2621-6-P THE UNIVERSITY OF MI CHI G AN COLLEGE OF ENGINEERING Department of Chemical and Metallurgical Engineering Progress Report AN INVESTIGATION OF METAL SPINNING B..Avitzur S. Floreen E. E. Hucke D. V. Ragone UMRI Project 2621 under contract with: SPINCRAFT, INC. MILWAUKEE, WISCONSIN Ref.: Contract No. DA-11-022-ORD-2542 Army Ballistic Missile Agency administered by: THE UNIVERSITY OF MICHIGAN RESEARCH INSTITUTE, ANN ARBOR March 1959

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TABLE OF CONTENTS List of Tables iv List of Figures v Abstract 1 Objective I. Introduction 2 A. The Process, General Description 3 B. The Procedure 5 1. The Analytical Solution 8 2. The Experimental Verification 9 II. Nomenclature 10 III. The Analytical Approach 13 A. The Process 13 B. The Process Variables 17 C. The Velocity Field 20 D. The Strain-Rates Field 23 E. The State of Stress 25 F. The Power 27 IV. Mathematical Description of the Process 30 A. The Sets of Axis and Their Transformation 30 B. The Roller 32 C. The Cone33 D. The Area of Contact 36 V. Solving the Power by the Deformation Theory 45 VI. Simplifying the Equations for Numerical Evaluation 49 A. Integrating the Power 50 B. Computing the Partial Power 52 C. Computing the Complementary Power 52 D. Evaluating the Strain-Rates Ratio i 55 VII. The Tangential Force 57 VIII. The Experimental Work 60 A. The Experimental Determination of the Nature of 60 Deformation B. Measuring the Power and Tangential Force 65 IX. The Results 70 X. Conclusions 72 XI. References 80 Appendices 1. Program for the IBM 650 Digital Computer 81 2. Comparing the Tangential Force for Shear and Bending 98 3o Evaluating the Yield Limit 101 iii

LIST OF TABLES Table I - Directional Cosines 30 II - Assignment List for the Results 86 III - General Assignment List 87 IV - Identification Number (To) Key 91 V - Distortion of Holes Pattern in Spun Cones 64 VI - Cincinnati Results of Recorded Forces 67 VII - Computed Weighted Power and Tangential Force under separate VIII - The Distribution of Strain Rates and - j cover iv

LIST OF FIGURES Figure Page 1 - The Cone and Mandrel 4 2 - The Roller 6 3 -The Positioning of the Roller 7 4 - The Process 14 5 -The Deformation 15 6 - Disc Analogy 16 7 -Displacing the Disc 18 8 -Viscous Flow Around an Obstacle 21 9 -The Zones of the Cone 35 10 - The Area of Contact 35 11 - Line BC 38 12 - The Boundary CDEF 48 13 - Feed Marks 49 14 - Approximating the Torus by a Cylinder 54 15 - Approximated Area of Contact 82 16 - Holes in the Original Disc 62 17 - A Cut of a Spun Cone 63 18 - Directions of Holes in the Cone 66 19 - The Shear and Bending Strains 99 20 - Comparing Shear Deformation to Bending 102 21 - Tangential Force vs Feed 76 22 - Tangential Force vs Included Angle 77 23 - Tangential Force vs "Round-off" Radius 78 24 - Tangential Force vs Roller's Radius 79 25 - Spincraft Experimental Set Up 68 26 - Specimen for Tensile Test 71 27 Flow Diagram for Computer Program 92 28 Stress Strain Curve 104 V~~~~~~~~~0

ABSTRACT The cone, the roller, and the geometry of the operation are described mathematically. A shear type of deformation is postulated, based on experimental evidence. The displacements, velocities, strain rates, and stress fields are computed for "Mises Material," and hence with Mises stress-strain rate law. The power consumed in the operation is computed from the strain rates and stress fields. The expression for the power is in a form that cannot be solved analytically. A numerical solution is therefore presented in graphical form, where the power and tangential force are plotted for a variety at process variables. The numerical solution is compared with actual measured power and forces. OBJECTIVE The purpose of this study is to relate the power requirement and the tangential force for mechanical spinning of cones to the process variables. 1

INTRODUCTION The aim of this study is to help toward better understanding of the process of spinning, and to develop a method for prediction of the power requirements. The power consumed in the process is of interest for the design of new equipment, for intelligent choice of the best machine for the job, and for the right choice of the process variables for a specified job. Furthermore, this power is the predominant factor dictating the interaction force between the roller and the cone. Therefore it can be later utilized for the determination of the forces. Spinning is a metal-shaping process widely used to fabricate pieces having rotational symmetry. In common spinning practice, a pattern having the final shape of the desired piece is mounted on a lathe. A flat sheet of metal is then clamped to the pattern, and while the pattern and sheet are revolving, the sheet is forced back over the pattern by pressing against the sheet with some type of spinning tool. Reduction in thickness of the sheet may or may not take place. Generally the spinning tool is a roller or a heavy wooden stick. The process can be done by hand when the spinning is done by a skilled craftsman who knows by experience how to lay the sheet against the pattern; or it can be done mechanically, in which case the forces are applied by some mechanical system. Many shapes and sizes can be spun; pieces up to 10 feet in diameter are common. Because of the nature of the process, very large reductions in thickness, up to 75% per pass in some cases, can be achieved. With suitable 2

equipment it is also possible to hot-spin. There do not seem to be any restrictions as to deformable materials which can be spun. Aluminum, brass, stainless steel, titanium, and super-alloys, for example, have all been successfully spun. Because of the simplicity of the operation, spinning offers some distinct economical advantages. Lenbridge has shown that spinning is more economical for producing a small number of pieces than deep drawing because of the low setup time and costs. The pattern used for spinning, for example, can often be made of wood, which saves a great deal of tooling expense. This study deals with the mechanical spinning of cones. It does not pertain to the "formability" problem but it does deal with the effect of the process variables on the power and tangential force for successful operation, where the variables are within their permissible range. A. THE PROCESS, GENERAL DESCRIPTION The raw material is a disc of diameter D and uniform thickness S. 0 The disc is mounted on a circular conical mandrel, which is clamped to the head of the spinning machine, and rotated. A forming roller is driven on tracks on the bed of the machine parallel to the side of the mandrel. The outer diameter D1 of the cone (Fig. 1) remains the same as the original diameter of the disc. The cone's thickness is now: S1 = So Sin ao 1l o o

So THE FINISHED CONE THE CONE MANDRELA| III THE BLANK MANDREL. THE OPERATION Li ROLLER /( F UNDEFORMED I REGION I Rf2 1 I iRi Rfl I do THE DIS- - ___ - PLACEMENTS AI IA'I II Fig. I - The cone

This type of spinning is called by different names like forming, hydrospinning, etc. Later on the displacements and strains will be analyzed in more detail. The roller is built from a hardened tool steel and mounted on a shaft with ball or roller bearings. There are many shapes of rollers; some are illustrated in Fig. 2. A common feature of all rollers is the "round-off" portionlof radius roo When thin discs are spun with high feeds, a land is added to the roller to smooth off the feed marks. It is sometimes found convenient to add a front as in shape III of Fig. 2, to push the flange forward The relative position of the roller's axis to the side of the mandrel may differ much in different setups, Some examples are shown in Figo 3. From this variety of possibilities, a common roller (shape 1) which is oriented with its axis parallel to the side of the mandrel was chosen to be analyzed. The changes in the roller's shape and its positioning might sometimes make the difference between a successful operation or failure, However, the strains and stresses during a successful operation do not differ much from case to case, Bo THE PROCEDURE This study can be divided into two parts: the analytical solutions, and the experimental verification. 5

LANDPo LAND LAND - yo FRONT SHAPE I SHAPE I SHAPE Fig. 2-The roller 6

A~ft+ N i N CP C 0 () O T U 0 0 V\ \ N K) ~\ L N N Iv 7

1. The Analytical Solution The geometry of the operation has been described mathematically. The equations of the cone and the roller have also been formulated. The boundaries of the area of contact between the roller and the cone have then been found. A shear type of deformation has been postulated, based on experimental evidence. Two solutions have been derived for the plastic work of deformation: one was based on the incremental theory (Mises stress-strain-rate law), the other, on the deformation theory (stress-strain law). Both solutions were computed for Mises Material, which implies: (1) no elastic deformation, and consequently no volumetric change; (2) no strain hardening. However, the deformation theory has an additional solution for the case of linear strain-hardening effect. a. The Incremental Theory Solution. - The velocity and strain-rate fields were computed using the geometric formulation above. An expression for the plastic work of deformation was then developed. The numerical values are computed using a digital computer. b. The Deformation Theory Solution. - The solution was simply derived by assuming that the deformation was reached by monotonically increased shear from zero to the final state in a finished cone. 8

2. The Experimental Verification Two different types of experiments were conducted. a. First Set of Experiments. - This set was designed to help choose the right type of deformations that take place during the process. Holes were drilled in the original discs and then plugged. Their direction was traced after the spinning. This series provides support for the choice of shear deformation in the analysis. b. Second Set of Experiments. - In this set of experiments the forces and power were measured. This set was run separately in two locations. At the Cincinnati Milling Machine Co., a dynamometer was used. Three components of the spinning force were recorded. A three-dimensional dynamometer at the arm of the roller is connected to a three-channel recorder. The actual forces for a diversified set of parameters was found. At Spincraft, Inc., the power requirement for the d-c motor driving the mandrel was recorded. The power requirement for a diversified set of parameters was found. 9

