ME 406 Manufacturing and Design 191-Term Project Instructors Dr. Ahmed Sarhan/Dr Numan Abu Dheir (Due on or before December 16 2019) Part of the vision of 2030 is to transform the Kingdom Oil-based economy to knowledge based added value economy. As a result local investors with large capitals are eager to find opportunities to take part in achieving this vision. You as a young engineer are targeting those local investors to start a jointventure in the manufacturing sector. You decided to come up with the idea of manufacturing Powder Metallurgy metal Porous Sliding Bearing as shown in Fig. 1. Porous-metal bearings are used widely in instruments and general machinery, in which their self-lubricating characteristics and load-carrying ability is very desirable. Most porous-metal bearings consist of either bronze or iron which has interconnecting pores. These voids take up 10% to 35% of the total volume. In operation, lubricating oil is stored in these voids and feeds through the interconnected pores to the bearing surface. When properly designed, they can be both economical and highly functional. Their manufacturing method consists of compacting the metal powder mixtures to the proper density. Subsequently, they are sintered for different duration subject to the temperatures. Sintered bearings are then sized to obtain the required dimensional characteristics. This is followed by inspection and impregnation with a lubricating oil. If the targeted production volume per year is 1M parts, then prepare your technical report and provide details for the following: 1- Basic overview for processes plan. (5pts) 2- Materials selection: What are the possible different powder and die materials for manufacturing the bearing. Justify your selection of the powder and die materials (10pts) 3- Design for the Porous Bearing (provide dimensions; Engineering Drawing 2D+3D), sizing of Porous-Metal Bearings and Bearing Clearances. (10 pts) 4- A design for the mechanical compacting die-punch (Sketch and Engineering Drawing 2D+3D). Show Press Fits and how the die will be fixed on machine and measures for die alignment. (25pts) 5- Force and power requirement for the press performing the compacting process and estimate the resulting green density at the proper porosity. (15pts) 6- Sintering process and its parameter and an estimate for the resulting properties. (10pts) 7- Heat treatment process (5pts) 8- Details for finishing processes with reasons and the expected dimensional tolerances of the part and surface finish expected. (5pts) 9- Detailed estimate for the cost of the part. (15 pts) 10- Providing a simulation showing the processing (compaction and/or sintering) will carry extra grade to the team. (extra 10 pts) Note: the report must be formatted to show all elements of technical reports. The report will be check for plagiarism. Make sure to provide references as needed. Both hard and soft copy of the report and the associated drawing and simulation files if any must be submitted on or before the due date. Figure 1
Purchase answer to see full attachment

ITS 832 Chapter 2 Information Technology in aGlobal Economy Introduction • • Educating Policy Managers and PolicyAnalysts Two practitioner types • • • Policy information analyst Needs addressed by two graduate programs • • • Policy informatics-savvy publicmanager Master of PublicAdministration (MPA) Master of Public Policy (MPP) Chapter focus • Role of policy informatics in preparing practitioners Practitioner Orientations to Policy Informatics • • • Two “ideal” types • • Policy informatics-savvy public managers Policy informatics analyst Manager • • Play important roles inimplementation Instrumental inusing policy informatics projects Analyst • • Execute policy informatics initiatives Other common rolenames • • • • Analyst Researcher Modeler Programmer Policy Informatics-savvy Manager • • • May not needto • • Build models Analyze big data Must be ableto • Lead personnel who build models and manage big data Necessary competencies • • • • • • Systems thinker Process orientations topublic policy Research methodologies Performance/Financial management Collaborative/communicative Social media, IT, and e-Governance awareness Policy Informatics Analyst • • Start with basic competencies of managers Additional necessarycompetencies • • • • Advanced research methods of IT applications Data visualizationand design Programming skills Modeling skills • Perhaps one of the most important Summary • • • • Two types of practitioners • • Managers Analysts Managers • • Leadership Interacting with actors Analysts • • Technical skills Responsible forexecuting requirements University ofVermont • Distinct programs to address each role
