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框架
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毕业设计
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6层框架住宅毕业设计结构计算书带CAD图,框架,住宅,毕业设计,结构,计算,cad
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某 办 公 楼 框 架 结 构 设 计专业：土木工程姓名：学号：指导教师： 前 言毕业设计是大学本科教育培养目标实现的重要阶段，是毕业前的综合学习阶段,是深化、拓宽、综合教和学的重要过程,是对大学期间所学专业知识的全面总结。本组毕业设计题目为某办公楼框架结构设计。在毕设前期，我温习了结构力学、钢筋混凝土、建筑结构抗震设计等知识，并借阅了抗震规范、混凝土规范、荷载规范等规范。在毕设中期，我们通过所学的基本理论、专业知识和基本技能进行建筑、结构设计。特别是在SARS肆掠期间，本组在校成员齐心协力、分工合作，发挥了大家的团队精神。在毕设后期，主要进行设计手稿的电脑输入，并得到老师的审批和指正，使我圆满的完成了任务，在此表示衷心的感谢。毕业设计的三个月里，在指导老师的帮助下，经过资料查阅、设计计算、论文撰写以及外文的翻译，加深了对新规范、规程、手册等相关内容的理解。巩固了专业知识、提高了综合分析、解决问题的能力。在进行内力组合的计算时，进一步了解了Excel。在绘图时熟练掌握了AutoCAD，以上所有这些从不同方面达到了毕业设计的目的与要求。框架结构设计的计算工作量很大，在计算过程中以手算为主，辅以一些计算软件的校正。由于自己水平有限，难免有不妥和疏忽之处，敬请各位老师批评指正。 二零零三年六月十五日 内容摘要本设计主要进行了结构方案中横向框架2、3、7、8轴框架的抗震设计。在确定框架布局之后，先进行了层间荷载代表值的计算,接着利用顶点位移法求出自震周期，进而按底部剪力法计算水平地震荷载作用下大小，进而求出在水平荷载作用下的结构内力（弯矩、剪力、轴力）。接着计算竖向荷载（恒载及活荷载）作用下的结构内力，。 是找出最不利的一组或几组内力组合。 选取最安全的结果计算配筋并绘图。此外还进行了结构方案中的室内楼梯的设计。完成了平台板，梯段板，平台梁等构件的内力和配筋计算及施工图绘制。关键词： 框架 结构设计 抗震设计 Abstract The purpose of the design is to do the anti-seismic design in the longitudinal frames of axis 2、3、7、8. When the directions of the frames is determined, firstly the weight of each floor is calculated .Then the vibrate cycle is calculated by utilizing the peak-displacement method, then making the amount of the horizontal seismic force can be got by way of the bottom-shear force method. The seismic force can be assigned according to the shearing stiffness of the frames of the different axis. Then the internal force (bending moment, shearing force and axial force ) in the structure under the horizontal loads can be easily calculated. After the determination of the internal force under the dead and live loads, the combination of internal force can be made by using the Excel software, whose purpose is to find one or several sets of the most adverse internal force of the wall limbs and the coterminous girders, which will be the basis of protracting the reinforcing drawings of the components. The design of the stairs is also be approached by calculating the internal force and reinforcing such components as landing slab, step board and landing girder whose shop drawings are completed in the end.Keywords : frames, structural design, anti-seismic design毕业设计进度计划安排:第一周 课题调研，选题、结构方案设计第二周 复习结构设计有关内容，借阅相关规范及资料第三周 导荷载、计算楼层荷载第四周 计算结构自震周期、水平地震作用大小第五、六周 水平地震力作用下的框架内力分析、计算第七、八周 竖向荷载作用下的框架内力计算第九周 框架内力组合分析、计算第十周 义务劳动周第十一周 楼层板配筋、框架梁配筋及构造第十二周 框架柱配筋、框架节点配筋及构造第十三周 TBSA计算及分析第十四周 楼梯计算，整理计算书第十五周 翻译科技资料，画图第十六周 打印论文，准备答辩目 录第一部分：工程概况1建筑地点、建筑类型、建筑介绍、门窗使用、地质条件 1柱网与层高 1框架结构承重方案的选择 2框架结构的计算简图 2梁、柱截面尺寸的初步确定 3第二部分：框架侧移刚度的计算5横梁、纵梁、柱线刚度的计算 5各层横向侧移刚度计算 6各层纵向侧移刚度计算12第三部分：重力荷载代表值的计算 13资料准备13重力荷载代表值的计算14第四部分：横向水平荷载作用下框架结构的内力和侧移计算23横向自振周期的计算23水平地震作用及楼层地震剪力的计算24多遇水平地震作用下的位移验算27水平地震作用下框架内力计算28第五部分：竖向荷载作用下框架结构的内力计算32计算单元的选择确定32荷载计算33内力计算40梁端剪力和柱轴力的计算45横向框架内力组合46框架柱的内力组合54柱端弯矩设计值的调整57柱端剪力组合和设计值的调整60第六部分：截面设计62框架梁62框架柱68框架梁柱节点核芯区截面抗震验算78第七部分：楼板设计82楼板类型及设计方法的选择82设计参数82弯矩计算83截面设计87第八部分：楼梯设计91设计参数91楼梯板计算91平台板计算93平台梁计算94第九部分：框架变形验算96梁的极限抗弯承载力计算96柱的极限抗弯承载力计算97确定柱端截面有效承载力Mc98各柱的受剪承载力Vyij的计算 99楼层受剪承载力Vyi的计算100罕遇地震下弹性楼层剪力Ve的计算 101楼层屈服承载力系数yi的计算101层间弹塑性位移验算 103第十部分： 科技资料翻译104科技资料原文 104原文翻译 113第十一部分：设计心得120参考资料 123第一部分：工程概况建筑地点：北京市建筑类型：六层办公楼，框架填充墙结构。建筑介绍：建筑面积约1000平方米，楼盖及屋盖均采用现浇钢筋混凝土框架结构，楼板厚度取120mm，填充墙采用蒸压粉煤灰加气混凝土砌块。门窗使用：大门采用钢门，其它为木门，门洞尺寸为1.2m2.4m，窗为铝合金窗，洞口尺寸为1.8m2.1m。地质条件：经地质勘察部门确定，此建筑场地为二类近震场地，设防烈度为8度。柱网与层高：本办公楼采用柱距为7.2m的内廊式小柱网，边跨为7.2m，中间跨为2.4m，层高取3.6m，如下图所示： 柱网布置图 框架结构承重方案的选择：竖向荷载的传力途径：楼板的均布活载和恒载经次梁间接或直接传至主梁，再由主梁传至框架柱，最后传至地基。根据以上楼盖的平面布置及竖向荷载的传力途径，本办公楼框架的承重方案为横向框架承重方案，这可使横向框架梁的截面高度大，增加框架的横向侧移刚度。框架结构的计算简图： 框架结构的计算简图 纵向框架组成的空间结构 横向框架组成的空间结构本方案中，需近似的按纵横两个方向的平面框架分别计算。 梁、柱截面尺寸的初步确定：1、梁截面高度一般取梁跨度的1/12至1/8。本方案取1/127200=600mm，截面宽度取6001/2=300mm，可得梁的截面初步定为bh=300*600。2、框架柱的截面尺寸根据柱的轴压比限值，按下列公式计算： （1）柱组合的轴压力设计值N=Fg E n注：考虑地震作用组合后柱轴压力增大系数。F按简支状态计算柱的负载面积。g E 折算在单位建筑面积上的重力荷载代表值，可近似的取14KN/m2。n为验算截面以上的楼层层数。 （2）AcN/uNfc 注：uN 为框架柱轴压比限值，本方案为二级抗震等级，查抗震规范可知取为0.8。 fc 为混凝土轴心抗压强度设计值，对C30，查得14.3N/mm2。3、计算过程：对于边柱：N=Fg E n=1.325.92146=2830.464（KN） AcN/uNfc=2830.464103/0.8/14.3=247418.18（mm2） 取700mm700mm 对于内柱：N=Fg E n=1.2534.56146=3628.8（KN） AcN/uNfc=3628.8*103/0.8/14.3=317202.80（mm2） 取700mm700mm梁截面尺寸（mm）混凝土等级横梁（bh）纵梁（bh）AB跨、CD跨BC跨C30300600250400300600柱截面尺寸（mm）层次混凝土等级bh1C307007002-6C30650650第二部分：框架侧移刚度的计算一、 横梁线刚度i b的计算：类别Ec（N/mm2）bh（mmmm）I0（mm4）l（mm）EcI0/l（Nmm）1.5EcI0/l（Nmm）2EcI0/l（Nmm）AB跨、CD跨3.01043006005.4010972002.2510103.3810104.501010BC跨3.01042504001.3310924001.6710102.5010103.341010二、 纵梁线刚度i b的计算：类别Ec（N/mm2）bh（mmmm）I0（mm4）l（mm）EcI0/l（Nmm）1.5EcI0/l（Nmm）2EcI0/l（Nmm）跨3.01043006005.410942003.8610105.7910107.711010其它跨3.01043006005.410972002.2510103.3810104.501010三、 柱线刚度i c的计算：I=bh3/12层次hc（mm）Ec（N/mm2）bh（mmmm）Ic（mm4）EcIc/hc（Nmm）147003.01047007002.001101012.7710102-636003.01046506501.448101012.4010101层A-5柱83003.01047007002.00110107.231010四、各层横向侧移刚度计算： (D值法)1、底层、A-2、A-3、A-7、A-8、D-2、D-3、D-4、D-7、D-8（9根）K=0.352c=(0.5+K)/(2+K)=0.362Di1=c12ic/h2=0.3621212.771010/47002=25112、A-1、A-4、A-6、A-9、D-1、D-5、D-6、D-9 （8根）K=3.38/12.77=0.266c=(0.5+K)/(2+K)=0.338Di2=c12ic/h2 =0.3381212.771010/47002 =23447、B-1、C-1、B-9、C-9 （4根）K=（2.5+3.38）/12.77=0.460c=(0.5+K)/(2+K)=0.390Di3=c12ic/h2 =0.3901212.771010/47002 =27055、B-2、C-2、B-3、C-3、C-4、B-7、C-7、B-8、C-8 （9根）K=（3.34+4.5）/12.77=0.614c=(0.5+K)/(2+K)=0.426Di4=c12ic/h2 =0.4261212.771010/47002 =29552、B-4、B-6、C-5、C-6 （4根）K=（3.34+3.38）/12.77=0.526c=(0.5+K)/(2+K)=0.406Di5=c12ic/h2 =0.4061212.771010/47002 =28165、B-5 （1根）K=3.34/12.77=0.262c=(0.5+K)/(2+K)=0.337Di6=c12ic/h2 =0.3371212.771010/47002 =23378D1=251129+234478+270554+295529+281654+23378 =9238102、第二层：、A-2、A-3、A-7、A-8、D-2、D-3、D-4、D-7、D-8 （9根）K=4.52/（12.42）=0.363c=K/(2+K)=0.154Di1=c12ic/h2 =0.1541212.41010/36002 =17681、A-1、A-9、D-1、D-5、D-6、D-9 （6根）K=3.382/（12.42）=0.273c=K/(2+K)=0.120Di2=c12ic/h2 =0.1201212.41010/36002 =13778、A-5 （1根）K=4.5/7.233=0.622c=（0.5+K）/(2+K)=0.428Di3=c12ic/h2 =0.428127.2331010/83002 =5392、A-4、A-6 （2根）K=（4.5+3.8）/（12.4*2）=0.318c=K/(2+K)=0.137Di4=c12ic/h2 =0.1371212.41010/36002 =15730、B-1、C-1、B-9、C-9 （4根）K=（2.5+3.38）2/（12.42）=0.474c=K/(2+K)=0.192Di5=c12ic/h2 =0.1921212.41010/36002 =22044、B-2、B-3、C-2、C-3、C-4、B-7、B-8、 C-7、 C-8 （9根）K=（3.34+4.5）2/（12.42）=0.632c=K/(2+K)=0.240Di6=c12ic/h2 =0.240*12*12.4*1010/36002 =27556、B-4、B-6 （2根）K=（3.342+4.5+3.38）/（12.42）=0.587c=K/(2+K)=0.227Di7=c12ic/h2 =0.2271212.41010/36002 =26063、C-5、C-6 （2根）K=（3.34+3.38）2/（12.42）=0.542c=K/(2+K)=0.213Di8=c12ic/h2 =0.2131212.41010/36002 =24456、B-5 （1根）K=3.342/（12.42）=0.269c=K/(2+K)=0.119Di9=c12ic/h2 =0.1191212.41010/36002 =13663D2=176819+137786+5392+157302+220444+275569+260632+244562+13663=7295303、第三层至第六层：、A-2、A-3、A-4、A-5、A-6、A-7、A-8、D-2、D-3、D-4、D-7、D-8 （12根）Di1=17681、A-1、A-9、D-1、D-9、D-5、D-6 （6根）Di2=13778、B-1、C-1、B-9、C-9 （4根）Di3=22044、B-2、B-3、B-4、B-5、C-2、C-3、C-4、B-6、B-7、B-8、C-7、C-8 （12根）Di4=27556、C-5、C-6 （2根）Di5=24456D3-6=1768112+137786+220444+2755612+244562 =7626004、顶层：、D-5、D-6 （2根）K=3.382/（12.42）=0.273c=K/(2+K)=0.120Di1=c12ic/h2 =0.1201212.41010/36002 =13778 、C-5、C-6 （2根）K=（3.382+3.34）/（12.42）=0.407c=K/(2+K)=0.169Di2=c12ic/h2 =0.1691212.41010/36002 =19404D顶=137782+194042=66364由此可知，横向框架梁的层间侧移刚度为：层次123456顶Di（N/mm）92381072953076260076260076260076260066364D1/D2=923810/7295300.7，故该框架为规则框架。五、各层纵向侧移刚度计算：同理，纵向框架层间侧移刚度为：层次123456顶层Di（N/mm）1035634935623942398942398942398942398103102D1/D2=1035634/9356230.7，故该框架为规则框架。第三部分：重力荷载代表值的计算一、资料准备：查荷载规范可取：、 屋面永久荷载标准值（上人）30厚细石混凝土保护层 220.03=0.66KN/m2 三毡四油防水层 0.4 KN/m220厚矿渣水泥找平层 14.50.02=0.29 KN/m2150厚水泥蛭石保温层 50.15=0.75 KN/m2120厚钢筋混凝土板 250.12=3.0 KN/m2V型轻钢龙骨吊顶 0.25 KN/m2（二层9mm纸面石膏板、有厚50mm的岩棉板保温层） 合计 5.35 KN/m2、1-5层楼面：木块地面（加防腐油膏铺砌厚76mm） 0.7 KN/m2120厚钢筋混凝土板 250.12=3.0 KN/m2V型轻钢龙骨吊顶 0.25 KN/m2 合计 3.95 KN/m2、屋面及楼面可变荷载标准值：上人屋面均布活荷载标准值 2.0 KN/m2楼面活荷载标准值 2.0 KN/m2屋面雪荷载标准值 SK=urS0=1.00.2=0.2 KN/m2 （式中ur为屋面积雪分布系数）、梁柱密度25 KN/m2蒸压粉煤灰加气混凝土砌块 5.5KN/m3二、重力荷载代表值的计算：1、第一层：（1）、梁、柱：类别净 跨（mm）截 面（mm）密 度（KN/m3）体 积（m3）数 量(根)单 重（KN）总 重（KN）横梁6500300600251.171729.25497.251700250400250.1794.2538.25纵梁6500300600251.172829.25819.003500300600250.63415.7563.00类别计算高度（mm）截 面（mm）密 度（KN/m3）体 积（m3）数 量（根）单 重（KN）总 重（KN）柱3600700700251.7643644.11587.6（2）、内外填充墙重的计算： 横墙： AB跨、CD跨墙：墙厚240mm，计算长度6500mm，计算高度3600-600=3000mm。 单跨体积：0.246.53=4.68m3 单跨重量：4.685.5=25.74KN 数量：17 总重：25.7417=437.58KN BC跨墙：墙厚240mm，计算长度1700mm，计算高度3600-600=3000mm。 单跨体积：（1.73-1.5*2.4）0.24=0.36m3 单跨重量：0.365.5=1.98KN 数量：2 总重：1.982=3.96KN 厕所横墙：墙厚240mm，计算长度7200-2400=4800mm，计算高度3600-120=3480mm。 体积：0.244.83.48=4.009m3 重量：4.0095.5=22.0495KN 横墙总重：437.58+3.96+22.0495=463.5895KN 纵墙： 跨外墙：单个体积：（6.53.0）-（1.82.12） 0.24=2.8656 m3数量：12总重：2.8656125.5=189.1296KN 厕所外纵墙：体积：6.53.0-1.82.1=15.72 m3总重：15.725.5=86.46KN 楼梯间外纵墙：体积：3.53.0-1.82.1=6.72 m3总重：6.725.5=36.96KN 门卫外纵墙：体积：3.53.0-1.22.4=7.62m3总重：7.625.5=41.91KN 内纵墙：单个体积：(6.53.0-1.22.4*2) 0.24=13.74m3 单个重量：13.745.5=75.57KN数量：12总重：75.5712=906.84KN 厕所纵墙：单个体积：0.24（3.6-0.12）4.93=4.1175m3 单个重量：4.11755.5=22.6463KN数量：2总重：22.64632=45.2926KN 正门纵墙：总重：（1.86.5-1.82.1）0.245.5=10.4544KN 纵墙总重：189.1296+86.46+36.96+41.91+906.84+45.2926+10.4544=1317.0466KN（3）、窗户计算（钢框玻璃窗）： 走廊窗户：尺寸：1800mm2100mm 自重：0.4KN/m2 数量：26 重量：1.82.10.426=39.312KN 办公室窗户：尺寸：1500mm2100mm 自重：0.4KN/m2 数量：2重量：1.52.10.42=2.52KN 总重：39.312+2.52=41.832KN（4）、门重计算： 木门：尺寸：1200mm2400mm 自重：0.15KN/m2 数量：26.25 重量：1.22.40.1526.25=11.34KN 铁门：尺寸：6500mm3000mm 自重：0.4KN/m2 数量：0.5重量：6.53*0.40.5=3.9KN 总重：11.34+3.9=15.24KN（5）、楼板恒载、活载计算（楼梯间按楼板计算）： 面积：48.441613+117.4176+30.24=777.3984（m2） 恒载：3.95777.3984=3070.7237KN 活载：2.0777.3984=1554.7968KN由以上计算可知，一层重力荷载代表值为G1=G 恒+0.5G活=（497.25+38.25）1.05+（819+63）1.05+1587.61.05+463.5895+1317.0466+41.832 +15.24+（3070.7237+1554.7968）0.5 =9618.5836KN注：梁柱剩上粉刷层重力荷载而对其重力荷载的增大系数1.05。2、第二层：（1）、梁、柱横梁：AB跨：300mm600mm 29.25KN18根=526.5KNBC跨：250mm400mm 4.25KN9根=38.25KN 纵梁：819+63=882KN 柱： 类别计算高度（mm）截 面（mm）密 度（KN/m3）体积（m3）数 量（根）单 重（KN）总 重（KN）柱3600650650251.5213638.0251368.9（2）、内外填充墙重的计算：横墙总重：463.5895KN纵墙： 比较第二层纵墙与第一层的区别有： 大厅：一层有铁门 二层A、B、B跨有内墙。 比较异同后，可得第二层纵墙总重为： 1317.0466+（3.06.55-21.82.1）0.245.5-3.9+（1.56.55-1.51.2）0.245.5+（1.53.55-1.51.2）0.245.5 =1317.0466+15.9588-3.9+10.593+4.653 =1344.3514KN（3）、窗户计算（钢框玻璃窗）： 第一类：尺寸：1800mm2100mm 自重：0.4KN/m2 数量：29 重量：1.82.10.429=43.848KN 第二类：尺寸：1500mm2100mm 自重：0.4KN/m2 数量：2重量：1.52.10.42=2.52KN 总重：43.848+2.52=46.368KN（4）、门重计算： 木门：尺寸：1200mm2400mm 自重：0.15KN/m2 数量：27.25 重量：1.22.40.1527.25=11.772KN（5）、楼板恒载、活载计算（楼梯间按楼板计算）： 面积：777.3984+11.166.96=855.072（m2） 恒载：3.95855.072=3377.5344KN 活载：2.0855.072=1710.144KN由以上计算可知，二层重力荷载代表值为G2=G 恒+0.5G活=（526.5+38.25）1.05+8821.05+1368.91.05+463.5895+1344.3514+46.368 +11.772+（3377.5344+1710.144）0.5 =9910.1918KN注：梁柱剩上粉刷层重力荷载而对其重力荷载的增大系数1.05。3、第三层至第五层：比较其与第三层的异同，只有B、B不同，可得三到五重力荷载代表值为：G3-5=9910.1918-10.593-4.653+（3.010.1-21.22.4）0.245.5=9927.3386KN4、第六层重力荷载代表值的计算：横梁：526.5+38.25=564.75KN 纵梁：882KN 柱：计算高度：2100mm截面：650mm650mm数量：36总重：0.650.652.12536=798.525KN 横墙：463.5895/2=231.7948KN 纵墙：（1344.3514+32.3928-10.593-4.653）/2=680.7491KN 窗重：46.368/2=23.184KN 木门重：门高2400mm，计算高度为门的1500mm以上，故系数=（2.4-1.5）/2.4=3/8则木门重：11.7723/8=4.4145KN 屋面恒载、活载计算： 恒载：855.0725.35=4574.6352KN 活载：855.0722.0=1710.144KN 雪载：855.0720.2=171.0144KN由以上计算可知，六层重力荷载代表值为G6=G 恒+0.5G活=（564.75+882+798.525）1.05+231.7948+680.7491+23.184+4.4145+4574.6352 +（1710.144+171.0144）0.5 =9753.4748KN注：梁柱剩上粉刷层重力荷载而对其重力荷载的增大系数1.05。5、顶端重力荷载代表值的计算：横梁：29.252=58.5KN纵梁：15.752=31.5KN柱：38.0254=152.1KN横墙：25.742=51.48KN纵墙：（3.03.55-1.22.4）0.245.5+（3.03.55-1.82.1）0.245.5=19.3248KN 木门：1.22.40.15=0.432KN 窗：1.82.10.4=1.512KN 楼板恒载、活载计算： 面积：4.27.2=30.24m2 恒载：30.245.35=161.784KN 活载：30.242.0=60.48KN雪载：30.240.2=6.048KN由以上计算可知，顶端重力荷载代表值为G顶=G 恒+0.5G活=58.5+31.5+51.48+19.3248+152.1+0.432+1.512+161.784+(60.48+6.048) 0.5 =543.1608KN集中于各楼层标高处的重力荷载代表值G i的计算结果如下图所示：第四部分：横向水平荷载作用下框架结构的内力和侧移计算一、横向自振周期的计算： 横向自振周期的计算采用结构顶点的假想位移法。按式Ge=Gn+1（1+3h1/2/H）将突出房屋重力荷载代表值折算到主体结构的顶层，即：Ge=543.16081+33.6/（3.65+4.7）=650.8153（KN）基本自振周期T1（s）可按下式计算：T1=1.7T （uT）1/2注：uT假想把集中在各层楼面处的重力荷载代表值Gi作为水平荷载而算得的结构顶点位移。T结构基本自振周期考虑非承重砖墙影响的折减系数，取0.6。uT按以下公式计算：VGi=Gk（u）i= VGi/D ij uT=（u）k注：D ij 为第i层的层间侧移刚度。 （u）i为第i层的层间侧移。 （u）k为第k层的层间侧移。 s为同层内框架柱的总数。结构顶点的假想侧移计算过程见下表，其中第六层的Gi为G6和Ge之和。