oAccepted 18 May 2012Available online 29 May 2012

Keywords:

Analytical

Journal

r th

ally

of

solution of the homogenous Reynolds equation thus a closed form expression for the denition of the

lubricant pressure is presented. The Reynolds equation is split into four linear ordinary differential

equations of second order with non-constant coefcients and together with the boundary conditions

they form four SturmLiouville problems with the three of them to have direct forms of solution and

eynoldnowa

ave ac

are based on the assumptions that one of the two terms in the left

[9]elec-ssureion.

Vogelpohl [1113] achieved to give closed forms of the functions

bearing, axial and circumferential. He also [1113] assumed a

Contents lists available at SciVerse ScienceDirect

journal homepage: www.els

Tribology Int

Tribology International 55 (2012) 4658distribution was dened in closed form. Cameron and Wood [14]E-mail address: chasalevris@sdy.tu-darmstadt.de (A. Chasalevris).partial solution of the Reynolds equation that corresponds to along bearing approximation and then used the technique ofseparation of variables for the solution of the homogenousReynolds equation. By adding the two solutions, the pressure

0301-679X/$ - see front matter & 2012 Elsevier Ltd. All rights reserved.

http://dx.doi.org/10.1016/j.triboint.2012.05.013

Abbreviations: ODE, Ordinary differential equation; BC, Boundary condition;

BVP, Boundary value problem; PSM, Power series method; FDM, Finite difference

method; CFD, Computational uid dynamicsn Corresponding author. Tel.: 49 6151 16 3461; fax: 49 6151 16 3668.side of Eq. (1.1) can be neglected. The rst term can be neglectedwhen the journal bearing is considered as a bearing with high for the pressure distribution along both directions of the journal@

@x

h3

6m@Px,y

@x 1R2

@

@yh3

6m@Px,y

@yO @h

@y2 @h

@t1:1

The approximate analytical solutions of the Reynolds equation

not many and have not been made recently. Kingsburydetermined the pressure distribution by an experimentaltrical analogy. Christopherson [10] determined the predistribution by utilizing the mathematical model of relaxattions of pressure distribution P(y,x) under approximate analyticalsolutions of Reynolds equation, in the form of Eq. (1.1) or insimilar forms, considering both the squeeze lm and the wedgelm effect. However, numerous works in the numerical solutionof the Reynolds equation have been made: ! !

that they satisfy Eq. (1.1) for the case of innite axial length or ofinnite short bearing. However they are regarded as approximatesolutions when they are used to determine the pressure distribu-tion in bearings of nite length.

The research contributions in the analytical calculation of thepressure distribution in a journal bearing with nite length are1. Introduction

The exact analytical solution of Rtion of journal bearings [1] is up toinvestigation. Most investigators hone of them to be confronted using the method of power series. The mathematical procedure is

presented up to the point that the application of the boundaries for the pressure distribution yields the

nal denition of the solution with the calculation of the constants. The current work gives in detail the

mathematical path with the help of which the analytical solution is derived, and ends with the pressure

evaluation and a comparison with past numerical solutions and an approximate analytical solution for

a nite bearing. The resultant pressure distribution presents slight differences compared to this of the

numerical solution and the approximate analytical solution in the values of maximum and minimum

pressure but also in the domain of lower values of pressure.

& 2012 Elsevier Ltd. All rights reserved.

s equation for lubrica-days a problem underhieved to dene func-

length to diameter ratio (long bearing, L/Db1 [2]) and the secondterm when the journal bearing is considered as a bearing withlow length to diameter ratio (short bearing L/Do1 [38]). Suchsolutions for pressure distribution give considerably simpliedmathematical expressions. They are exact solutions in the senseLubrication

HydrodynamicAn exact analytical solution of the Reynbearing lubrication

D. Sfyris a, A. Chasalevris b,n

a University of Aegean, Aegean 81400, Greeceb Darmstadt University of Technology, Darmstadt 64287, Germany

a r t i c l e i n f o

Article history:

