( ) ( )

++++= sfihfoff ErrE

rrErrE

rrpr 22

22

22

22

r

Through interference fit torque can be transmitted, which can be estimated

( ) ( ) sifshfoh ErrEErrE,

with a simple friction analysis at the interface. ( ) = coefficient of ( )

( )LdpFApNF

fff

ff

=== coefficient of friction( )

LdpTTorque

p

f

fff

2

2

= Abrasion Adhesion

211/19/2011 1

C.A.Coulomb 17811)Clearly distinguished between static & kinetic friction

2)Contact at discrete points.) p

3)Friction due to interlocking of rough surfaces

4)No adhesion5)f func(v)

11/19/2011 2

5)f func(v)

PLOUGHING Eff tPLOUGHING Effect

Assume n conical asperities of hard metal in contact with flat soft metal, vertically project area of contact: y p j

( )2*5.0 rnA =

HrnW )*5.0( 2= HnrhF )(=

cot2= Slope of real surfaces are always less than 10 (i.e. > 80), therefore <

0 1 11/19/2011 3

0.1.

ADHESION Theory

Two surfaces are pressed together under load W.

They deformed until area of contact (A) is sufficient to support y ( ) ppload W. A = W/H.

To move the surface sideway, must overcome shear strength of y, gjunctions with force F F = A s 4

Hs=11/19/2011

Aim: To reduce shear strength of interface

LUBRICATION Process by which the

friction in a moving contact is reduced. Six distinct form of l b i i lubrication are: Hydrodynamic Hydrostatic Elastohydrodynamic Mixed Boundary S lid fil

11/19/2011 5

Solid film

11/19/2011 6

FLUID FILM >5FLUID FILM BEARINGS

>5

Machine elements designed to produce smooth (low friction) motion between solid surfaces in relative motion and to generate a load support for mechanical and to generate a load support for mechanical components. Fluid between surfaces may be a gas, liquid or solid. Word film implies that fluid thickness (clearance) separating

the surfaces is several orders of magnitude smaller than other dimensions of bearing (width & length).

Successful design requires film thickness to be larger than the micro asperities on the surfaces, operation without contact of surfaces.

Operation principles of liquid film bearings are hydrodynamic, hydrostatic or combination.

11/19/2011 7

Hydrodynamic Hydrostatic

4/6/8 pockets

A i l diAxial coordinate Axial coordinateAxial coordinate

11/19/2011 8

Hydrodynamic HydrostaticRelative motion between two External source of pressurized Relative motion between two mechanical surfaces is utilized to generate pressure and levitate one surface relative to other

External source of pressurized fluid is required to levitate the one surface and separate it from other surface Costlyone surface relative to other

surface. Self-actingfrom other surface.. Costly

Load support is a strong function f l b

Load support is a weak function of lubricant viscosity. of lubricant viscosity.HDL provides an infinite bearing life

Infinite life if supporting ancillary equipments function welllife equipments function well

Able to damp the external vibrations.

Able to control the effect of external vibration.. Active control

Significant difference in static & kinetic friction coefficients

Almost same value of coefficient of frictions.

High relative speed generates much higher load capacity &

Very good control on the shaft position.g p y

destabilize the shaft-system.p

11/19/2011 9

Petroffs Petroff s Equation

Friction = Shear Stress * Area

( i i * /h)*

C is radial clearance

RLARNV 2 ; 2 ==F = (Viscosity* V/h)*Area

CRLRNF

F

2*2* force,Friction =

RLPCRLRN

iont of frictCoefficien

2/2*2*

WF,

=

=

Bearing

Stribeck

CR

PN

RLP 22

2

=Bearing characteristic number

11/19/2011 10Conclusion: Coefficient of friction is a function of speed, load and viscosity

Lubricant Viscosity

VI relates viscosity change at 37.8 0c and 98.90c.

Pennsylvanian oil~VI=100 gulf coast oil ~ VI=0 100*HL

U-LVI =gulf coast oil VI 0 H-L

1111/19/2011

Dynamic viscosity, 1cP = 1mPa.s

Kinematic viscosity, 1cS = cP/0.85 (g/cm3)

Variation with Temperaturep

More viscous oil is more susceptible to change in viscosity with temp.

