Introduction to Computer Aided Engineering with Ansys

Introduction Tradition every(prenominal)y, Engineers have role laboratory try outing equipment to test the structural behavior of materials. While this method is appreciated and is highly refreshing especially for linear cases the reliance on time consuming and expensive laboratory has hindered progress in the complexity of marked considered. However, the continual rapid advances in computer aided take aiming (CAE) over the eld have affected this argona signifi reartly.In many engineering disciplines, the application of advance delimited grammatical constituent tools has not only allowed the introduction of innovative, effective and efficient designs, but also the development of unwrap and more ideal design methods. (M. Mah cobblers lastren, 2007). In this duty assignment, an advance Finite element tool (Ansys parametric design Language) is manipulationd to canvas the design, material properties, linear focussing and modal psychoanalysis on components with linear isotr opic structural materials.The root of bounded element analysis (FEA) relies on the decomposition of the do main(prenominal) into a finite number of sub-domains (elements) for which the arrogant approximate solution is constructed by applying the variation or weighted residual methods (Erdogan Madenci. Ibrahim Guven, 2006). In effect, FEA reduces the problem to that of a finite number of un cognizes by dividing the domain into elements and by expressing the unknown field variable in basis of the assumed approximating functions within each element (M. Asghar Bhatti, 2005).These functions (also called interpolation functions) be defined in terms of the rate of the field variables at specific apexs, referred to as nodes. Nodes are usually located along the element boundaries, and they link adjacent elements. This assignment is a demonstration on how engineers l destination oneself numerical solutions to refine and validate design in the early stages of product design. For the t ask1 of this assignment, a sustain with structural isotropic material properties of youngs Modulus, E=200Gpa, v=0. 3 and .Will be analyze, both things are important to the design engineer, what is the utilise violence on the material that allow for cause it to begin to fail given the properties and geometry exposen in figure 1A below. At what point does it begin to fail (What point has the upper limit nervous strain). Having knowledge of these two factors, the engineer bequeath counterbalance to design the angle bracket to bear this consign without failure or if the load to be applied leave alone be reduce provided the design is not necessary a product or component that must bear such load.At every point in the design, the design engineer is inclined to make decisions that provide affect the overall functionality of the Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 tax assessor H. Wahyudi Page1 various components involve in the design. Com puter aided engineering , has make sure that the engineer will not pass through the cumbersome experience of conducting laboratory test to determine failure, rather few hours spend on the workstation ( computer system ) with a hightech finite element software, will not only save time, but the resources involve for every laboratory experiment.And with the integration of hot dog pretenseing software to FEA software, the engineer can actually stupefy the real components and conduct test that are closely related to how the system will perform in its application. Task2 of this assignment is to explore the effect of crease moment and torque and the corresponding, trim attempt and normal stress respectively. There are some designs that the engineer has to consider the effect at a particular point, element or component. For this task, we will consider the stress at point A cod to the effect of the bending moment and torque produce by the applied wedge.Task 3 is a modal analysis on a simply supported satisfying brick two natural frequencies are to be presented. In design, it is essential that the natural frequency of the system is known so as to find out if the system can perform effectively without failure repayable to resonance (vibration). For this the first natural frequency is important. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 tax assessor H. Wahyudi Page2 Task 1 come in1A Bracket stupefy compend steps 1. Pre affect Preprocessing involves, preparing the model for analysis, defining the type of analysis, discretization of the model into finite elements. For any analysis in the finite element method, this step is very essential as the result is dependent on this stage. 1. 1 set apart element type For this model, element type 8-node-plane82 is defined. And on the option, plane stress w/thk is selected. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 assessor H. Wahyudi Page3 foretell1 cov ering Element selection with option. . 2 Setting real Constant The thickness of the model is 10mm. Figure2 Showing accepted Constants with thickness 10mm. 1. 3 Material Models A linear elastic isotropic material is applied with a Youngs Modulus of elasticity of 200GPa and Poisson ratio of 0. 3 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 tax assessor H. Wahyudi Page4 Figure 3 Showing materials model with Youngs Modulus of elasticity of 200GPa. ( 1. 4 geometric Model The steps involve in the molding bracket to be analyze is shown.To model the geometry correctly, cardinal points are created, lines are created to join the key points, the lines are use to create force field, the two hardifications are drawn and subtracted from the area and so is the slot. 1. 4. 