stresses and deformation of a long hydrodynamic journal bearing

6
Computers d Strucrwes Vol. 48, No. 1, pp. 81-86, 1993 0045-7949/93 s6.lm + 0.00 Printed in Great Britain. 8 1993 Pergamoa Press Ltd STRESSES AND DEFORMATION OF A LONG HYDRODYNAMIC JOURNAL BEARING S. SINGH College of Technology, G. B. Pant University of Agriculture and Technology, Pantnagar-263 145 (U.P.), India (Receioed 23 April 1992) Abstract--The pressure in the lubricant film of a long hydrodynamic journal bearing is non-axisymmetric in nature. Therefore, Lame’s equations are not valid to calculate the stresses and deformation of the bearing. In this paper, a closed-form solution has been obtained to calculate the stresses and deformation of the bearing by means of Airy’s stress function approach. The bearing has been considered as a thick cylinder under plane strain loading. The pressure distribution on the bearing has been represented by Fourier series whose coefficients have been calculated by Simpson’s composite formula for numerical integration. The variation of hoop stress and deformation of the inner surface of the bearing has been plotted. Ao,...,H,, : k G h n ;(e) PO,...*% r rl,r2 u W i L rl v 4 Ue 6, Gl cr be ;:e, NOTATION stress function constants diametral clearance, mm journal diameter, mm eccentricity, mm modulus of elasticity, MPa modulus of rirciditv. MPa film thickness,mm journal speed, rpm an integer hydrodynamic pressure, MPa Fourier coefficients any radius, mm inner and outer radii of bearing respectively, mm journal speed, m/set journal load, N attitude angle, degree Airy’s stress function any angle, degree eccentricity ratio, 2e/c lubricant viscosity, mPa see Poisson’s ratio radial displacement, mm hoop displacement, mm radial strain hoop strain radial stress, MPa hoop stress, MPa shear stress, MPa polar coordinates INTRODUCTION The analysis of stresses and deformation of a thick cylinder under uniform pressure can be carried out by using the well-known Lame’s equations [ 11.However, when the pressure distribution is non-uniform then these equations become invalid and one has to resort to some numerical technique, for example the finite element method [2], to calculate the stresses and deformation. The pressure distribution in cylinders of high capacity hydraulic presses, in the lubricant film of hydrodynamic bearings and that which is due to flow past a cylinder is non-axisymmetric. Closed- form solutions for such problems are not readily available. In this paper, a closed-form solution is presented to calculate the stresses and deformations of a long hydrodynamic journal bearing by using the Airy stress function approach. The pressure distribution on the bearing has been represented by Fourier series whose coefficients have been calculated by adopting Simpson’s composite formula [3] for numerical inte- gration. The stress function constants have been evaluated by the Gauss-Jordan [4] method of solving non-homogeneous simultaneous equations. A nu- merical example has been solved to show the vari- ation of hoop stress and deformations of the inner surface of the bearing. LUBRICANT FILM PRESSURE The geometry of a hydrodynamic journal bearing is shown in Fig. 1. The pressure in the lubricant film of a long hydrodynamic journal bearing is [5] 3qUd P(e) = ~ 6 sin O(2 + 6 cos 0) c2 1 (2+8)(1+.5 cosQ2 . (1) The maximum pressure occurs at -36 c0se =- 2fE2’ (2) The pressure distribution on the journal and bear- ing is shown in Fig. 2. The pressure distribution is obviously highly non-axisymmetric. PROBLEM FORMULATION The bearing can be considered as a thick cylinder having axial symmetry. Further, the bearing can be 81

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Computers d Strucrwes Vol. 48, No. 1, pp. 81-86, 1993 0045-7949/93 s6.lm + 0.00

Printed in Great Britain. 8 1993 Pergamoa Press Ltd

STRESSES AND DEFORMATION OF A LONG HYDRODYNAMIC JOURNAL BEARING

S. SINGH

College of Technology, G. B. Pant University of Agriculture and Technology, Pantnagar-263 145 (U.P.), India

(Receioed 23 April 1992)

Abstract--The pressure in the lubricant film of a long hydrodynamic journal bearing is non-axisymmetric in nature. Therefore, Lame’s equations are not valid to calculate the stresses and deformation of the bearing. In this paper, a closed-form solution has been obtained to calculate the stresses and deformation of the bearing by means of Airy’s stress function approach. The bearing has been considered as a thick cylinder under plane strain loading. The pressure distribution on the bearing has been represented by Fourier series whose coefficients have been calculated by Simpson’s composite formula for numerical integration. The variation of hoop stress and deformation of the inner surface of the bearing has been plotted.

