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CAE REPORT

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  • 1. 1.0 INTRODUCTION The Engineering Design Process outlines the steps necessary in solving any type ofengineering problems. The engineering design process is a multi-step process including theresearch, conceptualization, feasibility assessment, establishing design requirements,preliminary design, detailed design, production planning and tool design, and finallyproduction. The sections to follow are not necessarily steps in the engineering design process,for some tasks are completed at the same time as other tasks. This is just a general summaryof each part of the engineering design process. The Engineering Design Process is not a linearprocess. Successful engineering requires going back and forth between the six main steps asshown in Figure 1. Figure 1 Engineering design process During the engineering design process, designers frequently jump back and forth betweensteps. Going back to earlier steps is common. This way of working is called iteration, and it islikely that our process will do the same. 1
  • 2. In this project, we have to use the finite element method (FEM). Its practical applicationoften known as finite element analysis (FEA)) which is a numerical technique for findingapproximate solutions to partial differential equations (PDE) and their systems, as well as(less often) integral equations. In simple terms, FEM is a method for dividing up a verycomplicated problem into small elements that can be solved in relation to each other. FEM isa special case of the more general Galerkin method with polynomial approximationfunctions. The solution approach is based on eliminating the spatial derivatives from thePDE. This approximates the PDE with a system of algebraic equations for steady state problems, a system of ordinary differential equations for transient problems. These equation systems are linear if the underlying PDE is linear, and vice versa.Algebraic equation systems are solved using numerical linear algebra methods. Ordinarydifferential equations that arise in transient problems are then numerically integrated usingstandard techniques such asEulers method or the Runge-Kutta method. In solving partial differential equations, the primary challenge is to create an equationthat approximates the equation to be studied, but is numerically stable, meaning that errors inthe input and intermediate calculations do not accumulate and cause the resulting output to bemeaningless. There are many ways of doing this, all with advantages and disadvantages. Thefinite element method is a good choice for solving partial differential equations overcomplicated domains (like cars and oil pipelines), when the domain changes (as during asolid state reaction with a moving boundary), when the desired precision varies over theentire domain, or when the solution lacks smoothness. For instance, in a frontal crashsimulation it is possible to increase prediction accuracy in "important" areas like the front ofthe car and reduce it in its rear (thus reducing cost of the simulation). Another example wouldbe in Numerical weather prediction, where it is more important to have accurate predictionsover developing highly nonlinear phenomena (such as tropical cyclones in the atmosphere,or eddies in the ocean) rather than relatively calm areas. 2
  • 3. 2.0 MODEL DESCRIPTION 2.1 TASK 1 3D CAD model as the dimension A = 25mm, B= 20mm, w= 1500 N/m , Poisson’s Ratio (v) = 0.29, Em = 200GPa, Es= 400MPa Figure 2 Original Beam Figure 3 Design 1 3
  • 4. Figure 4 Design 2Figure 5 Design 34
  • 5. 2.2 ANALYSIS  ORIGINAL BEAM (3D) Figure 6 Deformation of original beam Figure 7 Von-mises stress 5
  • 6. Figure 8 Translational displacement Figure 9 Principle stress 6
  • 7. Figure 10 Max & min nodal of Von-misesFigure 11 Max & min nodal of translational displacement 7
  • 8. Figure 12 Max & min nodal of principle stress Original Beam MESH: Entity Size Nodes 256 Elements 771 ELEMENT TYPE: Connectivity Statistics TE4 771 ( 100.00% ) 8
