THE UNIVERSITY OF MICHIGAN INDUSTRY PROGRAM OF THE COLLEGE OF ENGINEERING DYNAMICALLY LOADED JOURNAL BEARINGS OF FINITE LENGTH WITH AXIAL FEED Philip Go Kessel A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the University of Michigan Department of Engineering Mechanics 1964 May, 1964 IP-671

ACKNOWLEDGMENTS The author wishes to express his gratitude to Professor Jesse Ormondroyd, Chairman of the committee, who suggested the problem and under whose continued guidance the work was performed. Thanks are also due to Professor George Lo West Jr. for his assistance in the initial formulation of the problem and in arranging trips to the Newport News Shipbuilding Company, Newport News, Virginia and the U.S. Naval Engineering Experiment Station, Annapolis, Maryland to secure background data. These trips were sponsored by the University of Michigan Research Institute and the Engineering Mechanics Department, University of Michigan respectively. The author is also indebted to the other members of his committee for their many helpful suggestions. The use of the IoB.Mo 7090 computer at the University Computing Center is gratefully acknowledged. Finally the author wishes to express his deepest appreciation to his wife, Mary Joan, for her constant patience and understanding, without which this work would not have been possible.

TABLE OF CONTENTS Page ACKNOWLEDGMENTS............................................. ii LIST OF TABLES................................................ v LIST OF FIGURES.......................................... vi NOMENCLATURE.................... o............ o................... viii I INTRODUCTION o..... o................................... 1 Ao Statement of the Problem......................... 1 Bo Brief History of Hydrodynamic Lubrication...... 5 Co General Plan of Attack........................... 9 II MATHEMATICAL DESCRIPTION............................. 11 A. Formulation of Reynolds Lubrication Equation..... 11 B. Boundary Value Velocities and Film Thickness,..o. 15 C. Boundary Conditions.............................. 20 III THE PRESSURE EQUATION..................... o......... 22 Ao Solution of the Reynolds Equation................o o o 22 Bo Convergence of the Series........................ 31 Co Estimate of the Series Error..................... 34 Do Condition on the Inlet Pressure.................. 37 IV EQUATIONS OF MOTION OF THE JOURNAL................... 39 Ao Formulation of the Equations of Motion........... 39 B.o Evaluation of the Pressure Integrals............. 41 Co Simplification of the Equations of Motion........ 47 V RESULTS AND DISCUSSION. o o.............o o o o o..o o.........o o 53 Ao Physical Data of the Case Considered for Numerical Evaluation............................ 53 B.o Journal Orbits.o................................ 54 Co Pressure Distributions in the Bearing............ 66 VI SUMMARY.............................................. 78 A. Conclusions.................................. 78 B. Areas in Need of Further Study................... 80 iii

TABLE OF CONTENTS (CONT'D) Page Ao Derivation of the Reynolds Lubrication Equation......................................... 81 B. Solution of Reynolds Equation for the Infinite Length Bearing...o....................... 89 C. Evaluation of the Series Coefficients Am(z) and Bm(z).................................. 91 BIBLIOGRAPHY.................................................. 101 iv

LIST OF TABLES Table 4.1 Page y, 7, ac, 4, ( vs n fl.......................... 52 5.1 5.2 5.5 5.4 Characteristics of the Orbits for the Case of H2 = 0.0, b = Steady State Coordinates of n Characteristics of the Orbits for the Case of H2 = 0.0, b = Steady State Coordinates of n of the 4, K = = 0.85 of the 4, K = = 0.95 Journal Center 0o1796 and and Z = 90.... Journal Center 0.0969 and and i = 90,~., Characteristics of the Orbits of the Journal Center Considering Both the First and Second Harmonic Components of the Dynamic Load for b = 4, K = 0.1796 and Steady State Coordinates of n = 0.85 and = 90~................................... Characteristics of the Orbits of the Journal Center Considering Both the First and Second Harmonic Components of the Dynamic Load for b = 4, K = 0.0969 and Steady State Coordinates of n = 0.95 and 0 = 90 ~ o o........o. o.,. o o......................... 59 59 63 63 v

LIST OF FIGURES Figure Page 1.1 Stern Bearing Location............................... 2 1.2 Nature of Tailshaft Damage................. 5.......... 1.3 Actual Stern Tube Bearing............................ o 6 2.1 Physical Configuration................................... 12 2.2 Boundary Value Velocities and Film Thickness Configuration........................................ 16 4.1 Geometry for Journal Equations of Motion............. 40 5.1 Locus of the Journal Center for H1 = 0.8, H2 = 0.0, b = 4, p = 2700 and K = 0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85.oo...... 55 5o2 Locus of the Journal Center for HI = 0.8, H2 = 0.0, b = 4, 5= 300~ and K = 0.1796. Steady State Coordinates are 0 = 90~ and n = 0o85 o....o... 56 5o3 Locus of the Journal Center for HI = 0o8, H2 = 0.0, b = 4, P = 330~ and K = 0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85.......... 57 5.4 Locus of the Journal Center for HI = 0.8, H2 = 0.0, b = 4, 2 = 0~ and K = 0.1796. Initial Coordinates are 0 90~ and n = 0o85.o....... oo...... 60 5.5 Locus of the Journal Center for Hi = 0o8, H2 = 0o0, b = 4, p = 2700 and K = Oo0969. Steady State Coordinates are =0 90~ and n = 0.95........... 61 5.6 Locus of the Journal Center for Hi = 0.8, H2 = 0.0, b = 3, p = 270" and K = 0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85........... 64 5.7 Locus of the Journal Center for HI = 0.8, H2 = 0.0, b = 5, p = 2700 and K = 0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85........... 65 5.8 Pressure Variation Along the Length of the Bearing for t = 0.7300 in the Journal Orbit; po = 350o0psig., i = 74.0 inches, b = 4, p = 270~. Steady State Coordinates of the Orbits are n = 0.85 and 0 = 90~..... 67 vi

LIST OF FIGURES (CONTID) Figure Page 5.9 Pressure Variation at a Fixed Point (0 + O = 260~ and z = 0o0) in the Bearing for the Orbit Defined by HI = 0.8, H2 = 0.0, b = 4, P = 270~ and and K = 0.1796. Steady State Coordinates of the Orbit are 0 = 900 and n = 0.85.............o..o...... 68 5.10 Pressure Profile at t = 0.7850, n = 0.7951, z = 0.0 and 0 = 85o 47~ for the Orbit Defined by Hi = 0.8, H2 = 0.0, b = 3, p = 270~ and A = 0.0, Steady State Coordinates are n = Oo85 and 0 = 90 ~o....... ~.~.~..~ 71 5.11 Pressure Profile at t = 0.7300, n = 0.80835, z = 0.0, and 0 = 89.83~ for the Orbit Defined by HI = 0.8, H2 = Oo0, b = 4, p = 270~ and A = 0.0o Steady State Coordinates are n = 0.85 and 0 = 90~o.. 72 5.12 Pressure Profile at t = 0.6950, n = 0.8187, z = 0,0 and 0 = 88~650 for the Orbit Defined by H1 = 0.8, H2 = 0.0, b = 5, p = 270~ and A = 0o0. Steady State Coordinates are n = 0.85 and 0 = 90,,...0...0.0 o o.... 73 5.13 Pressure Profile at t = 0.7300, n = 08004, z = 0o0 and 0 = 89.82~ for the Orbit Defined by H1 = 0.8, H2 = 0,4, b = 4, p = 2700 and A 0.0, Steady State Coordinates are n = 0.85 and 9 = 90 90............ 00. 74 5014 Pressure Profile at t = 0o7300, n = 0.8020, z = 0.0 and 0 = 89.91~ for the Orbit Defined by Hi = 0.8, H2 = 0.4, b = 4, p = 270 and A = 11,25~o Steady State Coordinates are n = 0.85 and 0 = 900000.000000o 75 5.15 Pressure Profile at t = 0o7300, n = 0.8270, z = 0O0 and 0 = 89 96~ for the Orbit Defined by HI - 0.4, H2 = 0.2, b = 4, p = 270~ and A = 11.25~o Steady State Coordinates are n = 0~85 and 0 = 900..o..... 76 5.16 Pressure Profile at t = 0.7300, n = 0.9408, z = 0o0 and 0 = 90.07~ for the Orbit Defined by HI = 0.4, H2 = 0.2, b = 4, p = 270~ and A = 11,250. Steady State Coordinates are n = 0.95 and 0 = 90~,.00...00.. 77 Ao1 Auxiliary Coordinate System.......... o.....,...... o 82 vii

NOMENCLATURE Unless otherwise specified, the following symbols are used: Am(m=l,..o,oo) Bm(m=l,o o,oo) Cm(m=l,.o,o) D2 = r2 d2 dz2 functions of z functions of z integration constants differential operator D a + u a + v a + W a Dt at yx Ty z Fi(i=l,2,etc. ) F,F Gq;P(op=l,2,3) Hi(i=l,2) K L Nm(m=l, o,oo) 0 0o substantial derivative auxiliary symbols or functions dimensionless metric tensor ratio of the amplitudes of the first and second harmonic components of the dynamic load to the static load modified Sommerfeld number a characteristic length mass of the journal plus the mass of the propeller integration constants center of the bearing center of the journal function used for order of magnitude analysis general radial coordinate a Reynolds number critical Reynolds number Sommerfeld number Taylor's parameter viii R Re (Re)c S T

u U' = U/L V V' = V/L = 6L2 + 62 + -2 6x2 6y2 6Z2 velocity of a point on the journal surface tangential to the bearing wall in the circumferential direction dimensionless form of the velocity U velocity of a point on the journal surface normal to the bearing wall dimensionless form of the velocity V Laplacian operator WS WD WDlWD2 X,YZ Xf(i=l,2',3 ) a b c d e eN'et fi(i=1,2,etc.),fg gp(a p=l 2,3) 6 static load on the bearing dynamic load on the bearing amplitudes of the first and second harmonic components of the dynamic load body forces per unit mass corresponding respectively to the general coordinates body forces per unit mass corresponding respectively to the x,y,z Cartesian coordinates function of n, a = n/[l + (l-n2)/2 ] acceleration of the journal center number of propeller blades bearing clearance diameter of the journal journal eccentricity unit vectors normal and tangential to the bearing wall auxiliary symbols or functions metric tensor determinate of the metric tensor, g=lgrIl ix

h m(m=l,., 0oo) n p P0 oo t u'(~=1,2,3) film thickness at any point minimum film thickness length of the bearing series indici eccentricity ratio, n = e/c pressure at any point in the bearing supply pressure of the lubricating fluid radius of the journal radius of the bearing time coordinate velocity corresponding to the x Cartesian coordinate velocity components corresponding to the general coordinates ~1 dimensionless form of the general velocity components velocity corresponding to the y Cartesian coordinate velocity of any point on the journal surface velocity of the journal center velocity of any point on the journal surface relative to the journal center velocity corresponding to the z Cartesian coordinate auxiliary Cartesian coordinates 1/2 function of n, y = n/(l-n2) Cartesian coordinates functions of n ul(i=11223) v vB v0B/O VB/0 w y x,y,z o;,a x

angular location of the dynamic load with respect to the static load Euclidean Christoffel symbols rl(llim,0 &j(i,jcl,2,5) ~ ti(id,2.3) A function of n kronecker delta dimensionless small parameter for an order analysis, e = ho/L function of n general coordinates angular location around the journal with respect to the attitude angle phase angle of the second harmonic component of the dynamic load dynamic viscosity kinematic viscosity dimensionless pressure mass density of the lubricant attitude angle of the journal function of n angular velocity of the journal it p * (1) xi

I. INTRODUCTION A. Statement of the Problem The problem being considered here is that of finding a solution for the pressure distribution and simultaneous shaft loci of a 360o journal bearing subjected to dynamic loading. The bearing is lubricated by a circumferential source at one end of the bearing. The lubricating fluid flows out the other end of the bearing. The bearing is considered to be finite in length.. The bearing lubricant is water which is supplied at a constant rate and pressure. It is assumed that the bearing and journal surfaces will always remain parallel and that both are completely rigid. The surfaces are further assumed to be perfectly smootho The initial motivation for this problem stems from the stern tube bearing of ships, within which the tailshaft seemingly undergoes cavitation damage due to the dynamic loading of the propeller. It has been observed in several ships that the tailshaft is eroded at several definite positions around its periphery and within the confines of the stern bearing. (See Figure 11ol for the location of the stern bearing and Figure 1.2 for the nature of the tailshaft damage involvedo) The number of locations of damage and their positions around the periphery of the journal vary directly as the number of blades which the propeller haso If the number of blades is odd, either three or five, then there will be either three or five locations respectively of tailshaft damage and their positions will be directly in line with the propeller blades. If the number of blades is four, then there will be four locations of damage which are exactly 45~ offset from the line of the propeller blades. The -1

Figure 11. Stern Bearing Location,

Figure 12. Nature of Tailshaft Damage.

location along the length of the bearing where this damage usually occurs is slightly forward of the quarter way point of the bearing as measured from the propeller end of the bearing. From a survey made by the Bethlehem Steel Company, Shipbuilding Division,(l) possible explanations for an attack of this nature were (1) galvanic attack, (2) stray current leakage, (3) contact erosion with shaft idle and (4) cavitation erosion. The first of these was eliminated by direct electrical measurements which showed no current flow. The second was eliminated by the fact that if grounded DoCo leakage should attempt to leave the hull by jumping across to the shaft, then the damage should occur at the jumping off place, which would be the bearing and not the shaft. The third.was eliminated by the multiple locations of damage. This left the fourth, cavitation erosion, as the probable explanation. While the problem studied in this effort differs from the actual physical problem of stern tube bearings in that is was necessary to make certain assumptions to surpass some mathematical complexities, it is felt that it represents a reasonable initial model of the actual problem. It should be noted in passing that there are innumerable physical situations which correspond very closely to the problem studied hereo The three most important assumptions departing from the actual physical situation of stern tube bearings are (1) the shaft and bearing are to remain parallel at all times, (2) the bearing surface is completely smooth and (3) the propeller loading can be represented by at most the first two harmonic components.

