Lecture 19: Damping in the Euler-Lagrange Formulation

1. Lecture 19: Damping in the Euler-Lagrange Formulation

2. Let’s bring back our friend Figure 3.1 from Den Hartog f1 f2 k1 k3 k2 m1 m2 c1 c3 c2 y2 y1

3. We know how to write the differential equations for this using the Euler-Lagrange method y1 and y2 are obvious choices for generalized coordinates The energies The Lagrangian And we can go on to steps 5-8

4. For q1

5. For q2

6. This problem is nice in that linearization is unnecessary We have two types of forces: the applied forces f1 and f2 and the forces caused by the motion of the dampers For now we’ll treat them the same way — by looking at virtual work

7. Move block 1 f1 f2 k1 k3 k2 m1 m2 c1 c3 c2 f1 does work, as do the two dampers, c1 and c3 y2 y1

8. Move block 2 f1 f2 k1 k3 k2 m1 m2 c1 c3 c2 f2 does work, as do the two dampers, c2 and c3 y2 y1

9. The Euler-Lagrange equations Generalized forces We can rearrange these to make them look more familiar

10. Notice that the form of the spring terms and the damping terms are the same (except for the differentiation, of course) Lord Rayleigh noticed this and defined The Rayleigh dissipation function

11. The Rayleigh dissipation function (for this problem — a model) The modified Lagrange equations And you can see that it works

12. We form the Rayleigh dissipation function exactly as we form the spring potential energy find the dampers figure out how the dampers work write down the individual components of the RDF We add this to our methods of deriving dynamical equations

13. The ritual Find the kinetic and potential energies Find the Rayleigh dissipation function Figure out the generalized coordinates Use the method of virtual work to find the generalized forces Find the Euler-Lagrange equations

14. A general problem There may be forcing The initial conditions may be important To be general you need to find a homogeneous solution find a particular solution use the initial conditions to combine them

15. Simple example of how to solve once you have the equations The initial value problem The homogeneous equation Its solution where The particular equation Its solution

16. These are connected by the initial conditions

17. THE OVERHEAD CRANE WITH DAMPING

18. OVERHEAD CRANE add damping, c y1, f1 M q m (y2, z2)

19. The governing equations were We added the generalized forces Now we need the Rayleigh dissipation function

20. The damper works when the angle changes, but not when the cart moves So, the Rayleigh dissipation function for this problem is

21. We can linearize and revisit the problem for forced motion of the cart with harmonic forcing and initial conditions

22. It is my contention that once we introduce damping the exponential approach makes the most sense We haven’t looked at that in a while but it’s going to be more and more important as we go along so here we go . . . For the particular solution replace and take the real part at the end For the homogeneous solution, just use

23. The homogeneous problem Write the big difference Drop the forcing and replace each dot by an s We can cancel the time function, as usual

24. Write the algebraic equation for the constants in matrix form The determinant must vanish Again this is complicated, and it makes sense to put in numbers we’ll use the same numbers for the masses and the length as before m = 50, M =250 and l = 3

25. The nonzero roots are I want to choose c such that I have light damping The critical value of c = √(2207250) = 1485.68 . . . I will choose 150 N-m/s The nonzero values of s are then s = -0.2 ± 1.97079j Note that s2 = s1*

26. We can find the modal vectors in the usual way. The matrix from which we find them is The generic modal vector is (by direct substitution) Since modal vectors have no amplitude, I can factor out the s2

27. The oscillatory modal vector is the same for both, which makes sense, because there is only one “frequency” The modal vector for s = 0 is the same as it was before The general homogeneous solution becomes the “zero modal function”

28. Or, putting in for the modal vectors, The time derivative, which we’ll need for the initial conditions, is

29. This problem is simple enough that we can do the initial conditions by hand Suppose we start from rest with q0 = π/20 radians (= 9 °)

30. The positions From which The speeds From which Note that b0 is zero we’ll talk about this more later

31. The A equations Rewrite in matrix form Solve Note that

32. We can combine all of this to write the solution to the unforced initial value problem

33. We can actually put all the complex stuff in and sort out the result

34. The j times cosine terms cancel The wd times j sine terms cancel We are left with a purely real solution

35. Pick out the components The cart has a permanent offset of π/40 = 0.0785

36. We can plot the position (in red) and the angle (in blue)

37. What did we do? We found the equations of motion with forcing and damping We found the homogeneous solution by setting the forcing equal to zero the (complex) natural ”frequencies” the eigenvectors We applied a set of initial conditions found the constants in the general homogenous solution We did a lot of algebra to beat the answer into real form Note that we allowed everything to be complex and we got real results automatically because the initial conditions were real

38. We want to go on and work the entire problem We’ll be able to make use of much of what we have already done Specifically, the homogeneous solution will reappear. First we need to find a particular solution Our equations were For the particular solution replace and take the real part at the end

39. All the exponentials cancel and we are left with a matrix equation for the coefficients

40. invert to get where

41. Again this is complicated, and it makes sense to put in numbers we’ll use the same numbers as before m = 50, M =250, c = 150, and l = 3

42. We put it all together and we get the particular solution Actually taking the real part on screen would be remarkably tedious

43. I’m going to look at the results, and then we can go to Mathematica and get a feel for how we got them The particular solution for P0 = 100 and wf = 1 y is in red; q is in blue

44. We can start the system from rest The curious difference between the homogeneous initial value problem, the pure particular solution, and the actual case is in the parameter b0 which equals 1/3. This means that we have a steady motion of the cart

45. Most of the results are in the Mathematica notebook but the “resonant” result is perhaps interesting

46. But the pendulum has a large swing before the damping can kick in to limit it 10° is a smallish angle, the equations are valid and the result is probably correct

47. WE QUIT HERE. FEEL FREE TO GO ON.

48. Let’s look at some more examples

50. You can see from the diagram that the author of this example uses FBDs He also omits any force applied to the lathe from its operation To do this right we would have the lathe driven by an electric motor and there would be a load on the work piece on the lathe Motors are self-limiting in speed, so there’d be a maximum rotation rate and we could see oscillations as the cutting force is varied. This is more than we have the skills to do at the moment so we’ll follow along with the problem as posed