# MATERIAL MECHANICS, SE2126 COMPUTER LAB 4 MICRO MECHANICS. E E v E E E E E v E E + + = m f f. f f

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1 MATRIAL MCHANICS, S226 COMPUTR LAB 4 MICRO MCHANICS f m f f m T m f m f f m v v + + = +

2 PART A SPHRICAL PARTICL INCLUSION Consider a solid granular material, a so called particle composite, shown in Figure. One common example of such a material is concrete, which is a mix of construction aggregate (stone or sand) in a cement matrix. This lab will deal with a little more exotic material than that. You will do calculations on a glass bead/epoxy particle composite with constituent properties given below. ν Glass poxy Glass = 70 GPa = 5GPa = ν = 0.3 poxy Figure. An example of a particle composite. One can not without great difficulty analyse this material in its complete configuration, analytically or with the aid of finite element calculations. This difficulty is of course attributed to the particles and the fact that they give rise to an inhomogeneous material. A way around this problem is to determine some effective properties and model the composite as a homogeneous solid with the effective properties instead. The effective mechanical properties of the homogenized material are a kind of mix, or weighted average, of the constituents mechanical properties. There are several analytical techniques suggested to find these effective properties, more or less accurate. This homogenization can also be performed using a finite element approach and the volume average equations from micro mechanics shown below.

3 Ω Ω To be able to perform the analysis one need to define a representative volume element (RV) that is a model of the material on the micro-scale. The RV will be inhomogeneous and needs to give a good representation of the micro structure, but it will also need to be able to represent the material on the macro-scale in an average sense. It needs to be space filling, and have the same constituent orientation and volume fraction as the whole structure it represents. Figure 2 shows the RV for a material with spherical inclusions. Figure 2. Cut out of spherical inclusion in matrix - RV The finite element approach adopted in this lab is summarized in the following steps:. Apply prescribed displacements on the boundary of the RV. The displacements should correspond to homogeneous strain fields within th e RV. 2. Calculate the volume average of stresses and strains, and. 3. The volume averages can be thought of as the homogenized macroscopic stresses and strain and will give the effective properties using Hooke s law.

4 The relation between average strains and displacements on the boundary of the RV reads, where one needs to remember that shear strains like γ ij is equal to 2ε ij etc. For the particular case of a prescribed strain we then set, 0 on the boundary of the RV. Solving the boundary value problem gives the stresses σ k ij and strains ε k ij in each element k. Since the volume of each element V k is known, the average stresses and strains can be approximated by a discrete sum over all N elements where denotes the average stress in element k. The homogenized effective properties and ν can be found from Hooke s law. Here shown for macroscopic strain in the x-direction, but similar in the y and z-directions. This theory session sets the scene for the lab, now lets get started!

6 processing ability in this lab, but it is a good way to check if your finite element solver has done what you wanted it to do. To get some meaningful results you will need to work with something that is called lement Tables, which is just a tabular form of the results for each element. There will only be one result for each element and field variable, which might be a little surprising since the elements used in this analysis are 20 node brick elements and therefore should contain more than one value for each field variable (depends on the number of integration points in the element). What ANSYS does is that it averages the results in each element so you only get one unique element result for each field variable. A slight loss of accuracy, but we will have to live with that. If we know the volume average strains in two directions and the volume average stresses we can calculate the effective properties using Hooke s law. Verify these equations by considering Hooke s law with 0. To find the volume average results needed in the equations above go to ANSYS Main Menu > General Postproc > lement Table > Define Table to bring up the lement Table Data screen. Here you will need to Add the results you want to average. By highlighting one result set at a time from the list and clicking on apply you will create an element table for each of these results. Don t forget to add a table for the element volumes. When you are done adding results just close the lement Table Data screen. Now you will need to multiply each element result with the volume of that particular element. To do this, click on General Postproc > lement Table > Multiply to bring up

7 the Multiply lement Table Items screen. Multiply all your result tables with the volume table. This will create a new table that you will need to give a unique name so you know which results it contains. See picture below for explanation. Here the element table results for shear stress SXZ is being multiplied with the volume table VOLU. Notice the user label for the new table! When you have all your results multiplied with the volume you will need to sum them all up (this operation represents the integral over the volume). This is done by clicking on Sum of ach Item. Now you will need to divide each sum with the summed total volume, thus finalizing the volume average operation. You can use the result sheet at the end to record your values and to calculate the effective properties. For the second part just clear the model and reload the geometry using the pre-processor file again. This time you will solve the same model but with a shear strain loaded boundary instead. This is done by reading in the solver file PartA_ShearStrainXZ.txt that applies the following displacement field to the outer boundary. 2 2

