ASE 365 - Lecture 8 - Rotating unbalance How to get an EOM...

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Rotating unbalance k c x(t ) m e M- m ωt ω How to get an EOM for x(t)? FBD 2 f H f V z(t ) H f z m - = What is z(t)? For Newton’s 2nd law, z(t) must be an absolute (inertial) displacement. f S f D FBD 1 f H f V R A R B W ( 29 t e t x m x c kx f f f x m M H D S ϖ ϖ sin ) ( ) ( 2 - - - - = + - - = - t e m kx x c x M ϖ ϖ sin 2 = + + : EOM
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t e M m x x x M n n ϖ ϖ ϖ ζϖ sin 2 2 2 = + + : by Divide ( 29 ( 29 [ ] 2 2 1 2 2 2 2 2 ) / ( 1 ) / ( 2 tan ) / ( 2 ) / ( 1 ) ( ) sin( ) ( ) ( sin 2 n n n n p n n i G t i G A t x t A x x x n ϖ ϖ ϖ ϖ ζ φ ϖ ϖ ζ ϖ ϖ ϖ φ ϖ ϖ ϖ ϖ ϖ ζϖ - = + - = - = = + + - and with , which for : with Compare . ) ( ) sin( ) ( ) ( 2 φ ϖ φ ϖ ϖ ϖ ϖ and same with now So i G t i G e M m t x n p - = (ω/ωn)2 | G(iω)| 1 1 ω/ωn
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Harmonic base motion k c m x(t) y(t ) FBD fS fD m y y x x x n n n n 2 2 2 2 ϖ ζϖ ϖ ζϖ + = + + : Solve ky y c kx x c x m + = + + : EOM . ) ( ) ( ) ( ) ( t i p t i e i X t x Ae t y t y ϖ ϖ ϖ = = , use , harmonic For ( 29 ( 29 t i n n t i n n Ae i e i X i ϖ ϖ ϖ ζϖ ϖ ϖ ϖ ζϖ ϖ ϖ 2 2 2 2 ) ( 2 + = + + -
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φ ϖ ϖ ϖ ζϖ ϖ ζϖ ϖ ϖ ϖ ζϖ ϖ ζϖ ϖ ϖ ϖ ζϖ ϖ ϖ i n n n n n n n n e i X A i G i A i i A i i i X - = + = + - + = + + - + = ) ( ) ( ) / 2 1 ( / 2 ) / ( 1 / 2 1 2 2 ) ( 2 2 2 2 ( 29 + - = - - = - - - 2 2 3 1 1 2 1 ) / 2 ( ) / ( 1 ) / ( 2 tan / 2 tan ) / ( 1 / 2 tan n n n n n n ϖ ζϖ ϖ ϖ ϖ ϖ ζ ϖ ζϖ ϖ ϖ ϖ ζϖ φ and ( 29 A i G A i X e i X e i X t x n n n n t i t i p ) ( ) / 2 ( 1 ) / 2 ( ) / ( 1 ) / 2 ( 1 ) ( ) ( ) ( ) ( 2 2 2 2 2 ) ( ϖ ϖ ζϖ ϖ ζϖ ϖ ϖ ϖ ζϖ ϖ ϖ ϖ φ ϖ ϖ + = + - + = = = - where So
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Vibration isolation k c m x F(t)=F0cos ωt Force transmitted to base: x c kx f f F D S tr + = + = amplitude. s ' in interested are We . sinusoidal is and , sinusoidal are and so , that know We tr tr p F x c kx F x c kx t i G A t x + = - = ) cos( ) ( ) ( φ ϖ ϖ fS fD . so , Then . if easier is Analysis t i p p tr t i p t i e i AG c i k x c kx F e i AG t x e F t F ϖ ϖ ϖ ϖ ϖ ϖ ) ( ) ( ) ( ) ( ) ( 0 + = + = = = . when find so , when s ' equals when s ' t i tr tr t i tr e F t F F t F t F F e F t F F ϖ ϖ ϖ 0 0 0 ) ( cos ) ( ) ( = = = amplitude magnitude
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k c m x F(t)=F0cos ωt Force transmitted to base: kx x c f f F D S tr + = + = : of Magnitude tr F ( 29 2 2 2 2 0 2 ) / 2 ( ) / ( 1 ) / 2 ( 1 / 2 ) / ( 1 / 2 1 ) ( ) ( n n n n n n t i tr F i i kA e i AG k c i F ϖ ζϖ ϖ ϖ ϖ ζϖ ϖ ζϖ ϖ ϖ ϖ ζϖ ϖ ϖ ϖ + - + = + - + = + = ( 29 2 2 2 2 0
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