{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

SolSec 4.4

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

This preview shows pages 1–4. Sign up to view the full content.

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 = = = . . . . . . . . .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}