土木工程毕业设计目录

第一篇：土木工程毕业设计目录

绪论………………………………………………………..………………..…………….1 1.1 工程背景……………………………………………………………………………1

1.1.1 设计资料……………………………………………………………………..1 1.1.2 材料…………………………………………………………………………..1 1.2工程特点…………………………………………………………………………….1 1.3 本章小节…………………………………………………………………………….2 2 框架结构计算………………………………………………..............................................3 2.1 工程概况………………..…………………………………………………………...3 2.1.1 设计资料……………………………………….…………………………....3 2.2 梁柱截面、梁跨度及柱高度的确定……….………………………………………..4 2.3 何载计算……………….…….……………………..................................................7 2.3.1 屋面均布荷载……………………………………….………………………7 2.3.2 楼面均布恒载……………………………………………………………….7 2.3.3 屋面均布活载……………………………………………………………….7 2.3.4 楼面均布活荷载…………………………………………………………….7

2.4 梁柱自重……………….…….……………………...................................................8

2.5 墙体自重……………….…….……………………...................................................9

2.6 荷载总汇……………….…….…………………….................................................12 3 水平地震作用下框架的侧向位移验算…………………………...…………………….15 3.1 横向线刚度………………………………………………………...........................15

3.1.1 横向框架柱的侧移刚度D值……………………………………………..15 3.1.2 横向框架自振周期………………………………………………………...15 3.2 横向地震作用计算………………………………………………...........................17 3.3 横向框架抗震变形验算…………………………………………………………...18 3.4 水平地震作用下横向框架的内力分析…………………………………………...19 3.5 竖向荷载作用下横向框架的内力分析…………………………………………...20 3.5.1 荷载计算…………………………………………………………………...20 3.5.2 用弯矩分配法计算框架弯矩……………………………………………...22

3.6 梁端剪力及柱轴力的计算…………………………….…………………………26 4 内力组合计算…………………………………………………........................................28 4.1 框架梁内力组合………………………………………….....................................28 4.1.1 柱内力组合……………………………………………...........................32

4.1.2 截面设计……………………………………………................................33

4.2 内力计算…………………………………………………………………………...34

4.3 柱截面设计………………………………………………………………………..35

4.3.1 底层D柱截面设计………………………………………………………..35

4.3.2 本章小节…………………………………………………………………..39 5 楼板计算 ………………………………………………………………………………40 5.1 基本数据………………………………………………………………………….40 5.2 计算结果………………………………………………………………………….40 5.3 跨中挠度验算…………………………………………………………………….41 5.4 裂缝宽度验算…………………………………………………………………….42 6 楼板设计……………………………………………………………………………….46

6.1 基本数据…………………………………………………………………………..46

6.2 计算结果…………………………………………………………………………..46 6.3 跨中挠度验算……………………………………………………………………..47 6.4 裂缝宽度验算……………………………………………………………………...48 7 楼梯钢筋计算…………………………………………………………………………..52

7.1 荷载和受力计算…………………………………………………………………...52

7.2 配筋面积计算………………………………………………………………….......53

7.3 配筋结果…………………………………………………………………...............54 8 屋面板设计…………………………………………………………………................56

8.1 基本资料…………………………………………………………………...............56

8.2 计算结果…………………………………………………………………...............56

8.3 跨中挠度验算………………………………………………………………..........57

8.4 裂缝宽度验算………………………………………………………………..........58 结论…………………………………………………………………………………………..62

致谢…………………………………………………………………………………………..63 参考文献……………………………………………………………………………………..64

第二篇：土木工程毕业设计目录及摘要

延安市政府办公楼

摘 要

本工程为政府办公楼工程，采用框架结构，主体为六层，本地区抗震设防烈度为8度，近震，场地类别为II类场地。主导风向为西南，基本风压0.40KN/M，基本雪压0.25 KN/M。楼﹑屋盖均采用现浇钢筋混凝土结构。

本设计贯彻“实用、安全、经济、美观”的设计原则。按照建筑设计规范，认真考虑影响设计的各项因素。根据结构与建筑的总体与细部的关系。

本设计主要进行了结构方案中横向框架第12轴抗震设计。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。对楼板进行了配筋计算，本设计采用桩基础，对基础承台和桩进行了受力和配筋计算。

整个结构在设计过程中，严格遵循相关的专业规范的要求，参考相关资料和有关最新的国家标准规范，对设计的各个环节进行综合全面的科学性考虑。总之，适用、安全、经济、使用方便是本设计的原则。设计合理可行的建筑结构方案是现场施工的重要依据。

关键词： 框架结构，抗震设计，荷载计算，内力计算，计算配筋

I

Luoyang City West office building

ABSTRACT

This works for the city of Luoyang West office building, a framework structure for a six-storey main, in the region earthquake intensity of 8 degrees near Lan site classification as Class II venues.Led to the southwest direction, the basic Pressure 0.40 KN / M, basic snow pressure 0.25KN/M.Floor roof were using cast-in-place reinforced concrete structure.The design and implement “practical, security, economic, aesthetic,” the design principles.With the architectural design, design seriously consider the influence of the various factors.According to the structural and architectural detail and the overall relationship.The design of the main structure of the program horizontal framework 12-axis seismic design.In determining the distribution framework, the first layer of representative value of the load, Then use vertex from the displacement method for earthquake cycle, and then at the bottom of shear horizontal seismic load calculation under size, then calculated the level of load under the Internal Forces(bending moment, shear and axial force).Then calculate vertical load(constant load and live load)under the Internal Forces.Identify the most disadvantaged group or an internal force several combinations.Select the best safety results of the reinforcement and Mapping.In addition, the structure of the program indoor staircase design.Completion of the platform boards, boards of the ladder, platform beam component and the reinforcement of internal forces calculation and construction mapping.On the floor reinforcement calculation, the use of pile foundation design, foundation and pile caps for the force and reinforcement calculation.II

The whole structure of the design process, in strict compliance with the relevant professional standard, reference to relevant information and the latest national standards and specifications, and design of the various components of a comprehensive scientific considerations.In short, application, security, economic and user-friendly design is the principle.Design reasonable and workable structure of the program is the construction site of the important basis for the construction.KEY WORDS:Frame