NOMENCLATURE X,Y,Z - Rectangular coordinate system with: O - origin. R,GZ - Cylindrical polar coordinate system with origin 0 x,y,z - Rectangular coordinate system with: 0' - origin. X,Y,Z,-R,QZ,-x,yz, coordinates of a point in space in the corresponding coordinate system. ai - (where i = 1,2,3) - The three coordinates of the origin 0 in the x,y, z coordinates system. b - (where j = 1,2,3) - The three coordinates of the origin O'in the X,Y,Z coordinates system. a - The directional cosines of XY,Z axis in the x,y,z coordinate system and vice versa. r - "Round-off" radius of the roller. 0 p - The roller's radius. - Half the included angle of the cone; also the angle between Z and z axis. R - Instantaneous radius at which the roller touches the cone. 0 R - Cose's radius (R) at the bend. ~Min. R - The blank's outer radius. oMax. D - Outer diameter of the cone. S - Thickness of the blank. S - Thickness of the final cone. Ca - The cone's material yield limit at uniaxial tensile test. 0 k - Mises yield limit. N - The speed in rpm. F - The feed in ipr. t - Time t - Initial time n - No. of revolutions passed from initial time. U - Velocity in ipm. U -. Velocity vector. UR - Velocity in R direction in ipm U - Velocity in Q direction in ipm. UZ - Velocity in Z direction in ipm U - Principal part of UZ component of the velocity, zp UZC - Complementary part of UZ component of the velocity AUzp - Change in UZp. AUzc - Change in UZC 10

v - Velocity S.t - Circumferential velocity, SA - Feed velocity SR - Radial velocity, R.,.Z - Velocities in R,Q, and Z directions. G, G(R), G(R,G,Z), G(R,@,Z,n), etc. The function of the "round-off" surface of the roller. GR,G, GZGt, - The partial derivatives of G with respect to R,~,Z and t, respectively. Z = H(R,Q,n) - The function of the "round-off" surface expressing Z explicitly. A,B,C,D, E, F - Points on the boundaries of the area of contact. AB, BC, CD, DE, EF, FA - Boundary lines of the area of contact. RA, RB, RC RD RE, RF - Radius at points, A,B,C,D,E and F respectively. 0AB' gBC GCD, GDE,. EF, i FA - ~ as a function of R along lines AB, BC, CD, DE, EF, and FA,respectively. Zone 1, Zone 2, Zone 3 - Zones of the cone (Fig. 9). dv - Infinitesimal volume ds - Infinitesimal area - Strain, Eij - Strain components, where: i = R,Q, Z and j = R9,Z e - Hydrostatic portion of the strain, eij - Strain deviator tensor. - A dot on top represents time rate, and can be applied to any variable J,~iJ, e, eij - These are strain rate, strain-rate components, hydrostatic portion of the strain rate and strain-rate deviator tensor, respectively. [.,- Repeated index represent summation, like Ei e-RR + EQ Q + ZZ i, - Unit tensor called "Kronecker delta" and: ecqal 1 when i = J LJ equal 0 when i 4 j. a - Stress. a - Components of stress, S - Hydrostatic portion of the stress. Si - Components of the stress deviator.;c - Shear stress. 7 - Shear strain. v - Poisson's ratio, E - Young's modulus of elasticity. 7 - Shear angle - Displacement - Strain-hardening coefficient. i - Strain-rates ratio factor. h - Feed marks height. 11

t - Proportionality factor in Mises stress-strain-rate law. I - Strain function. J2 - The second invariant of stress, w - Work per unit volume. W - Power. W - Weighted power. Wp - Partial weighted power. Wc - Complementary weighted power, Wcl - Complementary weighted power consumed over areas 1 and 2. respectively. Wt - Power consumed through tangential motion. W - Power consumed through feed motion. t - Tangential force. FA - Feed force, FR - Radial force a, oa, aRAR - Experimental values of shear angles as prescribed in Fig. 18. 12

THE ANALYTICAL APPROACH A. THE PROCESS For mathematical representation, a model spin was formulated; The geometrical relations between the cone and the roller are shown in Fig. 4. In Fig. 5 the simplified displacements field is described. This displacement field is a "geometrically admissible" field. It is also close to the actual displacements field. The area of contact between the cone and the roller is that area where the outer surface of the cone and the roller are interfering with each other. This area takes the shape of the roller. The strain field, stress field, work, etc., were computed for the cone under the area of contact only. Although deformations exist outside of this region, they are ignored. This omission is discussed in more detail later. As Fig. 5 indicates, the radius of any material point remains unchanged during the whole process. Radial planes in the original disc remain radial planes in the deformed cone, so for visual realization one might conceive of the original disc as being constructed from concentric thin cylinders which are slipped axially to form the final cone (see Fig. 6). The above description does omit one crucial fact. The rings do not slip a full ring at a time. Only the portion of the ring directly affected by the roller is slipped, and as the roller advances it slips a bigger and bigger portion of the ring, until a full revolution is achieved, and the ring as a whole is displaced. 13

,THE CONE c THE ROLLER Iz ZONE 3 I o' B ZONE 2 ZONE I x THE BEND z I t 0o \lO "a F i._ -Th r____ -cs Fig. 4-The process 14

ty I I I I ro~o~so R z - l _l______________ Fig. 5 - The deformations 15

THE INTERMEDIATE FINAL DISC STAGE CONE Fig. 6- Disc analogy

Although at the completion of the deformation the cylinder is reshaped as a cylinder, during the operation the cylinders are distorted, and regain their shape only at the end. This distortion is the shear EQZ and is shown in y-z plane of Fig, 5. Fig. 7 illustrates ithe,way a disc is displaced if it is opened to form a strip. This model cone spinning differs in certain points from the actually spun cone. It will be more convenient tb discuss these differences after describing the experimental results of the first set of experiments. The justification for choosing this model will be given later. Meanwhile, the analysis proceeds for the chosen model. B. THE PROCESS VARIABLES To summarize the process in terms of its independent parameters, we classify the parameters in three groups: (1) the tool parameters, (2) the cone parameters5, and (3) the process parameters. The relative position of the tool toward the cone was fixed so that z axis of the tool was parallel to the side of the cone, and the tool feed is in z direction. Thus the positioning is not a variable, and is included in the standard unchanged conditions of the process. 17

crI NOILISOd -IVNIOI18 < in L0 3aOV-dsla C U NOIIISOd 1 NS19180 WU -c GV1dS II G-'a3s,dsa z NOILISOd -IlVNIO80 0: a GV —— 1-dSI H N Od -l v NI 91IS On < v N/ArJ~OI \ J-'12dSIG < NOIllSOd "IVNI9180 i- -, -,\ ~ - S i ~= —,' _ _ u <a z < I aJQ3VVdSI adSI 18

(a) The tool parameters, using only a tool of shape I in Fig. 2, are: r - the roller's "round-off" radius, and 0 p' the roller's radius (b) The cone parameters are: ao half the included angle, R - instantaneous radius at which the roller touches the cone, R - cone'sradius (R) at the bend, oMin RoMax - the blank's outer radius, S - the blank's thickness, and a - the cone's material yield limit at uniaxial tensial test. (c) The process parameters are: N - the speed in rpm, and F - the feed in ipr. Of these eight parameters, the cone angle (2o), the cone bend radius (RoMin) and outer radius (RoMax), as well as the material (ao) and the thickness (So), are determined by the designer of the cooe. The operation can be performed with a wide choice of the other parameters, r0, Po, N, and F. This study will enable the right choice of these last four parameters as far as the power is concerned. For a good description of the process, see part 1 of Ref. 2. 19

C. THE VELOCITY FIELD (U) Let the cone be considered in the R, 9, Z polar coordinate system. Each point of the rotating disc has rotational motion with velocity: U = 2itRN around the Z axis. No change in the radius R takes place during this process, and therefore: U = 0. Layers at O Z are identical with each other. The volume under the area of the instaneous contact with the roller has an additional component of the velocity (Uz) in the z direction. Let the velocity UZ be computed in terms of the other two components of the velocity U0 and UR and the geometry of the roller. Let the roller be described as the rigid body G(R,, Zt)zO. In this analysis, the disc is the flowing medium, and the roller is the rigid body (see Fig. 8). Stating that the component of the velocity of the flow, Normal to the Rigid Body, is the same for the medium as for the rigid body leads to: GR R + G' + G'Z + G = 0 R~ z t where G = G 9 G = =,Z G - H R G d9' tdZ R d= dZ- R - dt' Q = dt' dt 20

RIGID BODY /VELOCITY iVECTOR FLOWING, MEDIUM RIGID BODY \ Un-THE COMPONENT OF THE VELOCITY OF THE FLOW, NORMAL TO THE RIGID BODY Fig. 8-Viscous flow around an obstacle 21

or U' G = Un -G t where: U - the velocity vector of the medium ZG - the gradient of the surface of the rigid body. dR dQ dz - GR dt Gg -dt Gt - R dR dQ -. dt GZ GZ dt GZ dt GZ Knowing the values U = and U 2iN, dt R dt R one gets: U 2-N at z )Q +t, Evaluating the number of revolutions from time t = to, one gets: n = N-(t - to) and deriving with respect to t, dn dt and thus one continues and gets dz- U =2N + *dn = 2JtN aZ + NdZ = N.2,, z ] dt z - Q an dt ao on I- n Another approach, using Euler's method (the observer is stationary) follows. If the rigid body equation is F(R, 0, Z, n) = 0 22

it gives z as a function of R, 9 and n. z = H(R, @, n) n = f(t) = N(t-to) dz =, dR + 6z dQ + z dn aR 69 an dz _ z dR + z dO z dn U + d z dt R dt Q dt an dt inserting the values d = U = o d U 2iN and d N, one gets t U dt R dt the same expression for Uz as before. U =N [ 2 z+ z L 6Q zn J Summing up the velocity field under direct effect of the roller, one gets: UR = UQ = 2iRN (1) J6zz 6z t n where the derivatives - and a are yet to be defined from the geometry of the roller. D. THE STRAIN-RATES FIELD The rates of deformation3 in cylindrical polar coordinates assume the form: 23

u.A _ 1 _ ug e" U Z BR R R' eg R R a z z = 1 aUR + UQ UJ ( 1 UR aUZ 6z 6R aQZ ( au Rauz ^z= 2 (Z + R~ ) This description is of the general case where the "Euler Method" is used, which means that the observer is stationary. In the specific case of the velocity field of Eqs. (1), where U = 0 R and U = 2iRN, one can develop the strain-rates field further. Because layers at OSZ4SO are identical with each other, we can infer that no variable is dependent on z direction. And therefore a-, zz az and also U = 0, az az Because U, = O0 RR -6 O 24

UR 1 SU., 0 1 0 1 = R+R = -+- --- (2gRN) 0o -i R 6 R R RR 1B B )21 2aN 1U + U U (2gRN). 22-RN 2gN) - RQ __( =~ -+ u.) 2 R 1__ 1 6Uz e _.+ 1 U eez =2 (~as + aUz ~ = Uz 2, dz R69 = R - 6g And thus the strain-rates field is given in the form RZ = 2 aR E 1 aUz (2) 2R B( All other eij = 0 E. THE STATE OF STRESS The Levy-Mises stress-plastic-strain-rate law, with the use of stress deviators and strain deviators, is written: eij = / Sij () where Sij is the stress deviator tensor, e.. is the plastic-strain-rate deviator tensor, and J/ is the proportionality factor; and is a function of the strain rates. From the assumption that there is no plastic volumetric change; it follows that the plastic-strain-rate deviator is equal to the plastic strain rate itself.

