1 / 88

I’ll be running late Tuesday — dental appointment

I’ll be running late Tuesday — dental appointment. Lecture 10: More Forcing (plus). Things you should know. Scaling. Rotating imbalance as forcing. Vibration of a gear train (from DH Problem 31). Impulsive loading. Problem strategies: understanding and solving problems.

aldon
Download Presentation

I’ll be running late Tuesday — dental appointment

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. I’ll be running late Tuesday — dental appointment

  2. Lecture 10: More Forcing (plus) Things you should know Scaling Rotating imbalance as forcing Vibration of a gear train (from DH Problem 31) Impulsive loading Problem strategies: understanding and solving problems Den Hartog problems 41, 50, 56

  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 How to write problems independent of dimensions, minimizing the number of parameters The remaining parameters are dimensionless At this point, we have one of these: the damping ratio You had a bunch of these in fluids and heat transfer: Reynolds number, friction factor, Rayleigh number, Grashof number . . . I’ll go at this slowly in the context of a problem we are familiar with

  6. The generic problem m c f k Divide by m to get a partially nondimensional equation

  7. If I can make the equation fully nondimensional, then I can solve it once for all Scaling We’ve taken mass out and have only length and time to play with pickt = 1/wn

  8. suppose that we have a harmonic forcing select

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

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

  11. 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

  12. We need the derivatives for the initial conditions

  13. Here are the initial conditions in scaled form and we can solve these

  14. which completes the generic scaled problem

  15. We combine all this to get the scaled solution

  16. QUESTIONS? Now let’s look at more forcing mechanisms

  17. Rotating imbalance m w denote the offset distance by a0 M M includes the yellow mass and the blue rotating mass but not the black imbalance mass

  18. Centripetal force Suppose the system to be constrained from side to side motion We get a standard undamped equation

  19. Make the usual assumption for an undamped 1 DOF system solve for the amplitude This is the particular solution

  20. We don’t much care about the transient if we are running the unbalanced machine all the time m What is the force on the ground? w M

  21. All we have is a spring force f = kz where z denotes the particular solution we’ve just found. We get

  22. For w small compared to the natural frequency For w large compared to the natural frequency

  23. Plot with wn equal to unity (without loss of generality)

  24. Let’s look at this with dissipation in the underdamped case The acceleration is given by and it is harmonic at sine The particular solution is for an acceleration A

  25. so we have We care about the frequency-dependent part

  26. We can find the amplitude easily enough We see that this looks a lot like the undamped formula, as it must (In looking at the amplitude we miss the change in phase through resonance)

  27. z = 0.1

  28. We can plot the amplitude of the undamped and the damped on the same graph At this small damping ratio, there is little difference outside the resonance.

  29. QUESTIONS? Vibrations of a gear train: one degree of freedom model

  30. Label the outside gears 1 and 2 from left to right and the inside gears 3 and 4 from bottom to top Label the shafts 1 and 2 from left to right 2 4 2 1 3 1 Denote the diameters of the transmission gears by D3 and D4, respectively

  31. We’ll have k1 and k2 associated with the shafts The gear ratio n = D3/D4 (= 2) The gear box is “perfect” gear 4 turns at –n times the speed of gear 3

  32. We have equations of motion for all four gears, but the gears in the transmission have negligible inertia, so we can address them statically with free body diagrams The main gear equations are If we choose the proper common origin, q4 = -nq3

  33. Free body diagrams The transmission t2 4 F4/3 F3/4 3 t1

  34. Write the equilibrium equations for the two gears

  35. Plug in for the two torques Solve for q3

  36. substitute for q3 divide by the inertias

  37. subtract (4) from n times (3) This is now a one degree of freedom problem and we see that which is equivalent to the formula in Den Hartog

  38. QUESTIONS? What happens to our standard system if we hit it?

  39. Impulsive forcing m c k

  40. integrate from just before the interaction to after the interaction mean displacement mean force the system is at rest before the interaction, so change in momentum

  41. negligible We can solve these problems as homogeneous initial value problems the mass of the struck object not the mass of the striker generally supposed to be zero I’ll do that

  42. The momentum transferred depends on the nature of the collision The collision takes place quickly enough that the sum of the momentum of the struck mass and the striking mass is conserved If perfectly elastic, the momentum transferred is twice the hammer momentum Look at a scaled result: Dp = 1, m = 1,wn = 1, z = 0.1

  43. Mathematica code to solve the initial value problem

  44. maximum displacement is 0.8626 at t = 1.4780

  45. Mathematica code to find the amplitude and location of the peak We can extend the displacement picture by plotting out to two (or more) actual periods for various values of the damping ratio

  46. z= 0.2

  47. z= 0.4

  48. z= 0.6

  49. You can “undo” the scaling by multiplying the displacement by Dp/m dividing the time by wd

  50. Let me look at how this connects to what we might think of as an impulsive load Start with the differential equations and set the forcing equal to a delta function at t = t0 with zero initial conditions

More Related