SolSec3_2 - 3- 17Problems and Solutions for Section 3.2...

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Unformatted text preview: 3- 17Problems and Solutions for Section 3.2 (3.15 through 3.25) 3.15Calculate the response of an overdamped single-degree-of-freedom system to an arbitrary non-periodic excitation. Solution:From Equation (3.12): x t( )=F!( )h t"!( )d!t#For an overdamped SDOF system (see Problem 3.4) h t!"( )=12m#n$2!1e!$#nt!"()e#n$2!1t!"()!e!#n$2!1t!"()%&’(d"x t( )=F"( )t)12m#n$2!1e!$#nt!"()e#n$2!1t!"()!e!#n$2!1t!"()%&’(d"*x t( )=e!$#n2m#n$2!1F"( )t)e$#n"e#n$2!1t!"()!e!#n$2!1t!"()%&’(d"3- 183.16Calculate the response of an underdamped system to the excitation given in Figure P3.16. Plot of a pulse input of the form f(t) = Fsint. Figure P3.16 Solution:x t( )=1m!de"#!ntF$( )e#!n$sin!dt"$( )%&’(d$t)F t( )=Fsint( )t<*From Figure P3.16( )For t+*, x t( )=Fm!de"#!ntsin$e#!n$sin!dt"$( )( )d$t)x t( )=Fm!de"#!nt$12 1+2!d+!n2%&’(e#!nt!d"1( )sint"#!ncost%&’("!d"1( )sin!dt"#!ncos!dt{ }%&))+12 1+2!d+!n2%&’(e#!nt!d"1( )sint"#!ncost%&’(+!d"1( )sin!dt"#!ncos!dt{ }’(**For!>",:f(!)h(t"!)d!t#=f(!)h(t"!)d!$#+(0)h(t"!)d!$t#3- 19x t( )=Fm!de"#!ntsin$e#!n$sin!dt"$( )( )d$%&=Fm!de"#!nt’12 1+2!d+!n2"#$%e&!nt!d’1( )sin!dt’(( )"#$%’&!ncos!dt’(( )"#$%"#$%’!d’1( )sin!dt’&!ncos!dt)*+,+-.+/+"#+12 1+2!d+!n2"#$%e&!nt!d+1( )sin!dt’1( )"#$%+&!cos!dt’(( )"#$%"#$%+!d’1( )sin!dt’&!ncos!dt)*+,+-.+/+$%22Alternately, one could take a Laplace Transform approach and assume the under-damped system is a mass-spring-damper system of the form m!!x t( ) +c!x t( ) +kx t( ) =F t( )The forcing function given can be written as F t( ) =FH t( )!H t!"( )( )sint( )Normalizing the equation of motion yields !!x t( ) +2!"n!x t( ) +"n2x t( ) =fH t( )#H t#$( )( )sint( )where f=Fmand m, cand kare such that<!<1. Assuming initial conditions, transforming the equation of motion into the Laplace domain yields X s( ) =f1+e!"s( )s2+1( )s2+2#$ns+$n2( )The above expression can be converted to partial fractions X s( ) =f1+e!"s( )As+Bs2+1#$%&’(+f1+e!"s( )Cs+Ds2+2)*ns+*n2#$%&’(where A, B, C, and Dare found to be 3- 20A=!2"#n1!#n2( )2+2"#n( )2B=#n2!11!#n2( )2+2"#n( )2C=2"#n1!#n2( )2+2"#n( )2D=1!#n2( )+2"#n( )21!#n2( )2+2"#n( )2Notice that X s( )can be written more attractively as X s( ) =fAs+Bs2+1+Cs+Ds2+2!"ns+"n2#$%&’(+fe)*sAs+Bs2+1+Cs+Ds2+2!"ns+"n2#$%&’(=fG s( ) +e)*sG s( )( )Performing the inverse Laplace Transform yields x t( ) =fg t( ) +H t!"( )g t!"( )( )where g(t) is given below g t( ) =Acost( ) +Bsint( ) +Ce!"#ntcos#dt( ) +D!C"#n#d$%&’()e!"#ntsin#dt( )!dis the damped natural frequency,!d=!n1"#2. Let m=1 kg, c=2 kg/sec, k=3 N/m, and F=2 N. The system is solved numerically. Both exact and numerical solutions are plotted below 3- 21Figure 1 Analytical vs. Numerical Solutions Below is the code used to solve this problem % Establish a time vector t=[0:0.001:10]; % Define the mass, spring stiffness and damping coefficient m=1; c=2; k=3; % Define the amplitude of the forcing function F0=2; % Calculate the natural frequency, damping ratio and normalized force amplitude zeta=c/(2*sqrt(k*m));...
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SolSec3_2 - 3- 17Problems and Solutions for Section 3.2...

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