Received 3 February 2012

Received in revised form

15 May 2012

a b s t r a c t

The Reynolds equation fo

length is solved analytic

multiplicative form, a setlds equation for the nite journal

e pressure distribution of the lubricant in a journal bearing with nite

. Using the method of separation of variables in an additive and a

particular solutions of the Reynolds equation is added in the general

evier.com/locate/triboint

ernational

we present an analysis for evaluating the function g using againthe method of separation of variables, but now in a multiplicative

D. Sfyris, A. Chasalevris / Tribology International 55 (2012) 4658 47had extended the work of Christopherson [10] to show the effectof length to diameter ratio on eccentricity ratio, attitude angleand friction coefcient. In all cases, these solutions expressnatural phenomena in the oil lm on the basis of Reynoldsassumptions regarding lubrication, the most important assump-tion being that certain terms in the generalized NavierStokesequations for ow in a viscous uid may be neglected.

In the current work the Reynolds equation of the form ofEq. (1.1) is treated similarly to the path that Vogelpohl [1113]followed, but there are crucial differences that have to do with theway that the particular solution is obtained and with the solu-tions of the ordinary differential equations that are yielded duringthe procedure. In the beginning of the current work the Reynoldsequation is classied. As expected it turns out that it is an ellipticequation, which is reasonable since it denes a static problem(the time is not considered as an independent variable but as aparameter that gives the eccentricity and its rate of change). Thestrategy for obtaining the analytical solution is based in theapplication of the powerful method of separation of variables.The crucial step is splitting of the solution into two parts. The onesatises the homogeneous Reynolds equation; namely, Reynoldsequation without the second part that the pressure does notinterfere. For solving this part we assume a multiplicativeseparation of the independent variables so we obtain twoSturmLiouville problems. In this point, the contribution of thecurrent work is that the power series method is used in order toobtain the eigenfunctions of the one SturmLiouville problemwhile in the literature the corresponding confrontment of thisproblem is made with the linear approximation of the uid lmthickness function and the expression of the eigenfunctions usingBessel functions or normal sinusoidal functions. In the currentwork, the easy SturmLiouville problem is solved and theeigenvalues of it are incorporated in the functions dened withpower series.

The second part is a particular solution of the Reynolds

Nomenclature

P resulting pressure of the lubricantu pressure of the particular solutiong pressure of the homogeneous solutionx axial coordinate of the bearingy angular coordinate of the bearingyn angular location of zero pressureh uid lm thicknesshn approximate uid lm thicknessR journal radiusequation itself. For nding a particular solution we assume anadditive splitting of the independent variables and two SturmLiouville problems are obtained. The rst has to do with thepressure distribution along the circumferential coordinate and ithas a direct solution with a closed form expression taken from theliterature for ODE treatment. The other problem has to do withthe pressure distribution along the axial coordinate and theboundary conditions which are chosen to yield a trivial solution,without this the further progress of the solution is going to beproblematic. The current particular solution is also a contributionof the current work since it is actually a set of particular solutionsthat can be different from the solution of the innitely longbearing as used in [11]. The current particular solution yields thelong bearing pressure distribution as a sub case.

The paper is organised as follows: Section 2 contains the basicingredients of Reynolds equation together with the classicationform. The treatment with the Bessel functions and the approx-imating analytical solution is also given. The last section, Section5, deals with the boundary value problems. The resulting pressureis evaluated for a specic set of values of the physical andgeometrical journal bearing characteristics and is compared withthe analytical approximate solution for the nite bearing and twonumerical solutions, the one using the nite differences methodand the other with a very recent 3D-CFD analysis. The articleconcludes in Section 6 where also the forthcoming results of thefuture work are described.

The current work is considered to be the rst and initial stepfor the analytical confrontment of the analysis and design of theplain nite journal bearing. A future upcoming work that is basedin the analytical pressure denition presented in this paper, usesthe current exact analytical function for pressure distribution andafter analytical procedures gives the closed form denition of themain operational characteristics for the nite journal bearing,such as stiffness and damping coefcients, load carrying capacity,location of the maximum pressure, location of the minimum uidlm thickness etc. The fact that the uid lm forces can beexpressed analytically for the nite journal bearing is based in theevaluations presented in the current work. The general concept ofanalytically simulating other types of nite journal bearings, forexample non circular, or grooved can be reconsidered after thesolution presented in the current work.