Purchase answer to see full attachment

AE-418 NASTRAN PROJECT NAME:_______________ GRADE:_________ Part 1: Create Finite Element Model A finite element model using FEMAP (NxNastran) of the problem stated below must be created. The *.modfem file must be uploaded into CANVAS at submission time along with the report. Part 1 of the report must include a detailed step by step procedure showing how to create the finite element model. Consider the finite element representation of a stiffened panel as shown below in Figure 1. Figure 1: Stiffened Plate Finite Element Representation The 20 x 20 inches plate is 0.1 inches thick and is clamped along all four edges. A uniform pressure equal to 0.5 psi is applied to the top of the plate. Because of the centroidal axes of the stiffeners does not coincide with the mid-plane of the plate, you will have to account for this when you define the element properties for the stiffeners. Dimensions for the I-shape stiffeners are given in Figure 2 Figure 2: General Dimensions, Boundary Conditions and Loading Based on the results of the finite element model you created provide the answers at the following loactions  Component of the displacement vector at the center of the plate: Disp X = Disp Y = Disp Z =  Axial on the stiffener at the center of the plate Axial Stress =  Plate von Mises stresses at the center of the plate (@-t/2) stress = (@-t/2) stress = Part 2: Analytical Approximation for Stiffened Plate By looking in the literature or online find an analytical approximation solution to the response (deformations and stresses) of a stiffened panel. Clearly describe the assumptions and equations you will be using to find the analytical solutions to:  Component of the displacement vector at the center of the plate: Disp X = Disp Y = Disp Z =  Axial on the stiffener at the center of the plate Axial Stress =  Plate von Mises stresses at the center of the plate (@-t/2) (@-t/2) stress = stress = Part 3: Compare Part 1 and Part 2 Compare Part 1 and Part 2 and comment on your findings and proposed procedure to improve agreement between numerical (finite element) and analytical solutions. NASTRAN PROJECT Stiffened Plate Structural Analysis Pol Fontdegloria Balaguer AE 418 Embry-Riddle Aeronautical University, Daytona Beach, FL, 32114-3900 June 24, 2022 Introduction This report presents (1) finite element, and (2) analytical solutions for the structure presented in the problem statement. The mentioned structure consists of a stiffened rectangular panel of thickness 0.1 inches and equal sides equal to 20 inches. Constraints are defined as simply supported along all four edges of the structure. The model will be analyzed for a uniform pressure equal to 0.5 psi applied downwards vertically to the top of the plate. It must be stated that the reference axis has been shifted from the problem statement. Whereas the problem statement defines thickness along the z-axis, the model presented in the report defines it along the y-axis. In the problem statement the stiffeners go along the x-axis, whereas in the model it is along the z-axis. More detailed information will be shown in the following report. Another point that must be mentioned, is that the results are expected to differ due to the assumptions taken in the analytical solution, specially for the stress calculations. 