m,桩顶伸入承台50mm,钢筋保护层厚取70mm,则承台有效高度为h0=0.9-0.07=0.83m。计算柱顶荷载137 取平台及其上的土的平均重度桩顶的平均竖向力==837.76KN<1.2Ra=1224.96KN和593.78KN>0满足要求。单桩水平力设计值,其值远小于由公式估算的单桩水平承载力特征值,满足要求。相应于荷载效应应基本组合时作用于柱底的荷载设计值扣除承台及其上填土的自重后的桩顶竖向力设计值===1063.7KN和743.3KN137 承台受冲切承载力验算柱边冲切受冲切承载力高度影响系数计算冲跨比与系数的计算==7480KN>5395.5KN角桩向上冲切,,,,137 ==1658.3KN>=1063.7KN承台受剪切承载力计算受剪切承载力截面高度影响系数计算对于I-I斜截面(介于0.3~3之间)剪切系数对于II-II斜截面取剪切系数>3N=3899=2697KN从以上计算可知,该承台高度首先取决于I-I137 截面的受剪承载力,其次取决于沿柱边的受冲切承载力。承台受弯承载力计算选用20Φ18,=5098mm2,沿平行y轴方向均布布置。选用16Φ25,=7854mm2,沿平行x轴方向均布布置。137 结论毕业设计是土木工程专业本科培养计划中最后一个主要环节,也是最重要的综合性实践环节,通过毕业设计这几个多月的学习与实践,培养了我综合运用所学基础课、技术基础课及专业课知识和相应技能,也培养我查阅相关资料的能力,解决具体的土木工程设计问题所需的综合能力。培养了我的综合素质,增强了工程意识和创新能力。本设计完全根据新规范进行设计的,采用了当今最前沿的设计方法。主要目的是以满足各种功能为主,具有较强的适用性,在设计的全过程中采用了新工艺,新方法等,综合考虑各种因素,务实地做到了经济,并宜就地取材,运用准确的设计方法,这是本设计中一个最显著的特点。建筑设计合理,能够满足公共建筑需要,体现出了时代精神,具有很好的适用性。结构设计考虑了便于施工,安全,经济合理等因素,具有较好的安全性,适用性及耐久性。从建筑设计到结构设计都运用了科学设计方法,合理的设计理论。在设计过程中,对结构设计进行了大量的理论验算,然后进行比较,取其合理的计算方法。总之,通过本次设计,使我们对四年的知识又系统的学习一遍,把所有的零散的东西拼凑起来,在脑海中对我们专业知识有了更系统、更鲜明的认识,这是我们从学校进入工作单位的一个前提,对我们以后的工作有很大帮助。137 致谢毕业设计是对大学知识的一次汇总,同时也是搭建学习与工作的桥梁,经过这半年时间对高层旅馆的毕业设计,我感觉自己获益匪浅!!这不仅让我加深了对所学知识的了解,巩固了对所学知识的掌握程度,同时也让我熟悉了一个结构设计所需的全过程,更为重要的是我学会了如何在一个群体当中如何互相帮忙,互相探讨,共同进步更好地与团体相处,为以后工作奠定一个坚实的基础!在这半年毕业设计过程中我所取得的收获离不开王涛老师和牟晓梅老师孜孜不倦地教诲,耐人寻味的讲解以及同组同学的热心的帮忙,感谢他们在这半年来对我的关心与帮助,能让我顺利地完成毕业设计,使我在学到知识的同时体会到许许多多团体的力量,使我终生难忘!在这里我对所有帮助我、关心我的老师和同学们表示衷心的感谢!!2005年6月15日邵宪彪137 参考文献[1]混凝土结构设计算例[M],中国建筑工业出版社,2003[2]周祥广,公共建筑平面设计方案精选,东南大学出版社,2001[3]房屋建筑学(第二版)[M],中国建筑工业出版社,2002[4]建筑材料编写组.建筑材料(第四版)[M],中国建筑工业出版社,2001[5]梁兴文,史庆轩,童岳生编著,钢筋混凝土结构设计,北京,科学技术文献出版社,1999[6]混凝土小型空心砌块墙体结构构造(含2003年局部修改版),2004[7]包世华,方鄂华,高层建筑结构设计.清华大学出版社,2001[8]建筑世界杂志社.办公建筑,天大出版社,黑龙江科学技术出版社,1999[9]混凝土结构的耐久性[M],中国建筑工业出版社,1999[10]王振东,张景岱,钢筋混凝土及砌体结构,中国建筑工业出版社,2001[11]赵西安,钢筋混凝土高层建筑结构设计,北京,中国建筑工业出版社,1996[12]黄兴棣,钢筋混凝土高层建筑结构设计实例,上海,上海科学技术出版社,1998[13]方鄂华,钱稼茹,叶列平,高层建筑结构设,中国建筑工业出版社,2001[14]李国强,李杰,建筑结构抗震设计.中国建筑工业出版社,2002[15]吴湘兴,建筑地基基础,广州,华南理工大学出版社,2002[16]郑刚,基础工程,第一版,北京,中国建材工业出版社[17]原长庆,高层建筑结构设计,黑龙江科学技术出版,1999[18]梁兴文,史庆轩,土木工程专业毕业设计指导,科学技术出版社,2002137 附录一:抗震设计的发展摘要:1抗震设计思路发展历程;2现代抗震设计思路;3保证结构延性能力的抗震措施;4常用抗震分析方法关键词:结构设计抗震1.抗震设计思路发展历程随着建筑结构抗震相关理论研究的不断发展,结构抗震设计思路也经历了一系列的变化。最初,在未考虑结构弹性动力特征,也无详细的地震作用记录统计资料的条件下,经验性的取一个地震水平作用(0.1倍自重)用于结构设计。随着地面运动记录的不断丰富,人们通过单自由度体系的弹性反应谱,第一次从宏观上看到地震对弹性结构引起的反应随结构周期和阻尼比变化的总体趋势,揭示了结构在地震地面运动的随机激励下的强迫振动动力特征。但同时也发现一个无法解释的矛盾,当时规范所取的设计用地面运动加速度明显小于按弹性反应谱得出的作用于结构上的地面运动加速度,这些结构大多数却并未出现严重损坏和倒塌。后来随着对结构非线性性能的不断研究,人们发现设计结构时取的地震作用只是赋予结构一个基本屈服承载力,当发生更大地震时,结构将在一系列控制部位进入屈服后非弹性变形状态,并靠其屈服后的非弹性变形能力来经受地震作用。由此,也逐渐形成了使结构在一定水平的地震作用下进入屈服,并达到足够的屈服后非弹性变形状态来耗散能量的现代抗震设计理论。由以上可以看出,结构抗震设计思路经历了从弹性到非线性,从基于经验到基于非线性理论,从单纯保证结构承载能力到允许结构屈服,并赋予结构一定的非弹性变形性能力的一系列转变。2.现代抗震设计思路在当前抗震理论下形成的现代抗震设计思路,其主要内容是:(1)合理选择确定结构屈服水准的地震作用。一般先以具有统计意义的地面峰值加速度作为该地区地震强弱标志值(即中震的),再以不同的R(地震力降低系数)得到不同的设计用地面运动加速度(即小震的)来进行结构的强度设计,从而确定了结构的屈服水准。137 (2)制定有效的抗震措施使结构确实具备设计时采用的R所对应的延性能力。其中主要包括内力调整措施(强柱弱梁、强剪弱弯)和抗震构造措施。   现代抗震设计理念是基于对结构非弹性性能的研究上建立起来的,其核心是关系,关系主要指在不同滞回规律和地面运动特征下,结构的屈服水准与自振周期以及最大非弹性动力反应间的关系。其中R为弹塑性反应地震力降低系数,简称地震力降低系数;而为最大非弹性反应位移与屈服位移之比,称为位移延性系数;T则为按弹性刚度求得的结构自振周期。  在滞回曲线为理想弹塑性及弹性刚度始终不变的前提下,通过对不同周期,不同屈服水准的非弹性单自由度体系做动力分析,得到了有关弹塑性反应下最大位移的规律:对T大于1.0秒的体系适用“等位移法则”即非弹性反应下的最大位移总等于同一地面运动输入下的弹性反应最大位移。对于T在0.12-0.5秒之间的结构,适用“等能量法则”即非弹性反应下的弹塑性变形能等于同一地震地面运动输入下的弹性变形能。当“等能量原则”适用时,随着R的增大,位移延性需求的增长速度比“等位移原则”下按与R相同的比例增长更快。由以上规律我们可以看出,如果以结构弹性反应为准,把结构用来做承载能力设计的地震作用取的越低,即R越大,则结构在与弹性反应时相同的地震作用下达到的非弹性位移就越大,位移延性需求就越高。这意味着结构必须具有更高的塑性变形能力。规律初步揭示出不同弹性周期的结构,当其弹塑性屈服水准取值大小不同时,在同一地面运动输入下屈服水准与所达到的最大非弹性位移之间的关系。也揭示出了延性能力和塑性耗能能力是屈服水准不高的结构在较大地震引起的非弹性动力反应中不致发生严重损坏和倒塌的主要原因。让人们认识到延性在抗震设计中的重要性。   137 之所以存在上诉的规律,我们应该注意到钢筋混凝土结构的一些相关特性。首先,通过人为措施可以使结构具有一定的延性,即结构在外部作用下,可以发生足够的非线性变形,而又维持承载力的属性。这样就可以保证结构在进入较大非线性变形时,不会出现因强度急剧下降而导致的严重破坏和倒塌,从而使结构在非线性变形状态下耗能成为可能。其次,作为非线弹性材料的钢筋混凝土结构,在一定的外力作用下,结构将从弹性进入非弹性状态。在非弹性变形过程中,外力做功全部变为热能,并传入空气中耗散掉。我们可以进一步以单质点体系的无阻尼振动来分析,在弹性范围振动时,惯性力与弹性恢复力总处于动态平衡状态,体系能量在动能、势能间不停转换,但总量保持不变。如果某次振动过大,体系进入屈服后状态,则体系在平衡位置的动能将在最大位移处转化为弹性势能和塑性变形能两部分,其中,塑性变性能将耗散掉,从而减小了体系总的能量。由此我们可以想到,在地震往复作用下,结构在振动过程中,如果进入屈服后状态,将通过塑性变性能耗散掉部分地震输给结构的累积能量,从而减小地震反应。同时,实际结构存在的阻尼也会进一步耗散能量,减小地震反应。此外,结构进入非弹性状态后,其侧向刚度将明显小于弹性刚度,这将导致结构瞬时刚度的下降,自振周期加长,从而减小地震作用。   随着对规律认识的深入,这一规律已被各国规范所接受。在抗震设计时,对在同一烈度区的同一类结构,可以根据情况取用不同的R,也就是不同的用于强度设计的地震作用。当R取值较大,即用于设计的地震作用较小时,对结构的延性要求就越严;反之,当R取值较小,即用于设计的地震作用较大时,对结构的延性要求就可放松。   目前,逐步形成了一套“多层次,多水准性态控制目标”的抗震理念。这一理念主要含义为:工程师应该选择合适的形态水准和地震荷载进行结构设计。建筑物的性态是由结构的性态,非结构构件和体系的性态以及建筑物内容物性态的组合。目前性态水准一般分为:损伤出现、正常运作、能继续居住、可修复的、生命安全、倒塌。性态目标指建筑物在一定程度的地震作用下对所期望的性态水准的表述。对建筑抗震设计应采用多重性态目标,对一般结构、必要结构、对安全起控制作用的结构分别建议了相应的性态目标――基本目标(常遇地震下完全正常运作,少遇地震下正常运作,罕遇地震下保证生命安全,极罕遇地震下接近倒塌)、必要目标(少于地震下完全正常运作,罕遇地震下正常运作,极罕遇地震下保证生命安全)、对安全其控制作用的目标(罕遇地震下完全正常运作,极罕遇地震下正常运作)。