ME5554/ME5305 – Tutorial on a 2D Problem with Plasticity and Contact 1. The problem We want to study a 500x20x200mm3 aluminium flat bar loaded in 3-point bending. The testing rig uses 3 60mm steel rollers, as shown in the figure below, and the test is conducted by prescribing the displacement of the top roller.
Figure 1: geometry and boundary conditions.
The width-to-depth (200/20=10) of the cross section can be considered large enough to assume a state of plane stress is a good enough approximation. Hence, we can study the problem with a 2D model. 1.1 How to apply the actual load? Modelling the contact between rollers and beams allows an easier and more accurate representation of the actual load applied on the flat bar by the rollers, both in the middle and on the supports. An alternative method could be the application of a downward displacement only at one point at the mid span, and prescribe the vertical displacements only on two points at the supports, which would generate concentrated forces. In the case of a linear elastic analysis, these concentrated forces would result in infinite values of the stresses at such points in the exact mathematical solution. In the finite-element solution, which is an approximation of the exact mathematical one, the stress would not be infinite, but as the mesh is refined around the points of application of load and reactions, the stress at these points would always increase, and tend to infinity as the element size tends to zero. This normally leads to a relatively small over prediction of displacements, whereas for stresses and strains errors are limited to a small zone around the points of application of load and reactions. For an elasto-plastic problem, concentrated forces result in infinite values of the plastic strains in the exact mathematical solution. This is a bigger problem because, upon mesh refinement, unrealistic large plastic strains are found that, for a refinedPage 1 of 38
enough mesh would lead to an unrealistic collapse. Furthermore, for cases when the assumption of small displacements and strains could normally be made (geometrically linear problem), the actual FE solution would end up violating such assumptions. Alternative options could be of creating small volumes around the points of application of load and reactions, where the constitutive behaviour is assumed to be linear elastic. However, this can lead to excessive errors at the mid-span, where the plastic hinge would not be well captured. Other alternative options can be considered, e.g. with loads/displacement applied on small areas rather than points, and by constraining these small areas to the points of applications of loads and displacements, but these methods require experience and can still lead to errors. With experience, one can also consider the use of a mesh which is coarse enough around the point of applications of load and reactions, to avoid excessive stress concentration, yet not too coarse to have an excessive error in the solution. Again, with limited experience this can lead to big errors.. Therefore, with contact models available in most commercial codes, the actual model of the part of the rollers in contact with the flat bar is probably the most effective option. Disadvantages however exist and are (a) an increased computational cost (more iterations and sometimes smaller increments needed to achieve convergence) and (b) possible difficulty in achieving convergence in the equilibrium iterations of an increment, particularly when friction is used. 2 FE model 2.1 Parts Note that the rollers can be considered stiff enough, so that only half of each of them is modelled. Also, we can exploit symmetry. Therefore, three parts are created in ABAQUS, all of them started as “deformable, 2D planar, shell”: one half flat bar, as a 250x20mm^2 rectangle; one half roller and one quarter roller. The latter are made from a 30mm diameter circle and by suitably using “cut extrude” in the Part Module. All these operations are described in the other 2D and 3D tutorials on BBL, but an example of how to cut the circle into half is shown below. Click on “Create Cut: Extrude”:
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Sketch a rectangle with the top side containing the horizontal diameter:
Click on “Cancel procedure” and then on “Done” to obtain the half roller:
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It is convenient to right-click on the part, which is now the half roller, and click on “Copy”, to make the quarter roller with an analgous “Cut: Extrude” operation.
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2.2 Assembly In Assembly, double-click on Instances, select all the three of them, and for convenience tick on “Auto-offset from other instances”:
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To translate the half-roller in the right position, click on “Translate Instance”:
Then select the half-roller and click on “Done”. Click on the top point of the roller as “start point for the translation vector” and to the left-bottom vertex of the flat bar as the “end point”.
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Start point
End point
Then click on “Ok”. Again, click on the “Translate Instance” button. Select again the half-roller, click on “Done”, and now enter manually the X,Y values 0,0 as start point, follwed by “Enter”, and then “25,0” as end point, again followed by “Enter”.
