MECHANICAL
SolSec3_2

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

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3- 17 Problems and Solutions for Section 3.2 (3.15 through 3.25) 3.15 Calculate 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 ! 0 t # For an overdamped SDOF system (see Problem 3.4) h t ! " ( ) = 1 2 m # n \$ 2 ! 1 e ! \$# n t ! " ( ) e # n \$ 2 ! 1 t ! " ( ) ! e ! # n \$ 2 ! 1 t ! " ( ) % & ( d " x t ( ) = F " ( ) 0 t ) 1 2 m # n \$ 2 ! 1 e ! \$# n t ! " ( ) e # n \$ 2 ! 1 t ! " ( ) ! e ! # n \$ 2 ! 1 t ! " ( ) % & ( d " * x t ( ) = e ! \$# n 2 m # n \$ 2 ! 1 F " ( ) 0 t ) e \$# n " e # n \$ 2 ! 1 t ! " ( ) ! e ! # n \$ 2 ! 1 t ! " ( ) % & ( d "

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3- 18 3.16 Calculate the response of an underdamped system to the excitation given in Figure P3.16. Plot of a pulse input of the form f ( t ) = F 0 sin t . Figure P3.16 Solution: x t ( ) = 1 m ! d e " #! n t F \$ ( ) e #! n \$ sin ! d t " \$ ( ) % & ( d \$ 0 t ) F t ( ) = F 0 sin t ( ) t < * From Figure P3.16 ( ) For t + * , x t ( ) = F 0 m ! d e " #! n t sin \$ e #! n \$ sin ! d t " \$ ( ) ( ) d \$ 0 t ) x t ( ) = F 0 m ! d e " #! n t \$ 1 2 1 + 2 ! d + ! n 2 % & ( e #! n t ! d " 1 ( ) sin t " #! n cos t % & ( " ! d " 1 ( ) sin ! d t " #! n cos ! d t { } % & ) ) + 1 2 1 + 2 ! d + ! n 2 % & ( e #! n t ! d " 1 ( ) sin t " #! n cos t % & ( + ! d " 1 ( ) sin ! d t " #! n cos ! d t { } ( * * For ! > " , : f ( ! ) h ( t " ! ) d ! 0 t # = f ( ! ) h ( t " ! ) d ! 0 \$ # + (0) h ( t " ! ) d ! \$ t #
3- 19 x t ( ) = F 0 m ! d e " #! n t sin \$ e #! n \$ sin ! d t " \$ ( ) ( ) d \$ 0 % & = F 0 m ! d e " #! n t 1 2 1 + 2 ! d + ! n 2 " # \$ % e &! n t ! d 1 ( ) sin ! d t ( ( ) " # \$ % &! n cos ! d t ( ( ) " # \$ % " # \$ % ! d 1 ( ) sin ! d t &! n cos ! d t ) * + , + - . + / + " # 0 0 + 1 2 1 + 2 ! d + ! n 2 " # \$ % e &! n t ! d + 1 ( ) sin ! d t 1 ( ) " # \$ % + &! cos ! d t ( ( ) " # \$ % " # \$ % + ! d 1 ( ) sin ! d t &! n cos ! d t ) * + , + - . + / + \$ % 2 2 Alternately, 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 ( ) = F 0 H t ( ) ! H t ! " ( ) ( ) sin t ( ) Normalizing the equation of motion yields !! x t ( ) + 2 !" n ! x t ( ) + " n 2 x t ( ) = f 0 H t ( ) # H t # \$ ( ) ( ) sin t ( ) where f 0 = F 0 m and m , c and k are such that 0 < ! < 1 . Assuming initial conditions, transforming the equation of motion into the Laplace domain yields X s ( ) = f 0 1 + e ! " s ( ) s 2 + 1 ( ) s 2 + 2 #\$ n s + \$ n 2 ( ) The above expression can be converted to partial fractions X s ( ) = f 0 1 + e ! " s ( ) As + B s 2 + 1 # \$ % & ( + f 0 1 + e ! " s ( ) Cs + D s 2 + 2 )* n s + * n 2 # \$ % & ( where A , B , C , and D are found to be

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3- 20 A = ! 2 "# n 1 ! # n 2 ( ) 2 + 2 "# n ( ) 2 B = # n 2 ! 1 1 ! # n 2 ( ) 2 + 2 "# n ( ) 2 C = 2 "# n 1 ! # n 2 ( ) 2 + 2 "# n ( ) 2 D = 1 ! # n 2 ( ) + 2 "# n ( ) 2 1 ! # n 2 ( ) 2 + 2 "# n ( ) 2 Notice that X s ( ) can be written more attractively as X s ( ) = f 0 As + B s 2 + 1 + Cs + D s 2 + 2 !" n s + " n 2 # \$ % & ( + f 0 e ) * s As + B s 2 + 1 + Cs + D s 2 + 2 !" n s + " n 2 # \$ % & ( = f 0 G s ( ) + e ) * s G s ( ) ( ) Performing the inverse Laplace Transform yields x t ( ) = f 0 g t ( ) + H t ! " ( ) g t ! " ( ) ( ) where g ( t ) is given below g t ( ) = A cos t ( ) + B sin t ( ) + Ce ! "# n t cos # d t ( ) + D ! C "# n # d \$ % & ( ) e ! "# n t sin # d t ( ) ! d is the damped natural frequency, ! d = ! n 1 " # 2 . Let m =1 kg, c =2 kg/sec, k =3 N/m, and F 0 =2 N. The system is solved numerically. Both exact and numerical solutions are plotted below
3- 21 Figure 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)); wn=sqrt(k/m); f0=F0/m; % Calculate the damped natural frequency wd=wn*sqrt(1-zeta^2); % Below is the common denominator of A, B, C and D (partial fractions % coefficients) dummy=(1-wn^2)^2+(2*zeta*wn)^2; % Hence, A, B, C, and D are given by A=-2*zeta*wn/dummy; B=(wn^2-1)/dummy;

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• Spring '09
• J.G.Lee
• Sin, Impulse response

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