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Numerical Analysis and Parameter Study of a Mechanical Damper in Machine Tool

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NUMERICAL ANALYSIS AND PARAMETER STUDY OF A MECHANICAL DAMPER IN MACHINE TOOL By DONGKI WON A THESIS PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLOR IDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE UNIVERSITY OF FLORIDA 2004

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Copyright 2004 by Dongki Won

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ACKNOWLEDGMENTS The researcher would like to express his gratitude to Dr. Nam Ho Kim for his inspiration, support and guidance during this research. Thanks also go to other committee members Dr. John Ziegert and Dr. Raphael Haftka for their support and service on the authors advisory committee. Special thanks go Mr. Charles Stanislaus and the students in the design lab for their help in completing tasks associated with the research. Finally, the author wishes to thank his wife, parents, and other family members. Without their support and caring this work would not have been possible. iii

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TABLE OF CONTENTS page ACKNOWLEDGMENTS.................................................................................................iii LIST OF TABLES.............................................................................................................vi LIST OF FIGURES..........................................................................................................vii ABSTRACT.......................................................................................................................ix CHAPTER 1 INTRODUCTION........................................................................................................1 2 SIMPLIFIED MODEL.................................................................................................5 3 REVIEW OF ANALYTICAL APPROACH................................................................9 Calculation of the Normal Force and the Contact Pressure........................................10 Calculation of the Relative Displacement and the Work............................................16 4 BACKGROUND OF FINITE ELEMENTANALYSIS IN CONTACT PROBLEMS...............................................................................................................25 Contact Formulation in Static Problems.....................................................................26 The Lagrange Multiplier Method...............................................................................29 The Penalty Method....................................................................................................30 5 FIITE ELEMENT ANALYSIS..................................................................................32 Finite Element Model.................................................................................................32 Boundary Conditions..................................................................................................35 Calculation of Friction Work......................................................................................36 Finite Element Analysis Results.................................................................................37 Load Step 1..........................................................................................................37 Load Step 2 (Centrifugal Force + Vertical Force)...............................................38 6 PARAMETER STUDY..............................................................................................41 Determination of the Mesh Size.................................................................................41 iv

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Determination of the Start Angle................................................................................42 Parameter Study..........................................................................................................46 Change the Inner Radius of the Finger.......................................................................46 Change the Number of the Finger..............................................................................47 Final Results...............................................................................................................48 Comparison between the Analytical and Numerical Results.....................................50 7 CONCLUSION AND FUTURE WORK...................................................................53 APPENDIX A MATLAB CODE FOR THEORITICAL ANALYSIS...............................................55 B ANSYS INPUT FILE 01............................................................................................59 C ANSYS INPUT FILE 02............................................................................................61 D ANSYS INPUT FILE 03............................................................................................63 E ANSYS INPUT FILE 04............................................................................................65 F ANSYS INPUT FILE 05............................................................................................68 REFERENCES..................................................................................................................72 BIOGRAPHICAL SKETCH.............................................................................................73 v

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LIST OF TABLES Table page 2-1. Material properties used for the shank and finger........................................................8 3-1. First moment Q of various quantities.........................................................................13 5-1. The number of nodes and elements............................................................................33 6-1. The results of parameter study....................................................................................48 vi

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LIST OF FIGURES Figure page 1-1. Chatter mark.................................................................................................................1 2-1. Endmill and damper......................................................................................................5 2-2. Model.simplification.....................................................................................................6 2-3. Dimensions of geometry...............................................................................................7 3-1. Cross sectional area of the endmill system.................................................................11 3-2. Cross sectional area of the finger...............................................................................14 3-3. Cross sectional area of the finger...............................................................................15 3-4. Plot of work done by the friction force according to the change of the inner radius of finger........................................................................................................................22 3-5. Plot of work done by the friction force for varying start angles of the finger............23 3-6. Plot of the work done by the friction force for different numbers of fingers.............23 5-1. Solid 95, 20 node solid element..................................................................................32 5-2. FEA model of the endmill using 20 node cubic elements and 8 node contact elements with boundary conditions.........................................................................................33 5-3. 8-node contact and target element description...........................................................34 5-4. Force boundary conditions in each time step.............................................................35 5-5. Diagram explaining how to calculate the damping work...........................................36 5-6. FEA result for load step 1...........................................................................................37 5-7. FEA result for load step 2...........................................................................................39 5-8. Schematic diagram of endmill behavior according to the applied forces...................40 6-1. Damping according to the number of elements..........................................................41 vii

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6-2. Damping according to the position of the finger........................................................42 6-3. The maximum and the minimum values of the damping work for the given number of fingers..................................................................................................................43 6-4. Relative displacement of the two-finger case.............................................................44 6-5. Relative displacement of five-finger case..................................................................45 6-6. The first design variable.............................................................................................46 6-7. The second design variable.........................................................................................46 6-8. The result of the parameter study in which the inner radius of finger was changed..47 6-9. The result of the parameter study in which the number of fingers is changed...........48 6-10. The plot of the Table 6-1..........................................................................................49 viii

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Abstract of Thesis Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Master of Science NUMERICAL ANALYSIS AND PARAMETER STUDY OF A MECHANICAL DAMPER IN MACHINE TOOL By Dongki Won May, 2004 Chair: Nam Ho Kim Major Department: Mechanical and Aerospace Engineering When an endmill is used in high-speed machining, chatter vibration of the tool can cause undesirable results. This vibration increases tool wear and leaves chatter marks on the cutting surface. To reduce the chatter vibration, a layered-beam damper is inserted into the hole at the center of the tool. The friction work done by the relative motion between the tool and damper reduces chatter vibration. The purpose of this research is to design and optimize the configuration of the damper to obtain the maximum damping effect. The analytical method has been reviewed, which is based on the assumption of constant contact pressure and uniform deflection. For the numerical approach, nonlinear finite element analysis is employed to calculate the distribution of the contact pressure under the centrifugal and cutting forces. The analytical and numerical results are compared and discussed. ix

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In order to identify the effect of the dampers configuration, two design variables are chosen: the inner radius of the damper and the number of slotted dampers. During the parameter study and optimization, the inner radius is varied from 1.5mm to 3.5mm and the number of slotted dampers is varied from 2 to 10. Results show that the damping effect is maximum when the inner radius is 1.5mm and the number of slotted dampers is 5. However, this result depends on the operating condition. Thus, it is suggested to prepare a set of dampers and to apply the appropriate one for the optimum damping effect for a given operating condition. x

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CHAPTER 1 INTRODUCTION Milling is widely used in many areas of manufacturing. Traditionally, milling has been regarded as a slow and costly process. Therefore, many efforts have been made to improve the efficiency of milling. The main limitation of milling is caused by the vibration of the machine tool and workpiece. As the speed and the power of milling are increased, it is very important to control vibration of the tool. Two different kinds of vibration affect the cutting operation. The one is the self-excited (chatter) vibration at the high spindle speed, and the other is vibration at the critical natural frequency. This research is focused on the former, which produces a wavy surface during the milling operation, as shown in Fig.1-1. Figure 1-1. Chatter mark. Various methods of preventing chatter have been incorporated into machine tool systems. In 1989, Cobb [2] developed dampers for boring bars. These dampers are composed of two different types. The first type, which Cobb calls a shear damper, has two end caps that fit snugly around a boring bar. Between these are sandwiched an 1

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2 annular mass with plastic at each end, either in the form of ring or of several blocks at each end. This middle section of a mass and plastic pieces has a clearance from the boring bar, and is preloaded between the end caps with bolts. When the boring bar vibrates, the end caps transmit this vibration through the plastic pieces to the annular mass, which vibrates in tune. Since the plastic pieces do not slide on the faces of either one, a shear force is produced on the end faces of the plastic. The viscoelastic properties of the plastic provide damping for the system. The other type of damper proposed by Cobb is the compression damper in which the mass is again an annulus. This annulus is cut in half down its axis to form two half annuli. These are then bolted together around rings of plastic that are in contact with the boring bar. The bolts provide a preload on the plastic rings, and when the bar vibrates, the annular mass vibrates out of phase with it, compressing one side and then the other of the plastic rings. This alternate squeezing of the plastic creates damping, again by deforming the plastic, but in a compressive rather than shearing manner. In 1998, Dean [3] focused on increasing the depth of cut and increasing axis federates to improve the metal removal rate (MRR). While he does not present any original ideas on chatter reduction, his thesis refers to work done by Smith [6] in which a chatter recognition system was developed. This system uses a microphone to detect the frequency of chatter when it occurs. The system then selects a different speed(according to the parameters of the system) and tries to machine again. This process repeats itself until chatter no longer occurs. Much work in the field of structural damping has been done by Slocum [5]. In order to damp vibrations, Slocum uses layered beams with viscoelastic materials between

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3 the layers. Two cantilevered beams are stacked on top of each other, and a force is applied to the end of the top beam. It is known that beams experience an axial shear force when displaced in this manner. Slocums derivation calculates a relative displacement between corresponding points on the un-deformed beams. This is incorporated into a selfdamping structure by placing several small beams inside of a larger beam and injecting a viscoelastic material between them. This material bonds to each surface and thus, when there is a relative displacement, is deformed. The stretching of this material causes a dissipation of vibration energy and thus, damping. In 2001, Sterling [7] explored the possibility of deploying a damper directly inside of a rotating tool. To reduce the chatter vibration, a layered-beam damper, which Sterling calls a finger, is inserted into the hole at the center of the tool. Due to the high-speed rotation, the outer surface of the damper contacts with the inner surface of the tool. When chatter vibration occurs, which is a deflection of the tool, work is done in the contact interface due to the friction force and the relative motion. This work is dissipative and reduces chatter vibration. He developed an analytical model and performed an experiment for the layered beam damper. In this research, the work done by Sterling is further extended. Using finite element analysis, his analytical approach is compared to the numerical results. The objective of this research is to calculate the amount of friction work and to maximize its effect by changing the dampers configuration. The organization of thesis is as follows. In Chapter 2, a simplified model of the endmill is introduced that can be used for analytical study and numerical simulation. In Chapter 3, the analytical approach is reviewed that qualitatively estimates the damping

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4 work. Chapter 4 presents the background knowledge of finite element analysis in contact problems. Chapter 5 describes the numerical simulation procedure using the finite element method. Chapter 6 represents the parameter study according to the change of two design variables, followed by conclusions and future work at Chapter 7.

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CHAPTER 2 SIMPLIFIED MODEL The machine tool that we are considering in this research is a 4 long endmill, as shown in Fig.2-1. Most endmills are of the solid beam type as shown in Fig.2-1 (a). In this type of tool, the only available damping mechanism is structural damping, which is very small. Structural damping, which is a variant of viscous damping, is usually caused by internal material friction. When the damping coefficient is small, as in the case of structures, damping is primarily effective at frequencies close to the resonant frequency of the structure. (b) (a) Figure 2-1. Endmill and damper. (a) the original solid endmill and (b) the damper inserted model When a layered-beam damper (see Fig.2-1 (b)) is inserted into the hollow tool, the high-speed rotation causes a strong contact between the beam and tool. When chatter vibration occurs, it generates a relative motion between the beam and tool. Due to the contact force, this relative motion causes a friction force in the interface, which damps 5

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6 the vibration. In this research, this damping mechanism will be referred to as a mechanical damper. While the tool geometry is very important for cutting performance, the objective of this research is on the vibration of the tool. Thus, we want to simplfy the tool geometry so that the analytical and numerical studies in the following chapters will be convenient. The first step of simplifying the endmill model is to suppress unnecessary geometric details, while maintaining the endmills mechanical properties. The endmill model can be simplified as a cylinder because we are only interested in the contact surface, which is the inner surface of the tool. Fig.2-2 (b) illustrates the simplified model. =2722.71 rad/s (a) (b) F=100 N Figure 2-2. Model simplification. (a) the detailed tool model in which the damper is inserted and (b) the simplified model using hollowed cylinder. The simplified endmill model is composed of two-hollowed cylinders. The outer cylinder represents the endmill tool, and the inner cylinder represents the damper. For convenience, the outer part (tool) is denoted as a shank, while the inner part (damper) is denoted as a finger. As schematically illustrated in Fig.2-3, the outer radius R1 of the finger is 1.5 mm, the inner radius R2 of the finger is 4.7625 mm, the inner radius R3 of the shank is 4.7625 mm, and the outer radius R4 of the shank is 9.525 mm. The length of

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7 the endmill is 101.6mm. Because the gap between two contact surfaces is ignored, R2 is equal to R3. R1 will be varied from 1.0 mm to 3.5 mm during the parameter study in Chapter 6. Figure 2-3. Dimensions of geometry. Although Fig.2-3 shows only two fingers, the number of fingers can be altered to improve the damping performance. The parameter study in Chapter 6 will examine the effect of varying the number of fingers between 2 and 10. Because a damper with only one finger would have a lower contact pressure than the other cases, this case will not be considered. Next, the operating condition is also simplified. The applied force is assumed to be sequential. It is first assumed that the tool is rotated with a constant angular velocity. The constant angular velocity will generate a constant contact force at the interface. For this

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8 particular model, an angular velocity of 2,722.713rad/sec is used, which is equal to 26,000 rpm. In this initial state the endmill has not started cutting the surface. When the endmill starts cutting the surface, the tool undergoes a vertical force at the end of the endmill. To approximate the cutting process of the tool, a vertical force is applied at the tip. To accurately approximate the cutting force, the vertical force on the endmill needs to be measured and then an equal force needs be applied at the tip. However, since the objective of this research is vibration control, a representative force of 100N is applied. Thus, the damping work that will be calculated is not the actual magnitude, but rather a relative quantity. For simplicity, the same material properties are assumed for both the shank and finger even though the stiffness of the finger is actually slightly higher than that of the shank. The material properties used are listed in Table 2-1. Table 2-1. Material properties used for the shank and finger. Material Property Value Youngs Modulus 206780 MPa Mass Density 7.8210 -9 ton/mm 3 Friction Coefficient 0.15 In the following chapters analytical and finite element analysis will use the simplified model to determine the conditions for maximum damping of the endmill.

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CHAPTER 3 REVIEW OF ANALYTICAL APPROACH It would be beneficial to review an analytical model before starting the finite element analysis because it will provide a qualitative estimation of the numerical approach. Sterling [7], a former researcher, developed an analytical method that can estimate the amount of friction work during chatter vibration. In this chapter, his analytical approach is reviewed and the results will be compared with finite element analysis results in Chapter 5. The work done by the friction force that occurs between the inner surface of the endmill and the outer surface of the damper causes the damping effect that reduces the chatter vibration. According to the Coulomb friction model [9], the friction force and the damping work can be written as follows: fffFNWFU f (3.1) where f F is the friction force, is the friction coefficient, is the normal contact force, is the friction work, and Uis the relative displacement between the two contact surfaces. The normal force N is mainly caused by the centrifugal force created when the endmill is rotating. The relative displacement U is mainly caused by the vertical deflection of the tool when the endmill starts cutting. Therefore, we can divide the endmill system into two states. The first state is when the endmill is rotating without any cutting operation. In this case, only the centrifugal force is applied. The second state is N fW f f 9

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10 when the endmill starts cutting. In this state, the vertical force is added at the tip of the endmill. Both the centrifugal force and the vertical force are applied in this state. If we assume that there is no relative motion in the first state, then we can calculate the work done by the friction force during the second state. Calculation of the Normal Force and the Contact Pressure In this section, the normal force and pressure that are caused by the rotational motion of the tool will be calculated. There are three assumptions for the analytical method in this step. Those are listed as below. There is no angular acceleration, which means the angular velocity is constant. There is no relative motion between the two contact surfaces during the first step, which means there is no slip in the contact surface during the rotational motion. Contact occurs throughout the entire contact area during the second state. In the actual case, contact may not occur in some portions of the interface. For example there will be no contact near the fixed end or on the two sides where the neutral axis lies. All of these effects are ignored, and it is assumed that contact occurs throughout the entire area. Due to the second assumption, the contact pressure is calculated using the centrifugal force only and is assumed to remain constant. Now let us consider the simplified model, which was developed in Chapter 2 (Fig.2-3). The shank and the finger are hollowed cylinders. For simplicity, we only consider the case of a two-finger configuration. Figure 3-1 (a) shows the cross-sectional area and dimensions of the endmill system in which two fingers are inserted. Considering the symmetric geometry of the fingers, we can consider one finger, which is illustrated in Fig.3-1 (b). Since the finger can have an arbitrary location, represents the start angle of the finger. The point G indicates the first moment (centroid) of the fingers cross section, and R is the distance

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11 between the center of the tool and the centroid G. The normal force N and contact pressure can then be obtained as, cP (a) 22,cc M RNMRPA (3.2) where is the contact surface area, cA M is the mass of the finger, and is the angular velocity. The first step of calculating the contact pressure is to calculate the distance R, which is determined by the centroid of the fingers cross section. R1= 1 mm R4= 9.525 mm R2= R3= 4.7625 mmGR(b) = start angle Figure 3-1. Cross sectional area of the endmill system. (a) Cross sectional area and dimension of the original model in which two fingers are inserted. (b) Cross sectional area of the finger. G is the mass center of the cross section, and is the start angle. The centroid (first moment) of an assemblage of n similar quantities, 2 3 , situated at point , ,, for which the position vectors relative to a selected point O are , , has a point vector n 1P2r 2P3r 3P nP 1r nr r defined as

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12 11niini rr where is the th quantity (for example, this could be the length, area, volume, or mass of an element), ris the position vector of ith element, ii 1ni is the sum of all elements, and n 1i ni r is the first moment of all elements relative to the selected point O. In terms of x, y, and z coordinates, the centroid has coordinates 11niini x x 11niiniyy 11niinizz where is the magnitude of the ith quantity (element), ,, x yz are the coordinates of centroid of the assemblage, and ,ii i x yz are the coordinates of at which iP is concentrated. The centroid of a continuous quantity may be located though calculus by using infinitesimal elements of the quantity. Thus, for area A and in terms of x, y, z coordinates, we can write yzxzxyxdAQxAdAydAQyAdAzdAQzAdA

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13 whereQ, Q, Q are first moments with respect to the xy, yz, and xz planes, respectively. The following table indicates the first moments Q of various quantities xy yz xz about the coordinate planes. In Table 3-1 Q, Q, Q are the first moments with respect to xy, yz, xz planes, is the length, and mis the mass, respectively. Note that in two-dimensional work, e.q. in the xy plane, Qbecomes Q, and becomes xy yzyzQ xz L xy x yQ Table 3-1. First moment Q of various quantities. xyQ yzQ xzQ Dimensions Line zdL x dL ydL 2L Area zdA x dA ydA 3L Volume zdV x dV ydV 4L Mass zdm x dm ydm mL Now let us consider the case illustrated in Fig.3-2, which is a cross section of the finger. According to the figure, y can be expressed as xydAQyRAdA If we choose the polar coordinate system y is represented by, cosyr Using this polar coordinate system, the y of the centroid can be calculated by, 2121212132cos()1cos()312RRxRRRRRRrrdrydAQyAdArdrdrdrdrdrd d

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14 33212221332122212sin()sin()32sin()sin()3RRRRRRRR d y=cos Figure 3-2. Cross sectional area of the finger. If 0 332122212sin()3RRxRR Since the centroid of the cross section is always located along the symmetric line of the finger, it is convenient if the y-axis is chosen such that the centroid is located on the y-axis, as illustrated in Fig.3-3. Due to the symmetry, the integration of the domain can be done between and for angles, which provides a convenient formula. If we choose the polar coordinate system, from the Fig.3-3, y is represented by, cosyr

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15 d y=cos Figure 3-3. Cross sectional area of the finger. Using this polar coordinate system, the y of the centroid can be calculated by, 2121212132cos()1cos()312RRxRRRRRRrrdrydAQyRLdArdrdrdrdrdrd d 33212221332122212sin()sin()32sin()3RRRRRRRR 332122212sin()3RRRRR (3.3) This equation can be used for the general case which undergoes the centrifugal force. For the simplified model in Chapter 2, 2 24.7625 R mm 11.5 R mm

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16 Substituting these values into the above equation yields 2.1738yRmm21520[] Substituting this value into Eq.(3.2) yields the normal force and the contact pressure between two contact surfaces. In Eq., the contact areacan be calculated by cA4.7625 mm 22 [ (3.2) 2101.6c A RL which assumes that all parts of the surface are in contact. Using the material properties shown in Table 2-1, the mass M can be calculated by. 6227.82104.76251.5101.62.549910[]MVkg Therefore the contact pressure can be obtained from Eq.(3.2), as 222.5499102.17382722.7131520270.33]0.2703[]ccMRPAKPaMPa 2 It is noted that the contact pressure is calculated from the assumption that the whole surface is in contact with a constant pressure. cP Calculation of the Relative Displacement and the Work In this section the relative displacement that is caused by the vertical force (cutting force) will be calculated and the work that has been done by the friction force will be calculated. This state represents the one in which the endmill starts the cutting operation. The same assumptions are used as in the last case except for the second one because there is now a relative motion between the two contact surfaces. The assumptions are: There is no angular acceleration, which means the angular velocity is constant. There is no normal contact force change caused by the vertical force. Contact occurs throughout the entire contact area.

