1. ME 440 Intermediate Vibrations Th, Feb. 5, 2009 Section 2.2 and 2.6 © Dan Negrut, 2009 ME440, UW-Madison

2. Before we get started… Last Time: Started Chapter 2 – free response of a 1DOF system (Sec. 2.2) Discussed how to derive the EOM N2L, Energy Methods, Lagrange’s Equations and Hamilton Principle Mentioned the equivalent mass approach Today: HW Assigned: 2.106, 2.108, 2.109 (due on Feb. 12) Topics covered: How to solve EOMs once you obtain them Some analytical considerations regarding the nature of the response Undamped, underdamped, critically damped, and overdamped 2

3. Short Excursion: A Word on the Solution of Ordinary Differential Equations Classical analytic techniques Laplace transforms Numerical solution Usually found using MATLAB, or some other software package (Maple, EES, Sundials, etc.) MATLAB demonstrated later in this lecture (or next…) 3

4. ODE vs. IVP How is the concept of ODE related to that of IVP? Ordinary Differential Equation (ODE) Typically, has an infinite number of solutions Initial Value Problem (IVP) Is an ODE plus a set of initial conditions (ICs): The solution is indicated to assume at time T=0 a certain given value The solution for the IVP’s that we’ll deal with is UNIQUE It makes sense: only having the rate of change of a variable cannot tell you the value of the variable as a function of time 4

5. ODE: Infinite Number of Solutions ODE Problem: Initial Condition: y 0 =[-1000:50:1000] 5

6. ODE vs. IVP (Contd.) Example: consider a simple first order ODE It has an infinite number of solutions: However, if you specify an Initial Condition (IC) at time t=0, for instance, x(0) =2.5, then there is a UNIQUE solution that satisfies both the ODE and the imposed IC: Remember: 1. IVP = ODE + IC 2. IVP has a UNIQUE solution (unlike on ODE) 6

7. End short excursion. Back to regular business: Introducing  n &  EOM: 7 General form of the EOM for a one degree of freedom system: Compare the boxed forms of the EOM to understand how the natural frequency  n and damping ratio  are defined:

8. Some Quick Remarks EOM looks like (in standard form): 8 In Chapter 2, no applied external force present, that is, the right hand side is zero In fact, mass moves due to presence of nonzero initial conditions (ICs): You always have a set of two initial conditions One initial displacement and one initial velocity This is because we’re dealing with an IVP for a second order ODE

9. Motion Taxonomy In standard form, EOM looks like: 9 Motion taxonomy exclusively based on value of  : Undamped motion for  = 0Case 1 Overdamped motion if  >1Case 2A Critically damped motion if  =1Case 2B Underdamped motion if 0<  <1Case 2C

10. Case 1: Undamped Motion (  =0) Try a solution of this type: 10 Assume ICs at time t=0 are: That is, and are given to you (you know them) Based on chosen expression of x(t) and IC values, you get that That is,

11. Case 1: Undamped Motion (  =0) ~ Concluding Remarks ~ You can add the two harmonics to obtain only one harmonic and a phase angle that capture the response: 11 Solution could have been obtained by starting with an exponential form and substituting back into ODE This requires the solution of the Characteristic Equation (CE, see ME340): Then look for solution of the form (“exponential form”):

12. Short Excursion: HW Problem 2.35 Requires you to compute the natural frequency of a flywheel system (undamped system) In longitudinal direction With respect to torsional motion (vibration) In transversal direction In the end, it all boils down to computing the equivalent mass m eq and equivalent spring constant k eq, since then 12

13. Short Excursion Problem 2.35 (Cntd): Computing Equivalent Spring The tricky k eq is for transversal motion All information needed is in provided below 13

14. Short Excursion Problem 2.35 (Cntd): Computing Equivalent Spring You need the same info for the torsional motion and axial motion This info available in the collection of tables provided to you 14 Now Back To Original Business

