Seismic Analysis of Structures - III

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Seismic Analysis of Structures - III

  1. 1. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisChapters – 5 & 6Chapter -5RESPONSE SPECTRUMMETHOD OF ANALYSIS
  2. 2. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisIntroductionResponse spectrum method is favoured byearthquake engineering community because of: It provides a technique for performing anequivalent static lateral load analysis. It allows a clear understanding of thecontributions of different modes of vibration. It offers a simplified method for finding thedesign forces for structural members forearthquake. It is also useful for approximate evaluationof seismic reliability of structures.1/1
  3. 3. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… The concept of equivalent lateral forces for earth-quake is a unique concept because it converts adynamic analysis partly to dynamic & partly tostatic analysis for finding maximum stresses. For seismic design, these maximum stresses areof interest, not the time history of stress. Equivalent lateral force for an earthquake isdefined as a set of lateral force which willproduce the same peak response as thatobtained by dynamic analysis of structures . The equivalence is restricted to a single mode ofvibration.1/2
  4. 4. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… A modal analysis of the structure is carried outto obtain mode shapes, frequencies & modalparticipation factors. Using the acceleration response spectrum, anequivalent static load is derived which willprovide the same maximum response as thatobtained in each mode of vibration. Maximum modal responses are combined tofind total maximum response of the structure.1/3 The response spectrum method of analysis isdeveloped using the following steps.
  5. 5. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis The first step is the dynamic analysis while , thesecond step is a static analysis. The first two steps do not have approximations,while the third step has some approximations. As a result, response spectrum analysis iscalled an approximate analysis; but applicationsshow that it provides mostly a good estimate ofpeak responses. Method is developed for single point, singlecomponent excitation for classically dampedlinear systems. However, with additionalapproximations it has been extended for multipoint-multi component excitations & for non-classically damped systems.Contd…1/4
  6. 6. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Equation of motion for MDOF system undersingle point excitation(5.1) Using modal transformation, uncoupled sets ofequations take the form is the mode shape; ωi is the natural frequencyis the more participation factor; is themodal damping ratio.Development of the methodgx+ + = −&& & &&Mx Cx Kx MI22 ; 1 (5.2)i i i i i i i gz z z x i mξ ω ω λ+ + = − =&&&& LLTii Ti iφλφ φ=MIM1/5iφλi ξi
  7. 7. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Response of the system in the ith mode is(5.3) Elastic force on the system in the ith mode(5.4) As the undamped mode shape satisfies(5.5) Eq 5.4 can be written as(5.6) The maximum elastic force developed in the ithmodeContd…i i ix =φ zsi i i if = Kx = Kφ ziφ2i i iKφ = ω Mφ2si i i if =ω Mφ z2simax i i imaxf = Mφ ω z1/6
  8. 8. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Referring to the development of displacementresponse spectrum(5.8) Using , Eqn 5.7 may be written as(5.9) Eq 5.4 can be written as(5.10) is the equivalent static load for the ith modeof vibration. is the static load which produces structuraldisplacements same as the maximum modaldisplacement.Contd…( )max ,ii i d i iz Sλ ω ξ=max ii aSλ= = isi i ef M Pφ1 1max max− −= = ii si ex K f K P2a dS Sω=Pei1/7Pei
  9. 9. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Since both response spectrum & mode shapeproperties are required in obtaining , it is knownas modal response spectrum analysis. It is evident from above that both the dynamic &static analyses are involved in the method ofanalysis as mentioned before. As the contributions of responses from differentmodes constitute the total response, the totalmaximum response is obtained by combining modalquantities. This combination is done in an approximate mannersince actual dynamic analysis is now replaced bypartly dynamic & partly static analysis.Contd…Pei1/8
  10. 10. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Three different types of modal combination rulesare popular ABSSUM SRSS CQCContd…Modal combination rules ABSSUM stands for absolute sum of maximumvalues of responses; If is the response quantityof interestxmax1miix x==∑ (5.11)is the absolute maximum value ofresponse in the ith mode.maxix2/1
  11. 11. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis The combination rule gives an upper bound to thecomputed values of the total response for tworeasons: It assumes that modal peak responses occur atthe same time. It ignores the algebraic sign of the response. Actual time history analysis shows modal peaksoccur at different times as shown in Fig. 5.1;furthertime history of the displacement has peak value atsome other time. Thus, the combination provides a conservativeestimate of response.Contd…2/2
  12. 12. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis0 5 10 15 20 25 30-0.4-0.200.20.4Topfloordisplacement(m)t=6.150 5 10 15 20 25 30-0.4-0.200.20.4Time (sec)Firstgeneralizeddisplacement(m)t=6.1(a) Top storey displacement(b) First generalized displacement2/3Fig 5.1
  13. 13. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…0 5 10 15 20 25 30-0.06-0.04-0.0200.020.040.06Time (sec)Secondgeneralizeddisplacement(m)t=2.5(c) Second generalized displacementFig 5.1 (contd.)2/3
  14. 14. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis SRSS combination rule denotes square root of sumof squares of modal responses For structures with well separated frequencies, itprovides a good estimate of total peak response. When frequencies are not well separated, someerrors are introduced due to the degree ofcorrelation of modal responses which is ignored. The CQC rule called complete quadraticcombination rule takes care of this correlation.Contd…2max1(5.12)miix x== ∑2/4
  15. 15. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis It is used for structures having closely spacedfrequencies: Second term is valid for & includes the effectof degree of correlation. Due to the second term, the peak response may beestimated less than that of SRSS. Various expressions for are available; hereonly two are given :Contd…21 1 1(5.13)m m mi ij i ji i jx x x xρ= = == +∑ ∑∑i j≠2/5iρ j
  16. 16. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis(Rosenblueth & Elordy) (5.14)(Der Kiureghian) (5.15) Both SRSS & CQC rules for combining peak modalresponses are best derived by assumingearthquake as a stochastic process. If the ground motion is assumed as a stationaryrandom process, then generalized coordinate ineach mode is also a random process & thereshould exist a cross correlation betweengeneralized coordinates.Contd…( )( )222 211 4ijijij ijξ βρβ ξ β+=− +( )( ) ( )32 22 228 11 4 1ij ijijij ij ijξ β βρβ ξ β β+=− + +2/6
  17. 17. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Because of this, exists between two modalpeak responses. Both CQC & SRSS rules provide good estimates ofpeak response for wide band earthquakes withduration much greater than the period of structure. Because of the underlying principle of randomvibration in deriving the combination rules, thepeak response would be better termed as meanpeak response.Fig 5.2 shows the variation of with frquencyratio. rapidly decreases as frequency ratioincreases.Contd…2/7iρ jiρ jiρ j
  18. 18. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisFig 5.2Contd… 2/8
  19. 19. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisAs both response spectrum & PSDF representfrequency contents of ground motion, a relationshipexists between the two. This relationship is investigated for the smoothedcurves of the two.Here a relationship proposed by Kiureghian ispresentedContd…02.8( ) 2ln (5.16 b)2pωτωπ =  ÷ 2/9( )( )( )220,2 4(5.16 a)gffxDSpθθ θω ξω ξωωω ω π πτ ω+  = +   +    &&
  20. 20. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…0 10 20 30 40 50 6000.010.020.030.040.05Frequency (rad/sec)PSDFofacceleration(m2sec-3/rad)Unsmoothed PSDF from Eqn 5.16aRaw PSDF from fourier spectrum0 10 20 30 40 50 60 70 80 90 10000.0050.010.0150.020.025Frequency (rad/sec)PSDFsofacceleration(m2sec-3/rad)Eqn.5.16aFourier spectrum of El CentroUnsmoothed5 Point smoothedFig5.32/10Example 5.1 : Compare between PSDFsobtained from the smoothed displacement RSPand FFT of Elcentro record.