8 In the same manner as the previous part you will now need to find the effective shear modulus. Hooke s law for the shear modulus is a lot simpler, so you will only need two results to determine it. In linear isotropic elasticity there is a well known relation between the Young s modulus, the Poisson s ratio and the shear modulus. Check if this still applies between your effective properties? 2 Finally you will have a look at the Reuss bound for compliance and Voigt s bound for stiffness. These bounds define maximum and minimum values for the effective shear modulus and bulk modulus. You will only look at the bounds for the shear modulus here. The Reuss bound is; and the Voigt s bound is: Ω Ω Can you figure out how to calculate these bounds given what you know of the volume fractions of particles and the shear modulus of the particle and the matrix? A hint is that the equations are quite simple, and are similar in style to the Rule of mixtures equations. Did your previously calculated effective shear modulus fall with in the bounds? It should have

9 PART B HXAGONAL R.V.. Consider a unidirectional fibre composite with the fibres arranged in a hexagonal pattern, as shown in Figure 3a below. The representative volume element would, as it might be easy to guess, have a hexagonal shape with the fibre in the middle. A schematic figure of what the RV would look like is shown in Figure 3b. y x z Figure 3. a) The hexagonal pattern of the fibre composite. b) The representative volume element. Once again your task is to find the effective properties of this composite by using the micro mechanics approach. This time it will be much harder to do this from an analytical formula like the one derived from Hooke s law in part A. Why is that do you think? What is the major difference between the directional dependence of the RV in part A and in this part? The composite is a carbon fibre reinforced epoxy with properties given below. ν Carbon poxy Carbon = 250 GPa = 5GPa = ν = 0.3 poxy First you will need to get 7 files from the course home page. The pre-processor file contains the geometry and the mesh, as usual. The other 6 files correspond to the 6 possible strain loadings, all according to the name of the file. Before you read in the geometry into ANSYS you should open up that file in a text editor and change the value

10 of the parameter r, the radius of the fibre. Pick any value between 3 and 7 mm (you might get a warning about bad shapes of some elements ignore that). r = v f = Now save the file and read in the geometry in ANSYS in the usual way. Consider the following matrix equation for the volume averages of the stresses and strains. So if you apply each strain component one at a time you can look at the resulting stress components, and in that way calculate the corresponding column in the stiffness matrix. Since the definition and multiplication of the element results tables are somewhat time consuming those operations has been included in the solver text files for simplicity. All you will have to do after each solution is to click on Sum of ach Item to bring up the numerical results. Now solve the model by applying strain in the x-direction to the boundary, i.e. read in the solver file PartB_X.txt in ANSYS. Look at the summed results. For each solution you do, 5 of the 6 strain results are very close to zero, and from an analytical point of view they are exactly zero, and that is what you will assume them to be.

11 So, for example, for strain loading on the boundary in the z-directio, the matrix equation above would reduce to the following: Thus, by some easily performed calculations you have the entire third column of the stiffness matrix. The stresses multiplied by the element volumes are named according to; S_X_V for the stresses in the x-direction and S_YZ_V for the shear stress in the yz-plane. Similarly for the strains multiplied with the element volumes you will have; PS_X_V, and PS_YZ_V for the corresponding axial and shear strains. The height, or the length, of the RV along the z-axis will have an effect on the results. This is not really what you would want for an accurate model, but it can be good to have seen this behaviour. You will for example be able to see effects of this when you look at the longitudinal stresses after prescribing a longitudinal displacement on the boundary. Look at the edge between the fibre and the matrix, and you should see a ring-like stress gradient. Preferably, that should not be there. But to avoid this effect all together you would need to change the boundary conditions to something called periodic boundary conditions, and they are a little bit tricky to apply and more than we have time to do here. Now solve for all the different load cases, one at a time, and start to fill out the stiffness matrix with you calculated values. It will be sufficient if you record the values in GPa with only one decimal place (the program will give you the stresses measured in MPa). After a few solutions, can you see a pattern emerging in the stiffness matrix? What kind of symmetries would you expect given the geometry of the RV?

