同济大学：《建筑结构抗震》课程教学资源（课件讲稿）Chapter 4 Dynamic Response of MDOF System

2012/10/25 Chapter Dynamic Response of MDOF System 3.1 Introduction 4 1 Introduction 3.2 Free Vibration of SDOF Systems 4.2 Modal Analysis of MDOF Systems AA 3.7Response of Nonlinear SDOF Systems ( MDOF Systems Response to SDoF☐ Base Excitation Single-degree-of-freedom oscillators 自→2 DOF] OF SYSTEMS Mulli-degree-of-free dom system y g n degrees of freedom will hav ca1espeeeCanteewenemsondu 且 na 窗目目目 1

2012/10/25 1 Chapter 3 Dynamic Response of Structure 3.1 Introduction 3.2 Free Vibration of SDOF Systems 3.3 Impulsive Vibration of SDOF Systems 3.4 Harmonic Vibration of SDOF Systems 3.5 Response of Linear SDOF Systems 3.6 Spectrum of Seismic Response of SDOF 3.7 Response of Nonlinear SDOF Systems （*） No. 1 Chapter 4 Dynamic Response of MDOF System 4.1 Introduction 4.2 Modal Analysis of MDOF Systems 4.3 Mode Superposition Spectrum Analysis of MDOF Systems 4.4 No. 2 Instructor: Dr. C.E. Ventura No. 3 Instructor: Dr. C.E. Ventura No. 4 time, sec 0.0 0.2 0.4 0.6 0.8 Vibration Period T W g K  2 SDOF • Single-degree-of-freedom oscillators W K T Instructor: Dr. C.E. Ventura No. 5 First￾Mode Shape Third￾Mode Shape Second￾Mode Shape • Linear response can be viewed in terms of individual modal responses. MDOF Idealized Model Actual Building • Multi-story buildings can be idealized and analyzed as multi-degree-of-freedom systems. Instructor: Dr. C.E. Ventura No. 6 MDOF SYSTEMS • Multi-degree-of-freedom systems – Systems having n degrees of freedom will have: • n modes of vibration; • n natural frequencies. – Response will, in general, be some linear combination of response in these modes – Common approach is to treat the n-DOF system as n SDOF systems instead

2012/10/25 Natural Vibration Period Fundamental Vibration Period for Buildings 40 DOF Systems Example-3-storey building 4.2 Modal Analysis of MDOF Systems Newton's Second Law states: ndnaorsrngd .Axial deformations of the beam and columns neglected e the the floor levels en6-明明g+g0 nfo:8al份}份}-a

2012/10/25 2 Instructor: Dr. C.E. Ventura No. 7 Natural Vibration Period Every body has a vibration natural period that depends on its mass and stiffness. If the body has several degrees of freedom, it has many vibration periods, and the larger period is called the Fundamental Period of Vibration Instructor: Dr. C.E. Ventura No. 8 Fundamental Vibration Period for Buildings Instructor: Dr. C.E. Ventura 4.2 Modal Analysis of MDOF Systems No. 9 Instructor: Dr. C.E. Ventura No. 10 MDOF Systems • Example – 3-storey building Instructor: Dr. C.E. Ventura No. 11 Formulation of the Equation of Motion and Selection of the Dynamic Degrees of Freedom Idealized : • Beams and floors rigid • Axial deformations of the beams and columns neglected • Axial load on columns neglected • Concentrate the mass at the floor levels Instructor: Dr. C.E. Ventura No. 12 Formulation of the Equation of Motion and Selection of the Dynamic Degrees of Freedom •Newton’s Second Law states: If a force acts upon a body, the rate of change of the quantity of motion of the body (momentum = m*v) is equal to the applied force. pj - fSj - fDj = mjüj or mjüj + fDj + fSj = pj (t)                                  ( ) ( ) 0 2 1 0 2 1 2 1 2 1 2 1 p t p t f f f f u u m m S S D D   •For our Example: •In matrix form: mu dt dv f  m  

2012/10/25 …与Σ要 +For Damping fn=c盒，+G(位-，) ·石=+0 fm=c(,-) 。·=k1u1+k(u-u2） ·=kuu) 中份-好的 in matrix form: 份生 Equation of Motion: mutcu+ku=p(r) System:Translational Ground Motion 0n .ForN c+ku=-mlo( with ++6-0 ma+c0+ku =p(r) system: faaotou eneral Overview:MDOF Discrete systems dal Analysis Procedres of odal analysis Solutions to the equations of motions This method is applicable when: cally damped Response spectramethod 3