  21. 21. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Degree of freedom is sway degree of freedom. Sway d.o.f are obtained using condensationprocedure; during the process, desired responsequantities of interest are determined and storedinan array R for unit force applied at each swayd.o.f. Frequencies & mode shapes are determinedusing M matrix & condensed K matrix. For each mode find (Eq. 5.2) & obtain Pei(Eq. 5.9)Application to 2D framesiλ( )121(5.17)Nrirri NrirrWWφλφ===∑∑2/11
  22. 22. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Obtain ; is the modal peakresponse vector. Use either CQC or SRSS rule to find mean peakresponse.Example 5.2 : Find mean peak values of top dis-placement, base shear and inter storey drift between1st& 2ndfloors.Contd…( 1... )j ejR RP j r= = R j2341 23ω =5.06rad/s; ω =12.56rad/s;ω =18.64rad/s; ω = .5rad/s2/12Solution :[ ] [ ][ ] [ ];;T T1 2T T3 4φ = -1 -0.871 -0.520 -0.278 φ = -1 -0.210 0.911 0.752φ = -1 0.738 -0.090 -0.347 φ = 1 -0.843 0.268 -0.145
  23. 23. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisApproachesDisp (m)Base shear in termsof mass (m)Drift (m)2 modes all modes 2 modes all modes 2 modes all modesSRSS 0.9171 0.917 1006.558 1006.658 0.221 0.221CQC 0.9121 0.905 991.172 991.564 0.214 0.214ABSSUM 0.9621 0.971 1134.546 1152.872 0.228 0.223Time history 0.8921 0.893 980.098 983.332 0.197 0.198Table 5.1Contd…2/13
  24. 24. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Analysis is performed for ground motion applied toeach principal direction separately. Following steps are adopted: Assume the floors as rigid diaphragms & findthe centre of mass of each floor. DYN d.o.f are 2 translations & a rotation; centersof mass may not lie in one vertical (Fig 5.4). Apply unit load to each dyn d.o.f. one at atime & carry out static analysis to findcondensed K matrix & R matrix as for 2D frames. Repeat the same steps as described for 2DframeApplication to 3D tall frames3/1
  25. 25. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis3/2C.G. of mass line1CM2CM3CMLLLLgx&&θx(a)C.G. of mass line1CM2CM3CMLLLL Lgx&&θx(b)Figure 5.4:
  26. 26. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisExample 5.3 : Find mean peak values of top floordisplacements , torque at the first floor &at the base of column A for exercise forproblem 3.21. Use digitized values of the responsespectrum of El centro earthquake ( Appendix 5A ofthe book).Results are obtained following the steps ofsection 5.3.4.Results are shown in Table 5.2.Contd…1 2 34 5 6ω =13.516rad/s; ω =15.138rad/s; ω = 38.731rad/s;ω = 39.633rad/s ; ω = 45.952rad/s; ω =119.187rad/sX YV and V3/3Solution :
  27. 27. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisApproachesdisplacement (m)Torque(rad)Vx(N) Vy(N)(1) (2) (3)SRSS 0.1431 0.0034 0.0020 214547 44081CQC 0.1325 0.0031 0.0019 207332 43376Timehistory0.1216 0.0023 0.0016 198977 41205TABLE 5.2Contd… Results obtained by CQC are closer to those oftime history analysis.3/4
  28. 28. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Response spectrum method is strictly valid forsingle point excitation. For extending the method for multi supportexcitation, some additional assumptions arerequired. Moreover, the extension requires a derivationthrough random vibration analysis. Therefore, it isnot described here; but some features are givenbelow for understanding the extension of themethod to multi support excitation. It is assumed that future earthquake isrepresented by an averaged smooth responsespectrum & a PSDF obtained from an ensembleof time histories.RSA for multi support excitation3/5
  29. 29. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…3/6 Lack of correlation between ground motions attwo points is represented by a coherencefunction. Peak factors in each mode of vibration and thepeak factor for the total response are assumed tobe the same. A relationship like Eqn. 5.16 is establishedbetween and PSDF. Mean peak value of any response quantity rconsists of two parts:dS
  30. 30. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…212 ; 1.. (5.18)si i i i i ki kkz z z u i mξω ω β=+ + = =∑ &&&& &(5.19)i kkii iβ =φφ φTTMRM3/7• Pseudo static response due to thedisplacements of the supports• Dynamic response of the structure withrespect to supports.Using normal mode theory, uncoupleddynamic equation of motion is written as:
  31. 31. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis If the response of the SDOF oscillator tothen Total response is given by are vectors of size m x s (for s=3 &m=2)Contd…( ) ( ) ( )( ) ( )( ) ( ) ( )1 11 1 1(5.21)(5.22)(5.23)s mk k i ik is m sk k i ki kik i kr t a u t z tr t a u t zr tφφ β= == = == += += +∑ ∑∑ ∑ ∑βφT Ta u t z t3/8k kiu is z&&1(5.20)si ki kikz zβ==∑βφ and z
  32. 32. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Assuming to be randomprocesses, PSDF of is given by: Performing integration over the frequency rangeof interest & considering mean peak as peakfactor multiplied by standard deviation,expected peak response may be written as:Contd…{ }ϕ ϕ ϕ ϕ ϕ ϕ  Tβ 1 11 1 21 1 31 2 12 2 22 2 32T11 21 31 12 22 32φ = β β β β β β ( 5.24a)z = z z z z z z ( 5.24b)( ) ( ) ( )tr t ,u t and z( )r t(5.25)rrS = + + +T T T Tuu zz uz zua S a S a S S aβ β β βφ φ φ φ3/9
  33. 33. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…( )[( )....         l l l lL1 2 3 S1 2T T T Tuu uzβD βD zz βD βD zuT1 p 2 p 3 p S pTβD 1 11 11 1 21 21 1 s1 s1 m 11 1mij i j jE max r t = b b +bφ +φ φ + φ b ( 5.26)b = a u a u a u a u ( 5.27a)φ = φ β D φ β D φ β D ...φ β D ( 5.27b)D =Dω ,ξ i=1,..,s ; j=1,..,m ( 5.27c) and are the correlation matriceswhose elements are given by:,uu u zl l z zl( )∫l i j i ji jαu u u uu u -α1= Sω dω ( 5.28)σ σ3/10
  34. 34. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…( )∫ &&l i kj i ki kjα*u z j u uu z -α1= h Sω dω ( 5.29)σ σ( )∫ && &&l ki lj k lki ljα*z z i j u uz z -α1= hh Sω dω ( 5.30)σ σ( )( )&& && && &&i k i k g1 12 2uu u u u2 2coh i,k1S = S S coh i,k = S ( 5.31)ω ω( ) ( )&& && && && &&k l k l g1 12 2u u u u uS = S S coh k,l = coh k,l S ( 5.33)( )( )&& && &&i j i j g1 12 2u u u u u4 4coh i, j1S = S S coh i, j = S ( 5.32)ω ω3/11
  35. 35. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…( )ij i j jD =Dω ,ξ For a single train of seismic wave,that is displacement response spectrum for aspecified ξ ; correlation matrices can beobtainedif is additionally provided; can bedetermined from (Eqn 5.6). If only relative peak displacement is required,thirdterm of Eqn.5.26 is only retained. Steps for developing the program in MATLAB isgiven in the book.coh( i,j )( )j jDω ,ξu gS&&Example 5.4 Example 3.8 is solved for El centroearthquake spectrum with time lag of 5s.3/12
  36. 36. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…Solution :The quantities required for calculating theexpected value are given below:1 211 11 11 21 11 31 12 12 12 22 12 3221 11 21 21 21 31 22 12 22 22 22 321 1 1 1 1 1 11; ; ,0.5 1 0.5 1 1 1 1312.24rad/s ; 24.48rad/s1 1 11;1 1 130.0259 0.0259 0.0259 -TDw wβϕ β ϕ β ϕ β ϕ β ϕ β ϕ βϕ β ϕ β ϕ β ϕ β ϕ β ϕ βϕ     = = =     − −     = =  = =      =βφT Tφ φ ra( )11 21 31 112 22 32 21 21 1 1 22 10.0015 -0.0015 -0.00150.0129 0.0129 0.0129 0.0015 0.0015 0.0015( 12.24) 0.056m( 24.48) 0.011m05 10, 0 ; exp ; exp2 20D D D DD D D Dcoh i jωωρ ρω ωρ ρ ρ ρπ πρ ρ   = = = = == = = = = − −    = = = ÷  ÷       3/13
  37. 37. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis1 0.873 0.7650.873 1 0.8730.765 0.873 10.0382 0.0061 0.0027 0.0443 0.0062 0.00290.0063 0.0387 0.0063 0.0068 0.0447 0.00680.0027 0.0063 0.0387 0.0029 0.0068 0.04471 0.0008 0.0001 0.01420.0008 1 0  =      =   =llluuuzzz0.0007 0.0001.0008 0.0007 0.0142 0.00070.0001 0.0008 1 0.0001 0.0007 0.01420.0142 0.0007 0.0001 1 0.0007 0.00010.0007 0.0142 0.0007 0.0007 1 0.00070.0001 0.0007 0.0142 0.0001 0.0007 1         Contd…3/14
  38. 38. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Mean peak values determined are:Contd…1 21 2( ) 0.106 ; ( ) 0.099( ) 0.045 ; ( ) 0.022tot totrel relu m u mu m u m= == = For perfectly correlated ground motion1 0 00 1 0 null matrix0 0 11 1 1 0 0 01 1 1 0 0 01 1 1 0 0 00 0 0 1 1 10 0 0 1 1 10 0 0 1 1 1  = =       =      l lluu uzzz3/15
  39. 39. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… Mean peak values of relative displacement12RSA RHAu =0.078m ; 0.081mu = 0.039m ; 0.041m It is seen that’s the results of RHA & RSA matchwell. Another example (example 3.10) is solved for a timelag of a 2.5 sec.Solution is obtained in the same way and resultsare given in the book. The calculation stepsare self evident.3/16
