HW6soln - CE 573: Structural Dynamics HW#6 Solutions...

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CE 573: Structural Dynamics HW#6 Solutions Problem #1 Given : A popular method for limiting the effects of earthquakes on structures is to base isolate them; that is, to put in supports underneath the structure that isolate the main part of the structure from the ground. To understand why base isolation works so well, consider the following pair of structures. Structure (1) is unisolated, with and . Structure (2) is isolated using a base slab having a total mass and a series of rubber bearings having a total stiffness . (The building properties for (2) are the same as (1).) 3000 kg b m = 3 750 10 N/m b k 2000 kg i m = 3 50 i k Required : (a) For Structure (1), estimate the maximum base shear of the building using the pseudospectral acceleration design spectrum shown below. (b) For Structure (2), estimate the maximum “base” shear of the building with respect to the isolator slab. Use the same design spectrum as in (a), and model Structure (2) using the shear frame model. (c) Based on your results from parts (a) and (b), explain why base isolators are effective against earthquakes. Solution : (a). First need to determine the period of the unisolated structure: 3 rad sec 750 2 15.81 0.397 seconds. 3000 kg b b k T m π ω × == = = =
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From the design spectrum, we estimate the pseudospectral acceleration as 2 m s 25 pa S = . From this, we compute the base shear to be: () ( ) 2 3 m max s 3000 kg 25 75 10 N. bd p a Vk S m S == = = × Answer (b) The model for the building/base isolator slab is as follows: [] [] 3 3000 750 750 N kg; 10 . 2000 750 800 m MK ⎡⎤ == ⎢⎥ ⎣⎦ × (Note that this numbering scheme assumes that the building is DOF #1 while the isolator is DOF #2.) Based on these property matrices, we obtain the modal information as follows: >> kmat = [750 -750;-750 800]*10^3 kmat = 750000 -750000 -750000 800000 >> mmat = [3000 0; 0 2000] mmat = 3000 0 0 2000 >> [vecs,vals] = eig(kmat,mmat) vecs = -0.0144 -0.0113 -0.0138 0.0176 vals = 9.7620 0 0 640.2380 >> w1 = sqrt(vals(1,1)) w1 = 3.1244 >> a1 = vecs(:,1)/sqrt(vecs(:,1)'*vecs(:,1)) a1 = -0.7210 -0.6929 >> w2 = sqrt(vals(2,2)) w2 = 25.3029 >> a2 = vecs(:,2)/sqrt(vecs(:,2)'*vecs(:,2)) a2 = -0.5394 0.8420
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Now, for each mode, we need to perform a response spectrum analysis to determine the modal contribution to the “base” shear: Mode #1 : rad 1 sec 22 2.011 seconds. 3.124 T ππ ω == = From the formulas given with the response spectrum graph, this implies 2 m 2 s 1 20 9.945 1.0188 m. 2.011 pa pa d S SS = = We also need the participation factor {} []{} [] 1 11 1 : T T aM aMa Γ= >> onevec = [1;1]; gamma1 = a1'*mmat*onevec/(a1'*mmat*a1) gamma1 = -1.4084 Thus, we estimate the maximum displacement in this mode to be: {} {} ( ) ( ) 1,max 1 33 1,max 1 -0.7210 1.0346 * -1.4084 * 1.0188 m m. -0.6929 0.9942 750 750 1.0346 30.30 N 10 m 10 N. 750 800 0.9942 19.41 m d equiv ua S FK u ⎧⎫ = = ⎨⎬ ⎩⎭ ⎛⎞ ⎡⎤ ⇒= = × ⎜⎟ ⎢⎥ ⎣⎦ ⎝⎠ From equilibrium, the static equivalent force acting on the building (DOF #1) equals the “base” shear force with respect to the isolator slab. This is the first entry in the vector of equivalent forces, according to our numbering scheme. Thus, we estimate 3 1,max 30.30 10 N.
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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 University-West Lafayette.

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HW6soln - CE 573: Structural Dynamics HW#6 Solutions...

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