Structure,Seismic

Design，Load Calculation，Internal force calculation，Calculation reinforcement

III

符 号

Ec －混凝土弹性模量；

C20－表示立方体强度标准值为20N/mm2的混凝土强度等级； N－轴向力设计值； M－弯矩设计值； V－剪力设计值； A－构件截面面积； I－截面惯性矩；

Gk－永久荷载标准值； QK－可变荷载标准值；

G－永久荷载分项系数；

Q－可变荷载分项系数；

Fek－结构总水平地震作用标准值；

GE、Geq－地震时结构的重力荷载代表值、表值；

T－结构自振周期；

RE－承载力抗震调整系数；

－构件长细比；

fak－地基承载力特征值； H0－基础高度；

d－基础埋置深度，桩身直径；

－土的重力密度；

k－风荷载标准值；

Fn－结构顶部附加水平地震作用标准值；

u－楼层层间位移；

e－偏心距；

Asv－箍筋面积； B－结构迎风面宽度；

h0－截面有效高度；

IV

等效总重力荷载代

AS－受拉区、受压区纵筋面积。

目 录

前言 ··················································· 错误！未定义书签。第1章 工程概况 ································· 错误！未定义书签。第2章 结构布置及计算简图 ············· 错误！未定义书签。§2.1结构布置及梁,柱截面尺寸的初选 错误！未定义书签。

§2.1.1梁柱截面尺寸初选 ··········· 错误！未定义书签。§2.1.2结构布置 ··························· 错误！未定义书签。§2.2框架计算简图及梁柱线刚度 ································· 6

§2.2.1确定框架计算简图 ········································· 6 §2.2.2框架梁柱的线刚度计算 ································· 8

第3章 荷载计算 ······························································· 8 § 3.1荷载计算 ································································ 8

§3.1.1 屋面及楼面恒荷载计算 ····························· 9 §3.1.2 屋面及楼面活荷载计算 错误！未定义书签。§3.1.3 梁，柱，墙，门窗重力荷载的计算 ········· 9

第4章 框架梁柱的线刚度计算 ····································· 12 §4.1框架梁柱的线刚度计算 ········ 错误！未定义书签。

§4.1.1 框架梁柱的线刚度计算·························· 13 §4.1,2 梁柱的线平均刚度比 ······························ 13

第5章 横向水平荷载作用下的框架内力计算 错误！未定义书签。

§5.1重力荷载代表值计算 ··············· 错误！未定义书签。

§5.1.1墙自重 ··························································· 16 §5.1.2梁，柱，墙，板，门窗重力荷载标准值 ··· 19 §5.1.3集中与各楼层处的重力荷载代表值计算 ··· 21 §5.2水平地震作用下框架内力合侧移的计算 ··········· 23 §5.2.1横向自振周期 ··············································· 23 §5.2.2水平地震作用 ··················· 错误！未定义书签。

V

§5.2.3水平地震作用下框架内力计算错误！未定义书签。

§5.3横向凤荷载作用下框架结构内力和侧移计算 ··· 错误！未定义书签。

§5.3.1凤荷载标准值 ··················· 错误！未定义书签。

§5.3.2风荷载作用下的水平位移验算 ··············· 35 §5.3.2凤荷载作用下的框架结构的内力计算 ··· 35

第6章竖向荷载作用下内力计算 ··································· 42

§6.1框架结构的荷载计算 ······································· 42 §6.2.1板传荷载计算 ··········································· 42 §6.2恒荷载作用下框架的内力 ······························· 45

§6.2.1恒荷载作用下框架的弯矩计算 ··············· 49 §6.3活荷载作用下框架的内力 ······························· 55

§6.3.1活载作用下的框架内力 ··························· 55 §6.3.2活荷载作用下框架的内力计算 ··············· 55 第7章 框架结构的内力组合 ········································· 64

§7.1框架结构梁内力组合 ··········· 错误！未定义书签。

§7.1.1框架结构梁的内力组合 错误！未定义书签。§7.1.2梁端弯矩控制值 ······································· 74 §7.1.3梁端截面组合的剪力设计值调整 ··········· 86 §7.2框架结构柱的内力组合 ··································· 86 §7.2.1框架结构柱的内力组合 ··························· 86 §7.2.2柱端弯矩设计值调整 ······························ 102 §7.2.3柱端截面组合的剪力设计值调整 ·········· 104 第8章 框架柱正、斜截面配筋计算 ···························· 105

§8.1框架柱的截面设计 ·········································· 105

§8.1.1框架柱截面设计 ······································ 105 §8.1.2柱斜截面受弯承载力计算 ······················ 119 §8.2框架梁的正，斜截面配筋计算 ······················ 119

§8.2.1正截面受弯承载力计算 ·························· 119 §8.2.2截面受弯承载力计算 ······························ 128

VI

第9章板配筋计算 ·························································· 130

§9.1板的配筋 ·························································· 130

§9.1.1设计资料 ·················································· 130 §9.1.2荷载计算 ·················································· 130 §9.1.3板的计算 ·················································· 130 第10章 楼梯计算 ·························································· 138

§10.1楼梯的设计 ···················································· 138

§10.1.1楼梯设计资料 ········································ 138 §10.2踏步板（TB-1）计算 ···································· 138

§10.2.1荷载计算 ················································ 139 §10.2.2斜截面内力计算 ···································· 140 §10.2.3截面承载力计算 ···································· 140 §10.3楼梯斜梁(TL-1)计算 ································ 140

§10.3.1荷载计算 ················································ 140 §10.3.2斜截面内力计算 ···································· 141 §10.3.3截面承载力计算 ···································· 141 §10.4休息平台（PB-2）计算 ································ 142

§10.4.1荷载计算 ················································ 142 §10.4.2平台板弯矩计算 ···································· 142 §10.4.3平台板配筋计算 ···································· 142 §10.5平台梁计算 ···················································· 143

§10.5.1荷载计算 ················································ 143 §10.5.2内力计算 ················································ 144 §10.5.3正截面承载力计算.······························ 146

第11章 基础配筋计算 ·················································· 146

§11.1桩基础的设计及验算 ···································· 147 §11.1.1桩设计资料 ············································ 147 §11.1.2确定桩的规格数量及初定承台的尺寸 147 §11.1.3单桩竖向承载力特征值 ························ 147 §11.1.4确定桩的数量 ········································ 147 §11.1.5桩承台的设计 ········································ 147

VII

§11.1.6单桩承载力的验算 ································ 147 §11.1.7偏心荷载验算 ········································ 147 §11.1.8楼梯设计资料 ············· 错误！未定义书签。§11.2承台的结构计算 ············································ 149 §11.2.1承台高度验算 ········································ 149 §11.2.2柱对承台的冲切验算 ···························· 149 §11.2.3角柱对承台的冲切验算 ························ 150 §11.2.4承台抗剪承载力验算 ···························· 150 §11.3基础配筋计算 ················································ 150 §11.3.1承台配筋 ················································ 150 §11.3.2桩身配筋 ················································ 151 §11.4中桩基础的设计 ············································ 152 §11.4.1桩设计资料 ············································ 152 §11.4.2确定桩的规格数量及初定承台的尺寸 152 §11.4.3单桩竖向承载力特征值 ························ 152 §11.4.4确定桩的数量 ········································ 154 §11.4.5桩承台的设计 ········································ 154 §11.4.6单桩承载力的验算 ································ 154 §11.4.7偏心荷载验算 ········································ 154 §11.5中桩承台的结构计算 ···································· 155 §11.5.1承台高度验算 ········································ 155 §11.5.2柱对承台的冲切验算 ···························· 155 §11.5.3角柱对承台的冲切验算 ························ 155 §11.5.4承台抗剪承载力验算 ···························· 155 §11.6基础配筋计算 ················································ 156 §11.6.1承台配筋 ················································ 156 §11.6.2桩身配筋 ················································ 157 第12章 结论 ·································································· 158 参考文献 ·········································································· 160 致 谢 ················································································ 161 外文资料译文 ·································································· 162

