mel 311 lecture gear
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Design of Gears
R. K. Pandey, Ph.D.Associate Professor
Department of Mechanical Engineering
I.I.T. Delhi, New Delhi110 106, India
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Gears are used
to transmit
torque and
angular velocity
in wide varieties
of applications.
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Gearbox
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Straight Bevel Gear
Bevel Gear: teethare formed on a
conical surface, used
to transfer motionbetween non-
parallel andintersecting shafts.
Spiral Bevel Gear
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Worm gear set: consists
of a helical gear and a
power screw (worm), usedto transfer motion between
non-parallel and non-
intersecting shafts.Worm
Gear
Rack and Pinion set: a
special case of spur gears
with the gear having aninfinitely large diameter,
the teeth are laid flat.
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Gearset: Two
gears in mesh
Pinion: Smaller
of two gears
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Nomenclature /Terminology
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Nomenclature /Terminology
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Gear tooth
sizes for
various
diametral
pitches.
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Standard diametric pitch and
corresponding tooth size
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Fundamental law of gearing
Angular velocity ratio betweengears of a gearset must remain
constant throughout the mesh/ /
V out in in out m r r
VR (velocity ratio) =
(p / g) = (dg / dp) = (Ng / Np)
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Torque ratio or Mechanicaladvantage ( )
1/ / /A V in out out in
m m r r
Am
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Pressure angle ()
Angle between
the line of action
(common tangent)and the direction
of velocity at the
pitch point.
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Pressure angle ()
1. Standard values are14.50 , 200 and 250 .
2. 200 most commonly
used.
3. 14.50 is now obsolete.
14.50
200 250
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: /cCircular pitch p d N
: /dDiametral pitch p N d
: /Module m d N
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Interference and undercutting
The involute tooth form is onlydefined outside of the base circle.
Portion of tooth
below the basecircle will not be
involute and will
interfere with tip
of the tooth on the
mating gear.
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Interference and undercutting
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NB: Two mating gears must
have the same diametricpitch and pressure angle
().
( )dp
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Gear manufacturing:
1. Forming (casting, molding,
drawing, extrusion etc.)2. Machining (milling, shaping,
hobbing etc.)
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Free body diagrams of pinion and gear
F A l i (H li l G )
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Force Analysis (Helical Gears)
t= pressure angle (20o or 25o)
= helix angle (10, 20, 30, or 40o)
n = normal pressure angle
= helix angle
t = tangential pressure angle
tan n = tan t cos
Wr= W sin nWt = W cos
n
cos
Wa = W cos n sin
Where W = total force
Wr= radial component
Wt = tangential component (transmitted load)
Wa = axial component (thrust load)
Wr= Wt tan t
Wa = Wt tan
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American Gear ManufacturersAssociation (AGMA) has for manyyears been the responsible authority
for the dissemination of knowledgepertaining to the design and analysis
of gearing.
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The Lewis Formula
W. Lewis (1892)
was the first to
present a formulafor computing
the bending
stress in gear
tooth.
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1. The load is applied to the tip of a single
tooth.
2. The radial component of the load, Wr , isnegligible.
3. The load is distributed uniformly across the
full face width.4. Stress concentration in the tooth fillet is
negligible.
Assumptions made in deriving Lewis
equation
B di
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Bending stress
Lewis equation,
where
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Modifications in Lewis equation according to
AGMA standards (American Gear Manufacturers
Association)
Wt
tangential transmitted loadKa application factor (accounts shocks)
KV dynamic factor (account for internally generated vibration)
KS size factor (refer fatigue concept)
KI Idler factor
Pd
transverse diameteral pitch
F face width of the narrower member
Km load-distribution factor (accounts axial misalignment)
KB rim-thickness factor (gear has rim and spokes)
J geometry factor for bending strength
which includes root fillet stress concentration factorKf
Modified Lewis
equation
B di St M dif i F t
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Bending Stress Modifying Factors
Geometry factor J (Table 11-8 to 11-15)
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Geometry factor J
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Dynamic factor (Kv)
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Application factor, Ka
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Rim thickness factor, KB
KB = -2mB + 3.4 0.5 mB 1.2
KB = 1.0 mB 1.2
Backup ratio (mB)
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Load Distribution factor, Km
AGMA Bending Fatigue Strengths
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AGMA Bending Fatigue Strengths
for Gear Materials
Sfb is the corrected strength
KL is life factor
KT is the temperature factor
KR
is the reliability factor
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Reliability factor, KR
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Temperature factor, KT
AGMA recommends using temperature factorof 1 for operating temperatures (lubricant
temperature) up to 250oF. For higher
temperatures it can be estimated from:
KT=(460+TF)/620
This equation is valid for steel made gears.
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KL is life factor
AGMA bending fatigue strengths for selection of gear materials
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g g g g
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AGMA Bending FatigueStrength for the Steels
may be read from Fig.
