Rayleigh Quotient

# Rayleigh Quotient - CE 573 Structural Dynamics Example...

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CE 573: Structural Dynamics Example: Rayleigh Quotient Given: A six-story building is modeled as a shear-frame structure. Parameters are , psi, and . 2 66 lb-sec /in m = 6 30 10 E 4 882 in I = Required: Estimate the fundamental natural frequency 1 ω for this building using the Rayleigh quotient and the following approximations to mode #1: (a) linear displacement of each dof – ; (b) static displacement due to self-weight – {} . {} { } 123456 T linear φ = [][ ] 1 static KM g = Solution: 1. Property matrices – Mass matrix uses the lumped mass approximation: [] 2 lb-sec in 66 66 66 =. 66 66 66 m m m M m m m ⎡⎤ ⎢⎥ = ⎣⎦ As for the stiffness matrix, each floor has a stiffness ( )( ) () 64 3 lb in 33 24 30 10 psi 882 in 12 2* 108.89 10 180 in i EI k L × == = × . Thus, the overall stiffness matrix is 12 2 22 3 3 4 4 44 5 5 55 66 217.78 108.89 108.89 217.78 108.89 108.89 217.78 108.89 108.89 217.78 108.89 108.89 217.78 108.89 108.89 108.89 kk k k k K +− −+ −− 3 lb in 10 .

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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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Rayleigh Quotient - CE 573 Structural Dynamics Example...

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