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University of Engineering and Technology Peshawar, [email protected]: Introduction to Structural Dynamics and Earthquake Engineering LECTURE 9:MODAL ANALYSIS OF MDOF SYSTEMS SUBJECTED TO EARTHQUAKE LOADINGCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET2Modal decoupling of the EOMsIt is known that the equations of motion for a a MDOF with lumped mass system and undergoing only lateral displacement can be written as:[]{ } [ ]{ } [ ]{ } { } p(t) u k u c u m = + + & & &CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET3Letbe the modal matrix (matrix of mode shapes) in which the nth column is the nth mode shape of vibration (i.e. each column represents a particular mode shape) which means that[] { } []{ } q u = { } []{ } q u & & = { } []{ } q u & & & & = Substitutingthese in the equation given on previous slide [][]{ } [ ][]{ } [ ][]{ } { } p(t) q k q c q m = + + & & &Modal decoupling of the EOMsCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET4Modal decoupling of the EOMsPre-multiply both sides byOrWhere because of orthogonally properties of mode shapes (a matrixsaid to be orthogonal if where [I] is an identity matrix in which diagonal terms are 1 and off diagonal terms are 0 and therefore det[I]=1) the matrices are diagonal matrices(i.e., matrices in which off diagonal terms are zero) []T[] A[] [] [ ] I A AT=[ ] [ ] [] K and C , M[] [][]{ } [] [ ][]{ } [] [ ][]{ } [] { } p(t) q k q c q mT T T T = + + & & &[ ]{ } [ ]{ } []{ } { } P(t) q K q C q M = + + & & &CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET5Modal decoupling of the EOMsSince are diagonal matrices so the N coupled equations replaces by N uncoupled equation for SDOF systems [ ] [ ] [] K and C , M(t) P q K q C q Mn n n n n n n= + + & & &Where Mn= Generalized mass for the nth natural modeKn= Generalized stiffnes for the nth natural modeCn= Generalized damping for the nth natural modePn(t)= Generalized force for the nth natural modeCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET6Modal decoupling of the EOMsThe equation given on previous slide is for nth mode of MDOF of order N. All the independent equations for N modes in matrix form can be written asWhere M= Diagonal matrix of the generalized modal massesK= Diagonal matrix of the generalized modal stiffnessesC= Diagonal matrix of the generalized modal dampingsP(t) = Column vector of the generalized modal forces Pn(t)(t) P Kq q C q M = + +& & &CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET7Modal analysis for earthquake forcesThe uncoupled equations of motions for earthquake excitations can be written as{}= position vector={1} for shear structuresWhere {1} represents a unit vector matrix of dimensions N (the total DOF)[ ]{ } [ ]{ } []{ } { } [] { } ) (t p (t) P q K q C q MeffTeff = = + + & & &{ } []{} ) (t u m (t) pg eff& & =[ ]{ } [ ]{ } []{ } { } [] []{}[] []{ }structures shear for) ( 1) (t u mt u m (t) P q K q C q MgTgTeff& && & & & & = = = + + CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET8Modal analysis for earthquake forcesReplacingFor nth mode [] []{} L mT= [ ]{ } [ ]{ } []{ } { } { } ) (t u L (t) P q K q C q Mg eff& & & & & = = + + ) (t u L q K q C q Mg n n n n n n n& & & & & = + +) (t uMLqMKqMM 2qMMgnnnnnnnn n nnnn& & & & & = + + ) (2t uMLq q 2 q orgnnn n n n n n& & & & & = + + L and m has same unitsCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET9Participation factorsThe term Ln/Mnhas been given the name of participationfactor for the nth mode and is represented by n(Gamma)The participation factor is a measure (although rough) of how much the nth mode contributes to the displacement responseThe magnitude of the participation factor is dependent on thenormalization method used for the mode shapes.It can be seen that the normalization has an effect once in thenumerator and twice in the denominator.{ } []{}{ } []{ }nTnTnnnnmmML = = CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET10Participation factorsOnce the modal amplitudes {q} have been found the displacements of the structure are obtained fromand the effect of the normalization is cancelled out.What must be noted is that one should not compare one computer programs Participation Factors with those of another program unless they are both using the same normalization method{ } []{ } q u = CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET11Participation factorsThe participation factors are used to determine the number of mode shapes used to get a satisfactory estimate of the displacements of the structure..The displacements associated with the nth mode are given by{ } [ ]{ } (t) q (t) un n n =CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET12Effective weight of structure in nth mode, WnEffective weight of structure in nth mode=It shall be noted that the sum of the all effective weights for an excitation in a given direction, i.e. for a given {}, should equal the total weight of the structure. Note, this may be not be the casewhere rotational inertia terms also exist in the mass matrix.Many building codes require that a sufficient number of modesbe used in the analyses such that the sum of the effective weights is at least 90% of the weight of the structure. This provides a second measure on the number of modes required in the analysis.