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# Lecture 10

Lecture 10. Some comments: things you should know, and scaling. Problem strategies: understanding and solving problems. Den Hartog problem 50. Den Hartog problem 31. Den Hartog problem 41 (which you are handing in NOW). Den Hartog problem 56. Den Hartog problem 31 from “first principles”.

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## Lecture 10

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1. Lecture 10 Some comments: things you should know, and scaling Problem strategies: understanding and solving problems Den Hartog problem 50 Den Hartog problem 31 Den Hartog problem 41 (which you are handing in NOW) Den Hartog problem 56 Den Hartog problem 31 from “first principles” Rotating imbalance as forcing

2. Today’s office hours end at 5:00 Monday’s office hours will shift forward one hour for the rest of the semester Mondays 2:15-5:00

3. Some things engineers should know if you do, congratulations the mass of an object is its density times its volume (see Archimedes) the density of water is about 1000 kg/m3 and the density of air is about 1.23 kg/m3 the density of animals is a little less than that of water (most animals float) the density of steel is about 7800 kg/m3 and the density of aluminum is about 2700 kg/m3

4. A cubic foot of water weighs 62.4 lbs A gallon of water weighs about 7 lbs Gasoline is lighter (less dense) than water The pound is not a unit of mass — you have to divide by g (in the appropriate units!) Air drag is proportional to

5. Scaling We’ve taken mass out and have only length and time to play with pickt = 1/wn

6. Now we have a once for all problem with two dimensionless parameters: z and r The particular solution which we need to expand and sort out (we’ve done this before)

7. Now we need to map this back to the physical coordinates

8. The homogeneous solution (for the underdamped case) which we can also map back We can do initial conditions in either the scaled or physical world

9. We need the derivatives for the initial conditions

10. Here are the equations in scaled form and we can solve these

11. which completes the generic scaled problem

12. I leave it to you to see whether you like this or not. I discovered Tuesday night that maybe we don’t have quite as firm a grip on the material as I had hoped, so I want to spend some time on problems I’m going to start with the one that seemed to cause the most difficulty: #50

13. You have interpretations to make. The first thing to do is draw a picture. The picture is an interpretation of the first two sentences of the problem Let’s make the bar horizontal (it’s a harder problem if vertical) P0sin(wt) m c k

14. Now we are to find the equations of motion Is gravity important? NO P0sin(wt) m c k How many degrees of freedom? ONE q What’s a good variable?

15. How can we find the equations of motion? How about a torque balance around the hinge? P0sin(wt) m c k damping torque spring torque forcing torque

16. Inertial torque is Combine everything to get the equation of motion or divide by the inertia

17. The square of the natural frequency That turns out not to be what den Hartog is asking for He wants the damped natural frequency, although he does not say so We need to seek exponential solutions to the homogeneous equation

18. By small c he means that the argument of the square root is negative which agrees with the result in the back of the book

19. “Amplitude of the forced vibration” means the amplitude of the particular solution The easiest way to answer this question is probably to use complex notation

20. The square of the amplitude will be

21. At we’ll have displacement amplitude is 3l times that

22. QUESTIONS?

23. DH Problem 31 What do we know? Dimensions and materials of the two shafts Moments of inertia of the two outer gears The gear ratio 6/3 = 2 The moments of inertia of the shafts and the inner gears are negligible

24. What can/should we assume? The inner gears mesh perfectly, so that the rotation rate of the second is twice that of the first That all the gear angles are equal to zero in the neutral position

25. We can do this following the formulas

26. modify it according to the text I’ll derive this eventually

27. If you use the numerical formula following formula (13) you will get the answer in the back of the book

28. This seems like a difficult and confusing problem I solved it by copying from the book, but that’s not very satisfying although it’s acceptable . . . Let’s review the simpler problem on pp. 28-29 of Den Hartog Here’s an approximation of Figure 2.6

29. The shaft has a spring constant that we can find using the Appendix for now let’s just call it k. Each gear has a moment of inertia I There’s no damping The equations of motion This looks like a two degree of freedom problem, and indeed it is but . . . We can define the two degrees of freedom to be: the gears moving together the gears moving in opposite directions This is an example of mode “shapes”, of which more later ONLY THE LATTER IS INTERESTING

30. Divide 1A by I1 and 1B by I2 Take the difference 1A – 1B The natural frequency for this mode of oscillation is determined from

31. The formula for k is They don’t give us any numbers; this is just a symbolic exercise.

32. I’d like to do two more problems before we plunge into the intricacies of problem 31

33. DH Problem 41 (which you have done for homework, and for which the solution is available on line) m c f(t) k

34. They give you k = 20 lb/in, W = mg = 10 lb and a decay rate of 1% per cycle Part (a) The natural frequency is k/m, and the only issue here is one of units — we are in US customary (USC), and we need lengths in inches m = W/g = 10/32.2/12 = 0.02588 — not slugs, but USC mass anyway Natural frequency is then the square root of k/m. Putting in the numbers gives us 27.80 rad/sec

35. Part (b) They tell you that the amplitude decreases by 1 % for each full cycle This gives a decrement ratio of d = 1/0.99, the log of which is 0.01005 This is small enough that we can use the simplified formula (p. 40) z = c/c0 = ln(d)/2π = 0.00160 c = 2zwnm, which, with the numbers found so far, is 2 x 0.00160 x 27.8 x 0.02588 = 0.00230

36. Part (c) Equation 2.28a in den Hartog gives the response amplitude in terms of the forcing we’re driving at resonance, so this is zero So we have P0 = cwx0 = 0.00230 x 27.80 x 1 = 0.0639 lbs

37. Part (d) is most easily answered after we have done part (f) so I’ll come back to it. Part (e) is easy. The problem is linear, so doubling the force will double the eventual response

38. Part (f) We know that the transient dies out at exp(-zwnt), so we expect the amplitude to do the same That gives us amplitude = 2 - exp(-zwnt), or, in Den Hartog’s language 2 - exp(-ct/2m) (Recall that c = z/2wnm, so that zwn = c/2m)

39. Part (d) We can find the decay rate by differentiating the answer to part (f) c/2m = 0.0444 per second There are 27.8 rad/sec, hence 4.425 cycles per second decay rate is then 0.0444/4.425 ≈ 0.01 per cycle Note that this agrees with the initial information we were given

40. QUESTIONS?

41. DH Problem 56 “f is small” means we can linearize, which I’ll do in a couple of slides “the blade is stiff” means we can treat it as rigid

42. (This is really sort of a centrifugal pendulum)

43. Does gravity count in this problem?? NO Why not?

44. Draw our own picture and state any assumptions we are making b We also assume that the hinge is perfect, and that the only motion the blade can perform (except for the spin) is up and down

45. r denotes the radial distance from the rotation axis

46. we can linearize sinf —> f integrate on z

47. QUESTIONS?

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