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中南地质局综合办公楼设计 建筑工程专业 优秀毕业设计,建筑工程专业,优秀毕业设计,建筑工程,专业,优秀,毕业设计

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中南地质局综合办公楼的设计桥梁与结构工程学院 建筑工程2001级-01班 李小英 指导老师：卜乐奇 付涛 吴研摘要：本次毕业设计的题目是“中南地质局综合办公楼设计”。根据“长沙理工大学桥梁与结构工程学院2005届建筑工程毕业设计任务书”的要求，设计内容包括以下几方面：、建筑部分设计；、结构部分设计；、施工预算；、外文翻译。本项目建筑面积约3400平方米，采用钢筋混凝土框架结构。建筑物共六层，多为办公用房，层高均为3.6米。抗震设防烈度为7度。本设计提交的文件包括计算书和施工图两部分。计算书包括该设计的主要计算过程及相关说明，如结构选型、结构布置、框架内力计算等。施工图包括该设计的主要建筑施工图和结构施工图。关键词： 办公楼；建筑设计；结构设计；施工图预算1 课题内容1.1 题目中南地质局综合办公楼设计1.2 目的通过毕业设计培养我们综合运用所学知识，全面分析并解决工程实际问题的能力，从中掌握工程项目方案设计，施工图设计以及施工预算的具体步骤和方法，了解工程建设全过程中的有关常识，使我们毕业后能较快适应工作需要。1.3 建筑规模与要求此工程建设地点位于长沙市区。地段尺寸25m40m，平面形状L型。拟建项目的总建筑面积30005000平方米；投资标准：中等；建筑层数：58层；层高：办公室净高不小于2.6米；建筑总高度不大于24米。1.4 地质条件本工程地质条件分布均匀，自上而下为：人工填土，厚度1.32.1米，不宜作为持力层；新冲击粘土，厚度0.21.2米，不宜作为持力层；新冲击粉质粘土，厚度4.55.6米，是较好的持力层，承载力标准值270KPa;残积粉质粘土，厚度小于1米，是良好的持力层和下卧层，承载力标准值300KPa;强风化泥质粉砂岩（未钻透），是理想的持力层，承载里标准值350 KPa。地下水位：地表以下2米内无侵蚀性，稳定地下水位埋深为2.32.8米之间。抗震设防烈度按7度考虑，场地土属中硬土，二类场地，建筑物类别为二类。2 设计任务要求本次设计包括以下几方面内容：2.1 方案设计；2.2建筑设计；2.3结构设计；2.4施工图预算；2.5外文翻译。2.1 方案设计在进行方案设计的时候，首先根据建筑功能的要求，结合工程地质条件以及地形地貌特征，综合考虑经济技术、施工等条件，在初步确定结构方案的前提下，确定建筑平面布局，立面造型。由于本设计地段位于市区，要求立面造型美观。平面设计时，考虑地段的因素，整栋楼设置为坐南朝北向。房间布置上，基本按照设计任务书所要求的各房间面积指标及房间数进行设计，其中主要的包括普通办公室、机要办公室、财务办公室、会议室、接待室、计算机房、档案室、资料室等，并在各层设有男女厕所各一间，休息室。主入口设置在北面，位于道路旁，另在西面和南面设置两个侧入口，每层各设一主一副两部楼梯，并设消防栓，满足使用以及防火要求。门窗的设置方面，充分考虑采光及通风要求，每个房间均设有窗户，在内走廊设有高窗。屋顶采用坡屋顶形式，利于美观及排水。2.2 建筑设计 建筑设计主要是指建筑施工图的设计。主要有各层平面图，主要立面图及剖面图。1、 建筑平面图（除屋顶平面图外）是用一个假想的水平剖切面，在房屋的窗台上方剖开整栋房屋，移去剖切面上面的部分，将留下部分向水平投影面投影所得到的水平剖视图。建筑平面图包含被剖切到的断面、可见的建筑配件的设置情况。它是墙体砌筑、门窗安装和室内装修的重要依据。底层地面需反映出室外台阶、散水（或明沟）等，其他各层除应表示本层情况外，还需反映出下一层平面图未反映的可见建筑物构件。2、 建筑立面图是房间各个方向的外墙面以及按投影方向可见的构配件的正投影图。有定位轴线的建筑物，宜根据两端定位轴线号编注立面图名称。建筑物立面是用来反映房屋的体型和外貌、门窗形式和位置、墙面装饰材料等的图样。3、 建筑剖面图是假想用一个垂直于横向或纵向轴线的竖直平面剖切房屋所得正投影图，反映门窗洞口及窗台高度以及简要的结构形式和构造方式等情况。剖切面的剖切位置，应选在能反应房间全貌构造特征以及有代表性的部位，一般应显示楼梯，并在底层平面中表示。4、 节点详图反映该节点的具体做法和构造，一般有外墙防水，散水大样，女儿墙构造等。5、 门窗表：均选用中南地区标准图集或通用图集中的门窗型号。6、 建筑总说明：一般包括工程概况（工程名称、位置、建筑规模、建筑技术经济指标），对图纸上未能详细注写的用料、做法、部位或者可以统一说明的问题进行详细说明。2.3 结构设计结构设计包括各层结构平面图及基础，一榀框架、一根连续梁、一块现浇板的施工详图，并提供详细的结构设计计算书。1、 结构平面图即结构平面布置图，表示了墙、柱、梁等承重构件在平面图中的布置，是施工图中布置各层承重构件的依据。假想在建筑物底层室内地面下方做一水平剖切面，将剖切面下方的各构件向下做水平投影，即为基础平面图。楼层平面图是假想用一个紧贴楼面的水平剖切楼层后所得到的水平剖视图。2、 结构计算结构计算包括：确定框架计算简图、初估梁柱截面尺寸、荷载计算、内力计算、内力组合、配筋计算。在确定框架计算简图的时候应注意：框架柱的底层柱高应取基础顶面到二层楼面的高度，即基础顶面到室外地面的高度+室内外高差+底层层高。另外在计算梁的线刚度的时候，边跨梁应乘以1.5的系数，而中间跨应乘以2的系数。框架来年感的计算跨度，应为各柱的轴线距离。由于荷载计算时需计算梁柱的自重，所以应先估算出梁柱的截面尺寸，估算梁的节目尺寸时，应根据梁的跨度按不需验算饶度要求估算，估算柱的截面尺寸时，可根据底层柱的竖向荷载，综合考虑侧向刚度和轴压比要求在轴心受压基础上考虑1.21.4的放大系数。荷载计算分为恒荷载计算和活荷载计算，恒荷载楼板传给框架的荷载，梁柱自重等。而活荷载则是根据房间的功用不同来确定的，它是通过楼板传给框架的。内力计算同样也分为横荷载作用下的内力计算和活荷载下的内力计算。竖向荷载作用下的内力计算用分层法，水平荷载作用下的内力计算用D值法计算。对于活荷载应值得注意的是：按实际情况，应将活荷载分层分跨进行布置，但此种活载布置方法计算量大，不适合手算。所以，为了减轻计算量，本设计采用分跨布置的方法对活荷载进行最不利的布置。这种布置方法对负弯矩来说影响不大，但正弯矩的影响较大。可将计算出的正弯矩乘以1.1的放大系数，以减少分跨布置所产生的不利影响。内力组合。由于本设计考虑抗震，应先进行不考虑抗震时的内力组合，在进行考虑抗震的内力组合。又分为梁和柱的内力组合，组合时将各种荷载（恒荷载、活荷载、风荷载）作用下产生的内力乘以分项系数和组合系数得到各种荷载组合下的内力设计值。配筋。配筋时取最不利荷载组合时的内力设计值计算。同时还应考虑施工方便、经济节省和构造规定进行配筋。对于其他自选构件的计算，除一些构造要求有所不同外，原理与框架的计算相同。2.4 施工图预算施工图预算是指在施工设计阶段，当工程设计完成后，在单位工程开工之前，根据施工图纸计算的工程量，施工组织设计和国家规定的现行工程预算定额、单位估价表及各项费用的取费标准，建筑材料的预算价格，建设地区的自然的技术经济条件等资料，进行计算和确定单位工程或单项工程建设费用的经济文件。次设计只需进行建筑工程施工图预算。首先确定编制依据，及各部位的具体做法，然后计算工程量，再是使用定额，汇总各类费用。参考文献沈浦生主编。混凝土结构基本原理 。北京，高等教育出版社，2002.9沈浦生主编。混凝土结构设计 。北京，高等教育出版社，2003.11沈蒲生，苏三庆主编。高等学校建筑工程专业毕业设计指导。北京，中国建筑工业出版 2000.6梁兴文，史庆轩主编。土木工程专业毕业设计指导。北京，科学出版社，2002.7杨志勇主编。工民建专业毕业设计手册（本科、专科）。武汉，武汉工业大学出版社，2002.1同济大学，西安建筑科技大学，东南大学，重庆建筑大学编。房屋建筑学。北京，中国 筑工业出版社，2000.2吴德安主编。混凝土结构设计手册（第三版）。北京，中国建筑工业出版社，2003.3建筑结构荷载规范。北京，中国建筑工业出版社，2002.3刘建荣主编。建筑构造。成都，四川科学技术出版社，1991建筑制图标准（GBJ1-86）周起敬主编。混凝土结构构造手册（第二版）。北京，中国建筑工业出版社，2001.4The Design of Mid-south Geological Portion Complex Office Building Li XiaoyingAbstract: The total construction area of this design is about 3400, six floors, frame structure was adopted. The height of every floor is 3.6m, and the total height is 23.4m.The design strength of anti-earthquake is 7 degrees. This design include architectural design , structure design and construction budget.Key word:complex office building; architectural design ;structure design;cnstruction budget - 4 - 2005 届 毕 业 设 计（论 文）题目：中南地质局综合办公楼设计 班 级：工民建01-01班学 号： 200110030401姓 名： 李小英指导教师：卜乐奇 付涛 吴研2005年 6月摘 要本次毕业设计的内容是进行中南地质局综合办公楼的设计。根据“长沙理工大学桥梁与结构工程学院2005届建筑工程专业毕业设计任务”的要求进行，设计中应进行方案设计，拟定尺寸, 结构中一榀框架的受力及配筋的计算，基础、板、一根连续梁的计算，施工图的绘制。本次设计采用的是框架结构，共6层，全楼总高为23.4m，总建筑面积为3414m2。本次毕业设计的特点：按照毕业设计的任务书及指导书的要求，运用所学的课程，掌握房屋建筑的基本理论，基本知识，基本的设计方法，不断地提高分析问题，解决问题的能力。同时也不断提高运用计算机绘图的能力。在设计中我们力求做到：实事求是，运用公式正确无误，积极搜集查阅资料，理论依据充分，计算准确；按时独立完成设计任务，使之有一定的理论价值和实用价值，将理论与实际有机的结合起来，提高我们对工程实际问题的分析能力，锻炼我们实际动手的技能，掌握房屋结构设计及施工图绘制的步骤和方法。通过本次毕业设计不但提高了我们的设计能力和创新能力，拓宽了我们的知识面，增强了我们的自信心，而且使我们能更好的适应即将走上的工作岗位，为社会，为单位贡献自己的聪明才智目 录第一章 建筑设计说明书（1）第二章 结构设计说明书 （2）第三章 基本资料及结构选型（4）第四章 结构布置（5）第五章 框架结构荷载计算（7）第六章 框架内力计算（17）第七章 框架内力组合（36）第八章 框架梁柱截面配筋（44）第九章 框架柱基础尺寸确定及配筋计算（53）第十章 板的计算（57）第十一章 多跨梁的计算（58）第十二章 施工图预算第十三章 专业英语翻译致谢参考文献材料汇总表工程名称:中南地质局综合办公楼第1页共7页序号编号材料名称单位数量单价合价1ACF安拆费元4953.061.004953.062CLF材料费元31241.951.0031241.953CLFBC材料费补差元-1.831.00-1.834C0805球形执手锁把29.5820.00591.605C0809门碰珠只29.581.7050.296C1206瓷质面砖 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建筑工程专业毕业设计计算书 第一章 建筑设计说明书一、 建筑设计概况与主要技术经济指标1、 设计详细说明（1）、气象条件、 温度：最热月平均30.3，最冷月平均3.7；室外计算温度：夏季极端最高40.1，冬季极端最底9.5；、 相对湿度：最热月平均73%；、 主导风向：全年为西北风，夏季为东南风，基本风压0.35kn/m2、 雨雪条件：年降水量1450mm，日最大降水强度192mm/日；暴雨降水强度3.31mm/s，100mm2；基本雪压0.35kn/m2；（2）、工程地质条件 拟建场地各地层由上往下依次为：、 人工填土，厚度1.32.1m，不宜做为持力层；、 新冲击黏土，厚度0.41.2m，不宜做为持力层；、 冲击粉质黏土，厚度4.55.6m，是较好的持力层，承载力标准值270KPa；、 残积粉质黏土，厚度小于一米，是良好的持力层，承载力标准值300KPa；、 强风化泥质粉砂岩（未钻透），是理想的持力层，承载力标准值350KPa；地下水位：地表以下2米内无侵蚀性，稳定低水位埋深为2.3m2.8m之间。、 抗震设防烈度按7度考虑，设计基本本地加速度值为0.15g，场地土属中硬性，二类场地，建筑物类别为二类。（3）、施工技术条件 “三通一平”等施工现场准备工作已经做好，各种机具、材料能满足要求。2、 建筑主要技术经济指标（1）、根据要求及地形状况，本建筑设计为L形综合办公楼，建筑面积为3413.34平方米，占地面积487.52平方米。首层高3.6m，其余各层高3.6m主体为六层，总高为23.4m。建筑横总长为23.4m，纵向总长为33.6m。（2）、耐火等级：二级。二、 建筑内容1、 平面功能分析本建筑设计中间设有走廊，两侧均有采光及兼作通风的窗，轴线宽2.1m，满足工作人员的通行要求及疏散要求；办公楼底层设有两个入口，一个宽2.4m的正门，一个宽2.1m的侧门，均直接通往大厅；主体内各层都设有两楼梯，一电梯，各有单独的入口，具体尺寸及做法详见建筑施工图，这也满足了疏散的入口要求。2、 房间布置本办公楼设有接待室及会客室，满足休息要求；横向部分的房间呈南北向布置，且在南北两侧均设有用来通风及采光的适当宽度、1.5m高的窗户；纵向部分的房间呈东西向布置，且在东面设有宽度不等的窗户，西面设有半透明的玻璃幕墙，满足通风几采光的要求，且防止了西晒。3、 安全设施（1）、疏散要求方面：最远端房间的门距离楼梯距离为14.7m；两个楼梯至少可以并列通过三个人，合乎建筑规范要求。