LU

BR

ICA

TIO

N

AN

D

JO

UR

NA

L B

EA

RIN

GS

SC

OP

E•

Need

•Types of lubrication

•Viscosity

•Petroff’sequation

•Stable lubrication

•Thick-film lubrication

•Hydrodynam

ic theory

•Design considerations

•The relations of the variables

•Steady-state conditions in self-contained bearings

•Clearance

•Pressure fed bearings

•Load and m

aterials

•Bearing types

•Boundary lubricated bearings

Need

•The most important are revolute pair and sliding pair

•Objective of lubrication is to reduce friction, wear and heating

of machine parts which m

ove relative to each other.

•Lubricant is exactly that substance which does the above

when inserted between m

oving surfaces

•Lubrication is needed everywhere, for exam

ple, sleeve

bearings, antifriction bearing, cam and follower, gear teeth,

piston in cylinder, crank shaft and connecting rod bearings

•In applications requiring low load bearing capacity, nylon

bearings requiring no lubrication, a powder m

etallurgy

bearing with lubricant built-in, a bronze bearing with ring

oiling, wick feeding, or solid lubricant film

or grease

lubrication m

ay be satisfactory

•Journal=shaft and sleeve=

bushing

Types o

f lu

bri

cation

•Hydrodynamic

–Thick-film, fluid m

echanics, speed is necessary

•Hydrostatic

–Lubricant is introduced at high pressure, speed could be sm

all

•Elasto-hydrodynamic

–Occurs in rolling contact, like in gears and rolling element bearings, H

ertzian

contact theory and fluid m

echanics are required

•Boundary

–Insufficient surface

area, drop in velocity, lessening of lubricant quantity,

increase in bearing load, or increase in lubricant temperature lead to no thick

film

; partial metal to metal contact occurs which is mixed hydrodynam

ic-

boundary lubrication before gross metal-to-m

etal contact occurs;

fluid

mechanics and viscosity are less im

portant and chem

ical composition is more

important

•Solid-film

–Necessary when operation is to be at extrem

ely high temperaturesbecause

ordinary m

inerals oils degrade; graphite and m

olybdenum disulphide are often

used; composite bearing m

aterials are being researched because liquid lubricants

also proved to be environmentally non-sustainable

VIS

CO

SIT

Y

Fdu

Ady

τµ

==

Newton’s law of viscosity:

FU

Ah

τµ

==

µis the absolute viscosityor dynamic

viscosity. If shear rate is constant,

hU

dy

du

//

=

Then,

The units of absolute viscosity

SI System: Pa-s or N-s/m

2.

IPS system: lbf-sec/in2or psi-sec = 1 reyn

Conversion: 1 reyn= 6.89 * kPa-s= 6890 Pa-s.

The a

dora

ble

outc

asts

!

•CGS units have been banished but are still in

widespread use

•Force -Dyne (dyn), dynam

ic viscosity -poise (P),

kinem

aticviscosity –

stoke (St)

•Poise = dyn-sec/cm

2= 10-5*104N-sec/m

2= 0.1 Pa-s =

(1/68900) reyn

•Centi-poise (cP) = Z= 10-2*(10-5*104N-sec/m

2) =

0.001 Pa-s ⇒

µ(Pa-s) = 103Z (cP)

•µ(reyn) = 68900 Poise =6.89*106Z (cP)

Saybolt

univ

ers

al vis

cosim

ete

rof A

STM

•Kinem

aticviscosity is measured

•Measured is the time in seconds required for 60 m

l of

lubricant at a specified tem

perature to run through a

tube 17.6 m

m in a diameter and 12.25 m

m long.

•This m

easured tim

e is terms as Sayboltseconds.

•Using the Sayboltseconds, the kinem

aticviscosity of

the lubricant in centiStokes

(Stoke is cm

2/sec) can be

obtained by the following equation:

−=

tt

Zk

180

22

.0

Vis

cosity fro

m S

aybolt

vis

cosim

ete

r

()

6180

0.22

10

tt

υ−

=−

()

6180

0.22

10

tt

µρ

−

=

−

()

()

cSt

Zm

Since

k

62

10

sec

/,

−=

υ

The dynamic or absolute viscosity is density times the kinematicviscosity.

m2/sec.

