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Unformatted text preview: Problems and Solutions Section 8.2 (8.8 through 8.20) 8.8 Consider the bar of Figure P8.3 and model the bar with two elements. Calculate the frequencies and compare them with the solution obtained in Problem 8.3. Assume material properties of aluminum, a crosssectional area of 1 m, and a spring stiffness of 1 10 6 N/m. Solution: The finite element model for the twoelement bar is M u ( t ) + ( 29 = 0 where u ( t ) = 1 2 [ ] u 1 u 2 M = 12 4 1 1 2 = 2 211 1 As in problem 8.3, the spring adds a stiffness K to degree of freedom 2. The equation of motion is then Al 12 4 1 1 2 u ( t ) + 2 EA l 2 1 1 1 + Kl 2 EA u ( t ) = The natural frequencies can be found by eigenanalysis. Using the material properties of aluminum = 2700kg/m 3 , E = 7 10 10 Pa 1 = 129.0 rad/s 2 = 368.4 rad/s The solution obtained in problem 8.4 is 1 = 149.1 rad/s. 8.9 Repeat Problem 8.8 with a threeelement model. Calculate the frequencies and compare them with those of Problem 8.8. Solution: The finite element model of the 3 element rod for equal length elements is (from equation (8.25)) Al 18 4 1 1 4 1 1 2 u + 3 EA l 2 1 1 2 1 1 1 u = With the spring stiffness included, the global stiffness becomes K = 3 211 211 1 + 3 Solving for the natural frequencies gives 1 = 125.85 rad/s, 2 =333.1 rad/s, and 3 = 591.7 rad/s The natural frequencies predicted in 8.9 should be better than those predicted in 8.8. You can compare them to the results of 2 element model by using VTB8_2 and loading the file p8_3_10.con. 8.10 Consider Example 8.2.2. Repeat this example with node 2 moved to /2 so that the mesh is uniform. Calculate the natural frequencies and compare them to those obtained in the example. What happens to the mass matrix? Solution: (8.10, 8.11) The equation of motion can be shown to be Al 12 4 1 1 2 u + 2 EA l 2 1 1 1 u = w 1 = 1.16114 l E r = 8204.8 rad/s w 2 = 5.6293 l E r = 28663 rad/s The first natural frequency is slightly improved (closer to the distributed parameter true value) while the second natural frequency has become worse. Truth Example 8.22 Problem 8.10 Example 8.2.1 1.571 1 l E 1.643 1 l E 1.611 1 l E 1.579 1 l E 4.712 1 l E 5.196 1 l E 5.629 1 l E 5.167 1 l E The natural frequencies found using the 3 element model are much better than the 2 element model. 8.11 Compare the frequencies obtained in Problem 8.10 with those obtained in Section 8.2 using three elements....
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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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