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Unformatted text preview: Problems and Solutions Section 8.4 (8.34 through 8.43) 8.34 Refer to the tapered bar of Figure P8.13. Calculate a lumpedmass matrix for this system and compare it to the solution of Problem 8.13. Since the beam is tapered, be careful how you divide up the mass. Solution: The lumped mass at node 2 should be the total mass between x = .25 and x = .75. Therefore M 2 = 2700 π 4 .25 .75 ∫ η 1 2 + η 2 η 1 λ 2 ξ 2 + 2 η 1 η 2 η 1 λ ξ δξ = 26.5 likewise for node 3 M 3 = 2700 π 4 .75 1 ∫ η 1 2 + η 2 η 1 λ 2 ξ 2 + 2 η 1 η 2 η 1 λ ξ δξ = 7.289 The mass matrix is then M = 26.5 7.289 and the natural frequencies are ϖ 1 = 6670 rad/s and ϖ 2 = 13106 rad/s. For the distributed mass system ϖ 1 = 7414 rad/s and ϖ 2 = 20368 rad/s. The first natural frequency found by the distributed mass model is slightly better than the lumped mass model when compared to the three element distributed mass model derived in problem 13. 8.35 Calculate and compare the natural frequencies obtained for a tapered bar by using first, the consistentmass matrix (Problem 8.12), and second, the lumpedmass matrix (Problem 8.34). Solution: See solution for Problem 8.34. 8.36 Consider again the machine punch of Problem 8.16 and Figure P8.15. Calculate the natural frequencies of this system using a lumpedmass matrix and compare...
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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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