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Finite Element Method AERSP 301 Finite Element Method Jose Palacios July 2008 Slide 2 Today HW 5 has been uploaded (Stationary Principle, MOS FOS) Short assignment Due Thursday HWs 4 due Tuesday I will upload solutions Wednesday evening HWs 3 & 4 due no later than Friday Solutions will be uploaded Friday evening No Class on Thursday Friday during class, Exam Review Exam Time Equation Sheet Start Finite Element Method - Bars Slide 3 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 Slide 4 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) Slide 5 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. Slide 6 Axial deformations in bars Entire Bar: Displacements (us) must be continuous along length of bar (no jumps) Slide 7 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 Slide 8 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 Slide 9 Axial deformations in bars Global x-coordinate system is difficult to use so we transform to a non-dimensional local coordinate system (s-coordinate) such that: At x = x 1, s = 0 At x = x 2, s = 1 Transformation expressed as where (1-s) and (s) are mapping functions and dx = l ds where l is the length scaling for mapping Slide 10 Axial deformations in bars For individual element, express the displacements within the elements (in terms of values of nodes) For this, assume some displacement profile or shape within the element (in terms of element nodal displacements u 1 and u 2 ) An obvious choice is a linear profile wherein Such that at And at Slide 11 Axial deformations in bars This is the expression for the axial displacement, u, within an element (at every point within the element) in terms of local coordinates, s, and displacements at end points (nodes) u 1, u 2 : Define N 1 (s) = 1 s, and N 2 (s) = s displacement shape functions Shape functions are always in the local coordinate system Recall that The mapping functions are identical to the shape functions for this problem (this is not always the case) where M 1 and M 2 are mapping functions Slide 12 Axial deformations in bars Element Stiffness Matrix Recall, where Total Potential U Strain Energy W External Work Potential Look at U (Strain Energy) For the bar in extension: Then discretize the bar into a number of elements: Slide 13 Axial deformations in bars Element Stiffness Matrix Since strain energy, U, is a scalar quantity, the integral may be broken up Strain energy may be calculated for each individual element and summed to obtain the total strain energy of the entire structure: Consider a single element: Also Recall: Mapping function Assumed displacement within the element with shape f n s, N 1 (s) & N 2 (s) Slide 14 Axial deformations in bars Element Stiffness Matrix Then This partial derivative can be written in two forms: Introducing into the strain energy for the ith element: EA a function of s? Are u 1 and u 2 functions of s? How? Recall: Slide 15 Axial deformations in bars Element Stiffness Matrix Look at Each term in the matrix may be integrated individually. Slide 16 Axial deformations in bars Element Stiffness Matrix Individual terms are: This yields: or ELEMENT STIFFNESS MATRIX What if an element moves through space without elongation (Rigid Body Translation)? Can an element represent a constant strain condition? Slide 17 Axial deformations in bars Element Load Vector Now look at the External Work Potential, W: Recall Consider the integral term:: Discretize the bar using the finite element method Since Work is also a scalar and can be calculated for individual elements and added. Slide 18 Axial deformations in bars Element Load Vector Look at the contribution from a single element (putting integral in local coordinate system): Slide 19 Axial deformations in bars Element Load Vector Or in vector form: This is the contribution to the External Work Potential of the distributed force, f, over an element. What is the physical implication of representing: Distributed load is represented as two individual forces (one at each node) Slide 20 Axial deformations in bars Element Load Vector Now consider a constant distributed force, f = c Half of total load, cl, is placed at each node. What about other loading conditions? Could integrate exactly across element length, if convenient. Or, could use linear interpolation across the element Try these. Slide 21 Axial deformations in bars Global Load Vector and Global Stiffness Matrix So far, have obtained expressions for strain energy, U i, and external work potential, W i, for any individual element. Now add the contributions from individual elements together to obtain the strain energy, U, and external work potential, W, for the entire structure. This process is called ASSEMBLY. Look at the External Work Potential, W, due to the distributed force, f. Slide 22 Axial deformations in bars Global Load Vector and Global Stiffness Matrix The external work potential for element #1, W 1, can be written as: Similarly, the external work potential for element #2, W 2, can be written as: Slide 23 Axial deformations in bars Global Load Vector and Global Stiffness Matrix And, the external work potential for element #3, W 3, can be written as: The total external work potential due to the distributed force, f, is: Element #1 Element #2 Element #3 Slide 24 Axial deformations in bars Global Load Vector and Global Stiffness Matrix The external work potential due to the tip load, P, is W = q 4 P Similarly, there is a reaction force, R, acting at the root, x = 0 (because the bar is clamped at the end). The external work potential due to this reaction force, R, is W = q 1 R Slide 25 Axial deformations in bars Global Load Vector and Global Stiffness Matrix The total External Work Potential, W, is obtained by adding each load vector What if there were additional concentrated loads acting at some other locations on the bar? Global Load Vector (obtained from elemental load vectors and contributions of concentrated loads) Slide 26 Axial deformations in bars Global Load Vector and Global Stiffness Matrix Now, look at the strain energy, U, of the structure. The strain energy for element #1, U 1, can be written as: Slide 27 Axial deformations in bars Global Load Vector and Global Stiffness Matrix Similarly, the strain energy for element #2, U 2, can be written as: And, the strain energy for element #3, U 3, can be written as: Slide 28 Axial deformations in bars Global Load Vector and Global Stiffness Matrix The total strain energy of the structure: Global Stiffness Matrix (obtained from 2x2 elemental stiffness matrices)