VIII

IX

第三篇：土木工程毕业设计中英文摘要和目录

西南石油大学毕业设计（论文）

摘要

建筑名称：瑞宁家园1#住宅楼设计

位置：达州市

结构类型：钢筋混凝土框架结构

层数：六层

总建筑面积：7629.4m2

本住宅楼建筑总高度25.875米，共6层，层高5.1米，采用现浇钢筋混凝土框架结构，抗震等级为三级，抗震设防烈度为7度，整个设计包括建筑设计和结构设计两部分。

建筑设计部分，首先根据建筑的使用功能确定建筑的初步布局，包括建筑体形及室内空间布置等，然后结合使用的材料、做法进一步确定具体尺寸和平、立面风格等。具体来说，主要包括总平面设计、建筑平面设计（功能布局及布置思路、房间的布置、卫生间的设计、楼梯布置、柱网布置等）、立面设计、剖面设计（层数层高确定及楼梯设计）、建筑材料设计（屋面做法、楼面做法）、防火设计，并对主要设计规范、尺寸单位、荷载及钢筋混凝土作了附加说明。

结构设计部分，首先根据建筑设计、工艺使用要求、材料供应情况以及场地地质条件等确定结构布置方案，然后选取一榀典型框架进行设计计算。在明确计算简图的前提下，进一步确定框架承受的荷载，包括恒载、屋（楼）面活载和风荷载，在此基础上进行框架内力计算（采用D值法计算风荷载作用下的内力，采用弯矩二次分配法计算竖向荷载作用下的内力）、内力组合、梁、柱和基础等构件的配筋计算以及风荷载作用下的侧移验算。除此之外，还进行了楼梯设计（梯段板设计、平台板设计及平台梁设计）、雨篷的设计与计算等。

为进一步验证手算结果的精确度，采用PKPM 2012行业软件进行了电算，比较分析表明手算结果是可靠的。最后用AutoCAD2012进行绘图。

关键词：钢筋混凝土框架结构；图书馆建筑；建筑设计；结构设计

Abstract

Building Name： Bo Wen Library

Location：DaZhou City SiChuan

Struture System： R.C.frame

Number of Stories： 4 stories

Total floor area：7629.4m2

The present project is a 6-storey residential building with the floor height

5.1m and the total height 25.875m.The four-layer building is adopted the cast-in-place reinforced concrete frame structure, and the seismic fortification intensity is 7 with the anti-seismic level 3.The entire graduation project divides into two parts, that is, the architectural design part and the structural design part.In the architectural design part, the general layout is first determined according to the building function, including conformation and indoor spatial setting;then the dimension, plane style and elevation style are chosen considering the engineering materials and constructional details.Make it more specific,it includes the total plane design, architectural graphic design(the function layout and layout thinking, the room arrangement and the design of toilet, stair layout, column, etc), the facade design layout, section design(the number and the height of layers identity and stair design), building materials design(roofing approach, the floor practice)and fire prevention design.In addition, instructions for the main design norms and dimensions, load and reinforced concrete is given.In the structural design part, the array of structure is first determined according to architectural design, process demand, material supply and geological conditions;then a typical transverse frame is taken to be computed.After the extract of computation sketch, the external loads such as the dead load, living load on the roof(floor)and the wind load are calculated;then the internal force in the structure is obtained through the moment distribution method under vertical loading and the modified knick point method under horizontal loads.In the following, the combination of internal force is performed.On the above basis, the design of structural components such as beams, columns and foundations are respectively carried out, and at the same time the checking computation of the

slidesway under wind load is done.In addition, the design of the stair(ladder period of board design, smooth bedplate design and platform beam design), the canopy and other accessory components are also completed..To verify the accuracy of above results, numerical tests are performed with the software PKPM 2012, and the comparisons prove that the original results are reliable.Finally, the construction documents are completed using the software AutoCAD 2012.Keywords:Reinforced concrete frame;Multi-storey library building;Architectural design;Structural design.目录建 筑 设 计................1

1.1设计资料................11.1.1建筑环境和地点...........1

1.1.2建筑设计要求.............2

1.2 总平面设计..............3

1.2.1总平面设计要求...........3

1.2.2总图设计..........4

1.2.3方案特点..........5

1.2.4基地环境分析.............6

1.2.5交通组织..........7

1.2.6绿化景观设计.............8

1.2.7 变形缝的设计.............8

1.3 建筑平面设计............9

1.3.1设计指导思想及本方案平面特点..........10

1.3.2功能分区.........12

1.3.3房间的布置..............14

1.3.4卫生间的设计............16

1.3.5人流组织说明............17

1.3.6防火疏散组织与楼梯的布置.......17

1.3.7柱网的布置..............19

1.4 立面设计...............20

1.4.1立面设计要求............21

1.4.2方案造型设计............22

1.5 剖面设计...............25

1.5.1剖面设计基本要求........25

1.5.2屋面、楼面做法..........28

1.5.3楼梯选择.........29

1.6细部构造总说明.........31

1.6.1勒脚.............31

1.6.2散水.............31

1.6.3楼地面工程..............32

1.6.4屋面工程.........32

1.6.5门窗工程.........32

1.6.6基础工程.........33

1.7防火设计...............34

1.8其它说明...............35

2.结构设计...........36

2.1 板的设计...............36

2.1.1 设计资料.........36

2.1.2 楼盖的结构平面布置......36

2.1.3单向板计算..............37

2.1.4 双向板计算..............40

2.2 次梁计算...............47

2.3框架设计...............50

2.3.1 设计资料.........50

2.3.2 结构布置及结构计算简图.........52

2.3.3 荷载统计.........54

2.3.4 内力计算.........66

2.3.5 内力组合.........94

2.3.6截面设计........102

2.4 基础设计..............114

2.4.1基础资料........114

2.4.2荷载统计........115

2.4.3地基承载力验算.........117

2.4.4配筋计算........126

2.4.5联合基础设计...........134

2.5 楼梯设计..............138

2.5.1 设计资料........138

2.5.2 梯段板设计.............139

2.5.3平台板设计.............140

2.5.4平台梁设计.............141

2.6 雨篷设计..............143

2.6.1 设计资料........143

2.6.2 雨篷板抗弯强度计算.....143

2.6.3 雨篷梁计算.............145 参考文献.............150 致谢..........152 附录 博文图书馆图纸目录表...........153

