design of spur and helical gears - qltien.com

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...... Design of Spur and Helical Gears

Ratna Kumar Annabattula

208, Machine Design sectionDepartment of Mechanical Engineering

IIT Madras

Office hours: Wednesday, 3:00 p.m. - 4:00 p.m.email: [email protected]

Ratna Kumar (IIT Madras) Design of Gears July-Nov 2018 1 / 88

[email protected]

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Outline I...1 The Lewis Bending Equation...2 Surface Durability...3 AGMA Stress Equations...4 AGMA Strength Equations...5 Correction Factors...6 Finding St and Sc

Finding StFinding Sc

...7 ALL THE FACTORSDynamic factor Kv

Kv factorGEOMETRY FACTORS (I and J) or (ZI and YJ)

J(YJ)I(ZI)

Overload factors (Ko)Load-Distribution factor (Km(KH))


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Outline IIFinding Cp f

Finding Cpm

Finding Cma

Finding Ce

Size factor KsRim thickness factor KBElastic Coefficient Cp(ZE)Surface condition factor C f (ZR)Temperature factor KT(Yθ)Safety factors (SF and SH)Reliability factor (KR(YZ))Stress cycle factors (YN(ZN))Hardness ratio factor (CH(Zw))Flowchart of spur gear design for dendingFlowchart for spur gear design for wear

...8 Spur gear Problem. 1Ratna Kumar (IIT Madras) Design of Gears July-Nov 2018 3 / 88

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Outline III...9 Spur gear Problem 2

...10 Guidelines for design of a gear meshParallel Helical GearsForce Analysis - Helical Gears

...11 Bevel Gear - Force Analysis


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The Lewis Bending Equation

The Lewis Bending Equation I

..

Wilfred Lewis intorduced an equation in 1892, for estimating thebending stresses in gear teeth which still remains the basis formost of the gear design today.


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The Lewis Bending Equation II

From similar triangles,

t/2x

=ℓ

t/2=⇒ ℓ =

t2

4x

Using the concept of a rectangular cantilever beam

σ =MI/c

=6Wtℓ

Ft2,

where I/c is the section modulus.By rearranging the above equation,

σ =6Wtℓ

Ft2=

Wt

F1

t2/6ℓ


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The Lewis Bending Equation III

Substituting for the value of ℓ and using y = 2x/(3p), with p beingcircular pitch, the above equation can be written as

σ =Wt

Fpy

Here, y is called Lewis form factorIn the above equation, some times, it is preferred to havediametral pitch instead of circular pitch. Hence, by replacingp = π/P and y = πY

..σ =

WtPFY

, where Y =2xP3

.


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The Lewis Bending Equation IV

Figure 1 : Values of the Lewis Form Factor Y for a normal pressure angle20◦ and full-depth teeth.


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Dynamic Effects I

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Dynamic factor Kv has been defined as the reciprocal of that usedin previous AGMA standards. It is now greater than 1.0. Inearleir, AGMA standards it was less than 1.0.

Velocity factor in terms of the current AGMA standards is givenby C. G. Barth as (V is the pitch-line velocity in ft/min)

Kv =600 + V600

(cast iron, cast profile)

Kv =1200 + V1200

(cut or milled profile).


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Dynamic Effects II

AGMA equation for velocity factor

Kv =50 +

√V

50(hobbed or shaped profile)

Kv =

√78 +

√V

78(shaved or ground profile)


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Dynamic Effects III

The above equations for velocity factor in SI units (V is in m/sec)are given by

Kv =3.05 + V3.05

(cast iron, cast profile)

Kv =6.1 + V6.1

(cut or milled profile)

Kv =3.56 +

√V

3.56(hobbed or shaped profile)

Kv =

√5.56 +

√V

5.56(shaved or ground profile)


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Dynamic Effects IV

Effect of velocity factor on Lewis equation

..σ =

KvWtPFY

J

The metric version of the Lewis equation with velocity factor is givenby

..σ =

KvWt

FmY, J

where face width F and module m are in mm, Wt is in N leading tostress units of MPa.


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Dynamic Effects V

..

The above two equations form the basis for AGMAapproach to bending strength of the gear teeth.For spur gears, F should be 3 to 4 times the circular pitch asa general rule.


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Example-1

A stock spur gear is available having a module of 4 mm, a 44 mm face,16 teeth, and a pressure angle of 20◦ with full-depth teeth. Thematerial is AISI 1020 steel in as-rolled condition. Use a design factornd = 3.5 to rate the horse power output of the gear corresponding to aspeed of 25 rev/s and moderate applications.

..

Note: For AISI-1020 steel, the tensile strength (Sut) is 379 MPaand the yield strength (Sy) is 206 MPa.


