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# Seismic Analysis of Structures - II

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### Seismic Analysis of Structures - II

1. 1. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionChapters – 3 & 4Chapter -3RESPONSE ANALYSISFORSPECIFIED GROUND MOTION1
2. 2. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionIntroduction Time history analysis of structuures is carried outwhen input is in the form of specified time history ofground motion. Time history analysis can be performed using directintegration methods or using fourier transformtechniques. In the direct integration methods, there are manyintegration schemes; two most popular methods usedin earthquake engineering will be discussed here. In addition, time history analysis using FFT will bepresented. Before they are described, several concepts usedin dynamic analysis of structures under supportmotion will be summerised ( assuming that they arealready known to the students).1/1
3. 3. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion1/2Models for an SDOFSpring-mass-dashpot systemFig: 3.1 akcxmgx..Idealized single frameFig: 3.1bRigid beamLumped massAll membersareinextensiblegx gx.. ..
4. 4. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Equation of motion of an SDOF system can bewritten in three different ways: For MDOF system, equations of motion are writtenfor two cases; single & multi support excitations.Contd..             && & &&&& & &&&&&&gt t tg ggt tmx + cx + kx = -mx ( 3.1)mx + cx + kx = cx + kx ( 3.3)X = AX +f ( 3.4a)0x 0 1X= A = f = ( 3.4b-xx -k/m -c/mX = AX + Ff ( 3.5a)in which (3.5 )b            &&tgttgx0 0xX = ; F = ; f =xk/m c/mx1/3
5. 5. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion For single support single component excitationFor two component ground motion For three component ground motionContd..{ }   && && &&TTg g1 g21 0 1 0 - - - - - -I = ( 3.9a)0 1 0 1 - - - - - -X = x x ( 3.10a){ }     && && && &&TTg g1 g2 g31 0 0 1 0 0 - - - -I = 0 1 0 0 1 0 - - - - ( 3.9b)0 0 1 0 0 1 - - - -X = x x x ( 3.10b)1/4&& & &&gMX + CX + KX = -MIX ( 3.8)
6. 6. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Example 3.1: Determine for the following structures .Solution :I         1 1 2 3 1 2 3T Tu v u u u u uI = I =1 0 1 1 1 1 1Bracket frame3u2u1uShear building frameu2u3u1v1gx&&gx&&Fig3.4aFig3.4b1/5
7. 7. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion[ ]==010010001001001001TTIIContd..Case 1 Single componentCase 2 Two component11 & vuθ2θ1v2u2v1u13-D model of a shear building frameFig3.4c1/6
8. 8. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionExample 3.2 : Find the mass and stiffness matricesfor the two models of 3D frame shown in Fig 3.5.Solution :Contd..3u2u1uLxyC.M.Lk kgx&&Model-1                      4 -2 2 4 -1 3 0mK = k -2 3 -2 ; M = -1 4 -3 ; I = 062 -2 3 3 -3 6 1Fig3.5a1/7[ ] &&Teff gmP ( model1)= - 3 -3 6 x6
9. 9. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..gx&&θC.M. uC.R.vLk kModel-2For Model -2                    23 0 0.5L 1 0 0 0K = k 0 3 0.5L ; M = m 0 1 0 ; I= 10.5L 0.5L 1.5L L 00 06Fig3.5b1/8[ ] &&Teff gP ( model2)= -m 0 1 0 x
10. 10. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionExample 3.3 : All members are inextensible for thepitched roof portal; column & beam rigidities are K &0.5K obtain mass matrix & force vector.Solution:For Model 1Contd..        2.5 1.67 1M = m I =1.67 2.5 02ulmu1mLLgx&&2m3Lgx&&1u 2um2mmModel-1 Model-2Fig3.6a Fig3.6b1/9
11. 11. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion              &&eff g1.406 -0.156 1 1.25M = m I = P = -m x-0.156 1.406 1 1.25Contd..For Model 2CBm mDA E2m Instantaneous CentreαClB θDA Em2msecθmUnit acceleration givento u1 (model-1)Unit acceleration givento u1 (model-2)Fig3.6c Fig3.6d1/10
12. 12. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionFor multi support excitation, equation of motionContd..)11.3(0=++ggtgggssgssgtgggssgssgtgggssgssPXXKKKKXXCCCCXXMMMM&&&&&&(3.12)0 (3.13)(3.14)(3.15)( ) ( )( )tgt tss sg g ss sg g ss gt t tss ss ss sg g sg g sg gt t tss ss ss sg gss ss ss sg ss g sg ss gsg ss gX X rXM X M X C X C X K Xor M X C X K X M X C X K XM X C X K X K XM X C X K X M M r X C C r XK K r X= ++ + + + =+ + =− − −+ + =−+ + =− + − +− +&& && & &&& & && &&& &&& & && &1(3.16)0 (3.17)(3.18 )0 (3.18 )(3.19)ss s sg gs ss sg g gss sgss ss ss ss gK X K XX K K X rX aK r K bM X C X K X M rX−+ == − =+ =+ + =−&& & &&2/1
13. 13. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Example 3.4 : Find the r matrices for the two framesshown in Fig 3.7 & 3.8.Solution : For the rectangular framek k k1u1u52k2k2kxg12mFig3.72/2   ss3 -3K = k-3 9   sg0 0 0K =-2k -2k -2k     gg2k 0 0K = 0 2k 00 0 2k               -1ss sg1 10 0 0 1 1 11 12 6r = -K K = - =1 1 -2k -2k -2k 1 1 1k 36 6.. ..xg2..xg3u3 u4mu2
14. 14. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.. For the inclined leg portal frameUnit rotation givenat B1B(1)(2)(3) 3LD(5)C(4)BA EL(6) (7)KinematicD.O.F.Fig3.82/3238.4 12 0 6 20 6 012 48 12 0 0 0 03.60 12 38.4 6 12 6 6rr ruEI EIK KL L− −      = =      −   3 324 16 12 12 8 5.53 8 416 181 0 16 5.53 52 5.33 6.3312 0 12 0 8 5.33 8 412 16 0 12 4 6.33 4 3.69uuEI EIKL L− − − −      − −   = −   − − −   − − −   3 3119.56 10.51 4 810.51 129 5.33 22.30.0661 0.0054 4 80.0054 0.0082 5.33 22.310.2926 0.40740.654 0.1389us usgus usgEI EIK KL Lr K K−− −   = =   − −   − − −   = − =    − −    =  − 
15. 15. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Example 3.5 : Obtain the r matrix for d.o.fs 1, 2 & 3 forthe bridge shown in Fig 3.9.Solution: To solve the problem, following values areassumed2/4Simplified model of a cable stayed bridgeFig 3.960 m(3)Cables1l20 m(1)(9)240m20 m(2)(10)3(6)120m(7)450m30m(5)2(8)120m(4)1gxgx gx gxt dEI =1.25EI =1.25EI;10.8480 deck cableAE AEL  =  ÷ ÷   12Cosθ =13513Sinθ =( ) ( ) ( ); ;3 3 31AE 12EI 3EI 120m 12EI= = = 400ml 80120 80 120
16. 16. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion( )( )( )2 211 3i21 31i241 51 3i61 81 91 101222 11; 32 31 71 52162 51 72 41 82 92 10233 313.75EI 2AEk = + cosθ = 1.875m + 800mcos θl80AEk = 0; k = - cosθsinθ;l3 AE 3.75EIk = - cosθ;k = -2 l 80k =k = k = k = 0AEk = k k = -k ; k = - cosθ; k = 02Lk = k ;k = k ;k = k = k = 024EI 2AEk = +L120( )2 243 53 63 73sinθ = 800m 1+ sin θ ;k = k = k = k = 02/5Contd..
