Unit 4 helical gear

Helical Gears, Helix Angle, Parallel helical gears, Crossed helical gears, Herringbone gears, Left Hand & Right Hand Gear

1. Unit-4
Helical Gear
Dr. L.K. Bhagi
Associate Professor
School of Mechanical Engineering
Lovely Professional University

2. Helical Gears
Inhelicalgears,thetwomeshinggearsmaybe
mountedonparallelorintersectingshafts.The
teethonhelicalgeararecutatanangle(helix
angle)tothegearaxis.
Thehelixangleusuallyrangesbetween15ºand
30º.
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
2

3. Helical Gears
Spurgears,aremerelyhelicalgearswithazero
helixangle.
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
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4. Helical Gears
Spurgears,aremerelyhelicalgearswithazero
helixangle.
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5. Helical Gears
Spurgears,aremerelyhelicalgearswithazero
helixangle.
Carrymoreloadthanequivalent-sizedspurgears
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6. Helical Gears» Helix Angle
Thehelixangleofhelicalgearsisgenerally
selectedfromtherange6,8,10,12,15,20,30
degrees.
Thelargertheanglethesmootherthemotionand
thehigherspeedpossiblehoweverthethrust
loadingsonthesupportingbearingsalso
increases.
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7. Helicalgearsareclassifiedas
Parallelhelicalgears
Crossedhelicalgearsand
Herringbonegears
Allthehelicalgearsgeneratethrustloadsontheshafts
becauseofinclinedteeth;hence,thesemustbetaken
carewhiledesigningthemachines.
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Helical Gears» Classification

8. Helical Gears» Thrust & Radial Load
THRUSTloadisloadparalleltotheshaftofthegear.Itis
producedbyhelicalgearsbecausethehelixangle,notthe
pressureangle.Itisnotproducedbyspurgears,whichhave
straightteeththatareparalleltotheshaftaxis.
RADIALloadistheloadthattendstoseparatethegears.It
actsperpendiculartotheshaft.Thisiswhatisproducedby
thepressureangle.Bothspurgearsandhelicalgears
producethiskindofload.
29-11-2019
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9. Helical Gears» Parallel Helical Gears
•Referstowhentheshaftsareparalleltoeachother.
•Notethatmatinghelicalgears(onparallelshafts)
musthavethesamehelixanglebuttheopposite
hand.
29-11-2019
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10. Helical Gears» Crossed Helical Gears
Theshaftsarenon-parallel,andinthisconfigurationthe
gearsaresometimesknownas"skewgears"
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11. Helical Gears» Herringbone Gears
Doublehelicalgearsovercometheproblemofaxial
thrustpresentedbysinglehelicalgearsbyusingtwo
setsofteeththataresetinaVshape.Thisarrangement
cancelsoutthenetaxialthrust,sinceeachhalfofthe
gearthrustsintheoppositedirection,resultinginanet
axialforceofzero.
However,doublehelicalgearsaremoredifficultto
manufactureduetotheirmorecomplicatedshape.
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12. Helical Gears» Herringbone Gears
Toavoidaxialthrust,two
helicalgearsofoppositehand
canbemountedsidebyside,
tocancelresultingthrust
forces.
Herringbonegearsaremostly
usedonheavymachinery.
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13. Helical Gears»Left Hand & Right Hand
Gear
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14. Helical Gears»Advantages
❑Theangledteethengagemoregraduallythandospur
gearteethcausingthemtorunmoresmoothlyand
quietly.
❑Helicalgearsarehighlydurableandareidealfor
highloadapplications.
❑Helicalgearscanbeusedonnonparallelandeven
perpendicularshafts,andcancarryhigherloadsthan
canspurgears.
❑Atanygiventimetheirloadisdistributedover
severalteeth,resultinginlesswear.
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15. Helical Gears»Drawbacks
❑Oneofthedisadvantagesofthesegearsisthethrust
whichresultsalongthegearaxis,whichneedstobe
accommodatedbyusingadequatethrustbearings.
❑Thereisagreaterdegreeofslidingfrictionbetween
theteeth.Thisproducesgreaterwearduring
operation,andtheneedforlubricationsystems.
❑Thehelicalgearefficiencyislowerduetothecontact
betweenitsteeth,whichproducesaxialthrustand
generatesheat.Agreaterlossofenergyreduces
efficiency.
❑Highermanufacturingcostthanspurgears.
29-11-2019
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16. Helical Gears»Efficiency
Efficiencies of spur gear and helical gear compared?
a)efficiency of spur gear = efficiency of helical gear
b)efficiency of spur gear < efficiency of helical gear
c)efficiency of spur gear > efficiency of helical gear.
d)none of the above
29-11-2019
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17. Helical Gears»Efficiency
Efficiencies of spur gear and helical gear compared?
a)efficiency of spur gear = efficiency of helical gear
b)efficiency of spur gear < efficiency of helical gear
c)efficiency of spur gear > efficiency of helical gear.
d)none of the above
29-11-2019
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18. Helical Gears»Efficiency
Spurgearsarethegearsinwhichtheteethareparallel
tothegearaxis,whereasinhelicalgear,theteethare
notparalleltothegearaxis.
Becauseofthisarrangementofteethinhelicalgears,
theyhaveslidingcontactbetweentheteeth.
Duetotheslidingcontactbetweentheteethofhelical
gear,axialtrustofgearshaftalsoproducesandalso
producemoreheat.Thustheefficiencyofhelicalgear
islessthanthatofthespurgear.
29-11-2019
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19. Helical Gears»Strength
Forsametoothsize(module)andequivalentwidth,helicalgears
canhandlemoreloadthanspurgearsbecause...?helicalgear
toothiseffectivelylargersinceitisdiagonallypositioned
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20. Helical Gears»Strength
Forsametoothsize(module)andequivalentwidth,helicalgears
canhandlemoreloadthanspurgearsbecausethehelicalgear
toothiseffectivelylargersinceitisdiagonallypositioned.
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21. 29-11-2019
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SpurGears HelicalGears
Teetharecutparalleltotheaxisofthe
shaft
Teetharecutintheformofhelixon
thepitchcylinderbetweenmeshing
gears
Contactbetweenmeshingteethoccurs
alongtheentirefacewidthofthetooth
Contactbetweenmeshinggearsbegins
withapointontheleadingedgeofthe
toothandgraduallyextendsalongthe
diagonallineacrossthetooth
Loadapplicationissuddenresulting
intoimpactconditionsandgenerating
noiseinhighspeedapplications
Pickupofloadbythetoothisgradual,
resultinginsmoothengagementand
quietoperationevenathighspeeds
Helical Gears»Helical vsSpur Gear