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Solution-1

Allowable stress σall = Sy/nd = 206/3.5 = 58.86 MPa

The pitch diameter d = Zm = 16(4) = 64 mmThe pitch-line velocity is V = πdN = π(64/1000)(25) = 5.0265 m/sThe velocity factor Kv = (6.1+ V)/6.1 = (6.1+ 5.0265)/6.1 = 1.824

Form factor Y for 16 teeth is 0.296Lewis equation

Wt =FmYσall

Kv=

(44)(4)(0.296)(58.86)

1.824= 1681 N.

The transmittable power is H = WtV = 1681(5.0265) = 8450 W


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Solution-1

Allowable stress σall = Sy/nd = 206/3.5 = 58.86 MPaThe pitch diameter d = Zm = 16(4) = 64 mm

The pitch-line velocity is V = πdN = π(64/1000)(25) = 5.0265 m/sThe velocity factor Kv = (6.1+ V)/6.1 = (6.1+ 5.0265)/6.1 = 1.824


Wt =FmYσall

Kv=

(44)(4)(0.296)(58.86)

1.824= 1681 N.



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Solution-1

Allowable stress σall = Sy/nd = 206/3.5 = 58.86 MPaThe pitch diameter d = Zm = 16(4) = 64 mmThe pitch-line velocity is V = πdN = π(64/1000)(25) = 5.0265 m/s

The velocity factor Kv = (6.1+ V)/6.1 = (6.1+ 5.0265)/6.1 = 1.824


Wt =FmYσall

Kv=

(44)(4)(0.296)(58.86)

1.824= 1681 N.



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Solution-1

Allowable stress σall = Sy/nd = 206/3.5 = 58.86 MPaThe pitch diameter d = Zm = 16(4) = 64 mmThe pitch-line velocity is V = πdN = π(64/1000)(25) = 5.0265 m/sThe velocity factor Kv = (6.1+ V)/6.1 = (6.1+ 5.0265)/6.1 = 1.824


Wt =FmYσall

Kv=

(44)(4)(0.296)(58.86)

1.824= 1681 N.



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Solution-1


Form factor Y for 16 teeth is 0.296

Lewis equation

Wt =FmYσall

Kv=

(44)(4)(0.296)(58.86)

1.824= 1681 N.



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Solution-1



Wt =FmYσall

Kv=

(44)(4)(0.296)(58.86)

1.824= 1681 N.



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Example-2

Estimate the horse power rating of the gear in the previous examplebased on obtaining infinite life in bending.


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Surface Durability

Surface Failure I

Failure of surafces of gear teeth is called wear.Pitting (fatigue of high contact stresses), scoring (lubricationfailure) and abrasion (foreign material)Hertz theory to obtain the surface contact stress between twocylinders

pmax =2Fπbℓ

,

where pmax is the largest surface pressure, F is the force pressingthe two cylinders together, ℓ is the length of the cylinders and b isthe half-width given by

b =

[2Fπℓ

(1 − ν21)/E1 + (1 − ν22)/E2

1/d1 + 1/d2

]1/2Ratna Kumar (IIT Madras) Design of Gears July-Nov 2018 17 / 88

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Surface Durability

Surface Failure IIThe above equations for the notion used in gears may be obtainedby replacing the quantities as follows

F → Wt/ cosϕd → 2rℓ → face width F

pmax → σC(surface compressive stress)

Therefore, we get

..

σ2C =Wt

πF cosϕ1/r1 + 1/r2

(1 − ν21)/E1 + (1 − ν22)/E2, (1)

where r1 and r2 are the instantaneous radii of curvature of pinionand gear tooth profiles, respectively.


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Surface Durability

Surface Failure IIIThe equation for Hertzian stress doesn’t account for sliding butonly for rolling.In the gear contact only the contact at pitch point is under purerolling. Elsewhere, the motion is a combination of rolling andsliding.The radii of curvature of the tooth profiles at the pitch point are

r1 =dP sinϕ

2, r2 =

dG sinϕ2

To smplify the equation for compressive stress (Eq. 1), we definean elastic coefficient CP as

CP =

1

π(1−ν2P

EP+

1−ν2GEG

)1/2


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Surface Durability

Surface Failure IV

Hence,

..

σC = −CP

[KvWt

F cosϕ

(1

r1+

1

r2

)]1/2(2)

Negative sign is to indicate compressive stress.


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Surface Durability

Example-3

The pinion in Example-1 and Example-2 is to be mated with a50-tooth gear manufactured of ASTM No. 50 cast iron. Using thetangential load of 1700 N, estimate the factor of safety of the drivebased on the possibility of surface failure.

