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Therefore the weak form becomes 20 1 1 4 1 1 1 2 1 1

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Unformatted text preview: 2 1 For integration of first order: 0, 2 , 2. 2.0 1 1 2 00 1 1 2 00 1 0 2 2 For integration of 2nd order: 0.5773, , 0.5773, 0.5773 1 1 0.5773 2 1 0.333 1 0.333 1 1 2 0.5773 0.5773 0.5773 0.5773 0.5773 0.122 1.366 0.5773 1.488 1.488 0.86 0.86 1.0 0.86 0.86 0.666 0.86 1 1 2 1 1 2 1.366 0.5773 0.5773 0.5773 0.5773 0.122 0.666 Problem No 3 1 1 4 1 1 4 1 1 4 1 1 4 Part (i) , , 0.005 0.005 0 0.003 0 0 0.004 0 1 1 1 1 : 0 0 1 0 0 1 1 41 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0.005 1 4 0.005 1 Now at the center of the element: , 0.005 0.005 0 0.003 0 0 0...
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This test prep was uploaded on 02/26/2014 for the course EML 5526 taught by Professor Staff during the Spring '08 term at University of Florida.

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