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example_IV_4_02

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

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1 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 Ω . 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: ω 1 = 0.4912 k/m ; ω 2 = 1.4142 k/m ω 3 = 2.1667 k/m ; ω 4 = 2.6579 k/m ˜ P [ ] = φ (1) , φ (2) , φ (3) , φ (4) [ ] = 1 m 0.2280 0.5774 0.6565 0.4285 0.4285 0.5774 0.2280 0.6565 0.5774 0 0.5774 0.5774 0.6565 0.5774 0.4285 0.2280 1 2 3 4 f t ( ) = Me Ω 2 sin Ω t shaker
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2 The EOM’s for this model of the building are given by: M [ ] ˙ ˙ x + K [ ] x = f t ( ) where: f t ( ) = Me Ω 2 0 0 0 1 sin Ω t The uncoupled EOM’s become: ˙ ˙ q j + ω j 2 q j = ˆ f j t ( ) where x t ( ) = ˜ φ j ( ) q j t ( ) j = 1 4 and the modal forcings are given by: ˆ f j t ( ) = ˜ φ j ( ) T f t ( ) Explicitly we have: ˆ f 1 t ( ) = φ 4 1 ( ) Me Ω 2 sin Ω t = ˆ f 01 sin Ω t ˆ f 2 t ( ) = φ 4 2 ( ) Me Ω 2 sin Ω t = ˆ f 02 sin Ω t ˆ f 3 t ( ) = φ 4 3 ( ) Me Ω 2 sin Ω t = ˆ f 03 sin Ω t ˆ f 4 t ( ) = φ 4 4 ( ) Me Ω 2 sin Ω t = ˆ f 04 sin Ω t Solving for the particular response of the modal EOM’s (harmonic excitation): q jp t ( ) = ˆ f 0 j / ω j 2 1 − Ω 2 / ω j 2 sin Ω t = Q j sin Ω t
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