MAE 323: Chapter 6. Structural Models

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1 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 2011 Alex Grishin MAE 323 Chapter 6 1

2 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. Element DoF s are bending stress, strain, The element dissipates work by bending and linear extension exclusively. This means that only bending and compressive/tensile stresses and strains are calculated for derived quantities 2011 Alex Grishin MAE 323 Chapter 6 2

3 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 2011 Alex Grishin MAE 323 Chapter 6 3

4 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 2011 Alex Grishin MAE 323 Chapter 6 4

5 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 small 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 Alex Grishin MAE 323 Chapter

6 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 Alex Grishin MAE 323 Chapter 6 6

7 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) 2011 Alex Grishin MAE 323 Chapter 6 7

8 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 Alex Grishin MAE 323 Chapter 6 8

9 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) 2011 Alex Grishin MAE 323 Chapter 6 9

10 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 (this will be an important idea if you re ever asked to solve a matrix structural problem manually ). The number of canonical DoF s of an element solving Ndifferential equations may N be found by: c ndof = n pi (1) Where p i is the order of the ith differential equation divided by 2 The number of numerical DoF s will be equal to or larger than this depending on the number of independent spatial directions, m: i= 1 m N n ndof = n pij j= 1 i= 1 (2) 2011 Alex Grishin MAE 323 Chapter 6 10

11 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 u EA + b 2 x = 0 So, p 1 = 2/2=1 x And it has two nodes, n=2 So, by equation (1), ndof c =1*2=2 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 rigid body DoF s. It is less because the element can only dissipate energy in the direction of its length 2011 Alex Grishin MAE 323 Chapter 6 11

12 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? 2011 Alex Grishin MAE 323 Chapter 6 12

13 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 2011 Alex Grishin MAE 323 Chapter 6 13

14 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: 2011 Alex Grishin MAE 323 Chapter 6 14

15 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 2011 Alex Grishin MAE 323 Chapter 6 15

16 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 constitutes yet another DoF, quantities involving other deflection derivatives (often called derivedquantities) 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 taking derivatives of approximate quantities 2011 Alex Grishin MAE 323 Chapter 6 16

17 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 2011 Alex Grishin MAE 323 Chapter 6 17

18 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 r s σ xy 2011 Alex Grishin MAE 323 Chapter 6 18

19 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 Alex Grishin MAE 323 Chapter 6 19

20 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, Alex Grishin MAE 323 Chapter 6 20

21 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 Alex Grishin MAE 323 Chapter 6 21 r

22 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 Alex Grishin MAE 323 Chapter 6 22

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

23 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 2011 Alex Grishin MAE 323 Chapter 6 23

This usually manifests itself as elevated stress or strain levels in poorly

24 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 Alex Grishin MAE 323 Chapter 6 24

25 Structural Symmetry 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 2011 Alex Grishin MAE 323 Chapter 6 25

26 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 2011 Alex Grishin MAE 323 Chapter 6 26

27 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 2011 Alex Grishin MAE 323 Chapter 6 27

28 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 2011 Alex Grishin MAE 323 Chapter 6 28

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