Presentation on theme: "AERSP 301 Finite Element Method"— Presentation transcript:
1 AERSP 301 Finite Element Method Jose PalaciosJuly 2008
2 Today HW 5 has been uploaded (Stationary Principle, MOS FOS) Short assignmentDue ThursdayHW’s 4 due TuesdayI will upload solutions Wednesday eveningHW’s 3 & 4 due no later than FridaySolutions will be uploaded Friday eveningNo Class on ThursdayFriday during class, Exam ReviewExamTimeEquation SheetStart Finite Element Method - Bars
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
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 senseThe structure is subdivided (“discretized”) into a number of segments (or finite elements)
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 lengthU and W are related to displacements at a finite number of points which are the end-points of the elementSolution of the problem is then reduced to the system of an algebraic system for a finite set of “end point” or NODAL values.
6 Axial deformations in bars Entire Bar:Displacements (u’s) must be continuous along length of bar (no jumps)
7 Axial deformations in bars When the elements of the bar are reassembled:This GLOBAL system has 4 DOFs (q1…q4)Called GLOBAL degrees of freedom = Global nodal displacementsEach element has 2 DOFs:Called LOCAL DOFs or Element DOFs or Element nodal displacements
8 Axial deformations in bars Every local DOF corresponds to a nodal displacementLooking closely at a single element:u1, u2: local DOFsx1, x2: element nodal coordinatesl = x2 – x1 = length of element
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 = x1, s = 0At x = x2, s = 1Transformation expressed aswhere (1-s) and (s) are mapping functions and dx = lds where l is the length scaling for mapping
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 u1 and u2)An obvious choice is a linear profile whereinSuch that atAnd at
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)u1, u2:Define N1(s) = 1 – s , and N2(s) = s displacement shape functionsShape functions are always in the local coordinate systemRecall thatThe mapping functions are identical to the shape functions for this problem (this is not always the case)where M1 and M2 are mapping functions
12 Axial deformations in bars – Element Stiffness Matrix Recall , where – Total PotentialU – Strain EnergyW – External Work PotentialLook at U (Strain Energy)For the bar in extension:Then discretize the bar into a number of elements:
13 Axial deformations in bars – Element Stiffness Matrix Since strain energy, U, is a scalar quantity, the integralmay be broken upStrain energy may be calculated for each individual element and summed to obtain the total strain energy of the entire structure:Consider a single element:Mapping functionAlso Recall:Assumed displacement within the element with shape fns, N1(s) & N2(s)
14 Axial deformations in bars – Element Stiffness Matrix ThenThis partial derivative can be written in two forms:Introducing into the strain energy for the ith element:How? Recall:EA a function of s?Are u1 and u2 functions of s?
15 Axial deformations in bars – Element Stiffness Matrix Look atEach term in the matrix maybe integrated individually.
16 Axial deformations in bars – Element Stiffness Matrix Individual terms are:This yields:orELEMENT STIFFNESS MATRIXWhat if an element moves through space without elongation (Rigid Body Translation)?Can an element represent a constant strain condition?
17 Axial deformations in bars – Element Load Vector Now look at the External Work Potential, W:RecallConsider the integral term::Discretize the bar using the finite element methodSince Work is also a scalar and can be calculated for individual elements and added.
18 Axial deformations in bars – Element Load Vector Look at the contribution from a single element (putting integral in local coordinate system):
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)
20 Axial deformations in bars – Element Load Vector Now consider a constant distributed force, f = cHalf 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 elementTry these.
21 Axial deformations in bars Global Load Vector and Global Stiffness Matrix So far, have obtained expressions for strain energy, Ui, and external work potential, Wi, 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.
22 Axial deformations in bars Global Load Vector and Global Stiffness Matrix The external work potential for element #1, W1, can be written as:Similarly, the external work potential for element #2, W2, can be written as:
23 Axial deformations in bars Global Load Vector and Global Stiffness Matrix And, the external work potential for element #3, W3, can be written as:The total external work potential due to the distributed force, f, is:Element #1Element #2Element #3
24 Axial deformations in bars Global Load Vector and Global Stiffness Matrix The external work potential due to the tip load, P, is W = q4PSimilarly, 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 = q1R
25 Axial deformations in bars Global Load Vector and Global Stiffness Matrix The total External Work Potential, W, is obtained by adding each load vectorWhat if there were additional concentrated loads acting at some other locations on the bar?(obtained from elemental load vectors and contributions of concentrated loads)Global Load Vector
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, U1, can be written as:
27 Axial deformations in bars Global Load Vector and Global Stiffness Matrix Similarly, the strain energy for element #2, U2, can be written as:And, the strain energy for element #3, U3, can be written as:
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)