-
( ) ( ) ⎥⎥⎦⎤
⎢⎢⎣
⎡−
−
+++
−
+= 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
=Δ
=Δ
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
6η
-
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
2θ
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
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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
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Desired life.
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Various Types of ypClutches & Brakes
PP
PP
P
Mating frictional surfacestransmitting torqueP
P
transmitting torqueActuating mechanisms
Centrifugal P
P
Centrifugal Magnetic Hydraulic
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Hydraulic Pneumatic
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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.
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Simple construction.
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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
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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
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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.
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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
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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
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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
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To compensate reduced coefficient of friction, multiple disks are used.
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∫=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
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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