e 1. 3 ii A where e is the hydrostatic strain, e is the hydrostatic strain rate, and ie e= + e22 + e3 = 0 (volume constancy);..e = - 0, and 3 n ~ij = eij + ej = eij where c.. is the plastic-strain-rate tensor. Because "Mises Material" has no elastic strains, it follows that the plastic strains are the only existing ones. The stress-strain-rate relations become: Eij =/ Sij (4) Let a term I be defined such that - 1ij (5) | T 2 i3j ij Substituting (4) in (5), it follows that I= ijiJ /92 ijSi Substituting "Mises Yield Condition" 1 2 J2 = Si;Sj, = k in the preceding equation, one gets: 2 2 ijij 2 2 I: f2 k I k 26

and Mises "stress-strain-rate" law becomes eT Y; kl kl _ Eij..=+ Sij = + k Sik - -- i or = + k +i k~kl ekl The stress deviator field is therefore: k CRZ GRz = SRz = + G.2.2 Rz OZ - (7) k CQZ GQz = S +z = +. C RZ + eQz All other S = 0 The sign is -t be chosen from physical considerations. Whether it is plus or minus is not relevant in this study. F. THE POWER Let the rate of work per unit volume be w _ ai ei (8) The stress tensor can be separated into stress deviator and hydrostatic stress. 27

w = (i +S) = iji j i+ i = Siij + i i ii where:- S - the hydrostatic stress S =- o. 3 ll And because of the incompressibility, Eii = 0; w = Sijj (9) Substituting for i.. its value from Eq. (6), one gets 13 w = SijSij klkl k Substituting now the yield condition, J = 1SijSij = k, one gets: 2 - 2 get13 2k2 1 ^i- k |2 k l = + +k E-ijij It is now desired to replace k by the yield at uniaxial tensile test. For uniaxial stress: C- c - Co S = a S = a S = =S 5o 1 2 3 3o All other S.. = 0 Mises Yield condition gives 1 i 4_+ 1 _2 1 a2 ik2 J2 2 SijSi =2 9 +9 9 9 53 = k = o~28 28

It follows that the rate of work per unit volume is: W + a- 0 0 ijij (1 Defining work done on a system to be positive, it follows that: 6w E. With the particular strain-rate field in this case, 2.~ 2 22 i U I U2 If~ o''z)+: ao o' 6vZRZ, EZ 6 = a 0 a -2 1 2 The rate of work for the deformed cone is |W6 4 ) V = fi o t I U | z2 l'~21 SU z ( — ~U2' a21 w = fdV = A / 2 ff \J dz ds vo lume': Z ) 1 av, cll $Surface J volume Because the strains are independent on z direction, one can first integrate with respect to z and get: 0 - 6 /%~.. /~. z'ds J1 | V R )~(R ^. /surface G S f ~U 21 Uz2 W= II, R )~ as dfd@ (11) R~ 29 29

MATHEMATICAL DESCRIPTION OF TEE PROCESS A. THE SETS OF AXIS AND THEIR TRANSFORMATION Consider three coordinate systems: (1) (x,y,z) cartesian coordinate system, with the origin O. The axis z is the axis of cylindrical symmetry for the roller, and the origin is the center of the torical portion of the roller. (2) (XY,Z) cartesian coordinate system, with the origin 0. The axis Z is the axis of cylindrical symmetry for the cone. (3) (R,Q,Z) cylindrical polar coordinates with the same origin 0 and Z axis as the second cartesian system. The directional cosines for transformation from (xy,z) system to (X,Y,Z) system and vice versa can be represented in the following way. Table I - Directional Cosines x y z X cos 0 sin a 0 Y o 1 0 Z -sina 0 cosa 0 0 The transformation scheme is according to the following equation. x. =. +a X Xi a + aijXj j J'jixi 1J 3o

where: i = 1,2,3 denote the column number in Table I, j = 1,2,3 denote the raw number in Table I, ai = denote the coordinates of the origin 0 in the (x,y,z) coordinates system, al = a2 = 0 a = -Fn bj = denote the coordinates. of the origin O in the (X,Y,Z) coordinate system, bl =Fn Sin co, b2 = 0, b3 = Fn cos aCo X1 = x, xi = X= X1 = X, X? = Y, and. X = Z. The transformation is now getting this shape: x1 =x =a l alXX2 +a aX, =+ allX1 + a3X3 = X cos a, -:Z sin a X2 2 Y = a2 + a21X1 + a22X2 += a22X2 = Y x z = a + a3X = + a31+ Xl a33x = -Fn + Xn a + zco 3 511 2 3 3 3 + 511 = -Fn +3 Xsin + o And applying the second of Eqs. (12), one gets: 1. X1 =-X x cos x ro + Z sin ao + Fn sin ao X' Y = y X3 = Z = -xsin + z cos a + Fn cos 35 o The trahsformation from R, 0, Z system to (X,YZ) system is to be performed by: X $R cos 9 Y =R sin 0 Z = Z 351

and from (X,Y,Z) to (R,,Z) is performed through R = Y X -1 Y t Q = cos-1 = sin-1 = tn-l Y Z = Z The transformation from either system to any other system of the three is now given. x = X cos a - Z sin a = R cos @ cos - Z sin a y = Y = R sin ~ z = X sin ao + Z cos ao - Fn = R cos Q sin a + Z cos a - Fn 0 0 X = R cos O = x cos ao + z sin a + Fn sin a Y = R sin 0 = y (13) Z = Z = -x sin ao + z cos a + Fn cos a 0o BR = |x2 + y2 = + (x cos a + z cos a + Fn sin a) + y -1 X = sin-1 Y -1 Q=cos- sin =cos x Cos + z sin a~+Fn sinca |i+y2 fXY2+y 1 (x cosa +z sinaO + Fn sin a )2+y Z = z =-x sin a + z cos a + Fn cos a B.o o oO B. THE ROLTLER The roller is composed of a half torus and a cylinder. The half torus exists for z O. Its equations are: x +Y - Po + z - r = (14) 32

G =[(R cos Q cos - Z sin a) + R sin - po +[R cos Q sin a + Z cos a -Fn] - r = 0 (14) Let the second of Eqs. (14) be referred to as G = 0, or G(R,Q,Z) = 0. The cylindrical portion of the roller exists only for z O. Its equation is: x2 + y2 (r +po)2 = By transforming this equation to (R,QZ) axis, one gets: (R cos Q cos a - Z sin ao ) + R2sin2 - (r +p ) =0 R cos Q cos ao - Z sin ao = (P+ r)2 R2 sin2 Z sin R cos Q cos ao + O + r)2 R2 sin2 ] From geometrical considerations, the plus sign is to be chosen for that portion of the roller that is in touch with the cone. The equation of the cylindrical portion of the roller becomes x2 + y2 -(ro + P) = 0 Z = [R cos Q cos c + V(P + ro)2 R sin2 (15) C. THE CONE Let the cone be separated into three zones. Zone 1 is the already deformed portion of the cone. This zone is bounded by the bend on one end and by the cylinder of the same radius as spiral A is leaving. Spiral A is the trace that point A on the roller leaves on the cone. Zone 2 is the zone 533

being deformed; it is bounded by the speralic areas of radius as that of spiral A and as that of spiral B. Spiral B is the trace that point B on the roller leaves on the cone. Zone 3 is the yet undeformed disc from spiral B and up. Spiral A: R = F (n- 2 ) sin a - (r +p ) cos ZA =F (n ) cos + (r + po) sin a Spiral B: RB = F (n ) sin a - p cos a ZB= F (n - ) cos a + p sin a 2i oro o Zone 2 of the Cone The generating circle AB of the torus is: (x + p0)2 + z2 = r2 y = (17) 2 2 2 (R cos a% - Z sin a + p) +R sin a + Z cos a - Fn) -r = or: = O And thus, zone 2 is: 22 2 (R cos a - Z sin a + p )+ R sin a+ 2 +si Z cos a - F(n - )] r =0 0 0 0 )]2 Because the zone of interest is at the contact area between the cone and the roller, for which - 2ir, one can approximate - by 1 Thus, the approximate equation of the cone at the area of contact is: 54

ZONE 3 B ZONE 2 — ZONE I \/ / / /THE BEND-' SPIRAL B SPIRAL A Fig. 9-The cones zones Area A:(3+4+5+6) G= [J4Rcos ecosao-Zsinao)2+R2sin2e- po + + [Rcos sin ao+Z cos o -Fn2 -r2=0 B Z B D i EE/Area B:(+2) A Rreo B s: ( c+ 2) F. Z e sie area of cotcs +(r +p)+ Rsin2 t Fig. 10-The area of contact 35

(R cosa - Z sin o + +R sin a + Z cos a - F(nl - r = ____________^b~_____- i_ - r = 0 Z- 2Z'[ sin a0 (Rcos a + P) - cos aC (R sin a - F(n-l) + (R cos + p)2 +R sin a - F(n-)]2 - r = Z b + b -c 1,2 - r' For the area of contact Z = Z2 Z = b+ tb2- c where: b = P sin + F(n-l) cos a (18) c = (R cos ac + po) + [R sin a - F(n-1)] - r Zone 3 of the Cone For Zone 3 of the cone: Z = F(n - 2 ) cos ao + p sin a0 + ro F(n-l) cos a + p sin a + r (19) D. THE AREA,OF CONTACT The area of contact between the roller and the cone takes the shape of the roller. Let the area of contact be composed of two main areas. Area A will be that shaped like the torus, and Area B will be that shaped cylindrically (see Fig. 10). Let the boundary ABCDEFA be defined line by line. Line AB is the circle described by Eqs. (17). Line BC The contour of line BC can be found by deriving E = 0

on the equation of the torus G = O. However, because the determination of this line gets involved, it can be approximated by a circle passing through point B around the main axis of the torus (see Fig. 11). The plane of the circle can be described by: z = r cos a o o 2 2 2 And the circle x + y ( + sin ) In the (R,Q,Z) axis, line BC will become: ( R cos O sin a + Z cos - Fn - r cos a =0 > o o o o (R cos Q cos a - Z sin a ) + R sin (pO + r sin)a2= O Inserting Z = — co ( Fn + r cos a - R cos Q sin a ) from the zirst 0OS C~o 0 0 equation into the second equation, one gets [R cos Q cos a - (Fn + r cos a, - R cos Q sin a ) tan ]2 + R2 sin2 Q - o + ro sin a )2 0 This equation of line BC is written implicitly with respect to R and Q. Line CD For Line CD the torus G = 0 cuts Zone 3 of the cone Eq. (19). Repeating the equations once more, one gets: G = [(R cos cos a - Z sin a )2 + R sin Q - P + [R cos Q sin t0 + Z cos a - Fnl - r2 = 2 where: Z = F(n-l) cos a + p Sin a + r 57