  40. 40. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisCascaded analysis Cascaded analysis is popular for seismic analysisof secondary systems (Fig 5.5). RSA cannot be directly used for the total systembecause of degrees of freedom becomeprohibitively large ; entire system becomesnonclasically damped.4/1Secondary Systemxg....kcmxa = xf + xg.. .. ..Secondary system mountedon a floor of a building frameSDOF is to be analyzed forobtaining floor response spectrumxfFig 5.5
  41. 41. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…In the cascaded analysis two systems- primaryand secondary are analyzed separately; output ofthe primary becomes the input for the secondary. In this context, floor response spectrum of theprimary system is a popular concept forcascaded analysis.The absolute acceleration of the floor in the figureis Pseudo acceleration spectrum of an SDOF isobtained for ; this spectrum is used for RSA ofsecondary systems mounted on the floor.ax&&ax&&4/2
  42. 42. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…Example 5.6 For example 3.18, find the mean peakdisplacement of the oscillator for El Centro earthquake.for secondary system = 0.02 ; for the mainsystem = 0.05 ;floor displacement spectrum shown inthe Fig5.6 is usedSolution4/3ξ ξ0 5 10 15 20 25 30 35 4000.511.5Frequency (rad/sec)Displacement(m)Using this spectrum,peak displacement of thesecondary system withT=0.811s is 0.8635m. The time history analysisfor the entire system (withC matrix for P-S system) isfound as 0.9163m.Floor displacement responsespectrum (Exmp. 5.6)
  43. 43. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisApproximate modal RSA For nonclassically damped system, RSA cannotbe directly used. However, an approximate RSA can be performed. C matrix for the entire system can be obtained(using Rayleigh damping for individual systems& then combining them without coupling terms) matrix is obtained considering all d.o.f. &becomes non diagonal. Ignoring off diagonal terms, an approximatemodal damping is derived & is used for RSA.1200CCC =   φTCφ φ4/4
  44. 44. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisSeismic coefficient method Seismic coefficient method uses also a set ofequivalent lateral loads for seismic analysis ofstructures & is recommended in all seismic codesalong with RSA & RHA. For obtaining the equivalent lateral loads, it usessome empirical formulae. The method consists ofthe following steps:• Using total weight of the structure, baseshear is obtained byis a period dependent seismic coefficient(5.34)b hV W C= ×hC4/5
  45. 45. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…• Base shear is distributed as a set of lateralforces along the height asbears a resemblance with that for thefundamental mode.• Static analysis of the structure is carried outwith the force . Different codes provide different recommendationsfor the values /expressions for .( ) (5.35)i b iF V f h= ×(i = 1,2...... n)iF( )if hhC & ( )if h4/6
  46. 46. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis4/7 Distribution of lateral forces can be written asj j j1j j j1j j1j bj j1j jj bj jkj jj b kj jSa1F =ρ×W ×φ× ( 5.36)1j j j1 gF W ×φ= ( 5.37)∑F ΣW ×φW ×φF = V × ( 5.38)ΣW×φW ×hF = V ( 5.39)ΣW×hW ×hF = V ( 5.40)ΣW×h
  47. 47. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis4/8 Computation of base shear is based on first mode.Following basis for the formula can be put forward.( )( )iiiiaeb ib ba eia1bSaiV =ΣF =( ΣW×φ× )×λ ( 5.41)b ji j ji igiSV = W ( 5.42)gV≤ Σ V ( 5.43)S≤ Σ W i =1ton ( 5.44)gSV = ×W ( 5.45)g
  48. 48. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisSeismic code provisions All countries have their own seismic codes. For seismic analysis, codes prescribe all threemethods i.e. RSA ,RHA & seismic coefficientmethod.Codes specify the following important factors forseismic analysis:• Approximate calculation of time period forseismic coefficient method.• plot.• Effect of soil condition onhC Vs Ta& hSAor Cg g5/1
  49. 49. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…• Seismicity of the region by specifying PGA.• Reduction factor for obtaining design forcesto include ductility in the design.• Importance factor for structure. Provisions of a few codes regarding the first threeare given here for comparison. The codes include:• IBC – 2000• NBCC – 1995• EURO CODE – 1995• NZS 4203 – 1992• IS 1893 – 20025/2
  50. 50. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…IBC – 2000• for class B site,• for the same site, is given byhCAg0.4 7.5 0 0.08s1.0 0.08 0.4s (5.47)0.40.4sn nnnnT TATgTT + ≤ ≤= ≤ ≤ >1111.0 0.4s(5.46)0.40.4shTCTT≤= ≥5/3
  51. 51. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… T may be computed by can have any reasonable distribution. Distribution of lateral forces over the heightis given byiF1(5.49)kj ji b Nkj jjW hF VW h==∑21112 (5.48)Ni iiNi iiW uTg Fuπ ==   =   ∑∑5/4
  52. 52. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… Distribution of lateral force for nine story frame isshown in Fig5.8 by seismic coefficient method .( )1 1 1 1k={ 1; 0.5 T +1.5 ; 2 for T≤ 0.5s ; 0.5 ≤ T ≤ 2.5s; T ≥ 2. 5s ( 5.50)0 2 4123456789Storey forceStoreyT=2secT=1secT=0.4secW2WWWWWWWW9@3mFig5.85/5
  53. 53. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… NBCC – 1995• is given by• For U=0.4 ; I=F=1, variations of with Tare given in Fig 5.9.hCeh eC UC = ; C =USIF ( 5.51a);( 5.51b)RAS &g0 0.5 1 1.511.522.533.544.5Time period (sec)SeismicresponsefactorSFig5.95/6
  54. 54. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…• For PGV = 0.4ms-1, is given by• T may be obtained by• S and Vs T are compared in Fig 5.10 forv = 0.4ms-1, I = F = 1; (acceleration andvelocity related zone)    ∑∑1N 22i i11 Ni i1FuT = 2π ( 5.53)g FuAgnnn1.2 0.03≤ T ≤ 0.427sA= ( 5.52)0.512T > 0.427sgTA/gh vz = z5/7
  55. 55. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…0 0.5 1 1.5 2 2.5 3 3.5 400.20.40.60.811.21.4Time period (sec)A/gSSorA/gFig5.105/8
  56. 56. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…• Distribution of lateral forces is given by1t 1 b 1b 10 T≤ 0.7sF = 0.07T V 0.7 < T < 3.6s ( 5.55)0.25V T≥ 3.6s( )∑i ii b t Ni ii=1WhF = V -F ( 5.54)Wh5/9
  57. 57. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…  ÷ 1 c1e -3c1 c1A0≤ T ≤ TgC = ( 5.57)TAT≥ Tg T5/10 EURO CODE 8 – 1995• Base shear coefficient is given by• is given by• Pseudo acceleration in normalized form is givenby Eqn 5.58 in which values of Tb,Tc,TdsCeCesCC = (5.56)q′b c dT T Thard 0.1 0.4 3.0med 0.15 0.6 3.0soft 0.2 0.8 3.0 (A is multiplied by 0.9)are
  58. 58. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…5/11• Pseudo acceleration in normalized form,is given by  ÷ &&0nn bbb n ccc n dgnc dn d2nT1+1.5 0≤ T ≤ TT2.5 T≤ T ≤ TA= ( 5.58)T2.5 T≤ T ≤ TuTT T2.5 T≥ TT
  59. 59. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Rayleighs method may be used for calculating T. Distribution of lateral force is Variation of are shown inFig 5.11.∑∑i i1i b Ni i1i=1i ii b Ni ii=1WφF = V ( 5.59)WφWhF = V ( 5.60)Wh/ & /e go goc u A u&& &&5/12
  60. 60. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…0 0.5 1 1.5 2 2.5 3 3.5 400.511.522.53Time period (sec)A/ug0Ce/ug0Ce/ug0orA/ug0..........Fig 5.115/13
  61. 61. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… NEW ZEALAND CODE ( NZ 4203: 1992)• Seismic coefficient & design response curvesare the same.• For serviceability limit,is a limit factor.( ) ( )( )b 1 s 1b s 1C T = C T ,1 RzL T≥ 0.45 ( 5.61a)= C 0.4,1 RzL T≤ 0.45 ( 5.61b)sL1T6/1
  62. 62. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…• Lateral load is multiplied by 0.92.• Fig5.12 shows the plot of• Distribution of forces is the same as Eq.5.60• Time period may be calculated by usingRayleigh’s method.• Categories 1,2,3 denote soft, medium and hard.• R in Eq 5.61 is risk factor; Z is the zone factor;is the limit state factor.1bc vs T for µ =sl6/2
  63. 63. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…0 0.5 1 1.5 2 2.5 3 3.5 400.20.40.60.811.2Time period (sec)Category 1Category 2Category 3CbFig5.126/3
  64. 64. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd… IS CODE (1893-2002)• are the same; they aregiven by:aeSC vs T & vs Tg• Time period is calculated by empiricalformula and distribution of force is given by:∑2j jj b N2j jj=1WhF = V ( 5.65)Wha1+15T 0≤ T ≤ 0.1sS= 2.5 0.1≤ T ≤ 0.4s for hard soil ( 5.62)g10.4≤ T ≤ 4.0sT6/4
  65. 65. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…aa1+15T 0≤ T ≤ 0.1sS= 2.5 0.1≤ T ≤ 0.55s for medium soil ( 5.63)g1.360.55≤ T ≤ 4.0sT1+15T 0≤ T ≤ 0.1sS= 2.5 0.1≤ T ≤ 0.67s for soft soil ( 5.64)g1.670.67≤ T ≤ 4.0sT6/5For the three types of soil Sa/g are shown in Fig5.13Sesmic zone coefficients decide about the PGAvalues.