17. 17. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion( )( );  ÷ 83 93 103244 42 71 74deck54 64 84 94 10455 65 75 85 95 105366 76 86 96 106377 87 97 107886EIk = - = -24000m;k = 0;k = 24000mLAEk = = 320m;k = k k = -320m480k = k = k = k = k = 03.75EIk = ;k = k = k = k = k = 0803.75EIk = ;k = k = k = k = 080k = 320m;k = k = k = 07EIk =12( )298 88 10899 88 109 88 1010 887 2= ×400m× 120 ;k = k ;k = 00 12 77EI 8 2k = = k ;k = k ;k = k120 7 72/6Contd..
18. 18. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion     -0.781 -0.003 0.002 -0.218r = - -0.218 0.002 -0.003 -0.781-0.147 -0.0009 0.0009 0.1472/7 Using the above stifness coefficients,the condensedstifness matrix corresponding to the translational degeesof freedom is obtained.The first 3X3 sub matrix is the stifness matrix correspondingto the non support translational degrees of freedom.The coupling matrix between the support and non supporttranslational degree of freedom is the upper 3X4 matrix.Using them the above matrices the r matrix is obtained asContd..
19. 19. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Equation of Motion in state space&Z = AZ + f ( 3.20)in which             &&& -1 -1g ss ss ss ss0 0 IxZ = ; f = ; A = ( 3.21)-rx -K M -C Mx& t tZ = AZ + f ( 3.22)in which         &&tt-1tss ss g0xZ = ; f = ( 3.23)-M K xx2/8Example 3.6: Write equations of motion in statespace for example problem 3.4 using both relative &absolute motions.
20. 20. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion For relative motion of the structureContd..2/92 21 2k k k mω =1.9 ; ω =19.1 ; α = 0.105 ; β = 0.017m m m k               ss1 0 3 -3 0.156 -0.051C =α m+β k = km0 2 -3 9 -0.051 0.205      2 22 20 0 1 00 0 0 1A =-3ρ 1.5ρ -0.156ρ 0.026ρ3ρ -4.5ρ 0.051ρ -0.182ρ
21. 21. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion2/10 For absolute motion of the structure( )1 2 32000g g gx x x ρ   =    + + f( )( )       && && &&&& && &&1 2 31 2 3g g gg g g00kρ = ; f = -0.33 x + x + xm-0.33 x + x + xContd..
22. 22. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Both time & frequency domain solutions forSDOF system are presented first and then, they areextended to MDOF system. Two methods are described here: duhamel integration&Newmark’s - - methods.Duhamel integration treats the earthquake force as aseries of impulses of short duration shown in thefigure.In Newmark’s method, the equation of motion is solvedusing a step by step numerical integration technique. For both methods, a recursive relationship is derivedto find responses at K+1 time station given those at Ktime station.βt∆Response analysis2/11
23. 23. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..gx&&τ dτt τ−( ) ( )n-ξω t1 d 2 dx t = e C cos ω t +C sinω t ( 3.24)( ) ( ) ( )  & n-ξω tn 1 2 d d d 1 n 2 dx t =e -ξω C +C ω cos ω t+ -ω C -ξω C sin ω t ( 3.25)Fig3.10Duhamel Integration:Δt = t - t ( 3.34)k+1 k( )   ÷ k+1 kkF -FFτ = F + τ ( 3.35)Δt2/12
24. 24. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Responses at the tk+1 is the sum of following :• Response for initial condition• Response due to Fk between tk and tk+1• Response due to triangular variation of F Response at tk+1 clearly depends upon The constants etc can be evaluated from thethree response analyses mentioned above.Contd..(0) 0 (0) 0x x= =&&& &&& && &x = C x + C x + C F + C F ( 3.36)1 4k+1 k 2 k 3 k k+1x = D x +D x +D F +D F ( 3.37)1 4k+1 k 2 k 3 k k+12x = -x - 2ξω x - ω x ( 3.38)n nk+1 gk+1 k+1 k+1C ,C1 22/13
25. 25. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..k+1 k k+1 ( 3.49)q = Aq +HFin which               &&&i 4i i 42in 4 n 4x Cq = x H= D ( 3.50)x 1-2ξω D -ω Cm{ } { }{ }( )( )( )( )                             1 3 2 3 31 3 2 3 32 23n 1 3 n 2 3n 3n 1 3 n 2 3C +kC C +cC m CA = D +kD D +cD m D ( 3.51)-m C-ω C +kC -ω C +cC-2ξ ω m D-2ξω D +k D -2ξω D +cD2/14 Using expression for these constants, the respons atcan be written in recursive form ast k+1
26. 26. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion    ÷  ÷ ÷       ÷  ÷           ÷ ÷  ÷   ÷ ÷  ÷  ÷      -ξω Δt ξnC =e cosω Δt+ sinω Δt ( 3.39a)1 d d21-ξ-ξω Δt 1nC =e sinω Δt ( 3.39b)2 dωd2-ξω Δt1 2ξ 2ξ 1-2ξ ξnC = +e - 1+ cosω Δt+ - sinω Δt3 d dkω Δt ω Δt ω Δt 2n n 1-ξd         ÷ ÷  ÷  ÷  ÷  ÷        ( 3.40)2-ξω Δt1 2ξ 2ξ 2ξ -1nC = 1- +e cosω Δt+ sinω Δt ( 3.41)4 d dkω Δt ω Δt ω Δtn n dContd..2/15
27. 27. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion    ÷  ÷ ÷       ÷  ÷ ÷            ÷  ÷  ÷  ÷     ÷    ω-ξ ω Δ t nnD =e - sinω Δt ( 3.42)1 d21-ξ-ξ ω Δt ξnD =e cosω Δt- sinω Δt ( 3.43)2 d d21-ξω-ξω Δt1 1 1ξ nnD = - +e cosω Δt+ + sinω Δt (3 d dkΔt Δt 2 21-ξ Δt 1-ξ      ÷   ÷  ÷    3.44)-ξω Δt1ξ nD = 1-e cosω Δt+ sinω Δt ( 3.45)4 d dkΔt 21-ξContd..2/16
28. 28. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Newmark’s - method: With known displacement, velocity & acceleration at kthtime, it calculates the corresponding quantities at k+1thtime; Fk+1 is known. Two relationships are used for this purpose; they meanthat within time interval , the displacement isassumed to vary quadratically. Substituting these relationships in the equation ofmotion & performing algebraic manipulation,following recursive relationship is obtained.βt∆( )( ) ( )  ÷ & & && &&& && &&k+1 k k k+12 2k+1 k k k k+1x = x + 1-δ x Δt + x δΔt ( 3.52)1x = x + xΔt + -β Δt x +β Δt x ( 3.53)23/1
29. 29. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..k+1 N k N k+1 ( 3.66)q = F q +H F( )                 &&&2ii i Nixβ Δt1q = x H =δΔt ( 3.67)mαx 1( ) ( ) ( ) ( ) ( ) ( )( ) ( )        2 2 3 2 22 2n n n22 2N n n n2 2n n n1α -ω α Δt Δt -2ξω β Δt -ω β Δt α Δt -β α +γ Δt21F = -ω δΔt α -2ξω δΔt -ω δ Δt αΔt -δ α+γ Δtα-ω -2ξω -ω Δt -γ( 3.68)in which3/2( )221 2 n nt tα ξω δ ω β= + ∆ + ∆
30. 30. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion With known responses at kth time step, responsesat k+1 th time step are obtained.State space solution in time domain( ) ( )( ) ( )( )              ∫∫&&& &00k+1k+1kgggtA t-t At -As0 gttAtAΔt -Ask+1 k gtZ = AZ + f ( 3.69)0 10 xA = f = Z = ( 3.70)k c-x x- -m mZ t = e Z t + e e f s ds ( 3.71)Z = e Z + e e f s ds ( 3.72)3/3
31. 31. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Eqn (3.73) is preferred since it does not requireinversion. Once Z is known, displacement and velocity areknown.Second order differential equation is solved to find theacceleration .Contd..( )AΔtAΔtZ = e Z +Δte f ( 3.73)k+1 k gkAΔt -1 AΔtZ = e Z + A e -I f ( 3.74)k+1 k gkAtλt -1e =φe φ ( 3.75) The integration in Eqn (3.72) can be performed in twoways.3/4
32. 32. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion It obtains steady state solution of the equation ofmotion & hence, not strictly valid for shortduration excitations like, earthquake. However, under many cases (as mentioned before) agood estimate of rms & peak values of response maybe obtained. For obtaining the response, ground motion is Fouriersynthesized and responses are obtained by the use ofthe pair of Fourier integral (discussed before) .Frequency domain analysis( ) ( )( ) ( )∫∫&& &&&& &&α-iωtg g-ααiωtg g-α1x iω = x t e dt ( 3.76)2πx t = x iω e dω ( 3.77)3/5
