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example_IV_4_02

# example_IV_4_02 - Example IV.4.2 Say we reconsider the...

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Example IV.4.2 Say we reconsider the building problem. Here we will mount an eccentric shaker on the top floor of the building to simulate a modal test of the building. Find the response of the building as a function of the shaker frequency . How would the response change if the shaker were instead mounted on the second floor from the top? The natural frequencies and modal vectors are repeated here: ϖ = 0.4912 ; ϖ = 1.4142 ϖ = 2.1667 ; ϖ = 2.6579 1 2 3 4 5 1 f(t) = M e  2 σινΩτ σηακερ 1 2 3 4 f t ( ) = Me W 2 sinW t shaker

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The EOM’s for this model of the building are given by: M [ ] ú ú x + K [ ] x = f t ( ) where: f t ( ) = Me W 2 0 0 0 1 ì í ï ï î ï ï ü ý ï ï þ ï ï sinW t The uncoupled EOM’s become: ú ú q j + w j 2 q j = ö f j t ( ) where x t ( ) = ÷ f j ( ) q j t ( ) j =1 4 å and the modal forcings are given by: ö f j t ( ) = ÷ f j ( ) T f t ( ) Explicitly we have: ö f 1 t ( ) = f 4 1 ( ) Me W 2 sinW t = ö f 01 sinW t ö f 2 t ( ) = f 4 2 ( ) Me W 2 sinW t = ö f 02 sinW t ö f 3 t ( ) = f 4 3 ( ) Me W 2 sinW t = ö f 03 sinW t ö f 4 t ( ) = f 4 4 ( ) Me W 2 sinW t = ö f 04 sinW t Solving for the particular response of the modal EOM’s (harmonic excitation): q jp t ( ) = ö f 0 j / w j 2 1- W 2 / w j 2 sinW t = Q j sinW t where the modal amplitudes of response are: 2
Q 1 = f 4 1 ( ) Me W 2 w 1 2 1 1- W 2 / w 1 2 = f 4 1 ( ) MeH W/ w

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