2. Reynolds equation: classication and splitting of thesolutionof it and the crucial step where the unknown pressure P is splitinto two functions g and u. Section 3 is related with the evaluationof the particular solution of the Reynolds equation, u. In Section 4

O journal rotational speedm lubricant dynamic viscosityj0 attitude angle of the journalcr bearing radial clearancee journal eccentricity_e journal eccentricity rate of changee journal eccentricity ratio_e journal eccentricity ratio rate of changeW external bearing loadLb bearing/journal lengthc18 constants of integrationThe problem of the lubrication of journal bearings with nitelength is dened in this work as the calculation of the pressuredistribution of the Newtonian lubricant that is assumed to owunder laminar, isoviscous, and isothermal conditions in betweenthe rotating journal and the static bearing. The journal of radius Rand length Lb is assumed to be rotating with a constant rotationalspeed and to be constantly located in a point of eccentricity ewithrespect to the geometric centre of the bearing of radius Rcr andlength Lb after an application of a virtual vertical load W as shownin Fig. 1. The load is not used as a parameter in this work since noforces are evaluated or expressed and the unique inputs in thepressure evaluation are considered to be the eccentricity e and itsrate of change _e. The journal and the bearing are supposed to bein parallel (aligned bearing) and the uid lm thickness hbecomes a function of the unique independent parameter y fora time moment of constant e and _e which means that the function

3.1. Additive separation of variables and direct solution

D. Sfyris, A. Chasalevris / Tribology International 55 (2012) 465848for the uid lm thickness is h cre cosy and its timederivative is @h=@t _e cosy. The dynamic viscosity of the lubri-cant is assumed to be constant and equal to m through the entirecontrol volume (notied with shadow in Fig. 1) that is denedfrom the bearing and the journal surfaces. The attitude angle ofthe journal is dened as j0 with respect to the vertical coordinateaxis (see Fig. 1). The starting point is the equation of Reynoldswhich is expressed as

@

@x

h3

6m@Px,y

@x

! 1R2

@

@yh3

6m@Px,y

@y

!O @h

@y2 @h

@t2:1

After substituting the uid lm thickness function of Eq. (2.2)into Eq. (2.1) and performing the derivations one will arrive atEq. (2.3):

h cre cosy 2:2

cre cosy36m

@2Py,x@x2

3cre cosy2 e siny

6mR2@Py,x

@y

cre cosy3

6mR2@2Py,x

@y2eO siny2 _e cosy 2:3

Eq. (2.3) is the one that we are going to work with. This is anon homogenous linear partial differential equation of the secondorder for the unknown function P(x,y) with trigonometric coef-cients. Before embarking on the core of our analysis we classifyEq. (2.3). For doing so we need to evaluate the discriminant:

D B2AGh3=6mh3=6mR2 h6=36R2m2o0 2:4

Whatever the expressions of h, R, m are, it is Do0 so we speakabout an elliptic partial differential equation.

The crucial step for obtaining the solution is based on the

Fig. 1. Denition of the coordinate system and of the parameters of operation anddesign in a plain cylindrical journal bearing.following splitting of the solution. We assume that the unknownfunction can be written in the form::

Px,y ux,y|{z}particular

gx,y|{z}homogeneous

2:5

The function u(x,y) is a particular solution of Eq. (2.3) while thefunction g(x,y) describes the set of solutions for the homogeneousReynolds equation, namely, Eq. (2.3) without the right hand sideterms. In order to see more clearly that instead of seeking P(x,y)we can seek for the functions g(x,y) and u(x,y) one may write thedifferential operator of Reynolds equation as in the followingequation:

cre cosy36m

@2

@x23cre cosy

2e siny6mR2

@

@y cre cosy

3

6mR2@2

@y2

2:6In order to evaluate the function u(x,y) a particular solution ofEq. (2.3) is needed. For doing so we assume that u(x,y) can be splitin the following additive form:

ux,y jycx 3:1When the function u(x,y) is known at the boundary @L we

obtain from the above separation that ux,y9@L jy9@Lcx9@L.