33. 33. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion The FFT algorithm now available solves the integralusing their discrete forms. For linear systems, the well known responseequation in complex frequency domain is used: For earthquake excitation, the response due to thejth frequency component of excitation is given byContd..( )( )∑∑&& &&&& &&N-1-i 2πkr Ngk grr=0N-1i 2πkr Ngr gkk=01x = x e ( 3.78)Nx = x e ( 3.79)x( iω) = h( iω) p( iω)( ) ( ) ( )&& jiω tj j g jx t =h iω x iω e ( 3.80)3/6
34. 34. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Total response is given by∑N/2jj=0x( t) = x( t)Contd..in which ( ) ( )  -12 2n j n jjh iω = ω - ω + 2iξω ω The following steps may be used for programmingthe solution procedure.• Sample at an interval of (N).• Input in FFT.• Consider first N/2+1 values of the output• Obtain( )gx t&& t∆( 0... 1)grx r N= −&&( )eNT tπω =+ ∆  ÷ j jNh( iω ) ; ω =0... Δω23/7
35. 35. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..• Obtain .• Add to make input for IFFT.• IFFT of gives .&&j j gjx( iω)= h( iω) x ( iω)*jx ( iω)*jx ( iω) j = 0...N-1 ( )x t Since both frequency & time domain ( Duhamelintegration) can be used for the solution, thereexistsa relation between and . This relationships forms Fourier transforms pair ofintegral.)(ωh )(τh( ) ( )( ) ( )∫∫α-iωt-ααiωt-αh iω = h t e dt ( 3.83)1h t = h iω e dt ( 3.84)2π3/8
36. 36. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Example 3.7:Single bay portal frame in Fig 3.11 issubjected to Elcentro earthquake. Find withSolution:For Duhamel integralndΔt=0.02sω =12.24 rad/sω =12.23rad/s0.0150 0.0312 0.0257 0.00980.5980 0.9696 0.0146 0.00601.5153 3.4804 0.0436 1.0960− −      = − − =      − −   A H               n n0.9854 0.0196 0.0001 0.0001F = -1.4601 0.9589 0.0097 H = 0.0097-146.0108 -4.1124 -0.0265 0.97353/9m m,EI Lu,EI L60ok L3= 12ΕΙ10 rad/seckm=gx&&60o,EILFor New Mark’s method( )x t(0) 0; (0) 0; 0.05x x ξ= = =&Fig 3.11
37. 37. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..For state space solution            AΔt0 1 -0.0161-0.16045i -0.0161+0.16045iA =φ =-37.50 -1.2247 0.9869 0.98690.9926 0.0197e =-0.7390 0.9684 Responses for first few time steps are given in Table3.1 and 3.2.(in the book) Time history of responses are shown in Fig.3.123/10
38. 38. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion0 4 8 12 16 20 24 28 30-0.1-0.0500.050.1Time (sec)Displacement(m)0 4 8 12 16 20 24 28 30-0.1-0.0500.050.1Time (sec)Displacement(m)Newmark’s-β-methodFrequency domain FFT analysisFig3.12Contd..3/11
39. 39. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionResponse of MDOF system Two types of analysis are possible: direct & modalanalysis. For direct analysis damping matrix is generated usingRayleigh damping, For modal analysis, mode shapes & frequencies areused & modal damping is assumed the same for allmodes. Both time & frequency domain analyses using secondorder & state space equations can be carried out.Frequency domain analysis uses FFT algorithm.C =αM + βK4/1
40. 40. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionDirect analysis in time domain Equation of motion takes the form. The same two equations as used in SDOF are used byreplacing x by vector X Substituting those two equations in equation (3.85),the solution is put finally in the recursive form&& & &&MX +CX +KX = -MrX ( 3.85)k+1 k+1 k+1 gk+1( )( ) ( )  ÷ & & && &&& && &&k+1 k k k+12 2k+1 k k k k+1x = x + 1-δ x Δt + x δΔt ( 3.86)1x = x + xΔt + -β Δt x +β Δt x ( 3.87)2( )&&k+1 N k N gk+1Q = F Q +H X 3.924/2Contd..
41. 41. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..{ }{ }{ }                   2 2 2 2 22N2SΔt Δt Δt QΔt SΔt ΔtI- IΔt- Q+SΔt I - +4 4 4 2 4 4SΔt Δt Δt QΔt SΔt ΔtF = - I- Q+SΔt I - + ( 3.93)2 2 2 2 4 2QΔt SΔt-S - Q+SΔt - +2 4         2-1 -1 -1NΔt-T4ΔtS = G K Q = G C T = G Mr H = -T ( 3.94)2-T4/32G M C t K tδ β= + ∆ + ∆
42. 42. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionFrequency domain analysis using FFT Using the steps mentioned before for SDOF X(t)is obtained using FFT & IFFT Note that the method requires inversion of acomplex matrix Eq.(3.98)Contd..{ }( ) ( ){ }( ) ( ) ( )( )..........  &&&& && && && && && &&Tg 1 2 n gTg g 1 11 g1 12 g2 13 g3 2 21 g1 22 g2 23 g3j j gj-12j j jP =- m m m x ( 3.95 )P =-M rX =- m r x +r x +r x ,m r x +r x +r x , ( 3.96 )X iω =H iω P i ω ( 3.97 )H iω = K-M ω +i C ω ( 3.98 )4/4
43. 43. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Example 3.8 : For the portal frame shown in Fig 3.7 ,find the displacements by Newmark’s method:time delay = 5 s; k/m = 100; = 5%; duration = 40s• : Last 10s of record have zero values• : First 5s & last 5s have zero values• : First 10s have zero values1gx&&2gx&&3gx&&mk k k1u12u22k2k2k2mFig3.7ξ4/5Solution:xg2 xg3xg1.... ..u3 u4u5
44. 44. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionEquation of motion can be written as:1 212.25 / ; 24.5 / ; 0.816; 0.0027; 0.02rad s rad s t sω ω α β= = = = ∆ =Contd..1231 1 12 2 21 0 1 0 3 3 3 30 2 0 2 3 9 3 91 0 1 1 10 2 1 1 13gggx x xm m k kx x xxmxxα β −  −            + + +             − −                  =           && && &&&& && &&&&&&&&0.9712 0.0272 0.0193 0.0006 0.0001 0.00.0132 0.9581 0.0003 0.0192 0.0 0.00012.8171 2.7143 0.9281 0.0611 0.0096 0.00031.3572 4.1751 0.0302 0.8973 0.0002 0.0094281.7651 271.4872 7.1831 6.1402 0.0432 0.0343135.7433 417.5n−=−− − −−F091 3.0701 10.2531 0.0171 0.0602        − −  4/6
45. 45. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion[ ]0 0 0Tx =&Contd..0.0 0.0 0.00.0 0.0 0.00.0033 0.0033 0.0033;0.0032 0.0032 0.00320.3301 0.3301 0.33010.3182 0.3182 0.3182n− − −  − − −  − − −=  − − −  − − − − − −  HUsing recursive Eqn. (3.92), relative displacement,velocity and accelerations are obtained.Time histories of displacements, (u1 and u2) areshown in Fig 3.134/7
46. 46. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..0 5 10 15 20 25 30 35 40-0.04-0.0200.020.04Time (sec)Displacement(m)0 5 10 15 20 25 30 35 40-0.02-0.0100.010.02Time (sec)Displacement(m)Displacement u1Displacement u2Fig3.134/8
47. 47. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionExample 3.9: For the pitched roof portal frame shownin Fig 3.8, find displacements 4 & 5 for zero & 5s timedelay between the supports.Contd..1A E1Unit displacement givenat AUnit displacement givenat EFig3.81/ 232 / ; 5%EIrad smLξ = = ÷ 1 25.58 / ; 18.91 / ; 0.4311; 0.004; 0.02rad s rad s tω ω α β= = = = ∆ =4/9Solution:
48. 48. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionFor the first case , duration is 30s and excitation aresame at all supports. For the second case, duration = 35 s and excitationsare different at different supports.Contd..32.50 1.67;1.67 2.5016 10.5010.50 1290.2926 0.40740.654 0.139mEIL =    =    =  − MKr Time histories of displacements are shown for thetwo cases in Fig3.144/10
49. 49. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion4/110 4 8 12 16 20 24 28 32 36-0.1-0.0500.050.1Time (sec)Displacement(m)0 4 8 12 16 20 24 28 32 36-2-1012x 10-3Time (sec)Displacement(m)Without time delay for the d.o.f. 4Without time delay for the d.o.f. 5Contd..
