r r 2 r r ν r r p f f ( ) ( ) s E r r E o f h s f i...

52
( ) ( ) + + + + = s f i h f o f f E r r E r r E r r E r r p r ν ν δ 2 2 2 2 2 2 2 2 r Through interference fit torque can be transmitted, which can be estimated ( ) ( ) s i f s h f o h E r r E E r r E with a simple friction analysis at the interface. ( ) μ = coefficient of ( ) ( ) L d p F A p N F f f f f f π μ μ μ = = = μ coefficient of friction L d p T Torque f f f f 2 2 μ π = Abrasion Adhesion 2 11/19/2011 1

Transcript of r r 2 r r ν r r p f f ( ) ( ) s E r r E o f h s f i...

  • ( ) ( ) ⎥⎥⎦⎤

    ⎢⎢⎣

    ⎡−

    +++

    += sfihfoff ErrE

    rrErrE

    rrpr ννδ 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.

    ( ) μ = coefficient of ( )( )LdpF

    ApNF

    fff

    ff

    πμ

    μμ

    =

    ==μ coefficient of friction

    ( )LdpTTorque

    p

    f

    fff

    2

    2μπ

    μ

    =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 π=HrnW )*5.0( 2π= HnrhF )(=

    θπ

    μ cot2=Slope of real surfaces are always less than 10° (i.e. θ> 80°), therefore μ < 0 1

    11/19/2011 3

    π 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

    =μ11/19/2011

  • Aim: To reduce shear strength of interface

    LUBRICATIONProcess by which the friction in a moving contact is reduced. Six distinct form of l b i i lubrication are:

    HydrodynamicHydrostaticElastohydrodynamicMixed BoundaryS lid fil

    11/19/2011 5

    Solid film

  • 11/19/2011 6

  • FLUID FILM Λ>5FLUID FILM BEARINGS

    Λ>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

  • Petroff’s Petroff s Equation

    Friction = Shear Stress * Area

    ( i i * /h)*

    C is radial clearance

    RLARNV ππ 2 ; 2 ==

    F = (Viscosity* V/h)*Area

    CRLRNF ππη

    F

    2*2* force,Friction =

    RLPCRLRN

    iont of frictCoefficien

    ππημ

    μ

    2/2*2*

    WF,

    =⇒

    =

    Bearing

    Stribeck

    CR

    PN

    RLPηπμ 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.

    • Walther’s equation: Form the basis of ASTM viscosity temperature chart

    Tlt t)60l (l• Vogel’s equation: Most accurate; very useful in engineering calculations

    Tc logconstant)6.0log(log −=+ν

    engineering calculations

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

    12

    yp

    11/19/2011

  • Temperature RiseFriction, due to shear of lubricant film, generates heat (F×V)) 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

    Δ

    Δ=

    LNR

    tCmNRC

    LRNRP

    πη

    πππη

    ( )/860 3= mkgρ

    ( )

    ( ) 232

    2

    LNRtor

    LCmC

    NRtorP

    πη

    πη ( )1000

    /1760

    =

    =R

    CkgJCPo

    ( ) 222

    2

    ⎞⎛

    ⎟⎠⎞

    ⎜⎝⎛

    =Δ LCLCRNC

    torP

    πρ1000=

    C

    Nt η252=Δ( )( )

    228⎟⎠⎞

    ⎜⎝⎛=Δ

    CR

    CNtor

    Pρπη Nt η2.52=Δ

    11/19/2011 13

  • Nt η2.52=Δ Assume rotational speed = 900 rpmη p p

    η783=ΔtIn 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

    −−= βηηgrade 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 EquationA 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 hηη

    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 τττ +⎟⎞⎜⎛∂

    +=⎟⎟⎞

    ⎜⎜⎛ ∂

    ++

    dx, dy, and dz

    dzdxdzdydxx

    pdzdxdyy

    dzpdy .... :balanceForce ττ +⎟⎠

    ⎜⎝ ∂

    +=⎟⎟⎠

    ⎜⎜⎝ ∂

    ++

    u∂=ητflowNewtonianFor

    y∂=ητflowNewtonian For

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    =∂∂

    yu

    yxP η

    11/19/2011 18

    ⎟⎠

    ⎜⎝ ∂∂∂ yyx

  • u P∂∂ ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    =∂∂

    yu

    yxP η

    1yu :nintegratioOn Cy

    xP

    +∂∂

    =∂∂η ⎠⎝

    ∂∂∂ yyx

    21

    2

    2CyCy

    xPu ++∂∂

    =→ηx∂

    2;Uu 0,y :conditionsboundary Using ==

    21

    1

    2

    )( ,

    hPUUUuhy

    ∂−==

    η

    2

    121

    22 2)( ,

    yPyhy

    ChxP

    hUUCU

    ∂⎟⎞

    ⎜⎛

    =∂∂

    −=ηη

    ( ) 2212 UhyUU

    xPyhyu +−+∂∂

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛ −=⇒

    ηCheck !!!