第四篇：土木工程毕业设计中英文摘要和目录

山东科技大学毕业设计（论文）

摘要

建筑名称：博文图书馆

位置：青岛市经济技术开发区

结构类型：钢筋混凝土框架结构

层数：四层

总建筑面积：7629.4m2

本设计建筑名称为博文图书馆。图书馆不仅仅是一个阅读的场所，它更是市民交流的平台。作为城市地标，拟建建筑物要考虑与临边建筑风格相协调，同时也要突出自己的建筑风格。

本图书馆建筑建筑总高度25.875米，共4层，层高5.1米，采用现浇钢筋混凝土框架结构，抗震等级为三级，抗震设防烈度为7度，整个设计包括建筑设计和结构设计两部分。

建筑设计部分，首先根据建筑的使用功能确定建筑的初步布局，包括建筑体形及室内空间布置等，然后结合使用的材料、做法进一步确定具体尺寸和平、立面风格等。具体来说，主要包括总平面设计、建筑平面设计（功能布局及布置思路、房间的布置、卫生间的设计、楼梯布置、柱网布置等）、立面设计、剖面设计（层数层高确定及楼梯设计）、建筑材料设计（屋面做法、楼面做法）、防火设计，并对主要设计规范、尺寸单位、荷载及钢筋混凝土作了附加说明。

结构设计部分，首先根据建筑设计、工艺使用要求、材料供应情况以及场地地质条件等确定结构布置方案，然后选取一榀典型框架进行设计计算。在明确计算简图的前提下，进一步确定框架承受的荷载，包括恒载、屋（楼）面活载和风荷载，在此基础上进行框架内力计算（采用D值法计算风荷载作用下的内力，采用弯矩二次分配法计算竖向荷载作用下的内力）、内力组合、梁、柱和基础等构件的配筋计算以及风荷载作用下的侧移验算。除此之外，还进行了楼梯设计（梯段板设计、平台板设计及平台梁设计）、雨篷的设计与计算等。

为进一步验证手算结果的精确度，采用PKPM 2005行业软件进行了电算，比较分析表明手算结果是可靠的。最后用AutoCAD进行绘图。

关键词：钢筋混凝土框架结构；图书馆建筑；建筑设计；结构设计1

Abstract

Building Name： Bo Wen Library

Location：Economic and Technological Development Zone of QingdaoStruture System： R.C.frame

Number of Stories： 4 stories

Total floor area：7629.4m2

The building name of the design is Bo Wen Library.It is not only a reading place but a platform for communication.As a landmark of the city, The proposed building should be considered to be in accordance with the style of surrounding architecture, and also highlight its own style.The present project is a 4-storey library with the floor height 5.1m and the total height 25.875m.The four-layer building is adopted the cast-in-place reinforced concrete frame structure, and the seismic fortification intensity is 7 with the anti-seismic level 3.The entire graduation project divides into two parts, that is, the architectural design part and the structural design part.In the architectural design part, the general layout is first determined according to the building function, including conformation and indoor spatial setting;then the dimension, plane style and elevation style are chosen considering the engineering materials and constructional details.Make it more specific,it includes the total plane design, architectural graphic design(the function layout and layout thinking, the room arrangement and the design of toilet, stair layout, column, etc), the facade design layout, section design(the number and the height of layers identity and stair design), building materials design(roofing approach, the floor practice)and fire prevention design.In addition, instructions for the main design norms and dimensions, load and reinforced concrete is given.In the structural design part, the array of structure is first determined according to architectural design, process demand, material supply and geological conditions;then a typical transverse frame is taken to be computed.After the extract of computation sketch, the external loads such as the dead load, living load on the roof(floor)and the wind load are calculated;then the internal force in the structure is obtained through the moment distribution method under vertical

loading and the modified knick point method under horizontal loads.In the following, the combination of internal force is performed.On the above basis, the design of structural components such as beams, columns and foundations are respectively carried out, and at the same time the checking computation of the slidesway under wind load is done.In addition, the design of the stair(ladder period of board design, smooth bedplate design and platform beam design), the canopy and other accessory components are also completed..To verify the accuracy of above results, numerical tests are performed with the software PKPM 2005, and the comparisons prove that the original results are reliable.Finally, the construction documents are completed using the software AutoCAD 2004.Keywords:Reinforced concrete frame;Multi-storey library

building;Architectural design;Structural design.目录建 筑 设 计................1

1.1设计资料................1

1.1.1建筑环境和地点...........1

1.1.2建筑设计要求.............2

1.2 总平面设计..............3

1.2.1总平面设计要求...........3

1.2.2总图设计..........4

1.2.3方案特点..........5

1.2.4基地环境分析.............6

1.2.5交通组织..........7

1.2.6绿化景观设计.............8

1.2.7 变形缝的设计.............8

1.3 建筑平面设计............9

1.3.1设计指导思想及本方案平面特点..........10

1.3.2功能分区.........12

1.3.3房间的布置..............14

1.3.4卫生间的设计............16

1.3.5人流组织说明............17

1.3.6防火疏散组织与楼梯的布置.......17

1.3.7柱网的布置..............19

1.4 立面设计...............20

1.4.1立面设计要求............21

1.4.2方案造型设计............22

1.5 剖面设计...............25

1.5.1剖面设计基本要求........25

1.5.2屋面、楼面做法..........28

1.5.3楼梯选择.........29

1.6细部构造总说明.........31

1.6.1勒脚.............31

1.6.2散水.............31

1.6.3楼地面工程..............32

1.6.4屋面工程.........32

1.6.5门窗工程.........32

1.6.6基础工程.........33

1.7防火设计...............34

1.8其它说明...............35

2.结构设计...........36

2.1 板的设计...............36

2.1.1 设计资料.........36

2.1.2 楼盖的结构平面布置......36

2.1.3单向板计算..............37

2.1.4 双向板计算..............40

2.2 次梁计算...............47

2.3框架设计...............50

2.3.1 设计资料.........50

2.3.2 结构布置及结构计算简图.........52

2.3.3 荷载统计.........54

2.3.4 内力计算.........66

2.3.5 内力组合.........94

2.3.6截面设计........102

2.4 基础设计..............114

2.4.1基础资料........114

2.4.2荷载统计........115

2.4.3地基承载力验算.........117

2.4.4配筋计算........126

2.4.5联合基础设计...........134

2.5 楼梯设计..............138

2.5.1 设计资料........138

2.5.2 梯段板设计.............139

2.5.3平台板设计.............140

2.5.4平台梁设计.............141

2.6 雨篷设计..............143

2.6.1 设计资料........143

2.6.2 雨篷板抗弯强度计算.....143

2.6.3 雨篷梁计算.............145 参考文献.............150 致谢..........152 附录 博文图书馆图纸目录表...........153