1 Finite Element Model This section of the report presents an explanation of the model building process in FEMAP. To start with, a material must be defined. In this case, the material chosen is Al-2024-T3, which is the most used material for aircraft panel structures. Based on the military handbook [1] the properties of this material are Young’s Modulus (E) 10.7e6 psi, and Poisons Ratio (v) equal to 0.3. Having defined these values in the Material section, the Properties can be defined. Two properties have been used for the model, one for the plate, and one for the stiffeners. Therefore, using the Material defined previously, defining the element type as Plate, setting the thickness as 0.1 inches and selecting OK, the plate property will be created. On the other hand, the stiffener property is defined choosing the same Material, defining, the element type as Beam, and clicking on Shape to define the following properties for the cross section as shown in Figure 1. Drop down menu Shape and select IBeam. Select Orientation Direction (y) to Up. Select OK, define Y Neutral Axis Offset to -1 in both squares and select OK again. At this point both properties have been created. Y Neutral Axis Offset will ensure the I-Beam is attached to the plate at the top edge. Figure 1. Cross-Section Definition of I-Beam Stiffener. The following step is to create the geometry, this is done by creating the edges of the panel. In this case, a 20 inches per side square panel. The points used to create this geometry are noted as coordinates in Table 1. Points are selected under Geometry > Point. 2 Point 1 2 3 4 5 6 7 8 9 10 x-coordinate -10 -5 0 5 10 -10 -5 0 5 10 z-coordinate -10 -10 -10 -10 -10 10 10 10 10 10 Table 1. Geometry Points for FEMAP Model. Once the reference points have been created, by going to Geometry > Surface > Corners, and selecting Method > On Point, the desired surfaces that will act as the panels can be created. For this model, four surfaces are created following Table 2. It is important to select points in the order indicated to maintain the same surface orientation. Surface 1 2 3 4 Points 6-7-2-1 7-8-3-2 8-9-4-3 9-10-5-4 Table 2. Surfaces and Respective Corners for FEMAP Model. Once the surfaces have been created, the following step is to mesh the geometry. First, the geometry mesh size must be defined. To do so select Mesh > Mesh Control > Size on Surface, then proceed to select the four surfaces previously created. For this model, the mesh size is defined to be equal to 1. At this point the model will look like shown in Figure 2. Figure 2. Pre-mesh View of FEMAP Model. 3 After the mesh size has been determined, meshing of the panels and stiffeners must be done. This is done with a similar process for both Properties. For the panels, selecting Mesh > Geometry > Surface > Select All > OK. After making sure the plate property is selected, proceed. The model should now look like Figure 3. Figure 3. FEMAP Model after Meshing of Plate. For the meshing of the stiffeners, Mesh > Geometry > Curve, and select the five curves that go along the z-axis. After ensuring the selected property is the I-Beam and selecting OK. The orientation vector must be set up in the y-axis. After this the model should look like Figure 4. Figure 4. FEMAP Model after Meshing of Stiffeners. After the meshing is complete, it is important to make sure there are no coincident nodes. To do so, select Tools > Check > Coincident Nodes. After that is checked, it is important to define the Constraints for the model. In this case, the problem states a simply supported structure along all four edges of the panel. This means that the constraints are going to be different for the curves along the x-axis and z-axis. The constraints are shown in Table 3, marking with an ‘x’ the constraints selected for each curve type in the model. 