..

Material Properties

.

EP = 207 GPaνP = 0.292

EG = 100 GPaνG = 0.211


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Surface Durability

Solution-3 I

The elastic coefficient CP

CP =

1

π(1−0.2922207(109)

+ 1−0.2112100(109)

)1/2

= 150927.3

From the Example-1, we have pitch diameter of the pinion asdP = 64 mm.The pitch diameter of the gear is dG = 50(4) = 200 mm.Radii of curvature at the pitch points is

r1 =64 sin 20◦

2= 10.94 mm, r2 =

200 sin 20◦

2= 34.2 mm


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Surface Durability

Solution-3 II

Face width F = 44 mm and using Kv = 1.824 from Example-1,

σC = −CP

[KvWt

F cosϕ

(1

r1+

1

r2

)]1/2σC = −150927.3

[1.824(1700)

0.044 cos 20◦(

1

0.01094+

1

0.0342

)]1/2σC = −454 MPa.

The surface endurance strength of cast iron for 108 cycles is givenby

SC = 2.206HB MPa

The value of HB for ASTM no. 50 cast iron is 262. Hence,SC = 2.206(262) = 578 MPa.


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Surface Durability

Solution-3 III

Note that the contact stress is not a linear function of transmittedload. Hence, the factor of safety is defined as

nd =

(SC

σC

)2=

(578

454

)2≈ 1.62


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AGMA Stress Equations

AGMA I

Many assumptions and simplification go into the Lewis equation.It has to be corrected for various factors. Standards are set byAGMA

For bendingFor pitting resistance or contact stress

FOR BENDING:

..

σ = WtKoKvKsPd

FKmKB

J[US units] (3)

..

σ = WtKoKvKs1

bmKHKB

YJ[SI units] (4)


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AGMA II

Wt tangential load lbf(N)Ko is the overload factorKv is called the dynamic factor [Norton has 1

Kvwhere Kv < 1. New

standards have Kv > 1] �Ks is called the size factorPd is the transverse diametral pitchm is the moduleF(b) is the face width in(mm)Km(KH) is the load-distribution factorKB is the rim thickness factorJ(YJ) is the geometric factor for bending strength (includes theroot fillet stress concentration factor)


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AGMA IIIFOR PITTING:

..

σc = Cp

√WtKoKvKs

Km

dPFC f

I[US units] (5)

..

σc = ZE

√WtKoKvKs

KH

dw1bZR

ZI[SI units] (6)

Cp(ZE) is an elastic coefficient,√

lb f/in2(√

N/mm2)C f (ZR) is the surface condition factordP(dw1) is the pitch diameter of the pinion in(mm)I(ZI) is the geometry factor for pitting resistance


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AGMA Strength Equations

AGMA Strength Equations I

Allowable bending stress

..σall =

St

SF

YN

KTKR[US units] (7)

..σall =

St

SF

YN

YθTZ[SI units] (8)

St is the bending strength or allowable bending stress number,lbf/in2(N/mm2)YN is the stress cycle factor for bending stressKT(Yθ) are temperature factors


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AGMA Strength Equations

AGMA Strength Equations IIKR(YZ) are the reliability factorsSF is the AGMA factor of safety, which is a stress ratio

Allowable contact stress, σc,all

..σc,all =

Sc

SH

ZHCH

KTKR[US units] (9)

..σc,all =

Sc

SH

ZNZW

YθYZ[SI units] (10)

Sc is the allowable contact stress,lbf/in2(N/mm2)ZN is the stress cycle life factorCH(ZW) are the hardness ratio factors for pitting resistanceSH is the AGMA factor of safety.


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Correction Factors

Correction factors

15 factors need to be found to design a gear!


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Finding St and Sc Finding St

Finding St I

Depends on...1 Material...2 Material Hardness...3 Surface treatment...4 Grade. Higher grades are more expensive, have higher strengths.

We find these using a combination of tables, equations and charts.

FINDING St


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Finding St II

Figure 2 : Allowable bending strength for nitrided through hardened steelgears (i.e. AISI 4140, 4340), St. The SI equations are St = 0.568HB + 83.8MPafor grade 1 and 0.749HB + 110MPa for grade 2


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Finding St III

Figure 3 : Allowable bending fatigue strength. Norton uses S′f b for St �

.. Go to AGMA Process


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Finding St and Sc Finding Sc

Finding Sc

Figure 4 : Allowable surface fatigue strength. Norton uses S′f c for Sc �

.. AGMA Wear


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ALL THE FACTORS Dynamic factor Kv

Dynamic factor Kv I

Is present both in the bending stress and the surface stressequations equations 3,4,5,6. Note that Norton uses 1