（2）、防火要求方面，在各个楼道内均设有消防栓。三、 构造处理1、 墙面构造处理、 内外纵横墙为240mm厚，卫生间的标高比室内标高低30mm。、 防潮处理：墙身水平防潮处理，在标高为0.060m处铺设3%防水剂的细石混凝土厚60mm；垂直防潮处理，在所有外墙壁窗台标高以下采用防水砂浆。2、 门窗选用所有的向内开的门均采用普通的镶板木门，进大厅的正门采用玻璃门，另一个侧门则采用玻璃弹簧自动门；窗户均采用铝合金推拉窗，具体的构造做法详见建筑施工图内的门窗表。3、 楼地面及内外墙做法根据房间的功能不同，所采用的不同材料装饰。具体做法详见建筑施工图。4、 屋顶构造做法采用构造找坡做法，柔性防水屋面，详见建筑施工图中的节点构造详图，屋面排水采用女儿墙内排水，天沟采用现浇天沟。5、 楼梯构造做法防滑采用98ZJ401-29-1，起步采用98ZJ401-28-6，栏杆做法见建筑施工图。四、 其它补充1、 本设计中除标高以m为单位外，其它均以mm为单位。2、 所以未标注的门垛宽均为240mm。3、 室内标高为0.000，室外标高为-0.450。4、 所有的卫生器具、办公用品及家居均购买成品。5、 设计中的未尽事宜，请参照有关建筑规范施工。第二章 结构设计说明书一、 设计资料1、 建筑平面图如建筑施工图所示2、 基本风压W0=0.35KN/m23、 工程地质资料：自然地表下1.2m内为砂砾石（地基承载力标准值fk=240kp），其下层为圆砾石（地基承载力标准值fk=300kp），再下层为粘土层（地基承载力标准值fk=230kp）。4、 该工程所在的长沙市地震基本烈度7度，近震。场地土为类。二、 结构选型1、 结构体系选型：采用钢筋混凝土现浇框架结构体系（纵横向承重框架）体系。对于六层的办公楼，可采用钢筋混凝土框架结构、混合结构、底层框架或内框架砖房结构。该建筑要求布置灵活，同时考虑该建筑处于7度地震区，故选用框架 结构体系。由于结构承受纵横向水平地震作用，故选用纵横向承重框架体系。2、 其它结构选型（1）、屋面结构：采用现浇混凝土板作承重结构，屋面板按上人屋面的使用荷载选用。（2）、楼层结构：所有露面均采用现浇混凝土结构。（3）、楼梯结构：由于楼梯段水平投影长度不大于3米，故采用钢筋混凝土板式楼梯。（4）、天沟：采用现浇天沟。（5）、过梁：窗过梁以及门的过梁均采用钢筋混凝土梁，并采用可兼做过梁的框架梁做窗过梁。（6）、基础梁：因持力层较深，采用现浇钢筋混凝土基础梁。（7）、基础：因荷载不很大，地基承载力较大，采用钢筋混凝土柱下独立基础。三、 结构布置标准层楼面及屋面结构布置图详见结构施工图。四、框架结构计算1、 确定框架计算简图假定框架柱嵌固于基础顶面，框架梁与框架柱刚接。由于各层柱的截面尺寸不变，故梁跨等于柱截面行心轴线之间的距离。底层柱高从基础顶面算至二楼楼面，基础顶面标高根据地质条件、室内外高岔等定为-0.450m，二楼楼面标高为3.600m，故柱高为4.05m，其余各层的柱高从楼面算至上一层楼面（即层高），故均为3.0m。故可绘出计算简图如计算书内所示。多层框架为超静定结构，在内力计算之前，要预先估算梁、柱的截面尺寸及结构所才的材料强度等级，以求得框架中各杆的线刚度及相对线刚度。2、 梁柱的截面尺寸大致根据下列公式估算：框架梁：h=L/（1018） b=h/（23） 框架柱：b=H/（1520）3、 计算线刚度用公式I=EI/L计算，但在计算梁的刚度时，边梁要乘以1.5，中间梁要乘以2，这是因为考虑到楼板对所起的翼缘约束作用。4、 荷载计算把板划分为双向板计算。取荷载的标准值均查阅荷载结构规范而来。5、 梁主要验算正、斜截面的承载力，柱主要验算轴压比几两者均需验算草料强度。在验算时活荷载按满跨布置。6、 内力计算恒荷载和活荷载的内力计算均采用力矩分配发，活荷载选取三种情况来计算。风荷载和地震荷载均采用反弯点法，因为梁、柱相对线刚度接近于3。其中地震荷载的作用按底部剪力法计算。7、 内力组合各种框架情况下的内力求的后，根据最不利又是可能的原则进行内力组合。当考虑结构塑性内力重分布的有利影响时，应在内力组合之前对竖向荷载作用下的内力进行调幅。当有地震作用时，应分别考虑恒荷载和活荷载的组合及重力荷载代表值与地震作用的组合，并比较考虑这两种组合的内力，取最不利者。由于构件控制截面的内力值应区支坐边缘处，为此，应先计算各梁的控制截面处的（支坐边缘处的）内力值。为了简化起见，也可采用轴线处的内力值，这样计算得的钢筋用梁比需要的钢筋用梁略多一点。第三章 基本资料及结构选型一、 工程名称：中南地质局综合办公楼二、 建设地点：长沙市区三、 设计资料：基本风压基本雪压地面粗糙类型：C类极端最高温度：40.1极端最低温度：-9.5工程地质条件：场地土质分布均匀。自上而下为：人工填土层，厚度1.32.1m,不宜作为持力层；新冲击粘土层，厚度0.41.2m,不宜作为持力层；冲击粉质粘土，厚度4.55.6m,是较好的持力层，承载力标准值270KPa；残积粉质粘土,厚度1m,是良好的持力层和下卧层,承载力标准值300KPa；强风化泥质粉砂岩（未钻透）,是理想的持力层,承载力标准值350KPa；地下水位：地表以下两米内无侵蚀性，稳定地下水位埋深为2.32.8m之间该地区抗震设防烈度为7度建筑物类别：二类四、结构选型1、结构体系选型：采用钢筋混凝土现浇框架结构体系2、其他结构选型： 屋面结构：采用现浇钢筋混凝土板做承重结构，屋面按不上人屋面事业荷载选用 楼面结构：采用全部现浇钢筋混凝土板 楼梯结构：采用全部现浇钢筋混凝土板式楼梯 、天沟：采用现浇内天沟。、过梁：窗过梁以及带雨蓬的门过梁均采用钢筋混凝土梁，有些框架梁可兼作窗过梁使用。、基础梁：因持力层较深，且天然地表下土质较好，采用现浇钢筋混凝土基础梁。 、基础：因荷载不是很大，地基承载力较大 ，采用钢筋混凝土柱下单独基础。 第四章 结构布置一、 确定框架计算简图选一榀框架进行计算，假定框架柱嵌固于基础顶面，框架梁与柱刚接。由于各层柱的截面尺寸不变，故梁跨等于柱截面形心轴线之间的距离。底层柱高从基础顶面算至一层板底，基顶标高根据地质条件、室内外高差定为-0.45m，室外地坪至基础顶面的高度为0.25 m，底层楼层3.6m，故底层柱高为4.3m。其余各层的柱高为各层层高，故均为3.6m。由此可绘出框架的计算简图如下所示： 二、 初估梁柱截面尺寸多层框架结构为超静定结构，在内力计算之前，先估算梁柱的截面尺寸及结构所采用强度等级，以求得框架中各杆的线刚度及相对线刚度。 混凝土强度等级：梁柱均采用C30（E=2.95） 梁的截面尺寸按跨度大者计算，l=5400mm,h=(1/8-1/12)l=675mm-450mm，取h=600mmb=(1/2-1/3)l=300mm-200mm,取b=250mm 故框架梁的初选截面尺寸为bh=250mm600mm其惯性矩：=1/12250=4.5 柱的截面尺寸按层高确定，底层H=3600mm450mm250mm=4300mm b=(1/15-1/20)H=286mm-215mm,取b=400mm h=(1-2)b=573mm-286mm, 取h=400mm故框架柱的截面尺寸为bh=400mm400mm其惯性矩：=1/12400=2.133 板的截面尺寸由于=5400/3600=1.52，故按双向板设计 h=(1/30-1/40)=120mm-90mm, 取h=100mm三、 梁柱线刚度计算由iEI/l,得 梁：=1.54.5/5400E=12.495E =24.5/2100E=42.8E=1.54.5/5400E=12.495E 柱：=2.133/4300E=4.96E取为基准值1，则相对线刚度表示如下：第五章 框架结构荷载计算一、恒载标准值计算1、屋面 防水层（刚性）30厚细石混凝土防水 1.0 防水层（柔性）三毡四油铺小石子 0.4 波形石棉瓦 0.2 找平层：15厚水泥砂浆 0.015m20=0.3 保温层：80厚矿渣水泥 0.08m14.5=1.16 结构层：100厚现浇钢筋混凝土板 0.1m25=2.5 抹灰层：10厚混合砂浆 0.01m17=0.17合计 5.732、走廊及标准层楼面 小瓷砖地面（包括水泥粗砂打底） 0.55 结构层：100厚现浇钢筋混凝土板 0.1m25=2.5 抹灰层：10厚混合砂浆 0.01m17=0.17 合计 3.22 3、梁自重主梁 bh=250mm600mm 梁自重 250.25m（0.6-0.1）m=3.125 抹灰层：10厚混合砂浆 0.01m（0.6-0.1+0.25）m217=0.255 合计 3.38 基础梁 bh=250mm400mm 梁自重 250.25m0.4m=2.5 4、柱自重 bh=400mm400mm 柱自重 250.4m0.4m=4 抹灰重：10厚混合砂浆 0.01m0.4m417=0.27 合计 4.27 5、外纵墙自重 标准层 纵墙 （3.6-1.8-0.6）m0.24m18=5.184 铝合金窗 0.351.8m=0.63 贴瓷砖外墙面（包括水泥砂浆打底25mm）0. 5（3.6-1.8）m=0.9 水泥粉刷内墙面 0.36（3.6-1.8）m=0.648 合计 7.362底层 纵墙 （3.6-1.8-0.6-0.4）m0.24m18=3.456 铝合金窗 0.351.8m=0.63 贴瓷砖外墙面（包括水泥砂浆打底25mm）0.5（3.6-1.8）m=0.9 水泥粉刷内墙面 0.36（3.6-1.8）m=0.648 合计 5.634 6、内纵墙自重 标准层 纵墙 （3.6-0.6-0.9）m0.24m18=9.072 水泥粉刷内墙面 0.362.1m2=1.512 合计 10.584 7、内隔墙自重 标准层 内隔墙 （3.6-0.6）m0.24m9.8=7.056 水泥粉刷墙面 0.363m2=2.16 合计 10.216 底层 内隔墙 （3.6-0.6-0.4）m0.24m9.8=6.115水泥粉刷墙面 0.363m2=2.16 合计 8.275二、活载标准值计算1、 屋面和楼面活载标准值不上人屋面： 0.5楼面： 办公室、阅览室 2.0 楼梯、走廊 2.52、雪荷载 =0.80.35=0.28 屋面活荷载与雪荷载不同时考虑，两者中取大值。3、 竖向荷载下框架受荷总图 计算简图如下：1 A-B轴间框架梁 板传至梁上的三角形或梯形荷载等效为均布荷载 屋面板传荷载： 恒载：5.731.8m（1-2+）2=16.91 活载：0.51.8m（1-2+）2=1.48 楼面板传荷载： 恒载：3.221.8m（1-2+）2=9.51 活载：2.01.8m（1-2+）2=5.90 梁自重：3.38A- B轴间框架梁的均布荷载为：屋面梁 恒载=梁自重+板传荷载=3.38+16.91=20.29 活载=板传荷载=1.48 楼面梁 恒载=梁自重+板传荷载=3.38+9.51=12.89 活载=板传荷载=5.90B-C轴间框架梁 屋面板传荷载：恒载： 5.731.8m5/82=12.89 活载： 0.51.8m5/82=1.125楼面板传荷载： 恒载： 3.221.8m5/82=7.245 活载： 2.51.8m5/82=5.63 梁自重：3.38B- C轴间框架梁的均布荷载为： 屋面梁 恒载=梁自重+板传荷载=3.38+12.89=16.27 活载=板传荷载=1.125 楼面梁 恒载=梁自重+板传荷载=3.38+7.245=10.625 活载=板传荷载=5.63 C-D轴间框架梁同A-B轴间框架梁 A轴柱纵向集中荷载计算顶层柱 天沟自重 25（0.6+0.2-0.1 ）m0.1 m+（0.6+0.2）m（0.5+0.36）=2.44顶层柱恒载=天沟自重+梁自重+板传荷载 =2.443.6 m+3.38（3.6-0.4）m+16.91 3.6m+5.731.8m5/83.6m =74.76KN顶层柱活载=板传活载 =1.553.6m/2+0.51.8m3.6m5/8 =4.82KN标准层柱恒载=墙自重+梁自重+板传荷载 =7.362（3.6-0.4）m+3.38（3.6-0.4）m+9.973.6m/2+3.221.8m 3.6 m5/8 =65.36KN标准层柱活载=板传活载 =6.193.6m/2+2.01.8 m5/83.6m =19.24KN基础顶面恒载=底层外纵墙自重+基础梁自重 =5.364（3.6-0.4）m +2.5（3.6-0.4）m =26.03KNB轴柱纵向集中荷载计算 顶层柱恒载=梁自重+板传荷载 =3.38（3.6-0.4）m +16.913.6m +5.731.8m 3.6m 5/8+5.731.8m3.6m0.82+5.731.8m 0.9m 5/8 87.16KN 顶层柱活载=板传活载 = 1.553.6m/2 +0.51.8m 3.6m 5/8+2.01.8m3.6m0.82+2.01.8m 0.9m 5/8 =18KN标准层柱恒载=墙自重+梁自重+板传荷载 = 3.38（3.6-0.4）m+10.584（3.6-0.4）m+9.973.6m/2+3.221.8m 3.6 m5/8+3.221.8m 3.6 m5/8+3.221.8m 0.9m5/8 =96.9KN 标准层柱活载=板传活载 =6.193.6m/2+ 2.01.8m 3.6 m5/8+2.51.8m 3.6 m0.82+2.51.8m 0.9m5/8 =35.7KN基础顶面恒载=底层内纵墙自重+基础梁自重 =10.584（3.6-0.4）m +2.5（3.6-0.4）m =41.87KNC轴柱同B轴D轴柱同A轴 4、 风荷载计算作用在屋面梁和楼面梁节点处的风荷载标准值：风荷载标准值公式如下：K=zsz0 荷载规范规定，高度小于30m或高宽比小于1.5的房屋结构，允许不考虑风荷载的动力影响，即取=1.0。风荷载体型系数S，查荷载规范表7.3.1的30项次得：S=0.8-0=0.8风压高度变化系数z，查荷载规范表7.2.1，地面粗糙度类别为C类，并结合以内差法得：离地面高度 Z（m）3.67.210.814.418.021.6z 0.53 0.74 0.74 0.74 0.8 0.87 设计资料提供：0=0.35KN/m2。