SolSec 2.4 - 2 27 Problems and Solutions Section 2.4(2.37...

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2- 27 Problems and Solutions Section 2.4 (2.37 through 2.50) 2.37 A machine weighing 2000 N rests on a support as illustrated in Figure P2.37. 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 ! = 0.01, and calculate the transmitted force and the amplitude of the transmitted displacement. Figure P2.37 Solution: Given: Y = 0.2 cm, ! = 0.01, r = 1, mg = 2000N. The stiffness is computed from the static deflection and weight: Deflection of 5 cm implies: k = mg ! = mg 5cm = 2000 0.05 = 40,000 N/m Transmitted displacement from equation (2.70): X = Y 1 + (2 ! r ) 2 (1 " r 2 ) 2 + (2 ! r ) 2 # $ % & ( 1/2 = 10 cm Transmitted force from equation (2.77): F T = kYr 2 1 + (2 ! r ) 2 (1 " r 2 ) 2 + (2 ! r ) 2 # $ % & ( 1/2 = 4001N 2.38 Derive Equation (2.70) from (2.68) to see if the author has done it correctly. Solution: Equation (2.68) states: x p ( t ) = ! n Y ! n 2 + (2 "! b ) 2 ( ! n 2 # ! b 2 ) 2 + (2 "! n ! b ) 2 $ % & & ( ) ) 1/2 cos( ! b t " # 1 " # 2 ) The magnitude is: X = ! n Y ! n 2 + (2 "! b ) 2 ( ! n 2 # ! b 2 ) 2 + (2 "! n ! b ) 2 $ % & & ( ) ) 1/2
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2- 28 = ! n Y ( ! n " 4 )( ! n 2 + (2 #! b ) 2 ) ( ! n " 4 )(( ! n 2 " ! b 2 ) 2 + (2 #! n ! b ) 2 ) $ % & & ( ) ) 1/2 = ! n Y ( ! n " 2 )(1 + (2 # r ) 2 ) (1 " r 2 ) 2 + (2 # r ) 2 $ % & & ( ) ) 1/2 * = ! n Y 1 ! n 1 + (2 " r ) 2 (1 # r 2 ) 2 + (2 " r ) 2 $ % & ( ) 1/2 * X = Y 1 + (2 ! r ) 2 (1 " r 2 ) 2 + (2 ! r ) 2 # $ % & ( 1/2 This is equation (2.71). 2.39 From the equation describing Figure 2.13, show that the point ( 2 , 1) corresponds to the value TR > 1 (i.e., for all r < 2 , TR > 1). Solution: Equation (2.71) is TR = X Y = 1 + (2 ! r ) 2 (1 " r 2 ) 2 + (2 ! r ) 2 # $ % & ( 1/2 Show TR > 1 for r < 2 TR = X Y = 1 + (2 ! r ) 2 (1 " r 2 ) 2 + (2 ! r ) 2 # $ % & ( 1/2 > 1 1 + (2 ! r ) 2 (1 " r 2 ) 2 + (2 ! r ) 2 > 1 1 + (2 ! r ) 2 > (1 " r 2 ) 2 + (2 ! r ) 2 1 > (1 ! r 2 ) 2 Take the real solution: 1 ! r 2 < + 1 or 1 ! r 2 < ! 1 " ! r 2 > ! 2 " r 2 < 2 " r < 2
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2- 29 2.40 Consider the base excitation problem for the configuration shown in Figure P2.40. 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. Figure P2.40 Solution: The entire force passes through the spring. Thus the support sees the force F T = kX where X is the magnitude of the displacement. From equation (2.65) F T = kX = 2 !" n " b kY ( " n 2 # " b 2 ) 2 + (2 !" n " b ) 2 = 2 ! rkY (1 # r 2 ) 2 + (2 ! r ) 2 2.41 A very common example of base motion is the single-degree-of-freedom model of an automobile driving over a rough road. The road is modeled as providing a base motion
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