The first of these assumptions has the effect of making the film thickness a function of only angular displacement around the journal. It is believed that the two major consequences of this assumption are a change in the intensity of the pressures obtained and a slight shift along the length of the bearing of the region of minimum pressures developed in the lubricating filmo It is not felt however that the general pressure profile would be substantially altered by this assumption. These conjectures will be further explored in the presentation of the results. The second assumption definitely violates the actual bearing which is composed of staves (See Figure 1.3) spaced in the order of onehalf inch around the bearing periphery. The mathematical complexity of incorporating these effects however makes it necessary to assume a smooth bearing surface. Considering the third assumption, although the actual propeller loading is certainly composed of many harmonic components, the first two of these are known to represent the major portion of the propeller loading. Bo Brief History of Hydrodynamic Lubrication Although the field of Hydrodynamic Lubrication is a relatively old one, it was not until the latter part of the 1950's that solutions for finite length journal bearings considering only static loading became available. Suprisingly few papers on dynamically loaded journal bearings of either finite or infinite length appear in the literature. The initiation of Hydrodynamic Lubrication dates back to Tower,(2) who in 1883 seemingly by accident discovered this phenomena. He was engaged in an investigation of the friction characteristics of 157' partial

Figure 1.35 Actual Stern Tube Bearing.

railroad car journal bearings. In the course of his experiments it was necessary to drill a hole in the loaded region of the bearing. Plugging the hole with a cork, he noted that in subsequent experiments the cork was continuously forced out of the hole. Upon connecting a pressure gauge he found pressures in excess of 200 psi. while the unit bearing load was only 100 psi. Connecting more pressure gauges around the periphery of the bearing he found a very definite pressure profile. This of course confirmed his growing suspicion that a pressurized fluid film was being developed between the journal and bearing which actually supported the journal. In 1886 Reynolds(3) put Hydrodynamic Lubrication upon a sound mathematical basiso His work is noted to such a degree for its clarity, scope and understanding, that at present his analysis is basically still followed. Some of his integrations have been improved or extended, but his basic theory is still intact. Considering the case of steady loading and infinite length, Reynolds was able to obtain an approximate solution to his basic equation for a journal bearing in the form of a Fourier series. His results however, were limited to lightly loaded bearing; that is eccentricity ratios of less than 0.o5 Sommerfeld, (4) in 1904, through a series of clever mathematical substitutions succeeded in finding an exact solution to Reynolds equation for steady loading of infinite length journal bearingso The major shortcomings of his results were unrealistic attitude angles and negative pressures of a magnitude that a fluid would be incapable of withstandingo These results were a direct consequence of his assuming a complete oil film around the bearing.

Following Sommerfeld, the next most notable works were that of Harrison;(5) who in 1919 first treated the problem of dynamic loading and Swift,(6) who in 1937 presented a rather extensive treatment of dynamically loaded journal bearings of infinite lengtho Although the major portion of Swift's effort was devoted to stability of steady loads and alternating loads with no journal rotation, he did consider the case of a sinusoidal load on a rotating journal. His results indicated that at a frequency ratio of one-half (forcing frequency/journal frequency) the load capacity of a dynamically loaded journal bearing is zero. Below one-half the load capacity is less than an equivalent statically loaded bearing and above one-half is greater than an equivalent statically loaded bearing. The general journal orbits were approximately elliptical in shape becoming flatter with increasing frequency ratioo For frequency ratios less than one-half the ~major axis of the orbit was perpendicular to the load and above one-half parallel to the load. His results however were limited to frequency ratios of one or less and no pressure profiles were obtained. Burwell,(7) in 1947 extended Swift's results for the case of square wave loadingo His results indicated that for equal load amplitudes, the sinusoidal loading gave greater load capacity to the bearing. A major contribution was made by Tao,(8) who in 1959 succeeded in finding an exact and complete solution of Reynolds equation for statically loaded journal bearings of finite lengtho His method of approach was a multiple separation of variables technique which led to a form of Heun's equation. Fedor,(9) in 1960, by adding a sine series to the known solution for a statically loaded journal bearing of infinite length was able to

find a solution for a statically loaded journal bearing of finite length with circumferential feedo The analysis presented below will be based on Fedor s method. In 1961 Hays(10) presented a solution for a finite length journal bearing subjected to a sinusoidal loading considering only the squeeze film effect; that is without journal rotation. His method of approach was the assumption of a double sine-cosine series solution for the pressure function and numerical evaluation of the coefficients on a digital computer. To the best of the author's knowledge he also presented the first pressure profile for a dynamically loaded journal bearing. This distribution, as would be expected for a symmetrically loaded journal bearing, was in the form of a paraboloid; being symmetrical with respect to the length of the bearing and with respect to angular displacement around the journal. Co General Plan of Attack The initial phase of the problem deals with the formulation of Reynolds equation governing dynamic loading of finite length journal bearings. Although this equation is well known, the method chosen for its derivation is a relatively new one. Following the recent work of Elrod(ll) who considered the case of statically load journal bearings of finite length, the equation is formulated by a small parameter approacho The reason for this choice is that it eliminates the necessity of making the usual assumptions that the convective inertial terms of the Navier-Stokes equations are negligible, the fluid film curvature is negligible and the pressure is constant across the film thickness. The solution of the Reynolds equation is based on the recent work of Fedor,(9) who as mentioned above considered the case of a statically

-10 loaded finite length journal bearing with circumferential feed. To his solution an assumed series term is added to account for the dynamic loading, which when made to satisfy Reynolds equation and the given boundary conditions will yield the equation for the pressure distributiono This equation however will contain time indirectly in the form of two unknown velocity componentso These velocity components are the translational and rotational velocities of the journal center in its orbit about a steady state positiono These two unknown velocity components must then be solved for from the two scalar equations of motion of the journal mass center. These equations of motion will yield two simultaneous, non-linear, ordinary differential equations which must be evaluated numerically; in this case on a digital computer. In addition to yielding the two unknown velocity components, the path of the journal center will now be known. Thus for a given position in the journal orbit and the corresponding velocity components of the journal center, the pressure profile around and along the length of the bearing may then be evaluated.

IIo MATHEMATICAL DESCRIPTION A. Formulation of Reynolds Lubrication Equation If the assumptions of constant density and viscosity of the bearing lubricant are made, the Navier-Stokes equations and equation of continuity may be written respectively as Dup|p (261) Dtp = -+ pX + x u (2 1) Dt ax Dv Dt p Dw Dt 6x in which x,y,z are velocity components dynamic viscosity, ^ 6x2 r F = - + pY + tV v, ay = - P + pZ + ij2w, 6z: -.1 I -_ (2.2) (253) l + 2_ + Lw 0 O, (2~ 4) 6y 6z Cartesian co-ordinates, u,v,w are the corresponding, p is the density, p is the pressure, p. is the X,Y,Z are the body forces per unit mass, a2 a2 XD a a a 6+ 2 + -5 and =- u - v + w- o ~y2 az2 Dt at ax y az If the further assumptions are made that the flow is everywhere 6u 6v 6w laminar and the inertial terms at, t', ~ and the body forces XY,Z are neglible, then it can be shown (See Appendix A for the derivation.) by following the small parameter approach of Elrod(ll) that Equations (2.1) and (2o5) may be integrated directly for the velocity components u and w. Equation (2.2) yields ~. = 0 and thus the pressure is constant across the film thicknesso If the values of u and w are substituted into Equation (2.4) and then integrated with respect to y

r Ws (STATIC LOAD) I/j WD (DYNAMIC LOAD) / t^/y////N/////////////////// PROPELLER P END o 8 x h I z.,,,,, X O' r r 0000", / /.0, I - poo. l - tin e ll I As I II.. -t Figure 2.1. Physical Configuration.

across the film thickness h, the Reynolds equation, Equation (2o5), with first order correction terms may be written as a [h3(l- _) hap + a-[h3(i +.) 6p] ox d ax 6z d 6z = -6U L[h(l - )] + 6,h(l - ) aU + 12NV, (2.5) 6x 3d 3d Tx where h is the film thickness at any point, d is the diameter of the journal and V and U are the velocity components of any point on the journal surface normal and tangential respectively to the bearing. Justification of the five assumptions necessary for the derivation of Equation (2~5), namely, (1) constant density, (2) constant viscosity, (3) the flow is laminar, (4) t t = = = 0 and (5) X = Y = Z = 0 is now in order. The first of these (1) is very reasonable when it is noted that the compressibility of lubricating oils and water is in the order of one part in two thousand or less. Considering (2), the viscosity of lubricating oils or water is known to decrease with temperature. However, when it is noted.that the bearing is supplied by forced feed lubrication and that the hull of the ship, which is directly adjacent to the stern tube bearing, is immersed in the lake water, the temperature rise is very small. Considering assumption (3) and assuming for a moment that the journal and bearing are concentric, then the critical Reynolds number (Re)c for Couette flow between circular cylinders may be calculated from the Taylor parameter(12) T as follows, T = -2.0(r - 1)4(Re) ( r2 ) > 0 r c r2-r2 1

For numerical evaluation of the problem considered in this study the data used will be that of the SoS. John Go Munson of the Bradley Transportational Line. This is a single screw ship with a four bladed propeller. From this data rl = 9.2935 inches and r = 9.25 inches. This gives (Re)c = 1.810 x 105, The actual Reynolds number is given by (Re) _ r2 4.669 x 105 where v is the kinematic viscosity of water (v = 1o4 x 10-5 ft2/sec) and C = 11 rad/sec. This value however may be corrected(13) by a factor of (1.0 +.89n2) where n = e/c, e being the eccentricity of the journal and c the bearing clearance. Based on the minimum eccentricity encountered here the correction factor is lo7. This gives a new value for Reynolds number (Re) = 2.746 x 105o This value of (Re) is of course larger than (Re)c but not significantly and does not necessarily mean that the flow is turbulento At (Re)c for Couette flow between circular cylinders a secondary flow is induced which is also a laminar flow. An actual bound on (Re) before turbulent motion occurs has not yet been established but it is known that this secondary flow is actually more stable than the initial laminar flowo It is therefore not unreasonable to assume laminar flowo For purposes of mathematical analysis this is a mandatory assumptiono The validity of assumption (4) can be shown by a comparison of the magnitudes of the inertial terms versus the viscous terms, that is _u << v 2u This is equivalent to 1 << v The element of time can at ax2 t x2 be taken to be one half of the period correspond the period corresponding to the forcing frequency (a =44 rad/sec) and that of displacement (x 0.0012 in.).

-15 This corresponds to the maximum journal movement in this period of time. Considering these values 44<< 1.4 x 10 5 x 144 or i << 100 it (0.0012)2 Assumption (5) is obviously good. Considering the diameter of the journal, approximately two feet, the hydrostatic head will be in the order of 1.0 psi. This is less than 0.5% of the total pressure. For the bearing considered here, h/d = 0.0017. In all other known journal bearing uses, h/d is even less than this. Therefore, h/d << 1.0 and these terms may be neglected in Equation (2.5) which then can be written as a[h3 -P] + a-[h3 p-] = -6LU -h + 6ph 6U + 12pV. (2.6) xx x z z ax ax B. Boundary Value Velocities and Film Thickness Referring to Figure 2.2, the absolute velocity of point B may be written as vB = v' + B/O In terms of components parallel to the unit vectors eN and "e v, = d[e cos eN + e sin eG] + d x[e cosG + e sin or vO = [d- cosG + e sing eN + [ sin - e cos. ] es * Similarly vo,/0 = -rw sin 5 eN + rco cos 5 ~e"

-16 Figure 2.2. Boundary Value Velocities and Film Thickness Configuration.

-17 Therefore vB = [-rco sinb + de cosg + e sinG d_] e% vBy (dt dt + [rw cosb + d sing - e cosG es o dt dt., Separating vB into it scalar components and noting 6 is a small angle, therefore cos 6 1.0 and sin b -, ax the components of velocity U and V of a point on the journal tangential and normal respectively to the bearing may be written U = ro + da sing - e cosg d:, (2.7) dt dt V = ro ah + de- cosG + e sing d. (2.8) ax dt dt Referring to Figure 2.2 the film thickness h at any point B around.the journal may be written as h = AB = A - OB-= (r+c) - OB o It is necessary at this point to assume that the journal and bearing always remain parallel such that the film thickness is a function of only angular displacement around the journal. Now S = e sing and S = r sin b therefore e sing = r sin..

Further m = r sinb and m = OB sin, thus OB = r sin7/sinG. Noting that 7 =0-65 r r OB = sin(@-6) = - sinG sin~ [sinG sin6 - cosO cos5]. However sin6 = e sin, r and thus cos5 = (1 e2 r2 sin2)12 Therefore OB = 1r sinG 1/2' [ sin (1 - e2 sin2) 1/2 _ e. sinG cosQ], r2 r or OB = (r2 - e2 sin2)1/2 - e cosg. The film thickness h may thus be written as h = r + c - (r2 -e2 sin2G)1/2 + e cosG. Noting that e2 sin2 << r2, h = c + e cos = c(l + n cosg), (2.9) Substituting Equations (2.7) and (2.8) into Equation (2~6) and noting x = rG, therefore - - 1- ~ 6x r G 1 a -(h3 P) + (h3 P)] = - h e do S rh 1 de (10) 5G r dt 6G r d.t

-19 eh d _ h de + -. a< sing + - - cos9 r dt r dt +2o sin+ 2e sin + 2 d cos (2.10 cont'd) d~ dt dt From Equation (2.9) h = _ cn sinG o Substituting this result and Equation (2.9) into the right hand side of Equation (2.10) with the exception of the first term 1 [L (h ) + (h3 k)] - 6. r2 r 9 3~ 3z 3z o h + d e sinG(- + 2) 5~ dt r + de [ + - cos + 2 cosg] (2.11) dt r r Finally noting ~ << 2, ~ 2 cosg, cn e r r r r and 2cn d0 sing = - 2 d_ h dt dt S9 Equation (2.11) may we written 1 [1 (h5 3p) + A (h3 P)] = 6p r2 sG 3G9 z 3z (Co - 2 do) 3h + 2c dn cos (2.12) dt Gs dt The solution of Equation (2.12) together with the appropriate boundary conditions will yield or the pressure distribution around the bearing.