{ } []{}{ } []{}{ } []{ }gMLgmm mWnnnTnTnTnn2 = = CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET13Base Shear Force in the structure in the nth node,VbnThe base shear in nth mode can be determined using relationWhere Wn = Effective weight of structures in nth mode==gWggMLVnnnnnbnA A2CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET14Distribution of nodal (joint) forces in the structure from the base shear In many design codes the first step is to compute the modal base shear force and this is then distributed along the structures (shown on next slide) to each degrees of freedom.The distributed loads are assumed to give the same displacements in the structure as those generated by the exciting base shear.{ } []{ }nnbnnmLVf =CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET15Vbnm1nm2nm3nf1nf2nf3nBase shear acting in nth mode, VbnVbndistributed along the structure in each DOF= + + =n n n n bnf f f f V3 2 1Distribution of nodal (joint) forces in the structure from the base shear CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET16Response spectrum analysis: ExampleA 3 story R.C. building as shown below is required to bedesigned for a design earthquake with PGA=0.3g, and its elastic design spectrum is given by Fig 6.9.5 multiplied by 0.3). It is required to carry out the dynamic modal analysis by using the afore mentioned design spectrum . Take: Story height = 10ftTotal stiffness of each story = 250 kips/in. Weight of each floor = 386.4 kipsm1m2m3k1k2k3CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET17Develop mass and stiffness matrices[]=1 0 00 1 00 0 1mm=W/g = (386.4k) /(396.4 in/sec2)= 1.0 kip-sec2/in[ ] =250 250 0250 500 2500 250 500k[ ] [] = 2222250 250 0250 500 2500 250 500nnnnm kCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET18[ ] [] [ ] 0 det2= m k SettingnYields following valuessec /491 . 28sec /685 . 19sec /0360 . 7321radradradnnn===sec 221 . 0sec 319 . 0sec 893 . 0321===nnnTTTDetermine frequenciesCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET190250 250 0250 500 2500 250 500312111212121= nnnNormalized coordinates of first mode shape[ ] [] [ ]{ } 0121= m knSubstituting 1 and 51 . 9 41121= = nDetermine mode shapesCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET200149 . 200 250 0250 49 . 450 2500 250 49 . 4503121= First row gives802 . 1 0 250 49 . 45021 21= = 247 . 20 250 ) 802 . 1 ( 49 . 450 2500 250 49 . 450 25031 3131 21= = + = + Second row gives{ }==== 1.0000.8020.4457 2.247/2.247 1.802/2.247 1.000/2.242.2471.8021.0003121111Determine mode shapesCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET2105 . 387 250 250 0250 5 . 387 500 2500 250 5 . 387 500322212= 5 . 38722 =n015 . 137 250 0250 5 . 112 2500 250 5 . 1123222= 798 . 0 &45 . 03222 == ++=0.7980.4501.000322212Determine mode shapesCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET22++=++=0.4441.0000.801000 . 1250 . 2802 . 1
332331Similarly[]++++= = 0.4441.0000.8010.7980.4501.0001.0000.8020.4453 2 1 Determine mode shapesCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET230123-2 -1 0 1 2First mode shapeSecond mode shapeThird mode shapeMode shapesCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET24Calculate Modal mass and participation factors for each mode, n{ } []{}{ } []{ }nTnTnnnnmmML = = { } []{ }{ } []{ }1 1111 1 mmMLTT1= = For shear buildings {}={1}{ } []{ }{ } []{ }nTnTnnnnmmML 1= = CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET25=1111 0 00 1 00 0 1000 . 1802 . 0445 . 0T1L=1111.0000.802.445 01L/ft sec - kips26.964 /in. sec - kips 247 . 22 2= =1L{ } []{ }= =1111 0 00 1 00 0 113121111TT1m LCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET26{ } []{ }1 1 1 m MT==000 . 1802 . 0445 . 01 0 00 1 00 0 1000 . 1802 . 0445 . 0T1M=000 . 1802 . 0445 . 01.0000.802.445 01M/ftsec - kip 22.092 /in. sec - kip 841 . 12 2= =1MCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET27221 . 1841 . 1247 . 211= = = ML1Similarly355 . 0839 . 1652 . 022= = = ML2133 . 0839 . 1245 . 033= = = ML3CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET28Calculate effective weight of structure in each mode, Wn{ } []{}{ } []{}{ } []{ }gMLgmm mWnnnTnTnTnn2 = = ( )kips72 . 1059 4 . 386 *841 . 1247 . 221211= = = gMLW/in sec - kip 4 . 3862= g( )kips32 . 89 4 . 386 *839 . 1652 . 022222= = = gMLW( )kips61 . 12 4 . 386 *839 . 1245 . 023233= = = gMLWCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET29Calculate participating mass of the structure in mode n, PMnParticipating mass of the structure in nth mode= WWPMnn =*% 4 . 91 914 . 04 . 386 * 372 . 10591*1= = = =WWPM% 7 . 7 077 . 04 . 386 * 332 . 892*2= = = =WWPM% 09 . 1 0109 . 04 . 386 * 361 . 123*3= = = =WWPM00 . 