HW06_Solution - EML 4507 FEA & DESGN Fall 2010 HW06...

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Unformatted text preview: EML 4507 FEA & DESGN Fall 2010 HW06 Solution 18. Use FEM to solve the plane truss shown below. Assume AE = 10 6 N, L = 1 m. Determine the nodal displacements, forces in each element and the support reactions. Solution: Connectivity table Element LN #1 LN #2 AE L l 1 4 1 10 6 N 1.414 m 135 -.7071 2 4 2 10 6 N 1 m 180 -1 3 4 3 10 6 N 1.414 m -135 -.7071 Individual element stiffness matrix after applying transformation 2 2 2 2 2 2 2 2 cos cos sin cos cos sin cos sin sin cos sin sin [ ] cos cos sin cos cos sin cos sin sin cos sin sin EA L -- -- = -- -- k Element 1: 4 4 6 1 1 .354 .354 .354 .354 .354 .354 .354 .354 [ ] 10 .354 .354 .354 .354 .354 .354 .354 .354 u v u v -- -- = -- -- k Element 2: 4 4 6 2 2 1 1 0 [ ] 10 1 0 1 u v u v - = - k 1 2 3 4 1 2 3 10,000 N L L L x y Element 3: 4 4 6 3 3 .354 .354 .354 .354 .354 .354 .354 .354 [ ] 10 .354 .354 .354 .354 .354 .354 .354 .354 u v u v -- -- = -- -- k Global stiffness matrix after assembly 6 6 6 .354 .354 .354 .354 .354 .354 .354 .354 1 1 [ ] 10 .354 .354 .354 .354 .354 .354 .354 .354 .354 .354 1 0 .354 3.54 1.71 .016 10 .354 .354 .354 .354 .016 10 .707 s-- -- -- - = -- -- ---- --- K 1 1 2 2 3 3 4 4 u v u v u v u v Applying boundary conditions 6 6 6 .354 .354 .354 .354 .354 .354 .354 .354 1 1 10 .354 .354 .354 .354 .354 .354 .354 .354 .354 .354 1 0 .354 3.54 1.71 .016 10 .354 .354 .354 .354 .016 10 .707-- -- -- - -- -- ---- --- 1 1 2 2 3 3 4 4 10000 x y x y x y F F F F F F u v = - Striking rows and columns yields the following matrix equation:...
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This note was uploaded on 06/07/2011 for the course EML 4507 taught by Professor Kim during the Fall '11 term at University of Florida.

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HW06_Solution - EML 4507 FEA & DESGN Fall 2010 HW06...

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