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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 21. M material properties used for the shank and finger........................................... ........... 8 31. First moment Q of various quantities. ................................. ............... 13 51. The number of nodes and elements. ........................ ............ ....... 33 61. The results of parameter study ............ .................................... ..........................48 LIST OF FIGURES Figure page 11. Chatter m ark. ...................................................................1 21. Endm ill and dam per....................................... .................... ......... 22. M odel sim plifi cation......................................................... ................................... 6 23. D im tensions of geom etry .............................................................................. ............. 7 31. Cross sectional area of the endmill system........................................................11 32. Cross sectional area of the finger. ........................................ ......................... 14 33. Cross sectional area of the finger. ........................................ ......................... 15 34. Plot of work done by the friction force according to the change of the inner radius of fi n g er. ............................................................ ................ 2 2 35. Plot of work done by the friction force for varying start angles of the finger............23 36. Plot of the work done by the friction force for different numbers of fingers.............23 51. Solid 95, 20 node solid elem ent......................................... ............................ 32 52. FEA model of the endmill using 20 node cubic elements and 8 node contact elements w ith boundary conditions. ............................................................. .....................33 53. 8node contact and target element description. .................................. ...............34 54. Force boundary conditions in each time step. ................................. .................35 55. Diagram explaining how to calculate the damping work.................... ...............36 56. FE A result for load step 1............................................... ................................ 37 57. FEA result for load step 2 ....................................................................... 39 58. Schematic diagram of endmill behavior according to the applied forces...................40 61. Damping according to the number of elements ......................................................41 62. Damping according to the position of the finger......................................................42 63. The maximum and the minimum values of the damping work for the given number of fingers. .......................................................................................................... ..... 43 64. Relative displacement of the twofinger case ......................................................44 65. Relative displacement of fivefinger case. ............................................................45 66. The first design variable. ...... ........................... .........................................46 67. The second design variable............................................... .............................. 46 68. The result of the parameter study in which the inner radius of finger was changed..47 69. The result of the parameter study in which the number of fingers is changed..........48 610. The plot of the Table 61 ....... ........................... ........................................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 highspeed 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 layeredbeam 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. 11. Figure 11. 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 undeformed 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 layeredbeam damper, which Sterling calls a finger, is inserted into the hole at the center of the tool. Due to the highspeed 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.21. Most endmills are of the solid beam type as shown in Fig.21 (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 21. Endmill and damper. (a) the original solid endmill and (b) the damper inserted model When a layeredbeam damper (see Fig.21 (b)) is inserted into the hollow tool, the highspeed 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.22 (b) illustrates the simplified model. (a)(b) co=2722.71 rad/s F=100 N Figure 22. 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 twohollowed 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.23, 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 23. Dimensions of geometry. Although Fig.23 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 ^^49,$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 21. Table 21. Material properties used for the shank and finger. Material Property Value Young's Modulus 206780 MPa Mass Density 7.82x109 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.23). The shank and the finger are hollowed cylinders. For simplicity, we only consider the case of a twofinger configuration. Figure 31 (a) shows the crosssectional 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.31 (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 31. 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 31 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 twodimensional work, e.q. in the xy plane, Qx becomes Q, and Q becomes Q . Table 31. 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.32, 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) 8a  R ) sin(/7) + sin(a) R2) fa Figure 32. 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 yaxis is chosen such that the centroid is located on the yaxis, as illustrated in Fig.33. 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.33, y is represented by, y = r cos 2(R3 3(p R 2(R:3 3(P, 15 Y Ox X Figure 33. 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 21, the mass M can be calculated by. M = pV= 7.82x106 (4.762521.52)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 loadstress 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 xy plane. A represents the crosssectional 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) = (4.762541.54 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 xaxis, 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 +3xL22L3) 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 +3xL22 (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 34, 35 and 36 show the results of theoretical analysis. x 105 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 103 Figure 34. Plot of work done by the friction force according to the change of the inner radius of finger. Figure 34 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 35 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 35. 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 104 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 36. 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 36 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 multifield 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 uqdSU 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 01 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) [P1] 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 trialanderror procedure is required to find all contacting nodes. In the following, we discuss the solution of problem [P1] 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 [P1] is now transformed into the following saddlepoint problem: Find U and A such that H,(U, A) is stationary, i.e. [P2] =L 0 U (4.10) HL =0 BA Eq.(4.10) yields KUF+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'KUU F+P2P 2 2 The constrained minimization problem [P1] 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 51. 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 21, 20node solid elements in ANSYS (solid 95) are used to build the shank and finger. Figure 51 illustrates a 20node solid element that is used in ANSYS, and Fig.52 plots the finite element model of the endmill with boundary conditions. In Fig.52, 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. Table51 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 52. FEA model of the endmill using 20 node cubic elements and 8 node contact elements with boundary conditions. Table 51. 