4 Curve along x-axis z-axis Tx x Ty x x Tz x Rx x Ry x x Rz x Table 3. Constraints (reactions) for Various Edges of the Panel. The constraints are added by selecting Model > Constraint > Curve, selecting the curves that must be constrained for each case, and selecting OK. Then, selecting Arbitrary in CSys, and switching the coordinate system to Global Rectangular. This will ensure that the constraints are being set in the same coordinate system as the rest of the model. Check the boxes to coincide with Table 3 and select OK. After all constraints, the model should look like Figure 5. Figure 5. FEMAP Model after Constraints. Once all constraints have been modeled, we can proceed with the application of the loads. To do so, it is important to deactivate the I-Beam property first, this will allow for an easier selection of the panel elements only. Once this has been done, select Model > Load > Elemental. Select the drop-down menu Pick and select on Box. Select all the elements of the plate, which should be a total of 400. Select OK. Input a value of -0.5 for the Pressure box and proceed. Select any of the elements on the plate and click OK. Reactivate the I-Beam property, and the model should look like Figure 6. Figure 6. FEMAP Model after Loads. 5 Proceed by ensuring that the model is properly built by selecting File > Rebuild. Then, for the analysis, select File > Analyze, create a new set by selecting Create/Edit Set. Following, select New and make sure that the analysis program is set to NASTRAN, and the analysis type is Static. Proceed by selecting OK, Select, and OK to run the simulation. The results for the simulation are summarized further in this report. Although, Figure 7 presents a graphical visualization of what the output for Total Translation with a 1% deformation model and Contour activated. For the model to look like shown below, it is important to deactivate the geometry surfaces modeled at the beginning of the process. Figure 7. Total Translation Output of FEMAP Model. Analytical Solution In this section of the report, an analytical solution to verify Nastran model. The equations shown in this section have been derived from Theory of Plates and Shells. [2] More in particular section 30 of the book, where relations for a simply supported rectangular plate under uniform loading are derived. To start with, let us define some parameters for our problem. Following, Figure 8 shows a two-dimensional representation of the problem as well as the system of reference that will be used during this solution. For the sake of consistency, the reference axis has been stablished equal to the ones used for the finite element model. Figure 8. Schematic of Plate with Reference Axis. 6 Where a and b represent the total distance on the x-direction and z-direction respectively. Similarly, w and l, represent the distance between stiffeners in x-direction and z-direction respectively. Note that there are no stiffeners in the z direction, thus b will be equal to l. The different variables presented in the schematic above are stated in the following Table 4. Other values such as the thickness (t) of the plate are also noted. Variable a [in] b [in] w [in] l [in] t [in] q [psi] Value 20 20 5 20 0.1 0.5 Table 4. Plate Variables for Analytical Solution. Once all the variables have been obtained, the analytical solution proceeds in the following way. To start with, the maximum deformation of this rectangular plate must be found. Based on the previously mentioned book this will be defined by equation 141, found in page 117: 𝑞𝑎4 𝑤𝑚𝑎𝑥 = 𝛼 𝐷 Equation 1 Where α represents a numerical factor depending on the ratio b/a of the sides of the plate, and D is the stiffness of the plate. As for this case, the ratio b/a is found to be equal to 1. Thus, based on the table 8 found on page 120 of the book, the value of α for our case is equal to 0.00406. This leads to the following equation: 𝑤𝑚𝑎𝑥 = 0.00406 𝑞𝑎4 𝐷 Equation 2 At this point, all the variables in Equation 2 but the stiffness D are known. Thus, this variable must be found prior to performing any calculations. To find this variable, components for D must be averaged as shown in Equation 3. 𝐷= 𝐷𝑥 + 𝐷𝑧 2 Equation 3 7 Starting with Dx for its simplicity, it is defined as shown in Equation 4. 𝐷𝑥 = 𝐸′𝐼 𝐸 1 𝑏𝑡 3 𝐸𝑡 3 = ∙ ∙ = 𝑏 1 − 𝑣 2 12 𝑏 12(1 − 𝑣 2 ) Equation 4 The value used for Young’s Modulus (E) and Poisons Ratio (v) are the same ones used in the finite element model, thus 10.7e6 psi and 0.3 respectively. Thus, plugging in the values for all known variables, we obtain a Dx equal to 979.85 lb·in. Once this value has been found, the following is to find the value for Dz. Because stiffeners are found along the z-direction this component of the stiffness will present an extra step of complexity. First thing is to find the y distance to the centroid of the combination between plate and stiffeners. This can be found using Equation 5, with 0 starting as shown in Figure 9. 