We are looking for a solution where the independent variablescan be split in the above form. If we substitute Eq. (3.1) in Eq. (2.3)after some calculations Eq. (3.2) is obtained:

d2c xdx2

1R2

d2j ydy2

3e sinyR2cre cosy

djydy

6meO sinycre cosy3

12m _e cosycre cosy3

3:2

By inspecting the above equation one observes that the righthand side is a function of ywhile the left hand side is a function ofx only. So, the equality will be feasible only when both sides areequal to the same constant C. From the latter equation we obtaintwo equations for the functions j(y) and c(x). These are ordinarydifferential equations written as in Eqs. (3.3) and (3.4):

d2cxdx2

C 3:3

1

R2d2jydy2

3e sinyR2cre cosy

dj ydy

6meO sinycre cosy3

12m _e cosycre cosy3

C

3:4If one applies this operator to both sides of Eq. (2.5) he canimmediately verify that equivalently one may seek for thefunctions g(x,y) and u(x,y).

For treating the boundaries, it is assumed that P(x,y) equals toa known functionP(x,y) on the boundary of the lubricants controlvolume @L as shown in the following equation:

Px,y9@L Px,y 2:7

Then, the splitting according to Eq. (2.5) renders Eq. (2.8). Lateron, it is assumed that the pressure vanishes on the boundaryand that u(x,y) has to be equal to g(x,y) at the boundary, seeEq. (2.9).

ux,y9@Lgx,y9@L Px,y 2:8

ux,y9@L gx,y9@L 2:9

It is also worth stressing that in the solution adopted in thispaper, the Dirichlet boundary condition is throughout utilized.The pressure eld is assumed to be known at the boundaries andto be set equal to zero, so as to correspond to the physical statusthat the developed uid lm pressure is much higher from theatmospheric pressure. At any case the boundary pressure P(x,y)can be dened in correspondence to the demands that are setfrom the physical problem. If one is willing to tackle the Neumannproblem then he has to specify the directional derivative of thepressure along the boundary. The mathematical strategy forobtaining the solutions can be applied equally well to Dirichlet,Neumann or mixed boundary conditions, since the directionalderivative is a linear operator.

3. Evaluation of the particular solution

distribution of the circumferential direction would be negligible,even for high values of C. In the case that the boundaries ofEqs. (3.12) and (3.13) would be different (not zero) then theboundaries of the homogenous solution g(x,y) would be differentalso in order to yield the zero pressure in the ends of the bearing.In other words, the boundaries for the particular solution come toa correspondence with those of the homogenous solution so asthe resulting pressure P(x,y) to be bounded under the physicalexplanation for the lubricant ow.

The constants c3 and c4 that are incorporated in the circumfer-ential distribution can be dened using Eqs. (3.14) and (3.15):

j0 0 3:14

R 2p0 jydy 0, _e 0lim_e-0

R 2p0 jydy 0, _ea0

8>: 3:15

Eq. (3.15) is a two case formula, for the static (_e 0) and forthe dynamic problem ( _ea0). In the static case it expresses thesymmetry of pressure distribution j(y) to the horizontal axis(j0) of Fig. 2. The physical result agrees with the SommerfeldBC since it also yields j(p)0. In the dynamic case, Eq. (3.15)

D. Sfyris, A. Chasalevris / Tribology International 55 (2012) 4658 49Eq. (3.3) can be solved directly to give Eq. 3.5:

cx Cx2=2c1xc2 3:5The constants c1, c2 are arbitrary constants of integration.

For solving Eq. (3.4) one observes that we talk about a linearordinary differential equation with non-constant coefcientswhere the unknown function j(y) is not present explicitly. So,by setting z(y) as in Eq. (3.6), instead of Eq. (3.4) we may solve thefollowing linear ordinary differential equation dened in Eq. (3.7)of the rst order for the function z(y):

djydy

zy 3:6

1

R2dzydy

3e sinyR2cre cosy

zy 6meO sinycre cosy3

12m _e cosycre cosy3

C

3:7The generic set of solutions of the last equation is given in

Eq. (3.8) [20]:

zy eR3e siny=cr e cosydy c3

ZeR3e siny=cr e cosydy

6OmR2e siny

cre cosy3 12OmR

2 _e cosycre cosy3

C !!

