SolSec8.2

# SolSec8.2 - Problems and Solutions Section 8.2(8.8 through...

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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 cross-sectional area of 1 m, and a spring stiffness of 1 × 10 6 N/m. Solution: The finite element model for the two-element bar is M ú ú u ( t ) + Κ υ( τ 29 = 0 where u ( t ) = υ 1 υ 2 [ ] Τ u 1 u 2 M = ρΑλ 12 4 1 1 2 Κ = 2 ΕΑ λ 2 -1 -1 1 As in problem 8.3, the spring adds a stiffness K to degree of freedom 2. The equation of motion is then ρ Al 12 4 1 1 2 é ë ê ù û ú ú ú u ( t ) + 2 EA l 2 - 1 - 1 1 + Kl 2 EA é ë ê ù û ú u ( t ) = 0 The natural frequencies can be found by eigenanalysis. Using the material properties of aluminum ρ = 2700kg/m 3 , E = 7 × 10 10 Pa ϖ 1 = 129.0 rad/s ϖ 2 = 368.4 rad/s The solution obtained in problem 8.4 is ϖ 1 = 149.1 rad/s.

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8.9 Repeat Problem 8.8 with a three-element 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)) ρ Al 18 4 1 0 1 4 1 0 1 2 é ë ê ê ù û ú ú ú ú u + 3 EA l 2 - 1 0 - 1 2 - 1 0 - 1 1 é ë ê ê ù û ú ú u = 0 With the spring stiffness included, the global stiffness becomes K = 3 ΕΑ λ 2 -1 0 -1 2 -1 0 -1 1 + Κλ 3 ΕΑ Solving for the natural frequencies gives ϖ 1 = 125.85 rad/s, ϖ 2 =333.1 rad/s, and ϖ 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 ρ Al 12 4 1 1 2 é ë ê ù û ú ú ú u + 2 EA l 2 - 1 - 1 1 é ë ê ù û ú u = 0 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 ρ 1.643 1 l E ρ 1.611 1 l E ρ 1.579 1 l E ρ 4.712 1 l E ρ 5.196 1 l E ρ 5.629 1 l E ρ 5.167 1 l E ρ 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. Solution: See the solution for problem 8.10.

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8.12 As mentioned in the text, the usefulness of the finite element method rests in problems that cannot readily be solved in closed form. To this end, consider a section of an air frame sketched in Figure P8.13 and calculate a two-element finite model of this structure (i.e., find M and K ) for a bar with Solution: A ( x ) = π 4 η 1 2 + η 2 - η 1 λ 2 ξ 2 + 2 η 1 η 2 - η 1 λ ξ Two methods exist for creating a finite element model for this wing.
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