12 In chapter 5.4 of Gudmundsons book Material Mechanics several material symmetries are presented. Do any of them seem to fit your results? So what about those effective properties then? How do you find them now? If you consider the compliance matrix of a transversely isotropic material given below you will see that it is quite easy now. Remember that the compliance matrix S is just the inverse of the stiffness matrix C Record your values for the effective properties on the answer sheet at the end. There are several alternatives to performing this kind of analysis to get the effective properties. You will compare your results to two of them. The first is what is called the Rule of mixtures and the second is an empirical set of relations called the Halpin-Tsai equations. Rule of mixtures gives the following equations for the effective properties:

13 Halpin-Tsai predicts the same longitudinal stiffness and Poisson s ratio as Rule of mixtures, but for the other two, the equations are as follows: ; ; ; 2 ;

14 PART C DOS LNGTH MATTR? The most common model of a fibre composite is the one where the fibres are as long as the whole structure. The fibres go from end to end so to speak. In many common applications this is usually not true. One example is short fibre glass/polyester composite, that are very common in many applications today. In this part you will look at a unidirectional fibre composite with short fibres embedded in a matrix, see Figure 4 below. ν Glass Polyester Glass = 70 GPa = 2.5 GPa = ν = 0.3 Polyester Figure 4. A short fibre composite. The question you will investigate here is if this has any effect on the effective mechanical properties of the composite? You will only look at the longitudinal stiffness in this part. The model of the representative volume element can be very much simplified by using symmetries. First realize that the geometry is symmetric with respect to the longitudinal axis of the fibre, so an axi-symmetric 2D model is applicable. Secondly there is also symmetry with respect to the middle of the fibre, so a symmetry condition can be applied there as well, effectively cutting the model in half. A schematic of the model can be seen in Figure 5.

15 y R L/2 L f /2 x R/2 Figure 5. The model geometry. Rotational symmetry is imposed along the y-axis and ordinary symmetry along the x-axis. Download the single input file from the course webpage. Open up the input file and change the value of Lf, the fibre length, to a value between 20 and 80 mm. Lf = v f = Save the file and open it in ANSYS, this will initiate both the pre-processor and the solver. The loading in this part will be a prescribed displacement on the boundary in the y-direction. If you look at the solution for the stress in the y-direction, you will notice that there is a stress gradient at the end of the fibre. The stresses in the fibre needs some distance to build up, so the fibre is not used in an optimal way.

16 Because of this the estimated effective properties calculated by the Rule of mixtures are quite off. This can be corrected by assuming that there is an ineffective fibre length L i. The approach to estimating this is outlined below: Find the stress in the y-direction at the node located at position (0,0). Provided that the fibre is not too short, this is the stress that should appear in the whole fibre if it was used optimally. To do this, use the List Results > Nodal Solution. The node you are looking for has node number. Record the stress in the y-direction as 0,0 on the answer sheet. Now calculate the volume average of in the whole fibre. To do this use element tables, where you can multiply and sum element results. You will however need to select just the elements of the fibre; otherwise you will average the whole model instead of just the fibre. Type in the command: CMSL,S,FIBR. This will select only the predefined area component corresponding to just the fibre part of the model. Now you need to select all the elements in that area by typing: SLA. This selects the elements from an already selected area. Now calculate the volume average stresses of the fibre in the y-direction. Ω f The difference between the two stresses is essentially a measure of the effective fibre length, where denotes the so called ineffective stress. An average stress that is smaller than 0,0 can be seen as the stress 0,0 acting on the effective fibre length, and a vanishing stress acting on the remaining part of the fibre. Thus, 0,0 Now you solve this for the ineffective fibre length and record it on the answer sheet. What does that length measure actually mean?

17 That was all for this time, and for this course. Thanks for your participation. Good luck on the exam!

18 RSULTS FROM PART A SPHRICAL PARTICL INCLUSION Rp = v f = 2??

19 RSULTS FROM PART B HXAGONAL R.V.. r = v f = [ C] = To invert C you will probably need to use a numerical tool like Matlab or an advanced calculator. FM RoM Halpin-Tsai L L L T T T ν LT ν LT ν LT G LT G LT G LT

20 RSULTS FROM PART C DOS LNGTH MATTR? Lf = v f = 0,0

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