同济大学：《高层建筑结构》课程教学资源（教案讲稿）Chapter 05 Structural Wall System（1/2）

2015/10/25 Chapter 5 Structural Wall System 5.1Structural Wall System 5.1 Structural Wall System Classification: 5.2 Structural Analysis 5.3Capacity Calculation Steel Brace Shear Wall Steel Reinforced Concrete(SRC) Xiong Haibei,Tongji University,2015 5.1 Structural Wall System n the Sie f Structural Openings .On the Size of Openings .single wall with few openings .Wall with small openings Influence of Openings Plane Section Assumption 本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 1 Chapter 5 Structural Wall System 5.1 Structural Wall System 5.2 Structural Analysis 5.3 Capacity Calculation 5.4 Design of Structures with wall system 5.5 Case Study Xiong Haibei, Tongji University, 2015 5.1 Structural Wall System Classification :  On the Basis of Materials  Reinforced Concrete Shear Wall  Steel Plate Shear Wall (SPSW)  Steel Brace Shear Wall  Steel Reinforced Concrete (SRC) 5.1 Structural Wall System  On the Size of Openings  single wall with few openings  Wall with small openings  Coupled walls  Wall Frame On the Size of Structural Openings single wall 整体墙 小开口墙 Wall with 双肢墙 多肢墙 壁式框架 small openings Coupled wall Wall frame Influence of Openings Plane Section Assumption single wall Wall with small openings Coupled-walls Construction of RC-Shear Wall Colum Slab Beam Beam Colum Main reinforcement of the colum Stirrup Main reinforcement of the beam Longitudinal reinforcement of the wall horizontal reinforcement of the wall Strengthening rebar of the opening 本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 Reinforced Concrete Shear Wall 5.2Structural Analysis .Characteristic 5.2.1 structural analysis the wall pe,but weak at .3-Dimer selected as calculating sketch; .In some cases,the structure can be considered as .The lmtation of buildings'height are increased; wall frame system if the opening is large nough Arrangement of shear walls are much more sensitive to but the influence of rigidoe of itsjonsshould buldng'storsionn plan; be taken into account Main disadvantage:space limited Simed Cladon of ne s光ae&wwh调hoae黑8opg .Calculation model-looks like cantilever beam Out-of-plane forceand deformaion arenot ▣胆 single wall A.=YA 0 rae8orwrialamagyr =1-1.25√A1A 本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 2 Reinforced Concrete Shear Wall  Characteristic  Great stiffness within the wall plane, but weak at out-of￾plane;  Both the lateral loads and vertical loads are resisted by the shear walls ;  The limitation of buildings’ height are increased;  Arrangement of shear walls are much more sensitive to building’s torsion in plan;  Main disadvantage：space limited 5.2 Structural Analysis 5.2.1 structural analysis  3-Dimensional spatial model is selected as calculating sketch;  In some cases, the structure can be considered as wall frame system if the opening is large enough, but the influence of rigid zone of its joints should be taken into account; 5.2.2 Simplified Calculation of single wall Methodology： similar to the analysis of cantilever Beam  Calculation model－ looks like cantilever beam  Calculation assumption：  Single wall, overall height;  Fixed support at the bottom;  Lateral loads result both flexural deformation and shear deformation;  Out-of-plane force and deformation are not considered.  Single wall & Wall with small openings structures should be revised by using the coefficient of openings. 倒三角形分布荷载 均布荷载 顶部集中荷载 整体墙 Inverted triangular distributed load Uniformly distributed load Concentrated load at the top of a structure Integral Displacement of single wall —— non-uniform coefficient of shear, for rectangular section: ＝1.2 G=0.42E, H—— height of the wall V0 —— Base shear force. . ( ) ( ) ( ) 3 0 w 2 w w 3 0 w 2 w w 3 0 w 2 w w 11 V H 3 64 EI u 1 60 EI H GA 1 V H 4 EI u 1 8 EI H GA 1 V H 3 EI u 1 3 EI H GA          Under Inverted triangular distributed load Under Uniformly distributed load ,m Under Concentrated load at the top of the structure 倒三角形分布荷载 均布荷载 顶部集中荷载 整体墙 5.2.2 Simplified Calculation of single wall Aw —— equivalent cross-sectional area of the wall; Aop—— area of openings; Af —— gross vertical area of the wall; 0 —— coefficient of the openings; Iwi —— equivalent cross-sectional moment of inertia of the i segment along the vertical; hi ——height of the ith segment 。 . / w 0 0 op f n wi i i 1 w n i i 1 A A 1 1 25 A A I h I h           本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 总一缺一物 a8w Method:similar to the analysis of cantilever The deformation of wall contain two parts Integral deformation+Local deformatior Ma=Ma+Ma .Integral deformation:same to single wall sumpt::M-k then:M=(-)M Ma=Ma+Ma =M+-kMr立 Displacement of Wall with mll open =12置，受 H'GA 1号爱 3uEL 524smp8C sSimpired Callation of Coued Wa inuous connecting-rod meth on 本课件版权归作者所有，仅供个人学习使用，请勿转载。 3