结构顶点的假想侧移计算层次Gi（KN）VGi（KN）D i（N/mm）ui（mm）ui（mm）610404.290110404.290176260013.641265.98959927.338620331.628776260026.661252.68549927.338630258.967376260039.679225.68539927.338640186.305976260052.696186.00629910.191850096.497772953068.670133.3119618.583659715.081392381064.64064.64T1=1.7T （uT）1/2 =1.70.6(0.265989)1/2=0.526(s)二、水平地震作用及楼层地震剪力的计算：本结构高度不超过40m，质量和刚度沿高度分布比较均匀，变形以剪切型为主，故可用底部剪力法计算水平地震作用，即：1、结构等效总重力荷载代表值GeqGeq=0.85Gi=0.85（9618.5836+9910.1918+9927.33863+9753.4748+543.1608）=50666.3128（KN） 2、计算水平地震影响系数1查表得二类场地近震特征周期值Tg=0.30s。查表得设防烈度为8度的max=0.161=（Tg/T1）0.9max =（0.3/0.526）0.90.16 =0.0965 3、结构总的水平地震作用标准值FEkFEk=1Geq =0.096550666.3128 =4890.5658（KN）因1.4Tg=1.40.3=0.42sT1=0.526s，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 n=0.08T1+0.07=0.080.526+0.07=0.1121 F6=0.11214890.5658=548.2324KN 各质点横向水平地震作用按下式计算： Fi=GiHiFEk（1-n）/（GkHk） =4342.333（KN）地震作用下各楼层水平地震层间剪力Vi为 Vi=Fk（i=1，2，n）计算过程如下表：各质点横向水平地震作用及楼层地震剪力计算表层次Hi（m）Gi（KN）GiHi（KNm）GiHi/GjHjFi（KN）Vi（KN）26.3543.160813959.230.01773.82073.820622.79753.4748221403.880.2691168.0881241.908519.19927.3386189612.170.230998.7372240.645415.59927.3386153873.750.187812.0163052.662311.99927.3386118135.330.143620.9543673.61528.39910.191882254.590.100434.2334107.84814.79618.583645207.340.055238.8284346.676824446.29各质点水平地震作用及楼层地震剪力沿房屋高度的分布见下图： （具体数值见上表）三、多遇水平地震作用下的位移验算：水平地震作用下框架结构的层间位移（u）i和顶点位移u i分别按下列公式计算：（u）i = Vi/D iju i=（u）k各层的层间弹性位移角e=（u）i/hi，根据抗震规范，考虑砖填充墙抗侧力作用的框架，层间弹性位移角限值e1/550。计算过程如下表：横向水平地震作用下的位移验算层次Vi（KN）D i（N/mm）（u）i （mm）ui（mm）hi（mm）e=（u）i /hi61241.9087626001.62923.72336001/221052240.6457626002.93822.09436001/122543052.6627626004.00319.15636001/89933673.6157626004.81715.15336001/74724107.8487295305.63110.33636001/63914346.6769238104.7054.70547001/999由此可见，最大层间弹性位移角发生在第二层，1/6390，说明xal，则x=（VA+alq2/2）/（q1+q2）可得跨间最大正弯矩值：Mmax=MA+VAx-（q1+ q2）x2/2+alq2（x-al/3）/2若VA0，则Mmax=MA2）同理，三角形分布荷载和均布荷载作用下，如下图：VA= -（MA+MB）/l+q1l/2+q2l/4 x可由下式解得： VA=q1x+x2q2/l可得跨间最大正弯矩值：Mmax=MA+VAx-q1x2/2-x3q2/3l第1层AB跨梁： 梁上荷载设计值：q1=1.28.46=10.15 KN/m q2=1.2（14.22+0.57.2）=21.38 KN/m 左震： MA=270.47/0.75=360.63 KNm MB=-367.14/0.75=-489.52 KNmVA= -（MA-MB）/l+q1l/2+（1-a）lq2/2 =-（360.63+489.52）/7.2+10.157.2/2+21.387.2/3 =-30.22 KN0，故xal=l/3=2.4mx=（VA+alq2/2）/（q1+q2）=（205.03+1.221.38）/（10.15+21.38）=5.73m Mmax=MA+VAx-（q1+q2）x2/2+alq2（x-al/3）/2=-541.72+205.035.73-（10.15+21.38）（5.73）2/2+21.382.4（5.73-2.4/3）/2=241.98 KNm REMmax=0.75241.98=181.48 KNm 其它跨间的最大弯矩计算结果见下表： 跨间最大弯矩计算结果表层次123跨ABBCABBCABBCMmax281.79214.91234.07191.00206.00178.40层次456跨ABBCABBCABBCMmax151.44134.1581.5188.2029.2638.434、梁端剪力的调整：抗震设计中，二级框架梁和抗震墙中跨高比大于2.5，其梁端剪力设计值应按下式调整：V=Revb（M lb +M rb）/ln +V Gb（1）、对于第6层， AB跨：受力如图所示：梁上荷载设计值：q1=1.24.5=5.4 KN/m q2=1.2（19.26+0.57.2）=27.43 KN/m V Gb=5.47.2/2+27.434.8/2=85.27 KN ln=7.2-0.65=6.55 m左震：M lb=18.54/0.75=24.72 KNm M rb=-132.97/0.75=-177.29 KNm V=Revb（M lb +M rb）/ln +V Gb =0.751.2（24.72+177.29）/6.55+85.27 =91.71 KN右震：M lb=147.21/0.75=196.28 KNm M rb=6.4/0.75=8.53 KNm V=Revb（M lb +M rb）/ln +V Gb =0.751.2（196.28+8.53）/6.55+85.27 =92.09 KNBC跨：受力如图所示：梁上荷载设计值：q1=1.22.5=3.0 KN/m q2=1.2（9.63+0.5*3.6）=13.72 KN/m V Gb=3.02.4/2+13.721.2/2=11.83 KN ln=2.4-0.65=1.75 m左震：M lb= M rb=36.72/0.75=48.96 KNm V=Revb（M lb +M rb）/ln +V Gb =0.751.2248.96/1.75+11.83 =59.23 KN右震：M lb= M rb=57.21/0.75=76.28 KNm V=Revb（M lb +M rb）/ln +V Gb =0.751.2276.28/1.75+11.83 =87.33 KN（2）、对于第1-5层， AB跨：q1=1.28.46=10.15 KN/m q2=1.2（14.22+0.57.2）=21.38 KN/m V Gb=10.157.2/2+21.384.8/2=87.85 KN BC跨：q1=1.22.5=3.0 KN/m q2=1.2（7.11+0.53.6）=10.69 KN/m V Gb=3.02.4/2+10.691.2/2=10.01 KN剪力调整方法同上，结果见47页各层梁的内力组合和梁端剪力调整表。六、框架柱的内力组合：取每层柱顶和柱底两个控制截面，组合结果如下表： 横向框架A柱弯矩和轴力组合层次截面内力SGk调幅后SQk调幅后SEk（1）SEk（2）Re1.2（SGk+0.5SQk）+1.3SEk1.35SGk +1.0SQk1.2SGk+1.4SQkMmaxMM12NNminNmax6柱顶M58.84 16.43 -85.00 85.00 -22.53 143.22 95.86 93.61 143.22 -22.52 95.86 N195.37 51.84 -20.82 20.82 178.86 219.46 315.59 307.02 219.46 178.86 315.59 柱底M-36.41 -10.94 18.66 -18.66 -19.50 -55.89 -60.09 -59.01 -60.09 -19.50 -60.09 N233.39 51.84 -20.82 20.82 213.08 253.68 366.92 352.64 366.92 213.08 366.92 5柱顶M26.42 7.94 -125.30 125.30 -94.82 149.52 43.61 42.82 149.52 -94.82 43.61 N428.25 103.68 -58.62 58.62 374.93 489.24 681.82 659.05 489.24 374.93 681.82 柱底M-30.65 -9.21 61.72 -61.72 28.45 -91.91 -50.59 -49.67 -91.91 28.45 -50.59 N466.27 103.68 -58.62 58.62 409.14 523.45 733.14 704.68 523.45 409.14 733.14 4柱顶M30.65 9.21 -152.88 152.88 -117.3180.79 50.59 49.67 180.79 -117.33 50.59 N661.13 155.52 -115.13 115.13 552.75 777.25 1048.05 1011.1777.25 552.75 1048.05 柱底M-30.65 -9.21 101.92 -101.92 67.64 -131.10 -50.59 -49.67 -131.10 67.64 -50.59 N699.15 155.52 -115.13 115.13 586.97 811.47 1099.37 1056.7811.47 586.97 1099.37 3柱顶M30.65 9.21 -168.64 168.64 -132.7196.15 50.59 49.67 196.15 -132.70 50.59 N894.01 207.36 -187.91 187.91 714.71 1081.13 1414.27 1363.11081.13 714.71 1414.27 柱底M-31.00 -9.31 137.98 -137.98 102.44 -166.62 -51.16 -50.23 -166.62 102.44 -51.16 N932.03 207.36 -187.91 187.91 748.93 1115.35 1465.60 1408.71115.35 748.93 1465.60 2柱顶M30.16 9.08 -161.28 161.28 -126.0188.48 49.80 48.90 188.48 -126.02 49.80 N1126.9 259.20 -267.10 267.10 870.42 1391.26 1780.50 1715.21391.26 870.42 1780.50 柱底M-34.58 -10.48 197.12 -197.12 156.35 -228.03 -57.16 -56.17 -228.03 156.35 -57.16 N1164.9259.20 -267.10 267.10 904.64 1425.48 1831.83 1760.81425.48 904.64 1831.83 1柱顶M21.68 6.59 -149.94 149.94 -123.7168.67 35.86 35.24 168.67 -123.71 35.86 N1359.8 311.04 -357.58 357.58 1015.11712.40 2146.73 2067.21712.40 1015.12 2146.73 柱底M-10.84 -3.30 405.39 -405.39 384.01 -406.50 -17.93 -17.63 -406.50 384.01 -17.93 N1397.8311.04 -357.58 357.58 1049.31746.62 2198.06 2112.81746.62 1049.34 2198.06 横向框架B柱弯矩和轴力组合层次截面内力SGk调幅后SQk调幅后SEk（1）SEk（2）Re1.2（SGk+0.5SQk）+1.3SEk1.35SGk +1.0SQk1.2SGk+1.4SQkMmaxMM12NNminNmax6柱顶M-41.56 -11.57 -113.08 113.08 -152.86 67.64 -67.68 -66.07 -152.86 -152.86 -67.68 N246.59 69.12 -19.32 19.32 234.20 271.87 402.02 392.68 234.20 234.20 402.02 柱底M28.53 7.82 48.46 -48.46 76.44 -18.05 46.34 45.18 76.44 76.44 46.34 N284.61 69.12 -19.32 19.32 268.42 306.09 453.34 438.30 268.42 268.42 453.34 5柱顶M-22.61 -6.13 -174.88 174.88 -193.62 147.40 -36.65 -35.71 -193.62 -193.62 -36.65 N523.59 138.24 -60.81 60.81 474.15 592.73 845.09 821.84 474.15 474.15 845.09 柱底M24.86 6.77 116.59 -116.59 139.10 -88.25 40.33 39.31 139.10 139.10 40.33 N561.61 138.24 -60.81 60.81 508.37 626.95 896.41 867.47 508.37 508.37 896.41 4柱顶M-24.86 -6.77 -218.41 218.41 -238.37 187.53 -40.33 -39.31 -238.37 -238.37 -40.33 N800.59 207.36 -132.22 132.22 684.93 942.76 1288.16 1251.0684.93 684.93 1288.16 柱底M24.86 6.77 178.70 -178.70 199.65 -148.81 40.33 39.31 199.65 199.65 40.33 N838.61 207.36 -123.22 123.22 727.92 968.20 1339.48 1296.6727.92 727.92 1339.48 3柱顶M-24.86 -6.77 -262.83 262.83 -281.68 230.84 -40.33 -39.31 -281.68 -281.68 -40.33 N1077.6 276.48 -207.19 207.19 892.24 1296.26 1731.23 1680.2892.24 892.24 1731.23 柱底M25.20 6.80 215.04 -215.04 235.40 -183.92 40.82 39.76 235.40 235.40 40.82 N1115.6276.48 -207.19 207.19 926.45 1330.48 1782.55 1725.8926.45 926.45 1782.55 2柱顶M-24.31 -6.72 -256.95 256.95 -275.43 225.62 -39.54 -38.58 -275.43 -275.43 -39.54 N1354.6345.60 -295.57 295.57 1086.47 1662.83 2174.30 2109.41086.47 1086.47 2174.30 柱底M26.87 7.63 301.64 -301.64 321.72 -266.48 43.90 42.93 321.72 321.72 43.90 N1392.6345.60 -295.57 295.57 1120.69 1697.05 2225.62 2155.01120.69 1120.69 2225.62 1柱顶M-17.10 -5.01 -228.73 228.73 -240.66 205.37 -28.10 -27.53 -240.66 -240.66 -28.10 N1631.6 414.72 -393.38 393.38 1271.51 2038.60 2617.37 2538.51271.51 1271.51 2617.37 柱底M8.55 2.50 424.79 -424.79 422.99 -405.35 14.04 13.76 422.99 422.99 14.04 N1669.6 414.72 -393.38 393.38 1305.73 2072.82 2668.69 2584.11305.73 1305.73 2668.69 七、柱端弯矩设计值的调整：1、A柱：第6层，按抗震规范，无需调整。第5层，柱顶轴压比uN = N/Ac fc=489.24103/14.3/6502=0.080.15，无需调整。 柱底轴压比uN = N/Ac fc=523.45103/14.3/6502=0.0870.15。 可知，一、二、三层柱端组合的弯矩设计值应符合下式要求： Mc=cMb 注：Mc为节点上下柱端截面顺时针或逆时针方向组合的弯矩设计值之和，上下柱端的弯矩设计值可按弹性分析分配。 Mb为节点左右梁端截面顺时针或逆时针方向组合的弯矩设计值之和。 c柱端弯矩增大系数，二级取1.2。横向框架A柱柱端组合弯矩设计值的调整（相当于本层柱净高上下端的弯矩设计值）层次654321截面柱顶柱底柱顶柱底柱顶柱底柱顶柱底柱顶柱底柱顶柱底RE（Mc=cMb）143.22 60.09 149.52 91.91 180.79 199.89 199.89 216.62 216.62 240.19 247.36 508.12 REN219.46 366.92 489.24 523.45 777.25 811.47 1081.13 1115.35 1391.26 1425.48 1712.40 1746.62 2、B柱： 第6层，按抗震规范，无需调整。 经计算当轴力N=fc Ac=0.1514.36502/103=902.26 KN 时， 方符合调整的条件，可知B柱调整图如下：横向框架B柱柱端组合弯矩设计值的调整层次654321截面柱顶柱底柱顶柱底柱顶柱底柱顶柱底柱顶柱底柱顶柱底RE（Mc=cMb）152.8676.44 193.62 139.10 238.37 199.65 281.68 322.41 322.41 351.11 361.58 528.74 REN234.20 268.42 474.15 508.37 684.93 727.92 892.24 926.45 1086.47 1120.69 1271.51 1305.73 八、柱端剪力组合和设计值的调整：例：第6层：恒载SGk =（M上+M下）/h=（-54.84-36.41）/3.6=-25.35 活载SQk =（M上+M下）/h=（-16.43-10.94）/3.6=-7.6 地震作用SEk =（M上+M下）/h=（85.00+18.66）/3.6=28.79 调整：1.2（143.22+60.09）/3.6=67.77 横向框架A柱剪力组合与调整（KN）层次SGkSQkSEk1SEk2Re1.2（SGk+0.5SQk）+1.3SEk1.35SGk +1.0SQk1.2SGk+1.4SQkV=Revc（M bc +M lc）/hn 126-25.35 -7.60 28.79 -28.79 1.84 -54.31 -41.82 -41.06 67.77 5-15.85 -4.76 74.79 -74.79 56.51 -89.33 -26.16 -25.68 80.48 4-17.03 -5.12 70.78 -70.78 51.38 -86.64 -28.11 -27.60 126.89 3-17.12 -5.14 85.17 -85.17 65.32 -100.76 -28.25 -27.74 138.84 2-17.98 -5.43 99.56 -99.56 78.45 -115.70 -29.70 -29.18 152.27 1-6.92 -2.10 118.16 -118.16 108.03 -122.38 -11.44 -11.24 192.89 同理，横向框架B柱剪力组合与调整（KN）层次SGkSQkSEk1SEk2Re1.2（SGk+0.5SQk）+1.3SEk1.35SGk +1.0SQk1.2SGk+1.4SQkV=Revc（M bc +M lc）/hn 12619.47 5.39 44.87 -44.87 63.70 -23.80 31.67 30.91 76.43 513.19 3.58 80.96 -80.96 92.42 -65.45 21.39 20.84 110.91 413.81 3.76 110.31 -110.31 121.67 -93.43 22.40 21.84 146.01 313.91 3.77 132.74 -132.74 143.64 -115.21 22.55 21.97 201.36 214.22 3.74 155.16 -155.16 165.76 -136.80 22.94 22.30 224.51 15.46 1.60 139.05 -139.05 141.21 -129.94 8.97 8.79 227.32 第六部分：截面设计一、框架梁： 以第1层AB跨框架梁的计算为例。1、梁的最不利内力：经以上计算可知，梁的最不利内力如下： 跨间： Mmax=281.79 KNm 支座A：Mmax=406.29 KNm 支座Bl：Mmax=367.14 KNm 调整后剪力：V=182.70 KN2、梁正截面受弯承载力计算： 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：（1）、考虑跨间最大弯矩处： 按T形截面设计，翼缘计算宽度bf，按跨度考虑，取bf，=l/3=7.2/3=2.4m=2400mm，梁内纵向钢筋选II级热扎钢筋，（fy=fy，=310N/mm2），h0=h-a=600-35=565mm，因为fcm bf，hf，（ h0- hf，/2）=13.42400120（565-120/2）=1948.90KNm994.06 KNm属第一类T形截面。下部跨间截面按单筋T形截面计算：s=M/（fcmbf，h02）=281.79106/13.4/2400/5652=0.027=1-（1-2s）1/2=0.027As=fcmbf，h0/fy=0.02713.42400565/310=1582.58 mm2实配钢筋225、222，As=1742 mm2。=1742/300/565=1.0%min=0.25%，满足要求。梁端截面受压区相对高度：=fyAs/（fcmbf，h0）=3101742/13.4/2400/565min=0.3%，又As，/ As =1742/2724=0.640.3，满足梁的抗震构造要求。3、梁斜截面受剪承载力计算： （1）、验算截面尺寸： hw=h0=565mm hw/b=565/300=1.88V=182700N 可知，截面符合条件。 （2）、验算是否需要计算配置箍筋： 0.07fcmbh0=0.0714.3300565 =169669.5N182700N sv= nAsv1/bs=2*50.3/100/300=0.34%svmin=0.02fcm/fyv=0.0214.3/210=0.14% 加密区长度取0.85m，非加密区箍筋取8150。箍筋配置，满足构造要求。配筋图如下图所示：其它梁的配筋计算见下表：层次截面M（KNm）计算As，（mm2）实配As，（mm2）计算As（mm2）实配As（mm2）As/As，（%）配箍1支座A406.29 02472.86 425、222(2724)0.64 2.6加密区双肢8100,非加密区双肢8150Bl367.14 02234.57 425、222(2724)AB跨间281.79 0.10 1582.58 225、222(1742)支座Br226.77 02209.41 525(2454)0.804.8加密区四肢880非加密区四肢8100BC跨间214.91 0.05 1931.36 425(1964)2支座A361.0302197.36 525(2454)0.622.3加密区双肢8100非加密区双肢8150Bl335.2402108.44 525(2454)AB跨间234.07 0.09 1400.71 422(1520)支座Br202.110.31 1975.66 523(2077)0.874.3加密区四肢8100非加密区四肢8150BC跨间191.00 0.05 1724.41 424(1808)3支座A333.15 02027.69 523(2077)0.602.0加密区双肢8100非加密区双肢8150Bl318.29 01937.26 523(2077)AB跨间206.00 0.08 1162.12 323(1246)支座Br189.400.23 1851.39 522(1900)0.843.80加密区四肢8100非加密区四肢8150BC跨间178.40 0.04 1590.36 426(1593)4支座A278.59 01695.60 424(1808)0.541.6加密区双肢8100非加密区双肢8150Bl258.67 01574.36 424(1808)AB跨间151.44 0.07 891.01 225(982)支座Br145.15 0.16 1418.84 325(1473)0.853.0加密区双肢8100非加密区双肢8150BC跨间134.15 0.03 1211.71 420(1256)5支座A210.98 01284.13 325(1473)0.351.2加密区双肢8100非加密区双肢8150Bl196.86 01198.17 325(1473)AB跨间81.51 0.05 479.98 218(509)支座Br98.26 0.03 960.48 516(1005)1.132.4加密区双肢8100非加密区双肢8150BC跨间88.20 0.03 795.18 322(1140)6支座A147.21 0895.98 516(1005)0.310.8加密区双肢8100非加密区双肢8150Bl132.97 20.08620.12320.19120.26220.33520.35320.06320.12020.17120.21820.26420.2642轴压比：N/fcmbh0=2328.83*103/14.3/700/660=0.3535，故应考虑偏心矩增大系数。 1=0.5fcmA/N=0.514.37002/（1746.62103）=2.0061.0 取1=1.0 又l0/h15，取2=1.0 得=1+ l0212h0/1400eih2 =1+6.712660/1400/314.25 =1.068 轴向力作用点至受拉钢筋As合力点之间的距离 e=ei+h/2-as =1.068314.25+700/2-40 =645.62 mm 对称配筋： =x/h0=N/fcmbh0=1746.62103/14.3/700/660 =0.2645，故应考虑偏心矩增大系数。