Section 5.5

# Section 5.5 - 5- 53Problems and Solutions Section 5.5 (5.53...

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Unformatted text preview: 5- 53Problems and Solutions Section 5.5 (5.53 through 5.66) 5.53Design a Houdaille damper for an engine modeled as having an inertia of 1.5 kg.m2and a natural frequency of 33 Hz. Choose a design such that the maximum dynamic magnification is less than 6: XkM&lt;6The design consists of choosing J2and ca, the required optimal damping. Solution: From Equation (5.50), XkM!&quot;#\$%&amp;max=1+2Since XkM&lt;6,then 6&gt;1+2&gt;0.4Choose = 0.4. From Equation (5.49), the optimal damping is !op=12+1( )+2( )=0.3858The values of J2and caare J2=J1=0.4( )1.5 kg!m2/ rad( )=0.6 kg!m2/radca=2&quot;opJ2#p=2 0.3858( )0.6( )33( )2\$( )=95.98 N!m!s/rad5- 545.54Consider the damped vibration absorber of equation (5.37) with fixed at = 1/2 and fixed at = 0.25. Calculate the value of !that minimizesX/!. Plot this function for several values of 0 &lt; !&lt; 1 to check your design. If you cannot solve this analytically, consider using a three-dimensional plot of X/!versus rand !to determine your design. Solution: From equation (5.37), with = 0.5 and = 0.25, let f r,!( )=X&quot;4!2r2+r2#0.25( )24!2r21.25r2#1( )2+0.065r2#r2#1( )r2#0.25( )\$%&amp;2From equataions (5.44) and (5.45), with f=A1/2B1/2,!f!&quot;=becomes BdA!AdBSince B=4!2r21.25r&quot;1( )2+0.0625r2&quot;r2&quot;1( )r2&quot;0.25( )#\$%&amp;2and A=4!2r2+r2&quot;0.25( )2,then dA=!A!&quot;=8&quot;r2dB=!B!&quot;=8&quot;r21.25r2#1( )2So, 4!2r21.25r2&quot;1( )2+0.0625r2&quot;42&quot;1( )r2&quot;0.25( )#\$%&amp;2{ }8!r2( )=4!2r2+r2&quot;0.25( )2{ }8!r2( )1.25r2&quot;1( )20.0625r2&quot;r2&quot;1( )r2&quot;0.25( )2#\$%&amp;(=r2&quot;0.25( )21.25r2&quot;1( )25- 55Taking the square root yields 0.625r2!r2!1( )r2!0.25( )= r2!0.25( )1.25r2!1( )Solving for ryields r=0.4896, 0.9628Now take the derivative !f!r=becomes BdA=AdBSince B=4!2r21.25r2&quot;1( )2+0.0625r2&quot;r2&quot;1( )r2&quot;0.25( )#\$%&amp;2and A=4!2r2+r2&quot;0.25( )2,then dA!&quot;A&quot;#=8#2r+2r2\$0.25( )2r( )dB!&quot;B&quot;#=8#2r1.25r2\$1( )2+8#2r21.25r2\$2r( )( )2.5r( )+2 0.0625r2\$42\$1( )r2\$0.25( )%&amp;(0.125r\$2r( )r2\$0.25( )\$r2\$1( )2r( )%&amp;(Solving B dA = A dBfor yields r=0.4896!&quot;=0.1145!X#st=1.4279r=0.9628!&quot;=0.3197!X#st=6.3029To determine the optimal damping ratio, make a plot ofX/!versus rfor = 0.01, 0.1145, 0.3197, and 0.7. 5- 56The value of = 0.3197 yields the best overall response (i.e., the lowest maximum). 5- 575.55For a Houdaille damper with mass ratio = 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 equation (5.49), with = 0.25, !op=12+1( )+2( )=0.422From equation (5.48), r=22+=0.943The damper would be most effective at!=r!n=0.943!n, i.e., where the amplitude is greatest: 5- 585.56Consider again the system of Problem 5.53. If the damping ratio is changed to = 0.1, what happens toXk/M?...
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## Section 5.5 - 5- 53Problems and Solutions Section 5.5 (5.53...

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