Lecture 8: Some Alternate Excitations

1. Lecture 8: Some Alternate Excitations General forcing Impulsive loading Support motion (air bag sensor speed bump) First a bit of review so we know where we are

2. The story so far . . . We can reduce one degree of freedom systems to a single problem

3. You know how to figure out the natural frequency using energy You know how to find the damping ratio when it is less than unity

4. Mass-spring-damper system model m c ma(t) k which we have made independent of the mass by defining the natural frequency and the damping ratio

5. Look at setting up two examples

6. possible rotational damping torque applied here rotational spring Let the angle q be the single degree of freedom torque balance standard equation where

7. rotational damping applied torque let q denote the 1 DOF Balance the torque

8. linearize divide by I and rearrange standard form

9. Procedure Overview in Words You have an equation (an ode) and a set of initial conditions Find the homogeneous and particular solutions to the ode The particular solution takes care of the forcing term Use the SUM of the particular and homogeneous solution to satisfy the initial conditions

10. Procedure Overview in Symbols Find yH such that This gives you two arbitrary constants Find yP such that There is no arbitrary constant here — we “use up” the forcing amplitude

11. Combine the two solutions to deal with initial conditions Determines the two arbitrary constants in the homogeneous solution. These depend on the forcing amplitude through the particular solution as well as on the specific initial conditions

12. QUESTIONS?

13. We know how to find the particular solution for harmonic forcing Suppose that the forcing is not harmonic? We can find solutions for some other simple forcings: constant, A0 linear, A1t quadratic, A2t2 In fact, we can find the particular solution for any power but I don’t want to

14. constant linear quadratic In fact we can find the particular solution for any forcing

15. What does the function f have to be? We can find out by substituting into the differential equation

17. If f satisfies the homogeneous differential equation the integral vanishes If f(0) = 0, two more terms vanish If we get the right hand side f must be the solution to the homogeneous problem with the initial conditions

18. We aren’t going to use this much now but the basic idea of an integral solution will reappear QUESTIONS?

19. Impulsive forcing m c k

20. 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

21. 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

22. 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

23. Mathematica code to solve the initial value problem

24. maximum displacement is 0.8626 at t = 1.4780

25. 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

26. z= 0.2

27. z= 0.4

28. z= 0.6

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

30. Support Motion m c f k We know what to do if f is zero or a harmonic function, assuming the wall is fixed What do we do if f = 0, but the wall is not fixed?

31. m c k y+yW yW y denotes the relative motion of the mass with respect to the wall

32. force balance rearrange divide by m and so we have our standard problem the motion is forced inertially

33. Note that we can define y differently, as an absolute y m c k y yW This changes the governing equation

34. force balance rearrange divide by m Either formulation is fine — choose the one you fancy One may be better in terms of the meaning of the output The first formulation may make more sense for an air bag sensor

35. Air Bag Sensors MEMS devices: a cantilever beam with a mass some damping associated with the gas in the device These are basically accelerometers I want to spend some time discussing these

36. Model as a simple mass-spring system damping is not particularly important here I will put in a damping ratio of 0.1 eventually The air bag is triggered if the distance between the mass and the far wall is less than some critical distance, measured capacitively, perhaps A SPECIFIC DISPLACEMENT TRIGGERS THE AIR BAG

37. We can get the spring constant from beam bending theory The end mass puts a load on the beam and we can look at (This is also in Den Hartog, p. 429.)

38. We can analyze this using the support motion equation we just derived with y as the motion relative to the support Consider a constant deceleration and neglect the damping for clarity (I’ll put it back for numerical results) The vehicle is decelerating, so a is actually a negative number

39. particular solution homogeneous solution initial conditions

40. Plot the simple result peak response atwnt = π

41. What really happens? What can we say about designing one of these? Let’s take a look at a couple of frontal crashes

42. airbags?

44. The crash takes some time to happen — we can estimate how long from deceleration The air bag sensor takes some time to activate the air bag The air bag takes some time to deploy (I found the number 33 msec somewhere) First let’s look at the response with some damping, z = 0.1 This only reduces the homogeneous frequency by a factor of 0.995 It reduces the maximum deflection to 1.729a/wn2

45. Displacement vs. wt for a/wn2 = 1

46. For the designer Ignition takes place at t ≤ π/wn into the crash Maximum deflection is 1.729a/wn2 (for z = 0.1) So, wn has to be large, and the design deflection small And, the critical acceleration a must be larger than anything one would expect under normal operation

47. Questions arise about the timing of all this In a crash the car deforms, so it takes some time for the cabin to come to rest I’ve been able to find that airbags deploy in about 33 msec I don’t know if that includes the sensor reaction, but I think not

48. Crash deceleration criteria Crash threshold should be much higher than normal braking or minor parking lot encounters The following cannot happen, which is sort of too bad http://www.youtube.com/watch?v=u2zgr_Gg11U Crash deceleration data are not easy to come by typically tests are barrier tests at some impact speed accelerometer data are somewhere, but I’ve not done well searching

49. Deceleration from normal braking depends on the brake and tires — 1 g is pretty good (Federal standard asks for a minimum of an effective friction coefficient of 0.625 — deceleration at 0.625g.) Let’s say that we will set a threshold at 7g (I’ve seen suggestions of 7 g) Let’s set the sensor to trigger at 2 mm so we want a deflection of 2 mm at a deceleration of 7 g

50. Time to trigger will be (faster for higher speed crashes) which is comparable to statements I have found The bag takes 33 msec to deploy (an average value) Look at sensor performance for various deceleration values