dy

!3:8

All the integrals in Eq. (3.8) can be evaluated using directclosed form expressions. By evaluating the integrals that appearin the expression for z(y) one can integrate the outcome one timeand obtain the function j(y):

jy Z

zydyc4 3:9

The sum of j(y) and c(x) renders classes of particular solu-tions by choosing values for the constants C, c1, c2, c3, c4. Bychoosing C0 we obtain Eqs. (3.10) and (3.11):cx c1xc2 3:10

jy c41

22B18cre

2R2mO2c2r e2c3c2r e25=2

3e2c2r R

2m Ocrc3sinyc2re22cre cosy

12c2r e2 _eR2me26crR2m Oc3sinyc2r ee3cre cosy2

!3:11

3.2. Boundary conditions for the particular solution and evaluation

of the function

For the pressure distribution expressed from the particularsolution, the boundary conditions are setting the pressure equalto zero at the both ends of the bearing in axial direction, x7Lb/2and at the circumferential beginning y0 and end of the uidlm y2p. The boundary conditions in the axial direction aregiven in Eqs. (3.12) and (3.13):

cLb=2 0 3:12

cLb=2 0 3:13The constants c1 and c2 are then dened as zero and the

solution for c(x) becomes trivial. Since the particular solutionu(x,y) is expressed as in Eq. (3.1) the trivial solution c(x)0expresses the case of no variation of the pressure in the axialdirection (Long bearing approximation). In the case that Ca0, theboundaries of Eqs. (3.12) and (3.13) would yield a parabolic

distribution in the axial direction that in comparison to theincorporates the _e as a parameter and the fact that the expressionis analytical gives the possibility to solve for c3 and c4 so as theintegral

R 2p0 jydy to tend to become zero for small values _e.

The cases that are studied considering the values of _e extend up to_e=OR 0:001. In the dynamic case the result of the location ofzero pressure yn is in a good agreement with this of the ReynoldsBC but since djy=dy9y yna0 it cannot be said that the formulaof Eq. (3.15) corresponds to Reynolds BC. The benet of usingEq. (3.15) is that when _ea0, instead of predening the angle yn

and use it as a boundary of zero pressure it is let to be dened bythe pressure distribution through y during pressure calculation. Inthe static case the pressure distribution is evaluated with Eqs.(3.14) and (3.15) to give the Sommerfeld BC and these results arecompared with the literature in Section 5. The values of yn as afunction of _e are presented in Section 5 because yn is lightlyeffected also from the values of the pressure yielded by thehomogeneous solution g(x,y) that is added to j(y). The constants

Fig. 2. Pressure distribution along the angular coordinate y given from the

particular solution j(y).

c3 and c4 are given in Eqs. (3.16) and (3.17):

c4 6 _eR2m

ecre23:17

The current particular solution corresponding to the pressuredistribution developed in an innitely long bearing is shown inFig. 2 for some cases of the eccentricity rate of change _e.The current particular solution gives a further contribution inReynolds equation treatment because the additive separation ofvariables can yield the solutions of the long or of the short

WobtBy

mm

is aforsam

mm

1 f y 1 3e siny f y

ing transformation as in Eq. (3.6)) and then to be treated asEq. (3.7). Such an assumption would yield linear distribution ofthe pressure through the axial coordinate, since it would bem(x)c5xc6, thus the case of l0 is not accepted.

If la0 then the differential equation of Eq. (4.5) has no directclosed form of solution. Vogelpohl in [11] incorporates in detailthe treatment of the homogenous Reynolds problem with theassumptions made by Michell [5] and Dufng [15] for thelinearization of the uid lm thickness function so as to makeEq. (4.5) solvable. Sections 4.1 and 4.2 show how the currentproblem was treated in the past and how it is treated in the

c3 6R2m p 2cre2

c2r e2

p_e3cre3O Log cr ec2r e2p

Log cr

cp

e2c2r e2 p Log cr ec2r e2p

Log cr ec2r e2

p

a

D. Sfyris, A. Chasalevris / Tribology International 55 (2012) 465850Fig. 3. Pressure distribution along both angular and axial coordinates given from4. Evaluation of the homogeneous solution

The evaluation of g(x,y) of Eq. (2.5) is performed in this section.As claimed this should be the generic set of solutions for thehomogeneous Reynolds equation, namely:

cre cosy36m

@2gx,y@x2

3cre cosy2e siny

6mR2@gx,y

@y

cre cosy3

6mR2@2gx,y

@y2 0 4:1bearing, depending on the BC assumptions for c(x)and j(y)without the need of erasing the one of the two left hand termsin Eq. (2.1).