对重要性不同的建筑,如协助进行灾害恢复行动的医院等建筑,应该按较高的性态目标设计,此外,也可以针对甲方对建筑提出的不同抗震要求,选择不同的性态目标。   3.保证结构延性能力的抗震措施   合理选择了结构的屈服水准和延性要求后,就需要通过抗震措施来保证结构确实具有所需的延性能力,从而保证结构在中震、大震下实现抗震设防目标。系统的抗震措施包括以下几个方面内容:   (1)137 “强柱弱梁”:人为增大柱相对于梁的抗弯能力,使钢筋混凝土框架在大震下,梁端塑性铰出现较早,在达到最大非线性位移时塑性转动较大;而柱端塑性铰出现较晚,在达到最大非线性位移时塑性转动较小,甚至根本不出现塑性铰。从而保证框架具有一个较为稳定的塑性耗能机构和较大的塑性耗能能力。   (2)“强剪弱弯”:剪切破坏基本上没有延性,一旦某部位发生剪切破坏,该部位就将彻底退出结构抗震能力,对于柱端的剪切破坏还可能导致结构的局部或整体倒塌。因此可以人为增大柱端、梁端、节点的组合剪力值,使结构能在大震下的交替非弹性变形中其任何构件都不会先发生剪切破坏。   (3)抗震构造措施:通过抗震构造措施来保证形成塑性铰的部位具有足够的塑性变形能力和塑性耗能能力,同时保证结构的整体性。   这一系统的抗震措施理念已被世界各国所接受,但是对于耗能机构却出现了以新西兰和美国为代表的两种不完全相同的思路。首先,这两种思路都是以优先引导梁端出塑性铰为前提。   抗震研究者认为耗能机构宜采用符合塑性力学中的“理想梁铰机构”,即梁端全部形成塑性铰,同时底层柱底也都形成塑性铰的“全结构塑性机构”。其具体做法是通过结构分析得到各构件组合内力值后,对梁端截面就按组合弯矩进行截面设计;而对除底层柱底以外的柱截面,则用人为增大了以后的组合弯矩和组合轴力进行设计;对底层柱底截面则用增大幅度较小的组合弯矩和组合轴力进行截面设计。通过这一做法实现在大震下的较大塑性变形中,梁端塑性铰形成的较为普遍,底层柱底塑性铰出现迟于梁端塑性铰,而其余所有的柱截面不出现塑性铰,最终形成“理想梁铰机构”。为此,这种方法就必须取足够大的柱端弯矩增强系数。 另一方则认为取的柱弯矩增强系数过大,根据经验取了较小的柱弯矩增强系数,这一做法使结构在大震引起的非弹性变形过程中,梁端塑性铰形成较早,柱端塑性铰形成的相对较迟,梁端塑性铰形成的较普遍,柱端塑性铰形成的相对少一些,从而形成“梁柱塑性铰机构”。137    “理想梁铰机构”抗震措施的好处在于“理想梁铰机构”完全利用了延性和塑性耗能能力较好的梁端塑性铰来实现框架延性和耗散地震能量,同时因为除底层柱底外的其它柱端不出现塑性铰,也就不必再对这些柱端加更多的箍筋。但是这种思路过于受塑性力学形成理想机构概念的制约,总认为底层柱底应该形成塑性铰,这样就对底层柱底提出了较严格的轴压比要求,同时还要用足够多的箍筋来使柱底截面具有所需的延性,此外,底层柱底如果延性不够发生破坏很容易导致结构整体倒塌。这些不利因素使该方法丧失了很大的优势。   因此不需要被塑性力学的机构概念所限制,只要能在大震下实现以下的塑性耗能机构,就能保证抗震设计的基本要求:(1)以梁端塑性铰耗能为主;   (2)不限制柱端塑性铰出现(包括底层柱底),但是通过适当增强柱端抗弯能力的方法使它在大震下的塑性转动离其塑性转动能力有足够裕量。   (3)同层各柱上下端不同时处于塑性变形状态。   我国的抗震措施中对耗能机构的考虑也基本遵循了这一思路,采用了“梁柱塑性铰机构”模式。   抗震设计中我们为了避免没有延性的剪切破坏的发生,采取了“强剪弱弯”的措施来处理构件受弯能力与受剪能力的关系问题。值得注意的是,与非抗震抗剪破坏相比,地震作用下的剪切破坏是不同的。以梁构件为例,在较大地震作用下,梁端形成交叉斜裂缝区,该区混凝土受斜裂缝分割,形成若干个菱形块体,而且破碎会随着延性增长而加剧。由于交叉斜裂缝与塑性铰区基本重合,垂直和斜裂缝宽度都会随延性而增大。抗震下根据梁端的受力特征,正剪力总是大于负剪力,正剪力作用下的剪压区一般位于梁下部,但由于地震的往复作用,梁底的混凝土保护层可能已经剥落,从而削弱了混凝土剪压区的抗剪能力;交叉斜裂缝宽度比非抗震情况大,以及斜裂缝反复开闭,混凝土破碎更严重,从而使斜裂缝界面中的骨料咬合效应退化;混凝土保护层剥落和裂缝的加宽又会使纵筋的销栓作用有一定退化。可见,地震作用下,混凝土抗剪能力严重退化,但是试验发现箍筋的抗剪能力仍可以维持。   当地震作用越来越小时,梁端可能不出现双向斜裂缝,而出现单向斜裂缝,裂缝宽度发育也从大于非抗震情况到接近非抗震情况,抗剪环境越来越有利。此外,抗震抗剪要求结构构件应在大震下预计达到的非弹性变形状态之前不发生剪切破坏。因为框架剪切破坏总是发生在梁端塑性铰区,这就不仅要求在梁端形成塑性铰前不发生剪切破坏,而且抗剪能力还要维持到塑性铰的塑性转动达到大震所要求的程度,这就需要更多的箍筋。同时,在梁端塑性变形过程中作用剪力并没有明显增大,也进一步说明这里增加的箍筋不是用来增大抗剪强度,而是为了提高构件在发生剪切破坏时所达的延性。137    综上所述,与非抗震抗剪相比,抗震抗剪性能是不同的,其性能与剪力作用环境,塑性区延性要求大小有关。我们可以采取以下公式来考虑抗震抗剪的强度公式:   其中为混凝土抗剪能力,为箍筋抗剪能力,为由于地震作用导致的混凝土抗剪能力下降的折减系数,且随着剪力作用环境、延性要求而改变。我国的抗震抗剪强度公式也以上面公式为基础的,但是为设计方便,不同的烈度区取用了相同的公式,均取为0.6,与上面提到的混凝土抗剪能力随地震作用变化而不同的规律不一致,较为粗略。   延性对抗震来说是极其重要的一个性质,我们要想通过抗震措施来保证结构的延性,那么就必须清楚影响延性的因素。对于梁柱等构件,延性的影响因素最终可归纳为最根本的两点:混凝土极限压应变,破坏时的受压区高度。影响延性的其他因素实质都是这两个根本因素的延伸。如受拉钢筋配筋率越大,混凝土受压区高度就越大,延性越差;受压钢筋越多,混凝土受压区高度越小,延性越好;混凝土强度越高,受压区高度越低,延性越好(但如果混凝土强度过高可能会减小混凝土极限压应变从而降低延性);对柱子这类偏压构件,轴压力的存在会增大混凝土受压区高度,减小延性;箍筋可以提高混凝土极限压应变,从而提高延性,但对于高强度混凝土,受压时,其横向变形系数较一般混凝土明显偏小,箍筋的约束作用不能充分发挥,所以对于高强度混凝土,不适于用加箍筋的方法来改善其延性。此外,箍筋还有约束纵向钢筋,避免其发生局部压屈失稳,提高构件抗剪能力的作用,因此箍筋对提高结构抗震性能具有相当重要的作用。根据以上规律,在抗震设计中为保证结构的延性,常常采用以下措施:控制受拉钢筋配筋率,保证一定数量受压钢筋,通过加箍筋保证纵筋不局部压屈失稳以及约束受压混凝土,对柱子限制轴压比等。     按地震作用降低系数(“中震”的地面运动加速度与“小震”的地面运动加速度之比)来划分延性等级,“小震”取值越高,延性要求越低,“小震”取值越低,延性要求越高。对延性要求则并未按关系来取对应的,而是按抗震等级来划分,抗震等级实质又主要是由烈度分区来决定的。这就导致同一个R对应了不同的,从而制定了不同的抗震措施,这与关系是不一致的。137 另外,我国规定的“小震不坏,中震可修,大震不倒”的三水准抗震设防目标也存在一定的问题。该设防目标对甲类、乙类、丙类这三类重要性不同的建筑来说,并不都是恰当的。这种笼统的设防目标也不符合当今国际上的“多层次,多水准性态控制目标”思想,这种多性态目标思想提倡在建筑抗震设计中应灵活采用多重性态目标。甲类建筑指重大建筑工程和地震时可能发生严重此生灾害的建筑,乙类建筑指地震时使用不能中断或需要尽快修复的建筑,由于不同类别建筑的不同重要性,不宜再笼统的使用以上同一个性态目标(设防目标),此外,还应该考虑建筑所有者的不同要求,选择不同的设防目标,从而做到在性态目标的选择上更加灵活。   4.常用抗震分析方法   伴随着抗震理论的发展,各种抗震分析方法也不断出现在研究和设计领域。   在结构设计中,我们需要确定用来进行内力组合及截面设计的地震作用值。通常采用底部剪力法,振型分解反应谱法,弹性时程分析方法来计算该地震作用值,这三种方法都是弹性分析方法。其中,底部剪力法最简便,适用于质量、刚度沿高度分布较均匀的结构。它的大致思路是通过估计结构的第一振型周期来确定地震影响系数,再结合结构的重力荷载来确定总的水平地震作用,然后按一定方式分配至各层进行结构设计。对较复杂的结构体系则宜采用振型分解反应谱法进行抗震计算,它的思路是根据振型叠加原理,将多自由度体系化为一系列单自由度体系的叠加,将各种振型对应的地震作用、作用效应以一定方式叠加起来得到结构总的地震作用、作用效应。而对于特别不规则和特别重要的结构,常常需要进行弹性时程分析,该方法为直接动力分析方法。以上方法主要针对结构在地震作用下的弹性阶段,保证结构具有一定的屈服水准。   对结构抗震性能进行分析是抗震研究的一项重要内容,非线性时程分析,非线性静力分析是目前常用的几种抗震分析方法。其中针对结构非线性反应的非线性时程分析法(非线性动力反应分析),从建立在层模型或单列梁柱模型上的方法到建立在截面多弹簧模型上的方法,再到目前正在研究发展的建立在截面纤维滞回本构规律的纤维模型法,模拟的准确程度正在不断提高。其基本思路是通过一系列数值方法建立和求解动力方程从而得到结构各个时刻的反应量。但由于对地震特点和结构特性所做的假设,其结果存在不确定性,其主要价值是用来考察地震作用下普遍的而非特定的反应规律,以及对抗震设计后的结构进行校核分析,评估其抗震性能。非线性静力分析法137 是近年来得到广泛应用的一种结构抗震能力评估的新方法。这种方法从本质上说是一种静力非线性计算方法,但它将反应谱引入了计算过程和结果。其根本特征是用静力荷载描述地震作用,在地震作用下考虑结构的弹塑性性质。它的基本原理和步骤是先以某种方法得到结构在可能遭遇地震作用下所对应的目标位移,然后对结构施加竖向荷载的同时,将表征地震作用的一组水平静力荷载以单调递增的形式作用到结构上,在达到目标位移时停止荷载递增,最后在荷载中止状态对结构进行抗震性能评估,判断是否可以保证结构在该水平地震作用下满足功能需求。参考文献1.丰定国、王清敏、钱国芳、苏三庆编,工程结构抗震,地震出版社,20022.艾伦·威廉斯著,建筑与桥梁抗震设计,中国水利水电出版社,20023.王社良、曹照平,框架结构弹塑性性能试验研究,工程力学,1998:15(2)4.