𝑦̅𝑥 = 𝑁𝑥 ∙ 0.38 ∙ 1 𝑎𝑡 + 𝑁𝑥 ∙ 0.38 Equation 5 Figure 9. Schematic of Cell for Neutral Axis and Dx. Because the number of stiffeners in this case is 5, then the value for 𝑦̅𝑥 will be equal to 0.4871 inches. It must be mentioned that the value of 0.38 found in Equation 5 corresponds to the cross-sectional area of the stiffener. Due to the simplicity of the calculations to obtain this value, these have not been included in the report. Once the y distance for the neutral axis has been found, by using Equation 6, the equivalent stiffness Dz can be found. 𝐷𝑧 = 𝐷𝑧 𝑝𝑎𝑛𝑒𝑙 + 𝐷𝑧 𝑠𝑡𝑖𝑓𝑓𝑒𝑛𝑒𝑟 Equation 6 8 Where the equivalent stiffness in the z-axis is found by adding the equivalent stiffness provided by the panel, to the one provided by the stiffeners. Moreover, this mentioned equivalent stiffnesses can be found by using Equations 7. 𝐷𝑧 𝑝𝑎𝑛𝑒𝑙 = 𝐸 1 lb ∙ [ 𝑡 3 + 𝑡(𝑦̅𝑥 )2 ] = 280,054 2 (1 − 𝑣 ) 12 in Equation 7 (a) 𝐷𝑧 𝑠𝑡𝑖𝑓𝑓𝑒𝑛𝑒𝑟 = 𝐸 ∙ [ 𝑁𝑥 𝑒 𝑁𝑥 𝑐 lb ∙ ] ∙ [ 𝐼𝑥𝑥𝑠𝑡𝑖𝑓 + 0.38(1 − 𝑦̅𝑥 )2 ] = 1,231,975 2𝑤 𝑎 in Equation 8 (b) In Equation 7 (b), the value for the moment of inertia of the stiffener Ixx has been obtained from the cross section modeled previously in FEMAP. The value for this parameter is 0.229 in4. Values for Nx correspond to 2 and 3 for the edge and central stiffeners Leading for a total value Dz equal to 1,512,029 in/lb. With these values the value for equivalent stiffness can be found recalling Equation 3. This will result in an equivalent stiffness of D equal to 756,373 lb/in. Now that the value for equivalent stiffness has been found, by plugging in the known variables into Equation 1 the maximum displacement at the center of the panel can be obtained. This results in a value of wmax equal to 0.0004294 inches. Based on our model, this displacement will be found in the y-axis, thus will correspond to the y-displacement at the center of the plate. With the obtained value for equivalent stiffness D, we can proceed and obtain an equivalent thickness based on Equation 8. 3 𝑡𝑒𝑞 = √ 12𝐷(1 − 𝑣 2 ) 𝐸 Equation 9 Plugging in the values defined previously, the equivalent thickness obtained is equal to 0.9173 in. From now on, this value will be used to calculate the axial stress on the stiffener and the von Misses stress on the plate. To start with, the moment of x and z must be calculated. The moment of x on the edges of the constrained curves will be used to calculate the axial stress on the stiffener by simplifying to a pure 9 bending case in a two-dimensional situation. The equations for the bending moments at different positions of the plate can be found in the same book mentioned above. The equations are the following: (𝑀𝑥 ) = 𝛽𝑞𝑎2 | (𝑀𝑧 ) = 𝛽1 𝑞𝑎2 Equation 10 Timoshenko’s book presents multiple variations of these equations depending on the desired position in the plate. There is up to six variations of this equations, where the only parameter that changes is the value of the numerical variable β. As mentioned before, the first moment that we need is the moment of the x axis on the edge that goes along that same axis. Looking at tables attached in Theory of Plates and Shells the value of the numerical variable at that position is 0.0479. Therefore, the equation to obtain Mx of interest will show as follows: 𝑀𝑥 𝑥−𝑒𝑑𝑔𝑒 = 0.0479𝑞𝑎2 Equation 11 Plugging in the values for the known variables in Equation 11, we obtain a moment of the x-axis equal to 9.58 lb·in. As mentioned previously, this problem can now be reduced to a two-dimensional problem of pure bending for a beam, just as shown in Figure 10. Figure 10. FBD of 2-Dimensional Representation of Pure Bending in Beam. Based on the book Aerospace Structures I [3], the equation for axial stress in a beam due to pure bending is as follows. Furthermore, based on the values that have been found the derivation of this equation for our case is shown below. 