=1.00.80.530.35=0.15=1.00.80.740.35=0.21=1.00.80.740.35=0.21=1.00.80.740.35=0.21=1.00.80.80.35=0.22=1.00.80.870.35=0.24转化为集中荷载： 6层： Fw6k=0.243.6/2（3.6+1.8）=2.33KN5层： Fw5k=0.223.6/2（3.6+3.6）=2.85KN4层： Fw4k=0.213.6/2（3.6+3.6）=2.72KN3层： Fw3k=0.213.6/2（3.6+3.6）=2.72KN2层： Fw2k=0.213.6/2（3.6+3.6）=2.72KN 1层： Fw1k=0.153.6/2（3.6+3.6）=1.94KN5、 风荷载作用下的侧移验算横向2-6层D值计算 （KN/m）A轴柱 0.559 7557B轴柱 0.848 11464C轴柱 0.848 11464D轴柱 0.559 7557 横向底层D值计算： （KN/m）A轴柱 0.669 9044B轴柱 0.886 11977C轴柱 0.886 11977D轴柱 0.669 9144 风荷载作用下框架侧移计算水平荷载作用下框架的层间侧移按下式计算： 第一层的层间侧移值求出以后，就可以计算各楼板标高处的侧移值的顶点侧移值，各层楼板标高处的侧移值是该层以下各层层间侧移之和。顶点侧移是所有各层层间侧移之和。J层侧移 顶点侧移 框架在风荷载作用下侧移计算如下：层次/KN/KN（KN/m）/ m/h6 2.33 2.33 38042 0.0001 1/360005 2.85 5.18 38042 0.0001 1/360004 2.72 7.90 38042 0.0002 1/180003 2.72 10.62 38042 0.0003 1/120002 2.72 13.34 38042 0.0004 1/90001 1.94 15.28 42042 0.0004 1/9000 =0.0015m侧移验算： 层间侧移最大值：1/90001/550 （满足）6、 水平地震作用计算本建筑建筑高度为23.4m，且高度和刚度沿高度均匀分布，可采用底部剪力法计算地震作用。作用于屋面梁及各层楼面梁处的重力荷载代表值：=74.76+87.16+87.16+74.76=323.84（KN）=（65.36+19.24）2+（96.9+35.7）2=434.4（KN）框架自震周期计算： H=23.4m B=12.9m 得出 =0.491s多遇水平地震作用标准值及位移的计算 由设防烈度微度，场地土二类。查得， 则横向地震影响系数：=0.059 =0.4911.4=0.49 则 =0.059（323.84+434.45）0.85=125.16（KN） =0.1092125.16=13.67（KN） 则每层计算如下： 层次 或 6 323.84 21.6 6994.9431452.54 42.42 5 434.4 18 7819.20 32.14 4 434.4 14.4 6255.36 25.71 3 434.4 10.8 4691.52 19.28 2 434.4 7.2 3127.68 12.86 1 434.4 3.6 1563.84 6.43 层次/h 6 42.4242.42380420.00111/3273 532.1474.56380420.00201/1800 425.71100.27380420.00261/1385 319.28119.55380420.00311/1161 212.86132.41380420.00351/1029 16.43138.84420420.00331/1091顶点位移：0.0156/23.4=1/6661/550 （满足）第六章 框架内力计算为简化计算，考虑如下几种单独受荷情况： 恒载作用； 活荷载作用于A-B轴跨间； 活荷载作用于B-C轴跨间； 风荷载作用（从左向右或从右向左）； 横向水平地震作用（从左向右或从右向左）；对于、种情况，框架在竖向荷载作用下，采用力矩分配法进行计算，对于第、种情况，框架在水平荷载作用下，采用分层法进行计算。一、恒载作用下的内力计算1 、 恒载作用下引起的杆端弯矩计算（转化为设计值） 均布荷载引起的： 屋面梁的固端弯矩： 楼面梁固端弯矩： 恒载引起的不平衡弯矩： 则恒载作用下的M图如下所示：(单位：)恒载作用下的剪力图如下所示：（单位：KN）恒载作用下的N图：（单位：KN） 其中包括梁传来的荷载及梁自重。 梁自重设计值为： 1.20.40.4253.6=17.28（KN）二、活载作用下的内力计算 活载设计值=活载标准值1.4采用分层法，除底层外，其他各层柱的线刚度均乘0.9，且相应的传递系数为1/3。（底层为1/2）1、 顶层：由于屋面活载较小，不考虑活载最不利，将活载满跨布置。 2、 其他层为便于内力组合，将活荷载分跨进行布置。AB跨布置活荷载时： BC跨布置活载时 三、风荷载作用下的内力计算先求各柱V值：风荷载设计值=风荷载标准值1.4。用D值法进行计算。 6层 5层2.5311.1811.182.532.5311.1811.182.530.560.850.850.560.560.850.850.560.560.850.850.560.560.850.850.560. 650.980.980.651.442.192.191.44 4层 3层2.5311.1811.182.532.5311.1811.182.530.560.850.850.560.560.850.850.560.560.850.850.560.560.850.850.562.203.333.332.202.954.484.482.95 2层 1层2.5311.1811.182.532.5311.1811.182.530.560.850.850.560.670.890.890.670.560.850.850.560.670.890.890.673.715.635.633.714.606.106.104.60 求反弯点高度 层次 六层（m=6 n=6 h=3.6m）柱号0.60.251.00001.000.91.190.401.00001.001.441.190.401.00001.001.440.60.251.00001.000.9层次 五层（m=6 n=5 h=3.6m）柱号0.60.351.001.001.001.261.190.451.001.001.001.621.190.451.001.001.001.620.60.351.001.001.001.26层次 四层（m=6 n=4 h=3.6m）柱号0.60.401.001.001.001.441.190.451.001.001.001.621.190.451.001.001.001.620.60.401.001.001.001.44层次 三层（m=6 n=3 h=3.6m）柱号0.60.451.001.001.001.621.190.501.001.001.001.801.190.501.001.001.001.800.60.451.001.001.001.62层次 二层（m=6 n=2 h=3.6m）柱号0.60.501.001.001.001.801.190.501.001.001.001.801.190.501.001.001.001.800.60.501.001.001.001.80层次 一层（m=6 n=1 h=3.6m）柱号0.60.701.001.001.002.521.190.551.001.001.001.981.190.551.001.001.001.980.60.701.001.001.002.52风荷载作用下框架内力值柱上端弯矩：柱下端弯矩：梁端弯矩：先求每个节点柱端弯矩之和，然后按梁的线刚度进行分配。 从左向右吹的情况： 其M图如下所示：（单位：） V图如下所示：（单位：KN） N 图如下所示：（单位：KN） 从右向左吹风的情况与以上三图相反。 四、地震荷载作用下的内力计算 横向水平地震作用下的框架柱剪力和柱弯矩标准值的计算。A轴柱：层次h/m63.642.423804275570.198.060.60.97.2521.7653.674.563804275570.1914.710.61.2617.8533.1643.6100.273804275570.1919.050.61.4427.4341.1533.6119.553804275570.1922.720.61.6236.8144.9923.6132.413804275570.1925.160.61.845.2945.2913.6138.844204290440.2230.550.62.5276.9932.99B轴柱：层次h/m/KN/63.642.4238042114640.3113.151.191.4418.9428.4053.674.5638042114640.3123.111.191.6237.4445.7643.6100.2738042114640.3131.081.191.6250.3561.5433.6119.5538042114640.3137.061.191.8066.7166.7123.6132.4138042114640.3141.051.191.8073.8973.8913.6138.8442042119770.2838.881.191.9876.9862.99 得出从左向右时横向水平地震作用下的M图（单位：）N图（单位：KN）V图（单位：KN）如下所示： 五、与地震作用相结合的重力荷载代表值作用下的内力计算屋面梁的线荷载设计值=恒载+0.5雪荷载 =（20.29+0.50.80.353.6）1.2=24.95KN/m 楼面梁的线荷载设计值=恒载+0.5活荷载 =（12.89+0.52.03.6）1.2=19.795KN/m其中走廊=（12.89+0.52.53.6）1.2=20.87KN/m屋面梁固端弯矩： 楼面梁固端弯矩： 恒载引起的不平衡弯矩： 得出重力荷载代表值作用下的M图（单位：）N图（单位：KN）V图（单位：KN）如下所示： 第七章 框架内力组合各种荷载情况下的框架内力求得后，根据最不利又是可能的原则进行内力组合。当考虑结构塑性内力重分布的有利影响，应在内力组合之前对竖向荷载作用下的内力进行调幅。当有地震作用时，应分别考虑横荷载和活荷载代表值与地震荷载作用的组合，并比较二种组合内力，取最不利者。由于构件控制截面的内力值应取自支座边缘处，为此，在进行组合之前，应先计算各控制截面（支座边缘处）的内力值。 - 73 - 用于承载力计算框架梁荷载效应基本组合表 AB梁（非抗震）恒载活载活载活载左风右风 Mmax相应的V Mmin相应的V Vmax相应的M123456组合项目值组合项目值组合项目值6左M-25.42-1.45-1.45-1.451.7-1.71+2+3+4+0.6*6-30.791+2+3+4+0.6*6-30.79V46.833.173.173.17-0.430.4356.656.66中M27.081.711.711.710.54-0.541+2+3+4+0.6*532.53V6右M-68.34-5.92-5.92-5.92-0.610.611+2+3+4+0.6*5-86.471+2+3+4+0.6*5-86.47V-62.73-4.83-4.83-4.83-0.430.43-77.48-77.481左M-24.82-9.170.22-0.3511.6-11.61+0.7*2+0.7*4+6-43.081+2+3+4+0.6*6-41.08V28.7813.760.180.38-3.213.2141.8945.031中M13.586.48-0.29-0.642.92-2.921+2+0.6*521.81V1右M-41.97-20.88-0.761.69-5.765.761+2+3+0.6*5-66.871+2+0.6*5-66.31V-40.82-180.180.38-3.213.21-60.57-60.75用于承载力计算框架梁荷载效应基本组合表 BC梁（非抗震）恒载活载活载活载左风右风 Mmax相应的V Mmin相应的V Vmax相应的M123456组合项目值组合项目值组合项目值6左M-55.79-5.56-5.56-5.561.55-1.551+2+3+4+0.6*6-73.41+6-57.34V17.08-0.18-0.18-0.18-1.481.4817.4318.566中M-46.82-5.26-5.26-5.26001+2+3+4-62.6V6右M-55.79-5.56-5.56-5.56-1.551.551+2+3+4+0.6*5-73.41+0.7*（2+3+4）+5-69.02V-17.08-0.18-0.18-0.18-1.481.48-18.51-18.941左M-27.88-17.25-1.32.8914.8-14.81+0.7*2+0.7*4+6-43.081+2+3+0.6*6-55.31V11.169.5915.2-9.59-14.114.141.8944.411中M-22.02-7.171.8-7.17001+2+0.6*521.81V1右M-27.882.89-1.3-17.25-14.814.81+2+3+0.6*5-66.871+0.7*4+5-54.76V-11.069.5915.2-9.59-14.114.1-60.57-31.87用于承载力计算框架柱荷载效应基本组合表 A轴柱（非抗震）恒载活载活载活载左风右风Nmax相应的MNmin相应的MMmax相应的V123456组合项目值组合项目值组合项目值6上M18.691.451.451.45-1.71.71+2+3+4+0.6*624.061+516.991+2+3+4+0.6*624.06N121.593.173.173.17-0.430.43131.56121.16131.366下M-15.36-0.48-0.48-0.480.57-0.571+2+3+4+0.6*6-17.141+5-14.791+5-14.79N138.873.173.173.17-0.430.43148.64138.44138.44V9.46-0.54-0.54-0.540.43-0.437.589.899.891上M9.481.53-0.08-0.18-5.075.071+2+3+4+0.6*615.821+54.411+0.7*2+614.49N678.6971.884.073.55-11.6411.64765.17667.05740.651下M-15.36-4.580.030.211.82-11.821+2+3+4+0.6*6-26.81+5-3.541+0.7*2+6-30.39N695.9771.884.073.55-11.6411.64782.45684.33757.93V2.64-1.70.04-0.073.21-3.21-1.025.85-1.76用于承载力计算框架柱荷载效应基本组合表 