Ao,...,H,,

:

k G h n

;(e) PO,...*% r rl,r2 u W

i L

rl v

4 Ue 6, Gl cr be

;:e,

NOTATION

stress function constants diametral clearance, mm journal diameter, mm eccentricity, mm modulus of elasticity, MPa modulus of rirciditv. MPa film thickness,mm journal speed, rpm an integer hydrodynamic pressure, MPa Fourier coefficients any radius, mm inner and outer radii of bearing respectively, mm journal speed, m/set journal load, N attitude angle, degree Airy’s stress function any angle, degree eccentricity ratio, 2e/c lubricant viscosity, mPa see Poisson’s ratio radial displacement, mm hoop displacement, mm radial strain hoop strain radial stress, MPa hoop stress, MPa shear stress, MPa polar coordinates

INTRODUCTION

The analysis of stresses and deformation of a thick cylinder under uniform pressure can be carried out by using the well-known Lame’s equations [ 11. However, when the pressure distribution is non-uniform then these equations become invalid and one has to resort to some numerical technique, for example the finite element method [2], to calculate the stresses and deformation. The pressure distribution in cylinders of high capacity hydraulic presses, in the lubricant film of hydrodynamic bearings and that which is due to

flow past a cylinder is non-axisymmetric. Closed- form solutions for such problems are not readily available.

In this paper, a closed-form solution is presented to calculate the stresses and deformations of a long hydrodynamic journal bearing by using the Airy stress function approach. The pressure distribution on the bearing has been represented by Fourier series whose coefficients have been calculated by adopting Simpson’s composite formula [3] for numerical inte- gration. The stress function constants have been evaluated by the Gauss-Jordan [4] method of solving non-homogeneous simultaneous equations. A nu- merical example has been solved to show the vari- ation of hoop stress and deformations of the inner surface of the bearing.

LUBRICANT FILM PRESSURE

The geometry of a hydrodynamic journal bearing is shown in Fig. 1. The pressure in the lubricant film of a long hydrodynamic journal bearing is [5]

3qUd P(e) = ~

6 sin O(2 + 6 cos 0)

c2 1 (2+8)(1+.5 cosQ2 . (1) The maximum pressure occurs at

-36 c0se =-

2fE2’ (2)

The pressure distribution on the journal and bear- ing is shown in Fig. 2. The pressure distribution is obviously highly non-axisymmetric.

PROBLEM FORMULATION

The bearing can be considered as a thick cylinder having axial symmetry. Further, the bearing can be

81

82 s. SINOH

I ,- BEARING

LUBRIC

Fig. 1. Geometry of hydr~yn~ic jouruaf bearing.

treated as a plane strain concentric annular plate. The Airy stress function d, expressed in the two-dimen- sional polar coordinates should satisfy the bihar- monk equation

( a* 1 a 1 8 $i+;s+;Iae’

>(

a’d, 1 a&, 1 &#I ,,i+;,,+i2 =o.

r ae > (3)

The relations between the stress components, dis- placements and stress function # are expressed as

(0) JOURNAL PRESSURE

El?0 PRESSURE

(bl BEARING PREkSURE de9

Fig. 2. Hydrodynamic pressure distribution. Fig. 3. Actual and approximated hydrodynamic pressure.

a ia4

‘dB--dr rae ( > -- (6)

(7)

Q+T*-vo,)=~+~~. (8) r

The most general expression for the stress func- tion [6] satisfying eqn (3) is

d, = A,ln r + Bor2 + C,r2 In r f DOr2@

+ (A r /2)rB sin + A ; e - (E, /2)r@ cos e

+(B,r-‘+C,r3+D,rlnr)cosf3

+ c (A,? t i&r-” f Cnr”+‘+ D,,r”++2) n=2

+ f (E,r”+F,r-“+G,rn+2+H,r-“f2) n=2

x sin ne, (9)

where Ao,..., H, are undete~in~ coefficients. For the bearing subjected to the loading depicted

in Fig. 2(b) and Fig. 3, the stress function # should be such that the stresses obtained from it should satisfy the boundary conditions and displacements should be uniquely determined. We assume the stress function in series form as

+ f [A,r”$-B,r-“+C,r”t2$D,r-“‘2f II=2

xcosne + f [E,r”+F,r-“+G,r”+2 m=2

+H,r-“+2]sinnB.

