Lecture 14: Forced Damped Two Degree of Freedom Systems

1. Lecture 14: Forced Damped Two Degree of Freedom Systems For any system we need to build a model figure out the natural frequencies figure out the homogeneous solution(s) figure out how to get particular solutions We started with undamped, unforced 2DOF systems figured out the natural frequencies figured out the homogeneous solution(s) We added forcing Now we need to add damping

2. But first, a useful digression/review

3. Let’s think a little more about now to do a real problem Figure out what the problem is Draw a picture Write some equations Solve the equations and interpret the results

4. Figuring out what the problem is requires reading a paragraph and/or listening to a client You are looking for degrees of freedom, masses, springs, dampers, forcing . . . anything you can use to build a (mathematical/engineering) model Will you care about initial conditions or not? Did the paragraph or client say anything to imply they do? In a face to face discussion, ask. This can be the hardest part of the whole process You can’t answer a question if you don’t understand it.

5. You more or less know how to draw a picture Be neat and clear. It’s not just for you now. It’ll help you explain the meaning of your results later Some of the DH problems don’t have pictures — drawing pictures for them is a helpful exercise

6. Right now the only way we now how to write is equations is to use FBDs If you’ve drawn your picture correctly, the FBDs follow pretty directly The net force on each element is equal to its mass times its acceleration or we may have a torque balance with the rate of change of angular momentum The net force was zero in statics; in dynamics it equals the rate of change of momentum or the net torque, zero in statics, equals the rate of change of angular momentum When in doubt, put in all the forces and/or torques if something is irrelevant, then you can ignore it later if it’s important and you left it out, you may be doomed.

7. We’ve mostly looked at harmonic right hand side proportional to sin(wt+f), with the phase almost always equal to zero If the right hand side is harmonic, the left hand side must be harmonic This in turn means that the particular solution must be harmonic but it need not have the same phase In general

8. The right hand side goes like sin(wt); so must the left hand side. If there is dissipation, there will be sines and cosines on the left. You have to add them up to make the cosines vanish and the sines match This will determine the vectors a and b

9. For a one degree of freedom system we have the scalar equivalent of this Den Hartog §2.8 (pp 47-55) explores this in great detail I highly recommend it We are going to do a two degree of freedom problem with dissipation during today’s lecture

10. If the initial conditions are important If you need to know frequencies and decay rates Then you need the homogeneous solution(s) These satisfy the same differential equations, but with the right hand side equal to zero Now we have the comment I have made MANY times Homogeneous differential equations with constant coefficients have exponential solutions

11. So now let us review a 1DOF problem we have seen to identify the steps

12. Review DH problem 50 50. A weightless , stiff bar is hinged at one end. At a distance l from the hinge there is a mass m, at a distance 2l there is a dashpot c, and at a distance 3l there is a spring k and an alternating force P0sin(wt). Set up the differential equation. Assuming small damping (but not zero damping), calculate the natural frequency; the amplitude of forced vibration of the spring at the natural frequency and at half the natural frequency. P0sin(wt) m c k

13. We want a torque balance for this problem The forces (in the z direction) are: The corresponding lever arms are: P0sin(wt) So the torque is m k O c

14. The moment of inertial is ml2 so we have the torque balance We need to linearize this for small q P0sin(wt) m k c

15. We do the usual for the particular solution Eitherq = Acos(wt) + Bsin(wt) Or, replace sin(wt) by –jexp(jwt) and let q = Qexp(jwt) Let’s try the second approach, but first, we need to get rid of gravity by writing then the differential equation becomes

16. I don’t need the primes any more, so I’ll drop them and continue with the second approach becomes from which

17. There are more things that you need to do: Clarify the meaning of the particular solution Find the homogeneous solution, and satisfy the initial conditions if necessary becomes and you know what to do from this point on

18. Here’s what we did once more Figured out what the problem is Drew a picture Wrote some equations Solved the equations (and interpreted the results)

19. QUESTIONS? Let me start with a specific 2 DOF example with damping before we go to a general discussion

20. AUTOMOBILE SUSPENSION SYSTEM

21. A suspension has at least two aims: To keep the tires on the ground for control of the car To minimize the motion of the body for passenger comfort We looked at a one quarter model of a car earlier and we’ll stick with that We consider: one quarter of the body, the tire/axle system, a coil spring and a shock absorber, including the elastic behavior if the tire It’s a two degree of freedom system

