SolSec 4_3 - Problems and Solutions for Section 4.3 (4.34...

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Problems and Solutions for Section 4.3 (4.34 through 4.43) 4.34 Solve Problem 4.11 by modal analysis for the case where the rods have equal stiffness (i.e., k 1 = k 2 ), J 1 = 3 J 2 , and the initial conditions are x (0) = 0 1 ! " # $ T and ! x 0 ( ) = 0 . Solution: From Problem 4.11 and Figure P4.11, with k = k 1 = k 2 and J 1 = 3 J 2 : J 2 3 0 0 1 ! " # $ % !! + k 2 ( 1 ( 1 1 ! " # $ % = 0 Calculate eigenvalues and eigenvectors: J ! 1/2 = J 2 ! 1/2 1 3 0 0 1 " # $ $ $ % ! K = J ! 1/2 KJ ! 1/2 = k J 2 2 3 ! 1 3 ! 1 3 1 " # $ $ $ $ % ( det ! K ! ) I ( ) = 2 ! 5 k 3 J 2 + k 2 3 J 2 2 = 0 1 = 5 ! 13 ( ) k 6 J 2 ( * 1 = 1 , and 5 + 13 ( ) k 6 J 2 ( 2 = 2 ! 1 = 5 " 13 ( ) k 6 J 2 # 5 + 13 ( ) k 6 J 2 " k 3 J 2 " k 3 J 2 5 + 13 ( ) k 6 J 2 $ % ( ) ) ) ) ) ) ) v 11 v 12 $ % ( ) ) = 0 # v 11 = 1.3205 v 12 # v 1 = 0.7992 0.6011 $ % ( )
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! 2 = 5 + 13 ( ) k 6 J 2 " # 1 # 13 ( ) k 6 J 2 # k 3 J 2 # k 3 J 2 1 # 13 ( ) k 6 J 2 $ % ( ) ) ) ) ) ) ) v 2 11 v 22 $ % ( ) ) = 0 " v 21 = # 0.7522 v 22 " v 2 = # 0.6011 0.7992 $ % ( ) Now, P = v 1 v 2 ! " # $ = 0.7992 % 0.6011 0.6011 0.7992 ! " # $ Calculate S and S -1 : S = J ! 1/2 P = 1 J 2 0.4614 ! 0.3470 0.6011 0.7992 " # $ % S ! 1 = P T J 1/2 = J 2 1/2 1.3842 0.6011 ! 1.0411 0.7992 " # $ % Modal initial conditions: r 0 ( ) = S ! 1 " 0 ( ) = S ! 1 0 1 # $ % ( = J 2 1/2 0.6011 0.7992 # $ % ( ! r 0 ( ) = S ! 1 ! 0 ( ) = 0 Modal solution: r 1 t ( ) = 1 2 r 10 2 + ! r 10 2 1 sin 1 t + tan " 1 1 r 10 ! r 10 # $ % ( r 2 t ( ) = 2 2 r 20 2 + ! r 20 2 2 sin 2 t + tan " 1 2 r 10 ! r 20 # $ % ( r 1 t ( ) = 0.6011 J 2 1/2 sin 1 t + 2 # $ % ( = 0.6011 J 2 1/2 cos 1 t r 2 t ( ) = 0.7992 J 2 1/2 sin 2 t + 2 # $ % ( = 0.6011 J 2 1/2 cos 2 t r t ( ) = 0.6011 J 2 1/2 cos 1 t 0.7992 J 2 1/2 cos 2 t " # $ $ %
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Convert to physical coordinates: ! t ( ) = S r t ( ) = J 2 1/2 0.4614 " 0.3470 0.6011 0.7992 # $ % ( 0.6011 J 2 1/2 cos ) 1 t 0.7992 J 2 1/2 cos 2 t # $ % % ( ( t ( ) = 0.2774cos 1 t " 0.2774cos 2 t 0.3613cos 1 t + 0.6387cos 2 t # $ % % ( ( where 1 = 0.4821 k J 2 and 2 = 1.1976 k J 2 ,
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4.35 Consider the system of Example 4.3.1. Calculate a value of x (0) and ! x 0 ( ) such that both masses of the system oscillate with a single frequency of 2 rad/s. Solution: From Example 4.3.1, S = 1 2 1/ 3 1/ 3 1 ! 1 " # $ % S ! 1 = 1 2 3 1 3 ! 1 " # $ % From Equations (4.67) and (4.68), r 1 t ( ) = ! 1 2 r 10 2 + r 10 2 1 sin 1 t + tan " 1 1 r 10 ! r 10 # $ % ( r 2 t ( ) = 2 2 r 20 2 + r 20 2 2 sin 2 t + tan " 1 2
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SolSec 4_3 -