p(a)

2

MP~

1

The pressure distribution on the bearing can be represented by Fourier series [7] as

p(8) =p,/2 +p, cos 0 + q, sin 6

where

Stresses and deformation of a long hydrodynamic journal bearing 83

-“g*[n(n - l)r”-2E,-n(n + l)rene2F,

+n(n + I)r”G, -n(n - l)r-“HJcosnB (16)

+ f [p. cos n0 + q. sin n6], (11) 2Gu,= -Aor-1+2(1-2v)rBo+B,r-2cosfI n-2

+ F,r-‘sin0 + f [-nr”-‘A, n-2

p,=l 2n s = 0 p(6) cos n6 d0 (12)

q”= 1 s 2r = 0

p(8) sin ntJ dt7 (13)

forn=Oto co.

SOLUTION OF PROBLEM

Substituting eqn (10) in eqns (4)-(8), we get

+ i [-n(n - l)rne2A,, - n(n + l)rwnm2B, n-2

-(n - 2)(n + l)r”C,

- (n + 2)(n - 1)r -“DJcos n0

+ i [-n(n - l)r”-2E, - n(n + l)renV2F, n-2

-(n -2)(n + l)r”G,-(n +2)(n - l)r-“H&inn0

(14)

ug= -Aor-2+2Bo+2B,r-JcosB +2F,r-‘sin0

+ f [n(n - l)rnV2 A,, + n(n + l)r-“-2B,, n-2

+(n + 2)(n + l)r”C,

+ (n - 2)(n - 1)r -“D&OS n0

+ t [n(n - l)r”-2E, + n(n + l)renm2F, n-2

+(n + 2)(n + l)r”G,,

+ (n - 2)(n - I)r -“H,]sin nfJ (15)

T&x: -2B,r-‘sin8 +2F,r-3cos0

+ i [n(n - l)rnm2A, -n(n + l)renV2B, n-2

+n(n + l)r”C,-n(n - l)r-“DJsinnt7

+nr-“-‘B,-(n -2+4v)r”+‘C,

+ (n + 2 - 4v)r --)) + ‘D,]cos n0

+ i [-nr”-’ E,+nr-“-‘F, n-2

-(n -2+4v)r”+‘G,

+(n +2-4v)r-“+‘H,]sinne (17)

2Gu0=B,r-2sin0-F,r-2cos8

+ fj [nr”-’ A,+nr-“-‘B, n-2

+(n +4-4v)r”+‘C,

+(n -4+4v)r-“+‘D,]sinntl

-m$2 [nr”-‘E,, + nr+-IF,

+(n +4-4v)r”+‘G,

+(n -4+4v)r-“+‘HJcosnB. (18)

The constants A,, . . ..H. are to be determined from the boundary conditions given in eqns (19x22) below

cArA = -p(8) (19)

c,(r2) = 0 (20)

54(rl) = 0 (21)

td(r2) = 0. (22)

Substituting eqns (14) and (16) in eqns (19)-(22), and comparing coefficients, we get

[;I: ;]t} = { -;“) (23)

2B,ri’*p, (24)

2F,ri’=q, (25)

84 S. SINGH

[

u(L 1) u(l, 2) a(l,3) a(l,4)

~(2, 1) u(2,2) ~(2~3) u(2,4)

u(3, 1) u(3,2) u(3,3) u(3,4)

~(4, 1) ~(4~2) u(4,3) ~(4~4) I

Table 1. Fourier series coefficients (p(B) = p0/2 + p, cos 0 + Ii

q, sin 0 + C [p, cos n0 + q, sin &I]) n=Z

n p(n) 9(n)

0 O.l1208E+Ol O.OOOOOE + 00 I -0.21249E + 00 0.859lOE + 00 2 -0.31666E + 00 -0.244378 + 00

0.11373E + 00 0.6142SE - 01 = (26) ;: -0.10407E + 00 -0.14441E - 01

5 0.3728lE - 01 0.32558E - 02 - and 6 -0.416698 -01 -0.71309E 03

7 0.16416E - 01 0.15279E - 03

~(1, 1) u(l,2) u(l,3) u(1,4) En

~(2, 1) ~(292) u(2,3) u(2,4)