Kvfor

this. �Accounts for the transmission error

...1 Inaccuracies in tooth profile

...2 Vibration of the tooth during meshing

...3 magnitude of the pitch-line velocity

...4 dynamic unbalance

...5 wear and permanent deformation of contacting portions of the teeth

...6 Tooth friction, Gear shaft misalignment and the linear and angulardeflection of the shaft


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Dynamic factor Kv II

Kv =A +

√V

AV in

f tmin

(11)

Kv =A +

√200V

AV in m

s(12)

A = 50 + 56(1 − B);B = 0.25(12 −Qv)23 (13)

V is the pitch-line velocity πdpnp

12 , if dp (pitch diameter) .Qv is called the quality factor, of the least quality gear in mesh.Will be given to you.


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Kv factor

Figure 5 : Kv factor. Maximum velocity at the end point of the Qv curve isVmax

t = [A + (Qv − 3)]2 [ f tmin ] and Vmax

t =[A+(Qv−3)]2

200 [ms ]

.. AGMA Wear


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ALL THE FACTORS GEOMETRY FACTORS (I and J) or (ZI and YJ)

Geometry Factors I and J JUseful in a data sheet

Are present both in the bending stress (J(YJ)) and the surfacestress(I(ZI)) equations equations 3,4,5,6.


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Bending Strength Geometry FactorJ(YJ) I

Like the Lewis form factor, it accounts for the tooth form and thegeometry of the tooth profile. Includes the filleting at the root tip,which may change the Lewis form factor.Depends on where the loading is, the depth of the tooth (Full orstub) and the pressure angle �


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Bending Strength Geometry FactorJ(YJ) II

Figure 6 : Geometry factor (J(YJ)). For 20◦ pressure angle, full depthteeth. The loading is at the highest point of single tooth contact.



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Surface Strength Geometry Factor I(Z(I) I

We use equations (+ for external and - for internal)

I =cos

(ϕt

)sin

(ϕt

)2mN

mG

mG ± 1(14)

Load sharing factor mN is given by

mN =pN

0.95Z; Z Jis not the number of teeth (15)

mN = 1 for spur gearspN = pn cosϕn (16)

Z in Eq.15 is given by

Z =[(rP + a)2 − r2bP

]1/2+

[(rG + a)2 − r2bG

]1/2 − (rP + rG) sinϕt (17)

rP and rG are pitch radii. rbG and rbP are base circle radii.Ratna Kumar (IIT Madras) Design of Gears July-Nov 2018 41 / 88

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Surface Strength Geometry Factor I(Z(I) II

Surface strength geometry factor. Use + for external gears and −for internal gears

.. AGMA Wear


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ALL THE FACTORS Overload factors (Ko)

Overload factor (Ko)

Are present both in the bending stress and the surface stressequations equations 3,4,5,6. we make allowance for all externallyapplied loads in excess of the load Wt in a specific kind ofapplication.Called application factor in NORTON �

There is a table.

Figure 7 : Overload factor Ko. Called Application factor in NORTON(Ka)

.. Go to AGMA Wear


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ALL THE FACTORS Load-Distribution factor (Km(KH))

Load distribution factor (Km(KH)) I

Are present both in the bending stress and the surface stressequations equations 3,4,5,6.Corrects for the fact that the loads are not distributed uniformlyacross the line of contact.Depends on the face width F. Larger this is more chances of nonuniformity in the load.

Km = 1 + Cmc(Cp f Cpm + CmaCe) (18)

Cmc = 1 for uncrowned teeth and 0.8 for crowned teeth


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Load distribution factor (Km(KH)) II

Figure 8 : A crowned gear tooth


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Finding Cp f I

F(b) Cp f

F ≤ 1in F10d − 0.025

1 < F ≤ 17in F10d − 0.0375 + 0.0125F

17 < F ≤ 40in F10d − 0.119 + 0.0207F − 0.000228F2

b ≤ 25mm b10d − 0.025

25 < b ≤ 425mm b10d − 0.0375 + 4.92(10−4)b

425 < b ≤ 1000mm b10d − 0.1109 + 8.15(10−4)b−3.53(10−7)b2

Table 1 : Values of Cp f . If F10d < 0.05 then use F

10d = 0.05. d is the pinionpitch diameter


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Finding Cpm

For straddle mounted pinion with S1

S < 0.175 use Cpm = 1

straddle mounted pinion with S1

S ≥ 0.175 use Cpm = 1.1

Figure 9 : Definitions of S1 and S to find Cpm


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Finding Cma

Cma = A + BF + CF2 (F is face width in inches)

Figure 10 : Constants to calculate Cma. Note that F is in inches


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Finding Ce

Ce is 0.8 for gearing adjusted at the assembly, or when thecompatibility is improved by lapping or both. Ce = 1 for all otherconditions.