THE CIRCLE REPLACING LINE BC A Fig. 11- Line BC 38

Line DE At line DE, the torus cuts Zone 2 of the cone: G =[|(R cos O cos a - Z sin 2 + 2 sin2 - ] 2 + [R cos sin a0 + Z cos ao - Fn]2 r2 =0 u0 O 0 where: Z = b + b2 - c b = p sin ca + F(n-l) cos a c = (R cos a + ) + [R sin a - F(n-l)] - r 0 0 0 - Line EA is on the plane z = 0 From the transformation scheme (Eq. 13) z = R cos Q sin a + Z cos a - Fn = 0 0 ecos - Fn - Z cos ao R sin a o Substituting the right hand of the preceding equation for cos ~ in the equation of the cylinder, (R cos @ cos a - Z sin ca) + R sin 0 - (ro + ) 0 one gets: [(Fn - Z cos a0) cot a0 - Z sin a 2 +R [1 - ( ~ os ) ]a 0 = (r +'P)2 -o2 Fn-z cos a o [(Fn - Zos Ca0) cot a - Z sin a0 ( Fn- co ) 2 0 -(..... sin %o = (PO + )2 2 [Fn cot a - Z(cos cot aC + sin a )1 sin2 - (Fn - Z cos a ) =[('o + r)2:R21,sin 2 39

[Fn cos a - Z(cos2 a + sin a0)] (Fn - Z cos ao) =[(Pop + ro)2 R2 ].sin a2 (Fn cos a0 - Z) - (Fn - Z cos a)2 =[('Po + r)2 - R2] sin a (Fn)2(cos2 a - 1)-2Z (cos a - cos a ) Fn + Z2 (l-cos a ) =[(:P + r )2 2] sin2 a s2 n2 2 -(Fn)2 sin2 a + Z2 sin aO = [( p + r )2 - R ] sin2 a Z2= (.P + r0) -R + (Fn) Z - + 3('PO +r) -B2 + (Fn)2 Fn- cos (Fn - r )2 - 2 2 Cos = 0 -+Fn cos0 cos e =. o___ =. -- OSR sin a R sin ao Fn- J( P + r) - R + (Fn) cos a cos 9 P= -. --- -- lAE BR sin a Line FE Line FE is the intersection of Zone 2 of the cone with the cylindrical portion of the roller. The cone is: Z = b + Vb c where: b = P sin a + F(n-l) cos a 4o

c =(R cos a + p) + Rsin a -F(n-1) r 0 o 0 0 And the cylinder is::1R co — o a R (cos + cos in + sin Cao 0 0 Solving the equation of the cylinder for cos 0, one proceeds: (r~ + p )2 _ R2 sin a - R cos Q cos a0 (ro + po) - R + R cos2 = R cos ~ cos <a + Z sin a 0 - 2RZ cos Q sin a cos a o o 0 0 R2(1 - cos2 a ) cos2 +c 2 RZ sin cos cos - (Z sin ) -R2 + (ro + Po') = 0 cos Q = -RZ sin a cos ao + (RZ sin a cos a) -R2sino [(r + p ) -R-(Z sin a ) R2 sin2 a cos = -Z cos Q0o + (Z qos ac)2 -(p + r)2 + R + z sin a R sin a 0 2 2 2 -Z cos C + Z + (p + r ~c ~~Ros,. o _sin cx _ R' sin Q 41l

where: Z = b + b2 - c b = p sin a + F(n-l) cos ao c = (R cos aC + p)2 + [R sin aC - F(n-l)]2 r2 Line FA ~ = O After defining the equations of the line ABCDEFA, the intersection points A, B, C, D, E, F of these lines are to be defined. The radius (R) of these points will be determined. From either of the two equations of the lines passing through any of these points and its radius (R), the angle Q can later be computed. Point A: RA = Fn sin a - (r + p ) cos a Point B: RB = Fn sin a - p cos a (21) O O o Point C: The value of the radius RC can be solved from the intersection of the circle BC with Zone 3 of the cone. Equation (19) of Zone 3, transformed to (x,y,z) axis, will become: Z = F(n-l)cos a + P sin a + r =-x sin a + z cos a + Fn cos a 0 o o o o o The intersection of Zone 3 with the circle, for which z = r cos a, will give: F(n-l)cos a + Po sin a + r =-x sin a + r cos a + Fn cos a 0 0 o o0 0 0 o -F cos a + P sin a + r (l-cos a) = - x sin a 0 0o 0 0 0 o x x= i (F cos - p sin a r sin 2 ) sin a o o o x = F cot a0o - (p + r sin a ) 0 0 4

And from the equation of the circle BC, one gets 2 2 2 y = (P + r sin a ) -x o o o Point C in(x,y,z) axis is defined by x = F cot a - (P + r sin a ) o o o o Y = ( ro sin _x2 z = ro cos a From the transformation equation (13), one gets R cos Q = x cos a0 + z sin a + Fn sin a o o R sin 8 = y R = (x cos a + z sin a + Fn sin a2) + y 000 R = ( (x cos a + z sin a + Fn sin a ) + y2 = I 0 0 0 where: x =\ F cot a - (p0 + ro sin ao) y = (P + r sin a )2 x2 z = r cos a o o Point D RD is defined by the intersection of lines GED and CD Point E RE is defined by the intersection of either.two of the three lines: 0FE', ED and GAE. Point F RF = F(n-l) sin a - (P + r ) cos a. 43

Up to now, n indicated the position of the roller. For practical applications, a term whose physical meaning is easier to visualize should be used. Let the radius R of point A be used, and denoted by Ro: R = Fn sin a - (r +Po) cos 0 A 0 o 0 0 Fn = sin a R+o +(o +P) a This value of Fn is to be inserted wherever it appears. It is now appropriate to discuss the deflection of the neglected zones. 1. The Zone over Line BC, (R>RBC) The difference between a perfect disc and the actual shape of Zone 3 of the cone diminishes rapidly as R increases over RB. Therefore, neglecting the plastic deformation work on this zone is justified. 2. Lines CD, DE, EF The actual strain rate is finite and thus does not confirm with our picture (Fig. 12). The strain-rate field is as described except on that line and very close to it. Therefore the work under the area of contact is computed correctly. The work at the discussed zone (line CDEF) is only estimated. The velocity and strain-rates field From the equations of the roller, one can compute the values of the following terms: [ 6Z 6Z Uz = N.[2na.] (22) 44

. 1 a Uz bUz Eez - 2R These terms are long and bulky. Further, under the torus they are expressed as functions of R, Q, Z and n while z is implicitly deduced from the equation of the torus. Altogether, it is possible by the use of these equations and double integration by numerical methods to solve for the work of deformation. But as this requires much labor, some approximations will be made which will not affect the results appreciably and be less laborious. SOLVING THE POWER BY THE DEFORMATION THEORY Let a hypothetical cubic block of unit length be exposed to a pure shear. Clamps The Loadi Block I Let the load be k =, and the shear angle be 7. The distance traveled is g = 1.' tan 7 = tan 7, and the external work of deformation is a W = k tan 7 = r tan y. This is the work of deformation for a unit volume. The volume worked on the cone is v = 2WRN S. F sin a [ Mmn -] 45

and the relation between ao and the shear angle 7 is 7 = 90 ~-. One can now write the power for the spinning of a cone as W = 2rRN S F sin - tan 7 W = r Ca o RN S F sin a cot a 2 o O W = r S c S N F R cos a Let now the same approach be introduced to compute the power for spinning of strain-hardened materials. Let the strain-stress curve and the shear angle and displacement be described as below. The work per unit volume will be cos (25) 7=0 Let the stress-strain curve be approximated by the straight line (r=.. + b7; 46

the integral then becomes w= (IO+b b - C d7 7 + b S2 - o ifS' Cos7 2y~ cOs2y C OS2 7 =o 7 =o 7=o w =, o.tan 7 + b (7 tan 7y+ in cos 7') = ~ tan 7 + b -tan 7 + ln(cos y)] 3 And substitutin 7 = - o one gets: w = cot ao + b [(- - ao) cot ao + in (sin a ) This is the work per unit volume. To find the power, one multiplies the work per unit volume by the volume machined per minute. ~ W = 2nRN SO F sin ac f C cot ao + b [(2 - a) cot a + In (sin ca ) Summarizing the results for perfect plastic material and linearly strain-hardened material, one gets 2 W = r O S0 N F R cos a - for perfectly plastic material W = t a S NFR cos a + 2 a S NFR b [ ( - a )cos ac+ sin aCln(sina )] (24) For linearly strain-hardened material where: = b 7 I"~~~~~~~~~~~~~~~~~~~~ —b

x INFINITE SHEAR (STRAIN RATE) zi Fig. 12- The boundary CDEF 48

SIMPLIFYING THE EQUATIONS FOR NUMERICAL EVALUATION Repeating here Eq. (11) for the power of plastic deformation (.by the incremental theory), nW = a ~ tJ Ri( Ra * 0) 1dRdQ (11) {Y oR Let aweightedpower (W) be defined such that W = oN' and W = NW; (25) then the weighted power is: f %U2 * / 1 f J R 1|( a (dRdO (26) o R Let = (9) = - /. Z so that the equation for the RZ LReQ / R weighted power can be written: Wi R /ERZ2 + dRdQ = f RiT RZ dRdZ (27) Q R Q R The work of deformation done by the cylindrical portion of the roller will not be computed. The duty of this cylindrical portion is to smooth out the feed mark (Fig. 13). These feed marks are minor in size as the following example of a practical case shows. Fig. 13 5- -__ Feed Marks if 49

2 F2 h = ro- ro For example, if r =.250 in, and F =.040 in, one gets h =.250 -.2502 -.0202= oo8 in., and when reducing F to a half or F =.020 in., one gets h =.250 -.250 -.010 =.002 in. The amount of the displacement caused by the cylindrical portion of the roller is thus neglected. (The work of deformation caused by the cylindrical portion was computed for a number of cases, and found to be negligible. This part of the work is not included in the report). From here on the equations deal only with the torical portion of the roller. It is to be expected that ~ is a function of R and Q. For @ = 0, one gets 0 = O. For simplicity, and to be able to perform an integration, it will be assumed that T is independent of R and of Q. The constant value of S will be chosen arbitrarily as that of T at the point E. The distribution of T as well as that of ~RZ and ~Qz along line R = Rg and along line ED are given in Table VIII. INTEGRATING THE POWER Assuming a to be a constant, one can take the square root outside of the integral and Eq. (27) becomes: 50