Walthers equation: Form the basis of ASTM viscosity temperature chart

Tlt t)60l (l Vogels equation: Most accurate; very useful in engineering calculations

Tc logconstant)6.0log(log =+engineering calculations

ty.in viscosiincrease with increasesbtemp.ofunitshasb iscosity.inherent v givesk )/( += tbke

12

yp

11/19/2011

Temperature Rise Friction, due to shear of lubricant film,

generates heat (FV)) in lubricant oil and i th t t f l b i t increases the temperature of lubricant.

Assuming that total generated heat is i d b th il fl i th h b icarried by the oil flowing through bearing

( )( )22=

LRNRflowoilbyconvectedHeatgeneratedHeat

( )( ) ( )( ) 232

222

=

LNR

tCmNRC

LRNRP

( )/8603= mkg( )

( ) 2322

=

=

LNRtor

LCmC

NRtorP

( )1000

/1760

==

RCkgJCP

o

( ) 222

2

= LCLCRNCtor

P

1000=C

Nt 252=( )( )

228

=CR

CNtor

P Nt 2.52=

11/19/2011 13

Nt 2.52= Assume rotational speed = 900 rpm p p783=t In hydrodynamic lubrication, increase in viscosity increases load capacity but also increases friction. We require Reynolds equation.

SAE grade

Viscosity in mPa s 400c

Viscosity in mPa s 1000c

friction. We require Reynolds equation.

( )atat e

= grade mPa.s 40 c mPa.s 100 c

10W 32.6 5.5720W 62 3 8 81

25.525848 7809

4.36606 898220W 62.3 8.81

SAE 30 100 11.9SAE 40 140 14 7

48.780978.3000

109 6200

6.89829.3177

11 5101SAE 40 140 14.75W-20 38 6.9210W 30 66 4 10 2

109.620029.754051 9912

11.51015.41847 986610W-30 66.4 10.2

10W-40 77.1 14.410W 50 117 20 5

51.991260.369391 6110

7.986611.275216 051510W-50 117 20.591.6110 16.0515

11/19/2011 14

Reynolds Equation A basic pressure distribution equation for Fluid

Film Lub. In 1886, Reynolds derived for estimation of

pressure distribution in the narrow, converging gap between two surfaces gap between two surfaces.

Reynolds equation helps to predict hydrodynamic, squeeze, and hydrostatic film mechanisms.

33equation Reynolds'

( ) ( ) ( )

++

++=

+

02121

33

26 VVhWWz

hUUxz

Phzx

Phx h

11/19/2011 15

U1

y

No pressure development within the parallel surfaces.

U2=0 x

p p p

U1 U1 U1

11/19/2011 16

Pressure driven flow

11/19/2011 17

Small element ofFluid with sides

dzdxdzdydxppdzdxdydzpdy:balanceForce + += ++dx, dy, and dz

dzdxdzdydxx

pdzdxdyy

dzpdy .... :balanceForce + += ++u=flowNewtonianFory=flowNewtonian For

=

yu

yxP 11/19/2011 18

yyx

u P

=

yu

yxP

1yu :nintegratioOn Cy

xP +=

yyx

21

2

2CyCy

xPu ++=

x2;Uu 0,y :conditionsboundary Using ==

21

1

2

)( ,

hPUUUuhy

==

2

121

22 2)( ,

yPyhy

ChxP

hUUCU

=

=

( ) 2212 UhyUU

xPyhyu ++

= Check !!!

11/19/2011 19

:unit widthper direction -in x rate Flow

.0

dyuqh

x = ( )

212 213 hUU

xPhqx ++= Check !!!

direction-zin rate flowSimilarly

dh

( ).

30

hWWPhq

dywqz

++=

= ( )

212 21WW

zqz ++=

equationcontinuitymassusingderivedisequationReynolds

0)(

equationcontinuity massusingderivedisequation Reynolds

0 =++

qVVq zhx11/19/2011 20

zx

equation Reynolds'

( ) ( ) ( )

++

++=

+

02121

33

26 VVhWWz

hUUxz

Phzx

Phx h xxx

Stretching action

iU1 iiU1iiiU1

xx11/19/2011 21

equation Reynolds'

( ) ( ) ( )

++

++=

+

02121

33

26 VVhWWz

hUUxz

Phzx

Phx h xxx

Wedge action (inclined surfaces

2h1h

z11/19/2011 22

equation Reynolds'

( ) ( ) ( )

++

++=

+

02121

33

26 VVhWWz

hUUxz

Phzx

Phx h xxx

Squeeze action q(bearing surfaces move

di l

0h

tt

1h

tt

2h

perpendicular to each other)