11-25.
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Surface Durability Analysis
Surface fatigue failure due to
many repetitions of high contact
stresses may evaluated using the
expression for the surface
contact stress.
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Contact Stresses
Two bodies having curved
surfaces are pressed together then
point/line contact changes to areacontact and the stresses
developed in the two bodies willbe 3-D.
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Stresscomponents
below the
surface of
contactingspheres
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(a) Two cylinders held in contact by force F;
(b) Contact stress at face of contact of width 2b
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Stress
components
below the
surface ofcontacting
cylinders
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Plot of
shear
stressesbelow the
surfaces
for pointand line
contacts
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Surface Durability Analysis
maxp = 2W/( b l)
Where,
= largest surface pressure
W = force pressing the two cylinders
= length of cylinders
maxp
l
2W/( b l)
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1/ 22 2
1 1 1 1
1 2
(1 ) / (1 ) /2
1/ 1/
E EWb
l d d
1 222 2
1 1 1 1
1/ 1/
cos (1 ) / (1 ) /
t
c
d dW
F E E
maxp = 2W/( b l)
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Proper lubricating system
can minimize the surface
damage due to wear andcorrosion. But, surface
fatigue can occur even with
proper lubrication and itsthe most common mode of
gear failure and is
characterized by pitting and
spalling of the tooth
surface. The damage is
caused by repeated contact
stresses.
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1/ 21 22 2
1 1 1 1
1/ 1/
cos (1 ) / (1 ) /
t
c
d dW
F E E
Stress at concentrated line contact
(Buckingham equation)
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AGMA Surface Stress Equation
Cp elastic coefficient, (lb/in2)0.5
Wt
transmitted tangential load
Ca overload factor (same asKa)
Cv dynamic factor (same asKv)
Cs size factor (same asKs)
Cm load-distribution factor (same asKm)
Cf surface condition factord pitch diameter of thepinion
F face width of the narrowest member
I geometry factor
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Surface Geometry Factor (I)
This factor takes into account the radiiof curvature of the pinion/gear teeth
and the pressure angle. AGMS provides
equations for I for different type of
gears. Please refer pages 724/725 and
761/764 of the main text book forprocedures of calculation of Ifor spur
and helical gears, respectively.
AGMA Elastic coefficient C
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AGMA Elastic coefficient, CP
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AGMA recommends using
surface finish factor of 1 for gears
made by conventional methods.
Surface finish factor Cf
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For CH refer page
734-735 of Machine
Design book by
Norton.
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Safety factors against bending failure:
Nb pinion = Sfb/b pinion
Nb gear= Sfb/b gear
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Safety factor against surface failure:
Nc pinion-gear= (Sfc/c pinion)2
The safety factor against surface failure is found by
comparing the actual load to the load that would produce a
stress equal to the materials corrected surface strength.
Because surface stress is related to the square rot of theload, the surface fatigue safety factor can be calculated as
the quotient of the square of the corrected surface strength
divided by the square of the surface stress for each gear in
the mesh.
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Please go through thecase study-7C printed on pages
741-747 of the text book(Machine DesignAn Integrated
Approach written by R. L.
Norton. This will bring clarity
related to the gear design.
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Exercise:
Following figure shows photographic and
schematic views of a reducer (a helical
gear set). The reducer connects a steam
turbine and an alternator in a power plant.
Shafts of the turbine (corresponds to
pinion shaft) and alternator (correspondsto gear shaft) rotate at 8350 rpm and 1500
rpm, respectively.
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continued
A need arises to determine the fatigue
safety factors of the pinion and gear teeth
for the data mentioned below:Centre distance = 432 mm
Power to be transmitted = 3 MW
Plant operation= 3 shifts of 8 hoursExpected life of the reducer= 20 years
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continued
Assume involute teeth profiles and
number of teeth on the pinion and gear 20
and 111, respectively. AGMA standardfull depth teeth may be used in design.
Both pinion and gear are made of same
material and the mating teeth surfaceshave equal hardness 60 HRC.
continued
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continued
Design steps
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Step-1:Number of teeth on gear and pinion:
(Ng)=111, Np=20
Velocity ratio:
mG=Ng/Np=111/20=5.55
Design steps
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Step-2:
Torque on the pinion shaft:
Tp =P/p= 3.0 x 106 / (2 x x8350 / 60)
= 3430 N-m
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Step-3:
Output torque:
Tg= mG x Tp = 5.55 x 3430
= 19036.5 N-m
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Step-4:
Transmitted load will be same on
pinion and gear.
