同济大学：《建筑结构抗震》课程教学资源（课件讲稿）Chapter 6 Seismic Design of Reinforced Concrete Buildings（1/2）

2012/11/22 Chapter 6 Chapter 6 Bu Seismic Design of Structural System and m Grading for Reinforced Concrete 6Selsmic desgn of RCframes Buildings 6.1 Introduction RC structural system Frame structure Structural wall Dualsystem Mega structure 钢混土架买柱楼板配 梁柱节点钢筋连 1

2012/11/22 1 Chapter 6 Seismic Design of Reinforced Concrete Buildings Chapter 6 6.1 Introduction 6.2 Earthquake Damage in Reinforce Concrete Buildings 6.3 Structural System and Seismic Grading for Structures 6.4 Seismic design of RC frames 6.5 Seismic design of RC walls 6.6 Detailing 6.7 Dual system 6.8 Case Study 6.1 Introduction RC structural system Frame structure Structural wall Dual system Mega structure 钢筋混凝土框架 钢筋混凝土框架梁柱楼板配筋 梁柱节点钢筋连接

2012/11/22 梁柱节点钢筋 6.21

2012/11/22 2 梁柱节点钢筋 连接 梁柱节点钢筋 连接 一、现行设计能实现强柱弱梁吗 6.2 Earthquake Damage in Reinforce Concrete Buildings Sources: China Southwest Architecture Design Institute Co., Ltd Con’d 现行设计能实现强柱弱梁吗 Sources: China Southwest Architecture Design Institute Co., Ltd Weak Column

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2012/11/22 3 Con’d 现行设计能实现强柱弱梁吗 Sources: China Southwest Architecture Design Institute Co., Ltd 如何考虑框架、框剪结构中填充墙对整体结构的刚度贡献 Con’d Sources: China Southwest Architecture Design Institute Co., Ltd 如何考虑框架、框剪结构中填充墙对整体结构的刚度贡献 Con’d Sources: China Southwest Architecture Design Institute Co., Ltd 怎样合理设计框架、框剪结构中的填充墙 Con’d Sources: China Southwest Architecture Design Institute Co., Ltd 建议提高竖向构件的最低配筋水准 Con’d Sources: China Southwest Architecture Design Institute Co., Ltd 箍筋设置问题 Damage regarding to stirrups 建筑材料、施工管理问题 Con’d Sources: China Southwest Architecture Design Institute Co., Ltd

2012/11/22 Key lessons from 5.12 earthquake From the earthquake of RC Frames,wo ceg2sesgnorRctane was worse than other ederior an No-dc and bord failres

2012/11/22 4 十六、重视角柱、加腋梁柱的抗震设计 Photo by Prof. Xiong Haibei, 2008.7 Photo by Prof. Xiong Haibei, 2008.7 Photo by Prof. Xiong Haibei, 2008.7 Key lessons from 5.12 earthquake From the earthquake vulnerability of RC Frames, we learnt: • Vulnerability of corner columns was worse than other exterior and interior columns • Column’s failure were more occurred and more severe than beam’s. • All elements must be detailed so that they can respond to strong earthquakes in a ductile mode. • Non-ductile modes such as shear and bond failures must be avoided. • A high degree of structural redundancy should be provided. 6.3 Structural System and Seismic Grading for Structures Conceptual Design for RC frame structures  Height  Structural System  Seismic Fortification Intensity Grading  Requirements in Design  Configuration  Calculation  Details

2012/11/22 The appropriate maximum height for R/C buildings(m) Seismic grading for reinforced concrete buildings Table6.1.1inGB50011-2010) (able6.1.2inGB50011-2010) Notes:the heightn()is thevaluelistedn be continued (Table6.1.2inGB50011-2010) 6.4 Seism design of RC frames 5 Earthquake action and responses The Min Value of story shear force Three kinds of Cakulation method Base Shear Method all be qu thcbn0thhartoreoferentareyof