第五篇：土木工程毕业设计

土木工程毕业设计-某多层办公楼的设计

一、设计资料(1)设计标高:室内设计标高±0.000,室内外高差450mm.(2)墙身做法:采用加气混凝土块,用M5 混合砂浆砌筑,内粉刷为混合砂浆 底,纸筋灰面,厚 20mm,“803”内涂料两度。外墙采用贴面砖,1:3 水泥砂浆底厚20mm。(3)楼面作法:楼板顶面为水磨石地面,楼板底面为15mm厚白灰砂浆天花抹 面,外加V 型轻钢龙骨吊顶。(4)屋面作法:现浇楼板上依次铺 20mm 厚水泥砂浆找平层、300mm 厚水泥 珍珠制品隔热找平层、20mm厚水泥砂浆找平层和SDC120 复合卷材,下 面依次为15mm厚白灰砂浆天花抹面和V 型轻钢骨龙吊顶。(5)基本风压:ω o =0.3KN/m 2(地面粗糙度属C 类)。(6)基本雪压:S 0 =0.3KN/m 2。(7)抗震设防烈度:八度(0.2g)第二组,框架震等级为二级。地质条件 : 全套CAD 图纸,计算书,联系153893706 由上至下: 人工添土:厚度为1m 粉质粘土:厚度为7m,地基承载力特征值为500KPa 中风化基岩:岩石饱和单轴抗压强度标准值为3.6MPa 建筑场地类别为Ⅱ类;无地下水及不良地质现象。活荷载:上人屋面活荷载2.0KN/m 2 ,办公室楼面活荷载2.0KN/m 2 ,走廊楼面活荷 载2.5KN/m 2 ,档案室楼面活荷载2.5KN/m 2。

二、结构布置及结构计算简图的确定 结构平面布置如图1 所示。各梁柱截面尺寸确定如下: 主梁:取h=1/9l=1/9×7200=800mm,取h=800mm,取b=350mm, 次梁:取h=1 /16l=1/16×7200=450mm,取h=500mm,取b=250mm, 柱子:取柱截面均为b×h=600×600mm,现浇板厚为100mm。取③轴线为计算单元,结构计算简图如图2 所示??根据地质资料,确定基础顶面 离室外地面为500mm,由此求得底层层高为4.7m。各梁柱构件的线刚度经计算 后列于图 2 中,其中在求梁截面惯性矩时考虑到现浇楼板的作用,取 I=2I 0(I 0 为不考虑楼板翼缘作用的梁截面惯性矩)。梁柱均采用C30 混凝土。由于后面梁 的组合弯矩过大,相对受压区高度超过了界限受压区高度,所以改用 C40 混凝 土,因为梁柱刚度成比例增加,不会引起后面分配系数的改变。地震作用下的弹 性侧移将更小,一定符合要求。图2.框架计算简图 注:图中数字为各杆件的线刚度,单位为1.0e10 N*mm 框架刚度计算: 1.边跨框架梁线刚度: i b =2Ec×I。/l =2×3.0×10 4 ×350×800 3 /(12×7200)=12.44×10 10 N·mm 2.中间跨框架梁线刚度: i b =2Ec×I。/l =2×3.0×10 4 ×350×800 3 /(12×2400)=37.33×10 10 N·mm 3.二~五层框架柱线刚度: ic=Ec×I。/l =3.0×10 4 ×600×600 3 /(12×3600)=9.00×10 10 N·mm 4.首层框架柱线刚度: ic=Ec×I。/l =3.0×10 4 ×600×600 3 /(12×4700)=6.89×10 10 N·mm 在计算内力时,柱的线刚度取框架实际柱线刚度的0.9 倍。即:首层框架柱线刚度:ic=0.9×6.89×10 10 = 6.20×10 10 N·mm 二~五层框架柱线刚度:ic=0.9×9.00×10 10 = 8.10×10 10 N·mm