Ii

F, I

~(3, 1) u(3,2) ~(393) ~(394) G,

u(4, 1) ~(4~2) ~(4~3) u(4,4) H, J

8 -0.22152E - 01 -0.32257E - 04 9 0.92877E - 02 0.67629E - OS

10 -0.13779E - 01 -0.14283E - 05 > 11 0.59870E - 02 0.363708 - 06

12 -0.93370E - 02 -0.1978SE - 06 13 0.41416E - 02 --0.2403SE - 06 14 -0.66379E - 02 0.12694E - 07 15 0.29708E - 02 0.38633E - 08

(27) Knowing the constants A,,, , H,,, the stresses

for n = 2,3,4, . . ,

where

u(1, 1) = -n(n - l)rye2

a(l,2) = -n(n + l)r;“-2

a@, 3) = -(n + l)(n - 2)r;

u(l,4) = -(n + 2)(n - 1)r;”

u(2, 1) = n(n - l)rye2

u(2,2) = n(n + 1)r;

u(2,3) = -(n + l)nr;“-2

a(2,4) = -n(n - 1)r;”

a(3, 1) = -n(n - I)$2

u(3,2) = -n(n + l)rTfle2

u(3,3) = -(n - 2)(n + l)rl

u(3,4) = -(n + 2)(n - 1)r;”

u(4, 1) = n(n - l)r?-2

a(4,2) = (n + l)nr2

a(4,3) = -(n + l)nr;“-2

u(4,4) = -n(n - 1)r;“.

and displacements can be calculated from eqns

(14H18).

Table 2. Actual and approximated hydrodynamic pressure

0 Actual pressure Approximated pressure (degree) WW NW

O.OOOOOE + 00 0.23396E - 01 0.0 5.0

10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 65.0 70.0 75.0 80.0 85.0 90.0 95.0

100.0 105.0 110.0 115.0 120.0 125.0 130.0 135.0 140.0 145.0 150.0 155.0 160.0 165.0 170.0 175.0 180.0

0.88980E - 01 0.8430lE - 01 0.178OSE + 00 0.17lllE+00 0.26730E + 00 0.26797E + 00 0.3568OE + 00 0.36128E + 00 0.44663E + 00 0.44796E + 00 0.53682E f 00 0.53394E + 00 0.6274lE + 00 0.62496E + 00 0.71838E + 00 0.71972E + 00 0.80969E i- 00 0.81256B + 00 0.90124E + 00 0.90139E + 00 0.99287E + 00 0.99019E + 00 0.10843E + 01 0.10829E + 01 0.11753E + 01 O.l1772E+Ol 0.12653E + 01 0.12677E + 01 0.13538E + 01 0.13530E + 01 O.l44OOE+01 0.1437lB + 01 0.1523OE + 01 0.15224E + 01 0.16016E + 01 0.16044E+01 0.16743E + 01 0.16762E + 01 0.17396E + 01 0.1737SE + 01 0.17953E + 01 0.17922E + 01 0.18392E + 01 0.18400E + 01 0.18687E + 01 O.l872SE+Ol 0.18809E + 01 0.18818Et01 0.1873OE + 01 O.l8689E+Ol 0.18417E + 01 0.18388E + 01 0.17843E f 01 0.17877E + 01 0.16982E + 01 0.17034E + 01 0.1581SE + 01 0.15796E + 01 0.14333E + 01 0.14257E + 01 0.12537E + 01 0.12528E + 01 0.1044SE+Ol 0.1055lE+Ol 0.80903E + 00 0.8 146OE + 00 0.552lOE + 00 0.53508E + 00 0.27999E + 00 0.26394E + 00 0.48676E - 06 0.6874SE - 01

Stresses and deformation of a long hydrodynamic journal bearing

Table 3. Values of constants for stress function 4

n A” & Cl Dzl 0 -0.11634EfO4 0.36356s + 00 - - 1 - -0.288JSE + 04 - - 2 0.312%E + 01 O.l4092E+07 - 0.80599E - 03 -0.38248E + 04 3 -0.41538E - 02 -0.335808 + 07 0.1559933 - 05 0.63430E + 04 4 0.29795E - 04 0.35664E + 08 -O.l3116E-07 - 0.58372E f 05

z -O.l1626E-06 0.1701OE - 08 -0.1972433 0.40738E + + 09 10 -0.86064E 0.557788 - - 10 12 - 0.5943OE 0.3003lE + + 06 07

; -0.98315E- 0.20964E - 11 12 -0.3354OE 0.10312E + + 13 11 -O.l1278E- 0.515538 - 14 15 -0.14286E 0.475208 + + 08 10 9 -0.14623E - 14 -0.10489E + 14 0.80175E - 18 0.14262E + 11

