MAE 323: Lecture 6. Modeling Topics: Part I. Modeling Topics Alex Grishin MAE 323 Lecture 6 FE Modeling Topics: Part 1

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1 Modeling Topics 1

2 Common element types for structural analyis: oplane stress/strain, Axisymmetric obeam, truss,spring oplate/shell elements o3d solid ospecial: Usually used for contact or other constraints What you need to know before selecting an element: owhat is the dominant mechanism by which work is dissipated? owhat DoF s and derived quantities does the element support? owhat special inputs (if any are required) owhat parent shape(s) does the element support? These considerations make some elements more suitable for certain problems than others 2

3 Beams and Shells θ z1 u z1 1 z 3 θ y1 θ x1 u y1 y X-X x ux1 θ z2 z u z2 2 Cross section X-X The user must input cross sectional properties. θ x2 θ y2 y u y2 u x2 v 1,y v 2 1 θ 1 2 θ 2 x u u 1 2 The traditional Euler-Bernoulli beam element has two nodes, with a third optional orientation node in three dimensions. This is a reduced continuum element. Supported nodal DoF s are: displacement, and slope. The displacements usually include longitudinal extension. Element DoF s are bending stress, strain, The element dissipates work by bending, torsion and linear extension exclusively. This means that only bending, torsion and compressive/tensile stresses and strains are calculated for derived quantities 3

4 Beam and Shells z x y u z1 θ z1 1 u z1 θ z1 4 t θ y1 u y1 u x1 θ x1 θ y1 u y1 u x1 θ x1 2 s r u z θ z 1 1 θ y1 u y1 u x1 θ x1 u z1 θ z1 3 θ y1 u y1 u x1 θ x1 1 t 3 s r 2 Traditional (Mindlin) shell elements come in rectangular (4-node) shapes, as well as triangular (3-node) shapes. These elements are also dominated by bending deformation. Nodal and element DoF s are the same as beams but with an additional spatial dimension Again, the user must supply cross-sectional information 4

5 Beam and Shells z,u z x C P x,u x z,u z C P θ y x,u x Kirchoff Assumption: Straight lines normal to the mid-plane before bending remain straight and normal to midplane after bending. Suitable for thin shells z,u z x C P x,u x z,u z γ xz C P x,u x Mindlin Assumption: straight lines normal to midplane before bending remain straight BUT not necessarily normal after bending. This is due to transverse shear. Suitable for thick shells 5

6 Beam and Shells Today, some commercial FE software products offer fully parametric beam and shell elements like those shown below. The advantage of these types of elements over traditional small-deflection beams and shells is that they carry transverse shear components (and so are better at modeling thick beams and shells), and they more accurately represent large rotations and large curvatures They also offer a fully integrated solution through the cross-section. This is helpful when modeling composites or assessing the affects of cross section warping. The disadvantage of these element types is that they may experience shear locking and/or localized spurious deflections. Also, care must be taken to reproduce Euler- Bernoulli (predominantly bending) behavior 4 4 t t s r 1 2 r s

7 2D Continuum Elements s r s r All the above isoparametric elements (whose shape functions we have already seen) can be used to model plane stress, plane strain, and axisymmetric problems in structural mechanics. The work dissipation mechanism is governed by the elastic internal strain energy relationship (Chapter 4, equation 3). Nodal DoF s are displacement x and y displacement. Element DoF s are stress and strain At this point, the reader should notice a relationship between the number of nodes an element has and the number of shape functions. When we solve the algebraic system equations, the solution vector represents shape function coefficients, which in turn represents solution values at nodal locations. Because of this, an element is said to have nodal degrees of freedom. 7

8 3D Continuum Elements s s t r t r All the above isoparametric elements (whose shape functions are obtained the same way as the 2D elements) may be used to model structural problems over any 3D continuum. Nodal and element DoF s are the same as for plane stress and strain, but with an additional spatial dimension The energy dissipation mechanism is again supplied by the elastic strain energy density relationship (Hooke s Law) 8

9 Spring Elements y z 1 u z1 u y1 U y2 ux x1 2 u x2 u z2 Type 1: Longitudinal Spring 1 2 u z1 u z2 u y2 u y1 u x1 u x2 Type 2: Coincident-Node Spring We have already seen spring elements (Chapter 2). They may be defined in two spatial dimensions or three. The longitudinal spring is usually fully defined by two nodal locations and a longitudinal stiffness, k (as in Chapter 2). The second type of spring (the Coincident-Node spring) is fully defined by two stiffness components in 2D (three in 3D) and two coincident nodal locations. This type of spring may also have either translational OR rotational DoF s (the latter are identical to the former except that the stiffness values are defined in terms of force x distance divided by degree or radian) The energy dissipation mechanism is supplied by Hooke s Law. 9

10 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s In Chapter 2, the student (hopefully) got a feel for what nodal degrees of freedom are when we spoke of spring, truss and beam elements. In that chapter, we saw that they are simply the primary variable (the one we want to solve for) at nodal locations. If we were to count the total number of nodal DoF s an element has, we would see a pattern that looks something m*n, where m is the spatial dimension and n is the number of nodes. For example, if we have a two-dimensional spring element, it appears that we should have 2 x 2 = 4 nodal DoF s. However, in the element coordinate system, this reduces to 1 x 2 =2 nodal DoF s (spring elements don t use the transverse, or y-element direction). Furthermore, a two-dimensional beam element appears to have (with extension) 6 DoF s instead of 4! So, it seems that the DoF count per node for an element depends on : - coordinate reference -degree and form of the governing equation(s) 10

11 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s In an attempt to make this all a little less bewildering, we ll introduce some general principles. It helps to make a distinction between what we ll call canonical DoF s and numerical DoF s. Numerical DoF s will refer to the total number of DoF s per element that we use in constructing the algebraic system. This will usually determine the matrix system size during solution. Canonical DoF s will refer to the theoretical minimum number of DoF s needed to solve the problem, usually for a single characteristic differential equation in a preferred coordinate system (this will be an important idea if you re ever asked to solve a matrix structural problem manually ). This preferred coorinate system is usually obvious in 1 dimension (along the domain), but canget more complicated in two or more dimensions (more on this later) The number of canonical DoF s of an element solving Ndifferential equations may N be found by: c ndof = n p (1) i Where p i is the order of the ith differential equation divided by 2 and the index i ranges over the number of independent differential equations defined over the element and n is the number of nodes. i= 1 11

12 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s The number of canonical degrees of freedom, ndof c is important in physics and relates to the idea of generalized coordinates. This notion helps us to understand what an element can do when it is used to approximate a structure. Example 1: A bar/truss element utilizes a single differential equation: 2 d u EA 2 dx + b x = 0 So, p 1 = 2/2=1 And it has two nodes, n=2 So, by equation (1), ndof c =1*2=2 12

13 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s Example 2: A canonical beam element utilizes a single differential equation: 4 d v EI = 0 So, p 1 = 4/2=2 4 dx And it has two nodes, n=2 So, by equation (1), ndof c =2*2=4 In the previous two examples, it is instructive to compare this number to m*n (the number of spatial dimensions times the number of nodes), because this is the number of total free-particle DoF s(the cumulative number of DoFseach node would have in space individually). In example 1, It is less because the element can only dissipate energy in the direction of its length. In example 2, it is equal to m*n simply by accident (the order of the governing differential equation). The governing differential equation places restrictions on whether a node can move in a given spatial direction, thus implying a preferential coordinate orientation -a set of coordinates in which motion is prescribed by a unique differential equation along each axis. In structural mechanics, this coordinate system will always coincide with principle coordinates 13

14 Element vs Nodal Degrees of Freedom Element name Dimension # DOFs per node where these elements are found when these elements are used Solid ( 2D elastic continuum --pl. stress, strain, axisymmetry) 2D 2 (x and y translation) everywhere applications involving fully elastic domains in only two spatial dimensions Solid (2D elastic continuum) 3D 3 (x,y,z translations) everywhere Applications involving fully elastic domains Plates/Shells 2D Canonical 3 (1 transverse deflection and 2 inplane rotations) textbooks and specialized code Thin structural members under transvers load or moments with two distinct principle curvatures Plates/Shells 3D general 6 (x, y, z translations and x,y,z rotations. Extension and torsion added to canonical) commercial code thin structural members with two distinct principle curvatures, but also extension and surface-normal torsion Beams 1D canonical 2 (1 transverse translation and 1 rotation) textbooks and specialized code long-hand calculations involving structural bending problems of thin members (transverse loads and moments only) Beams 2D general 3 (x,y translations and 1 rotation. Usually, extension is added to canonical behavior) textbooks and commercial code hand calcs and commercial applications involving structural bending problems of thin structures Beams 3D general 6 (x, y, z translations and x,y,z rotations. Extension and Torsion added to canonical) commercial code commercial applications involving thin members subjected to bending, extension, and torsion Springs/bars/trusses 1D canonical 1 (axial translation) textbooks and specialized code hand calcs involving structural members with extension only Springs/bars/trusses 2D general 2 (x and y translation. Same behavior as canonical) textbooks and commercial code hand calcs and commercial applications involving extension in multiple directions Springs/bars/trusses 3D general 3 (x,y, z translations. Same behavior as canonical) commercial code commercial applications involving extension in multiple directions 14

15 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s Thus, this simple assessment of ndof c tells us something very important about truss/bar elements. Without any further calculations or evidence, we can immediately deduce what will happen in the following situation: y x u x =0 1 2 u y =0 F y A truss element fixed at node 1 with a transverse load at node 2 We already know from the previous example that this element cannot dissipate energy in the y-direction (only along it s length). So, what does this mean in terms of the finite element solution? 15

16 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s To reinforce this line of reasoning, we can always check it by actually solving the finite element system! We utilize equation 12b of Chapter 2: F 2 2 c cs c cs 1 1x 2 2 v F cs s cs s 1 1y = 2 2 L c cs c cs u F 2 2x 2 2 cs s cs s v F 2 2 y EA u 16

17 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s Using the technique we learned there (slide 12 of Chapter 2) of striking out DoF s that are set to zero: F 2 2 c cs c cs 1 1x 2 2 v F cs s cs s 1 1y = 2 2 L c cs c cs u cs s cs s v F 2 y EA u EA c 2 cs 0 = 2 L cs s Fy But recall that c and s are the direction cosines the element makes with the coordinate axes: 17

18 Element vs Nodal Degrees of Freedom Canonical DoF s and Numerical DoF s So, we have: E A L = Fy This equation has no solution (k is singular)! On the other hand, if we had used a beam element (slide 28 of Chapter 2) and set rotation at node 1 equal to zero, we wouldget a solution. So, if we need a line element that carries a transverse load, use beams (this is the difference between trusses and frames)! This is a major theme of ROM (Reduced Order Modeling - using reduced continuum elements) We ll leave the student with a final thought (and exercise). Count the DoF s of the beam of slide 28 in Chapter 2 and compare to equation (1). Then compare this number to the rigid-body DoF s m*n and give an explanation 18

19 Element vs Nodal Degrees of Freedom So far, we have discussed nodal DoF s, which usually represent the primary variable(s) at nodal locations. However, just as the slope of the transverse deflection of beam elements may constitute yet another DoF, quantities involving other deflection derivatives (often called derivedquantities) may also constitute DoF s of their own. The most notable examples of such quantities are stress and strain. However, in continuum elements, numerical integration results in an odd phenomenon. Recall that in the FEM, we are actually solving integral equations and the integrals are performed numerically with quadrature rules. This produces some error. This error gets amplified when we take derivatives*, which we must do to calculate stresses and strains (see slides 15 and 16 of Chapter 4). This error is not great (and diminishes with increased order of shape functions used), but is enough to give stress and strain contours unphysical and spurious patterns *This is usually the case when estimating derivatives approximately 19

20 Element vs Nodal Degrees of Freedom As an example, consider a two-element (2 nd order Serendipity) plane stress model with an applied uniform shear stress at the free end: y s r s r v=-1 x Now, let s look at the internal shear stress. Before we continue, students should be able to predict what the analytical result would be 20

21 Element vs Nodal Degrees of Freedom But look at the shear stress contours that result if one simply plots the stresses as calculated naively from equation (8) of Chapter 5 (with second degree isoparametric shape functions): 0 r 0 N 0 u σ = C s 0 N v s r σ rs 21

22 Element vs Nodal Degrees of Freedom Specifically, let s focus on the shear stress distribution at s =1/ 3 σ ( s = 1/ 3) xy The location s = 1/ 3 happens to coincide with one of the Gauss points of a two-point rule The figure marks the x-locations of the same Gauss rule for both elements. Note that the averagestress at Gauss point locations is closer to the analytical result. As we refine the elements over this domain, these points would converge faster to the analytical result than other points. r 22

23 Element vs Nodal Degrees of Freedom The fact that derived quantity averages are closer to the analytical solution at Gauss points reflects the superconvergentproperties of such sites. Even though taking derivatives of approximate quantities amplifies their error, it turns out that Gauss points of a rule 1 order less than that used to perform the integration have the same error as the displacements!* Most commercial FE code calculates and stores stress and strains at these Gauss point locations OR at element centroids. Such locations may be thought of as providing the element DoF s (and are often referred to as such) And because these sites provide such increased accuracy for derived quantities, the value of these quantities at other locations is usually extrapolated from these sites *R. D. Cook, D. S. Malkus, and M. E. Plesha, Concepts and Applications of Finite Element Analysis, 3rd ed. New York, NY, USA: John Wiley & Sons,

24 Element vs Nodal Degrees of Freedom Thus, to obtain smoother, more accurate stress and strain contour plots, most modern FE codes extrapolate these quantities according to a scheme like the one shown below for 2 nd order isoparametric elements D 4 4 s q A B p C Here, r and s are the usual isoparametric coordinates, such that Gauss Points 1, thru 4 lie at r, s = ± 1/ 3 To make things a little easier, let a new parametrization, p, q range from ±1 over the Gauss points Thus, the two coordinate systems are related by a scale factor of sqrt(3). In other words: p q = = 3r 3s 3 24 r

25 Element vs Nodal Degrees of Freedom To find element stresses anywhere within the domain, use: σ 4 p = i= 1 N So, if we wanted to calculate σ x (say)at point A (node 1: p=q=-sqrt(3)), we would have: p i Where σ i is any stress component at the four Gauss point locations, and the N i are now bilinear shape functions in p and q given by: σ ( 1 p)( 1 q) ( 1+ p)( 1 q) ( 1+ p)( 1+ q) ( 1 p)( 1+ q) 1 N( p, q) = 4 σ = 1.866σ 0.500σ σ 0.500σ i xa x1 x2 x3 x 4 25

26 Element vs Nodal Degrees of Freedom Similar extrapolation formulas can be used for triangular, as well three dimensional tetrahedral and hexahedral elements Once the extrapolation is performed for each element, coincident stress/strain values at corner nodes are averaged. In commercial systems, such plots are often referred to as averaged, or nodal stresses (unmodified stress values at Gauss points are referred to as element stresses). With schemes such as this, only corner node stresses are calculated and stored. Below is an averaged shear stress plot of the problem on slide 17 Even with just two elements, the stress plot is now much closer to the analytical value. This plot at least now starts to make physical sense 26

27 Element vs Nodal Degrees of Freedom Unaveraged element stresses and strains also provide a good measure of mesh quality. This usually manifests itself as elevated stress or strain levels in poorly shaped elements, as shown below Ω 3 Ω 2 Poorly shaped element Leads to spurious strain result Ω 4 Ω 1 27

28 The Patch Test and Energy Error In the last lecture, we learned about the element shape metrics. Most commercial FE software comes with some such metric to alert the user when element aspect ratios, lengths, or angles become so distorted that the solution accuracy may suffer. There are also ways the user may check element quality directly One way is to conduct a patch test. This is done by selecting a group of suspect elements and applying either a constant strain or constant stress condition to them. If the element aspect ratios are within acceptable limits (and the underlying formulation is sound), the elements should respond by re-producing the constant strain or stress field exactly. Wherever this is not the case, the user can conclude that the elements may be excessively distorted. 28

29 The Patch Test and Energy Error p y x Example showing a test for constant σy u x = 0 u y = 0 29

30 Implementing the patch test in Mechanical APDL Below is an example of apply a constant σ y condition on an element patch p y u x =0 x u y =0 30

31 Implementing the patch test in Mechanical APDL Results of the constant stress patch test: y x Problem elements 31

32 Implementing the patch test in Mechanical APDL Below is an example of applying a constant strain condition to the element patch. This is done by applying a bi-linear displacement field to all nodes Displacement Gradient y Y X x 32

33 Implementing the patch test in Mechanical APDL The resulting strain in the x-direction y x 33

34 The Patch Test and Energy Error Most commercial FE codes also have the capability of plotting the strain energy or strain energy density. This is usually done on an unaveraged (or element) basis. This corresponds to element strain energy, which is proportional to the square of the stress. Because of this, it is easy to see differences between adjacent elements. If the differences are large, that may indicate element problems 34

35 Structural Symmetry And Modeling Boundary Conditions We have already encountered two types of symmetry found in structural elastostatic problems: planar and axisymmetry. Both of these types of symmetry exploit the fact that a domain s loads, boundary conditions, and geometry admit a planar section in which all other parallel planar sections yield an identical structural response. Other types of symmetry one can encounter in structural problems involve boundary conditions. These are: Symmetry Boundary Condition Anti-Symmetry Boundary Condition Cyclic Symmetry Boundary Condition 35

36 Structural Symmetry osymmetry Boundary Condition A symmetry boundary condition is a surface on which primary solution variables (displacement in structural models) are set to zero normal to the surface. Such a boundary condition usually reflects a theoretical dividing line through a domain both sides of which behave identically. The figure below shows a planar model (one type of symmetry), which is identically loaded at opposite ends. The geometry and loading admit a plane of symmetry which can be used to cut the model in half u x =0 y Half-symmetry model x 36

37 Structural Symmetry oanti-symmetry Boundary Condition An anti-symmetry boundary condition is a surface on which primary solution variables (displacement in structural models) are set to zero within the plane of the surface. Such a boundary condition also reflects a hypothetical dividing line through a domain, but this time, it usually represents a half of structure the other half of which deforms in the opposite direction along the symmety plane. The figure below shows a planar model whose loading suggests a plane of symmetry parallel to it y u x =0 Half-anti- symmetry model x 37

38 Structural Symmetry ocyclic Symmetry Boundary Condition A cyclic symmetry boundary condition can be imposed on a structure if it has repeated or identical features (or parts) around a central axis of revolution. S 2 θ S 1 y θ S = θ S 1 2 r x 38

39 More About Shell/Plate Elements 39

40 DoF s at each node: ux,uy,uz,θx, θy, θz Solid (tet) elements Shell elements Shells are a planar extension of beam elements. They are used to model thin structures whose primary energy dissipation mechanism involves bending in more than 1 dimension A tubular section modeled with shell elements 40

41 Solid Shell Elements ANSYS also supports a special kind of shell (Solsh190) that actually has the same geometry (node configuration) as a linear hexahedral or prism element It is referred to as a solid shell element. This works by extrapolating displacement and rotation DoF results from the midplane (or neutral axis) to 8 (or six) corner nodes The advantage of this type of element is that it eliminates the need for the analyst to extract mid-plane geometry from solid CAD data in order to define the shell/plate surfaces. It is still a shell element (dominated by bending), but it has full 3D solid element geometry. DoF s at each node: ux,uy,uz 41

42 Below is an example of the type of model for which the solid shell option makes sense. It has several thin members in bonded contact A typical member 42

43 This special solid shell is implemented in Workbench by first inserting a sweep meshing method. Then select Automatic Thin next to Src/Trg Selection Sweep Method Automatic Thin Solid shell elements 43