三、荷载计算 1.恒载计算(1)屋面框架梁线荷载标准值: SDC120 复合卷材 0.15KN/m 2 20mm厚水泥砂浆找平层 0.4 KN/m 2 300mm厚水泥砂浆珍珠制品隔热找平层 1.2 KN/m 2 20mm厚水泥砂浆找平层 0.4 KN/m 2 100mm厚现浇混凝土楼板 2.5 KN/m 2 15mm厚白灰砂浆天花抹面 0.26 KN/m 2 V 型轻钢龙骨吊顶 0.12 KN/m 2 屋面恒载 5.03 KN/m 2 框架梁自重 0.35×0.80×25=7.00 KN/m 梁侧粉刷2×(0.80-0.1)×0.02×17=0.48KN/m 作用在顶层框架梁上的线荷载为: g 5AB1 =g 5CD1 =7.00+0.48=7.48 KN/m g 5BC1 =7.48KN/m g 5AB2 =g 5CD2 =5.03×3.6=18.12 KN/m g 5BC2 =5.03×2.4=12.07KN/m(2)楼面框架梁线荷载标准值: 水磨石地面 0.65 KN/m 2 100mm厚现浇混凝土楼板 2.5 KN/m 2 15mm厚白灰砂浆天花抹面 0.26 KN/m 2 V 型轻钢龙骨吊顶 0.12 KN/m 2 楼面恒载 3.53 KN/m 2 边跨填充墙自重: 0.20×(3.6-0.80)×7=3.92KN/m 墙面粉刷:(3.6-0.80)×0.02×2×17=1.90KN/m 作用在中间层框架梁上的线荷载为: g AB1 =g CD1 =7.48+3.92+1.90=13.30KN/m g BC1 =7.48 KN/m g AB2 =g CD2 =3.53×3.6=12.71 KN/m g BC2 =3.53×2.4=8.47 KN/m(3)屋面框架节点集中荷载标准值: 边柱连系梁自重: 0.35×0.80×7.2×25=50.40KN 粉刷: 2×0.02×(0.80-0.10)×7.2×17=3.43KN 次梁自重: 0.25×0.50×7.2×25×0.5=11.25 KN 粉刷: 2×0.02×(0.50-0.10)×7.2×17×0.5=0.98KN 1.5m高女儿墙自重: 1.5×7.2×0.20×7=12.10 KN 粉刷: 2×0.02×1.2×7.2×17=5.88 KN 连系梁传来屋面自重: A、D 点相同:(3.6×3.6+1.8×1.8)×5.03=81.49 KN 顶层边节点集中荷载: G 5A =G 5D =165.53KN 中柱连系梁自重: 0.35×0.80×7.2×25=50.40KN 粉刷: 2×0.02×(0.80-0.10)×7.2×17=3.43 KN 次梁自重: 0.25×0.50×7.2×25×0.5=11.25KN 粉刷: 2×0.02×(0.50-0.10)×7.2×17×0.5=0.98 KN 连系梁传来屋面自重: B、C 点相同:[0.5×1.2×1.2+2.4×1.2+0.5×1.8×1.8+1.8×3.6] ×2× 5.03??117.70 KN 顶层中间节点集中荷载: G 5B =G 5C =183.76KN(4)楼面框架节点集中荷载标准值: 边柱连系梁自重: 0.35×0.80×7.2×25=50.40KN 粉刷: 2×0.02×(0.80-0.10)×7.2×17=3.43 KN 次梁自重: 0.25×0.50×7.2×25×0.5=11.25KN 粉刷: 2×0.02×(0.50-0.10)×7.2×17×0.5=0.98 KN 钢窗自重: 2.1×2×1.8×0.45=3.40KN 窗下墙体自重: 0.2×1.00×6.6×7=9.24KN 粉刷: 2×0.02×1.00×6.6×17=4.49 KN 窗边墙体自重: 1.8×(6.6-2.1×2)×0.2×7=6.05 KN 粉刷: 2×0.02×1.8×(6.6-2.1×2)×17=2.94 KN 框架柱自重: 0.60×0.60×3.6×25=32.4 KN 粉刷: 1.8×0.02×3.6×17=2.20 KN 连系梁传来楼面自重: A、D 点相同:(3.6×3.6+1.8×1.8)×3.53=57.19 KN 中间层边节点集中荷载: G A =G D =183.97KN 中柱连系梁自重: 50.40 KN 粉刷: 3.43 KN 次梁自重: 11.25 KN 粉刷: 0.98 KN 内纵墙自重: 6.6×(3.6-0.80)×0.2×7=25.87KN 粉刷: 6.6×(3.6-0.80)×2×0.02×17=12.57KN 扣除门洞重加上门重:-2.1×1.0×(7×0.2+17×2×0.02-0.2)=-3.95KN 框架柱自重: 32.4KN 粉刷: 2.20KN 连系梁传来楼面自重: B、C 点相同: [0.5×1.2×1.2+2.4×1.2+0.5×1.8×1.8+1.8×3.6] ×2×3.53=82.60 KN 中间层中节点集中荷载: G B =G C =217.75KN(5)恒载作用下的结构计算简图如图3 所示, 其中:G 5A = G 5D =165.53KN G 5B = G 5C =183.76 KN G A = G D =183.97 KN G B = G C =217.75 KN g 5AB1 = g 5BC1 = g 5CD1 =7.48 KN/m g 5AB2 = g 5CD2 =18.12 KN/m g 5BC2 =12.07KN/m g AB1 = g CD1 =13.30KN/m g BC1 =7.48KN/m g AB2 = g CD2 =12.71KN/m g BC2 =8.47 KN/m 2.屋、楼面活荷载计 屋、楼面活荷载作用下的结构计算简图如图4 所示。图中各荷载值计算如下 P 5AB = P 5CD =3.6×2.0=7.20 KN/m P 5BC =2.4×2.0=4.80KN/m P 5A =P 5D =(1.8×1.8+3.6×3.6)×2.0=32.40KN P 5B = P 5C = [1.2×1.2+4.8×1.2+1.8×1.8+3.6×3.6 ]×2.0=46.8 KN P AB =P CD =2.0×3.6=7.2 KN/m 走廊:P BC =2.5×2.4=6.00KN/m P A = P D =(1.8×1.8+3.6×3.6)×2.0=32.40KN P B = P C = 〔1.2×1.2+4.8×1.2〕×2.5+〔1.8×1.8+3.6×3.6〕×2.0=50.40KN 3.风荷载计算 风压标准值计算公式为: ω=β z ·μ s ·μ z ·ω o 因结构高度H=18.600m<30m,可取β z =1.0;对于矩形平面μ s =1.3;μ z 可查荷 载规范得到,当查得得μ z <1.0 时,??μ z =1.0。将风荷载换算成作用于框架每 层节点上的集中荷载,计算过程如表 1 所示。表中 A 为一榀框架各层节点的受 风面积。计算结果如图5 所示。风荷载计算 表1 层次 β z μ s Z(m)μ z ω o(KN/m 2)A(m 2)P w(KN)五层 1.0 1.3 19.1 0.78 0.3 24 7.30 四层 1.0 1.3 15.5 0.74 0.3 28.8 8.31 三层 1.0 1.3 11.9 0.74 0.3 28.8 8.31 二层 1.0 1.3 8.3 0.74 0.3 28.8 8.31 一层 1.0 1.3 4.7 0.74 0.3 33.2 9.58 4.地震作用计算(1)计算框架梁柱的线刚度: 框架梁线刚度:边跨:i b1 =12.44×10 4 KN·M 中跨:i b2 =37.33×10 4 KN·M 框架柱线刚度:首层:i c1 =6.20×10 4 KN·M 其他层:i c2 =8.1×10 4 KN·M(2)本设计仅考虑水平地震作用即可,并采用基底剪力法计算水平地震作用 力。