Chapter 2 Gear Drive-3

8/10/2019 Chapter 2 Gear Drive-3

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2.7.1 Design principle of contact fatigue strength

To avoid fatigue pitting of gear drive, satisfying

H HP Contact stress at

critical point

Allowable

contact stress

n

H 2 2

1 2

1 2

1

1 1( )

F

L

E E

21

211

n

E

FZ

L

1. Contact stress: Based on Hertz formula

1,2

Poisson ratio of materials;

E1, E2Modulus of elasticity of materials;

+ for outside engagement;

- for inside engagement;

1,2Curvature radii at the contact point;


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--Converting gear engagement to Hertz contact of two cylinder

Variable during engagement.A1

A2

B1

B2

Base circle

1

2

Which point is the critical point

So, different contact points have different values of H.

What is the curvature radius on an involute

spline? Constant or variable?

Low boundary D can be regarded as the

critical point, where His maximum.

C

Pitch point C

D

When only a pair of teeth in mesh,

pitting failures often occur at C.

It will be easier to describe pitch point C by Hertz formula,

so we just take the pitch pointC as the critical point.
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Two radii of the cylinders =

At the pitch point C

''dCN sin2

111

''d

CN sin2

222

d1'

d2'

Dia. of engaging circle;'angle of engagement, for

standard gear which is equal to

(pressure angle on pitch circle);

Ratio of teeth number

'd

'd

z

zu

1

2

1

2

Dia. of circle

'

d'd

cos

cos11

Then, 2 1

1 2 1 2 1

1 1 2( 1)

cos tan

u

d u '

Curvature radii at pitch point C nH E

FZ

L
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nnc KFF cos

2

1

1

d

K T

Length of contact line L

If considering more than one pair

of teeth in mesh, then

bWidth of face;

2

Z

bL

ZContact ratio coefficient;

Z0.850.92, the more teeth, the less value.

Calculated loading Fnc

n

EH

FZ

b

Substituting all of equations above into Hertz formula, then we have

ubd

uKT

'ZZ

2

1

1

2E

)1(2

tancos

2

MPa

)1(22

1

1

HEubd

uK TZZZ

+ for outside engagement;- for inside engagement;
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1

H E H 2

1

2 ( 1)KT u

Z Z Z bd u

Coefficient of local area at pitch point:

Influence of tooth profile curvature radius

(Fig. 2-18)

Coefficient of

elasticity

2N mm

Table 2-15

Yes!

Are the contact

stress on pinion andon gear equal?

Coefficient of contact ratio

b

FZ

n

EH

Fig. 2-18 ZH Helical angle /()

Z0.850.92, the more number of

teeth, the less value of Z.


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Table 2-15 Coefficient of elasticity, ZE

Pinion GearZE/

Material E1/Nmm-2 Material E2/Nmm

-2

Steel 2.06105

Steel 2.06

105

189.8Cast steel 2.02105 188.9

Nodular cast iron 1.73105 181.4

Grey iron 1.18-1.26105 162-165.4

Cast steel2.06

105Cast steel 2.02105 188


Grey iron 1.18-1.26105 161.4

Nodular cast iron1.73105


Grey iron 1.18105 156.6

Grey iron1.18-

1.26105Grey iron 1.18-1.26105 143.7-146.7

2N mm


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2. Allowable contact stress HP

Contact fatigue limit by experiments

Life coefficient of contact strength, (Fig. 2-32)

Minimum safety coefficient of contact strength,

(Table 2-17)

Hlim

HP N

Hmin

Z

S

Contact fatigue limit Hlim (Fig. 2-23 - Fig. 2-26)

MLLow quality of heat treatment and material properties

MQMedium quality of heat treatment and material properties

MEExcellent quality of heat treatment and material properties

Table 2-17 Recommendation of minimum safety coefficient

Requirements of

application

Minimum safety coefficient

SFmin SHmin

Very high reliability 2.00 1.50-1.60

High reliability 1.60 1.25-1.30

Medium reliability 1.25 1.00-1.10

Low reliability 1.00 0.85


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Fig.2-23 Hlim , Steel and cast

steel, Normalizing heat treatment

Steel, Normalizing

Cast steel, Normalizing

Fig.2-24 Hlim , Hardening &

Tempering heat treatment

Alloy steel

Steel

Alloy cast steel

Cast steel


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Fig.2-25 Hlim , Cast iron

Nodular cast ironMalleable cast-iron Grey iron


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Fig.2-26 Hlim , Hardening & Carburizing