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17 Because the centrifugal force is dominant in this endmill system, the change of the normal contact force due to the vertical force is ignored in this step. The first step of calculating the frictional work is to obtain the second moment of inertia of the finger. The following definitions of the second moment, or moment of inertia are analogous to the definitions of the first moment of a plane area, which were given in the previous section. The derivation of load-stress formulas for beams may require solutions of one or more of the following equations: 22xyxy I ydA I xdA I xydA (3.4) where is an element of the plane area dA A lying in the x-y plane. A represents the cross-sectional area of a member subjected to bending and/or torsional loads. The integrals in the above equations are commonly called moments of inertia of the area A because of the similarity with integrals that define the moment of inertia of bodies in the field of dynamics. From Eq.(3.4), if we choose the polar coordinate system, x and y are represented as, cosxr sinyr and x I y I xy I are 21212132442132442134422211sinsincossincos81cossincossincos81sincossinsin8RxRRyRRxyRIrdrdRRIrdrdRRIrdrdRR (3.5) If 0 the Eq.(3.5) can be written as

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18 212121324421032442103442101sinsincos81cossincos81sincossin8RxRRyRRxyRIrdrdRRIrdrdRRIrdrdRR 2 (3.6) For the simplified model in Chapter 2, 11.5 R mm 24.7625 R mm Substituting those values into the above equation yields, 444414.76251.5sincos4.76251.588xI Thus, the moment of inertia of the upper finger is 4444344.76251.5200[]88x 4 I RRmm The moment of inertia of the lower finger can be obtained in the same way because the two fingers are symmetric about the x-axis, and the two values would be the same. Now let us calculate the relative displacement and the friction work. It is well known that beams undergo internal shear deformations along their axes during bending. Members of a composite beam that are not securely fixed together will slide over each other in proportion to their distance from the neutral axis of entire composite beam. It is known that for a cantilevered beam with a point load at the end, the vertical deflection at any point in the beams neutral surface is 32326F 3 x xLLEI where F is the force on the end of the beam, E is the beams elastic modulus, I is the beams moment of inertia, x is the position along the length of the beam measured from the free end, and L is the length of the beam.

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19 If a composite beam is bent, all members will have the same deflection at the tip. Therefore it can be written that 32332323266fsssffFF 3 x xLLxxLLEIEI (3.7) where, subscript s means the shank, and f means the finger. The external force F at the tip of the endmill must equal the sum of the forces required to deflect each member. s fFFF (3.8) (3.8) If there are only two fingers inside the shank Eq.(3.7) can be written as 121122ffs s sffffFFFEIEIEI The above equation can be written as follows 111 f sss f fFEIFEI and 122211 f fffffFEIFEI Substituting these equations into yields 111211111 22 f ssfffsfffffffFEIFEIFFFFFEIEI Solving above equation for 1 f F yields 1111122fffssffffFEIFEIEIEI This equation can be reduced to 1111fffn s sfiiFEIFEIEI fi The same operation for s F and 2 f F yields

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20 1sssn s sfiFEIFEIEI ifi 2221fffn s sfiiFEIFEIEI fi The normal stress at any point in a cantilevered beam is given by Fxc I where is the stress, F is the force on the free end, x is the position along the beam measured from the free end, c is the distance from the beams neutral axis to the point of interest, and I is the area moment of inertia of its cross section. The axial strain in the beam is then FxcEEI where is the axial strain and is the Youngs Modulus of the material. c is the sum of the quantity (d+y) where d is the distance of the neutral axis of the finger from the neutral axis of the composite beam and y is the perpendicular distance from the point in question to the neutral axis of the finger. E For any point in any component member of the composite beam, the change in position of the point related to x=0 (the free end of the beam) can be written as 0xaxialdx Substituting the previously obtained equations into this integral for both the fingers and the shank, the following equations are obtained. The value of d of the finger can be

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21 represented by sinR where is the start angle. Note that in the equation for the shank, d is zero because its neutral axis lies on the neutral axis of the composite beam. 222211()(0)()()22Lsssaxialnnxssssfifi s sfiiiFEIFxdyyFyLxdxLxEIEIEIEIEIEI fi 222211()()()(()22LfififaxialnnxfifissfifissfifiiiFEIFxdydyFdyLxdxLxEIEIEIEIEIEI ) The relative displacement can then be obtained by subtracting the above two equations 22222211()()()()222faxialsaxialnn 1n s sfifissfifissfiiiFdyLxFyLxFdLxEIEIEIEIEIEI fii The work done through this displacement is by friction. The amount of work done is equal to the integral solved over the entire length of the beam of the frictional force (friction coefficient times the normal force, which is the force/unit length at a point times the differential length) multiplied by the relative displacement. Writing this equation (assuming the pressure, P, is uniform over the entire area of the finger) 2201()()2axialLfsaxialnssfifiiFdLxPPEIEI dx The work done by friction for a displacement of the end of the beam by a specified force, F, is then 3113n s sfiidWLPFEIEI fi (3.9)

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22 The MATLAB code used for calculating this friction work is implemented in appendix A. Figure 3-4, 3-5 and 3-6 show the results of theoretical analysis. 1 1.5 2 2.5 3 3.5x 10-3 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 7 7.2 7.4x 10-5 Inner Radius of Damper Finger (m)Work(J) Figure 3-4. Plot of work done by the friction force according to the change of the inner radius of finger. Figure 3-4 is the work according to the change of the inner radius of the finger. As the radius increases the work decreases because the mass decreases. When the radius is 1.5mm and the number of finger is 2, the work done by the friction force is7.1182. Figure 3-5 is the work calculated for different values of the start angle when the number of damper fingers is two and the inner radius is 1.5 mm. When the start angle is 0 it has its maximum value and this value gradually decreases as the angles decrease. Even when the number of fingers is held constant, the damping values differ for different positions of the finger. This result is reasonable because the relative displacement is zero at the neutral axis. In Chapter 6 this effect will be discussed in more detail. 510[]J

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23 0 20 40 60 80 100 120 140 0 1 2 3 4 5 6 7 8x 10-5 Start Angle of Damper Finger (degree)Work(J) Figure 3-5. Plot of work done by the friction force for varying start angles of the finger. The radius of the finger is 1.5 mm and the number of the finger is 2. 0 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 1x 10-4 Number of Damper FingerWork(J) Figure 3-6. Plot of the work done by the friction force for different numbers of damper fingers. The inner radius of the finger is 1.5 mm and the start angle is chosen to have the maximum damping in each number of the finger.

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24 Figure 3-6 shows the work for different numbers of fingers. The inner radius of the finger is 1.5 mm and the start angle is chosen to have the maximum damping for the number of fingers being used. The work increases gradually as the number of fingers is increased. Finite element analysis will be done for these analytical results in the following chapter.

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CHAPTER 4 BACKGROUND OF FINITE ELEMENTANALYSIS IN CONTACT PROBLEMS Since the material properties of the endmill is linear elastic, the classical finite element can be used without difficulty. Major concern is the contact constraints in the interface. Contact problems are highly nonlinear and require significant computer resources to solve for. Contact problems present two significant difficulties. First, the regions of contact are generally unknown until running the model. Depending on the loads, material, boundary conditions, and other factors, surfaces can come into and go out of contact with each other in a largely unpredictable and abrupt manner. Second, most contact problems need to account for friction. Frictional response can be chaotic, making solution convergence difficult. In addition to these two difficulties, many contact problems must also address multi-field effects, such as the conductance of heat and electrical currents in the areas of contact. In this chapter the general procedure of performing the contact analysis using FEA is discussed. Contact problems are characterized by contact constraints which must be imposed on contacting boundaries. To impose the contact constraints, two basic methods are available: the Lagrange multiplier method [9] and the penalty method [9]. Other constraint methods based on the basic methods have been proposed and applied. The augmented Lagrangian method [9] and the perturbed Lagrangian method [9] are two examples. 25

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26 Contact Formulation in Static Problems To introduce the basic constraint methods, we consider static mechanical problems subjected to contact constrains on known contacting boundaries. For small displacements, the total potential energy of the structure can be written as, 1()2TddddSueCeubuqU TF (4.1) where is the total potential energy of the body, is the engineering strain, is the elastic modulus, is the body force, is the surface traction. By using standard element procedures, the discretized form of Eq.(4.1) can be obtained as follows: e C b q 1()2TUUKUU TFc (4.2) where U is the global displacement vector, K is the global stiffness matrix, and F is the global load vector. The virtual work due to the contact load is calculated as (4.3) 1()tLncnWw where indicates that the quantity in the parentheses is evaluated in association with a hitting node n, t is the total number of hitting nodes at time t, and n L cw the virtual work due to the concentrated contact force at hitting node n and is calculated as 21()tciwf ti uu N (4.4) where t i f is the components of the contact force at a target point in the directions of t. The virtual displacement iN 2 u at the target point can be evaluated by using equation (4.5) 21Nnnnu 2,u

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27 where N denotes the total number of target nodes on the target segment, n denotes the shape function associated with contact node n on the contact segment andis the displacement of target node n. 2,nu To obtain a matrix expression for 2 1 u u the following notation is used: 1112,12,12,12,2,2,123123123...TNNNcuuuuuuuuuu 12121210000000000100000...000010000000NcNN Q Then Eq.(4.4) can be written as tTTtTtccciicwfuQNur c where 123tTtttiiiittTtcciNNNfNrQN where tdenotes the ijN j th component of the boundary unit vector t. iN We realize that tis a nodal force vector contributed by the contact at the associated contacting node. It will be convenient to distinguish, in the evaluation of t, between the contribution of normal contact forces and the contribution of tangential friction forces. Thus we write cr cr tttccnrrr cf where 112,3ttTtcncttTtcfcJJffsummationonJrQNrQN

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28 The penetration of the hitting node can be calculated as 211ttttp xxN where t and t are the position vectors of the hitting node and target point, respectively. 1x 2x Expressing t and tin terms of the displacementsuand u, respectively, we have the following discrete form of the kinematic contact condition: 1x 2x 1 2 2121110ttttttTccpp 1 xxNuuNNQu (4.6) where 1212121112233,,TtTp x xxxxxNXX This means that for small displacements the configuration of the contact system may be considered unchanged after the displacements. Thus, we can approximatet, by in Eq.(4.6). Furthermore, we assume only one load step. This means that we need to move only one step in the time domain and 1N 01N p and tin Eq.(4.6) can be replaced by and p p0 p respectively, where denotes any initial penetration (or gap) and p0 p any penetration after deformation. Therefore, the discretized kinematic contact condition can now be written as 010oTccp p NQu (4.7) Eq.(4.7) applies to a single contacting node. If there are contacting nodes, there will be contact constraint equations, each taking the form of Eq.(4.7). Those contact constraint equations can be assembled to obtain )1(LL L L

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29 00 PQUP (4.8) where 12011000120{,,...,},,...,LTLTcIITLppppppPQNQp Since the contacting boundaries are known, all the contacting nodes can be identified and Eq.(4.8) can be established explicitly. Based on the above preparations, the contact problem can be stated as follows: Minimize in Eq.(4.2) subject to the constraints in Eq.(4.8) [P-1] U It must be observed that the mechanical contact condition should be satisfied automatically by assuming that all the contact nodes are actual contacting nodes. In general, actual contacting nodes are not known a priori and an iterative trial-and-error procedure is required to find all contacting nodes. In the following, we discuss the solution of problem [P-1] with alternative constraint methods. L The Lagrange Multiplier Method In the Lagrange multiplier method, the function to be minimized is replaced by the following function: 01(,)()2TTTLUUKUUFQUP (4.9) where is an unknown vector which contains as many elements as there are constraint equations in Eq.(4.8). The elements in are known as Lagrange multipliers.

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30 The constrained minimization problem [P-1] is now transformed into the following saddle-point problem: Find and such that U (,)L U is stationary, i.e. [P-2] 00LL U (4.10) Eq.(4.10) yields 000T KUFQHQUP (4.11) Combining Eq.(4.11), we obtain LLL KUF (4.12) where 00TLLL KQKQUFPUU By solving Eq.(4.12), we can obtain the displacement and the Lagrange multiplier The elements in are interpreted as contacting forces at the corresponding contacting nodes. U The Penalty Method In the penalty method, the potential energy of the structure is penalized when a penetration occurs on the contact surface. The following penalty potential is added to the structural potential:

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31 12Tp PP whereis a diagonal matrix with elements ,ii which is the penalty parameters and is a vector of penetration.. P The function to be minimized is now replaced by 2122ppTT 1 UKUUFPP (4.13) The constrained minimization problem [P-1] is then transformed into the following unconstrained minimization problem: Find such that U p is minimized. To find its minimum, is held stationary by invoking the following condition: L 0p U (4.14) (4.14) Substituting Eq.(4.13) and (4.8) into we can obtain p p KUF (4.15) where 0TpTPKKQQFFQP The solution of Eq.(4.15) gives the displacement The contacting forces are then calculated as U C FP where the penetration P is a function of the displacement vector U

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CHAPTER 5 FIITE ELEMENT ANALYSIS In this chapter, the finite element analysis procedure and results of the endmill system are presented. Even though the cutting process is dynamic, a static finite element analysis is performed with the centrifugal force and the cutting force at the tip. Thus, the friction work obtained must be interpreted as a qualitative measure. Since the simulation condition is the same as that of the analytical method in Chapter 3, it is still valid to compare the results of finite element analysis with the analytical results. Finite Element Model Figure 5-1. Solid 95, 20 node solid element The first step of finite element analysis is to build a computational model. The simplified geometry for the endmill in Chapter 2 is used in the FEA. Using the same 32

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33 material properties listed in Table 2-1, 20-node solid elements in ANSYS (solid 95) are used to build the shank and finger. Figure 5-1 illustrates a 20-node solid element that is used in ANSYS, and Fig.5-2 plots the finite element model of the endmill with boundary conditions. In Fig.5-2, the shank is modeled using two elements through the radial direction, and fingers are modeled inside the shank. The cutting force F=100N is distributed to 4 nodes at the tip. Table5-1 shows the number of elements used in the endmill finite element model. Figure 5-2. FEA model of the endmill using 20 node cubic elements and 8 node contact elements with boundary conditions. Table 5-1. The number of nodes and elements. Node 24826 Element 6480 SOLID95 4320 TARGE170 1080 CONTA174 1080

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34 Because the contact area is curved, higher order elements must be used to prevent inaccurate representation of the surface. Since the contact element is defined on the surface of solid elements, the consistent order of the element must be used for the structure and contact surface. The counter part of SOLID95 structural element is 8-node contact element, as illustrated in Fig.5-3. The contact elements are defined between the shank and finger. Figure 5-3. 8-node contact (CONTA174) and target (TARGE170) element description. ANSYS offers CONTA174 and TARGE170 for contact and target elements, respectively. CONTA174 is used to represent the contact and sliding between 3-D target surfaces and a deformable surface, defined by this element. The element is applicable to 3-D structural and coupled thermal-structural contact analysis. This element is located on the surfaces of 3-D solid or shell elements with midside nodes. It has the same geometric characteristics as the solid or shell element face with which it is connected. Contact occurs when the element surface penetrates one of the target segment elements on a specified target surface. Coulomb and shear stress friction is allowed.

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35 TARGE170 is used to represent various 3-D target surfaces for the associated contact elements. The contact elements themselves overlay the solid elements describing the boundary of a deformable body and are potentially in contact with the target surface, defined by TARGE170. This target surface is discretized by a set of target segment elements (TARGE170) and is paired with its associated contact surface via a shared real constant set. Any translational or rotational displacement, temperature, and voltage on the target segment element can be imposed. Forces and moments on target elements can also be imposed. Boundary Conditions According to the forces applied to the endmill, we can divide the analysis procedure into two steps. The first step is when the endmill starts rotating. In this step, only the angular velocity of 2,722.713 [ is applied without considering the cutting force. /sec]rad Force Vertical Force =100 [N] Rotational Velocity =2722.713 [rad/s] Time Step 1 Time Step 2 Time (sec) Figure 5-4. Force boundary conditions in each time step. Each time step is divided into 5 steps.

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36 The second step is when the endmill starts the cutting process. At this time a vertical force of 100 N is applied at the tip of the endmill. Figure 5-4 illustrates the force boundary conditions in each step. Each time step is divided into 5 substeps in order to improve the convergence of nonlinear analysis. Calculation of Friction Work Even if the structure is linear elastic, the contact constraints make the problem nonlinear. In ANSYS, a Newton-Raphson iterative method is employed to solve the nonlinear system of equations. All default parameters in ANSYS are used in nonlinear analysis. Contact Surface Slide (Relative Displacement) Friction Stress Damping = Dot Product of Friction Force and Slide (Relative Displacement) Figure 5-5. Diagram explaining how to calculate the damping work. After finishing FEA, the damping work which is done by the friction force is calculated. The input commands to ANSYS for obtaining the frictional work are listed in Appendix D. Figure 5-5 shows a schematic procedure of the calculation. First, the friction stress and the relative slide in each contact element are obtained for each load step. By assuming a constant stress within an element, the friction force can be obtained by

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37 multiplying the friction stress with the element area. The dot product of the friction force vector and the relative displacement vector in each substep and in each element yields the friction work. Finite Element Analysis Results Load Step 1 Figure 5-6 is the results at load step 1 when only the angular velocity is applied. In Chapter 3, the analytical method estimates the contact pressure to be 0.27[ ] M Pa ( a ) ( b ) ( c ) ( d ) Figure 5-6. FEA result for load step 1 when only the angular velocity is applied. (a) is the von misses stress (MPa), (b) is the contact pressure (MPa), (c) is the frictional stress (MPa), and (d) is the slide (mm). MX indicates where the maximum value and MN means the minimum value.

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38 Whereas, the maximum contact pressure from FEA is 0.58[ ] M Pa as shown in Fig.5-6 (b). Since the actual contact occurs only half of the contact surface, the higher contact pressure from FEA is expected. In addition, the contact pressure is not constant: maximum at the top and the bottom surface and zero at both sides. Even though we assume that there is no relative motion between the contact surfaces during load step 1 in Chapter 3, there is a relative motion due to the diameter change. However, the relative motion in load step 1 should not be counted because it is not related to the chatter vibration. Load Step 2 (Centrifugal Force + Vertical Force) In the load step 2, a vertical force (cutting force) is applied on top of the centrifugal force. Figure 5-7 shows the results of load step 2. The frictional work is calculated by dot producing the friction force (Fig.5-7 (c)) and the relative slide (Fig.5-7 (d)). Since the load step is divided into five sub steps, the friction work at each sub step must be summed. The total friction work during load step 2 is 3.3426. By comparing with the analytical results in Chapter 3, the finite element analysis estimates about 50% of the analytical result. The friction work calculated from FEA is less than that from the analytical approach because the actual contact area is small in FEA. There is no contact on two sides: the neutral axis lies and the fixed end. In addition, the contact pressure is not constant. That is why the analytical result is about two times higher than the FEA result. Another interesting observation is that most of the relative displacement occurs on the bottom side of the finger (see Fig.5-7 (d)). That happens because the vertical force is applied to the top side of the endmill. 510[]J 57.118210[]J

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39 ( a ) ( b ) ( c ) ( d ) Figure 5-7. FEA result for load step 2 when both ngular velocity and vertical force are applied. (a) is the von misses stress (MPa), (b) is the contact pressure (MPa), (c) is the frictional stress (MPa), and (d) is the slide (mm). MX indicates where the maximum value and MN means the minimum value. Figure 5-8, a schematic diagram of endmill bending according to the applied force, explains in more detail. This figure is exaggerated; the real deformation is very small. Fig.5-8 (a) shows the initial state. When the endmill starts rotating, the centrifugal force is applied and the deformed shape looks like Fig.5-8 (b). Due to the mass conservation, the length of the shank reduces as the diameter increases. Whereas, the length of the finger is not reduced because the cylindrical finger is cut along its neutral axis. When the vertical load is applied on top of the centrifugal force, the shank and the finger are deformed, as illustrated in Fig.5-8 (c). Due to the difference in geometric center, the deflected shape has the largest slide at the bottom surface.

