9._FEM_-_Hw_6

9._FEM_-_Hw_6 - AERSP 301 Finite Element Method Jose...

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AERSP 301 Finite Element Method Finite Element Method Jose Palacios

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Axial deformations in bars Axial deformations in bars We were able to write expressions for the strain energy, U, and external work, W, for a bar under axial loading. For the 1-DOF and 2-DOF spring problems, the next step was to apply the stationary principle. – For the 1-DOF system, gave the equilibrium equation. Likewise, for the 2-DOF system, gave the (system of two) equilibrium equations. As in HW #5, Problem 2
Axial deformations in bars Axial deformations in bars For the bar, though being a continuum structure – it has an infinite number of degrees of freedom. – Displacement at every x will vary, u = u(x). Second issue – if you were applying the stationary principle, differentiating is a problem because the expressions for U and W have integrals. The finite element method represents continuum structures (with an infinite number of degrees of freedom) into a system with a finite number of degrees of freedom. – Represents structure in an approximate sense The structure is subdivided (“discretized”) into a number of segments (or finite elements)

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Axial deformations in bars Axial deformations in bars Variation of displacement (u) within each element is assumed to be very simple, such that U and W may be easily integrated across the element length U and W are related to displacements at a finite number of points which are the end-points of the element Solution of the problem is then reduced to the system of an algebraic system for a finite set of “end point” or NODAL values.
Axial deformations in bars Axial deformations in bars Entire Bar: Displacements (u’s) must be continuous along length of bar (no jumps)

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Axial deformations in bars Axial deformations in bars When the elements of the bar are reassembled: This GLOBAL system has 4 DOFs (q 1 …q 4 ) Called GLOBAL degrees of freedom = Global nodal displacements Each element has 2 DOFs: – Called LOCAL DOFs or Element DOFs or Element nodal displacements
Axial deformations in bars Axial deformations in bars Every local DOF corresponds to a nodal displacement Looking closely at a single element: u 1 , u 2 : local DOFs x 1 , x 2 : element nodal coordinates l   = x 2 – x 1 = length of element

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Axial deformations in bars
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9._FEM_-_Hw_6 - AERSP 301 Finite Element Method Jose...

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