22. From Gillispie, TD Fundamentals of Vehicle Dynamics (1992) We can fit this to a general 2 DOF vertical model

23. The general 2 DOF system in a vertical orientation c2 k2 z2 Start adjusting it to make it look like Gillespie f2 m2 c3 k3 f1 z1 m1 c1 k1

24. c2 k2 z2 f2 m2 c3 k3 f1 z1 m1 c1 k1 Finally add ground motion zG

25. Gillespie Us z2 m2 k3 c3 z1 m1 k1 zG

26. A spring tire model

27. Typical shock absorbers

28. Free Body Diagrams m2 m1 Be careful with the arrows and the signs I suppose z to be positive up

29. Collect the forces on each mass and write Newton’s law which we can rearrange into differential equations with left and right hand sides

30. The differential equations, supposing z positive up We want to know both the (damped) natural frequencies and the response to rough roads This means a homogeneous and a particular solution I’ll do the homogeneous solution first — this is new

31. Start with the homogeneous solution by setting zG to zero Seek exponential solutions: replace each dot by an s and the variable by a constant Z

32. Put in matrix form Now we have the matrix eigenvalue problem again This is quite messy in general, so let us put in some realistic numbers

33. Some numbers from Gillispie, TD Fundamentals of Vehicle Dynamics (1992) 5.0 L Mustang front suspension (year unspecified) Tire spring constant: 1198 lb/in Spring spring constant: 143 lb/in Weight: 957 lb “The 40% damping ratio is reasonably representative of most cars” It’s not clear to me exactly what this represents, but I’m guessing it refers to a damping ratio with respect to the body mass-spring spring system

34. If we are to do this correctly, we need to guess the effective weight of the wheel-axle-tire segment: 75 lb, say Standard US customary gravity acceleration is 386 in/s2; we take mass to be weight in pounds divided by 386 We take c = 2zwm ≈ 2z√(km), where we use k from the spring spring and m of the vehicle. In our case this will be 2 x 0.4 x √(143 x 2.479) = 15.06 lb-sec/in

35. Put these numbers into our picture z2 m2 = 2.479 lb-sec2/in m2 k3 c3 k3 = 143 lb/in c3 = 15.06 lb-sec/in z1 m1 = 0.194 lb-sec2/in m1 k1 = 1198 lb/in k1 zG

36. The determinant with the parameters substituted in is expanding It has four roots, which come in complex conjugate pairs: -39.3025 ± j70.4975, -2.5494 ± j6.9412 rad/sec

37. The eigenvectors are complex We find them in the same way as before The eigenvalues are complex conjugates so, too, are the eigenvectors

38. If we normalize the eigenvectors, which allows us to get a better picture of the relative amplitudes we have We can focus on the real parts and they tell us that the high frequency motion is concentrated in mass 1, the tire the low frequency motion is concentrated in the body The axle bounces at a higher frequency and is more strongly damped the driver will not be aware of this

39. The general homogeneous solution will be where the constant coefficients will be complex The complex solution looks magic, and I do not have a nice simple physical explanation Let’s move forward cautiously and see how this works in practice trusting to the ritual

40. We find the coefficients by selecting an initial condition an impulsive loading can represent a RR track, say A unit impulse from the ground will give initial conditions substituting the homogeneous solution gives me four inhomogeneous algebraic equations There’s a lot of algebra involved, so I used Mathematica

41. find the constants assign them The general homogeneous solution

42. y looks complex, but it is actually real as we can see below fixes roundoff error I plot this response on the next slide

43. The response to the unit impulse tire motion body motion

44. QUESTIONS? On to forced motion

45. THE PARTICULAR SOLUTION We care about the “steady state” response to a rough road, which we can model as a harmonic forcing We suppose the ground motion to be harmonic at frequency w zG = Z0sin(wt) And seek the particular solution I will address some general issues before putting the numbers in

46. The response will also be harmonic at the same frequency — we have a choice as to how to represent it Trigonometrically: Exponentially with the accompanying change

47. The differential equations, supposing z positive up Divide each by its appropriate m

48. Now, if I use the trigonometric approach I will have to write

49. If we substitute these into equations (1) and (2) we get a pair of equations— each has a coefficient multiplying the sine and one multiplying the cosine Each of these four coefficients must vanish independently This is pretty messy, and I will go to Mathematica to sort this out

50. The four coefficients that must vanish can be collected as coefficients multiplying the constants to be determined