B轴柱（非抗震）恒载活载活载活载左风右风Nmax相应的MNmin相应的MMmax相应的V123456组合项目值组合项目值组合项目值6上M-12.56-0.36-0.36-0.36-2.162.161+2+3+4+0.6*6-14.941+5-14.721+0.7*（2+3+4）+5-15.48N104.246.016.016.01-1.051.05122.9103.19115.816下M11.410.120.120.121.44-1.441+2+3+4+0.6*610.911+512.851+0.7*（2+3+4）+513.1N121.526.016.016.01-1.051.05140.18120.47133.09V16.180.130.130.131.48-1.4815.6817.6617.931上M-11.41-1.810.27-0.6-10.2810.281+2+3+0.6*6-6.781+4+0.6*5-18.181+0.7*（2+4）+5-23.38N730.94143.9682.91-55.86-35.5635.56979.15653.74757.051下M11.410.7-0.090.211.9-11.91+2+3+0.6*64.881+4+0.6*518.751+0.7*（2+4）+523.94N748.22143.9682.91-55.86-35.5635.56996.43671.02871.47V3.350.7-0.160.2214.1-14.1-4.5712.0318.09横向水平地震作用与重力荷载代表值组合效应 AB梁层次截面内力重力荷载重力荷载水平地震水平地震Mmax相应的VMmin相应的VVmax相应的M1234组合项目值组合项目值组合项目值6左M-22.76-18.9721.76-21.76（1+4）*0.75-33.54（1+4）*0.85/0.8V25.62-5.55.533.07中M45.566.89-6.89V右M-51.78-43.15-7.967.96（1+2）*0.7539.34（1+3）*0.75-44.81（1+3）*0.85/0.8V-55.53-5.55.5-64.841左M-23.42-19.5278.28-78.28（2+3）*0.7544.07（1+4）*0.75-76.28（1+4）*0.85/0.8V34.47-22.1622.1660.17中M31.6418.24-18.24（1+3）*0.7537.41V右M-39.9-33.25-41.441.4（2+4）*0.756.11（1+3）*0.75-60.98（1+3）*0.85/0.8V-45.59-22.1622.16-72.41横向水平地震作用与重力荷载代表值组合效应 BC梁层次截面内力重力荷载重力荷载水平地震水平地震Mmax相应的VMmin相应的VVmax相应的M1234组合项目值组合项目值组合项目值6左M-42.3-35.2520.44-20.44（1+4）*0.75-47.06（1+4）*0.85/0.8V21.21-19.4719.4743.22中M-33.7900V右M-42.3-35.25-20.4420.44（1+3）*0.75-47.06（1+3）*0.85/0.8V-21.21-19.4719.47-43.221左M-26.45-22.04106.38-106.38（2+3）*0.7563.26（1+4）*0.75-99.62（1+4）*0.85/0.8V17.95-101.31101.31126.71中M33.7900V右M-26.45-22.04-106.38106.38（2+4）*0.7563.26（1+3）*0.75-99.62（1+3）*0.85/0.8V-17.95-101.31101.31-126.71横向水平地震作用与重力荷载代表值组合效应 A轴柱层次截面内力重力荷载地震作用地震作用Nmax相应MNmin相应MMmax相应N123组合项目值组合项目值组合项目值6上M19.77-28.2928.29（1+3）*0.7536.05（1+2）*0.75-6.82（1+3）*0.7536.05N125.33-7.157.1599.3694.5499.36下M-18.59.43-9.43（1+3）*0.75-20.95（1+2）*0.75-7.26（1+3）*0.75-20.95N142.61-7.157.15112.32108.37112.32V-10.6310.48-10.48（1+3）*0.85-17.94-0.12-15.831上M12.62-42.8942.89（1+3）*0.844.41（1+2）*0.8-24.26（1+3）*0.844.41N730.68-125.5125.5684.94484.14684.94下M-12.62100.09-100.09（1+3）*0.8-90.17（1+2）*0.869.98（1+3）*0.8-90.17N747.96-125.5125.5698.77497.97698.77V-3.5139.72-39.72（1+3）*0.85-36.7528.9734.58横向水平地震作用与重力荷载代表值组合效应 B轴柱层次截面内力重力荷载地震作用地震作用Nmax相应MNmin相应MMmax相应N123组合项目值组合项目值组合项目值6上M-15.34-36.9236.92（1+3）*0.7516.19（1+2）*0.75-39.2（1+2）*0.75-39.2N113.36-18.1618.1698.6471.471.4下M13.8924.62-24.62（1+3）*0.75-8.05（1+2）*0.7528.88（1+2）*0.7528.88N130.64-18.1618.16111.684.3684.36V14.5617.1-17.1（1+2）*0.8526.9123.7523.751上M-16.94-96.0696.06（1+3）*0.863.3（1+2）*0.8-90.4（1+2）*0.8-90.4N793.82-407.06407.06960.7309.41309.41下M13.89100.07-100.07（1+3）*0.8-68.94（1+2）*0.891.17（1+2）*0.891.17N811.1-407.06407.06974.53323.23323.23V4.0450.54-50.54（1+2）*0.8546.3943.6643.66 第八章 框架梁柱截面配筋框架柱的配筋计算 混凝土强度：C30 fc=14.3N/mm2, ft=1.43 N/mm2ftk=2.01 N/mm2钢筋的强度：HPB235 fy=210N/mm fyk=235N/mm2HRB400 fy=360N/mm fyk=400N/mm2A轴柱：（1）、轴压比验算底层柱 Nmax=748.22kN轴压比 满足要求。则A轴柱的轴压比满足要求。（2）、截面尺寸复核取h0=400mm-35mm=365mm Vmax=36.75Kn因为：hw/b=365mm/400mm=0.91336.75kN 满足要求（3）、正截面受弯承载力的计算柱同一截面分别承受正反向弯矩，故采用对称配筋A轴柱：一层： =1fcbh0b=14.3N/mm2400mm365mm0.518=1081.48kN NNb,为大偏心受压。选用M大，N大的组合为最不利。 最不利组合：M=26.8KNm N=782.45KNM=90.17 KNm N=698.77KN 在弯矩中没有由水平荷载产生的弯矩，柱的计算长度=1.0H=3.6m 第一组：M=26.8KNm N=782.45KNei=e0+ea=34mm+20mm=54mm1.0 取=1.0因为 l0/h=3.6/0.4=915, 取=1.0 NNb,为大偏压。所以按构造配筋，最小总配筋率=0.5% =0.005400mm400mm=800mm2 每侧实配320（As=As=941mm2）第二组：M=90.17 KNm N=698.77KNei=e0+ea=129mm+20mm=149mm1.0 取=1.0因为 l0/h=3.6/0.4=915, 取=1.0 NNb,为大偏压。 =0.005400mm400mm=800mm所以按构造配筋，每侧实配320（As=As=941mm2）六层：最不利组合：M=30.39KNm N=757.93KN ei=e0+ea=40mm+20mm=60mm1.0 取=1.0因为 l0/h=3.6/0.4=915, 取=1.0 NNb,为大偏压。 0所以按构造配筋，每侧实配320（As=As=941mm2）（4）、垂直于弯矩作用平面的受压承载力验算一层：Nmax=782.45KNl0/b=3.6m/0.4m=9 查表得=0.99满足要求（5）、斜截面受剪承载力计算A轴柱：一层最不利组合：M=90.17KNm N=698.77KN V=36.75KN因为剪跨比，所以=3.0因N所以N=686.4KN 按构造配箍，取复式箍410250B轴柱：一层：=1081.48KN最不利组合：M=4.88KNM N=996.43KN M=91.17KNM N=323.23KN 第一组：M=4.88KNM N=996.43KN在弯矩中没有水平荷载产生的弯矩，柱的计算长度l0=1.0H=3.6m取=1.0因为l0/b=3.6m/0.4m=9 15所以=1.0 按构造配筋：由于最小总配筋率=0.005400mm400mm=800mm每侧实配320（As=As=941mm2） 第二组：M=91.17KNM N=323.23KN 取=1.0因为l0/b=3.6m/0.4m=9 15所以=1.0 按构造配筋：由于最小总配筋率=0.005400mm400mm=800mm每侧实配320（As=As=941mm2） 六层：最不利组合：M=28.88KNM N=84.36KN 取=1.0因为l0/b=3.6m/0.4m=9 Vb)504.96(Vb)504.96(Vb)504.96(Vb)0000 实配钢筋28350283502835028350注：所有钢筋均为级钢裂缝宽度验算取=71.01 KNm计算： mm =0.218mm用于正常使用极限状态验算的基本组合表（AB梁） 层恒载活载活载活载左风右风123456组合项目值组合项目值6左-16.05-0.82-0.82-0.82-1.71.71+2+3+4+0.6*5-19.53中27.081.711.711.71-0.460.461+2+3+4+0.6*632.49右-55.79-4.95-4.95-4.950.61-0.611+2+3+4+0.6*6-71.01 1左-19.06-6.420.18-0.27-11.611.61+（2+4）*0.7+5-35.34中13.586.48-0.250.59-2.282.281+2+4+0.6*622.02右-33.81-16.82-10.595.76-5.761+2+3+0.6*6-55.09用于正常使用极限状态验算的基本组合表（BC梁） 层恒载活载活载活载左风右风123456组合项目值组合项目值6左-52.37-5.32-5.32-5.321.25-1.251+2+3+4+0.6*6-69.08中-46.82-5.26-5.26-5.26001-46.82右-52.37-5.32-5.32-5.32-1.251.251+2+3+4+0.6*5-69.08 1左-25.65-15.33-4.340.9711.98-11.981+2+4+0.6*6-52.51中-22.02-7.171.8-7.17001+3-20.22右-25.650.97-4.34-15.33-11.9811.981+2+4+0.6*5-55.09 框架梁的裂缝宽度验算见下表层计算公式 梁AB 梁BC6左中右左中右19.5332.4971.0169.0846.8269.0865.89109.61239.57233.06157.96233.060.0080.0080.0080.0080.0080.008-1.38-0.390.4180.3680.020.3681616161616162.12.12.12.12.12.10.060.100.2180.2100.1420.210层计算公式 梁AB 梁BC1左中右左中右69.0846.8269.0852.5120.2252.51233.06157.96233.06177.1668.22177.160.0080.0080.0080.0080.0080.0080.3990.0660.3990.178-1.2940.1781616161616162.12.12.12.12.12.10.210.1420.210.1590.0610.159 第九章 框架柱基础尺寸确定及配筋计算1、 B轴柱基础的确定B轴柱承受的上部结构两组最不利内力：M=91.17KNM N=323.23KN V=43.66KNM=68.94 KNm N=974.53KN V=46.39KN 该框架结构层数不多，地基土较均匀且柱距较大，可选择独立基础，根据地质报告，基础埋深需在冲击粉质粘性土层上，取C25混凝土（fc=11.9N/mm2，ft=1.27N/mm2）；钢筋采用HRB335（fy=300N/mm2）2B轴柱独立基础的计算选择基础埋深 d=2400mm修正后的地基承载力特征值：地基承载力设计值 根据设计资料提供以及土力学可得； 承载力修正系数： 重度计算：人工填土 新冲击粘土 冲击粉质粘土 =（16+20+27.2）/2.4=26.3 假设b3m,故只对基础埋深进行修正。 =325+1.626.3（2.4-0.5）=404.95KPa 求基础底面积（地面尺寸）=974.53/404.95-20（2.4+0.50.75）=2.79 由于偏心荷载不大，基础底面积初步增大10%A=（1+0.1）A=1.12.79=3.07取方形基础取b=1.8m, l=1.8m A=1.8m1.8m=3.24持力层强度验算 当有时： =974.54+2062.78=1308.14KN =68.94+46.391=115.33 l/6=0.5m 符合要求1.2=1.2404.95=485.94KPa 故持力层强度满足要求。 当有时： =323.23+2062.78=656.83KN =91.17+43.661=134.83 l/6=0.5m 符合要求1.2=1.2404.95=485.94KP 故持力层强度满足要求。