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（2/2）

Chapter 5 5.3 Capacity Calculation Structural Wall System 5.3.1 Flexural Capacity Calculation 5.1 Structural Wall System 5>5 5.4 Design of Structural Wall n Hai Tongji University, 5.3 Capacity Calculation 5.3.1 Flexural Capacity Calculation 5.3.1 Flexural Capacity Calculation Calculation Assumption: 4.The principlef themeRCcomn Large Eccentric Compression Large Eccentric Compression Calculation Sketch N=,6.x-nh-15 MM+M. 本课件版权归作者所有，仅供个人学习使用，请勿转载。1

1 Chapter 5 Structural Wall System 5.1 Structural Wall System 5.2 Structural Analysis 5.3 Capacity Calculation 5.4 Design of Structural Wall Xiong Haibei, Tongji University, 2015 5.3 Capacity Calculation 5.3.1 Flexural Capacity Calculation  Large eccentric compression  Small eccentric compression  Large eccentric tension  Small eccentric tension 1 b y s cu f 1 E      b    b    w 0 s M h e a N 2   w 0 s M h e a N 2   β1 is a factor to adjust the compresion strength of concrete. If strength of concrete C80, β1=0.74 Linear interpolation for strength between C50 and C80. 5.3 Capacity Calculation 5.3.1 Flexural Capacity Calculation  Large eccentric compression  Small eccentric compression 1 b y s cu f 1 E      b    b    β1 is a factor to adjust the compression strength of concrete. C80, β1=0.74; linear interpolation between C50 and C80. / 0   x h c c cu c f concrete   A B’ B C D E y f Es Rebar 5.3.1 Flexural Capacity Calculation Calculation Assumption: 1. Plane section assumption; 2. Consider the contribution of vertical distributing reinforcements ; 3. To guarantee safety, the vertical distributing reinforcements in compressive region are not considered in calculation because they may buckle for their small diameter; 4. The principle of design is the same as RC column.  Large Eccentric Compression Calculation Sketch ' ' ( .) ( )( ) ( ) ( ) sw 1 c w yw w0 w 0 yw sw w0 y s w0 s w0 yw sw w 0 w s y w0 s A N f b x f h 1 5x h f Ah x N M 1 1 fA h a 2 h fA MM M M M A fh a             Establish the Equilibrium Equation: First, set the vertical distributing reinforcements to get Mw. And then calculate M0.  Large Eccentric Compression α1 --- Ratio of the stress of compressive region and the strength of concrete The strength of concrete C80, α1=0.94 Linear interpolation for strength between C50 and C80. 本课件版权归作者所有，仅供个人学习使用，请勿转载

Small Eccentric Compression Small Eccentric Compression N■afAx+A,-a,A, Ne=a/h.s(h)+f,(h.-o) c=,+冬-d Large Eccentric Tension Large Eccentric Tension Calculation Sketch N=-ax+人是-l5国 w.4-之-网》 国上，速胜大物粒层国食力学为 5.3.2 Shear Resistance Calculation 5.3.2 Shear Resistance Calculation .Shear strength of eccentric compression: 1.No earthquako: ysa05h。+013w+冬 2.Earthquake: 北s是005A人，-01Bw+÷ z-0++0w+08 2.Earthquake:0.8/0.85 北≤n50点-0w+a8 本课件版权归作者所有，仅供个人学习使用，请勿转载。2

2 Calculation Sketch----Not consider the contribution of both the tensile and the compressive reinforcements, just like the analysis of small eccentric compressive column.  Small Eccentric Compression ' ' ' ()( ) 1cw y s s s 1 c w w0 y s w0 s w 0 s 1 s y b 1 N fbx f A A x Ne f b x h f A h a 2 h ee a 2 f                    Establish equilibrium equation; Establish compatibility equation of deformation, because the stress of one side of the rebar does not yield.  Small Eccentric Compression Calculation Sketch  Large Eccentric Tension ' ' ( .) . / ( )( ) ( ) ( ) sw 1 c w yw w0 w0 yw sw sw 1 c w yw sw w0 yw yw sw w0 y s w0 s w0 yw sw w 0 w s y w0 s A N f b x f h 1 5x h fA N N x 0 A fb 15f A h f f Ah x N M 1 1 fA h a 2 h fA MM M M M A fh a                      Large Eccentric Tension 5.3.2 Shear Resistance Calculation  Shear strength of eccentric compression: 1. No earthquake： 2. Earthquake： (. . ) . sw sh w t w w0 yh w0 1 A A V 05fb h 013N f h  0 5 A s    [ (. . ) . ] . sw sh w t w w0 yh w0 RE 1 1 A A V 0 4 f b h 0 1N 0 8 f h   0 5 A s      Shear strength of eccentric tension: 1. No earthquake： 2. Earthquake：*0.8/0.85 (. . ) . sw sh w t w w0 yh w0 1 A A V 05fb h 013N f h  0 5 A s    [ (. . ) . ] . sw sh w t w w0 yh w0 RE 1 1 A A V 0 4 f b h 0 1N 0 8 f h   0 5 A s     5.3.2 Shear Resistance Calculation 本课件版权归作者所有，仅供个人学习使用，请勿转载