50. 50. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..0 5 10 15 20 25 30 35 40-0.0500.05Time (sec)Displacement(m)0 5 10 15 20 25 30 35 40-4-2024x 10-3Time (sec)Displacement(m)With time delay for the d.o.f. 4With time delay for the d.o.f. 5Fig3.144/12
51. 51. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion The excitation vector fg is of size 2nX1 (single point) The excitation vector fg is of size 2nX1 (multi pointexcitation). The time domain analysis is performed in the sameway as for SDOF system. In frequency domain, state space solution can beperformed as given belowg gp rx= − &&0(3.99)g gx = − &&fr0(3.100)gg =   fp5/1Contd..State Space Direct Analysis
52. 52. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..(3.101)g= +& &z Az fin which1 10; ; (3.102)g g ggxandx− −    = = = = −    − −    0&&&IA f z p rxpKM CM By using FFT of , jthcomponent of isobtained. jth frequency component of response is given by:)( ωifgjgf( ) ( ) ( )( ) \$1(3.103)(3.104)j j gjji i iiω ω ωω ω−= = − z H fH I A By using IFFT, may be obtained (as before)( )z t5/2
53. 53. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Example 3.9: Find the displacement, responsescorresponding to d.o.f 1,2 and 3 using frequencydomain ordinary & state space solutions for the beamshown in Fig 3.153316; 48 ; 0.6 ; 2%s sEIk m C mmLξ= = = =5/3A pipeline supported on soft soil (Exmp. 3.10)(1) (2) (3)2m m (4) 2msk scgx&&L L(5) (6) (7)Wave propagationsk scgx&&sc skgx&&
54. 54. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion0.0 0.0 0.0 1.0 0.0 0.0 00.0 0.0 0.0 0.0 1.0 0.0 00.0 0.0 0.0 0.0 0.0 1.0 0112.0 16.0 16.0 1.622 0.035 0.035 132.0 80.0 32.0 0.070 0.952 0.070 116.0 16.0 112.0 0.035 0.035 1.622 1gm             = =   − − − − −    − − −  − − − − −     A f gx&&Contd.. For solution of second order differential equation,FFT & IFFT are used as before.5/41 2 356 16 8 0.813 0.035 0.01716 80 16 0.035 0.952 0.0358 16 56 0.017 0.035 0.8138.1 / , 9.8 / , 12.2 /0.1761 0.0022m mrad s rad s rad sω ω ωα β− −      = − − = − −      − −   = = == =K CSolution:
55. 55. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion0 5 10 15 20 25 30-0.1-0.0500.050.1Time (sec)Displacement(m)0 5 10 15 20 25 30-0.2-0.100.10.2Time (sec)Displacement(m)Displacement for d. o. f. 1Displacement for d. o. f. 2Contd..Fig3.165/5 Time histories of displacements are shown inFig 3.16 - 3.17
56. 56. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionFig3.1715 20 25 30 35 40 45-0.1-0.0500.050.1Time (sec)Displacement(m)15 20 25 30 35 40 45-0.2-0.100.10.2Time (sec)Displacement(m)Displacement for d. o. f. 1Displacement for d. o. f. 2Contd..5/6
57. 57. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Excitation load vector Pg is of the form Generally is set to zero (in most cases) The solution procedure remains the same. and are of the order of n x s; s is thenumberof supports. Solution requires time histories of and .Solutions can be obtained both in time and frequencydomains.Solution for absolute displacements( ) (3.105)g sg g sg g= − + &P K x C xsgCsgksgCgx gx&5/7
58. 58. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionModal analysis In the modal analysis, equation of motion is decoupledinto a set of N uncoupled equations of motion . Normal mode theory stipulates that the response is aweighted summation of its undamped mode shapes.( )....∑&& & &&&& & &&Ti i i i i i i gT 2 ii i i i i i i iis2i i i i i ik gkk=1Ti kik Ti iX =φz ( 3.106)mz + c z +k z = -φ Mrx ( 3.111)kk =φ Kφ ω = c = 2ξ ωm ( 3.112)mz +2ξω z +ω z = - λ x i =1 m ( 3.113)φ Mrλ = ( 3.114)φ Mφ5/8
59. 59. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion For s=1, equation 3.113 represents the equation forsingle point excitation. Each SDOF system can be solved in time orfrequency domain as describes before.Example 3.11: For the cable stayed bridge shownbefore , find the displacement responses of d.o.f 1,2,3for Elcentro earthquake; 5s time delay is assumedbetween supports.Contd..684 0 149 20 0 00 684 149 0 20 0149 149 575 0 0 60m m−      = =      −   K M5/9Solution:
60. 60. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..[ ][ ][ ]     1 2 3T1T2T3-0.781 -0.003 0.002 -0.218r = - -0.218 0.002 -0.003 -0.781-0.147 -0.0009 0.0009 0.147ω = 2.86rad/s ω = 5.85rad/s ω = 5.97rad/sφ = -0.036 0.036 -0.125φ = 0.158 0.158 0φ = -0.154 0.154 0.030 First modal equation        &&&&&& &&&&&g1Tg22 11 1 1 1 1 Tg31 1g4xxφ Mrz + 2ηω z + ω z = -xφ Mφx5/10
61. 61. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion[ ]121341.474 0.008 0.0061 1.474gggggxxpxx   = −     &&&&&&&&0.025t∆ =• will have first 30s as the actual record & the last15s as zeros.• will have first 10s as zeroes followed by 30s of’actual record & the last 5s as zeros.• Time histories of generalized loads are shown inFig3.18.• Time histories of generalized displacement areshow in Fig 3.191gx&&3gx&&Contd..5/11
62. 62. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion0 5 10 15 20 25 30 35-1-0.500.51Time (sec)Secondgeneralizedforce(g)0 5 10 15 20 25 30 35-1-0.500.51Time (sec)Firstgeneralizedforce(g)First generalized forceSecond generalized force (pg2)Fig3.18Contd..5/12
63. 63. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..0 5 10 15 20 25 30 35 40 45-0.04-0.0200.020.04Time (sec)Displacement(m)0 5 10 15 20 25 30 35 40 45-0.04-0.0200.020.04Time (sec)Displacement(m)First generalized displacement (Z1)Second generalized displacement (Z2)Fig3.195/13
64. 64. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..Solution Z1 Z2 Z3Time historyrms(m)peak(m)rms(m)peak(m)rms(m)peak(m)0.00910.03690.0048 0.0261 0.0044 0.0249Frequencydomain0.0090.03680.0049 0.0265 0.0044 0.02501z2z3z5/14Table 3.4 Comparison of generalized displacement
65. 65. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion One index which is used to determine number ofmodes required is called mass participation factor. Number of modes m to be considered in theanalysis is determined byi r irrimMλ φρ =∑Contd..11miiρ=≈∑5/15 Number of modes depends upon the nature ofexcitation, dynamic characteristics of structure &response quantity of interest.
