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

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Unformatted text preview: 2- 27Problems and Solutions Section 2.4 (2.37 through 2.50) 2.37A 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!=mg5cm= 20000.05= 40,000 N/m Transmitted displacement from equation (2.70): X= Y1+(2!r)2(1"r2)2+(2!r)2#$%&’(1/2= 10 cm Transmitted force from equation (2.77): FT= kYr21+(2!r)2(1"r2)2+(2!r)2#$%&’(1/2= 4001N 2.38Derive Equation (2.70) from (2.68) to see if the author has done it correctly. Solution: Equation (2.68) states: xp(t) = !nY !n2+(2"!b)2(!n2#!b2)2+(2"!n!b)2$%&&’())1/2cos(!bt"#1"#2) The magnitude is: X = !nY !n2+(2"!b)2(!n2#!b2)2+(2"!n!b)2$%&&’())1/22- 28= !nY (!n"4)(!n2+(2#!b)2)(!n"4)((!n2"!b2)2+(2#!n!b)2)$%&&’())1/2= !nY(!n"2)(1+(2#r)2)(1"r2)2+(2#r)2$%&&’())1/2*= !nY1!n1+(2"r)2(1#r2)2+(2"r)2$%&’()1/2*X = Y1+(2!r)2(1"r2)2+(2!r)2#$%&’(1/2This is equation (2.71).2.39From 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= XY=1+(2!r)2(1"r2)2+(2!r)2#$%&’(1/2Show TR > 1 for r< 2TR= XY= 1+(2!r)2(1"r2)2+(2!r)2#$%&’(1/2>11+(2!r)2(1"r2)2+(2!r)2> 1 1 +(2!r)2>(1"r2)2+(2!r)21 > (1!r2)2Take the real solution: 1!r2< +1 or 1!r2<!1"!r2>!2"r2<2"r<22- 292.40Consider 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 FT = kXwhere Xis the magnitude of the displacement. From equation (2.65) FT=kX=2!"n"bkY("n2#"b2)2+(2!"n"b)2=2!rkY(1#r2)2+(2!r)22.41A 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 displacement of y(t) = (0.01)sin (5.818t) m. The suspension provides an equivalent stiffness of k= 4 x 105N/m, a damping coefficient of...
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SolSec 2.4 - 2- 27Problems and Solutions Section 2.4 (2.37...

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