  66. 66. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis6/60 0.5 1 1.5 2 2.5 3 3.5 400.511.522.53Time period (sec)Hard SoilMedium SoilSoft SoilSpectralaccelerationcoefficient(Sa/g)Variations of (Sa/g) with time period TFig 5.13Contd…
  67. 67. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…Example 5.7: Seven storey frame shown in Fig 5.14is analyzed withFor mass: 25% for the top three & rest 50% of liveload are considered.1 2 3T = 0.753s ; T = 0.229s ; T = 0.111sR = 3; PGA = 0.4g ; for NBCC, PGA≈ 0.65gSolution: First period of the structure falls in the fallingregion of the response spectrum curve. In this region, spectral ordinates are differentfor different codes.-3 7 -2-1Concrete density = 24kNm ; E = 2.5×10 kNmLive load = 1.4kNm6/7
  68. 68. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis6/8A Seven storey-building frame for analysisFig 5.145m 5m 5m7@3mAll beams:-23cm × 50cmColumns(1,2,3):-55cm × 55cmColumns(4-7):-:-45cm × 45cmContd…
  69. 69. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…Table 5.3: Comparison of results obtained by different codesCodesBase shear (KN)1st Storey Displacement(mm)Top Storey Displacement(mm)SRSS CQC SRSS CQC SRSS CQC3 all 3 all 3 all 3 all 3 all 3 allIBC 33.51 33.66 33.52 33.68 0.74 0.74 0.74 0.74 10.64 10.64 10.64 10.64NBCC 35.46 35.66 35.46 35.68 0.78 0.78 0.78 0.78 11.35 11.35 11.35 11.35NZ420337.18 37.26 37.2 37.29 0.83 0.83 0.83 0.83 12.00 12.00 12.00 12.00Euro 8 48.34 48.41 48.35 48.42 1.09 1.09 1.09 1.09 15.94 15.94 15.94 15.94Indian 44.19 44.28 44.21 44.29 0.99 0.99 0.99 0.99 14.45 14.45 14.45 14.456/9
  70. 70. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd…0 2 4 6 8 10 12 14 161234567Displacement (mm)NumberofstoreyIBCNBCCNZ 4203Euro 8IndianComparison of displacements obtained by different codesFig 5.156/10
  71. 71. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisLec-1/74
  72. 72. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisChapter - 6Inelastic SeismicResponse of Structures
  73. 73. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisIntroduction Under relatively strong earthquakes, structuresundergo inelastic deformation due to currentseismic design philosophy. Therefore, structures should have sufficientductility to deform beyond the yield limit. For understanding the ductility demand imposedby the earthquake, a study of an SDOFsystem in inelastic range is of great help.The inelastic excursion takes place when therestoring force in the spring exceeds or equal tothe yield limit of the spring.1/1
  74. 74. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis For this, nonlinear time history analysis of SDOFsystem under earthquake is required; similarly,nonlinear analysis of MDOF system is useful forunderstanding non-linear behaviour of MDOFsystem under earthquakes. Nonlinear analysis is required for other reasonsas well such as determination of collapse state,seismic risk analysis and so on. Finally, for complete understanding of theinelastic behavior of structures, concepts ofductility and inelastic response spectrum arerequired.The above topics are discussed here.Contd.. 1/2
  75. 75. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisNon linear dynamic analysis If structure have nonlinear terms either in inertiaor in damping or in stiffness or in any form ofcombination of them, then the equation of motionbecomes nonlinear. More common nonlinearities are stiffness anddamping nonlinearities. In stiffness non linearity, two types of nonlinearityare encountered :• Geometric• Material (hysteretic type) Figure 6.1 shows non hysteric type non linearity;loading & unloading path are the same.1/3
  76. 76. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..fLoadingLoadingUnloadingUnloadingxx∆Fig.6.11/4
  77. 77. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd.. Figure 6.2 shows hysteric type nonlinearity;experimental curves are often idealised as(i) elasto plastic; (ii) bilinear hysteretic ;(iii) general strain hardeningyxfxyfyxfxyfyffxyxyfyxfxVariation of force with displacement under cyclic loadingIdealized model of forcedisplacement curveIdealized model of forcedisplacement curveFig.6.21/5
  78. 78. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Equation of motion for non linear analysis takesthe formand matrices are constructed for thecurrent time interval. Equation of motion for SDOF follows as Solution of Eqn. 6.2 is performed in incrementalform; the procedure is then extended for MDOFsystem with additional complexity. and should have instantaneous values.Contd..mΔx + c Δx + k Δx = -mΔx (6.2)gt t&& & &&1/6(6.1)K gt t+ + = −&& & &&M x C x x Mr x∆ ∆ ∆ ∆KtCtCt Kt
  79. 79. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis and are taken as that at the beginning ofthe time step; they should be taken as averagevalues. Since are not known, It requires aniteration. For sufficiently small , iteration may beavoided. NewMark’s in incremental form isused for the solutiontc tkxx &∆∆ &t∆Method−βContd..( )( )& && &&& && &&k22k kΔx = Δt x + δ Δt Δx ( 6.3)ΔtΔx = Δt x + x +β Δt Δx ( 6.4)21/7
  80. 80. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..( )( )  ÷       ÷  ÷    && & &&& & &&&& & &&& & & &&k k2k kt t 2g t k t kk+1 k k+1 k k+1 1 1Δx = Δx - x - x ( 6.5)βΔt 2ββ Δtδ δ δΔx = Δx - x + Δt 1- x ( 6.6)βΔt β 2βkΔx = Δp ( 6.7)δ 1k = k + c + m ( 6.8a)βΔt β Δtmδ m δΔp = -mΔx + + c x + + Δt -1 c x ( 6.8b)βΔt β 2β 2βx = x +Δx x = x +Δx x && &&1 k= x +Δx ( 6.9)1/8
  81. 81. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis81 For more accurate value of acceleration, it iscalculated from Eq. 6.2 at k+1th step. The solution is valid for non hysteretic nonlinearity. For hysteretic type, solution procedure ismodified & is first explained for elasto - plasticsystem. Solution becomes more involved becauseloading and unloading paths are different. As a result, responses are tracked at every timestep of the solution in order to determine loadingand unloading of the system and accordingly,modify the value of kt.Contd.. 1/9
  82. 82. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisElasto-plastic non linearity For material elasto plastic behaviour, is takento be constant. is taken as k or zero depending uponwhether the state is in elastic & plastic state(loading & unloading). State transition is taken care of by iterationprocedure to minimize the unbalanced force;iteration involves the following steps.Elastic to plastic statetctk( ) ( )0(6.10)eex a x∆ = ∆1/10
  83. 83. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysisp e t(Δ x) for( 1- a )Δp withk =0e pΔ x =( Δ x ) +( Δ x )Contd.. Use Eq. 6.7, findPlastic to plastic stateEq. 6.7 with Kt=0 is used ; transition takes place ifat the end of the step; computation is thenrestarted.Plastic to elastic stateTransition is defined byis factored (factor e) such thatis obtained for with&x < 0&x = 0a(Δ x ) ( )1- eΔp tk≠ 0aΔ x =( Δ x ) +Factored Δ x&x = 0x∆1/11
  84. 84. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisExample 6.1 Refer fig. 6.3 ; ; findresponses at t=1.52 s & 1.64s given responses att= 1.5s & 1.62s ; m=1kgSolution:sradn /10=ωContd.. 2/1xxfmcgx..SDOF system with non-linearspring0.15mg0.0147m xxfForce-displacement behaviourof the springFig . 6.3