W - 1 I R; RZ dRdQ R ddQ.aUz One can perform the integration IR aR dR by substituting R = U and bUz a- dR = dv. (R JUz. But a close check of the equation reveals that, while is changing dR appreciably for every minor change in R, R itself can be regarded as constant R = R. So: JR dR BE Ro aBR dR = R J, The weightedpower becomes W = R B AUzdQ (28) Recalling that U = N [2-t a + a6Z -Q + Let: UZp = 2iN N (29) Uz N ZC a (N2 and thus: U Jz = iJZP + tU JUzp = 2tNN A Z bo aTzc = N A z Let: The totalweightedpower W1 = Wl + Wc 51

where: the partial power is W = A?- Ro A UdzpQ -p N 9 ZP and the complementary power is W =- R' J A UzcdQ 0 B. COMPUTING THE PARTIAL POWER W R (i + __ _ z p N ozo'p = Ri / Ro 2 A ^ aQ p = 2x Ro * + AS where: the total A Z = F cos a, and therefore W = 2Tj R -1 +' F coso (30) p 0 0 The similarity of this partial power to the power computed by the deformation theory is to be noticed. For V= O0 W is the work of deformation p computed by the deformation theory, Eq. (20). C. COMPUTING THE COMPLEMENTARY POWER The complementary power is W = AUd = T dR (51) c'N ~O 0 u ff o J n The value of 7- can be evaluated from Eq. (14), G = 0, of the torus. However, an approximate expression that will be as good for numerical evaluation can be developed. 52

Let the torus be of infinite radius (P') (see Fig. 14). o The torus which is now a cylinder as described by the circle of radius r is (a - z) + (b x)2 r 2 o a z= + r 2.(b -x)2 z =a - (b - )2 Z =a + r - (b - x)-a+ 2 (b - R cos )2 Z =Fn cos aC + P sin aO + r2 -[R cosQ -(Fn sin - P Cos a ) (2) Deriving Z with respect to n, one gets z = F co o + cosQ -(Fn sin ao-Qo cosa,)j. F sin ao nr 2 _ tR cos 9 - (Fn sin aGo-Po cos a )32 )z bz bz One notices that: A (R ) (R And therefore: A - = F sin a 5 R _ cose -(Fn sina0o- cos a ) ro2-[Rucos -(Fn sin o-Po cos ao)]2 R1 cosQ -(Fn sin o-p-o cos Co) where subscripts.ui denoteupper bound of integration and 1 denotes lower bound of integration. 35

b/ / /PFn cos ao a/ Fn cos ao po sin ao a= Fn cos cro + os in ao _ Fig 14-Approximating the torus by a cylinder 54

And so, Eq. (31) for the complementary power becomes Wc = R F sin c o. +s o r2-uoso - (Fn sin a. - Cos a) Boundary of the area r -[RucosQ-(Fn sin cos o)2 of contact RIcosQ -(Fn sin ao-P cos ) Csl (33a) 2-[RlcosQ-(Fn sin a -.P cos ao)] These integrals can be transformed to take the form of elliptic integrals, for which there is a standard procedure for a solution.4 However because the boundaries of integration have to be found by a numerical method (namely, a computer), it was found advantageous to do the integration on the computer also. D. EVALUATING THE STRAIN-RATES RATIO Let the approximated torus be used [Eq. (32]: Z =Fn cos a0 + po sin ac + ro -[R cos @ -(Fn sin a -p cos a ) The velocity is U_ = 1 dz = F cos a - cR cos -(Fn sin a -pocos C)][-R sin 92t-F sin (4) N N dt 0~ lr2-rR cos Q -(Fn sin aO-p cos a )c And evaluating 1 Uzz and 1 U, oe on tinues: N aR N ag 55

(Fn sin a-p cos c) = / r -[4R cos Q-(Fn sin ao-po cos ca)]2 /r2-(R cos- )2 = Thus the derivative of -- from Eq. (34) with respect to R becomes: N 1 UZ RF sin R s 2 2i sin ] + (R ) 2 sin 2(C,)2 (R cos Q - $)2(2) R sin Q + F sino, )cos 1 oYz 1 N aR = ()3 cos O (2JR sin 0 + F sin o)[( + (R cos - ) ] + (R cos 0 - D) 2g sin 0 (T)2 1 a _Z 1 f r2 cos ~ (2iRR sin 9 + F sin ao) + 2g sin ()2(R cos Q - ).N TR 0 0 And the derivative of Uz with respect to 0 becomes: N a Zum =f lfR sin Q[tR sn + sin o] + RF cos Q -] 2iR cos (R cos - p)2 r[2iR sin 9 + F sin a0] R sin 0 (71)2 Z ~= i* (- sin Q [2n R sin 0 + F sin aol (r)2 +(R cos 0 - )2 + (R cos 0 - p) 2iR cos 0 (')2) 1. a1 Z = frO2in [2rKR sin 0 + F sin a0o]+ 2r cos Q (R cos e-S) (f)2 56

Recalling that by definition: 2 2 /1 az =!(z /)= (R RQ RZ \Uz/)R one gets: 2t cos 0 (RCos 0 - ) (Rj)2 - r2 sin Q[2R sin + F sin a 2i sin 0 (R cos Q-P) (()2 + rO2 cos 9 [2tR sin s + F sin a0 (35) where: = Fn sin ab - po cos ao (fi= f ro2 -[R cos Q - (Fn sin ao - po cos Co)]2 THE TANGENTIAL FORCE The numerical results will be presented in a graphical form. The power is a linear function of the following parameters: yield limit (%o), thickness (So), and speed in RPM (N). It will therefore be better to introduce a weighed tangential force and plot its dependence on the other process variables. The number of graphs will thus be reduced, and the real tangential force and power can be found from those plots and computed for a variety of materials, thicknesses and speeds. First it will be shown that the greatest portion of the power is absorbed through the tangential component of the force acting between the roller and the cone. Let the force be resolved to its three following components: Tangential Force (t), Radial Force (FR) and Feed Force (FA) (see Fig. 1). The power consumed 57

can be composed of the total power delivered by these three components of the force separately by each cone. In general, the power is the force multiplied by the velocity directed parallel to this force. (The dot product of the force by the velocity). This can be expressed as W = t.St + FA.SA + FR.SR The roller does not move in the radial direction, or SR = 0. It follows that W = t.St + FA.SA = Wt + WA where:t = t.St - The "tangential power" where: WA = FASA - The "feed power" The "circumferential velocity" is St = 2iRN, and the feed velocity is SA = F. N The ratio for the "feed power" to the total power is: WA WA _ 1 Wt = icR t WA+Wt + 1+ 2 t WA F SA A typical case follows (see Table VI reading M10). F =.028 inch/R R = 1 inch FA = 1640 lb t = 100 lb WA= 1 = 1 = 1 =.068 2, 100 1 + 13.7 14.7 1 +-.028 1640 58

The power consumed by the "feed force" is about 6.8%3 of the total power. Let the feed power be neglected and the work consumed be regarded as done entirely by the tangential force. One now gets: W = t,St = 2'R'N't And the tangential force is t = W OS W'lb. 2itRN 2'5 irR Let a weighed tangential force (t) be introduced such that: t''SOo Equation (36) summarizes the relations between the power, weighed power, tangential force, and weighed tangential force. *1 t -0So a-o.So t 2rtRN 2 IcR W W w - t t' 2rtRN-OcS0 2 g' rtR a S (36) W = Q NW = 2rtRNoSot = 2JtRNt W = 2 Sj Rt' 2 F R t = So 0 S 59

THE EXPERIMENTAL WORK Two types of experiments were conducted. One was designed to study the nature of deformation that takes place during the process. The other was designed to measure the forces between the tool and the work during the process. In usual practice during many spinning operatings, the flange (Zone 3) of the cone is distorted. This distortion is very undesirable because a highly distorted flange cannot be worked into a cone and the operation stops or the cone is cracked. This distortion is discussed in Ref42,5,and 6, and some explanations of the factors governing this distortion are given. It was experienced in this study that the predominant factor governing distortion is the thickness S1 of the cone. The thickness of all cones spun and analyzed in this study was kept uniform and according to the equation S1 = So sin ac, which will be called the "sine law." In this case the flange is always straight and undistorted. A. THE EXPERIMENTAL DETERMINATION OF THE NATURE OF DEFORMATION The study of the nature of the deformations took two tracks. One was a study of the microstructure of the deformed cone, the other, of the general pattern of the deformation. The microstructure study showed that the outer surface of the cone was more deformed than the side facing the mandrel. At the same time higher temperature existed at the outer surface and hence more recrystallization occurred on this side of the cone. These phenomena are discussed in Ref. 5 (p. 6). 60

The general pattern of the deformations was first studied by putting grid lines on the original disc and then following the direction they took after deformation. The results of this study are given in Ref. 5. This method showed only the displacement pattern over the surface, ignoring the depth. Later a better technique was used, as follows. In the original disc.0125-in. holes were drilled and plugged with "sculp" metal (see Fig. 16). After the cones were spun, the metal was carefully cut and filed until the holes were revealed. From the direction of the plugs, a three-dimensional deformation picture was constructed. A typical picture of the holes is Fig. 17. This cone was spun and checked by Cincinnati Milling Machine Co. as part of a study conducted there. A top view on the radial line of holes, shows the shear ciQ of the cone (see Fig. 18). The angle ac indicate by how much the outer surface slipped over the inner surface of the cone. A listing of a, AR and a for a variety of cones is given in Table V. L6aR was measured at different radial distances on each cone. a. was considered to be zero in the analytical study and compared to (90-o0), which is the main shear, it is seen to be small enough. No explanation has been found of the odd values of tc4, which prevents inclusion of this distortion in the analytical approach. The analytical approach assumed a pure shear cRZ which means that ca=O. This assumption does not hold. For big ao, it seems that the deformation is closer to pure bend. For smaller ao, the deformation is closer to pure shear 61

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A cut of a spun cone:::::'-: (courtesy of Cincinnati Milring Machine Co,) 63

