MIT2_092F09_sol4

MIT2_092F09_sol4 - 2.092/2.093 FINITE ELEMENT OF SOLIDS AND...

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9 2 2 analytically in HW #1. Using, E=1.0×10 N/m , A=0.0025m , a=2m, and R 5 = 300 kN, we found that: 1 + 1 0 0 ⎥ ⎡ U 3 R 3 0 EA 1 1 ⎥ ⎢ 0 + 1 U 5 = R 5 = 300 kN a 2 2 2 2± U 6 R 6 0 1 1 2 2 + 1 2 2 2 2 U 3 0 U = 5 0.1903 m U 6 0.0497 1 1 1 R 1 62.13 0 R 2 ⎥ ⎡ 0 62.13 EA R = 1 ⎥ ⎢ 0 1 0.1903 = 62.13 kN ⎥ ⎢ 4 a 2 2 R 7 0.0497 237.5 1 1 0 R 8 0 2 2 1 0 0 For numbering of R i , see HW #1. These analytical linear elastic results can now be compared to the results calculated numerically by ADINA. 1 1 2 2 2 2 2 2 2 2 2 2 2.092/2.093± F INITE E LEMENT OF S OLIDS AND F LUIDS F ALL 2009± Homework 4- solution
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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_sol4 - 2.092/2.093 FINITE ELEMENT OF SOLIDS AND...

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