-20 Co Boundary Conditions One of the major difficulties encountered in solving the Reynolds equation is adequately defining the boundary conditions. There are three general approaches adapted in all of the bearing literature. The first of these is the classical Sommerfeld approach, namely p(@) = p(@ + 2it) and 6P(g) = 6P(@ + 2X) This approach in general yields negative pressures over a considerable portion of the bearing and unrealistic attitude angles. The second method adopted is the same as the first except the pressure is set equal to zero for the entire negative region. The third approach is to let p = 0 for all 9 > Q9 and aP = 0 for @ = 91 This condition of course requires both the pressure and pressure gradient to be zero at the beginning of the negative region. This last method in general yields results that compare more favorably with practice, in particular with respect to the attitude angle. The approach used in this study is that of Sommerfeld for the specific reason that the major questions in this study are (1) do negative pressures occur, (2) if so, are they of sufficient magnitude to allow cavitation of the lubricating fluid and (3) if cavitation is possible what is the location of the areas of possible cavitation around and along the length of the journalo Accordingly the boundary conditions adapted are p(O,z) = p(2t,,z), (2.13) _p (O,z) = _p (2r,z), (2ol4) 59 6~

-21p(G,') =, (2.15) p(Q, - i) = Po ~ (2d16) 2

III. THE PRESSURE EQUATION A. Solution of the Reynolds Equation In order to determine a solution to Equation (2.12) first consider the infinite bearing; thus ( ) = 0. Equation (2.12) may then be written i-(h3 aP-) = 6tr2(c - 2 d1) ah + 12ir2c n cos9. (351) @9 39 dt 69 dt The solution to Equation (3.1) satisfying the first two boundary conditions, Equations (2.13) and (2.14), is a well known solution (See Appendix B.) which may be written as 6 = (a) -2 d)[n(2+n cos@) sing c2 dt (2+n2)(l+n cos )2 + ~2 dn 1 ] + arbitrary constant. (3.2) c2n dt (l+n cosg)2 The arbitrary constant may be absorbed into the infinite series accounting for the finite length of the bearing. For the total pressure function, in view of Equation (3.2) let P( Z + 2 d n(2+n cosG), sinG ] 0(Zc2 (2 cc2 dt -2 (2+n2)(l+n cos@)2 + 6r2 d 1 c2n dt (l+n cos9)2] 00 PO - Z Am(z) sin mi - Z Bm(z) cos m9. (353) m —1 m=l1 Equation (35.53) satisfies the first two boundary conditions, Equation (2.13) and (2.14). The first term represents the circumferential source function. The next two terms represent the infinite length -22 -

-23 bearing and the last two terms represent finite length of the bearing. The first as may readily be shown by substitution. to satisfy Equation (5353) by substituting equating the coefficients of sin mg and this is done, and for convenience let a correction to account for the three terms satisfy Equation (5.5) The last two terms can be made r them into the equation and L cos mg equal to zero. If Am(z) = Am and Bm(z) = Bm the following recurrence relations are obtained. For m = 1 2(D2-1) A1 + n(D2+2-) A2 = 0, 2(D2-1) B1 + n(D2+2) B2 = 0, and for m > 1 2(D2-m2)Am + n(D2 m2-m+2)Am.l + n(D2-m2+m+2)Am+l = 0 2(D2-m2)Bi + n(Dm2-m+2)Bm_1 + n(D2 -m2+m+2)Bm+1 = 0 where D2 = r2 d2 dz2 (3.4) (3.5) (3.6) (357) Considering Equations (306) and (357), not only are they three term recurrence relations, but they are also second order differential equations. However following the work of Fedor(9) assume A2 = -7A1 (3.8) and B2 = -4B1 (3~9) where 7 and * are positive quantities to be determined. Substituting

Equation (3.8) into Equation (3.4), it becomes [(2-7n)D2 - 2(1+7n)]Al = 0. Letting U2 = 2(1+yn) (3.10) (2-yn) then D2A - CA1 = 0 o (3.11) Thus we have a second order, linear, homogeneous equation with constant coefficients. In view of the third boundary condition, Equation (2.15), the solution of Equation (3.11) must be even in z. Accordingly the solution of Equation (35.11) is A1 = C1 cosh _z (35.12) r where C1 is a constant of integration. In view of Equation (3.8) A2 = -7C1 cosh w- o (3513) r Substituting Equation (35.9) into Equation (35.5), it becomes [(2 -mn)D2 - 2(l+4n)]Bi = 0 o Letting 2 = 2(1+fn) (3 14) (2-pn) then D2B1 - 2B1 = 0. (3o15) Again in view of the third boundary condition, Equation (2.15), the solution of (3515) must be even in z. Accordingly the solution of (3515) is B1 = N1 cosh z (3.16) r

-25 where N1 is a constant of integration. In view of Equation (3.9) B2 = -rN1 cosh Z. (5.17) r Knowing Ai. A2, B1, and B2, all of the Am and Bm may be evaluated from the general recurrence relations, Equations (35.6) and (3.7) respectively. This is done for the first ieight terms of each series in Appendix C. A bound for this particular number of terms will be established later on. Now applying the third boundary condition, Equation (2.15) p(G, -) = 0, to Equation (3.3) v6r2(-2 d0)[n(2+n cosG) sinG + 6.r dn[ 1 (5.18) c2 dt (2+n2)(l+n cosg)2 c2n dt (l+n cosg)2 - ~S Am(r) sin mg - i Bm(2) cos mg = 0. m m=1l This requires that the first two terms of Equation (3518) be expanded into Fourier series. Considering the first term 6ur2 K rn(2+n cos~) sinG C2 (X-2 dt)[(2+n2)(l+n cos)2] (3019) -= _6 (-2 d) 1 [ n sing s n sing ] 2 dt (2+n2) (l+n cos (l+n cosg)2 Now n sinG 2a sing (l+n cosG) (l+2a cQo + a2) 9 where 8, a;; (3.20) It is known that(14) sing ("a)I-4l sin mg (1+2a cosQ + a-) ml)

-26 which is uniformly convergent for 0 < a < 1. Accordingly n sing (l+n cosg) n sinG (l+n cosG)2 00 =2 Z m=l (-l)m-la sin mg (3.21) Now a n sins I = n co)J 9 6n (l+n cosg) therefore from Equation (3.21) n sinG (l+n cosG)2 6 00 = n - [2 Z 6a m=l (-l)m- sin m (1)m a sin mG ~n or n sinG (l+n cosQ)2 2 (-1) m1m a sin mG = m=l ( n2) /2 m=l (1 -n2)1/2 (3522) Substituting Equations (3521) and (3.22) into Equation (3519) -6.-r2(2 d)[. n(2+n cosQ) sinG] c2 dt (2+n2)(l+n cosG)2 2 c2 00 12pr, 2 (2 d0) ( 1, ) Z c2 dt 2+n2 m=l (3023) mS. (.l)m-lam(l + m ) sin (1-n2)l/2 Considering the second term of Equation (3518) it is noted n2. 1 2n3 sinG 9[ (l+n cosG)2' (l+n cos)3 n [ n sinG = (~n sinG + n2 sing cosG) an (l+n cosg)2 (l+n cos)3 (3.24) (3.25) Now and n2 n sin 2n sin (-n sin + n2 sing cosQ) S~2 (l+n cosG) ( l.+n cos)5, + (+n cos,) 862 (l+n cosg) (l+n cosg)3 (3.26)

-27 By adding and substracting a term in Equation (3.24) it may be written as n2 a[ 1 = 2n3 sin 6G (l+n cosg)2 (l+n cosQ)3 (-n sinG + n2 sinG cosQ) (1+n cosg)3 +(-n sinG + n2 sinG cosG) (1 + n cosg)3 (3.27) In view of Equations (35.25) and (3.26), Equation (3.27) may be written n2 a[ 1 1 n - (1+n cosg)2 ~) 2 n sinG =~ -(l+n cosG) n n sinG + n (l+n cosq)2] (3.28) From Equation (3.22) n [ n sin9 ] an (l+n cosG)2 00 =2 Z m=l m2 mn2 m-l m [(1n2) + (1_n2)/2](-l) a sln mG (3.29) Considering Equation (3.21) n sinG [~ (l+n cosg) 00, (-1) mm amcos mg m —l (3.30) and therefore 00 82 [ n sin ] = - 2 892 (1+n cosg) m=l (.,)m-lm2apsin mg. (3.31) Substituting Equations (3.29) and (3531) into Equation (3528) n2 [ 1 ] -2 G~ (l+n cosg)2 m=l (.l)m-lm am[m - m (1-n2) - -1n2.] sin mG. (1.-n (3232) (5.52) Integrating Equation (3.22) with respect to G n2[(+n cos 2] = 2 Z ( m=l m n2 (-l)m-amm- ( m - (1 n2 -os mgn (3.33)

-28 From Equation (5533) the second term of Equation (3518) may thus be written as 6[ir2 dn[ 1 nc2 dt (l+n cos@)2 (3-4) v~ dn {2 Z ( -l[m a m _ n2} 2 r= n 1-ia 2 - m —w - ]. cos m@} rin5 c2 d-t m=l (ln2) (1-n2)3/2 Although termwise integration of a Fourier series is quite legitimate, in this case Equation (5 32), the termwise differentiation of a Fourier series requires more caution. The necessary criteria is that the resultant series must converge uniformly in the interval in questionn It is thus necessary to establish the uniform convergence of the series in Equations (3522), (3529), (3530), and (3531)o Considering the series in Equation (35.22) 2 (-l)m-lmam 2 E 2)1-... sin mg m=l (1-n2)1/2 this series is uniformly convergent on the interval 0 < a < r if 0 < r < lo um(a) = maim and. M mrm, noting that sin mg is bounded by one. Then on the stated interval jal < r and so |um(a)l < Mm' Since the series Z mrm is convergent, the uniform convergence follows by virtue of the Weierstrauss M-testo The series in Equation (3530) 2 Z (-l)m-lmam cos mg m=l is uniformly convergent on the interval 0 < a < r if 0 <r <. Noting that cos mg is bounded by one the proof is identical to that for Equation (3.22) directly above.

-29 Considering the series in Equation (3531) C00 2 Z (-l)m-lm2am sin m, m=l this series is uniformly convergent on the interval 0 < a s r if 0 < r < 1. For let um(a) = m2am and Mm = m2rm, noting that sin mG is bounded by one. Then on the stated interval |a| < r and so |um(a)| < Mm. Since the series Z m2rm is convergent, the uniform convergence again follows by virtue of the Weierstrauss M-testo The series in Equation (3.29) 2 0 ~ m2 in2 m- lam 2 m [ n + n2/ ] (-1)m-lam sin mg m=l (1-n2) (1-n2)3/2 is uniformly convergent on the interval 0 < a < r if 0 < r < 1. Noting that sin mi is bounded by one and taking advantage of the fact that the sum of two uniformly convergent series is itself uniformly convergent, the proof of the uniform convergence of the individual terms of Equation (35.29) is identical respectively to that of Equations (3531) and (3.22) above. Substituting the results of Equations (3.23) and (3534) into Equation (3.18) 12 (c-2 )( _ 2) Z (-1) lam[1 + in ] sin mg c2 dt 2+n2 m=l (1-n2)1/2 + 12rdn2 (-l)m-lam[m- m _ n2 cos m n5c2 dt m-l (1-n2) (l-n2)3/2 - Z Am(-) sin m9 - Z Bm(2) cos m9 = 0. (3535) m -d ml 2

-30 Equating coefficients of sin mg and cos mg in Equation (3535) equal to zero Am(Q) = 12A (a2 (Dc2 dt)( l2-n)(l)m lam[l + 1 (3~36) 2 C2 dt 2+n2 (1-n2)1/2 and Bm() -12}lr2 dn ( ml)m lam[m _ m n2 n3c2 m n- (3~37) (2) nc2 dt (1-n2) (1.-n2)3/2 Equations (3536) and (3537) in addition to yielding sufficient boundary conditions to evaluate the constants of integration for all of the Am(z) and Bm(z) also permit the evaluation of y and r in Equations (358) and (359) respectively. Considering Equations (3o12) and (3536) for Ai (i) C cosh ay2 (a) -2 )( ) a[l + 1 (58) 2r c dt 2+n- (n)1/1 and from Equations (3.13) and (3356) for A2(-) -C cosh - - -1 2r(-2 dn) (3i 29) Dividing Equation (35.539) by (53538) and substituting Equation (53.20) for a r n[2 + (l-n 2)1/2] ( [1 + (l1n2)l/2]2 From Equations (3516) and (3537) for B1(2) N1 cosh I 122r2 dn l 1 n2 3 (5 41) 2r nTcT- dt (l-n2) (l —n2)3/2 and from Equations (5317) and (53537) for B2(4) 12c.r2 dn 2 2 n2 -1rNl cosh = - nc2 d- a[2 - (n (in2)3/2 ~ (5342)

Dividing Equation (3542) by (3541) and again substituting Equation (3520) for a = n[l + 2(1-n2 )1/2] (343 [1 + (1-n2)1/2]2 It is to be noted finally that the last boundary condition, Equation (2.16), is satisfied by the chosen form of p(@,z). Now since Equation (353), which is repeated for convenience, p(~,z) 2 po(1- ) 6 r2 )[ n(2+n cos9) sing 3 ~(2 i d dt (2+n2)(l+n cosg)2 6 r2 d 1 C + nt (l+n IosZ) m- Am(z) sin mG - 1 Bm(z) cos mg nc dt (l+n cosg)2 m=l mM l satisfies Reynolds Equation (531) and meets all prescribed boundary conditions, it is a solution of the problem under consideration. Bo Convergence of the Series Now the question of convergence comes up for the series co00 Z Am(z) sin mg m=l and 00 Z Bm(Z) cos m o m=l The first of these has been shown(9) to be convergent in the following manner. Making the approximations (m-l)(m+2) (m-1)2 and (m+l) (m -2) (m+L) 2