    11/19/2011 19

  • :unit widthper direction -in x rate Flow

    .0

    dyuqh

    x = ∫

    ( )212 21

    3 hUUxPhqx ++∂∂

    −=η Check !!!

    direction-zin rate flowSimilarly

    dh

    ( )

    .

    30

    hWWPhq

    dywqz

    ++∂

    −=

    = ∫

    ( )212 21

    WWz

    qz ++∂=

    η

    equationcontinuitymassusingderivedisequationReynolds

    0)(

    equationcontinuity massusingderivedisequation Reynolds

    0 =∂∂

    +−+∂∂ qVVq zhx

    11/19/2011 20

    ∂∂ zx

  • equation Reynolds'

    ( ) ( ) ( )⎭⎬⎫

    ⎩⎨⎧ −++

    ∂∂

    ++∂∂

    =⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    +⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    02121

    33

    26 VVhWWz

    hUUxz

    Phzx

    Phx hηη ⎭⎩ ∂∂⎠⎝ ∂∂⎠⎝ ∂∂ xxx ηη

    Stretching action

    iU1 iiU1iiiU1

    xx11/19/2011 21

  • equation Reynolds'

    ( ) ( ) ( )⎭⎬⎫

    ⎩⎨⎧ −++

    ∂∂

    ++∂∂

    =⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    +⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    02121

    33

    26 VVhWWz

    hUUxz

    Phzx

    Phx hηη ⎭⎩ ∂∂⎠⎝ ∂∂⎠⎝ ∂∂ xxx ηη

    Wedge action (inclined surfaces

    2h1h

    z11/19/2011 22

  • equation Reynolds'

    ( ) ( ) ( )⎭⎬⎫

    ⎩⎨⎧ −++

    ∂∂

    ++∂∂

    =⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    +⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    02121

    33

    26 VVhWWz

    hUUxz

    Phzx

    Phx hηη ⎭⎩ ∂∂⎠⎝ ∂∂⎠⎝ ∂∂ xxx ηη

    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 hηη ⎭⎩ ∂∂⎠⎝ ∂∂⎠⎝ ∂∂ xxx ηη

    ( )⎭⎬⎫

    ⎩⎨⎧∂∂

    +=⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂ h

    xUU

    zP

    zh 21

    3 61:Ition Simplifica η

    xz

    ( )⎭⎬⎫

    ⎩⎨⎧∂∂

    +=⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂ h

    xUU

    xPh

    x 213

    6:IItion Simplificaη ⎭⎩∂⎠⎝ ∂∂ xxx η

    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 stiffnessOil 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 mechanismHigh 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: θηδ rd

    drdphq .

    12

    3

    −=Ref to slide 20

    If flow is axisymmetrical, and radial flow rate

    η dr12

    y ,is constant, then flow rate:

    23 dphQ

    If film thickness is constant then on

    πη

    2..12

    rdrpQ −=

    If film thickness is constant, then on integration:

    )(log 6 1

    3

    CrQph +−=η

    π

    11/19/2011 27

  • 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

    ( )drrdpRpWo

    i

    R

    Ris ∫ ∫+=

    πθπ

    2

    0

    2 .

    Substituting expression of p and rearranging iR 0

    ⎟⎞

    ⎜⎛ iR1

    2

    ⎟⎞

    ⎜⎛

    ( )⎟⎟⎟⎟

    ⎠⎜⎜⎜⎜

    ⎝⎟⎠⎞⎜

    ⎝⎛

    −=

    o

    o

    i

    os

    RR

    RRpWlog.2

    1 .

