ASE 365 - HW 8 Solutions

ASE 365 - HW 8 Solutions - Solutions: Homework Set 8 1. For...

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Unformatted text preview: Solutions: Homework Set 8 1. For Problem 7.2, draw free body diagrams of the three masses. Then Newton’s second law gives X F x 1 = m 1 ¨ x 1 =- k 1 x 1 + k 2 ( x 2- x 1 ) X F x 2 = m 2 ¨ x 2 =- k 2 ( x 2- x 1 ) + k 3 ( x 3- x 2 )- ( k 5 + k 6 ) x 2 X F x 3 = m 3 ¨ x 3 =- k 3 ( x 3- x 2 )- k 4 x 3 EOMs: m 1 m 2 m 3 ¨ x 1 ¨ x 2 ¨ x 3 + k 1 + k 2- k 2- k 2 k 2 + k 3 + k 5 + k 6- k 3- k 3 k 3 + k 4 x 1 x 2 x 3 = For Lagrange’s equations, we need kinetic energy: T = 1 2 3 X i =1 m i ˙ x 2 i = 1 2 [ ˙ x 1 ˙ x 2 ˙ x 3 ] m 1 m 2 m 3 ˙ x 1 ˙ x 2 ˙ x 3 potential energy: V = 1 2 k 1 x 2 1 + k 2 ( x 2- x 1 ) 2 + k 3 ( x 3- x 2 ) 2 + k 4 x 2 3 + k 5 x 2 2 + k 6 x 2 2 = 1 2 [ x 1 x 2 x 3 ] k 1 + k 2- k 2- k 2 k 2 + k 3 + k 5 + k 6- k 3- k 3 k 3 + k 4 x 1 x 2 x 3 and non-conservative virtual work: δW nc = 3 X i =1 Q i,nc δx i = 0 from which the non-conservative generalized forces are all zero. Lagrange’s equa- tions give the same EOMs as above. For the masses and spring stiffnesses given in Problem 7.21, the eigenvalues are { λ 1 , λ 2 , λ 3 } = { 1 . 3409 , 1 . 8250 , 6 . 3342 } k/m and the eigenvectors (not mass-normalized) are ˆ U = [ ˆ u 1 ˆ u 2 ˆ u 3 ] = 1 1 1 . 6591 . 1750- 4 . 3342 2 . 0710- . 2693 . 4483 1 The modal masses m i = u T i M u i , with the eigenvectors above, are { m 1 , m 2 , m 3 } = m { 10 . 0125 , 1 . 1757 , 20 . 1872 } . The mass-normalized eigenvectors, formed by dividing each one above by the square root of its modal mass, are U = [ u 1 u 2 u 3 ] = 1 √ m . 3160 . 9223 . 2226 . 2083 . 1614- . 9647 . 6545- . 2484 . 0998 2. For Problem 7.6, draw free body diagrams of masses m 1 , m 2 and m 3 . Applying Newton’s second law to these three, and its rotational analogue to m 3 , yields m 1 ¨ x 3 = c 1 ( ˙ x 1- ˙ x 3 ) + k 1 ( x 1- x 3 )- k 3 x 3 m 2 ¨ x 4 = c 2 ( ˙ x 2- ˙ x 4 ) + k 2 ( x 2- x 4 )- k 4 x 4 m 3 ¨ x C =- c 1 ( ˙ x 1- ˙ x 3 )- c 2 ( ˙ x 2- ˙ x 4 )- k 1 ( x 1- x 3 )- k 2 ( x 2- x 4 ) I C ¨ θ = c 1 ( ˙ x 1- ˙ x 3 ) a- c 2 ( ˙ x 2- ˙ x 4 ) b + k 1 ( x 1- x 3 ) a- k 2 ( x 2- x 4 ) b Since x C = ( bx 1 + ax 2 ) /L and θ = ( x 2- x 1 ) /L , the equations of motion become m 3 b/L m 3 a/L- I C /L I C /L m 1 m 2 ¨ x 1 ¨ x 2 ¨ x 3 ¨ x 4 + c...
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This note was uploaded on 09/18/2011 for the course ASE 365 taught by Professor Staff during the Spring '10 term at University of Texas at Austin.

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ASE 365 - HW 8 Solutions - Solutions: Homework Set 8 1. For...

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