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SolSec 4_6 - Problems and Solutions for Section 4.6(4.67...

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Problems and Solutions for Section 4.6 (4.67 through 4.76) 4.67 Calculate the response of the system of Figure 4.16 discussed in Example 4.6.1 if F 1 ( t ) = δ ( t ) and the initial conditions are set to zero. This might correspond to a two-degree-of-freedom model of a car hitting a bump. Solution: From example 4.6.1, with F 1 ( t ) = δ ( t ), the modal equations are !! r 1 + 0.2 ! r 1 + 2 r 1 = 0.7071 ! ( t ) !! r 2 + 0.4 ! r 2 + 4 r 2 = 0.7071 ! ( t ) Also from the example, ! n 1 = 2 rad/s " 1 = 0.07071 ! d1 = 1.4106 rad/s ! n 2 = 2 rad/s " 2 = 0.1 ! d2 = 1.9899 rad/s The solution to an impulse is given by equations (3.7) and (3.8): r i ( t ) = ˆ F m i ! di e " # i ! ni t sin ! di t This yields r ( t ) = 0.5012 e ! 0.1 t sin1.4106 t 0.3553 e ! 0.2 t sin1.9899 t " # $ % & The solution in physical coordinates is x ( t ) = M ! 1/2 P r ( t ) = .2357 ! .2357 .7071 .7071 " # $ % & 0.167 e ! 0.1 t sin1.4106 t ! 0.118 e ! .02 t sin1.9899 t " # $ % & x ( t ) = 0.0394 e ! 0.1 t sin1.4106 t + 0.0279 e ! 0.2 t sin1.9899 t 0.118 e ! 0.1 t sin1.4106 t ! 0.0834 e ! 0.2 t sin1.9899 t " # $ % &
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4.68 For an undamped two-degree-of-freedom system, show that resonance occurs at one or both of the system’s natural frequencies. Solution: Undamped two-degree-of-freedom system: M !! x + K x = F ( t ) Let F ( t ) = F 1 ( t ) 0 ! " # $ % & Note: placing F 1 on mass 1 is one way to do this. A second force could be placed on mass 2 with or without F 1 . Proceeding through modal analysis, I !! r + ! r = P T M " 1/2 F ( t ) Or, !! r 1 + ! 1 2 r 1 = b 1 F 1 ( t ) !! r 2 + ! 2 2 r 2 = b 2 F 1 ( t ) where b 1 and b 2 are constants from the matrix P T M -1/2 . If F 1 ( t ) = a cos ω t and ω = ω 1 then the solution for r 1 is (from Section 2.1), r 1 ( t ) = ! r 10 ! 1 sin ! 1 t + r 10 cos ! 1 t + b 1 a 2 ! 1 t sin ! 1 t The solution for r 2 is r 2 ( t ) = ! r 20 ! 2 sin ! 2 t + r 20 " b 2 a ! 2 2 " ! 1 2 # $ % & ( cos ! 2 t + b 2 a ! 2 2 " ! 1 2 t sin ! 1 t If the initial conditions are zero,
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r 1 ( t ) = b 1 a 2 ! 1 t sin ! 1 t r 2 ( t ) = b 2 a ! 2 2 " ! 1 2 cos ! 1 t " cos ! 2 t ( ) Converting to physical coordinates X ( t ) = M -1/2 Pr ( t ) yields x 1 ( t ) = c 1 r 1 ( t ) + c 2 r 2 ( t ) x 2 ( t ) = c 3 r 1 ( t ) + c 4 r 2 ( t ) where c i is a constant from M -1/2 P . So, if the driving force contains just one natural frequency, both masses will be excited at resonance. The driving force could contain the other natural frequency ( ω = ω n2 ), which would cause r 1 and r 2 to be r 1 ( t ) = b 1 a ! 1 2 " ! 2 2 cos ! 2 t " cos ! 1 t ( ) r 2 ( t ) = b 2 a 2 ! 2 t sin ! 2 t and x 1 ( t ) = c 1 r 1 ( t ) + c 2 r 2 ( t ) x 2 ( t ) = c 3 r 1 ( t ) + c 4 r 2 ( t ) so both masses still oscillate at resonance. Also, if F 1 ( t ) = a 1 cos ω 1 t + a 2 cos ω 2 t where ω 1 = ω n1 and ω 2 = ω n2 , then both r 1 and r 2 would be at resonance, so x 1 ( t ) and x 2 ( t ) would also be at resonance.
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4.69 Use modal analysis to calculate the response of the drive train system of Problem 4.44 to a unit impulse on the car body (i.e., and location
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