prob_07_02_01 - 2 y 1 x 1 y 3 x 3 y 1 x 2 y 3 x 3 y 2 =...

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Problem 7.2-1 Given: Show that det(J) = 2A for a three-node triangle of arbitrary shape, where det(J) is given by Eq. 7.2-4. Suggestion: Place the triangle in the first quadrant, drop lines from vertices to the x axis and work with trapezoidal areas thus created. Solution: 1 3 2 x y A y 1 y 3 2 x 3 x 1  y 2 y 3 2 x 2 x 3  y 1 y 2 2 x 2 x 1  = 2A y 1 y 3  x 3 x 1  y 2 y 3  x 2 x 3  y 1 y 2  x 2 x 1  = y 1 y 3  x 3 x 1  y 2 y 3  x 2 x 3  y 1 y 2  x 2 x 1  expand x 1 y 2 x 2 y 1 x 1 y 3 x 3 y 1 x 2 y 3 x 3 y 2 Thus: 2A x 1 y 2 x
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Unformatted text preview: 2 y 1 x 1 y 3 x 3 y 1 x 2 y 3 x 3 y 2 = From Eq. 7.2-3 J x 21 y 31 x 31 y 21 = x 2 x 1 y 3 y 1 x 3 x 1 y 2 y 1 = J x 2 x 1 y 3 y 1 x 3 x 1 y 2 y 1 = expand J x 1 y 2 x 2 y 1 x 1 y 3 x 3 y 1 x 2 y 3 x 3 y 2 = Compare 2A with |J| gives: J 2 A =...
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This note was uploaded on 01/11/2011 for the course MAE 5020 taught by Professor Folkman during the Fall '10 term at Utah State University.

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