Rayleigh Damping

# Rayleigh Damping - CE 573: Structural Dynamics Example:...

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CE 573: Structural Dynamics Example: MDOF Damped Forced Vibration Given: A six story shear frame model building has the following mass and stiffness properties: , psi and for each column. The property matrices are: 2 66 lb-sec /in m = 6 30 10 E 4 882 in I = [] 2 3 lb-sec lb in in 66 217.78 108.89 66 108.89 217.78 108.89 66 108.89 217.78 108.89 ;1 66 108.89 217.78 108.89 66 108.89 217.78 108.89 66 108.89 108.89 MK ⎡⎤ ⎢⎥ −− == ⎣⎦ 0 × Required: (a) Assuming that 1 0.04 ξ = and 3 0.04 = , determine the damping ratios for all of the modes using the Rayleigh damping model. (b) Suppose that floor #1 is subjected to a rectangular impulse force of amplitude lb for a duration of 0.5 seconds, with all other floors not receiving any forcing. Using the damping ratios from (a), determine the response of the building for seconds. (Assume the building is at rest initially.) 3 200 10 o F 0.5 t > (c) Repeat part (a), but now assume that 2 0.04 = and 4 0.04 = . What effect would you predict this assumption has on the response as compared to part (b)? Solution: (a) The general solution for the damping ratios under the Rayleigh damping model is 1 1 1 ; 2 . 1 2 r r r jj s j s s ω α ξβω β ⎛⎞ =+ = ⎜⎟ ⎝⎠ The calculation proceeds as follows: >> mmat = diag([66 66 66 66 66 66])

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mmat = 66 0 0 0 0 0 0 66 0 0 0 0 0 0 66 0 0 0 0 0 0 66 0 0 0 0 0 0 66 0 0 0 0 0 0 66 >> k = 2*12*(30*10^6)*882/(12*15)^3 k = 1.0889e+005 >> kmat = [2*k -k 0 0 0 0; . .. -k 2*k -k 0 0 0; . .. 0 -k 2*k -k 0 0; . .. 0 0 -k 2*k -k 0; . .. 0 0 0 -k 2*k -k; . .. 0 0 0 0 -k k] kmat = 217780 -108890 0 0 0 0 -108890 217780 -108890 0 0 0 0 -108890 217780 -108890 0 0 0 0 -108890 217780 -108890 0 0 0 0 -108890 217780 -108890 0 0 0 0 -108890 108890 >> dmat = inv(mmat)*kmat; >> [eigvecs,eigvals] = eig(dmat) eigvecs = 0.5187 -0.3678 -0.1327 0.5507 -0.4565 -0.2578 0.3678 -0.5507 -0.2578 -0.1327 0.5187 0.4565 -0.2578 -0.4565 -0.3678 -0.5187 -0.1327 -0.5507 -0.5507 -0.1327 -0.4565 0.2578 -0.3678 0.5187 -0.1327 0.2578 -0.5187 0.4565 0.5507 -0.3678 0.4565 0.5187 -0.5507 -0.3678 -0.2578 0.1327 eigvals = 1.0e+003 * 2.1296 0 0 0 0 0 0 0.8298 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 3.6974 0 0 0 0 0 0 5.1741 0 0 0 0 0 0 6.2214 >> omega = sort( diag(sqrt(eigvals)) ) omega = 9.7919 28.8068 46.1474 60.8062 71.9311 78.8757
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## Rayleigh Damping - CE 573: Structural Dynamics Example:...

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