𝜎𝑎𝑥𝑖𝑎𝑙 = 𝑀𝑥 𝑥−𝑒𝑑𝑔𝑒 ℎ 𝑀 𝑦= 𝐼 𝐼𝑥𝑥 2 Equation 12 10 Where h is the height of the I-beam defined previously in this report, and y is the distance from the point of interest to the neutral axis. Due to symmetry, the neutral axis is found at the center of the cross section. In this case, analyzing the top of the beam, the distance y will be equal half the height of the beam. By plugging in all the known variables, the axial stress at the edges of the beam will be equal to 41.834 psi. Finally, the problem statement requests to obtain the von Misses stress at the center of the plate. To do so, the first step is to obtain the principal stresses in the plate, as those stresses will be used in Equation 13 to obtain the von Misses stress as follows. 𝜎𝑣 = √𝜎12 − 𝜎1 𝜎2 + 𝜎22 Equation 13 These principal stresses can be found knowing the stresses in the x and z direction, as well as the shear stress. Because this problem is a plate, and to simplify this problem, the assumption taken is that it will not carry shear stresses, just like it has been done for most of the analytical solutions during this class. Knowing this, the principal stresses equation is simplified to the following two-dimensional case. 𝜎1 , 𝜎2 = 𝜎𝑥 + 𝜎𝑧 𝜎𝑥 − 𝜎𝑧 ± 2 2 Equation 14 Recalling to Timoshenko’s book, the x and z stresses can be found by combining the following sets of equations. 𝜎𝑥 = 𝐸𝑦 1 1 ( +𝑣 ) 2 1 − 𝑣 𝑟𝑥 𝑟𝑧 | 𝜎𝑧 = 𝐸𝑦 1 1 ( +𝑣 ) 2 1 − 𝑣 𝑟𝑧 𝑟𝑥 Equation 15 𝑀𝑥 = 𝐷 ( 1 1 +𝑣 ) 𝑟𝑥 𝑟𝑧 | 1 1 𝑀𝑧 = 𝐷 ( + 𝑣 ) 𝑟𝑧 𝑟𝑥 Equation 16 By combining Equation 15 and Equation 16, the following expression is obtained to find the x and z stresses as a function of the moment about the same axis and the equivalent stiffness found previously. Once again, y in these equations represents the vertical distance to the point of interest from the neutral axis. The equations are as shown below. 11 𝜎𝑥 = 𝐸𝑦 𝑀𝑥 1 − 𝑣2 𝐷 | 𝜎𝑧 = 𝐸𝑦 𝑀𝑧 1 − 𝑣2 𝐷 Equation 17 As mentioned previously, the calculations for the von Misses stress will be conducted using a new plate of equivalent thickness equal to 0.9173 inches. Furthermore, because the plate that is being analyzed has equal sides, the moments of x and z for it will be equal, therefore both equations above can be merged into a single equation. 𝜎𝑥 = 𝜎𝑧 = 𝐸𝑦 𝑀𝑐𝑒𝑛𝑡𝑒𝑟 1 − 𝑣2 𝐷 Equation 18 Furthermore, due to this relation between x and z stresses that equals them, Equation 13 simplifies to the following expression. 𝜎𝑣 = 𝐸 𝑀𝑐𝑒𝑛𝑡𝑒𝑟 𝑡𝑒𝑞 1 − 𝑣2 𝐷 2 Equation 19 Recalling Equation 10 and based on the value for β found on Theory of Plates and Shells equal to 0.0209. The value obtained for the moment about both axis at the center is 4.18 lb·in. Therefore, plugging all the known values into Equation 19, von Misses stress equals to 29.803 psi. 12 Results & Discussion In the previous sections, two different methods to analyze a simply supported stiffened plate model under a uniform pressure have been presented. Following, the results obtained through both methods are presented side by side and compared. Table 5 displays the results that have just been mentioned. x – displacement [in] y – displacement [in] z – displacement [in] stiffener axial stress top [psi] von Mises stress [psi] Finite Element Model 0 0.0003917 0 49.504 32.855 Analytical Solution 0 0.0004294 0 41.834 29.803 % Error 0 8.77 0 15.49 9.29 Table 5. Results for both Methods Starting by analyzing the different displacements, the only axis that results in a displacement is the y-axis. This is totally understandable as it is the only direction along which a load is applied. Also, due to the structure being symmetrical about the z-axis and x-axis, no displacement on the center of the plate will be experimented. Then, focusing on the results of the y-displacement, the Finite Element Model provides a result of 0.0003917 inches as shown in Figure 11, whereas the analytical solution ends up providing a solution of 0.0004294 inches for this same parameter. This translates in a percentage error of 8.77% with the analytical solution providing a larger deflection. In case of choosing one of the values, the analytical solution would provide a safer solution. Figure 11. Four Central Elements of Plate (Y-Displacement). Following, the second parameter requested by the problem statement s the axial stress on the stiffener situated at the center of the plate. The Finite Element Model results in a value of 49.504 psi as shown in Figure 12. On the other hand, the analytical solution provides a result of 41.834 psi, for a total percentage error of 15.49%, the largest error presented on this report. 