SolSec 4_5 - Problems and Solutions for Section 4.5 (4.56...

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Problems and Solutions for Section 4.5 (4.56 through 4.66) 4.56 Consider the example of the automobile drive train system discussed in Problem 4.44. 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 4.44, the equation of motion with damping is 75 0 0 0 100 0 0 0 3000 ! " # # # $ % !! x + 10,000 1 1 0 1 3 2 0 2 2 ! " # # # $ % x = 0 x 0 ( ) = 0 and ! x 0 ( ) = 0 0 1 ! " $ % T m/s Other calculations from Problem 4.44 yield: ! 1 = 0 " 1 = 0 rad/s 2 = 77.951 2 = 8.8290 rad/s 3 = 362.05 3 = 19.028 rad/s v 1 = 0.1537 0.1775 0.9721 # $ % % % ( ( ( v 2 = ) 0.8803 ) 0.4222 0.2163 # $ % % % ( ( ( v 3 = 0.4488 ) 0.8890 0.0913 # $ % % % ( ( ( Use the summation method to find the solution. Transform the initial conditions: q 0 ( ) = M 1/2 x 0 ( ) = 0 ! q 0 ( ) = M 1/2 ! x 0 ( ) = 0 0 54.7723 ! " # $ T Also, 1 = 2 = 3 = 0.1. d 2 = 8.7848 rad/s d 3 = 18.932 rad/s The solution is given by q t ( ) = c 1 + c 2 t ( ) v 1 + d i e ! i # i t sin di t + $ i ( ) v i 2 3 % where i = tan " 1 di v i T q 0 ( ) v i T ! q 0 ( ) + i i v i T q 0 ( ) % ( ) * i = 2,3 Eq. (4.114) d i = v i T ! q 0 ( ) di cos i # i i sin i i = 2,3 Thus, 2 = 3 = 0 d 2 = 1.3485 d 3 = 0.2642
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Now, q 0 ( ) = c 1 v 1 + d i sin ! i v i i = 2 3 " ! q 0 ( ) = c 2 v 1 + # $ i % i d i sin i + di d i cos i ( ) i = 2 3 " v i Pre-multiply by v 1 T : v 1 T q 0 ( ) = 0 = c 1 v 1 T ! q 0 ( ) = 53.2414 = c 2 So, q t ( ) = 53.2414 v 1 ! 1.3485 e ! 0.8829 t sin 8.7848 t ( ) v 2 + 0.2648 te ! 1.9028 t sin 18.932 t ( ) v 3 The solution is given by x t ( ) = M ! 1/2 q t ( ) x t ( ) = 0.9449 t 1 1 1 " # $ $ $ % ! ! 0.1371 ! 0.05693 0.005325 " # $ $ $ % e ! 0.8829 t sin 8.7848 t ( ) + 0.01369 ! 0.002349 0.0004407 " # $ $ $ % e ! 1.9028 t sin 18.932 t ( ) m The following Mathcad session illustrates the solution without the rigid body mode (except for x 1 which shows both with and without the rigid mode) The read solid line is the first mode with the rigid body mode included.
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4.57 Consider the model of an airplane discussed in problem 4.47, Figure P4.46. (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 4.47, with damping 3000 0 0 0 12,000 0 0 0 3,000 ! " # # # $ % !! x + C ! x + 13455 13455 0 13,455 26910 13,455 0 13,455 13,455 ! " # # # $ % x
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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SolSec 4_5 - Problems and Solutions for Section 4.5 (4.56...

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