MIT2_092F09_sol8

# MIT2_092F09_sol8 - 2.092/2.093 FINITE ELEMENT ANALYSIS OF...

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ω 2 0.3029 1 = 1.7753 , φ = Note: T T M = δ 2 ij , K = ω i δ 0 ij 1 .6739 i j i j 2.092/2.093± F INITE E LEMENT A NALYSIS OF S OLIDS AND F LUIDS F ALL 2009± Homework 8-solution Instructor: Prof. K. J. Bathe Assigned: TA: Seounghyun Ham Due: Problem 1 (20 points): a) static correction p Δ R=R ( M i r i ) i=1 where p=1. 10 1 0 ⎤ ⎡ 0.3029 9.0825 Therefore Δ M 1 r 1 = ⎥ ⎢ 3.029 = 0 0 2 ⎦ ⎣ 0.6739 4.0825 s Calculate K Δ U = Δ R using Gauss elimination. s 2.1498 U 1 0.3029 2.1498 Δ U = and U= = 1.7062(1-cos 1.7753t) + . 0.4832 U 2 0.6739 0.4832 b) 4 -1 1 0 10 K= , M= , R= -1 4 0 2 0 0 0 U=0 ; U=0 Considering the eigenproblem, K = ω 2 M Session 23 Session 25

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2 0.9531 ω 2 = 4.2247 , φ = 2 0.2142 Using U= Φ X where Φ = 1 2 ω 2 0 T 10 3.029 X + 1 X= Φ = (1) 0 ω 2 2 0 -9.531 The generalized solution for (1) is 3.029 Asin ω 1 t+Bcos ω 1 t+ 2 x 1 ω 1 Asin ω t+Bcos ω t+1.7062 X= = = 1 1 x 2 -9.531 Asin ω t 2 t+Bcos ω Asin ω +Bcos
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## This note was uploaded on 02/24/2012 for the course MECHANICAL 2.092 taught by Professor Klaus-jürgenbathe during the Fall '09 term at MIT.

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MIT2_092F09_sol8 - 2.092/2.093 FINITE ELEMENT ANALYSIS OF...

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