5.3.2 Shear Resistance Calculation 5.4 Design of Structural Wall V.=nV. .9 degree soismic design: Layout Vertical Layout Principle the imtegral lateral stiffmess.but not too d up in themec than 8m and th of-p Bottom Strengthening Region Limit of Axial-Load Ratio 1oth nt of shear wall, 50m 10 本课件版权归作者所有，仅供个人学习使用，请勿转载。3

3  Shear force of the wall’s strengthening region should be amplified--- “Strong-Shear-Weak-Bending” 1. no earthquake or shear wall Ⅳ: no adjustment 2. shear wall Ⅰ, Ⅱ, Ⅲ: * amplification coefficient 1.6、1.4、1.2 successively。 3. 9 degree seismic design: V V w vw w  . wua w w M V 11 V M  5.3.2 Shear Resistance Calculation 5.4 Design of Structural Wall  Plane layout  Vertical layout  Axial-load ratio  Boundary element  Strengthening region at the bottom  Minimum ratio of reinforcement of wall  Construction requirement of tie beam Layout  Principle：  Increase the integral lateral stiffness, but not too high;  Try to make the stiffness center superpose upon the centroid to reduce eccentricity and avoid torsion;  Avoid short-width wall;  The length of wall should be shorter than 8m and the height-to-width should be lager than 2;  Wall should be arranged in two ways along the main axis and the axis of wall should align to the axis of frame.  It’s inappropriate to set the frame beam onto the tie beam. Vertical Layout  Principle：  The shear wall should be arranged continuously from bottom to top;  Openings should be lined up in the same place. Irregular openings should be strengthened;  Pay attention to the situation that the shear wall is set upon beams. These beams are frame-supported beams, so their seismic intensity should be upgraded. ;  Try to avoid weak layer. The shear force of weak layer should multiply the amplification coefficient 1.15;  The out-of-plane stiffness should be controlled. Bottom Strengthening Region  Purpose： To ensure draw ability after plastic hinges appear in the shear walls, the strengthening region at the bottom should be reinforced.  Principle：  1/8 of the total height of shear wall， When H>150m, 1/10  Or reinforce the two stories at the bottom Limit of Axial-Load Ratio  principle：  Increase wall’s ductility to make the shear walls at the bottom form plastic hinges when facing rare earthquake, avoiding brittle failure.  Axial pressure N based on representative value of gravity load. (different from columns) Axial-load ratio Ⅰ（9 degree） Ⅱ（7,8 degree） Ⅱ N/fcA 0.4 0.5 0.6 本课件版权归作者所有，仅供个人学习使用，请勿转载

Boundary Member Confined Boundary Membe Restraining boundar nb .Constructing boundary membe irst stor Confined Boundary Member 学 omm (I) 。 Longitudina Ordinary boundary member Ordinary boundary member 质玉打郭力烟的狗态达蝶将外 本课件版权归作者所有，仅供个人学习使用，请勿转载。4

4 Boundary Member  Restraining boundary member  Constructing boundary member  Principle  Restraining boundary member：the ends of Ⅰ,Ⅱ shear wall’ bottom-strengthening region and the first story above;  Constructing boundary member：the rest ends of Ⅰ,Ⅱ shear wall. The ends of Ⅲ,Ⅳ shear wall and non-seismic design wall. Confined Boundary Member  Length of wall: lc  Volume stirrup ratio:  Characteristic value of stirrup: v Item Ⅰ（9 degree） Ⅰ（7,8 degree） Ⅱ v 0.20 0.20 0.20 lc (embedded column) 0.25hw 0.20hw 0.20hw lc (flanking column or column at the end of wall) 0.20hw 0.15hw 0.15hw c v v yv f f     Diameter of stirrup: 8mm;  Stirrup spacing value: 100mm（Ⅰ） 150mm（Ⅱ）  Longitudinal reinforcement: range—shaded area A; Area—1.4%， 1.2%， 1.0%； （ special Ⅰ,Ⅰ,Ⅱ shear wall respectively ）  Diameter：  616, 614; Confined Boundary Member 约束边缘构件 约束边 缘构 件截 面及 配筋 Ordinary boundary member  Length of wall, minimum of stirrup, maximum stirrup spacing Ordinary boundary member 本课件版权归作者所有，仅供个人学习使用，请勿转载

