SolSec 5.4 - 5- 29 Problems and Solutions Section 5.4 (5.37...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
5- 29 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 Eq. (5.21), the steady-state vibration will be zero when a a m k = 2 ω Choosing µ = 0.2 yields mm m a aa == () ( ) = = µ 0 2 100 20 20 20 8000 2 2 . k g k N/m a With damping of ζ = 0.001 and ζ a = 0.0015, the values of c and c a are ck m m a a ( ) ( ) = ( ) ( ) = 2 2 0 001 2000 100 0 894 2 2 0 0015 8000 20 1 2 ζ .. kg/s kg/s From Eq. (5.32), X km Fc F j KM j C a = + −+ ωω 2 00 2 det Since MC K = = = 100 0 02 0 2 0944 1 2 12 10 000 8000 8000 8000 , the denominator is –6.4 × 10 7 -1.104 × 10 6 j , so the value of X is X F c Fj j C a = + 2 2 det
Background image of page 1

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

View Full DocumentRight Arrow Icon
5- 30 Using Window 5.4, the magnitude is X 375 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 5.36 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 mm cc kk F a a a == 100 20 0 8944 1 2 2000 8000 100 0 kg kg kg/s kg/s N/m N/m N 20 rad/s .. ω Using Eq. (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.
Background image of page 2
5- 31 5.39 Derive Eq. (5.35) for the damped absorber from Eqs. (5.34) and (5.32) along with Window 5.4. Also derive the nondimensional form of Eq. (5.37) from Eq. (5.35). Note the definition of ζ given in Eq. (5.36) is not the same as the ζ values used in Problems 5.37 and 5.38. Solution: Substituting Eq. (5.34) into the denominator of Eq. (5.32) yields X F km c j mk m k k m mc j aa a a a 0 2 22 2 = () + −+ [] +−+ ωω ω Referring to Window 5.4, the value of 0 F X can be found by noting that Akm Bc Amk m k m k Bk m m c a a a 1 1 2 2 2 2 =− = + + Since 2 2 2 2 2 1 2 1 0 B A B A F X + + = then X F c m k m k k m m c a a a a a 2 0 2 2 2 2 2 2 2 2 2 = + +− + which is Eq. (5.35) To derive Eq. (5.37), substitute cm k m m m p a a a a == = 2 2 ζω µ ,, , and then multiply by k 2 to get Xk F kk m k m ap dr p 0 2 2 2 2222 2 2 2 2 2 2 222 4 14 = + +−− ζ ωω ω µω µω ζωω Substituting r pp a p = ωω ω βω 2 , , and yields
Background image of page 3

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

View Full DocumentRight Arrow Icon
5- 32 Xk F mr k r r m m r m mr r pp p p dr pppp p p p p 22 0 2 24 22 222 2 2 2 4 2 2 2 22 2 4 14 = () + [] +− ωβω ω ζωω ωω β µ ζ Canceling m 2 and 8 p yields
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

Page1 / 15

SolSec 5.4 - 5- 29 Problems and Solutions Section 5.4 (5.37...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online