2012/11/22 5 The appropriate maximum height for R/C buildings (m) (Table 6.1.1 in GB50011-2010) Notes: the height in () is the value listed in GB 50011-2008 Structure Types System Seismic Fortification Intensity 6 7 8 (0.2g) 8 (0.3g) 9 Frame System 60 50 (55) 40 (45) 35 24 (25) Frame-Wall System 130 120 100 80 50 Structural Wall System 140 120 100 80 60 Frame supported Wall System 120 100 80 50 N.A Frame- Tube System 150 130 100 90 70 Tube in Tube System 180 150 120 100 80 Slab-Column and Wall System 80(40) 70(35) 55(30) 40 N.A Seismic grading for reinforced concrete buildings (Table 6.1.2 in GB50011-2010) Types of structure Seismic fortification intensity 6 7 8 9 Fram structure Height (m) ≤24 >24 ≤24 >24 ≤24 >24 ≤24 Frames 4th 3rd 3rd 2nd 2nd 1st 1st Large span frames 3rd 2nd 1st 1st Wall￾Frame structure Height (m) ≤60 >60 ≤24 25~60 >60 ≤24 25~60 >60 ≤24 25~60 Frames 4th 3rd 4th 3rd 2nd 3rd 2nd 1st 2nd 1st Structural walls 3rd 3rd 2nd 2nd 1st 1st Structural wall structure Height (m) ≤80 >80 ≤24 25~80 >80 ≤24 25~80 >80 ≤24 25~60 Structural walls 4th 3rd 4th 3rd 2nd 3rd 2nd 1st 2nd 1st be continued Types of structure Seismic fortification intensity 6 7 8 9 Frame - supported wall structure Height (m) ≤80 >80 ≤24 25~ 80 >80 ≤24 25~ 80 Struc￾tural walls General 4th 3rd 4th 3rd 2nd 3rd 2nd Streng￾thening 3rd 2nd 3rd 2nd 1stI 2nd 1st Frames that supporting walls 2nd 2nd 1st 1st Framed-tube structure Frame 3rd 2nd 1st 1st Tube 2nd 2nd 1st 1st Tube in tube structure Exterior tube 3rd 2nd 1st 1st Interior tube 3rd 2nd 1st 1st Slab-column￾wall structure Height (m) ≤35 >35 ≤35 >35 ≤35 >35 Columns 3rd 2nd 2nd 2nd 1st Walls 2nd 2nd 2nd 1st 2nd 1st (Table 6.1.2 in GB50011-2010) Seismic design flow chart 6.4 Seismic design of RC frames Earthquake action and responses  22 Three kinds of Calculation method • Base Shear Method • Response Spectrum Method • Time History Analysys Method To get the lateral force to every storey of the building struture, then To get the shear force of every storey of the building struture The Min Value of story shear force  Determination of story shear force  (GB 50011-2010, 5.2.5) The horizontal seismic shear force at each floor level of structure shall be comply with the requirement of the following equations:   n j i Veki  Gj 6 7 8 9 Structures with obvious torsion effect or fundermental period is less than 3.5s 0.008 0.016 (0.024) 0.032 (0.048) 0.064 Structures with fundermental period greater than 5s 0.006 0.012 (0.018) 0.024 (0.036) 0.048

2012/11/22 DValue methods the shear force of the k column in-t floor texural stitness for beam and column e the position of the point f contra-dexure in columr M。= ) 。-形h-的 5.12 M-Curve Three Principles of seismic design Lesson from this Diagran strong shear-weak fexure D Lesson from this Curve Lesson from these picture M-eCu 6

2012/11/22 6  the shear force of the k-th column in i-th floor where Vik is the horizontal shear force induced in column k on i th story; Dik is relative shear stiffness of column k on i th story; n i k ik ik ik V D D V   1 (5.7)  n k Dik 1 is summation of relative shear stiffness of total columns on i th story. D Value methods 2 c h 12i D  Determine the internal forces of lateral-force-resisting frames :  Determine flexural stiffness for beam and column;  Calculate D value;  Determine shear force for each column;  Determine the position of the point of contra-flexure in column;  Calculate column moments, then derive the beam moments based on equilibrium of beam to column joint, and distribution of beam stiffness / M kl  Vik h (5.11) ( ) / M V h h ku  ik  (5.12) 6.4 Seismic design of RC frames 6.4.1 Principle of seismic design Three Principles of seismic design strong shear - weak flexure strong column - weak beam strong joints - weak members M-φ Curve  Lesson from this Diagram M-φ Curve of a beam  Lesson from this Curve Ductility and compress ratio  Lesson from these picture

2012/11/22 6.4.2 Shear resistance of members 6.4.2 Shear resistance of members ion ratio:definc al she What's ditt ent from To dirtemin size of a member's section we should: bh≥fyE,',Bf) 6.4.2 Shear resistance of members 6.4.2 Shear resistance of members r六 M27s'10.20A.=5ym'/A. 5.46 6.4.3 strong shear-weak flexure Shear strength ofbeams ralfailureand prevent s n is I 7

2012/11/22 7 6.4.2 Shear resistance of members Members in a Frame: BEAM, COLUMN, and JOINT What's different from beam and column? A member subjected to an axial force of 0.1fc Ag or less can be treated as a beam (Ag = the gross area of the section, fc = design compression strength for concrete). To dirtermin size of a member's section we should: ( , , ) 0 RE c c bh  f  V  f • Shear compression ratio : defined as nominal shear stress divided by design compression strength of concrete (cylinder strength), or V/f cbh0 , to quantify a nominal shear stress across a beam section • To ensure that premature diagonal compression failure not occur before the onset of yielding of shear reinforcement, the diagonal compression principal stress should be limited. 6.4.2 Shear resistance of members For Beam：If ratio of span to depth (跨高比) is greater than 2.5, For Column and Wall：If shear span ratio (剪跨比) is greater than 2, the design shear force should satisfy the following equation: RE c c RE c c c c RE bh V f V f V f bh       / 0.20 5 / (0.20 ) 1 0 0     6.4.2 Shear resistance of members (0.15 ) 1 bh0 V f c c RE    For Beam：If ratio of span to depth (跨高比) is NOT greater than 2.5, For Column and Wall：If shear span ratio (剪跨比) is NOT greater than 2, the design shear force should satisfy the following equation: (0.30 ) 1 j c j j RE Vj  f b h  For Joint：  (5.46) 6.4.2 Shear resistance of members  To ensure ductile flexural failure and prevent brittle shear failure, a principle of “strong shear—weak flexure” should be followed in seismic design of reinforced concrete beams. Gb n r b l b vb V l M M V     6.4.3 strong shear—weak flexure For beams in grade 1 frames, is 1.3. For beams in grade 2 frames, is 1.2. For beams in grade 3 frames, is 1.1. vb vb vb Gb n r bua l bua V l M M V    1.1 For beams in grade 1 and earhtquake intensity 9,the beam and lintel may not be adjusted,but it should satisfy the following requirement. Shear strength of beams