Pa-s.

Density in kg per m

3.

Diffe

rent te

mpera

ture

s o

f re

levance

T1= Inlet temperature

T2= Outlet tem

perature

Tf= T

av= Film tem

perature which is also the average

temperature; Also the temperature at which to

Ts=T1=for both bearings, external sump type and self-

contained type

T∞=Ambient temperature

PETR

OFF’S

EQ

UA

TIO

N:

•Petroffwas the first person to explain the phenomenon of bearing friction

•He assumed that the shaft is concentric hence his theory is not good for actual

bearing which is eccentric when it is running

•Nevertheless, petroff’sbearing theory explains the useful (i) dimensionless groups

(ii) gives an equation for coefficient of friction that is good even for eccentric

bearings

•Petroff’stheory is developed for lightly loaded bearings

∆+

=2T

TT

fi

∆−

=2T

TT

fi

fT

Contd

.2

UrN

hc

πµ

τµ

==

23

24

()(

)(2

)()

rN

rlN

TAr

rlr

cc

πµ

πµ

τπ

==

=

2

()(2

)()

2T

fWr

frlPr

rflP

==

=2

2Nr

fPc

µπ

=2

rN

Sc

Pµ

=

2

22

22

rN

rf

Sc

Pc

µπ

π

=

=

τ=shear

stress,

S=

Som

merf

eld

num

ber; r=radius of the journal; c=radial

clearance between the bearing and journal; r/c=radial clearance ratio; N=speed in

rps; P=pressure in the bearing; W=load on the bearing; T=torque; f=coefficient of

friction; h=film thickness; U=relative velocity of bearing and journal; l=length of

bearing; µ=coefficient of dynamic viscosity

Petr

off’s

equation

(1883)

Contd

.

2j

mlc

Ulc

rNρ

ρπ=

=&

()

()

00

00

2

f

loss

b

UA

TT

QUA

TT

∞

∞

−=

−=

&

gen

pj

pQ

mC

Tlc

rNC

Tρπ

=∆

=∆

&&

()

2

32

22

24

j

gen

jj

jj

Nrl

frc

rc

QTN

fWrN

rWN

rWN

cr

cW

r

µπ

ππ

π

=

==

=

&

23

3

00

16

j

f

Nlr

TT

UAc

µπ

∞=

+(

)0

02

fgen

pj

p

UA

TT

QT

mC

lcrN

Cρπ

∞−

∆=

=& &

area

density

Average velocity

Combined radiation & convention

universal heat transfer coefficient

Bearing surface area

Because the bushing temperature is half-way

between the film temperature and ambient

temperature

Heat generated due to work

done on the lubricant by the

journal

Sta

ble

Lubrication

()

60.36210

jN P

µ−

≥

Thic

k F

ilm

Lubrication

ce=

ε

Hydro

dynam

ic T

heory

Hydro

dynam

ic lubrication e

quations:

0dp

Fp

dxdydz

dxdz

dydxdz

pdydz

dx

yττ

τ

∂

=

++

−+

−=

∂

∑

dp

dx

yτ∂=∂

u yτ

µ∂

=∂

2

2

dp

u

dx

yµ∂

=∂

⇒

Substituting,

1

1u

dpy

Cy

dx

µ∂

=+

∂integrating

Integrating once more,

2

12

1 2

dp

uy

Cy

Cdx

µ=

++

0;

0;

yy

h

uu

U

==

==−

⇒

Som

merf

eld

’sconclu

sio

n

()

12

2

;0

2

1 2Uhdp

CC

hdx

dp

Uu

yhy

ydx

h

µ

µ

=−

−=

⇒=

−−

Uu

yh

=−

When the pressure is

maximum, dp/dx=0, hence

0h

Qudy

=∫

Volume flow rate:

3

212

Uh

hdp

Qdx

µ=−

−

If the lubricant is incompressible, then the volume flow rate issame for all cross-

sections. Hence

0=

dx

dQ

Contd

.