  • 9. ELEMENT QUALITY: Criterion Good Poor Bad Worst Average 771 ( Stretch 0 ( 0.00% ) 0 ( 0.00% ) 0.410 0.587 100.00% ) 771 ( Aspect Ratio 0 ( 0.00% ) 0 ( 0.00% ) 3.865 2.234 100.00% ) Materials.1 Material Steel Youngs modulus 2e+011N_m2 Poissons ratio 0.29 Density 7860kg_m3 Coefficient of thermal expansion 1.17e-005_Kdeg Yield strength 4e+008N_m2 Static Case Boundary Conditions Figure 1 STRUCTURE ComputationNumber of nodes : 256Number of elements : 771Number of D.O.F. : 768Number of Contact relations : 0Number of Kinematic relations : 0Linear tetrahedron : 771 9
  • 10. RESTRAINT Computation Name: Restraints.1 Number of S.P.C : 90 LOAD Computation Name: Loads.1 Applied load resultant : STIFFNESS ComputationNumber of lines : 768Number of coefficients : 12642Number of blocks : 1Maximum number of coefficients per bloc : 12642Total matrix size : 0 . 15 Mb SINGULARITY ComputationRestraint: Restraints.1Number of local singularities : 0Number of singularities in translation : 0Number of singularities in rotation : 0Generated constraint type : MPC CONSTRAINT ComputationRestraint: Restraints.1Number of constraints : 90Number of coefficients : 0Number of factorized constraints : 90Number of coefficients : 0Number of deferred constraints : 0 FACTORIZED ComputationMethod : SPARSENumber of factorized degrees : 678Number of supernodes : 135Number of overhead indices : 3954Number of coefficients : 26211Maximum front width : 72Maximum front size : 2628 10
  • 11. Size of the factorized matrix (Mb) : 0 . 199974 Number of blocks : 1 Number of Mflops for factorization : 1 . 356e+000 Number of Mflops for solve : 1 . 082e-001 Minimum relative pivot : 2 . 764e-002 Minimum and maximum pivot Value Dof Node x (mm) y (mm) z (mm) 2.8078e+008 Tx 256 5.5259e+000 6.5176e+000 2.6446e+001 1.2646e+010 Ty 221 1.4819e+001 5.8076e+000 1.1835e+002 Minimum pivot Value Dof Node x (mm) y (mm) z (mm) 3.1157e+008 Ty 256 5.5259e+000 6.5176e+000 2.6446e+001 8.5879e+008 Ty 240 8.8304e+000 1.2999e+001 1.0319e+002 8.8355e+008 Tz 225 1.0611e+001 1.9616e+001 5.6535e+001 8.8865e+008 Tx 244 6.1865e+000 5.9668e+000 8.2072e+001 8.9038e+008 Tz 243 1.0395e+001 1.3270e+001 9.3997e+001 8.9809e+008 Tz 226 1.0087e+001 1.9574e+001 6.5662e+001 9.0152e+008 Ty 251 1.0388e+001 1.3258e+001 5.6463e+001 9.0499e+008 Tx 227 9.4759e+000 1.9519e+001 7.4732e+001 9.1072e+008 Tx 226 1.0087e+001 1.9574e+001 6.5662e+001 Translational pivot distribution Value Percentage 10.E8 --> 10.E9 3.0973e+000 10.E9 --> 10.E10 9.6313e+001 10.E10 --> 10.E11 5.8997e-001 DIRECT METHOD ComputationName: Static Case Solution.1Restraint: Restraints.1Load: Loads.1Strain Energy : 2.495e-005 J 11
  • 12. Equilibrium Applied Relative Components Reactions Residual Forces Magnitude Error Fx (N) -2.2500e+002 2.2500e+002 9.9476e-013 2.3287e-014 Fy (N) -2.4980e-014 8.9084e-013 8.6586e-013 2.0270e-014 Fz (N) 4.5157e-014 7.1487e-013 7.6003e-013 1.7792e-014 Mx (Nxm) -1.0037e-014 -7.6279e-014 -8.6316e-014 1.3471e-014 My (Nxm) -1.6875e+001 1.6875e+001 1.5277e-013 2.3842e-014 Mz (Nxm) 2.8125e+000 -2.8125e+000 5.0626e-014 7.9011e-015 Static Case Solution.1 - Deformed mesh.2 Figure 2On deformed mesh ---- On boundary ---- Over all the model 12
  • 13. Static Case Solution.1 - Von Mises stress (nodal values).2 Figure 33D elements: : Components: : AllOn deformed mesh ---- On boundary ---- Over all the model Global Sensors Sensor Name Sensor Value Energy 2.495e-005J 13
  • 14. 2.3 TASK 2  DESIGN 1 Figure 13 Design 1 Figure 14 Deformation of Design 1 14
  • 15. Figure 15 Von-mises stress of Design 1Figure 16 Translational displacement of Design 1 15
  • 16. Figure 17 Principle stress of Design 1 DESIGN 2 Figure 18 Design 2 16
  • 17. Figure 19 Deformation of Design 2Figure 20 Von-mises stress of Design 2 17
  • 18. Figure 21 Translational displacement of Design 2 Figure 22 Principle stress of Design 2 18
  • 19.  DESIGN 3 Figure 23 Design 3 Figure 24 Deformation of Design 3 19