66. 66. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion The approach provides a good estimate of responsequantity with few number of modes.)(tRMode acceleration approach[ ]( )( ) ( )  ∑ ∑∑&& && &&& && &&& &i i g i i i2 2i im mii g i i i i2 2i=1 i=1i imi i i i2i=1 i1 1z = -λ x - z +2ξω z ( 3.117)ω ωφ 1R( t)= -λ x - z +2ξω z φ ( 3.118)ω ω1=R t - z +2ξω z φ ( 3.119)ω is the quasi static response for whichcan be proved as belowgMrx− &&( ) ( )( ) ( )&&&&gT TgKX t = -Mrx t ( 3.120)φ Kφ z t = -φ Mr x t ( 3.121)6/1
67. 67. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion The solution is obtained by:• Find quasi static response for .• Find :The second quantity is obtained from( )112mi i i iiiz zξω φω=+∑ && &&)(tRContd..( ); ∑ ∑&& &&n ni g i gi i i i2 2i=1 i=1i i-λ x λ xz = R t =φ z = - φ ( 3.122,123)ω ωgMrx− &&( )  ÷ &&&& & i gi i i i2 2i i-λ x1z +2ξω z = - z ( 3.124)ω ω6/2 The contribution of the second part of the solutionfrom higher modes is small; first part contributingmaximum to the response consists of contributionfrom all the modes.
68. 68. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionComputation of internal forces Determination of internal forces may beobtained in two ways:Using the known displacements and member properties.Using the mode shape coefficients of the internal forces. For the latter , any response quantity of interest isobtained as a weighted summation of mode shapes;the mode shape coefficients correspond to thoseof the response quantities of interest.The method is applicable where modal analysis ispossible;number of modes to be considered dependsupon the response quantity of interest.6/3
69. 69. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd..6/4 These coefficients are obtained by solving the MDOFsystem for a static load given below & findingthe response quantity of interest. For the first approach member stiffness matrix ismultiplied by displacement vector in memberco-ordinate system. Rotations which are condensed out are regeneratedfrom the condensation relationships. The second approach is widely used in softwarewhich deals with the solutions of framed structurein general.2; 1 (3.125)i i i i mω= = LφP M
70. 70. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion z(t) is expressed as a weighted summation of modeshapes. Equation of motion can be then written as2n uncoupled equations are then obtained asThe initial condition is obtained fromState space analysis( )Z t =φq ( 3.126)&&g-1 -1 -1gφq = Aφq+ f ( 3.127)φ φq = φ Aφq+ φ f ( 3.128)( )&i i i giq =λ q + f i=1.......2n ( 3.129)-10 0q =φ Z ( 3.130)6/5
71. 71. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion In frequency domain, Eq 3.129 is solved using thesame FFT approach withContd..( ) ( )11 (3.131)jh iω ω λ−= −Example 3.12: For the frame shown in Fig3.20, findthe base shear for the right column; k/m = 100; 5%ξ =k kkk2k 2k2k2kgx&&4u3u2u1um2m2m2mFig3.20Solution:2 2 0 02 4 2 00 2 6 40 0 4 8k−  − − = − − − K1 0 0 00 2 0 00 0 2 00 0 0 2m   =   M1 25.06rad/s; 12.57rad/sω ω= =3 418.65rad/s; 23.85rad/sω ω= =6/6
72. 72. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion[ ][ ][ ][ ]1.0 0.871 0.520 0.2781.0 0.210 0.911 0.7521.0 0.738 0.090 0.3471.0 0.843 0.268 0.145TTTT= − − − −= − −= − − −= − −1234φφφφContd..1.500 1.140 0.0 0.01.140 3.001 1.140 0.00.0 1.140 4.141 2.2800.0 0.0 2.280 5.281mα β−  − − = + = − − − C M K0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.00.0 0.0 0.0 0.0 0.0 1.0 0.0 0.00.0 0.0 0.0 0.0 0.0 0.0 1.0 0.00.0 0.0 0.0 0.0 0.0 0.0 0.0 1.02.0 1.0 0.0 0.0 1.454 0.567 0.0 0.02.0 2.0 1.0 0.0 1.134 1.495 0.567 0.00.0 1.0 3.0 2.0 0.0 0.567 2.062 1.1340.0 0.0 2.=− −− −− −A0 4.0 0.0 0.0 1.134 2.630           − −  6/7
73. 73. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionTime histories of the base shear obtained by modesimulation, mode acceleration and modal state spaceanalyses are shown in Fig 3.21.Contd..1 23 41.793 1.571 ; 1.793 1.571 ;1.157 1.461 ; 1.157 1.461i ii iλ λλ λ= − + = − −= − + = − −0 5 10 15 20 25 30-0.1-0.0500.050.1Time (sec)Shear(intermsofmassm)0 5 10 15 20 25 30-0.1-0.0500.050.1Time (sec)Shear(intermsofmassm)Mode acceleration methodState space method× 102× 1026/8Fig 3.21
74. 74. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Computational steps for MATLAB programmingare given in the book in section 3.5.7 for alltypes of analyses. Steps are given in following sections:Contd.. Computation of basic elements required for alltypes of analysis. Time domain analysis covering direct analysis,modal analysis, mode acceleration approach,state space analysis (modal and direct). Frequency domain analysis covering allcases considered for the time domain analysis6/9
75. 75. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionLec-1/74
76. 76. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionChapter -4FREQUENCY DOMAINSPECTRAL ANALYSIS
77. 77. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionIntroduction Spectral analysis is a popular method for findingseismic response of structures for ground motionsmodeled as random process. Since the analysis is performed in frequency domainit is known as frequency domain spectral analysis. The analysis requires the knowledge of randomvibration analysis which forms a special subject. However, without the rigors of the theory of randomvibration, spectral analysis is developed here. It requires some simple concepts which will beexplained first.1/1
78. 78. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionStationary random processx2 (t)x1 (t)t1t2x3 (t)x4 (t)∞∞∞∞∞1( )x t 1( )x tFig 4.1Mean of is2 21 1[{ ( ) ( )} ]X E x t x tσ = −Sample2 21[ ( ) ( )]ixi i iT x t x t dtσ = −∫Ergodic process 2 2x xiσ σ=1/2
