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Unformatted text preview: Chapter 8 Problems and Solutions Section 8.1 (8.1 through 8.7) 8.1 Consider the one-element 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 free-free 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 = 1-1-1 1 = 6 2 1 1 2 -1 = 6 2 1-1-1 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 clamped-free rod of length l /2 has the same natural frequency as the second mode of a free-free rod of length l . 8.3 Consider the system of Figure P8.3, consisting of a spring connected to a clamped-free 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 clamped-free 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