22. 29-11-2019
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SpurGears HelicalGears
Usedforparallelshaftsonly Crossedhelicalgearsarealsousedon
shaftswithcrossedaxes
Speedislimitedtoabout20m/s Usedinautomobile,turbinesandhigh
speedapplicationsupto50m/s
Imposesradialloadonly Imposesradialandaxialthrustloads
Contactratioislow Contactratioishigh
Helical Gears»Helical vsSpur Gear

23. Helical Gears» Geometry
Spurgearshavethediametralandcircularpitches.
Helicalgeargeometryrequiresadditionalpitches.
Figureshowsaportionofthetopviewofahelicalrack.
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24. Helical Gears» Geometry
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25. Helical Gears» Geometry
TheangleΨisthehelixangle.
Thetransversecircularpitch(p)orcircularpitchismeasuredonaplane
normaltotheshaftaxis(A-Aplane).
Thenormalcircularpitch(p
n)isthedistancebetweencorrespondingpointsof
adjacentteeth,measuredonaplaneperpendiculartothehelix(B-Bplane).
Theaxialpitch(p
a)isthedistancebetweencorrespondingpointsofadjacent
teeth,measuredonaplaneparalleltotheshaftaxis.Forsmoothtransferof
load,thefacewidthofhelicalgear(b)isusuallymadeatleast20%longer
thantheaxialpitch(p
a).
29-11-2019
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26. Helical Gears» Geometry
So,
Diametralpitchspurgear Circularpitchspurgear
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26p
p
n
=cos cospp
n
= )d(diameter circlepitch per
(z) teeth of No.

=P (z) teeth of No.
d
p

=
 =Pp

27. Helical Gears» Geometry
So,normalandtransversediametralpitchesare
andnormal(m
n)andtransverse(m)module
29-11-2019
Design of Machine Elements II MEC306
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27=Pp =
nn
pP 

 coscos
PP
pp
n
n
== cos
P
P
n
= cos
11
mm
n
= cosmm
n
=

28. The Module of a Gear
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28
isdefinedasinverseofdiametralpitch

29. Helical Gears» Geometry
Foraxialpitch(p
a)
Therelationbetweenthenormalpressureangle(α
n)inthe
normalplaneandtransversepressureangle(α)is
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
29a
p
p
=tan tan
p
p
a
= 


tan
tan
cos
n
=

30. Helical Gears» Geometry
Thepitchcirclediameterofthehelicalgear(d),isrelatedtothe
numberofteeth(z),normalmodule(m
n)andhelixangle(ψ)as
29-11-2019
Design of Machine Elements II MEC306
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30

cos
zm
zm
zzp
z
n
=====
P
d
d
p cos
zm
n
=d

31. Helical Gears» Geometry
Centredistancebetweenthematinggearsis
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31cos2
z
cos2
z
22
2121 nn
mmdd
a +=+= ( )
cos2
zz
21
+
=
n
m
a

32. Helical Gears» Geometry
Centredistancebetweenthematinggearsis
29-11-2019
Design of Machine Elements II MEC306
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32cos2
z
cos2
z
22
2121 nn
mmdd
a +=+= ( )
cos2
zz
21
+
=
n
m
a

33. Helical Gears» Virtual Number of
Teeth
Thevirtualnumberofteethforahelicalgearmaybedefinedas
thenumberofteeththatcanbegeneratedonthesurfaceofa
cylinderhavingaradiusequaltotheradiusofcurvatureatapoint
atthetipoftheminoraxisofanellipseobtainedbytakinga
sectionofthegearinthenormalplane.
29-11-2019
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34. Helical Gears» Virtual Number of
Teeth
Theshapeofthetoothin
thenormalplaneisnearly
thesameastheshapeofa
spurgeartoothhavinga
pitchradiusequaltoradius
ofcurvaturer’atpointPof
theellipsesis
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34b
a
r
2
=

35. Helical Gears» Virtual Number of
Teeth
Theequationofanellipsewithitscentreatoriginofanxysystem
withaandbasthesemi-majorandsemi-minoraxesis
Alsotheformulaforradiusofcurvatureis
Bythesetwoequationsandbyputtingx=0andy=b,weget
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351
2
2
2
2
=+
b
y
a
x 2
2
2
3
2
1
dx
yd
dx
dy














+
= b
a
2
=

36. Helical Gears» Virtual Number of
Teeth
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37. Helical Gears» Virtual Number of
Teeth
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Design of Machine Elements II MEC306
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37b
a
r
2
= ellipse theof axismajor Semi
cos2
−
=

d
a axisminor Semi
2
−==
d
b 
2
cos2
d
r=

38. Helical Gears» Virtual Number of
Teeth
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Design of Machine Elements II MEC306
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38
2
cos
2
d
rd ==
Inthedesignofhelicalgears,an
imaginaryspurgearis
consideredintheplaneA-Awith
apitchcircleradiusr’and
modulem
n.Itiscalledvirtual
spurgear.

39. Helical Gears» Virtual Number of
Teeth
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Design of Machine Elements II MEC306
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39nn
m
d
p
r
z












=

=
2
cos2
2
2
Thenumberofteethz’onthis
imaginaryspurgeariscalledthe
virtualnumberofteeth.

2
2
cos
cos
nn m
d
m
d
z =








=

40. Helical Gears» Virtual Number of
Teeth
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40
Thedesignofhelicalgearsisbasedonvirtualnumberofteeth(z’)
andzistheactualnumberofteeth..



3
22
cos

cos
cos
cos
z
m
zm
m
d
z
n
n
n
=
==

41. Helical Gears» Tooth Proportions
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41
Asageneralguide,toothproportionsshouldbebasedonanormal
pressureangleof20°.Inhelicalgears,thenormalmodulem
n
shouldbeselectedfromstandardvalues.
m
n(inmm)=1,1.25,1.5,2,2.5,3,4,5,6,8and10
Thestandardproportionsoftheaddendumandthededendumare,
addendum(h
a) =m
n
dedendum(h
f) =1.25m
n
clearance(c) =0.25m
n(Tipandrootclearance)

42. Helical Gears» Tooth Proportions
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Theaddendumcirclediameterd
aisgivenby
Similarly,dedendumcirclediameterd
fisn
n
aa
m
zm
hdd 2
cos
2 +=+=
 





+= 2
cos
z
md
na n
n
ff
m
zm
hdd 5.2
cos
2 −=−=
 





−= 5.2
cos
z
md
nf

43. Helical Gears» Tooth Proportions
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Trailing edge of the tooth
Leading edge of the tooth
Leading edge of the tooth should be advanced ahead of the
trailing edge by a distance greater than the circular pitch
From the ΔA
1A
2C
So, px  tantan
1
2
bx
b
x
CA
CA
=== 

tan
bor tan
p
pb 

44. Helical Gears» Tooth Proportions
29-11-2019
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And
Therefore,
This is the minimum face width
Manyauthoritiesrecommendedthatthefacewidthbeatleast
twicetheaxialpitchtoobtaintruehelical-gearaction.