    22π

    ⎟⎟⎟⎟

    ⎠⎜⎜⎜⎜

    ⎝⎟⎠⎞⎜

    ⎝⎛−

    =

    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

    load

    6

    8

    10

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

    ratio

    ( )drrdpRpWoR

    Ris ∫ ∫+=

    π

    θπ2

    0

    2 .iR 0

    )/1l (1

    6

    30 phQ sπ= 1

    2CQ =)/1log(6 1rQ

    η )/1log( 12 r

    CQ11/19/2011 29

  • flow vs ratio1

    2CQ =

    180

    200

    220

    240 )/1log( 12 r

    Q

    120

    140

    160

    180

    w

    60

    80

    100

    120

    flow

    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 η

    244

    0 1

    .

    ωηπ⎟⎟⎞

    ⎜⎜⎛

    ⎟⎟⎞

    ⎜⎜⎛

    −=

    =

    RRP

    PQP

    i

    sh=

    02

    0

    )(

    R

    rAh

    rF ωη

    00

    12

    ωη⎟⎠

    ⎜⎝

    ⎟⎟⎠

    ⎜⎜⎝

    =Rh

    Pf

    443 ⎞⎛ ⎞⎛

    ∫=⇒=0

    3

    0

    2

    2iR

    ff drrhPrFP πωηω

    24

    00

    402

    0

    30 1

    2)/log(61 ωηππη ⎟

    ⎜⎜

    ⎛⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛−+=

    RR

    hRP

    RRhP is

    it

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

    11/19/2011 31

    power loss.

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

    244

    0230 11 ωηππ ⎟

    ⎟⎞

    ⎜⎜⎛

    ⎟⎟⎞

    ⎜⎜⎛

    −+=RRPhP i

    0001

    2)/log(6ωη

    η ⎟⎠⎜⎝

    ⎟⎟⎠

    ⎜⎜⎝

    +=Rh

    PRR

    P si

    t

    ( )⎟⎟⎟⎞

    ⎜⎜⎜⎛

    ⎞⎛

    −= o

    i

    os RR

    R

    RpW1

    .2

    2

    2π0

    2301 h

    ChCPt +=( )⎟⎟⎠

    ⎜⎜⎝

    ⎟⎠⎞⎜

    ⎝⎛

    i

    oos

    RR

    plog.2

    0h

    d/523605*2π

    ( ) Pa 824,585.01)2log(.2

    1.0*1000

    rad/s5236.0605*2

    22 =⇒−=

    =⇒=

    ss PP π

    ωπω

    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 61η ⎭⎩∂⎠⎝ ∂∂ xzz η

    dhUP32

    2 6 η=

    ∂dxhz 32∂

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

    hU

    dzdp η

    L/2 zat 0p using 4

    3 223 ±==⎟⎟

    ⎞⎜⎜⎝

    ⎛−=

    Lzdxdh

    hUp η

    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 cosθ

    θeLdhU ⎟⎞⎜⎛η3equation followingin h of expression Using

    rCedθRdxLz

    dxdh

    hUp ==⎟⎟

    ⎞⎜⎜⎝

    ⎛−= εη ;;

    43 2

    3

    ( )

    4sin

    cos13 22

    32 ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛−⎟

    ⎠⎞

    ⎜⎝⎛−

    +=

    LzRC

    Upr

    θεθε

    η

    11/19/2011 34

  • Load capacity of short journal bearing θbearing

    centres of line of direction in component Load

    θ

    ( )222

    2

    3

    0 1

    22

    cos)...(2

    2 ε

    εηθθ θπ

    θ−

    =⇒∫ ∫=− rC

    LUWdzRdpWL

    L φ

    2 3

    centres of line lar toperpendicu component Load

    επηπ LULW

    ( ) 232

    2220 14

    sin)...(ε

    επηθθπ

    −=⇒∫ ∫=

    − rrr C

    LUWdzRdpWL

    2/1

    ( )2/1

    22222

    322 1116

    14 ⎭⎬⎫

    ⎩⎨⎧

    +⎟⎠⎞

    ⎜⎝⎛ −

    −=+=⇒ ε

    πε

    επηθ

    rr C

    LUWWW( )14 ⎭⎩ ⎠⎝− πεrCεπφφ

    21tantan −=⇒= Wrε

    φφθ 4

    tantan =⇒=W

    11/19/2011 35

  • Locking of Journal Locking of Journal Position

    valueMaxW ===

    01

    φε

    valueMaxW =

    02

    0

    =

    =

    =

    W

    πφ

    ε

    20

    10

  • Lesser the attitude angle, better the stability of bearing.