-32 for large m, then the recurrence relation for Am(z), Equation (3o6), may be written as 2[D2_m2]Am(z) + n[D2-(ml)2]Aml(Z) + n[DK2(m+l)2]Aml(z) = 0 (3s44) Letting (D2Km2)Am(z) = Fm (3~ 45) where Fm = Fm(m,n,z) and substituting into Equation (3544) yields the following second order difference equation with constant coefficients 2Fm + nFm+l + nFml = 0 (346) Letting Fm = Ux, then from Equation (3546) UX-1[U2 + 2 U + 1] =0 (3~47) or - 1 + (l-n2)1/2 U1,2 = n Now in order to insure Fm is finite at n = 0 it is noted that n U12 = ~ 1 + (l-n2) 2 also satisfies Equation (3047)0 Accordingly F p ~ n m - n m -1 + (1-n2)1!2] T21' + (1-n2)1/2 ~ In order to insure F. = 0 at n = 0 it follows that T1 = 0. Letting F0 = f(z), then Fm = (1)m2)/ f(z)[ (-l)mam f(z) (3048) 1+ (1-n2)1/2 =(1 (~

-33Substituting Equation (35.48) into Equation (35.45) (D2-m2)Am(z) = (-l)mam f(z). (3.49) Since Am(z) is an even function the complementary solution is Am(z) = Cm cosh mz o r A particular solution by variation of parameters is Am(z) = (-l)mam(r) f f(zl) sinh[m(z-zz)] dz or replacing the integral by its maximum value M, then the solution to Equation (35.49) may be written Am(z) = Cm cosh mz - (-)mamM (550) r m2 Evaluating Equation (3550) at z = - it follows that Am(") + (-l)mamM Cm = — - m cosh - Equation (3550) may then be written as Am( )cosh m- (-l)mamM cosh mz Am(z) =r[ r - csh m2 m2 cosh m Therefore the inequality exists < Am(')cosh - amM cosh m z IAm(z)I +) --- = sh m CO2 coshh mi 2r 2r or more strongly Am(z)l <_- +)l a OM 2 M2~~~~~~~

-34 Letting Rm = IAm() I + amM then it follows from the Cauchy ratio test that Z Rm is absolutely m=l convergent since 0 < a < 1 for 0 < n < 1 and lim Am+1(2) m Iand and lim am+lM /amM m o (m+1)2 m2 Now | sin mi| <- 1 and IAm(Z) < ER comparison test the series = a 1 for large m and therefore by the 00 Z Am(z) sin m@ m=l absolutely converges for - I< z < A 2 - _2' 2 2 0 <n<l 2r 0 < n < 1. The proof for.00 Z Bm(z) cos mg m=l is identical since the recurrence relation is identical and Icos m~] < 1. Co Estimate of the Series Error An upper bound on the error involved in a finite number of terms of the series

-35 00 Am(z) sin mg m=l can be determined in the following manner. A little reflection on this series will reveal that the Am(z) are attenuated at z = 0 and reach a maximum at z = A- where the axial flow in the bearing will attain its 2 greatest importance. This is further substantiated by both the mathematical form (see Appendix C) and the numerical evaluation of the Am(z). Accordingly the following inequality may be written that Z IAm(z)l|sin mg| < Z IAm(i) | |sin m@| m=l- m=l 2 or 00 00 Z I'Am(z)I Z i Am) Considering Equation (35.536) which is rewritten for convenience Am() = En (-1)m-l am[1 + (3.36)'(2+n2) (1=-(n2)1/2 where F1 - c X-2 dt it is noted that Am(r) is the combination of the two geometric terms am and mam a d (am) da By virtue of the known sum of the geometric series morm and its derivative co mrm-1 = 1 m= (.1-r)2

-36 it follows that 00 Z IAm()l Fl + a -1] (5351) m=l 2 (2+n2) (l-a) (l-n2)l/2(l-a)2 Considering the series o00 Z Bm(z) cos m ~ m=l it is again noted that Z |Bm(z)!Icos mj < z IBm(T)Iicos B( |os m m=l m=l or,v!~(z)] < z IBm()1 o m=l m=l Rewriting Equation (3.37) for convenience F2m-lm| m n2 ______ Bm(X) = f(-) a m - ( m - n2 ] (557) where F 12-lr2 dn F2 2 dt then again by virtue of the known sum for the geometric series and its derivative it follows that m=l 2 n5 (l-a)2 (1-a)2 (1-n2) (1-a)(ln2)3/2 1] 52) It is to be noted that the error involved with the series Bm(z) is approximately twice as great as that for Am(z) for any given number of terms. A typical maximum error for the series 00 Z Bm(z) cos mg m=l

-37 considering the first eight terms and corresponding to a steady state position of n = 0.80 is 3.5 percent. However in the actual evaluation, the boundary condition for a supply pressure of 50.0 psig was checked with a maximum error of 0.2 percent for eight terms of the series. The error interior to the ends of the bearing would of course be even less than this. Do Condition on the Inlet Pressure In order to insure that the oil film is continuous, the condition is imposed that the axial pressure gradient at the end of the bearing is negative. It follows then from Equation (3.3) that (P) = - P - E. c sin m - E m cos mg ~ < 0. (3~53) 6z Z = A- I m=l oz m=l 6z z = If the first term is taken from each series, an equation can be found that imposes a minimum value for the feed pressure po. Greater accuracy on this lower bound may be found by including more terms of the two series. From Equation (3512) (M ) I2_ = Cl(r) sinh -, where from Equation (3536) - 12ytr2(o2 d)( 1 )( + 1a 1 2 dt 2+n2 (1-n2)1/2) cosh a_ 2r From Equation (3.16) (Bl) = Nl(r~) sinh A

-38 where from Equation (3537) N1 =! r2 dn 1- 1 _n2 a n3c2 dt (1-n2) (1-n2)3/2 cosh ~ Substituting these results into Equation (5.53) and noting Equation (3520) for a o > 1 - 2 d))( 1 )( n )tanh 2)(sin ) 1 0 c2 dt 2+n2 (1-n2)1/22r - (d)((1_n2)3/2)()(tanh 2)(cos 9) B (354) Now 9 is the angle at z = ~ where the tendency is the greatest for positive gradients to exist. Thus the angle 9 in question is the angle which maximizes the right hand side of Equation (3554). Differentiating the right hand side of Equation (3554) with respect to 9 and equating to zero gives G = tan -[ (^)an e t n-. n (3n55) tandn 2+n2 tanh d- d

IVo EQUATIONS OF MOTION OF THE JOURNAL A. Formulation of the Equations of Motion Referring to Figure 4.1 the displacement of the journal center with respect to a fixed reference point, in this case the center of the bearing, may be written as Y' 5= e eN ~ The velocity of point O is then -~ de -h.b do -_ vo =- eN + e e It follows that the acceleration of point 0' may be written as aO, = [d2 - e -- )2] eN + [e d- + 2 de - e a0 dt2 dt dt2 dt dt Considering the scalar equations of motion Z FN = maN and Z Ft = mat, then for the normal direction 2/2 2it WD cos(Z + P) + f f p(9,z) cosGrdgdz + Ws cos~ (4.1) -2/2 0 = M [d2e - e(d-)2] dt2 dt and for the tangential direction 2/2 2rt -WD sin(0 + P) + f J p(O,z) sinGrdgdz - W sin0 -2/2 0 - M [e d +2 d d] (4.2) dt2 dt dt -39

eN Figure 4.1. Geometry for Journal Equations of Motion.

where M represents the mass of the journal plus the mass of the propeller. The pressure function p(g,z) is defined by Equation (353) which for convenience is repeated below: i( z 6= r2p.2 dr n(2+n cos~) sinG z) = 2 2 + dt)[(2+n2)(l+n cosg)2 n dn [ 1 ] c2n dt (l+n cosG)2 00 00 - Z Am(z) sin mg - Z Bm(z) cos mg. m=l m=l (4.3) B. Evaluation of the Pressure Integrals Considering the evaluation of the integrals of the pressure function in Equation (4.1), then for the first term of Equation (4.35) 2/2 2it z _/2 / 2t f f po(1 -) cosgrdgdz = por f [(2 - z) sing] dz = 0. -2/2 0 o -i/2 i 0 The integral of the second term of Equation (43.5) may be written as 2/2 2n 6(r2 do n(2+n cosG) sinG cosrdd z f f -— (co-2 -) cosrddz -2/2 0 c2 dt) (2+n2)(l+n cosg)2 f/2 (2+n cosQ)sincosdQ (2+n cos (2+n cos)sinGcosGQd F f {f 2 d (1+n.cos)2 z -/2 0- (l+n cosG) )2 (4.4) 5) where 6pr3 (o2 do) n t2 a d-(2+n2) Letting G = P + 2it in the second term of the right hand side of Equation (4.5), then the right hand side of Equation (4 5) becomes F / { T (2+n cosQ)sinecosQd (2+n cos))sinGcocossdd ( 4 -2/2 0 (l+n cosG)2 _ (l+n cos3)2

-42Replacing P = -y, inverting the limits of the second term in (4.6) and substituting this result into Equation (4.5) it follows that / 2 6 r 2 ) )[n(2+n cosG)sin cos9rddz -^/2 0 c dt (2+n2)(l+n cos@)2 2 /2 r (2+n cosg)singcosid@' (2+n cos7)sin7cosydy d -2/2 0 (l+n cos@)2 0 (l+n cosy)2 = 0. Considering the third term of Equation (4.3) then Y/2 21 2/2 2n 6~ dn [ 1 ] cosirdidz -,/2 0 nc2 dt (l+n cos@)2 6pr3 dn 2/2 1 sing n d@ ]2 d --- * -.. —--- J. dz nc2 dt -2/2 (1-n2) (l+n cosm) (1-n2) (l+n cos) 0 6~r dn f2 -2n [ 2 ll 1-n 0_)~ - /{ -- [ L 2 tan l(F tan )] dz nc2 dt _-/2 (1.n2) (in2)l/2 l+n 2 O Evaluating the limits with respect to @ and integrating with respect to z 2/2 2i 6Fr2 dn 1 f fd [ 1 ] cosgrdgdz -2/2 0 nc t (l+n cosQ)2 12mnir3 1 dn (4.8) c2 (1-n2)3/2 dt ~ In view of the orthogonality relationship 2i f cos rx sin sx dx =0, 0 where r and s are integers then the integral of the fourth term

of Equation (4.3) becomes Q/2 2r 00 f f [- Z Am(z) sin mg] coserdedz = 0. (4.9) -,/2 0 mdl From the additional orthogonality relationship that ~2<~ 0, r s 2 cos rx cos sx dx = i t, r = s. 0 then the integral of the last term of Equation (4.3) becomes Q/2 2 2 Q/2 J [f - Bm(z)cos m@] coserdedz = -rx f Bl(z)dz. (4.10) -2/2 0 m- l -2/2 From Equation (3.16) Bi(z) = N1 cosh _z r and thus Equation (4.10) may be written /J 2 [- Bm(z)cos me] cosgrdadz = 2 N1 sinh. (411) -/22 0 m= — 1 2r However from Equation (3.41) 1 12 r2 dn n n2 cosh { n3c2 dt l+(l-n2)1/2 (1-n2) (1-n2)5/2 2r and Equation (4.11) may be written as i/2 2,t C f f [- Z Bm(z)cos mg] cosGrdgdz -2/2 0 m(4.12) 24 r4n d [ 1 1[ n2 tanh - n2a2^ dt 1+(l-n2)l/2 "(1-n2) (l1-n2)5/2 2r

Considering next the evaluation of the integrals of the pressure function in Equation (4,2), then for the first term of Equation (453) Q/2 2Tn. f -Y/2 0 Po ( - )sinGrd.dz = - por 2I [( - ) s ] z -2/2 2 2 (4.13) The integral of the second term of Equation (4.3) may be written as 2/2 2it 6,r2 d n(2+n cosg)sinG f f (cw-2 -)[ (2+n2)+n cos)2 sinGrdGdz -I/2 0 c2 dt (2+n2)(l+n cos~)2 /2 2Ic =6 r(-2( dI t)( n f f c2 dt (2+n2) -Q/2 0 [(2+n cosg)sin2g] dGdz (l+n cosQ)2 (4.14) Letting 20 0 (2+n cosG)sin2G (l+n cosG)2 2 [1 + (l+n cosg)]sin2G 0 (l+n cosO)2 2i sin2 d Q (l+n cosg) + 2 2 sin2G f dG 0 (l+n cosG)2 and making the substitution that u = cosO and du = -sinGdO, then = - f l-u2 du 0 l+nu 2it F lU2 du__ (1-u2 du2 0 (l+nu)2 Making the further substitution that z =! + nu and dz = ndu, then 1 21t (n2_l)+2zz2 dz F =n2 - -0 -- dz 1 2 r (n2-1)+2z-z2 n2 05 dzz2

-45 The solution of these integrals may be written as T; 1 1 2-) z -Z 2t + / 0 dz l(n2-1) + 2z z2 27K + (n2-1) f - 0 z dz V(n2-1)+2z'z2 _- _ (n2-1) + 2z-z2 z 20 + f 0 dz z (n2-l)+2z-z2 2T -f 0 dz V (n2-l)+2zoz2 or 1 n [ (1 - z) (n2-l) +2z-z2 2-A + n2 f 0 dz z ~(n2-l)+2z-z2 - 1 [( 1 ) (n2-l)+2z-z2 n2 sin-1 2z+2(n2-1)} ] - z V'42 However z = (l+n cos~) and therefore - 2 n2singcosg nF (l+n cosg) n2 sinln+cosQ t (1-n2)1/2 l1+ c~sg 0 2 (1-n2)1/27 Substituting this value of F back into Equation (4.14) and integrating with respect to z then Q/2 ^2 f6r2 di n(2+n cosg) sinG sinrdgd z J J - 2 ( ci-)t (2 sin(rdGdz -o/2 0 c dt (2+n2)(l+n cosg) c2 td/ n (2+n2)(1-n2)1/2 (4.15)

