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MDOF Forced Vibration

MDOF Forced Vibration - CE 573 Structural Dynamics Example...

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CE 573: Structural Dynamics Example: Solving the Forced Vibration Problem Given: A three story building is modeled using shear frames. The property matrices and modal information for this structure are: [ ] [ ] [ ] 2 kip-sec kips in in rad rad rad 1 2 3 sec sec sec 0.1035 0 0 405.57 349.63 0 0 0.0776 0 ; 349.63 802.76 453.13 ; 0 0 0.0518 0 453.13 453.13 0.5382 0.6329 0.1736 15.067 ; 73.463 ; 131.86 ; 0.5881 - M K ω ω ω = = = = = Φ = 0.2770 -0.6920 0.6038 -0.7230 0.7007 It is subjected to a constant force ( ) 15 kips F t = on its top story, and it has initial conditions { } { } in sec 0 0 ( 0) 0 ; ( 0) 0 0 2 u t u t ⎧ ⎫ ⎪ ⎪ = = = = ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ ± . .5 Required: Determine the forced vibration response of this building. Solution: 1. Compute all of the required modal quantities: >> mmat = diag([0.1035 0.0776 0.0518]) mmat = 0.1035 0 0 0 0.0776 0 0 0 0.0518 >> kmat = [405.57 -349.63 0; -349.63 802.76 -453.13; 0 -453.13 453.13] kmat = 405.5700 -349.6300 0
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-349.6300 802.7600 -453.1300 0 -453.1300 453.1300 >> dmat = inv(mmat)*kmat dmat = 1.0e+004 * 0.3919 -0.3378 0 -0.4506 1.0345 -0.5839 0 -0.8748 0.8748 >> [eigvecs, eigvals] = eig(dmat) eigvecs = 0.1736 0.6329 0.5382 -0.6920 -0.2770 0.5881 0.7007 -0.7230 0.6038 eigvals = 1.0e+004 * 1.7387 0 0 0 0.5397 0 0 0 0.0227
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