As shown in Fig. 2 the pressure distribution of the particularsolution becomes zero at the angle yn that is a function of theeccentricity rate of change. The pressure distribution of theparticular solution is affected from _e and obtains much highermaximum values even for small values of _e such as _e=OR 0:001,while the domain with negative pressure has an increment inits minimum pressure. The particular solution u(x,y) is presentedin Fig. 3.the particular solution u(x,y).R2 f yR2 cre cosy f y

l 4:5

The primes denote ordinary derivative with respect to thefunctions arguments. For treating Eq. (4.4) we distinguish thefollowing cases:

(a) If l0 Eq. (4.4) becomes as in Eq. (4.6) which can be solved togive Eq. (4.7).

m00x 0 4:6

mx c5xc6 4:7

(b) If l40 then lk2 and the solution is as in Eq. (4.8):mx c5 coskxc6 sinkx 4:8

(c) If lo0 then lk2 and one will nally arrive to thefollowing solution:

mx c5ekxc6ekx 4:9

Eq. (4.5) can be very easily solved if l0 so as to makeEq. (4.5) a rst order differential equation (using the correspond-curfunction of xwhile the right is a function of y only. So, in orderthis equality to be feasible both sides should be equal to thee constant, say l. We thus obtain Eqs. (4.4) and (4.5):00xx l 4:4

00 0B00xx

1

R2f 00yf y

1

R23e siny

cre cosyf 0yf y 4:3

y inspecting the latter equation one realizes that the left handfolloy f ymx: 4:2hen the function g(x,y) is known at the boundary @L we

ain from the above separation that g(x,y)9@L f(y)9@L m(x)9@L.using this assumption in Eq. (4.2) one will arrive at thewing outcome:We assume that the independent variables of the functiong(x,y) can be separated in the multiplicative form:

gx,

e2r e2

3cre3Oarc cos cre

Log cr e

c2r e2p

Log cr e

c2r e2p

rc cos cre Log c2r e2p

ep

cr ep

Log

c2r e2

pe

p cr e

p

3:16rent paper.

Two ways are presented in this paper in order to continue withsolving Eq. (4.9) rewritten as Eq. (4.10). The rst way is toapproximate the trigonometric coefcients, introduced by theterm 3h

0/h in Eq. (4.10), with a linear function of rst order as

Michell [5] and Dufng [15] did, so as to obtain a solution usingBessels functions. This approximation is presented in Section 4.1with the use of a linear function for h. The second way is to usethe power series method and to give the exact solution ofEq. (4.10) as a sum of innite series. This way is presented indetail in Section 4.2.

f 00y 3h0

hf 0ylR2f y 0 4:10

4.1. Linear approximation of the uid lm thickness and use of

Bessels functions for the denition of f(y)

Michell in [5] used the formula hconst.y in order forEq. (4.10) to become as Eq. (4.11) and its solution to be feasibleusing Bessel functions:

f 00y 3yf 0ylR2f y 0 4:11

D. Sfyris, A. Chasalevris / Tribology International 55 (2012) 4658 51Fig. 4. The trigonometric function of uid lm thickness h and the linearDufng in [15] used also a linear varying uid lmthickness; he used the transformation f(y)w(y)/H, with H2h3/m; and using these transformations Eq. (4.10) was written asEq. (4.12). Since H2const.y2a2y2, this yields H00 0 and then theeigenfunctions of Eq. (4.12) can be of the form wi sinliy.

w00ylwy H00

Hwy 4:12

In the analysis of this section the trigonometric function h isapproximated with the linear function h* that is dened inEq. (4.13) and shown in Fig. 4.