1 seduce key points development table 1 below Table 1 key points for bracket KP. No 1 2 3 4 5 6 X 0 30 50 74 74 cxxx Y 0 0 36 50 25 50 Z 0 0 0 0 0 0 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 tax assessor H. Wahyudi Page5 7 8 130 0 85 85 0 0 Figure4 key points mapped for bracket 1. 4. 2 Create cable (PreprocessorModelingLineStraight line join the keypointsFigure 5 screening lines created from the key points. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page6 Figure 6 Arc created employ Larc,3,4,5,25 ( Line arc joining keypoints,3, 4 at center 5 and radius 25mm. ) 1. 4. 3 Create area-(preprocessormodelingcreateAreasArbitraryBy lines ) select all lines Figure 7 created area from lines. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page7 1. 4. 4 Create two circles gird1 x =15, y=15, radius=7. 5 Circle2x=40,y=62. , radius=7. 5 Cut out the circle from the main area using Preprocessormodeling Operate Boolean Subtract (Select the big area and click apply and then the two circles) Figure 8 showing subtracted circular areas. 1. 4. 5 Create the s lot- first create the two circles, then the rectangle, use Boolean subtraction operation to cut out the slot. Circle1 x=87. 5, y=67. 5, radius=7. 5 Circle 2 x=112. 5, y=67. 5, r=7. 5 Rectangle coordinates (87. 5, 60) & (112. 5, 75) Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. WahyudiPage8 Figure 9 showing model with slot 1. 4. 6 mesh This is a key part of the finite element method. The model is discretized into finite element. This process is necessary as the solution is solved for each element and then a spheric solution is guarded by combining the result for each element. This involves finding the stiffness matrix for each element, the gist matrix for each element, and then obtaining both global stiffness and lunge matrix. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page9 Figure 10 Meshed Model of the bracketFigure11 polish mesh model at the slot, circles and arc. Nangi Ebughni Okoria - Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page10 2. 0 Processing ( closure) To obtain the solution for the model, the type of analysis, constraints (displacement constraints), and the load will be define. This is like defining the bounds conditions. 2. 1 bourne Conditions (All DOF= 0 at the two circles) Figure 11 Boundary condition (0 displacements to all DOF at the two circles) Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page11 2. Boundary Condition (apply pressure at the slot) Figure 12 Pressure of 19. 26 MPa is applied on the slot 2. 3 Solve the built model to obtain the solution Figure 2. 3 The step use to solve the current Load step Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page12 3. 0 Post processing In this stage, the result will be listed, plotted and study. deform shape to illustrate result has been obtained in the Postprocessor Phase . TASK1B TASK 1B Maximum Load applied without causing pay offing Analytical solution of task1 Free luggage compartment Diagram of Bracket W WA L1 0L2 10 47. 5 72. 5 90 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page13 In this analysis, we are going to consider the effect of the uniformly distributed load to act at ? of the width of the bracket h= 35/2=17. 5mm. First we analyze the system for the shear force, v and bending moment, M. The shear force and bending moment is plotted against x. W is the distributed load along the 25mm slot. is the distribution reaction load along the 10mm length from the center of the circle. Sum of vertical forces decents zero Where F is the force due to W and (i. . I). For the Boundary Condition is the force due to . ) 0 ? x ? 10 Mx WA x V Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page14 . (2) 10 ? x ? 47. 5 M wA 10 V FA x . (3) (4) Nan gi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page15 47. 5 ? x ? 72. 5 V 25w 10 M x (5) 0 (6) Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page16 .. (7) (8)Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page17 Shear coerce & Bending Moment Diagram Graph of x against shear force v 0 0 -5 -10 10 X-Axis 47. 5 72. 5 90 V-Axis -15 v -20 -25 -30 Figure1B. Shear Force Diagram (Graph) Graph of x against bending Moment M 1600 1 cd 1200 Axis Title 1000 800 600 400 200 0 M 0 0 10 125 47. 5 1062. 5 72. 5 1375 90 1375 Figure 1C Bending Moment draw. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page18 From the shear force and bending moment plat, it can be observe that at x=47. the shear force is maximal and the bending moment is maximum at the region , however the shear force at this region is zero. So using x=47. 5 as the point where the stress will begin to be maximum (initiate) value, the value of w and F can be obtained there as followed. mm note that we are using 17. 5 on the assumption that the uniformly distributed load acts at the center of the bracket. Shear stress, Note that this is the shear stress due to the effect of the shear force when the bracket is fully restrained at the two circles. form stress, Note that the normal stress above is due to the bending moment, M.Now, in other to find the value of w, Von mises failure bill is applied. First , we calculate the first and second principal stress, since the bracket is subject and to be analyze under plane stress condition. Von Mises stress, , =2. 11075w Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page19 Now by Von Mises Stress Failure Criteria, where is the yield strength of the material use for analysis. Since this uniformly distributed load acts at the s lot of 25mm, the force that is been applied due to this uniformly distributed load, .For the purpose of analysis of the bracket as presented in the assignment using ansys APDL, this force could be applied as a pressure Task1B ( II) Where will the stress initiate From the shear force diagram and bending moment diagram above, the stress will initiate ate x=47. 