MIT2_092F09_sol7

# MIT2_092F09_sol7 - 2.092/2.093 FINITE ELEMENT ANALYSIS OF...

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x T 10 + ω 2 1 x= Φ =3.029 (1) 0 The generalized solution for (1) is 3.029 x =Asin ω 1 t+Bcos ω t+ = Asin ω 1 1 1 t+Bcos ω t+1.7062 2 1 ω 1 0 0 From U= U = 0 , x=0 and x=0 Using these initial conditions, x 1 = 1.7062(1-cos ω 1 t) = 1.7062(1-cos 1.7753t) 2.092/2.093± F INITE E LEMENT A NALYSIS OF S OLIDS AND F LUIDS F ALL 2009± Homework 7-solution Instructor: Prof. K. J. Bathe Assigned: TA: Seounghyun Ham Due: Problem 1 (20 points): 4 -1 1 0 10 K= , M= , R= -1 4 0 2 0 0 0 U=0 ; U=0 Considering the eigenproblem, K φ = ω 2 M 0.3029 2 T T T T 2 ω 1 = 1.7753 , 1 = Note: i M j = δ ij , K = ω δ j i ij i 0.6739 0.3029 Using U= Φ X where Φ = = 1 0.6739 Session 16 Session 19

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0.3029 0.5168(1-cos 1.7753t) Therefore, U= Φ X = 1.7062(1-cos 1.7753t)= 0.6739 1.1498(1-cos 1.7753t)
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MIT2_092F09_sol7 - 2.092/2.093 FINITE ELEMENT ANALYSIS OF...

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