为基底剪力,先要计算结构各层的总重力荷载代表值: 顶层: G5= g 5AB1 ×16.8+(3.6+7.2)×(g 5AB2 + g 5CD2)×0.5+ g 5BC2 ×2.4×0.5+ G 5A + G 5B + G 5C + G 5D +4×G 柱/2+0.3×7.2×16.8×0.5 =7.48×16.8+10.8×18.12×2×0.5+12.07×2.4×0.5+(165.33+183.76)×2+4 ×0.5×(32.4+2.20)+0.3×7.2×16.8×0.5=1121.36KN G4 = G3= G2= g AB1 ×14.4+ g BC1 ×2.4+ 0.5×g AB2 ×(3.6+7.2)×2+0.5×2.4× g BC2 +G A +G B + G C + G D +4×G 柱×0.5+[(3.6+7.2)×(p AB + p CD)×0.5+ p BC ×2.4× 0.5+P A + P B + P C + P D ] ×0.5 =13.30×14.4+7.48×2.4+10.8×12.71+0.5×2.4×8.47+(183.97+217.75)× 2+2×34.6+[10.8×0.5×14.8+1.2×6.0+(32.4+46.8)×2] ×0.5 =1352.30KN G1= G2+4×(2.35-1.8)×G 柱/3.6=1352.30+4×0.65×34.6/3.6 =1373.44 KN 质量重力荷载代表值见图6。图6.重力荷载代表值 G1 G2 G4 G3 G5(3)用“D 值法”计算柱的侧向刚度,框架各构件线刚度如图7: 其中:i 5 = i 4 = i 3 = i 2 =8.1×10 10 N*mm i 1 =6.2×10 10 N*mm i 51 =i 41 =i 31 =i 21 =i 11 =12.44×10 10 N*mm i 52 =i 42 =i 32 =i 22 =i 12 =37.33×10 10 N*mm 除底层外,其余柱的D 值: 边柱 由K=(i 51 + i 41)/(2 i 5)=1.536 α =K/(2+K)=0.434则D 5 边=12α i c /h j 2 =0.414×12×8.1×10 4 /3.6 2 =32550KN/m 中柱 由K=(i 51 + i 52 + i 41 + i 42)/(2 i 5)=6.144 α =K/(2+K)=0.754 则D 5 中=α ×12 i c / h j 2 =56550KN/m 底层柱的D 值: 边柱 由K= i 11 / i 1 =2.006 α =(0.5+K)/(2+K)=0.626 故D 1 边=α ×12 i c / h j 2 =21083 KN/m 中柱 由K=(i 11 + i 12)/ i 1 =8.027 得α =(0.5+K)/(2+K)=0.850 故D 1 中=α ×12 i c / h j 2 =28628 KN/m 从而得各层的侧向刚度如下: K 1 =2(D 1 边+ D 1 中)=2×(21083+28628)=99422KN/m K 2 = K 3 = K 4 = K 5 =2(D 边+ D 中)=2×(32550+56550)=178200KN/m T u 图7.框架各构件线刚度 i51 i5 i52 i51 i5 i5 i5 i41 i4 i4 i42 i4 i41 i4 i3 i2 i21 i3 i31 i32 i22 i2 i2 i3 i31 i21 i3 i2 i1 i11 i1 i12 i1 i11 i1(4)用顶点位移法求结构基本周期T 1 : 将各层的重力荷载代表值当作水平力,产生的楼层剪力为: V 5 =G 5 =1121.36KN V 4 = G 4 +G 5 =1352.30+1121.36=2473.66KN V 3 = V 4 +G 3 =3825.96KN V 2 = V 3 +G 2 =5178.26KN V 1 = V 2 +G 1 =6551.70KN 求顶点位移,则: U 1 = V 1 / K 1 =0.0659m U 2 = U 1 + V 2 / K 2 =0.0659+0.0291=0.0950m U 3 = U 2 + V 3 / K 3 =0.0950+0.0215=0.1165m U 4 = V 4 / K 4 + U 3 =0.1165+0.0139=0.1304m U 5 = V 5 / K 5 + U 4 =0.1304+0.0063=0.1367m= U T 故T 1 =1.8 =0.6655s 由于要考虑填充墙对基本周期的折减,故取折减系数为0.6, 则T 1 =0.6×0.6655=0.3993s 由于T 1 <1.4 T g =1.4×0.4=0.56s,故不考虑顶层附加作用。查抗震设计规范得:α max=0.16 由于T 1 =0.3993s< M 2=“131.33” ×6.41×7.2 12 边·l边2=“1” 5 P BA AB 各杆的固端弯矩为: ×6.0=“3.75KN/m” 8 5BC=“5” 中=“g” m KN ×7.2=“6.41” 64 5AB=“M” 边=“g” C B A(b)6.41KN 3.75KN 7.2KN D 6.0KN 图30.二~四层活载分布图及计算简图 28 姓名:余雪祯 某多层办公楼设计 重庆大学本科学生毕业设计 ??二~四层,其荷载分布情况如图30.所示: 140 215 645 22.34 0.73-0.72 CB-1-22.34-4.85-7.45-0.16 1.13-0.24-0.73 8.39 27.69 B5-1.44 BC 606-3.73 0.05 1.47 0.07-0.12 2.26 10.91 A5 16.78-27.69 1=“α” 22.35-5.01 29.52-24.51 12.43-12.43 图29.活荷载作用下顶层弯矩分配法计算过程 故其弯矩分配过程如图29.所示: ×3.00×1.2 6 中·l中2=“1” 3=“188.30KN” ×4.8=“3.00” 3.00KN 4.8KN(a)图28.顶层活荷载分布图 27 并利用结构对称性,取二分之一结构,如图(b)所示。图示(a)结构内力较复杂,可将梯形荷载等效为均布荷载计算,同样利用弯矩分配法计算, 对顶层,其荷载分布情况如图28.所示: 由于顶层活载大小和标准层不一样,所以分开算。活载分布按满布荷载法,活载作用下的内力计算也采用分层法。(1)2.楼面活载作用下的内力计算: 图27.恒载作用下的轴力图 26 图26.恒载作用下的剪力图 恒载作用下的内力简图为:剪力图(图26.)、轴力图(图27.)。(6)+V 柱+V +G 2C=“G” +N 1C=“G” G GC=“217.75” N2C V2 V1 N1C 底层中柱轴力 底层中柱 2D=“G” 1D=“G” N 25 GD=“183.97” N2D N1D 底层边??轴力 底层边柱 3C=“G” 二层中柱轴力 N3C 二层中柱 3D=“G” N3D 二层边柱轴力 二层边柱 4C=“G” 24 N4C 三层中柱轴力 三层中柱 4D=“G” N4D 三层边柱轴力 三层边柱 5C=“G” 四层中柱轴力 N5C 四层中柱 23 5D=“G” 四层边柱轴力 N5D 四层边柱 顶层中柱轴力柱G=“34.60” G5C=“183.76” 五层中柱: 顶层边柱轴力 G5D=“165.53” 22 五层边柱 柱子轴力计算:(5)V  ×8.47×2.4×1.2=“0” V·2.4-7.48×2.4×1.2-8.47KN 64.71 7.48KN 底层中间跨 对首层中间跨: 同理:V ×12.75×1.8×7.2-58.06+91.83=“0” ×7.2-13.30×7.2×3.6-12.71×3.6×3.6-91.83 12.71KN 13.30KN 58.06 底层边跨 对首层边跨: 57.86 二层中间跨 ??二层中间跨: 21 ×12.75×1.8×7.2-64.69+89.57=“0” 89.57 64.69 二层边跨 对二层边跨: 三层中间跨 对三层中间跨: ×12.71×1.8×7.2-64.40+89.65=“0” 89.65 64.40 三层边跨 对三层边跨: V×2.4-7.48×2.4×1.2-20 57.19 四层中间跨 对四层中间跨: ×12.71×1.8×7.2-65.82+89.20=“0” 89.20 65.82 四层边跨 对四层边跨: ×12.07×2.4×1.2=“0” ×2.4-7.48×2.4×1.2-顶层中间跨 73.13 12.07KN 对顶层中间跨: 19 ×18.12×1.8×7.2-44.29+91.37=“0” ×7.2-7.48×7.2×3.6-18.12×3.6×3.6-顶层边跨 91.37 18.12KN 44.29 对顶层边跨: 按简支跨计算: 剪力计算:(4)图25.