Introduction to Computer Aided Engineering with Ansys

Introduction Traditionally, Engineers hold up gived laboratory testing equipment to test the structural behavior of materials. period this method is appreciated and is highly acceptable especially for running(a) cases the credit on time consuming and expensive laboratory has hindered progress in the complexity of designed meeted. However, the continual rapid advances in computer aid engineering (CAE) over the years have affected this area signifi gougetly.In some(prenominal) engineering discip pull backs, the application of advance finite factor tools has not lone(prenominal) allowed the introduction of innovative, returnive and efficient designs, but also the development of wear break and more accurate design methods. (M. Mah fetch upren, 2007). In this assignment, an advance Finite fraction tool (Ansys parametric design Language) is used to see the design, material properties, linear nervous strain and modal digest on components with linear isotropous structural materials.The basis of finite component synopsis (FEA) relies on the decom mystify of the ambit into a finite number of sub-do master(prenominal)s (elements) for which the systematic approximate dis authorizer is constructed by move overing 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 cognises by dividing the domain into elements and by expressing the unknown world variable in terms of the assumed approximating functions within severally element (M. Asghar Bhatti, 2005).These functions (also called interpolation functions) are defined in terms of the values of the orbital cavity variables at specific signifys, referred to as nodes. Nodes are usually located along the element boundaries, and they connect adjacent elements. This assignment is a demonst dimensionn on how engineers use numerical solutions to refine and validate design in the earliest stages of product des ign. For the task1 of this assignment, a angle hold with structural isotropic material properties of preteens Modulus, E=200Gpa, v=0. 3 and .Will be analyze, twain things are important to the design engineer, what is the utilize mash on the material that leave behind cause it to begin to fail given the properties and geometry shown in descriptor 1A below. At what point does it begin to fail (What point has the maximal puree). Having knowledge of these two factors, the engineer exit decide to design the support to bear this bill without reverse or if the send to be employ result be reduce provided the design is not necessary a product or component that must bear such dilute.At any point in the design, the design engineer is inclined to make decisions that pass on affect the boilersuit functionality of the Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 tax assessor H. Wahyudi Page1 various components feign in the design. Computer aided engineering , has made sure that the engineer will not pass by means of the cumbersome experience of conducting laboratory test to determine failure, rather fewer 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 CAD molding software to FEA software, the engineer jackpot actually mannikin the real components and conduct test that are well related to how the system will perform in its application. task2 of this assignment is to explore the effect of divagation moment and torque and the corresponding, gazump adjudicate and conventionalism filtrate respectively. There are some designs that the engineer has to consider the effect at a single-valued functionicular point, element or component. For this task, we will consider the try out at point A due to the effect of the digression moment and torque produce by the utilis e upshot.Task 3 is a modal analysis on a simply supported truehearted brick two indwelling 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 cigaret perform effectively without failure due to ringing (vibration). For this the first natural frequency is important. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 assessor H. Wahyudi Page2 Task 1 foresee1A square square bracket Model Analysis pure tones 1. Pre bear upon Preprocessing involves, preparing the flummox for analysis, defining the type of analysis, discretization of the exemplar 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 Define element type For this model, element type 8-node- insipid82 is defined. And on the option, plane hear w/thk is selected. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 tax assessor H. Wahyudi Page3 aim1 showing divisor selection with option. . 2 Setting real Constant The thickness of the model is 10mm. approximate2 Showing really Constants with thickness 10mm. 1. 3 Material Models A linear e uttermost(a)ic isotropic material is applied with a new-fangleds Modulus of elasticity of 200GPa and Poisson ratio of 0. 3 Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 assessor H. Wahyudi Page4 visit 3 Showing materials model with Youngs Modulus of elasticity of 200GPa. ( 1. 4 Geometric Model The steps involve in the modeling bracket to be analyze is shown.To model the geometry correctly, fundamental out points are drawd, lines are created to join the key points, the lines are use to create area, the two circles are drawn and subtracted from the area and so is the slot. 1. 4. 1 acquire key points apply table 1 below Table 1 key points for bracket KP. No 1 2 3 4 5 6 X 0 30 50 74 74 130 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 assessor H. Wahyudi Page5 7 8 130 0 85 85 0 0 phase4 key points mapped for bracket 1. 4. 2 make banknote (PreprocessorModelingLineStraight line join the keypointsFigure 5 showing 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 using L expelling,3,4,5,25 ( Line arc join keypoints,3, 4 at warmheartedness 5 and spoke 25mm. ) 1. 4. 3 do area-(preprocessormodelingcreateAreasArbitraryBy lines ) select all lines Figure 7 created area from lines. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 tax assessor H. Wahyudi Page7 1. 4. 