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40 (a)(b)(c) Figure 5-8.Schematic diagram of endmill behavior according to the applied forces. (a) shows the initial state, in (b) the angular velocity is applied and in (c) the vertical force is added.

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CHAPTER 6 PARAMETER STUDY In this chapter, a parameter study is performed to find the maximum value of the damping work. Two design variables, the inner radius of the finger and the number of the fingers, are changed. The results of the parameter study are discussed with respect to the results of the analytical approach. Determination of the Mesh Size It is very important to determine the proper mesh size. A fine mesh will usually give an accurate result, but it requires a large amount of computational cost. Since the finite element analysis needs to be repeated 45 times during the parameter study, computational cost is an important issue. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2x 104 3.15 3.2 3.25 3.3 3.35x 10-5 Element NumberWork(J) Figure 6-1. Damping according to the number of elements. Number of fingers = 2, inner radius R1=1mm, and start angle = 0. 41

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42 Figure 6-1 shows the damping work according to the change of the number of elements. The work gradually increases until the element number reaches 4200, and then it converges on 3.28. Based on these results the number of elements is chosen to be 4200. 510[]J Determination of the Start Angle. Before starting the parameter study, a start angle must be chosen for different configurations. This is because the damping value changes with starting angle even if the same number of fingers is used. 503.2810[]StartAngleWorkJ5452.0110[]StartAngleWorkJ5900.9510[]StartAngleWorkJ51352.0110[]StartAngleWorkJ502.8610[]StartAngleWorkJ503.4010[]StartAngleWorkJ5302.1610[]StartAngleWorkJ5602.8610[]StartAngleWorkJ5903.4110[]StartAngleWorkJ5182.8210[]StartAngleWorkJ5363.4010[]StartAngleWorkJ5543.7010[]StartAngleWorkJ Figure 6-2. Damping according to the position of the finger

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43 Figure 6-2 illustrates the difference of damping work according to the fingers position. The magnitude of damping work is different for different start angle. Thus, it is necessary to evaluate the maximum and minimum values together. Figure 6-3 is the plot of the minimum and maximum values for the same number of fingers when the inner radius is 1.0 mm. The results show a large difference between the minimum and maximum values when the number of fingers is small. However the difference is reduced as the number of fingers is increased. If we compare this result with the analytical result, shown in Fig.3-6, the FEA estimated damping work converges to a third of analytical damping work. Possible explanations are from the fact that not all portion of the fingers are in contact and the relative slide is large at the bottom surface. 1 2 3 4 5 6 7 8 9 10 0 0.5 1 1.5 2 2.5 3 3.5 4x 10-5 Number of FingerWork(J) Maximum ValueMinimum Value Figure 6-3. The maximum and the minimum values of the damping work for the given number of fingers.

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44 Let us consider the cases where the number of fingers is 2 and 5 more in detail. Figure 6-4 shows the relative displacement for the two-finger case. Figure 6-4 (a) and (b) show the case where the start angle of the finger is. Fig.6-4 (c) and (d) show the result when the start angle is0. In both cases the pressure between the contact surfaces is almost the same for each position of the finger. But the relative displacement is different. 90 ( a ) To p ( b ) Botto m ( c ) To p ( d ) Botto m Figure 6-4. Relative displacement of the two-finger case. (a), (b) is when the start angle of the finger is90, and (c), (d) is0. MX indicates where the maximum value occurs and MN means the minimum value. The figure shows that most sliding occurs in the bottom part of the contact surface. That is because the force is applied at the top surface. If we use this relative displacement

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45 for calculating the frictional work, the work calculated when the start angle is0is bigger than. 90 ( a ) To p ( b ) Botto m ( c ) To p ( d ) Botto m Figure 6-5. Relative displacement of five-finger case. (a), (b) is when the start angle of the finger is 18, and (c), (d) is MX indicates where the maximum value occurs and MN means the minimum value. 54 The five-finger case is very similar to the two-fingered case. Figure 6-5 shows the five-finger case. Parts (a) and (b) are for a start angle of 18 and parts (c) and (d) are for a start angle of If we compare Fig.6-5 (b) with Fig.6-5 (d), the latter has almost twice greater sliding region than the former. This is because the location of the finger differently affects the damping work. However, Fig.6-3 shows that these effects 54

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46 disappear as the number of fingers increases. This means that when the number of fingers is small the work done by the friction force depends on the start angle. Parameter Study To find the maximum value of damping, a parameter study was done for two design variables. Figure 6-6 shows the first design variable, which is the inner radius of the finger, and Fig.6-7, the second design variable, the number of fingers. Figure 6-6. The first design variable: the inner radius (R1) of the finger (varied from 1.5 mm to 3.5 mm). Figure 6-7. The second design variable: the number of fingers (from 2 to 10). Change the Inner Radius of the Finger. First the radius is changed from 1.0mm to 3.5mm for the two-finger case. Figure 6-8 shows the result.

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47 1 1.5 2 2.5 3 3.5 2.5 2.6 2.7 2.8 2.9 3 3.1 3.2 3.3 3.4 3.5x 10-5 Inner Radius of Damper Finger (mm)Work(J) Figure 6-8. The result of the parameter study in which the inner radius of finger was changed. Its obvious that the work done by the friction force is gradually reduced, since the mass of the finger is reduced when the radius is reduced. When the radius is 1.0mm the damping is and when the radius is 3.5mm it is Change the Number of the Finger Next the number of fingers is changed from 2 to 10 for the case where the inner radius is 1.5 mm. The damping work shows a minimum value for the four-finger case, and a maximum value for the five-fingered case. The start angle of the finger (position of the finger) is chosen to have the maximum damping. 53.2810[]J 52.5310[]J

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48 1 2 3 4 5 6 7 8 9 10 0 0.5 1 1.5 2 2.5 3 3.5 4x 10-5 Number of FingerWork(J) Figure 6-9. The result of the parameter study in which the number of fingers was changed. Final Results The parameter study is repeated for all cases, and the results are as follows. Table 6-1. The results of parameter study 10 -5 [J] 2 3 4 5 6 7 8 9 10 1.0 mm 3.28 3.41 3.21 3.69 3.55 3.54 3.54 3.48 3.51 1.5 mm 3.34 3.43 3.24 3.72 3.58 3.55 3.57 3.52 3.56 2.0 mm 3.36 3.41 3.24 3.69 3.54 3.47 3.52 3.45 3.50 2.5 mm 3.31 3.31 3.18 3.57 3.36 3.33 3.39 3.35 3.42 3.0 mm 3.08 3.10 3.02 3.32 3.10 3.08 3.12 3.04 3.16 3.5 mm 2.53 2.62 2.65 2.83 2.61 2.63 2.65 2.60 2.68

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49 1 1.5 2 2.5 3 3.5 2 3 4 5 6 7 8 9 10 2 2.5 3 3.5 4x 10-5 Inner Radius of Damper Finger (mm) Number of FingerWork(J) Figure 6-10. The plot of the Table 6-1. In order to find the configuration that yields the maximum damping work, the first design variable (R1) is changed by six different values and the second design variable (number of finger) is changed by nine different values. Table 6-1 shows the results in 69 matrix. In each configuration, the start angle is chosen such that the maximum damping work can be occurred. Figure 6-10 plots the response surface of the damping work. Even if the second design variable is discrete, a continuous surface is plotted for illustration purpose. It is noted that the local peak when the number of fingers is five is maintained throughout all different radii. The general trend of the response surface is consistent. Based on the response surface, we can conclude that the damping work has its maximum value when the inner radius is 1.5 mm and the number of fingers is 5. However, the large

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50 difference in maximum and minimum damping values in this configuration, as shown in Fig.6-3, may reduce the significance of this choice of design. As a conclusion, the effect of damping work increases as the number of fingers is increased and the inner radius is decreased. Comparison between the Analytical and Numerical Results Figure 6-9 shows the analytical and numerical results of friction work as a function of the number of fingers. The friction work estimated from finite element analysis is less than half of the friction work obtained from the analytical method. The possible explanations of such discrepancy are the assumption of constant contact pressure. 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 1 1.2x 10-4 Number of FingersWork(J) Analytical ResultsNumerical Results Figure 6-9. Plot of the analytical and numerical resutls. In reality, the contact pressure is not constant and same portion of the finger does not contact with the shank. And most relative motion occurs at the bottom part of the

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51 endmill. That is because the vertical force is applied at the top and the nonlinearity associated with the centrifugal force contributes the asymmetry between the top and bottom fingers. During the nonlinear analysis ANSYS automatically update the geometry and refer to the deformed configuration, which means the body force is calculated at the deformed geometry. Even though the analytical and numerical results show the difference, the general trends of both results are very similar each other. In order to explain the general trends of friction work, consider the analytical explanation of the contact pressure: 2cc M RPA where c M A and is constant. Value R which is the distance between the rotational center and the mass center of the finger, can be calculated by 332122212sin()3RRRRR If we assume that the friction work is proportional to the contact pressure, then the friction work is proportional to R which increases as the number of fingers increases. If the number of the finger increases, which means the angle goes to zero, R is converges to 3321222123 R RR R R Therefore, the maximum value of the contact pressure can be calculated by 332221222123cccRRMRMPAARR

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52 As a conclusion, the friction work increases along with the number of fingers, but its effect is reduced as the number of fingers increases.

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CHAPTER 7 CONCLUSION AND FUTURE WORK The goal of this research was to design the mechanical damper which is inserted into the endmill. Through the nonlinear finite element analysis and parameter study, the trend of the damping effect is identified. The results from the analytical method were used to verify the FEA result. Although the number and the inner radius of finger were varied in finite element analysis, only the two-finger case was considered in the comparison. The FEA results were smaller than those of the analytical approach because some regions did not contact and the contact pressure was not constant. A parameter study was carried out by changing the inner radius and the number of fingers. The inner radius was varied from 1.0 mm to 3.5 mm, and the number of fingers was varied from 2 to 10. The results show general trends of the damping work according to the change of the two design variables. As the inner radius decreased and the number of finger increased, the damping work increased. The Maximum value was when the inner radius was 1.5 mm and the number of fingers was 5. The parameter study also showed that when the number of the fingers is small the damping work is affected by the position of the finger, but that this dependence disappears as the number of finger increases. 53.7210[]J Recommendations for Future Research As discussed in Chapter 6, the damping work depends on the start angle. However, in practice the endmill is continuously rotating. Thus, it is recommended to perform a 53

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54 series of static FEA by rotating the endmill by on cycle and to calculate the integrated damping work. This procedure will provide more accurate estimation of the real damping work.

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APPENDIX A MATLAB CODE FOR THEORITICAL ANALYSIS %------------------------------------------------------------------------------------% Matlab function to calculate the work done by the friction force % % By Dongki Won %------------------------------------------------------------------------------------function cal_work %------------------------------------------------------------------------------------% Material Properties, radius and applied forces. %------------------------------------------------------------------------------------E=2.0678e11; %Modulus of Elasticity : 2.0*10^11[MPa] rho=7820; %Mass Density : 7820[kg/m^3] mu=0.15; %Friction Coefficient : 0.15 r1=1.5e-3; r2=4.7625e-3; r3=r2; r4=9.525e-3; L=101.6e-3; omega=2722; F=100; %------------------------------------------------------------------------------------% Calculate the work according to the chage of the inner raius of finger. %------------------------------------------------------------------------------------n=2;m=1; for r1=1.0e-3:0.5e-3:3.5e-3 ang(1)=0; for i=1:n ang(i+1)=ang(1)+(360/n)*i; end ang=ang*(pi/180); II=(1/4)*((r4)^4-(r3)^4)*pi; for i=1:(n-1) tmp1=sin(ang(i+1))*cos(ang(i+1))-sin(ang(i))*cos(ang(i))-(ang(i+1)-ang(i)); tmp1=tmp1*((r1)^4-(r2)^4); tmp1=tmp1*(1/8); II=II+tmp1; end work=0; for i=1:n alpha=(ang(i+1)-ang(i))/2; 55

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56 d1=(2*sin(alpha).*(r2^3-r1^3))/(3*alpha*(r2^2-r1^2)); d2=d1*sin((ang(i)+ang(i+1))*0.5); V=(ang(i+1)-ang(i))*(r2^2-r1^2)*L*0.5; P=abs(V*rho*(omega^2)*d1/L); work=work+abs((1/3)*(L^3)*mu*P*F*d2/(E*II)); end y1(m)=work; m=m+1; end x1=[1.0e-3:0.5e-3:3.5e-3]; plot(x1,y1,'b-o'); grid; xlabel('Inner Radius of Damper Finger (m)'); ylabel('Work(J)'); [tmp,n]=size(x1); clc fprintf('\nCalculate the work according to the chage of the inner raius of finger.\n'); fprintf('The Number of Finger=%3.0f\nStart Angle=%3.0f [Degree]\n',n,0); for i=1:n fprintf('Inner Raius of Finger =%16.9e [m]\t\tWork=%16.9e [J]\n',x1(i),y1(i)); end %------------------------------------------------------------------------------------% Calculate the work according to the chage of the start angle of finger. %------------------------------------------------------------------------------------n=2;m=1;r1=1.5e-3; atmp=(360/n)/4; for a=0:atmp:atmp*3 ang(1)=a; for i=1:n ang(i+1)=ang(1)+(360/n)*i; end ang=ang*(pi/180); II=(1/4)*((r4)^4-(r3)^4)*pi; for i=1:(n-1) tmp1=sin(ang(i+1))*cos(ang(i+1))-sin(ang(i))*cos(ang(i))-(ang(i+1)-ang(i)); tmp1=tmp1*((r1)^4-(r2)^4); tmp1=tmp1*(1/8); II=II+tmp1; end work=0; for i=1:n alpha=(ang(i+1)-ang(i))/2; d1=(2*sin(alpha).*(r2^3-r1^3))/(3*alpha*(r2^2-r1^2)); d2=d1*sin((ang(i)+ang(i+1))*0.5); V=(ang(i+1)-ang(i))*(r2^2-r1^2)*L*0.5; P=abs(V*rho*(omega^2)*d1/L);

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57 work=work+abs((1/3)*(L^3)*mu*P*F*d2/(E*II)); end y2(m)=work; m=m+1; end x2=[0:atmp:atmp*3]; figure; plot(x2,y2,'b-o'); grid; xlabel('Start Angle of Damper Finger (degree)'); ylabel('Work(J)'); [tmp,n]=size(x2); fprintf('\nCalculate the work according to the chage of the start angle of finger.\n'); fprintf('Number of Finger=%5.0f\nInner Radius of the Finger=%16.9e [m]\n',2,r1); for i=1:n fprintf('Start Angle =%16.9e [Degree]\t\tWork=%16.9e [J]\n',x2(i),y2(i)); end %------------------------------------------------------------------------------------% Calculate the work according to the change of the number of finger. %------------------------------------------------------------------------------------m=1;clear ang for n=2:10 clear ang atmp=(1+(-1)^(n-1))/2; ang(1)=((360/n)/4)*atmp; for i=1:n ang(i+1)=ang(1)+(360/n)*i; end ang=ang*(pi/180); II=(1/4)*((r4)^4-(r3)^4)*pi; for i=1:(n-1) tmp1=sin(ang(i+1))*cos(ang(i+1))-sin(ang(i))*cos(ang(i))-(ang(i+1)-ang(i)); tmp1=tmp1*((r1)^4-(r2)^4); tmp1=tmp1*(1/8); II=II+tmp1; end work=0; for i=1:n alpha=(ang(i+1)-ang(i))/2; d1=(2*sin(alpha).*(r2^3-r1^3))/(3*alpha*(r2^2-r1^2)); d2=d1*sin((ang(i)+ang(i+1))*0.5); V=(ang(i+1)-ang(i))*(r2^2-r1^2)*L*0.5; P=abs(V*rho*(omega^2)*d1/L); work=work+abs((1/3)*(L^3)*mu*P*F*d2/(E*II)); end y3(m+1)=work;

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58 m=m+1; end x3=[1:10]; figure; plot(x3,y3,'b-o'); axis([0,10,0,11e-5]); grid; xlabel('Number of Damper Finger'); ylabel('Work(J)'); [tmp,n]=size(x3); fprintf('\nCalculate the work according to the chage of the number of finger.\n'); fprintf('Inner Radius of the Finger=%16.9e [m]\n',r1); for i=1:n fprintf('Number of Finger =%3.0f\t\tWork=%16.9e [J]\n',x3(i),y3(i)); end

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APPENDIX B ANSYS INPUT FILE 01 !-----------------------------------------------------------------------------------------! ANSYS input file to determine the mesh size according to the change of elemnt number. The results are stored in mesh_size,txt. ! Dongki Won !-----------------------------------------------------------------------------------------/SHOW,JPEG *SET,rad_in,1.5 Inner radius of the finger *SET,rad_out1,4.7625 Outer radius of the finger *SET,rad_out2,9.525 Outer radius of the shank *SET,f_num,2 The number of fingers *SET,f_ang,360/f_num The angle of one finger *SET,st_ang,(f_ang/4)*0 Start angle of the finger *DO,ii,1,8 *IF,ii,EQ,1,THEN Element Number is 864 *SET,s_mesh,6 *SET,f_mesh,6 *SET,l_mesh,12 *SET,st_mesh,2 *SET,ft_mesh,2 *ELSEIF,ii,EQ,2 Element Number is 1536 *SET,s_mesh,8 *SET,f_mesh,8 *SET,l_mesh,16 *SET,st_mesh,2 *SET,ft_mesh,2 *ELSEIF,ii,EQ,3 Element Number is 2400 *SET,s_mesh,10 *SET,f_mesh,10 *SET,l_mesh,20 *SET,st_mesh,2 *SET,ft_mesh,2 *ELSEIF,ii,EQ,4 Element Number is 4200 *SET,s_mesh,14 *SET,f_mesh,14 *SET,l_mesh,25 *SET,st_mesh,2 59

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60 *SET,ft_mesh,2 *ELSEIF,ii,EQ,5 Element Number is 6480 *SET,s_mesh,18 *SET,f_mesh,18 *SET,l_mesh,30 *SET,st_mesh,2 *SET,ft_mesh,2 *ELSEIF,ii,EQ,6 Element Number is 8640 *SET,s_mesh,18 *SET,f_mesh,18 *SET,l_mesh,30 *SET,st_mesh,3 *SET,ft_mesh,3 *ELSEIF,ii,EQ,7 Element Number is 11500 *SET,s_mesh,18 *SET,f_mesh,18 *SET,l_mesh,40 *SET,st_mesh,3 *SET,ft_mesh,3 *ELSEIF,ii,EQ,8 Element Number is 19200 *SET,s_mesh,20 *SET,f_mesh,20 *SET,l_mesh,60 *SET,st_mesh,3 *SET,ft_mesh,3 *ENDIF tmp1=s_mesh*l_mesh*2*(1+st_mesh) tmp2=f_mesh*l_mesh*2*(1+ft_mesh) num_elem=tmp1+tmp2 /INPUT,endmill_model,inp /INPUT,endmill_work,inp /OUTPUT,mesh_size,txt,,APPEND *VWRITE,num_elem,work (F7.0,3X,F16.9,3X) /OUTPUT *ENDDO

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APPENDIX C ANSYS INPUT FILE 02 !-----------------------------------------------------------------------------------------! ANSYS input file to determine the start angle of the finger. The results are stored in start_angle,txt. ! Dongki Won !-----------------------------------------------------------------------------------------/SHOW,JPEG *SET,rad_in,1 Inner radius of the finger *SET,rad_out1,4.7625 Outer radius of the finger *SET,rad_out2,9.525 Outer radius of the shank *SET,s_mesh,14 *SET,l_mesh,30 *SET,st_mesh,2 *SET,ft_mesh,2 *DO,ii,8,10 *SET,f_num,ii *SET,f_ang,360/f_num *IF,ii,EQ,2,THEN *SET,f_mesh,14 *ELSEIF,ii,EQ,3 *SET,f_mesh,9 *ELSEIF,ii,EQ,4 *SET,f_mesh,7 *ELSEIF,ii,EQ,5 *SET,f_mesh,6 *ELSEIF,ii,EQ,6 *SET,f_mesh,5 *ELSEIF,ii,EQ,7 *SET,f_mesh,4 *ELSEIF,ii,EQ,8 *SET,f_mesh,4 *SET,s_mesh,16 *SET,ft_mesh,2 *ELSEIF,ii,EQ,9 *SET,f_mesh,4 *SET,s_mesh,18 !*SET,ft_mesh,2 61

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62 *ELSEIF,ii,EQ,10 *SET,f_mesh,3 *SET,s_mesh,15 !*SET,ft_mesh,3 *ENDIF *DO,jj,0,3 *SET,st_ang,(f_ang/4)*jj /INPUT,endmill_model,inp /INPUT,endmill_work,inp /OUTPUT,start_angle,txt,,APPEND *VWRITE,f_num,jj,st_ang,work (F7.0,3X,F7.0,3X,F16.9,3X,F16.9,3X) /OUTPUT *ENDDO *ENDDO

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APPENDIX D ANSYS INPUT FILE 03 !-----------------------------------------------------------------------------------------! ANSYS input file to do the parameter study as the change of two design variables, the number of the finger(f_num) and the inner radius of the finger (rad_in). The results are stored in parameter_study,txt. ! Dongki Won !-----------------------------------------------------------------------------------------/SHOW,JPEG *SET,rad_out1,4.7625 Outer radius of the finger *SET,rad_out2,9.525 Outer radius of the shank *SET,s_mesh,14 *SET,l_mesh,25 *SET,st_mesh,2 *SET,ft_mesh,2 *DO,ii,2,10 *SET,f_num,ii *SET,f_ang,360/f_num *IF,ii,EQ,2,THEN *SET,f_mesh,14 *SET,st_ang,(f_ang/4)*0 *ELSEIF,ii,EQ,3 *SET,f_mesh,9 *SET,st_ang,(f_ang/4)*3 *ELSEIF,ii,EQ,4 *SET,f_mesh,7 *SET,st_ang,(f_ang/4)*2 *ELSEIF,ii,EQ,5 *SET,f_mesh,6 *SET,st_ang,(f_ang/4)*3 *ELSEIF,ii,EQ,6 *SET,f_mesh,5 *SET,st_ang,(f_ang/4)*2 *ELSEIF,ii,EQ,7 *SET,f_mesh,4 *SET,st_ang,(f_ang/4)*3 *ELSEIF,ii,EQ,8 *SET,f_mesh,4 63

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64 *SET,s_mesh,16 *SET,st_ang,(f_ang/4)*2 *ELSEIF,ii,EQ,9 *SET,f_mesh,4 *SET,s_mesh,18 *SET,st_ang,(f_ang/4)*3 *SET,ft_mesh,2 *SET,l_mesh,30 *ELSEIF,ii,EQ,10 *SET,f_mesh,3 *SET,s_mesh,15 *SET,st_ang,(f_ang/4)*2 *SET,ft_mesh,2 *SET,l_mesh,25 *ENDIF *DO,jj,1,3.5,0.5 *SET,rad_in,jj /INPUT,endmill_model,inp /INPUT,endmill_work,inp /OUTPUT,parameter_study,txt,,APPEND *VWRITE,f_num,rad_in,work (F7.0,3X,F16.9,3X,F16.9,3X) /OUTPUT *ENDDO *ENDDO

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APPENDIX E ANSYS INPUT FILE 04 !-----------------------------------------------------------------------------------------! ANSYS input file to solve the endmill model. This input file is used in endmill_run.inp and endmill_start.inp ! Dongki Won !-----------------------------------------------------------------------------------------PARSAV /CLEAR PARRES /FILNAME,endmill,1 /UNITS,MPA /TITLE,Damping Design of Endmill /prep7 !--------------------------------------------------------------------! Element Type and Material Properties !--------------------------------------------------------------------ET,1,SOLID95 !20-Nodes Cubic Element MP,EX,1,206800 !Modulus of Elasticity : 206780[MPa] MP,DENS,1,7.82e-9 !Mass Density : 7.82e-6[kg/mm^3] MP,NUXY,1,0.29 !Poisson's Ratio : 0.29 MP,MU,1,0.15 !Friction Coefficient : 0.15 !--------------------------------------------------------------------! Create Model Geometry !--------------------------------------------------------------------!= Shank:contact surface is 3, 8 K,1,0,0,0 K,2,0,0,1 K,11,rad_out2,0,0 K,12,rad_out1,0,0 K,13,rad_out1,0,101.6 K,14,rad_out2,0,101.6 A,11,12,13,14 VROTAT,1,,,,,,1,2,360,2 != Finger:contact surfaces are 13,19,25,31,37 *DO,i,1,f_num,1 CYLIND,rad_in,rad_out1,101.6,,st_ang+f_ang*(i-1),st_ang+f_ang*i *ENDDO 65

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66 !--------------------------------------------------------------------! Generating Mesh !--------------------------------------------------------------------LSEL,S,LENGTH,,101.6 LESIZE,ALL,,,l_mesh vsel,S,volu,,1,2,1,1 LSEL,R,RADIUS,,rad_out1 LSEL,A,RADIUS,,rad_out2 LESIZE,ALL,,,s_mesh LSEL,S,LENGTH,,rad_out1 LESIZE,ALL,,,st_mesh vsel,S,volu,,3,3+f_num,1,1 LSEL,R,RADIUS,,rad_out1 LSEL,A,RADIUS,,Rad_in LESIZE,ALL,,,f_mesh LSEL,S,LENGTH,,rad_out1-rad_in LESIZE,ALL,,,ft_mesh AllSEL MSHAPE,0,3D != hexahedral-shaped elements,3D MSHKEY,2 != Map meshing VMESH,ALL !--------------------------------------------------------------------! Designating Contact Pairs !--------------------------------------------------------------------ET,2,TARGE170 ET,3,CONTA174 ASEL,S,,,3,8,5,1 NSLA,S,1 ESLN,0 TYPE,2 ESURF,,TOP ASEL,S,,,13,13+(f_num-1)*6,6,1 NSLA,S,1 ESLN,0 TYPE,3 ESURF,,TOP FINISH /SOLU !--------------------------------------------------------------------! Apply Boundary Conditions and Sove !--------------------------------------------------------------------NSEL,S,LOC,Z,0

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67 D,ALL,ALL ALLSEL SOLCONTROL,ON Load Step 1 OMEGA,0,0,2722.713,0 TIME,1 NROPT,UNSYM NSUBST,5,10,2 AUTOTS,OFF OUTRES,BASIC,ALL KBC,0 LSWRITE,1 Load Step 2 CSYS,1 NSEL,S,LOC,Z,101.6 NSEL,R,LOC,X,rad_out2 NSEL,R,LOC,Y,90 F,ALL,FY,-25 *DO,kk,1,2 NSEL,S,LOC,Z,101.6 NSEL,R,LOC,X,rad_out2 NSEL,R,LOC,Y,90+(360/(s_mesh*4))*((-1)**kk) F,ALL,FY,-25 *ENDDO *DO,kk,1,2 NSEL,S,LOC,Z,101.6 NSEL,R,LOC,X,rad_out2 NSEL,R,LOC,Y,90+(360/(s_mesh*4))*2*((-1)**kk) F,ALL,FY,-12.5 *ENDDO CSYS,0 ALLSELL TIME,2 NROPT,UNSYM NSUBST,5,10,2 AUTOTS,OFF OUTRES,ALL,ALL KBC,0 LSWRITE,2 LSSOLVE,1,2,1 FINISH

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APPENDIX F ANSYS INPUT FILE 05 !-----------------------------------------------------------------------------------------! ANSYS input file to calculate the work done by the frictional force. This input file is used in endmill_run.inp and endmill_start.inp ! Dongki Won !-----------------------------------------------------------------------------------------/POST1 SET,1,1 ESEL,S,ENAME,,174 *GET,emin,ELEM,0,NUM,MIN *GET,emax,ELEM,0,NUM,MAX *GET,aa,ELEM,emin,VOLU aa=aa*0.25 !-------------------------------------------------------------------------------! Read Friction Stress of R and S direction in each elements !-------------------------------------------------------------------------------ETAB,taur1,SMISC,5 ETAB,taur2,SMISC,6 ETAB,taur3,SMISC,7 ETAB,taur4,SMISC,8 ETAB,taus1,SMISC,9 ETAB,taus2,SMISC,10 ETAB,taus3,SMISC,11 ETAB,taus4,SMISC,12 !-------------------------------------------------------------------------------! Calculate Friction Force of R and S direction by multiplying area !-------------------------------------------------------------------------------SMULT,rforce1,taur1,,aa SMULT,rforce2,taur2,,aa SMULT,rforce3,taur3,,aa SMULT,rforce4,taur4,,aa SMULT,sforce1,taus1,,aa SMULT,sforce2,taus2,,aa SMULT,sforce3,taus3,,aa SMULT,sforce4,taus4,,aa !-------------------------------------------------------------------------------! Read Slide(Relative Displacement) of R and S direction in each elements !-------------------------------------------------------------------------------ETAB,tasr1,NMISC,17 68

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69 ETAB,tasr2,NMISC,18 ETAB,tasr3,NMISC,19 ETAB,tasr4,NMISC,20 ETAB,tass1,NMISC,21 ETAB,tass2,NMISC,22 ETAB,tass3,NMISC,23 ETAB,tass4,NMISC,24 !-------------------------------------------------------------------------------! Calculate the work done by the friction force in each elements. !-------------------------------------------------------------------------------VDOT,work1,rforce1,sforce1,,tasr1,tass1 VDOT,work2,rforce2,sforce2,,tasr2,tass2 VDOT,work3,rforce3,sforce3,,tasr3,tass3 VDOT,work4,rforce4,sforce4,,tasr4,tass4 SABS,1 SSUM *GET,work1,SSUM,,ITEM,work1 *GET,work2,SSUM,,ITEM,work2 *GET,work3,SSUM,,ITEM,work3 *GET,work4,SSUM,,ITEM,work4 work=work1+work2+work3+work4 SABS,0 !-------------------------------------------------------------------------------! Save the slide value for the next step. !-------------------------------------------------------------------------------ETAB,tasr1tmp,NMISC,17 ETAB,tasr2tmp,NMISC,18 ETAB,tasr3tmp,NMISC,19 ETAB,tasr4tmp,NMISC,20 ETAB,tass1tmp,NMISC,21 ETAB,tass2tmp,NMISC,22 ETAB,tass3tmp,NMISC,23 ETAB,tass4tmp,NMISC,24 !-------------------------------------------------------------------------------! Repeat the procedure until reaches to the last set !-------------------------------------------------------------------------------ww=0 *DO,uu,1,2 *DO,vv,1,5 *IF,uu,EQ,1,AND,vv,EQ,1,CYCLE SET,uu,vv ww=ww+1 !-------------------------------------------------------------------------------! Read Friction Stress of R and S direction in each elements !-------------------------------------------------------------------------------ETAB,taur1,SMISC,5

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70 ETAB,taur2,SMISC,6 ETAB,taur3,SMISC,7 ETAB,taur4,SMISC,8 ETAB,taus1,SMISC,9 ETAB,taus2,SMISC,10 ETAB,taus3,SMISC,11 ETAB,taus4,SMISC,12 !-------------------------------------------------------------------------------! Calculate Friction Force of R and S direction by multiplying area !-------------------------------------------------------------------------------SMULT,rforce1,taur1,,aa SMULT,rforce2,taur2,,aa SMULT,rforce3,taur3,,aa SMULT,rforce4,taur4,,aa SMULT,sforce1,taus1,,aa SMULT,sforce2,taus2,,aa SMULT,sforce3,taus3,,aa SMULT,sforce4,taus4,,aa !-------------------------------------------------------------------------------! Read Slide(Relative Displacement) of R and S direction in each elements !-------------------------------------------------------------------------------ETAB,tasr1,NMISC,17 ETAB,tasr2,NMISC,18 ETAB,tasr3,NMISC,19 ETAB,tasr4,NMISC,20 ETAB,tass1,NMISC,21 ETAB,tass2,NMISC,22 ETAB,tass3,NMISC,23 ETAB,tass4,NMISC,24 SMULT,tasr1tmp,tasr1tmp,,-1 SMULT,tasr2tmp,tasr2tmp,,-1 SMULT,tasr3tmp,tasr3tmp,,-1 SMULT,tasr4tmp,tasr4tmp,,-1 SMULT,tass1tmp,tass1tmp,,-1 SMULT,tass2tmp,tass2tmp,,-1 SMULT,tass3tmp,tass3tmp,,-1 SMULT,tass4tmp,tass4tmp,,-1 SADD,tasr1,tasr1,tasr1tmp SADD,tasr2,tasr2,tasr2tmp SADD,tasr3,tasr3,tasr3tmp SADD,tasr4,tasr4,tasr4tmp SADD,tass1,tass1,tass1tmp SADD,tass2,tass2,tass2tmp SADD,tass3,tass3,tass3tmp SADD,tass4,tass4,tass4tmp

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71 !-------------------------------------------------------------------------------! Calculate the work done by the friction force in each elements. !-------------------------------------------------------------------------------VDOT,work1,rforce1,sforce1,,tasr1,tass1 VDOT,work2,rforce2,sforce2,,tasr2,tass2 VDOT,work3,rforce3,sforce3,,tasr3,tass3 VDOT,work4,rforce4,sforce4,,tasr4,tass4 SABS,1 SSUM *GET,work1,SSUM,,ITEM,work1 *GET,work2,SSUM,,ITEM,work2 *GET,work3,SSUM,,ITEM,work3 *GET,work4,SSUM,,ITEM,work4 work=work+work1+work2+work3+work4 SABS,0 !-------------------------------------------------------------------------------! Save the slide value for the next step. !-------------------------------------------------------------------------------ETAB,tasr1tmp,NMISC,17 ETAB,tasr2tmp,NMISC,18 ETAB,tasr3tmp,NMISC,19 ETAB,tasr4tmp,NMISC,20 ETAB,tass1tmp,NMISC,21 ETAB,tass2tmp,NMISC,22 ETAB,tass3tmp,NMISC,23 ETAB,tass4tmp,NMISC,24 *ENDDO *ENDDO FINISH

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REFERENCES 1. Boresi, A.P., Schmidt, R.J. and Sidebottom, O.M., Advanced Mechanics of Materials, fifth Edition, John Wiley & Sons, Inc., NY, 1993. 2. Cobb, W.T., Design of Dampers for Boring Bars and Spindle Extensions, Masters Thesis, University of Florida, 1989. 3. Dean, M.R., Improving Milling Efficiency Through Chatter Suppression and Increased Axis Feedrates, Masters Thesis, University, 1998. 4. Schmitz, T., Predicting High-Speed Machining Dynamics by Substructure Analysis, Annals of the CIRP, 49(1), 2000, pp. 303-308. 5. Slocum, A.H., Precision Machine Design, Prentice Hall, Inc., Englewood Cliffs, NJ, 1992. 6. Smith, K.S., Chatter, Forced Vibrations, and Accuracy In High Speed Milling, Masters Thesis, University of Florida, 1985. 7. Sterling, Robert A., Damping of Long Endmills, Masters Thesis, University of Florida, 1989. 8. Tlusty, J., Manufacturing Processes and Equipment, Prentice Hall, Inc., Upper Saddle River, NJ, 2000. 9. Zhong, Z.H., Finite Element Procedures for Contact-Impact Problems, NY, Oxford University Press, 1993 72

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BIOGRAPHICAL SKETCH Dongki Won was born May 5, 1972. In February 2000, he received a bachelors degree with high honors in mechanical engineering from the Dongguk University in Seoul, Korea. In September 2001, he began graduate study in the Department of Mechanical and Aerospace Engineering at the University of Florida. At the time of writing, he is in his last semester at the University of Florida, after which he will start working on a PhD in the Department of Mechanical and Aerospace Engineering at the University of Florida. 73


Permanent Link: http://ufdc.ufl.edu/UFE0004942/00001

Material Information

Title: Numerical Analysis and Parameter Study of a Mechanical Damper in Machine Tool
Physical Description: Mixed Material
Copyright Date: 2008

Record Information

Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
System ID: UFE0004942:00001

Permanent Link: http://ufdc.ufl.edu/UFE0004942/00001

Material Information

Title: Numerical Analysis and Parameter Study of a Mechanical Damper in Machine Tool
Physical Description: Mixed Material
Copyright Date: 2008

Record Information

Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
System ID: UFE0004942:00001


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NUMERICAL ANALYSIS AND PARAMETER STUDY
OF A MECHANICAL DAMPER IN MACHINE TOOL















By

DONGKI WON


A THESIS PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
MASTER OF SCIENCE

UNIVERSITY OF FLORIDA


2004

































Copyright 2004

by

Dongki Won















ACKNOWLEDGMENTS

The researcher would like to express his gratitude to Dr. Nam Ho Kim for his

inspiration, support and guidance during this research. Thanks also go to other committee

members Dr. John Ziegert and Dr. Raphael Haftka for their support and service on the

author's advisory committee. Special thanks go Mr. Charles Stanislaus and the students

in the design lab for their help in completing tasks associated with the research. Finally,

the author wishes to thank his wife, parents, and other family members. Without their

support and caring this work would not have been possible.
















TABLE OF CONTENTS

page

A C K N O W L E D G M E N T S ................................................................................................. iii

LIST O F TA B LE S .................. .......... .. ............................. ....... ....... vi

L IST O F F IG U R E S .... ...... ................................................ .. .. ..... .............. vii

ABSTRACT ........ .............. ............. ...... ...................... ix

CHAPTER

1 IN TRODU CTION ................................................. ...... .................

2 SIM P L IF IE D M O D E L ...................................................................... .....................5

3 REVIEW OF ANALYTICAL APPROACH......................................9

Calculation of the Normal Force and the Contact Pressure.......................................10
Calculation of the Relative Displacement and the Work................ ..................16

4 BACKGROUND OF FINITE ELEMENTANALYSIS IN CONTACT
P R O B L E M S ...............................................................................................................2 5

Contact Formulation in Static Problem s................................ ....................... 26
The Lagrange M ultiplier M ethod ........................................ .......................... 29
T he P penalty M ethod ............ .............................................................. .......... .... 30

5 FIITE ELEMENT ANALYSIS ...........................................................................32

F inite E lem ent M odel ...................................................... .. .... .. .... ........ 32
B boundary Conditions .................. ...................................... .. ........ .... 35
C alculation of F riction W ork ........................................................... .....................36
Finite Elem ent A analysis R esults..............................................................................37
L o ad Step 1 .............. ...... ............. .. .................................... .. ............... 3 7
Load Step 2 (Centrifugal Force + Vertical Force)................... ..... ............38

6 PA R A M E TE R STU D Y ...................................................................... ..................4 1

D eterm nation of the M esh Size ..................................................................... .. .... 41










D eterm nation of the Start A ngle..................................................................... ......42
P aram eter Stu dy .................................................................................. 46
Change the Inner Radius of the Finger. .......... .........................................................46
Change the N um ber of the Finger .......................................................... ..... .......... 47
Final R results ..................... ......... ... ..... .........................48
Comparison between the Analytical and Numerical Results ...................................50

7 CONCLUSION AND FUTURE WORK ........................................ ............... 53

APPENDIX

A MATLAB CODE FOR THEORITICAL ANALYSIS............................ .................55

B A N SY S IN P U T F IL E 0 1 ................................................................... ....................59

C AN SY S INPU T FILE 02 ......................................................................... 61

D A N SY S IN PU T FILE 03 .................................................... .............................. 63

E AN SY S IN PU T FILE 04 ........................................................................ ............... 65

F A N SY S IN PU T FILE 05 ................................................... ............................... 68

REFERENCES ..................... .... ........... ...... .. ......... ........ 72

B IO G R A PH IC A L SK E T C H ...................................................................... ..................73




























v
















LIST OF TABLES

Table page

2-1. M material properties used for the shank and finger........................................... ........... 8

3-1. First moment Q of various quantities. ................................. ............... 13

5-1. The number of nodes and elements. ........................ ............ ....... 33

6-1. The results of parameter study ............ .................................... ..........................48
















LIST OF FIGURES


Figure page

1-1. Chatter m ark. ...................................................................1

2-1. Endm ill and dam per....................................... .................... .........