79. 79. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionFourier series & integral Fourier series decomposes any arbitrary functionx(t) into Fourier components.( )  ÷ ∑∫ ∫∫α0k kk=1T T2 2k 0T T- -2 2T2kT-2a 2πkt 2πktx( t )= + a cos +b sin ( 4.1)2 T T2 2πkt 2a = x( t ) cos dt a = x t dt ( 4.2)T T T2 2πktb = x( t ) sin dt ( 4.3)T T1/3  ∑ ∫   T2ok kT2a Δωx( t )= + x( t ) cos(ω t ) dt cos( ω t )+2π k=1 -∞(4.4)ωω ωπ ∆ ∑ ∫   ∞T2k kT2x( t ) sin( t ) dt sin( t )k=1 -
80. 80. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion( ) ( )∫ ∫∫ ∫α αω=0 ω=0α α-α -αx( t )= 2 A(ω )cos( ωt )dω+2 B( ω )sin( ωt )dω ( 4.7a)x( t )= Aω cosωtdω+ B ω sin ωtdω ( 4.7b)Contd. The complex harmonic function is introducedto define the pair of Fourier integral.∫αiωt-α( 4.10)x( t )= x(ω ) e dω, 2 /T T dω π ω→ ∞ ∆ = → . It can be shown that (book)1/4∫α-iωt-α1x(ω )= A( ω )-i B( ω )= x( t ) e dt ( 4.9)2π
81. 81. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. Discrete form of Fourier integral is given by  ÷   ÷ ∑∑2πkrN-1 -iNk rr=02πkrN-1 iNr kk=01x(ω )= x e ( 4.12)Nx( t )= x e ( 4.13) FFT & IFFT are based on DFT. From , Fourier amplitude is obtained.xk Ak( )2 2k k k0 0NA = 2 c + d k =1...... ( 4.14)2A = c ( 4.15)For the Fourier integral to be strictly valid∫α-αx( t ) dt <α ( 4.11)1/5
82. 82. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion In MATLAB, , is divided by N/2 (not N), then Parsavel’s theorem is useful for finding meansquare valueNote equation 4.18 pertains to Eq 4.1 & Eq 4.19pertains to Eqs 4.12 – 4.13.(X(t) is divided by N not N/2 as in MATLAB)rxContd.( )2 2k k k00A = c + d ( 4.16)cA = ( 4.17)2∑∫∑ ∑∫T 22 2 20k k0T N-1 N-122 2r kr=0 K=00a1 1x( t ) dt = + ( a +b ) ( 4.18)T 4 21 1x( t ) dt = x = x ( 4.19)T N1/6
83. 83. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionCorrelation functions Random values taken across the ensembleare different than those would takealthough and are the same. How these two sets of random variables aredifferent is denoted by auto correlation function Obviously, mean square value of theprocess & the two sets are perfectly correlated. Similarly, cross correlation between two randomprocess is given by1x( t )2 1x( t = t +τ )21E [x( t ) ] 22E [x( t ) ]XXR (τ = 0) =[ ] yx xyxy 1 1 ( )R (τ )=E x( t )y( t +τ ) ; R τ =R ( -τ ) ( 4.21)1/7xx 1 1R (τ )= E[x( t )x( t + τ ) ] ( 4.20)
84. 84. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionE [x2] = σ2+ m2xR(τ)2mτo − σ2+ m20τoxyR (τ)σx σy + mx mymx my− σx σy + mx myFig 4.2Fig 4.3Contd.1/8
85. 85. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionPSDFs & correlation functions Correlation functions & power spectral densityfunctions form Fourier transform pairsFrom equation 4.23, It followsωω∫∫∫∫∞-iwτxx xx-∞∞-iwτxy xy-∞∞iwτxx xx-∞∞iwτxy xy-∞1S (ω)= R ( τ) e dτ ( 4.22)2π1S (ω)= R ( τ) e dτ ( 4.23)2πR (τ)= S ( ω) e d ( 4.24)R (τ)= S ( ω) e d ( 4.25)yx xyS (ω) = comp.conjugS ( ω)1/9
86. 86. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion An indirect proof of the relationships may begiven asContd.[ ]  ∫∫∞2 2xx x xx-∞αxy xy-αR ( 0)=r = S (ω) dω =E x ( 4.26a)R ( 0)= S (ω) dω =E xy ( 4.26b) Eq. 4.26a provides a physical meaning of PSDF;distribution of mean square value of the processwith frequency. PSDF forms an ideal input for frequency domainanalysis of structures for two reasons:1/10
87. 87. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion A second order random process is uniquelydefined by its mean square value. Distribution of mean square value with frequencyhelps in ascertaining the contribution of eachfrequency content to the overall response.Contd.S(ω)dω ωω Fig 4.41/11
88. 88. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion1/12ContdFig 4.5(b)Fig 4.5(a)kS2 Tπkω ωS = | x | at kth frequencykω ωkSCompactedordinatesS at closer frequenciesS(ω)ωωkkth ordinate of the power spectral density functionFig 4.5(c)
89. 89. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Concept of PSDF becomes simple if ergodicity isassumed;for a single sample, mean square value For large T , ordinates become more packed; kthordinate is divided by ; sum of areas willresult in variance; smooth curve passing throughpoints is the PSDF. This definition is useful in the development ofspectral analysis technique for single pointexcitation and widely used in the randomvibration analysis of structures.ωdContd.( )∑ ∑∫T 2 N-122 2 2 20x k k kk=00a1 1r = x( t ) dt = + a +b = x ( 4.27a)T 4 21/13
90. 90. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion It is difficult to attach a physical significance tocross PSDF; however some physical significancecan be understood from the problem below.For different values of ǿ, degree of correlationvaries; the responses will be different.Contd.1P1 =A sin ωt P2 =A sin (ωt + φ) For random excitations & ,the responsewill be obviously different depending upon thedegree of correlation & hence cross PSDF.1p( t ) 2p( t )1/14
91. 91. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion PSDF matrix is involved when more than onerandom processes are involvedPSDF matrix( ) ( ) ( )( ) ( )( ) ( ) ( )1 1 2 21 2 2 11 1 2 2 1 211 1 2 2 1 222 2 2 2 21 1 2 2 1 2 1 2 2 1 2 12 21 21 2 2 12 21 2 1 2[ ] (4.28)(4.30)yy x x x xx x x xyy x x x x x xxy a x a x a axE y E a x a x a a x x a a x xS d a S d a S da a S d a a S dS d a S a S a a Sα α αα α αα αα αααω ω ω ω ω ωω ω ω ωω ω ω ω− − −− −− = + =      = + + +   = ++ += + +∫ ∫ ∫∫ ∫∫ ( ) ( ){ }2 11 1 2 2 1 2 2 12 11 1 1 2 12 21 2 1 2 2 1 1 22 1 2 2 2yy(4.31)(4.32 )S (4.32 )x xx x x xyy x x x x x x x xx x x xa a S dS S aS a S a S a a S a a S a a aS S abααω ω ω− +    = + + + =     =∫TxxaS a2/1