15. Case 2: Damped Motion EOM 15 Trial Solution Characteristic Equation Roots

16. Case 2, Possible Subcases… Case A: Overdamped (“Supercritical”) Nature of Motion: Aperiodic 16 Case B: Critical Damping Nature of Motion: Aperiodic Case C: Underdamped (“Subcritical”) Nature of Motion: Periodic

17. Subcase 2A:  > 1 Solution can be found to assume form 17 Constants A 1 and A 2 found based on initial conditions at time t=0: Then, one gets

18. Subcase 2A:  > 1 (Contd) Bringing the expression of the solution x(t) to a “nicer” form This is only cosmetics… Use trick of the trade. Express A 1 and A 2 using two yet unknown variables B 1 and B 2 as follows: 18 Next, recall the definition of the hyperbolic sine and cosine: Then x(t) can be equivalently expressed as

19. Subcase 2A:  > 1 (Contd) An overdamped system does not oscillate, see picture next slide Such a system dissipates energy due to presence of damper As such, it should come to rest Theoretically, takes infinite amount of time to reach rest For mass-spring-damper system, plot on next slides displays time evolution of The generalized coordinate x(t) (upper plot) Time derivative, that is, the velocity of the mass (lower plot) Results obtained for (units used: SI) m = 2 c = 12 k = 8 19

20. Overdamped Response clear m = 2; c = 12; k = 8; posIC = 2.; velIC = 5.; Amat = [ 0 1 ; -k/m -c/m] Cmat = eye(2) sys = ss(Amat, [], Cmat, []) icConds = [posIC, velIC] initial(sys, icConds) Matlab Code: 20

21. Subcase 2B:  = 1 Characteristic Equation has a double root. Solution assumes form 21 Use initial conditions to get constants A 1 and A 2. Final form of x(t): Note that for critical damping, m, c, and k should be such that the following condition holds (it leads to  =1):

22. Subcase 2B:  = 1 (Cntd) An overdamped system does not oscillate, see picture on next slide Parameters used (units used: SI) m = 2 c = 8 k = 8 Since damper is present in system, energy dissipation occurs It takes infinite amount of time to come to rest However, it gets within any neighborhood of the equilibrium configuration faster then any overdamped system you compare against For mass-spring-damper system, plot on next slide displays The generalized coordinate x(t) (upper plot) Time derivative, that is, the velocity of the mass (lower plot) 22

23. Critical Damping Response clear m = 2; c = 8; k = 8; posIC = 2.; velIC = 5.; Amat = [ 0 1 ; -k/m -c/m] Cmat = eye(2) sys = ss(Amat, [], Cmat, []) icConds = [posIC, velIC] initial(sys, icConds) Matlab Code: 23

24. Subcase 2C:  < 1 Solution can be found to assume form 24 Nomenclature:  d – Damped Natural Frequency Don’t rush like last time to get A 1 and A 2 based on initial conditions… Rather, used the same trick with the B 1 and B 2 variables to massage the solution x(t) a bit

25. Subcase 2C:  < 1 (Contd) Solution then is equivalently expressed as 25 This is a better time to use ICs to get the constants B 1 and B 2 associated with your solution… Solution can also be equivalently expressed as

26. Quick Remarks Note that we get the undamped case by setting in expression of x(t) on previous slide the value of the damping ratio to be  =0 Kind of intuitive, no damping corresponds to case when   0 For the underdamped solution, amplitude of successive oscillations is decreasing exponentially. Specifically, like In typical structural systems,  ¿ 0.2. For all purposes,  d ¼  n Periods of successive cycles are the same; i.e., we have periodic motion with the following attributes: 26

27. Quick Remarks (Contd) Graphical representation of system response, underdamped case 27

28. Underdamped Example clear m = 2; c = 2; k = 8; posIC = 2.; velIC = 5.; Amat = [ 0 1 ; -k/m -c/m] Cmat = eye(2) sys = ss(Amat, [], Cmat, []) icConds = [posIC, velIC] initial(sys, icConds) Matlab Code: 28

29. Root Locations 29