基础高度验算 地基净反力： 当有时：=1.35256.39-1.35333.6/3.24=271.11KPa =1.35179.65-1.35333.6/3.24=167.47KPa 当有时：=1.35154.36-1.35333.6/3.24=133.33KPa =1.3564.59-1.35333.6/3.24=12.14KPa取基础高度：h=800mm, =760mm. 基础分两级 下阶 取 柱边截面： =1.07 故柱边基础高度满足要求。 变阶处截面： =0.94 故变阶处高度满足要求。 配筋计算 计算基础的长边方向：截面：柱边地基净反力值：=12.14+（3+0.4）（271.11-12.14）/6=260.39KPa =260.39 III-III截面：柱边地基净反力值：=12.14+（3+1.2）（271.11-12.14）/6=374.70KPa =209.24 比较与，应按配筋，现于平行于l方向2m内宽度范围内配取复式箍1812120（1890）符合构造要求。 计算基础的长边方向：II-II截面： =109.15 IV-IV截面： =61.18比较与，应按配筋，现于平行于b方向3m内宽度范围内配取复式箍810160（552.68）符合构造要求。第十章 板的计算1 设计资料取三层楼面上的一块板进行计算，计算简图如下：楼面活荷载：q=2.5KN/M2材料选用： 混凝土强度等级C30 HRB235（f=210N/MM2） 2 板，梁的截面尺寸的确定 板厚：h=100mm 框架梁截面尺寸：bxh=250x600mm 连系梁截面尺寸：bxh=250x600mm 3. 板的计算 采用塑性理论计算 荷载计算：水磨石地面（包括水泥粗砂打底） 0.65KN/M2 100厚钢筋混凝土板 250.1=2.5KN/M2 10厚混合砂浆抹灰 0.0117=0.17KN/M2合计：3.32KN/M2 设计值：1.23.32=3.984KN/M2 活载：1.42.5=3.5KN/M2 总荷载：p=3.984+3.5=7.48KN/M2 板：ly=5.4，lx=3.6 n= ly/ lx=1.5 取=1/n2=0.44，=2 采用分离式配筋方式，跨中钢筋不弯起也不截断Mx=lymx=5.4mxMy=lxmy=lxmx=0.443.6mx=1.58mxMx=Mx”= lymx =lymx=25.4mx=10.8mxMy=My”= lxmy =lxmy =lxmx=20.443.6mx=3.17mx 区格板四周与梁连接，内力折减系数为0.8 2 Mx 2My +Mx +My+ Mx”+ My”=plx2(3 ly lx)/12 25.4mx+21.58mx+210.8mx+23.17mx=0.87.483.63.6(35.4-3.6)/12 mx =2.01KNm/m 故 mymx0.442.01=0.88KNm/m mx= mx”=mx=22.01=4.02KNm/m my= my”=my=2x0.88=1.76 KNm/m 取截面有效高度hx=80mm,hy=70mm,即可近似按计算钢筋截面面积 x方向：跨中：As=2.01106/(0.9521080)=126mm2 选用8150支座：As=4.02 106/(0.9521080)=256mm2 选用8100 y方向：跨中：As=0.88106/(0.9521070)=63mm2 选用8200支座：As=1.76106/(0.9521070)=126mm2 选用8150 配筋图见施工图第十一章 连续梁的计算1 三层楼面的一根连系梁进行计算梁尺寸：bxh=250mm600mm 计算跨度取支座中心线间距离l=l0=3.6m 净跨l0=3.6-0.3=3.3m2荷载计算板传来1.23.323(1-20.252+0.253)=10.64KN/M梁自重 1.2250.25(0.6-0.1)=3.75KN/M10厚混合砂浆抹1.20.01(0.6-0.1) 2+0.25 17=0.26KN/M2恒载： G=10.64+3.75+0.26=14.65KN/M活载： Q=板传活载1.423x(1-2x0.252+0.253)=7.48KN/M总荷载：G+Q=14.65+7.48=22.13KN/M 3内力计算弯矩计算：截面端支座边跨跨中离端第二支座离端第二跨跨中中间支座中间跨跨中弯矩系数m1/161/14-1/111/16-1/141/16M=m（G+Q）l2 (KN.m)-24.7920.49-26.0724.79-20.4924.79剪力计算：截面端支座内侧Ain离端第二支座 中间支座外侧Bex内侧Bin外侧Cex内侧Cin剪力系数V0.50.550.550.550.55V=V（G+Q）l0(KN)36.5240.1740.1740.1740.171配筋计算：跨中截面按T形截面进行承载力计算翼缘宽度 bf=l/3=3.6/3=1.2mV504.97V0.7 bh0ft（KN）0.72505651.43=141.39V141.39V箍筋肢数，直径构造配箍（2，8200）2，8200Tests on a Half-Scale Two-Story Seismic-Resisting Precast Concrete BuildingThis paper describes experimental studies on the seismic behavior and design of precast concrete buildings. A half-scale two-story precast concrete building incorporating a dual system and representing a parking structure in Mexico City was investigated. The structure was tested up to failure in a laboratory under simulated seismic loading. In some of the beam-to-column joints, the bottom longitudinal bars of the beam were purposely undeveloped due to dimensional constraints.Emphasis is given in the study on the evaluation of the observed global behavior of the test structure. This behavior showed that the walls of the test structure controlled the force path mechanism and significantly reduced the lateral deformation demands in the precast frames. Seismic design criteria and code implications for precast concrete structures resulting from this research are discussed. The end result of this research is that a better understanding of the structural behavior of this type of building has been gained results of simulated seismic load tests of a two story precast concrete building constructed with precast concrete elements that are used in Mexico are described herein. The structural system chosen in the test structure is the so called dual type, defined as the combination of structural walls and beam-to-column frames. Connections between precast beams and columns in the test structure are of the window type. This type of construction is typically used in low- and medium rise buildings in which columns are connected with windows at each story level. These windows contain the top and bottom reinforcement. Fig. 1 shows this type of construction for a commercial building in Mexico City. In most precast concrete frames such as those shown in Fig, 1, longitudinal beam bottom bars are not fully developed due to constraints imposed by the dimensions of file columns in beam-to-column joints. In an effort to overcome this deficiency, and as described later, some practicing engineers in Mexico design these joints by providing hoops around the hooks of that reinforcement in order to achieve its required continuity. However, this practice is not covered in the ACI Building Code (ACI318-02), nor in the Mexico City Building Code (MCBC, 1993). Part of this research was done to address this issue. The objectives of this research were Io evaluate the observed behavior of a precast concrete structures in the laboratory and to propose the use of precast structural elements or precast structures with both an acceptable level of expected seismic performance and appealing features from the viewpoint of construction Emphasis is given in this paper on the global behavior of the test structure. In the second part of this research which gill be presented in a companion paper, the observed behavior of connections between precast elements in the test structure, as well as the behavior of the precast floor system will be discussed in detail. Structural and non structural damages observed in buildings during past earthquakes throughout the world have shown the importance of controlling lateral displacement in structures to reduce building damage during earth- quakes. It is also relevant to mention that there are several cases of structures in moderate earthquakes in which the observed damage in non-structural elements in buildings was considerable even though the structural elements showed little or no damage. This behavior is also related Io excessive lateral displacement demands in the structure. To minimize seismic damage during earthquakes, the above discussion suggests the convenience of using a structural system capable of controlling lateral displacements in structures. A solution of this type is the so-called dual system. Studies by Paulay and Priestley4 on the seismic response of dual systems have shown that the presence of walls reduce the dynamic moment demands in structural elements in the frame subsystem. Also in conjunction