Considering the integral of the third term in Equation (4-3) then it follows directly that /2 2 6(r2d n 1;f f - [ 2 —] singrdgd.z -/2 0 nc dt (l+n cos~)2 6cr2 dn nc2 dt Q/2 1 2t f E 1 ] dz 0 o 5 /2 n(l+n cos@) 0 -19/2 (4.16) From the orthogonality relationship 2n f sin rx sin sx dx = 0 0, r A s, r = s > 0 then the integral of the fourth term of Equation (4.3) becomes Q/2 2it f f [- Z Am(z)sin mg]sinGrdgdz -e/2 0 m=l r/2 = -rT f -/2 Al(z)dz (4.17) From Equation (3.12) A1 = C1 cosh r r and thus Equation (4.17) may be written 9/2 2ir oo Jf [- E Am(z)sin mg] singrdgdz = - 2r2T- C sinh a y -Q/2 0 m=l However from Equation (35.538) 1 =,1 [l21pr2(2 dt n n q cosh — e (1) m -2a (2r2)e1/2)] 2r and Equation (4.18) may be written as (4.18) Q/2 2it f f [-../2 0 00 Z Am(z)sin m~] singrdadz m=l 24r= _, do_ n cd o2 ~^(c-2 d -t tanh 2 =Cc2 d (2+n2)'(1-n2)1./2 2r (4.19)

-47 In view of the orthogonality relationship that 2x J cos rx sin sx dx =0, 0 where r and s are integers, then the integral of the last term in Equation (4.3) becomes 0/2 2r 00 f [- Z Bm(z)cos mO] sinGrdgdz = 0. (4.20) -X/2 0 mel C. Simplification of the Equations of Motion If it is noted that 1 1 [i- 1 n2 1 [l+(l-n2)/2 1 (n2 - (1-n2)3/2 ] (1-n2)32 and the results of Equations (4.4), (4.7), (4.8), (4.9) and (4.12) are substituted into Equation (4.1), then the equation of motion for the normal direction becomes WD cos (0+) + Ws coso +. [2 tanh l - = M[- - e( ) ]. (4.21) c2 2r 1-n2)2dt dt2 dt Similarly, substituting the results of Equations (4.13), (4.15), (4.16), (4.19) and (4.20) into Equation (4.2), then the equation of motion for the tangential direction becomes -WD sin(1+p) - WS sin0 12tr35r [2r tanh "- l](o-2 di) ni c2 2r dt) (2+n2)(1-n2)/2 = M[e d2 + 2 dt dt (4.22) -dt2 d-t dt]

Considering Equations (4.21) and (4.22) the inertial terms can be shown to be negligible. Since the bearing is going to oscillate about a steady state position with a maximum radial travel equal to the minimum film thickness, which is in the order of 0.001 inches, and the circular frequency of the dynamic load is four times the angular velocity of the journal for a four-bladed propeller, which in this case is 11.0 radians/ second, then the maximum acceleration expected will be a = r2 = 0001[4 x 112 0.16. 12 This as a percentage of gravity is 0.16 x 100 = 0.5% o 32.2 The inertial forces are therefore in the order of 1/200 of the gravity forces and shall be neglected. The dynamic load will be assumed to be of the form WD = WD sin bcot + WD sin 2b(ot - A) (4.23) where WD1 and WD2 represent the amplitudes of the first and second harmonic components of the propeller loading, b is the number of blades and A a phase angleo It should be noted that these are termed the first and second harmonic components in terms of the blade number and are actually not the first and second harmonics in the usual sense. Utilizing the fact that the inertial forces are negligible, substituting Equation (4.23) for WD and introducing the dimensionless constants Hi = WD, H 2 (4.24) w ~S wS

-49 and the constant K = 12.r3 (4.25) c2WS Equations (4.21) and (4.22) may be written respectively as [{H1 sin bot + H2 sin 2b(wt-A)} cos3 + 1] cosO -[{Hi sin bot + H2 sin 2b(wt-A)} sinp] sing 2r ti 1 dn + K[| tanh 2 - 1] ( 2)3/2 dt4.26) 2r r dt n and. -[{Hi sin but + H2 sin 2b(cnt - A)} cosp + 1] sin0 [{Hi sin but + H2 sin 2b(Lut-A)} sink] cos0 - K[ - tanh 1-l(w2 d.n 0 (4~27) - K[2 tanh r - " -2 dt (2+n2)(1n2)l/2 04o27 It is convenient at this point to make the change of variable = (ln2)2 (428) In terms of the variable y it is noted that 1 dn = dy (l1-n2)3/2 dt dt and n y(l+y2) (2+n2) (1-n2)1/2 (2+3y2) Substituting these relations into Equations (4.26) and (4.27) and.solving for the time derivativesof the dependent variables y and ~, then Equations (4.26) and (4.27) become respectively

-50 dy = [{H1 sin bwot + H2 sin 2b(wt-A)} cosp + 1] coso dt ~ K[l - 2r tanh ] CQ 2r [{Hi sin bwt + H2 sin 2b(cwt-A)} sinp] sing K[l - 2r tanh y ~l 2r (4.29) and do_ [{H1 dt 2 - sin bct + H2 sin 2b(wt- A)} cosp + 1] sing 2K[ 1 - 2 tanh_] Y(1+y2) ~a 2r (2+3y2) [{HI sin bot + H2 sin 2b(wt-A)} sina] coso 2K[1 - tanh ( 2) c~e Y, 2r' (2+3y2) (4~30) In terms of the new variable y Equation (3o10) for 1/2 -2 = 2[7y + (l+y2) ] 2(1+y2)/2- -y 2 becomes (4.31) where from Equation (3.40) 7 = y[l + 2(1+y)1/2] [1 + (l+2)1/2]2 (43.2) Similarly Equation (3.14) for ~2 2 _ 2[*y + (1+y2)1/2] 2(1+y )/2 - becomes (4.33) where from Equation (3543) y[2 + (++y2) 1/2] [1 + (1+y2)l/2]2

-51 Substituting Equation (4.32) into Equation (4o31) 2{ 1+2(l+y )1/ + (l+y ) l+(+y )1/2 ] 1 (4o ). 2(~+r2) /2{~+(~+Y) /2} - 2{+2(1+y2)1/2} and substituting Equation (4~34) into Equation (4.5533) 2[Y{2+(l+y2)l/2} + (l+y2)l/2{1+(1+y2)1/2}2] 1/2 L 2(l+y2)l/2{l+( l+y2) 1/2}2 {2+(l+y2)l/2} 4-36 Values of y, 7, Oa, and. verses n are given in Table 4.1. Considering Equations (4.35) and (4.36) for oC and, it is seen that these expressions are quite complex for computational work. These may be approximated by polynomials of the form oc 0.o002204y - o04980'2 + 0o6518y + 0o7843 (4.37) and 0,o008050y3 - 0.1215y2 + 0.6154y + 0.8332, (4.38) where for 0.6 < n < 1.0 or 0.75 < y <oo the maximum error involved in either expression is 0.68 per cent. The solution of the two simultaneous, non-linear, first order differential equations given by Equations (4o29) and (4.30) will give the path of the journal in its orbit about a steady state position and the velocity components corresponding to the translational and rotational motion of the journal center in this orbit. Thus for a given point (n, 0) in this orbit, and the corresponddn do ing velocity components (dt d) the pressure may be evaluated around the circumference and along the length of the bearing from Equation (4.3)o

-52 TABLE 4.1 y, 7,) (X'', / vs n n y 7 c 0.00 0.0000 0.0000 1.C 0.10 0.1005 0.0753 1oC 0.20 0.2041 0.1520 o C 0.30 0.3145 0.2321 1.C 0.40 0.4364 0.3176 1.C 0.50 0.5574 0.4115 1.1 0.60 0.7500 0.5185 1.2 0.70 0.9802 0.6466 1.3 0.80 1.55553333 0.8125 1.5 0.85 1.6136 0.9214 1.7 0.90 2.0647 1.0633 1.9 0.95 350424 1.2756 2.3 0.96 35.4286 1.3359 2.5~ 0.97 35.9900 1.4080 2,7. 0.98 4.9247 1.4990 35.0 0.99 7.0179 1.6280 3o 6 1.00 oo 2.0000 oo )000 )056 )229 )527 )970 L593;460.702 634'120 371 691 223 314 497 676 __ 0. 0000 0.0751 0.1510 0.2285 0.3085 0.3923 0. 4815 0. 5785 0.6875 0.7484 0. 8171 0. 8962 0. 9141 0.o 9329 0.9530 0.9749 1. 0000 __ 1.o 0000 1. 0056 1.0227 1.0519 1. 0942 1.1516 1.2274 1.3273 1. 4622 1. 5492 1.6566 1. 7955 1. 8290 1.8652 1. 9048 1. 9488 2. 0000

V. RESULTS AND DISCUSSION A. Physical Data of the Case Considered for Numerical Evaluation Numerical results for the simultaneous location of the journal center and the corresponding pressure distribution in the bearing will be presented under separate headings as they are actually two distinct steps in the numerical evaluation. The data considered for numerical results of this problem is from the stern tube bearing of the So S. John Go Munson of the Bradley Transportational Line. The pertinent characteristics of the bearing are: X= 11.0 radians/second, r = 9.25 inches, = 74.0 inches, c = 0.0435 inches, b =4, p 1.4 x 107 pound-second/inch2, Po = 30.0 (pound/inch2) gage o The actual static load on the bearing is unknown due to the effect of journal deflection. If the effect of this deflection is ignored the static load can be shown to be approximately 7 x 104 pounds. This static load however would yield a steady state eccentricity of 0.999966 and a minimum film thickness of 1.47 x 10-5 incheso This is considerably below the limit for which a lubricating film would be developed. It is felt that the journal actually rests on the after end of the bearing due to its static deflection and allows a film to be developed in the remainder of the bearing. For this reason the steady state eccentricity of the Journal will be assumed and the value of WS is then fixed. -53

-54 With respect to the location of the dynamic load and the amplitude ratios of its first and second harmonic components little is known. From the sparse information available it appears that the ranges of p and H1 are respectively 270 < P < 360 and 0 < H1 < loOo No information is available on H2 although it is certainly considerably less than HI. Bo Journal Orbits The solutions of the two simultaneous differential equations, Equations (4.29) and (4.30), describing the journal orbits were obtained numerically by the Runge-Kutta fourth order method onan I.BoM. 7090 digital computer. The step size used in the numerical integration was t = 0.005 seconds. As a check on the accuracy of the integration a step size of t = 0.001 seconds indicated a discrepancy of only one part in 5 x 106 for a total of 0.5 seconds. The integration was treated as an initial value problem utilizing the static loading position as initial values for n and 0. The integration was then continued until the journal orbit repeated itself within an error of 0.5 percent. In all cases considered here this occurred by at least the second time around the orbit. It should be noted that due to the assumption of a complete oil film, the steady state position for 0 from Equations (4.29) and (4.30) will always be ( = 90~O Figure 5.1, 5.2 and 5.3 show the orbits obtained considering only the first harmonic component of the dynamic load for three values of P (27005 300~, 3300) and using a steady state position of n = 0.85 and 0 = 90~. These figures illustrate that it takes 9.0 cycles in terms of the dynamic loading frequency (44.0 radians/second) or 2.25 cycles in terms of the journal frequency (11.0 radians/second) to complete

-55 ~ INDICATES THE START OF THE ORBIT Figure 5.1. Locus of the Journal Center for Hi 0.8, H2 = 0.0, b = 4, P = 270~ and K = 0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85.

-56 990 0 INDICATES THE START OF THE ORBIT Figure 5.2. Locus of the Journal Center for Hi = 0.8, H2 = 0.0, b = 4, P = 300~ and K = 0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85.

-57 * INDICATES THE START OF THE ORBIT Figure 5.3. Locus of the Journal Center for Hi = 0.8, H2 = 0.0, b = 4, P = 330~ and K = 0.1796. Steady State Coordinates are 0 = 90' and n = 0.85.

-58 one orbit of the journal center. The orbits are almost symmetrical about a line connecting the mid-point of the first cycle and the end of the fifth cycle. This shall be designated as 0mean in the discussion. It is seen that Omean is very sensitive to a change in the value of P locating the dynamic load, decreasing at a much faster rate than P is increasing. This amounts to a rapid conversion from one velocity (Ed) to the other velocity (dt) and as shall be seen below considerably effects the magnitudes of the pressures developed in the bearing. The principal characteristics of the above orbits along with other orbits investigated for H2 = 0.0 and a steady state position of n = 0.85 and 0 = 90~ are summarized in Table 5o.1. These results show that the size of the orbits vary directly as the amplitude ratio H1 but than Omean is almost independent of H1. It should be noted that the conditions of 0 = 90~ and B = 0~ are a singular point of Equations -.29) and (4o50) and represent an unstable condition. Figure 5. 4 shows the orbit obtained for these conditions, which although appearing similar in form to the preceding orbits continues to grow in sizeo Considering again the case of H2 = 0.0, the effect of increasing the eccentricity of the initial conditions is shown in Figure 5..5. Although the orbits are elongated in the 0 direction with a corresponding decrease in the n direction the general shape of the orbits is maintained. The former point is to be expected as resistance to motion in the n direction increases with increasing eccentricity.