Table V - Distortion of Holes Pattern in Spur Cones. (original thickness So =.081") See Fig. 18 Cone Cone Radius Average Number Mat. Angle,c R A c | 1 90~-a Degrees Degrees Inch inch Degrees Degrees Degrees 4 3,50 1-A-1-R-.5 Al 31.5 14.5 6 25 65 4.50 ~ 1-A-l-R-l Al 31.5 4 4.5 13 22 68 4.5 1.75 5 2.75 4 6 7-A-1-R-.5 Brass 31.5 4.5 2 12 18 72 7-A I I 1-RI - 0 t - 9-A —R — Brass 31.5 5375 5.5 4.25 3.5 17 33 4.75 1.75 l-A-20-R-1 Al 42.5 0 7 35 55 1-A-20-R-5 Al 42.5 0 2 45 45 9-A-3-JJ-R-1 Brass 54 3.25 2 3 23 67 3.75 2 4.25 3 4.75 2 64

than to bending. It is felt that by assuming pure shear, a computed value for the power will be fairly close to the actual one (see Appendix 2), B. MEASURING THE POWER AND TANGENTIAL FORCE This set of experiments was done in two separate places, one at Cincinnati Milling Machine Company, and the other at Spincraft, Incorporated. Cincinnati Milling Machine Co. was conducting a study of the spinning operation for quite a long time. Equipment, already constructed, consisted of a three-dimensional dynamometer connected to a 3-channel recorder. The roller is mounted on the dynamometer, and thus the force between the roller and the cone is measured. It was agreed that Cincinnati Milling Machine Co. would run aset of tests designed especially to check the analytical approach developed in this study. The results are plotted in Figs. 21, 22, 23, and 24. In this study only the tangential force was analyzed. At Spincraft the power consumption of thed-c motor,which drives the main spindle, was measured. This was done by recording the voltage and current through the motor (see Fig. 25). The data and experimental results are listed in Table IX. The analytical results, compared with the experimental data.will be given in the next report. One can write the power consumed by the motor, to be: Wt = Wo+ where: Wt is the power measured at load, Wo is the power measured at no load, W is the power of deformation consumed by the cone, and h is the efficiency factor. 65

en Q C3.a c: _) U a. Q T- ~ ~6C w (I:~~~~~~~~~~~~~I 0 w0 IL) /CL ~ ~ 66 U-J K~~~~~~ 0z I~~~~~~~~~~~J~Q a~~~~~~~~-J ~ ~ ~ ~ ~ ~ L

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iL -- iiiitiiii"~ iii ^r~~~~~~ \&~~~~~~~~~~~~~~~~~ C \ ~*o ^**^' C> I.................r~ - -. C o o) cr E 0 OD ro ^ Id ^ | --'~ h-. cr L;O C ( Z a. OD_ ^ rr) ro:p <n. O _ a: co o z (n (f) w a. Lu E:n,. cr -----------— y cc-~ — >00, C,;S ct~~~~~~~~~~~~~~i

Table IX with Spincraft experimental data and the plots of Spincraft experimental data against the theoretical answei. are not included in this report. These results will follow at a later date. Table VII - computed weighted power and tangential force - and Table VIII the distribution of strain rates and R r are provided under separate cover. 69

The same equation can be transformed to: W t,-wo The efficiency ) for the experiment at Spincraft was arbitrarily chosen. For the analytical solution, a value for the uniaxial yield limit is required. From each cone spun in this set of tests, a specimen was cut as Fig. 26 shows, and tested on a tensile test machine. A typical stressstrain curve is shown in Fig. 28. An average stress was chosen to be the yield limit (see Appendix 3). RESULTS The nature of the deformations has been discussed in Ref. 5. Table V gives the deformations as had been found by the holes technique. The effects of each of the parameters, Feed F, cone included angle 2ao, cone radius Ro, roller "round-off" radius ro, and roller radius p on the power and tangential force, are given in the following graphs. Knowing those five parameters for an actual case, one finds from the suitable graph the value of theweighted tangential force. By a simple computation, the tangential force and power are obtained: t = a So t W = 2xRN t = 2iRNaoSo t 70

t _(-3/16!-1 1/4I 3/8" 1-1/8" min 4" l Fig. 26-Specimen for tensile test 71

where t - the tangential force, Co - yield limit of the actual cone at uniaxial tensile test, So - original disc thickness, t - weightedtangential force as found from the graph, R - radius of cone where the force t is applied, and N - speed of cone in rpm. CONCLUSIONS (1) The aim of this study was to find the power consumed in the mechanical spinning of cones. The information gathered about the process gave the geometrical picture, and some vague idea about the general pattern of the displacements. For the analytical study, a displacement field was postulated, which gave the strainrates field and the stress-deviator field. The strain-rate field does satisfy automatically the compatability conditions. Because of the very complicated boundaries, which require numerical methods for determination, no attempt was made to check the stresses for satisfaction of the equations of equilibrium. No load-boundary conditions are given, and therefore no steps were taken to satisfy any load distribution. Only the geometrical boundary conditions are satisfied. Although the above approach is far from giving a rigorous solution for the stress and strain fields, it does give a fairly good approximation of the required work of deformation. 72

(2) In addition to the solution by the incremental theory, a solution by the deformation theory was given. It is seen that as the radius (R) of the cone increases, the incremental-theory solution approaches the solution given by the deformation theory. This is explained by the following reasoning. For the incremental theory, in the numerical solution, it was assumed that the ratio V between the two nonzero strain-rate components was constant. This assumption gave a fairly good value for the power. By assuming5 to be a constant, a "radial load" was postulated. A "radial load" is one in which each component of strain (and stress) increases with time at the same rate as the others. For this load it has been proved that the incremental theory is in complete agreement with the deformation theory. In the deformation-theory solution, it was assumed that cRg = 0. As the radius of the cone Ro grows bigger, the maximum angle 0 of point D on the area of contact gets smaller. The circle R = constant approaches the cord on which eRQ = 0. It is therefore to be expected, as R grows and ERg approaches zero, that the solution according to the incremental theory will approach that of the deformation theory. The deformation-theory solution is therefore a good quick way to find the approximate power requirements. (3) The effect of the parameters on the tangential force can be summarized as follows. 73

For any set of parameters, the cone is spun from a minimum Ro at the bend to a maximum Ro at the outer radius. As the radius gets bigger, the tangential force is decreased, but never gets less than the tangential force predicted by the deformation theory. Let this minimum tangential force be called the "most efficient process force." Whenever a change of parameters brings the tangential force closer to the minimum, it will be regarded as making the process more efficient. (a) The tangential force is linearly dependent on the yield limit aO and on the blank's thickness, So as long as the thickness is much less than the minimum diameter spun. (b) Increasing the feed (F) will decrease the efficiency. (c) The minimum tangential force is proportional to coso. The more deformation is introduced (smaller oa), the worse is the efficiency. (d) As the roller round-off radius r0 approaches zero, the efficiency approaches its best value. As ro gets bigger than about 1 in. in practical conditions, a further increase does not change the efficiency very much. (e) As the roller radius po approaches zero, the efficiency approaches its best value (see Ref. 2, Part II, p. 255). Here for values of 6"< p <10" the changes in efficiency are very small. Each one of the effects mentioned above is more noticeable as R gets o smaller. For very big R, the efficiency is approaching its best value for any value of the other parameters. 74

(4) For the numerical evaluation of the expression for the power, many approximations were made. The original expression could have been evaluated for the exact form of the roller. This would have required a much more complicated program handled by a bigger machine than the IBM 650. This study was mainly conducted to find a way to evaluate the power and to prove the soundness of this way. The shorter and simpler program proved it. (5) In the set of tests to find the forces, a dynamometer was used in one case and a power measurement in the other. The tangential force measured by the dynamometer showed a wide spread or irreproducibility. The most likely reasons for that spread are: (a) the conditions might not have been under full control, or (b) the dynamometer might be inaccurate. At any rate, if further tests are to be run, this source of trouble should first be located. The dynamometer readings were accepted only at the smaller radius Ro = 1 in. As the roller advanced, it started to rub on the flange. In regular operation this rubbing is undesirable and can be eliminated. On any further test, care should be taken to get rid of this rubbing, and have a record of the power for the full length of the operation. 75

THE EFFECT OF THE FEED (F) ON THE WEIGHTED TANGENTIAL FORCE t' --- Predicted by the simplified deformation theory Predicted by the incremental theory 0 Cincinnati experimental data o= 7 in., ro=.250in., ao=35~.026 wgi gi5l. iN.l iN ~~~~~~~~I.OIC0~~~~~~~- 10 F. I I I'iwi~''1'" =*~~ - 17:-.;u,:.,o'_ — ~7 _: -- -I -- K i -tj o 2 0.010.020.030.040.050 FEED PER REVOLUTION (ipr) The tangential force t= o *So- t'(lb) The power w 2irR(Nt=2~roo-So Ro N t( 1^n) Fig. 21-Tangential force vs. feed 76 W ~ ~....~_~.. ~, -.4 —~ z~~~~T 0 0I 0. 02 14 0.0.5 FEDPR EOUIOjir ThXagnialfret=Oo~S't(b The powe~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~r ~~~~~~~~w=2rot2oos-oN.' b!. L~~~~l

THE EFFECT OF THE INCLUDED ANGLE (2ao) ON THE WEIGHTED TANGENTIAL FORCE t' - Predicted by the simplified deformation theory Predicted by the incremental theory o Cincinnati experimental data P = 7 in., ro =.250 in..026 i |!,-!-:'.:. t.i T -I J- 4;,.. 00-;;! —-^l':- — 1;;- -t i'i::...... "^fc t i'' *I *:'<>>- ii Mi l i: -. - - _- t-; — J,r rl-J! Wi. -J t: JA71:-.-I-^I^i h&^1 4 __U____L __ _O1 i ji^': N.: -'i:F=.028(ipr): —:;-':":. "-_ -'....^...;t.TI' i —':, ii —'-~-T O -0- Ro N. I,,.0 1, —:: _:T. z T i'.':-;t' i~': —:: ^I77_171^iN iN7ii7 0 10~ 20~ 30~ 40~ 50~ 60~ 70~ 80~ 90~ HALF THE INCLUDED ANGLE ao The tangential force t1croSo *S t (Ib) The power ~w-2xRoN t^7rCToo IR -c N -1 - Fig. 22- Tangential force vs. included angle 77 L ~ ~~- ri -—: -l. 4,, —

THE EFFECT OF THE "ROUND-OFF" r0 ON THE WEIGHTED TANGENTIAL FORCE t' ---— Predicted by the simplified deformation theory -Predicted by the incremental theory 0 Cincinnati experimental data Po 7 in.,a= 35~.0262 1_7 =t+ *0S.020 +'-i-,' i -iiii-: t7 i- — F-l-i- it: z _, - i — —'-,l *' i t-" ~-"''-i'- _ k-,.4 t~ ttit" ~~~-;i t ti -, - i -, t —- I,I- * I i -4- t L: ^ 0i —I.-' tIt! I i L I I I I I I 1 - I. 5 1f, —i — 1.. -I —.030 ( Ii I -: T ep-.r ] —- _ _..- -I J ~".010-::. L1Ro I I Fig. 23-Tangentil force vs. Round off rdius 78 0 I 3 ROUND -OFF r0 (IN) i-1 T-I-1 i ~ — i ttr 7-'-t ~ ~7.I:-4 ___..:- _;I i +~" ii 77 Z t — -— l-i i-i t —; -f-L~ ~~~t4_ -c-,-l —:-1-0

THE EFFECT OF THE ROLLER'S RADIUS po, ON THE WEIGHTED TANGENTIAL FORCE t' ---- Predicted by the simplified deformation theory Predicted by the incremental theory o Cincinnati experimental data ao =35~ ro =.250 in. I I I -II i I 2 3 4 5 6 7 8 II 12.020 I - I ~THE ROLLER'S RADIUS p(i N.) The tangential force t0 t'(b) Fig. 24- Tangential force vs. rollers radius 9 I I I I I I t F 24 -Tngt f vs ro roius Lum L 011 All o 2 3- 4 6 7 8 9 10 11 12 THE ROLLER'S RADIUS pr IN.) The tangential force t=a-0so-t'(lb) The power wit= 2vRoNt=2-r arO so Ro N -t ( r 79

REFERENCES 1. Lengbridge, J., "Economics of Spinning and Drawing,' Tool Eng., 30, 89-94 (1953). 2. Reichel, H., "Roll Spinning of Cone Shaped Aluminum Parts," Fertigungstechnik, 8, Part 1, No. 5, 181-184 (April, 1958); Part 2, No. 6, 252-260 (June, 1958). 3. Rouse, H. (Editor), Advanced Mechanics of Fluids (John Wiley and Sons, Inc., New York, 1959), p. 204. 4. Von Karman, T., and Biot, M. A., Mathematical Methods in Engineering (McGraw-Hill Book Co., New York, 1940), pp. 119-1530. 5. Avitzur, B., Carleton, W. D., Floreen, S., Hucke, E. E., and Ragone, D. V., Mechanically Spun Cones, Univ. of Michigan Eng. Res. Inst. Report 2621-4-P, Ann Arbor, June, 1958. 6. Feola, J. N., Experimental Analysis of Shear Deformation, unpublished M. S. thesis, Syracuse University, January, 1955. 7. Perlic, A. S. Smith, J. W., and Van Zoeren, H. R., Internal Translator (IT), A compiler for the 650, Univ. of Mich. Statistical Res. Lab., January, 1957. 80

APPENDIX 1 PROGRAM FOR THE IBM 650 DIGITAL COMPUTER It now remains to solve numerically the following sets of equations. From the deformation theory W = v.,taOSo NFRo cos a W = 2R F Ro cos a 0~~~~~~S ~(20) t = aoS F cos a t = cos Cx o 0 And from the incremental theory: W = 2t F Ro l +' cos Co + fl + RF in,o(f Rucos-(Fn sina -p0cosa ) RlcosQ-(Fn sinao-p os )os 0 ~J iro-RucosQ-(Fn siO CosO -rRccoso-(Fn sin si c-pOcosaCO) Boundary of the area of contact where: - 2i cos(R cosQ-p)(WT)2-ro2 sing [2tR sing + F sina] 2n sino(R coso-p)(~f )2-rO2 cos [21R sinR + F sinao] P = Fn sin a0 -'Po cos a FT* = I r02 -tR cosQ - (Fn sin aO-pO cos a 0)]2 W= aoS N W.1 s W t:_ 2C t = a, S t' 0 0 81

R /A A A F Fig. 15 - Approximated area of contact 82

For the integration, the boundaries of the area of contact are to be computed. Because the cylindrical portion of the roller is disregarded, area B is omited, Let the remained of the area of contact be approximated by the configuration of Fig, 15. The integration will be done for Area 1 and for Area 2c For area 1: Ru = RB = Fn sin aO-Po cos a C R1= RA = R u = GE Q= 0 O1 =0 For area 2: RB = = Fn sin ar-p cos ao R = RED(Q) = R of line ED as a function of ~ u D Q1 = E For simplicity (and to save drum locations on the computer), let R1 of area 2 be linearly dependent Q. a - i)h R =R + (RB - R) I o uB u 1 To solve for QE and QD, one solves the equation of the torus: G = 0 where: Z = b + b2b = o sin aO + F(n-l) cos a~ C = (R cos O + po)2 + [R sin o-F(n-l)]2 r2 83

for 9 when R = R and when R = RB, respectively. Newton's method of successive approximations will be used. The equation G(R,Q) = 0 is to be solved for Ro, for example. Thus G(Ro, ) = 0 is given; it is necessary to solve for 9. According to Newton's method: _ G(Ro 0o) 0 o Go(RoG,) where g0 is an approximation of the exact solution. 91 is now closer to the exact solution of 9 than 0 is. If Qo is replaced now by 01 and the computation is repeated, the solution approaches the exact value more and more closely. Repetition of the operation yields any desired accuracy. Let: G(cosQ) = [(R cosQ coso-Z sinao)2+R2(1-cos2g)-po] +[R cosQ sinao+ Zcosa-F 4 -r2 = 0 where: Z = b + b2 - C b = po sin ao + F(n-l) cos a C = (R cos a0 + po)2 + [R sin ao - F(n-1)]2 -ro2 6[G(coso)) ]2 2 a (cos)] =2(R cosQU0- Z sinao) + R2( - cos2) - po. (R cosQ cosao -Z sina o) R cos a - 2 cos Q + 2 [R cos Q sin ao + Z cos ao - Fn] R sin a6 0

The value of cos 0 will be solved by Newton's method for any desired R. cos Q = cos - G(cos - ~o a[G(cos Q,)1 6(cos i) Then the value of Q itself will be computed. After solving for QE and QD, the program will solve for the value of i at these two points. The valve of T at D will be used as a representative value for an average of i. The integration of the terms under the integral is done by Simpson's rule for numeric integration. As Fig. 12 shows, some work is done along the boundary CDEF, which is not included in the work of deformation under the area of contact. A direct computation of this work seems difficult. To account for this work, it was assumed that the additional work is proportional to the value of egZ and it has been added to the correcting factor under the root ( (1 + ). Let the equation for the main power be written as: W P 2aSFR il + a.v' cos aC P o 0~o where a is a multiplying factor to account for the work at the boundary CDEF. The "IT" language for an IBM 650 was used in this solution with the library sub-routines of The University of Michigan Statistical Research Laboratory. Before presenting the program, the assignment list is given: 85

Table II Assignment list for the results C5 - Ro radius at point A C6 - p roller radius C7 - r Roller "round-off" radius C8 - a Half included angle C9 - F Feed per revolution C10 - half the number of intervals used in Simpson integration Y1 - Wl Complementary weighted power at area 1 Y2 - W2 Complementary weighted power at area 2 Y9 - W9 Principal weighted power Yll - Wll Total weighted power by the deformation theory Y15 - t Weighted tangential force by deformation theory Y14 - tWeighted tangential force by incremental theory Y19 - W19 Totalweighted power by the incremental theory 10 - Identification number for the set of parameters (see Table (4).) I1 - Percent of Y1 of total power I5 - Percent of Y2 of total power I4 - Percent of Y4 total power I9 - Percent of Y9 of total power 117 - Percent of power by incremental theory to power by deformation theory Cll0 - Cos D E C111 - Cos "D C113 - Strain-rates ratio v-at point E C114 - Strain-rates ratio T at point D 86

Table III - General Assignment List C1 -R C2 - 0 coordinates position in space c3 - z C4 - N - Speed in rpm C5 - R - Radius at point A C6 - Po - Roller radius C7 - ro -Roller "round-off" radius C8 - a - Half the included angle of the cone C9 - F - Feed in ipr C10 - Half the number of intervals used in Simpson integration Cll - Sin 9 C12 - cos 9 C13 - sin aO C14 - cos a C23 - PO Sin a0 C26 - F sin a o C27 - n - The number of revolution passed from time t = 0 C28 - F-n C32 - R cos 9 cos a - Z sin a 0 0o C33 - (R cos 9 cos ao - Z sin ) + sin2 9 C37 - R cos sin a + Z cos a - Fn o C42 - (ro + Po) cos o C46 - ro + p0 87

C50 - 9E, at the point E C55 - F(n-l) cos a 056 - Po sin a +F(n-l) cos a = b C57 - F(n-l) C61 - C62 C63 Temporary variables used to compute the complementary power. C64 c69 - p cos a C72 - RB = Fn sin a - p0 cos aG C73 - R cos Q - (Fn sin - p cos ) o o o C74 - r 2 C75 - r2 - [R cos Q - (Fn sin a - p cos a)]2 C76 - F sin a0 + 2rR sin 9 C77 - 2[BR cos 9 - (Fn Sin a - po cos a )][C75 C78 - (C77 * Cll) + C74*C12*C76 C79 - (C77 - C12) - C74 * Cllw C76 c80 - C79/C78 C83 - C = (R cos a + po)2 + [R sin a - F(n-1)]2 r c94 - (c56.c56) - c83 C95 - C96 - y Multiplying factors C97 -J

1 -Clll C111 C100- cill Cll C101 - GED(R,Q,Z) The function G = 0 along line ED E~jD C102 - rGED(RQ,Z)J 6(cos @) C103 - C101/C102 C110 - cos GE C1ll - cos GD C113 - j at point E, strain rates ratio at point E C114 - at point D, strain rates ratio at point D C121 - p sin a + F(n-l) cos a + r - W - Complementaryweightepower at area 1 Y2 - W2 - Complementary weightedpower at area 1 Y2 - W2 - Complementary weighted power at area 2 Y9 - W9 - Principal weightedpower Yll - Wll - Totalweighted power by the deformation theory Y13 - t -Weighted tangential force by the deformation theory Y14 - t -Weighted tangential force by the incremental theory Y19 - W19 - Totalweighted power by the incremental theory IO - Identification number for the set of parameters (Table 4) I1 - Percent of Y1 of total power I2 - Alphabetic variable - "NOT" I3 - Alphabetic variable - "GOOD" t 14 - Data for machine decission I5 - Percent of Y2 of total power 16 - Data for machine decission. 89

I8 - Data for machine decission I9 - Percent of Y9 of total power 117 - Percent of power by incremental theory to power by deformation theory 90

