MATERIAL MECHANICS, SE2126 COMPUTER LAB 3 VISCOELASTICITY N t i Gt () G0 1 i ( 1 e τ = α ) i= 1 k a k b τ
PART A RELAXING PLASTIC PAPERCLIP Consider an ordinary paperclip made of plastic, as they more commonly are today. The basic principle behind paperclips is that the papers force the two parts to displace relative each other and they respond by clamping the papers together. This can only be achieved if there is some part of the paperclip that can spring back at the papers, see the dashed area in Figure 1 encircling the connection parts between the two triangles. Figure 1. A paperclip with the connecting parts encircled with dashed lines. The plastic that the paperclips are made from is a type of polystyrene, maybe more recognized in its solid foam state as the packaging material Styrofoam. The completely solid state is a cheap plastic with fairly good mechanical properties for simple applications. One of the drawbacks are that polystyrene is much more susceptible to creep and relaxation than metals are. Imagine clamping a few papers together with a polystyrene paperclip and then leaving it like that for a while. The purpose of this part is to investigate the effect of time on the clamping ability of the paperclip. Assume that the material can be described with a Generalized Maxwell model according to Figure 2. The combined effect of the springs and the dashpot will represent the
relaxation modulus E(t). A simple analysis of the model in Figure 2 shows how the material behaves over time. First consider an instantaneous loading where the dashpot does not have time to deform, i.e. it is infinitely stiff. This corresponds to the instantaneous stiffness E 0. When time goes to infinity the dashpot will relax and spring b will return to its equilibrium position, and therefore not adding anything to the overall stiffness of the system. The remaining stiffness will only be controlled by spring a, this is called E. k a k b τ 0 Figure 2. Schematic illustration of the Generalized Maxwell model with only one Maxwell element. To define an isotropic material one needs to define one more elastic constant, usually Poisson s ratio ν. But any two of Lame s five constants are equally good, for instance the shear modulus G, and the bulk modulus K. The familiar relationship between E and G is given below. 2 1 So there is a simple relationship between the shear modulus and the Young s modulus. This means that the equation in the course book for E(t) can easily be rewritten to an equation for the relaxation modulus in shear G(t) instead, called a Prony series.
1 1 / Since we only have one Maxwell element we set N = 1 and get 1 / 0 0 1 This is how ANSYS deals with viscoelastic formulations and we see that besides defining the instantaneous response G 0 we need to define how much the material will relax, given by α, and the characteristic relaxation time τ. Similar to the shear response this can also be done for the bulk modulus K(t). We will not let K change with time in this lab, but it can be useful to understand why we do not use the standard elastic constants E and ν. Any deformation (strain state) of a solid body can be divided into two parts, the volume preserving part (incompressible), and the volume changing part (volumetric). The shear modulus relates to the incompressible strains and the bulk modulus is related to the volumetric strains, and since many materials show different viscoelastic response to these two types of deformations there is an advantage to this formulation. Ok, that should be enough theory for now, let s apply it! Start ANSYS via the ANSYS Product Launcher and MAKE SURE the working directory is a folder on the C: hard drive and in the TEMP directory! Since the loading and the geometry is symmetric only one of the connecting parts will be modelled. Save the geometry input file PartA_Preprocessor.txt from the course web
page and open it up in ANSYS via File > Read Input from Your model should look like in Figure 3 below. The loading applied to this model will be in the form of prescribed displacements on the two circular areas seen in Figure 3. 1 26 Figure 3. Geometry of the paperclip with the two displacement loaded areas shown. Area 1 will be given a displacement of 1 mm in the positive z- direction, and Area 26 will be fixed in all three directions. You will investigate two things in this part. First you will look at how the clamping force changes with time. Secondly you will look at how the paperclip would behave if you removed the clamped papers after a while. Save the file PartA_Solver.txt from the course web page in your working directory. Open it up and have a look at it. You will notice that there are four different load steps in this part. First there is a load step where the displacement is applied as a linear ramping. This is solved over a very short time period compared to the relaxation time of the material. Secondly there is a much longer load step where the displacement is kept constant, corresponding to just leaving the papers clamped for a long while. The time the solution has to solve for is governed by the value of the variable MultipleOfRelaxTime defined at the top of the file. Make sure it is set to 4.
The third load step is proceeded by the end-of-file-command /eof, which means that anything after that line will be ignored. Leave it like that for now. You will use the last two load steps later. Now you should be ready to solve the model by reading the solver text file. This will take a little while since it is a full 3D model with a viscoelastic material model. (This is a good time to familiarize yourself with later parts of this document so you are prepared for what is to come ) When the model is solved you can view the results as usual under ANSYS Main Menu > General Postproc > Plot Results. Compare the von Mises equivalent stress in the first load step and the last. What has happened? Since you have given Area 1 a fixed displacement there will be a reaction force there. It is this force that clamps the papers together. There are several ways of evaluating this force. Each node in a FE-model carries a set of reaction forces. They are usually minimized so to maintain equilibrium, but not at constraints. There will, and should, be a non zero sum of reaction forces in the direction of the applied deformation on Area 1. To find this sum you will need to first deselect all other nodes in the model, or rather select just the nodes on the constrained surface. Start by giving the command ASEL, S, AREA,, 1 in the command line to select only Area 1. Then give the command NSLA, S, 1 to select all the nodes associated with that area. Now, when ever you list nodal results, ANSYS will only list the active nodes. You can list the nodal reaction forces from General Postproc > List Results > Reaction Solu and there choose to list the reaction forces in the z-direction. This will list the reaction force at each node, but at the end of the list there is a summation over all nodes and that is the value you want. Now do this for all available solver output times. You can change at what time you are viewing the results by General Postproc > Read Results By Pick. To avoid having to click back and forth between listing results and picking the next time step you can issue the PRRSOL, FZ command to bring up the list of nodal results.