2-2. M odel sim plifi cation......................................................... ................................... 6

2-3. D im tensions of geom etry .............................................................................. ............. 7

3-1. Cross sectional area of the endmill system........................................................11

3-2. Cross sectional area of the finger. ........................................ ......................... 14

3-3. Cross sectional area of the finger. ........................................ ......................... 15

3-4. Plot of work done by the friction force according to the change of the inner radius of
fi n g er. ............................................................ ................ 2 2

3-5. Plot of work done by the friction force for varying start angles of the finger............23

3-6. Plot of the work done by the friction force for different numbers of fingers.............23

5-1. Solid 95, 20 node solid elem ent......................................... ............................ 32

5-2. FEA model of the endmill using 20 node cubic elements and 8 node contact elements
w ith boundary conditions. ............................................................. .....................33

5-3. 8-node contact and target element description. .................................. ...............34

5-4. Force boundary conditions in each time step. ................................. .................35

5-5. Diagram explaining how to calculate the damping work.................... ...............36

5-6. FE A result for load step 1............................................... ................................ 37

5-7. FEA result for load step 2 ....................................................................... 39

5-8. Schematic diagram of endmill behavior according to the applied forces...................40

6-1. Damping according to the number of elements ......................................................41









6-2. Damping according to the position of the finger......................................................42

6-3. The maximum and the minimum values of the damping work for the given number
of fingers. .......................................................................................................... ..... 43

6-4. Relative displacement of the two-finger case ......................................................44

6-5. Relative displacement of five-finger case. ............................................................45

6-6. The first design variable. ...... ........................... .........................................46

6-7. The second design variable............................................... .............................. 46

6-8. The result of the parameter study in which the inner radius of finger was changed..47

6-9. The result of the parameter study in which the number of fingers is changed..........48

6-10. The plot of the Table 6-1 ....... ........................... ........................................49
















Abstract of Thesis Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Master of Science

NUMERICAL ANALYSIS AND PARAMETER STUDY
OF A MECHANICAL DAMPER IN MACHINE TOOL

By

Dongki Won

May, 2004

Chair: Nam Ho Kim
Major Department: Mechanical and Aerospace Engineering

When an endmill is used in high-speed machining, chatter vibration of the tool can

cause undesirable results. This vibration increases tool wear and leaves chatter marks on

the cutting surface. To reduce the chatter vibration, a layered-beam damper is inserted

into the hole at the center of the tool. The friction work done by the relative motion

between the tool and damper reduces chatter vibration. The purpose of this research is to

design and optimize the configuration of the damper to obtain the maximum damping

effect.

The analytical method has been reviewed, which is based on the assumption of

constant contact pressure and uniform deflection. For the numerical approach, nonlinear

finite element analysis is employed to calculate the distribution of the contact pressure

under the centrifugal and cutting forces. The analytical and numerical results are

compared and discussed.









In order to identify the effect of the damper's configuration, two design variables

are chosen: the inner radius of the damper and the number of slotted dampers. During the

parameter study and optimization, the inner radius is varied from 1.5mm to 3.5mm and

the number of slotted dampers is varied from 2 to 10.

Results show that the damping effect is maximum when the inner radius is 1.5mm

and the number of slotted dampers is 5. However, this result depends on the operating

condition. Thus, it is suggested to prepare a set of dampers and to apply the appropriate

one for the optimum damping effect for a given operating condition.














CHAPTER 1
INTRODUCTION

Milling is widely used in many areas of manufacturing. Traditionally, milling has

been regarded as a slow and costly process. Therefore, many efforts have been made to

improve the efficiency of milling. The main limitation of milling is caused by the

vibration of the machine tool and workpiece. As the speed and the power of milling are

increased, it is very important to control vibration of the tool.

Two different kinds of vibration affect the cutting operation. The one is the self-

excited (chatter) vibration at the high spindle speed, and the other is vibration at the

critical natural frequency. This research is focused on the former, which produces a wavy

surface during the milling operation, as shown in Fig. 1-1.














Figure 1-1. Chatter mark.

Various methods of preventing chatter have been incorporated into machine tool

systems. In 1989, Cobb [2] developed dampers for boring bars. These dampers are

composed of two different types. The first type, which Cobb calls a shear damper, has

two end "caps" that fit snugly around a boring bar. Between these are sandwiched an









annular mass with plastic at each end, either in the form of ring or of several blocks at

each end. This middle section of a mass and plastic pieces has a clearance from the

boring bar, and is preloaded between the end caps with bolts. When the boring bar

vibrates, the end caps transmit this vibration through the plastic pieces to the annular

mass, which vibrates in tune. Since the plastic pieces do not slide on the faces of either

one, a shear force is produced on the end faces of the plastic. The viscoelastic properties

of the plastic provide damping for the system. The other type of damper proposed by

Cobb is the compression damper in which the mass is again an annulus. This annulus is

cut in half down its axis to form two half annuli. These are then bolted together around

rings of plastic that are in contact with the boring bar. The bolts provide a preload on the

plastic rings, and when the bar vibrates, the annular mass vibrates out of phase with it,

compressing one side and then the other of the plastic rings. This alternate squeezing of

the plastic creates damping, again by deforming the plastic, but in a compressive rather

than shearing manner.

In 1998, Dean [3] focused on increasing the depth of cut and increasing axis

federates to improve the metal removal rate (MRR). While he does not present any

original ideas on chatter reduction, his thesis refers to work done by Smith [6] in which a

chatter recognition system was developed. This system uses a microphone to detect the

frequency of chatter when it occurs. The system then selects a different speed(according

to the parameters of the system) and tries to machine again. This process repeats itself

until chatter no longer occurs.

Much work in the field of structural damping has been done by Slocum [5]. In

order to damp vibrations, Slocum uses layered beams with viscoelastic materials between









the layers. Two cantilevered beams are stacked on top of each other, and a force is

applied to the end of the top beam. It is known that beams experience an axial shear force

when displaced in this manner. Slocum's derivation calculates a relative displacement

between corresponding points on the un-deformed beams. This is incorporated into a

selfdamping structure by placing several small beams inside of a larger beam and

injecting a viscoelastic material between them. This material bonds to each surface and

thus, when there is a relative displacement, is deformed. The stretching of this material

causes a dissipation of vibration energy and thus, damping.

In 2001, Sterling [7] explored the possibility of deploying a damper directly inside

of a rotating tool. To reduce the chatter vibration, a layered-beam damper, which Sterling

calls a finger, is inserted into the hole at the center of the tool. Due to the high-speed

rotation, the outer surface of the damper contacts with the inner surface of the tool. When

chatter vibration occurs, which is a deflection of the tool, work is done in the contact

interface due to the friction force and the relative motion. This work is dissipative and

reduces chatter vibration. He developed an analytical model and performed an

experiment for the layered beam damper.

In this research, the work done by Sterling is further extended. Using finite element

analysis, his analytical approach is compared to the numerical results. The objective of

this research is to calculate the amount of friction work and to maximize its effect by

changing the damper's configuration.

The organization of thesis is as follows. In Chapter 2, a simplified model of the

endmill is introduced that can be used for analytical study and numerical simulation. In

Chapter 3, the analytical approach is reviewed that qualitatively estimates the damping






4


work. Chapter 4 presents the background knowledge of finite element analysis in contact

problems. Chapter 5 describes the numerical simulation procedure using the finite

element method. Chapter 6 represents the parameter study according to the change of two

design variables, followed by conclusions and future work at Chapter 7.














CHAPTER 2
SIMPLIFIED MODEL

The machine tool that we are considering in this research is a 4" long endmill, as

shown in Fig.2-1. Most endmills are of the solid beam type as shown in Fig.2-1 (a). In

this type of tool, the only available damping mechanism is structural damping, which is

very small. Structural damping, which is a variant of viscous damping, is usually caused

by internal material friction. When the damping coefficient is small, as in the case of

structures, damping is primarily effective at frequencies close to the resonant frequency

of the structure.


(a) (b)













Figure 2-1. Endmill and damper. (a) the original solid endmill and (b) the damper
inserted model

When a layered-beam damper (see Fig.2-1 (b)) is inserted into the hollow tool, the

high-speed rotation causes a strong contact between the beam and tool. When chatter

vibration occurs, it generates a relative motion between the beam and tool. Due to the

contact force, this relative motion causes a friction force in the interface, which damps









the vibration. In this research, this damping mechanism will be referred to as a

mechanical damper.

While the tool geometry is very important for cutting performance, the objective of

this research is on the vibration of the tool. Thus, we want to simply the tool geometry

so that the analytical and numerical studies in the following chapters will be convenient.

The first step of simplifying the endmill model is to suppress unnecessary

geometric details, while maintaining the endmill's mechanical properties. The endmill

model can be simplified as a cylinder because we are only interested in the contact

surface, which is the inner surface of the tool. Fig.2-2 (b) illustrates the simplified model.


(a)(b) co=2722.71 rad/s





F=100 N








Figure 2-2. Model simplification. (a) the detailed tool model in which the damper is
inserted and (b) the simplified model using hollowed cylinder.

The simplified endmill model is composed of two-hollowed cylinders. The outer

cylinder represents the endmill tool, and the inner cylinder represents the damper. For

convenience, the outer part (tool) is denoted as a shank, while the inner part (damper) is

denoted as a finger. As schematically illustrated in Fig.2-3, the outer radius R1 of the

finger is 1.5 mm, the inner radius R2 of the finger is 4.7625 mm, the inner radius R3 of

the shank is 4.7625 mm, and the outer radius R4 of the shank is 9.525 mm. The length of









the endmill is 101.6mm. Because the gap between two contact surfaces is ignored, R2 is

equal to R3. R1 will be varied from 1.0 mm to 3.5 mm during the parameter study in

Chapter 6.


Figure 2-3. Dimensions of geometry.

Although Fig.2-3 shows only two fingers, the number of fingers can be altered to

improve the damping performance. The parameter study in Chapter 6 will examine the

effect of varying the number of fingers between 2 and 10. Because a damper with only

one finger would have a lower contact pressure than the other cases, this case will not be

considered.

Next, the operating condition is also simplified. The applied force is assumed to be

sequential. It is first assumed that the tool is rotated with a constant angular velocity. The

constant angular velocity will generate a constant contact force at the interface. For this


^^4-9,$2SOM4
7 '62S MM


R2V4, ,625MM
R1_1 ,5000 -\u
k%









particular model, an angular velocity of 2,722.713rad/sec is used, which is equal to

26,000 rpm. In this initial state the endmill has not started cutting the surface. When the

endmill starts cutting the surface, the tool undergoes a vertical force at the end of the

endmill. To approximate the cutting process of the tool, a vertical force is applied at the

tip. To accurately approximate the cutting force, the vertical force on the endmill needs to

be measured and then an equal force needs be applied at the tip. However, since the

objective of this research is vibration control, a representative force of 100N is applied.

Thus, the damping work that will be calculated is not the actual magnitude, but rather a

relative quantity.

For simplicity, the same material properties are assumed for both the shank and

finger even though the stiffness of the finger is actually slightly higher than that of the

shank. The material properties used are listed in Table 2-1.

Table 2-1. Material properties used for the shank and finger.
Material Property Value
Young's Modulus 206780 MPa
Mass Density 7.82x10-9 ton/mm3
Friction Coefficient 0.15

In the following chapters analytical and finite element analysis will use the

simplified model to determine the conditions for maximum damping of the endmill.














CHAPTER 3
REVIEW OF ANALYTICAL APPROACH

It would be beneficial to review an analytical model before starting the finite

element analysis because it will provide a qualitative estimation of the numerical

approach. Sterling [7], a former researcher, developed an analytical method that can

estimate the amount of friction work during chatter vibration. In this chapter, his

analytical approach is reviewed and the results will be compared with finite element

analysis results in Chapter 5.

The work done by the friction force that occurs between the inner surface of the

endmill and the outer surface of the damper causes the damping effect that reduces the

chatter vibration. According to the Coulomb friction model [9], the friction force and the

damping work can be written as follows:

Ff = u x N
(3.1)
Wf=F xUf

where Ff is the friction force, /u is the friction coefficient, N is the normal contact force,

Wf is the friction work, and Uf is the relative displacement between the two contact

surfaces. The normal force N is mainly caused by the centrifugal force created when the

endmill is rotating. The relative displacement Uf is mainly caused by the vertical

deflection of the tool when the endmill starts cutting. Therefore, we can divide the

endmill system into two states. The first state is when the endmill is rotating without any

cutting operation. In this case, only the centrifugal force is applied. The second state is









when the endmill starts cutting. In this state, the vertical force is added at the tip of the

endmill. Both the centrifugal force and the vertical force are applied in this state. If we

assume that there is no relative motion in the first state, then we can calculate the work

done by the friction force during the second state.

Calculation of the Normal Force and the Contact Pressure

In this section, the normal force and pressure that are caused by the rotational

motion of the tool will be calculated. There are three assumptions for the analytical

method in this step. Those are listed as below.

* There is no angular acceleration, which means the angular velocity is constant.

* There is no relative motion between the two contact surfaces during the first step,
which means there is no slip in the contact surface during the rotational motion.

* Contact occurs throughout the entire contact area during the second state. In the
actual case, contact may not occur in some portions of the interface. For example
there will be no contact near the fixed end or on the two sides where the neutral
axis lies. All of these effects are ignored, and it is assumed that contact occurs
throughout the entire area.

Due to the second assumption, the contact pressure is calculated using the

centrifugal force only and is assumed to remain constant. Now let us consider the

simplified model, which was developed in Chapter 2 (Fig.2-3). The shank and the finger

are hollowed cylinders. For simplicity, we only consider the case of a two-finger

configuration. Figure 3-1 (a) shows the cross-sectional area and dimensions of the

endmill system in which two fingers are inserted. Considering the symmetric geometry of

the fingers, we can consider one finger, which is illustrated in Fig.3-1 (b). Since the

finger can have an arbitrary location, 0 represents the start angle of the finger. The point

G indicates the first moment centroidd) of the finger's cross section, and R is the distance









between the center of the tool and the centroid G. The normal force N and contact

pressure Pc can then be obtained as,


N = MRo 2,


P MR) 2
P-


(3.2)


where Ac is the contact surface area, M is the mass of the finger, andw is the angular

velocity. The first step of calculating the contact pressure is to calculate the distance R,

which is determined by the centroid of the finger's cross section.

(a) (b)





06 = start angle


-R1= 1 mm


R2= R3= 4.7625 mm
- R4= 9.525 mm


Figure 3-1. Cross sectional area of the endmill system. (a) Cross sectional area and
dimension of the original model in which two fingers are inserted. (b) Cross
sectional area of the finger. G is the mass center of the cross section, and 0 is
the start angle.

The centroid (first moment) of an assemblage of n similar quantities, A,, A2, A3,

..., A situated at point P], P2, P,,..., Pn for which the position vectors relative to a


selected point O are rl, r2, r3, ..., rn has a point vector r defined as







12



A=1
r=
n
A,


where A, is the i th quantity (for example, this could be the length, area, volume, or mass

n
of an element), r is the position vector of i th element, i A, is the sum of all n elements,
1=1

n
and Vr A, is the first moment of all elements relative to the selected point 0. In terms of
1=1

x, y, and z coordinates, the centroid has coordinates

n n n

X= y= Z= -
n n n

1=i 1=i 1=1


where A, is the magnitude of the ith quantity (element), x, y, z are the coordinates of


centroid of the assemblage, and x, y,, z are the coordinates of P, at which A, is


concentrated.

The centroid of a continuous quantity may be located though calculus by using

infinitesimal elements of the quantity. Thus, for area A and in terms of x, y, z

coordinates, we can write


xdA Q

SdA A

ydA Q

\ dA A

fzdA Q

SdA A









where Qy, Qy, Q. are first moments with respect to the xy, yz, and xz planes,

respectively. The following table indicates the first moments Q of various quantities A

about the coordinate planes. In Table 3-1 Qx, Qyz, Qx are the first moments with respect to

xy, yz, xz planes, L is the length, and m is the mass, respectively. Note that in two-dimensional

work, e.q. in the xy plane, Qx becomes Q, and Q becomes Q .

Table 3-1. First moment Q of various quantities.
A Qx Qyz Qxz Dimensions
Line fzdL fxdL fydL
Area fzdA fxdA fydA
Volume fzdV xdV fydV L
Mass zdm xdm ydm mL

Now let us consider the case illustrated in Fig.3-2, which is a cross section of the

finger. According to the figure, y can be expressed as


SydA Q
y=R=
fdA A

If we choose the polar coordinate system y is represented by,

y = r cos

Using this polar coordinate system, the y of the centroid can be calculated by,


x fydA 2I Ca s(0)rdr
y A \dA 2 drdrdO

1 R2
r3 cos(O)drdO
3 f g _
S R
yr2 2 drdO
2J I -a
L^JIr










- R3) sin(/7) + sin(a)
-R2) 8-a
- R ) sin(/7) + sin(a)
-R2) f-a


Figure 3-2. Cross sectional area of the finger.


If a= 0,


2 (R23- R13) sin(f)
3(R2 -R,2) f


Since the centroid of the cross section is always located along the symmetric line of

the finger, it is convenient if the y-axis is chosen such that the centroid is located on the

y-axis, as illustrated in Fig.3-3. Due to the symmetry, the integration of the domain can

be done between -a and a for angles, which provides a convenient formula. If we

choose the polar coordinate system, from the Fig.3-3, y is represented by,

y = r cos


2(R3
3(p R
2(R:3
3(P,






15


Y








Ox



X



Figure 3-3. Cross sectional area of the finger.

Using this polar coordinate system, the y of the centroid can be calculated by,


y RfydA Jc J2 TrCOs(O)rdrdO
L fdA RR rdrd
S J r cos(O)rddO
1 0 R1



2 r2 3 R, cos()drdo
3 JR,


2 JR 1

2(R23 -R,3) sin(a) sin(-ca)
3( -R,2) (-a)

2(R23 -R,3) sin(a)



.R = 2(R3 -R,3) sin(a) (3.3)
3(R2 2 ) a

This equation can be used for the general case which undergoes the centrifugal


force. For the simplified model in Chapter 2, a = R2 = 4.7625mm, R, = 1.5mm .
2









Substituting these values into the above equation yields y = R = 2.1738mm. Substituting

this value into Eq.(3.2) yields the normal force and the contact pressure between two

contact surfaces. In Eq.(3.2), the contact area Ac can be calculated by

Ac = 7rR2L = x 4.7625 x101.6 = 1520 [mm2]

which assumes that all parts of the surface are in contact. Using the material properties

shown in Table 2-1, the mass M can be calculated by.


M = pV= 7.82x10-6 (4.76252-1.52)x101.6

= 2.5499x10 2 [kg]

Therefore the contact pressure can be obtained from Eq.(3.2), as

MRo)2 2.5499 x102 x2.1738 x 2722.7132
S A 1520

=270.33 [KPa]= 0.2703 [MPa]

It is noted that the contact pressure Pc is calculated from the assumption that the whole

surface is in contact with a constant pressure.

Calculation of the Relative Displacement and the Work

In this section the relative displacement that is caused by the vertical force (cutting

force) will be calculated and the work that has been done by the friction force will be

calculated. This state represents the one in which the endmill starts the cutting operation.

The same assumptions are used as in the last case except for the second one because there

is now a relative motion between the two contact surfaces. The assumptions are:

* There is no angular acceleration, which means the angular velocity is constant.

* There is no normal contact force change caused by the vertical force.

* Contact occurs throughout the entire contact area.









Because the centrifugal force is dominant in this endmill system, the change of the

normal contact force due to the vertical force is ignored in this step. The first step of

calculating the frictional work is to obtain the second moment of inertia of the finger. The

following definitions of the second moment, or moment of inertia are analogous to the

definitions of the first moment of a plane area, which were given in the previous section.

The derivation of load-stress formulas for beams may require solutions of one or more of

the following equations:

SI = y 2dA
I = x2dA (3.4)

I = xydA

where dA is an element of the plane area A lying in the x-y plane. A represents the

cross-sectional area of a member subjected to bending and/or torsional loads. The

integrals in the above equations are commonly called moments of inertia of the area A

because of the similarity with integrals that define the moment of inertia of bodies in the

field of dynamics. From Eq.(3.4), if we choose the polar coordinate system, x and y are

represented as,

x= r cosO, y = r sinO

andlx, ly, I are


,I = 2 r3 sin2 OdrdO = (R 4 R1) ( -- sin / cos/ + sin a cos a)


f I f = 23 cos2 OdrdO = (4 R R,4 )( -a+sin Scos8 -sinacosa) (3.5)

I R2 r3 sin0 cosOdrdO= b(R -R4 sin2 sin2 a
x Ja Jj1


If a = 0 the Eq.(3.5) can be written as









x = f 2 r3 sin2 0drdO = -(R4 R,(a sina cosa)

I= R2 r3 cos2 OdrdO=(R4 -R,4)(a+sin acosa) (3.6)

I = f 2r3 sin O cosOdrd = (R4 R4 )sin2


For the simplified model in Chapter 2, a = ;r, R = 1.5mm, R2 = 4.7625mm.