End point
Then Click on “Ok” and you shoulf find the half-roller in the right place, as below:
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For the quarter-roller, first rotate the instance by clicking on “Rotate Instance”:
Then select the quarter-roller, click on “Done”, select any of the verteces of the halfroller (e.g. the top point, but this is not important) as the centre of rotation, insert “180” as angle of rotation (this is important), click “Enter” and then click “Ok”. Choosing the top point of the quarter roller for the centre of rotation and one would find the result in the figure below. Page 8 of 38
Then repeat the procedure to translate instances, applying it to the quarter-roller and using as start point the bottom point of the roller, and for end point the top-right vertex of the flat bar:
Start point
End point
Click “Ok” and you should find the result below:
To set up the meshes directly on the instances of the assembly instead of in the parts, expand the Instances by clicking on “+”, Page 9 of 38
select all instances in the assembly, right-click an choose “Make Independent”.
2.3 Assembly To mesh the model, the datum planes shown below are used. They are at distances of 20, 40 and 200 (mm) from the left end.
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They have been created by clickin gon “Datum plane”, choosing principal plane YZ as the one to offset, and then using an offset value equal to: X coordinate of left end + distance desired from the left hand. The X coordinate of the left end can be found by using the “Query” button shown below, which is very useful in a large number of situations, selecting “point” and picking up the point of interest. Notice that ‘prospective view’ has been toggled off in the figure. Prospective can be toggled on or off using the button also shown below.
Prospective on/off
Query
Point to query Coordinates of the point
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Using the above datum points, and following methods well described in the other tutorials on BBL, partitions are made and all instances are meshed, using planestrain elements with incompatible modes and without reduced integration. Notice only that in order to select and “seed” in a different way different lines of the model, it is useful to use the “Seed edge” button and the different selection modes shown in the figure below:
Selection modes (self-explanatory) Seed Edges
The meshes created are shown below: .
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2.4 Material model We will consider an aluminum, and will use for it an elastoplastic model with nonlinear isotropic hardening. Let us assume we are given as material parameters the proof strength 𝜎𝑃 , that is the yield stress in correpondence of a plastic strain of 0.2%=0.002 (in a uniaxial test) and the utimate tensile strength (UTS) and the corresponding strain 𝜀𝑈𝑇𝑆 . We will consider the following data: . 𝜎𝑃 = 280 MPa
UTS = 330 Mpa
𝜀𝑈𝑇𝑆 = 0.06 = 6%
The material behaviour in a uniaxial test can be well approximated with the RambergOsgood law: 𝜀=
𝜎 𝜎 𝑛 + 0.002 ( ) 𝐸 𝜎𝑃
where 𝐸 is the Young’s modulus and 𝑛 is a parameter to be determined. For any aluminium the Young’s modulus is 70GPa (or very close to it). Parameter 𝑛 can be determined by enforcing that for 𝜎 = UTS one has 𝜀 = 𝜀𝑈𝑇𝑆 : 𝜀𝑈𝑇𝑆
UTS UTS 𝑛 = + 0.002 ( ) 𝐸 𝜎𝑃
In this case it must be: 310 330 𝑛 0.06 = + 0.002 ( ) 70000 280 which leads to: 𝑛=
1 330 ln [0.002 (0.06 − 70000)] 330 ln (280)
= 20.2
The input data needed for an elastoplastic model with nonlinear isotropic hardening in ABAQUS is a list of stress values and the corresponding plastic strains during the tensile test. One can use the equation: 𝜎
𝑛
𝜎
𝜀𝑝 = 0.002 (𝜎 ) = 0.002 (280)
𝑛=20.2027128
𝑃
which leads to the following table:
stress 0.0001 10 20 30 40 50
plastic strain 1.1245E-133 1.16014E-32 1.40002E-26 5.05424E-23 1.6895E-20 1.53321E-18 Page 13 of 38
total strain 1.42857E-09 0.000142857 0.000285714 0.000428571 0.000571429 0.000714286
𝜎
𝜀 = 𝜀𝑝 + 𝐸
60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 310 320 330
6.09927E-17 1.37336E-15 2.03882E-14 2.20191E-13 1.85023E-12 1.26902E-11 7.36039E-11 3.70843E-10 1.65732E-09 6.67953E-09 2.46038E-08 8.37371E-08 2.65719E-07 7.92143E-07 2.23279E-06 5.98313E-06 1.53141E-05 3.75931E-05 8.88225E-05 0.000202625 0.00044752 0.000959275 0.002 0.004063672 0.008060619 0.015633776 0.029690988 0.055285714
0.000857143 0.001 0.001142857 0.001285714 0.001428571 0.001571429 0.001714286 0.001857143 0.002000002 0.002142864 0.002285739 0.002428655 0.002571694 0.002715078 0.002859376 0.003005983 0.003158171 0.003323307 0.003517394 0.003774053 0.004161805 0.004816417 0.006 0.008206529 0.012346333 0.020062348 0.034262416 0.06
What should be inserted in ABAQUS are the first two columns. The third one is useful to check that for a total strain of 6%=0.06 one has a stress equal to the UTS. Likewise, from the second column one can check that for the platic strain equal to 0.002 the stress is equal to the proof stress. Notice that to have this precisely the more precise value in of 𝑛 = 20.2027128 was used in Excel, otherwise . With 𝑛 = 20.2 the values are nearly the same. One can notice that the plastic strain is nonzero for values very small of the stresses. However, this is just a numerical feature of the Ramberg-Osgood equation, as for values of the stress lower than about 230 the plastic strain is less than 1% of the total strain. Hence, it is a good approximation to assume that the elastic law is valid until a stress of 220, which can be considered as the initial yield stress, and then the rest of the table is used. Hence we are going to input the following table: 220 230 240 250 260
0 3.76132E-05 8.88597E-05 0.000202687 0.00044761
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270 280 290 300 310 320 330
0.000959369 0.002 0.004063285 0.00805911 0.01562946 0.029680234 0.055261077
To this end, in ABAQUS we create a material model Aluminium. We first choose ‘Mechanical’ → ’Elasticity’ → ‘Elastic’ and input 70000 and 0.33 as Young’s modulus and Poisson’s ratio. We then choose ‘Mechanical’ → ’Plasticity’ → ‘Plastic’, and the copy and past the Excel table above into the one in the ABAQUS field below:
with the “Hardening’ option left ‘Isotropic’ (default one) and all the other choices also left as by default. You can then save. For the roller, another elastic material model that you can name ‘High-srength steel’ should be made, with Young’s modulus equal to 70000 (MPa) and Poisson’s ratio equal to 0.27. We can reasonably assume that the the high strength of the steel is sufficient to avoid plasticity, and therefore that no plasticity is included. This is reasonable if in the experiment no residual deformation of the roller is found. To actually assign the material models to the right parts, the usual steps as in the previous tutorials can be used.
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2.5 Modelling contact To model contact, one needs to define slave and master surfaces that are in ‘potential contact’ with each other. To this end, double click on ‘Interactions’:
Name the ‘interaction’ (to be created) as you wish (here ‘Left-roller’ is chosen) and click on Surface-to-surface contact (Standard).
and on ‘Continue’. Select the top edge of the roller as first ‘master surface’: Page 16 of 38
Click on ‘Done’. Now the slave surface needs to be defined. To this end, click on ‘Node region’ to choose slave type and then click on the part of the beam bottom surface that is I contact with the roller:
Click on ‘Done’ and the ‘Edit interaction’ window appears. Leave the default options and click right below on the ‘Create interaction property’ button as shown below:
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Give a name (‘Unilateral-contact’ here) and click on ‘Contact’, if not yet selected, and then on ‘Continue’:
Choose ‘Mechanical’ → ‘Normal behaviour’ → and then leave the default optinon of ‘Hard contact’. Then click ‘Ok’. Do the same operation for the quarter roller. It is better to have the roller edge always as ‘master surface’ because it is likely to have a coarser mesh. This is because it is more accurate to thave the master surface coinciding with the one having the coarser mesh (see manual for more explanation of why this is recommended). Page 18 of 38
2.5 Create a step and apply boundary conditions Create a step with ‘Static Analysis’, here named ‘Load’ with a period of 1 (s), an initial increment of 0.05 (s) and a maximum increment of 0.05 (s). Notice that here the response is assumed rate-independent and that the analysis is static so that (timedependent) inertia forces are not included. Therefore, the time is really a ‘pseudo’ time that is just a parameter that shows the evolution of loading. In other words, whether the load is applied in 1 or in 20 seconds does not change the final results. You can change the time period and the maximum and initial time increments accordingly if you wish to compare the results with an experiment more easily. For example, if in the experiment the displacement is applied with a rate of 0.1 mm/s, then you can apply 60mm of displacements in 600s. Hence the period would be 600 s and the minimum and maximum time increments can be scaled down to 0.05*600=30s. As boundary conditions, fix the bottom edge of the left roller (zero displacements U1 and U2). Since we are using solid elements, which only have translational degrees of freedom, there is no need to constrain rotations, so you can ignore U3. For top edge of the roller at the right-end , apply zero displacement in the horizontal direction (U1) and -60mm in the bottom direction (U2):
Apply symmetry boundary conditions as shown below:
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2.6 Run Now you can create and submit a job and run the analysis in the usual way.
3 Post-processing In this section, a few tips for post-processing of results are given. 3.1 Resize legend so that numbers are readable Click on ‘Viewport’, then on ‘Viewport Annotation Options’ and then on ‘Set Font’, choosing 14 instead of 8. You can also tick ‘Triad’, ‘Title block’ and ‘State block’ to make the other annotations readable but what matters really is that the legend with the contour plot values can be read. Otherwise the colours have only a very qualitative meaning.
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The result is below and numbers can be read (in the screen shot as well as in figures that can be exported, as shown below).
To export the actual Viewport into a jpg figure, first of all make sure that the current work directory is the one where you want the file in the end. To check and possibly change the work directory, on the main menu go on ‘File’ → ‘Set Work Directory’ and change it to the right one is not already so. To export the picture, click on ‘File’ and then on ‘Print …’, as shown below:
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Then tick ‘File’ instead of ‘Printer’, choose a name (here ‘VMises’) and choose ‘TIFF’ (easier to paste into a Word file).
Click ‘OK’ and you should fine the file VMises_stress.tif in the work directory.
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If you copy and paste the file into a Word file, as was done below, you can see the the legend is clearly well readable even if the figure is reduced.
3.2 Scaled or actual deformation For a geometrically linear analysis, the deformation is automatically scaled by the program so that it can be easily be appreciated ‘with naked eye’. To change this, one can click on ‘Options’ → ‘Common’, then tick ‘Uniform’ instead of ‘Auto-compute’ and insert the value 1 to have the actual deformation in scale 1:1.
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For a geometrically nonlinear analysis (where displacements may be very large), the default option is that the deformation is shown in the actual size (that is in scale 1:1).
3.2 Plotting the load-displacement curve In this case, displacement control is applied. Hence, the load measured by the load cell in the experiment is the total reaction of the centre roller. However, in this case because of symmetry, the total force to be compared with the experimentally measured load is twice the reaction of the quarter of roller in the centre. In turn, the reaction force at the roller is determined as the sum of all nodal reactions of the nodes of the roller edge, as these are the nodes where the a downward displacement was applied. On these nodes, at each increment of the analysis the same displacement (equal to the prescribed one) but different nodal reactions are computed by the FE model. In summary, the numercally computed load-displacement curve is obtained as a XY dataset where the X-data are the displacements (at each increment) of any one (arbitrarily chosen) node on the mid-roller edge while the load is the sum of all nodal reactions. Both the displacement and the load turn out to be negative in the reference system of the model, so these should be changed sign before comparing them with the loaddisplacement curve experimentally measured. Practically it is quite useful in ABAQUS to use node sets. In this case, when the boundary conditions were applied at the top edge of the mid roller (zero horizontal displacement and prescribed (negative) vertical displacement), automatically the program creates a node set with all nodes on the top edge where such boundary condition is applied. To determine what this set is, when visualising the results, in the left pane one can ‘expand’ the ‘Output Database’, by clicking on the related [+] sign Page 24 of 38
(which then changes to [-]), and then likewise expand ‘Jobs-1.odb’ and “Node-sets’. Clicking on each one of the node sets, the related nodes will be highlighted in red in the model. Here, we can find that node-set 29 is associated with the nodes of the top edge of te mid roller, as shown in the figure below:
Expanded
Expanded
Node set 29
Now that we know that node set 29 contains the nodes of interests we can proceed as follows. Click on ‘Create XY Data’ as shown below:
Choose ‘ODB field output’ and click on ‘Continue’ Page 25 of 38
In the drop-down list next to ‘Position’, click on ‘Unique Nodal’. This is because we are interested in the results at the nodes of the FE model, and not at the integration points of elements.
