chopra讲学结构动力学2009 5lecture3_W

1、Estimating Seismic Demands for Performance-Based Engineering of BuildingsHarbin Institute of Technology Harbin, ChinaApril 30-May 2, 2009Anil K. ChopraUniversity of California, BerkeleySeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.1LECTURE 3Estimating Roof D

2、isplacementSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.2Organization SDF system estimation of roof displacment Deformation of inelastic SDF systemsCurrent practiceImproved methodsSeismic Demands forPerformance-Based EngineeringHarbin Institute of Technolog

3、yLecture 3.3MPA: Practical Application Roof displacement required for each pushover analysis (MPA and FEMA) Estimated fromurn = GnfrnDn Deformation Dn of nth-”mode” inelastic SDF system from Inelastic design spectrum Empirical equations for inelastic deformation ratioSeismic Demands forPerformance-B

4、ased EngineeringHarbin Institute of TechnologyLecture 3.4PART ISDF-System Estimate of Roof DisplacementSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.5How Well Can Roof Displacement for Pushover Analysis of Buildings Be Estimated from SDF System Analysis? Bas

5、ic premise in all pushover analyses(ur)SDF = G1fr1D1 Compare(ur )SDFand(ur )MDFdetermined by nonlinear RHASeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.6Response Statisticsr rr u* =(u)SDF (u)MDFCompute median value of u*r SDFBias = Median u*r SDF-1.0Seismic

6、Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.7r )Estimate of Median Displacement, (u*SDFSAC BuildingElastic SystemsInelastic SystemsBoston 9-story0.8290.860Boston 20-story0.7830.721Seattle 9-story0.8210.944*Seattle 20-story0.7410.947L.A. 9-story0.9121.19*L.A. 20-st

7、ory0.8811.19* Some collapsesSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.8Estimate of Median Displacement: Elastic Systems SDF system underestimates median roof displacement by9 to 26% over 6 SAC buildingsSeismic Demands forPerformance-Based EngineeringHarb

8、in Institute of TechnologyLecture 3.9SDF Estimate of Roof Displacement: Elastic SystemsMedian estimate = 91% of exact Underestimated for 17 out of 20 excitations Underestimated by as much as 40% (up to 47%)Seismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.10SAC

9、Buildings: Elastic Systems Roof Displacement Example of dominant first modeSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.11SAC Buildings: Elastic Systems Roof Displacement Example of first mode not dominantSeismic Demands forPerformance-Based EngineeringHarb

10、in Institute of TechnologyLecture 3.12Estimates of Median Displacement: Inelastic Systems SDF-system Estimate of Roof DisplacementUnderestimates by 14-28% for Boston buildingsUnderestimates by 6% for Seattle buildingsOverestimates by 19% for Los Angeles buildingsSeismic Demands forPerformance-Based

11、EngineeringHarbin Institute of TechnologyLecture 3.13SDF Estimate of Roof Displacement: Inelastic SystemsMedian = 1.19 exact Alarmingly small: 0.63 (as low of 0.31) Surprisingly large: 1.65 (as large as 2.15) Worse because of “collapsed” casesSeismic Demands forPerformance-Based EngineeringHarbin In

12、stitute of TechnologyLecture 3.14SDF & MDF System ResponsesCASE 1: Peak response occurs before yielding-induced drift SDF-system is highly accurateSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.15SDF & MDF System ResponsesCASE 2: Permanent drift in SDF respon

13、se is smaller than in exact response SDF-system underestimates by 37%Seismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.16SDF & MDF System ResponsesCASE 3: Permanent drift in SDF response is larger than in exact response SDF-system overestimates by 65%Seismic Dem

14、ands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.17Observations: 6 SAC Buildings Using the SDF-system method, the median roof displacement is:Underestimated by up to 28%Overestimated by up to 19% For individual excitations, the SDF-system estimate can be alarmingly small

