Chapter 5 Chapter 5 5.1 Bulling asfionandmic forification 5.2 Seismic conceptual design Earthquake Effect and 5.3 Calculation Methodoogy ofsmio Seismic Design principles Configuration Conceptual Design Simpliciry Regularity quirem nts Symmetn Integral Construction Redundancy Objective 贤 度度 -tesdh design methodology 2 Structural detall 器敲盘 Return period (yr)50 475 2475
1 Chapter 5 Earthquake Effect and Seismic Design principles 5.1 Building classification and seismic fortification 5.2 Seismic conceptual design 5.3 Calculation Methodology of seismic action 5.4 Seismic Check of structural members and structural lateral deformations Chapter 5 Review Seismic Conceptual Design Simplicity Regularity Symmetry Integral Construction Redundancy 3 Review Configuration n Configuration characteristics and their effects n Configuration Regularities and Irregularities n Essentials of structural systems for seismic resistance Requirements n Structural Materials n Structural Systems n Connections n Details Review Objective Level 1 Level 2 Level 3 minor earthquake level moderate earthquake level major earthquake level 1) Elastic force 2) Elastic deformation (Stage 1) Structural detailing 1) Elastoplastic deformation (Stage 2) “two-stages and three-levels seismic design methodology” Review Return period (yr) 50 475 2475 Review
Chapter 5 ng00 2.The woanmn Calculation Methods , Seismic Influnce Coefficient Curve When 0.1 <T,<T, When T,T.57, Characteristic Pariods (s)
2 Chapter 5 Building classification and seismic fortification Review 5.3 Calculation Methodology of seismic action Based on the National Code of Seismic design for the Buildings (GB 50011-2010) 5.3.1 General Requirements 1. Seismic calculations should be performed in two directions of the structures; 2. The elements located in the diagonal with the main resistant systems ( angle larger than 15 degrees), should be considered in two directions; 3. If the system with obviously unsymmetrical distribution of mass and stiffness, the torsion effects should be considered; 4. the structures with large spans, long cantilever beams in the 8 or 9 degree areas, or high-rise buildings in the 9 degree areas should be considered of vertical earthquake effects. Calculation Methods n The Response Spectrum Method (RSM) for MDOF System n The Base Shear Method (BSM) for regularity structures with the height is less than 40m n The Time history method (THM) for dynamic analysis of structures, used for the buildings like: in the site of 7,8 and I,II, and with height large than 100m, or in the site of 8 and III, and with the height large than 80m, or in the site of 9 and with the height large than 60m, Seismic Influnce Coefficient Curve GB 50011-2010, Fig. 5.1.5 Seismic Influence Coefficient amax Rare Eq. 0.28 0.50/0.72 0.90/1.20 1.40 Frequently Eq. 0.04 0.08/0.12 0.16/0.24 0.32 Intensity 6 7 8 9 Characteristic Periods（s） 3 0.35 0.45 0.65 0.90 rd Group 2 0.30 0.40 0.55 0.75 nd Group 1 0.25 0.35 0.45 0.65 st Group Site Category I 0 I 1 II III IV 0.20 0.25 0.30 Seismic Influnce Coefficient Curve 2 max g g i T T When T T 5 a h a g = × < < （ ） ： T g 2 1 max i [ 0.2 ( 5 )] When : 5Tg T 6.0 a h h a g T T g s = × × × - < < 2 max 0 Ti When .1 a = h a ： < < Tg