10 0.374OlE - 16 0.395058 + 15 -0.20814E - 19 -0.52857E + 12 11 -0.28654E - 18 -0.44953E + 16 0.16135E - 21 0.59294E + I3 12 0.80256E - 20 0.18816E + 18 - 0.45632s - 23 -0.24506E + 15 13 -0.646738 - 22 -0.22775E + 19 0.37067E - 25 0.29329E + 16 14 0.18996E - 23 0.10096E + 21 -0.10962E - 26 -0.12872E + 18

85

15 -6115670E - 25 -O.l26llE+22

Ftl 0.11667E + 05 0.108748 + 07

n

1 2 3 4

:

: 9

10 11 12 13

E” -

0.2415lE+Ol - 0.224348 - 02

0.41344E - 05 -O.l0153E-07

0.29llOE - 10 -0.91505E - 13

0.30528E - 15 -0.10648E - 17

0.38742E - 20 -0.17406E -22

0.17006E - 24 0.375328 - 26

-0.181368 +07 0.494888 + 07

-0.172248 + 08 0.697I8E + 08

-0.312178 +09 0.15016E + 10

-0.76374E + 10 0.40924E + 11

-0.27308E + 12 0.39872E + 13 0.13217E + 15

0.90952E - 29 0.15936E + 19

G. H.

- -0.621978 -03

0.842508 - 06 -0.182OOE -08

0.48710E - 11 -0.147298 - 13

0.47983E - 16 -0.16422E - 18

0.58380E - 21 -0.2156lB - 23

0.980178 - 26 -0.96695E - 28 -0.21512E - 29

-0.295GE + 04 0.34258E + 04

-0.80999E + 04 0.26226E + 05

-0.1017lE + 06 0.44229E + 06

-0.20802E + 07 0.10385E + 08

-0.54754E + 08 0.36019E + 09

-0.51928E + 10 -0.1702lE + 12

f: -0.363248 -0.20378E - - 29 31 -0.19306E -0.16405E + + 16 15 0.209628 O.l1828E-34 - 32 0.24613E 020723E + + 12 13

Table 4. Hoop stress and displacements of inner surface of bearing

(de&e, (;;a) Cllk, U0

(mm)

0.0 0.91274E + 01 10.0 0.10624s + 02 20.0 O.l1123E+02 30.0 0,1%98E c 02 40.0 0.93936E + 01 50.0 0.73515E + 01 60.0 0.47426E + 01 70.0 0<17900E+01 80.0 -0.1213lE + 01 90.0 -0.39714E + 01

100.0 -0.61293E + 01 110.0 -0.7384lE + 01 120.0 -0.7457OE + 01 130.0 -0.61842E + 01 140.0 -0.35828E + 01 150.0 0.15891E + 00 160.0 0.45097E + 01 170.0 0.88t85E + 01 180.0 0.12074E + 02 190.0 0.13375E + 02 200.0 0.12914E + 02 210.0 O.l1128E+02 220.0 0.83287E + 01 230.0 0.49750E + 01 240.0 0.14322E + 01 250.0 -0.19158E + 01 260.0 -0.47510E + 01 270.0 -0.68309E i- 01 280.0 -0.79616E + 01 290.0 -0.80768E + 01 300.0 -0.71594E + 01 310.0 -0.53186E + 01 320.0 -0.27288E + 01 330.0 0.3574lE + 00 340.0 0.36017E + 01 350.0 0.666868 + 01 360.0 0.91274E + 01

-0.21052E - 02 -0.135828 -02 -0.2569CJE - 02 -0.72713E -03 -0.27017E -02 -0.20243E - 04 - 0.24908E - 02 0.68217E - 03 -0.195908 -02 0.13018E - 02 -O.l1629E-02 0.17693E - 02 -O.l87lOE-03 0.20314E - 02 -0.86282E - 03 0.20554E - 02

0.18697E - 02 0.18343E - 02 0.27166E - 02 0.13878E - 02 0.32985E - 02 0.76222E - 03 0.35354E - 02 0.27584E - 04 0.33835E - 02 -0.72908E - 03 0.28442E - 02 -0.1412OE - 02 O.t%99E - 02 -0.19288E - 02 0.861338 - 03 -0.22027E - 02

