# SolSec 4_4 - Problems and Solutions for Section 4.4(4.44...

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Problems and Solutions for Section 4.4 (4.44 through 4.55) 4.44 A vibration model of the drive train of a vehicle is illustrated as the three-degree- of-freedom system of Figure P4.44. 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. 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 ! " # # # \$ % !! 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 Calculate eigenvalues and eigenvectors: M ! 1/2 = 0.1155 0 0 0 0.1 0 0 0 0.0183 " # \$ \$ \$ % ! K = M ! 1/2 KM ! 1/2 = 133.33 ! 115.47 0 ! 115.47 300 ! 36.515 0 ! 36.515 6.6667 " # \$ \$ \$ % det ! K ! " I ( ) = 3 ! 440 2 + 28,222 = 0 1 = 0 # 1 = 0 rad/s 2 = 77.951 2 = 8.8290 rad/s 3 = 362.05 3 = 19.028 rad/s

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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 mode 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 The solution is given by: q t ( ) = c 1 + c 4 t ( ) v 1 + c 2 sin ! 2 t + " 2 ( ) v 2 + c 3 sin 3 t + 3 ( ) v 3 where i = tan " 1 # i v i T q 0 ( ) v i T q 0 ( ) \$ % ( ) i = 2,3 c i = v i T ! q 0 ( ) i cos i = 2,3 Thus, 2 = 3 = 0, c 2 " 1.3417, and c 3 = 0.2629 So, q 0 ( ) = c 1 v 1 + c i sin i v i i = 2 3 " ! q 0 ( ) = c 4 v 1 + i c i cos i v i i = 2 3 " Premultiply by v 1 T ; v 1 T q 0 ( ) = 0 = c 1 v 1 T ! q 0 ( ) = 53.2414 = c 4 So, q t ( ) = 53.2414 t v 1 + 1.3417sin 8.8290 t ( ) v 2 + 0.2629sin 19.028 t ( ) v 3 Change to q( t ): x t ( ) = M ! 1/2 q t ( ) x t ( ) = 0.9449 t 1 1 1 " # \$ \$ \$ % + ! 0.1364 ! 0.05665 0.005298 " # \$ \$ \$ % sin8.8290 t + 0.01363 ! 0.02337 0.0004385 " # \$ \$ \$ % sin19.028 t m
4.45 Calculate the natural frequencies and normalized mode shapes of 4 0 0 0 2 0 0 0 1 ! " # # # \$ % !! x + 4 1 0 1 2 1 0 1 1 ! " # # # \$ % x = 0 Solution: Given the indicated mass and stiffness matrix, calculate eigenvalues: M ! 1/2 = 0.5 0 0 0 0.7071 0 0 0 1 " # \$ \$ \$ % ( ! K = M ! 1/2 KM ! 1/2 = 1 ! 0.3536 0 ! 0.3536 1 ! 0.7071 0 ! 0.7071 1 " # \$ \$ \$ % det ! K ! " I ( ) = 3 ! 3 2 + 2.375 ! 0.375 = 0 1 = 0.2094, 2 = 1, 3 = 1.7906 The natural frequencies are: ! 1 = 0.4576 rad/s 2 = 1 rad/s 3 = 1.3381 rad/s The corresponding eigenvectors are: v 1 = ! 0.3162 ! 0.7071 ! 0.6325 " # \$ \$ \$ % v 2 = 0.8944 0 ! 0.4472 " # \$ \$ \$ % v 3 = 0.3162 ! 0.7071 0.6325 " # \$ \$ \$ % The relationship between eigenvectors and mode shapes is u = M ! 1/2 v The mode shapes are: u 1 = ! 0.1581

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SolSec 4_4 - Problems and Solutions for Section 4.4(4.44...

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