MDOF Forced Vibration

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 123 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 - MK ωω ω ⎡⎤ ⎢⎥ == ⎣⎦ === Φ = 0.2770 -0.6920 0.6038 -0.7230 0.7007 It is subjected to a constant force ( ) 15 kips Ft = on its top story, and it has initial conditions {} {} in sec 00 (0 ) 0 ; ) 0 02 ut ⎧⎫ ⎪⎪ ⎨⎬ ⎩⎭ ± . . 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
-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 >> w1 = sqrt(eigvals(3,3)); phi1 = eigvecs(:,3) phi1 = 0.5382 0.5881 0.6038
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online