{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ASE 365 - HW 8 Solutions

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

This preview shows pages 1–3. Sign up to view the full content.

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 = 0 0 0 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 0 . 6591 0 . 1750 - 4 . 3342 2 . 0710 - 0 . 2693 0 . 4483 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The modal masses m 0 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 0 . 3160 0 . 9223 0 . 2226 0 . 2083 0 . 1614 - 0 . 9647 0 . 6545 - 0 . 2484 0 . 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 1 c 2 - c 1 - c 2 - ac 1 bc 2 ac 1 - bc 2 - c 1 c 1 - c 2 c 2 ˙ x 1 ˙ x 2 ˙ x 3 ˙ x 4 + k 1 k 2 - k 1 - k 2 - ak 1 bk 2 ak 1 - bk 2 - k 1 ( k 1 + k 3 ) - k 2 ( k 2 + k 4 ) x 1 x 2 x 3 x 4 = 0 0 0 0 The matrices are not symmetric. For Lagrange’s equations, we need kinetic energy: T = 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern