FEA Solution Procedure
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1 EA Soltion Procedre (demonstrated with a -D bar element problem) EA Procedre for Static Analysis. Prepare the E model a. discretize (mesh) the strctre b. prescribe loads c. prescribe spports. Perform calclations (solve) a. generate stiffness matrix (k) for each element b. connect elements (assemble K) c. assemble loads (into load vector R) d. impose spports conditions e. solve eqations (KDR) for displacements. Postprocess
2 Deflection of a Bar Element If we fix the left end of a bar (with constant cross section) it s end deflection is given by: P δ δ AE P If the left end is NOT fixed, the relationship between force and deflection is given by: AE AE ( ) ( ) Deflection of a Bar Element These two eqations can be conveniently expressed in matrix form as: AE The different parts are known as: the elemental stiffness matrix the elemental displacement vector d the elemental force vector r This form allows s to easily combine the eqations from all elements of a strctre. k AE
3 The -D Bar Problem Calclate: P E.g. - deflections, - strains, - stresses, and - internal forces. a. Generate Elemental Eqations,,,,,,,,,, Element nmber Node nmber
4 b. & c. Combine Eqations The global set of eqations is obtained by smming the eqations for each force. In matrix form, the reslt is: P R E A Combined Eqations The different parts are known as: - the global stiffness matrix (K) - the global displacement vector (D) - the global force vector (R) I.e.: KDR
5 Combined oads Note also that the internal forces are balanced by the external applied loads and reactions: If there are no external loads on a node, the internal forces mst balance to. P R,,,,,,,,,, d. Spport Conditions In this example we have one spport condition:. If we set in the global set of eqations, then the first colmn of the stiffness matrix is not necessary, and the bottom eqations can be written as: P
6 Spport Conditions As well, the reaction at the bondary can be solved sing the top row eqation: AE A E R Bt this mst be done after, and have been calclated. e. Solving for Deflections The global matrices (with bondary condition rows/colmns removed) are solved for deflections (U). UK - In practice, the compter does not actally calclate K -, bt solves for U directly, sing some techniqe sch as Gassian Elimination. 6
7 .a. Solving for Strain Once we know the deflections, we can calclate the strain for each element. In a one dimensional problem, the strain is given by: or the bar, ε ε d dx.b. Stress and Internal orce Below the yield stress, for a onedimensional problem, stress is given by: The force in each bar can be calclated by: or by: σ Eε σa AE( ) 7
8 What abot Planar (-D) problems? The eqation for a bar element with an arbitrary orientation in planar space is obtained by transforming the local element coordinate system to the global coordinate system. What abot Planar (-D) problems? Mathematically this is done by mltiplying the elemental stiffness eqation by a rotation matrix: 8
9 Varying cross-section A bar element with varying cross-section does not have constant strain, therefore δp/ae can not be sed. We cold develop a new eqation for a bar with non-niform cross-section, bt instead, we approximate the soltion with a set of constant cross-section bar elements. Distribted oad A bar element with distribted loading does not have constant strain. (δp/ae can not be sed.) We cold develop a new eqation for a bar with distribted loading, bt in the inite Element Method, we approximate the soltion with a set of bar elements with loaded nodes in between. 9
10 ormal Procedre for Bar and Beam Elements Another way to calclate the stiffness matrix! Bar Element ormal Method We have shown how to obtain the elemental stiffness eqations for a bar element sing the direct method. We can also obtain these eqations throgh a more general, formal procedre.
11 Bar Element ormal Method or most elements a general formla is sed to calclate k, where B is the strain-displacement matrix and E is the material property matrix. To obtain B for a bar, we mst first find (x). Bar Element ormal Method et s assme that (This is correct for a bar with constant cross-section and no distribted loads.) N is called the shape fnction matrix.
12 Bar Element ormal Method The axial strain is given by: Ths ε x ( - )/. E is simply the elastic modls E (a scalar) dv is A dx, ths Beam Element Direct Method We start with the shape fnctions.
13 Beam Element Direct Method There are two degrees of freedom (displacements) at each node: v and θ z. Each shape fnction corresponds to one displacement eqal to one and all the others eqal to zero. Note that everything we do in this corse assmes that the displacements are small. Beam Element Direct Method Using standard beam deflection formlae and statics, we solve for one colmn of k at a time. E.g., to solve for colmn of k:
14 Beam Element Direct Method The reslt is: which operates on d [v, θ z, v, θ z ] T. Beam Element ormal Method The formal beam element stiffness matrix derivation is mch the same as the bar element stiffness matrix derivation. We start with the formla: The commonality is that d T kd/ gives the strain energy.
15 Beam Element ormal Method Beam Element ormal Method Stress is given by: σ x My / I
16 Beam Element w/axial Stiffness If we combine the above bar and beam stiffness matrices, we get a general beam stiffness matrix with axial stiffness. Uniformly Distribted oads can be represented by eqivalent loads at intermediate nodes. Axially: 6
17 Uniformly Distribted oads aterally: Orientating Element in -D Space Transformation matrices are sed to transform the eqations in the element coordinate system to the global coordinate system, as was shown for the bar element in -D planar space. 7
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