⇒=

−=

012

2

3

dx

dp

h

dxd

dx

dh

U

dx

dQ

µ

3

6d

hdp

dh

Udx

dx

dx

µ

=−

33

6h

ph

ph

Ux

xz

zx

µµ

∂∂

∂∂

∂−

=−

∂∂

∂∂

∂

2r

rN

fc

cPµ

φ

=

Norm

ally difficult to solve this partial differential equation analytically. Sommerfeld

gave a solution in 1904, which can be expressed as

Eff

ective journ

al speed

2

jN

rS

cP

µ

=

2j

bW

NN

NN

=+

−

fW

NN

=

()

22

12

jb

Wb

W

jj

NN

NN

Nr

Sc

PN

N

µ

+−

==

+−

S

Desig

n C

onsid

era

tions

•There are two groups of variables.

•The first group of variables are independent variables and

either given or under the control of the designer.

1) The viscosity, µ

2) The load per unit of projected bearing area, P

3) The speed N

4) The bearing dim

ensions r, c, β

•In the second group are the dependent variables. The

designer can not control these except indirectly by changing

one or more of the first group. These are:

1)

The coefficient of friction f

2)

The temperature rise ∆T

3)

The volume flow rate of oil Q

4)

The minim

um film thickness ho

The d

imensio

nle

ss v

ari

able

s a

nd c

hart

s o

f R

aym

ondi&

Boyd

variable

ess

dim

ensionl

rise

eTem

peratur

pressure

film

maxim

um

of

Position

ratio

pressure

film

Maxim

um

ratio

Flow

variable

Flow

degrees

, thickness

film

minim

um

of

Position

variable

thickness

film

Minim

um

ariable

friction v

of

t coefficien

max

0

=∆=

=

=

=

==

=

P

TCP

PQQrcNl

Qch

crf

p

Ps

θφ

Fir

st:

Contd

.

Absolu

te v

iscosity:

Viscosity with respect to average

temperature

for

various

SAE

grades of lubricants from 10 to 70.

You can use the table 12-1 in

conjunction

with

the

following

equation as an alternative to this

chart.

.lubricants

of

grades

60

-SAE

to

10

-SAE

for

1-12

table

from

obtained

be

may

&0

127

8.1

0 µµµ b

Cin

Te

oT

b

+

=

()

0.120

11

2

c

sj

Trfc

PQ

QrcNl

Q

∆=

−

The value of

“P”

must be substituted

in MPa

Anoth

er

exam

ple

: Itera

tive D

ete

rmin

ation o

f A

vera

ge T

em

pera

ture

of th

e F

ilm

12-11 of Shigley(6ed)

Contd

.

12-21 (Shigley, 6ed)

12-11 of Shigley(6ed)

12-21 (Shigley, 6ed)

Contd

.

Matlab

code for

the p

recedin

g e

xam

ple

:

pause off

%Iterative programmefor the determ

ination of the film

temperature in journal bearning

%from Raymondiand Boyd charts

% This is for SAE30; for other grades the viscosity temperature relation must be appropriately substituted

% The design is for minimum radial clearance scenario

% this computer program is valid only for l/d=1, 1/2 or 1/4

W=3000 %

load on the bearing, in Newtons

N=8 %

journal speed in rps

Ts=60 %

sump temperature in degree centrigrade

lbyd=1 %

the l/dratio

jdia=80 %

nominal diameter of the journal in mm

jtol=-0.01 %

unilateral tolerance (deviation) on journal diameter, m

m;must be a signed quantity

bdia=80.08 %

nominal diameter of the sleeve, in mm

btol=0.03 %

unilateral tolerance (deviation) on bearing diameter, mm; must be a signed quantity

% In minimum clearance design, we look for maximum shaft (journal) and minimum hole (sleeve)

% The radial clearance c is then equal to the cmin, the m

inimum clearance of the fit

%seeking maximum shaft

dmax=max(jdia, jdia+jtol)

%seeking the minimum sleeve diameter

bmin=min(bdia, bdia+btol)

%radial clearance

c=(bmin-dmax)/2

r=jdia/2

%clearance ratio

clratio=r/c

Contd

.%length of the bearing

l=lbyd*jdia

% load per unit projected area of the journal, MPa

P=W/(jdia*l)

