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Unformatted text preview: Chapter 8 Problems and Solutions Section 8.1 (8.1 through 8.7) 8.1 Consider the oneelement model of a bar discussed in Section 8.1. Calculate the finite element of the bar for the case that it is free at both ends rather than clamped. Solution: The finite element for a rod is derived in section 8.1. Since u 1 is not restrained equations (8.7) and (8.11) are the finite element matrices. 8.2 Calculate the natural frequencies of the freefree bar of Problem 8.1. To what does the first natural frequency correspond? How do these values compare with the exact values obtained from methods of Chapter 6? Solution: K = 111 1 = 6 2 1 1 2 1 = 6 2 111 1 1,2 = 0, 12 E r l 2 and the corresponding eigenvectors are x 1 = 1 1 [ ] T / 2 and x 2 = 1 1 [ ] T / 2 Therefore, 1 = 0, w 2 = 12 E r l 2 The first natural frequency corresponds to the rigid body mode, or pure translation. From the solution to problem 6.8, 1 = 0, w 2 = p 2 E rl 2 The first natural frequency is predicted exactly while the second is 10.2% high. A point of interest is that, due to symmetry, the first mode of a clampedfree rod of length l /2 has the same natural frequency as the second mode of a freefree rod of length l . 8.3 Consider the system of Figure P8.3, consisting of a spring connected to a clampedfree bar. Calculate the finite element model and discuss the accuracy of the frequency prediction of this model by comparing it with the method of Chapter 6. Solution: The finite element for the clampedfree rod is given by (8.14) as Al 3 u 2 ( t ) + EA l u 2 ( t ) = 0 The spring has the effect of adding stiffness K at u 2 . Thus, Al 3 u 2 ( t ) + EA l + K ae u 2 ( t ) = 0 From (1.16) = 3( Kl + EA ) r Al Next consider the first natural frequency as predicted from the distributed...
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This note was uploaded on 11/28/2010 for the course ME 4440 taught by Professor Hill during the Winter '09 term at Detroit Mercy.
 Winter '09
 Hill

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