SolSec 5.5 - 5- 44 Problems and Solutions Section 5.5 (5.53...

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5- 44 Problems and Solutions Section 5.5 (5.53 through 5.66) 5.53 Design a Houdaille damper for an engine modeled as having an inertia of 1.5 kg-m 2 and a natural frequency of 33 Hz. Choose a design such that the maximum dynamic magnification is less than 6: 6 0 < M Xk The design consists of choosing J 2 and c a , the required optimal damping. Solution: From Eq. (5.50), Xk M 0 1 2 =+ max µ Since Xk M 0 6 < , then 4 . 0 2 1 6 > + > Choose µ = 0.4. From Eq. (5.49), the optimal damping is ζ µµ op = + () + = 1 21 2 0 3858 . The values of J 2 and c a are J J rad c a 04 15 06 == =⋅ ( ) ζω π .. / . / kg m rad kg m 2 J 2 0.3858 0.6 33 2 95.98 N m s/rad 22 op 2 p
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5- 45 5.54 Consider the damped vibration absorber of Eq. (5.37) with β fixed at β = 1/2 and µ fixed at µ = 0.25. Calculate the value of ζ that minimizes st X δ . Plot this function for several values of 0 < < 1 to check your design. If you cannot solve this analytically, consider using a three-dimensional plot of st X versus r and to determine your design. Solution: From Eq. (5.37), with β = 0.5 and µ = 0.25, let fr X rr r st , . .. . () = +− [] 40 2 5 4 1 25 1 0 065 1 0 25 22 2 2 2 2 2 2 From Eqs. (5.44) and (5.45), with f A B = 12 / / , = f 0 becomes AdB BdA Since Brr r =− 4 1 25 1 0 0625 1 0 25 2 2 2 . and Ar r =+ 2 5 2 2 ., then dA A r dB B = = = 8 8 1 25 1 2 2 . So, 4 1 25 1 0 0625 4 1 0 25 8 4 0 25 8 1 25 1 0 0625 1 0 25 0 25 1 2 2 2 2 2 2 2 2 2 2 2 2 ζζ r r r r r r r . . . {} −− 25 1 2 2 r Taking the square root yields
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5- 46 0 625 1 0 25 0 25 1 25 1 22 2 2 2 .. . . rr r r r −− () Solving for r yields r = 0 4896 0 9628 ., . Now take the derivative = f r 0 becomes AdB BdA = Since Brr r =− +− [] 4 1 25 1 0 0625 1 0 25 2 2 2 2 ζ . and Ar r =+ 40 2 5 2 2 then dA A r dB B r r r r r r =+− ζζ 820 2 5 2 8 125 1 8 2 25 2 0 0625 4 1 0 25 0 125 2 0 25 1 2 2 2 2 2 . . . . ( ) 2 r Solving B dA = A dB for ζ yields r X r X st st =→ = = 0 4896 0 1145 1 4279 0 9628 0 3197 6 3029 . . δ To determine the optimal damping ratio, make a plot of st X versus r for ζ = 0.01, 0.1145, 0.3197, and 0.7.
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5- 47 The value of ζ = 0.3197 yields the best overall response (i.e., the lowest maximum).
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5- 48 5.55 For a Houdaille damper with mass ratio µ = 0.25, calculate the optimum damping ratio and the frequency at which the damper is most effective at reducing the amplitude of vibration of the primary system. Solution: From Eq. (5.49), with µ = 0.25, ζ µµ op = + () + = 1 21 2 0 422 . From Eq. (5.48), 943 . 0 2 2 = + = µ r The damper would be most effective at ωω ω == r nn 0 943 . , i.e., where the amplitude is greatest:
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5- 49 5.56 Consider again the system of Problem 5.53. If the damping ratio is changed to ζ = 0.1, what happens to 0 M Xk ? Solution: If ζ op = 0.1, the value of µ becomes 01 1 21 2 002 006 096 0 8 589 5 589 2 . .. . ., . = + () + += = =− µµ µ Clearly µ = 5.589 is the physical solution. The maximum value of 0 M Xk would be Xk M 0 1 2 1 358 =+ = max .
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SolSec 5.5 - 5- 44 Problems and Solutions Section 5.5 (5.53...

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