Scroll down in the list of variables, and expand displacements (‘U’) and reaction forces (‘RF’), clicking on the black triangle as shown below. Then select the vertical components, by ticking RF2 and U2.
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Select vertical components
Scroll down
Now click on ‘OK’. Then click on the tab ‘Elements/Nodes’ and then on ‘Node sets’ and, within the window that appears, on ‘SET-29’, as shown below:
Now click on ‘Save’ and then on ‘OK’. Close the window by clicking on the top-right corner button.
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All the nodal forces and displacements (X data) in direction 2 of Node set 29 are now stored, each one against time (Y data) for each increment. However, we need a XY dataset with the load against the displacement (of anyone of the nodes as they have the same displacement). To this end we need to suitably operate with these data. Hence, click again on ‘XY data’ and choose ‘Operate with XY data’.
Click on ‘Continue’. In the dialog window that appears, scroll down in the operator field and click on ‘Combine(X,X)’ that will appear in the top field.
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At this point, within the round brackets next to Combine, type ‘-‘ (minus):
minus
Click on anyone of the displacements ‘U:U2 ….’ and then on ‘Add to Expression’
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Type a comma and then a ‘-2’ and a *, that is type ‘,-2*’ (2 is because of symmetry, as otherwise only half of the total load would be computed).
With the cursor blinking after the ‘*’, scroll down in the list of operators and choose ‘Sum((A,A,…))’ then click on the first of the list of reaction forces, press and hold the Shift key and click the last of the reaction forces, so that all of them are selected as below. At this point click on ‘Add to Expression’.
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Scroll down
You should see as below:
Click on ‘Save As …’ and choose a name for the curve. Here it is called ‘Loaddisplacement’:
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Click on ‘OK’. You can now click on ‘Plot Expression’ and close the window. You should see the plot below.
Clicking on ‘XY Data Manager’
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The you can select all the XY data except ‘Load-Displacement’ and delete them (click on ‘Detele’ and then on ‘OK’). This is practical to avoid keeping all the data that were only needed to combine the XY data. Deleting these data is particularly useful if the analysis is run again after making some changes and then one wants to plot a new load-displacement curve, because the displacements and rotations from the previous and current analyses could be mixed up.
The curve can now be seen and edited by going on XY Data, right-clicking on ‘Loaddisplacement’ and choosing ‘Edit…’ Page 33 of 38
In older versions of ABAQUS one could select and copy all the XY data and paste them onto an Excel sheet. In more recent versions it seems that these data cannot be copied and pasted, so a report should be exported. To this end, click on ‘Report’ in the main menu and on ‘XY…’
Select ‘Load-displacement’ and click on ‘OK’.
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This will add the XY data to the report in the file ‘abaqus.rpt’. This is a text file that can be opened with any text editor, for example with the standard ‘Write’ of Windows. In this case the following file appears:
This file can be copied and pasted on Excel, where with standard techniques such as clicking on ‘Data’ in the main menu, then on ‘Text to Columns’, choosing ‘Delimited’, then ‘Next’, and ticking on ‘Space’ as one additional delimeter, ‘Next’ then ‘Finish’.
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The result below is found
Clicking again on a contour plot, one can view the coponents of strains against time by clicking again on ‘Create XY data’ (as earlier shown). Choose ‘ODB field output’, untick ‘RF’ and ‘U’ in the variables and tick the variable of interest, for example E11 in this case for the strain in direction x. This time, clicking on ‘Elements/Nodes’ you can click on ‘Edit Selection’ and pick the node which is of interest. In this case, the top right point in the flat bar (on the axis of symmetry), is chosen as shown below.
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Make sure that the point is the one in the beam and not in the roller. To this end you can disactivate the option ‘Select the Entity Closer to the Screen’.
With this optin disactivated, when you click on the node you can check whether the selection is correct or change clicking on ‘Next’ or ‘Previous’ as shown below:
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Once the selection is correct, you can click on ‘OK’. The following graph should appear when you click on ‘Plot’.
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