Modal Combinations

# Modal Combinations - CE 573 Structural Dynamics Example...

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CE 573: Structural Dynamics Example: Response Spectrum Analysis using Modal Combinations Given: A four story building is modeled using shear frames. The mass and stiffness properties are expressed in terms of kg and N/m; property matrices are given below. Each story is subjected to a rectangular pulse expressed in terms of 3 12 10 m 6 41 0 k 500 o F = kN. Assume that the pulse lasts for seconds. 0.15 d t = [] 3 3 3 3 66 66 6 24 10 2 1.5 18 10 kg; 12 10 12 10 12 10 8 10 32 2 4 2 8 10 16 10 8 10 N/m. 23 81 0 1 21 0 0 0 0 m m M m m kk kk k K kkk ⎡⎤ × ⎢⎥ × == × ⎣⎦ × ×− × −− − × × × −− −× × × Required: Estimate the maximum response of the top story using the SRSS rule. Solution: 1. Obtain the modal information: >> mmat = diag([24 18 12 12]*10^3); >> kmat = [12 -8 0 0; -8 16 -8 0; 0 -8 12 -4; 0 0 -4 4]*10^6;

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>> dmat = inv(mmat)*kmat; >> [vecs, vals] = eig(dmat); >> omega = sort( diag( sqrtm(vals) ) ) omega = 6.6721 17.1954 27.9496 40.0105 >> phimat = [vecs(:,3) vecs(:,4) vecs(:,2) vecs(:,1)] phimat = -0.3370 -0.4999 -0.5021 -0.1779 -0.4604 -0.3064 0.4236 0.5876 -0.5378 0.0909 0.6048 -0.7634 -0.6207 0.8049 -0.4502 0.2008 >> modalkmat = phimat'*kmat*phimat modalkmat = 1.0e+007 * 0.0651 0.0000 0.0000 0.0000 0 0.4601 -0.0000 0.0000 0.0000 0 1.2579 0.0000 0.0000 0.0000 0.0000 2.3135 2. Modal static responses:
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## This note was uploaded on 05/09/2008 for the course CE 573 taught by Professor Whalen during the Fall '05 term at Purdue.

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Modal Combinations - CE 573 Structural Dynamics Example...

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