5. This is because at this point the shear force, v is maximum and the moment, M is maximum amid 47. 5 to 72. 5. Note that for this calculation, the assumption use is that since the material is a linear isotropic material, the load is linearly proportional to the stress. Nangi Ebughni Okoria- Cume42-09/10-00089. February 2012- MED 305-assigt1 Assessor H. Wahyudi Page20 Figure1B. II showing that the stress will initiate at 47. 5, this also where the maximum stress exist. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page21 Task1C Maximum divagation Figure1. 1C Nodal Displacement plot showing maximum Deflectionof 0. 136653mm The nodal plot above shows that the maximum deflection at the undecomposed end of the bracket is 0. 136653mm. I have included deformed shape plot of the bracket to better show how the bracket deformed.Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page22 Figure1. 2C Deformed cause & un-deformed shaped of the bracket Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page23 Figure 1. 3C Deformed shaped of bracket. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page24 Task 1D Maximum stress The maximum von Mises stress obtained is259. 676MPa. The Von Mises stress failure measure is use for this analysis. Figure1. 1D Maximum Von-mises StressVon Mises Failure Criterion The von Mises Criterion (1913), also known as the maximum deviance energy criterion, octahedral shear stress theory, or Maxwell-Huber- Hencky-von Mises theory, is often utilise to estimate the yield of ductile materials. The von Mises criterion states that failure occurs when the energy of distortion reaches the same energy for yield/failure in uniaxial tension. Mathematically, this is expressed as, Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page25 In the cases of plane stress, s3 = 0. The von Mises criterion reduces to,This equation represents a principal stress ellipse as illustrated in the adjacent figure, Figure 1. 2D Illustration of Von Mises Theory. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page26 Figure 1. 2D Showing position of maximum Von-Mises Stress Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page27 Task1E Discussion Of result 1E. 1 Discussion on nodal displacement Figure1. 1E Nodal Displacement temporary hookup From the nodal displacement plot abo ve, it can be observed that the deflection on the left side of the bracket after the circle.The minimum deflection is on the first circle from the right. This is to say that the displacement at this circle is fully restrained, meaning all DOF is zero. The Blue part of the plot shows that there is no deflection. to a fault a closer look shows that at the right end of the bracket, the displacement is maximum. The plot shows that maximum deflection occurs at the uppermost right node of the bracket. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page28 Figure 1. 3E Displacement Vector plot showing the steering of the deflection and how the bracket deflect.IE. 2 Discussion of Maximum Stress Distribution Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page29 Figure1. E1Arrow diagram the stress at different locations Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Asse ssor H. Wahyudi Page30 Figure 1. E2 stress distribution contour plot. Fig. 1. E1 and Fig. 1. E2 shows that the bracket will experience maximum stress around x= 47. 5 mm, this is to say at the stress is maximum. This is in accordance with the manual calculation obtained in Task1B above. Also comparing Figure 1. E1&1.E2 and the bending moment & shear force diagram shown in figure1B and figure 1C above of task 1B, one could conclude that the assumption used for the manual calculation is correct since the min stress on the model is at the 2nd circle. Also the stress at the top circle is minimal and is increasing from zero to the maximum value of stress at x=47. 5. This result plotted above is when P=19. 26MPa, though this value is slightly high than the12. 32MPa obtained from the manual calculation the result is similar. The pressure is less at Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. WahyudiPage31 12. 32 because the assumption use for the calculation was the uniformly distributed load was acting at the center of the slot. In the application of this bracket, one will be careful not to use a pressure greater than 12. 32MPa on it as this may result to yielding. The design engineer ensured that the applied force on the bracket does not initiate a stress greater than the yield strength of the material. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page32 Task2 Analysis of a prise Arm For the assignment component no2, a lever ramp up is to be analyzed using ansys.The analysis will be conducted to determine the Von-Misses stress at element A as shown in fig. 2. 1 below. A force acts on the components at the 38cm component shown. Figure2. 1 Showing a component of lever arm analyzed in this assignment Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page33 2A Analysis using Ansys Parametric designs Language (Mechanical APDL). Ste ps in the analysis Preprocessor 2A-1 Define Element type Element Type AddSolid10 node solid 187ok Figure2A. Element type 2A-2 Material Model Material PropsMaterial Model StructuralLinearElasticIsotropic Young Modulus of 206 X103 N/mm2 is applied. And poison ratio v=0. 3. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page34 Figure 2B material properties 2A-3 Geometric model Steps in Modeling the Geometry are as followed 2A-3. 