Ch2-4 to2-9_Oct11_2010

Ch2-4 - Correction Slide#44 in Ch2-1toCh2-3 As 2n Bs As f 0 cos t 2 Bs 2n As Bs sin t 0 2 2 n Sign of this term was incorrect in the previous

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Correction: Slide #44 in Ch2-1toCh2-3.ppt   22 0 ( 2 )cos 2 sin 0 s n s n s s n s n s A B A f t B A B t     Sign of this term was incorrect in the previous presentation. Correct it in your notes 1
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2.4 Base Excitation One way to excite the motion of a system is to apply a force to its mass. Another way is to apply displacement or force on the base (as opposed to the mass) Important class of vibration analysis: Preventing excitations from passing from a vibrating base through its mount into a structure/machine/etc Vibration isolation: Vibrations in a vehicle: Mainly excited by road conditions Disk drive in a laptop: rotating disk at a high speed m k 2
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FBD of Base Excitation m c(x y) () k x y FBD m k x(t) c base y(t) System Model =- ( - )- ( - )= + + = + F k x y c x y mx mx cx kx cy ky Engineering System 3
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Base Excitation Example: For a car excited by an event with a period of The steady-state solution is just the superposition of the two individual particular solutions (system is linear). T V Assume: ( ) sin( ) and plug into eqn. of motion: + + = + + + = cos( ) + sin( ) harmonic forcing functions y t Y t mx cx kx cy ky mx cx kx c Y t kY t 0 0 22 +2 + =2 cos( ) + sin( ) s c f f n n n n x x x Y t Y t    V T   4
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Particular Solution for the sine Term With a sine for the forcing function, the eqn of motion:     2 0 0 2 2 2 2 22 0 2 2 2 2 +2 + = sin cos sin sin( ) 2 ( ) 2 () ( ) 2 n n s ps s s s s ns s nn s x x x f t x A t B t X t f A f B      Use rectangular form to make it easier to add the cos term 5 2 2 1 ( ) cos sin , tan p s s s ss s x t A t B t B X A B A    
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Particular Solution for the cosine Term With a cosine for the forcing function, the particular solution:     2 0 22 0 2 2 2 2 0 2 2 2 2 +2 + = cos cos sin cos( ) () ( ) 2 2 ( ) 2 n n c pc c c c c nc c nn c x x x f t x A t B t X t f A f B      6
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Magnitude X/Y Now add the sine and cosine terms to get the magnitude of the full particular solution: 2 00 and use 2 and p pc ps c n s n x x x f Y f Y             2 2 2 2 22 2 2 2 2 2 2 2 2 (2 ) ( ) 2 ( ) 2 Using (plug into above), 1 (2 ) this term becomes (1 ) 2 c s n n n n n n n n ff XY rr r        2 2 1 (2 ) ) 2 Xr Y Displacement Transmissibility Ratio 7
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Relative Magnitude plot of X/Y vs Frequency Ratio 0 0.5 1 1.5 2 2.5 3 -20 -10 0 10 20 30 40 Frequency Ratio r X/Y (dB) =0.01 =0.1 =0.3 =0.7 Displacement Transmissibility n r   2 2 22 1 (2 ) (1 ) 2 Xr Y rr  m k x(t) c base y(t) Attenuation Amplification 8
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Potentially severe amplification at resonance r ≈1 Attenuation for r > sqrt(2): Isolation Zone If r < sqrt(2), transmissibility decreases with damping ratio: Amplification Zone If r >> 1, then transmissibility increases with damping ratio X p ~2Y /r Observations from the Relative Displacement Transmissibility Plot 9
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This note was uploaded on 11/01/2010 for the course ME 455 taught by Professor Centinkaya during the Fall '10 term at Clarkson University .

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Ch2-4 - Correction Slide#44 in Ch2-1toCh2-3 As 2n Bs As f 0 cos t 2 Bs 2n As Bs sin t 0 2 2 n Sign of this term was incorrect in the previous

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