-59 TABLE 5.1 CHARACTERISTICS OF THE ORBITS OF THE JOURNAL CENTER FOR THE CASE OF H2 = 0.0, b = 4, K = 0.1796 AND STEADY STATE COORDINATES OF n = 0.85 AND 0 = 90~ Hi n max nmin Omax Omin Omean 0.8 0~ 0,00~ 0.8 2700 0.8804 0.8025 96.20~ 83588~ 84151~ 0.8 300~ 0.8790 0.8054 99160~ 82.29~ 21o06~ 0.8 330~ 0.8750 0.8138 100.760 79850 7.88~ 0.4 2700 0.8664 0.8272 93~15~ 86.90~ 84.73~ 0.4 300~ 0.8655 0.8284 94.48~ 85.85~ 21.11~ o.4 330~ 0.8630 0.8323 95.47~ 84,75~ 7.90~ TABLE 5.2 CHARACTERISTICS OF THE ORBITS OF THE JOURNAL CENTER FOR THE CASE OF H2 = 0.0, b = 4, K = 0o0969 AND STEADY STATE COORDINATES OF n = 0.95 AND 0 = 90~ Hi i nmax nmin Omax Omin Omean 0.8 00 - - - o0.00 0.8 270 0 o 9611 0.9309 96. 32 83.79~ 65.490 0.o 8 300~ 0o 9606 0.9321 99.280 82.160 8.13 0.8 330~ 0.9591 0.9356 101.180 79.820 3.35 0.o 4 2700 0.9561 0.9409 93.26 ~ 86.80~ 65 860 0.4 300~ 0.9557 0.9413 94, 590 85.740 8.24~ 0.4 330~ 0.9548 0.9429 95~55~ 84.69~ 3,51~

-60 * INDICATES THE START OF THE ORBIT Figure 5.4. LoCUs of the Journal Center for H1 = 0.8. H2 - 0.0, b = 4, and K - 0.1796. Initial Coordinates are = 90 and n 0.85.

-61 n Figure 5.5. Locus of the Journal Center for H1 = 0.8, H2 = 0.0, b - 4, p = 270 0 a nd K 0.0969. Steady State Coordinates are -90 an n m 0.95.

-62 Table 5.2 summarizes the principal characteristics of orbits investigated for steady state conditions of 0 = 90~ and n = 0.95 with H2 = 0~0. It is again noted that Omean is essentially independent of H1, depending largely on p. A comparison of Tables 5.1 and 5.2 shows that 0mean also depends on the steady state value of n but to a much smaller extent than $o Including the second harmonic component of the dynamic load at various phase angles A produces no substantial changes in the general shape of the journal orbitso Principal characteristics of the orbits investigated are summarized in Tables 5.3 and 5.4. The variations of Omean verses P, the amplitude ratios HI and H2, and for differential initial eccentricities follow the same patterns as those for the case of H2 = 0o0o It is also seen that Omean is essentially independent of Ao Figure 5.6 shows a typical journal orbit obtained if the blade number is changed to b = 35 In this case it takes only 7.0 cycles in terms of the dynamic loading frequency (3350 radians/second) or 2533 rotations of the journal to complete one orbit. If the blade number is changed to b = 5 it now takes 11.o0 cycles in terms of the loading frequency (55~0 radians/second) or 2.2 rotations of the journal to complete one orbito A typical orbit obtained for this case is shown in Figure 5.7.

-63 TABLE 5.3 CHARACTERISTICS OF THE ORBITS OF THE JOURNAL CENTER CONSIDERING BOTH THE FIRST AND SECOND HARMONIC COMPONENTS OF THE DYNAMIC LOAD FOR b = 4, K = 0.1796 AND STEADY STATE COORDINATES OF n = 0.85 AND 0 = 90~ HI H2 A 1 nmax nmin Omax Omin Omean 0.8 O.4 0.00~ 270~ 0.8805 0.7885 97.74~ 82.37~ 84.15~ 0.8 0.4 3.75~ 270~ 0.8815 0.7903 97.480 82.520 83.59~ 0.8 0.4 7.50~ 270~ 0.8819 0.7940 96.88~ 83.01~ 83.23~ 0.8 O. 4 11.25~ 2700 0.8817 0.7999 96.100 83.70~ 83.18~ 0.8 0.4 11.25~ 300~ 0.8799 0.8049 99.26~ 81.95~ 28.09~ 0.4 0.2 0.000 270~ o.8665 0.8216 93.81~ 86.26~ 84.42~ 0.4 0.2 3.75~ 270~ 0.8671 0.8223 93.70~ 86.330 83.86~ 0.4 0.2 7.50~ 270~ 0.8673 0.8238 93.44~ 86.540 83.45~ 0.4 0.2 11.25~ 270~ 0.8672 0.8262 93.10~ 86.85~ 83.25~ TABLE 5.4 CHARACTERISTICS OF THE ORBITS OF THE JOURNAL CENTER CONSIDERING BOTH THE FIRST AND SECOND HARMONIC COMPONENTS OF THE DYNAMIC LOAD FOR b = 4, K = 0.0969 AND STEADY STATE COORDINATES OF n = 0.95 AND 0 = 90~ Hi H2 A P nmax nmin Smax Omin Omean 0.8 0.4 0.00~ 270~ 0.9612 0.9251 97.86~ 82.29~ 64.11~ 0.8 O. 4 3.75~ 270~ 0.9615 0.9258 97.60~ 82.440 62.88~ 0.8 0.4 7.50~ 270~ 0.9617 0.9274 97.01~ 82.93~ 62.22~ 0.8 0.4 11.25 ~ 270~. 9616 0.9298 96.22~ 83.62~ 62.33 ~ 0.8 0.4 11.25~ 300~ 0.9609 0.9315 99.38~ 81.87~ 11.09~ 0.4 0.2 0.00~ 2700 0.9561 0.9387 93.93~ 86.17~ 64.55~ 0.4 0.2 3.75 2700 0.9563 0.9389 93.82 ~ 86.23 ~ 63.33 ~ 0.4 0.2 7.50 ~ 270~ 0.9564 0.9395 93.56~ 86.45~ 62.69~ 0.4 0.2 11.25~ 270~ 0.9564 0.9404 93.22~ 86.75~ 62.80~

* INDICATES THE START OF THE ORBIT. Figure 5.6. Locus of the Journal Center for H1 = 0.8, H2 = 0.0, b = 3, p = 270~ and K =0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85.

-65 0 INDICATES THE START OF THE ORBIT Figure 5.7. Locus of the Journal Center for HI - 0.8, H2 0.0, b = 5, a = 270~ and K 0.1796. Steady State Coordinates are 0 = 90~ and n = 0.85.

Co Pressure Distribution in the Bearing Having obtained the journal orbits, then for any point (n, i) in these orbits and the corresponding velocity components (in, 2 ) the dt dt pressure profile around and along the length of the bearing may be determined from Equation (4.3). This again was accomplished on the I.B.M. 7090 digital computer including the first eight terms of each series accounting for the finite length of the bearing. Two of the most important questions to be answered about the pressure distribution in the bearing are how it varies around and along the length of the bearing for a complete orbit at a given initial eccentricity position. These two points were investigated rather extensively of which the results are indicated in Figures 5o8 and 5.9. Figure 5o8 shows that the region of minimum pressure encountered along the length of the bearing occurs slightly past the middle of the bearing toward the propeller endo The particular plot is for the point in a complete orbit where the greatest minimum pressure occurso All other positions in the orbit follow a similar patterno It is of course obvious that pressures of the negative magnitude obtained could not actually occur without rupture of the fluid filmo Figure 5.9 shows the variation of pressure at a fixed point in the bearing for a complete cycle. Again the point selected for representation is that where the greatest negative pressure occurs during a complete orbit. In studying the numerical results of the pressure variation around the journal for many different initial conditions it is found that the pressure profiles are all of a very similar design; being an

-67 a. H=: 0.8; H2. 0.0; A = 0.0; 8 39.835; e = 170~; n = 0.808 0.0 - 2.0 -4.0 b. HI. 0.8; H2 = 0.4; A = 89.82~; @ = 170~; c. HI " 0.8; H2 w 0.4; A =0 89.91~; @ = 170~; = 0.0; n = 0.800 = 11.25; n m 0.802 - j h'7 m / -8 A 0 0.. -I0 -12.0.0.0,,o_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __b_ I'l tlI I IL I I -14 -16. - 1i -0.5 -0.25 0.0 Z/t 0.25 0.5 Figure 5.8. Pressure Variation Along the Length of the Bearing for t = 0.7300 in the Journal Orbit; po = 30.0 psig, A = 74.0 inches, b = 4, p = 270~. Steady State Coordinates of the Orbits are n = 0.85 and 0 = 90~.

0 -100.0 -200.0 -300.0 OD CI CO a. -4000 -500.0 t (Sec) Figure 5.9. Pressure Variation at a Fixed Point (0 + ~ = 260~ and z - 0.0) in the Bearing for the Orbit Defined by HI = 0.8, H2 = 0.0, b = 4, p = 270~ and K = 0.1796. Steady State Coordinates of the Orbit are 0 = 90~ and n = 0.85.

-69 eight lobed pattern with eight regions each of positive and negative pressures o This pattern although fluctuating considerably in size remains essentially fixed in space during a complete orbit. It oscillates approximately + 5~ about a mean position. It is further found, as indicated in Figure 5.9, that there is one extreme value of negative pressure encountered in each orbit. This extreme value of negative pressure occurs slightly past the middle point of the orbit; that is just past the mid-points of the fourth, fifth and sixth cycles for three, four and five blades respectively. The angular position around the bearing where this occurs is defined by ( + = 170~ The numerical results also show that this extreme value occurs dn at an optimum condition of a large positive value of (-) and a large negative value of (-)O The relative magnitudes of these two velocidt ties are approximately equalo The pressure profiles shown below will all be for the position z = 0.0 along the length of the bearing and at the point in the respective journal orbit where the largest negative pressure occurso Figures 5o10, 5o11 and 5.12 show the pressure distributions around the bearing for three, four and five bladed propellers respectively using the parameters H2 = 0.0, 9 = 270~ and a steady state position of n = 0.85 and 0 = 90~o It is seen that the three bladed propeller produces the largest values of both positive and negative pressureso Figures 5o13 and 5,14 illustrate the effect of including the second harmonic of the dynamic loading at different phase angles Ao

-70 A considerable reduction in the magnitudes of both positive and negative pressures is obtained for an appropriate value of A, in this case A = 11.25~. The effect of the initial eccentricity on the magnitudes of pressures developed in the bearing is illustrated in Figures 5.15 and 5o16. Both positive and negative pressures are considerably reduced with increasing eccentricity. Thus, although the static portion of the pressure increases with increasing eccentricity, the total pressure is greatly reduced. This is to be expected in view of the previous discussion of the journal orbits. As the initial eccentricity is increased there is a conversion from the velocity (i-) to the velocity (d-) dt dt and the more important one in terms of pressure levels appears to be the radial velocity (dn), dt

-71 Figure 5.10. Pressure Profile at t = 0.7850, n = 0.7951, z = 0.0 and, = 85.47~ for the Orbit Defined by HI = 0.8, H2 = 0.0, b = 3, p = 270~ and A = 0.0 Steady State Coordinates are n = 0.85 and, = 90~.

-72 Figure 5.11. Pressure Profile at t = 0.7300, n = 0.8083, z = 0.0, and 0 = 89.83~ for the Orbit Defined by H1 = 0.8, H2 = 0.0, b = 4, P = 270~ and A = 0.0. Steady State Coordinates are n = 0.85 and 0 = 90~.

-73 Figure 5.12. Pressure Profile at t = 0.6950, n = 0.8187, z = 0.0 and 0 - 88.65~ for the Orbit Defined by Hi = 0.8, H2 = 0.0, b = 5, A - 270~ and A = 0.0. Steady State Coordinates are n = 0.85 and =90~.

Figure 5.13. Pressure Profile at t = 0.7300, n = o.8004, z = 0.0 and 0 = 89.82~ for the Orbit Defined by H1 = 0.8, H2 = 0.4, b = 4, P = 270~ and A = 0.0. Steady State Coordinates are n = 0.85 and 0 = 90.

-75 Figure 5.14. Pressure Profile at t = 0.7300, n = 0.8020, z = 0.0 and 6 = 89.91~ for the Orbit Defined by H = 0.8, H2 = 0o4, b = 4, p = 270~ and A = 11.25~. Steady State Coordinates are n = 0.85 and ) = 90~.

-76 Figure 5.15. Pressure Profile at t = 0.7300, n = 0.8270, z = 0.0 and 0 = 89.96~ for the Orbit Defined by HI = 0.4, H2 = 0.2, b = 4, p = 2700 and A = 11.25~. Steady State Coordinates are n = 0.85 and 0 = 90~.

-77 Figure 5.16. Pressure Profile at t = 0.7300, n = 0.9408) z = 0.0 and 0 = 90.07~ for the Orbit Defined by H = 0.4, H2 = 0.2, b = 4, P = 270~ and A = 11.25~. Steady State Coordinates are n = 0.95 and 0 = 90~.

VI. SUMMARY Ao Conclusions The journal orbits are found to depend primarily on four factors, these being (1) the amplitude ratios of the dynamic load H1 and H2, (2) the frequency of the dynamic load b X, (3) the initial steady state eccentricity position no and (4) the angular location of the dynamic load p. Considering the first of these it is found that the. size of the journal orbits varies directly with the amplitude ratios H1 and H2. With respect to the second, the frequency of the dynamic load is found to effect the general shape of the orbitso In terms of the loading frequency it takes exactly 2b + 1 cycles to complete one journal orbito The third and. fourth factors are found to influence the mean attitude of the journal orbito An increase in either p or no decreases Omean considerably, with f being the more important factor. As Omean of the journal orbit decreases there is a rapid change from a translational velocity (-n) to a rotational velocity (dp) This results in long narrow dt dt journal orbits nearly symmetrical about Omeano The position along the length of the bearing.for extremum values of pressure, either positive or negative, occurs slightly past the middle of the bearing toward the after endo The general profile of the pressure distribution around the bearing is independent of all parameters. It is felt that this is. a..result of the assumption of a complete oil film around the bearing and the nature of the sine-cosine form of the solution for the pressure distribution. -78

For given amplitude ratios Hi and H2 the magnitudes of the pressures developed in the bearing are found to depend mostly on the relative values of the velocities (-i) and. (d)o It appears that of these an two quantities the more important is the radial velocity (d-)< It would take considerable more data to firmly establish this. The difficulty encountered here is that it takes nearly ten minutes of computer time to completely evaluate one journal orbit and the corresponding pressure profile around and along the length of the bearingo If the above conjecture is true concerning the velocity (d) than the greatest reduction of dt pressure magnitudes will be obtained for a minimum value of Omean' a maximum value of the initial steady state eccentricity and an appropriate value for the phase angle A of the second harmonic componento With respect to the physical problem motivating this study the possibility of cavitational damage has been firmly established. It is felt that with the proper choice of the physical parameters involved in this problem all negative pressure regions around the circumference of the bearing could be raised above the vapor pressure of the fluid except one and this one occurring only once every complete journal orbito If this were the case then for a propeller having four blades, as it takes exactly 2o25 revolutions of the journal for one complete orbit, there would be four distinct areas of cavitation damage exactly 90~ apart around the circumference of the journal. If the number of propeller blades is changed to three, it takes exactly 2o33 revolutions of the journal for a complete orbit. This would result in three distinct areas of cavitation damage exactly 120 apart around the journal.