1 Create Key points using the table below Table 2- Table Key Point No 1 2 3 4 X 0 0 50 50 Y 0 19 19 12. 5 Z 0 0 0 0 Nangi Ebughni Okoria- Cume42-09/10-00089. February 2012- MED 305-assigt1 Assessor H. Wahyudi Page35 5 6 7 8 355 355 455 455 12. 5 19 19 0 0 0 0 0 Figure 2A-3. 1 temporary hookup of Key points Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page36 2A-3. 2 Create straight Line between the following key points Kp1&Kp2 Kp2&Kp3 Kp3&Kp4 Kp4& Kp5 Kp5&Kp6 Kp6& Kp7 Kp7&Kp8 Kp8&Kp1. Figure 2A-3. 2 Line Plot 2A-3. 3 Create Line tenia PreprocessorModeling Createlinesline Fillet First fillet is created between lines KP3 &KP4 and line KP4& KP5 fillet radius is 3. mm, click apply. Second Fillet is created between line KP4 & KP5 and KP5 & KP6, fillet radius 3. 2mm, click Ok. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page37 Figure 2A-3. 3 Plot of section to show Fillet 2A-3. 4 Create area The area is created by selecting all the lines PreprocessorModelingCreateAreaArbitraryByline Figure 2A-3. 4 Plot of created area Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page38 2A-3. Create an extrusion This is to convert the 2D area created to a 3D solid Cylinder PreprocessorModelingOperateExtrudeAreaabout Axis Please note that I selected the about bloc because we want the extrusion to be alike revolving the area 360o around the axis to be selected. The selected line joining KP1 & KP2 is use as the axis of rotation as this is the center line drawn when the lever arm is dissected into two equal halve from the origin. Figure2A-3. 5 Extrude area about axis Kp1& Kp8 (360o revolution) Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H.Wahyudi Page39 2A-3. 6 create the end point of the arm. Solid cylinder command is use to create this end part. After creating this Volume all the Volumes are added together to form one complete component. Table 3 Features for end part of lever arm Attributes WP X WP Y Radius Depth Part1 405mm 0mm 10mm 380mm Part2 405mm 0mm 10mm -80mm Figure 2A-3. 6 Complete Model Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page40 2A-4 Meshing Figure 2A-4 mesh plot of lever arm. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H.Wahyudi Page41 2A-5 Apply Boundary Conditions The first border condition app lied is to fully restrain the left end of the lever arm. Displacement on area is used, and the area at the left end of the lever arm is picked. All Degree of freedom (ALL DOF) is set to zero. Lastly, the second boundary condition is applied. A force of 1890N in the negative Y-direction is applied to the right end of the lever arm. (Note that 1890N is use because my passport No. is A3543390A and the last two digits on my passport no is 90 respectively). 2A-6 Solve the analysis The current load step is solved and result obtained.To view the obtain result, under postprocessor, click load result and then nodal solution, stress, Von Mises stress. The result is plotted. 2A-7 Refined Mesh for better result, the mesh is smooth at the lines to minimum size of 1 as shown in Figure 2A-7 below. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page42 Figure 2A-6 Refined Mesh Plot Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-ass igt1 Assessor H. Wahyudi Page43 2A-7 Von-Mises stress at Element A The Von Mises stress obtained at A is 866. 984N/mm2Figure2A-8 Von Mises Plot displaying maximum stress obtained at A to be 866. 984 N/mm2. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page44 Task2B Analytical Solution Y A Z 35. 5cm B F=1890N Figure2B-1 Free Body Diagram of the lever arm 38cm From Figure2B-1 above, the force on the 38cm cylinder, will cause a torque about element A. C The plane line from will be the axis upon which it will act. T V=1890N Ansys result 1 M 2nd result ( change position of F) Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. WahyudiPage45 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page46 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page47 No3 Modal Analysis of a simply supported rectangular beam Task3A Finite Element Model Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page48 Figure 3A. 1 Geometric Model of beam. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page49 Figure3A-2 Mesh Plot of beam Nangi Ebughni Okoria- Cume42-09/10-00089. February 2012- MED 305-assigt1 Assessor H. Wahyudi Page50 Task3B Boundary Condition The boundary condition is applied as followed, on the left side all DOF is set to zero whereas on the right side only the vertical is set to zero ( i. e. Fy=0). Figure 3B-1 Boundary Conditions on the beam. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page51 Task 3C Procedure 3. 1. 1 Element Type Solid Brick 8-node 45 (Solid45) 3. 1. 2 Material properties Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page52Geometric Modeling Create rectangle Nangi Ebughni Okoria- Cume42-09/10 -00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page53 Operate Extrude for a length of 5cm which is equal to 0. 05m Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page54 isometric line view of model geometry Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page55 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page56 First Frequency Mode shape Deformed shaped Nangi Ebughni Okoria- Cume42-09/10-00089. February 2012- MED 305-assigt1 Assessor H. Wahyudi Page57 Def + Undeformed Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page58 2nd mode shape Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page59 3rd Result Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page60 fourth mode shape Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page61 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page62