SolSec 4.5 - Problems and Solutions for Section 4.5 (4.51...

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Problems and Solutions for Section 4.5 (4.51 through 4.60) 4.51 Consider the example of the automobile drive train system discussed in Problem 4.43. Add 10% modal damping to each coordinate, calculate and plot the system response. Solution: Let k 1 = hub stiffness and k 2 = axle and suspension stiffness. From Problem 4.41, the equation of motion with damping is 75 0 0 0 100 0 0 0 3000 10 000 11 0 13 2 02 2 0 00 0 0 1 + −− = () = () = [] ˙˙ , ˙ xx x0 x and m/s T Other calculations from Problem 4.41 yield: λω 2 3 12 3 77 951 8 8290 362 05 19 028 0 1537 0 1775 0 9721 0 8803 0 4222 0 2163 0 4488 0 8890 0 0913 == = = =− rad/s rad/s rad/s 2 3 .. . . . . . . . . . vv v Use the summation method to find the solution. Transform the initial conditions: qx 0 0 0 0 0 54 7723 M M T / / ˙ ˙ . Also, . 1 . 0 3 2 1 = = = ζ rad/s 932 . 19 rad/s 7848 . 8 3 2 = = d d ω The solution is given by qv v tc c t d e t i t di ii () =+ ++ 14 1 3 3 2 ζω ωφ sin where φ i di i T i T iii T i = () + () = tan ˙ , 1 0 23 vq (Eq. (4.114)) di i i T di i i = = ˙ cos sin , 0 Thus, 2642 . 0 3485 . 1 0 3 2 3 2 = = = = d d Now, v v 0 0 2 3 41 2 3 + = = cd d i i i i di i i sin ˙ sin cos
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Pre-multiply by v 1 T : vq 11 14 00 0 53 2414 T T c c () == ˙ . So, qv v v te t t e t tt =+ + −− 53 2414 1 3485 8 7848 0 2648 18 932 1 0 8829 2 1 9028 3 . . sin . . sin . .. The solution is given by xq x tM t e t e t () = () () = + +− 12 0 8829 1 9028 0 9449 1 1 1 0 1371 0 05693 0 005325 8 7848 0 01369 0 002349 0 0004407 18 932 / . . . . sin . . . . sin . m The following Mathcad session illustrates the solution without the rigid body mode The read solid line is the first mode with the rigid body mode included.
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4.52 Consider the model of an airplane discussed in problem 4.44, Figure 4.36. (a) Resolve the problem assuming that the damping provided by the wing rotation is ζ i = 0.01 in each mode and recalculate the response. (b) If the aircraft is in flight, the damping forces may increase dramatically to ζ i = 0.1. Recalculate the response and compare it to the more lightly damped case of part (a). Solution: From Problem 4.44, with damping 3000 0 0 0 12 000 0 0 0 3 000 53 820 53 820 0 53 820 107 640 53 820 0 53 820 53 820 , , ˙˙ ˙ ,, , ++ −− = xx x 0 C q q0 0 0 0 200 00 17 94 4 2356 26 91 5 1875 11 2 33 () = [] == . ˙ .. T m 0 rad/s rad/s rad/s 2 λω vvv 123 0 4082 0 8165 0 4082 0 7071 0 0 7071 0 5774 0 5774 0 5774 = = =− .
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SolSec 4.5 - Problems and Solutions for Section 4.5 (4.51...

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