with shake table tests conducted on a cast-in-place reinforced concrete dual system. Bertero5 has shown the potential of the dual system, in achieving excellent seismic behavior n this investigation, the dual system is applied to the case of precast concrete structures.DUCTILITY DEMAND IN DUAL SYSTEMSIn order to develop a base for a later analysis of the observed seismic response of the test structure studied in this project a simple analytical model is used to evaluate the main features of ductility demands in dual systems. Fig 2 shows the results of a simple approach to analyze the lateral load response iii a dual system. The lateral load has been normalized in such a manner that the combination of maximum lateral resistance in both subsystern i.e. walls and frames-leads to a lateral resistance of the global system equal to unity b is also assumed that both subsystems have the same maximum lateral resistance. In the first case (Fig 2a), it is assumed that the wall and frame subsystems have global displacement ductility capacities equal to 4 and 2 respectively. In the second case (Fig. 2b), the frame subsystem response is assumed to be elastic, and the lateral stiffness of the wall subsystem is taken to be 4 times that of the frame subsystem.As shown in Fig 2, the lateral deformation compatibility of the combined system is controlled by the lateral deformation capacity of the wall subsystem. In the first case Fig 2ak an elastic-plastic envelope for the lateral global response of the dual system is assumed, and the corresponding displacement ductility (u) is equal to 33.For the second case (Fig. 2b) with an elastic behavior of the frame subsystem, this ductility is equal to 25. These simple examples illustrate that in the analyzed cases, due to the higher flexibility in the frame subsystems as compared to those of the wall subsystern, in a dual system, the ductility demands in the frame subsystem result in smaller ductility values than those of the wall subsystem. This analytical finding was verified in this study from the experimental studies conducted on the test structure. This verification is later discussed in the paper It is of interest to note that results of the type shown in Fig. 2 have been also found by Bertero in shake table tests of a dual system. DESCRIPTION OF TEST STRUCTUREThe test structure used in this investigation is a two-story precast concrete building, representative of a low-rise parking structure located in the highest seismic zone of Mexico City. The prototype was constructed at one-half scale. For the sake of simplicity, ramps required in a parking structure have not been considered in the selected prototype structure. Their use, requiring large openings in the floor system, would have required a very complex model of the floor system for both linear and nonlinear analysis of the structure.A detailed description of the dimensions, materials, design procedures, and construction of the test structure can be found elsewhere.6 A summary of this information is given below. The dimensions and some characteristics of the test structure are shown in Fig. 3. The longitudinal and transverse are shown in Fig3a. Also, the exterior (longitudinal) frame containing the wall (Column Lines 1 and 3) are termed the lateral frame (see Fig, 3b), and the internal (longitudinal) frame with the single tee (Column Line 2) are termed the central frame. Doable tees spanning in the longitudinal direction are supported by L-shaped precast beams in the transverse direction as shown in Fig3a. The structure uses precast frames and precast structural walls, the latter elements functioning as the main lateral load resisting system. Fig. 4 shows an early phase of the construction of the test structure. As can be seen, the windows in the columns and walls are left in these elements for a later assemblage with the precast beams.The unfastened design base shear required by the Mexico City Building Code (MCBC, 1993)2 is 0.2WT, where WT is the total weight of the prototype structure, assuming a dead load of 5,15 KPa (108 psi) and a live load of 0.98 KPa (20.5 psi). The prototype structure was designed using procedures of elastic analyses and proportioning requirements of the MCBC, In these analyses, the gross moment of inertia of the members in the structure was considered and rigid offsets (distances from the joints to the face of the supports) were assumed for all beams in the structure except for beams in the central frame, which had substandard detailing as will be described latch. Results from these analyses indicated that the structural walls in the test structure would take about 65 percent of the design lateral loads. A review of the nominal lateral resistance of the structure using the MCBC procedures showed that this resisting force was about 1.3 times the required code lateral resistance (0,2Wr), This is one of several factors, later discussed, that contributed to the over-strength of the structure.The longitudinal reinforcement in all the structural elements of the test structore was deformed bars from Grade 420 steel. Table 1 lists the concrete compressive cylinder strengths for different members of the prototype structure. Fig. 5 shows typical reinforcing details for precast beams spanning in the direction of the applied lateral load (see Fig. 3). Figs. 6 and 7 show reinforcing details for the columns, and for the structural wails and their foundation, respectively. It should be mentioned that the test structure was designed with the requirements for moderately ductile structures specified by the MCBC. According to these provisions, the test structure did not require special structural walls with boundary elements such as those specified in Chapter 21 of AC1 318 02.The precast two-story columns were connected to the precast foundation by unthreading them in a grouted socket type connection. The reinforcing details of the foundation, as well as its design procedure and behavior in the test structure are discussed in the companion paper? Tae beam-to-cadmium joints in file test structure were cast-in-place to enable positioning