HW05_Solution - EML 4507 FEA & Design Fall 2010 HW05...

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Unformatted text preview: EML 4507 FEA & Design Fall 2010 HW05 Solution 2) Three rigid bodies, 2, 3 and 4, are connected by six springs as shown in the figure. The rigid walls are represented by 1 and 5. A horizontal force F 3 = 1000 N is applied on Body 3 in the direction shown in the figure. Find the displacements of the three bodies and the forces (tensile/compressive) in the springs. What are the reactions at the walls? Assume the bodies can undergo only translation in the horizontal direction. The spring constants (N/mm) are r 1 = 500, r 2 = 400, r 3 = 600, r 4 = 200, r 5 = 400, and r 6 = 300. Solution: Element equilibrium equations: (1) 1 1 (1) 2 2 1 1 500 1 1 f u u f - = - , (2) 2 2 (2) 4 4 1 1 400 1 1 f u u f - = - (3) 2 2 (3) 3 3 1 1 600 1 1 f u u f - = - , (4) 1 1 (4) 3 3 1 1 200 1 1 f u u f - = - (5) 3 3 (5) 4 4 1 1 400 1 1 f u u f - = - , (6) 4 4 (6) 5 5 1 1 300 1 1 f u u f - = - Nodal equilibrium equations: (1) (4) 1 1 1 (1) (2) (3) 2 2 2 (3) (4) (5) 3 3 3 3 (2) (5) (6) 4 4 4 (6) 5 5 f f R f f f f f f F f f f f R + = + + = + + = + + = = Substituting element equations in the above nodal equilibrium, the structural matrix equation can be obtained as F 3 1 2 3 4 5 6 1 2 3 4 5 1 1 2 3 4 5 5 7 5 2 5 15 6 4 2 6 12 4 1000 100 4 4 11 3 3 3 u R u u u u R -- --- ---- = --- - Applying boundary condition 1 5 u u = = (rigid wall) by deleting first and fifth rows and columns, the global matrix equation can be obtained as...
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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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HW05_Solution - EML 4507 FEA & Design Fall 2010 HW05...

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