2015/10/25 3 5.2.3 Simplified Calculation of a Wall with Small Openings Method: similar to the analysis of cantilever  The deformation of wall contain two parts: Integral deformation + Local deformation  Integral deformation: same to single wall  Local deformation: caused by moment of one part of the wall Calculation Diagram Moments of Small-Opening Wall Total deformation＝ Integral deformation＋local deformation ' '' ' '' ' '' assumpt then ( ) ( ) Zi Zi Zi Zi Zi Zi Zi Zi Zi Zi i i PZ PZ i M M M M k M M 1 k M M M M I I kM 1 k M I I              ： ： Local moment Cross-section positive stress Integral moment Shear and Axial Force of Shear Wall  Shear force : distributed by the mean value of area and moment of inertia;  Axial force: caused by integral bending of the wall. ' i i Zi i i PZ i Zi Zi i i i 1 A I V 2 A I kM y N N A A I                 Local moment Cross-section positive stress integral moment Displacement of Wall with small openings u The deformation of wall with small openings is larger than the single wall’s. ×1.2 . . ( ) . ( ) . ( ) 3 0 w 2 w w 3 0 w 2 w w 3 0 w 2 w w 11 V H 3 64 EI u 1 2 1 60 EI H GA 1 V H 4 EI u 1 2 1 8 EI H GA 1 V H 3 EI u 1 2 1 3 EI H GA             Under Inverted triangular distributed load Under Uniformly distributed load Under Concentrated load at the top of the structure 双肢墙 倒三角形分布荷载 均布荷载 顶部集中荷载 5.2.4 Simplified Calculation of Coupled Wall  Method: Continuous connecting-rod method  Assumption： Shear wall is simplified as walls and coupling beams Point of contra-flexure is in the middle of coupling beams; All walls have the same stiffness and deformations ; Bending deflection and shear deformation are considered when calculating coupling beams and walls. coupling beams are regarded as uniformly distributed connecting-rods along the wall 5.2.4 Simplified Calculation of Coupled Wall Calculation method: continuous connecting-rod Size of the structure Serialization assumption Fundamental system 本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 Continuous deformation shear-balance equation: Continuous deformation shear-balance equation: 6,(x)+6,(x)+6,(x)=0 6(x)+6,(x)+6,(x)=0 -relative displacement caused by bending of wal 62(x) 6(x)=-2c0.(x)/ Compatible displacement equation Continuous deformation shear-balance equation: 6(x)+6,(x)+d,(x)=0 From:6x)+6,(x)+6(x)=0 (x) relative displacement caused by tion and shea 6)(x)ha' 0-f(r(x)) 3E ssumption:mx)=2crx） r()d solving the equotion) the total moment of goe历ciem he f ofhe:N-空 6H2 Inner force of wall 本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 4 Continuous deformation shear-balance equation: 1 2 3    （x x x 0 )+ （ )+ （ ) 1 （x)——relative displacement caused by bending of wall ( ) 1 m   （x 2c x )  1 2 3    （x x x 0 )+ （ )+ （ ) 2 （x) ( ) ( ) H x 2 1 2 x 0 1 1 1 x x dxdx E A A     （   ) Continuous deformation shear-balance equation: ——relative displacement caused by axial deformation of wall 3 （x) ——relative displacement caused by bending deflection and shear deformation of connecting-rod ( ) 3 3 0 b x ha x 3EI  （ ) 1 2 3    （x x x 0 )+ （ )+ （ ) Continuous deformation shear-balance equation: Compatible displacement equation '' '' '' From ( ) abtained ( ) ( ) ( ) Differential ( ) ( ) ( ) ( ( )) assumption ( ) ( ) solving the equ otion : ( 1 2 3 H x 3 m 0 1 2 x 0 b 3 m 0 1 2 b m x x x 0 1 1 1 x ha 2c x x dxdx 0 E A A 3EI 1 1 1 2ha 2c x x 0 E A A 3EI f x m x 2c x m                             : （ )+ （ )+ （ ) ： ： ： ) ( ) 2 1 0 2 V      Stiffness coefficient of coupling beam D: Stiffness ratio of coupling beam and wall: ( ) 0 2 b 3 2 2 1 1 2 2 2 2 1 1 2 1 2 I c D a 6H D h I I 6H D 2hsc 2cA A s A A           the total moment of the floor's connecting rods is: corresponding shear force is: / the end moment of coupling beam is : the axial th i i bi i bi bi 0 i m m h the V m h 2c M V a   =（ ) =（ ) = force of the wall is and ( ) ( ) th i bj n 1 i1 Pi j 1 2 j i n 2 i2 Pi j 1 2 j i i N V I M M m I I I M M m I I            n j=i ： = Inner force of wall 本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 Simplified Calculation of multiple Walls 60E IVH IV H u=3E Characteristic of multiple Walls Characteristic of multiple Walls stiffness of coupled beams mF刀m) D=Icla De D,阳 moment distribution coefficient mle walls israted total moment of rods:m=h beam's shear force: V=m(5)h/2c beam's end moment:M=d ↓Displacement axial force of walls:N,=∑Wy-yg-） oment ofa piece of wall.is distributed according =H 60 EI flexual moment IVH u二3Ela 本课件版权归作者所有，仅供个人学习使用，请勿转载。 5