Alloys steel,

carburizing

Flame or induction

Hardening

HV150 55 60 65HRC


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Life coefficient ZNFig. 2-32

Rotational speed r/min

Design life in hours h

Number of load applications

per revolution

Number of load cycle N

natN 60

Hlimobtained by experiments of infinite load cycles. If gear is serviced

in a finite life, the contact fatigue limit should be increased, ZN > 1

One-sided load, H is fluctuating Two-sided load, H is reverse and

repeated.

Driving Gear

a =2 a =1

Mating with more than one

gearHow to decidea

a =1a =1

Driving gear


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1Pitting allowable: Structural steel, H & T steel, Nodular cast iron (pearlite, bainite),

malleable cast-iron (pearlite) ;

2No Pitting: Structural steel, H & T steel, Flame/induction hardening steel, nodular cast

iron, nodular cast iron (pearlite, bainite), malleable cast-iron (pearlite) ;

3Grey iron, nodular cast iron (ferrite), Nitriding steel, H&T steel, Carburizing steel;

4Carbonitriding H&T steel, carburizing steel.

ZN

Fig. 2-32 Life coefficient


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MPa)1(2

HP2

1

1HEH

ubd

uKTZZZ

Checking formula of contact fatigue strength for gear drive:

for checking

2H E 13

1

HP d

2 1( ) mm

Z Z Z KT ud

u

for design

Maybe HP1doses not equals to HP2so the smaller one should be taken

account in design.

Order that,

1

d

d

b

Face width coefficient

(Table 2-14 )

Calculating formula of contact fatigue strength for gear drive:

ArrangementFace width coefficient

Soft surface Hard surface

symmetrical 0.8-1.4 0.4-0.9

asymmetrical 0.6-1.2 0.3-0.6

Cantilever 0.3-0.4 0.2-0.25

Table 2-14 Face width coefficient, d


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Attentions

The contact fatigue strength mainly depends on d,the Dia. of pitch circle.

d, Contact fatigue strengthH ,

bshould be proper: if b too much, unbalanced load

Module mhas nothing with the contact fatigue strength. d1m z1

For the two engaged gears, H1H2.

Find exact face width coefficient din table 2-14.

But HP1 HP2

So in gear drive, the contact strength may be different.

A smaller HP means poor contact strength.

We should design the contact strength by the smaller HP.

MPa)1(2

HP2

1

1HEH

ubd

uKTZZZ


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First calculating d1

2H E 13

1

HP d

2 1( ) mm

Z Z Z KT ud

u

Choosing z1 Calculatingm = d1/z1 Calculatinga = m(z1+z2)/2

1. Module mshould be basic value by recommendation , and m 1.5 for power

transmission

2. d1and d2should be re-calculated accurately by module m, to three decimal

places

3. Center distance a should be an integer, easy to manufacture and examine;

4. If a is not an integer, it is should be rounded up by modifying the tooth

profile.

Attentions


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Measures to upgrade the contact fatigue strength

Increasing d,dia. of pitch circle or a, center distance

Increasing bor d

Positive modified tooth profile

Hlim

HP

Hmin

NZ

S

Better materials

Upgrading precision of gear drive

Increasing hardness by advanced heat treatment

Dose the width of pinion equals to width of gear?

HP2

1

1HEH

)1(2

ubd

uKTZZZ

b1= b2+ (5~10) mm b2= dd1

b1=b2 b1>b2


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2.7.2 Design principle of bending fatigue strength

a) Mechanics modelCantilever

b) Critical section 300tangential line

F FP

To avoid tooth breakage by bending fatigue, satisfying Fn

Bending stress at

critical pointAllowable

bending stress

Rectangular section:

width (tooth width ) SF, length b

1. Bending stress F

c) Load position to produce the maximum bending stress

Upper boundary of meshing area, D

300

M N

SF

B

D

SFb

E

Base

circle


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For the error of manufacturing and assembling, for the gear drive with medium

precision, contact ratio can be regarded as 1.

