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SolSec8.1

# SolSec8.1 - Chapter 8 Problems and Solutions Section...

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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 = ΕΑ λ 1 -1 -1 1 Μ = ρΑλ 6 2 1 1 2 Μ -1 Κ = 6 Ε ρλ 2 1 -1 -1 1 λ 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, ϖ 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, ϖ 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 .

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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 ρ Al 3 ú ú u 2 ( t ) + EA l u 2 ( t ) = 0 The spring has the effect of adding stiffness K at u 2 . Thus, ρ Al 3 ú ú u 2 ( t ) + EA l + K æ è ö ø u 2 ( t ) = 0 From (1.16) ϖ = 3( Kl + EA ) r Al
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