92. 92. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion For single variable Eq (4.32b) can be extended to establish twostochastic vectorsContd.2y xS = a S ( 4.33)( ) ( )( ) ( ) ( )1 1 2 2 1 2 2 1n×mn×1 m×1Tyy xxn×m 1 n×r 2m×1 r×1T T T Tyy x x x x x x x xxy xxy t = A x t ( 4.34)S = AS A ( 4.35)y t = A x t +B x t ( 4.36)S = AS A +BS B +AS B +BS A ( 4.37)S = AS ( 4.38) PSDF of the derivatives of the process isrequired in many cases. It can be shown (book):&&& &2x x2 4x x xS =ω S ( 4.48)S =ω S = ω S ( 4.49)2/2
93. 93. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion It is assumed that p(t) is an ergodic process; then infrequency domain From (4.51a), it is possible to write Using Eq(4.19), mean square value may be written asSISOx(ω)= h( ω) p( ω) ( 4.51a)&&2 2 -1n ngh(ω)=( ω - ω + 2iξω ω) ( 4.51c)p(ω)= -x( ω) ( 4.51d)( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )* * *2 2 2xω x ω =h ω h ω p ω p ω ( 4.53a)xω = h ω p ω ( 4.53b)∑ ∑ ∑∫∑∫T N-1 N-1222 2r Kr=0 K=00T2 2201 1x( t ) dt = x = x = x(ω ) ( 4.54)T N1x( t ) dt = h(ω ) p( ω ) ( 4.55)T2/3
94. 94. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionα→TContd.∫ ∫T∞ 22 2x p001x( t ) dt =r = h(ω ) S dω ( 4.56)T∫ ∫ω ω2x p0 02x p*x pS(ω) dω = h( ω) S dω ( 4.57)S(ω)= h( ω) S ( 4.58)S(ω)= h( ω)S h( ω) ( 4.59)2/4
95. 95. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion( ) ( )( ) ( )( )( )∗∗&& &&&&& &&&&&& &&& &&&&&2x xxx x xx x24x x2 2xx x xx x2x x4x xTxx x xx xT2 2xx x xx xxω = iωx ω ( 4.60)S =ω S ( 4.61)S = iωS S = -iωS ( 4.62)xω = -ω x ω ( 4.63)S =ω S ( 4.64)S = -ω S S = -ω S ( 4.65)S =ω S ( 4.66)S =ω S ( 4.67)S = iωS ; S = iωS ( 4.68)S = -ω S ; S = -ω S ( 4.69a)& &xx xxS + S = 0 ( 4.69b)Contd.2/5
96. 96. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionExample 4.1: For the problem in example3.7, findthe rms value of the displacement;Solution: Digitized values of PSDF of Elcentro are given inAppendix 4A(book) can be obtained using aboveequations.0ω = 12.24 rad/s ; Δω = 0.209 rad/sContd.2 2 -1n n2x ph(ω)=( ω - ω +2iξω ω) ( 4.51c)S(ω)= h( ω) S ( 4.58)Xh( w ) & S2/6
97. 97. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. PSDF of Elcentro earthquake, absolute square ofand PSDF of displacements are shown in theFigs 4.8-4.10 .h(ω )0 20 40 60 80 100 120 140 16000.0050.010.0150.02Frequency (rad/sec)PSDFofacceleration(m2sec–3/rad)Fig 4.82/7
98. 98. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.0 4 8 12 16 20 2402468 x 10-4Frequency (rad/sec)|h(w)|20 5 10 15 20 250246Frequency (rad/sec)PSDFofdisplacement(m2sec/rad8 x 10-4Fig 4.10Fig 4.92/8
99. 99. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Single point excitation, P(t) is given by Using Eqns.4.35 & 4.71, following equation can bewritten Using Eqns.4.35 & 4.72( ) ( ) ( )&&gP( t ) = -MIx ( 4.70)xω = H ω P ω ( 4.71)MDOF system( ) ( )*Txx ppS = Hω S H ω ( 4.72)(4.74)&&gT Tpp xS = MII M S2/9(4.75)&&gT T *Txx xS =HM II M H S&& &&g gx xS = - HMIS (4.76)x
100. 100. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionExample 4.2 : For example 3.8, find PSDFs of thedisplacement & for the same excitationsat all supports.Solution:1u 2uContd.               2 -1T1 21 0M = m0 21 0 3 -3C = 0.816 m + 0.0027 k0 2 -3 93 -3K = k-3 9H = [K - Mω + iCω]k= 100; I = {1 1}mω = 12.25rad/s; ω = 24.49rad/s2/10
101. 101. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Using Eqn.4.75, is obtained; PSDFs of &are shown in Fig.4.11XXSContd.1u 2u0 5 10 15 20 2500.20.40.60.80.11.21.4Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)(10-4)0 5 10 15 20 2500.511.522.533.5Frequency (rad/sec)-For u1For u2Fig 4.112/11u1u2σ = 0.0154mσ = 0.0078mPSDFofdisplacement(m2sec/rad)(10-5)
102. 102. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. For multipoint excitation:&& &&&& &&g gg gT T *Tx x xx x xS =HMrS r M H ( 4.77)S = -HMrS ( 4.78)&&gxS is of size s x s; r is of size n x s If all displacements are not required, then a reducedis used of size m x n & is given byH(ω) XXS&&gT T *Txx m×n x n×mS =H MrS r M H (4.79) Without the assumption of ergodicity, Eqns 4.75-4.78 can be derived (1-3).3/1
103. 103. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.Example4.2: (a) For example 4.2, if a time lag of 5s isintroduced between supports, find the PSDFs of u1 &u2 ; b) For example 3.9, find PSDFs of the degrees offreedom 4 & 5 for correlated and partially correlatedexcitations (with time lag 5s).3/2Solution: For example 4.2     ÷ ÷       ÷  ÷       && &&ijs1 2xg 1 1 xg 1 22 11 1 11r =1 1 13rωcoh( i, j)= exp -2πv1ρ ρ5ω 10ωS =ρ 1 ρ S ρ = exp - ρ = exp -2π 2πρ ρ 1PSDFs & cross PSDFs are shown in Fig4.12 (a-d)
104. 104. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.0 5 10 15 20 25012345x 10-5Frequency (rad/sec)0 5 10 15 20 25 3000.20.40.60.811.2x 10-5Frequency (rad/sec)Displacement u1Displacement u2Fig 4.123/3 Rms values of u1 & u2 are 0.0089m (0.0092m) &0.0045m (0.0048m)PSDFofdisplacement(m2sec/rad)PSDFofdisplacement(m2sec/rad)
105. 105. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion3/40 5 10 15 20 25012345x 10-5Frequency (rad/sec)0 5 10 15 20 25 3000.20.40.60.811.2x 10-5Frequency (rad/sec)Displacement u1Displacement u2Contd.PSDFofdisplacement(m2sec/rad)PSDFofdisplacement(m2sec/rad)
106. 106. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion For the problem of example 3.9, required matrices& values of other parameters are given below:3/51 5.58rad/sω = 2 18.91rad/sω =   319.56 10.5EIK =10.5 129L3EI= 2mLα = 0.431β =0.004   0.479 0.331r =-0.131 0.146C =αM+βK   2.5 1.67M= m1.67 2.5 Using Eq 4.77, PSDFs are obtained & areshown in Figs.4.13 & 4.14.