Now you can draw your graph of the clamping force versus time at the end of this report. The removal of the paperclip will now be analysed. Will there be any effect of the relatively long time the paperclip was stuck to the papers or will it return to its original shape right away? Open up the solver text file again. Now you will remove the end-of-file-command /eof so that the solver will solve for the last two load steps as well. In the third load step the displacement control of Area 1 will be removed, hence simulating the removal of the paperclip. This is done over a very short time period, just like in the first load step. In the fourth and last load step the paperclip is free of any loading while the solution time advances an equal amount of time as it did in the second load step. You will have to change the time the paperclip was stuck to the papers by changing the value of the variable MultipleOfRelaxTime. Pick any value between 1 and 10. MultipleOfRelaxTime: When this is done, save the solver file (do not forget to save!). Clear the last analysis by using File > Clear & Start New and just reload the geometry file and solve with the updated solver file. How much deformation remains just after unloading, i.e. in the first sub step of load step 4? How much deformation remains at the last sub step in load step 4, i.e. after a period of time since the loading was removed? Should there be any difference, and if it is, why? Remember that in the fourth load step there is absolutely no external load whatsoever on the paperclip. Record these values on your answer sheet.
PART B CREEPING SANDWICH BEAM Sandwich beam constructions are becoming more and more common today. A sandwich structured composite is a special class of composite materials that is fabricated by attaching two thin and stiff skins to a lightweight but thick core. The core is usually of much lower stiffness than the skin material. It is not uncommon for the core to be some kind of polymer foam, and the skin to be a fibre composite of some kind. What you will be looking at in this part is a sandwich structured beam made from carbon reinforced epoxy with a polymer foam core. As investigated in part A of this lab, polymers are susceptible to relaxation and creep. Consider the beam in Figure 4 below, loaded by a force couple at one end resulting in a bending moment. h h/20 Figure 4. Sandwich beam. The instantaneous elastic deformation would be something like what is shown in Figure 5 below. This would be true even if the core material is viscoelastic and at least moderately stiff in comparison to the carbon fibre. In this case the initial stiffness of the core material is 20 times less than the carbon fibre. Figure 5. Schematic view of the instantaneous elastic deformation.
Imagine that the sandwich beam in Figure 4 was subjected to the load for a sufficiently long time for the core material to experience creep. What do you think would happen to it? How would the deformation change do you think? Don t just guess, try and reason your way to an answer. What is it that will happen and how will that effect the deformation? Use the answer sheet at the end and sketch a rough figure representing the beam after a long while. Don t solve the model before you have done this! That would be cheating! (Do not use up the whole sheet, as you will need to draw a graph there as well.) Now download the files PartB_Preprocessor.txt and PartB_Solver.txt from the course webpage. Read in the geometry file in ANSYS and then solve the model with the solver file. The model is a 2D plane stress model with PLANE183 elements. Have a look at the solution at the end time. Does it look anything like you suspected it would? Do you understand why it looks like it does? Creep loading is characterized by a creep compliance that increases with time. The situation at a certain time could then maybe approximately be captured by an elastic solution with parameters that reflects the current value of the creep compliance. It would therefore be tempting to try and figure out what those parameters are. This would possibly save a lot of modelling time since an elastic solution is almost instantaneous in comparison to the viscoelastic solution time. In the current problem we assumed that the only part of the stiffness that changed with time was the part related to shear (volume preserving deformation). In other words, if we hope to model this we need to take this into account. We can not just change the E- modulus and hope for the best As already said there are several ways in which one can describe the same isotropic solid. We know from the definition of the Prony series how to find the shear modulus as a
function of time. So given that we want the bulk modulus to be constant in time we can find an expression for the Poisson ratio as a function of time 3 2 6 2 Thus the E-modulus will also be a function of time as 2 1 Ok, so say we would like to find the response of the structure at time t = 3τ. Can we just replace the initial elastic constants with E(3 τ ) = 602MPa and ν (3 τ ) = 0.485 and solve the elastic problem? Find the viscoelastic solution that corresponds to t 3 and record the maximum value of the displacement vector sum, and when this is done you should clear the model. You will now evaluate the elastic solution to the sandwich problem and see if it is possible to find a similar solution with this simplified method. In order to solve the elastic problem with different core stiffness you need to change a few things. First look in the solver file. There is a section in there that has been commented with the use of!, remove these. This will trigger an elastic solution instead of a viscoelastic time dependent solution. (Save the updated file under a new name!) But before you solve you will need to remove the viscoelastic material formulation. In the Preprocessor > Material Props > Material Models there are two materials defined. Material number 1 is the core material. First remove the Prony: Shear Response formulation, and then change the initial (elastic) constants to the values given above. Did the elastic solution look anything even remotely close to the viscoelastic solution? If not, why is that do you think? How does the elastic theory depend on the load history?