2012/10/25 3 Instructor: Dr. C.E. Ventura No. 13 Formulation of the Equation of Motion and Selection of the Dynamic Degrees of Freedom • fs1 = fs1 b + fs1 a • fs1 = k1u1 + k2 (u1 -u2 ) • fs2 = k2 (u2 -u1 )   columns c j h EI k 3 12                       2 1 2 2 1 2 2 2 1 u u k k k k k f f s s •In matrix form: •Story Stiffness = Instructor: Dr. C.E. Ventura No. 14 Formulation of the Equation of Motion and Selection of the Dynamic Degrees of Freedom ( ) 1 1 1 2 1 2 f c u c u u D       ( ) 2 2 u2 u1 f c D                           2 1 2 2 1 2 2 2 1 u u c c c c c f f D D   Equation of Motion: •In matrix form: •For Damping: m ( ) .. . u cu ku  p t Instructor: Dr. C.E. Ventura No. 15 Planar Systems: Translational Ground Motion •uj t (t) = uj (t) + ug (t) •u t (t) = u(t) + ug (t)1 •Displacements related by: •For N masses: •Equation of Dynamic Equilibrium fI + fD + fS = 0 •For a linear system: fS = ku fD = ců •Inertia forces related to total acceleration of mass fI = müt •1= vector of order N with each element = 1 Instructor: Dr. C.E. Ventura No. 16 Planar Systems: Translational Ground Motion The new equations of motion have now become: mü+ ců + ku = -m1üg (t) Comparing: mü+ ců + ku = -m1üg (t) with mü+ ců + ku = p(t) The ground motion can be replaced by the effective earthquake forces: peff(t) = -m1üg (t) Instructor: Dr. C.E. Ventura No. 17 General Overview: MDOF Discrete systems • Procedures to formulate equations of motions • Solutions to the equations of motions – Numerical solution of eigenproblems – Modal superposition method • Classically damped systems • Nonclassically damped systems – Solution in the frequency domain – Rayleigh-Ritz method – Direct integration of equation of motion – Response spectra method Instructor: Dr. C.E. Ventura No. 18 Modal Analysis • Dynamic analysis using modal superposition method for classically damped systems also called: – Modal analysis – Classical modal analysis – Classical mode superposition method – Classical mode displacement superposition method • This method is applicable when: – The system is classically damped – The system is linear

2012/10/25 Modal Analyais dal Analysis Overview of the method mii+ci+ku=plt) .The re se in each mode can be c 0=立a.0=pq Substitute u(t)in equation of motion: 立m.0+立c40+4g.0= 之'ma0+立'e.o)+立4.e)=0 Modal Analysis Modal Analysis .Because of orthogonality of modes We define: -Generalined modal mass: M,=m响 Generalized modal stiffness: K.■kp Since the system is elassically damped c=02r Modal Analysis Modal Analysis beunoupdde .Divide the uncoupled equations by M..+C.A.+K.9.-P.0) Or in matrix form: Mu+Ca+Ku=P( Psdiagomalmtrofrthegerenledmoalarce