15、(31-82% of exact) OR surprisingly large (145-215% of exact) Not included are the few cases where the first- “mode” system collapsed whereas the building did not.Seismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.18SDF-System Estimate of Roof Displacement Inapprop

16、riate for individual ground motions Reasonable for estimating median displacement due to ground motion ensemble or design spectrum Biased estimate; Develop correction factorSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.19PART IIDeformation of Inelastic SDF S

17、ystems Current PracticeSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.20SDFATC-40toCaEpsaticmitayteSpDe:cturrum=MGferDthod1. Plot elastic design spectrum in A-D formatSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.21AT

18、C-40 Capacity Spectrum Method2. Plot the demand diagram and capacity diagram together. Intersection point gives deformation demand.Seismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.22ATC-40 Capacity Spectrum MethodAnalysis of equivalent linear systemsIterative p

19、rocedure does not always convergeEven if it converges, deformation may be inaccurateSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.23Evaluation of ATC-40 ProcedureSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.24Discre

20、pancy in ATC-40 ProcedureSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.25FEMA 440 Method to Estimate D Has rectified both flaws of the ATC-40 methodLack of convergence in some casesLarge errors in many cases Improved procedures for equivalent linearization o

21、f inelastic systemsSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.26Benefit of Equivalent Linearization Detour?Deformation of inelastic systems available fromInelastic design spectrumEquations for inelastic deformation ratioSeismic Demands forPerformance-Base

22、d EngineeringHarbin Institute of TechnologyLecture 3.27Inelastic Design SpectraSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.28D from Inelastic Design Spectrum Classical= 1 Tn 2Dm Ry 2p A Graphical: Capacity-Demand-DiagramMethodSeismic Demands forPerformance

23、-Based EngineeringHarbin Institute of TechnologyLecture 3.29Capacity-Demand-Diagram Method Attractive graphical feature of FEMA 440 Equivalent Linear System or ATC-40 Capacity Spectrum Method can be retained without the equivalent linearization detour. Achieved in the CDD Method by using the inelast

24、ic design spectrum to define demand.Seismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.30Inelastic Demand Diagram Inelastic design spectrum plotted in A-D format Deformation from spectrum: Plot Ay vs D for constant mD = mDy = m (Tn 2p )2 AySeismic Demands forPerf

25、ormance-Based EngineeringHarbin Institute of TechnologyLecture 3.31Capacity-Demand-Diagram Method Plot capacity and demand diagrams in A-D format Yielding branch of capacity diagram intersects the demand diagram for several m The deformation is given by the one intersection point where m from the tw

26、o diagrams matchesSeismic Demands for Performance-Based EngineeringHarbin Institute of TechnologyLecture 3.32Capacity-Demand-Diagram MethodSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.33Deformation of Inelastic SDF Systems Capacity-demand-diagram method usi

27、ng the inelastic design spectrum has been developed Graphical procedure gives same deformation as the classical method using R-m-Tn relations Capacity-demand-diagram method retains graphical feature of ATC-40 methodConstant m-demand diagramElastic demand diagram in ATC-40 Demand diagram used is diff

28、erentSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.34PART IIIDeformation of Inelastic SDF SystemsImproved MethodsSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.35Improved Methods to Estimate D FEMA-440 Equivalent Line

29、arization Method FEMA-440 Improvements for C1 and C2 Equations for inelastic deformation ratio considering all hysteresis featuresSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.36Bilinear Systems: Definitionsumuouyu00kakmfyfofsf Post-yield stiffness: a k Yiel

30、d-strength reduction factor:Ry =fof yum Ductility factor:m = uySeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.37Inelastic Deformation Ratio Systems with known ductility m:Cm = umuo Systems with known strength:CR= umuoSeismic Demands forPerformance-Based Engin

