This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 5 35Problems and Solutions Section 5.4 (5.37 through 5.52) 5.37A machine, largely made of aluminum, is modeled as a simple mass (of 100 kg) attached to ground through a spring of 2000 N/m. The machine is subjected to a 100N harmonic force at 20 rad/s. Design an undamped tuned absorber system (i.e., calculate maand ka) so that the machine is stationary at steady state. Aluminum, of course, is not completely undamped and has internal damping that gives rise to a damping ratio of about = 0.001. Similarly, the steel spring for the absorber gives rise to internal damping of about a= 0.0015. Calculate how much this spoils the absorber design by determining the magnitude Xusing equation (5.32). Solution: From equation (5.21), the steadystate vibration will be zero when !2=kamaChoosing = 0.2 yields ma=m=0.2( )100( )=20 kgka=ma!a2=20( )20( )2=8000 N/mWith damping of = 0.001 and a= 0.0015, the values of cand caare c=2!km=2 0.001( )2000( )100( )=0.894 kg/sca=2!akama=2 0.0015( )8000( )20( )=1.2 kg/sFrom equation (5.32), X=ka!ma"2( )F+ca"FjdetK!"2M+"jC( )Since M=10020!"#$%&C=2.09441.21.21.2!"#$%&K=10,000800080008000!"#$%&the denominator is 6.41071.104106j, so the value of Xis 5 36X=kama!2( )F+ca!Fj( )detK"!2M+!jC( )Using Window 5.4, the magnitude is X=3.75!10"5mThis is a very small displacement, so the addition of internal damping will not affect the design very much. 5.38Plot the magnitude of the primary system calculated in Problem 5.37 with and without the internal damping. Discuss how the damping affects the bandwidth and performance of the absorber designed without knowledge of internal damping. Solution: From Problem 5.37, the values are m=100 kgma=20 kgc=0.8944 kg/sca=1.2 kg/sk=2000 N/mka=8000 N/mF=100 N!=20 rad/sUsing Equation (5.32), the magnitude of Xis plotted versus with and without the internal damping (c). Note that Xis reduced when X< F/k= 0.05 m and magnified when X> 0.05 m. The plots of the two values of Xshow that there is no observable difference when internal damping is added. In this case, knowledge of internal damping is not necessary. 5 375 385.39Derive Equation (5.35) for the damped absorber from Eqs. (5.34) and (5.32) along with Window 5.4. Also derive the nondimensional form of Equation (5.37) from Equation (5.35). Note the definition of given in Equation (5.36) is not the same as the values used in Problems 5.37 and 5.38. Solution: Substituting Equation (5.34) into the denominator of Equation (5.32) yields XF=ka!ma"2( )+ca"j!m"2+k( )!ma"2+ka( )#$%&+k!m+ma( )"2( )ca"#$%&jReferring to Window 5.4, the value of XFcan be found by noting that A1=ka!ma"B1=ca"A2=!m"2+k( )!ma"2+ka( )!maka"2B2=k!m+ma( )"2( )ca"Since XF=A12+B12A22+B22then X2F2=ka!ma"2( )2+ca2"2!m"2+k( )!ma"2+ka( )!maka"2#$%&2+k!m+ma( )"2#$%&2ca2"2which is Equation (5.35) To derive Equation (5.37), substitute ca=2!...
View
Full
Document
This note was uploaded on 02/15/2010 for the course ME 3222 taught by Professor Tang during the Spring '10 term at UConn.
 Spring '10
 Tang

Click to edit the document details