SolSec 2.4 - 2 27 Problems and Solutions Section 2.4(2.31...

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2- 27 Problems and Solutions Section 2.4 (2.31 through 2.42) 2.31 A machine weighing 2000 N rests on a support. The support deflects about 5 cm as a result of the weight of the machine. The floor under the support is somewhat flexible and moves, because of the motion of a nearby machine, harmonically near resonance ( r =1) with an amplitude of 0.2 cm. Model the floor as base motion, and assume a damping ratio of ζ = 0.01, and calculate the transmitted force and the amplitude of the transmitted displacement. Solution: Given: Y = 0.2 cm, = 0.01, r = 1, mg = 2000N Deflection of 5 cm: k = mg 5cm = 05 . 0 2000 = 40,000 N/m Transmitted displacement: X = Y 12 2 22 2 + −+ () ( ) / r rr = 10 cm Transmitted force: F T = kYr 2 2 2 + ( ) / r = 4001N 2.32 Derive Equation 2.61 from 2.59. Solution: Equation (2.59) x p ( t ) = ω n Y ωζ ωω nb n b 2 2 2 2 + / cos( ωθ θ b t −− ) The magnitude is: X = n Y n b 2 2 2 2 + / = n Y ( ( ) ) ( )(( ) ( ) / nn b b n b + 42 2 422 2 2 2 2 = n Y ( ) ( ) / n r + 2 = n Y 11 2 2 2 n r + ( ) / X = Y 2 2 + ( ) / r
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2- 28 2.33 From the equation describing Figure 2.11, show that the point ( 2 , 1) corresponds to the value TR > 1 (i.e., for all r < 2 , TR > 1). Solution: Equation (2.62) TR = Y X = 12 2 22 2 + −+ () ( ) / ζ r rr Show TR > 1 for r < 2 TR = Y X = 2 2 + ( ) / r > 1 2 2 2 2 ) 2 ( ) 1 ( ) 2 ( 1 r r r + + > 1 1 + 2 2 2 2 ) 2 ( ) 1 ( ) 2 ( r r r + > 1 > 2 2 ) 1 ( r Take the real solution: 11 2 2 2 2 2 −< + −> < < r r r or
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2- 29 2.34 Consider the base excitation problem for the configuration shown in Figure P2.34. In this case the base motion is a displacement transmitted through a dashpot or pure damping element. Derive an expression for the force transmitted to the support in steady state.
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This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 2.4 - 2 27 Problems and Solutions Section 2.4(2.31...

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