Seismic Influnce Coefficient Curve How to simplify the real building toamodel? The Power Index of the curvilnear decrease sectio In two main direction.Some time sdir.And Ydi 7=0.9+8i of declined slope %=1+80流205 BSM Assumption tical and F1-8 G Efficient Weight RSM 1.Without Torsion vibration For SDOM.G-G .For MDOM.G G,? and Based on G:Represent Weigkt G,=Gu+Lw w(T Bunch or balls
3 Seismic Influnce Coefficient Curve The Adjusting coefficient of declined slope . . . 0 05 0 9 0 3 6 x g x - = + + .0 .0 1 0 5 0 2 4 32 x h x - = + + .0 .0 . 2 0 5 1 0 8 1 6 x h x - = + + The Power Index of the curvilinear decrease section: The damping adjusting factor h2 ³ 0.55 and How to simplify the real building to a model? In two main direction. Some time is X dir. And Y dir. 框架梁 框架梁 框架梁 框架梁 框架柱 PKPM 3D model Simplification of regular structure. 质量 刚度 X 向 X 向 Kx m 质量 刚度 Y 向 Y 向 Ky m BSM Assumption： Regularity in vertical and plan； Height less than 40m； First mode shape is in dominant; High modes effects are concerned。 n n Ek n Ek n j j j i i i Ek eq F F F G H GH F F G d d a D = = - = å= (1 ) 1 1 Geq Efficient Weight n For SDOM, n For MDOM, Geq i =0.85 G n i å Geq 1 =G G i ? Gi Represent Weight Gi =Gki+ L ji ki Live Load combination factor roof loads: 0.0 or 0.5 or 1.0 For floor loads: 0.5 or o.8 hook loads: 0.0 or 0.3 ( Table 5-4) ji RSM q Important and dominant； q Based on Elastic response spectra analysis q Used for elastic structure q Plan regularity q Bunch or balls 1. Without Torsion vibration
Combination of modes Combination of modes ·SRSs ·cac F.=aX.G. sa-oas,s. 8561+2以 P0-T+4k6+广方 ar ofthndtmode With Torsion vibration Only under X dir.Eq. 入 工x,0, 旧 ++8 Bunch of plates Fo=.X,G Fn=《,y,YG =a,yrpG .Undor two directions excited How many modes should be composed?Tota? Combination of two dirction t Sa=5:+0.53,厅 how角any Sm=5+(0.85S,)月 地0
4 Combination of modes § SRSS å å å = = = = = 2 1 2 1 Ek j n i ji i n i ji i j ji j j ji i S S X G X G F X G g a g Combination of modes § CQC å å= = = m j m k S Ek jk S j S k 1 1 r T j k T T j k T T jk l z z l l z z l l r 2 2 2 1.5 (1 ) 4 (1 ) 8 (1 ) - + + + = － coupling factor of i-th and j-th mode . rjk With Torsion vibration q Plan irregularity q Torsion should be concerned carefully q Bunch of plates What’s different between regular and irregular structure? O n ly und e r X d ir. E q . ( ) O n ly und e r Y d ir. E q . ( ) W ith a n a n g le o f θ to X D ir. E q . c o s sin n ji i i 1 tj n 2 2 2 2 ji ji ji i i i 1 n ji i i 1 tj n 2 2 2 2 ji ji ji i i i 1 tj x j yj X G X Y r G Y G X Y r G g f g f g g q g q = = = = = + + = + + = + å å å å tji j tj i ji i Yji j tj ji i Xji j tj ji i F r G F Y G F X G a g j a g a g 2 = = = tj g 2 2 2 2 ( 0 .85 ) ( 0 .85 ) Ek y x Ek x y S S S S S S = + = + n Under two directions excited Combination of two direction excited 1 0.85 How many modes should be composed? The numbers should satisfy that the 90% mass are participant in the vibration. According to some rules, based on the codes or on the required of engineering needed. Total? How many?
