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HOMEWORK 1 1. Derive equation of motion of SDOF using energy method 2. Find amplitude A and tanΦ for given x0, v0 3. Find natural frequency of cantilever, l=400mm, Φ=5mm, E=2e11Pa, m=2.7kg. Confirm with SW Simulation 4. Work with exercises in chapter 19 – blue book 1 RING Energy method Ring.SLDASM 2 ωn, x0, v0 fully define free undamped vibration 3 MODULE 02 DAMPED VIBRATION Inman (3rd edition) section 1.3 4 SINGLE DEGREE OF FREEDOM SYSTEM WITHOUT DAMPING stiffness mass Notice that mass and stiffness are completely separated. & kx 0 mx& This is the equation of motion of a single degree of freedom system with no damping. It states that inertial forces are equal and opposite to stiffness forces. 5 SINGLE DEGREE OF FREEDOM SYSTEM WITH DAMPING Notice that mass, damping and stiffness are completely separated. Viscous damper, damping force is proportional to velocity 6 mx Equation of motion We assume solution in the form of: 7 x ae t x a e t x a 2 e t Characteristic equation Critical damping Determinant of characteristic equation Damping ratio (modal damping) n k m 8 underdamped 1 1 9 Equation is linear, therefore the sum of solutions is also solution underdamped 1 or, using Euler’s relations: Damped natural frequency Two forms of solution of free under- damped vibration 10 underdamped 1 ωd- natural frequency of damped oscillations A and Ф are calculated from initial conditions Inman p. 24 11 overdamped 1 This is a non-oscillatory motion 12 critically 1 Damping is critical Damping iswhen: critical when: c ζ= =1 2 km 13 FREE VIBRATIONS WITH VISCOUS DAMPING Depending on the sign of ζ we have three cases ζ<1 system is under damped, motion is oscillatory with an exponential decay in amplitude ζ=1 system is critically damped, at most one overshot of system resting position is possible ζ>1 system is over damped, motion is exponentially decaying 14 nt atop Ae abottom Aent Damped vibration amplitude decay.xls 15 m =10kg k = 1000N/m c = 20Ns/m SDOF damped.SLDASM 16 Linear damper 20Ns/m Modal damping 10% 17 0.62, 56.48 t1 = 0.62s a1 = 56.48 t2 =1.23s a2 = 30.30 1000 n 10 10 1.23, 30.30 Ae 10 x 0.62 56.48 1.86 10 x1.23 Ae 30.30 e6.1 1.86 6.1 ln1.86 0.62 0.10 18 swing arm.SLDASM 19 L1=0.2m m =0.56kg Assume that beam mass is negligible kL=2000N/m cL=10Ns/m L2=0.1m Is this system under damped, critically damped or over damped? 20 System is underdamped, modal damping is 7.4% 21 Linear damper 10Ns/m Modal damping 7.4% 22 ζ = 5% ζ = 100% ζ = 10% ζ = 200% ζ = 50% 23 TYPICAL VALUES OF DAMPING RATIO SYSTEM Damping ratio (% of critical damping) METALS 0. 1% CONTINUOUS METAL STRUCTURES 2% - 4% METAL STRUCTURES WITH JOINTS 3% - 7% ALUMINUM/STEEL TRANSMISSION LINES ~0.4% SMALL DIAMETER PIPING SYSTEMS 1% - 2% LARGE DIAMETER PIPING SYSTEMS 2% - 3% SHOCK ABSORBERS 30% RUBBER ~5% LARGE BUILDINGS DURING EARTHQUAKE 1%-5% PRE-STRESSED CONCRETE STRUCTURES 2% - 5% REINFORCED CONCRETE STRUCTURES 4%-7% TIMBER 5% - 12% 24 MODAL SUPERPOSITION METHOD Structure vibrating in a given mode can be considered as the Single Degree of Freedom (SDOF) system. Structure can be considered a series of SDOF. For linear systems the response can be found in terms of the behavior in each mode and these summed for the total response. This is the Modal Superposition Method used in linear dynamics analyses. A linear multi-DOF system can be viewed as a combination of many single DOF systems, as can be seen from the equations of motion written in modal, rather than physical, coordinates. The dynamic response at any given time is thus a linear combination of all the modes. There are two factors which determine how much each mode contributes to the response: the frequency content of the forcing function and the spatial shape of the forcing function. Frequency content close to the frequency of a mode will increase the contribution of that mode. However, a spatial shape which is nearly orthogonal to the mode shape will reduce the contribution of that mode. 25 MODAL SUPERPOSITION METHOD The response of a system to excitation can be found by summing up the response of multiple SODFs. Each SDOF represents the system vibrating in a mode of vibration deemed important for the vibration response. 