! 2 = k a ! = 125, Problems and Solutions Section 5.3 (5.27 through 5.36)

Transcription

1 5 3 Problems and Solutions Section through 536) 57 A motor is mounted on a platform that is observed to vibrate excessively at an operating speed of 6000 rpm producing a 50N force Design a vibration absorber undamped) to add to the platform Note that in this case the absorber mass will only be allowed to move mm because of geometric and size constraints The amplitude of the absorber mass can be found from equation 5) and used to solve for : X a = 000 m = = 50 From equation 51), = 15,000 N/m! = =! = 15,000 ) 6000 &, 60 = 0317 kg

2 Consider an undamped vibration absorber with β = 1 and µ = 0 Determine the operating range of frequencies for which Xk /! 05 From equation 54), with β =! p = 1ie, =! p ) and µ = 0, Xk = ) )! a 1! & a &, ) 1! & a,! 0 1 ) = & a 1! & a 4! & a 1 For Xk = 05, this yields 05!! Solving for the physical solution gives 4 01!!! & = = 0 Solving for! & gives! = 0955, 1813 &

3 5 5 Comparing this to the sketch in Figure 515, the values for which Xk! 5 are 0955! 1051 and! 1813

4 Consider an internal combustion engine that is modeled as a lumped inertia attached to ground through a spring Assuming that the system has a measured resonance of 100 rad/s, design an absorber so that the amplitude is 001 m for a measured) force input of 10 N The amplitude of the absorber mass can be found from equation 5) and used to solve for : X a = 001m = = 100 = 10,000 N/m Choose ω = ω n = 00 rad/s From equation 51), =! = 10, = 05 kg 530 A small rotating machine weighing 50 lb runs at a constant speed of 6000 rpm The machine was installed in a building and it was discovered that the system was operating at resonance Design a retrofit undamped absorber such that the nearest resonance is at least 0 away from the driving frequency By observing Figure 515, the values of µ = 05 and β = 1 result in the combined systems natural frequencies being 81 above the driving frequency and 18 below the driving frequency since! = a p = 1 and ω = ω p ) So the absorber should weigh and have stiffness = µm = 05) 50 lb) = 15 lb! a! = 15 lb) 4448 N/lb) 1 981& & ) = 4 ) 10 6 N/m = 1,800 lb/in

5 A 3000kg machine tool exhibits a large resonance at 10 Hz The plant manager attaches an absorber to the machine of 600 kg tuned to 10 Hz Calculate the range of frequencies at which the amplitude of the machine vibration is less with the absorber fitted than without the absorber For Xk = 1, equation 54) yields ) 1 µ! a! p &!!, ) 1!!, µ! a! p & = 1!! Since µ m = = 0, this becomes! & = 0, For Xk = 1, equation 54) yields Since =! p, 1 µ! a! ),! p &! p & 1! ),! µ! a =!! p &! 4! a!! p &! µ 1)! ), a!! p &! = 0! & 4 3!! = 0 1! & = 099,1534 The range of frequencies at which Xk > 1 is 0 <! < 099 and <! < 1534

6 5 8 Since ω a = ω p, 0 < ω < 6958 rad/s and 859 < ω < rad/s

7 A motorgenerator set is designed with steadystate operating speed between 000 and 4000 rpm Unfortunately, due to an imbalance in the machine, a large violent vibration occurs at around 3000 rpm An initial absorber design is implemented with a mass of kg tuned to 3000 rpm This, however, causes the combined system natural frequencies that occur at 500 and 3000 rpm Redesign the absorber so that ω 1 < 000 rpnd ω > 4000 rpm, rendering the system safe for operation The mass of the primary system can be computed from equation 55) Since! = a p = 1 and! 1 & = & = 06944, then 1) 06944)! 1 1) 1 µ ) µ = m µ = = kg ) 1 = 0 & By increasing µ to 055 and decreasing β to 089, the design goal can be achieved The mass and stiffness of the absorber should be = µm = 055) 14876) = 818 kg & 60 ) /! a! p = 818) 089), 3000 = 639,600 N/m 533 A rotating machine is mounted on the floor of a building Together, the mass of the machines and the floor is 000 lb The machine operates in steady state at 600 rpnd causes the floor of the building to shake The floormachine system can be modeled as a springmass system similar to the optical table of Figure 514 Design an undamped absorber system to correct this problem Make sure you consider the bandwidth To minimize the transmitted force, let ω a = ω = 600 rpm Also, since the floor shakes at 600 rpm, it is assumed that ω p = 600 rpm so that β = 1 Using equation 56) with µ = 01 yields! n = 08543, So the natural frequencies of the combined systere ω 1 = 516 rpnd ω = 703 rpm These are sufficiently enough away from 600 rpm to avoid problems Therefore the mass and stiffness of the absorber are

8 5 30 = µm = 01) 000 lbm) = 00 lbm slug ),! a = 00 lbm) lbm & & = 5,541 lb/ft

9 A pipe carrying steam through a section of a factory vibrates violently when the driving pump hits a speed of 300 rpm see Figure P534) In an attempt to design an absorber, a trial 9kg absorber tuned to 300 rpm was attached By changing the pump speed it was found that the pipeabsorber system has a resonance at 07 rpm Redesign the absorber so that the natural frequencies are 40 away from the driving frequency The driving frequency is 300 rpm 40 above and below this frequency is 180 rpnd 40 rpm This is the design goal The mass of the primary system can be computed from equation 55)! = a p = 1 and! 1 & = & = 04761, then Since 1) 04761)! 1 1) 1 µ ) µ = m µ = 9 = kg ) 1 = 0 & By increasing µ to 09 and decreasing β to 085, the design goal can be achieved The mass and stiffness of the absorber should be = µm = 09) 15611) = 1405 kg & 60 ) /! a! p = 1405) 085), 300 Note that µ is very large, which means a poor design = 10,00 N/m

10 A machine sorts bolts according to their size by moving a screen bacnd forth using a primary system of 500 kg with a natural frequency of 400 cycle/min Design a vibration absorber so that the machineabsorber system has natural frequencies below 160 cycles/min and above 30 rpm The machine is illustrated in Figure P535 Using Equation 56), and choose by trial and error) β = 04 and µ = 001, the design goal of ω 1 < 160 rpnd ω > 30 rpm can be achieved The actual values are ω 1 = 1598 rpnd ω = 4004 rpm The mass and stiffness of the absorber should be = µm = 001) 500) = 5 kg! a! [ = 5) 0), 400 & 60 ) / = N/m

11 A dynamic absorber is designed with µ = 1/4 and ω a = ω p Calculate the frequency range for which the ratio Xk / < 1 From Equation 54), with β =! p = 1 and µ = 05, Xk = ) = & a 4 )! a 1! & a! 5 & a 1! & a &, ) 1! & a 1,! 05 1 ) For Xk = 1, this yields! & 4 15!!! = 0, 1118 & = 0 For Xk = 1, this yields! & a 4 35 & a & = 09081, 1557 a! = 0 Comparing this to the sketch in Figure 515, the values for which Xk < 1 are

12 09081 <! < 1118 and! >