1=0.5fcmA/N=0.514.37002/（2198.06103）=1.5941.0 取1=1.0 又l0/h15，取2=1.0 得=1+ l0212h0/1400eih2 =1+6.712660/1400/31.49 =1.674 ei=1.67431.49=52.71mmbfcmbh0及Ne0.43fcmbh02. 因为N=2198.06KN0.8%3、柱斜截面受剪承载力计算：以第1层A柱为例，查表可知：框架柱的剪力设计值V c=192.89KN剪跨比=3.993，取=3轴压比n=0.353考虑地震作用组合的柱轴向压力设计值 N=1746.62KN192890N故该层柱应按构造配置箍筋。柱端加密区的箍筋选用4肢10100。查表得，最小配筋率特征值v=0.09，则最小配筋率vmin=vfcm/fyv=0.0914.3/210=0.6%柱箍筋的体积配筋率v=（Asvili）/s/Acor=78.56508/100/650/650=1.0%0.6%,符合构造要求。注：Asvi、li为第i根箍筋的截面面积和长度。 Acor为箍筋包裹范围内的混凝土核芯面积。 s为箍筋间距。 非加密区还应满足s10d=200mm,故箍筋配置为410150，柱的配筋图如下图所示：其它各层柱的配筋计算见下表：柱A柱层次123截面尺寸700700650650650650组合一二一二一二M（KNm）508.12 -17.93 240.19 57.16 216.62 51.16 N（KN）1746.62 2198.06 1425.48 1831.83 1115.35 1465.60 V（KN）192.89 152.27 138.84 e0（mm）290.92 8.16 168.50 31.20 194.22 34.91 ea（mm）23.33 23.33 21.67 21.67 21.67 21.67 l0（m）4.74.73.63.63.63.6ei（mm）314.25 31.49 190.17 52.87 215.89 56.58 l0/h6.71 6.71 5.54 5.54 5.54 5.54 11.0 1.0 1.0 1.0 1.0 1.0 21.0 1.0 1.0 1.0 1.0 1.0 1.068 1.674 1.070 1.253 1.062 1.236 e（mm）645.62 362.71 488.48 351.25 514.28 354.93 0.2640.2510.197计算As=As（mm2）668.41 980.00 0845.00 0.8%0.95%0.8%0.95%0.8%偏心判断大小大小大小配箍加密区4肢10100，非加密区4肢10150加密区4肢10100，非加密区4肢10150加密区4肢10100，非加密区4肢10150柱A柱层次456截面尺寸650650650650650650组合一二一二一二M（KNm）199.89 50.59 149.52 50.59 143.22 60.09 N（KN）811.47 1099.37 489.24 733.14 219.46 366.92 V（KN）126.89 80.48 67.77 e0（mm）246.33 46.02 305.62 69.00 652.60 163.77 ea（mm）21.67 21.67 21.67 21.67 21.67 21.67 l0（m）3.63.63.63.63.63.6ei（mm）268.00 67.69 327.29 90.67 674.27 185.44 l0/h5.54 5.54 5.54 5.54 5.54 5.54 11.0 1.0 1.0 1.0 1.0 1.0 21.0 1.0 1.0 1.0 1.0 1.0 1.050 1.198 1.041 1.147 1.020 1.072 e（mm）566.40 366.09 625.71 389.00 972.76 483.79 0.1430.0860.0390.065计算As=As（mm2）2.21 845.00 121.49 845.00 459.67 0.8%0.95%0.8%0.95%0.8%偏心判断大小大小大大配箍加密区4肢8100，非加密区4肢8150加密区4肢8100，非加密区4肢8150加密区4肢8100，非加密区4肢8150柱B柱层次123截面尺寸700700650650650650组合一二一二一二M（KNm）528.74 14.04 351.11 43.90 322.41 40.82 N（KN）1305.73 2668.69 1120.69 2225.62 926.45 1782.55 V（KN）227.32 224.51 201.36 e0（mm）404.94 5.26 313.30 19.72 348.00 22.90 ea（mm）23.33 23.33 21.67 21.67 21.67 21.67 l0（m）4.74.73.63.63.63.6ei（mm）428.27 28.59 334.97 41.39 369.67 44.57 l0/h6.71 6.71 5.54 5.54 5.54 5.54 11.0 1.0 1.0 1.0 1.0 1.0 21.0 1.0 1.0 1.0 1.0 1.0 1.050 1.742 1.040 1.323 1.036 1.300 e（mm）759.68 359.80 633.37 339.76 667.99 342.94 0.1980.1980.163计算As=As（mm2）1093.34 980.00 525.14 845.00 571.83 845.00 实配单侧选420（1256）选420（1256）选420（1256）s0.82%0.8%0.95%0.8%0.95%0.8%偏心判断大小大小大小配箍加密区4肢10100，非加密区4肢10150加密区4肢10100，非加密区4肢10150加密区4肢10100，非加密区4肢10150柱B柱层次456截面尺寸650650650650650650组合一二一二一二M（KNm）238.37 40.33 193.62 40.33 152.86 46.34 N（KN）684.93 1339.48 474.15 896.41 234.20 453.34 V（KN）146.01 110.91 76.43 e0（mm）348.02 30.11 408.35 44.99 652.69 102.22 ea（mm）21.67 21.67 21.67 21.67 21.67 21.67 l0（m）3.63.63.63.63.63.6ei（mm）369.69 51.78 430.02 66.66 674.36 123.89 l0/h5.54 5.54 5.54 5.54 5.54 5.54 11.0 1.0 1.0 1.0 1.0 1.0 21.0 1.0 1.0 1.0 1.0 1.0 1.036 1.258 1.031 1.201 1.020 1.108 e（mm）668.00 350.14 728.35 365.06 972.85 422.27 0.1210.0840.041计算As=As（mm2）364.20 845.00 379.29 845.00 503.36 845.00 实配单侧选420（1256）选420（1256）选420（1256）s0.95%0.8%0.95%0.8%0.95%0.8%偏心判断大小大小大小配箍加密区4肢8100，非加密区4肢8150加密区4肢8100，非加密区4肢8150加密区4肢8100，非加密区4肢8150三、框架梁柱节点核芯区截面抗震验算：以第1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，因为节点两侧梁不等高，计算时取两侧梁的平均高度，即hb=(600+400)/2=500mmhb0=(565+365)/2=465mm二级框架梁柱节点核芯区组合的剪力设计值Vj按下式计算：Vj=(jbMb)1-(hb0-as,)/(Hc-hb)/ (hb0-as)注：Hc为柱的计算高度，可采用节点上、下柱反弯点之间的距离，即 Hc=0.543.6+0.654.7=5.0 m Mb为节点左右梁端逆时针或顺时针方向组合弯矩设计值之和，即Mb=（367.14+226.77）/0.75=791.88 KNm 可知，剪力设计值Vj=(jbMb)1-(hb0-as,)/(Hc-hb)/ (hb0-as) =1.2791.881031-（465-35）/（5000-500）/（465-35） =1998.67 KN 节点核芯区截面的抗震验算是按箍筋和混凝土共同抗剪考虑的，设计时，应首先按下式对截面的剪压比予以控制： Vij0.30jfcmbjhj/RE 注：j为正交梁的约束影响系数，楼板为现浇，梁柱中心重合，可取1.5。 bj、hj分别为核芯区截面有效验算宽度、高度。 为验算方向柱截面宽度。 bj=bc=700mm, hj=700mm 可知，0.30jfcmbjhj=0.301.514.3700700/0.75=4204200NVj=1998670N，满足要求 节点核芯区的受剪承载力按下式计算： Vj1.1jftbjhj+0.05jNbj/bc+fyvAsvj(hb0-as,)/s/RE注：N取第2层柱底轴力N=1120.69KN和0.5fcmA=0.514.37002=3503.5KN二者中的较小值，故取N=1120.69KN。 该节点区配箍为410100，则 1.1jftbjhj+0.05jNbj/bc+fyvAsvj(hb0-as,)/s/RE=1.11.51.5700700+0.051.51120.69103+210478.5（465-35）/100/0.75=2107125NVj=1998670N 故承载力满足要求。 其它框架梁柱节点核芯区截面抗震验算见下表：层次123节点边节点中节点边节点中节点边节点中节点hb（mm）600500600500600500hb0（mm）565465565465565465Hc（m）5.4153.63.63.063.24 Mb（KNm）541.72791.88481.37716.47444.2676.92Vj（KN）1078.731998.67958.561808.3885.051707.73bj=bc（mm）700700650650650650hj（mm）7007006506506506500.30jfcmbjhj/RE（KN）4204.24204.23625.13625.13625.13625.1配箍4101004101004101004101004101004101001.1jftbjhj+0.05jNbj/bc+fyvAsvj(hb0-as,)/s/RE（KN）2137.62107.11971.81864.91941.41845.1结论合格合格合格合格合格合格层次456节点边节点中节点边节点中节点边节点中节点hb（mm）600500600500600500hb0（mm）565465565465565465Hc（m）2.633.061.842.522.592.52 Mb（KNm）371.45538.43281.31393.49196.28253.57Vj（KN）740.11358.35560.49992.69391.08639.7bj=bc（mm）650650650650650650hj（mm）6506506506506506500.30jfcmbjhj/RE（KN）3625.13625.13625.13625.13625.13625.1配箍4810048100481004810048100481001.1jftbjhj+0.05jNbj/bc+fyvAsvj(hb0-as,)/s/RE（KN）1745.21687.31729.51663.31714.81659.9结论合格合格合格合格合格合格第七部分： 楼板设计一、楼板类型及设计方法的选择：对于楼板，根据塑性理论，l02/l0151460N 满足要求。第九部分： 框架变形验算多遇水平地震作用下框架层间弹性位移验算以在第四部分中给出，在此不再赘述。现考虑罕遇水平地震作用下框架层间弹塑性位移计算。一、 梁的极限抗弯承载力计算：计算时，采用构件实际配筋和材料的强度标准值，可近似地按下式计算：Mbu=Asfyk（h0-as，） 注：为钢筋强度标准值计算过程和结果见下表：层次支座实配As（mm2）fyk（N/mm2）h0（mm）as，（mm）Mbu（KNm）1A353433556535627.46 Bl353433556535627.46 Br321733536535355.64 2A307933556535546.68 Bl307933556535546.68 Br282733536535312.52 3A286533556535508.68 Bl286533556535508.68 Br265533536535293.51 4A229133556535406.77 Bl229133556535406.77 Br190033536535210.05 5A171633556535304.68 Bl171633556535304.68 Br130533536535144.27 6A125633556535223.00 Bl125633556535223.00 Br7603353653584.02 二、 柱的极限抗弯承载力计算：根据抗震规范，当柱轴压比小于0.8时，其极限抗弯承载力可按下式计算，并且计算时采用构件的实际配筋和材料强度标准值：Mcu=Asfyk（hc0-as，）+0.5Nhc（1-N/bchcfcmk）注：fcmk为混凝土弯曲抗压强度标准值 N为考虑地震组合时相应于设计弯矩的轴力bc、hc、hc0为柱截面的宽度、高度、有效高度。 