.. Go to AGMA Wear


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ALL THE FACTORS Size factor Ks

Size factor Ks

Are present both in the bending stress and the surface stressequations equations 3,4,5,6.Takes into account the non uniformity of material properties dueto the size.

Ks = 1.192

(F√

YP

)0.0535US units (19)

Ks = 0.904(bm√

Y)0.0535

SI units (20)

If Ks < 1 set Ks = 1

.. Go to AGMA Wear


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ALL THE FACTORS Rim thickness factor KB

Rim thickness factor KB I

Are present in the bending stress equations 3,4When the rim thickness is not large enough to provide support tothe tooth root, faille occurs in the rim rather than at the toothroot fillet.

KB = 1.6 ln(2.242mB

)[For mB < 1.2] (21)

KB = 1 For mB ≥ 1.2 (22)

mB =tR

ht(23)


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ALL THE FACTORS Rim thickness factor KB

Rim thickness factor KB II

Figure 11 : Backup ratio and KB determination



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ALL THE FACTORS Elastic Coefficient Cp(ZE)

Elastic Coefficient Cp(ZE)

Are present in the surface stress equations equations 5,6Accounts for the fact that the materials of the pinion and the gearcan be differentEquation

Cp =

√√√ 1

π(1−ν2P

EP+

1−ν2GEG

) (24)

Where νP and νG are the Poisson’s ratio of the pinion and thegear, respectivelyWhere EP and EG are the Young’s modulus of the pinion and thegear, respectively


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ALL THE FACTORS Surface condition factor C f (ZR)

Surface condition factor C f (ZR)

Not establishedWhen detrimental surface finish is known to exist then C f > 1, sayabout 1.5.


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ALL THE FACTORS Temperature factor KT(Yθ)

Temperature factor KT(Yθ)

Accounts for higher operating temperaturesOccurs in Equations 7, 8, 9,10If oil or gear blank temperature up to 250◦F (120◦)C useKT(Yθ) = 1. For higher temperatures, use a value > 1.

.. Go to AGMA Process .. AGMa Wear


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ALL THE FACTORS Safety factors (SF and SH)

Safety factors (SF and SH) I

Occurs in Equations 7, 8, 9,10

SF =StYN/(KTKR)

σall=

fully corrected bending strengthbending stress (25)

SH =ScZNCH/(KTKR)

σc=

fully corrected contact strengthcontact stress (26)

NOTE: SF ∝Wt (proportional to load, since stress is proportionalto the load (σbending ∝Wt))

It so happens that σc ∝√

Wt. So, SH ∝√

Wt

So Compare SF to S2H while deciding the cause of failure (Bending

or Pitting) �If SF = 3 and SH = 2, then SF < S2

H, this means there is a threat offailure of the gear tooth from bending


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ALL THE FACTORS Safety factors (SF and SH)

Safety factors (SF and SH) II

If SF = 5.73 and SH = 1.69, then SF > S2H, this means is a threat of

failure of the gear tooth from wear or pittingThis relation ship σc ∝

√Wt is valid for when the contact is linear

or helical (Spur gear or helical gear)Crowned teeth then σc ∝ (Wt)

13 which means that we should

compare SF to S3H

.. Go to AGMA Process .. AGMA Wear


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ALL THE FACTORS Reliability factor (KR(YZ))

Reliability factor (KR(YZ)) I

Occurs in Equations 7, 8, 9,10Accounts for statistical distributions of material fatigue failureKR = 0.658 − 0.0759ln(1 − R) if 0.5 < R < 0.99

KR = 0.5 − 0.109ln(1 − R) if 0.99 < R ≤ 0.9999

Reliability KR(YZ)

0.9999 1.50.999 1.250.99 1.000.9 0.850.5 0.7

Table 2 : Reliability factors. Use this only for the values of reliability in thetable. For other reliability use the equation


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ALL THE FACTORS Reliability factor (KR(YZ))

Reliability factor (KR(YZ)) II

.. Go to AGMA Process .. AGMA Wear


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ALL THE FACTORS Stress cycle factors (YN(ZN))

YN(ZN) IOccurs in Equations 7, 8(YN), 9,10(ZN) .. Go to AGMA ProcessThe St and Sc choices are for a life of 107 cycles

We need YN(ZN) to modify for other values of life .. AGMA WearNote that if the cycles is slightly > 107 then the mating gear may be experiencing slightly fewerthan 107 cycles

