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HW06_Solution

# HW06_Solution - EML 4507 FEA DESGN Fall 2010 HW06 Solution...

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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 θ 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 θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ - - - - = - - - - 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 0 1 0 0 0 0 0 [ ] 10 1 0 1 0 0 0 0 0 u v u v - = - k 1 2 3 4 1 2 3 10,000 N L L L x y

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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 0 0 0 0 .354 .354 .354 .354 0 0 0 0 .354 .354 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 [ ] 10 0 0 0 0 .354 .354 .354 .354 0 0 0 0 .354 .354 .354 .354 .354 .354 1 0 .354 3.54 1.71 .016 10 .354 .354 0 0 .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 0 0 0 0 .354 .354 .354 .354 0 0 0 0 .354 .354 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 10 0 0 0 0 .354 .354 .354 .354 0 0 0 0 .354 .354 .354 .354 .354 .354 1 0 .354 3.54 1.71 .016 10 .354 .354 0 0 .354 .354 .016 10 .707 - - - - - - - - - - - - - - - × - - - × 1 1 2 2 3 3 4 4 0 0 0 0 0 0 0 10000 x y x y x y F F F F F F u v                                           =                                     -          Striking rows and columns yields the following matrix equation: 6 4 6 4 1.71 10 .016 0 10000 .016 .71 10 u v    ×       =       - ×       Solving for the displacements gives: 4 4 { } {0 14.1}mm u v = -
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HW06_Solution - EML 4507 FEA DESGN Fall 2010 HW06 Solution...

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