SolSec 4.6 - Problems and Solutions for Section 4.6 (4.61...

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Problems and Solutions for Section 4.6 (4.61 through 4.67) 4.61 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 ) ( 7071 . 4 4 . ) ( 7071 . 2 2 . 2 2 2 1 1 1 t r r r t r r r δ = + + = + + & & & & & & Also from the example, rad/s 9899 . 1 1 . rad/s 2 rad/s 4106 . 1 07071 . rad/s 2 d2 2 2 d1 1 1 = = = = = = ω ζ n n The solution to an impulse is given by equations (3.7) and (3.8): t e m F t r di t di i i ni i sin ˆ ) ( = This yields r () . sin . . sin . . . t et t t = 5013 1 4106 3553 1 9899 1 2 The solution in physical coordinates is xr x .. . sin . . sin . . sin . . sin . . sin . . sin . / . . tMP t t e t t tt == = + −− 12 1 2 2357 2357 7071 7071 5013 1 4106 3553 1 9899 1182 1 4106 08374 1 9899 3544 1 4106 2512 1 9899 9899 t
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4.62 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: ) ( t K M F x x = + & & Let F () t Ft = 1 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, ) ( 2 / 1 t M P I T F r r = Λ + & & Or, ) ( ) ( 1 2 2 2 2 2 1 1 1 2 1 1 t F b r r t F b r r = + = + ω & & & & where b 1 and b 2 are constants form 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), t t a b t r t r t r 1 1 1 1 10 1 1 10 1 sin 2 cos sin ) ( + + = & The solution for r 2 is rt r tr ba t tt 2 20 2 22 0 2 2 2 1 2 2 2 2 2 1 2 1 ˙ sin cos sin =+ + ωω If the initial conditions are zero, t t a b t r t t a b t r 2 1 2 1 2 2 2 2 1 1 1 1 cos cos ) ( sin 2 ) ( = =
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Converting to physical coordinates X ( t ) = M -1/2 Pr ( t ) yields ) ( ) ( ) ( ) ( ) ( ) ( 2 4 1 3 2 2 2 1 1 1 t r c t r c t x t r c t r c t x + = + = where c i is a constant form 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 () t t a b t r t t a b t r 2 2 2 2 1 2 2 2 2 1 1 1 sin 2 ) ( cos cos ) ( ω = = and ) ( ) ( ) ( ) ( ) ( ) ( 2 4 1 3 2 2 2 1 1 1 t r c t r c t x t r c t r c t x + = + = so both masses still oscillate at resonance. Also, if F 1 ( t ) = a 1 cos w 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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4.63 Use modal analysis to calculate the response of the drive train system of Problem 4.41 to a unit impulse on the car body (i.e., and location q 3 ). Use the modal damping of Problem 4.51.
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SolSec 4.6 - Problems and Solutions for Section 4.6 (4.61...

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