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ASE 365 - HW 7 Solutions

ASE 365 - HW 7 Solutions - Solutions Homework Set 7 1...

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Unformatted text preview: Solutions: Homework Set 7 1. Problem 5.1. Draw FBDs of the two disks. Consider rotation, due to external moments M 1 and M 2 and torques in the two shafts. Equations of motion: I 1 ¨ θ 1 + GJ 1 L 1 + GJ 2 L 2 θ 1- GJ 2 L 2 θ 2 = M 1 I 2 ¨ θ 2- GJ 2 L 2 θ 1 + GJ 2 L 2 θ 2 = M 2 These can be placed in matrix form: I 1 I 2 ¨ θ 1 ¨ θ 2 + GJ 1 L 1 + GJ 2 L 2- GJ 2 L 2- GJ 2 L 2 GJ 2 L 2 θ 1 θ 2 = M 1 M 2 2. Problem 5.12. For free vibration, with I 1 = I 2 = I , GJ 1 = GJ 2 = GJ , and L 1 = L 2 = L , EOMs become I 1 1 ¨ θ 1 ¨ θ 2 + GJ L 2- 1- 1 1 θ 1 θ 2 = The eigenvalue problem becomes 2- 1- 1 1 Θ 1 Θ 2 = λ Θ 1 Θ 2 where λ = ω 2 IL GJ . Characteristic equation: 2- λ- 1- 1 1- λ = λ 2- 3 λ + 1 = 0 Eigenvalues: λ 1 = 3- √ 5 2 , λ 2 = 3 + √ 5 2 . Natural frequencies: ω 1 = 0 . 6180 r GJ IL , ω 2 = 1 . 6180 r GJ IL . Modal vectors satisfy 2- λ i- 1- 1 1- λ i u 1 i u 2 i = and are u 1 = u 11 1 1 . 6180 , u 2 = u 12 1- . 6180 . 1 3. Problem 5.18. For free vibration, let θ ( t ) = U η ( t ) in EOMs, and multiply EOMs by U T to diagonalize both matrices: U T I 1 1 U ¨ η + U T GJ L 2- 1- 1 1 U η = Solutions of the resulting modal EOMs are η i = A i cos ω i t + B i sin ω i t , with ω i ’s found for Problem 5.12. Evaluating at t = 0: θ 1 (0) θ 2 (0) = 1 . 5 = U η (0) = U A 1 A 2 so A 1 A 2 = U- 1 θ 1 (0) θ...
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ASE 365 - HW 7 Solutions - Solutions Homework Set 7 1...

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