With the proposed method above we would predict an exponentially increasing compliance, and this is clearly wrong! We can find the true behaviour of the compliance by looking at the solution for the full problem again. So clear the model and solve it again with the full viscoelastic formulation. When the solution is finished, bring up the TimeHist Postpro. Click on the green plus sign to add the xy shear stress. You will have to click on a position in the model where you want to investigate this. Pick a position pretty much in the middle (it doesn t have to be exact, but at least make sure it is in the core material and between the supports). The entry SXY_# should appear in the list. Now also add the elastic xy shear strain in the same way (this time you can enter the node number in the box instead of trying to click on the same node). You can now calculate a shear compliance as the ratio of shear strain to shear stress as using the calculator function below the variable list. The drop down list above the CLEAR button will let you use the previously defined list entries in your calculations. To finalize you calculation, click on ENTER. When you have done this you can plot the compliance over time by selecting it in the variable list and click on the graph data button (the one that looks like a graph). Was it even remotely exponential in shape? Draw this graph in the answer section for this problem.
PART C RUNNING ON FIRM GROUND? One important aspect of working as an engineer is the question of physical models. Imagine that you are hired by a company that makes running shoes. They want to get a better understanding of how the ground responds to the shoe when running, see Figure 7a. For asphalt this is easy, a good enough engineering approximation is to say that it is infinitely stiff. But what if you run on softer ground, maybe a rained upon and sawdust covered running track? They have settled on a spring and damper model of the ground, shown in Figure 7b, and your job is to find a good set of model parameters k and c. ( ) = ( ω ) p t p sin t 0 a) b) () = ( ω ) F t F sin t 0 uy ( t) k c Figure 7. a) 2D model of ground and the pressure load exerted by a runner. b) 1D spring and damper model. In a real life situation you might have been given a set of test results with the displacement as a function of time for a few different running paces. Since we don t have time to put on our running shoes and head up into the forest behind Mechanical Engineering and do some real measurements we will simulate the tests with a few FE model solutions instead.
The displacement response of the model in Figure 7b is given as the solution to the first order differential equation 1 1 that has the solution 1 1 sin 1 cos The period of human walking is about 1.5 seconds for each cycle. A cycle is actually two steps, so the time period of one foot pressed to the ground is about half of that. This is for walking, and this time will decrease to about a tenth of that for really fast sprinters. (This is not to say that during a sprint the period of running also decreases to a tenth, but rather to accommodate for the fact that the sprinter is airborne for a while in each step.) The ground is here modelled as a Generalized Maxwell model with parameters given in the table below. G 0 43.5 MPa G 0 τ 0.15 s Now download the files PartC_Preprocessor.txt and PartC_Solver.txt from the course homepage. Open up the solver file and look at the parameter section in the beginning of the file. There is a parameter named VelocityMultiplier that controls the input to the solver. It ranges from 1 for walking, to 10 for fast running. A running person applies more load during a single step than the same person would while walking, and to accommodate for that, without having to do a dynamic analysis, a factor of dynamic loading has been introduced. It varies linearly with VelocityMultiplier from 1 for walking, to 2 for fast running. So the amplitude of the applied force will be as below.
8 9 9 In the solution for the displacement there is a parameter ω that also will change with increased pace. This is the angular frequency of the time dependent term in F(t). In this case that term is an ordinary sine function and the angular frequency is related to the loading time t load as ω = π. t load Now if you look at the analytic solution for u y (t) you can see that if you have two values for the displacement for two different times you could find k and c with a moderate amount of effort. But since you are probably running out of time you will not have to do this, unless you want of course. What you need to do instead is to draw the graph of the displacement versus time for a few different values of VelocityMultiplier. Look at the graphs and try to explain the differences. Open up the model by reading the geometry file into ANSYS. The model is 2D axisymmetric with PLANE183 elements suitable for this type of analysis. Now solve the model for some running paces by executing the solver file. When the solution is done, open up the TimeHist Postpro and by graphical picking you should plot the displacement in the y-direction for upper leftmost node in the model. This is the node that sits on the top of the symmetry line x = 0. Make sure you make a note on each graph of how fast the person is running. That is all for this time, see you next time, bye!
RESULTS FROM PART A RELAXING PLASTIC PAPERCLIP F z max ( Fz ) 1 t τ MultipleOfRelaxTime: δ unloading = mm δ unloading + time = mm
RESULTS FROM PART B CREEPING SANDWICH BEAM Yes, this page is supposed to be empty...
RESULTS FROM PART C RUNNING ON FIRM GROUND? u y u y t t u y u y t t