Substituting those values into the above equation yields,

I =(4.76254 -1.54 (-sin cosz) = (4.76254-1.54

Thus, the moment of inertia of the upper finger is


I= (R34 R44) (4.76254 1.54) 200 [mm4]

The moment of inertia of the lower finger can be obtained in the same way because the

two fingers are symmetric about the x-axis, and the two values would be the same.

Now let us calculate the relative displacement and the friction work. It is well

known that beams undergo internal shear deformations along their axes during bending.

Members of a composite beam that are not securely fixed together will slide over each

other in proportion to their distance from the neutral axis of entire composite beam. It is

known that for a cantilevered beam with a point load at the end, the vertical deflection at

any point in the beam's neutral surface is

6E(-(-x3 +3xL2-2L3)
6El

where F is the force on the end of the beam, E is the beam's elastic modulus, I is the

beam's moment of inertia, x is the position along the length of the beam measured from

the free end, and L is the length of the beam.









If a composite beam is bent, all members will have the same deflection at the tip.

Therefore it can be written that

-F Ff 2L3)
6= (-x3 +3xL2 -23 (-x3 +3xL2-2 (3.7)
6E I, 6E fi

where, subscript s means the shank, andfmeans the finger. The external force F at the tip

of the endmill must equal the sum of the forces required to deflect each member.

F = F + Ff (3.8)

If there are only two fingers inside the shank Eq.(3.7) can be written as

F F,, Ff2
EsJI EfIJfl Ef21f2

The above equation can be written as follows

FEIJ F 1Ef2 f2
F = f and Ff = f
Efhfl f2 EfJI1

Substituting these equations into (3.8) yields

F,,EI FF Ef 2 f2
F= 1+F,, +F, f s +Ff, + 1 22
fl f2 E p l l E p I
Efll Ifl fl

Solving above equation for Ff, yields

FEfIf
EI +EflI +Ef2f2

This equation can be reduced to

FEfjlf
Fs=
E=1
flfn
EJI, + EfIf


The same operation for F] and Ff2 yields











n



FEf2f2
F1==


Ff2 n
EjI + E,If


The normal stress at any point in a cantilevered beam is given by

Fxc
I

where a is the stress, F is the force on the free end, x is the position along the beam

measured from the free end, c is the distance from the beam's neutral axis to the point of

interest, and I is the area moment of inertia of its cross section. The axial strain in the

beam is then

a Fxc
E El

where E is the axial strain and Eis the Young's Modulus of the material. c is the sum of

the quantity (d+y) where d is the distance of the neutral axis of the finger from the neutral

axis of the composite beam andy is the perpendicular distance from the point in question

to the neutral axis of the finger.

For any point in any component member of the composite beam, the change in

position of the point related to x=O (the free end of the beam) can be written as





Substituting the previously obtained equations into this integral for both the fingers and

the shank, the following equations are obtained. The value of d of the finger can be









represented by Rx sin 0, where 0 is the start angle. Note that in the equation for the

shank, dis zero because its neutral axis lies on the neutral axis of the composite beam.

L Fxxx (d+y) FE,I (0 + y) 2 Fxyx(L 2)
F d( = (+)L( x L ) =
i- EI EI, +I E 2EI 2K E(J, + E fi2



=L Fxxx(d+y) FEflfi (d+y) ) F x(d +y)x(L -2)
facal I x E 2E I "E
El E,+Ef f 2E ,2 EI +jE EI.f



The relative displacement can then be obtained by subtracting the above two equations

fFx(d+y)x(L2 -x2) Fxyx(L -x2) Fxdx(L -x2)
2KEIZ + EfEIf 2 EI, + EfIf, 2 EI, + Efi)
1-=1 1- 1=1

The work done through this displacement is by friction. The amount of work done

is equal to the integral solved over the entire length of the beam of the frictional force

(friction coefficient times the normal force, which is the force/unit length at a point times

the differential length) multiplied by the relative displacement. Writing this equation

(assuming the pressure, P, is uniform over the entire area of the finger)

LP( Fxdx(L2 -_X2) d
pPp(3 9.") = P2P + idx
2 AEI +Z E 2I
1=1

The work done by friction for a displacement of the end of the beam by a specified

force, F, is then

1 d
W = -L3pPF (3.9)
3 KESI + EfIf,
1=1











The MATLAB code used for calculating this friction work is implemented in

appendix A. Figure 3-4, 3-5 and 3-6 show the results of theoretical analysis.


x 10-5
7.4 ---------------------------- --------

7.2 -



6.8 -

6.6 --



6 \
6.4 -- -----------

6.2---------- ---------- ---------- ---------r-\-------

6-- --------------------------- ----------- -\ -
\\ I\

5.8---

5.6 --

5.4
1 1.5 2 2.5 3 3.5
Inner Radius of Damper Finger (m) x 10-3

Figure 3-4. Plot of work done by the friction force according to the change of the inner
radius of finger.

Figure 3-4 is the work according to the change of the inner radius of the finger. As

the radius increases the work decreases because the mass decreases. When the radius is

1.5mm and the number of finger is 2, the work done by the friction force


is7.1182x 10 [J]. Figure 3-5 is the work calculated for different values of the start angle


when the number of damper fingers is two and the inner radius is 1.5 mm. When the start

angle is 0 it has its maximum value and this value gradually decreases as the angles

decrease. Even when the number of fingers is held constant, the damping values differ for

different positions of the finger. This result is reasonable because the relative

displacement is zero at the neutral axis. In Chapter 6 this effect will be discussed in more

detail.



































0 20 40 60 80 100
Start Angle of Damper Finger (degree)


120 140


Figure 3-5. Plot of work done by the friction force for varying start angles of the finger.
The radius of the finger is 1.5 mm and the number of the finger is 2.


x 10-4






0.16 ------ I ----L ---- L__ _
0.8 -







0.4-----



0.2


1 2 3 4 5 6
Number of Damper Finger


7 8 9 10


Figure 3-6. Plot of the work done by the friction force for different numbers of damper
fingers. The inner radius of the finger is 1.5 mm and the start angle is chosen
to have the maximum damping in each number of the finger.






24


Figure 3-6 shows the work for different numbers of fingers. The inner radius of the

finger is 1.5 mm and the start angle is chosen to have the maximum damping for the

number of fingers being used. The work increases gradually as the number of fingers is

increased. Finite element analysis will be done for these analytical results in the

following chapter.














CHAPTER 4
BACKGROUND OF FINITE ELEMENTANALYSIS
IN CONTACT PROBLEMS

Since the material properties of the endmill is linear elastic, the classical finite

element can be used without difficulty. Major concern is the contact constraints in the

interface. Contact problems are highly nonlinear and require significant computer

resources to solve for. Contact problems present two significant difficulties. First, the

regions of contact are generally unknown until running the model. Depending on the

loads, material, boundary conditions, and other factors, surfaces can come into and go out

of contact with each other in a largely unpredictable and abrupt manner. Second, most

contact problems need to account for friction. Frictional response can be chaotic, making

solution convergence difficult. In addition to these two difficulties, many contact

problems must also address multi-field effects, such as the conductance of heat and

electrical currents in the areas of contact. In this chapter the general procedure of

performing the contact analysis using FEA is discussed.

Contact problems are characterized by contact constraints which must be imposed

on contacting boundaries. To impose the contact constraints, two basic methods are

available: the Lagrange multiplier method [9] and the penalty method [9]. Other

constraint methods based on the basic methods have been proposed and applied. The

augmented Lagrangian method [9] and the perturbed Lagrangian method [9] are two

examples.









Contact Formulation in Static Problems

To introduce the basic constraint methods, we consider static mechanical problems

subjected to contact constrains on known contacting boundaries. For small displacements,

the total potential energy of the structure can be written as,


H(u) = eCedQ u bdfJ u-qdS-U F (4.1)


where H is the total potential energy of the body, e is the engineering strain, C is the

elastic modulus, b is the body force, q is the surface traction. By using standard element

procedures, the discretized form of Eq.(4.1) can be obtained as follows:


H(U) = UKU_ U'F (4.2)
2

where Uis the global displacement vector, K is the global stiffness matrix, and F is the

global load vector.

The virtual work due to the contact load is calculated as

tL
wi= Z(swY)" (4.3)
n=1

where ( )" indicates that the quantity in the parentheses is evaluated in association with a


hitting node n, 'L is the total number of hitting nodes at time t, and 8we the virtual work

due to the concentrated contact force at hitting node n and is calculated as

'w5 =t f, (Su2 -_ u'). N, (4.4)

where '/ is the components of the contact force at a target point in the directions of tN.

The virtual displacement (u2 at the target point can be evaluated by using equation

N
(u2 = (fu2 n (4.5)
n=\









where N denotes the total number of target nodes on the target segment, denotes the

shape function associated with contact node n on the contact segment and 3u2'" is the

displacement of target node n.

To obtain a matrix expression for 'u2 8u1, the following notation is used:

U 1 1 1 uI u 2,1 ,1 ,1 2,N 2,N ,N T
1 2 3 1 2 3 1 2 3

-10 0 0 0 0 0 0 0 N 0 0
Q = 0 -1 0 0 o 0 0 o 0 ... 0 O 0
0 0-1 0 0 0 0 0 0 N

Then Eq.(4.4) can be written as

C= <(QC "//)=
3w, = 3u(Q =ur

where


S I={ 'N2 'N,3
r QT ttf


where tNj denotes the j th component of the boundary unit vector tN,.

We realize that tre is a nodal force vector contributed by the contact at the

associated contacting node. It will be convenient to distinguish, in the evaluation of trc,

between the contribution of normal contact forces and the contribution of tangential

friction forces. Thus we write

t = tr + tr

where


r Q f summation on =2,3
trcf Q j &f summation on J= 2,3









The penetration of the hitting node can be calculated as

tp= (tx2- txl)t N


where tx' and tx2 are the position vectors of the hitting node and target point,

respectively.

Expressing tx' and tx2 in terms of the displacements u' and u2, respectively, we

have the following discrete form of the kinematic contact condition:

tp= (x2 -t X)t N, +(u2 U t
(4.6)
=Nt TQcu, p 0

where

p= trxZ--
p=' I X
1 _[ 1 2 [2 }I


This means that for small displacements the configuration of the contact system may be

considered unchanged after the displacements. Thus, we can approximate t N, by 0N, in

Eq.(4.6). Furthermore, we assume only one load step. This means that we need to move

only one step in the "time" domain and Tp and tp in Eq.(4.6) can be replaced by p and

p, respectively, where p denotes any initial penetration (or gap) and p any penetration

after deformation. Therefore, the discretized kinematic contact condition can now be

written as

p=o TQcu+ p= 0 (4.7)

Eq.(4.7) applies to a single contacting node. If there are L(L > 1) contacting nodes,

there will be L contact constraint equations, each taking the form of Eq.(4.7). Those

L contact constraint equations can be assembled to obtain









P =QU + P =0 (4.8)

where

p= {pl, p..., PL T
L
I=1
0 {0 1 0 2 ... 0 L T


Since the contacting boundaries are known, all the contacting nodes can be identified and

Eq.(4.8) can be established explicitly.

Based on the above preparations, the contact problem can be stated as follows:


Minimize n(U) in Eq.(4.2) subject to the constraints in Eq.(4.8) [P-1]


It must be observed that the mechanical contact condition should be satisfied

automatically by assuming that all the L contact nodes are actual contacting nodes. In

general, actual contacting nodes are not known a priori and an iterative trial-and-error

procedure is required to find all contacting nodes. In the following, we discuss the

solution of problem [P-1] with alternative constraint methods.

The Lagrange Multiplier Method

In the Lagrange multiplier method, the function to be minimized is replaced by the

following function:


,L(U, A) = UTKU UTF + AT (QU +0 ) (4.9)
2

where A is an unknown vector which contains as many elements as there are constraint

equations in Eq.(4.8). The elements in A are known as Lagrange multipliers.









The constrained minimization problem [P-1] is now transformed into the following

saddle-point problem:

Find U and A such that H,(U, A) is stationary, i.e. [P-2]


=-L 0
U (4.10)
HL =0
BA

Eq.(4.10) yields

KU-F+QHA= (4.11)
QU+0P=0

Combining Eq.(4.11), we obtain

KLUL = FL (4.12)

where


KF =_
KL- K QT
Q 0

FL {


UL {


By solving Eq.(4.12), we can obtain the displacement U and the Lagrange multiplier A .

The elements in A are interpreted as contacting forces at the corresponding contacting

nodes.

The Penalty Method

In the penalty method, the potential energy of the structure is penalized when a

penetration occurs on the contact surface. The following penalty potential is added to the

structural potential:









1
;T = poaP
P2

where a is a diagonal matrix with elements a, which is the penalty parameters and P is

a vector of penetration..

The function to be minimized is now replaced by

in = H + ~z
1 U U1 F +(4.13)
= U'KU-U F+-P2P
2 2

The constrained minimization problem [P-1] is then transformed into the following

unconstrained minimization problem:


Find U such that Hn is minimized.


To find its minimum, n, is held stationary by invoking the following condition:

an
I-= 0 (4.14)
dU

Substituting Eq.(4.13) and (4.8) into (4.14), we can obtain

KPU = F (4.15)

where

Kp =K+Q aQ
F, = F QTa0p

The solution ofEq.(4.15) gives the displacement U. The contacting forces are then

calculated as

F = aP


where the penetration P is a function of the displacement vector U.














CHAPTER 5
FIITE ELEMENT ANALYSIS

In this chapter, the finite element analysis procedure and results of the endmill

system are presented. Even though the cutting process is dynamic, a static finite element

analysis is performed with the centrifugal force and the cutting force at the tip. Thus, the

friction work obtained must be interpreted as a qualitative measure. Since the simulation

condition is the same as that of the analytical method in Chapter 3, it is still valid to

compare the results of finite element analysis with the analytical results.

Finite Element Model


Figure 5-1. Solid 95, 20 node solid element

The first step of finite element analysis is to build a computational model. The

simplified geometry for the endmill in Chapter 2 is used in the FEA. Using the same










material properties listed in Table 2-1, 20-node solid elements in ANSYS (solid 95) are

used to build the shank and finger. Figure 5-1 illustrates a 20-node solid element that is

used in ANSYS, and Fig.5-2 plots the finite element model of the endmill with boundary

conditions. In Fig.5-2, the shank is modeled using two elements through the radial

direction, and fingers are modeled inside the shank. The cutting force F=100N is

distributed to 4 nodes at the tip. Table5-1 shows the number of elements used in the

endmill finite element model.

AN
ELEMEIDTS
JUN 14 2003
19:47:56
F PLOT NO. 1
CMEG
























Figure 5-2. FEA model of the endmill using 20 node cubic elements and 8 node contact
elements with boundary conditions.

Table 5-1. The number of nodes and elements.
Node 24826
Element 6480
SOLID95 4320
TARGE170 1080
CONTA174 1080












Because the contact area is curved, higher order elements must be used to prevent

inaccurate representation of the surface. Since the contact element is defined on the

surface of solid elements, the consistent order of the element must be used for the

structure and contact surface. The counter part of SOLID95 structural element is 8-node

contact element, as illustrated in Fig.5-3. The contact elements are defined between the

shank and finger.

Targe Segmen EleriMt


Associated Target Surfaces L K
P N SN ode Quadrilateral
TSHAPQUAL
Contact Elements


I' V J
C-

I ,
'*'.. *lT
....... "" /
Surface of Solid/Shell Element aace-tAo-Srf
/ OCatack Element
GONTA173 or CONTA174

Figure 5-3. 8-node contact (CONTA174) and target (TARGE170) element description.

ANSYS offers CONTA174 and TARGE170 for contact and target elements,

respectively. CONTA174 is used to represent the contact and sliding between 3-D

"target" surfaces and a deformable surface, defined by this element. The element is

applicable to 3-D structural and coupled thermal-structural contact analysis. This element

is located on the surfaces of 3-D solid or shell elements with midside nodes. It has the

same geometric characteristics as the solid or shell element face with which it is

connected. Contact occurs when the element surface penetrates one of the target segment

elements on a specified target surface. Coulomb and shear stress friction is allowed.









TARGE170 is used to represent various 3-D target surfaces for the associated

contact elements. The contact elements themselves overlay the solid elements describing

the boundary of a deformable body and are potentially in contact with the target surface,

defined by TARGE170. This target surface is discretized by a set of target segment

elements (TARGE170) and is paired with its associated contact surface via a shared real

constant set. Any translational or rotational displacement, temperature, and voltage on the

target segment element can be imposed. Forces and moments on target elements can also

be imposed.

Boundary Conditions

According to the forces applied to the endmill, we can divide the analysis

procedure into two steps. The first step is when the endmill starts rotating. In this step,

only the angular velocity of 2,722.713 [rad sec] is applied without considering the

cutting force.


Force

Vertical Force
=100 [N]



Rotational Velocity
=2722.713 [rad/s]





Time Step 1 Time Step 2 Time (sec)


Figure 5-4. Force boundary conditions in each time step. Each time step is divided into 5
steps.









The second step is when the endmill starts the cutting process. At this time a

vertical force of 100 N is applied at the tip of the endmill. Figure 5-4 illustrates the force

boundary conditions in each step. Each time step is divided into 5 substeps in order to

improve the convergence of nonlinear analysis.

Calculation of Friction Work

Even if the structure is linear elastic, the contact constraints make the problem

nonlinear. In ANSYS, a Newton-Raphson iterative method is employed to solve the

nonlinear system of equations. All default parameters in ANSYS are used in nonlinear

analysis.







-0
'Contact Surface


Slide
Friction Stress
(Relative Displacement)


Damping = Dot Product of Friction Force and Slide (Relative Displacement)



Figure 5-5. Diagram explaining how to calculate the damping work.

After finishing FEA, the damping work which is done by the friction force is

calculated. The input commands to ANSYS for obtaining the frictional work are listed in

Appendix D. Figure 5-5 shows a schematic procedure of the calculation. First, the friction

stress and the relative slide in each contact element are obtained for each load step. By

assuming a constant stress within an element, the friction force can be obtained by











multiplying the friction stress with the element area. The dot product of the friction force


vector and the relative displacement vector in each substep and in each element yields the


friction work.


Finite Element Analysis Results

Load Step 1

Figure 5-6 is the results at load step 1 when only the angular velocity is applied. In


Chapter 3, the analytical method estimates the contact pressure to be 0.27[MPa].


NODAL SOLUTION
STEP=1
SUB =5
TIME=1
SEQV (AVG)
DMX =.0004
SMN =.0532
SMX =6.1193


AN

.0532
.7272f
1.4012
2.07521
2.7493
3.4233
4.0973
4.7713
5.4453
6.1193


Figure 5-6. FEA result for load step 1 when only the angular velocity is applied. (a) is the
von misses stress (MPa), (b) is the contact pressure (MPa), (c) is the frictional
stress (MPa), and (d) is the slide (mm). MX indicates where the maximum
value and MN means the minimum value.


NODAL SOLUTION (b)
STEP=1 0
SUB =5 .0646
TIME=1
CONTPRES (AVG) .1293
RSYS=O .1939
DHX =.0003 .2586
S-< =.5819 .3232
.3879
.4525
.5171
.5818
S / /.s









Whereas, the maximum contact pressure from FEA is 0.58 [MPa], as shown in

Fig.5-6 (b). Since the actual contact occurs only half of the contact surface, the higher

contact pressure from FEA is expected. In addition, the contact pressure is not constant:

maximum at the top and the bottom surface and zero at both sides. Even though we

assume that there is no relative motion between the contact surfaces during load step 1 in

Chapter 3, there is a relative motion due to the diameter change. However, the relative

motion in load step 1 should not be counted because it is not related to the chatter

vibration.

Load Step 2 (Centrifugal Force + Vertical Force)

In the load step 2, a vertical force (cutting force) is applied on top of the centrifugal

force. Figure 5-7 shows the results of load step 2. The frictional work is calculated by dot

producing the friction force (Fig.5-7 (c)) and the relative slide (Fig.5-7 (d)). Since the

load step is divided into five sub steps, the friction work at each sub step must be

summed. The total friction work during load step 2 is 3.3426 x10 5[J]. By comparing

with the analytical results 7.1182 x10 5[J] in Chapter 3, the finite element analysis

estimates about 50% of the analytical result. The friction work calculated from FEA is

less than that from the analytical approach because the actual contact area is small in

FEA. There is no contact on two sides: the neutral axis lies and the fixed end. In addition,

the contact pressure is not constant. That is why the analytical result is about two times

higher than the FEA result. Another interesting observation is that most of the relative

displacement occurs on the bottom side of the finger (see Fig.5-7 (d)). That happens

because the vertical force is applied to the top side of the endmill.