恒载作用下弯矩图 18 值。调幅后的弯矩,跨中弯矩为调幅后的端弯矩的平均值减去跨中弯矩组合 的折减系数如图25.所示,括号内为 弯矩调幅:将梁各固端弯矩值乘以0.85(3)故梁的跨中弯矩即等于梁端弯矩的平均值减去跨中弯矩组合值。=“891.03” 边中=“=”    *M 9.46 4.07 5.39 * 179.39 93.21 86.18 标准层: 11.18 5.79 181.35 132.88 48.47 顶层: 跨中弯矩组合:(其它层中跨)×1.2×8.47×1.2×(顶层中跨)=“1” KN·M ×1.2×12.07×1.2× M中=“1” 对于:(其它层边跨)4=“245.26” ×12.71×2×(顶层边跨)=“12.71×2×3+” ×18.12×2× ×13.30×7.2(中跨)=“1” ×7.48×2.4(五层边跨)=“1” ×7.48×7.2 ql 17 将固端弯矩简化为简支端计算,同时将荷载叠加,如图24.所示: 跨中弯矩计算:(2)12.40-5.82 柱A、D: 柱B、C: 24.79-11.64 图23.一层传至柱底的弯矩 一层传至柱底的弯矩为(图23.):-16.10-5.12+0.97=“-20.25”-12.39+0.75=“-11.64”-80.62+4.49=“-76.13” 27.55-2.59-0.17=“24.79”-63.53-5.18+0.75-0.35=“-68.31” 35.98+11.15-3.38-0.23=“43.52” 图22.一层弯矩传递图 109.12-2.59+1.50=“108.03” 一层(图22.):-76.85+8.78=“-68.07”-66.89-10.04+1.47-0.64=“-76.10” 图21.二层弯矩传递图 33.45+11.15-6.55-0.42=“37.63” 33.45+11.99-6.55-0.42=“38.47” 107.47-5.02+2.93=“105.38”-15.37-5.37+1.91=“-18.83”-15.37-5.12+1.91=“-18.58” 16 二层(图21.): 107.47-4.84+2.85=“105.48” 图20.三层弯矩传递图-15.37-5.12+1.85=“-18.64”-76.85+8.54=“-68.31” 33.45+11.15-6.31-0.40=“37.89”-66.89-9.68+1.43-0.62=“-75.76” 三层(图20.):-76.85+9.57=“-67.28” 107.47-5.73+3.20=“104.94” 33.45+11.15-7.47-0.45=“36.68” 33.45+15.23-7.47-0.45=“40.76” 图19.四层弯矩传递图-66.89-11.45+1.60-0.69=“-77.43”-15.37-5.12+2.08=“-18.41”-15.37-6.06+2.08=“-19.35” 四层(图19.): 109.26-3.38+1.83-0.28+0.06=“107.49”-45.70-6.76+0.92-0.56=“-52.10” 45.70+11.15-4.39-0.36=“52.10” 图18.顶层弯矩传递图-91.06+4.84+0.18=“-86.04”-18.19-5.12+1.05+0.04=“-22.22” 顶层(图18): 点重新分配,从而得到竖向荷载作用下的整个结构的弯矩图。传递,并将叠加后的各个不平衡的节 将各层分层法求得的弯矩图叠加,柱子按 35.98 27.55-12.39 109.12-16.10-80.62 71.43 图17.底层弯矩图-63.53 15 底层(图17.): 33.45-15.37 107.47 67.66-76.85-66.89 图16.标准层弯矩图 标准层(图16.):-18.19 45.70-45.70 109.26-91.06 80.19 图15.顶层弯矩图(顶层)顶层(图15.): 各层弯矩图为: 72.85-3.06-12.12 B下-6.13 097 583-72.85 1.64-0.27-1.64 126-15.74 B上 2.82-24.24 106.36 24.73 194 24.68 2.81 A下 232 5.64 49.46 3.67 32.23 A上 303-106.36 465-0.36-0.55 0.06-0.28 0.13 0.08 图14.底层弯矩分配法计算过程 底层除柱子的线刚度不同外,其它同二至四层,故其弯矩分配过程如图14.所示: 69.79-15.17 123 566-69.79 0.93-0.20-0.93 1.65-23.31 23.08 189-11.66 30.10 3.30 283 5.06 46.16 434-0.31 图13.标准层弯矩分配法计算过程 14 弯矩分配法计算过程如图13.所示: ×12.77×1.2 g=“0.4s,故” ×24.62×7.2 ×8.47=“12.77” BC2=“7.48+” + BC1 ×12.71=“24.62” 57 AB2=“13.30+” AB1 同理,先将梯形分布荷载及三角形分布荷载化为等效均布荷载。利用结构对称性取二分之一结构计算。二至四层荷载作用如图12.:-3.62 2.64 81.09-7.24 102.04 30.92-2.64-0.09-0.88 4.10-0.57 0.14-0.03-0.02-27.03-17.60-81.09-102.04 61.84 40.20 8.19-0.44 0.27 5.33 0.17-13.52 图11.顶层弯矩分配法计算过程 13 弯矩分配法计算过程如图11.所示: ×15.09×1.2 5EB=“1” 5BE=“1” ×23.62×7.2 5BA=“1” 图示结构内力可用弯矩分配法计算,并可利用结构对称性取二分之一结构计算, ×12.07=“15.02” 5BC2=“7.48+” 5BC1 ×18.12=“23.62KN/m” 5AB2=“7.48+” 5AB1     Q , 7  ABQ)(图10.所示: 梯形分布荷载及三角形分布荷载化为等效均布荷载。等效均布荷载的计算公式如 图中,梁上分布荷载由矩形和梯形两部分组成,可根据固端弯矩相等的原则先将 倍。刚度取框架柱实际线刚度的0.9 中的柱子线 所示。图9 中取出顶层进行分析,结构计算简图如图9 顶层:由图3 恒载作用下的内力计算采用分层法。

四、内力计算 ]×0.3=“7.02” 雪C=“[1.2×1.2+4.8×1.2+1.8×1.8+3.6×3.6” 雪B=“P” ×0.3=“4.86KN” 雪D=“(1.8×1.8+3.6×3.6)” 雪A=“P” 雪BC=“0.3×2.4=0.72” 雪CD=“0.3×3.6=1.08” 雪AB=“P” 5.屋面雪荷载(分布同屋面活荷载)位移,肯定能够满足。由于风荷载远小于水平地震作用,故不必验算风荷载作用下框架的弹性水平满足要求。550 887 =“1” 19.1 10 21.53 H U=“1” 顶点位移验算: 满足要求;450 787 3.6 4.576 二层: 524 4.7 8.962 层间位移验算:底层:-3 +1.406×10 +2.783×10 +3.839×10 +4.576×10 K F EK=“6455.17” 地震作用下的水平位移的弹性验算: F5 F3 F4 F2 F1 图8.地震力的水平作用 11 故地震的水平作用力如图8.所示: 同理: ×891.03=“75.53KN” 76150.26  j E=“0.85×0.16×(1121.36+1352.30×3+1373.44)” 1G 1Geq=“0.85α” FEK=“α” max=“0.16” α>