  85. 85. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis85&&&t = 1.5s; x = 0.01315m; x = 0.1902m/s;2 -1x = 0.46964 m/s ; f = 1.354 ; c =0.4Nsm ;x t-1k =100Nmt&&-1k = 10140NmΔx = - 0.00312ggΔp = 37.55N&& -1Δx = 0.0037m; Δx = - 0.01ms ; Δf =k Δxt( f ) =1.7243N> 0.15mgx t+Δt( f ) + eΔxk = 0.15mg( e = 0.3176)x t t2/2
  86. 86. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..( )& &&&&&&&& X +ΔX = 0t 1-1kΔx = 1-e Δp; k =0; Δx =0.00373m; Δx =-0.00749mst2-1x = x +Δx +Δx = 0.01725m x = 0.1827mst 1t+Δt 2 t+ΔtP -c x -ftk+1 k+1 x( k+1) -1x = = 0.279msk+1 mAt t =1.625s ; x >0 ; k = 10040;Δp =-0.4173; Δx =0 .000042;Δx = -0.061;( )( )&& & & &&&; e =-6.8;Δx = eΔx =-0.000283;1-5kΔx = 1-e Δp; k =100; Δx = Δx +Δx =-4.44×10 ; Δx =-0.061;t 12 2x = x +Δx =0.0298; x = x +Δx= -0.033t tt+Δt t+Δtx from Eqn =3.28; f = f +kΔx = 1.4435Nxt tt+Δt t+Δt 22/3
  87. 87. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisSolution for MDOF System Sections undergoing yielding are predefined andtheir force- deformation behaviour are specifiedas shown in Fig 6.4.0.5kk1.5kk0.5mmmm3yx2yx1yx3pV2pV1pV0.5kk1.5kkxxxFig.6.4 For the solution of Eqn. 6.1, state of the yieldsection is examined at each time step.2/4
  88. 88. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Depending upon the states of yieldsections, stiffness of the members are changed &the stiffness matrix for the incremental equation isformed. If required, iteration is carried out as explained forSDOF. Solution for MDOF is an extension of that of SDOF.Contd..( )      ÷     && & &&t t 2g t k t kKΔx = Δp ( 6.11)δ 1K =K + C + M ( 6.12a)βΔt β ΔtMδ M δΔp =-M r Δx + + C x + +Δt -1 C x ( 6.12b)βΔt β 2β 2β2/5
  89. 89. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisExample 6.2: Refer to Fig 6.5; K/m = 100; m = 1 kg;find responses at 3.54s. given those at 3.52s.Solution:Contd..                        ÷  ÷    &&&&& & &&1.44977 0.15mgf = 0.95664 < 0.15mg and x > 0k0.63432 0.15mg10260 symδ 1K = K + C + M = -124 10260t t 2βΔt β( Δt) 0 -124 10137Δx = 0.5913gMδ M δΔp = -MIΔx + + C x + + Δt -1 C xg t tk kβΔt β 2β 2β          32.6224= 18.02568.43760.0032-1Δx = K Δp = 0.00180.00092/6
  90. 90. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysisk/2k/2mmmk/2k/2k/2 k/23m3m3m1x2x3xyx xyf0.15m gyf =0.01475myx =3 storey frameForce displacementcurve of the columnContd..Fig.6.52/7
  91. 91. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..                                     0.0032 0.16KΔx = -0.0014 ; Δf = Δx = -0.072-0.0009 -0.045e( 0.16)1.60977 0.15mg1f = f +Δf = 0.88664 f + eΔf = f + e ( -0.07) =≤ 0.15mgk+1 k k k 20.58932≤ 0.15mg e ( -0.045)3e =0.136;1                                         e =1; e =12 3eΔx e Δx1 1 1 1Δx = Δx = e Δx + e ( Δx - Δx ) = e Δxe 1 1 1 12 2 2 2eΔx + e ( Δx - Δx )+ e ( Δx - Δx ) e Δx1 1 12 2 3 3 2 3 3e0.000435 0.13581= -0.000965 e = 0.68932-0.001865 3.07e32/8
  92. 92. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis                            ÷      & & && & & &0.0028 0.00324 0.02009Δx = 0.0027 ; Δx = Δx + Δx = 0.0018 ; x = x + Δx = 0.0083312 2 k+1 k0.0026 0.00074 0.01140.-0.0509 0δ δ δΔx = Δx - x + Δt 1- x = -0.0406 ; x = x +Δx =k k k+1 kβΔt β 2β-0.0524( )                                       && &.13610.070.0165e( 0.16) 0.00 0.0218 1.47151Δf = e ( -0.07) + -0.005 = -0.075 ; f = f +Δf = 0.8822 k+1 k-0.005 -0.05 0.584e ( -0.045)3-2.289-1x = M P -C x -F = -1.7tk+1 k+1 k+1 k+1     018-2.2825Contd.. 2/9
  93. 93. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Bidirectional interaction assumes importanceunder:• Analysis for two component earthquake• Torsionally Coupled System For such cases, elements undergo yieldingdepending upon the yield criterion used. When bidirectional interaction of forces onyielding is considered, yielding of a crosssection depends on two forces. None of them individually reaches yield value;but the section may yield.Bidirectional Interaction3/1
  94. 94. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisIf the interaction is ignored, yielding in twodirections takes place independently. In incremental analysis, the interaction effect isincluded in the following way. Refer Fig 6.6; columns translate in X and Ydirections with stiffness and .Contd..eyikexik     ∑ ∑ ∑ ∑i i i iex ex ye ey ey xex y ey xθex ex ey eyθ ex y ey xK 0 K eK = 0 K K e ( 6.13a)K e K e KK = K ; K = K ; K = K e + K e ( 6.13b)3/2
  95. 95. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..xeyeDDColm. 1Colm. 2Colm. 3Colm. 4CRYXC.M.Fig.6.63/3
  96. 96. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisTransient stiffness remaining constant overis given byThe elements of the modification matrix aret∆pKContd..t e pK =K -K ( 6.14)22yi xi yixipxi pyi pxyi pyxii i i2 2i exi xi eyi yixi exi xi yi eyi yiyixixi yi2 2pxi pyiB B BBK = ; K = ; K =K = ( 6.15)G G GG =K h +K h ( 6.16a)B =K h ; B =K h ( 6.16b)VVh = ; h = ( 6.16c)V V3/4tK
  97. 97. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis When any of the column is in the full plastic statesatisfying yield criterion, . During incremental solution changes as theelements pass from E-P, P-P, P-E; the changefollows E-P properties of the element & yieldcriterion. Yield criterion could be of different form; mostpopular yield curve istk =0tkContd..φ    ÷  ÷ ÷  ÷   2 2yixiipxi pyiVV= + ( 6.19)V V3/5
  98. 98. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis For , curve is circular ; ,curve is ellipse; shows plastic state,shows elastic state, is inadmissible. If , internal forces of the elements arepulled back to satisfy yield criterion; equilibriumis disturbed, corrected by iteration. The solution procedure is similar to that forSDOF. At the beginning of time , check the states oftheelements & accordingly the transient stiffnessmatrix is formed.pyipxi VV = pyipxi VV ≠1=iφ 1<iφ1>iφ1>iφContd.. 3/6
  99. 99. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisIf any element violates the yield condition at theend of time or passes from E-P, then aniteration scheme is used.If it is P-P & for any element, then anaverage stiffness predictor- corrector schemeis employed.The scheme consists of : is obtained with for the time internal Δt&incremental restoring force vector is obtained.1>iφContd..1U∆ taK1 111(6.21)(6.22 )(6.22 )tai ii iforce tolerance adisplacement tolerance b++∆ = ∆∆ − ∆ ≤∆ − ∆ ≤F K UF FU U3/7( )1 K = K +Kta tt02
  100. 100. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis After convergence , forces are calculated &yield criterion is checked ; element forces arepulled back if criterion is violated. With new force vector is calculated &iterationis continued. For E-P, extension of SDOF to MDOF is done. For calculating , the procedure as given inSDOF is adopted.taKContd..1(6.23)i iiϕ′ =F FpU∆3/8
  101. 101. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis If one or more elements are unloaded fromplastic to elastic state, then plastic workincrements for the elements are negative When unloaded, stiffness within , is taken aselastic.Example 6.3: Consider the 3D frame in Fig 6.8;assume:t∆Contd..1(6.25)(6.26)pi i pipi i ei iFw UU U K F−∆ = ∆∆ = ∆ − ∆( ) ( ) ( )( ) ( )( )px py p 0 p p oBA Dop o x y o B C o C oC Ax y P AD =3.5m; h = 3.5m; M = M = M = M ; M = M =1.5MkM = 2M ; k = k = k = k ; k = k =1.5k ;k = 2k ; = 50min which m = m = m = 620kg and V =152.053/9