Figure 12 : The Strength-Cycle factors YN and ZN for bending and pittingrespectively


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ALL THE FACTORS Hardness ratio factor (CH(Zw))

Hardness ratio factor (CH(Zw)) I

Occurs in Equations 9,10Pinion is generally subjected to more cycles than the gear since ithas fewer teethIf both are hardened by the same process, then one can actuallyharden the pinion alone a little more so that the pinion will workharden the gear as it makes contact with itApplies only to the GEAR not the pinion.If both gear and pinion are through hardened

CH = 1.0 + A′(mG − 1) (27)

A′ = 8.98 × 10−3HBP

HBG− 8.29 × 10−3 [1.2 ≤ HBP

HBG≤ 1.7] (28)


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ALL THE FACTORS Hardness ratio factor (CH(Zw))

Hardness ratio factor (CH(Zw)) IIHBP and HBG are the Brinell Hardness numbers of the pinion andthe gears, respectively. (10mm ball at 3000kg load)mG is the gear ratioHBPHBG

< 1.2, A′ = 0

HBPHBG

> 1.7, A′ = 0.00698

If surface hardened pinions with hardness of 48 HRC or harder arerun with through hardened then,

CH = 1 + B′(450 −HBG) (29)

B′ = 0.00075e−0.0112 fp , where fp is the surface finish of the pinionexpressed as root mean square roughness Ra in µinCHARTS ARE ALSO AVAILABLE

.. AGMA Wear


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ALL THE FACTORS Flowchart of spur gear design for dending

Spur Gear Bending based on ANSI/AGMA2001-D04

..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

Wt =33000H

V


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

Kv =A +

√V

AV in ft/min

Kv =A +

√200V

AV in m/sec

A = 50 + 56(1 − B);B = 0.25(12 −Qv)23

V is the pitch-line velocity πdpNp/12.

Qv is called the quality factor, of the least quality gear in mesh. Will be given to you.


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

Ks = max1.192 (

F√

YP

)0.0535, 1


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

Load distribution factor (Km)

Km = Cm f = 1 + Cmc(Cp f Cpm + CmaCe)

.. Go To Km


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

Rim thickness Factor (KB) .. Go to KB


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

Geometry Factor (J) .. Go to J


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

..

Bending endurance strength

.σall =

St

SF

YN

KTKR


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

..


.σall =

St

SF

YN

KTKR

Gear bending strength or allowable bending stress number St


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

..


.σall =

St

SF

YN

KTKR

Stress Cycle Factor YN


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

..


.σall =

St

SF

YN

KTKR

Temperature Factor KT


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

..


.σall =

St

SF

YN

KTKR

Reliability Factor KR


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..

dP =ZP

Pd, V =

πdN12

, σ = WtKoKvKsPd

FKmKB

J

..


.σall =

St

SF

YN

KTKR

Safety Factor SF


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ALL THE FACTORS Flowchart for spur gear design for wear

Spur Gear Wear based on ANSI/AGMA2001-D04

..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2Tangential Load

Wt =33000H

V


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2Overload Factor Ko


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2Dynamic Factor Kv


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2Size Factor Ks


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2Load Distribution Factor Km


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2Surface condition Factor C f is taken as 1


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2Geometry Factor I


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2

..

Contact endurance strength

.σC,all =

SC

SH

ZNCH

KTKR


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2

..


.σC,all =

SC

SH

ZNCH

KTKR

Gear contact strength or allowable contact stress number SC


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2

..


.σC,all =

SC

SH

ZNCH

KTKR

Stress Cycle Factor ZN


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2

..


.σC,all =

SC

SH

ZNCH

KTKR

Temperature Factor KT


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2

..


.σC,all =

SC

SH

ZNCH

KTKR

Reliability Factor KR


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2

..


.σC,all =

SC

SH

ZNCH

KTKR

Safety Factor SH


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..

dP =ZP

Pd, V =

πdN12

, σC = CP

(WtKoKvKs

Km

dPFC f

I

)1/2

..


.σC,all =

SC

SH

ZNCH

KTKR

Hardness Ratio Factor CH


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Spur gear Problem. 1

Spur gear Problem 1

A 17-tooth 20◦ pressure angle spur pinion rotates at 1800 rev/min andtransmits 4hp to 52 tooth disk gear. The diametral pitch is 10teeth/in, the face width 1.5 in and the quality standard is No. 6. Thegears are straddle mounted with bearings immediatelyadjacent (Cpm = 1). The pinion is a grade 1 steel with a hardness of240 HB tooth surface and a through hardened core. The gear is steel,through hardened also, grade 1 material, with a HB of 200, toothsurface and core. Poisson’s ratio is 0.3, JP = 0.3, JG = 0.4, and Young’sModulus is 30Mpsi. The loading is smooth because of motor and theload. Assume a pinion life of 108 cycles and a reliability of 0.9, anduse YN = 1.3558N−0.0178, ZN = 1.4488N−0.023. The tooth profile isuncrowned. This is a commercial enclosed unit gear. Find Factor ofsafety in bending, contact and examine the safety factors to commenton possible failure mode.


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Spur gear Problem. 1

Spur gear Problem 1 - Data

Number of teeth on pinion ZP = 17

Pressure angle ϕ= 20◦

Speed of rotation NP = 1800 rev/minPower transmitted H = 4 hpNumber of teeth on gear ZG = 52

Diametral pitch P = 10 teeth/inFace width F = 1.5 inQuality standard number Qv = 6

Gears are straddle mounted with bearing immediately adjacentPinion material: Grade 1 steel with HB = 240 with through and surface hardenedGear material: Grade 1 steel with HB = 200

Poisson’s ratio ν= 0.3

JP = 0.3, JG = 0.4, E = 30(106) psi.Smooth loading because of motor and loadPinion life: 108 cycles and a reliability of 0.9The load cycle factors: YN = 1.3558N−0.0178 and ZN = 1.4488N−0.023

Tooth profile is uncrownedThe gear set is a commercial enclosed unit.


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Spur gear Problem 2

Spur gear problem 2

A speed reducer has 20◦ full-depth teeth. The gear set has 22 and 60teeth with a diametral pitch of 4 and a face width of 31

4 in. The pinionshaft speed is 1145 rev/min. The expected life is for 5 years at 24 hourper day service. The transmission accuracy is 6.The gears are throughhardened and nitrided and are of grade 1.The nitriding results in thepinion and gear core hardness of 250 HB and 390 HB for the pinionand gear case. Estimate the power rating. The source of power issmooth and there could be some moderate shock in the gear. Assumethat the reliability is 0.99. [1hp=33, 000 lbf ft

min ]


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.

Spur gear Problem 2

Problem 3 IA pair of open gears having pressure angle 20◦ is designed to transmit 180hp [1hp=33,000 lbf ft

min ] with apinion speed of exactly 1000 rev

min . The distance between the two shafts is known to be 12.5 inches. Thegear is supposed to rotate at a speed of exactly 500 rev

min in a direction that is opposite to that of thepinion. The gear is made of grade 1, through hardened steel, while that of pinion is made of grade 2nitrided through hardened steel. The hardness of the gear and the pinion is 300 HB and 325 HB,respectively. The quality of mesh is 6. The face width of a tooth in the gear and the pinion is 4.17 in.The reliability of the design is 0.99 for a billion revolutions of the gear.If the following is known about the gear set, is it a good design? Is it a good design, if the life expected istwo orders of magnitude smaller?

The total number of teeth in the pinion must be greater than 27 and less than 95The pinion must be straddle mounted at the center of a simply supported shaft.The overload factor is 1.5The rim thickness factor is 1.0The Ce = 1

The Cma = 0.3153

The elastic coefficient, Cp = 2300

The surface finish factor, C f = 1

The geometry factor for pitting resistance, I = 0.1071

The Strength life factors for bending is given by, YN = 1.6831N−0.0323 and is 0.8618 for the gearThe Strength life factor for contact is, ZN = 2.466N−0.0566 and is 0.7727 for the gearThe temperature factor, KT = 1

The theoretical allowable bending strength for the gear is 35990 psiThe theoretical allowable contact strength for the pinion is 147725 psi

HINT: It would help if you calculated Kv, Ks, Km, J, YN for pinion, ZN for pinion, Allowable bending strength

for pinion, allowable contact strength for gear to organize your thoughts.


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Guidelines for design of a gear mesh

Design of a Gear Mesh I

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A priori decisions

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Function: load, speed, reliability, life, Ko

Unquantifiable risk: design factor nd

Tooth system: ϕ, ψ, addendum, dedendum, root fillet radiusGear ratio mG, ZP, ZG

Quality number Qv


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Design of a Gear Mesh II

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Desgin decisions

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Diametral pitch Pd

Face width FPinion material, core hardness, case hardnessGear material, core hardness, case hardness

Four design decision categoriesTotal 8 decisions to makeMore number of decisions compared to other m/c elements


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Design of a Gear Mesh III

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Steps for Design

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Choose a diametral pitch.Examine implications on face width, pitch diameters andmaterial properties. If not satisfactory, return to pitchdecision for change.Choose a pinion material and the associated core and casehardness. If not satisfactory, return to pitch decision anditerate until no decisions are changed.Choose a gear material and the associated core and casehardness. If not satisfactory, return to pitch decision anditerate until no decisions are changed.


