SolSec 4.3

# SolSec 4.3 - Problems and Solutions for Section 4.3 (4.31...

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Problems and Solutions for Section 4.3 (4.31 through 4.40) 4.31 Solve Problem 4.10 by modal analysis for the case where the rods have equal stiffness (i.e., , 3 ), 2 1 2 1 J J k k = = and the initial conditions are x (0) = 01 0 0 [] () = T and ˙ . x Solution: From Problem 4.10 and Figure 4.30, with : 3 and 2 1 2 1 J J k k k = = = Jk 2 30 21 11 0 + = ˙˙ θθ Calculate eigenvalues and eigenvectors: JJ KJ K J k J KI k J k J k J k J −− = == ⇒− () =− + = = ⇒= + 12 2 2 2 2 2 2 2 1 2 1 3 0 2 3 1 3 1 3 1 5 33 0 51 3 6 3 6 // ˜˜ det , and λλ λ λω 2 22 ωλ 1 2 11 12 11 12 1 3 6 3 63 3 3 6 0 1 3205 0 7992 0 6011 = + + = = k J k J k J k J k J v v vv . . . v 2 2 11 22 21 22 2 3 6 3 3 3 6 2 0 0 7522 0 6011 0 7992 = + = k J k J k J k J k J v v . . . v Now, P = = 0 7992 0 6011 0 6011 0 7992 .. Calculate S and S -1 :

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SJ PJ SP J J T == −− 12 2 11 2 2 0 4614 0 3470 0 6011 0 7992 1 3842 0 6011 1 0411 0 7992 // .. Modal initial conditions: rS S J 00 0 1 0 6011 0 7992 0 2 1 () = () = = θ / . . ˙ ˙ Modal solution: rt rr t r r t r r 1 1 2 10 2 10 2 1 1 1 110 10 2 2 2 20 2 20 2 2 2 1 210 20 () = + + + + ω ˙ sin tan ˙ ˙ sin tan ˙ J t J t J t J t 1 22 2 0 6011 2 0 6011 0 7992 2 0 6011 () =+ = = . sin . cos . sin . cos π Jt 0 6011 0 7992 2 12 1 2 2 . cos . cos / Convert to physical coordinates: ωω tS r tJ t tt + 2 2 1 2 2 0 4614 0 3470 0 6011 0 7992 0 6011 0 7992 0 2774 0 2774 0 3613 0 6387 / / / . cos . cos . cos . cos . cos . cos where , 1976 . 1 and 4821 . 0 2 2 2 1 J k J k = =
4.32 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 S = = 1 2 13 13 11 1 2 31 1 // From Equations (4.67) and (4.68), rt rr t r r t r r 1 1 2 10 2 10 2 1 1 1 110 10 2 2 2 20 2 20 2 2 2 1 22 0 20 () = + + + + ω sin tan ˙ sin tan ˙ Choose x (0) and ˙ x (0) so that r 1 ( t ) = 0. This will cause the frequency 2 to drop out. For r 1 ( t ) = 0, its coefficient must be zero. 0 or 0 2 10 2 10 2 1 1 2 10 2 10 2 1 = + = + r r r r Choose 0 20 10 = = r r & . Let r 20 = 0 r and 2 / 3 20 = & as calculated in Example 4.3.1. So, r 00 3 2 [] / T and ˙ . r Solve for x (0) and ˙ x 0 : xr 1 12 0 32 05 15 0 () = () = = S S / . . ˙ ˙

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4.33 Consider the system of Figure P4.33 consisting of two pendulums coupled by a spring.
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## This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 4.3 - Problems and Solutions for Section 4.3 (4.31...

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