TOPIC 6 Structural Dynamics III Analysis of Elastic …

1 TOPIC 6 Structural Dynamics IIIA nalysis of Elastic MDOF SystemsTOPIC 6 Structural Dynamics IIIA nalysis of Elastic MDOF Systems Equations of Motion for MDOF Systems Uncoupling of Equations through use of Natural Mode Shapes Solution of Uncoupled Equations Recombination of Computed Response Modal Response Spectrum Analysis (By Example) Use of Reduced Number of ModesuxuyrzMajority of Massis in FloorsTypical Nodal DOFM otion isPredominantlyLateralPlanar Frame with 36 Degrees of Freedom12345678910111213141516 Planar Frame with 36 Static Degrees of Freedombut with only THREE Dynamic DOFu1u2u3f11f21f31V1=1 kipfff112131 Development of Flexibility MatrixDevelopment of Flexibility Matrixf12f22f32V2=1 kip fff122232 Development of Flexibility Matrixf13f23f33V3=1 kip fff132333 UfffVfffVfffV= + + 112131112223321323333 UfffffffffVVV= 111213212223313233123

2 FVU=KUV=KF= 1 Concept of Linear Combination of Shapes (Flexibility)m1m3m2k1k2k3V1(t), u1(t)V2(t), u2(t)V3(t), u3(t)Kkkkkkkkkk= + + 11112222300 Mmmm= 123000000 Utututut()()()()= 123 VtVtVtVt()()()()= 123 Idealized Structural Property MatricesNote: Damping to be shown laterKKKKrrFnnnmmnmmnmn,,,,LNMOQPLNMOQP= LNMOQPKr K r Fnn nnm mn,,+=0Kr K rmn nmm m,,+=0 Static CondensationMassless DOFDOF with mass21 Condensed Stiffness MatrixStatic Condensation(continued)rKKrKr K K K r FKKKKrFKrFKK KK Kmmmmnnnnnnmmmmnnnnnnm mm mnnnnnnnnm mm mn= = === ,,,,,,,,,,,,,, 1111ns21 RearrangePlug intoSimplifymmmutututkkkkkkkkkutututVtVt Vt12312311112222312312300000000 + + + = () () ()()()()()()()MU tKU tV t ()()()

3 +=mu tk u tk u tV tmu tku tku tk u tk u tV tmu tk u tk u tk u tV t111112122111222232332223333 ()()()() ()()()()()() ()()()()()+ = ++ = ++=Coupled Equations of Motion for Undamped Forced VibrationDOF 1 DOF 2 DOF 3 MutKut ()()+=0 Assume utt()sin= KM =20 Thenhas three (n) solutions: 11121311= , 21222322= , 31323333= ,Natural Mode ShapeNatural FrequencySolutions for System in Undamped Free Vibration ()sinutt= 2KM =2 For a SINGLE ModeKM =2 For ALL Modes[] = 12 3 Where: 2122232= Solutions for System in Undamped Free VibrationNote: Mode shape has arbitrary scale.

4 Usually TMI= 110i=.orMODE 1 MODE 3 MODE 2 11 21 31 32 22 12 33 23 13 Mode Shapes for Idealized 3-Story FrameUYYY= + + 112131112223321323333 UYYY= 111213212223313233123UY= KUV=Concept of Linear Combination of Mode Shapes(Change of Coordinates)KY V = TTKYV=Mode ShapeModal Coordinate[] = 12 3 TKkkk= 123** TMmmm= 123** TCccc= 123**Generalized MassGeneralized DampingGeneralized StiffnessOrthogonality Conditions TVtVtVtVt()()()()**= 123 Generalized ForceMuCuKuVt ()++=yu =My Cy Ky Vt ()++= TTTTMyCyKyVt ()

5 ++=mmmyyycccyyykkkyyyVtVtVt1231231231231 23123123** ()()() + + = MDOF Equation of Motion:Transformation of Coordinates:Substitution:Premultiply by T:Using Orthogonality Conditions: Uncoupled Equations of Motion are:Development of Uncoupled Equations of Motion ug ,ur1 + = +++=)()()()( )()()()()()()(3,2,1,3,2,1,tututuMtuMtutu tutututuMtFRRRgRgRgRgI Move to RHS as)()(tuMRtVgEFF =Earthquake Loading for MDOF System)()(*tuMRtVgT = 111=Rm1=2m2=3m3=1u1u2u3M=+ 2131 011=Ru1u2u3m1=2m2=3m3=1M= 231 Modal Earthquake Loading ()utg ()utgmy cy k y V t1111111** * * ()++ =my cy k y V t3333333** * * ()++ =my cyk yV t2222222** * * ()++ = ()/**yyyVtm1111121112++= ()/**yyyVtm2222222222++= ()