The same is seen to be true if the blade number is changed to five. It now takes exactly 2.2 revolutions of the journal for a complete orbit resulting in five distinct areas of cavitation damage 72~ apart. If the conjecture concerning the existence of one negative pressure region for a complete journal orbit is true then the location along the length of the bearing where this would occur also agrees closely with the observed damage. Bo Areas in Need of Further Study To get a more exact understanding of the actual physical problem there are two particular areas of investigation which would be of considerable valueo The first of these is a more exact determination of the real propeller loading either analytically or experimentally, the latter being perhaps the more realistic approach. The second would be to attempt to treat the journal as a deformable body. This of course makes the film thickness a function of both ~ and z and would quite likely effect the velocities (ad) and (d_) considerably. dt

APPENDIX A. Derivation of the Reynolds Lubrication Equation Reynolds equation for dynamically loaded journal bearings may be derived by following the same approach as that of Elrod(ll) who considered the case of static loading. It is necessary only to change the boundary conditions to account for the radial motion of the journal due to the dynamic loading. Accordingly the work that follows is that of Elrod with the exception of that involving the radial velocity Vo The complete Navier-Stokes equations and equation of continuity for steady flow of an incompressible fluid and neglecting body forces Xi can be written respectively in general coordinates as vgu + 2ui i u + r ( ra u++T ri u rTTu vs r ^u1 i ^ pi ^ ~ ^ r ^ 6 + r_ r - ri rT )u aeF A6 aa ^Ad ac A ap a af T ad Cd YT -... + r- ) - - — a - (A.I) a (g1/2 U) =0 (A.2) where the Euclidean Christoffel symbols are Ir (~ l2,3) = 1 gia(~ +~ _.g -.) (A.ia) a? s~12~ 2 3 ~Ce or ri (Sl, ) = 2. Gi( +....) (A.3b) and the metric tensor is gap= L yi yi (A.4a) ^p ~ 6 1=1S&!

-82 or ga5 = -(cofactor of g,, in g). (AK4b) Further - =.gai Gap - 9 G (A.5) where L is some characteristic length and the velocity components ui may be written as ui d i dt Referring to Figure A.1 the following dimensionless variables are introduced; rl = 5 2 = Y and 3 = rR (Ao6) L L h(l,s2) Figure Avl Auxiliary Coordinate System. Introducing the dimensionless velocities ui = L2u= * V

-83 the boundary conditions become ui = 0 *X when 3 = -1 (A.7a) and 1 L2U' v u3 L2V V U2 = when 3 = 0 (A 7b) where U' = U and V' = - L h Now the transformations inverse to Equations (A.6) are yl = R sinG = (r-~3h) sing = (r-^3h) sin(L I), r y2 = L~2, y3 = r-R cosg = r - (r- 3h) cos ( r-) o In view of these relations it follows that (Ao7c) = L(r-3h) cosG - r = 3 h_ sinh, 6 2. 3 h sin, 3y1 - = - h sing ay2 = o, a3 9 y2 -0 asl ay2 _ L (A.8) ay5 ~y5 = L (r-3h) r sinG + 53 ah cos9, ael = 3 chh cos9, <62 3

-84 Considering Equations (A.4a), (A.4b), (A.5) and (Ao8) the following relations are obtained: L2 (1-3 h)+(3 h )2 (~3)2 ~h 6h L 61 662.3 h 6h L2 glT (3)2 ah h_ L' l 6g2 1 + (3 h ) —2 L 6~2 3 h 6h L2 6e2 g3 h.h L2 al ~3 h 6h L2 ae2 h)2 L |(A.9) L2gP = Gc = 1 (l-.3 h)2 r 0 0 1 3 ln h (1-_3 h)2 r 3 a ln h L2 (3 ln h2 h2 + c(l-3 h1)2 r + (53 a ln hi)2 a62 Ao 10) 3 a ln h (1-3 h)2 r~.53 ln h 6e2 and 1h2 L4h2 (Aoll) Introducing as a dimensionless small parameter e = ho/L, where ho represents the minimum film thickness, it is noted from Equation (A 10).that all GP are f(E0) except 553 which is &(e~2)o Therefore GC5 = C[exp(-25353ln e)] (A12) 55~~~~~~~~~(.2

-85From Equation (A.9) all derivatives of Gop are @'(e2) except Gil which is ((e). Accordingly XG^ = ~erexp(2 - o )1n e].(A13) Considering Equations (A.12) and (A.13), then from Equation (A.3b) ri = +2[exp(-2 6T + - 6161) in e] (A.l4) + e[exp(-2 6a3+ 2-6SCB)lni + 6[exp(-286 6 + 2-6 61)lne] l Picking the lowest power of e for each rIa is is noted that iA 5, a 1, 1: r e2) i / 3, a = 1, or P = 1, or a 1: rL = (e) i = 3, a = 1, or ( J 1: r = ) i = 3, C =: r( = (-1 up Introducing the dimensionless pressure.-t P(-)2 ho and. substituting it and u1 into Equation (A.1) it may be written as Gc[' + r + r u i + r a r + (air + ri rT - r Tr- )ua] - r -o* Cra+ Lap- - T a Ca T =(o - u* + roa ) - c | - 0 a (A.15)

With i A 3 and retaining terms of Equation (Ao15) of the lowest powers of E, that is 6(e-2) and er(E1) then -Glr3'u* 11 60 2ui 2i 6(~5)2 15 r3 au* s+ i3 *~ - e-2Gi --- = 0. Waca With respect to the given accuracy (A 16) 1r 1 il 1 1 Gl(_ h) = -Gil h= (e), r i -3 (e2) a53 ah as — = 0, ^~ r3 1 =G3 [a 33 2 63 - G (e2 a|~ as a -o, 33 r11 = G3 11 2 6G11. h(L) = e(1) r h e In view of these relations, Equation (A1o6) may be written 2 32ui (L) i I h 6(e:)2 - 2Gil h u r 6a3 h r u. r -F3 - E-2[Gil a 0 z + Gi2 T2 + Gi3 ] =0 0 s33 (Ao17) With i = 3 correct in terms of and 3(e3) Equation (Aol5) becomes aJT (A. 18) From Equation (Ao 17) 2 1 a u* c(a3)2 1 h 3u* r at 1h - h aU* _ G11(h )2 t r 6a5 ho 6a1 (A 19)

and a2u2 W(e:)2 h au2_ r 3 G22 ( (A.20) where Gl1 1 (1~-3 h)2 r = 1 + 23 h + r and G22 = 1 Retaining terms of f(E) in Equations (Ao19) and (Ao20) 1 and h ul h h ru2 r ~3 (1 + 2|3 h h 2 ~T (Ao21) a2u2 ( g3)2 (h)2 ho 6a2 (Ao22) In view of Equation (Ao18), with respect to the given accuracy, Equations (Ao21) and (Ao22) may be integrated directly to give U1 -L2U u *... v + l] 2 (1 + 3) + 3h[ 3(L2 r 2v - L) 36 + ( 3)2( 2U 2v + + (3)3 f1] 18 (Ao23) where h 2 fl = (ho) fi hr and 2 u,. = f2[3 (1+53) 2 + _(3 r 12 + (~3)2 4 + 63I (A. 24)

-88 where 2 f2 = (h) __ e ho c2 Equation (Ao2) in view of Equation (Aol8) and considering boundary conditions (Ao7a) and (Ao7b) may be integrated with respect to ~3 to give f [j (1-3 hl ) hu1] d.3 + f [h(lh53 ) hut] d. -= r [ (-3 h) hu d (A2) 0o 35 r * Substituting Equations (A.23) and (A.24) into Equation (A.25) and performing the required integration gives __ [(h ) { 2(h)(._ 2 -) h) L2U_} b~1 ho_ ho 2r 6~1 r v -U + h-[ (L-) {(L) (1 + ) T-}] 12 h L -V (A.26) 2 h ho ho 2r 2 ho v Finally in terms of the more conventional variables g1 x 2 =z p and noting Equations Ao then Equation (Ao26) may be written and noting Equations (Ao7c), then Equation (Ao26) may be written L_[h3(1- hL) _p] + a[h35(l + h_) Lp] (Ao27) 6x 2r 6x 6z 2rr z = -6U a-[h(l h)] + 6ph(l ^_) a u + 12pV'o8x 6r 6r 6x This of course represents the general Reynolds lubrication equation for finite length, dynamically loaded, journal bearings with first order correction terms; that is ((e)o

-89 Bo Solution of Reynolds Equation for the Infinite Length Bearing Considering Equation (3ol) with z variation neglected the partial derivatives become total derivatives and Equation (3ol) becomes L_(h35 p) = 6tr2(wc-2 d) dhL + 12r2c dn cosd o (Bol) dG d9 dt dG dt Letting F1. 6 (2 d) d F2 6 (2 dn ~ c2 dt 2 c2 dt noting Equation (2o9) that dh h = c(l+n cosG) Y d- en sinG, dG substituting these results into Equation (Bl1) and integrating with respect to G, Equation (Bol) becomes F1[c E + F2[ sins + F F3[ 1 ] (Bo2) dG (l+n cosg)2 (1+n cosG)3 c3(l+n cosG)3 where F3 is a constant of integration. Noting that the pressure must be continuous around the bearing and letting a 1 n Equation (Bo2) becomes P^ P O F it — dG + F t singdg p()-p(~-~) = o = (FT1 f, (5 - cosG)2 2 (a + cosg)3 +o3F 4 d( G.J (B35) + - (( + cos) )

-90 or 0 = F 2 tan - 1 tan } I5-+1 1 sin( i (2-_1) (a + cosg) -T + c(F2[2(S + cosG)2 ] t 2F + 1 1 + F3[ I(2 5/ tan1 a +1t - 3a sinG 2(K1-)2 (t + cos9) sinG T 2(a-21)(a + cosg)2 _r Evaluating Equation (Bo4) the constant of integration F3 -2F1(a2-1) F =.. 3 (2-2+1) Considering now Equation (Bo3) as an indefinite integral (B 4) is (B 5) P(G) = C —2F1 / d + FsinGdG + o —F3 (5 + cosG)2 (5 + cosg) de — + os-) + F4 (a + cosG)3 4 (B.6) where F4 is an arbitrary constant, performing the integration again as in Equation (Bo4) and substituting Equation (Bo 5) for F39 Equation (Bo6) becomes p(G) = Fi[ s(i + )] + (2a2+1)(a + cos@) a + cosG a F 1 - --- — ] + F4 2(5 + cosQ)3 (Bo7) Substituting the expressions for F1, F2 and a into Equation (Bo7) p(G) 6 c2 ( d-2 [ n( 2+n cosG)s2in. 6ir2 adn+ 1cntn + 1Ft-) ---- ] + arbitrary constant. c2 dt n(l+n cosg)2 (B.8)

-91 Co Evaluation of the Series Coefficients Am(z) and Bm(z) Considering the Am(z), and for convenience letting Am(z) = Am, from Equation (356) the general recurrence relation for m > 1 is 2(D2 -m2)Am + n(D2 m2-m+2)Am-l + n(D2-m2+m+2)Am+l = 0 (Col) where D2 r2 d2 dz2 From Equations (3512) and (3513) Al and A2 may be written as A1 = C1 cosh az (Co2) r and A2 = -Y7C cosh Uz (Co5) r where a and 7 are defined by Equations (3510) and (3540) respectively as 1/2 2(l+yn)l 1/2 2 n 2+(l-n2)1 2} 2-n' Y {l+(l-n2)1/2}2 and C1 is a constant of integrationo In view of the boundary condition, Equation (2.15), all of the Am must be even in z. The complementary solution for all Am will therefore contain only one term which shall be represented as the first term of each Am and Cm is the corresponding constant of integration. For m = 3, then from Equation (Col) D2A3 = - (D2-4)A2 - (D2-4)A1 o Substituting Al and A2 and solving for A3 A3 = 5 + ( 1 - )( ) Cl cosh (C.4) n a~2 — ~

-92 For m = 4, then from Equation (C.1) D2A4 - 4A4 = - (D2_9)A3 - (D2-10)A2 Substituting A2 and A3 and solving for A4 A4 = C4 cosh 2z r - 9C + 7( -2o ) 2n [ (a2-4) 2(a2-9)(27 _ 1)] C1 cosh a (C5) n - r For m = 5, then from Equation (Col) D2A5 - 10A5 = 2- (D2-16)A4 - (D2-l8)A3. Substituting A3 and A4 and. integrating A5 becomes A5 = C5 cosh J z - 4 C4 cosh 2z _ 9 C3(1 _ 8) + F C1 cosh cZ r n r 5 n (c2-1O) r (c.6) where F1 4(a2-16)( a29)(27 - 1) =n a2 n 2- n(2-16)(2-10) - (218)(27 -1) n -24 n (a2 -4 (C.7) For m = 6, then from Equation (Col) D2A6 18A6 = -2(D2-25)A5 -(D2-28)A4 o n Substituting A4 and A5 and solving for A6 A6 = C6 cosh 8 z r - 15 C5 cosh IT z 12 C4(1 - Z7) cosh 2z 4n r 7 n- r 4 C 10 F2 +n 7% - + ( ) C1.cosh az r (C.8) where F2 2= (a228) ( 9)( 27 1) _ n a2 n 22-10 2 a2-25 - n(2 —O) F1 o (C.9)