the longitudinal reinforcement of the framing beams. The beam top reinforcement was placed in sum on top of the precast beams. Fig. 8 shows typical reinforcing details for the joints in the double tees of the central frame. Since these tees and their supporting L-shaped beams in Axes A or C (see Fig. 3) had the same depth, the hooked bottom longitudinal bars in the double tees could not pass through the full depth of the column because of interference with the bottom bars from file transverse beam (see Fig. g).As a result, these hooked bars possessed only about 55 percent of the development length required by Chapter 21 of ACI 318-02. In an attempt to anchor these hooked bars, some designers in Mexico provide closed hoops around the hooks, as shown in Fig. 8. The effectiveness of this approach was also studied in the companion paper.3 Beam to-column joints in the lateral frames of the test structure had transverse beams that were deeper than the longitudinal beams. This made it possible for the top and bottom bars of the longitudinal beams to pass through the full joint, and, therefore, these bars achieved their required development length.Cast-in-place topping slabs in the test structure were 30 mm (1.18 in.) thick and formed the diaphragms in January-February 2005 Fig. 3. Plan and elevation of test structure: (a) Plan; (b) Lateral frame; (c) Transverse frame. Dimensions in mm. Note: 1 mm - 0.0394 in. the structural system. Welded wire reinforcement (WWR) was used as reinforcement for the topping slabs. The amount of WWR ill the topping slabs was controlled by the temperature and shrinkage provisions of the MCBC. which are similar to those of AC 318-02.It is of interest to mention that the requirements for shear strength in the diaphragms given by these provisions, which are similar to those of ACI 318-89, did not control the design. A wire size of 6 x 6 in. 10/10 led to a reinforcing ratio of 0.002 in the topping slab. The strength of the WWR at yield and fracture obtained from tests were 400 and 720 MPa(58 and 104 ksi),respectively.TEST PROGRAM AND INSTRUMENTATIONTest ProgramThe test structure was subjected to simulate seismic loading in the longitudinal direction (see Fig. 3a). Quasitatic cyclic lateral loads FI and F2 were applied at the first and second levels of the structure, respectively (see Fig. 3b). The ratio of F2 to FI was held constant throughout the test., with a value equal to 2.0. This ratio represents an inverted triangular distribution of loads, which is consistent with the assumptions of most seismic codes including the MCBC.The test setup is shown in Fig. 9.The structure had Hinges A, B, and Cat each slab level as shown in Fig. 9b.The purpose of the hinges was to avoid unrealistic restrictions in the structure by allowing the ends of the slabs to rotate freely during lateral load testing. As can be seen in Fig 9, the lateral loads were applied by hydraulic actuators that work in either tension or compression.When the actuators worked in compression, they applied the loads directly at one side of the structure. However, when the actuators applied tensile loads at one side of the test structure, they were convened to compression loads at the other side by means of four high strength reinforcing bars for each actuator see D32 (1 in) reinforcing bars in Fig. 9. Both ends of these bars were attached to 50 mm (2 in.) thick steel plates At each of the floor levels, two of these plates were part of Hinge A, and the other two plates at the actuator side were part of Hinge B (see Fig. 9b).As can be seen in Fig 9b before the application of tensile loads in the actuators, the latter end of the plates left a clear space with the end of the slab. This space at zero lateral load was about 50 mm (2 in.), and it allowed for beam elongation of the structure which occurs during the formation of plastic hinges in the beams? For the case of compressive loads on the actuators acting on the transverse beam at the side of Hinge B (see Fig. 9b), the system also allowed 50 mm (2 in.) of beam elongation. These particular features of the test setup allowed the application of compressive loads at points in the slabs with no special reinforcement at these points. If tensile forces had been applied to loading points in the slabs, it is very likely that these loading points would have required unrealistic, special reinforcing details that are not represent five of those in an actual structure. The gravity load was represented by 53 steel ingots, acting at each level of the structure, with the layout shown in Fig. 9a, The weight per unit area of the ingots per level was 2.79 KPa (58.3 psf), which added to the self-weight of tile slab, leading to a total floor dead load of 5.37 KPa (112 psf). This is 88 percent of the code required gravity load for seismic design (MCBC, 1993). It was not possible to apply the remaining 12 percent of gravity load due to space limitations in the slabs. The total weight of the structure, omitting the weight of the foundation, was 284.2 KN (63.9 kips). Transverse displacements in the structure (perpendicular to the loading direction) were precluded by using steel ball hearings installed in the lateral faces of beam to-column joints of the second level at Column Lines A1and A3 (see Fig. 9a). These ball bearings were supported by a steel frame. As shown in Fig. 10, the precast columns and wails were fixed to the strong floor using steel beams supported by the foundation and anchored to the floor. The lateral loading history used in the test structure was based on force control during elastic response of the structure, followed by displacement control during yield fluctuations, using the lateral roof displacement of the structure A.The target lateral load or top displacement was typically reached by incremental loading of both actuators in which the bottom actuator was fore-contrived to half the load value of the top actuator. A cycle of lateral load of about 0.75Vg was initially applied, where VR is the nominal theoretical base shear strength computed using the ACI 318 02 provisions. The value of VR, 198 KN (44.5 