2012/11/22 Shear strength of beams Shear strength of columns Principle of strong shear and weak flexure ≤0m+125亭 537 For colurrns in grade I frames.is 1.5.the others are 1.4 6六器做+子 Shear strength of columns shear resistance of columns oroh 1】 六0A+以子-02n 6.4.5 strong column-weak beam 品 ΣM,=n∑M, (Ft ∑M.=12∑M 8

2012/11/22 8 Considering shear strength reduction in the diagonal plan of beam under the repeated reversal loading, the shear strength in diagonal plan should be checked as the follows (0.42 1.25 ) 1 0 h0 s A V f bh f sv t yv RE b    ) 1 1.05 ( 1 0 h0 s A V f bh f sv t yv RE b      Shear strength of beams Principle of strong shear and weak flexure To ensure ductile flexural failure and prevent brittle shear failure n b c t c vc H M M V    (5.37) Shear strength of columns For columns in grade 1 frames, is 1.5, the others are 1.4 For columns in grade 2 frames, is 1.3. 1.2 For columns in grade 3 frames, is 1.2. 1.1 For columns in grade 4 frames, is 1.1. 1.1 vc vc vc vc   n t cua b cua H M M V  1.2 For columns in grade 1 and earhtquake intensity 9, the columns may not be adjusted, but it should satisfy the following requirement. Shear strength of columns shear resistance of columns 0.056 ) 1 1.05 ( 1 0 h 0 N s A V f b h f c sv t c c yv RE col       0.2 ) 1 1.05 ( 1 0 h 0 N s A V f b h f c sv t c c yv RE col       Design value of bending moments of members around a beam-column joint should satisfy the following requirements:  M c  c M b For frame structures in grades 1,2,3 and 4, is 1.7,1.5,1.3 and 1.2; For other types of frame, is 1.4 in grade 1,1.2 in grade 2,1.1 in grade 3and 4.  c  c  c  c 6.4.5 strong column—weak beam Mc,up Mc,low Mb,r Mb,l Vc,up Vb,l Vb,r Nc,up Nc,low Vc,low Vc,low Vc,low Beam Column  M c   M bua 1.2 For frame structures in grade 1 and earhtquake intensity 9, the design value of combined bending moment of menbers should satisfy the following requirement

2012/11/22 6.4.5 strong ioints-weak members (1)Design of beam-column joints (2)Design value of shear force -会 登-） (3)Seismic shearstrength checking of joint core 6.4.6 Lateral defection evaluation (b)Shear strength checking of beam-column joint ment should be limited -nA+05,g+L,52 64切 2） 548 Au.=[o (2cnderre 5 Aoygaeeaaleestcphc 651 9

2012/11/22 9 6.4.5 strong joints-weak members (1) Design of beam-column joints  The design of beam-column joints is primarily aimed at: (i) preserving the integrity of the joint so that the strength and deformation capacity of the connected beams and columns can be developed and substantially maintained. (ii) preventing significant degradation of the joint stiffness due to cracking of the joint and loss of bond between concrete and the longitudinal column and beam reinforcement or anchorage failure of beam reinforcement.                c b b o s b s b jb j H h h a h a M V / / 0 1 (5.41)               c b b o s b s bua j H h h a h a M V / / 0 1 1.15 (5.42) (2) Design value of shear force (b) Shear strength checking of beam-column joint             s h a f A b b V f b h N b s yv svj c j j t j j j RE j 0 1.1 0.5 1    5.47 (3) Seismic shear strength checking of joint core          s h a V f b h f A b s j t j j yv svj RE j 0 0.9 1   5.48 6.4.6 Lateral deflection evaluation • Drift, or inter-story displacement should be limited (causes non-structural damage and human discomfort and secondary stress in the main structure) • To provent the structure collapsed. (1) Checking elastic floor drifts under frequently accured earthquake, (2) Checking plastic fl oor drifts under r are accured earthquake, A simplified equation to calculate elastic-plastic story drift is the follows: u p   p  h (5.50) u p  pue (5.51) ue  e h