{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
5- 37
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 4
5- 39 X 2 k 2 F 0 2 = k 2 !
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern