Identification of needIdentification of need

Problem formulationDesign is an iterative process

Mechanism/Synthesisiterative process

AnalysisAnalysis requires mathematical model

Verification

of system/component.

PresentationPresentation

11/2/2011 1

Can we increase speed of Jute Flyer ??

Flyer Spinning Machine

Current speed 4000 rpm

Target speed 6000 rpm

Bobbin

11/2/2011 2

Bobbin

Can we increase speed of Jute Flyer ??

Flyer Spinning Machine•Increase rotational speedp

• Constraints: Stress < ??

Deflection < ???Deflection ???

11/2/2011 3

Increase operating speed wharve assembly

Bearing life must be at least 3 years least 3 years The wharves must be lighter than the current wharveswharvesTemperature rise must be within 5°C.Cost of new wharve assembly ≤ 1.5 times cost of existing assemblyof existing assembly

11/2/2011 4

Total Product Development…

Stress concentration factor, surface finish factor, size factor, etc.

Design Factors: “Factor of Safety”FOS is a ratio of two quantities that have same units:

Strength/stress ; Critical load/applied loadLoad to fail part/expected service loadMaximum cycles/applied cyclesMaximum cycles/applied cyclesMaximum safe speed/operating speed.

Necessary to calculate one or more factors Necessary to calculate one or more factors of safety to estimate likelihood of failure.

StressDeformationWear

11/2/2011 6

Design Factors: “Factor of Safety”Material properties (FSM ) 1.0,1.1,1.3Stress (FSS ) 1.0, 1.2, 3.0St ess ( SS ) 0, , 3 0Geometry (FSG ) 1.0, 1.2Failure analysis (FSFA ) 1 0 1 2 1 5Failure analysis (FSFA ) 1.0, 1.2, 1.5Environmental factors (FSE ) 1.0, 1.3, 1.6Danger to Personnel (FS ) 1 0 1 6Danger to Personnel (FSD ) 1.0, 1.6FOS is deterministic. Often data are statistical and there is a need to use Probabilisticand there is a need to use Probabilistic approach.

11/2/2011 7

Analysis of Steam yTurbine CouplingShrinkage stresses and power transmission capacity with a short transmission capacity with a short circuit factor of 4.4 at rated speed of 3000 rpm3000 rpm

11/2/2011 8

Press Fit δrsδrsBase-line

Pressure pf is caused by interference between h ft & h b P

δrhδrhshaft & hub. Pressure increases radius of hole and decreases radius of

rf

rh

and decreases radius of shaft.

rf

rfr

pf

frf

pf

11/2/2011 9

( ) ( )strain ntialCircumfere −==

−+=

drdr rrr σνσδθθδε θθ

( )t iR di l −∂−∂∂

+ drr

Erdr

rrr

rr σνσδδδδ

θ

θ

θ

( )

( )( ) 02

sin2 balance Force

strain Radial

=⎟⎠⎞

⎜⎝⎛−−++=

=∂

=∂=

dzdrddzrddzddrrd

Erdrr

rrr

rrr

θσθσθσσ

ε

θ

θ

11/2/2011 10

( )( )2 ⎠⎝

rrr θ

( ) ( )22

222 stress ntialCircumfere

io

iooiooii

rrpprrrrprpσ

−−−−

=θ rf

( ) ( )22

222 stress Radial

io

iooiooiir rr

pprrrrprpσ−

−+−=

CASE I: Internally Pressurized (Hub)-

( )( )2( )( )

( )( )2

22

22 1 stress ntialCircumfere

fo

off

rrrrrp

σ−

+=θ

( )fo

off

rrrrp

σ−

+= 22

22

max,θ

( )( )22

22 1 stress Radial

fo

offr rr

rrrpσ

−

−= fr pσ −=max,

( ) ⎞⎛ 22( )Er

rh

f

rh σνσδε θθ

−==strain ntialCircumfere

f

r

fo

off

rrrrr

Ep

hδ

νεθ =⎟⎟⎠

⎞⎜⎜⎝

⎛+

−

+= 22

22

max,

11/2/2011 11

CASE II: Externally Pressurized (shaft)-

ff rpσ ⎟⎟⎞

⎜⎜⎛

−= 2 2θ

( ) ( )22

222 stress ntialCircumfere

io

iooiooii

rrpprrrrprpσ

−−−−

=θ

fr

ifff

pσ

rrrpσ

−=

⎟⎠

⎜⎝ −

max,

22max,θ( ) ( )22

222 stress Radial

io

iooiooiir rr

pprrrrprpσ−

−+−=

( )⎟⎟⎞

⎜⎜⎛ +

−=2

2 1stressntialCircumfere iff

rrrpσθ rf

( )⎟⎟⎞

⎜⎜⎛ −

−=

⎟⎠

⎜⎝ −

=

22

22

1stressRadial

stress ntialCircumfere

i

ifff

rrrpσ

rrrpσθ rf

⎟⎟⎠

⎜⎜⎝ −

−= 22 stressRadialif

ffr rrrpσ

( )Er

rs

f

rs σνσδε θθ

−==strain ntialCircumfere

f

rs

if

fi

s

f

rrrrr

Ep

sδ

νεθ =⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

+−= 22

22

max,

11/2/2011 12

−= h δδδceinterferenTotal

⎥⎤

⎢⎡ +

+++ sfihfo

rsrh

rrrr ννδ

δδδ2222

r ceinterferenTotal

( ) ( ) ⎥⎥⎦⎢

⎢⎣

−−

++−

=s

s

ifs

f

h

h

foh

fff ErrEErrE

prδ 2222ror

Ex: A wheel hub is press fitted on a 105 mm diameter solid shaft. The hub and shaft material is AISI 1080 steel (E = 207 GPa). The hub’s outer diameter is 160mm The radial interference between shaft and outer diameter is 160mm. The radial interference between shaft and hub is 65 microns. Determine the pressure exercised on the interface of shaft and wheel hub.

⎤⎡ 2222

( ) ( )⎥⎥⎦⎤

⎢⎢⎣

⎡

−

++

−

+= 22

22

22

22

r :materials same of made areshaft and hub Ifif

fi

fo

foff

rrrr

rrrr

Epr

δ

( )⎥⎥⎦⎤

⎢⎢⎣

⎡

−= 22

2

r2 :solid isshaft If

fo

off

rrr

Epr

δ ANS: pf =73 MPa

I i !!!!( )⎦⎣ f

11/2/2011 13Iterations !!!!

4 59 4 34 4 5796 4 50 4 582 4 5847 4 59484.59,4.34,4.5796,4.50, 4.582,4.5847……………4.5948

( ) /22∑ ∑−=

Ndd iidσ

EX. NOMINAL SHAFT DIA. 4.5mmNUMBER OF SPECIMEN 34

1−=

Ndσ

NUMBER OF SPECIMEN 344.58mm0 0097

d6

4.5294

0 09870.0097dσ 0.0987

11/2/2011 14

Probabilistic Approach to Design

Ex: Tensile tests on 9 pieces of hot rolled steel. Measured ultimate tensile 34.24

67.468==

MPaMPa

s

s

σμ

Measured ultimate tensile strength data are: 433, 444, 454, 457, 470, 476, 481 493 and 510 MPa ( ) 1

2

34.2467.468

21

=⎟⎠⎞

⎜⎝⎛ −

−exf

x

s

481, 493, and 510 MPa. Find the values of mean, and std. dev. Assuming

l di t ib ti fi d

( )234.24

=

∫∞+

exfπ

normal distribution find the probability density function.

( ) 1=∫∞−

dxxf

NOTE: Reliability is probability that machine element will perform intended function satisfactorily.

11/2/2011 15

y

Ex: Consider a structural member subjected to Ex: Consider a structural member subjected to a static load that develops a stress σ

sσσ

Variation in load !!!!!

Variation in Area !!!FSFF ,, ( )f QPP <= 0 failure ofy Probabilit

s,,σσMargin σ−= SQ

fPR −=1y Reliabilit

NOTE: Addition or subtraction of subtraction of normal distribution provides normal provides normal distribution.

11/2/2011 16

21

1 ⎟⎟⎞

⎜⎜⎛ −

−QQ

( ) 2

21 ⎟

⎠⎜⎝= QS

Q

eS

Qfπ

QQ

Z

QSQQZ −

=

+ 11

variablenormalLet

2 1

Z

Z

Q

dZeR ∫=∞+ −

21

21

0

2

π ∫=∞−

−0 221

21 z z

dZeFπ

QSQwhere −=0 Z ALGEBRAIC MEAN STD. DEVIATION

FUNCTIONS Q103040 =−=Q

CQCxQCQ

==

CxC

C

+

0C x

σσ1086 22 =+=Qs

Q

xyQyxQxCQ

=±=+=

yxyxxC

±+

2222

22

xy yx

yx

x

σσ

σσ

σ

+

+

1100 0at

−=

Z

Q

xQyxQ

1==

xyx

y

1 2

22222

x

yxy

x

yx

σ

σσ +1

10 −==Z

11/2/2011

Z-Table provides probability of failure

11/2/2011 18

Value of normal variable provide the probability of f il failure.

I th tIn the present case Probability of failure is 0.1587 & reliability is0.1587 & reliability is .8413.

Selecting stronger materialSelecting stronger material (mean value of strength = 50 units!!!!)

0

)Material properties (FSM ) 1.0,1.1,1.3Stress (FSS ) 1.0, 1.2, 3.0 1 9( S )Geometry (FSG ) 1.0, 1.2Failure analysis (FSFA ) 1.0, 1.2, 1.5

1.9

11/2/2011 19

Failure analysis (FSFA ) 1.0, 1.2, 1.5

:arebar tensilea of Stress andStrength :Ex( ) ( )MPaMPaS y 15,184 & 32,270 == σ

2

dzeRz

243.22

211design ofy Reliabilit −−

∞−∫−=π

R = 1-0.0075 ???? Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980.

Prob: A steel bar is subjected to compressive load. Statistics of load are (6500, 420) N. Statistics of area are (0.64, 0.06) m2. Estimate the statistics of stress.Ans: (10156 1156 4) Pa

11/2/2011 20

Ans: (10156, 1156.4) Pa.

Example: Stress developed in a machine element is given by:

( )( )223 344/ LLkdP +=σGiven P = (1500, 50) N, Strength = (129, 3) MPa, L1=(150, 3) mm, L2=(100, 2) mm. Assume std. dev. of d is 1.5% mean

( )( )21 344/ LLkdP +=σ

value of d. k = 0.003811.Determine distribution of d if the maximum probability

of machine-element-failure is 0.001

ALGEBRAIC MEAN STD. DEVIATIONFUNCTIONS Q

lfd i tiSt d d CQCxQCQ

==

CxC

C 0C x

σσ

⎟⎞

⎜⎛ ∂

2

:by expressed isfunction complex aofdeviation Standard

φ xyQyxQxCQ

=±=+=

yxyxxC

±+

2222

22

xy yx

yx

x

σσ

σσ

σ

+

+

∑ = ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=n

i xiix1

2σφσμ

φ

xQyxQ

yQ

1==

xyx

y

1 2

22222

x

yxy

x

yx

y

σ

σσ +

21

Example: Stress developed in a machine element is given by:

( )( )223 344/ LLkdP +=σGiven P = (1500, 50) N, Strength = (129, 3) MPa, L1=(150, 3) mm, L2=(100, 2) mm. Assume std. dev. of d is 1.5% mean

( )( )21 344/ LLkdP +=σ

value of d. k = 0.003811.Determine distribution of d if the maximum probability

of machine-element-failure is 0.001

∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

= =ni xi

ix12

2

:by expressed isfunction complex a ofdeviation Standard σφσφ⎠⎝ i μ

2/1

22

2

22

1

22

22

21 LLdP LLdP⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=σ σσσσσσσσσ Statistically independent

( ) ( ) ( ) ( )2/1

22

32

2

32

2

42

2

3

21

002.085216003.0170430015.04136355022724dd

dd

ed

LLdP

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

⎥⎦⎢⎣ ⎠⎝ ∂⎠⎝ ∂⎠⎝ ∂⎠⎝ ∂

σσ

[ ] 2/13 290472614204183012291.11 e

d

dddd

+++=

⎥⎦⎢⎣ ⎠⎝⎠⎝⎠⎝⎠⎝

σσ

31136200

d=σσ 22

( )( )22

21

3 344/ LLdkP +=σ Strength = (129, 3)( )( )3

21

34087000d

=σ

Strength (129, 3)

Std. dev. of d is 1.5% mean value of d

d

( )3408700061290 3de( )113620063

340870006129009.32

12

2 ⎥⎤

⎢⎡

⎟⎞

⎜⎛+

−−=−=

e

deZ

( ) 1103121136

63

222

3

⎟⎞

⎜⎛

⎟⎞

⎜⎛

⎥⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

+d

e

( )

m0010;m66860

11031417482.11363000 332 ⎟

⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛+

σddd

m001.0;m 6686.0 == σσd11/2/2011 23

GEARSGEARS

Elements required to transmit power between t a s t po e bet eerotating shafts. ~ Different rotational speeds.Rotates about axis

Often gears are treated as pitch cylinder which rollRotates about axis.

Consists of gearwheel

pitch cylinder which roll together without slip.

Positive drive by meshingand teeth. Positive drive by meshing teeth.

Torque RatioGenerally gear pair acts as a speed g

pAm

ωω

= ratio, Torqueacts as a speedreducer aiming torque amplification

g

Gear Pair:Smaller – PINIONq p

at output shaft. Larger -- Gear

Internal & External gearingInternal & External gearing

Tooth Profile ωTooth Profileg

pAm

ωω

=

Involute tooth form: Locus of aform: Locus of a point on a line rolling on its base circle.

Velocity ratio does ynot change due to inaccuracies in

t di tcenter distance.

Tooth curves of the matingTeeth need to be tangentTeeth need to be tangent to each other.

Line of action is tangent toBoth pinion & gear base

c Circles.

On changing center distance

c

On changing center distanceLine of action still remainsTangent to both base circlesTangent to both base circlesBut slope changes. Pressure angle ??

⎟⎟⎞

⎜⎜⎛

= − bI

R1cosφ ⎟⎠

⎜⎝ I

I Rφ

Backlash: Difference between tooth space and tooth thickness.• Prevents jamming of teeth.

• Compensates for thermal expansion of teeth.

Any limit on Torque ratio ?????

Velocity ratio

Normally speed reduction for a single pair of spur gear < g p p g7:1.

Size of gear wheel increases Gear box size.

For high speed reduction. T hTwo stage or three stage construction are preferred.

Helical 10:1

Compound Gear Train: At least one shaft carries two gears.

Internal 4-8Bevel 1-8Cylindrical worm 3-80g Cylindrical worm 3 80

Velocity ratio in Spur Gear

Ex: Motor speed 1440 rpm. R i d d t t it Required speed to transmit a load of 10 kN is 100 rpm.

( ) n1ratioreductionOverall

ratioreduction Stage =

1440

( )ratioreduction Overall

795.3100

1440=

Pinion 18 teeth (20°)

11/2/2011 30

Spur Gears: Teeth Spur Gears: Teeth parallel to axis of rotation. Suitable to transmit

ti b t ll l motion between parallel shafts.

Spur Gear Drive11/2/2011 31

S GSpur Gears

Basic gear.Pressure angle is Pressure angle is measure of inclination of each teeth of each teeth.

Larger α greater strength & wear strength & wear resistance.

Diametrical Pitch: N in Diametrical Pitch: N in each inch of gear’s pitch diameter.pitch diameter.

11/2/2011 32

Classification of Gears based on Classification of Gears based on Position of Shaft Axes

Rolling Gears.1. Parallel axes.2. Intersecting axes

Quieter operation. Axial force. Herrringbone gears. Manufacturing l itcomplexity.

Rolling Cross Axis GearsMixture of rolling & Sliding at contact surfaces.

Lubrication requires ub a o quattention.

Worm GearsRelatively low efficiency

Worm Gears

Hypoid bevel

Helical crossed

Hypoid bevel gears

Helical crossed axis gears

Rolling Vs sliding

Rolling

Sliding

Rolling Vs slidingRolling Vs slidingWith involute

fil fAs contact moves

profile of gears, only one contact position

towards or away from pitch point, lidi position

experience pure rolling.

sliding occurs.

g

c

After few degrees rotationAfter few degrees rotation

Gear

Intersecting h ft

Non intersecting Parallel shaft non parallel shaftshaft

Spur Gear Helical Gear Crossed Hypoid WormSpur Gear Helical Gear Crossed Helical Gear

Hypoid Gear

WormGear

Single Helical Gear

Double Helical Gear

11/2/2011 37BevelGear

Spur Gear Nomen l t eNomenclature

Pitch circles: Imaginary tangent circles.

Pinion: Smaller.

Circular pitch: Sum of tooth thickness & width of spacethickness & width of space.

Addendum: Radial distance between top land and pitch i lcircle.

Backlash: Difference between tooth space and ptooth thickness.

Module: m=Dp/ZP=DG/ZG

11/2/2011 38Basic spur gear Geometry.

Pitch and Base CirclesTooth system

φ, deg

Addendum

Dedendum

Cross belt Full depth

20 1m 1.25m1.35m

22.5 1m 1.25m1.35m

25 1m 1.25m1.35m

Stub 20 .8m 1m

c

ModulesPreferred 1 1 25 1 5 2 2 5 3 4 5 6 8 10 12 16 20 25 32 40

11/2/2011 39

Preferred 1,1.25,1.5,2,2.5,3,4,5,6,8,10,12,16,20,25,32,40Second choice 1.125,1.375,1.75,2.25,2.75,3.5,4.5,5.5,7,9,11,14,18,22,28,36,45

Catalogue BModule MM

Bore diameter Pounds Module

MM

Bore diameter PoundsMM of cutter MM of cutter

HBIGM01 1mm 27mm or 1” 95.20 HBIGM05 5 27 or 1 155.40HBIGM01.25 1.25 27 or 1 95.20 HBIGM05.5 5.5 27 or 1 172.20HBIGM01.5 1.5 27 or 1 98.00 HBIGM06 6 27 or 1 193.20HBIGM01.75 1.75 27 or 1 100.80 HBIGM07 7 32 or 1-1/4 329.00

HBIGM02 2 27 or 1 103.60 HBIGM08 8 32 or 1-1/4 392.002 27 or 1 103.60 8 32 or 1 1/4 392.00HBIGM02.25 2.25 27 or 1 103.60 HBIGM09 9 32 456.40HBIGM02.5 2.5 27 or 1 107.80 HBIGM10 10 32 512.40HBIGM02 75 2 75 27 or 1 110 60 HBIGM11 11 40 560 00HBIGM02.75 2.75 27 or 1 110.60 HBIGM11 11 40 560.00

HBIGM03 3 27 or 1 119.00 HBIGM12 12 40 616.00HBIGM3.25 3.25 27 or 1 123.20 HBIGM14 14 40 665.00

3 5 27 1 120 40 16 40 980 00HBIGM03.5 3.5 27 or 1 120.40 HBIGM16 16 40 980.00HBIGM03.75 3.75 27 or 1 123.20 HBIGM18 18 50 1190.00

HBIGM04 4 27 or 1 124.60 HBIGM20 20 50 1358.00

HBIGM04.25 4.25 27 or 1 147.00

HBIGM04.5 4.5 27 or 1 142.80

HBIGM04.75 4.75 27 or 1 152.60

All sizes shown above are in 20o PA =Presure Angle (Note: 14.1/2o PA is also supplied)

11/2/2011 40

BSS Cutter Cuts Teeth (European Catalogue B

Number Cuts Teeth pCutter No)

1 135 - RACK 82 55 134 72 55 - 134 73 35 - 54 64 26 - 34 55 21 - 25 46 17 - 20 37 14 - 16 2

11/2/2011 41

Contact RatioContact Ratio

Base pitchaction ofLength

=

)r

(2

recess ofLength approach ofLength _p

bgπ

+

)Z

(2g

gπ

11/2/2011 42

Contact RatioContact Ratio

rrb b22*

p*

aLength

sinr caLength

−=

= φ

caba

rrb bpop

**cbLength

aLength

−=

φsinrcb or, p22 −−= bpop rr

iSi il l 22 φ

bgogbpop

bgog

rrrr

rr

sin)rr( ab action, ofLength

sinracSimilarly

gp2222

g22

φ

φ

+−−+−=

−−=

g

bgogbpop

ZCrrrr

/r2sin

ratioContact bg

2222

πφ−−+−

=

11/2/2011 43

Ex: For φ=20°, ZP=19, Zg=37, and m=4; Find Gear Ratio, φ , P , g , ; ,circular pitch, base pitch, pitch diameters, center distance, addendum, dedendum, whole depth, outside diameters, and contact ratio If center distance is increased by 2% what will contact ratio. If center distance is increased by 2% what will be new pressure angle and new contact ratio.

1937 RatioGear =

mp or, Zd pitch Circular

19

cg

g == ππ

⎟⎟⎞

⎜⎜⎛

− pr coscos 1 φ

φr

Zmp

p

gc

41 0Add d

)(r C dist,center Nominal

ddiaPitch ;cosppitch Base

g

gb

+=

== φ ⎟⎟⎠

⎜⎜⎝

=p

pnew r02.1

cosφ

a2ddmm5bm 1.25 b Dedendum,

mm4am1.0 a Addendum,

+==⇒×=

=⇒×=

New rpa2d d pop +=

bgogbpop Crrrr φsinratioContact

2222 −−+−

bpratioContact =

11/2/2011 44

Ex: A gear pair (ZP=23, φ=20°, Zg =24, m=1.75, F=10.0 mm) h d l 2 d l dhas center distance equal to 42 mm. Find nominal and running contact ratios.

55ppitchCircular;2324RatioGear =→== mmpmπ

125.41)(rCdist,center Nominal

42d diaPitch ;1662.5cosppitch Base

5.5ppitch Circular ;23 RatioGear

g

gb

c

→+=

=→=→=

→

p

ggc

c

r

mmdZmmmp

mmpm

φ

π

75.432d dmm1875.2bm 1.25 b Dedendum, ;mm75.1a m1.0 a Addendum,

5.)(Cd s ,ce eNo

pop

g

=→+==⇒×==⇒×=

→

op

p

da

bgogbpop Crrrr φsinCR

2222 −−+−=

ocos125.411 ⎟⎞

⎜⎛ φφφ

bpCR =

o06.2342

cos125.41cos 1 =→⎟⎠⎞

⎜⎝⎛= −

newnew φφφ

Nominal contact ratio = 1 6 Nominal contact ratio = 1.6,

Contact ratio after assembly = 1.14 45

Design of Spur Gear

t thdB di

strength Elasticstatic,

≤

≤bσ

strengthenduranceBendingdynamic, ≤bσ

strengthenduranceSurface≤σ strengthenduranceSurface≤contactσ

Core Case Core Case

Hardness; Strength > 1 GPa; g

Related to wear. Indirectly Lubrication.

11/2/2011 46

Stresses in Spur GearsStresses in Spur GearsTwo modes of failures:

Fatigue: fluctuating bending stresses at root of tooth.

Keep stress state within

Tooth system

φ, deg

Addendum

Dedendum

Keep stress state within modified Goodman line for material.. Infinite life

Full depth

20 1m 1.25m1.35m

Surface fatigue (Pitting)Repeated surface contact stresses materials do not

22.5 1m 1.25m1.35m

stresses … materials do not exhibits endurance limit

In the present course – Only

25 1m 1.25m1.35m

p ytwo pressure angle and Full depth teeth.

Stub 20 .8m 1m

11/2/2011 47

Bending Stresses W

Wr

g

ModulusSectionMoment

b =σ

Wt

6/2tFlWt

b =σ

AssumptionsCompression due to radial Components of gear

t th f W WCompression due to radial component of force is negligible.Teeth do not share load. Uniformly

tooth force Wt , Wrand Wa

distributed load across the face width.Greatest force is exerted at tip ( )

nceCircumfereπ

Nm =

Greatest force is exerted at tipLewis Eq.

mYFWt

b =σ( )

( )knessTooth thic knessTooth thic

SpacingknessTooth thic

π

π

m

NNm

+=

+=

2222 = m

mtl

π

11/2/20112

knessTooth thic ππ

m=2056.0

4=⇒ Y

No. of Teeth

Form factor Y

No. of Teeth

Form factor Y

No. of Teeth

Form factor YTeeth factor Y Teeth factor Y Teeth factor Y

1213

0.2450 261

2122

0.3280 331

5060

0.4090 42213

1415

0.2610.2770 290

222426

0.3310.3370 346

6075100

0.4220.4350 44715

1617

0.2900.2960.303

262830

0.3460.3530.359

100150300

0.4470.4600.47217

1819

0.3030.3090.314

303438

0.3590.3710.384

300400Rack

0.4720.4800.485

20 0.322 43 0.397

WtEx: A gear set is connected to 5 H P motor 1440 rpm Pinion mYF

Wtb =σEx: A gear set is connected to 5 H.P. motor, 1440 rpm. Pinion

(m=3mm, F=38 mm, number of teeth =16, pressure angle 20 degrees) is made of AISI 1020 steel (yield strength = 206 MPa) on milling machine Find factor of safety

49

on milling machine. Find factor of safety.

V = 3.6191 m/s ; Wt = 1031 N; Stress = 30.6 MPa

L d i ti d t t t tiLoad variation due to contact ratio

11/2/2011 50

Ex: Pinion shaft passes 15kW at 2500 rpm. For φ=25°, ZP=14, m=4, and Gear Ratio=3.5, determine transmitted P , , ,loads on gear teeth. Find pitch diameters, mean and alternative components of transmitted load.

T

Z

p

g

3.57)60/25002/(15000

49145.3

==

=×=

π

mmZmd

mNT

T

g

p

56144

.55.2003.575.3

3.57)60/25002/(15000

=×==

=×=

π

( )NWWl dR di l

NdTWloadTangential

mmZmd

ppt

pp

954t

20462//,

56144

==

=×==

φ NWWloadRadial tr 954tan, == φ47% load

2tWloadMean =

2

2tWloadeAlternativ =

11/2/2011 51

2

Homework: What will the pressure angle be if the center distance of a 20° pressure angle gear pair is increased by 7% 7%.

Homework: Find out the mean and alternating load components of a gear-set that transmits 50 kW at 1600 components of a gear set that transmits 50 kW at 1600 pinion rpm. φ=25°, ZP=23, Zg=57, and m=4.

F ti B di F il f S GFatigue Bending Failure of Spur Gear

Surface Finish Gear sizeR li bili S iReliability Stress concentrationRotation in one direction or bothTemperature11/2/2011 52

Temperature

AGMA EqLewis

AGMA introduced velocity factor in terms of pitch line velocity (m/s) in Lewis equation.

YFWK

AGMA

tvb =σ

Eq.Lewis( )profilecastironcastVKv ,

05.305.3 +

=

( )

V

profilemilledorCutVKv

56301.6

01.6

+

+=

mYF

( )

( )V

profileshapedorHobbedVKv

565

56.356.3

+

+=

Useful for preliminary estimation of gear size.

( )profilegroundorShavedVKv 56.556.5 +

=

For V = 15 m/s KFor V = 15 m/s , Kv

• Cast iron, cast profile = 5.2

• Cut or milled profile = 3 5• Cut or milled profile = 3.5

• Hobbed or shaped profile = 2.1

Sh d d 1 3• Shaved or ground = 1.311/2/2011 53

Ex: Find out the power rating of milled profiled spur No. of Form gear (AISI material, yield strength = 210MPa) for data: φ=20°, ZP=16, F=36mm, m=3.0, N = 20 rps. Assume factor of safety = 3.0.

Teeth factor Y14 0.277

Ans: Allowable bending stress = 70 MPa.

Pitch line velocity V=3.0 m/s.

1516

0.2900.296

Kv = 1.5 , Form factor Y = 0.296

Tangential load = 1492 N.

P ti 4 475 kWPower rating = 4.475 kW.

( )profilemilledorCutVKv 01601.6 +

=01.6

WK tvσmYF

tvb =σ

11/2/2011 54

AGMA Bending mBatv

b KKKJF

WK=σAGMA Bending

Stress EquationmBab JmF

loading ofpoint angle, pressureon depends FactorGeometrybendingAGMAJ =

11/2/2011 55

11/2/2011 56

mBatv

b KKKJmF

WK=σ

Driven Machines Power Source Uniform Light shock Moderate shock Heavy shock

Application factor K

JmF

Application factor, KaUniform (Electric motor, turbine) Light shock (Multicylinder)

1.00

1.20

1.25

1.40

1.50

1.75

1.75

2.25(Multicylinder) Moderate shock (single cylinder)

1.30

1.70

2.00

2.75

Character of operation Driven machines

Uniform Generator, conveyor, electric hoist,

M di M hi t l i Medium Machine tool, mixer, pump,

Heavy Press, shear, mill, drilling,

11/2/2011 57

d di ib i fmBa

tvb KKK

JmFWK

=σLoad distribution factor Km

Face width, mm Km

JmF

< 50150

1.61.7

250>500

1.82.0

R

BB

tmwheremK

mK

=≥=

<≤+−=

2101

2.1m0.5 4.32factor thicknessRim B

tBBB h

mwheremK =≥= 2.1 0.111/2/2011 58

Summary of Previous Lecture

Lewis Eq.mYF

Wtb =σ

AGMA Bending Stress Eq. tv KKKWKσ mBatv

b KKKJmF

=σ

11/2/2011 59

Ex: A gear pair (ZP=23, φ=20°, Zg =24, m=1.75, F=10.0 mm) gtransmits 8 N.m torque from crankshaft (rotational speed 8000 rpm) of single cylinder IC engine to wheels. Bore diameter of pinion is 17 mm, and bore dia of gear is 20 mm. Use AGMA pinion is 17 mm, and bore dia of gear is 20 mm. Use AGMA bending stress formula to determine the maximum bending stress. Assume gears are grounded.

Given: F = 10 mm m =

Btv

b KKKWK=σ

Given: F = 10 mm, m = 1.75, Wt = 8000/(23*1.75*0.5)

mBab KKKJmF

=σ Driven Machines

Load distribution Power Source Uniform Light shock Moderate shock Heavy shock Application factor, Ka Uniform 1.00 1.25 1.50 1.75

Load distribution factor Km

Face width mm

Km

(Electric motor, turbine) Light shock (Multicylinder)

1.20

1.40

1.75

2.25

width, mm

< 50 1.6

Moderate shock 1.30 1.70 2.00 2.75 11/2/2011 60

6.1 0.2 == ma KK ( )gearsgroundVK 56.5 +=ma ( )gearsgroundKv 56.5

( )80002540

mm 25.4075.1*23 ==

Nd

d p

π ( )

565

/86.1660

800025.4060

+

→==

V

smNd

V p ππ

875.3575.1*25.1*2 =−= pproot dd3185.1

56.556.5

=+

=VKv

( ) 43759509375.375.1*25.2 ==t

pproot

Boredtmmh

( )12.1

4375.95.0

=⇒>

=−=

BB

pprootR

Km

Boredt

mBatv

b KKKJmF

WK=σ

11/2/2011 61

J = 0.35

J = 0.26

KKKWK tv=σ W

22.4=vmBa KKKK

loadingtipforMPa

KKKJmF

b

mBab

6.368=⇒

=

σ

σMPa

mYFWLewis t

b 75==σ

AISI 1020 steel (yield 11/2/2011 62loadingHPSTCforMPab 8.273=⇒σ

AISI 1020 steel (yield strength = 206 MPa)

FoS = 0.75

( )butfinish SaK MPain =

Finishing a b

Endurance StrengthFinishing method

a b

Ground 1.58 -0.085

finishTrsLee kkkkkSS '=Machined or cold drawn

4.51 -0.265

Hot rolled 57.7 -0.718

=1.0 mmdd ee 51 if24.1 107.0 ≤= − Probability of

survival, %Reliability factor, kr

Hot rolled 57.7 0.718

0 for b

mmdd ee 51 if51.1 157.0 >= −survival, % factor, kr

5090

1.00 897bending

ltde 808.0=909599

0.8970.8680.814g load

99.999.99

99 999

0.7530.7020 65999.999

99.99990.6590.620

Temperature FactorTemperature Factor

Temperature KT Temperature KT

20°C 1.00 300°C 0.97550°C 1.01 350°C 0.943100°C 1.02 400°C 0.900100 C 1.02 400 C 0.900150°C 1.025 450°C 0.843200°C 1 02 500°C 0 768200°C 1.02 500°C 0.768250°C 1.0 550°C 0.672

NOTE: Initially increase in temperature causes the redistribution of stress-strain profiles at notches or stress concentration features, hence increases the fatigue strength.

11/2/2011 64

features, hence increases the fatigue strength.

( )butfinish SaK MPain =( )38051.4 265.0= −

finishK

Endurance Strength( )

934.0=finish

finish

K

finishTrsLee kkkkkSS '=Finishing method

a b

Machined or 4.51 -0.265

=1.0 de24.1 107.0= −

Probability of Reliability

cold drawn

0 for b

mmde 512.79 if ≤≤ltde 808.0=

survival, % factor, kr

5090

1.00 897

bending

( )( )mmd

mmd

e

e

65.225708.1808.0

=

=909599

0.8970.8680 814g load

99 0.814

0.1=sk( )( )( )( )( )( ) MPaMPaS e 159934.01897.0113805.0 =×=

Fatigue failure criteria for Fatigue failure criteria for fluctuating stresses

12

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

ut

m

e

a

SSσσ1

=+e

a

ut

m

FoSSSσσ

⎠⎝ ute

1tan ==m

a

eut FoSSS

σσθ

m

KKKJmF

WKmBa

tvba 22

==σσ

MPa

JmF

a 1372

27422

==σ

1=+yt

m

e

a

SSσσ

2

1137137 yte

FoS1

159137

380137

=+ 0.82

AGMA Strength Equation

Through Through Hardened Steel

MPab 8.273=σ

MPaHGradeMPaHGrade

Bball

Bball

3.88533.01113703.02

,

,

+=→

+=→

σσ

34817.2282

=→=→

B

B

HrequiredGradeHrequiredGrade

11/2/2011 67

Nitrided through hardened steel gears (AISI Nitrided through hardened steel gears (AISI 4140, 4340): Max hardness 340 HB.

MPaHGrade 11074902 +=→σ Yσ

Strength at 107 cycles and 0 99 Reliability MPaHGradeMPaHGrade

Bball

Bball

8.83568.01110749.02

,

,

+=→

+=→

σσ

RT

N

F

ballballcorrected KK

YS

,,,

σσ =

Strength at 10 cycles and 0.99 Reliability.

Material Designation Heat Treatment Allowable Bending Stress Bending Stress Number (MPa)

ASTM A48 Gray cast iron

Class 20Class 30

As CastAs Cast

3558Gray cast iron Class 30

Class 40As CastAs Cast

5890

Bronze Sand castHe t T e ted

39163Heat Treated 163

ASTM A536 ductile

Grade 60Grade 80

AnnealedQuenched & Temp

151-227151-227

11/2/2011 68Grade 100Grade 120

Quenched & TempQuenched & Temp

186-275213-275

bll Yσ

RT

N

F

ballballcorrected KK

YS

,,,

σσ =

Material properties (FSM ) 1.0,1.1,1.3Stress (FSS ) 1.0, 1.2, 3.0

√

√St ess ( SS ) 0, , 3 0Geometry (FSG ) 1.0, 1.2Failure analysis (FSFA ) 1 0 1 2 1 5

√√

XFailure analysis (FSFA ) 1.0, 1.2, 1.5Environmental factors (FSE ) 1.0, 1.3, 1.6Danger to Personnel (FS ) 1 0 1 6

X

X

X

√

Danger to Personnel (FSD ) 1.0, 1.6 X

11/2/2011 69

Correction Factors( )

( )⎩⎨⎧

≤≤<<−−

=999909901l109050

99.05.01log0759.0658.0RR

RRK e

R ( )⎩⎨ ≤≤−− 9999.099.01log109.05.0 RRe

R

Yσ

RT

N

F

ballballcorrected KK

YS

,,,

σσ =

Contact StressesTwo rolling surfaces under compressive load experience “contact stresses”.contact stresses .

Ball and roller bearingsCams with roller followerS h l l hSpur or helical gear tooth contact

PinionPinion

Gear

11/2/2011 71

Contact (Elastic) Stresses

Compressive load elastic deformation of surfaces deformation of surfaces over a region surrounding the initial point of contact. pStresses are highly dependent on geometry of dependent on geometry of the surfaces in contact as well as loading and material

Stress concentration near contact region is very high Stress

gproperties. is very high. Stress

concentration factor ????

11/2/2011 72

????

R1R1

Cylindrical

R2

Spherical

R2

Spherical

On varying radii of curvature: sphere-plane, sphere-in-cup, cylinder-on-plane and cylinder in troughplane, and cylinder-in-trough.Zero areas Infinite stress. Material will elastically deform and contact ygeometry will change.

Contact stresses…

1

dbdb <<

2db <<

Deformation b will be

High stress

Deformation b will be small compared to dimensions of two bodies g

concentrationbodies.

11/2/2011 74

h lSpherical contact⎤⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

2

max 1brpp

∫ ∫b

drrdpF2

ispatchcontactonloadappliedTotal θπ

∫ ∫= drrdpF0 0

ispatch contact on load appliedTotal θ

∫ ⎥⎤

⎢⎡

⎟⎞

⎜⎛b

drrrpF2

12ispatchcontactonloadappliedTotal π ∫⎥⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

−= drrb

pF0

max 12ispatch contact on load appliedTotal π

[ ]b2

( )0

max222 2i dpFb π∫

[ ]∫ −= drrrbbpF

0

22max2 or π

51=K r

( )

3max

max222

2or

2 assumingon

bpF

dtttbpFtrb

b

π

π

=

−==− ∫

5.1=tK r

max2

32 or

3 or

pbF

bF

π=

Cylindrical ContactR1

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−=

22

max 1ay

bxpp

LPressure variation along Y-axis is

li ibl

⎥⎦⎢⎣ ⎠⎝⎠⎝ ab

⎥⎤

⎢⎡

⎟⎠⎞

⎜⎝⎛−=

2

max 1 xpp

Lnegligible,

⎥⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

max bpp

R2YispatchcontactonloadappliedTotal

XZ

∫⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

b

dxbxpLF

0

2

max 12

ispatch contact on load appliedTotal

11/2/2011 76

Z⎦⎣

Cylindrical Contact…2

max 12 dxbxpLF

b

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−= ∫

Cylindrical Contact…

( )2

2

0

2il dbFθb

b

θθ

π

⎥⎦⎢⎣ ⎠⎝

∫

∫

( )0

2max cos2 sinlet

LbF

dbpFθb x

π

θθ== ∫

max2 or pLbF =

Stress concentration factor = 4/π

max2

32 pbF contactspherical π=

max2 pLbF contactlcylindricaπ

= How to determine b ???211/2/2011 77

How to determine b

Assume pmax = σy and find value of b.

5.1 Fb contactspherical=

max

2 Fb

pb

contactlcylindrica

π

=maxpL

bπ

=

Criterion b << d1 needs to be checked.

11/2/2011 78

For axi-symmetric point load For axi symmetric point load Timoshenko & Goodier suggested:

( )F

EGzyν

ρ

1

)1(2x

2

222

⎫⎧

+=++=

( )

( )

zG

Fz ρ

νρπ

δ 14 3

2

⎪⎫⎪⎧⎭⎬⎫

⎩⎨⎧ −

+=

( )yxE

F ν

νπ

δ 10

)1(24

221⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+

−+

+

=Y

( )rF

Eπνδ

ν

21

)1(22

1−

=

+

XZπ

Ref: S. Timoshenko and J.N.Goodier, Theory of elasticity, 2nd

Edition, McGraw Hill.

11/2/2011 79

( ) ( )dd

brpb 2 2max

21 /11

1sphere of Deflection

−−∫ ∫ θνθδ

π

( ) ( )

( )d

brp

drrdr

pE

r

b 2max

21

0 0

max

1

11

/121

2

,

−−

=

∫

∫ ∫

νδ

θπ

θδ

( )

( ) ( ) drbrp

drrr

pE

b2

21

0

max

1

11

/11or

22

or

−=

=

∫

∫νδ

ππ

δ

( ) ( ) drbrpE 0

max1

1 /1or −= ∫δ

( ) ( )coscos1sinbassumingon22

1 θθθνδθπ

dbpr −== ∫( ) ( )

( ) ( )12cos1or

coscos sinb assumingon

221

0max

11

θθνδ

θθθδθ

π

db

dbpE

r

+−

==

∫

∫

( ) ( )

( ) 2sin1or

12cos2

or

221

0max

1

11

θθνδ

θθδ

π

pb

dpE

⎥⎤

⎢⎡ +

−

+= ∫

22 pbF π=

( )

( )1or

22 or

max

21

1

0max

1

11

πνδ

θδ

pb

pE

−=

⎥⎦⎢⎣+=

max3pbF π=

11/2/2011 8022

or max1

1δ pE

( )22

1max

21

1πνδ p

Eb −

=22 bF22 1E

max2

3pbF π=

FEb

21

11

83 νδ −

=OEb 1

1 8 O

1 OCOB −=δ

22

2211

or

or,

bRR

ACOAR

−−=

−−=

δ

δ

2

11

111

11or,

or,

bR

bRR

⎟⎟⎟⎞

⎜⎜⎜⎛

⎟⎟⎞

⎜⎜⎛

−−=

=

δ

δ

2

111

1

,

b

R

⎟⎞

⎜⎛

⎟⎞

⎜⎛

⎟⎞

⎜⎛

⎟⎟⎠

⎜⎜⎝

⎟⎠

⎜⎝

AC2

111 termsnegligible

2111or,

b

RbR ⎟

⎟

⎠⎜⎜

⎝⎟⎟

⎠⎜⎜

⎝+⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=δ

2 B2

1

2

11 2 or,

RbR=δ F

ERb

1

21

13 175.0 ν−= 81

( )21b πν1hfD fl tiTwo spherical contacting surface

( )

( ) ( )⎥⎤

⎢⎡ −−

−=

22

21

max2

22

11

41

b

pE

b

ννπδ

πνδ

( )11sphere of Deflection

21

1πνδ pb −

= ( ) ( )⎥⎦

⎢⎣

+=2

2

1

1max4 EE

bpδ22 max

11δ p

E

+=22

22radii, geometricofin termspresentedbecan deflectionTotal

Rb

Rbδ

( ) ( )⎥⎦

⎤⎢⎣

⎡ −+

−=+

2

22

1

21

max2

2

1

2

21

11422

or

22gp

EEbp

Rb

Rb

RR

ννπ

22 bF( ) ( )⎥⎦

⎤⎢⎣

⎡ −+

−

⎥⎤

⎢⎡

+=

⎦⎣

2

22

1

21max

2121

11114

or EE

pb ννπ max2

3pbF π=

⎥⎦

⎢⎣

+21 22 RR

( ) ( )⎥⎤

⎢⎡ −

+−

=22

213 115.1 Fb νν

⎟⎞

⎜⎛ −− 22 114

contacts lcylindricaFor

F νν

⎥⎦

⎢⎣

+

⎥⎦

⎤⎢⎣

⎡+

=21

21 21

214 EE

RR

b

11/2/2011⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎠⎞⎜

⎝⎛ +

=2

2

1

1

21

1111

4EE

RRL

Fb νν

π

Surface/Contact Stresses Surface/Contact Stresses in Spur Gears

Surface failure of gear tooth occurs due to very high local y gcontact stresses. Maximum contact pressure at the contact point between two cylinders is given by:

( ) ( )[ ]=max

2Lb

Fpπ

( ) ( )[ ]⎟⎠⎞⎜

⎝⎛ +

−+−=

21

2221

21

11/1/12

dd

EELFbwhere νν

πAs per nomenclature of gear design: F = W, L=F, W = Wt /cosφ , d1=dp*sinφ

11/2/2011 83

⎠⎝ 21 dd /cosφ , d1 dp sinφ

As per nomenclature of gear d i F W L F W W /

=max2

LbFp

πdesign: F = W, L=F, W = Wt / cosφ , d1=dp*sinφ( ) ( )[ ]

⎟⎠⎞⎜

⎝⎛ +

−+−= 2

221

21

11/1/12

dd

EELFbwhere νν

π

2W

⎟⎠

⎜⎝

+21 dd

( ) ( )[ ]νν

π

/1/12 22

max

−+−

=

ggpp EEWbh

Fbp

FbWp t

πφcos/2

max =( ) ( )[ ]

φπ

sin111 ⎟

⎠⎞

⎜⎝⎛ +

=

gp

ggpp

ddF

bwhere Fbπ

( ) ( )[ ]ννφ

/1/12

cos/222max

+= t

EEW

Wp ( ) ( )[ ]

φ

ννφπ

π

i111

/1/1cos

2 22

⎟⎠⎞

⎜⎝⎛ +

−+− ggppt

dd

EEF

WF

84

φsin⎠⎝ gp dd

Surface contact compressive ⎥

⎥⎤

⎢⎢⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛ + gp

dddd

compressive stress( ) ( )

⎥⎥⎥⎥

⎦⎢⎢⎢⎢

⎣

−+−

⎟⎠

⎜⎝=

ggpp

gpt

EE

ddF

Wp/1/1cossin

222

2max ννφφπ

( ) ( ) ⎥⎥⎥⎤

⎢⎢⎢⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛ +

= gp

gp

tc

dddd

W222

2σ

⎥⎦⎢⎣

( ) ( )⎥⎥⎥

⎦⎢⎢⎢

⎣

−+− ggppc EEF /1/1cossin 22 ννφφπ

⎤⎡⎤⎡ dd( ) ( )[ ] ⎥

⎥⎦

⎤

⎢⎢⎣

⎡ +

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+−=⇒

g

pg

ggppp

tc d

ddEEdF

Wφφννπ

σcossin2

/1/11

222

Z

Z

2cossin

111 g

22P ZICLet

+=

⎤⎡⎟⎞

⎜⎛

⎟⎞

⎜⎛

=φφ

νν Z211 p g

g

g

P

pZ

EE

+

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟⎠

⎞⎜⎜⎝

⎛ − ννπ

tPc d

WC=⇒σp

Pc dIF11/2/2011 85

equationresistance pittingAGMA c vmat

P CCCWC=σqp g c vmap

P dIF

L d di ib i D i M hi Load distribution factor Cm

F idth C

Driven Machines

Power Source Unifor Light Moderate Heavy Face width, mm

Cm

50 1 6

m shock shock shock Application factor, Ca Uniform 1.00 1.25 1.50 1.75

< 50150

1.61.7

(Electric motor, turbine) Light shock

1.20

1.40

1.75

2.25

250>500

1.82.0

g(Multicylinder) Moderate shock

1.30

1.70

2.00

2.75

shock

( ) ( ) 3/21225.0 ;15650200

v

B

v QBBAwhereVA

C −=−+=⎟⎟⎞

⎜⎜⎛ +

= ( ) ( )5.0;5650 vv Qwhe eA

C ⎟⎠

⎜⎝

11/2/2011 86

Calculation of Factor Cv

200B

v AVA

C ⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

( ) ( ) 3/21225.0 and 15650 vQBBA

A

−=−+=

⎟⎠

⎜⎝

AGMA Qv

Tolerance Cv for 16.86 m/s velocityQv910

15 μm10 μm

1.341.23

1112

7 μm5 μm

1.131

11/2/2011 87

Ex: A gear pair (ZP=23, φ=20°, Zg =24, m=1.75, F=10.0 mm) gtransmits 8 N.m torque from crankshaft (rotational speed 8000 rpm) of single cylinder IC engine to wheels. Bore diameter of pinion is 17 mm, and bore dia of gear is 20 mm. Using AGMA pinion is 17 mm, and bore dia of gear is 20 mm. Using AGMA pitting resistance formula to determine the maximum contact stress. Assume gears’ quality = 9, E = 2.e5 MPa, ν=0.3

equation resistance pittingAGMA c = vmap

tP CCC

dIFWCσ

1.34 1.6 2.0 18700011

122

=⇒

⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛ −

+⎟⎟⎞

⎜⎜⎛ −

= pgp

P

p

CCνν

π

0.0821I Z

Z

2cossin g =⇒

+=

⎥⎥⎦⎢

⎢⎣

⎟⎟⎠

⎜⎜⎝

+⎟⎟⎠

⎜⎜⎝ gP

ZI

EE

φφ

π

Ans: 1334 MPaZ2 p + gZ

200B

vVA

C ⎟⎟⎞

⎜⎜⎛ +

=

11/2/2011 88

( ) ( ) 3/21225.0 and 15650 v

v

QBBA

A

−=−+=

⎟⎠

⎜⎝

Contact Stress vs. BrinellHardness

Ncall Z,σσ =

RTFcallcorrected KKS,,σ =

223741.2

120022.2

,

,

GradeMPaHB

GradeMPaHB

call

call

+=

+=

σ

σ

,call

Stess: 1334 MPa

Effect of Brinell Hardness on allowable contact stress for through-hardened steel.

11/2/2011 89

g

AlternativesReliability < 0.99

( )( )⎨

⎧ <<−−=

99.05.01log0759.0658.0 RRK e

R ( )⎩⎨ ≤≤−− 9999.099.01log109.05.0 RR

Ke

R

11334

RT

N

F

callcallcorrected KK

ZS

,,,

σσ =8996.01

12001334

<⇒< RR

KK

R < 0.9585

11/2/2011 90

Pitting resistance stress cycle g yfactor, ZN

Allowable contact stress is based on 10 million load cycles with reliability of 0.99.

Heat Treatment Grade 1 (MPa) Grade 2 Grade 3I d ti 1172 1310InductionCarburizedNitrided

1172 12061240

131013441551

--

1896

Strength of Steel

Increase in

lNN 60560 1051146621334

Increase in module.Increase in face width.

Zσ

cyclesNN 6056.0 1051.1466.212001334

×=⇒= −

11/2/2011 91RT

N

F

callcallcorrected KK

ZS

,,,

σσ =

Ex: Motor speed 1440 rpm. Required speed to transmit a load of 10 kW is 100 rpm Design transmit a load of 10 kW is 100 rpm. Design gears.

( ) n1ratioreductionOverall

ratioreduction Stage =

79531440

( )ratioreduction Overall

795.3100

=

Pinion 18 teeth (20°)Pinion 18 teeth (20 )

equationLewiswithstartingif75 MPa≤σ equationLewiswith startingif75 MPa≤σWt=σ

mFm 159 ≤≤W10000

92mYFb =σ( )( ) smTm

WWp

t /60144010000

π=

No. of Teeth

Form factor Y

No. of Teeth

Form factor Y

No. of Teeth

Form factor YTeeth factor Y Teeth factor Y Teeth factor Y

1213

0.2450 261

2122

0.3280 331

5060

0.4090 42213

1415

0.2610.2770 290

222426

0.3310.3370 346

6075100

0.4220.4350 44715

1617

0.2900.2960.303

262830

0.3460.3530.359

100150300

0.4470.4600.47217

1819

0.3030.3090.314

303438

0.3590.3710.384

300400Rack

0.4720.4800.485

20 0.322 43 0.397

mWt 37.710*75 6m=3.0 mm

Modules( )( )mm

mmYF

Wtb 309.012

37.710*75 6 =⇒=σ Select m = 3.0 mmPCD = 54 mm

93

ModulesPreferred 1,1.25,1.5,2,2.5,3,4,5,6,8,10,12,16,20,25,32,40

Is dp = 54 mm justifiable ? Easily available bearings 10, 12, 15 mm.Gears to be on stepped shaft 15 mmGears to be on stepped shaft 15 mm.Tooth height = 2.25*3mm.Mi i i hi k 1 2*hMinimum rim thickness =1.2*ht

dp ≥ 4 6.7 mmp

dp = 54 mm is justifiable.V = 4 m/sV = 4 m/s

11/2/2011 94

AGMA Bending mBatv

b KKKJF

WK=σAGMA Bending

Stress EquationmBab JmF

Teeth on on Gear = 68

J = 0.3

F 36 F= 36 mm

NW 250010000== NWt 2500

4==

016 +VprofilemilledorCut

11/2/201167.1

01.601.6

=+

=VKvMPaK ba 310;5.1 == σ

Checking contact stressChecking contact stressW

1

equation resistance pittingAGMA c = vmap

tP CCC

dIFWCσ

1.123 1.6 1.5 18700011

122

=⇒

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟⎠

⎞⎜⎜⎝

⎛ −= p

gpP C

EE

Cνν

π

0.1271I Z

cossin g =⇒=

⎥⎦⎢⎣⎟⎠

⎜⎝

⎟⎠

⎜⎝ gP

I

EE

φφAns: 977 MPaZ2 p + gZ Ans: 977 MPa

( ) d15650200 ⎟

⎞⎜⎛ +

B

BAVA

C ( )

( ) 10assuming12250

and 15650;

3/2 =−=

−+=⎟⎟⎠

⎜⎜⎝

=v

QQB

BAA

C

11/2/2011 96

( ) 10assuming1225.0 == vv QQB

2237412 GradeMPaHB+=σMPaHGrade Bball 113703.02 , +=→σ

281 HB 307 HB

MPab 310=σAns: 977 MPa

223741.2, GradeMPaHBcall +=σ

Ans: 977 MPa

11/2/2011 97

Helical GearsTeeth are angled (15°, 23°, 30°, 45° )w. r. t. axis of rotation

High contact ratio Smooth engagement Less noiseLess noise

Axial load

Increase in helix Increase in helix angle increases smoothness. Z

98

smoothness.

ψ3cos, ZZteethVirtual =′

H li l GHelical Gears

Axial load

φsinn

WForceRadialWForceContactW =

φtan nψφ

φcoscos

sin

nnt

nnr

WForceTangentialWWForceRadialW==

==

Two pressure anglesφφψ

tancos n=ψφ sincos nna WForceThurstW ==

Two pressure anglesTransverse pressure angle (φ)Normal pressure angle (φ )

11/2/2011 99

Normal pressure angle (φn)

For parallel axes, p ,Meshing of

opposite hand Helical Gears Helical Gears is essential.

Axial force = Separation force ψφ sincos nna WW =p

pp = ψcos

ca

cn

pppp

= ψψ

tancos

100apF >

( )ψππ 2cos/5022 dR B( )

ψπ

ψππ

cos/

cos/5.022

nc

e

mZdm

dpRZ ==′ B

ψψ

ψ 22 coscos/

cos n

n

n mmZ

mdZ ==′

dcos2

d a ellipse, of axismajor -Semi =ψ

a2

2d b ellipse, of axisminor -Semi =

ψ3cos, ZZteethVirtual =′

ba

eRradius Effectivecurvature,ofRadius =ψ

101

Ex: A pair of helical gears (Ψ=25°, Zp=22, Zg=44) have normal pressure angle=20° and normal module = 3 Fi d i l t di t & t 3mm. Find nominal center distance & transverse pressure angle, φ. Compare with spur gear.

Geometric entity in transverse plane

Helical gear Spur gear

M d l ( ) 3/ (25°) 3 31 3Module (mm) 3/cos(25°)=3.31 3

Dp (mm) 3*22/cos(25°)=72.823 66

Dg (mm) 3*44/cos(25°)=145.646 132

φ (°) tan(20°)/cos(25°)=21.88° 20°

C (mm) 0.5*(Dp+Dg)=109.234 99

ψ3cos, ZZteethVirtual =′

11/2/2011 102

ψcos

Axial Thrust Force

φsinn

WForceRadialWForceContactW

===

ψφφ

coscossin

nnt

nnr

WForceTangentialWWForceRadialW==

)()/()(

mrsradWdtransmittebetoPower

pω=

ψφ sincos nna WForceThurstW ==)()( p

EL iWB di tZZ ′Eq. , Lewis

mYFWstressBending

z

tb

′

=σ

WKψ3cos

Z =

11/2/2011 103 Eq.AGMA mBa

z

tvb KKK

JmFWK

′

=σ

Pinion teeth = 18, Gear teeth = 68, helix angle = 20helix angle = 20.Virtual teeth = 21.6, 81.6 17.1

3.035.0RatioJ ==

11/2/2011 104

BEARINGSMechanical elements which

1. allow relative motion between t l t (i two elements (i.e. shaft & housing).

2 Bear load2. Bear loadThrust loadRadial loadCombined load

11/2/2011 105

1. Dry Contacts 2. Chemical Films3. Lamellar Solids

4 P i d L b i t Fil5. Elastomers 6. Flexible Strips

4. Pressurized Lubricant Film

7. Rolling Elements 8. Magnetic FieldField

11/2/2011 106

Comparison in three types of bearingComparison in three types of bearing11/2/2011 107

Rolling Element Rolling Element Bearing

“Rotation is always easier than linear motion”. Lo friction & moderate l bricant req irementsLow friction & moderate lubricant requirements

are two important advantages of rolling bearing.

If you can buy it, don’t make it!B i l tiBearing selection….~ 20,000 Varieties of bearings. Conventional bearing steel to ceramics, with (out) cage (brass/polymers). Pin.Smallest bearings – few grams. Largest 70 Tonnes

Frequent questions~ 20,000 Varieties of bearings classification.c ass cat oHow do I select a bearing for given applicationapplication.How do I treat combined load.B i lif (106 l ?? 3000 )Bearing life (106 cycles ?? 3000 rpm).Mathematical formulation ?Any requirement of lubrication (How to we incorporate)p )

11/2/2011 109

Bearing ClassificationClassification

11/2/2011 110

http://www.skf.com/portal/skf/home/products?newlink=first&lang=en

11/2/2011 111

Deep Groove Ball Bearing (DGBB): Both rings possess deep grooves. Bearing can support high Bearing can support high radial forces as well as axial forces. There are single-row & d bl G d l& double row DGBB. Widely used in industry. Angular Contact Ball

Bearing (ACBB):Cage/Separator: Ensures uniform spacing and prevents mutual contact of

Bearing (ACBB):Raceways are so arranged that forces are transmitted prevents mutual contact of

rolling elements. from one raceway to other under certain contact angle-angle between line of action angle between line of action of the force & radial plane. Due to CA, ACBB are better

d h h lsuited to sustain high axial loads than DGBB.

Ball Cylindrical roller Angular contact ball Tapered roller Sphericalroller

Ball Cylindrical roller11/2/2011 113

Ball Cylindrical roller

Cylindrical Roller Bearing

Higher coefficient of friction because of small diameter rollers anddiameter rollers and rubbing action against each other

11/2/2011 114

Cylindrical roller bearings

11/2/2011 115

Shield: Profiles sheet steel discs pressed into the grooves of outer ring and forming gap-type seals grooves of outer ring and forming gap type seals with the inner-ring shoulders. Nomenclature with Z

S l Oft d f l ti bb B i l d Seals: Often made of elastic rubber. Bearings sealed on both sides are grease filled and in –normal working conditions the grease filling lasts the entire service life of g gthe bearings. Nomenclature with R

117

Selection of bearing t peSelection of bearing type

Cylindrical & Needle roller– Pure Radial LoadCylindrical roller thrust, ball thrust, four point y , , pangular contact ball bearings– Pure Axial loadTaper roller, spherical roller, angular contact p , p , gbearings– Combined loadCylindrical roller, angular contact ball bearing–y , g gHigh speedDeep groove, angular contact, and cylindrical eep g oo e, a gu a co tact, a d cy d caroller bearing– High running accuracy

11/2/2011 118

Designation – International Organization for Standardization

Each rolling bearing is designed by a code that clearly indicates construction, dimensions, tolerances and bearing clearance. , , g

05618 2Z

Multiply by 5 to get bore in mmd<10mm 618/8 (d=8mm)

00 = 10mm01= 12mmd<10mm… 618/8 (d=8mm)

d>500 mm… 511/530 (d=530mm)02 = 15mm03 = 17mm11/2/2011 119

0 Double row angular contact ball bearings1 Self aligning ball bearings1 Self-aligning ball bearings2 Spherical roller bearings3 Taper roller bearingsp g4 Double row deep groove ball bearings5 Thrust ball bearings

BS i6 Single row deep groove ball bearings7 Single row angular contact ball bearings8 Cylindrical roller thrust bearings

BoreSeries

8 Cylindrical roller thrust bearingsHK needle roller bearings with open endsK Needle roller and cage thrust assembliesgN Cylindrical roller bearingsA second and sometimes a third letter are used to identify the configuration of the flanges e g NJ NU NUP; double or configuration of the flanges, e.g. NJ, NU, NUP; double or multi-row cylindrical roller bearing designations always start with NN. QJ Four-point contact ball bearings

11/2/2011 120

ACBB

SRB TBB

SABB TRB DGBB DGBB ACBB CRTB

In order of

In increasing order

In order of increasing outside

11/2/2011 121bearing diameter

Examples of basic codes

11/2/2011 122

Rolling Element BearingsLoad Direction Misalignment

CapacityRadial Axial Both High Med Low

Deep groove ball y y y

Cylindrical Roller y Some types

y

Needle y yNeedle y y

Taper Roller y y y y

Self Aligning Ball y y ySelf Aligning Ball y y y

Self Aligning Spherical Roller

y y y

Angular contact ball

y y y

Thrust ball y yThrust ball y y

123

Suffix

61804 61804-2Z 61804-2RS161804 61804-2Z 61804-2RS1

11/2/2011 124

Basic Dynamic Load Rating: CRadial load (thrust load for thrust bearings) which a groupRadial load (thrust load for thrust bearings) which a group of identical bearings with stationary outer rings can theoretically endure one million revolutions of inner ring.y g

Static Load Rating: C0

Radial load causing permanent deflection greater thanRadial load causing permanent deflection greater than 0.01% of ball dia. ------ > 0.3 μm – 2320 N

11/2/2011 125

E i l t l dEquivalent load

P = V X Fr + Y Fa

V Rotation factor

X Radial factor

F Applied radial loadFr Applied radial load

Y Thrust factor

Fa Applied thrust load

11/2/2011 126

Bearing type Inner ring Single row Double row eRotating Stationary Fa/VFr > e Fa/VFr ≤ e Fa/VFr > ea r a r a r

Deep groove ball

Fa/C0 V V X Y X Y X Y.014 1 1.2 0.56 2.30 1 0 0.56 2.30 .19

ball bearing

.028

.056

.084

1.991.711.55

1.991.711.55

.22

.26

.28.11.17.28

1.451.311.15

1.451.311.15

.3

.34

.38.42.56

1.041.00

1.041.00

.42

.44

Angular 20 1 1.2 .43 1.0 1 1.09 .70 1.63 .57Angular contact ball bearing

20253035

1 1.2 .43.41.3937

1.0.87.7666

1 1.09.92.7866

.70

.67

.6360

1.631.441.241 07

.57

.68

.809535

40.37.35

.66

.57.66.55

.60

.571.07.93

.951.14

Self aligning

1 1 .4 .4 cot

1 .42 cot

.65 .65 cot

1.5 tanaligning

ball bearing

cotα cotα cotα tanα

Deep Groove Ball Bearing

RSH Sheet steel reinforced contact seal of acrylonitrile-butadiene rubber (NBR) on one side of the bearing. L stand for low friction.

Example: Assume radial and axial loads on a bearing are 7500N and 4500N respectively. Rotating shaft dia = p y g70 mm. Select a single row deep groove ball bearing.

Bearing type Inner ring

Single row eR t ti /Rotating Fa/VFr > e

Deep groove

Fa/C0 V X Y014 1 0 56 2 30 19groove

ball bearing

.014

.028

.056084

1 0.56 2.301.991.711 55

.19

.22

.2628.084

.11

.17

1.551.451.31

.28

.3

.34.28.42.56

1.151.041.00

.38

.42

.44

129Fa/Fr = 0.6; Fa/C0=4500/31000 X = 0.56, Y= 1.37; P=10365Fa/C0=4500/68000 X = 0.56, Y= 1.65; P=11625

Rolling Element Bearings-Rolling Element BearingsLoad Calculation: Tabular Approach

Load ratingC > P x fn x fL x fdn L d

Where C = radial dynamic rating P = calculated effective radial loadP calculated effective radial loadfn = speed (rpm) factorfl = Life (hours) factorlfd = dynamic or service factor

Load classification FactorfUniform 1.0

Light shock 1.5Moderate shock 2.0

11/2/2011 130

Heavy shock 3.0

11/2/2011 131

Example 1: Radial load = 4448 N, Speed = 1000 rpm, p p pShaft dia. 30 mm; Desired life= 30 000 hours, No Shock loading.

C > P x fn x fL x fdfd = 1.0; P = 4 448 N, fn= 2.78; fl= 3.42

C > 42 290 N=> C > 42, 290 N

11/2/2011 132

Revisiting example discussed in slide 129Example: Assume radial and axial loads on a bearing are 7500N and 4500N respectively Shaft dia = 70 mm Select a deep groove ball bearing 4500N respectively. Shaft dia 70 mm. Select a deep groove ball bearing. Consider shaft rotates at 1000 rpm and expected bearing life = 30000 hours

Fa/Fr = 0.6; Fa/C0=4500/31000 P=10365Fa/C0=4500/68000 P=11625

C > P x fn x flfn= 2.78; fl= 3.42

Case 1: C = 98.55 kN

C 2 C 110 53 kNCase 2: C = 110. 53 kN

M b d ti f lif 1300 h11/2/2011 133

May be a good option for life = 1300 hr

Summarizing previous lectureRolling element bearings/Rolling bearingsbearings

Ball bearingRoller bearing

Antifriction bearings Starting ~Running friction.

Roller bearingFive Components

Inner ringException for Taper roller bearing

Inner ringOuter ringRolling elements

ConeCup

Finite Life of bearingRolling elementsCage/Retainer/SeparatorS l/Shi ld

Finite Life of bearingCost of dis-assemblingMaintenance free b i f Fi it LifSeal/Shield

11/2/2011 134

bearings for Finite Life.

E i l t D i L dEquivalent Dynamic Load

Equivalent load: P = V X Fr + Y Fa

eFr ≤= /FwhenF Pringinner ofrotation Assuming

ar

e is a dimensionlessratio, indicating axial

load lower than a

eFFYFXP rar

r

>+= /Fwhen a

arcertain limit does not

affect total load

V l f d d t & t ti l dValue of e depends on arrangement & static load capacity (CO ) of bearing

11/2/2011 135

Fa/C0 Fa/VFr > e eSingle/Double row Deep

X Y

Single/Double row Deep groove ball bearing

.014

.0280.56 2.30

1.99.19.22

rFVP

.056

.08411

1.711.551 45

.26

.283

1

.11

.17

.28

1.451.311.15

.3

.34

.38.42.56

1.041.00

.42

.44 a

FVFe

rFV

X & Y Factors depend upon bearing geo., numberof balls and size of ballsof balls and size of balls.

11/2/2011 136

Fixed (Ball) vs Floating (Roller) Bearings

Predominately Used as FloatingUsed as Floating

Bearing

• Variations of shaft due to the thermal expansion are accommodated between the inner ring & the roller set.

Summarizing previous lectureSummarizing previous lecture…Considered an example: speed of inner ring

1000 t d b i lif 30000 = 1000 rpm, expected bearing life = 30000 hours 1.8e09 cycles.13000 hours 7 8e08 cycles13000 hours 7.8e08 cycles.Life should be infinite after 1.0e6 cycle ?

L-10 LifeRating life corresponding to 10% failure probability (Reliability ??)Dynamic load rating (catalogue C reading) is the l d hi h 90% ( li bilit 0 9) f f load which 90% (reliability=0.9) of a group of identical bearings will sustain for minimum of 106

cyclescycles.11/2/2011 138

L-10 LifeL 10 Life

( ) 10 3322116 aaaa LPLPLPC ===

10bearings ballfor 3a =

Load Rating bearingsroller for 3

10

a

a

⎞⎛

= Load Rating Factors

Speed60000,1000

PC

hoursin life Bearinga

⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒

p⎠⎝C = 110530 N, P = 11625 N, Speed = 1000 rpm Life = 14,326 hours.C=39700 N, P=10365 N, Speed = 1000 rpm Life = 936.5 hours

11/2/2011 139

Load Rating Factors 32

1 ⎤⎡Load Rating Factors1

1log48.4 ⎥⎦⎤

⎢⎣⎡=

Ra e

( ) 10 3322116 === LPLPLPC aaaa

Application FactorEl t i t t b 1 0 1 4

1061 ⎟

⎟⎠

⎞⎜⎜⎝

⎛= cyclesin

aPCaL

a Electric motor, turbo-compressor

1.0-1.4

Reciprocating 1.5-2.0

bearings ballfor 3=

⎟⎠

⎜⎝

a

aP f machinesImpact machines like hammer

2.5-3.5

bearingsroller for 3

10=a

V-belt drive 2.0Single ply belt drive 3.03Multi ply belt drive 3.5Chain drives 1.5

11/2/2011 140

Example 2: Radial load = 2 224 N, Speed = 1500 rpm

D i d lif 8 h /d 5 d / k f 5 Li h Sh kDesired life= 8 hours/day, 5 day/weeks for 5 years, Light Shock loading. For shaft dia of 25 mm.

C 2224*1 5*(10400*1500*60/106)1/C > 2224*1.5*(10400*1500*60/106)1/a

C > 32, 633 N for BALL BEARINGSC > 25, 978 N for ROLLER BEARINGS

11/2/2011 141

Types Conrad type

Bearing SelectionConrad typeFilling notch type

Radial + Thrust Deep groove ball bearingDeep groove ball bearing

Single row angular contact ball bearing i t th t l d i di ti can resist thrust load in one direction

only.

11/2/2011 142

Example 3: A radial load of 3000N combined with thrust load of 2500N is to be carried on a 6214 ball bearing for 70 mm dia rotating shaft at 1000 rpm. Determine equivalent radial load to be rotating shaft at 1000 rpm. Determine equivalent radial load to be used for calculating fatigue life. Compare life of 6214 bearing with that for a 7214 (nominal contact angle 30°)

Step 1: C0 for 6214 is 45kN and 7214 is 60 kN. C for 6214 is 63.7 kN and 7214 is 71.5 kNStep 2:

Bearing type Single row, Fa/VFr > e eg yp g , a r eDeep groove ball bearing Fa/C0 X Y

.056 0.56 1.71 .26Angular contact ball bearing 30 .39 .76 0.8

Step 3: Equivalent radial load for 6214 bearing is 5955N & for 7214 bearing radial load is 3070.

11/2/2011 143

Step 4: Life for 6214 will be 20,400 hours and for 7214, p , ,life=210,550 Hours

000,1000ChoursinlifeBearing3

⎟⎞

⎜⎛=

Speed60Phoursin lifeBearing ⎟

⎠⎜⎝

=

11/2/2011 144

( )

∑=

=

1

cos

cosmax

ψψ

ψ

ψ

ψ

WF

WW n

Fr

∑−=

=1

cosψψ

ψ ψWFr

max

max

).06.4/(

).37.4/(

WZF

WZF

llr

ballr

=

= loadunder element rolling of Deflection

max).06./( Wrollerr

b illf1 11 bearing, ballfor 1.5 n

; W == nKδ

( ) ; 2coszoneloadofextent angular of50%

11 =

−rdC δψ

bearingroller for 1.11n =

ring ofshift radial =rδ

11/2/2011 146

Failure of Four Row Failure of Four Row Cylindrical Roller Bearing

Two large roller bearings – (ID = 865 mm, OD = 1180 mm) failed in a cold

lli ill R 35 00 000 hrolling mill. Rs. 35,00,000 eachone bearing failed within 105 hours (installed on 05/01/03 and failed completely on 10/01/03) andon 05/01/03 and failed completely on 10/01/03), andother failed within 300 hours of operation (installed on 05/01/03 and removed on 24/01/03 due (to detection of excessive vibration and metal particles). Expected life of bearings was approximately Expected life of bearings was approximately 40,000 operating hoursSurvival rate 0.5% and 1.0%.

Failed Bearing

Bearing subjectedsubjected to normal load

Constant direction load. Quarter of the outer race is under load.

O t i ith ki I t IVOuter ring with marking I to IV.

First time mounting zone I along the direction of the load.

After a period of approximately 1000

ti h ( 2operating hours (≅ 2 months), outer ring is turned 90° Iturned 90 .

Conclusion: Rated bearing life = 4.* Life of one load zone.

I

Co c us o a ed bea g e e o o e oad o eExpected life of each load zone = 10,000 operating hours

Hole in line of maximum Hole

load.

Four holes of 3/8” 10 UNC 3B of 45mm depth were drilled and tapped to drilled and tapped to facilitate the handling of outer race.

I

• With new arrangement, noWith new arrangement, no complaint was received.

III III

II

IV

W II IVW

II

(a) Earlier arrangement (b) New arrangement

Homework: A single row cylindrical roller bearing N 205 ECP is subjected to pure radial load of 2800 N and rotational speed = 1500 rpm. Estimate the bearing life for reliability = 0.99.

Homework: Select a suitable deep groove ball bearing for a shaft of 30 mm dia rotating at 2000 rpm. Bearing needs to support a radial load of 2000 N and axial load of 400 N.

Drawback of Angular contact bearings ??11/2/2011 153

g g

T d t B k t b k F t fTandem arrangement Back to back Face to face

Back to back for rigid shaft mounting ( primary factor).g g ( p y )Requires inner rings to be clamped.

Face to face for shaft misalignment (primary concern). Requires outer rings to be clamped.

Bearing type

Single row Double row eFa/VFr > e Fa/VFr ≤ e Fa/VFr > e

α(°)

X Y X Y X Y

Angular contact

2025

.43

.411.0.87

1 1.09.92

.70

.671.631.44

.57

.68ball bearing

303540

.39

.3735

.76

.6657

.78

.6655

.63

.6057

1.241.0793

.80

.951 1440 .35 .57 .55 .57 .93 1.14

Self aligning

.4 .4 cotα

1 .42 cotα

.65 .65 cotα

1.5 tanαaligning

ball bearing

cotα cotα cotα tanα

Representative Bearing Design LivesT f A li ti D i Lif Type of Application Design Life

(thousands of hours)

Instruments and apparatus for infrequent use 0.1-0.5

Machine used intermittently, but reliability is of great importance

8-14 great importance Machine for 8-hours service, but not every day 14-20

Machines for 8 hour service every working day 20 30Machines for 8-hour service, every working day 20-30

Machine for continuous 24-hour service 50-60

Machine for continuous 24-hour service where reliability is of extreme importance.

100-200

NOTE: SKF recommends min load of 0 02 C to be imposed on roller11/2/2011 157

NOTE: SKF recommends min load of 0.02 C to be imposed on roller

bearings, while 0.01 C to be applied on ball bearing.

Variable Loadingaaaa LPLPLP

1

⎟⎞

⎜⎛ +++

Often bearings are subject to variable loading:

LLLLPLPLPP

321

332211

b ib llf3

......⎟⎟⎠

⎞⎜⎜⎝

⎛

++++++

=

loading:Bearing operates at 1000 rpm and applied load of 500 N

a

a

bearingsroller for 3

10bearingsballfor 3

=

=

applied load of 500 N for 100 hours, then bearing operates at 1200 rpm and 250 N

rotationsofNumberL,L,L 321

thenlifeexpectedLIF,...

3

=1200 rpm and 250 N for 250 hours….

In such situation it is d i bl t fi d

( )( )

aaaa LLPLPLPP1

332211 ...

thenlife,expectedLIF

⎟⎟⎞

⎜⎜⎛ +++

=

=

advisable to find an equivalent load using

( )

( ) aaaa fPfPfPP

LLLLP

1

332211

321 ...

+++=

⎟⎟⎠

⎜⎜⎝ +++

11/2/2011 158

g ( )fPfPfPP 332211 ...+++=

Example: A ball bearing is run at four piecewise load and speed conditions Find equivalent loadconditions. Find equivalent load.

Time Speed, Product, Rotation Applied Time fraction

Speed, rpm

Product, column 1*2

Rotation fraction

Applied load, kN

0.1 1000 100 0.0333 4

0.2 2000 400 0.1333 3

0.3 3000 900 0.3 2

0.4 4000 1600 0.5333 1

( ) 313333 fPfPfPfPP +++( ) 3

43

433

323

213

1 fPfPfPfPP +++=

( ) NP 20541066368 319 =×= ( ) NP 2054106636.8 =×=

11/2/2011 159

HomeworkA ball bearing (C = 85 kN) supports a shaft that rotates at 1000 rpm. A radial load varies in such a way that 50, 30 and 20 percent of the time the load is 3, 5, and 7 kN respectively. Estimate L-10 lifelife.

( ) 31

43

433

323

213

1 fPfPfPfPP +++= ( )44332211 ffff

( )∫2

32iπ

θθ dP11/2/2011 160

( )∫0

3max1 2sin θθ dP

Bearing ClearanceBearing Clearance

dD dD

Bearing Mounting Bearings are mounted on shaft and housing with transition to Interference fit. If interference fits exceed the internal radial If interference fits exceed the internal radial clearance, the rolling elements become preloaded.

C2, C3, C4 as bearing suffix.

High operating temperature environment requires larger bearing clearance.

11/2/2011 162

IT Grade 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Lapping

Honing

Super finishing

C li d i l i diCylindrical grinding

Diamond turning

Plan grinding

Broaching

Reaming

Boring TurningBoring, Turning

Sawing

Milling

Planing, Shaping

Extruding

Cold Rolling DrawingCold Rolling, Drawing

Drilling

Die Casting

Forging

Sand Casting

Hot rolling, Flame cutting

Nominal Sizes (mm)

over 1 3 6 10 18 30 50 80 120 180 250

inc. 3 6 10 18 30 50 80 120 180 250 315

IT Grade International tolerance grade of industrial processes.

1 0 8 1 1 1 2 1 5 1 5 2 2 5 3 5 4 5 61 0.8 1 1 1.2 1.5 1.5 2 2.5 3.5 4.5 6

2 1.2 1.5 1.5 2 2.5 2.5 3 4 5 7 8

3 2 2.5 2.5 3 4 4 5 6 8 10 12

4 3 4 4 5 6 7 8 10 12 14 164 3 4 4 5 6 7 8 10 12 14 16

5 4 5 6 8 9 11 13 15 18 20 23

6 6 8 9 11 13 16 19 22 25 29 32

7 10 12 15 18 21 25 30 35 40 46 527 10 12 15 18 21 25 30 35 40 46 52

8 14 18 22 27 33 39 46 54 63 72 81

9 25 30 36 43 52 62 74 87 100 115 130

10 40 48 58 70 84 100 120 140 160 185 21010 40 48 58 70 84 100 120 140 160 185 210

11 60 75 90 110 130 160 190 220 250 290 320

12 100 120 150 180 210 250 300 350 400 460 520

13 140 180 220 270 330 390 460 540 630 720 81013 140 180 220 270 330 390 460 540 630 720 810

14 250 300 360 430 520 620 740 870 1000 1150 1300

A company X, decided to design air-circulators for paintp y , g pshops. Length of 2-m and diameter of 0.6-m was designedfor rotor of air-circulator. Company X wanted to design

it bl b i t d th ti 2/3suitable bearings to reduce the power consumption on 2/3.

On studying the air-circulator it was found that rotor length could be reduced from 2-m to 1.4-m by relocating the drive-could be reduced from 2 m to 1.4 m by relocating the drivemotor. Reduction in length of rotor itself fulfilled the requirements.