26 TIME AND FREQUENCY RESPONSE ANALYSIS 27 TIME RESPONSE ANALYSIS In Time Response analysis the applied load is an explicit function of time, mass and damping properties are both taken into consideration and the vibration equation appears in its full form: Where: [ M ] mass matrix [ C ] damping matrix [ K ] stiffness matrix [ F ] vector of nodal loads [ d ] unknown vector of nodal displacements Dynamic time response analysis is used to model events of short duration. A typical example would be analysis of vibrations of a structure due to an impact load or acceleration applied to the base (called base excitation). Results of Time Response analysis will capture both the response during the time when load is applied as well as free vibration after load has been removed. 28 TIME RESPONSE ANALYSIS F(t) Force excitation or base excitation is a function of time. Solution is performed in time domain i.e. data of interest (displacement, stresses) are computed as functions of time. F(t) F(t) 5 12 10 4 8 3 6 2 4 1 2 0 0 0 2 4 6 8 10 0 2 time Examples of load time history in Time Response analysis. 4 6 8 10 time 29 FREQUENCY RESPONSE ANALYSIS Force excitation or base excitation is not directly a function of time, rather it is a function of the excitation frequency. Solution is performed in frequency domain i.e. data of interest (displacement, stresses) are computed as functions of frequency. Force amplitude does not have to be constant time Example of load time history that can be used in Frequency Response analysis. 30 TIME RESPONSE DUE TO IMPULSE LOAD force force time time Δt Different ways to illustrate a unit impulse Impulse load An impulse applied to a SDOF is the same as applying the initial conditions of zero displacement and initial velocity v0 = F Δt /m (Δt here is duration of “square” impulse, Δt is very short) 31 TIME RESPONSE DUE TO IMPULSE LOAD F t Fˆ A d md md v0 0 To model impulse load we set: x 0= 0 v 0 = F Δt /m 32 TIME RESPONSE DUE TO IMPULSE LOAD Analytical solution of unit impulse excitation problem. Response of a system due to an impulse at t = 0 Where Fˆ is force impulse Inman p. 195 33 TIME RESPONSE DUE TO IMPULSE LOAD Impulse load response.xls 34 TIME RESPONSE DUE TO IMPULSE LOAD Impulse 4.77Ns Impulse load definition A force is considered to be an impulse if its duration Δt is very short compared with the period T=1/f Here Δt = 0.0075, T=0.03 35 TIME RESPONSE DUE TO IMPULSE LOAD The critical damping for SDOF is: ccr = 2 km = 4000 Ns m To define damping as 5% of critical damping we can either enter it as 200 in the Spring-Damper Connector window or as 0.05 in the Global Damping window. Damping definition. Damping can be defined explicitly (left) or as a fraction of critical damping (right). The entries in both windows define the same damping. 36 TIME RESPONSE DUE TO IMPULSE LOAD Location for Time History graph Results of Dynamic Time analysis; displacement time history 37 In class exercise to demonstrate: Equivalence of explicit damping and modal damping (study 01, study 02) Equivalence of short impulse load to initial velocity (study 03, study 04) SDOF damped 38 01 02 Explicit damping 80Ns/m Equivalent modal damping 0.02 Results are identical 39 03 04 Impulse 4.77Ns Initial velocity 0.477m/s Results are identical except for the very beginning (can’t be seen in these graphs) 40 HOMEWORK 2 1. SDOF m=10kg, k=1000N/m is critically damped. Find a combination n of x0 and v0 that will make the SDOF cross the resting position. Prepare plot in Excel 2. SDOF m=10kg, k=1000N/m performs damped vibration. After 10s the displacement amplitude is 5% of the original amplitude. Find linear damping 3. Swing arm problem 4. SDOF m=10kg, k=1000N/m is at rest. At t=0 it is subjected an impulse 10Ns. What is the amplitude of displacement, velocity and acceleration? 41