Ch4-1(1-2)c - 1 MAE351 Mechanical Vibrations Lecture 15...

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1 1 MAE351 Mechanical Vibrations Lecture 15 (Chap. 4.1-4.2) 2 Chapter 4. Multiple DOF Systems (adapted from B&K Technical Note BA767412)
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2 3 Continuous System in Multiple DOF 4 4.1 Two Degrees of Freedom x 1 x 2 k 1 k m 1 m 2 2 FBD x 1 x 2 m 1 m 2 k 1 x 1
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3 5 Equations of Motion   11 2 1 2 () mx t kx t k x t x t    22 2 2 1 Rearranging: 1 2 1 2 2 21 () ( ) () () 0 mx tk k x x t   6 Initial Conditions • Two coupled, second-order, ordinary differential equations with constant coefficients • Needs 4 constants of integration to solve • Thus, 4 initial conditions on positions and velocities are required to solve velocities are required to solve:
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4 7 Solution by Matrix Methods 111 222 () x tx t t, t x t     xxx  11 2 2 22 2 0 0 mk k k , k       MK Mx Kx 0 I. C. 10 10 20 20 (0) , (0) x x x x xx 8 Solution form: ( ) ; unknown jt te xuu 0     - - M Ke u0 u 0 ODE Algebraic Eq: 2 algebraic eqs in 3 unknowns: Condition for Solution: 1 1) If det[- ] , [- ] exists (trivial) M KM K  0u 0 21 2) For (nontrivial) [- ] should not exist. or  :1 eq. in 1 unknown
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5 9 Characteristic eq. & eigenvectors 2 2 112 2 2 22 2 - 0 0 mkk k MK km k   Calculate   and calculate the corresponding vectors u 1 and u 2 2 11 2 () u0 2 2 2 0 0 Recall : mk k k , k      10 Two simultaneous equations 0 ( ) 0 0 xa y ab y xb y   There are two possibilities: 1) ( ) 0 0; 0 (trivial solution) 2) ( ) 0 or : - two equations are identical - one equation in two unknowns y x   - one equation in two unknowns 0 (non-trivial solution) - only the ratio, not the values, btw and is determined x y a y xy 
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6 11 Numerical example (4.1.5 & 4.1.6) m =9kg =1kg k =24 N/m and =3 1 9 kg, 2 1kg, 1 24 N/m and 2 3 N/m • Characteristic equation becomes 4 -6 2 +8=( 2 -2)( 2 -4)=0 2 =2 and 2 =4 =2 and =4 or 1,3  2 rad/s, 2,4   2 rad/s 12 Computing eigenvectors u 22 For =2, [- ] 930 27 9(2) 3 0 MK uu u u    u0 11 11 11 12 1 12 12 11 12 Let , then 30 33 ( 2 ) 0 u u      u From both equations: Note: only the direction (ratio), not the magnitude can be determined! 2 1 1 0 two identical equations (one eqn in 2 unknowns). Suppose satis  u 2 1 1 1 fies [ ] , so does , arbitrary: ( ) ( ) a a a u
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7 13 Likewise for the second value of 2 : hav w then let 4 = For 21 2 u u 0 0 ) 4 ( 3 3 3 ) 4 ( 9 27 ) (- have we let 4, For 22 21 2 1 22 2 2 u u K M u 0 u 3 1 or 0 3 9 22 21 22 21 u u u u Note that the other equation is the same 14 What to do about the magnitude!
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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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Ch4-1(1-2)c - 1 MAE351 Mechanical Vibrations Lecture 15...

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