13 Figure 12. Central Stiffener on the Plate (Axial Stress). Contrary to the previous case, the FEA model presents a larger value for axial stress, thus this is the value that must be chosen for safety purposes. As mentioned previously, this is the parameter with the largest percentage error. This is most likely since the analytical solution for the stiffener is a simplified twodimensional model that only accounts for the reactions of the plate on the edge of contact with the stiffener. If the stiffener is to be assumed to carry most of the load this is a reasonable assumption. However, it is important to account that the plate element has been neglected for this analysis. This reasoning would make more understandable this higher percentage error. The final parameter that has been analyzed in this report is the von Misses stress at the center of the plate. As shown in Figure 13 and Figure 14, the value for the FEA is 32.855 psi. If we recall the analytical solution, the result is slightly lower than that, being 29.803 psi, which leads to a percentage error of 9.29%. Figure 13. Full Plate (von Misses Stress). 14 Figure 14. Four Central Elements (von Misses Stress). As in the previous case, if we where to choose a value to proceed for our design, the chosen one would be the computational one, as it presents a larger value which will ensure a lower failing probability. In this case, the percentage error could come from the fact that we are using equations that have been derived for plates. Plates usually present a much lower thickness to width and length ratio. When the equivalent thickness method is used, the plate goes from having a thickness to width/length ratio of 0.005 to a 0.459, which represents an increase of over 900%. This could signify that the assumptions made for plates can not longer be applied, as there might be some shear flow that has not been accounted for. Conclusion When designing or analyzing a structure, it is usually faster and simpler to create a finite element model that will give you more detailed information of the overall performance. However, it is key to be able to support your computational solution with an analytical solution or hand calculations. Even if you make assumptions to be able to perform your analytical solution, it will still give you an overall vision of the problem and allow for verification of the FEA. On top of that, having two different methods that lead to a similar solution help you better understand the simplifications that analytical solutions require, and gives you a wider understanding of how the theoretical methods learned in the classroom must be applied in real life scenarios. Having two solutions derived from different methods also allows you to choose whichever one will give a higher safety factor. For this case, the analytical solution gave a safer solution when it came to displacements, whereas the computational method was the safer choice when it came to structure stresses, which would lead to ultimate failure of the structure. We can conclude then, that if our structure was to be constrained by space limitations, the analytical solution would be the ideal one, whereas if our structure was to be limited by the weight, like in most aerospace structures, the finite element model would be the proper solution. 15 References [1] U.S. Dept. of Defense. (1998). 3-71. Table 3.2.3.0 (b1). In Metallic materials and elements for aerospace vehicle structures. [2] Timošenko Stepan P., & Woinowsky-Krieger, S. (1996). Alternate Solution for Simply Supported and Uniformly Loaded Rectangular Plates. In Theory of plates and shells (pp. 113–124). essay, McGraw-Hill. [3] Radosta Frank, J. (2017). In Aerospace Structures I. essay, Embry-Riddle Aeronautical University, Aerospace Engineering Department. 16
Purchase answer to see full attachment