ASE 365 - Lecture 6

ASE 365 - Lecture 6 - 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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becomes EOM so , and Then t X t x t X t x p p ϖ cos ) ( sin ) ( 2 - = - = t kA t X k t X c t X m cos ) cos ( ) sin ( ) cos ( 2 = + - + - case. undamped the in do they as cancel simply not do terms dependent - time the case, damped the For ??? ???
? about How t X t X t x s c p ϖ sin cos ) ( + = ) sin cos ( ) ( ) cos sin ( ) ( 2 t X t X t x t X t X t x s c p s c p + - = + - = , Then or , becomes EOM t kA t X t X k t X t X c t X t X m s c s c s c cos ) sin cos ( ) cos sin ( ) sin cos ( 2 = + + + - + + - t kA t kX X c X m t kX X c X m s c s c s c cos sin ) ( cos ) ( 2 2 = + - - + + + -

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. and for , equations 2 the solve , From s c s c s c s c s c s c s c X X kX X c X m kA kX X c X m t kA t kX X c X m t kX X c X m 0 cos sin ) ( cos ) ( 2 2 2 2 = + - - = + + - = + - - + + + - ϖ balance" harmonic of method "
response. the and excitation the between lag phase the is where , try important, is response the of amplitude the since Or, φ ϖ ) cos( ) ( - = t X t x p ) sin sin cos (cos ) cos( ) ( ) sin cos cos (sin ) sin( ) ( ) sin sin cos (cos ) cos( ) ( 2 2 t t X t X t x t t X t X t x t t X t X t x p p p + - = - - = - - = - - = + = - = , , Then t kA t t kX t t X c t t X m cos ) sin sin cos (cos ) sin cos cos (sin ) sin sin cos (cos 2 = + + - - + - : EOM in Insert . and for solve and Rearrange X

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. and : Problem t t dt d t t dt d ϖ cos ) (sin sin ) (cos ! ! : Consider t i t i t i e e dt d t i t e + = ) ( sin cos ic m k kAe t x k ic m kA X kAe Xe
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This note was uploaded on 09/18/2011 for the course ASE 365 taught by Professor Staff during the Spring '10 term at University of Texas at Austin.

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ASE 365 - Lecture 6 - SDOF harmonic response x k m F(t) = F...

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