6 /**yyyVtm3333323332++= Simplify by Dividing Through by m* and defining iiiicm=**2 Development of Uncoupled Equations of Motion(Explicit Form)MODE 1 MODE 2 MODE 3 MODE 1 MODE 2 MODE 3 Definition of Modal Participation FactorTypical Modal Equation: ()**ywywyVtmiiiiiiii++=22 recallVtMrvtiiTg*() ()= Right hand side = iTigMrmvt* ()Modal Participation Factor PiConcept of Effective Modal MassFor each Mode iMPmiii=2*In Previous Development, We have Assumed: TCccc= 123** Rayleigh Proportional Damping Wilson Discrete Modal Damping Development of a Modal Damping Matrix Two Methods Described Herein:MASSPROPORTIONALDAMPERSTIFFNESSPR OPORTIONALDAMPERCMK=+ Rayleigh Proportional DampingCMK=+ For Modal Equations to Be Uncoupled:2 nnnTnC=Using Orthogonality Conditions.

7 22 nnn=+ nnn=+122 Rayleigh Proportional Damping TMI=AssumesMode 1 FrequenciesRayleigh Proportional Damping (Example) , Damping RatioMASSSTIFFNESSTOTALTYPE = = Critical in Modes 1 and 3 Modes 1 & 2 .36892 & 3 .41487 & 4 .43871 & 5 .45174 Factors (5% each indicated mode)Rayleigh Proportional Damping (Example)5% Damping in Modes 1 & 2, 1 & 3, 1 & 4, or 1 & , Radians/secModal Damping Ratio1,21,31,41,5 MODESMASSPROPORTIONALDAMPERSTIFFNESSPROP ORTIONALDAMPERCMK=+ Rayleigh Proportional DampingWilson DampingDirectly Specify Modal Damping Values i* TCcccmmm= = 123111222333222** TnnnnCc= LNMMMMMMOQPPPPPP= 2222112211 () TcC =11 CMMiiiTiin= = 21 FORMATION OF EXPLICIT DAMPING MATRIXFROM WILSON MODAL , Radians per secondModal Damping RatioWilson Damping (Example)

8 5% Damping in Modes 1 and 2, 310% in Mode 5, Zero in Mode 4 Wilson Damping (Example)5% Damping in all , Radians per secondModal Damping RatioSolution of MDOF Equations of Motion Explicit (Step by Step) Integration of Coupled Equations Explicit Integration of FULL SET of Uncoupled Equations Explicit Integration of PARTIAL SET of Uncoupled Equations Modal Response Spectrum AnalysisTime, tForce, V(t)t1t3t2t0Vt2Vt11 Time, = VVtttt2221 Computed Response for Piecewise Linear LoadingEXAMPLE of MDOF Response of Structure Responding to1940 El Centro Earthquakek3=180 k/inu1(t)u2(t)u3(t)k1=60 k/ink2=120 k/inm1= k-s2/inm2= k-s2/inm3= k-s2/inAssume WilsonDamping with 5%critical in each Component of 1940 El CentroEarthquake.

9 Maximum acceleration = gM= s / in2k3=180 k/inu1(t)u2(t)u3(t)k1=60 k/ink2=120 k/inm1= k-s2/inm2= k-s2/inm3= k-s2/inK= 60600601801200120300kip / inForm Property Matrices:Example (Continued) 2221 096 6212 4= ..seck3=180 k/inu1(t)u2(t)u3(t)k1=60 k/ink2=120 k/inm1= k-s2/inm2= k-s2/inm3= k-s2/in = 1 0001 0001 000064406012570 3000 6762 ..KM =2 Solve Eigenvalue Problem:Example (Continued)Normalization of Modes using TMI=vs = 1 MODE 2 MODE 3T= secT= secT= secExample (Continued)Mode Shapes and Periods of Vibration = rad/sec = rad/sec = rad/seck3=180 k/inu1(t)u2(t)u3(t)k1=60 k/ink2=120 k/inm1= k-s2/inm2= k-s2/inm3= k-s2/inMMT*.

10 Sec /== 1801245523102kipin n= 45898314 1370 Generalized Mass:Example (Continued)k3=180 k/inu1(t)u2(t)u3(t)k1=60 k/ink2=120 k/inm1= k-s2/inm2= k-s2/inm3= k-s2/inCompute Generalized Loading:)()(*tvMRtVgT =Vvtng*.. ()= 25661 2542 080 Example (Continued) ()/**yyyVtm1111121112++= ()/**yyyVtm2222222222++= ()/**yyyVtm3333323332++= k3=180 k/inu1(t)u2(t)u3(t)k1=60 k/ink2=120 k/inm1= k-s2/inm2= k-s2/inm3= k-s2/in .. ()yyyvtg1110 45821 01 425++= .. ()yyy vtg2220 98396 60 511++= .. ()yy y vtg33 31 457212 40 090++=Write Uncoupled (Modal) Equations of Motion:Example (Continued)Modal Participation Factors1 4250 5110 1 9110 ModeModeMode123 Modal Scaling i.