2012/10/25 4 Instructor: Dr. C.E. Ventura No. 19 Modal Analysis • Overview of the method: – The equations of motions, when transformed to modal coordinates, become uncoupled. – The response in each mode can be computed independently of the other modes by solving an SDOF system with the vibration properties of that mode. – Modal responses are combined to obtain the total response. Instructor: Dr. C.E. Ventura No. 20 Modal Analysis • Equation of motion: • Transfer to modal equations: – Displacements in terms of modal coordinates: – Substitute u(t) in equation of motion: – Premultiplying each term by : mu  cu  ku  p(t) ( ) ( ) ( ) 1 t q t t r N r u  r  Φq   ( ) ( ) ( ) ( ) 1 1 1 q t q t q t t r N r r r N r r r N r m r c k  p         ( ) ( ) ( ) ( ) 1 1 1 q t q t q t t r N r r T r n N r r T r n N r r T  n m  c  k  p            T n Instructor: Dr. C.E. Ventura No. 21 Modal Analysis • Because of orthogonality of modes: • Since the system is classically damped: 0 0   r T n r T n     k m r  0 T n c n  r n  r n  r Instructor: Dr. C.E. Ventura No. 24 Modal Analysis • We define: – Generalized modal mass: – Generalized modal stiffness: – Generalized modal damping: – Generalized modal forces: n T Mn  n m n T Kn  n k n T Cn  n c P (t) (t) T n  n p Instructor: Dr. C.E. Ventura No. 25 Modal Analysis • The modal equations will be uncoupled and reduces to: • Or in matrix form: M is the diagonal matrix of the generalized modal mass K is the diagonal matrix of the generalized modal stiffness C is diagonal matrix of the generalized modal damping P(t) is diagonal matrix of the generalized modal forces M q C q K q P (t) n n  n n  n n  n   Mu  Cu  Ku  P(t) Instructor: Dr. C.E. Ventura No. 26 Modal Analysis • Divide the uncoupled equations by : • Solve the above equation for the modal coordinate similar to an SDF system. Mn n n n n n n n n M P t q q q ( ) 2 2        q (t) n

2012/10/25 Modal Analyeis Modal Analysis Element forees: is:u(r) m.0) 0=立a.=立a.0 0=2x.0 Modal Analysis:Summary (cont'd) d the stiffnes Compute n-modal displacements) l displacements r .Determine the natural frequenciesand mode shapes Combine the contributions of all the modes to determine the total response. solving for modal coordinates( itl-Story Structures moda 4.3 The Response Spectrum Method of Analysis for MDOF Systems 5

2012/10/25 5 Instructor: Dr. C.E. Ventura No. 27 Modal Analysis • Displacements: – Contribution of the nth mode to the displacement is: – Combining these modal contributions gives the total displacement:       N n N n n n n t u t q t 1 1 u( ) ( )  ( ) (t) q (t) un  n n u(t) Instructor: Dr. C.E. Ventura No. 28 Modal Analysis • Element forces: – Determine the contributions of the individual modes to the element force : • From modal displacements using element stiffness properties or, • By static analysis of the structure subjected to the equivalent static forces associated with the nth-mode response defined as: – The total element force considering contributions of all modes is: r (t) n (t) un ( ) ( ) ( ) 2 t t q t n kun  mn n f     N n n r t r t 1 ( ) ( ) Instructor: Dr. C.E. Ventura No. 29 Modal Analysis: Summary • Define the structural properties: – Determine the mass matrix m and the stiffness matrix k. – Determine the modal damping ratios • Determine the natural frequencies and mode shapes . • Compute the response in each mode by uncoupling the equations of motion and solving for modal coordinates . n n q (t) n Instructor: Dr. C.E. Ventura No. 30 Modal Analysis: Summary (cont’d) • Compute n-modal displacements un (t) . • Compute the element forces associated with the n-modal displacements rn (t) . • Combine the contributions of all the modes to determine the total response. Instructor: Dr. C.E. Ventura No. 31 • Individual modal responses can be analyzed separately. Reference: A. K. Chopra, Dynamics of Structures: A Primer, Earthquake Engineering Research Institute • For typical low-rise and moderate-rise construction, first-mode dominates displacement response. • Total response is a combination of individual modes. Total Roof Displ. Resp. Multi-Story Structures Instructor: Dr. C.E. Ventura 4.3 The Response Spectrum Method of Analysis for MDOF Systems No. 32

2012/10/25 What is the Response Spectrum Method(RSM)? A way to solve the equations of motion. e凡esp g them to be ed as SDOF gp8DfYaeoReM3gtando5ynne wenraeform o ncoupled Equations of Motion Concept of Linear Combination of Mode Shapes MDOF Equation of Motion: 刷+Cn+u=) J=Φr MD+C0+KDy=F() Mode Shape 44g Concept of Linoar Combination of Mode Shapes odal Responses using Response Spectrum Maximum modal displacement Ms(T 0=立.Y0 faMm产s.(Ta.tq-on Modal base shears VhaMa小+ n=1 l。-tmi M。=◆mtn Mode shape Response of SDOF s)Systemresponse from specrum graph