Minimum Dimension of Shear Wall Minimum Dimension of Shear Wall 1.Nen-sismic: V.≤0.25Bfbh。 Shrar-span ratias?5 K50.15gA Distributing reinforcements 诗 nings should bo roir 200 Design of Coupling Beam Shear Capacity 1.non-seismic action Ks07iAw+人n section b d be adjusted of tle beams when 5≤六a2A+人 M≤f,A,(ho-a) 飞038h+09以n年如} 本课件版权归作者所有，仅供个人学习使用，请勿转载。5

5 Minimum Dimension of Shear Wall  Strength of concrete ≥C20；  Thickness of shear wall Seismic intensity region Embedded column Non-embedded column Ⅰ,Ⅱ Strengthening region at the bottom H/16 200 h/12 200 The rest H/20 160 h/15 180 Ⅲ, Ⅳ Strengthening region at the bottom H/20 160 H/20 160 The rest H/25 160 H/25 180 Non￾seismic design all H/25 160 H/25 180 1. Non-seismic action： 2）Seismic action： Shear-span ratio>2.5： Shear-span ratio≤2.5： . V 0 25 f b h w c   c w w0 (. ) w c c w w0 RE 1 V 0 20 f b h    (. ) w c c w w0 RE 1 V 0 15 f b h    c c w0 M V h   Minimum Dimension of Shear Wall Distributing reinforcements  Lateral and vertical reinforcement： sw sw w A b s   Shear wall Seismic intensity Minimum ratio of reinforcement Maximum spacing value Minimum diameter Normal height Ⅰ, Ⅱ, Ⅲ 0.25% 300 8 Normal height Ⅳ, non-seismic design 0.20% 300 8 B height Special Ⅰ Strengthening region: 0.40% The rest: 0.35% 300 8 Temperature -stress￾increase region Seismic and non-seismic 0.25% 200 —— Anchoring and overlapping of reinforcement of shear walls Process of openings  Lapping length≥ 1.2laE ;  Don’t lap on the same place;  Spacing between lap joints ≥500mm;  Openings should be reinforced. Design of Coupling Beam  principle：  Similar to the design of RC beam;  According to the design of double-tendon section beam;  Moment and shear force should be adjusted： Ideal elastic：6,7 degree： ×0.8， 8,9 degree： ×0.5，  Not consider the function of tie beams when meeting rare earthquake.  Flexure capacity： ' ( ) M y s b0  fA h a  1. non-seismic action： 2. seismic action： Span-to depth ratio>2.5: Span-to-depth ratio<=2.5: . sv w t b b0 yv b0 A V 07fbh f h s   ( . ) sv b t b b0 yv b0 RE 1 A V 0 42 f b h f h  s   (. . ) sv b t b b0 yv b0 RE 1 A V 0 38 f b h 0 9 f h  s   Shear Capacity 本课件版权归作者所有，仅供个人学习使用，请勿转载

Design of Coupling Beam Design of Coupling Beam .Strong-shear-weak-bending on-selsmic action 2.sismic action: 5=n.5+S+ Shear-span ratio52.5: .Misa Short-length Shear Wall 本课件版权归作者所有，仅供个人学习使用，请勿转载。6

6  Minimum dimension： 1. non-seismic action： 2. seismic action： Shear-span ratio>2.5： Shear-span ratio≤2.5： . V 0 25 f b h w c   c w w0 (. ) w c c w w0 RE 1 V 0 20 f b h    (. ) w c c w w0 RE 1 V 0 15 f b h    Design of Coupling Beam  Strong-shear-weak-bending Ⅰ,Ⅱ,Ⅲ,Ⅳ seismic： 9 degree： l r b b b vb Gb n M M V V l     . l r bua bua b Gb n M M V 11 V l    Design of Coupling Beam Reinforcement of coupling beams Short-length Shear Wall  Definition: ratio of length to thickness is 5 to 8;  Advantages: Better ductility compare with normal wall  Disadvantages: weaker in stiffness, and more drift than normal wall  Solutions: Be careful in detail design  The seismic rate is suggested to be higher, in order to make the wall with a sufficient ductility under rare earthquake, otherwise, it will be collapsed due to over large displacement  The ratio of short-length wall is suggested lower than normal wall. If the short-length wall is without boundary columns, the ratio should be 0.1 lower further  The shout-length wall is suggested to work with boundary columns and or Connor walls  The design shear force should be modified by the factor 1.4 or 1.2 for rate 1 or 2, except at the bottom strengthening zone  It is not allowed that all walls are short-length wall. Details of 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