计算过程和结果见下表：柱号层次As（mm2）fyk（N/mm2）hc0（mm）as，（mm）N（KN）hc（mm）bc（mm）fcmk（N/mm2）Mcu（KNm）A柱13768335660402198.0670070022783.38 23768335610401831.8365065022720.09 33768335610401465.60 65065022719.98 43768335610401099.3765065022719.86 5376833561040733.1465065022719.74 6376833561040219.4665065022719.57 B柱13768335660402668.6970070022783.55 23768335610402225.6265065022720.22 33768335610401782.5565065022720.08 43768335610401339.4865065022719.93 5376833561040896.4165065022719.79 6376833561040453.3465065022719.65 三、 确定柱端截面有效承载力Mc：节点A6：因MbuMcu，223.00 KNm719.57 KNm， 所以，Mc6u=Mbu=223.00 KNm节点A5：因MbuMcu，304.68 KNm719.57+719.74 KNm， 所以，Mc6l=Mbuk6/（k5+k6）=304.68/2=152.34 KNmMcu6l=719.57 KNm 取较小值152.34 KNm Mc5u=Mbuk5/（k5+k6）=304.68/2=152.34 KNm Mcu5u=719.74 KNm 取较小值152.34 KNm节点A4：Mc5l=203.38 KNm Mc4u=203.38 KNm节点A3：Mc4l=254.34 KNm Mc3u=254.34 KNm节点A2：Mc3l=273.34 KNm Mc2u=273.34 KNm节点A1：Mc2l=627.461.0/（1.0+1.03）=309.09 KNm Mc1u=627.461.03/（1.0+1.03）=318.37 KNm柱底A0：Mc1l=Mcu1l=783.38 KNm节点B6：Mc6u=223.00+84.02=307.03 KNm节点B5：Mc6l=224.22 KNm Mc5u=224.22 KNm节点B4：Mc5l=308.41 KNm Mc4u=308.41 KNm节点B3：Mc4l=401.10 KNm Mc3u=401.10 KNm节点B2：Mc3l=429.60 KNm Mc2u=429.60 KNm节点B1：Mc2l=（627.46+355.64）1.0/（1.0+1.03）=484.29 KNm Mc1u=（627.46+355.64）1.03/（1.0+1.03）=498.82 KNm柱底B0：Mc1l=Mcu1l=783.55 KNm四、 各柱的受剪承载力Vyij的计算：理论依据：第i层第j根柱的受剪承载力计算公式为：Vyij=（Mciju + Mcijl）/Hni注：Hni为第i层的净高，可由层高H减去该层上、下梁高的1/2求得。 可得：Vy6A =（223.00+152.34）/（3.6-0.6）=125.11 KN Vy5A =（152.34+203.38）/（3.6-0.6）=151.57 KN Vy4A =（203.38+254.34）/（3.6-0.6）=152.57 KN Vy3A =（254.34+273.34）/（3.6-0.6）=175.89 KN Vy2A =（273.34+309.09）/（3.6-0.6）=194.14 KN Vy1A =（318.37+783.38）/（4.7-0.6/2）=250.40 KN Vy6B =（307.02+224.22）/（3.6-0.6）=177.08 KN Vy5B =（224.22+308.41）/（3.6-0.6）=177.54 KN Vy4B =（308.41+401.10）/（3.6-0.6）=236.50 KN Vy3B =（401.10+429.60）/（3.6-0.6）=276.90 KN Vy2B =（429.60+484.29）/（3.6-0.6）=304.63 KNVy1B =（498.82+783.55）/（4.7-0.6/2）=291.45 KN五、 楼层受剪承载力Vyi的计算：将第i层各柱的屈服承载力相加即得Vyi=Vyij 则Vy6=（Vy6A+Vy6B）2=（125.11+177.08）2=604.38 KN Vy5=（151.57+177.54）2=658.22 KN Vy4=（152.57+236.50）2=778.14 KN Vy3=（175.89+276.90）2=905.58 KN Vy2=（194.14+304.63）2=997.54 KN Vy1=（250.40+291.45）2=1083.70 KN六、 罕遇地震下弹性楼层剪力Ve的计算：8度水平地震影响系数最大值max=0.9，此时可用0.9/0.16的比值剩以多遇地震作用下层间地震弹性剪力Vi求出Ve。Vi的计算结果在第四部分已经算出，则Ve6=1241.9080.9/0.16=6985.73 KNVe5=2240.6450.9/0.16=12603.63 KNVe4=3052.6620.9/0.16=17171.22 KNVe3=3673.6150.9/0.16=20664.08 KNVe2=4107.8480.9/0.16=23106.65 KNVe1=4346.6760.9/0.16=24450.05 KN七、 楼层屈服承载力系数yi的计算：该建筑共有9榀横向框架，故：y6=9Vy6/Ve6=9604.38/6985.73=0.779y5=9Vy5/Ve5=9658.22/12603.63=0.470y4=9Vy4/Ve4=9778.14/17171.22=0.408y3=9Vy3/Ve3=9905.58/20664.08=0.394y2=9Vy2/Ve2=9997.54/23106.65=0.389y1=9Vy1/Ve1=91083.70/24450.05=0.399以上计算部分可总结如下表：层次柱Mcu（KNm）Mcl（KNm）Vyij（KN）Vyi（KN）Vei（KN）yi1A223.00 152.34 125.11 604.38 6985.73 0.779 B307.02 224.22 177.08 C307.02 224.22 177.08 D223.00 152.34 125.11 2A152.34 203.38 151.57 658.22 12603.63 0.470 B224.22 308.41 177.54 C224.22 308.41 177.54 D152.34 203.38 151.57 3A203.38 254.34 152.57 778.14 17171.22 0.408 B308.41 401.10 236.50 C308.41 401.10 236.50 D203.38 254.34 152.57 4A254.34 273.34 175.89 905.58 20664.08 0.394 B401.10 429.60 276.90 C401.10 429.60 276.90 D254.34 273.34 175.89 5A273.34 309.09 194.14 997.54 23106.65 0.389 B429.60 484.29 304.63 C429.60 484.29 304.63 D273.34 309.09 194.14 6A318.37 783.38 250.40 1083.70 24450.05 0.399 B498.82 783.55 291.45 C498.82 783.55 291.45 D318.37 783.38 250.40 可知，y，min=y2=0.389，第二层薄弱层，但与相邻层比较：0.8y平均=0.8(0.394+0.399)/2=0.3170.389说明仍属于比较均匀的框架。查表得弹塑性位移增大系数p=1.665八、 层间弹塑性位移验算（第二层）：ue=Ve/D=23106.65103/729530=31.67 mm由于计算中D值采用纯框架刚度，并未考虑填充墙的刚度，而在计算基本周期T1时考虑了非结构填充墙的影响系数0.6，使得T1减小而Ve增大，二者不协调。由于Ve与T1成正比，故可近似用0.6将ue进行折减，得ue=0.631.67=19.00 mm则弹塑性层间位移up=pue=1.66519.00=31.64mmph=3600/50=72mm故满足要求。第十部分： 科技资料翻译一、科技资料原文：Structural Systems to resist lateral loadsCommonly Used structural SystemsWith loads measured in tens of thousands kips, there is little room in the design of high-rise buildings for excessively complex thoughts. Indeed, the better high-rise buildings carry the universal traits of simplicity of thought and clarity of expression.It does not follow that there is no room for grand thoughts. Indeed, it is with such grand thoughts that the new family of high-rise buildings has evolved. Perhaps more important, the new concepts of but a few years ago have become commonplace in today s technology.Omitting some concepts that are related strictly to the materials of construction, the most commonly used structural systems used in high-rise buildings can be categorized as follows:1. Moment-resisting frames.2. Braced frames, including eccentrically braced frames.3. Shear walls, including steel plate shear walls.4. Tube-in-tube structures.5. Tube-in-tube structures.6. Core-interactive structures.7. Cellular or bundled-tube systems.Particularly with the recent trend toward more complex forms, but in response also to the need for increased stiffness to resist the forces from wind and earthquake, most high-rise buildings have structural systems built up of combinations of frames, braced bents, shear walls, and related systems. Further, for the taller buildings, the majorities are composed of interactive elements in three-dimensional arrays.The method of combining these elements is the very essence of the design process for high-rise buildings. These combinations need evolve in response to environmental, functional, and cost considerations so as to provide efficient structures that provoke the architectural development to new heights. This is not to say that imaginative structural design can create great architecture. To the contrary, many examples of fine architecture have been created with only moderate support from the structural engineer, while only fine structure, not great architecture, can be developed without the genius and the leadership of a talented architect. In any event, the best of both is needed to formulate a truly extraordinary design of a high-rise building.While comprehensive discussions of these seven systems are generally available in the literature, further discussion is warranted here .The essence of the design process is distributed throughout the discussion.Moment-Resisting FramesPerhaps the most commonly used system in low-to medium-rise buildings, the moment-resisting frame, is characterized by linear