SolSec 4.4 - 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

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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 11 0 13 2 02 2 00 0 0 1 + −− = () = () = [] ˙˙ , ˙ xx 0 x0 x and m/s T Calculate eigenvalues and eigenvectors: M KM K M = == 12 0 1155 0 0 10 0 0 0 0183 133 33 115 47 0 115 47 300 36 515 0 36 515 6 6667 / // . . . ˜ .. det ˜ , KI () =− + = λλ λ λω 32 2 3 440 28 222 0 77 951 8 8290 362 05 19 028 rad/s rad/s rad/s 2 3 vv v 13 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: qx q x 0 0 0 0 0 0 0 54 7723 12 () = () = () = () = [] MM T // , ˙ ˙ . The solution is given by: qv v v tc c t c t c t () =+ ++ 14 2 2 23 3 3 3 sin sin ωφ where φ ω i ii T i T i i T i i ci = = = = tan , ˙ cos , 1 0 0 0 vq Thus, 2629 . 0 and , 3417 . 1 , 0 3 2 3 2 = = = c c So, v v 0 0 11 2 3 41 2 3 = = cc i i i i i sin ˙ cos Premultiply by T v 1 ; 00 0 53 2414 T T c c == ˙ . So, v v tt t t + 53 2414 1 3417 8 8290 0 2629 19 028 123 . . sin . . sin . Change to q ( t ): xq x tM t t t () = () () = + +− 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 400 020 001 41 0 12 1 01 1 0 + −− = ˙˙ xx Solution: Given the indicated mass and stiffness matrix, calculate eigenvalues: MK M K M = ⇒= = 12 05 0 0 0 0 7071 0 00 1 1 0 3536 0 0 3536 1 0 7071 0 0 7071 1 // / . . ˜ . .. . det ˜ ., , . KI () =− + = == = λλλ λ λλ 32 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 3 2 1 = = = ω The corresponding eigenvectors are: vvv 123 0 3162 0 7071 0 6325 0 8944 0 0 4472 0 3162 0 7071 0 6325 = = .
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This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 4.4 - 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

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