 sincostantantan
nn
mmmp
=== 

sin
n
m
b

45. 29-11-2019
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45
Problem4.1:Apairofparallelhelicalgearsconsistsofa20teeth
pinionmeshingwitha40teethgear.Thehelixangleis25°andthe
normalpressureangleis20°.Thenormalmoduleis3mm.Calculate:
(a)Transversemodule
(b)Transversepressureangle
(c)Axialpitch
(d)Pitchcirclediameterofthepinionandthegear
(e)Centredistance
(f)AddendumandDedendumcirclediamtersofthepinion
Helical Gears» Numerical Problem 4.1

46. Helical Gears» Numerical Problem»
Prob. 4.1 solution
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TransverseModule
TransversePressureAngle
AxialPitchmm
m
m
n
31.3
25cos
3
cos
=

==
 =


== 88.21
25cos
20tan
cos
tan
tan 



n ()
mm
mp
p
a
3.22
25tan
3
tantan
====





47. Helical Gears» Numerical Problem»
Prob. 4.1 solution
29-11-2019
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Pitchcirclediameterofpinionandgear
Centredistance()
mm
mz
d
np
p
2.66
25cos
320
cos
=

==
 mm
dd
a
gp
3.99
2
4.1322.66
2
=
+
=
+
= ()
mm
mz
d
ng
g
4.132
25cos
340
cos
=

==


48. Helical Gears» Numerical Problem»
Prob. 4.1 solution
29-11-2019
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48
AddendumandDedendumcirclediametersofpinionmm
z
md
na
2.722
25cos
20
32
cos
=






+=






+=
 mm
z
md
nf
7.585.2
25cos
20
35.2
cos
=






−=






−=


49. 29-11-2019
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49
Problem4.2:Apairofhelicalgearsconsistsofa25teethpinion
meshingwitha50teethgear.Thenormalmoduleis4mm.Findthe
requiredvalueofhelixangle,ifthecentredistanceisexactly165mm.
Helical Gears» Numerical Problem 4.2

50. Helical Gears» Numerical Problem»
Prob. 4.2 solution
29-11-2019
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50
Helixanglefromequation( )
cos2
zz
21
+
=
n
m
a ( )
cos2
50254
165
+
= ( )
818.1
165
50254
cos2 =
+
= 909.0cos= =619.24

51. 29-11-2019
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51
Problem4.2:Apairofparallelhelicalgearsconsistsofa20teeth
pinionandvelocityratio3:1.Thehelixangleis15°andthenormal
moduleis5mm.Calculate:
(i)Pitchcirclediameterofthepinionandthegear
(ii)Centredistance
Helical Gears» Numerical Problem 4.3

52. Helical Gears» Numerical Problem»
Prob. 4.3 solution
29-11-2019
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52
(i)Pitchcirclediameters()
mm
mz
d
np
p
53.103
15cos
520
cos
=

==
 ()
mm
mz
d
ng
g
58.310
15cos
560
cos
=

==


53. Helical Gears» Numerical Problem»
Prob. 4.3 solution
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53
(ii)Centredistance( )
cos2
zz
21
+
=
n
m
a ( )

+
=
15cos2
60205
a mm 06.207=a

54. 29-11-2019
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54
Problem4.4:Twentydegreefulldepthinvoluteprofiled19-tooth
pinionand37-toothgearareinmesh.Ifthemoduleis5mm,thecentre
distancebetweenthegearpairwillbe
(a) 140 mm.
(b) 150 mm
(c) 280 mm
(d) 300 mm
Helical Gears» Numerical Problem 4.4

55. 29-11-2019
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55
Problem4.5:Ifα=helixangle,andp
c=circularpitch;thenwhichone
ofthefollowingcorrectlyexpressestheaxialpitchofahelicalgear?
(a.) (b) (c) (d)
Helical Gears» Numerical Problem 4.5tan
c
p cos
c
p cos
c
p sin
c
p

56. 29-11-2019
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56
Problem 4.6:
Assertion(A):Shaftssupportinghelicalgearsmusthaveonlydeep
grooveball-bearings.
Reason(R):Helicalgearsproduceaxialthrusts.
(a)BothAandRareindividuallytrueandRisthecorrectexplanation
ofA.
(b)BothAandRareindividuallytruebutRisnotthecorrect
explanationofA
(c)AistruebutRisfalse
(d)AisfalsebutRistrue
Helical Gears» Numerical Problem 4.6

57. 29-11-2019
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57
Problem4.7:
Assertion(A):Whiletransmittingpowerbetweentwoparallelshafts,
thenoisegeneratedbyapairofhelicalgearsislessthanthatofan
equivalentpairofspurgears.
Reason(R):Apairofhelicalgearshasfewerteethincontactas
comparedtoanequivalentpairofspurgears.
(a)BothAandRareindividuallytrueandRisthecorrectexplanation
ofA
(b)BothAandRareindividuallytruebutRisnotthecorrect
explanationofA
(c)AistruebutRisfalse.
(d)AisfalsebutRistrue
Helical Gears» Numerical Problem 4.7

58. 29-11-2019
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58
Helical Gears» Force Analysis

59. F a
F t
F r
F  n
  

60. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
60
ResultantforceFactingonthe
toothofhelicalgear
F
t=tangentialcomponent
F
r=radialcomponent
F
a=axialcomponent
Helical Gears» Force Analysis
A
B
D
C
FromΔABCnr
FF sin= n
FBC cos=
E

61. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
61
Helical Gears» Force Analysis
A
B
D
C
FromΔBDCsinBCF
a
= sincos
n
F= cosBCF
t
= coscos
n
F=
DividingF
aandF
t


tan
coscos
sincos
==
n
n
t
a
F
F
F
F tan
ta
FF=
E

62. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
62
Helical Gears» Force Analysis
A
B
D
C
DividingF
randF
t



cos
tan
coscos
sin
n
n
n
t
r
F
F
F
F
== 







=


cos
tan
n
tr
FF
E

63. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
63
Helical Gears» Force Analysis
A
B
D
C
FromΔEDCt
r
F
F
=tan
E
and


costancos tan
cos
tan
n
==
==
t
r
t
rr
n
F
F
F
F
BC
F 


tan
tan
cos
n
=

64. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
64
Helical Gears» Force Analysis
The tangential component is calculated from the relationship2
d
M
F
t
t
=
M
t = transmitted torque (N-mm)
d = pitch circle diameter (mm)
Thefollowinginformationisrequiredinordertodecidethedirection
ofthreecomponents
(i)Whichisthedrivingelement?Whichisdrivenelement?
(ii)Ispinionrotatinginclockwiseoranticlockwise?
(iii)Whatishandofhelix?Isitrighthandedorlefthanded?

65. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
65
Helical Gears» Force Analysis

66. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
66
Helical Gears» Force Analysis

67. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
67
Problem4.8:Apairofparallelhelicalgears,5kWpower
at720rpmissuppliedtothepinionAthroughitsshaft.The
normalmoduleis5mmandthenormalpressureangleis
20°.Thepinionhasright-handteeth,whilethegearhasleft-
handteeth.Thehelixangleis30°.Thepinionrotatesinthe
clockwisedirectionwhenseenfromtheleftsideofthe
page.Determinethecomponentsofthetoothforceand
drawafree-bodydiagramshowingtheforcesactingonthe
pinionandthegear.
Helical Gears» Numerical Problem 4.8

68. Helical Gears» Numerical Problem»
Prob. 4.8 solution
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
68
Power(kW)=5
n
A=720rpm
z
A=20
z
B=30
m
n=5mm
ψ=30°
α
n=20°

69. Helical Gears» Numerical Problem»
Prob. 4.8 solution
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
69
Componentsoftoothforce
F
t=tangentialcomponent
F
r=radialcomponent
F
a=axialcomponent
Asweknow()






=
2
A
At
t
d
M
F ()
A
At
n
kWPower
M
2
)(1060
6

= cos
nA
A
mZ
d=

70. Helical Gears» Numerical Problem»
Prob. 4.8 solution
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
70
So,( )
NF
t
6.1148
47.115
56.663142
== ()
()
NmmM
At
56.66314
7202
51060
6
=

=
 ()
mm
mZ
d
nA
A
47.115
30cos
520
cos
===


71. Helical Gears» Numerical Problem»
Prob. 4.8 solution
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
71
So,( )
NF
t
6.1148
47.115
56.663142
== tan
ta
FF= () NFF
ta
14.66330tan6.1148tan === 







=


cos
tan
n
tr
FF NFF
n
tr
73.482
30cos
20tan
6.1148
cos
tan
=








=








=



72. Helical Gears» Numerical Problem»
Prob. 4.8 solution
29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
72
Free-bodydiagramofforces
Thedirectionofthreecomponentson
thegearatthepoint-2willbeopposite
tothatofthepinion.

73. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
73
Helical Gears» Beam Strength
ForBeamstrength,thehelicalgearisconsideredtobe
equivalenttoaformativespurgear.
The pitch circle diameter of this gear is d´, the number of teeth
is z´and the module m
n.
The beam-strength of the spur gear is given by,
Beamstrength(S
b)isthemaximumvalueofthetangential
forcethatthetoothcantransmitwithoutbendingfailure.YmbS
bb
=

74. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
74
Helical Gears» Beam Strength
ForBeamstrength,thehelicalgearisconsideredtobe
equivalenttoaformativespurgear.
The pitch circle diameter of this gear is d´, the number of teeth
is z´and the module m
n.
The beam-strength of the spur gear is given by,
Beamstrength(S
b)isthemaximumvalueofthetangential
forcethatthetoothcantransmitwithoutbendingfailure.YmbS
bb
=

75. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
75
Helical Gears» Beam Strength
The beam-strength of the spur
gear is given by,
S
b=(S
b)
n=beamstrength
perpendiculartothetooth
element
m=m
n=normalmoduleYmbS
bb= cos
b
b= YmbS
bb
= ()


cos
Ybm
S
bn
nb
=

76. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
76
Helical Gears» Beam Strength
Y= Lewis form factor based on
virtual number of teeth z´.YbmS
bnb
= ()


cos
Ybm
S
bn
nb
=
S
bis the component of (S
b)
nin
the plane of rotation. Thus,()cos
nbb
SS=
Inordertoavoidthefailureofgear
toothduetobending,thebeam
strengthshouldbemorethanthe
effectiveforcebetweenthe
meshingteeth.effb
FS

77. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
77
Helical Gears» Effective Load on Gear
Tooth
Whengearsrotateatverylowspeedthetangentialload(F
t)can
beconsideredasactualforcebetweentwomeshingteeth.
Butwhengearsrotateathighspeedthenitisnecessaryto
considerthedynamicforceresultingfromtheimpactbetween
matingteeth.()
d
M
d
M
F
tt
t
2
2







=

78. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
78
Helical Gears» Effective Load on Gear
Tooth
Thedynamicforceisinduceddueto:
(i)Inaccuraciesoftoothprofile
(ii)Errorsintoothspacing
(iii)Elasticityofparts

79. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
79
Helical Gears» Effective Load on Gear
Tooth
Therearetwomethodstoaccountfordynamicload:
(i)Approximateestimationbyvelocityfactor(C
v)inthe
preliminarystagesofgeardesignand
(ii)PrecisecalculationbyBuckingham’sequationinthefinal
stagesofgeardesign

80. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
80
Helical Gears» Effective Load on Gear
Tooth
Inthepreliminarystages,theeffectiveloadF
effbetweentwo
meshingteethisgivenby
C
s= service factor and C
v= velocity factor
Thevelocityfactorforhelicalgearsis(v>20m/s)
Where,visthepitchlinevelocityinm/sv
ts
eff
C
FC
F= v
C
v
+
=
6.5
6.5

81. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
81
Helical Gears» Effective Load on Gear
Tooth
Inthefinalstagesofgeardesign,thedynamicloadiscalculated
byequationderivedbyEarleBuckingham.Thedynamicloadis
givenby,( )
( )
t
t
d
FCebv
FCebv
F
++
+
=


2
2
cos21
coscos21
F
d=dynamicload; v=pitchlinevelocity
e=sumoferrorsbetweentwomeshingteeth
b=facewidthoftooth;F
t=tangentialforceduetoratedtorque

82. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
82
Helical Gears» Effective Load on Gear
Tooth
Theeffectiveloadisgivenby
Where,F
disthedynamicloadoradditionalloadduetodynamic
conditionsbetweentwomeshingteeth.( )
dtseff
FFCF +=

83. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
83
Helical Gears» Wear Strength
Tocalculatewearstrength,pairofhelicalgearsisconsideredto
beequivalenttoaformativepinionandaformativegearina
planeperpendiculartothetoothelement.Thewearstrengthof
thespurgearisgivenby
Wearstrengthisthemaximumvalueofthetangentialforcethat
thetoothcantransmitwithoutpittingfailure.KdbQS
pw
=

84. 29-11-2019
Design of Machine Elements II MEC306
(Dr. L K Bhagi)
84
Helical Gears» Wear Strength