hn cre2ey=p 4:13

With the use of h* the solution of Eq. (4.10) is feasible usingBessel functions and the general solution is presented in Eq. (4.14)approximation of it, h*, as a function of the angular coordinate y.for both cases of positive and negative value of l.

f y c7

BesselJ 1,icr ekpR2e ikRy crp ep2ey c8

BesselY 1,icr ekpR2e ikRy crp ep2ey , l k

2

c7 BesselJ 1,cr ekpR

2e kRy crpep2ey c8

BesselY 1,cr ekpR2e kRy crp ep2ey , lk

2

8>>>:

4:14Since no imaginary solution can be accepted, only the case for

lk2 is accepted. The solution of Eq. (4.10) given by Eq. (4.15)will form a boundary value problem with results plotted togetherwith those from the exact solution given by the Power SeriesMethod, presented in what follows:

f y c7BesselJ 1, cr ekpR2e kRy

crpep2ey

c8BesselY 1, cr ekpR2e kRy

crpep2ey

4:15

4.2. The use of the method of power series for the denition of f(y)

The method of the Power Series [18,19] is used in this sectionin order to dene a solution for Eq. (4.10). The rst step is toconvert Eq. (4.10) from a linear ODE with trigonometric coef-cients to a linear ODE with polynomial coefcients. For thisreason the transformation cosy x is used and we are lookingfor a function f y ~f x. The rst and the second derivative of~f x are dened in Eqs. (4.16) and (4.17) correspondingly:

df

dy d

~f

dxsiny f 0x

1x2

q4:16

d2f

dy2 d

2 ~f

dx2sin2y d

~f

dxcosy ~f 00 x1x2 ~f 0 xx 4:17

By substituting the last two expressions into Eq. (4.10) thefollowing differential equation in Eq. (4.18) is obtained:

ax3bx2axb ~f 00 x4ax2bx3a ~f 0 x3agx3bg~f x 04:18

The constants a, b, and g are dened as

a e=3, b cr=3, g 2lR2 4:19

4.2.1. Study of ~f x in the ordinary pointsFor all the ordinary points [18,19] of the differential equation

in Eq. (4.18) we assume that the solution can be written in theform of Eq. (4.20):

~f x X1n 0

dnxn 4:20

The rst and the second derivative of the function ~f x aredened in Eq. (4.21) and (4.22) correspondingly:

~f0x

X1n 1

dnnxn1 4:21

~f00x

X1n 2

dnnn1xn2 4:22

By substituting the last three expressions to the differentialequation one will nd out that the constants that appear inthe power series expansion are determined by the followingformulas:

2 2 2

an an23gadn1bn dnan 4an3adn1

D. Sfyris, A. Chasalevris / Tribology International 55 (2012) 465852bn23nb2bdn2 X1

n 0xn

!2a3agd1 0

d1 0an2an23gadn1bn2dnan24an3adn1bn23nb2bdn2 0 4:23

4.2.2. Study of ~f x in the singular pointsThe differential equation of Eq. (4.18) is now written in the

form of Eq. (4.24):

~f 00 xx1x1axb f 0xxaxb3ax1x1 ~f x3daxb 0 4:24The term multiplied with the higher order derivative has the

following roots as in Eq. (4.25):

x 1, x1, xb=a 4:25The rst corresponds to the value y0, while the second to the

value yp. The third one is not accepted because it yieldscosyo1. So, we have two singular points, in the valuesx71. For them, a separate analysis has to be performed.For the singular point x1 we have that it is a normal singularpoint since analytic functions A0(x), A1(x) exist such that Eq.(4.26) is to be satised:

x1xaxb3ax1x1 x1x1axbA1xx123gaxb x1x1axbA0x 4:26

The series expansions of the functions A0(x), A1(x) are as in Eq.(4.27):

A1x X1n 0

x12

n1 12

X1n 0

x12

n1

3X1n 0

a

ab x1 n1

A0x 3gX1n 1

x12

n14:27

So, the indicative equation of this normal singular point isgiven in Eq. (4.28) with the two roots of Eq. (4.29):

pl l212l 4:28

l1 0, l2 1

24:29

For this singular point the solutions will be of the form of Eq.(4.30).

