1 HOMEWORK 1 1.Derive equation of motion of SDOF using energy method 2.Find amplitude A and tanΦ for given x 0, v 0 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

2 RING Ring.SLDASM Energy method

3 ω n, x 0, v 0 fully define free undamped vibration

4 MODULE 02 DAMPED VIBRATION Inman (3 rd edition) section 1.3

5 SINGLE DEGREE OF FREEDOM SYSTEM WITHOUT DAMPING mass stiffness 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. Notice that mass and stiffness are completely separated.

6 Viscous damper, damping force is proportional to velocity SINGLE DEGREE OF FREEDOM SYSTEM WITH DAMPING Notice that mass, damping and stiffness are completely separated.

7 Equation of motion We assume solution in the form of:

8 Critical damping Damping ratio (modal damping) Characteristic equation Determinant of characteristic equation

9 Equation is linear, therefore the sum of solutions is also solution

10 Two forms of solution of free under- damped vibration Damped natural frequency or, using Euler’s relations:

11 Inman p. 24 A and Ф are calculated from initial conditions ω d - natural frequency of damped oscillations

12 This is a non-oscillatory motion

13 Damping is critical when:

14 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

15 Damped vibration amplitude decay.xls

16 SDOF damped.SLDASM m =10kg k = 1000N/m c = 20Ns/m

17 Linear damper 20Ns/m Modal damping 10%

, , t1 = 0.62sa1 = t2 =1.23sa2 = 30.30

19 swing arm.SLDASM

20 m =0.56kg Assume that beam mass is negligible k L =2000N/m c L =10Ns/m Is this system under damped, critically damped or over damped? L 2 =0.1m L 1 =0.2m

21 System is underdamped, modal damping is 7.4%

22 Modal damping 7.4%Linear damper 10Ns/m

23 ζ = 5% ζ = 10% ζ = 200% ζ = 100% ζ = 50%

24 SYSTEMDamping ratio  (% of critical damping) METALS0. 1% CONTINUOUS METAL STRUCTURES2% - 4% METAL STRUCTURES WITH JOINTS3% - 7% ALUMINUM/STEEL TRANSMISSION LINES~0.4% SMALL DIAMETER PIPING SYSTEMS1% - 2% LARGE DIAMETER PIPING SYSTEMS2% - 3% SHOCK ABSORBERS30% RUBBER~5% LARGE BUILDINGS DURING EARTHQUAKE1%-5% PRE-STRESSED CONCRETE STRUCTURES2% - 5% REINFORCED CONCRETE STRUCTURES4%-7% TIMBER5% - 12% TYPICAL VALUES OF DAMPING RATIO

25 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. MODAL SUPERPOSITION METHOD

26 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.

27 TIME AND FREQUENCY RESPONSE ANALYSIS

28 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.

29 TIME RESPONSE ANALYSIS 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. time F(t) Examples of load time history in Time Response analysis. F(t)

30 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. Example of load time history that can be used in Frequency Response analysis. time Force amplitude does not have to be constant

31 Different ways to illustrate a unit impulse time force Impulse load An impulse applied to a SDOF is the same as applying the initial conditions of zero displacement and initial velocity v 0 = F Δt /m (Δt here is duration of “square” impulse, Δt is very short) ΔtΔt TIME RESPONSE DUE TO IMPULSE LOAD

32 To model impulse load we set: x 0 = 0 v 0 = F Δt /m TIME RESPONSE DUE TO IMPULSE LOAD

33 Analytical solution of unit impulse excitation problem. Response of a system due to an impulse at t = 0 Inman p. 195 is force impulseWhere TIME RESPONSE DUE TO IMPULSE LOAD

34 TIME RESPONSE DUE TO IMPULSE LOAD Impulse load response.xls

35 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 = , T=0.03 TIME RESPONSE DUE TO IMPULSE LOAD

36 The critical damping for SDOF is: 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. TIME RESPONSE DUE TO IMPULSE LOAD

37 Results of Dynamic Time analysis; displacement time history Location for Time History graph TIME RESPONSE DUE TO IMPULSE LOAD

38 SDOF damped 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)

39 01 Explicit damping 80Ns/m 02 Equivalent modal damping 0.02 Results are identical

40 04 Initial velocity 0.477m/s 03 Impulse 4.77Ns Results are identical except for the very beginning (can’t be seen in these graphs)

41 HOMEWORK 2 1. SDOF m=10kg, k=1000N/m is critically damped. Find a combination n of x 0 and v 0 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?