周锡元等,建筑结构的隔震、减震和振动控制,建筑结构学报,2002:25.周云、徐彤,耗能减震技术的回顾与展望,力学与实践,2000:20137 附录二:EarthquakeresistancedesigndevelopmentAbstract:1,earthquakeresistancedesignmentalitydevelopmentcourse;2,modernearthquakeresistancedesignmentality;3,guaranteestructureductilityabilityearthquakeresistancemeasure;4,commonlyusedearthquakeresistanceanalysismethodKeyword:Structuraldesignearthquakeresistance1.EarthquakeresistancedesignmentalitydevelopmentcourseAlongwiththeconstructionstructureearthquakeresistancecorrelationfundamentalresearchunceasingdevelopment,thestructureearthquakeresistancedesignmentalityhasalsoexperiencedaseriesofchanges.Atfirst,inhasnotconsideredthestructureelasticdynamiccharacteristic,alsodoesnothavethedetailedearthquakefunctionrecordingstatisticaldataunderthecondition,theexperiencetakesanearthquakelevelfunction(0.1timeofdeadweight)touseinthestructuraldesign.Isunceasinglyrichalongwiththegroundmovementrecording,thepeoplethroughthesingledegreeoffreedomsystemelasticreactionspectrum,firsttimefromonmacroscopicsawtotheearthquaketheresponsewhichcausestotheelasticstructurealongwiththestructurecycleandthedampingratiochangeoveralltendency,haspromulgatedthestructureinundertheearthquakegroundmovementstochasticdriveforcedoscillationdynamiccharacteristic.Butatthesametimealsodiscoveredthecontradictionwhichisunabletoexplain,atthattimethestandardtookthedesignobviouslywassmallerthanwiththegroundmovementaccelerationthefunctionwhichobtainedaccordingtotheelasticreactionspectruminthestructuregroundmovementacceleration,thesestructuresmajoritybynomeansappearedthemajordamageactuallyandcollapse.Afterwardsalongwithtostructurenon-linearperformance137 unceasingresearch,thepeoplediscoveredwhendesignstructuretakestheearthquakefunctiononlyisentrustswithstructurebasicallytosubmitthesupportingcapacity,whenhasabiggerearthquake,thestructurewillenterafteraseriesofcontrolsspotsubmitstheinelasticitystrainedcondition,andwilldependonittosubmittheafterinelasticdistortionabilitytoundergotheearthquakefunction.Fromthis,alsoformedgraduallyhascausedthestructuretoenterinunderthecertainlevelearthquakefunctionsubmits,afterandachievedenoughsubmittedtheinelasticitystrainedconditiontocomethediffusionenergythemodernearthquakeresistancedesigntheory.Mayseebyabove,thestructureearthquakeresistancedesignmentalityhasexperiencedfromtheelasticitytothenon-linearity,frombasedonexperiencestobasedonthenon-linearizedtheory,submitsfromthepureguaranteestructural-load-carryingcapacitytothepermissionstructure,andentrustswiththestructurecertaininelasticitydeformabilityabilityaseriesoftransformations.2.ModernearthquakeresistancedesignmentalityModernearthquakeresistancedesignmentalityformswhichunderthecurrentearthquakeresistancetheory,itsmaincontentis:(1)Thereasonablechoicedeterminationstructuresubmitsthestandardearthquakefunction.Firstbyhasthestatisticalsignificancethegroundpeakvalueaccelerationtotakegenerallythislocalearthquakestrongandtheweaksymbolizedthevalue(namelycentershakes),then(seismicforcestep-downratio)obtainsthedifferentdesignbydifferentR(namelyslightlytoshake)withthegroundmovementaccelerationcarriesonthestructuretheintensitydesign,thushaddeterminedthestructuresubmitsthestandard.(2)WhichformulatestheeffectiveearthquakeresistancemeasuretocausetheductilityabilitywhichthestructuretrulyhaswhendesigntouseRcorresponds.Mainlyincludestheendogenicforceadjustmentmeasure(strongcolumnweakbeam,strongcutsweakiscurved)andtheearthquakeresistancestructuremeasure.Themodernearthquakeresistancedesignideaisbasedonestablishestothestructureinelasticperformanceresearchin,itscoreistherelations,therelations137 mainlyrefersdifferentlyisstagnatingundertheruleandthegroundmovementcharacteristic,thestructuresubmitsthestandardandtheselfoscillationcycleaswellasthebiggestinelasticdynamicresponserelations.Ristheballplasticityrespondstheseismicforcestep-downratio,theabbreviationseismicforcestep-downratio;Butforthebiggestinelasticityrespondedthedisplacementwithsubmitsratioofthedisplacement,iscalledthedisplacementductilitycoefficient;Tthenforpressesthestructureselfoscillationcyclewhichtheelasticrigidityobtains.Instagnatesthecurvefortheidealballplasticityandtheelasticrigiditythroughoutundertheinvariablepremise,throughtothedifferentcycle,differentlysubmitsthestandardinelasticsingledegreeoffreedomsystemtomakethemechanicalanalysis,obtainedundertherelatedballplasticityresponsethebiggestdisplacementrule:Isbiggerthan1.0secondssystemstoTtobesuitable"theequipotentiallinetomovetheprinciple"isundertheinelasticresponsebiggestdisplacementalwaysisequaltoundertheidenticalgroundmovementinputtheelasticreactionbiggestdisplacement.RegardingTin0.12-0.5secondbetweenstructure,suitable"andsoontheenergyprinciple"isundertheinelasticresponseballplasticenergyofdeformationisequaltoundertheidenticalearthquakegroundmovementinputtheelasticenergyofdeformation.When"andsoontheenergyprinciple"issuitable,alongwithRincreasing,thedisplacementductilitydemandrateofrisecomparedto"andsoonthedisplacementprinciple"underaccordingtogrowsquicklywiththeRsameproportion.Wemayseebytheaboverule,iftakethestructureelasticreactionas,usesforthestructuretodothebearingcapacitydesigntheearthquakefunctiontotakelowly,namelyRbigger,thenstructureinwithelasticreactiontimethesameearthquakefunctionissuestoinelasticdisplacementbigger,thedisplacementductilitydemandishigher.Thismeantthestructuremusthavethehigherplasticdeformationability.Therulepromulgatesinitiallythedifferentelasticcyclethestructure,whenitsballplasticitysubmitsthestandardvaluesizenotatthesametime,submitsthestandardundertheidenticalgroundmovementinputwithbetweenthebiggestinelasticdisplacementrelationswhichachievedAlsopromulgatedtheductility137 abilityandtheplasticityconsumesenergytheabilityissubmitsthestandardnothighstructurenottosendintheinelasticdynamicresponsewhichthebigearthquakecausedhasthemainreasonwhichthemajordamageandcollapsed.Letthepeoplerealizetotheductilityintheearthquakeresistancedesignimportance.Thereasonthathastheappealtherule,weshouldnotetoreinforcedconcretestructuresomerelatedcharacteristics.First,mayenablethestructurethroughtheartificialmeasuretohavethecertainductility,namelythestructureunderexteriorfunction,mayhavetheenoughnon-lineardistortion,butalsomaintainsthesupportingcapacitytheattribute.Likethismayguaranteethestructurewhenentersthebignon-lineardistortion,cannotappearbecausetheintensitytodroptheseriousdestructionsuddenlywhichbutcausesandtocollapse,thuscausesthestructurepossiblytoconsumeenergyintounderthenon-linearitystrainedcondition.Next,tookthenon-lineelasticmaterialthereinforcedconcretestructure,underthecertainexogenicprocess,thestructurewillentertheinelasticconditionfromtheelasticity.Intheinelasticdistortionprocess,theexternalforceactingbecomestheheatenergycompletely,andspreadstointheairthediffusiontofall.Wemayfurthernothavethedampedoscillationbythesingleparticlesystemtoanalyze,whenelasticscopevibration,theforceofinertiaandtheelasticresiliencyalwaysareatthedynamicalequilibriumcondition,thesystemenergyinthekineticenergy,thepotentialenergydoesnotstopthetransformation,butthetotalquantitymaintenanceisinvariable.Ifsomevibrationoversized,afterthesystementerssubmitsthecondition,thensysteminpositionofequilibriumkineticenergyinbiggestdisplacementplacetransformationforelasticpotentialenergyandplasticenergyofdeformationtwoparts.Amongthem,theplasticdeformationperformancefallsthediffusion,thusreducedthesystemalwaysenergy.Maythinkfromthisus,underearthquakemutualaction,structureinvibrationprocess,afterifenterssubmitsthecondition,willfallthepartialearthquakesthroughtheplasticdeformationperformancediffusiontolosetothestructurethecumulativeenergy,thuswillreducetheearthquaketorespond.Atthesametime,theactualstructureexistencedampingalsocangoastepfurtherthediffusionenergy,reducesthe137 earthquaketorespond.Inaddition,afterthestructureenterstheinelasticcondition,itslateralrigidityobviouslywillbesmallerthantheelasticrigidity,thiswillcausethestructureinstantrigiditythedrop,theselfoscillationcycletolengthen,thuswillreducetheearthquakefunctionAlongwithtoruleunderstandingthorough,thisrulehasbeenacceptedbythevariouscountries"standard.Whenearthquakeresistancedesign,tointheidenticalintensityareaidenticalkindofstructure,mayusedifferentRaccordingtothesituation,alsoisdifferentusesintheintensitydesignearthquakefunction.WhenRvaluebig,namelyusesintheearthquakefunctionwhichdesignscomparingthehour,isstrictertothestructureductilityrequest;Otherwise,whenRvaluesmall,whennamelyusesinearthquakefunctionwhichdesignstobebig,tothestructureductilityrequestmayrelax.Atpresent,graduallyhasformedaset"multi-level,multi-standardsconditioncontrolgoal"earthquakeresistanceidea.Thisideamainmeaningis:Engineershouldchoosetheappropriateshapestandardandtheearthquakeloadcarriesonthestructuraldesign.Thebuildingconditionisbythestructurecondition,non-structuralunitandsystemconditionaswellasbuildingcontentsconditioncombination.Atpresenttheconditionstandardgenerallydividesinto:Thedamageappears,thenormaloperation,cancontinuetolive,mayrepair,thesafety