% For finding the average temperature of the film

, Tavor Tf, a trial and error procedure is needed

% maximum possible value in the Fig 12-11 is 145 degree centrigrade

%iteration loop begins

deltaT=42 %

2*(145 -Ts)

Tav=Ts+deltaT/2 % This is the trial # 1 for average temperature

dis=Tav-Ts %

or you may assign any value greater than 1

iteration=0

while (abs(dis)>0.2)

iteration=iteration+1

% From Table 12-1 for SAE30, for the current trial of Tav

mu0=0.0971 %

base viscosity, mPa-s

bval=1360 %

degree centrigrade

mu=mu0*exp(bval/(1.8*Tav+127)) %

absolute or dynamic viscosity in milli-Pacal-second, mPa-s.

Contd

.

%temperature rise in the bearing deltaT

deltaT=2*(Tav-Ts)

%sommerfeldnumber

S=((clratio)^2)*(m

u*0.001*N)/(P*10^6)

%From Fig. 12-21, for SAE oils, the second estimate of temperature increase, for l/d=1, dT2 is

dT2=(P/0.12)*(0.349+6.009*S+0.0475*(S^2))

%discrepancy in the estimation of temperature increase

dis=deltaT-dT2

% Now the next trial for Tavcan be made as Tav-dis/2; sign willbeautomatically taken care of

Tav=Tav-dis/2

deltaT=dT2

pause

end

% Now discrepancy being less than one, we can consider the latest trail of Tavas the valid

%hence the T1 and T2 can be estimated as well

T1=Tav-deltaT/2 %

inlet temperature, same as sump temperature Ts

T2=Tav+deltaT/2 %

outlet temperature

iteration

Pro

ble

ms w

ith l/d

ratio o

ther

than 1

, ½

, ¼

or

infinity

•With l/d

ratio falling in between the

marked

values, it is possible to use an equation to

determine at the interm

ediate value of l/d:

()

1

3

12

14

11

11

21

41

21

48

31

11

11

41

12

424

ll

ll

ly

yd

dd

dd

yl

ll

lld

yy

dd

dd

∞

−

−−

−+

−−

=

−

−−

+−

−

Ste

ady-s

tate

conditio

ns i

n s

elf-c

onta

ined b

earings (

als

o c

alled a

s p

illo

w

blo

ck b

earings a

nd p

edesta

l bearings) (

)loss

bQ

UAT

T∞

=−

&

()

fb

bT

TT

Tα

∞−

=−

The value of alpha is usually

taken as 1 otherwise,

for

better accuracy, Table 12-2

can be referred.

Contd

.

()

00

1 1

loss

f

f

b

UA

QT

T

TT

T

α α α

∞

∞

=−

+ +=

+

&

Solving the preceding equation for Tb and substituting in the heat loss rate

gives the following equation in proper variables

Anoth

er

exam

ple

:

•A full journal bearing has a shaft journal w

ith a

diameter of 30 m

m and a unilateral tolerance of

-0.035 m

m. The busing bore is 50 m

m in length.

The bearing load is 2.75 kN

and the journal

rotates

at 1120 rpm. Analyze the

minim

um

clearance assem

bly and find the minim

um film

thickness, the coefficient of friction, and the total

oil flow if the average viscosity is 60 m

Pa-s.

Solu

tion:

Contd

.

Contd

.

exam

ple

•A journal bearing ha a shaft diameter of 75 m

m

with a

unilateral tolerance of -0.02 mm. The

bushing bore has a diameter of 75.10 m

m w

ith a

unilateral tolerance of 0.06 m

m. The bushing is 36

mm and supports a load of 2 kN. The journal speed

is 720 rpm. For the minim

um clearance assem

bly

find the minim

um film

thickness, the heat loss

rate, and the

maxim

um lubricant pressure for

SAE20 and SAE40 lubricant operating at a an

average film

tem

perature of 60oC.

Contd

.

Contd

.

Fig

. 12.2

6

Fig

. 12.2

7

Fig

. 12.3

2

Fig

. 12.3

3

Fig

. 12.3

4

Fig

. 12.3

5

Fig

. 12.3

7

Fig

. 12.3

8

Fig

. 12.3

9