Introduction Traditionally, Engineers have used laboratory testing equipment to test the structural behavior of materials. While this method is appreciated and is highly acceptable especially for linear cases the reliance on time consuming and expensive laboratory has hindered progress in the complexity of designed considered. However, the continual rapid advances in computer aided engineering (CAE) over the years have affected this area significantly.

In many engineering disciplines, the application of advance finite element tools has not only allowed the introduction of innovative, effective and efficient designs, but also the development of better and more accurate design methods. (M. Mahendren, 2007). In this assignment, an advance Finite element tool (Ansys parametric design Language) is used to analyze the design, material properties, linear stress and modal analysis on components with linear isotropic structural materials.

The basis of finite element analysis (FEA) relies on the decomposition of the domain into a finite number of sub-domains (elements) for which the systematic 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 unknowns by dividing the domain into elements and by expressing the unknown field variable in terms of the assumed approximating functions within each element (M. Asghar Bhatti, 2005).

These functions (also called interpolation functions) are defined in terms of the values of the field variables at specific points, referred to as nodes. Nodes are usually located along the element boundaries, and they connect adjacent elements. This assignment is a demonstration on how engineers use numerical solutions to refine and validate design in the early stages of product design. For the task1 of this assignment, a bracket with structural isotropic material properties of young’s Modulus, E=200Gpa, v=0. 3 and .

Will be analyze, two things are important to the design engineer, what is the applied force on the material that will cause it to begin to fail given the properties and geometry shown in figure 1A below. At what point does it begin to fail (What point has the maximum stress). Having knowledge of these two factors, the engineer will decide to design the bracket to bear this load without failure or if the load to be applied will 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 will affect the overall functionality of the Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1: Assessor: H. Wahyudi Page1 various components involve in the design. Computer aided engineering , has made 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 CAD modeling software to FEA software, the engineer can actually model 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 bending moment and torque and the corresponding, shear stress 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 due to the effect of the bending moment and torque produce by the applied force.

Task 3 is a modal analysis on a simply supported solid 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 due to resonance (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 Figure1A: Bracket Model Analysis steps 1. : Preprocessing 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: Define 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 Figure1: showing Element selection with option. . 2: Setting real Constant: The thickness of the model is 10mm. Figure2: Showing Real Constants with thickness 10mm. 1. 3: Material Models: A linear elastic isotropic material is applied with a Young’s 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 Figure 3: Showing materials model with Young’s 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, key points are created, 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: Create key points using 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 Figure4: key points mapped for bracket 1. 4. 2: Create Line (Preprocessor>>Modeling>>Line>>Straight line: join the keypoints

Figure 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 Larc,3,4,5,25 ( Line arc joining keypoints,3, 4 at center 5 and radius 25mm. ) 1. 4. 3: Create area-(preprocessor>>modeling>>create>>Areas>>Arbitrary>>By 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 Circle1: x =15, y=15, radius=7. 5 Circle2:x=40,y=62. , radius=7. 5 Cut out the circle from the main area using Preprocessor>>modeling>> 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. Wahyudi

Page8 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 global solution is obtained by combining the result for each element. This involves finding the stiffness matrix for each element, the force matrix for each element, and then obtaining both global stiffness and force matrix. Nangi Ebughni Okoria- Cume42-09/10-00089. , February 2012- MED 305-assigt1: Assessor: H. Wahyudi Page9 Figure 10: Meshed Model of the bracket

Figure11: Refined 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 (Solution): 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 boundary conditions. 2. 1: Boundary 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 analyzed. Deformed shape 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 Diagram of Bracket W WA L1 0

L2 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 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: Assessor: 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 Shear Force & 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 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 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. Normal 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 criterion 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 slot 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 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 right 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 shaped & 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 Stress

Von 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: Discussion Of result 1E. 1: Discussion on nodal displacement 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 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. Also 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 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. E1:Arrow 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 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 higher 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. Wahyudi

Page31 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 lever Arm For the assignment component no2, a lever 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: Assessor: H. Wahyudi Page33 2A: Analysis using Ansys Parametric designs Language (Mechanical APDL). Steps in the analysis Preprocessor 2A-1: Define Element type Element Type>> Add>>Solid>>10 node solid 187>>ok Figure2A. : Element type 2A-2: Material Model Material Props>>Material Model>> Structural>>Linear>>Elastic>>Isotropic 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 Plot 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 Fillet Preprocessor>Modeling> Create>lines>line 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 Preprocessor>Modeling>Create>Area>Arbitrary>Byline 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 Preprocessor>Modeling>Operate>Extrude>Area>about Axis Please note that I selected 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 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 boundary 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: Refined Mesh: for better result, the mesh is refined 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-assigt1: Assessor: H. Wahyudi Page43 2A-7: Von-Mises stress at Element A The Von Mises stress obtained at A is 866. 984N/mm2

Figure2A-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. Wahyudi

Page45 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 Page52

Geometric 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 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 4th 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