kips), can be assumed to correspond to first yield in the structure.This parameter was computed using measured material properties, a strength reduction factor (1) equal to unity and common assumptions for flexural strength of reinforced concrete sections. For the lateral force of 0.75VR, the corresponding lateral roof displacement was defined as 0.75y. Using this value and assuming elastic behavior, the calculated value of the displacement at first yielding, Ny, was equal to 4.5 mm (0.18 in.). After the cycle at 0.75y, the structure was subjected to three cycles for each of the maximum prescribed roof displacements 2y, 4y, 6y, 13y and 20y-which correspond to roof drift ratios, Dr, of 0.003, 0.006, 0.009, 0.020 and 0.031, respectively. The roof drift ratio can be defined as: Dr=A/H (1) where H is the height of the structure measured from the fixed ends of first story columns. This value is equal to 2920 nun (115 in.).The complete cyclic lateral loading history applied to the structure is shown in Fig. 11. This loading history is expressed in terms of both displacement ductility, A/Ny, and roof drift ratio, Dr. In addition, the peaks of lateral displacements in Fig. 11 are related to the measured base shear force, V, expressed as the ratio V/VR.InstrumentationA detailed description of the instrumentation of the structure can be found in the report by Rodriguez and Blandon? Lateral displacements of the structure were measured with linear potentiometers placed at each level of the structure as shown in Fig. 12. Beam elongation, discussed later, was evaluated from measurements using this instrumentation. Twelve pairs of potentiometers were used for measuring the average curvatures in critical sections of two columns of the structure. In addition, nine pairs of potentiometers were instrumented in vertical sections of beams in central and lateral frames, and in sections of a wall base. Electrical resistance strain gauges were placed in the longitudinal reinforcement of some beams, columns and wails of the test structure, as well as in the hoops of beam to-column joints with substandard reinforcing details. Some of the measurements from this instrumentation and from the potentiometers in the beams are discussed in the companion paper. Here, the observed experimental response of the beam to-column joints with substandard reinforcing details is evaluated.EXPERIMENTAL RESULTSThe applied lateral loading history in the test structure is shown again in Fig.13. in this case, the peaks of lateral displacement are related to the most important events observed during testing, such as first yielding of the longitudinal reinforcement, first cracking in walls and topping slabs, loss of concrete cover, and buckling of longitudinal reinforcement.Fig. 14 shows the measured base shear force, V, versus roof drift ratio hysteretic loops. The ordinate of this graph represents the measured base shear expressed in a dimensionless form as the ratio V/VR. As shown in Fig. 14a, first yielding of the longitudinal reinforcement occurred at wails in the critical sections at foundation faces, at a roof drift ratio of about 0.0030, corresponding t9 a base shear force equal to about 1.4VR. The maximum measured base shear was equal to 549KN(123.4 kips) or 2.77VR, corresponding to a roof drift ratio of 0.020.Fig. 14b shows envelopes of interstory drift ratio, dr, measured during testing in the two levels of the structure. The results show that the two levels had similar values of interstory drifts, which is due to the significant contribution of the structural walls to the global response of the structure. This feature of the displacement profile of the structure suggests that for a given Dr Value, interstory drift values would be similar in magnitude.Cyclic lateral loading was terminated at a roof drift ratio of 0.034 corresponding to a base shear force of 462KN (103.8 kips) or 2.3Vn. At this level of lateral displacement, the buckling of longitudinal reinforcement at the fixed ends of the first-story walls was excessive, and it led to significant out-of-plane displacements of the walls along with wide cracks in the topping slab-to-wall joints as discussed in the companion paper? Fig. 14a also shows some of the most important events observed during testing. A summary of relevant damage observed during testing of die structure is described below. Wall damage was initiated by the loss of concrete cover and buckling of longitudinal reinforcement at the fixed ends of the first-story walls. These events occurred at a D, value equal to about 0.020. Buckling of the longitudinal reinforcement occurred immediately after the loss of concrete cover in these critical sections. Figs. 15 and 16 illustrate the cracking pattern and damage observed at the end of testing for the lateral frame and for the central frame, respectively. Fig. 17 shows an overall view of the damage of the lateral frame at file end of testing, and Fig. 18 provides a closer look at the buckling of longitudinal reinforcement at the fixed end of a first-story wall at the end of testing. Figs. 15 through 18 indicate that the observed damage at the end of testing in the columns and beams to-column, and beam column joints was significantly less than that of the walls.The formation of plastic hinges in the critical sections of the structural elements, such as at the fixed ends of the first story columns and wails, and in beam ends at wall faces, was observed during testing, especially after reaching the maximum base shear. Evidence of buckling of the longitudinal reinforcement was also observed in some of these sections of columns and beams (see Figs. 15 and 16), foremen of some beams, columns and wails of the test structure, as well as in the hoops of beam to-column joints with substandard reinforcing details. Some of the measurements from this instrumentation and from the potentiometers in the beams are discussed in the companion paper. Here, the observed experimental response of the beam to-column joints with substandard reinforcing details is evaluated. 一个未完工的二层预制混凝土结构物的抗震测试这篇文章是关于地震和预制混凝土建筑物设计的试验性的研究。墨西哥市里一个带有双重系统