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 8node contact element, as illustrated in Fig.53. 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 aacetAoSrf / OCatack Element GONTA173 or CONTA174 Figure 53. 8node 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 3D "target" surfaces and a deformable surface, defined by this element. The element is applicable to 3D structural and coupled thermalstructural contact analysis. This element is located on the surfaces of 3D 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 3D 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 54. 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 54 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 NewtonRaphson 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 55. 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 55 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 56 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 56. 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.56 (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 57 shows the results of load step 2. The frictional work is calculated by dot producing the friction force (Fig.57 (c)) and the relative slide (Fig.57 (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.57 (d)). That happens because the vertical force is applied to the top side of the endmill. AN AH NODAL SOLUTION (a) NODAL SOLUTION (b) STEP2 .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 57. 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 58, 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.58 (a) shows the initial state. When the endmill starts rotating, the centrifugal force is applied and the deformed shape looks like Fig.58 (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.58 (c). Due to the difference in geometric center, the deflected shape has the largest slide at the bottom surface. (a) (b) Figure 58. 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 61. Damping according to the number of elements. Number of fingers = 2, inner radius Rl=lmm, and start angle a = 0. 42 Figure 61 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 62. Damping according to the position of the finger Figure 62 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 63 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.36, 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 63. 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 64 shows the relative displacement for the twofinger case. Figure 64 (a) and (b) show the case where the start angle of the finger is 90'. Fig.64 (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 =.265E03 fI 0 *B33E041 .167E03 .250E03 .333E03 .417E03 .500E03 .C83E03 .667E03 .750E03( (c) Top NODAL SOLUTION STEP=2 SUB =5 TIME=2 CONTSLID (AVG) RSYS=0 DMX =.027511 SMX =.754E03 .838E04 .168E03 .251E03 .335E03 ,419E13 .503E03 .586E03 .670E03 ,754E03 NODAL SOLUTION STEP=2 SUB =5 TIME=2 CONTSLID (AVG) RSYS=0 DMX =.026878 SMX =.265E03 NODAL SOLUTION STEP=2 SUB =5 TIME2 CONTSLID (AVG) RSYS=0 DMX =.027511 SMX =.754E03 Figure 64. Relative displacement of the twofinger 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 .833E04 .167E03 .250E03 .333E03 .417E03 .500E03 .E83E03 *667E03 .750E03 (d) Bottom 0 .B38E041 .168E03 .251E03 .335E03 .419E03 .502E03 .586E03 .670E03 .754E03 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 TIME2 CONTSLID (AVG) RSYS=0 DMX =.027544 SMX =.987E03 0 .110E031 .219E03 .329E03 .439E03 .549E03., .65@E03 .76 E03 .87BE03 .987E03 (c) Top NODAL SOLUTION STEP2 SUB =5 TIME=2 CONTSLID (AVG) RSYS=0 DMX =.027544 SMX =.987E03 NODAL SOLUTION STEP=2 SUB =5 TIME=2 CONTSLID (AVG) RSYS=0 DMX =.027613 SMX =.773E03 0M .859E04 .172E03 .258E031 .344E03 .420E03 .516E03 .601E03 .687E03 .773E03 NODAL SOLUTION STEP=2 SUB =5 TIME=2 CONTSLID (AVG) RSYS=0 DMX =.027613 SMX =.773E03 Figure 65. Relative displacement of fivefinger 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 fivefinger case is very similar to the twofingered case. Figure 65 shows the fivefinger 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.65 (b) with Fig.65 (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.63 shows that these effects (b) Bottom AN 0 .110E03 .219E03 .329E03 .43)E03 .543E03 .65SE03 .768E03 .876E03 .987E03 (d) Bottom 0 .85.E041 .17;E03 .25EE03 .344E03 .431E03 .51FE03 .601E03 .687E03 .773E031 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 66 shows the first design variable, which is the inner radius of the finger, and Fig.67, the second design variable, the number of fingers. Figure 66. The first design variable: the inner radius (R1) of the finger (varied from 1.5 mm to 3.5 mm). Figure 67. 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 twofinger 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 68. 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 fourfinger case, and a maximum value for the fivefingered 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 69. 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 61. The results of parameter study x105 [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 105 49 x 105 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 610. The plot of the Table 61. 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 61 shows the results in 6x9 matrix. In each configuration, the start angle is chosen such that the maximum damping work can be occurred. Figure 610 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.63, 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 69 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 104 Numerical Results S0.6 0.2 . 0.. 0 .2  ^        1 2 3 4 5 6 7 8 9 10 Number of Fingers Figure 69. 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 twofinger 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=1.5e3; r2=4.7625e3; r3=r2; r4=9.525e3; L=101.6e3; omega=2722; F=100; %  % Calculate the work according to the chage of the inner raius of finger. %  n=2;m=l; for rl=1.0e3:0.5e3:3.5e3 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:(n1) 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^A3rlA3))/(3*alpha*(r2A2rlA2)); d2=dl *sin((ang(i)+ang(i+1))*0.5); V=(ang(i+ )ang(i))*(r2A2rlA 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=[1.0e3:0.5e3:3.5e3]; plot(xl,yl,'bo'); 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=1.5e3; 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:(n1) 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).*(r2A3rlA3))/(3*alpha*(r2A2rlA2)); d2=dl *sin((ang(i)+ang(i+1))*0.5); V=(ang(i+l)ang(i))*(r2A2rlA2)*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,'bo'); 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(n1))/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:(n1) 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 3rlA3))/(3*alpha*(r2 2rl^2)); d2=dl *sin((ang(i)+ang(i+1))*0.5); V=(ang(i+l)ang(i))*(r2A2rlA2)*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,'bo'); axis([0,10,0,11e5]); 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 ! 20Nodes Cubic Element MP,EX, 1,206800 ! Modulus of Elasticity : 206780[MPa] MP,DENS,1,7.82e9 ! Mass Density : 7.82e6[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*(i1),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 outlrad in LESIZE,ALL,,,ft mesh AllSEL MSHAPE,0,3D != hexahedralshaped 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 num1)*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 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," Master's Thesis, University of Florida, 1989. 3. Dean, M.R., "Improving Milling Efficiency Through Chatter Suppression and Increased Axis Feedrates," Master's Thesis, University, 1998. 4. Schmitz, T., "Predicting HighSpeed Machining Dynamics by Substructure Analysis," Annals of the CIRP, 49(1), 2000, pp. 303308. 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," Master's Thesis, University of Florida, 1985. 7. Sterling, Robert A., "Damping of Long Endmills," Master's 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 ContactImpact 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. 