AN -AH-
NODAL SOLUTION (a) NODAL SOLUTION (b)
STEP-2 .0182 STEP=2 a
21 S5 SUB =5 39011
TIME=2 TIIM=2
SEQV (AVC) 6.1542 CONTPRES (AVG) 780;
DMX =.0281 9.2222 RSYS=0 1.1703
SM =.012 1.2902 DX -.0275 1 5604
SMB =27 6302 15.352 SM =3.511
15.35 2SYS0 .000
18.426231 M 2.34000
21.4942i 2.7307
24 5622 .3.120-)
27.630211 3.511





NODAL SOLUTION NODAL SOLUTION

SUB =5 055 SUB .0001
TINe=2 l f l=2
CONTSFR (AVG) s ( a CONTSLID (AVG) .0t02
RSYS=O 0 1755 RSYSO .0003
DNX =.0275 2341 DMX =.0275 .0003
SM =.5266 =.0008
.29S =.000 .0004
.3511 .0005
.4096 .0006
.4691 .0007
.5266 .1C,08






Figure 5-7. FEA result for load step 2 when both ngular velocity and vertical force are
applied. (a) is the von misses stress (MPa), (b) is the contact pressure (MPa),
(c) is the frictional stress (MPa), and (d) is the slide (mm). MX indicates
where the maximum value and MN means the minimum value.

Figure 5-8, a schematic diagram of endmill bending according to the applied force,


explains in more detail. This figure is exaggerated; the real deformation is very small.


Fig.5-8 (a) shows the initial state. When the endmill starts rotating, the centrifugal force


is applied and the deformed shape looks like Fig.5-8 (b). Due to the mass conservation,


the length of the shank reduces as the diameter increases. Whereas, the length of the


finger is not reduced because the cylindrical finger is cut along its neutral axis. When the


vertical load is applied on top of the centrifugal force, the shank and the finger are


deformed, as illustrated in Fig.5-8 (c). Due to the difference in geometric center, the


deflected shape has the largest slide at the bottom surface.









(a)






(b)
















Figure 5-8. Schematic diagram of endmill behavior according to the applied forces. (a)
shows the initial state, in (b) the angular velocity is applied and in (c) the
vertical force is added.
















CHAPTER 6
PARAMETER STUDY

In this chapter, a parameter study is performed to find the maximum value of the

damping work. Two design variables, the inner radius of the finger and the number of the

fingers, are changed. The results of the parameter study are discussed with respect to the

results of the analytical approach.

Determination of the Mesh Size

It is very important to determine the proper mesh size. A fine mesh will usually

give an accurate result, but it requires a large amount of computational cost. Since the

finite element analysis needs to be repeated 45 times during the parameter study,

computational cost is an important issue.


x 105
3.35



3.3 -- -




3.25



3.2 -




3.15
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Element Number x104


Figure 6-1. Damping according to the number of elements. Number of fingers = 2, inner
radius Rl=lmm, and start angle a = 0.







42



Figure 6-1 shows the damping work according to the change of the number of


elements. The work gradually increases until the element number reaches 4200, and then


it converges on 3.28 x105 [J]. Based on these results the number of elements is chosen


to be 4200.


Determination of the Start Angle.

Before starting the parameter study, a start angle must be chosen for different


configurations. This is because the damping value changes with starting angle even if the


same number of fingers is used.


Start Angle = 0
Work = 3.28 x 10 [J]









Start Angle = 0
Work = 2.86 x 10 [J]









Start Angle = 0
Work = 3.40 x 10 [J]


Start Angle =45
Work = 2.01x 10 [J]









Start Angle =30
Work = 2.16x10 5[J]









Start Angle=18
Work = 2.82 x10 [J]


Start Angle =90
Work = 0.95 x 10 [J]









Start Angle = 60
Work = 2.86 x 10 [J]









Start Angle=36
Work = 3.40 x10 [J]


Start Angle =135
Work = 2.01x 10 [J]









Start Angle = 90
Work = 3.41x 10[J]









Start Angle =54"
Work = 3.70 x 10 [J]


Figure 6-2. Damping according to the position of the finger










Figure 6-2 illustrates the difference of damping work according to the finger's

position. The magnitude of damping work is different for different start angle. Thus, it is

necessary to evaluate the maximum and minimum values together. Figure 6-3 is the plot

of the minimum and maximum values for the same number of fingers when the inner

radius is 1.0 mm. The results show a large difference between the minimum and

maximum values when the number of fingers is small. However the difference is reduced

as the number of fingers is increased. If we compare this result with the analytical result,

shown in Fig.3-6, the FEA estimated damping work converges to a third of analytical

damping work. Possible explanations are from the fact that not all portion of the fingers

are in contact and the relative slide is large at the bottom surface.


x 105


3.5 -------- -- -- -- ----

3 ----- ----------- I--- --- ----- ----- -----I------

Maximum Value
2.5 J---- Minimum Value

S2
2- ^- 7------ ------

1.5-
1, /--- - ---



0.5--- -------------------

0
1 2 3 4 5 6 7 8 9 10
Number of Finger


Figure 6-3. The maximum and the minimum values of the damping work for the given
number of fingers.








44



Let us consider the cases where the number of fingers is 2 and 5 more in detail.


Figure 6-4 shows the relative displacement for the two-finger case. Figure 6-4 (a) and (b)


show the case where the start angle of the finger is 90'. Fig.6-4 (c) and (d) show the result


when the start angle is 0 In both cases the pressure between the contact surfaces is


almost the same for each position of the finger. But the relative displacement is different.




(a) Top (b) Bottom


NODAL SOLUTION
STEP=2
SUB =5
TIME=2
CONTSLID (AVG)
RSYS=0
DMX =.026878
SMX =.265E-03


fI


0
*B33E-041
.167E-03
.250E-03
.333E-03
.417E-03
.500E-03
.C83E-03
.667E-03
.750E-03(


(c) Top


NODAL SOLUTION
STEP=2
SUB =5
TIME=2
CONTSLID (AVG)
RSYS=0
DMX =.027511
SMX =.754E-03


.838E-04
.168E-03
.251E-03
.335E-03
,419E-13
.503E-03
.586E-03
.670E-03
,754E-03


NODAL SOLUTION
STEP=2
SUB =5
TIME=2
CONTSLID (AVG)
RSYS=0
DMX =.026878
SMX =.265E-03


NODAL SOLUTION
STEP=2
SUB =5
TIME-2
CONTSLID (AVG)
RSYS=0
DMX =.027511
SMX =.754E-03


Figure 6-4. Relative displacement of the two-finger case. (a), (b) is when the start angle

of the finger is 90', and (c), (d) is 0 MX indicates where the maximum value
occurs and MN means the minimum value.


The figure shows that most sliding occurs in the bottom part of the contact surface.


That is because the force is applied at the top surface. If we use this relative displacement


0
.833E-04
.167E-03
.250E-03
.333E-03
.417E-03
.500E-03
.E83E-03
*667E-03
.750E-03


(d) Bottom


0
.B38E-041
.168E-03
.251E-03
.335E-03
.419E-03
.502E-03
.586E-03
.670E-03
.754E-03













for calculating the frictional work, the work calculated when the start angle is0 is bigger


than 90.


(a) Top


7B1--


NODAL SOLUTION
STEP=-2
SUB =5
TIME-2
CONTSLID (AVG)
RSYS=0
DMX =.027544
SMX =.987E-03


0
.110E-031
.219E-03
.329E-03
.439E-03
.549E-03.,
.65@E-03
.76 -E-03
.87BE-03
.987E-03|


(c) Top


NODAL SOLUTION
STEP--2
SUB =5
TIME=2
CONTSLID (AVG)
RSYS=0
DMX =.027544
SMX =.987E-03


NODAL SOLUTION
STEP=2
SUB =5
TIME=2
CONTSLID (AVG)
RSYS=0
DMX =.027613
SMX =.773E-03


0M

.859E-04
.172E-03
.258E-031
.344E-03
.420E-03
.516E-03
.601E-03
.687E-03
.773E-03


NODAL SOLUTION
STEP=2
SUB =5
TIME=2
CONTSLID (AVG)
RSYS=0
DMX =.027613
SMX =.773E-03


Figure 6-5. Relative displacement of five-finger case. (a), (b) is when the start angle of

the finger is 18', and (c), (d) is 54'. MX indicates where the maximum value
occurs and MN means the minimum value.


The five-finger case is very similar to the two-fingered case. Figure 6-5 shows the


five-finger case. Parts (a) and (b) are for a start angle of 18 and parts (c) and (d) are for a


start angle of 54'. If we compare Fig.6-5 (b) with Fig.6-5 (d), the latter has almost twice


greater sliding region than the former. This is because the location of the finger


differently affects the damping work. However, Fig.6-3 shows that these effects


(b) Bottom


AN


0
.110E-03|
.219E-03
.329E-03
.43)E-03
.543E-03
.65SE-03
.768E-03
.876E-03
.987E-03|


(d) Bottom


0
.85.E-041
.17;E-03
.25EE-03
.344E-03
.431E-03
.51FE-03
.601E-03
.687E-03
.773E-031









disappear as the number of fingers increases. This means that when the number of fingers

is small the work done by the friction force depends on the start angle.

Parameter Study

To find the maximum value of damping, a parameter study was done for two

design variables. Figure 6-6 shows the first design variable, which is the inner radius of

the finger, and Fig.6-7, the second design variable, the number of fingers.


Figure 6-6. The first design variable: the inner radius (R1) of the finger (varied from 1.5
mm to 3.5 mm).


Figure 6-7. The second design variable: the number of fingers (from 2 to 10).

Change the Inner Radius of the Finger.

First the radius is changed from 1.0mm to 3.5mm for the two-finger case. Figure 6-

8 shows the result.











x 105
3.5 ----

3.4



3.2 -



3-
3.1 -----^ ^ -------- ----------

3.2----------- -------------------------------------------



2.9 -

2.8-

2.7
2.73 ----------- ----------- -- -- -- -- -T- -- -- -- -\T- -- -- ---




2.6-------

2.5
1 1.5 2 2.5 3 3.5
Inner Radius of Damper Finger (mm)






Figure 6-8. The result of the parameter study in which the inner radius of finger was
changed.




It's obvious that the work done by the friction force is gradually reduced, since the

mass of the finger is reduced when the radius is reduced. When the radius is 1.0mm the


damping is 3.28 x10 5[J], and when the radius is 3.5mm it is 2.53 x10 5[J].


Change the Number of the Finger

Next the number of fingers is changed from 2 to 10 for the case where the inner

radius is 1.5 mm. The damping work shows a minimum value for the four-finger case,

and a maximum value for the five-fingered case. The start angle of the finger (position of

the finger) is chosen to have the maximum damping.


































2 3 4 5 6
Number of Finger


7 8 9 10


Figure 6-9. The result of the parameter study in which the number of fingers was changed.

Final Results

The parameter study is repeated for all cases, and the results are as follows.


Table 6-1. The results of parameter study


x10-5 [J]


2 3 4 5 6 7 8 9 10
1.0 mm 3.28 3.41 3.21 3.69 3.55 3.54 3.54 3.48 3.51
1.5 mm 3.34 3.43 3.24 3.72 3.58 3.55 3.57 3.52 3.56
2.0 mm 3.36 3.41 3.24 3.69 3.54 3.47 3.52 3.45 3.50
2.5 mm 3.31 3.31 3.18 3.57 3.36 3.33 3.39 3.35 3.42
3.0 mm 3.08 3.10 3.02 3.32 3.10 3.08 3.12 3.04 3.16
3.5 mm 2.53 2.62 2.65 2.83 2.61 2.63 2.65 2.60 2.68


x 10-5







49






x 10-5
4

3.51



2.5




10
9
8
7
6 3.5
5 3--
4 \ I----<. 2.5
4 2.51.5
3 2
Number of Finger 2 1.5
Inner Radius of Damper Finger (mm)



Figure 6-10. The plot of the Table 6-1.

In order to find the configuration that yields the maximum damping work, the first

design variable (R1) is changed by six different values and the second design variable

(number of finger) is changed by nine different values. Table 6-1 shows the results in 6x9

matrix. In each configuration, the start angle is chosen such that the maximum damping

work can be occurred. Figure 6-10 plots the response surface of the damping work. Even

if the second design variable is discrete, a continuous surface is plotted for illustration

purpose. It is noted that the local peak when the number of fingers is five is maintained

throughout all different radii. The general trend of the response surface is consistent.

Based on the response surface, we can conclude that the damping work has its maximum

value when the inner radius is 1.5 mm and the number of fingers is 5. However, the large









difference in maximum and minimum damping values in this configuration, as shown in

Fig.6-3, may reduce the significance of this choice of design.

As a conclusion, the effect of damping work increases as the number of fingers is

increased and the inner radius is decreased.

Comparison between the Analytical and Numerical Results

Figure 6-9 shows the analytical and numerical results of friction work as a function

of the number of fingers. The friction work estimated from finite element analysis is less

than half of the friction work obtained from the analytical method. The possible

explanations of such discrepancy are the assumption of constant contact pressure.


x 10-4








Numerical Results

S0.6-----






0.2 .
0..

0 .2 -- -^- - - - - - - -





1 2 3 4 5 6 7 8 9 10
Number of Fingers


Figure 6-9. Plot of the analytical and numerical resutls.

In reality, the contact pressure is not constant and same portion of the finger does

not contact with the shank. And most relative motion occurs at the bottom part of the









endmill. That is because the vertical force is applied at the top and the nonlinearity

associated with the centrifugal force contributes the asymmetry between the top and

bottom fingers. During the nonlinear analysis ANSYS automatically update the geometry

and refer to the deformed configuration, which means the body force is calculated at the

deformed geometry. Even though the analytical and numerical results show the

difference, the general trends of both results are very similar each other.

In order to explain the general trends of friction work, consider the analytical

explanation of the contact pressure:

AMRo 2
c A


MA
where- ando is constant. ValueR, which is the distance between the rotational center
A

and the mass center of the finger, can be calculated by


R 2(R23 R3) sin(a)
R-
3 3( 2 -R,2 a

If we assume that the friction work is proportional to the contact pressure, then the

friction work is proportional toR, which increases as the number of fingers increases. If

the number of the finger increases, which means the angle a goes to zero, R is converges

to

R 2(R23 -R3)
R=
3( 2 -R,2)

Therefore, the maximum value of the contact pressure can be calculated by

MRco2 2(R23 -R3) MC2
c A 3 (R22 -R,2) A






52


As a conclusion, the friction work increases along with the number of fingers, but

its effect is reduced as the number of fingers increases.














CHAPTER 7
CONCLUSION AND FUTURE WORK

The goal of this research was to design the mechanical damper which is inserted

into the endmill. Through the nonlinear finite element analysis and parameter study, the

trend of the damping effect is identified.

The results from the analytical method were used to verify the FEA result.

Although the number and the inner radius of finger were varied in finite element analysis,

only the two-finger case was considered in the comparison. The FEA results were smaller

than those of the analytical approach because some regions did not contact and the

contact pressure was not constant.

A parameter study was carried out by changing the inner radius and the number of

fingers. The inner radius was varied from 1.0 mm to 3.5 mm, and the number of fingers

was varied from 2 to 10. The results show general trends of the damping work according

to the change of the two design variables. As the inner radius decreased and the number

of finger increased, the damping work increased. The Maximum value was 3.72 x 105 [J]

when the inner radius was 1.5 mm and the number of fingers was 5. The parameter study

also showed that when the number of the fingers is small the damping work is affected by

the position of the finger, but that this dependence disappears as the number of finger

increases.

Recommendations for Future Research

As discussed in Chapter 6, the damping work depends on the start angle. However,

in practice the endmill is continuously rotating. Thus, it is recommended to perform a






54


series of static FEA by rotating the endmill by on cycle and to calculate the integrated

damping work. This procedure will provide more accurate estimation of the real damping

work.
















APPENDIX A
MATLAB CODE FOR THEORITICAL ANALYSIS

%-------------------------------------------------------------------------------------
% Matlab function to calculate the work done by the friction force
%
% By Dongki Won
%---------------------------------------------------------------- ---------------------
function cal work
%---------------------------------------------------------------- ---------------------
% Material Properties, radius and applied forces.
%---------------------------------------------------------------- ---------------------
E=2.0678el 1; %- Modulus of Elasticity : 2.0*10A11[MPa]
rho=7820; %- Mass Density : 7820[kg/mA3]
mu=0.15; %- Friction Coefficient : 0.15
rl=1.5e-3;
r2=4.7625e-3;
r3=r2;
r4=9.525e-3;
L=101.6e-3;
omega=2722;
F=100;
%---------------------------------------------------------------- ---------------------
% Calculate the work according to the chage of the inner raius of finger.
%---------------------------------------------------------------- ---------------------
n=2;m=l;
for rl=1.0e-3:0.5e-3:3.5e-3
ang(1)=0;
for i=l:n
ang(i+l)=ang(1)+(360/n)*i;
end
ang=ang*(pi/180);
II=(1/4)*((r4)A4-(r3)A4)*pi;
for i=l:(n-1)
tmpl=sin(ang(i+l))*cos(ang(i+l))-sin(ang(i))*cos(ang(i))-(ang(i+l)-ang(i));
tmpl=tmp l*((r l)^4-(r2)A4);
tmpl=tmpl*(1/8);
II=II+tmp 1;
end
work=0;
for i=l:n
alpha=(ang(i+ 1)-ang(i))/2;









dl=(2*sin(alpha).*(r2^A3-rlA3))/(3*alpha*(r2A2-rlA2));
d2=dl *sin((ang(i)+ang(i+1))*0.5);
V=(ang(i+ )-ang(i))*(r2A2-rlA 2)*L*0.5;
P=abs(V*rho*(omegaA2)*dl/L);
work=work+abs((1/3)*(LA3)*mu*P*F*d2/(E*II));
end
yl(m)=work;
m=m+1;
end
xl=[1.0e-3:0.5e-3:3.5e-3];
plot(xl,yl,'b-o');
grid;
xlabel('Inner Radius of Damper Finger (m)');
ylabel('Work(J)');
[tmp,n]=size(xl);
clc
fprintf('\nCalculate the work according to the chage of the inner raius of finger.\n');
fprintf('The Number of Finger=%3.0f\nStart Angle=%3.0f [Degree]\n',n,0);
for i=l:n
fprintf('Inner Raius of Finger =% 16.9e [m]\t\tWork=%16.9e [J]\n',x (i),yl(i));
end
%-----------------------------------------------------
% Calculate the work according to the chage of the start angle of finger.
%-----------------------------------------------------
n=2;m=l;rl=1.5e-3;
atmp=(360/n)/4;
for a=0:atmp:atmp*3
ang(1)=a;
for i=l:n
ang(i+l)=ang(1)+(360/n)*i;
end
ang=ang*(pi/180);
II=(1/4)*((r4)A4-(r3)A4)*pi;
for i=l:(n-1)
tmpl=sin(ang(i+l))*cos(ang(i+l))-sin(ang(i))*cos(ang(i))-(ang(i+l)-ang(i));
tmpl=tmpl*((rl)A^4-(r2)A4);
tmpl=tmpl*(1/8);
II=II+tmp 1;
end
work=0;
for i=l:n
alpha=(ang(i+ 1)-ang(i))/2;
dl=(2*sin(alpha).*(r2A3-rlA3))/(3*alpha*(r2A2-rlA2));
d2=dl *sin((ang(i)+ang(i+1))*0.5);
V=(ang(i+l)-ang(i))*(r2A2-rlA2)*L*0.5;
P=abs(V*rho*(omegaA2)*dl/L);









work=work+abs((1/3)*(LA3)*mu*P*F*d2/(E*II));
end
y2(m)=work;
m=m+l;
end
x2=[0:atmp:atmp*3];
figure;
plot(x2,y2,'b-o');
grid;
xlabel('Start Angle of Damper Finger (degree)');
ylabel('Work(J)');
[tmp,n]=size(x2);
fprintf('\nCalculate the work according to the chage of the start angle of finger.\n');
fprintf('Number of Finger=%5.0f\nInner Radius of the Finger=%16.9e [m]\n',2,rl);
for i=l:n
fprintf('Start Angle =%16.9e [Degree]\t\tWork=% 16.9e [J]\n',x2(i),y2(i));
end
%-----------------------------------------------------
% Calculate the work according to the change of the number of finger.
%-----------------------------------------------------
m=l;clear ang
for n=2:10
clear ang
atmp=(l+(-1)A(n-1))/2;
ang(l)=((360/n)/4)*atmp;
for i=l:n
ang(i+l)=ang(1)+(360/n)*i;
end
ang=ang*(pi/180);
II=(1/4)*((r4)A4-(r3)A4)*pi;
for i=l:(n-1)
tmpl=sin(ang(i+l))*cos(ang(i+l))-sin(ang(i))*cos(ang(i))-(ang(i+l)-ang(i));
tmpl=tmp l*((rl )A4-(r2)A4);
tmpl=tmpl*(1/8);
II=II+tmp 1;
end
work=0;
for i=l:n
alpha=(ang(i+ 1)-ang(i))/2;
dl=(2*sin(alpha).*(r2 3-rlA3))/(3*alpha*(r2 2-rl^2));
d2=dl *sin((ang(i)+ang(i+1))*0.5);
V=(ang(i+l)-ang(i))*(r2A2-rlA2)*L*0.5;
P=abs(V*rho*(omegaA2)*dl/L);
work=work+abs((1/3)*(LA3)*mu*P*F*d2/(E*II));
end
y3(m+l)=work;






58


m=m+1;
end
x3=[1:10];
figure;
plot(x3,y3,'b-o');
axis([0,10,0,11e-5]);
grid;
xlabel('Number of Damper Finger');
ylabel('Work(J)');
[tmp,n]=size(x3);
fprintf('\nCalculate the work according to the chage of the number of finger.\n');
fprintf('Inner Radius of the Finger=%16.9e [m]\n',rl);
for i=l:n
fprintf('Number of Finger =%3. Of\t\tWork=%16.9e [J]\n',x3(i),y3 (i));
end















APPENDIX B
ANSYS INPUT FILE 01

1--------------------------------------- --------------------------
ANSYS input file to determine the mesh size according to the change of
Selemnt number. The results are stored in meshsize,txt.