-93 For m = 7, then from Equation (C.1) D2A7 - 28A7 = - 2(D2-36)A6 - (D2-40)A5 o Substituting A5 and A6 and solving for A7 A7 = C7 cosh 28 z r 1 768 7 _ + 24[ 7(l - n2 36- C6 cosh lOn \/8 z r -5 (1 52) C cosh 3 n5. r 144 + n c4 cosh 2z 1288(3 10 C4 cosh 2z - 12-8(8 - 1) - 72(1 - )] C3 1 +.1 [. (a2-28) 2( - 36) n a2-18 - ( -4)F a2-10 az Cl cosh - r (C.10) where F1 and F2 are defined by Equations (Co7) and (C.9). For m = 8, then from Equation (Col) D2A8 - 40A8 = - 2(D2-49)A7 - (D2-54)A6o n Substituting A6 and A7 and integrating A8 becomes A8 = C8 cosh r-O z r ~-7C cosh. 7 _ 1[36(1 2n 7 r 22 62 cosh 62 10n C6 cosh FlV z r + [26(l 52) + 33] C5 cosh z 6 n n2 n r - [15 {768(1 7 + 44} - 600(1 - 7)] c4 cosh 2z 56 4n.7 n n n 7 n2 r (C. 11) 1[ 40 7 {288( 2n L. n 10) - 72(1 n2 _ 8 ) + 26(3 )] n2 n n - 1 [2(2-49 (a2-4o) n a2-28 { 2(a2536 n )218 - (a 2 )F1} + ( -5)F2]C cosh ~z. a2.18 r From Equation (3.36) A () F(-i)m a [1 + Am(') = F(-l)m- ai[l + - ('-n 2)1_/2] (C.12)

where F = 12r2(-2 d)( do 2) c2 dt 2+n2 all of the constants Cm may be evaluated. For m = 1 from Equations (Co2) and (Col2) C1 = Fa[l + + (!-n2) I/ coh 2r For m = 3 from Equations (Co4) and (Col2) (c.13) (c.14) C3 = Fa3[1 + I.; - X Fa(2. - (1_n2)!/2 n 1(2 )_4 1 (1-n2)l/2 * (Col5) For m = 4 from Equations (Co5) and (Col2) C4 = 1 { -Fa4[l+ + cosh ~ { (1-n2)2 2n r + [2( 2 )(27 - 1) y1-(2- )10 cosh } (C.16) For m = 5 from Equations (Co6) and (Co12) C5 = -1- - {Fa5[1+ 5 1 05 osh- (1-n2)1/2 2r 4 + - C4 cosh - n r + 9 C3-(1 ) - 1 C1 cosh (2} e For m = 6 from Equations (C 8) and (C. 12) (Co17) C6 - h1 {-Fa6 [1+ cosh ---- 2r 6 (1-n2)l/2 (i-n2)1/2 + 15 C5 cosh 4n 2r (C.18) 12 4(1 ) cos -4 10 7 n2 r n F2 (a2 -18) C1 cosh 2r}

-95 For m = 7 from Equations (Co10) and (C.12) Coh 1 — {Fa7[Jl + — 7 ] 7 cosh _8 (1-n2)1/2 2r + 36 C6 cosh 18 lOn 2r + 5( - ) C5 cosh 5 n2 2r 1768(1 7 ) + 1 24 7 n n n n C4 cosh ~ r + 1[288(3 0) - 72(1 ~)]C3 n. n2 1,1 [ (c2 -28) - 2(C2-36) n c(2-18 F2 -(t240) F1] c2 _10 C1 cosh a} 2r (c 19) For m = 8 from Equations (Co11) and (Co12) C8 = 1- [Fa [ cosh' 2r + 8 ] 7 (1-n2)1/2 2n C7 cosh - 28 2r + 1[36(1 62 )]c6 22 -10n2 cosh -i8 2r - [26(1 - 52) 33]C 6 n n2 n cosh-,l' I 2r ~ l[z5 I.{768(. +6 4n - 7n + -i[ 7 {288(3 40 2n. n2 - )+144} n2 n -600(1 - 7)]C4 cosh 7 n2- r - 72(1 - 8)} + 26(3 n2 n - 10)] C n2 (C.20) + 1 [2(a2-49) { (c-40 ) n a2 {28 -2. ) F2 n 2 182 - ( 240)F} a2 -10 + (a2-54)F2]C cosh 2r Considering the Bm(z), and for convenience letting Bm(z) = Bm, from Equation (357) the general recurrence relation for m > 1 is 2(D2-m2)Bm+ (D2 m2m+2)Bm1 + n(D2m2+m+2)Bm+1 = 0 (C.21) B1 and B2 d2 where again D2 = r2 d2 ~ dz2 From Equations (3o16) and (3o17) may be written as B1 = N1 cosh r r (C.22)

and B2 = -4N1 cosh z (C.23) where ~ and. are defined by Equations (35.14) and (3.43) respectively as: [2(1 + n)]l/2 [n {1 + 2(1-n2)1/2}]1/2 2 ^tn {1 + (1-n2)l/2} 2 and N1 is a constant of integration. In view of the bouncary condition, Equation (2.15), all of the Bm must be even in z. The complementary solution for all Bm will therefore contain only one term which shall be represented as the first term of each Bm and Nm is the corresponding constant of integration. For m = 35 then from Equation (Co21) D2B3 - (D2-4) B2 - (D2-4) B1, Substituting B1 and B2 and solving for B3 B = N + (n - 1)( ) N1 cosh r (C.24) For m = 4, then from Equation (Co.21) D2B4 - 4B4 = - (D2-9) B5 (D210) B2 Substituting B2 and B3 and solving for B4 B4 = N4 cosh - i + [ - _( )(L - l)]Nl cosh L. (C.25) r 2n (~2-4) n -2 n r For m = 5, then from Equation (C.21) D25 - 10B5 = - 2(D2-16)B4 (D2-l8) B3

Substituting and B4 and integrating B5 becomes B5 = N5 cosh - N csh 5' -'r n -2z 9 (1 8) r 5T 3 n1 + 3 N1 cosh z'Z ( 2-1o) r (C.26) where F3 = 4( 2 6)( 29) n 1) 3 n2,2 n - 2( 216) n 2 10 - (218)(228 n - 1)(L ) E2 (C 27) For m = 6, then from Equation (C.21) D2B6 - 18B6 = - 2(D2-25)B5 - (D2-28)B4 0 n Substituting B4 and B5 and solving for B6 B6 = N6 cosh -8l z r 15 cosh. z. 12 4 7 ) - - -' N4(1 -) 2z cosh r 4 + - N3(3 n 10) + F4 N1 cshz n ) +(8N cosh (C.28) where F4 = 2(-2 28) (2-9)(2_ - 1) _- (2-28)( 20) n (2 n 2_-4 - 2(2-25)F3 n 12-1o (C.29) For m = 7, then from Equation (C.21) D2B7 - 28B7 = - n(D2-36)B6 - (D2-4o)B5. Substituting B5 and B6 and solving for B7 B7 = N7 cosh 2 z r _ 36N6 lOn cosh 8 z.- 5(1 r 3 - 52)N c n r 1+ 768( 7_.) + 3144N4 24 7n n2 n cosh 2z -1 288 -1) r - n2 - 72(1)N 1 + - ) (: -28) 2( 2-36)F4 n t2-18 ( 2-)F3]N1 cosh 4z 2-'10 r (C.30)

-98 where F3 and F4 are defined by Equations (Co27) and (C.29)o For m = 8, then from Equation (Co21) D2B8 - 40B8 = - 2(D2 49)B7 (D2-54)B6 oa~ ~~~ns - (oD =-n(B. - o Substituting B6 and B7 and integrating B8 becomes B8 = N8 cosh 4__ z r -7 N7 cosh z 2n r - 1[36(1 62)]N6 cosh 1 z 22 li0n2 r 1[(l - 52) + 3]N5 cosh - - 6 n n2 n r - 1ll5 768 [40 2n {n2 40 2n n2 7+ 4144 600 ( n n 7 -) - 72(l -'n2 - 7n)]N4 cosh 2z ny- r (C 31) 216 + 26(3 n -n2 N 1 [2 - g -4 )[2( )-49 (~ -4o) n 2Q28 {- 2-( 2J) F4 n 2-18 - 22-0) + ( 2- F4 2 _18' coshL- r From Equation (3537) Bm() =F(-l)m am[m m - n2) 2(~) = ((-1) -n~[. (1_=2) (L-n2)3/2] where - 12tr2 dn Fnc2 dt n3c2 t (C.32) (C.33) all of the constants Nm may be evaluatedo For m = 1 from Equations (C.22) and (Co32) N1 = Fa[l - 1 n2 1 (1-n2) (1-n2)3/2 cosh - 2r (C.34) For m = 3 from Equations (C 24) and (C3.2) N = Fa3[5- "3~ (1-n2) n2 (l-n2)3/2- n )2 )[ + 2.(n2)(C.5) (Co35)

-99For m = 4 from Equations (C.25) and (C.32) 1 ___1 n2 N4- = r Fa4[4 - 4 n2 ] + 9 cosh QL (1-n2) (1-n2)3/2 2n + 2 1) - t( 10 cosh 2r 1 ~ (C.36) For m = 5 from Equations (0C26) and. (C,52) N5 - 1. Fa5[5 - 5 2 _-n ] + n N4 coshn 2r + [ n( () - ( 2-)N1 cosh 2. (C367) From m = 6 from Equations (Co.28) and (Co32) N65 = 1 _ -Fa6[6 - 6 n2 + 1 N4 cosh cosh l[8 (1-n2) (1-n2)3/2 4n 2r + o N4( - ) cosh (3 - ) -1 F N cosh J (C.,8) For m = 7 from Equations (Co.50) and. (C052) cosh Fr L (l-n2) (l-n2)5/2 lOn 2r + N3(1 - )N5 cosh (C3 - St7n n2 (+ 28 88 2-10) 72(1 8 1 _()F4 a (72 )F — N1 cos2r (C039) 7 2 ~.28) r n2 1)3 2r (~2-28) n t2-18)F 2-()F10 c os

For m = 8 from Equations (C3.5) and (Co32) _ _ _ _ _ _ _ - 8 N8 - 1 -Fa8[8- 8 cosh 0 i^ (1-rn - 2r n2 (17n2)3/2 + L N7 cosh./.F8 2n 2r 5) + 3] N5 cosh 2-r n2 n 2r + [36(l 1n ) N6 22 iOn2 6 cosh a- - 1[26(1 2r 6 n + [15 36 4n {768(1 7) + 14 6oo(1 )- )]N4 cosh 7n n2 n 7' r (c.4o) + [ L {i7(3 _- ) - 72(1- )} 40 2n n2 nn n2 + 216(3 N i( - )]N n n' + 1 [2( )2( 2_40) n ^2-28 -( -3)F4 - n ~2-18 2' io')F 3} + ( 2-5)4 ]N cosh *2 (~2_8'4] oor -r

BIBLIOGRAPHY 1. Mosher, Lo M Tailshaft Inspection Report Shipbuilding Division, Bethlehem Steel Corporation, Quincy, Mass,, 1957. 2. Tower, B. "First Report on Friction Experiments " Proc. Insto Mecho Engrs. (London), Vol. 34, (1883) 632. 3. Reynolds, Oo "On the Theory of Lubrication and Its Application to Mr. Beauchamp Tower's Experiments, Including an Experimental Determination of the Viscosity of Olive Oil." Trans. Roy. Soc. (London), Volo 177 (pto 1), (1886) 157. 4. Sommerfeld, Ao "Zur Hydrodynamischen Theorie der Schmiermittelreibung." Z, Math Phyo, Vol. 50, (1904) 97. 5. Harrison, W. J "'The Hydrodynamical Theory of the Lubrication of a Cylindrical Bearing Under Variable Load and of a Pivot Bearing." Trans. Cambridge Phil. Soc., Vol. 22, (1919) 373. 6. Swift, H. Wo "Fluctuating Loads in Sleeve Bearings." Jo Insto Civil Engrs. (London), Vol. 5, (1937) 161-95. 7. Burwell, Jo T. "The Calculated Performance of Dynamically Loaded Sleeve Bearings." Trans. Am. Soc. Mecho Engrso, Vol. 69, (1947) A-231-45. 8. Tao, L. N. "General Solution of Reynolds Equation for a Journal Bearing of Finite Widtho " Quart. Appl. Math., Vol. 17-18, (1959-61) No. 2, 129-36. 9o Fedor, J. Vo "A Sommerfeld Solution of Finite Length Journal Bearings with Circumferential Groves." Trans. Am. Soc. Mech. Engrso, Volo 82, Series D, (1960) 321-26. 10. Hays, D. Fo "Squeeze Films: A Finite Journal Bearing with a Fluctuating Loado" Trans. Am. Soc. Mech. Engrso, Vol. 83, Series D, (1961) 579-88. 11. Elrod, Ho Go "A Derivation of the Basic Equations for Hydrodynamic Lubrication with a Fluid Having Constant Propertieso" Quart. Apple Math., Volo 17-18, (1959-61) No 4, 349-59. 12. Lin, C. C. The Theory of Hydrodynamic Stability. Cambridge University Press, Great Britian, (1955) ChapO 2. 135 Barwell, Fo T. Lubrication of Bearings. Butterworths Scientific Publications, London, (1956) 166-67. 14. Dwight, H. B. Tables of Integrals and Other Mathematical Data. The MacMillan Company, New York, No Yo, (1949) 85. -101