~f 1x X1n 0

dnx1n

~f 2x x112

X1n 0

enx1n 4:30

After a similar procedure as the one described for the ordinarypoints, one will be led to the following formulas of Eq. (4.31), forthe calculation of the constants of the above power series inEq. (4.30):

d1 01na2gn12dn1n1bna3gn3n2dn12nabandn1 0 4:31For the coefcients of the other independent solution one will

be led to the following formulas:e1 0This section starts with two boundary value problems yieldedby the homogeneous solution dened in detail in Section 4. Therst has to do with the distribution of the pressure g(x,y) inthe axial direction of the bearing m(x) and the second with thedistribution in the circumferential direction f(y). The boundaryvalue problem for m(x) is dened from the ODE in Eq. (5.1) andthe BCs of Eqs. (5.2) and (5.3).

m00xk2mx 0 5:1

mLb=2 1 5:2

mLb=2 1 5:3The explanation for the BCs in Eqs. (5.2) and (5.3) is that the

resulting pressure at both ends of the bearing P(7Lb/2,y) has tobe equal to zero (the atmospheric pressure is set zero) and thusu(7Lb/2,y)g(7Lb/2,y), that means j(y)m(7Lb/2)f(y).

The general solution of Eq. (5.1) is given in Eq. (5.4).The general solution in Eq. (5.4) is substituted in the BCs ofEqs. (5.2) and (5.3) and one can dene the constants c5 and c6 andobtain the solution. The eigenfunctions mn(x) are written as inEq. (5.5) with kn to be the eigenvalues of the problem ofEqs. (5.15.3).

mx c5ekxc6ekx 5:4

mnx eLbkn=2

1eLbkn eknx e

Lbkn=2

1eLbkn eknx, 5:5

The boundary value problem for f(y) is treated in the twofollowing Sections 5.1 and 5.2 for the cases that f(y) is denedusing Bessel functions (Section 4.1) and the Power Series Methoda13

43g6nn2

en1

1

413ab12ag12bg32a4bn

12a4bn2enab 32n

en1 4:32

For the normal singular point x1 we have the followingindependent solutions of Eqs. (4.33) and (4.34):

~f 1x X1n 0

dnx1n 4:33

~f 2x 9x193=2X1n 0

enx1n 4:34

For the denition of the constants in Eqs. (4.33) and (4.34), theformulas of Eqs. (4.35) and (4.36) are used correspondingly:

d1 01na33g3nn2dn13ag2bn2a3bgn3n2dn1n2aban2bndn1 0 4:35

e0 e1 01

23agna 1

2n

en1 a

11

23g8n2n2

b 943g3nn2

en

1452n4a2an4b2n

en1 0 4:36

5. Boundary conditions for the pressure distribution andevaluation of the resulting pressure(Section 4.2) correspondingly.

5.1. Boundary value problem for f(y) using Bessel functions

The boundary value problem consists of the ODE of Eq. (5.6)and the BCs of Eqs. (5.7) and (5.8):

f 00y 3hn0

hnf 0yk2R2f y 0 5:6

f 0 0 5:7

f p 0 5:8The BCs in Eqs. (5.7) and (5.8) express that in the beginning of

the oil lm (y0) the pressure P(x,0) has to be zero. Thusj(0)m(x)f(0). Since j(0) is chosen to be zero (see section ofthe particular solution) then f(0)0. Also, since f(y) is thehomogenous solution there is no inuence from _e (this meansy*p) and the pressure yielded from the homogeneous problembecomes also zero at yy*p. The general solution of Eq. (5.6) isgiven in Eq. (5.9):

f y c7BesselJ 1,crekpR

2ekRy

=crpep2ey

c ekpR

D. Sfyris, A. Chasalevris / Tribology International 55 (2012) 4658 53c8BesselY 1, r 2e kRy =crpep2ey

5:9The substitution of the general solution in the boundary

conditions gives the system of equations in Eq. (5.10):

c7 BesselJ 1, cr ekpR2e

=crpep

c8 BesselY 1, cr ekpR2e

=crpep

0

c7 BesselJ 1,cr ekpR

2e kRp

=crpep

c8 BesselY 1, cr ekpR2e kRp

=crpep

0

8