horizontal and vertical members connected essentially rigidly at their joints. Such frames are used as a stand-alone system or in combination with other systems so as to provide the needed resistance to horizontal loads. In the taller of high-rise buildings, the system is likely to be found inappropriate for a stand-alone system, this because of the difficulty in mobilizing sufficient stiffness under lateral forces. Analysis can be accomplished by STRESS, STRUDL, or a host of other appropriate computer programs; analysis by the so-called portal method of the cantilever method has no place in todays technology.Because of the intrinsic flexibility of the column/girder intersection, and because preliminary designs should aim to highlight weaknesses of systems, it is not unusual to use center-to-center dimensions for the frame in the preliminary analysis. Of course, in the latter phases of design, a realistic appraisal in-joint deformation is essential.Braced FramesThe braced frame, intrinsically stiffer than the moment resisting frame, finds also greater application to higher-rise buildings. The system is characterized by linear horizontal, vertical, and diagonal members, connected simply or rigidly at their joints. It is used commonly in conjunction with other systems for taller buildings and as a stand-alone system in low-to medium-rise buildings.While the use of structural steel in braced frames is common, concrete frames are more likely to be of the larger-scale variety.Of special interest in areas of high seismicity is the use of the eccentric braced frame.Again, analysis can be by STRESS, STRUDL, or any one of a series of two or three dimensional analysis computer programs. And again, center-to-center dimensions are used commonly in the preliminary analysis. Shear wallsThe shear wall is yet another step forward along a progression of ever-stiffer structural systems. The system is characterized by relatively thin, generally (but not always) concrete elements that provide both structural strength and separation between building functions.In high-rise buildings, shear wall systems tend to have a relatively high aspect ratio, that is, their height tends to be large compared to their width. Lacking tension in the foundation system, any structural element is limited in its ability to resist overturning moment by the width of the system and by the gravity load supported by the element. Limited to a narrow overturning, One obvious use of the system, which does have the needed width, is in the exterior walls of building, where the requirement for windows is kept small.Structural steel shear walls, generally stiffened against buckling by a concrete overlay, have found application where shear loads are high. The system, intrinsically more economical than steel bracing, is particularly effective in carrying shear loads down through the taller floors in the areas immediately above grade. The sys tem has the further advantage of having high ductility a feature of particular importance in areas of high seismicity.The analysis of shear wall systems is made complex because of the inevitable presence of large openings through these walls. Preliminary analysis can be by truss-analogy, by the finite element method, or by making use of a proprietary computer program designed to consider the interaction, or coupling, of shear walls.Framed or Braced TubesThe concept of the framed or braced or braced tube erupted into the technology with the IBM Building in Pittsburgh, but was followed immediately with the twin 110-story towers of the World Trade Center, New York and a number of other buildings .The system is characterized by three dimensional frames, braced frames, or shear walls, forming a closed surface more or less cylindrical in nature, but of nearly any plan configuration. Because those columns that resist lateral forces are placed as far as possible from the cancroids of the system, the overall moment of inertia is increased and stiffness is very high.The analysis of tubular structures is done using three-dimensional concepts, or by two- dimensional analogy, where possible, whichever method is used, it must be capable of accounting for the effects of shear lag.The presence of shear lag, detected first in aircraft structures, is a serious limitation in the stiffness of framed tubes. The concept has limited recent applications of framed tubes to the shear of 60 stories. Designers have developed various techniques for reducing the effects of shear lag, most noticeably the use of belt trusses. This system finds application in buildings perhaps 40stories and higher. However, except for possible aesthetic considerations, belt trusses interfere with nearly every building function associated with the outside wall; the trusses are placed often at mechanical floors, mush to the disapproval of the designers of the mechanical systems. Nevertheless, as a cost-effective structural system, the belt truss works well and will likely find continued approval from designers. Numerous studies have sought to optimize the location of these trusses, with the optimum location very dependent on the number of trusses provided. Experience would indicate, however, that the location of these trusses is provided by the optimization of mechanical systems and by aesthetic considerations, as the economics of the structural system is not highly sensitive to belt truss location.Tube-in-Tube StructuresThe tubular framing system mobilizes every column in the exterior wall in resisting over-turning and shearing forces. The termtube-in-tubeis largely self-explanatory in that a second ring of columns, the ring surrounding the central service core of the building, is used as an inner framed or braced tube. The purpose of the second tube is to increase resistance to over turning and to increase lateral stiffness. The tubes need not be of the same character; that is, one tube could be framed, while the other could be braced.In considering this system, is important to understand