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ASE 365 - Lecture 7 - SDOF harmonic response x k m F(t = F...

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SDOF harmonic response EOM: A = F/k represents excitation amplitude in displacement units (static response). k c m x F(t) = F cos ω t (or F sin ω t)
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Recap We’ve solved the ODE ( 29 ( 29 [ ] . ) / ( 1 ) / ( 2 tan ) / ( 2 ) / ( 1 ) ( 2 2 1 2 2 2 n n n n i G ϖ ϖ ϖ ϖ ζ φ ϖ ϖ ζ ϖ ϖ ϖ - = + - = - and where
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Complex plane representation ) ( ) ( φ ϖ - = t i p kXe t kx : units force For ) ( ) ( φ ϖ ϖ - = t i p cXe i t x c : Also ) ( 2 ) ( φ ϖ ϖ - - = t i p mXe t x m : And mω2 X X kX F ωt ωt- φ Re I m φ Re I m F ωt X ωt- φ kX mω2 X X
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Frequency response plots ω/ωn | G(iω) | 1 1 ω/ωn φ 1 90 o 180 o
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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
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