,tocollapse.Theconditiongoalreferstothebuildinginunderthecertaindegreeearthquakefunctiontoconditionstandardindicatingwhichexpected.Shouldusethemultipleconditiongoaltotheconstructionearthquakeresistancedesign,Tothegeneralstructure,theessentialstructure,toweresafethecontrolactionthestructuretosuggestseparatelythecorrespondingconditiongoal-basicgoal(oftenmetunderearthquakecompletelynormallytooperate,littlemetunderearthquakenormaloperation,metunderearthquaketoguaranteerarelysafety,metunderearthquakeclosetocollapseextremelyrarely),theessentialgoal(isshortunderearthquakecompletelynormallyoperates,metunderrarelyearthquakenormaloperation,metunderearthquaketoguaranteeextremelyrarelysafety),(metunderearthquaketothesafeitscontrolactiongoalcompletelynormallyto137 operaterarely,metunderextremelyrarelyearthquakenormaloperation).Totheimportantdifferentconstruction,likeassiststocarryonthedisastertorestoremotionconstructionandsoonhospital,shouldaccordingtothehighconditiongoaldesign,inaddition,alsomayaimatthepartyofthefirstpartthedifferentearthquakeresistancerequestwhichproposedtotheconstruction,choosesthedifferentconditiongoal.3.GuaranteesthestructureductilityabilitytheearthquakeresistancemeasureAfterreasonablychosethestructuretosubmitthestandardandtheductilityrequest,neededthroughtheearthquakeresistancemeasuretoguaranteethestructurehadtheductilityabilitytrulywhichneeded,thustheguaranteestructureshook,underthebigquakeinthecenterrealizestheearthquakeresistancetogarrisonthegoal.Systemearthquakeresistancemeasureincludingfollowingseveralaspectscontent:(1)"Strongcolumnweakbeam":Artificialincreasesthecolumntobeoppositeinbeam"santi-curvedability,causesthereinforcedconcreteframeunderthebigquake,thebeamendplastichingeappearsearly,inachievedwhenbiggestnon-lineardisplacementtheplasticrotationisbig;Butthecolumnendplastichingeappearslate,inachievedwhenbiggestnon-lineardisplacementtheplasticrotationissmall,evensimplydoesnotappeartheplastichinge.Thustheguaranteeframehasastablerplasticitytoconsumeenergytheorganizationandthebigplasticityconsumesenergytheability.(2)"Strongcutsweaklyiscurved":Theshearingfailurebasicallydoesnothavetheductility,oncesomespothastheshearingfailure,thisspotonthoroughlywillwithdrawfromthestructureearthquakeresistanceability,alsopossiblywillcausethestructureregardingthecolumnendshearingfailurethepartorthewholecollapses.Thereforemayartificialincreasethecolumnend,beamsection,thepitchpointcombinationshearingforcevalue,enablesthestructuretheinelasticitytodistortunderthebigquakeitsanycomponentallnottobeableinturntohavetheshearingfailurefirst.(3)Earthquakeresistancestructuremeasure:Throughtheearthquakeresistance137 structuremeasureguaranteedformstheplastichingethespottohavetheenoughplasticdeformationabilityandtheplasticityconsumesenergytheability,atthesametimeguaranteesthestructuretheintegrity.Thissystemearthquakeresistancemeasureideahasbeenacceptedbythevariouscountries,butregardingconsumedenergytheorganizationtoappearactuallytakeNewZealandandUSasrepresentative"stwokindofquitesamenotlessthanmentalities.First,thesetwokindofmentalitiesallaretakefirstguidethebeamendtoleavetheplastichingeasthepremises.Resistsearthquakestheresearchertothinkconsumesenergytheorganizationsuitablytouseconformstointhemechanicsofplasticity"theidealbeamarticulationorganization",namelythebeamendformstheplastichingecompletely,atthesametimethefirstfloorbaseofcylinderalsoallformstheplastichinge"theentirestructureplasticorganization".Itsconcreteprocedureisobtainsvariouscomponentscombinationendogenicforcevalueafterthestructureanalysis,pressesthecombinationbendingmomenttothebeamendsectiontocarryonthesectiondesign;Buttobesidesfirstfloorbaseofcylindercolumnsection,thenchoosesapersonforajobforhasincreasedthelatercombinationbendingmomentandthecombinationaxlestrengthcarriesonthedesign;Tofirstfloorbaseofcylindersectionthenwithincreasesthescopesmallcombinationbendingmomentandthecombinationaxlestrengthcarriesonthesectiondesign.Throughthisprocedurerealizationinthebigquakebigplasticdeformation,thebeamendplastichingeformsiscommon,thefirstfloorbaseofcylinderplastichingeappearslateYuthebeamendplastichinge,butotherallcolumnssectiondoesnotappeartheplastichinge,finallyforms"theidealbeamarticulationorganization".Therefore,thismethodmusttaketheenoughbigcolumnendmomentenhancementcoefficient.Othersthoughttakescolumnbendingmomentenhancementcoefficientoversized,accordingtoexperiencedhastakenthesmallcolumnbendingmomentenhancementcoefficient,thisprocedurecausedthestructureintheinelasticdistortionprocesswhichthebigquakecaused,thebeamendplastichingeformed137 