2012/10/25 6 Instructor: Dr. C.E. Ventura 33 What is the Response Spectrum Method (RSM)? The Response Spectrum is an estimation of maximum responses (i.e., acceleration, velocity and displacement) of a family of SDOF systems subjected to a prescribed ground motion. The RSM utilizes the response spectrum to give the structural designer a set of possible forces and deformations a real structure would experience under earthquake loads. For SDF systems, RSM gives quick and accurate peak response without the need for a time-history analysis. For MDF systems, a true structural system, RSM gives a reasonably accurate peak response, again without the need for a full time-history analysis. Instructor: Dr. C.E. Ventura A way to solve the equations of motion. • This will be done by a transformation of coordinates from Normal Coordinates (displacements at the nodes) to Modal Coordinates (amplitudes of the natural Mode shapes). • Because of the Orthogonality Property of the natural mode shapes, the equations of motion become uncoupled, allowing them to be solved as SDOF equations. • After solving, we can transform back to the normal coordinates. 34 Instructor: Dr. C.E. Ventura Uncoupled Equations of Motion 35 Instructor: Dr. C.E. Ventura Concept of Linear Combination of Mode Shapes (Transformation of Coordinates) 36 Instructor: Dr. C.E. Ventura Concept of Linear Combination of Mode Shapes    N n 1 n n {u(t)} { }Y (t) 37 Mode shape Response of SDOF Instructor: Dr. C.E. Ventura X n.M ax L n M n S d Tn  n     n   F i.n.M ax L n M n S a Tn  n     n   V b.n.M ax L n 2 M n S a Tn  n     n  Maximum modal displacement Modal forces Modal base shears Ln , Mn , and φn are system parameters determined from the Modal Analysis Method. L n  n T L  m1 n M n  n T m  n M   n S a Tn  n    System response from spectrum graph. Modal Responses using Response Spectrum 38

2012/10/25 Modal Responses of a ten-storey frame building Modal Participation Factor, r.-L/M. -Φrm/rmΦ. It dependson how the mode shapes are scaled 脑瑞器 Concept of Effective Modal Mass,m' Effective modal masses and modal heights m.=L2/M. 0 Modal mass Modal heigh e modal mass is independent of fective modal masses and modal heights nse Spectrum Method steps Modal mass T (or frequencies w,=2m/T)and or themode,and repeat fo Modal height

2012/10/25 7 Instructor: Dr. C.E. Ventura Modal Responses of a ten-storey frame building Period SA (g) 1st Mode T = 1.70 s V = 0.75 Sa 2nd Mode T = 0.65 s V = 0.13 Sa 3rd Mode T = 0.34 s V = 0.05 Sa 1 2 39 Instructor: Dr. C.E. Ventura Modal Participation Factor, Γn Γn = Ln /Mn = {Φn } T[m]{1} ∕ {Φn } T [m]{Φn } It is a measure of the contribution of each mode to the total Response of the system to the given type of excitation It depends on how the mode shapes are scaled 40 Instructor: Dr. C.E. Ventura Concept of Effective Modal Mass, m* n Practical value of Effective Modal Mass: Use enough modes in the analysis to provide a total effective mass not less than 90% of the total structural mass (valid for base shear calculation only) • The sum of the effective modal mass for all modes is equal to the total structural mass. • The value of effective modal mass is independent of mode shape scaling. m* n = Ln 2 ∕ Mn = Γn 2 {Φn } T [m]{Φn } 41 Instructor: Dr. C.E. Ventura 42 Effective modal masses and modal heights Modal mass Modal height We use this information to compute modal base shears and modal base overturning moments Instructor: Dr. C.E. Ventura 43 Effective modal masses and modal heights Modal mass Modal height We can use this information to compute modal base shears and modal base overturning moments Instructor: Dr. C.E. Ventura Response Spectrum Method steps 44 Solution steps: - Determine mass matrix, m - Determine stiffness matrix, k - Find the natural periods, Tn (or frequencies ωn=2π/Tn ) and mode shapes φn of the system - Compute peak response for the nth mode, and repeat for all modes. - Combine individual modal responses for quantities of interest (displacements, shears, moments, stresses, etc)

2012/10/25 Physical meaning of mode shapes DOF with uniform mass and stiffness ular buildin Building with Podiu lative movement mode shapes 5DOF with uniform mass and stiffness 5 DOF-Base lsolated Relative movemont mode shapes 5DOF-Base Isolated