  102. 102. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysisfind Initial stiffness & stiffness at t = 1.38s, giventhat t = 1.36sContd.. 3/102kkyx3.5m1.5k1.5k3.5m3.5mAB CD3 D frameFor column ADisplacement (m)0.00467m152.05 NForce(N)Force-displacementcurve of column A
  103. 103. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis1033/11                                                          & &&& &&& &&U UU 0.00336 0.13675 -0.16679x xxU = 0.00037 U = 0.00345 U = -0.11434y y y0.00003 0.00311 -0.06153θ θ θk k kV 627.27xF = V = 70.ykVθ k    &&V =102.83 V =10.10A x AyV =154.24 V =19.56Bx By888 ; x =-0.08613ggkV =158.66 V =15.15Dx Dy773.51V = 211.54 V = 26.08Cx CySolution:Forces in the columns are pulled back (Eq. 6.23)& displacements at the centre
  104. 104. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..( )φφ φ φ φ         ÷  ÷ ÷  ÷   ∑ ∑ ∑22ex exi 0 ey eyi 0θ 0e2 2yixiipxi pyiA B C Dt et t t2KDK = K = 6k ; K = K = 6k ; K = = 3k 3.54186000 0 54250K = 0 186000 5425054250 54250 1139250VV= +V V=0.462 ; =0.465; =0.491; = 0.488K = Kδ 1K = K + C + M= K +1βΔt β( Δt)              ÷  ÷         & &&&&4g t k t k638.6 sym0 638.65.425 5.425 1379.760000M= ×1016282Mδ M δΔp = -MΔx + + C U + + Δt -1 C U = 286βΔt β 2β 2β6123/12
  105. 105. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis With the e values calculated as above, theforces in the columns are pulled back                       -1tk+1 kAxBxk+1 kDxCx0.0025 476.1ΔU = K Δp = 0.000001 and ΔF = K ΔU= 10.20.000001 181.20.0059U = U +ΔU = 0.00040.0001VV =183.891103.36V = 275.84F = F +ΔF = 81.09 ; ;V = 275.84954.75V = 367.79φ φ φ φ φφ φ φ φ    ÷  ÷ ÷  ÷   =AyByDyCy2 2yixii A B C Dpxi pyiA B D CA B D C= 13.52V = 20.27V = 20.27V = 27.03VV= + & = 1.47 ; = 1.47 ; = 1.47 ; =1.47V V1 1 1 1e = 0.824 e = = 0.824 e = =0.824 e = = 0.824Contd.. 3/13
  106. 106. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis5 3220.254 0.830.00122 10 ; is calculed as 0.83 10 in which .0.00124 0.8316268.57(1 ) 285.76631.480.291670.007950.00;xee xxi ix yixAxBxixipxiUe = etcUeK xe eKhhVhV− −   ∆   ∆ = × ×    ∆        ∆ = − ∆ =    = = ====∑∑;U ep p20.000590.0003953; ;0.000390.00530.000290.00398yAyByiyiyDxD pyiyCxChhVhhh Vhh=======Contd.. 3/14
  107. 107. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis1072 222 2246.56 18.12246.56 18.12; ; ;246.56 18.12246.56 18.121.9721.314;1.3140.98; ; ;2xA yAxB yBxi exi xi yi eyi yixD yDxC xCABi exi xi eyi yiDCyixipxi pyi p p pxyi pyxii iB BB BB K h B K hB BB ByGGG K h K hGGBB DK K K K K KG Gθ= == == == == ==== +=== = = =∑ xi yiiB BG=Contd.. 3/15
  108. 108. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..35211.016.06 185 100.29 53.96 062110000 10 0.16 63.85( )0 0.54 126.60.2917; 0.29170.00260.0001 ;0.00tt t tPyi i Pxi ixp ypPyi Pxipsymsymt tK x K ye eK Kδβ β−= − ×= + + + × −∆ ∆= = = =            ∆ = ∆ = ∆   ∑ ∑∑ ∑2 2KK K C M = K M =U K p U0.002598 0.00009830.002598 0.0001018;0.002602 0.00009830.002602 0.0001018x py         = ∆ =            U3/16
  109. 109. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..2 29.78151.91311.54228.043;11.37227.74510.98303.4711.002; 1.002; 1.0; 1.00ex px pxy pxpipxy ey py pyi iAyAxByBxDyDxCyCxyixiipxi pyiA C DK K K UK K K UVVVVVVVVVVV Vϕϕ ϕ ϕ ϕ− − ∆   ∆ =   − − ∆   ========   = + ÷  ÷ ÷  ÷   = = = =BV3/17Because yield condition is practically satisfied,no further iteration is required.
  110. 110. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisMulti Storey Building frames For 2D frames, inelastic analysis can be donewithout much complexity. Potential sections of yielding are identified &elasto–plastic properties of the sections aregiven. When IMI = Mp for any cross section, a hinge isconsidered for subsequent & stiffness matrixof the structure is generated. If IMI > Mp for any cross section at the end ofIMI is set to Mp, the response is evaluated withaverage of stiffness at t and (IMI = Mp ).t∆t∆tt ∆+4/1
  111. 111. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis At the end of each , velocity is calculated ateach potential hinge; if unloading takes place atthe end of , then for next , the sectionbehaves elastically. ( ).t∆t∆t∆~t small∆Contd..Example 6.4Find the time history of moment at A & the force-displacement plot for the frame shown in Fig 6.9under El centro earthquake; ; compare theresults for elasto plastic & bilinear back bone curves.• Figs. 6.10 & 6.11 are for the result of elasto-plastic case Figs 6.12 & 6.13 are for the resultof bilinear case• Moment in Fig 6.12 does not remain constantover time unlike elasto-plastic case.st 2.0=∆4/2
  112. 112. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis4/3kkkk3m3m3m 1.5kA1.5kmmm1x2x3xk = 23533 kN/mm = 235.33 × 103kgiKdK0.1diKK =0.01471m Displacement (m)346.23kNForce(kN)FrameForce-displacementcurve of columnContd..Fig.6.9
  113. 113. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis4/4-600000-400000-20000002000004000006000000 5 10 15 20 25 30Time (sec)Moment(N-m)-400000-300000-200000-10000001000002000003000004000000 0.005 0.01 0.015 0.02 0.025 0.03 0.035Displacement (m)Force(N)Contd..Fig.6.10Fig.6.11
  114. 114. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis4/5-500000-400000-300000-200000-1000000100000200000300000400000500000-0.005 0 0.005 0.01 0.015 0.02 0.025 0.03Displacement(m)ShearForce(N)-800000-600000-400000-20000002000004000006000008000000 5 10 15 20 25 30Time (sec)Moment(N-m)Contd..Fig.6.13Fig.6.12
  115. 115. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis For nonlinear moment rotation relationship,tangent stiffness matrix for each obtainedby considering slope of the curve at thebeginning of If unloading takes place, initial stiffness isconsidered. Slopes of backbone curve may be interpolated ;interpolation is used for finding initial stiffness. If columns are weaker than the beams, then top& bottom sections of the column becomepotential sections for plastic hinge. During integration of equation of motion isgiven byt∆t∆Contd..4/6tK(6.27)t e p= −K K K
  116. 116. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Non zero elements of Kp are computed usingEqns. 6.15 & 6.16 and are arranged so that theycorrespond to the degrees of freedom affected byplastification. The solution procedure remains the same asdescribed before. If 3D frame is weak beam-strong column system,then problem becomes simple as the beamsundergo only one way bending.The analysis procedure remains the same asthat of 2D frame.Contd.. 4/7
  117. 117. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis For 2D & 3D frames having weak beam strongcolumn systems, rotational d.o.f are condensedout; this involves some extra computationaleffort. The procedure is illustrated with a frame asshown in the figure (with 2 storey).Contd..• Incremental rotations at the member endsare calculated from incrementaldisplacements.• Rotational stiffness of member is modifiedif plastification/ unloading takes place.• The full stiffness matrix is assembled &rotational d.o.f. are condensed out.4/8
  118. 118. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Elasto-plastic nature of the yield section isshown in Fig 6.16. Considering anti-symmetry :Contd.. 4/9pθM1, M2Mp1 = Mp2 = Mp3θMoment-rotation relationshipof elasto-plastic beamfig. 6.16