early,thecolumnendplastichingeformedrelativelylate,thebeamendplastichingeformediscommon,thecolumnendplastichingeformedrelativelyfewsomewhat,thusformed"thebeamcolumnplastichingeorganization"."Theidealbeamarticulationorganization"theearthquakeresistancemeasureadvantagelayin"theidealbeamarticulationorganization"tousetheductilityandtheplasticitycompletelyconsumesenergytheabilitygoodbeamendplastichingetorealizetheframeductilityandthediffusionearthquakeenergy,atthesametimebecausedidnotappeartheplastichingebesidesthefirstfloorbaseofcylinderothercolumnends,alsodidnotneedagaintothesecolumnendCanadamorestirrups.Butthiskindofmentalitytooreceivesthemechanicsofplasticitytoformtheidealorganizationconcepttherestriction,alwaysthoughtthefirstfloorbaseofcylindershouldformtheplastichinge,likethisproposedtothefirstfloorbaseofcylinderthestrictaxispressescomparedtotherequest,meanwhilemustuseenoughmanystirrupstocausetheductilitywhichthebaseofcylindersectionhasneeds,inaddition,firstfloorbaseofcylinderiftheductilityinsufficientlyhasthedestructiontobeveryeasytocausethestructurewholetocollapse.Thesedisadvantagefactorscausedthismethodtolosetheverybigsuperiority.Thereforedoesnotneedtolimitbythemechanicsofplasticityorganizationconceptbelow,solongascanrealizetheplasticityunderthebigquaketoconsumeenergytheorganization,canguaranteetheearthquakeresistancedesignthebasicrequest:(1)Consumesenergybythebeamendplastichingeprimarily.(2)Doesnotlimitthecolumnendplastichingetoappear(includingfirstfloorbaseofcylinder),butenableitthroughthesuitableenhancementcolumnendanti-curvedabilitymethodtohavetheenoughallowanceunderthebigquakeplasticrotationtoitsplasticrotationability.(3)Theendatthesametimeisnotattheplasticdeformationconditionwiththelevelvariouscolumnsabout.Inourcountry"searthquakeresistancemeasuretoconsumedenergytheorganizationconsiderationalsobasicallytofollowthismentality,hasused"the137 beamcolumnplastichingeorganization"thepattern.Inearthquakeresistancedesignweinordertoavoidnothavingtheductilityshearingfailuretheoccurrence,took"strongcutweaklyiscurved"themeasuretoprocessthecomponenttobendtheabilitywithtocuttheabilitytherelationalquestion.Theworthnotingis,withmustresistsearthquakesanti-cutsthedestructiontocompare,undertheearthquakefunctionshearingfailureisdifferent.Takethebeamcomponentastheexample,underthebigearthquakefunction,thebeamendformsalternatelytheslantingcrackarea,thisareaconcretetheslantingcrackdivision,isformedcertaindiamondsblockbody,moreoverdestroyscangrowalongwiththeductilitybutintensifies.Asaresultofalternatelytheslantingcrackandtheplastichingeareabasicsuperposition,verticalandtheslantingcrackopeningcanbutincreasealongwiththeductility.Undertheearthquakeresistanceactsaccordingtothebeamendthestresscharacteristic,thepositiveshearalwaysisbiggerthanthenegativeshear,underthepositiveshearfunctioncutspressestheareatobelocatedlowerpartgenerallybeam,butasaresultoftheearthquakemutualaction,thebreastconcreteprotectorpossiblyalreadyflaked,thusweakenedtheconcretetocutpressestheareaanti-tocuttheability;Alternatelyslantingcrackwidthrationon-earthquakeresistancesituationbig,aswellastheslantingcrackmakeandbreak,theconcreteisrepeatedlymoreserious,thuscausesintheslantingcrackcontactsurfacetheaggregatelinkingeffectdegeneration;Concreteprotectorflakingandthecrackwidencanenabletheverticalmusclethecotterfunctiontohavethecertaindegeneration.Obviously,undertheearthquakefunction,theconcreteanti-cutstheabilityseriousdegeneration,buttheexperimentaldiscoverystirrupanti-cuttheabilitystilltobepossibletomaintain.Whentheearthquakefunctionmoreandmorehour,thebeamendpossiblydoesnotappearthedoublesynclinecrack,butappearsthesinglesynclinecrack,thecrackopeninggrowthalsofromisbiggerthanthenon-earthquakeresistancesituationtoapproachthenon-earthquakeresistancesituation,anti-cutstheenvironmenttobemoreandmoreadvantageous.Inaddition,resistsearthquakes137 