# SolSec3_6 - 3 57Problems and Solutions Section 3.6(3.43...

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Unformatted text preview: 3- 57Problems and Solutions Section 3.6 (3.43 through 3.44) 3.43A power line pole with a transformer is modeled by m!!x+kx=!!!ywhere xand yare as indicated in Figure 3.23. Calculate the response of the relative displacement (x – y) if the pole is subject to an earthquake base excitation of (assume the initial conditions are zero) !!y t( )=A1!tt"#\$%&’(t(2tt>2t)*+,+Solution: Given: m!!x+kx=!!!y!!y=A1!tt"#\$%&’(t(2tt>2t)*+,+x( )=!x( )=The response x(t) is given by Eq. (3.12) as x t( )=t!F"( )h t#"( )d"where h t¡"( )=1m¢nsin¢nt¡"( )for an undamped system For !t!2t,3- 58x t( )=A1!"t#\$%&’(1m)n#\$%&’(sin)nt!"( )d"t*x t( )=Am)n21!tt+1t)nsin)nt!cos)nt+,-./For t>2t, x t( )=A1!"t#\$%&’(1m)n#\$%&’(sin)nt!"( )d"2t*x t( )=Am)n21t)nsin)nt!sin)nt!2t( )( )!cos)nt!cos)nt!2t( )+,-./Find y(t) when!t!2t, !!y t( )=A1!tt"#\$%&’!y t( )=At!A2tt2+C1y t( )=A2t2!A6tt3+C1t+C2Using IC's yields C1= C2= 0. Find y(t) when t> wt: !!y t( )=!y t( )=C3y t( )=C3t+C4Using IC's yields C3= C4=0. The relative displacement =0....
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## This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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SolSec3_6 - 3 57Problems and Solutions Section 3.6(3.43...

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