-0.338728 - 03 -0.21863E - 02 -O.l4661E-02 -0,18724E-02 -0.23550E - 02 -O.l3004E-02 -0.28763E - 02 -0.55609E - 03 - 0.29774E - 02 0.25095E - 03 -0.2669lE - 02 O.l0143E-02 -0.200818-02 O.l64llE-02 -0.10909E -02 0.20605E - 02 -0.3593OE - 04 0.222998 - 02

0.102828 - 02 0.213668 - 02 0.19783E - 02 0.17977E - 02 0.27084E - 02 0.12567E - 02 0.31395E - 02 0.57809E - 03 0.3228lE -02 -O.l5992E-03 0.29680E - 02 -0.8739lE - 03 0.23923E - 02 -0.14838E - 02 O.l5684E-02 -0.19215E -02 0.59167E - 03 -0.2138OE - 02

- 0.42422E - 03 -0.21090E - 02 -0.13606E -02 -0.18383E - 02 -0.21052E - 02 -0.13582E - 02

86 S. SINGH

ZERO STRESS

SCALE lmm : 1Mpn

ZERO DISPLACEMFNT

-3 SCALElmm : 2X13mi-n

Fig. 4. Hoop stress at inner surface of bearing. Fig. 5. Displacements of inner surface of bearing.

Table 5. Maximum and minimum values of hooo stress and disulacements

Quantity Max. value

q,, MPa 0.13405E + 02 a,, mm 0.35381E - 02 ue, nun 0.22324E - 02

0 (degree)

192 111 241

Min. value

-0.81554E +Ol -0.29915E - 02 -0.22323E - 02

tl (degree)

286 196 154

NUMERICAL EXAMPLE

Let us assume that the bearing parameters are q = 15 mPa set, $ = 60 degrees, c = 0.06 mm, L = 0.36215, d = 60 mm, and n = 2000 rpm.

The material properties are G = 0.84E + 05 MPa and v = 0.30. Taking lubricant film pressure to be zero at 0 = 0 and 180 degrees, the pressure distri- bution on the bearing is plotted in Fig. 3. The Fourier coefficients pO, . . . , q, were obtained from eqns (12) and (13) by evaluating the integrals by Simpson’s composite formula and taking 36 subintervals. The maximum value of n was taken as 15. The Fourier coefficients are given in Table 1. The pressure calcu- lated from eqn (11) has been compared with that of eqn (1) in Table 2 and shown by solid dots in Fig. 3.

Taking r, = 30.06mm and r2 = 40mm, the con- stants&, . . , H, were calculated from eqns (23)-(27) by using Gauss-Jordan method of solving non- homogeneous simultaneous equations. These con- stants are given in Table 3.

A computer program was developed to evaluate eqns (14x18). The hoop stress and displacements of the inner surface of the bearing so evaluated at 0 = 10 degree step are given in Table 4 and plotted in Figs 4 and 5. The calculations were made on an IBM PC/XT computer.

DISCUSSION OF RESULTS

The maximum hoop stress due to uniform maxi- mum film pressure of 1.881 MPa, according to Lame’s equations, would have been 6.762 MPa, and is tensile in nature. From the present analysis, the

maximum and minimum values of hoop stress and displacements are as given in Table 5.

The maximum tensile hoop stress is larger by 98.24% from that of the value given by Lame’s equations.

CONCLUSIONS

A closed-form solution has been presented to calculate the stresses and deformations of a long hydrodynamic journal bearing by using the Airy’s stress function approach. An example has been solved, which shows that the results of the present analysis are quite offset from those of the values obtained from Lame’s equations. This will enable the designers to calculate the hoop stress and defor- mations more rigorously which is essential for an accurate analysis of hydrodynamic journal bearings for precision machinery.

5.

6.

7.

REFERENCES

S. Singh, Strength of Materials, 4th Edn. Khanna Publishers, Delhi (1988). 0. C. Zienkiewicz, The Finite Element Method, 3rd Edn. McGraw-Hill (1977). E. Kreyszig, Advanced Engineering Mathematics, 2nd Edn. John Wiley, New York (1967). S. D. Conte and C. de Boor, Elementary Numerical Analysis: An Algorithmic Approach, 2nd Edn. McGraw- Hill, Kogakusha, Tokyo (1972). M. F. Spotts, Mechanical Design Analysis. Prentice- Hall, Englewood Cliffs, NJ (1964). S. P. Timoshenko and J. N. Goodier, Theory of Elasticity, 3rd Edn. McGraw-Hill New York (1970). M. Abramowith and I. A. Stegun, Handbook of Mathematical Functions. Dover, New York (1964).