2015/10/25 5 3 0 eq 3 0 eq 3 0 eq 11 V H u 60 EI 1 V H u 8 EI 1 V H u 3 EI    双肢墙 倒三角形分布荷载 均布荷载 顶部集中荷载 Displacement Coupled wall Inverted triangular distributed load Uniformly distributed load Concentrated load at the top of the strusture Simplified Calculation of multiple Walls Characteristic of multiple Walls 1. k+1 walls； 2. k coupled beams； stiffness of coupled beams： 3. integral coefficient expression of multiple walls is related with axial deformation coefficient; 4. coupled beams’ end moments are calculated on the basis of distribution coefficient / 0 2 3 D I c a i bi i i  0 bi bi bi 2 bi i I I 3 EI 1 GA a    moment distribution coefficient )] / i i i i i k i i i 1 i i i i m m D D 1 r r 1 4 B B               = （ ) : [1+1.5 (1- Characteristic of multiple Walls total moment of rods beam's shear force / beam's end moment axial force of walls: ( ) moment of a piece of wall, , is distr th i i th bi i th bi bi 0 n th i blj blj 1 l=i i m m h i V m h 2c i M V a i N V V j       ： = j （ ) ： =（ ) ： = = ibuted according to the flexual moment M ( ) n j ij k 1 Pi l l i l l i 0 j ij k 1 pi 0 j l i I M m I I V V I            3 0 eq 3 0 eq 3 0 eq 11 V H u 60 EI 1 V H u 8 EI 1 V H u 3 EI    Displacement 倒三角形分布荷载 均布荷载 顶部集中荷载 多肢墙 Coupled wall Inverted triangular distributed load Uniformly distributed load Concentrated load at the top of the strusture 本课件版权归作者所有，仅供个人学习使用，请勿转载

2015/10/25 5.2.6 Simplified Calculation of Wall Frame 圓 eam: w=a,-h/4 2=a,-14 oss-sectional centric axis as the aods of beams and column: =,-614 12=92-b/4 图5,14刚长发 Rigidity Wall Column Frame Dmethod Wall beam: Wall colum k-(c+c). Height of Point of Contra-flexure Displacement relative story displacement 4动 y=a++男++y Displacement oftop of the wall 图系16装性反有点位置 本课件版权归作者所有，仅供个人学习使用，请勿转载。 6

2015/10/25 6 5.2.6 Simplified Calculation of Wall Frame Method：similar to frame structure Assumption：  Link；  the cross area of coupling beam and wall is regarded as rigid zone;  Choose cross-sectional centric axis as the axis of beams and columns;  coupling beams do not fixed at boundary of the openings, so the length of rigid zone is not the dimension of the cross;  Displacement: similar to D method. 壁式框架 倒三角形分布荷载 均布荷载 顶部集中荷载 Length of Rigid Zone beam： / / b1 1 b b2 2 b l a h 4 l a h 4     / / c1 1 c c2 2 c l c b 4 l c b 4     column： Rigidity ' ( ) 12 21 c c m m m 12i 2     ' , 1 1 2 2 k c i k ci   Wall beam： Wall column： ' ( ) c c 1 k c c i 2   Wall Column Frame D method Height of Point of Contra-flexure 0 1 2 3 y a sy y y y      Displacement i i V u D   Shear force of i story j wall-column j ji i j D V V D   i relative story displacement Displacement of top of the wall i u u  本课件版权归作者所有，仅供个人学习使用，请勿转载