107. 107. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. Rms values for d.o.f 4 & 5 are0.0237( corr.) & 0.0168( partially corr.)0.0005( corr.) & 0.0008( partially corr.)Fig 4.130 2 4 6 8 10 12 14 16 18 2002468x 10-4Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)0 5 10 15 20 25 3000.511.5233.5x 10-4Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)Without time lagWith time lag3/6
108. 108. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.0 2 4 6 8 10 12 14 16 18 2001234x 10-7Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)0 5 10 15 20 25 3000.511.52x 10-7Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)Without time lagWith time lagFig 4.143/7
109. 109. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionPSDFs of absolute displacements Absolute displacement is obtained by addingground displacement to the relative displacement PSDF of is obtained using Eqn. 4.37Example 4.4: For example 4.3, find the PSDFs ofabsolute displacement of d.o.f 4 & 5.Solution: Let & representx =Ix +rx ( 4.80)a gax( ) ( ) ( )&&a g g gg g g gT T Tx xx x xx x x2g g2 Tx x x xx x xS = S +rS r +IS r +rS I ( 4.81)xω = -HMrx ω =HMrω x ω ( 4.82)S =HMrω S and S = S ( 4.83)x &&gx[ ]   && && &&T T4 5 g g1 g2x = x x x = x x3/8
110. 110. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. Using Eqn. 4.83 Using Eqn. 4.81, is obtained.The PSDFs of & are shown in Figs 4.15(a-b). rms values are given as 0.052m & 0.015mrespectively.       &&&& &&g g gg gg1 4 g2 4gg1 5 g2 52 -2x x x x11 21x x ij12 22x x x xx xx x x xS =HMrω S =HMrω S ( 4.84)c cS = S c = coh( i,j) ( 4.85)c cS SS = ( 4.86)S SxaS4x aS 5x aS3/9
111. 111. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.0 2 4 6 8 10 12 14 16 18 2000.20.40.60.811.2x 10-3Frequency (rad/sec)0 2 4 6 8 10 12 14 16 18 2002468x 10-5Frequency (rad/sec)For d. o. f. 4For d. o. f. 5Fig 4.153/10PSDFofdisplacement(m2sec/rad)PSDFofdisplacement(m2sec/rad)
112. 112. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionPSDF of member forces Consider the frame in Fig 3.7; x is the vector ofdyn. d.o.f. & θ as that of the condensed out d.o.f.Consider column i-j; displacement at the ends ofcolumn are PSDF matrix of the member end forces areTθθ xxTxθ xx θx xθθ = Ax ( 4.87)S = AS A ( 4.88)S = AS & S = S ( 4.89)  Ti i j jδ = x θ x θ ( 4.90)f = Kδ ( 4.91)TffδδS = KS K ( 4.92)4/1
113. 113. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. If are in local coordinates, thenExample 4.5 : For the example 3.10, find PSDFs ofdisplacemens1 & 2; also find the PSDF of bendingmoment at the centre.Solution:δT T Tffδδ δδδ = Tδ ( 4.93)S = KS K = KTS T K ( 4.94)               1 2 356 -16 8 0.813 -0.035 0.017K = -16 80 -16 m C = -0.035 0.952 -0.035 m8 -16 56 0.017 -0.035 0.811ω =8.0rad/s; ω =9.8rad/s and ω =12.0rad/s4/2
114. 114. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion Following the same procedure , PSDFs of 1& 2are obtained and are shown in Fig4.16(a-b). For finding PSDF of bending moment, at thecenter is required& are zero as is zero at the center. Displacement vector isContd.θ[ ][ ]T1 2 3x = x x x3θ = -1 0 1 x = Ax4LθθSxθS θδ[ ]T21 1 2δ = x ,θ ,x ,θ4/3
115. 115. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.0 2 4 6 8 10 12 14 16 18 2002468x 10-3Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)Displacement for d. o. f. 1Displacement for d. o. f. 2Fig 4.16 (a-b)4/4
116. 116. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. Using modified stiffness for pinned end,is eliminated ; are abs.displacements.PSDF of bending moment obtained using Eqn.4.92 is shown in Fig 4.16c. rms values ofare 0.0637m, 0.1192m & 1.104m .1θ 1 2x & x1 2x ,x & B.M.4/50 2 4 6 8 10 12 14 16 18 2000.20.40.60.811.21.41.61.8Frequency (rad/sec)PSDFofB.M.xmassm(gm)2sec/rad)Fig 4.16(c)
117. 117. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionModal spectral analysis Advantages of modal analysis have beendescribed before. For multipoint excitation, the ithmodal equationcan be written asr is the influence coefficient matrix of size mxn.hi(w) for each modal equation can be easilyobtained.&& & &&&& & &&2 Ti i i i i i i i i i gT2 ii i i i i i g iiTi i imz +2ξ ωmz +mω z = -j Mrx ( 4.95)-j Mrz +2ξ ω z +ω z = x = p i=1.....r ( 4.96)mm = j Mj4/6
118. 118. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motioniΖ( ω) ip(ω) can be related to byContd.( ) ( ) ( )( )( )&&i i iTi giizω = h ω p ω i =1.....r ( 4.97a)-j Mrxωpω = ( 4.97b)m Elements of PSDF matrix of z are given by&&i j g*i j T T Tz z i x ji jh hS = j MrS r M j i =1....r, j=1....r ( 4.98)mm Using modal transformation ruleTxx zzx=φzS =φS φ ( 4.99)φ is m×r4/7
119. 119. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.Example 4.6: For problem example 3.11,find PSDF ofd.o.f 1(tower top) and d.o.f 2(centre of deck). It isassumed that a uniform time lag of 5s between thesupports exists.684 0 149 20 0 00 684 149 , 0 20 0149 149 575 0 0 60m m−      = =      −   K M0.781 0.003 0.002 0.2180.218 0.002 0.003 0.7810.147 0.009 0.0009 0.147− − −  = − − −  − − r[ ][ ][ ]1 2 32.86rad/s; 5.85rad/s ; 5.97rad/s0.036 0.036 0.1250.158 0.158 00.154 0.154 0.030TTTω ω ω= = == − −== −123φφφSolution4/8
120. 120. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. rms values of displacement for d.o.f 1 & 2are:rms modal directd.o.f1 0.021m 0.021md.o.f2 0.015m 0.015mvalue PSDFs are calculated using equations 4.98-4.99 and shown in Fig 4.17 a-c.4/9It is seen that the values obtained by modal anddirect analyses are the same because the numberof modes taken are equal to the number of d.o.f.(ie all modes are considered).
121. 121. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion0 1 2 3 4 5 6 7 8 9 100123456x 10-4Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)0 5 10 15 20 25 30 35 40 45 50-6-4-2024x 10-5Frequency (rad/sec)Real PartCrossPSDFbetweenu1andu2(m2sec/rad)Fig 4.17b4/10Contd.Fig 4.17a
122. 122. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.4/110 5 10 15 20 25 30 35 40-10123x 10-20Frequency (rad/sec)Imaginary PartCrossPSDFbetweenu1andu2(m2sec/rad)Fig 4.17c
123. 123. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motion4/120 1 2 3 4 5 6 7 8 9 1001x 10-4Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)0 1 2 3 4 5 6 7 8 9 100123456x 10-4Frequency (rad/sec)PSDFofdisplacement(m2sec/rad)Top of the left towerCentre of the deckContd.Fig 4.18
124. 124. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionSpectral analysis (state space) When state space equation is used, is obtainedas:  )g g*Tzz f f-1S =HS H ( 100a)H= Iω- A ( 100b)zzSg gf fS is the PSDF matrix of vector. contains and terms; addition of thesetwo terms becomes zero. For modal spectral analysis, eigen values &eigen vector of matrix A are obtained. Same equations as used before are utilized tofind PSDF of responses.gf& &xx xxS +S = 0&xxS &xxSzzS4/13
125. 125. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd.Example4.7: For the exercise problem 3.12, find thePSDF matrices of top & first floor of displacements;ground motion is perfectly correlated.0.0 0.0 0.0 0.0 1 0.0 0.0 0.00.0 0.0 0.0 0.0 0.0 1 0.0 0.00.0 0.0 0.0 0.0 0.0 0.0 1 0.00.0 0.0 0.0 0.0 0.0 0.0 0.0 12.0 1.0 0.0 0.0 1.454 0.567 0.0 0.02.0 2.0 1.0 0.0 1.134 1.495 0.567 0.00.0 1.0 3.0 2.0 0.0 0.567 2.062 1.1340.0 0.0 2.0 4.0 0.0=− −− −− −−A0.0 1.134 2.630           −   Using eqns 4.100(a-b), PSDFs are calculatedand are shown in fig.4.19.4/14Solution
126. 126. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionContd. rms values are as below:rms modal direct statespaced.o.f1 0.0903m 0.0907m 0.0905md.o.f4 0.0263m 0.0259m 0.0264mvalue Steps for developing a program for spectralanalysis of structures with multi-support excitationsusing MATLAB are given. Steps cover all types of methods ie;modal , directand state space formulations.The program can easily make use of the standardMATLAB routines.4/15
127. 127. T.K. DattaDepartment Of Civil Engineering, IITResponse Analysis for specified ground motionLec-1/74