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SolSec 4.4 - Problems and Solutions for Section 4.4(4.41...

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Problems and Solutions for Section 4.4 (4.41 through 4.50) 4.41 A vibration model of the drive train of a vehicle is illustrated as the three-degree- of-freedom system of Figure P4.41. Calculate the undamped free response [i.e. M ( t ) = F ( t ) = 0, c 1 = c 2 = 0] for the initial condition x (0) = 0, ˙ x (0) = [0 0 1] T . Assume that the hub stiffness is 10,000 N/m and that the axle/suspension is 20,000 N/m. Assume the rotational element J is modeled as a translational mass of 75 kg. Torque convertor inertia J 75 kg m 2 /rad M ( t ) Clutch torque F ( t ) Wind and road load Equivalent inertia of transmission m 2 100 kg Vehicle Equivalent inertia m 3 3000 kg Rotational Hub damping Tire damping x 1 ( t ) x 2 ( t ) x 3 ( t ) Hub stiffness Axle and suspension stiffness Translational Solution: Let k 1 = hub stiffness and k 2 = axle and suspension stiffness. The equation of motion is 75 0 0 0 100 0 0 0 3000 10 000 1 1 0 1 3 2 0 2 2 0 0 0 0 1 + = ( ) = ( ) = [ ] ˙˙ , ˙ x x 0 x 0 x and m/s T Calculate eigenvalues and eigenvectors: M K M KM = = = 1 2 1 2 1 2 0 1155 0 0 0 0 1 0 0 0 0 0183 133 33 115 47 0 115 47 300 36 515 0 36 515 6 6667 / / / . . . ˜ . . . . . . det ˜ , . . . . K I ( ) = + = = = = = = = λ λ λ λ λ ω λ ω λ ω 3 2 1 1 2 3 440 28 222 0 0 0 77 951 8 8290 362 05 19 028 rad/s rad/s rad/s 2 3 v v v 1 3 2 0 1537 0 1775 0 9721 0 4488 0 8890 0 0913 0 8803 0 4222 0 2163 = = = . . . . . . . . .
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Use the mode summation method to find the solution. Transform the initial conditions: q x q x 0 0 0 0 0 0 0 54 7723 1 2 1 2 ( ) = ( ) = ( ) = ( ) = [ ] M M T / / , ˙ ˙ . The solution is given by: q v v v t c c t c t c t ( ) = + ( ) + + ( ) + + ( ) 1 4 1 2 2 2 2 3 3 3 3 sin sin ω φ ω φ where φ ω ω φ i i i T i T i i T i i c i = ( ) ( ) = = ( ) = tan , ˙ cos , 1 0 0 2 3 0 2 3 v q v q v q Thus, 2629 . 0 and , 3417 . 1 , 0 3 2 3 2 = = = c c φ φ So, q v v q v v 0 0 1 1 2 3 4 1 2 3 ( ) = + ( ) = + = = c c c c i i i i i i i i i sin ˙ cos φ ω φ Premultiply by T v 1 ; v q v q 1 1 1 4 0 0 0 53 2414 T T c c ( ) = = ( ) = = ˙ . So, q v v v t t t t ( ) = + ( ) + ( ) 53 2414 1 3417 8 8290 0 2629 19 028 1 2 3 . . sin . . sin . Change to q ( t ): x q x t M t t t t t ( ) = ( ) ( ) = + + 1 2 0 9449 1 1 1 0 1364 0 05665 0 005298 8 8290 0 01363 0 02337 0 0004385 19 028 / . . . . sin . . . . sin . m
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4.42 Calculate the natural frequencies and mode shapes of 4 0 0 0 2 0 0 0 1 4 1 0 1 2 1 0 1 1 0 + = ˙˙ x x Solution: Given the indicated mass and stiffness matrix, calculate eigenvalues: M K M KM = = = 1 2 1 2 1 2 0 5 0 0 0 0 7071 0 0 0 1 1 0 3536 0 0 3536 1 0 7071 0 0 7071 1 / / / . . ˜ . . . . det ˜ . . . , , . K I ( ) = + = = = = λ λ λ λ λ λ λ 3 2 1 2 3 3 2 375 0 375 0 0 2094 1 1 7906 The natural frequencies are: rad/s 3381 . 1 rad/s 1 rad/s 4576 . 0
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