THM k THM 三一峰 Tabh512-2eoa辅emp0 aik vele【Gmk力 Vertical Earthquake Action Vertical Earthquake Action To High rise buildie F=G m 0.752 1=0.100.20 Base Shear Force nd structura a-2型 Strength-Capacity The Minimum Base Shear Force Displaceme 00602/02
5 THM 1. Dynamic analysis 2. Elastic THM 3. Elasto-plastic THM Table 5.1.2-2 acceleration peak velue（cm/s2） Intensity 6 7 8 9 Frequently Eq. 18 35/55 70/110 140 Rare Eq. 125 125/220 400/510 620 -0.20 -0.10 0.00 0.10 0.20 10 12 14 16 18 20 22 24 26 28 30 时间(s) 加速度(g) 结构反应记录 一层反应 -0.20 -0.10 0.00 0.10 0.20 10 12 14 16 18 20 22 24 26 28 30 时间(s) 加速度(g) 地面振动记录 台面输入 Responses record Ground motion record Vertical Earthquake Action To High rise building § Regularity in vertical and plan； § First mode shape is in dominant; § Vertical Eq. response spectra almost same to Horizontal Eq. § But, the peak value is 65% of H.Eq. v v,max v v i 1 v,max H,max 0.65 E k eq i i i n E k i i F G GH F F GH a a a = = = = å Geq i =0.75 G n i å Gi Gj Gn Hi Hj FEvk Fvn Fvj Vertical Earthquake Action To Long cantilever beam and the other large span structures Regularity in plan； First mode shape is in dominant; F G vi =lEv i lEv =0.10~0.20 Gi Gj Gn Hi Hj FEvk Fvn Fvj 5.4 Seismic Check of structural members and structural lateral deformations n Strength —— Capacity n Displacement —— Ductility Base Shear Force i n Eki j j V G l = The Minimum Base Shear Force > å 类别 6 7 8 9 Obviously Torsion and T1 less than 3.5s 0.008 0.016/0.024 0.032/0.048 0.064 T1 less than 3.5s only 0.006 0.012/0.024 0.024/0.036 0.048 j i = F n Eki j V = å
D Value methods forces of lateral the shear force of o. M.-FAW M-F(h-W) 6.12) Combination of responses to withstand the most Without earthquake action S=YoSck+WoloSok+W.Y.S. (S=YaSor +YaSp 5.13 是 With earthquake action aS■7aS+Ψe￥S+YaSw 5.14 S-YSa +YOSm +YnSm+W.7S. S<RIYE SsR性E 516) Combination of responses S=YGSGE+YBSEM+YeSEM earthquake action S=a+Sx+Sx+w,风 S<C/Ym 6
6 n 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 å= = 1 (5.7) å= n k Dik 1 is summation of relative shear stiffness of total columns on i th story. D Value methods Determine the internal forces of lateralforce-resisting frames : Ø Determine flexural stiffness for beam and column; Ø Calculate D value; Ø Determine shear force for each column; Ø Determine the position of the point of contra-flexure in column; Ø 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 ku = Vik h - h (5.12) Combination of responses Without earthquake action With earthquake action RE G GE Eh EhK Ev EvK w w wK S R S S S S S g g g g y g < / = + + + S R S S S S G GK Q Q QK w w wK < =g +y g +y g n Structures must be designed to withstand the most unfavorable combination of design loads. (1) G GE Eh Ehk S = g S + g S (5.13) (2) G GE W W Wk Eh Ehk S = g S + y g S + g S (5.14) (3) G G Q Q S = g S + g S (5.15) S R RE £ / g (5.16) • For some structures, when vertical seismic action is considered, the following combination of loads should be checked: G G E Eh Ehk Ev Evk S = g S + + g g S S Combination of responses Without earthquake action —— Displacement Check With earthquake action C/ GE EhK EvK w wK RE S S S S S S y g = + + + < GK Q QK w wK S S S S S C = + + y y <
Jsed for design of elements Displacement Check Elastic drift and Elasto-plastic drift requirement Duel syste 1/00 1/168 ahe鱼te,chear wall 1/1000 1/120 Stse frame 1/2s0 Elasto-plastic drift E-P Displacements under rare occurred ，=,4 Where 4 -Bade-Patic arder R.Fq 出，≤，☑圆 ed? Designa bulding? What kind of Building? How many? How to do that?