4 Create two circles Circle1 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 slot- 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 engage 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 global solution is deemed by unite the result for each element. This involves finding the stiffness matrix for each element, the deplume matrix for each element, and then obtaining both global stiffness and agitate matrix. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page9 Figure 10 Meshed Model of the bracketF igure11 elegant battle 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 (Solution) To obtain the solution for the model, the type of analysis, constraints (displacement constraints), and the charge up will be define. This is like defining the saltation conditions. 2. 1 bourn Conditions (All DOF= 0 at the two circles) Figure 11 limit 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, diagramted and analyzed. Deformed normal to illustrate result has been obtained in the Postprocessor Phase. TASK1B TASK 1B Maximum Load applied without causing yielding Analytical solution of task1 Free Body plot 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. offset 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 effect of the circle. Sum of vertical forces equals 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 Asses sor H. Wahyudi Page14 . (2) 10 ? x ? 47. 5 M wA 10 V FA x . (3) (4) Nangi 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 lop Force & Bending heartbeat plot 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 1400 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 diagram. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page18 From the shear force and bending moment diagram, it can be observe that at x=47. the shear force is maximum and the bendi ng 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. conventionality 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 quantity is applied. First , we weigh the first and import 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 try Failure Criteria, where is the yield stre ngth of the material use for analysis. Since this uniformly distributed load acts at the slot of 25mm, the force that is been applied due to this uniformly distributed load, .For the inclination 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 between 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 Deflection Figure1. 1C Nodal Displacement plot showing maximum Deflectionof 0. 136653mm The nodal plot above shows that the maximum deflection at the even up 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 shape & 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 criterion is use for this analysis. Figure1. 1D Maximum Von-mises StressVon Mises Failure Criterion The von Mises Criterion (1913), also known as the maximum distortion energy criterion, octahedral shear stress theory, or Maxwell-Huber-Hencky-von Mises theory, is often used 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 following 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 password Of result 1E. 1 Discussion on nodal displacem ent Figure1. 1E Nodal Displacement Plot From the nodal displacement plot above, 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 conjecture that the displacement at this circle is fully restrained, meaning all DOF is zero. The gritty part of the plot shows that there is no deflection. Also a circumferent 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 direction 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. E1cursor diagram the stress at different locations Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor 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(a) 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 second circle. Also the stress at the top circle is borderline 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 reasonably higher than the12. 32MPa obtained from the manual calculation the result is similar. The pressure is less at Nangi Ebughni Okoria- Cume42-09/10-0008 9. , 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 great 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 open Arm For the assignment component no2, a prize arm 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 Asse ssor H. Wahyudi Page33 2A Analysis using Ansys Parametric designs Language (Mechanical APDL). Steps in the analysis Preprocessor 2A-1 Define Element type Element fictional character 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 embitter 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 headstone 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 Plot of Key points Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page 36 2A-3. 2 Create straight person 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 Fillet PreprocessorModeling Createlinesline Fillet First fillet is created between lines KP3 &KP4 and line KP4& KP5 fillet radius is 3. mm, click apply. flash 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 entertain note that I selec ted the about axis 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 divide 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- Cume42-09/10-0008 9. , February 2012- MED 305-assigt1 Assessor H.Wahyudi Page41 2A-5 Apply Boundary Conditions The first terminal point condition applied 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 lithe Mesh for better result, the mesh is refined at the lines to minimum surface 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-assigt1 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 horizontal 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- Cume4 2-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page47 No3 Modal Analysis of a simply supported orthogonal 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- Cume42-09/1 0-00089. , 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 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 relative 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 3r d Result Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1 Assessor H. Wahyudi Page60 quaternary 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