0tt = 1tt = 2tt =

Can ca

High lofor shdurat

210

210

ttthhh

>

arryoadshortion210

ttt

equation Reynolds'

( ) ( ) ( )

++

++=

+

02121

33

26 VVhWWz

hUUxz

Phzx

Phx h xxx

( )

+=

h

xUU

zP

zh 21

3 61:Ition Simplifica

xz

( )

+=

h

xUU

xPh

x 213

6:IItion Simplifica xxx 11/19/2011 24

H d St ti B i (HSB)Hydro-Static Bearings (HSB) Completely removal of wear and reduction of Completely removal of wear and reduction of coefficient of friction to 1/500. Surfaces can be separated by full fluid film even at Surfaces can be separated by full fluid film even at zero speed. No problem with micro roughness and waviness. No problem with micro roughness and waviness.

Zero friction at zero speed.Useful feature for large size telescopes and radars.Useful feature for large size telescopes and radars.

High stiffness Oil film thickness varies as cube root of load.

Wh t b i i b d H d t ti h i

3/1Wh Why not every bearing is based on Hydrostatic mechanism

High pressure supply Reliability & life of high pressure oil lines are always in doubt.11/19/2011 25

Elementary 1-D Analysis

Assume a shaft of radius Ro is located oco-axially with a circular recess of radius Ri.

Assume all the oil in recess is at the supply pressure Ps.

11/19/2011 26

R f t

Elemental flow rate: rddrdphq .

12

3

=Ref to slide 20

If flow is axisymmetrical, and radial flow rate

dr12y ,

is constant, then flow rate:

23 dphQ

If film thickness is constant then on

2..12 rdrpQ =

If film thickness is constant, then on integration:

)(log 6 1

3

CrQph +=

11/19/2011 27

6

Using two boundary conditions to find unknown values of C1 and Q

is Rr

RR

rR

pp = 00

0

Rregion in the log

log

Load carrying capacity:iRlog

( )drrdpRpW oi

R

Ris +=

20

2 .

Substituting expression of p and rearranging iR 0

iR12 ( )

=o

o

i

os

RR

RRpWlog.2

1 .

22

=

1

21

11log.2

1

r

rCW

11/19/2011 28

iR 1r

20

22load vs ratio

16

18

20

C1 = 10

10

12

14

l

o

a

d

6

8

10

.1 .2 .3 .4 .5 .6 .7 .8 .94

ratio

( )drrdpRpW oRR

is += 20

2 .iR 0

)/1l (1

6

30 phQ s= 1

2CQ =)/1log(6 1rQ )/1log( 12 rCQ11/19/2011 29

flow vs ratio1

2CQ =

180

200

220

240 )/1log( 12 r

Q

120

140

160

180

w

60

80

100

120

f

l

o

w

0

20

40 C2 = 10

.1 .2 .3 .4 .5 .6 .7 .8 .9

ratio

Generally we require high load capacity but low flow rate Generally we require high load capacity but low flow rate.

11/19/2011 30

Power loss: consists of pumping power and friction losses.

+=PQP

PPP fht =0h

UAF

244

0 1

.

=

=RRP

PQP

i

sh =02

0

)(

R

rAh

rF

00

12

=Rh

Pf

443 == 0 3

0

2

2iR

ff drrhPrFP

24

00

402

0

30 1

2)/log(61

+=

RR

hRP

RRhP is

it

000 )g( i Generally we require high load capacity, low flow rate and low power loss

11/19/2011 31

power loss.

Example: W = 1000 N, =5 rpm, R0=100 mm, Ri=50 mm, =0.01 Pa.s. Optimize minimum film thickness for minimum power loss

244

0230 11

+= RRPhP i

0001

2)/log(6

+=Rh

PRR

P si

t

( )

= oi

os RR

R

RpW1

.2

2

20

2301 h

ChCPt +=( )

i

oos

RR

plog.2

0h

d/523605*2

( ) Pa 824,585.01 )2log(.21.0*1000rad/s5236.060

5*2

22 ====

ss PP

sC

C

/N.m10*404.0

)N/(s.m 10*614.226

2

2111

==

11/19/2011 32

( ) sC /N.m10404.02micronh losspowero 8.26min =

Short Static Short Static Hydrodynamic Bearing

( )