7 Used for design of elements Material Structural elements Force condition γRE Steel Column，beam, brace, connection, bolting, welding forcing stability 0.75 0.80 Reinforced Concrete beam Column with ratio of compression less than 0.15 Not less than 0.15 Share wall The other elements bending Eccentric compression E.C. E.C. Shearing, E. tension 0.75 0.75 0.80 0.85 0.85 Masonry Confined masonry wall, Masonry wall and Reinforced masonry wall shearing Shearing 0.9 1.0 RE g Displacement Check Elastic drift and Elasto-plastic drift requirements Steel frame 1/250 1/50 tube in tube , shear wall 1/1000 1/120 Duel system 1/800 1/100 RC Frame 1/550 1/50 Structure E-P supported system 1/1000 1/120 [ ] y qp [ ] q 39 p Target：No Collapse p The peak value of acceleration is about 4～6 times than that of frequently occurred earthquake. p The elements will be in elastic-plastic behavior p Using development of plastic hinges to dissipate the energy. p restrict the limitation of the drift of soft/weak story. Elasto-plastic drift under rare occurred Earthquake u h p p D £ [q ] [ ] p q 40 E-P Displacement： p p e Du =h Du Where ： Δup—— Elasto-Plastic displacement under R. Eq.； Δue—— Elastic displacement under R. Eq. ηp—— Amplified coefficient of E-P 。 How many modes should be composed? The numbers should satisfy that the 90% mass are participant in the vibration. According to some rules, based on the codes or on the required of engineering needed. Total? How many? Design a building? What kind of Building? How to do that?
Flow Chart of seismic design Flow Chart of seismic design → 用手子 翻 Flow Chart of seismic design Flow Chart of seismic design 嗣 Flow Chart of seismic design Flow Chart of seismic design ar乳rene 8
8 Flow Chart of seismic design simplified m4 m3 m2 m1 K4 K3 K2 K1 calculation of E. action Floor 1 Floor 2 Floor 3 Floor 4 C-F Point D value Shear force distribution Response combination Lateral Force Moment diagram under lateral load Flow Chart of seismic design Vertical Loads Flow Chart of seismic design Moment diagram under vertical loads combined Moment diagram Flow Chart of seismic design Design of Element Flexural Strength beam Shear Strength Flexural Strength Shear Strength The strength of Shear shall be stronger than fexual beams shall be stronger than columns Column Joint shear strength joint shall be stronger than connected beams and columns 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 Flow Chart of seismic design The strength of Shear shall be stronger than fexual Drift Flow Chart of seismic design 0 6 12 18 24 30 36 42 48 54 60 66 0 1/500 ETABS SATWE 1/542 0 6 12 18 24 30 36 42 48 54 60 66 0.8 1 1.2 1.4 ETABS SATWE 1.2 1.4
-m w-054x1J6xa270-66N 后413子0卫N a=08x09x270=61 Fa=0.08x0,429x÷2700=61.8 F,=0sx0429←x10=-61,4N 5=0x026x120=562N 鞋 。0sas（}2mo-2N 绿 eoaoe6 .=008T+0.0=008x0.4665+001-=0.04732 2G.-=270x5+2700x10+180x15=6750 =0-004723048=2,97 5=200t-a072k33048=2.9则 atal sesm sof th