Dongki Won


/SHOW,JPEG


*SET,radin,1.5
*SET,rad_outl,4.7625
*SET,radout2,9.525
*SET,f num,2
*SET,fang,360/f num
*SET,stang,(f ang/4)*0

*DO,ii,1,8
*IF,ii,EQ,1,lTHEN
*SET,s mesh,6
*SET,f mesh,6
*SET,l mesh,12
*SET,st mesh,2
*SET,ft mesh,2
*ELSEIF,ii,EQ,2 !
*SET,s mesh,8
*SET,f mesh,8
*SET,l mesh,16
*SET,st mesh,2
*SET,ft mesh,2
*ELSEIF,ii,EQ,3 !
*SET,s mesh,10
*SET,f mesh,10
*SET,l mesh,20
*SET,st mesh,2
*SET,ft mesh,2
*ELSEIF,ii,EQ,4 !
*SET,s mesh,14
*SET,f mesh,14
*SET,l mesh,25
*SET,st mesh,2


SInner radius of the finger
SOuter radius of the finger
SOuter radius of the shank
SThe number of fingers
SThe angle of one finger
SStart angle of the finger


Element Number is 864






element Number is 1536





element Number is 2400






element Number is 4200









*SET,ft mesh,2
*ELSEIF,ii,EQ,5
*SET,s mesh,18
*SET,f mesh,18
*SET,l mesh,30
*SET,st mesh,2
*SET,ft mesh,2
*ELSEIF,ii,EQ,6
*SET,s mesh,18
*SET,f mesh,18
*SET,l mesh,30
*SET,st mesh,3
*SET,ft mesh,3
*ELSEIF,ii,EQ,7
*SET,s mesh,18
*SET,f mesh,18
*SET,l mesh,40
*SET,st mesh,3
*SET,ft mesh,3
*ELSEIF,ii,EQ,8
*SET,s mesh,20
*SET,f mesh,20
*SET,l mesh,60
*SET,st mesh,3
*SET,ft mesh,3
*ENDIF


SElement Number is 6480





SElement Number is 8640





SElement Number is 11500





SElement Number is 19200


tmpl=smesh*lmesh*2*(1+st mesh)
tmp2=f mesh*lmesh*2*(1+ft mesh)
num_elem=tmp 1+tmp2
/INPUT,endmillmodel,inp
/INPUT,endmillwork,inp
/OUTPUT,meshsize,txt,,APPEND
*VWRITE,numelem,work
(F7.0,3X,F16.9,3X)
/OUTPUT
*ENDDO















APPENDIX C
ANSYS INPUT FILE 02

1--------------------------------------- --------------------------
ANSYS input file to determine the start angle of the finger.
The results are stored in startangle,txt.

Dongki Won
1--------------------------------------- --------------------------
/SHOW,JPEG

*SET,radin, 1 Inner radius of the finger
*SET,rad_outl,4.7625 Outer radius of the finger
*SET,radout2,9.525 Outer radius of the shank
*SET,s mesh,14
*SET,l mesh,30
*SET,st mesh,2
*SET,ft mesh,2

*DO,ii,8,10
*SET,f num,ii
*SET,fang,360/fnum
*IF,ii,EQ,2,THEN
*SET,f mesh,14
*ELSEIF,ii,EQ,3
*SET,f mesh,9
*ELSEIF,ii,EQ,4
*SET,f mesh,7
*ELSEIF,ii,EQ,5
*SET,f mesh,6
*ELSEIF,ii,EQ,6
*SET,f mesh,5
*ELSEIF,ii,EQ,7
*SET,f mesh,4
*ELSEIF,ii,EQ,8
*SET,f mesh,4
*SET,s mesh,16
*SET,ft mesh,2
*ELSEIF,ii,EQ,9
*SET,f mesh,4
*SET,s mesh,18
!*SET,ft mesh,2






62


*ELSEIF,ii,EQ,10
*SET,f mesh,3
*SET,s mesh,15
!*SET,ft mesh,3
*ENDIF
*DOjj,0,3
*SET,stang,(f ang/4)*jj
/INPUT,endmill_model,inp
/INPUT,endmillwork,inp
/OUTPUT, startangle,txt,,APPEND
*VWRITE,f num,jj,stang,work
(F7.0,3X,F7.0,3X,F 16.9,3X,F16.9,3X)
/OUTPUT
*ENDDO
*ENDDO















APPENDIX D
ANSYS INPUT FILE 03

1--------------------------------------- --------------------------
ANSYS input file to do the parameter study as the change of two design
variables, the number of the finger(f num) and the inner radius of the finger
(radin). The results are stored in parameterstudy,txt.

Dongki Won
1--------------------------------------- --------------------------
/SHOW,JPEG

*SET,rad_outl,4.7625 Outer radius of the finger
*SET,radout2,9.525 Outer radius of the shank
*SET,s mesh,14
*SET,l mesh,25
*SET,st mesh,2
*SET,ft mesh,2

*DO,ii,2,10
*SET,f num,ii
*SET,fang,360/fnum
*IF,ii,EQ,2,THEN
*SET,f mesh,14
*SET,stang,(f ang/4)*0
*ELSEIF,ii,EQ,3
*SET,f mesh,9
*SET,stang,(f ang/4)*3
*ELSEIF,ii,EQ,4
*SET,f mesh,7
*SET,stang,(f ang/4)*2
*ELSEIF,ii,EQ,5
*SET,f mesh,6
*SET,stang,(f ang/4)*3
*ELSEIF,ii,EQ,6
*SET,f mesh,5
*SET,stang,(f ang/4)*2
*ELSEIF,ii,EQ,7
*SET,f mesh,4
*SET,stang,(f ang/4)*3
*ELSEIF,ii,EQ,8
*SET,f mesh,4









*SET,s mesh,16
*SET,stang,(f ang/4)*2
*ELSEIF,ii,EQ,9
*SET,f mesh,4
*SET,s mesh,18
*SET,stang,(f ang/4)*3
*SET,ft mesh,2
*SET,lmesh,30
*ELSEIF,ii,EQ,10
*SET,f mesh,3
*SET,s mesh,15
*SET,stang,(f ang/4)*2
*SET,ft mesh,2
*SET,l mesh,25
*ENDIF
*DOjj, 1,3.5,0.5
*SET,radin,jj
/INPUT,endmill_model,inp
/INPUT,endmillwork,inp
/OUTPUT,parameter study,txt,,APPEND
*VWRITE, fnum,rad in,work
(F7.0,3X,F 16.9,3X,F16.9,3X)
/OUTPUT
*ENDDO
*ENDDO
















APPENDIX E
ANSYS INPUT FILE 04

1--------------------------------------- --------------------------
ANSYS input file to solve the endmill model. This input file is used
in endmillrun.inp and endmill_start.inp

Dongki Won
1--------------------------------------- --------------------------
PARSAV
/CLEAR
PADRES

/FILNAME,endmill,
/UNITS,MPA
/TITLE,Damping Design of Endmill
/prep7
1---------------------------------------- --------------
SElement Type and Material Properties
1---------------------------------------- --------------
ET,1,SOLID95 !- 20-Nodes Cubic Element
MP,EX, 1,206800 !- Modulus of Elasticity : 206780[MPa]
MP,DENS,1,7.82e-9 !- Mass Density : 7.82e-6[kg/mmA3]
MP,NUXY, 1,0.29 !- Poisson's Ratio : 0.29
MP,MU,1,0.15 !- Friction Coefficient: 0.15
1---------------------------------------- --------------
SCreate Model Geometry
1---------------------------------------- --------------
!= Shank:contact surface is 3, 8
K,1,0,0,0
K,2,0,0,1
K,11,rad out2,0,0
K,12,radoutl,0,0
K,13,radoutl,0,101.6
K,14,rad out2,0,101.6
A,11,12,13,14
VROTAT,1,,,,,,1,2,360,2

!= Finger:contact surfaces are 13,19,25,31,37
*DO,i,l,f num,l
CYLIND,radin,rad_outl,101.6,,stang+f ang*(i-1),stang+f ang*i
*ENDDO











SGenerating Mesh
1---------------------------------------- --------------
LSEL,S,LENGTH,, 101.6
LESIZE,ALL,,,l mesh

vsel,S,volu,,1,2,1,1
LSEL,R,RADIUS,,rad outl
LSEL,A,RADIUS,,rad out2
LESIZE,ALL,,,s mesh
LSEL,S,LENGTH,,rad outl
LESIZE,ALL,,,st mesh

vsel,S,volu,,3,3+f num,1,1
LSEL,R,RADIUS,,rad outl
LSEL,A,RADIUS,,Rad in
LESIZE,ALL,,,f mesh
LSEL,S,LENGTH,,rad outl-rad in
LESIZE,ALL,,,ft mesh
AllSEL

MSHAPE,0,3D != hexahedral-shaped elements,3D
MSHKEY,2 != Map meshing
VMESH,ALL
1---------------------------------------- --------------
SDesignating Contact Pairs
1---------------------------------------- --------------
ET,2,TARGE170
ET,3,CONTA174
ASEL,S,,,3,8,5,1
NSLA,S,1
ESLN,0
TYPE,2
ESURF,,TOP
ASEL,S,,,13,13+(f num-1)*6,6,1
NSLA,S,1
ESLN,0
TYPE,3
ESURF,,TOP
FINISH

/SOLU
1---------------------------------------- --------------
SApply Boundary Conditions and Sove

NSELS,LOC,Z,0
NSEL,S,LOC,Z,O









D,ALL,ALL
ALLSEL
SOLCONTROL,ON

! Load Step 1
OMEGA,0,0,2722.713,0
TIME, 1
NROPT,UNSYM
NSUBST,5,10,2
AUTOTS,OFF
OUTRES,BASIC,ALL
KBC,0
LSWRITE,

! Load Step 2
CSYS,1
NSEL,S,LOC,Z,101.6
NSEL,R,LOC,X,rad out2
NSEL,R,LOC,Y,90
F,ALL,FY,-25
*DO,kk, ,2
NSEL,S,LOC,Z,101.6
NSEL,R,LOC,X,rad out2
NSEL,R,LOC,Y,90+(360/(smesh*4))*((-1)**kk)
F,ALL,FY,-25
*ENDDO
*DO,kk, ,2
NSEL,S,LOC,Z,101.6
NSEL,R,LOC,X,rad out2
NSEL,R,LOC,Y,90+(360/(smesh*4))*2*((-1)**kk)
F,ALL,FY,-12.5
*ENDDO
CSYS,0
ALLSELL
TIME,2
NROPT,UNSYM
NSUBST,5,10,2
AUTOTS,OFF
OUTRES,ALL,ALL
KBC,0
LSWRITE,2
LSSOLVE,1,2,1


FINISH
















APPENDIX F
ANSYS INPUT FILE 05

1--------------------------------------- --------------------------
ANSYS input file to calculate the work done by the frictional force.
This input file is used in endmillrun.inp and endmill_start.inp

Dongki Won
1--------------------------------------- --------------------------
/POST1
SET, 1,1
ESEL,S,ENAME,, 174
*GET, emin,ELEM,0,NUM,MIN
*GET,emax,ELEM,0,NUM,MAX
*GET,aa,ELEM,emin,VOLU
aa=aa*0.25
1--------------------------------------- ---------------------
SRead Friction Stress of R and S direction in each elements

ETAB,taurl,SMISC,5
ETAB,taur2,SMISC,6
ETAB,taur3,SMISC,7
ETAB,taur4,SMISC,8
ETAB,taus1,SMISC,9
ETAB,taus2,SMISC,10
ETAB,taus3,SMISC,11
ETAB,taus4,SMISC,12
1--------------------------------------- ---------------------
SCalculate Friction Force of R and S direction by multiplying area
1--------------------------------------- ---------------------
SMULT,rforcel ,taurl,,aa
SMULT,rforce2,taur2,,aa
SMULT,rforce3,taur3,,aa
SMULT,rforce4,taur4,,aa
SMULT, force ,taus 1,,aa
SMULT, sforce2,taus2,,aa
SMULT, sforce3,taus3,,aa
SMULT, sforce4,taus4,,aa
1--------------------------------------- ---------------------
SRead Slide(Relative Displacement) of R and S direction in each elements

ETAB,tasrl-,NMISC,-17
ETAB, tasr l,NMISC,17










ETAB,tasr2,NMISC,18
ETAB,tasr3,NMISC,19
ETAB,tasr4,NMISC,20
ETAB,tassl,NMISC,21
ETAB,tass2,NMISC,22
ETAB,tass3,NMISC,23
ETAB,tass4,NMISC,24
1--------------------------------------- ---------------------
SCalculate the work done by the friction force in each elements.

VDOT,workl,-rforcel,-sforce -,,tasrl-,tass1
VDOT,work2,rforce2,sforce2,,tasr2,tass2
VDOT,work3,rforce3, sforce3,,tasr3,tass3
VDOT,work4,rforce4,sforce4,,tasr4,tass4

SABS,1
SSUM
*GET,workl, SSUM,,ITEM,workl
*GET,work2, SSUM,,ITEM,work2
*GET,work3, S SUM,,ITEM,work3
*GET,work4, SSUM,,ITEM,work4
work=work 1 +work2+work3 work4
SABS,0
1--------------------------------------- ---------------------
SSave the slide value for the next step.
1--------------------------------------- ---------------------
ETAB,tasrltmp,NMISC,17
ETAB,tasr2tmp,NMISC, 18
ETAB,tasr3tmp,NMISC, 19
ETAB,tasr4tmp,NMISC,20
ETAB,tassltmp,NMISC,21
ETAB,tass2tmp,NMISC,22
ETAB,tass3tmp,NMISC,23
ETAB,tass4tmp,NMISC,24
1--------------------------------------- ---------------------
SRepeat the procedure until reaches to the last set
1--------------------------------------- ---------------------
ww=0
*DO,uu, 1,2
*DO,vv,1,5
*IF,uu,EQ,1 ,AND,vv,EQ, 1, CYCLE
SET,uu,vv
ww=ww+l
----------------------------------------------------
Read Friction Stress of R and S direction in each elements
---------ETAB,taurl,SMISC,5
ETAB,taurl,SMISC,5










ETAB,taur2,SMISC,6
ETAB,taur3,SMISC,7
ETAB,taur4,SMISC,8
ETAB,taus 1,SMISC,9
ETAB,taus2,SMISC,10
ETAB,taus3,SMISC,11
ETAB,taus4,SMISC,12
------------------------------------------------------
Calculate Friction Force of R and S direction by multiplying area
----------------------------------------------------
SMULT,rforcel ,taurl,,aa
SMULT,rforce2,taur2,,aa
SMULT,rforce3,taur3,,aa
SMULT,rforce4,taur4,,aa
SMULT,sforcel,tausl,,aa
SMULT,sforce2,taus2,,aa
SMULT, sforce3,taus3,,aa
SMULT, sforce4,taus4,,aa
----------------------------------------------------
Read Slide(Relative Displacement) of R and S direction in each elements
----------------------------------------------------
ETAB,tasrl,NMISC, 17
ETAB,tasr2,NMISC, 18
ETAB,tasr3,NMISC, 19
ETAB,tasr4,NMISC,20
ETAB,tassl,NMISC,21
ETAB,tass2,NMISC,22
ETAB,tass3,NMISC,23
ETAB,tass4,NMISC,24
SMULTtasr,tasrtmp,tasrtmp,,- 1
SMULT,tasr2tmp,tasr2tmp,,- 1
SMULT,tasr3tmp,tasr3tmp,,- 1
SMULT,tasr4tmp,tasr4tmp,,- 1
SMULT,tassltmp,tassltmp,,-1
SMULT,tass2tmp,tass2tmp,,- 1
SMULT,tass3tmp,tass3tmp,,- 1
SMULT,tass4tmp,tass4tmp,,- 1
SADD,tasrl,tasrl,tasrltmp
SADD,tasr2,tasr2,tasr2tmp
SADD,tasr3,tasr3,tasr3tmp
SADD,tasr4,tasr4,tasr4tmp
SADD,tass ,tass ,tass tmp
SADD,tass2,tass2,tass2tmp
SADD,tass3,tass3,tass3tmp
SADD,tass4,tass4,tass4tmp










-------------------------------------------------------
Calculate the work done by the friction force in each elements.
----VDOT,work,--- rforcesforce--,,tasr,tass
VDOT,work2,rforce2,sforce2,,tasr2,tass2
VDOT,work3,rforce2,sforce3,,tasr3,tass3
VDOT,work4,rforce4, sforce4,,tasr4,tass4
VDOT,work4,rforce4,sforce4,,tasr4,tass4
SABS,1
SSUM
*GET,workl, SSUM,,ITEM,workl
*GET,work2, SSUM,,ITEM,work2
*GET,work3, S SUM,,ITEM,work3
*GET,work4, SSUM,,ITEM,work4
work=work+work 1 +work2+work3 work4
SABS,0
----------------------------------------------------
Save the slide value for the next step.
--------ETAB,tasrtmp,NM --SC,17
ETAB,tasr2tmp,NMISC, 18
ETAB,tasr3tmp,NMISC, 19
ETAB,tasr4tmp,NMISC,20
ETAB,tasr4tmp,NMISC,20
ETAB,tass2tmp,NMISC,22
ETAB,tass3tmp,NMISC,23
ETAB,tass4tmp,NMISC,24
ETAB,tass4tmp,NMISC,24
*ENDDO
*ENDDO
FINISH
















REFERENCES

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Materials, fifth Edition, John Wiley & Sons, Inc., NY, 1993.

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3. Dean, M.R., "Improving Milling Efficiency Through Chatter Suppression and
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4. Schmitz, T., "Predicting High-Speed Machining Dynamics by Substructure
Analysis," Annals of the CIRP, 49(1), 2000, pp. 303-308.

5. Slocum, A.H., Precision Machine Design, Prentice Hall, Inc., Englewood Cliffs,
NJ, 1992.

6. Smith, K.S., "Chatter, Forced Vibrations, and Accuracy In High Speed Milling,"
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7. Sterling, Robert A., "Damping of Long Endmills," Master's Thesis, University of
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8. Tlusty, J., Manufacturing Processes and Equipment, Prentice Hall, Inc., Upper
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9. Zhong, Z.H., Finite Element Procedures for Contact-Impact Problems, NY, Oxford
University Press, 1993















BIOGRAPHICAL SKETCH

Dongki Won was born May 5, 1972. In February 2000, he received a bachelor's

degree with high honors in mechanical engineering from the Dongguk University in

Seoul, Korea. In September 2001, he began graduate study in the Department of

Mechanical and Aerospace Engineering at the University of Florida. At the time of

writing, he is in his last semester at the University of Florida, after which he will start

working on a PhD in the Department of Mechanical and Aerospace Engineering at the

University of Florida.