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### EE312Chapter1&2

Course: EE 312, Spring 2008
School: CUNY City
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CUNY City - EE - 312
CUNY City - EE - 312
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CE 573: Structural Dynamics Fourier Series Example0 t 2 sec 0 lb F (t ) = , F (t + 5) = F (t ) 200 lb 2 t 5 sec rad T = 5 seconds =2 /5 sec .Constant term:ao = = 1 5 sec 1 2 sec 1 5 sec F (t ) dt = ( 0 lb ) dt + ( 200 lb ) dt 5 sec 5 sec
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Purdue - CIVIL ENGI - CE 573
CE 573: Structural Dynamics Characteristics of Structural Dynamic Systems Restoring Forces &quot;Inertial Forces&quot;: Newton's 2nd Law and D'Alembert's Principle Damping Dynamic Forces
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CE 573: Structural Dynamics Numerical Evaluation of Dynamic Response Forcing Interpolation ExampleGiven: Mass of tank m = 0.2533 kip-sec2 / in , stiffness k = 10 kips/in, undamped natural period T = 1.0 sec ( = 6.283 rad/sec ), and = 0.05 . Tan
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CE 573: Structural Dynamics Equivalent Viscous Damping Energy Dissipation in GeneralED =cycleFD du =cycleFD (t ) du dt ED = FD (t ) u (t ) dt dt cycle Energy Dissipation for Viscous DampingFd (t ) = cx (t ) = cX cos( t )FD
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CE 573: Structural Dynamics Response to Arbitrary Dynamic Excitationmu + ku = F (t ) u (t ) = 1 t F ( ) sin (t - ) d m 0(In all the examples, assume that the system starts at rest.) Example #1: Step Force u (t ) = Fo t 1 t Fo sin (t - ) d
Purdue - CIVIL ENGI - CE 573
CE 573: Structural Dynamics Structural Control Devices Supplemental DampersIdea: Add damping to the structural system by inserting dampers directly into the structure.Advantages: Fairly cheap and easy to install good for seismic retrofitting. D
Purdue - CIVIL ENGI - CE 573
CE 573: Structural Dynamics Example: Solving the Forced Vibration ProblemGiven: A three story building is modeled using shear frames. The property matrices and modal information for this structure are: 0 0 0 0.1035 405.57 -349.63 0 kip-sec2 ;
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CE 573: Structural Dynamics Example: Solving the Free Vibration ProblemGiven: A three-story building is modeled as shown. The dead load per unit length for each horizontal girder varies with each floor; you may assume that the weight of each column
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CE 573: Structural Dynamics Example: Response Spectrum Analysis using Modal CombinationsGiven: A four story building is modeled using shear frames. The mass and stiffness properties are expressed in terms of m = 12 103 kg and k = 4 106 N/m; prope
Purdue - CIVIL ENGI - CE 573
Problem 10.1:Given: A uniform beam of flexural stiffness EI = 109 lb-in 2 and length 300 in has one end fixed and theother simply supported.Required: Determine the system stiffness matrix considering three beam segments and the nodal coordinates
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CE 573: Structural Dynamics HW#2 SolutionProblem #1Given: A viscously damped structure is set into free vibration with an initial velocity. The resultingdamped oscillations are shown above.Required: (a) Determine the logarithmic decrement, the
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CE 573: Structural Dynamics Earthquake Response Spectrum Analysis: SDOF Case Recall: Spectral Displacement and Response Spectrumx (t ) = 1 t F ( )e - (t - ) sin [ d (t - ) ] d S d max x(t ) = S d ( m, c, k , F (t ) ) . m d 0 Earthquake Respon
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Journal of Earthquake Engineering, Vol. 8, No. 2 (2004) 175207 c Imperial College PressTHE VERTICAL-TO-HORIZONTAL RESPONSE SPECTRAL RATIO AND TENTATIVE PROCEDURES FOR DEVELOPING SIMPLIFIED V/H AND VERTICAL DESIGN SPECTRAYOUSEF BOZORGNIA Applied T
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CE 573: Structural Dynamics Solutions to HW#4Problem #1Given: A response spectrum is a plot of the maximum dynamic load factor (DLF) as a function of the(nondimensional) duration of the loading. See Figures 4.4 and 4.5 in the text for examples. W
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University of California, Berkeley Civil and Environmental EngineeringInstructor: Stephen Mahin Spring Semester 2006CEE 227 Earthquake Engineering Design Reading Assignments FEMA 356 Generally Topic Earthquake Engineering Borzorgnia and BerteroS
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University of California at Berkeley Civil and Environmental EngineeringInstructor: Stephen A. Mahin Spring Semester 2006CEE 227 - Earthquake Resistant DesignProblem 4 - Conceptual Design (due in Discussion Section, Feb. 16) In this problem, you
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Jeffrey Hunt CE 227 Problem 8 Part (a) The zip code for the site was chosen as 94704-1209LOCATION 37.870925 Lat. -122.268424 Long. The interpolated Probabilistic ground motion values, in %g, at the requested point are: 10%PE in 50 yr 5%PE in 50 yr 2
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CE 573: Structural Dynamics Response to Harmonic Forcing: Damped Systems Complementary Solution + Particular Solutionmu + cu + ku = Fo sin ( t + ) u + 2 u + 2u =Complementary solution:uc (t ) = e - t [ A cos d t + B sin d t ]Fo sin ( t +
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FEMA Program to Reduce Earthquake Hazards in Steel Moment-Frame Structures A New Paradigm for Design and Evaluation of Steel Moment Frame Buildings Stephen MahinNishkian Professor of Structural Engineering University of California at Berkeley Chair,
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Statistical Evaluation of Approximate Methods for Estimating Maximum Deformation Demands on Existing StructuresSinan D. Akkar1 and Eduardo Miranda, M.ASCE2Abstract: A statistical study is presented to evaluate the accuracy of five approximate metho
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Seismic Design Based on the Yield DisplacementMark Aschheim,a) M.EERIAlthough seismic design traditionally has focused on period as a primary design parameter, relatively simple arguments, examples, and observations discussed herein suggest that th
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CE 573: Structural Dynamics Logarithmic Decrement Matlab Example&gt; u1 = udata(2) u1 = 96.5749 &gt; u2 = udata(10) u2 = 83.9971(Located from the plot.)(Located from the plot.)&gt; delta = log(u1/u2) delta = 0.1395 u = log k , k = 1. uk +1 &gt;
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CE 227 Homework #2 SolutionsAdpated from: Troy MorganProblem 4 Conceptual DesignIn this problem, we investigate various structural configurations for our building. I've presented a sketch of possible solutions here, but obviously yours may (and
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CE 573: Structural Dynamics Numerical Evaluation of Dynamic Response Newmark Methods ExampleGiven: Mass of tank m = 0.2533 kip-sec2 / in , stiffness k = 10 kips/in, undamped natural period T = 1.0 sec ( = 6.283 rad/sec ), and = 0.05 . Tank star
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CE 573: Structural Dynamics HW#5 Supplemental ProblemGiven: Another common model used for buildings is the portal frame model, which tries to account forflexibility of both horizontal and vertical members by allowing for rotations at connections.