9 [Example 4.6] One 3-story reinforce concrete building is located at site of the second group in category I1 . The stifiness of the beams is ¥ and two columns have the same cross-section . The weight of each floor, three mode shapes and corresponding periods are shown in Figure 4.3. The seismic zone of intensity for design is 7. Tg=0.3s. x=0.05. Draw the figures of the moment under the horizontal seismic actions by mode analysis method. G3= 2700kN T1=0.4665 T2=0.2086 1/3 T3=0.1348 2/3 1 G2= 2700kN G1= 1800kN 2/3 1 1/4 2/3 -1 -3/4 Participating factor 1.364 3 1 2700 3 2 1800 1 2700 3 1 2700 3 2 1800 1 2700 2 2 2 1＝ ＝ ÷ ø ö ç è æ ÷ + ´ ø ö ç è æ ´ + ´ ´ + ´ + ´ g Seismic actions 2700 66.3kN 3 1 F 0.054 1.364 11＝ ´ ´ ´ ＝ F 0.054 1.364 1 1800 132.6kN 2700 132.6kN 3 2 F 0.054 1.364 13 12 ＝ ＝ ＝ ＝ ´ ´ ´ ´ ´ ´ The seismic actions of mode 2 and mode 3 can be obtained in a similar way. Solution: Given: Mode 1, T1=0.4665 > Tg Seismic influence coefficient 0.08 0.054 0.4665 0.3 T T 0.9 max 0.9 1 g 1＝ ＝ ÷ ´ ＝ ø ö ç è æ ÷ ÷ ø ö ç ç è æ a a The moments of the frame can be drawn according to the actions above and we’ll finally get the figure of the moment after conbinating modes by SRSS method, shown in Figure 4.4 and Figure 4.5. 2700 61.8kN 3 2 F22＝0.08´ 0.429´ ´ ＝ F23＝0.08´ 0.429´ (－1)´1800＝－61.8kN F31＝0.08´ 0.26´1´ 2700＝56.2kN 2700 42.1kN 4 3 F 0.08 0.26 32＝ － ÷´ ＝－ ø ö ç è æ ´ ´ 1800 9.4kN 4 1 F33＝0.08´ 0.26´ ´ ＝ 2700 61.8kN 3 2 F21＝0.08´ 0.429 ´ ´ ＝ 414.4 -42.1 56.2 132.6 132.6 66.3 9.4 (c) 40.9 11.5 29.4 11.5 29.1 40.9 422.6 342.8 745.9 11.8 (a) 497.3 61.8 165.8 331.5 165.8 -61.8 61.8 (b) 77.3 77.3 414.4 331.5 77.3 77.3 77.3 77.3 11.8 40.9 29.1 29.4 183.3 183.3 526.1 342.8 422.6 342.8 183.3 342.8 526.1 765.4 497.3 Figure 4.4 Moments of Each Seismic Actions Figure 4.5 Moment of the Frame Structure after Modes Conbined [Example 4.7] Draw the figures of the moment under the horizontal seismic actions in Example 4.6 by base shear method. Equivalent values of the whole gravity loads Geq＝0.85´ (1800 + 2700 ´ 2)＝6120kN Seismic influence coefficient 0.08 0.054 0.4665 0.3 0.9 1＝ ÷ ´ ＝ ø ö ç è æ a Total seismic actions FEk＝a1Geq＝0.054 ´ 6120＝330.48kN Because T1 ＝0.4665＞1.4Tg＝1.4 X0.3=0.42 d n＝0.08T1 + 0.01＝0.08´0.4665+ 0.01＝0.04732 å= ´ + ´ + ´ n k 1 GkHk＝2700 5 2700 10 1800 15＝67500 (1 0.04732) 330.48 62.97kN 67500 2700 5 F1＝ ´ - ´ ＝ ´ (1 0.04732) 330.48 125.94kN 67500 2700 10 F2＝ ´ - ´ ＝ ´ (1 0.04732) 0.04732 330.48 141.58KN 67500 1800 15 F3 ´ = ú û ù ê ë é ´ - + ´ ＝ The moments of the frame according to the actions above is shown in Figure 4.6(a). According to Table 5.7 , top additional seismic action coefficient
Case Study Wuhan University,2012.10
10 413.11 176.98 62.97 413.11 413.11 747.5 413.11 125.94 141.58 511.3 334.4 176.98 511.3 334.4 747.5 422.6 342.8 183.3 183.3 526.1 342.8 422.6 342.8 183.3 342.8 526.1 765.4 Figure 4.6 Moment of the Frame Structure by Base Shear Method Compared with Moment calculated by Mode Analysis Method in EP. 4.6 (a) Base Shear Method (b) Mode Analysis Method Calculation book of a frame under a rolling rock attack Wuhan University,2012.10 Case Study