  119. 119. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis4/10( )( )                       122 21122 22kl klk -k - -2 2Δkl-k 2k 0Δ2K = ( 6.28a)kl kl kl kl θ-α + 0.672 2 2 6 θkl kl kl- 0α +1.332 6 2( )( )( )( )( )( )                            -1Δ Δ Δθ θ θΔ-11-1θ 22-112-11Δ2K = K -K K K ( 6.28b)3α +0.67 16K = ( 6.29a)1 3α +1.33kl3α +0.67 11 -13θ = Δ ( 6.29b)1 3α +1.331 0l3α +0.67 11 -1 -1 -1 1 -13kK = k - ( 6.30)1 3α +1.33-1 2 1 0 1 0l
  120. 120. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd.. Equation of motion for the frame is given by:The solution requires to be computed at timet; this requires to be calculated. Following steps are used for the calculation&& & &&Δt gΔ0MΔx +CΔx +K Δx = -MIΔx ( 6.31)C =αK +βM ( 6.32)4/11tK∆21 & αα& & &x = x +Δx ; x = x + Δx ( 6.33a)i i-1 i-1 i i-1 i-1M =M +ΔM ; M =M + ΔM ( 6.33b)1i 1i-1 1i-1 2i 2i-1 2i-1
  121. 121. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis & are obtained using Eqn. 6.29bin which values are calculated as: & are then obtained; and hence& & are calculated fromand , is obtained using ( Eq. 6.30). If Elasto-plastic state is assumed, thenfor at the beginning of the timeinterval; for unloading are obtainedby (Eq.6.28a.)11 −∆ iθ 12 −∆ iθαContd..1i-1 2i-11 2c c1i-1 2i-11i-1 2i-11i-1 2i-1r l r lα = & α =6EI 6EIM Mr = ; r =θ θ4/1211 −∆ iM 12 −∆ iMiM1 iM2 1α 2α iM1tK∆iM2021 == ααPΜ=Μ=Μ 2121 & αα
  122. 122. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..Example 6.5: For the frame shown in Fig 6.17, findthe stiffness matrix at t = 1.36 s given the responsequantities in Table 6.14/13kkkk3m3m3m5m1θ2θ3θ4θ5θ6θk k1∆2∆3∆E = 2.48 × 107kN/m2Beam 30 × 40 cmColumn 30 × 50 cmFrame13524650KN-mMθY = 0.00109 rad θForce-displacement curveFig. 6.17
  123. 123. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis1234/14JointTimeStepx θ Msec m m/s m/s2rad rad/s rad/s2kNm1 1.36 0.00293 0.0341 -1.2945 0.00109 0.013 -0.452 503 1.36 0.00701 0.0883 -2.8586 0.00095 0.014 -0.297 -23.185 1.36 0.00978 0.1339 -3.4814 0.00053 0.009127 -0.098 42.892 1.36 0.00293 0.0341 -1.2945 0.00109 0.013 -0.452 -504 1.36 0.00701 0.0883 -2.8586 0.00095 0.014 -0.297 23.186 1.36 0.00978 0.1339 -3.4814 0.00053 0.009127 -0.098 -42.89x&x&&θθ&θ&&Table shows that sections 1 & 2 undergo yielding;recognising this, stiffness matrices are given below:x&x &&x &θ &&θTable 6.1Contd..
  124. 124. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis              1231423456Δ1.067Δ-1.067 2.133 symΔ0 -1.067 2.133θ0.8 -0.8 0 2.4Κ = 4.83×10 ×θ 0.8 0 -0.8 0.8 4θ0 0.8 0 0 0.8 3.2θ0.8 -0.8 0 0.4 0 0 2.4θ0.8 0 -0.8 0 0.4 0 0.8 4θ0 0.8 0 0 0 0 0 0.8 3.2     14Δ 230.4451 symΔΚ = 4.83×10 × -0.6177 1.276Δ0.2302 -1.0552 1.811ΔContd..4/15
  125. 125. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Push over analysis is a good nonlinear static(substitute) analysis for the inelastic dynamicanalysis. It provides load Vs deflection curve from rest toultimate failure. Load is representative of equivalent static loadtaken as a mode of the structure & total loadis conveniently the base shear. Deflection may represent any deflection & canbe conveniently taken as the top deflection.Push over analysis 5/1
  126. 126. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis It can be force or displacement control dependingupon whether force or displacement is given anincrement. For both , incremental nonlinear static analysis is‘performed by finding matrix at the beginningof each increment. Displacement controlled pushover analysis ispreferred because, the analysis can be carriedout up to a desired displacement level. Following input data are required in addition tothe fundamental mode shape(if used).tKContd.. 5/2
  127. 127. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Assumed collapse mechanism Moment rotation relationship of yieldingsection. Limiting displacement. Rotational capacity of plastic hinge.Contd.. Displacement controlled pushover analysis iscarried out in following steps: Choose suitable Corresponding to , find1δ∆1∆δ5/3rrφδδ ×∆=∆ 11
  128. 128. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Obtain ; obtain At nth increment, At the end of each increment , moments arechecked at all potential locations of plastichinge. For this, is calculated from condensationrelationship. If , then ordinary hinge is assumedat that section to find K for subsequentincrement.Contd..1∆Κ=∆ δp∑∆= iBn VBV 1 1n iδ∆ = ∆∑1BV∆nθPMM =||5/4
  129. 129. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd.. Rotations at the hinges are calculated at eachstep after they are formed. If rotational capacity is exceeded in a plastichinge, rotational hinge failure precedes themechanism of failure. is traced up to the desired displacementlevel.iB 1iV VsΔExample 6.6 Carry out an equivalent static nonlinear analysisfor the frame shown in Fig 6.19.5/5
  130. 130. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisCrosssectionLocation b (mm) d (mm) (kNm) (rad) (rad)C1 G,1st, 2nd400 400 168.9 9.025E-3 0.0271C2 3rd,4th, 5th& 6th300 300 119.15 0.0133 0.0399B1 G,1st, 2nd400 500 205.22 6.097E-3 0.0183B2 3rd,4th, 5th& 6th300 300 153.88 8.397E-3 0.0252yMyθmaxθContd..3m3m3m3m3m4m 4m3m3myMyθ cθFrameMoment rotationcurve for beamsFig.6.19Table 6.25/6
  131. 131. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..D (m)Base shear(KN) Plastic Hinges at section0.110891 316.825 10.118891 317.866 1,20.134891 319.457 1,2,30.142891 320.006 1,2,3,40.150891 320.555 1,2,3,4,50.174891 322.201 1,2,3,4,5,60.190891 323.299 1,2,3,4,5,6,70.206891 324.397 1,2,3,4,5,6,7,80.310891 331.498 1,2,3,4,5,6,7,8,90.318891 332.035 1,2,3,4,5,6,7,8,9,100.334891 333.11 1,2,3,4,5,6,7,8,9,10,110.350891 334.185 1,2,3,4,5,6,7,8,9,10,11,120.518891 342.546 1,2,3,4,5,6,7,8,9,10,11,12,130.534891 343.207 1,2,3,4,5,6,7,8,9,10,11,12,13,140.622891 346.843 1,2,3,4,5,6,7,8,9,10,11,12,13,14,151.448699 307.822 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,161.456699 308.225 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17Table 6.3 Solution is obtained by SAP2000.5/7
  132. 132. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..0.914310.75480.53450.31200.19880.0833Fig.6.200 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20100200300400Baseshear(kN)Roof displacement (m)Fig.6.215/8
  133. 133. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..2 13 3 34516 175678 812111310 914 1515Fig.6.225/9
  134. 134. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisDuctility & Inelastic spectrum A structure is designed for a load less thanthat obtained from seismic coefficient methodor RSA (say, for The structure will undergo yielding, if it issubjected to the expected design earthquake. The behavior will depend upon the forcedeformation characteristics of the sections. The maximum displacements & deformationsof the structure are expected to be greater thanthe yield displacements.)43(./ −≈RRVB6/1
  135. 135. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd.. How much the structure will deform beyondthe yield limit depends upon its ductility;ductility factor is defined asxmμ = ( 6.35a )xyFor explainingductility , two SDOFsare considered withelasto – plasticbehavior & the other acorrespondingelastic system shownin Fig 6.23.fxyfyx oxofStiffness kElasticElasto-plasticmxFig. 6.236/2