Fn

Tangential

component

Radial

component

Bending stress F

Shearing stress t

Pressure stress y

Very small, ignoring them,

and adding some coefficient

to compensate

Ignoring the friction force

Fn

FncosF

FnsinF

F

F

y

t

So, all loads are acting on only one pair of teeth, and consider the addendum as theacting point of load.

E

Base

circle


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Bending stress:

W

MF

n F FcosF hW

F

1 FF 2

1

2cos

cos / 6

KT h

d bS

mSF

1 F

2

1

2 6 cos

cos

KT

bd m

Fn

FncosF

F

SF

hF

mhF

F1

2

1

6 cos2

cos

mKT

bd m

1Fa

1

2KTY

bd m

, , scaling coefficients

YFa

SF b

Base

circle


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Coefficient of tooth profile YFa

F only depend on the tooth profile

So,YFa depends on number of teeth and

modification coefficient, independent with module m.

FFa 2

6 cos

cosY

Y

Fawith respect to modification coefficient:

YFa with respect to number of teeth:

Find exact YFafrom Fig. 2-20.

z

YFa

xYFa

Fi 2 20 C ffi i t f t th fil Y
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Fig. 2-20 Coefficient of tooth profile YFaLimitation of undercutting

Limitation of topping

C id i h i t t d t t ti t


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Considering shearing stress, pressure stress and stress concentration at

the fillets, coefficient of stress modification YSaFig. 2-21

Coefficient of stress modification


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1

F Fa Sa

1

2 MPa

KTY Y Y

bd m

Considering contact ratio, coefficient of contact ratio Y

Y0.650.85A larger number of teeth and a larger

contact ratio, a smaller Y

Then, the bending stress at tooth root

Dose the bending stress at tooth root of pinion equal to gear?

1F1 Fa1 Sa1

1

2KTY Y Y

bd m

1F2 Fa 2 Sa 2

1

2KTY Y Y

bd m

F1 F2

Fa1 Fa 2

Y Y

z1


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Only when ,

the bending strength of the pinion equals to that of the gear.

Bending fatigue limit by experiments

Life coefficient of bending strength (Fig. 2-32)

Minimum safety coefficient of bending

strength, Table 2-17

If under reverse and repeated bending stress , Flim should be multiplied by 0.7.

Flim ST

FP

Fmin

N

Y

YS

Stress modification coefficient of test gear, YST=2

Bending fatigue limit Flim

(Fig. 2-28 -- 2-30)

2. Allowable bending stress,FP

Does the allowable

bending strength of the

pinion equal to that ofthe gearSo, the bending strength of the pinion does

NOT equal to that of the gear.

FP1 FP2

Fa1 Sa1 Fa2 Sa2

FP1 FP2

Y Y Y Y

F1 F2


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Fig.2-28 Flim , Cast steel and structure Steel , Normalizing heat treatment

Structure steel ,

Normalizing heat treatment

Cast steel ,

Normalizing heat treatment

Fi 2 29 C i


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Fig.2-29 Flim , Cast iron

Nodular cast iron Malleable cast-iron

Grey iron


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Fig.2-30 Flim , Hardening & tempering

Alloy steel

Carbon steel

Steel,

Hardening & tempering

Cast steel,

Hardening & tempering

Alloy steel

Carbon steel


29/41


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Fig.2-31 Flim , Hardening

Steel,

Case Hardening


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3. Attentions of bending strength for gear drive

MPa2

FPSaFa

d

2

1

3

1F

YYY

zm

KT

Module meffects the bending strength most

m Bending strengthTooth widthSF Sectional area F

Different bending stresses on mating teeth: F1F2

For standard gear YFa1YSa1YFa2YSa2 So, F1F2

Because F1F2and the pinion will have a larger number of cycles of load,

the material of pinion should have better properties and harder surface.

Comparing

Just considering the larger value to design module

But, for checking, bending strength of the two gear should be calculatedseparately.