1 = PMMost of the code requires that such number of modes shall be considered so that PM 0.9. In our case, indeed, the consideration of just the first mode would have been sufficient as PM1 0.9CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET30==gWggMLVnnnnnbnA A2Calculate base shear in the structure in each mode, VbnMode 1: Tn1=0.893 seckipsggT gW Vn1b 8 . 640 3 . 0 *1*8 . 1* 72 . 1059 .A11 1===kipsgggW V2b 6 . 72 3 . 0 *71 . 2* 32 . 89 3 . 0 *A : sec 0.319 T For2 2 n2= == =kipsgggW V3b 2 . 10 3 . 0 *71 . 2* 61 . 12 3 . 0 *A : sec 0.221 T For3 3 n3= == =Values of A for each Tncan be determined from Fig. 6.9.5 given on next slideCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET310.221 sec0.319 sec0.893 secsec 221 . 0sec 319 . 0sec 893 . 0321===nnnTTTCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET32Nodal forces acting on the structure in each mode, fn{ } []{ }nnbnnmLVf =[] []==3121111131211132111321
mLVfffmLVfffbnnnbnnnFirst mode==2 . 2857 . 2289 . 126000 . 1802 . 0445 . 012 0 00 12 00 0 12) 12 * 247 . 2 (8 . 640312111fffCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET33Nodal forces acting on the structure in each mode, fn{ } []{ }nnbnnmLVf =[]++==0.7980.4501.00012 0 00 12 00 0 12) 12 * 652 . 0 (6 . 7232221222322212mLVfffbSecond mode++=9 . 881 . 503 . 111322212fffCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET34Nodal forces acting on the structure in each mode, fn{ } []{ }nnbnnmLVf =[]++==0.4441.0000.80112 0 00 12 00 0 12) 12 * 245 . 0 (2 . 1033231333332313mLVfffbThird mode++=5 . 186 . 414 . 33332313fffCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET35{ } { }++++= =5 . 186 . 414 . 339 . 881 . 503 . 1112 . 2857 . 2289 . 1263 2 1 n n n nf f f fNodal forces acting on the structure in each mode, fnCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET36Nodal forces acting on the structure in each mode, fn640.8 kipsMode 1285.2 k228.7 k126.9 kMode 272.6 kips88.3 k50.1 k111.3 k10.2 kipsMode 318.5 k41.6 k33.4 kCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET37Combination ofModal MaximaThe use of response spectra techniques for multi-degree of freedom structures is complicated by the difficulty of combiningthe responses of each mode.It is extremely unlikely that the maximum response of all the modes would occur at the same instant of time.When one mode is reaching its peak response there is no way ofknowing what another mode is doing.The response spectra only provide the peak values of the response, the sign of the peak response and the time at which the peak response occurs is not known.CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET38Combination ofModal MaximaThereforeand, in general The combinations are usually made using statistical methods.{ } []{ }max maxq u { } []{ }max maxq u CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET39Combined Response roLet rnbe the modal quantity (base shear, nodal displacement, inter-storey drift, member moment, column stress etc.) for mode n .The r values have been found for all modes (or for as many modes that are significant). Most design codes do not require all modes to be used but many do require that the number of modes used is sufficient so that the sum of the Effective Weights of the modes reaches, say, 90% of the weight of the building. Checking the significance of the Participation Factors may be useful if computing deflections androtations only.CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET40Absolute sum (ABSSUM) methodThe maximum absolute response for any system response quantity is obtained by assuming that maximum response in each mode occurs at the same instant of time. Thus the maximum value of the response quantity is the sum of the maximum absolute value of the response associated with each mode. Therefore usingABSSUM method This upper bound value is too conservative. Therefore, ABBSSUM modal combination rules is not popular is structural design applications=Nnno or r1CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET41Square-Root-of-the Sum-of-the-Squares (SRSS) methodThe SRSS rule for modal combination, developed in E.Rosenblueths PhD thesis (1951) isThe most common combination method and is generally satisfactory for 2-dimensional analyses is the square root of the sum of the squares method. The method shall not be confused with the root-mean-square of statistical analysis as there is no n denominator. 2 / 112=Nnnoor rCE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET42Modal combination ofnodal forces:kips Vb 6 . 723 2 . 10 6 . 72 8 . 640 = + + =ABBSUM ruleSRSS rulekips Vb 645 2 . 10 6 . 72 8 . 6402 2 2= + + =CE-412: LECTURE 9 Dr. Mohammad Javed, C.E.D, NWFP UET43Home Assignment No. 7A 3 story R.C. building as shown below is required to bedesigned for a design earthquake with PGA=0.25g, and its elasticdesign spectrum is given by Fig 6.9.5 (Chopras book) multiplied by 0.25). It is required to carry out the dynamic modal analysis by using the afore mentioned design spectrum . Take: Story height = 10ftTotal stiffness of each story = 2000 kips/ft. Mass of each floor = 5000 slugsm1m2m3k1k2k3