  136. 136. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis means that the strength of the SDOFsystem is halved compared to the elastic system. With the above definitions, equation of motionof SDOF system becomes:2=YRYR yfContd.. An associated factor, called yield reductionfactor, is defined as inverse of :Y YY0 0f xf = = ( 6.35b)f xym my0 y 0 yxx x μ= × =μ f = ( 6.36)x x x R6/3
  137. 137. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis&& yy2yy n 0 yf( x,x)f( x,x)= ; x( t)=μ( t)x ;ffa = =ω x fmContd..&& & & &&&&&& & &2n n y y gg2 2n n y nyx +2ξω x +ω x f( x,x)=- x ( 6.37)xμ+2ξω x +ω x f( μ,μ)=- ω ( 6.38)a depends upon .,, yn fξωµ6/4
  138. 138. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Time history analysis shows the following : For , responses remain within elasticlimit & may be more than that for . For , two counteracting effects takeplace (i) decrease of response due todissipation of energy (ii) increase of responsedue to decreased equivalent stiffness. Less the value of , more is the permanentdeformation at the end .1=Yf1<Yf1<YfYfContd.. is known if for a & can becalculated.µ mx Yf 0x6/5
  139. 139. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Effect of time period on areillustrated in Fig 6.24. For long periods, &independent of ; . In velocity sensitive region, may besmaller or greater than ; not significantlyaffected by ; may be smaller or largerthan . In acceleration sensitive region, ;increases with decreasing ; ;for shorter period, can be very high(strength not very less).Yf, , ,m o Yx x fµgoom xxx ≈≈YfYR=µmxoxµYRContd..om xx >TfY & YR>µµ6/6
  140. 140. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd..01m g yx x f =0m gxx0.125yf =0.25yf =0.5yf =Disp.sensitiveVel.sensitiveAcc.sensitive0.01 0.05 0.1 0.5 1 5 10 50 1000.0010.0050.010.050.10.515100.0010.0050.010.050.10.51510Ta=0.035Tf=15Tb=0.125Tc=0.5Td=3Te=10Spectral Regionsx0/xg0orxm/xg0T (sec)n0.5yf =0.25yf =0.125yf =1yf =Disp.sensitiveVel.sensitiveAcc.sensitive0.01 0.050.1 0.5 1 5 10 50 1000.10.51510Spectral Regionsxm/x0 0.10.51510T (sec)nTa=0.035Tf=15Tb=0.125Tc=0.5Td=3Te=10Normalized peakdeformations for elasto-plasticsystem and elastic systemRatio of thepeak deformationsFig. 6.246/7
  141. 141. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Inelastic response spectrum is plotted for : For a fixed value of , and plots ofagainst are the inelastic spectra or ductilityspectra & they can be plotted in tripartite plot. Yield strength of the E-P System. Yield strength for a specified is difficult toobtain; but reverse is possible by interpolationtechnique.µ ξ YYY AVD ,,nTInelastic response spectra2y y y n y y n yD = x V =ω x A = ω x ( 6.39)y yf = mA ( 6.40)µ6/8
  142. 142. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis For a given set of & , obtain response forE-P system for a number of . Each solution will give a ; , ismaximum displacement of elastic system. From the set of & , find the desired &corresponding . Using value, find for the E-P system. Through iterative process the desiredand are obtained .nT ξYfµ oo xKf = oxf µ µYfContd..Yf µµYf6/9
  143. 143. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisContd.. For different values of , the process isrepeated to obtain the ductility spectrum.nT0 0.5 1 1.5 2 2.5 300.210.40.60.81µ =1.5248(sec)nTfy/w=Ay/gFig. 6.256/10
  144. 144. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis From the ductility spectrum, yield strengthto limit for a given set of & can beobtained. Peak deformation .µ nT ξ2m Y Y nx =μ x =μ A ω1=µf nTµContd..0.01 0.050.1 0.5 1 5 10 50 1000.050.10.51Ta=0.035Tf=15Tb=0.125Tc=0.5Td=3Te=10fy2015102Tn(sec)Ry0.120.1950.378µ=4µ=2µ=1.5µ=0.0 If spectrum foris known ,it is possibleto plot vs. fordifferent values of . The plot is shown inFig. 6.26.Fig. 6.266/11
  145. 145. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Above plot for a number of earthquakes areused to obtain idealized forms of & .f nTContd..1 T < Tn a-1/2f = ( 2μ -1) T < T < T ( 6.41)y n cb-1μ T > Tn c0.01 0.050.1 0.5 1 5 10 50 1000.050.10.51Ta=1/33Tf=33Tb==11//82TcTe=10fyTn (sec)8µ=4µ=2µ=0.21µ=Tc1.5µ=Fig. 6.277/1
  146. 146. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisConstruction of the spectra As , idealized inelastic designspectrum for a particular can be constructedfrom elastic design spectrum. Inelastic spectra of many earthquakes whensmoothed compare well with that obtained asabove. Construction of the spectrum follows the stepsbelow : Divide constant A-ordinates of segmentby to obtain .YfYR 1=µcb − 12 −= µYR cb −7/2
  147. 147. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Similarly, divide V ordinates of segmentsby ; toget ; D ordina-tes of segmentsby to get ;ordinate by toget . Join & ; drawfor; take as the same; join . Draw for)( dc − µ=YR dc −)( ed −µ=YR ed −ffµfeµgoY xD = sTn 33>aa &bagoY xA &&= sTn331<Contd..Natural vibration period Tn (sec) (logscale)goV= &Ta=1/33 sec Tf=33 secTb=1/8 sec Te=10 secaabbccddeeff Elastic designspectrumInelastic designspectrumPseudo-velocityVorVy(logscale)/V µA=xgoA=αAxgoD=xgoD/µD/µD=xgoA/v2µ-1αVαDx.Illustration of theMethodFig. 6.287/3
  148. 148. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisExample 6.7 :Construct inelastic design spectrum fromthe elastic spectrum given in Fig 2.22.The inelastic design spectrum is drawn & shownin Fig 6.28b.2=µContd..Inelastic design spectrum for µ = 2Fig. 6.28b0.01 0.02 0.05 0.1 0.2 0.3 0.50.7 1 2 3 4 567 10 20 30 50 701000.0010.0020.0030.0040.0050.0070.010.020.030.040.050.070.10.20.30.40.50.7123457107/4
  149. 149. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of AnalysisDuctility in multi-storey frames For an SDOF, inelastic spectrum can providedesign yield strength for a given ; maximumdisplacement under earthquake is found as For multi-storey building , it is not possiblebecause It is difficult to obtain design yield strength ofall members for a uniform . Ductility demands imposed by earthquake onmembers widely differ. Some studies on multi - storey framesare summarized here to show how ductilitydemands vary from member to memberwhen designed using elastic spectrum foruniform .µyxµµµ7/5
  150. 150. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Shear frames are designed following seismiccoefficient method ; is obtained usinginelastic spectrum of El centro earthquakefor a specified ductility & storey shearsare distributed as per code. Frames are analysed assuming E-P behaviourof columns for El centro earthquake. The storey stiffness is determined usingseismic coefficient method by assumingstorey drifts to be equal.BYVContd.. 7/6
  151. 151. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis Results show thatFor taller frames, are larger in upper & lowerstories; decrease in middle storeys.Deviation of storey ductility demands from thedesign one increases for taller frames.In general demand is maximum at the firststorey & could be 2-3 times the designStudy shows that increase of base shear bysome percentage tends to keep the demandwithin a stipulated limit.µµµContd..µ7/7
  152. 152. T.K. DattaDepartment Of Civil Engineering, IIT DelhiResponse Spectrum Method Of Analysis

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