MEL 311 Lecture Gear

8/10/2019 MEL 311 Lecture Gear

1/91

Design of Gears

R. K. Pandey, Ph.D.Associate Professor

Department of Mechanical Engineering

I.I.T. Delhi, New Delhi110 106, India


2/91

Gears are used

to transmit

torque and

angular velocity

in wide varieties

of applications.


3/91

Gearbox


4/91


5/91

Straight Bevel Gear

Bevel Gear: teethare formed on a

conical surface, used

to transfer motionbetween non-

parallel andintersecting shafts.

Spiral Bevel Gear


6/91

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.


7/91

Gearset: Two

gears in mesh

Pinion: Smaller

of two gears


8/91

Nomenclature /Terminology


9/91

Nomenclature /Terminology


10/91

Gear tooth

sizes for

various

diametral

pitches.


11/91

Standard diametric pitch and

corresponding tooth size


12/91

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)


13/91

Torque ratio or Mechanicaladvantage ( )

1/ / /A V in out out in

m m r r

Am


14/91

Pressure angle ()

Angle between

the line of action

(common tangent)and the direction

of velocity at the

pitch point.


15/91

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


16/91

: /cCircular pitch p d N

: /dDiametral pitch p N d

: /Module m d N


17/91

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.


18/91

Interference and undercutting


19/91

NB: Two mating gears must

have the same diametricpitch and pressure angle

().

( )dp


20/91

Gear manufacturing:

1. Forming (casting, molding,

drawing, extrusion etc.)2. Machining (milling, shaping,

hobbing etc.)


21/91


22/91


23/91


24/91

Free body diagrams of pinion and gear

F A l i (H li l G )


25/91

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


26/91

American Gear ManufacturersAssociation (AGMA) has for manyyears been the responsible authority

for the dissemination of knowledgepertaining to the design and analysis

of gearing.


27/91

The Lewis Formula

W. Lewis (1892)

was the first to

present a formulafor computing

the bending

stress in gear

tooth.


28/91


29/91

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


30/91

Bending stress

Lewis equation,

where


31/91

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


32/91

Bending Stress Modifying Factors

Geometry factor J (Table 11-8 to 11-15)


33/91

Geometry factor J


34/91

Dynamic factor (Kv)


35/91

Application factor, Ka


36/91


37/91

Rim thickness factor, KB

KB = -2mB + 3.4 0.5 mB 1.2

KB = 1.0 mB 1.2

Backup ratio (mB)


38/91

Load Distribution factor, Km

AGMA Bending Fatigue Strengths


39/91

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


40/91

Reliability factor, KR


41/91

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.


42/91

KL is life factor

AGMA bending fatigue strengths for selection of gear materials


43/91

g g g g


44/91

AGMA Bending FatigueStrength for the Steels

may be read from Fig.

11-25.


45/91

Surface Durability Analysis

Surface fatigue failure due to

many repetitions of high contact

stresses may evaluated using the

expression for the surface

contact stress.


46/91

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.


47/91


48/91

Stresscomponents

below the

surface of

contactingspheres


49/91

(a) Two cylinders held in contact by force F;

(b) Contact stress at face of contact of width 2b


50/91

Stress

components

below the

surface ofcontacting

cylinders


51/91

Plot of

shear

stressesbelow the

surfaces

for pointand line

contacts


52/91

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)


53/91

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)


54/91

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.


55/91

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)


56/91

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


57/91

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


58/91

AGMA Elastic coefficient, CP


59/91

AGMA recommends using

surface finish factor of 1 for gears

made by conventional methods.

Surface finish factor Cf


60/91


61/91


62/91

For CH refer page

734-735 of Machine

Design book by

Norton.


63/91

Safety factors against bending failure:

Nb pinion = Sfb/b pinion

Nb gear= Sfb/b gear


64/91

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.


65/91

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.


66/91

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.


67/91

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


68/91

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


69/91

continued

Design steps


70/91

Step-1:Number of teeth on gear and pinion:

(Ng)=111, Np=20

Velocity ratio:

mG=Ng/Np=111/20=5.55

Design steps


71/91

Step-2:

Torque on the pinion shaft:

Tp =P/p= 3.0 x 106 / (2 x x8350 / 60)

= 3430 N-m


72/91

Step-3:

Output torque:

Tg= mG x Tp = 5.55 x 3430

= 19036.5 N-m


73/91

Step-4:

Transmitted load will be same on

pinion and gear.

Wt = Tp/(dp/2) = 3430/(0.1319/2)

= 52 x 103N

Step 5:


74/91

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


75/91

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:


76/91

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


77/91

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


78/91

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:


79/91

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


80/91

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


81/91

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:


82/91

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


83/91

Step-15: Pitting geometry factor:

Ipg= cos/{ (1/p +1/g) dpmN pg}

= .935/(50.67 x .1319 x 0.63)

= 0.222


84/91

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.


85/91

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


86/91

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


87/91

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


88/91

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


89/91

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.


90/91

Step-21:

Safety factor against surface failure:

Nc Pinion = (1369/ 1090.5)2 = 1.57

O.K.


91/91

Thank you foryour kind

attention

MEL 311 Lecture Gear

Documents

Transcript of MEL 311 Lecture Gear