+=

h

xUU

zP

zh 21

3 61 xzz dhUP

32

2 6 =dxhz 32

6 dhUdp 0zat 0dp/dzcondition pressuremax using 6 3 === zdxdh

hU

dzdp

L/2 zat 0p using 4

3 223 ==

= Lz

dxdh

hUp

11/19/2011 33

4 dxh

Film thickness, h, depends on geometry of tribo-pair. ForFilm thickness, h, depends on geometry of tribo pair. For example, in journal bearing h = Cr + e cos

eLdhU 3equation followingin h of expression Using

2 rC

edRdxLzdxdh

hUp ==

= ;;

43 2

3

( ) 4sin

cos13 22

32

+=Lz

RCUp

r

11/19/2011 34

Load capacity of short journal bearing bearing

centres of line of direction in component Load

( )222

2

3

0 1

22

cos)...(2

2

= = rC

LUWdzRdpWL

L

2 3

centres of line lar toperpendicu component Load

LUL W( ) 23

2

2220 14

sin)...(

= =

rrr C

LUWdzRdpWL

2/1

( )2/1

22222

322 1116

14

+

=+=

rr C

LUWWW ( )14 rC

21tantan == Wr 4tantan == W11/19/2011 35

Locking of Journal Locking of Journal Position

valueMaxW ===

01

valueMaxW =

02

0

===

W

20

10

Lesser the attitude angle, better the stability of bearing.

80

90eccentricity ratio vs. attitude angle

60

70

e

40

50

t

u

d

e

a

n

g

l

e

10

20

30

A

t

t

i

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 10

10

11/19/2011 37

Eccentricity ratio

Friction force in Journal BearingFriction force in Journal Bearing

Petroff equation (explained on slide 10)--- inaccurate

dAFhdxdp

hU ;

2

=+= h = Cr + e cos

dzRdhUF

L

L.0

2/

2/ 0

+=

2/1

21

2

=

C

ULRF ( )2/1

22222

31116

14

+

=

rC

LUW1 rC

If 0, F Petroff solution( )14 rC

11/19/2011 38

Raimondi & Boyd Method

hPhPh loadingstaticfor equation Reynolds'

33

xhU

zPh

zxPh

x =

+

6

R 2 N

dft tiCR Snumber Sommerfeld s

NpN

=

secondfor rotation sN

Dimensionless numberDimensionless number

11/19/2011 39

L/D S (R/C)f Q/(RCLNs ) Qs / Q p/pmax1 .631 .2 74 12.8 3.596 .28 .529

.264 .4 63 5.79 3.99 .497 .484

.121 .6 50.6 3.22 4.33 .68 .415

.045 .8 36.2 1.7 4.62 .842 .313

.019 .9 26.5 1.05 4.74 .919 .247

0.5 2.03 .2 75 40.9 3.72 .318 .506

.779 .4 61.5 17 4.29 .552 .441

.319 .6 48 8.1 4.85 .73 .365

.092 .8 33.3 3.26 5.41 .874 .267

.031 .9 23.7 1.6 5.69 .939 .206

0.25 7.57 .2 75.2 153 3.76 .33 .489

2.83 .4 60.9 61.1 4.37 .567 .415

1.07 .6 46.8 26.7 4.99 .746 .334

11/19/2011 40

.261 .8 31 8.8 5.6 .884 .24

.074 .9 21.9 3.5 5.91 .945 .18

Question: Estimate friction coefficient Question: Estimate friction coefficient and minimum film thickness of a full cylindrical hydrodynamic bearings cylindrical hydrodynamic bearings required for shaft of 50 mm diarotating at 1500 rpm Assume applied rotating at 1500 rpm. Assume applied load is 5 kN, bearing length is 50 mm radial clearance is 25 microns mm, radial clearance is 25 microns, and lubricant viscosity is 25 mPa.s (at

bi t t t ) ambient temperature). mkg /860 3= ( )t( )CkgJC mkgP o/1760

/860

= ( )at

at e=

11/19/2011 41

Clutches & Brakes Difference between coupling, clutch &

brake. Maximize friction (uniform) Smooth, gradual connection/disconnection, g Minimize Wear (if mechanical)

Design targets: Design targets: Required friction torque. Actuating Mechanism Actuating Mechanism

Heat dissipation. Desired life

11/19/2011 42

Desired life.

Various Types of ypClutches & Brakes

PP

PP

P

Mating frictional surfaces transmitting torqueP

P

transmitting torque Actuating mechanisms Centrifugal P

P

Centrifugal Magnetic Hydraulic

11/19/2011 43

Hydraulic Pneumatic

ClutchesCone clutch is installation. However, disk clutch a number f d t l t hof advantages over cone clutch.