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Program for Power and Tangential Force 0007 IO,C5*..C10,C86 READ F C13=Q22EKC8Q F C14=Q21EKC8Q F C23=C6*C13 F C69=C6*C14 F C46=C6+C7 F C26=C9*C13 F C42=C46*C14 F C27= ( C5+C42)/C26 C28=C9*C27 F C57=C9*(C27-1.) F C55=C57*C14 F C56=C55+C23 F C121=C56+C7 F C72=(C28*C13)-C69 F Y11=6.2831853*C5*C9*C 14 F C1=C72 F 16=1 F U006 C12=.99 F C83=( ( C6+C1*C14)*C6+C1*C14)+( ( C57-C1*C3 ) *C57-C*C13 ) -C7*C7F C94=(C56-C56)-C83 F G22IFC94VO. F C94=0. F u022 C3=C56+Q20EKC94Q F 0015 C32=( C1*C12*C14)-C3^C13 F C33=Q20EK((C (3C3C32)+C1C1*(1.C12*C12))Q F C37=( C11C12*C13 )+(C3C14 )-C28F C101=( (C33-C6)*(C33-C6))+(C37* C37)-C7*C7 F C102=2.*( ( ( C33-C6)*(C32*C*C1C4 )-C1*C1*C12)/C33)+C1*C13*C37 F C103=C101/C102 F C12=C12-C103 F G15 I FAC103WC86 F C(110+I6)=C12 F C73=(C1*C12 )-C72 F Cll1=Q20EK1.-C12*C12)Q F C74=C7*-C7 F C75=C74-C73*C73 F 93

C76=C26+6.283*C1*C1 1 F C77=6.283*C73*C75 F C78= C77*C11)+C74*C12*C76 F C79=(C77*C12 )-C74*C11C76 F C80=C79/C78 F C(113+I6)=C80*C80 F 0011 TIOTC(113+I6)TC(110+I6) F I6=I6-1 F C1=C5 F G14IFI8U1 F G6IFI6UO F 0014 G3IFAC114WC95 F 0020 Y9=Y11*(Q20EK(1.+C96*C114)Q) F G19 F 0003 ATI2TI3 F G20 F 0019 G18IFI4U1 F C12=Q20EK( (l.-C12*C12)/C12*C12 10a F C50=Q23EKC12Q F C3=Q27EKO.KC50KC1OK1K2K9Q F 0002 C12=Q21EKC3Q F C61= C5*C12 )-C72 F C62=( C7*C7)-C61*C61 F C62=AC62 F 0004 C63=C61/(Q20EKC62Q) F C61=C72*C12-1. F C62=tC7*C7)-C61*C61 F C62=AC62 F 0005 C64=C61/(Q20EKC62Q) F Y1=C5*C9*C13*C64-C63 F G2 F 0009 Y1=Y1*(Q20EK(1,+C96*C114)Q) F C100=Q20EK( (1.-C111*C111)/C111 *Cll11)Q F C101=Q23KC100Q F C3=Q27EKC50KC101KC10K2K10K16QF 0010 C12=Q21EKC3Q F C61=C5-C72+ ( C72-C5)*C12*(C3-C5 U)/(C101-C50) F C62=- C7*C7)-C61*-61 F 94

C62=AC62 F C63=C61/(Q20EKC62Q) F C61=C72*C12-1. F C62=(C7*C7)-C61-C61 F C62=AC62 F C64=C61/(Q20EKC62Q) F Y2=C5*C9*C13*C64-C63 F G10 F 0016 Y2=Y2*(Q20EK( 1.+-96*C114 )) F 0018 Y19=AY1lY2+2+AY9 F I17=100**Y19/Y11 F Y13=. 5*Y1 i/1.73205*3 14159*C5F Y14=. 5*Y19/1.73205'3.14159*C5F G23IFI4U1 F I1=100.*Y1/Y19 F I5=100,*Y2/Y19 F 19=100.*Y9/Y19 F 0012 TIOTY1TY2TY9TY1S F 0008 TIOTI1TI5TI9 F 0023 TIOTC5TC6TC7TC8 F 0001 TIOTC9TC1OTC86TC96 F 0021 TIOTC95TI4TI8TI17 F 0013 TIOTY11TY9TY13TY14 F 0017 TIOTC50TC101 F G7 F H FF 95

COMPARING THE STRAIN RATES 0007 I0,C5...C10,C86 READ F C13=Q22EKC8Q F C14=Q21EKC8Q F C23=C6*C13 F C69=C6*C14 F C45=C6+C7 F C26=C9*C13 F C42=C464C14 F C27= ( C5+C42 )/C26 F C28=C9*C27 F C57=C9*(C27-1.) F C55=C57*C14 F C56=C55+C23 F C72=(C28*C13)-C69 F I6=5 F C43=(C72-C5)/5. F 0001 C1=C72-I6*C43 F 0006 C12=.99 F C83=((C6+C1*C14)*C6+C1*C14)+(( C57-C*C13) *C57-C1*C13 ) -C7*C7F C94=(C56*C56)-C83 F G22IFC94VO. F C94=0. F 0022 C3=C56+Q20EKC94Q F 0015 C32=(C1*C12*C14)-C3*C13 F C33=Q20EK( (C32*C32)+C1*C1*( 1C12*C12))Q F C37=( CC12*C13 )+( C3*C14 )-C28F C101=((C33-C6)*(C33-C6))+(C37* C37)-C7*C7 F C102=2.*(( C33-C6)*(C32*C1*C14 )-C1*C1*C12 )/C33)+C*C13*C37 F C103=C101/C102 F C12=C12-C103 F G15IFAC103WC86 F C(105+I6)=C12 F C73=(C1*C12)-C72 F C 1=Q20EK (1.-C12-C12 ) F C74=C7*C7 F C75=C74-C73*C73 F C76=C26+6.283*CClC11 F C77=6.283*C73*C75 F C78=(C77C11 )+C74*C12*C76 F C79=(C77*C12)-C74*C11*C76 F C80=C79/C78 F C(113+I6 ) =C80*C80 F 96

0002 TIOTC(113+I6)TC79TC78TC( 105+6 0002 ) F I6=I6-1 F I7=I6+1 F G3IFI7UO F G1 F 0003 I6=1 F C43=(1.-C105)/5. F C1=C72 F C12=lo F 0004 C73= ( C1*C12)-C72 F C11=Q20EK( 1-C12*C12)Q F C74=C7*C7 F C75=C74-C73*C73 F C76=C26+6.283*C1*C11 F C77=6 283*C73*C75 F C78=(C77*C11)+C74*C12*C76 F C79=(C77*C12)-C74*C11*C76 F C80=C79/C78 F C(118+I6)=C80*C80 F 0005 TIOTC(118+I6)TC79TC78TC12 F C12=C12-C43 F I6=16+1 F G4IFC12WC105 F G7 F H FF cos 0 = C12 V = C119 C120 C121 C122 C123 C72 = RB e cos 0 = C105 = C113 (os @ = C106 7 = c114 C5 = R - \ rcos 0 = 0107 \ \ (fcos Q = C108 c\ \. = C117 \(cos 9 = c110 9 = c118 97

APPENDIX 2 COMPARING THE TANGENTIAL FORCES FOR SHEAR AND BENDING The type of deformation that takes place can be assumed as bending or shear, as will be explained immediately. The actual deformation can be anything betweenthese two types (see Fig. 19). A. BEND Each point on the central surface remains at its previous radial distance (R) from the axis Z. The straight lines AB and CD remain straight and perpendicular to the surface. AB becomes A'B' and CD becomes C'D'. The length A'B' = AB sin a B. SHEAR Shear was described earlier (Fig. 5). Each point remains at its previous distance (R) from the center (Z axis). The straight lines AB and CD remain straight with no change in length. However, in this case they are no longer perpendicular to the surface, but parallel to the Z axis. Let the tangential force be computed with the deformation theory. For shear deformation the tangential force was found to be: t S~ o F cos a 98

A-I: cLOYD C Di RECTION a)BEND - 1S DIRECTION b) SHEAR Bf B A: Fig. 19-The shear a bending strains 99

For pure bending the steps are as follows: Let the power be: W = V fd = Va 0 where: V is the volume deformed per minute a the yield limit under uniaxial tension 0 0 the effective logaritmic strain Let 0. be the circumferential logarithmic strain at the middle surface 0S be the width avg. logarithmic strain 1 be the length logarithmic strain (see Fig. 19) The middle surface has no circumferential change in length, and therefore 0 =0 The thickness is reduced from S to S = S sin a o o o0 and therefore: 0S = in S in (sin a) From the volume constancy, one gets K0 = -S-0e = - The definition of the effective strain in the absence of shear strains 2 j1 [(0 01) - +2 ( 1 2s& + )OS - 0) 2 in this case reduces to 0 i | 1(0 + 0S2 + 2) + p S = n (sin c ) The volume deformed per minute is: V = 2jRN S AR = 29RN S F sin a 0 0 0 o1 o 100

And thus: W= 2iRN S F sin a % in (sin a ) O o O o And the tangential force t = - = 2 S ~ F sin a -ln (sin a ) c 2rtRN For comparison purposes let aweighted tangential force be defined such that tt f S'a 0 o The bending assumption gives tI = 2F sin a in (sin a ) o O The shear assumption gives t' = F cos a S o Figure 20 describes the weighted tangential force for a feed of.100 ipr for half included angle from a = 0 to a = 90~. o o APPENDIX 3 EVALUATING THE YIELD LII[T In the analytical approach, Mises material was considered. The characteristics of the material were thus presented by a single value of the yield limit (a ). Actual materials, however, have an elastic range as well as a strain-hardening effect. The most convenient way to present the characteristics of an actual material is by its tensile test recording of a uniaxial stress 101

O0 C 0 ~~c~~~~~ o Pb 0~ 0 C0 4-'-6 / / - 00c C 0 c t-o~~~~~0 o,c o(9 E IC)'IO Boo 0 cc-o X / - 0 0 0 0 Cc C 0 c 50 0 0 - 4-.-./ 0 ^ o b' 0 0 000 0 0 102

piotted against the strain. For the check of the numerical answer against the 1\ experimental power consumed, a representative value for oa is to be chosen from the tensile test recording. First, the question of the elastic portion of the deformation was considered. Steel with modulus of elasticity of 30,000,000 /inch and with extremely high yield limit of aO = 80,000/0'/,,will serve as an example. For this particular material, the strain at yield with uniaxial stress will be 8o ooo Eyield = 8000 < 003 yie30,000,000 Let the shear at 45~ to the principal axis be found for this case. The shear stress T = ao 2 (l + 3) 1+5 OQs1 (?0022 The shear strain is = (1.0022 E 2 E Transferring the radiansto degrees, one gets maximum shear in the elastic range of.13~ for original angle of 90~ The shear strain at yield is much smaller than the minimum shear (5~) considered in this study. Considering this elastic deformation as part of the plastic deformation has no effect on the numerical results. This is clearly illustrated in Fig. 28 where a stress-strain curve of an actual tensile test is recorded at constant speed of the chart. From the actual recording an average value was picked up and denoted to be the representative yield limit ao. The value of ao is neither far from the minimum nor from the maximum value of the stress on the chart. 103

z 4~I/~ ~~~ ICO I 1' I 0 4O C a, c.'0 4U) (10 L) I OD 10L

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