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Section 5.4

# Section 5.4 - 5 35 Problems and Solutions Section 5.4(5.37...

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5- 35 Problems and Solutions Section 5.4 (5.37 through 5.52) 5.37 A 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 100-N harmonic force at 20 rad/s. Design an undamped tuned absorber system (i.e., calculate m a and k a ) 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 ζ = 0.001. Similarly, the steel spring for the absorber gives rise to internal damping of about ζ a = 0.0015. Calculate how much this spoils the absorber design by determining the magnitude X using equation (5.32). Solution: From equation (5.21), the steady-state vibration will be zero when ! 2 = k a m a Choosing μ = 0.2 yields m a = μ m = 0.2 ( ) 100 ( ) = 20 kg k a = m a ! a 2 = 20 ( ) 20 ( ) 2 = 8000 N/m With damping of ζ = 0.001 and ζ a = 0.0015, the values of c and c a are c = 2 ! km = 2 0.001 ( ) 2000 ( ) 100 ( ) = 0.894 kg/s c a = 2 ! a k a m a = 2 0.0015 ( ) 8000 ( ) 20 ( ) = 1.2 kg/s From equation (5.32), X = k a ! m a " 2 ( ) F 0 + c a " F 0 j det K ! " 2 M + " jC ( ) Since M = 100 0 0 20 ! " # \$ % & C = 2.0944 1.2 1.2 1.2 ! " # \$ % & K = 10,000 8000 8000 8000 ! " # \$ % & the denominator is –6.4 × 10 7 -1.104 × 10 6 j , so the value of X is

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5- 36 X = k a m a ! 2 ( ) F 0 + c a ! F 0 j ( ) det K " ! 2 M + ! jC ( ) Using Window 5.4, the magnitude is X = 3.75 ! 10 " 5 m This is a very small displacement, so the addition of internal damping will not affect the design very much. 5.38 Plot 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 kg m a = 20 kg c = 0.8944 kg/s c a = 1.2 kg/s k = 2000 N/m k a = 8000 N/m F 0 = 100 N ! = 20 rad/s Using Equation (5.32), the magnitude of X is plotted versus ω with and without the internal damping (c). Note that X is reduced when X < F 0 /k = 0.05 m and magnified when X > 0.05 m. The plots of the two values of X show that there is no observable difference when internal damping is added. In this case, knowledge of internal damping is not necessary.
5- 37

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5- 38 5.39 Derive 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 ζ given in Equation (5.36) is not the same as the ζ values used in Problems 5.37 and 5.38. Solution: Substituting Equation (5.34) into the denominator of Equation (5.32) yields X F 0 = k a ! m a " 2 ( ) + c a " j ! m " 2 + k ( ) ! m a " 2 + k a ( ) # \$ % & + k ! m + m a ( ) " 2 ( ) c a " # \$ % & j Referring to Window 5.4, the value of X F 0 can be found by noting that A 1 = k a ! m a " B 1 = c a " A 2 = ! m " 2 + k ( ) ! m a " 2 + k a ( ) ! m a k a " 2 B 2 = k ! m + m a ( ) " 2 ( ) c a " Since X F 0 = A 1 2 + B 1 2 A 2 2 + B 2 2 then X 2 F 0 2 = k a ! m a " 2 ( ) 2 + c a 2 " 2 ! m " 2 + k ( ) ! m a " 2 + k a ( ) ! m a k a " 2 # \$ % & 2 + k ! m + m a ( ) " 2 # \$ % & 2 c a 2 " 2 which is Equation (5.35) To derive Equation (5.37), substitute c a = 2 ! m a " p , k a = m a " a 2 , and m a = μ m , then multiply by k 2 to get
5- 39 X 2 k 2 F 0 2 = k 2 !

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