Large frictional area in relatively small area.

More effective heat More effective heat dissipation.

Simple construction.

11/19/2011 44

Simple construction.

Aim: Obtain expression of axial force F necessary to produce a certain torque with pressure distribution.

Uniform pressureUniform pressure

or

d2=ir

drrpF 2

( )222 ior rrpdrrpF o == ri

( )2ro ( )33322 ior rrpfdrrprfT i == 11/19/2011

Friction based brakesFriction mat. f Pmax

(MPa)Tmax(C)( ) ( C)

Dry OilMolded 0.25-0.45 0.06-0.09 1-2 230Molded

Woven 0.25-0.45 0.08-0.1 0.35-0.69 230

Sintered 0.15-0.45 0.05-0.08 1-2 300

Cast iron/Hard_steel

0.15-0.25 0.03-0.06 0.69-0.72 260

Wet or dry brakes

11/19/2011 46

Laws of Wear Wear Volume proportional

to sliding distance (L) True for wide range of

conditions Wear Volume proportional

to the load (N) D ti i Dramatic increase

beyond critical load Wear Volume inversely Wear Volume inversely

proportional to hardness of softer material

HNLkV

31= Transition from mild wear to severe

depends on relative speeddepends on relative speed, atmosphere, and temperature.

11/19/2011

Two methods to estimate pressure distribution are:

Uniform wear

Uniform pressurep

( )UtpAkNLkVvolumeWear 11or

( )

UpUkVHH

VvolumeWear ==33

,

1 =ir

drrpF 2 rprp

pUwHp

Atwratewear

===

meanratearConstat we3

, 1

ooii rprp =meanratear Constat we

= or

drrpF 2 r=ir

oo drrpF 2 = oi

r

roo drrrpfT 2

i

11/19/2011

Comparison( ) iioiwearuniform rrrrpfT at pressuremax assuming22max = ( )332fT ( )33max 32 iopressureuniform rrpfT = ( )2 33 rrT ( )( )32 22 == rrr rrTTRatio ioi iowearuniformpressureuniform( )( )1132 2

3 =RRRatio

1 451.5

1.551.6

Torque ratio vs dimension ratio

( )13 R1 2

1.251.3

1.351.4

1.45

T

o

r

q

u

e

r

a

t

i

o

11/19/2011 491.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2

1.11.15

1.2

R

Friction mat

f Pmax(MPa)

Value of ri to maximize Tmat. (MPa)

Dry Oil

M ld d 0 25 0 06 1 2( )22max ioiwearuniform rrrpfT = Molded 0.25-0.45 0.06-0.09 1-2Woven 0.25- 0.08- 0.35-

0=rdTd

Woven0.45 0.1 0.69

Sintered 0.15-0 45

0.05-0 08

1-2ird

or

0.45 0.08

Cast iron/Hard steel

0.15-0.25

0.03-0.06

0.69-0.72( )NrrrpfT ioi 22max =

3o

ir =Most of automotive clutches operate

Hard_steel

ost o auto ot e c utc es ope atewet. The oil serves as an effective coolant during clutch engagement. To compensate reduced coefficient

11/19/2011 50

To compensate reduced coefficient of friction, multiple disks are used.

= or

drrpF 2Disk Brake

ir

oo drrpF 2

F F= or

drdrpF 2

F F=ir

ii drdrpF 1

ddrrrpfTor

ii = 2

ddrrrpfTir

ii 1

= or

oo drrrpfT 211/19/2011 51

ir

oopf

Ex: Two annular pads, ri=98mm, ro=140 mm, subtend an angle of 100, have a coefficient of friction of 0.35, and are

t t d b i f h d li li d 36 i actuated by a pair of hydraulic cylinders 36 mm in diameter. Torque requirement is 1500 N.m. Determine max contact pressure, actuating force and hydraulic pressure. contact pressure, actuating force and hydraulic pressure.

CASE I: Uniform wear

ddrrrpfTor

ii = 2 NdrdrpF or

ii 180072

==

pf

irii

1

09801401500 22

pir

ii

1

pi 2098.014.0

180*100*098.0*35.0

21500

=

Papi 2506649=PF 17690959

11/19/2011 52( ) Paphydraulic 17690959018.0 2 ==