31、eeringHarbin Institute of TechnologyLecture 3.38Limiting Values of um/uo Tn = :Equal Deformation RuleCm = CR = 1 Tn = 0 :Equal Force Rule.For a = 0Lm =mLm = m1+ (m -1)aL= 11 +Ry -1L= RRy aR Limiting values apply to all excitations and systems Seismic Demands for Performance-Based EngineeringHarbin I

32、nstitute of TechnologyLecture 3.39Ground Motion Ensembles Nine ensembles, 170 ground motions LMSR, LMLR, SMSR, SMLR (4x20)Four combinations of large (M=6.6-6.9) or small (M=5.8-6.5) magnitude and small (R=13-30 km) or large (R=30-60 km) distance NEHRP Site Classes B, C, or D (3x20)M = 6.0-7.4R 120 k

33、m Near-Fault (2x15)Fault normal components M = 6.2-6.9 R 9 kmFault parallel componentsSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.40Influence of Earthquake Magnitude and Distance100110Period, Tn (sec)0.10.5 0.01LMSR LMLR SMSR SMLR503020105321yR =4 Median C

34、R-Tn plots, similar for LMSR, LMLR, SMSR, andMedian CRSMLR ensemblesCm -Tnsimilarplots are alsoSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.41Influence of NEHRP Site Classn100110Period, T(sec)0.10.5 0.01LMSR BC D503020105321yR =4Median CR Median Cm-Tnand CR

35、-Tn plots similar for site classes B, C, and D. Close to LMSR, although spectra differSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.4220.48.02.440.430.160.034fTeTdTcTbTaaTf0.1eb1dcS p e c t r a lR e g i o n s AccelerationVelocityDisplacementsensitivesensitiv

36、esensitive10Pseudovelocity, V/goMedian Response Spectrum: LMSR Ensemble0.1110Natural vibration period, Tn (sec)Seismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.43Median Response SpectrumPeriod, Tn (sec)32.521.510.5000.511.5LMSR NFFN NFFP2.52Pseudoacceleration,

37、A/go Spectral shapes of near-fault motions are much different than far-fault motions Tc = 0.85,0.61sec(0.43 forLMSR)Seismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.44ncT /T1001010.10.01110LMSR NFFN NFFP100nPeriod, T (sec)1001010.10.01110LMSR NFFN NFFP100CR for

38、 Near-Fault Ground MotionsMedian CR In acceleration- sensitive region, CR-Tnplots differ for near- and far- fault motionsMedian CR Similar when CR is plotted against Tn/ TcSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.45Estimating Deformation of Inelastic Sy

39、stems Deformation of elastic system from elastic design spectrum Need equation for Cmfor structures with known m Need equation for CR for structures with known fy or RySeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.46Empirical Equation for CR a Td-1CR = 1+ (L

40、-1)-1 + c n R Rb Tyc Numerical coefficients: a, b, c, d Function of Ry and TnTc Satisfies limits LR at Tn = 0 and 1 at Tn = L= 11 +Ry -1for all systems & excitations RRy a Seismic Demands for Performance-Based EngineeringHarbin Institute of TechnologyLecture 3.47Empirical Equation for CRncT /TncT /T

41、1001010.10.011001010.10.01111010LMSR NFFNProposed CR100100y(b) a=10%, R =6y(a) a=0%, R =6CR Same equation and coefficients for near- and far-fault motionsSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.48Empirical Equation for CRncT /TncT /T1001010.10.01100101

42、0.10.0111LMSR LMLR SMSR SMLRProposed CR1010100100y(b) a=10%, R =6y(a) a=0%, R =6CR Same equation and coefficients, independent of earthquake magnitude and distanceSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.49Empirical Equation for CRncT /TncT /T1001010.10

43、.011001010.10.01111010B C DProposed CR100100y(b) a=10%, R =6y(a) a=0%, R =6CR Same equation and coefficients for site classes B, C, and DSeismic Demands forPerformance-Based EngineeringHarbin Institute of TechnologyLecture 3.50Inelastic Deformation Ratio Empirical equations for Cmand CR available for bilinear systems Func