ch8sol-HW (1)

ch8sol-HW (1) - Chapter 8 Multi Degree-of-Freedom...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 8 Multi Degree-of-Freedom Vibration: Introductory Topics 1. The two degree-of-freedom system in Figure 8.38 undergoes translational motion. (a) Derive matrix equation of motion. Figure 8.38: Two degree-of-freedom system. Solution: matrix is symmetric, we need only derive three coe¢ cients. First, we apply a unit force to mass m 1 : The resulting free-body diagram is shown below. Note that the forces in the vertical direction 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 CHAPTER 8 MULTI DOF VIBRATION Since k 2 is unstretched, m 1 and m 2 displace as a rigid body, thus F 1 = k 1 x 1 1 = k 1 x 1 ; x 1 = 1 k 1 ; f 11 = f 21 = 1 k 1 : Next, we apply a unit force to mass m 2 : The resulting free-body diagram is shown below. This is a static problem, and the forces on each mass must sum to zero. Then, we have k 1 x 1 = k 2 ( x 2 x 1 ) k 2 ( x 2 x 1 ) = F 2 = 1 ; k 1 x 1 = 1 : Thus, mass m 1 will displace by a distance x 1 = 1 =k 1 : Then from the second equation, x 2 = k 2 x 1 + F 2 k 2 = 1 k 1 + 1 k 2 ; since F 2 = 1 and x 1 = 1 =k 1 : So, the ±exibility in±uence coe¢ cient matrix becomes: [ f ] = 2 4 1 k 1 1 k 1 1 k 1 1 k 1 + 1 k 2 ± 3 5 ; and 8 < : x 1 x 2 9 = ; = 2 4 1 k 1 1 k 1 1 k 1 1 k 1 + 1 k 2 ± 3 5 8 < : F 1 F 2 9 = ; : (b) Derive the sti/ness in±uence coe¢ cients We apply a unit displacement to each degree-of-freedom and observe the resulting force. We draw a free-body diagram for motion in the positive x 1 and x 2 directions.
Background image of page 2
3 First, we apply a unit displacement to the x 1 coordinate and force x 2 = 0 . P 1 and P 2 are the forces necessary to maintain static equilibria. The result is P 1 = ( k 1 + k 2 ) k 11 and P 2 = ± k 2 k 21 : Applying a unit displacement to x 2 and setting x 1 = 0 gives us P 1 = ± k 2 k 12 ; and P 2 = k 2 k 22 : [ k ] = 2 4 k 1 + k 2 ± k 2 ± k 2 k 2 3 5 : To see that [ f ] and [ k ] are inverses, we take the matrix product [ f ][ k ] : [ f ][ k ] = 2 4 ( k 1 + k 2 ) k 1 ± k 2 k 1 k 2 + k 2 k 1 k 2 + k 2 k 1 k 2 + k 2 k 1 + k 2 k 2 3 5 = 2 4 1 0 0 1 3 5 : Since the matrix product [ f ][ k ] is the identity matrix, [ k ] is the inverse of [ f ] ; or, [ f ] 1 = [ k ] : (d) Write the matrix equation of motion. We already have the [ k ] matrix from part (b) of this problem. The [ m ] matrix can be written as [ m ] = 2 4 m 1 0 0 m 2 3 5 :
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/26/2012 for the course 650 443 taught by Professor Benaroya during the Fall '11 term at Rutgers.

Page1 / 12

ch8sol-HW (1) - Chapter 8 Multi Degree-of-Freedom...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online