2012/10/25 8 Instructor: Dr. C.E. Ventura 45 Physical meaning of mode shapes Examine the behaviour of a simple 5 story building that represents these cases Regular building Building with Podium Base-Isolated building Instructor: Dr. C.E. Ventura 46 5 DOF with uniform mass and stiffness 1 0 1 1 2 3 4 5 Mode 1 1 0 1 1 2 3 4 5 Mode 2 1 0 1 1 2 3 4 5 Mode 3 1 0 1 1 2 3 4 5 Mode 4 1 0 1 1 2 3 4 5 Mode 5 Modal Periods (in sec): T 1  2.001 T 2  0.685 T 3  0.435 T 4  0.338 T 5  0.297 Modal Frequencies (in Hz): f1  0.500 f2  1.459 f3  2.300 f4  2.955 f5  3.370 Modal Masses (as a function of "m"): MM1  4.398 MM2  0.436 MM3  0.121 MM4  0.038 MM5  0.008 Modal Heights (as a function of "H"): MH1  3.513 MH2  1.204 MH3  0.764 MH4  0.594 MH5  0.521 Modal Participation Factors: 1  1.252 2  0.394 3  0.208 4  0.116 5  0.053 Total mass: MMm  1295.337 Instructor: Dr. C.E. Ventura 47 Modal Expansion (5 DOF with uniform mass and stiffness) 1 2 3 4 5Vector 1 2 3 4 5 Mode 1 1 2 3 4 5 Mode 2 1 2 3 4 5 Mode 3 1 2 3 4 5 Mode 4 1 2 3 4 5 Mode 5 = + + + + 1 2 3 4 5Vector 1 2 3 4 5 Mode 1 1 2 3 4 5 Mode 2 1 2 3 4 5 Mode 3 1 2 3 4 5 Mode 4 1 2 3 4 5 Mode 5 = + + + + 1 2 3 4 5Vector 1 2 3 4 5 Mode 1 1 2 3 4 5 Mode 2 1 2 3 4 5 Mode 3 1 2 3 4 5 Mode 4 1 2 3 4 5 Mode 5 = + + + + Deformation Modal Contributions Instructor: Dr. C.E. Ventura 48 Relative movement mode shapes 5 DOF with uniform mass and stiffness Slow Instructor: Dr. C.E. Ventura 49 5 DOF – Base Isolated 1 0 1 1 2 3 4 5 Mode 1 1 0 1 1 2 3 4 5 Mode 2 1 0 1 1 2 3 4 5 Mode 3 1 0 1 1 2 3 4 5 Mode 4 1 0 1 1 2 3 4 5 Mode 5 Modal Periods (in sec): T 1  5.866 T 2  0.900 T 3  0.482 T 4  0.351 T 5  0.299 Modal Frequencies (in Hz): f1  0.170 f2  1.111 f3  2.074 f4  2.845 f5  3.341 Modal Masses (as a function of "m"): MM1  4.994 MM2  0.006 MM3  0.000 MM4  0.000 MM5  0.000 Modal Heights (as a function of "H"): MH1  3.048 MH2  38.558 MH3  0.512 MH4  38.503 MH5  0.501 Modal Participation Factors: 1  1.037 2  0.046 3  0.012 4  0.004 5  0.002 Total mass: MMm  1295.337 Instructor: Dr. C.E. Ventura 50 Relative movement mode shapes 5 DOF – Base Isolated Slow

2012/10/25 movement mode shape 5 DOF-with Podium of soreyshear-beam type buding Natural vibration modesof building Mode shapes of the system: 001 1 k =201s=06sT3=044-0.34T-0.2 00011 Assume a damping ratio of 5%for all modes Spectrum 9

