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Unformatted text preview: Chapter 8 Multi DegreeofFreedom Vibration: Introductory Topics Problems for Section 8.2 &The Concepts of Sti/ness and Flexibility 1. The two degreeoffreedom system in Figure 8.38 undergoes translational motion. (a) Derive the &exibility in&uence coe¢ cients. (b) Derive the sti/ness in&uence coe¢ cients. (c) Find the inverse of the &exibility matrix and show that this equals the sti/ness matrix. (d) Write the matrix equation of motion. Figure 8.38: Two degreeoffreedom system. Solution: (a) Derive the &exibility in&uence coe¢ cients The &exibility matrix for this system has four elements. Since the &exibility in&uence coe¢ cient matrix is symmetric, we need only derive three coe¢ cients. First, we apply a unit force to mass m 1 : The resulting freebody diagram is shown below. Note that the forces in the vertical direction are excluded because the normal force exerted by the &oor is simply the gravitational force. 1 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 ; and therefore we &nd f 11 = f 21 = 1 k 1 : Next, we apply a unit force to mass m 2 : The resulting freebody 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 ; from which we &nd 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 degreeoffreedom and observe the resulting force. We draw a freebody diagram for motion in the positive x 1 and x 2 directions. 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 : Thus, the matrix of sti/ness in&uence coe¢ cients is: [ k ] = 2 4 k 1 + k 2 ¡ k 2 ¡ k 2 k 2 3 5 : (c) Find the inverse of the &exibility matrix and show that this equals the sti/ness matrix 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 m 2 3 5 : Thus, we may write the matrix equations of motion as...
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This note was uploaded on 09/24/2010 for the course MECHANICAL 14:650:443 taught by Professor Benaroya during the Spring '10 term at Rutgers.
 Spring '10
 Benaroya

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