  • 20. Figure 25 Von-mises stress of Design 3Figure 26 Translational displacement of Design 3 20
  • 21. Figure 27 Principle stress of Design 3 21
  • 22. 2.4 MORPHOLOGICAL CHART DESIGN Original beam Design 1 Design 2 Design 3A (mm) 25 25 25 25B (mm) 20 20 20 20w (N/m) 1500 1500 1500 1500Poisson’s 0.29 0.29 0.29 0.29ratio (v)Em (GPa) 200 200 200 200Es (MPa) 400 400 400 400Von Misses 8.32x107 9.89x107 2.82x107 1.26x107Stress (Nm2)Translational 0.000426 0.000507 0.000644 0.000277Displacement(mm)Principal 1.09x107 1.31x107 1.91x107 1.07x107Stress (Nm2) 22
  • 23. 3.0 DISCUSSION From the analysis above, I had done the finite element analysis (FEA) for originaljust in 3D. The Von-Mises Stress obtained is 8.32 x107 Nm2 , Translational displacement is0.000426 mm, and Principle Stress is 1.09x107 Nm2 for 3D beam by using CATIA. A beam is a structural element that is capable of withstanding load primarily byresisting bending. The bending force induced into the material of the beam as a result of theexternal loads, own weight, span and external reactions to these loads is called a bendingmoment. In order to increase the stiffness of the bracket without compromising thedimensions, we have to design 3 concepts to overcome the problem. However, only one outfrom three of my design was successfully done. By increasing the stiffness of the beam, thedeflection of the rectangular support bracket must be reduced. From the result obtained, the 1st design introduced us a poor result than the originalbeam. The value of Von Misses Stress = 9.89x107 Nm2, Translational Displacement =0.000507 mm and Principal Stress= 1.31x107Nm2. All the value is much greater than theoriginal beam, thus this design cannot promote to increase the stiffness of the bracket. For the 2nd design, the value of the Von Misses Stress = 2.82x107 Nm2, TranslationalDisplacement = 0.000644 mm and Principal Stress = 1.91x107x107 Nm2. This design alsogives the poor result than the original beam and design 1. However, the third design was successfully obtained the exact value to increase thestiffness due to its value for Von Misses Stress = 1.26 x107 Nm2, TranslationalDisplacement = 0.000277 mm and Principal Stress = 1.07x107 Nm2. This design is the bestamong the other design because have less value than the original beam. So, this design canbe promote to increase the stiffness and reduced the deflection. 23
  • 24. 4.0 CONCLUSIONS As the conclusion, based on the translational displacement, the best design goes to the rd3 design; where the translational displacement of the beam is smaller than the original beamwhich is 0.000277 mm rather than 0.000426 mm. Thus this design can overcome theproblem.5.0 APPENDIX (a) ADDITIONAL STRESS CONTOUR PLOT The program provides two types of stress contour plots: a von Mises-Hencky effective stress plot and a Factor of Safety on Material Yield plot. In both of these plots the maximum stress plotted is limited to the yield strength of the cable component. Thus, where stress concentrations occur, the maximum stress reported by the program will never exceed the yield strength. The following stress contours are available: 1. Stress xx - contour; 2. Stress yy - contour; 3. Stress xy - contour; 4. Stress zz - contour; 5. Major Principal - contour; 6. Minor Principal - contour; 7. Stress 1/ Stress 3 - contour; 8. Max. Shear - contour; 9. P - mean stress. contour; 10. q - shear stress. is the second stress invariants; 11. q/p - ratio of q/p as defined above; 12. Pore pressure - pore water pressure; 13. Yield zone - PISA actually plots the the value of the yield zone function f for plastic models. 24
  • 25. (b) REFERENCES i. http://en.wikipedia.org/wiki/Beam_(structure) ii. http://en.wikipedia.org/wiki/Finite_element_analysis iii. http://answers.yahoo.com/question/index?qid=20081120131225AAbKEvM iv. http://www.newport.com/Fundamentals-of Vibration/140234/1033/content.aspx v. http://cedb.asce.org/cgi/WWWdisplay.cgi?115570 25

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