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CE 573: Structural Dynamics MDOF Earthquake Response Spectrum ExampleGiven: A three story building is modeled using shear frames. The mass and stiffness properties are as indicated. Assume = 0.02 for all modes. Each story has a height of 14 ft. Re
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CE 573: Structural Dynamics Example on Orthogonality and Modal MatricesGiven: A three-story building is modeled as shown. The dead load per unit length for each horizontal girder varies with each floor; you may assume that the weight of each column
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CE 573: Structural Dynamics Example: MDOF Damped Forced VibrationGiven: A six story shear frame model building has the following mass and stiffness properties: m = 66 lb-sec 2 / in , E = 30 106 psi and I = 882 in 4 for each column. The property ma
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CE 573: Structural Dynamics Example: Rayleigh QuotientGiven: A six-story building is modeled as a shear-frame structure. Parameters are m = 66 lb-sec 2 / in , E = 30 106 psi, and I = 882 in 4 . Required: Estimate the fundamental natural frequency
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Purdue - CIVIL ENGI - CE 573
CE 573: Structural Dynamics Transmissibility
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Characterizing Near Fault Ground Motion For The Design And Evaluation Of BridgesPaul SomervilleABSTRACT Near-fault ground motions are different from ordinary ground motions in that they often contain strong coherent dynamic long period pulses and p
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CE 573: Structural Dynamics HW#6 Supplemental ProblemsProblem #1Given: A popular method for limiting the effects of earthquakes on structures is to base isolate them; that is, to put in supports underneath the structure that isolate the main part
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CE 573: Structural Dynamics HW#1 Supplemental ProblemsProblem #1Given: A concrete slab (mass = 20,000 kg) forms the floor of a manufacturing facility. To preventexcessive vibrations from reaching the equipment placed on this floor, the slab is su
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CE 573: Structural Dynamics HW#5 SolutionsGiven: Another common model used for buildings is the portal frame model, which tries to account forflexibility of both horizontal and vertical members by allowing for rotations at connections. In this pro
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CE 573: Structural Dynamics Example on Natural Frequencies and ModesGiven: A two-story building is modeled as a shear-frame structure. Parameters are m = 35 103 kg and k = 17.5 106 N/m. Required: Determine the natural frequencies and natural vibr
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CE 573: Structural Dynamics Numerical Evaluation of Dynamic Response Elastoplastic Behavior ExampleGiven: Mass of frame m = 0.5 kip-sec2 / in , stiffness k = 20 kips/in, and = 0.02 . Tank starts at rest and is subjected to pulse F(t) as shown. A
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CE 573: Structural Dynamics HW#2 Supplemental ProblemsProblem #1Given: A viscously damped structure is set into free vibration with an initial velocity. The resultingdamped oscillations are shown above.Required: (a) Determine the logarithmic de
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CE 573: Structural Dynamics Solutions to HW#3Problem #1m Y(t) y(t)v = constantCenter of MasskcFramegh yo y (x)RWheelxGiven: A car can be (crudely) modeled as a single degree of freedom system having the entire car's mass m lumpe
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CE 573: Structural Dynamics HW#4 Supplemental ProblemGiven: A response spectrum is a plot of the maximum dynamic load factor (DLF) as a function of the(nondimensional) duration of the loading. See Figures 4.4 and 4.5 in the text for examples. We w
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CE 573: Structural Dynamics General Periodic Response Theorymu + cu + ku = F ( t ) ao + an cos ( n t ) + bn sin ( n t )n =1 n =1 N NComplementary Solution: uc (t ) = e - t [ A cos D t + B sin D t ]. Particular Solution:For constant term: u
Purdue - CIVIL ENGI - CE 573
CE 573: Structural Dynamics HW#3 Supplemental Problemv = constantm Y(t) y(t)Center of MasskcFramegh yo y (x)RWheelxGiven: A car can be (crudely) modeled as a single degree of freedom system having the entire car's massm lumped a
Purdue - CIVIL ENGI - CE 573
CE 573: Structural Dynamics Example: Inverse Vector Iteration for Higher ModesGiven: A six-story building is modeled as shown. Parameters are m = 66 lb-sec 2 / in , E = 30 106 psi, and I = 882 in 4 . Required: Use inverse vector iteration to find