    80

    90eccentricity ratio vs. attitude angle

    60

    70

    e

    40

    50

    tude

    ang

    le

    10

    20

    30Atti

    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 ;

    2τητ

    ⎞⎛

    ∫=+=h = Cr + e cosθ

    dzRdhUF

    L

    L.0

    2/

    2/ 0θη

    π∫ ∫ ⎟

    ⎠⎞

    ⎜⎝⎛ +=

    2/1

    21

    2

    ε

    πη

    −=

    C

    ULRF( )

    2/12

    2222

    31116

    14 ⎭⎬⎫

    ⎩⎨⎧

    +⎟⎠⎞

    ⎜⎝⎛ −

    −=⇒ ε

    πε

    επη

    rCLUW

    1 εrC

    If ε 0, F Petroff solution

    ( )14 ⎭⎩ ⎠⎝− πεrC

    11/19/2011 38

  • Raimondi & Boyd Method

    hPhPh ∂⎟⎞⎜⎛ ∂∂⎟⎞⎜⎛ ∂∂

    loadingstaticfor equation Reynolds'33

    xhU

    zPh

    zxPh

    x ∂∂

    =⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂

    +⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛∂∂

    ∂∂ 6

    ηη

    R 2 N⎞⎛

    dft tiCR Snumber Sommerfeld s

    NpNη

    ⎟⎠⎞

    ⎜⎝⎛=

    secondfor rotation sN

    Dimensionless numberDimensionless number

    11/19/2011 39

  • L/D S ε φ (R/C)f Q/(RCLNs ) Qs / Q p/pmax

    1 .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=ρ ( )tβ( )CkgJC

    mkg

    Po/1760

    /860

    =

    ρ ( )atat e

    −−= βηη

    11/19/2011 41

  • Clutches & BrakesDifference between coupling, clutch & brake.

    Maximize friction (uniform)Smooth, gradual connection/disconnection, gMinimize Wear (if mechanical)

    Design targets:Design targets:Required friction torque.

    Actuating MechanismActuating MechanismHeat dissipation.Desired life

    11/19/2011 42

    Desired life.

  • Various Types of ypClutches & Brakes

    PP

    PP

    P

    Mating frictional surfacestransmitting torqueP

    P

    transmitting torqueActuating 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 π2

    ( )222 ior

    rrpdrrpFo

    −== ∫ ππri∫

    ( )2ro

    ∫ ( )33322 io

    r

    rrpfdrrprfTi

    −== ∫ ππ

    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 WearWear 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 inverselyWear 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 11

    or

    ( )

    UpUkVHH

    VvolumeWear ==33

    ,

    1

    ∫=ir

    drrpF π2ωω rprp

    pUwHp

    Atwratewear

    =

    ∝⇒==

    meanratearConstat we3

    , 1

    ωω ooii rprp =meanratear Constat we

    ∫=or

    drrpF π2 r∫=ir

    oo drrpF π2∫=o

    i

    r

    roo drrrpfT π2

    i

    11/19/2011

  • Comparison( ) iioiwearuniform rrrrpfT at pressuremax assuming22max −= π( )332fT ( )33max 3

    2iopressureuniform rrpfT −= π

    ( )2 33 rrT ( )( )3

    222 −

    −==

    rrrrr

    TT

    Ratioioi

    io

    wearuniform

    pressureuniform

    ( )( )1

    132

    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

    Torq

    ue ra

    tio

    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 −= π

    Molded 0.25-0.45

    0.06-0.09

    1-2

    Woven 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 −=π

    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 π2Disk Brake

    ∫ir

    oo drrpF π2

    F F∫∫=or

    drdrpF θθ2

    F F∫∫=ir

    ii drdrpF θθ1

    θθ

    ddrrrpfTor

    ii ∫∫=2

    θθ

    ddrrrpfTir

    ii ∫∫1

    ∫=or

    oo drrrpfT π211/19/2011 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

    θθ

    ddrrrpfTor

    ii ∫∫=2

    NdrdrpFor

    ii 180072

    == ∫∫ θθ

    θ

    pfir

    ii ∫∫1

    09801401500 22 ⎟⎞⎜⎛ −⎞⎛ π

    pir

    ii

    1

    ∫∫θ

    pi 2098.014.0

    180*100*098.0*35.0

    21500

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛ −⎟⎠⎞

    ⎜⎝⎛=

    π

    Papi 2506649=

    PF 1769095911/19/2011 52( ) Paphydraulic 17690959018.0 2 == π