anti-cutstherequeststructuralunittobesupposedunderthebigquaketoestimateachievedinfrontofinelasticitystrainedconditiondoesnothavetheshearingfailure.Becausetheframeshearingfailurealwaysoccursinthebeamendplastichingearea,thisnotonlyrequestsinthebeamendtoforminfrontoftheplastichingenottohavetheshearingfailure,moreoveranti-cutstheabilityalsotohavetomaintaintotheplastichingeplasticrotationachievedthebigquakerequestsdegree,thisneedmorestirrups.Atthesametime,affectstheshearingforceinthebeamendplasticdeformationprocessnotobviouslytoincrease,alsofurtherexplainedhereincreasesthestirrupisnotusesfortoincreasetheshearingstrength,theductilitywhichbutisforenhancethecomponentwhentohaveshearingfailurereaches.Insummary,withmustresistsearthquakesanti-cutscompares,resistsearthquakesanti-cutstheperformanceisdifferent,itsperformanceandtheshearingforcefunctionenvironment,theplasticareaductilityrequestsizeconcerns.Belowwemayadopttheformulatoconsiderresistsearthquakesanti-cutstheintensityformula.Anti-cutstheabilityfortheconcrete,anti-cutstheabilityforthestirrup,forbecausetheearthquakefunctioncausestheconcreteanti-cutsthereductioncoefficientwhichtheabilitydrops,alsoalongwiththeshearingforcefunctionenvironment,theductilityrequestbutchanges.Ourcountry"searthquakeresistanceshearingstrengthformulaalsotakeaboveformulaasfoundation,butforthedesignconvenient,thedifferentintensityareahasusedthesameformula,takesis0.6,mentionstheconcretewiththeaboveanti-tocuttheabilitybutthedifferentrulenottobeinconsistentalongwiththeearthquakefunctionchange,issketchy.Theductilitytoresistsearthquakessaidisanextremelyimportantnature,wemustwanttoguaranteethestructurethroughtheearthquakeresistancemeasuretheductility,thenmustclearlyaffecttheductilityfactor.Regardingcomponentandsoonbeamcolumn,theductilityinfluencefactorfinallymayinduceintothemostbasictwopoints:Concretelimitpressurestrain,timedestructioncompressionzonealtitude.Theinfluenceductilityotherfactorsessenceallisthese137 twobasicfactorsextending.Likeispulledthesteelbarratioofreinforcementinabigway,concretecompressionzonehighlybigger,theductilityisworse;Compressionsteelmore,theconcretecompressionzonehighgoespastslightly,theductilityisbetter;Theconcretegoespasthigh,thecompressionzonehighgoespastlowly,ductilitybetter(butifconcreteintensityexcessivelyishighpossiblycanreduceconcretelimitpressurestrainthustoreduceductility).Tothepillarthiskindofbiascomponent,theaxialcompressionexistencecanincreasetheconcretecompressionzonealtitude,reducestheductility;Thestirrupmayenhancetheconcretelimitpressurestrain,thusenhancestheductility,whenregardinghighstrengthconcrete,bearing,itstransversestrainfactorcommonconcreteisobviouslysmall,thestirruprestraintfunctioncannotfullydisplay,thereforeregardingthehighstrengthconcrete,isnotsuitableforwithtoaddthestirrupthemethodtoimproveitsductility.Inaddition,thestirrupalsohastherestraintlongitudinalreinforcement,avoidsithavingthepartialbucklingtolosesteadily,enhancesthecomponentanti-tocuttheabilitythefunction,thereforethestirruptoenhancesthestructureearthquakeresistanceperformancetohavethequitevitalrole.Belowaccordingtotheaboverule,intheearthquakeresistancedesignfortheguaranteestructureductility,usesthemeasurefrequently:Thecontrolispulledthesteelbarratioofreinforcement,guaranteedthecertainquantitycompressionsteel,throughaddsthestirruptoguaranteetheverticalmusclenotpartialbucklinglosessteadyaswellastherestraintbearingconcrete,comparesandsoontothepillarlimitaxispressure.("Centershakes"groundmovementaccelerationaccordingtotheearthquakefunctionstep-downratiowith"slightlyshakes"ratioofthegroundmovementacceleration)todividetheductilityrank,"slightlyshakes"thevaluehigh,ductilityrequestlower,"slightlyshakes"thevaluelowly,theductilityrequestishigher.Toductilityrequestthenpressesbynomeansrelatestakesthecorrespondence,butisdividesaccordingtotheearthquakeresistancerank,theearthquakeresistancerankessencemainlyisdecidedbytheintensitydistrict.ThiscausedidenticalRtocorresponddifferently,thushasformulatedthedifferentearthquakeresistance137 measure,thiswiththerelationswasinconsistent.Moreover,ourcountrystipulates"slightlyshakesdoesnotgobad,centershakesmayrepair,bigquakenotbutactually"threestandardsearthquakeresistancesgarrisonsthegoalalsotohavethecertainproblem.Shouldgarrisonthegoaltothearmorclass,thesecondgradeclass,thethirdkindofthesethreekindofimportantdifferentconstructionsaidthat,notallisappropriate.Thiskindgeneralgarrisonsthegoalnottoconformtonowinternationalon"multi-level,themulti-standardsconditioncontrolgoal"thethought,thiskindofconditionsgoalthoughtadvocatedshouldnimblyusethemultipleconditiongoalintheconstructionearthquakeresistancedesign.Thearmorclassbuildingreferstowhenthesignificantarchitecturalengineeringandtheearthquaketheconstructionwhichpossiblyhasseriousthistolivethedisastertheconstruction,thesecondgradeclassbuildingreferstowhenearthquakeusescannotseverorneedsrepairassoonaspossibleabove,becausethedifferentcategoryconstructiondifferentimportance,notsuitablegeneralusestheidenticalindividualityconditiongoal(togarrisongoal),inaddition,butalsoshouldconsiderthebuildingowner"sdifferentrequest,choosesdifferentlygarrisonsthegoal,thusachievesintheconditiongoalchoiceismorenimble.4.CommonlyusedearthquakeresistanceanalysismethodIsfollowingtheearthquakeresistancetheorydevelopment,eachearthquakeresistanceanalysismethodalsounceasinglyappearsintheresearchandthedesigndomain.Inthestructuraldesign,weneedtodetermineusesfortocarryontheendogenicforcecombinationandthesectiondesignearthquakevirtualvalue.Usuallyusesthebaseshearingforcelaw,inspiresthedecompositionreactionspectralmethod,theelastictimeintervalanalysismethodcalculatesthisearthquakevirtualvalue,thesethreemethodsallaretheelasticityanalysismethods.Amongthem,thebaseshearingforcelawissimplest,issuitableforthequality,therigidityalongdistributestheevenstructurehighly.Itsapproximatementalityisfirstinspiresthecyclethroughtheestimatestructuretodeterminetheearthquakeinfluencecoefficient,therecombinationstructuregravityloaddeterminedthetotalhorizontalearthquakefunction,thencarriesonthestructural137 designaccordingtocertainwayassignmenttoeachlevel.Tocomplexstructuresystemthenusessuitablyinspiresthedecompositionreactionspectralmethodtocarryontheearthquakeresistancecomputation,itsmentalityisaccordingtoinspirestheprincipleofsuperposition,changesintothemulti-degreeoffreedomsystemaseriesofsingledegreeoffreedomsystemthesuperimposition,inspireseachkindthecorrespondingearthquakefunction,thefunctioneffectsuperimposesbycertainwayobtainsthestructuretotalearthquakefunction,thefunctioneffect.singledegreeoffreedomsystemthesuperimposition,inspireseachkindthecorrespondingearthquakefunction,thefunctioneffectsuperimposesbycertainwayobtainsthestructuretotalearthquakefunction,thefunctioneffect.Carriesontheanalysistothestructureearthquakeresistanceperformanceisresistsearthquakestheresearchanimportantcontent,thenon-lineartimeintervalanalysis,thenon-linearstaticanalysisisthepresentcommonlyusedseveralearthquakeresistanceanalysismethod.Inwhichinviewofthestructurenon-linearityresponsenon-lineartimeintervalanalyticmethod(non-lineardynamicresponseanalysis),fromtheestablishmentinthelevelmodeloronthesinglerowbeamcolumnmodelmethodtotheestablishmentonthesectionmulti-springmodelmethod,isstudyingthedevelopmenttoatpresenttheestablishmenttostagnatethisconstructionruleagaininthesectiontextilefiberthetextilefibermodeling,thesimulationaccuratedegreeisenhancingunceasingly。Thusitsbasicmentalityisobtainsthestructureeachtimereactingweightthroughaseriesofnumericalmethodestablishmentandthesolutionpowerequation.Butbecausedoestotheearthquakecharacteristicandthestructurecharacteristicbutthesupposition,itsresultexistenceuncertainty,itsmainvalueisusesfortoinspectundertheearthquakefunctionuniversalthenon-specificresponserule,aswellasafterresistsearthquakesthedesignthestructuretocarryontheexaminationanalysis,appraisesitsearthquakeresistanceperformance。Non-linearstaticanalyticmethodwastherecentyearsobtainsthewidespreadapplicationonestructureearthquakeresistanceabilityappraisalnewmethod。Thismethodfromessentiallywillsayisonestaticnonlinearcomputationmethod,butitresponded137 thespectrumhasintroducedthecomputationprocessandtheresult.Itsbasiccharacteristiciswiththestaticloaddescriptionearthquakefunction,considersthestructureundertheearthquakefunctiontheelastoplasticitynature。Itsbasicprincipleandthesteparefirstobtainthestructurebysomemethodthegoaldisplacementwhichtocorrespondunderthepossiblebitterexperienceearthquakefunction,thentostructureinflictionverticalloadatthesametime,willattributetheearthquakefunctionagroupofhorizontalstaticloadtoaffectbythemonotoneincreasingformtothestructurein,inachievedwhengoaldisplacementwillstoptheloadincreasingprogressively,finallywillstoptheconditionintheloadcarryingontheearthquakeresistanceperformancetothestructuretoappraise,willjudgewhethermightguaranteethestructurewillmeetthefunctionneedunderthishorizontalearthquakefunction.Reference1.Fengdingguo,Wangqingmin,Qianguofang,Susanqing,Engineeringstructureearthquakeresistance,EarthquakePublishinghouse,20022.Aylen•Williams,Constructionandbridgeearthquakeresistancedesign,ChineseWaterconservationWaterandelectricityPublishinghouse,20023.Wangshelian,Caozhaoping,Portalframeconstructionelastoplasticityperformancetestresearch,Engineeringmechanics,1998:15(2)4.Zhouxiyuan,Theconstructionstructureseparatesshakes,theabsorptionofshockandthevibrationcontrol,Constructionstructurejournal,2002:25.Zhouyun,Xubin,Recallingenergyshockabsorptiontechnologyandprospects,Mechanicsandpractice,2002:20137'