Wt = Tp/(dp/2) = 3430/(0.1319/2)
= 52 x 103N
Step 5:
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Step-5:
Velocity factor (Kv):
Pitch line velocity (Vt)
= (dp/2)p= (0.1319/2) x (2 x x 8350/60)
= 57.66 m/s
Kv = { 78/ [78+ (200 x Vt)0.5]}0.5
=0.648
St 6
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Step-6:
Various factors:
Size factorKs = 1.0
Rim thickness factorKB = 1.0 (solid-disk gears)
Load distribution factorKm= 1.8
Application factorKa = 1.25 (Moderate shock)
Idler factorKI= 1.0 (non idler case)
Bending geometry factorJpinion = 0.428
Step 7:
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Step-7:
Pinion-tooth bending stress:
bp = {(Wtxpd)/ (FxJ)} x (Ka.Km.Ks.KB.KI/Kv)
= {(52050 x 151.63)/(.190 x 0.428)}
x (2.25/0.65)
= 335.95 MPa
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Step-8: Gear tooth bending stress:
bg= {(Wtxpd)/ (FxJ)} x (Ka.Km.Ks.KB.KI/Kv)
={(52050 x 151.63)/(.190 x 0.61)} x(2.25/0.65)
=235.71 MPa
h f i
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Step-9: Length of action:
Zpg= {(rp+ap)2(rp cos)2}0.5 +{(rg+ag)2
(rgcos)2}0.5 -Cpgsin
= { (0.0723)2(0.06595 x cos(20.7))2}0.5 +
{ (0.3723)2(0.366 x cos(20.7))2}0.5 -0.432
x sin (20.7)
=0.0377 + 0.14620.1527 = 0.0311 m
Step-10:
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Step-10:
Transverse contact ratio:
mpg=pdxZpg / (3.14 x cos (20.7)) =1.6
Step-11: Axial contact ratio:
mF= Fxpdx tan/3.14 = 0.190 x 151.63 x
0.28/3.14 =2.57px =ptcot = 0.0207 x cot 15.67 = 0.0737 m
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Step-12: Normal pressure angle and
helix angle:
n = 200 , b = 14.70
St 13 Mi i l th f th li f t t
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Step-13: Minimum length of the lines of contact
for mesh:
nrpg= Fractional part of mppg= 0.6
na=Fractional part of mF = 0.57
Lmin pg= {mp pgF(1-na)(1-nr pg)px}/ cosb
={1.6 x 0.19 - (1- 0.57) (1- 0.6) x 0.0737}/cos 14.7
= 0.301 m
m N pg=F/Lmin pg= .190/.301 = 0.63
Step-14:
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Step 14:
Radii of curvature of teeth:
p = {{ 0.5 [(rp+ap)+(Cpg-rg-ag)]}2 -(rp cos)
2}0.5
={0.00435 - .003805} 0.5 = 0.0233 m
g= Cpgsin -p = 0.432 sin (20.7)- 0.0233= 0.129 m
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Step-15: Pitting geometry factor:
Ipg= cos/{ (1/p +1/g) dpmN pg}
= .935/(50.67 x .1319 x 0.63)
= 0.222
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Step-16: The elastic coefficient:
Cp = {3.14 x [(1-2p)/Ep +(1-
2g)/Eg]}
-0.5
= 191.63
Material of pinion and gear is same.
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Step-17: Surface stress at mesh:
c p = Cp { (WtCaCmCs Cf)/ (F IpgdpCv)}0.5
= 191.63 {(52050 x1.25 x1.8x1.0x1.0)/(0.19x.222x0.1319x.65)}0.5
=1090.5 MPa
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Step-18: Corrected bending-fatigue strength:
Sfb = 6235 + 174HB0.126HB2
=6235 + 174 x 6000.126 x (600) 2
= 65275 x 6890= 450 MPa
This value needs to be corrected for certain
factors. Service life = 20 years continuousrun.
Operating temperature= 700C
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p g p
No. of cycles during service =8350 x 20 x 365 x 24 x60
=8.77 x1010
Life factorKL= 1.3558 (8.77 x1010) -0.0178 = 0.86
Temperature factorKT= 1.0Reliability factorKR = 1.0
Corrected bending fatigue strength:Sfb = (KLxSfb)/ (KTxKR)
=(0.86 x 450)/(1x1) =387 MPa
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Step-19: Corrected surface-fatigue strength:
Sfc = 27 000 + 364 xHB= 27000 + 364 x
600 =245400x6890= 1690 MPa
CL = Life factor = 1.44(8.77 x10
10
)
-0.023
=0.81CT=1.0
CR=1.0
CH=1.0Sfc = CL x CHx Sfc /(CTxCR) = 0.81 x 1690
= 1369 MPa
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Step-20:
Safety factor against bending failure:
Nb pinion = 387/335.95= 1.15 O.K.
Nb gear= 387/ 235.71 = 1.64 O.K.
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Step-21:
Safety factor against surface failure:
Nc Pinion = (1369/ 1090.5)2 = 1.57
O.K.
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Thank you foryour kind
attention
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