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# SolSec 4_1 - Problems and Solutions for Section 4.1(4.1...

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Unformatted text preview: Problems and Solutions for Section 4.1 (4.1 through 4.16) 4.1 Consider the system of Figure P4.1. For c 1 = c 2 = c 3 = 0, derive the equation of motion and calculate the mass and stiffness matrices. Note that setting k 3 = 0 in your solution should result in the stiffness matrix given by Eq. (4.9). Solution: For mass 1: m 1 !! x 1 = ! k 1 x 1 + k 2 x 2 ! x 1 ( ) " m 1 !! x 1 + k 1 + k 2 ( ) x 1 ! k 2 x 2 = For mass 2: m 2 !! x 2 = ! k 3 x 2 ! k 2 x 2 ! x 1 ( ) " m 2 !! x 2 ! k 2 x 1 + k 2 + k 3 ( ) x 2 = So, M !! x + K x = m 1 m 2 ! " # # \$ % & & !! x + k 1 + k 2 ’ k 2 ’ k 2 k 2 + k 3 ! " # # \$ % & & x = Thus: M = m 1 m 2 ! " # # \$ % & & K = k 1 + k 2 ’ k 2 ’ k 2 k 2 + k 3 ! " # # \$ % & & 4.2 Calculate the characteristic equation from problem 4.1 for the case m 1 = 9 kg m 2 = 1 kg k 1 = 24 N/m k 2 = 3 N/m k 3 = 3 N/m and solve for the system's natural frequencies. Solution: Characteristic equation is found from Eq. (4.9): det ! " 2 M + K ( ) = ! " 2 m 1 + k 1 + k 2 ! k 2 ! k 2 ! " 2 m 1 + k 2 + k 3 = ! 9 " 2 + 27 ! 3 ! 3 ! " 2 + 6 = 9 " 4 ! 81 " 2 + 153 = Solving for ω : ! 1 = 1.642 ! 2 = 2.511 rad/s 4.3 Calculate the vectors u 1 and u 2 for problem 4.2. Solution: Calculate u 1 : ! 2.697 ( ) 9 ( ) + 27 ! 3 ! 3 ! 2.697 + 6 " # \$ \$ % & ’ ’ u 11 u 21 " # \$ \$ % & ’ ’ = " # \$ % & ’ This yields 2.727 u 11 ! 3 u 21 = ! 3 u 11 + 3.303 u 21 = or, u 21 = 0.909 u 11 u 1 = 1 0.909 " # \$ % & ’ Calculate u 2 : ! 6.303 ( ) 9 ( ) + 27 ! 3 ! 3 ! 6.303 + 6 " # \$ \$ % & ’ ’ u 12 u 22 " # \$ \$ % & ’ ’ = " # \$ % & ’ This yields ! 29.727 u 12 ! 3 u 22 = ! 3 u 12 = 0.303 u 22 = or, u 12 = ! 0.101 u 22 u 2 = ! 0.101 1 " # \$ % & ’ 4.4 For initial conditions x (0) = [1 0] T and ! x (0) = [0 0] T calculate the free response of the system of Problem 4.2. Plot the response x 1 and x 2 . Solution: Given x (0) = [1 0] T , ! x ( ) = 0 0 ! " # \$ T , The solution is x t ( ) = A 1 sin ! 1 t + " 1 ( ) u 1 + A 2 sin ! 2 t + " 2 ( ) u 2 x 1 t ( ) x 2 t ( ) # \$ % % & ’ ( ( = A 1 sin ! 1 t + " 1 ( ) ) 0.101 A 2 sin ! 2 t + " 2 ( ) 0.909 A 1 sin ! 1 t + " 1 ( ) + A 2 sin ! 2 t + " 2 ( ) # \$ % % & ’ ( ( Using initial conditions, 1 = A 1 sin ! 1 " 0.101 A 2 sin ! 2 1 # \$ % & = 0.909 A 1 sin ! 1 + A 2 sin ! 2 2 # \$ % & = 1.642 A 1 cos ! 1 " 0.2536 A 2 cos ! 2 3 # \$ % & = 6.033 A 1 cos ! 1 + 2.511 A 2 cos ! 2 4 # \$ % & From [3] and [4], ! 1 = ! 2 = " / 2 From [1] and [2], A 1 = 0.916, and A 2 = ! 0.833 So, x 1 t ( ) = 0.916sin 1.642 t + ! / 2 ( ) + 0.0841sin 2.511 t + ! / 2 ( ) x 2 t ( ) = 0.833sin 1.642 t + ! / 2 ( ) " 0.833sin 2.511 t + ! / 2 ( ) x 1 t ( ) = 0.916cos1.642 t + 0.0841cos2.511 t x 2 t ( ) = 0.833 cos1.642 t ! cos2.511 t ( ) 4.5 Calculate the response of the system of Example 4.1.7 to the initial condition x (0) = 0, ! x (0) = [1 0] T , plot the response and compare the result to Figure 4.3....
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SolSec 4_1 - Problems and Solutions for Section 4.1(4.1...

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