clearly the difference between the shear and the flexural components of deflection, the terms being taken from beam analogy. In a framed tube, the shear component of deflection is associated with the bending deformation of columns and girders (i.e, the webs of the framed tube) while the flexural component is associated with the axial shortening and lengthening of columns (i.e, the flanges of the framed tube). In a braced tube, the shear component of deflection is associated with the axial deformation of diagonals while the flexural component of deflection is associated with the axial shortening and lengthening of columns.Following beam analogy, if plane surfaces remain plane (i.e, the floor slabs),then axial stresses in the columns of the outer tube, being farther form the neutral axis, will be substantially larger than the axial stresses in the inner tube. However, in the tube-in-tube design, when optimized, the axial stresses in the inner ring of columns may be as high, or even higher, than the axial stresses in the outer ring. This seeming anomaly is associated with differences in the shearing component of stiffness between the two systems. This is easiest to under-stand where the inner tube is conceived as a braced (i.e, shear-stiff) tube while the outer tube is conceived as a framed (i.e, shear-flexible) tube.Core Interactive StructuresCore interactive structures are a special case of a tube-in-tube wherein the two tubes are coupled together with some form of three-dimensional space frame. Indeed, the system is used often wherein the shear stiffness of the outer tube is zero. The United States Steel Building, Pittsburgh, illustrates the system very well. Here, the inner tube is a braced frame, the outer tube has no shear stiffness, and the two systems are coupled if they were considered as systems passing in a straight line from the “hat” structure. Note that the exterior columns would be improperly modeled if they were considered as systems passing in a straight line from the “hat” to the foundations; these columns are perhaps 15% stiffer as they follow the elastic curve of the braced core. Note also that the axial forces associated with the lateral forces in the inner columns change from tension to compression over the height of the tube, with the inflection point at about 5/8 of the height of the tube. The outer columns, of course, carry the same axial force under lateral load for the full height of the columns because the columns because the shear stiffness of the system is close to zero. The space structures of outrigger girders or trusses, that connect the inner tube to the outer tube, are located often at several levels in the building. The AT&T headquarters is an example of an astonishing array of interactive elements:1. The structural system is 94 ft (28.6m) wide, 196ft(59.7m) long, and 601ft (183.3m) high.2. Two inner tubes are provided, each 31ft(9.4m) by 40 ft (12.2m), centered 90 ft (27.4m) apart in the long direction of the building.3. The inner tubes are braced in the short direction, but with zero shear stiffness in the long direction.4. A single outer tube is supplied, which encircles the building perimeter.5. The outer tube is a moment-resisting frame, but with zero shear stiffness for the center50ft (15.2m) of each of the long sides.6. A space-truss hat structure is provided at the top of the building.7. A similar space truss is located near the bottom of the building8. The entire assembly is laterally supported at the base on twin steel-plate tubes, because the shear stiffness of the outer tube goes to zero at the base of the building.Cellular structuresA classic example of a cellular structure is the Sears Tower, Chicago, a bundled tube structure of nine separate tubes. While the Sears Tower contains nine nearly identical tubes, the basic structural system has special application for buildings of irregular shape, as the several tubes need not be similar in plan shape, It is not uncommon that some of the individual tubes one of the strengths and one of the weaknesses of the system.This special weakness of this system, particularly in framed tubes, has to do with the concept of differential column shortening. The shortening of a column under load is given by the expression=fL/EFor buildings of 12 ft (3.66m) floor-to-floor distances and an average compressive stress of 15 ksi (138MPa), the shortening of a column under load is 15 (12)(12)/29,000 or 0.074in (1.9mm) per story. At 50 stories, the column will have shortened to 3.7 in. (94mm) less than its unstressed length. Where one cell of a bundled tube system is, say, 50stories high and an adjacent cell is, say, 100stories high, those columns near the boundary between .the two systems need to have this differential deflection reconciled. Major structural work has been found to be needed at such locations. In at least one building, the Rialto Project, Melbourne, the structural engineer found it necessary to vertically pre-stress the lower height columns so as to reconcile the differential deflections of columns in close proximity with the post-tensioning of the shorter column simulating the weight to be added on to adjacent, higher columns.二、原文翻译：抗侧向荷载的结构体系常用的结构体系若已测出荷载量达数千万磅重，那么在高层建筑设计中就没有多少可以进行极其复杂的构思余地了。确实，较好的高层建筑普遍具有构思简单、表现明晰的特点。这并不是说没有进行宏观构思的余地。实际上，正是因为有了这种宏观的构思，新奇的高层建筑体系才得以发展，可能更重要的是：几年以前才出现的一些新概念在今天的技术中已经变得平常了。如果忽略一些与建筑材料密切相关的概念不谈，高层建筑里最为常用的结构体系便可分为如下几类：1 抗弯矩框架。2 支撑框架，包括偏心支撑框架。3 剪力墙，包括钢板剪力墙。4 筒中框架。5 筒中筒结构。6 核心交互结构。7 框格体系或束筒体系。特别是由于最近趋向于更复杂的建筑形式，同时也需要增加刚度以抵抗几力和地震力，大多数高层建筑都具有由框架、支撑构架、剪力墙和相关体系相结合而构成的体系。而且，就较高的建筑物而言，大多数都是由交互式构件组成三维陈列。将这些构件结合起来的方法正是高层建筑设计方法的本质。其结合方式需要在考虑环境、功能和费用后再发展，以便提供促使建筑发展达到新高度的有效结构。这并不是说富于想象力的结构设计就能够创造出伟大建筑。正相反，有许多例优美的建筑仅得到结构工程师适当的支持就被创造出来了，然而，如果没有天赋甚厚的建筑师的创造力的指导，那么，得以发展的就只能是好的结构，并非是伟大的建筑。无论如何，要想创造出高层建筑真正非凡的设计，两者都需要最好的。 虽然在文献中通常可以见到有关这七种体系的全面性讨论，但是在这里还值得进一步讨论。设计方法的本质贯穿于整个讨论。设计方法的本质贯穿于整个讨论中。抗弯矩框架抗弯矩框架也许是低，中高度的建筑中常用的体系，它具有线性水平构件和垂直构件在接头处基本刚接之特点。这种框架用作独立的体系，或者和其他体系结合起来
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