FP1

Sa1Fa1

YY

FP2

Sa2Fa2

YYwith


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If one-sided load, F is fluctuating; If two-sided load, F is reverse and

repeated.

Teeth number of pinion

Given that Dia. of pitch circle dgiven a, i

z1 YFaYSa

m

F

F F

z1

m

smooth movement

h to save time of manufacturing

Tipunder satisfying the

bending strength, teeth

number of pinion can be

as large as possible

Soft surface, closed gear drive: z1=20~40

Hard surface, closed gear drive or open gear drive: z1=17~25.

If two-sided load, how to decide F?


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Measures to improve the bending strength

m

Increasing bproperly

Increasing the precision grade of gear drive, and the radius of

fillet at tooth root.

A better material

Upgrading heat treatment and increasing the hardness of

tooth surface .

F

FP

Positive modification tooth profile to increase the tooth width

MPa2

FPSaFa

1

1F YYY

bmd

KT

the most significant geometrical factor

1. Choose material and heat treatment


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Design of spur gear

2.Compute HP, and FP

3. choose: z1, z2, d,

Find K, ZE, ZH

Closed?

3. Choose: z1, z2,d,

Find K, YFa, YSa

2H E 13

1

HP d

2 1( )

Z Z Z KT ud

u

Fa Sa13

2

d 1 FP

2 Y Y YKTm

z

4. Calculate

standard value11 zdm 4. Rounded up to standard value

5. Compute a, b1, b2, d1, d2,

da1, da2, df1, df2

5. Computer a, b1, b2, d1, d2, da1, da2,df1, df2

6. Find Yfa, Ysa, Check bending fatigue strength

6. Find ZE, ZH, check contact fatigue strength

Just considering

bending strength.

Increasing

module 10%.

Y

N

Soft?NY

1F Fa Sa FP

1

2KTY Y Y

bd m

1H E H HP2

1

2 ( 1)KT uZ Z Z

bd u


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3. choosez1, z2, dFind K, ZE, ZH

Pinion: z1=31, z2=115, K=1.5, ZE=188.9 (Table 2-15),

ZH= 2.5, Z=0.87,d=0.8 (Table 2-14 )

52 2H E 1 331

HP2 d

2 1 2.5 188.9 0.87 2 1.5 2.18 10 3.7 1( ) ( ) 84.7mm537 0.8 3.7

Z Z Z KT udu

4. Calculate module, standard value 1 1 85 31 2.83m d z

3m Table 2-4

5. Compute a, b1, b2, d1, d2, da1, da2, df1, df2(2.3 )

1 1 3 31 93mmd mz 2 2 3 115 345mmd mz

2 84.7 0.8 68mmb 1 2 5 73mmb b

1 2 2 93 345 2 219 mma d d

1 1 2 93 2 1 3 0.25 3 85.5mmf ad d h c

2 2 2 345 2 1 3 0.25 3 337.5mmf ad d h c

1 22 2 2 219 337.5 2 0.25 3 99mm

a fd a d c

2 12 2 2 219 85.5 2 0.25 3 351mma fd a d c

Y Fi 2

20


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6. Find YFa, YSa, Check bending fatigue strength

Pinion: YFa1=2.45, YSa1=1.65;

Gear: YFa2=2.2, YSa2=1.72.

5

1F1 Fa1 Sa1 FP1

1

2 2 1.5 2.18 102.45 1.65 0.85 118MPa

68 93 3

KTY Y Y

bd m

Y0.650.85

Fa2 Sa2F2 F1 FP2

Fa1 Sa1

111MPaY Y

Y Y

YFa, Fig. 2-20

YSa, Fig. 2-21

So, this design is safe.

Homework 8


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Problem:

1. A double speed reducer with spur gears, working in

unidirectional situation, endures moderate shock.

2. Service life: 10 years.

3. Speed ratio of the first stage gear drive i=3.7.

4. Inputting rotational speed n1=745r/min.

5. Transmission power P=17KW.6. Hard surface is recommended.

7. Try to design the first stage gear drive.

Homework-8

Tips: For hard surface gear drive, First design by Bending strength, thencheck by Contact strength.

Chapter 2 Gear Drive-3

Documents

Transcript of Chapter 2 Gear Drive-3