2012/10/25 9 Instructor: Dr. C.E. Ventura 51 5 DOF – with Podium 1 0 1 1 2 3 4 5 Mode 1 1 0 1 1 2 3 4 5 Mode 2 1 0 1 1 2 3 4 5 Mode 3 1 0 1 1 2 3 4 5 Mode 4 1 0 1 1 2 3 4 5 Mode 5 Modal Periods (in sec): T 1  2.001 T 2  0.724 T 3  0.523 T 4  0.407 T 5  0.341 Modal Frequencies (in Hz): f1  0.500 f2  1.382 f3  1.913 f4  2.457 f5  2.929 Modal Masses (as a function of "m"): MM1  3.316 MM2  0.990 MM3  0.634 MM4  0.056 MM5  0.004 Modal Heights (as a function of "H"): MH1  3.555 MH2  0.102 MH3  0.941 MH4  0.114 MH5  1.509 Modal Participation Factors: 1  1.327 2  0.658 3  0.441 4  0.169 5  0.046 Total mass: MMm  1295.337 Instructor: Dr. C.E. Ventura 52 Relative movement mode shapes 5 DOF – with Podium Slow Instructor: Dr. C.E. Ventura RSM – a sample calculations of a 5-storey shear-beam type building. 1 0 0 0 0 0 1 0 0 0 m = 0 0 1 0 0 X 100 kips/g 0 0 0 1 0 0 0 0 0 1 2 -1 0 0 0 -1 2 -1 0 0 k = 0 -1 2 -1 0 X 31.54kip/in. 0 0 -1 2 -1 0 0 0 -1 1 Typical storey height is h=12 ft. 53 (Example from Chopra’s book) Instructor: Dr. C.E. Ventura No. 54 Natural vibration modes of a 5-storey shear building. Assume a damping ratio of 5% for all modes T1 = 2.01s T2 = 0.68s T3 = 0.42s T4 = 0.34s T5 = 0.29s Mode shapes φn of the system: 54 Instructor: Dr. C.E. Ventura No. 55 Obtain values from Response Spectrum: T1 = 2.01s T2 = 0.68s T3 = 0.42s T4 = 0.34s T5 = 0.29s 55 Instructor: Dr. C.E. Ventura . . . Results: 56

2012/10/25 Time-historyna(roof Modal Combinations to estimate peak resps AA 0 ends on wh odal Combinations. Modal Combinations... a)Sum of the absolute values: b)um of there leads to very conservative results modal values at the Results can be conservative or unconservative. 飞-2 5天·J The response is estimated as: odal Combinations.. )Complete quadratic combination (CQC): Modal combinations-for 5 DOF example: orrect base shear is 73.278 kips (from time-history analysis) d in BSSUM:Summation of absolute values of individual mod P Vs.1 kipsgrossly over-estimated are spread out Pau( Where r=P./Pn and must be equal to or less than 1.0. ces are cosely spa 10

2012/10/25 10 Instructor: Dr. C.E. Ventura Time-history analysis results (roof displ. & base overturning moment):57 Instructor: Dr. C.E. Ventura 58 Modal Combinations to estimate peak response: • Modal maxima do not occur at the same time, in general. • Any combination of modal maxima may lead to results that may be either conservative or unconservative. • Accuracy of results depends on what modal combination technique is being used and on the dynamic properties of the system being analysed. • Three of the most commonly used modal combination methods are: Instructor: Dr. C.E. Ventura 59 Modal Combinations…. a) Sum of the absolute values: • leads to very conservative results • assumes that maximum modal values occur at the same time • response of any given degree of freedom of the system is estimated as Instructor: Dr. C.E. Ventura 60 Modal Combinations….. b) Square root of the sum of the squares (SRSS or RMS): •Assumes that the individual modal maxima are statistically independent. •SRSS method generally leads to values that are closer to the “exact” ones than those obtained using the sum of the absolute values. •Results can be conservative or unconservative. •Results from an SRSS analysis can be significantly unconservative if modal periods are closely spaced. •The response is estimated as: Instructor: Dr. C.E. Ventura 61 Modal Combinations…. c) Complete quadratic combination (CQC): •The method is based on random vibration theory •It has been incorporated in several commercial analysis programs •A double summation is used to estimate maximum responses, Similar equations can be applied for the computation of member forces, interstorey deformations, base shears and overturning moments. ρ is a cross-modal coefficient (always positive), which for constant damping, z, is evaluated by Where r = ρn / ρm and must be equal to or less than 1.0. Instructor: Dr. C.E. Ventura 62 Modal combinations – for 5 DOF example: ABSSUM: Summation of absolute values of individual modal responses Vb ≤ Σ |Vbn| = 98.41 kips → grossly over-estimated SRSS: Square root of sum of squares Vb = (Σ Vbn 2 ) 1/2 = 66.07 kips →good estimate if frequencies are spread out CQC: Vb = (ΣΣVbi ρinVbn) 1/2 = 66.51 kips → good estimate if frequencies are closely spaced Correct base shear is 73.278 kips (from time-history analysis)