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Problem-4

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Design a 4:1 spur-gear reduction for a 100-hp, theree phasesquirrel-cage induction motor running at 1120 rev/min. The loadis smooth , providing a reliability of 0.95 at 109 revolutions of thepinion. Gearing space is meager. Use Nitralloy 135M, grade 1material to keep the gear size small. The gears are heat-treatedfirst and then nitraded.


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Guidelines for design of a gear mesh Parallel Helical Gears

Helical Gears I


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Helical Gears II

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Nomenclature

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ψ: Helix anglept: Transverse circular pitchpn: Normal circular pitch pn = pt cosψpx: Axial pitch px =

pttanψ

Pn: Normal diametral pitch Pn = Ptcosψ

ϕn, ϕt: normal and tangential pressure angle

cosψ =tanϕn

tanϕt


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Helical Gears III

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Condition to avoid interference

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Smallest number of teeth ZP for mG = ZG/ZP = 1

ZP =2k cosψ3 sin2 ϕt

(1+

√1+ 3 sin2 ϕt

)

For a given gear ratio m = NG/NP, the smallest pinion

ZP =2k cosψ

(1+ 2mG) sin2 ϕt

[mG +

√m2

G + (1+ 2mG) sin2 ϕt

]

Largest gear teeth number

ZG =Z2

P sin2 ϕt − 4k2 cos2 ψ4k cosψ − 2NP sin2 ϕt


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Problem-1

A stock helical gear has a normal pressure angle of 20◦, a helix angle of25◦ and a transverse diametral pitch of 6 teeth/in and has 18 teeth.Find:

...1 The pitch diameter

...2 The transverse, the normal and the axial pitches

...3 The normal diametral pitch

...4 The transverse pressure angle


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Guidelines for design of a gear mesh Force Analysis - Helical Gears

Force Analysis-Helcial Gears I

Figure 13 : Tooth faces acting on a right-hand helical gear


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Force Analysis-Helcial Gears II

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The forces are applied on pitch plane and in the center ofthe gear face.The theree components of the total toothe force:

Radial component: Wr = W sinϕn (30)Tangential component: Wt = W cosϕn cosψ (31)

Axial components: Wa = W cosϕn sinψ (32)


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Force Analysis-Helcial Gears III.Problem-2..

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In the Figure, a 1-hp motor runs at 1800 rev/min in the clockwise direction, as viewed from the positive xaxis. Keyedto to the motor shaft is an 18-tooth helical pinion having a normal pressure angle of 20◦, ahelix angle of 30◦, and a normal diametral pitch of 12 teeth/in. The hand of helix is shown to be righthanded. Make a three-dimensional sketch of the motor shaft and pinion, and show the forces acting onthe pinion and the bearing reactions at A and B. The thrust to be taken out at A.

ϕt = tan−1tanϕn

cosψ = tan−1 tan 20◦

cos 30◦ = 22.8◦ (33)


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Force Analysis-Helcial Gears IV

Pt = Pn cosψ = 12 cos 30◦ = 10.39 teeth/in (34)

Hence,dP = 18/10.39 = 1.732 in

Pitch line velocity:

V =πdN12

=π(1.732)(1800)

12= 816 ft/min

Transmitted load:

Wt =33000H

V=

33000(1)

816= 40.4 lbf


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Force Analysis-Helcial Gears V

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Solution

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Wr = Wt tanϕt = (40.4) tan 22.8◦ = 17.0 lbfWa = Wt tanψ = (40.4) tan 30◦ = 23.3 lbf

W =Wt

cosϕn cosψ =40.4

cos 20◦ cos 30◦ = 49.6 lbf


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Force Analysis-Helcial Gears VI

The forces Wt,Wr and Wa act in the −y, z and −x directions,respectively at point C.Given that the thrust load to be taken by A and hence,FA = Wa = 23.3 lbfConsidering moment equilibrium about z axis at support A

Wa(dP/2) −Wr(13) + FyB(10) = 0

=⇒ FyB =

17(13) − 23.3(1.732/2)

10= 20.1 lbf

Taking force equilibrium in y direction, we get

−FyA + Fy

B −Wr = 0

=⇒ FyA = 20.1 − 17.0 = 3.1 lbf


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Force Analysis-Helcial Gears VII

Now considering moment equilibrium about y axis at A, we get

FzB(10) −Wt(13) = 0

=⇒ FzB = 40.4(13)/10 = 52.5 lbf

Considering the force equilibrium in z direction leads to

FzA = 2.1 lbf

Torque applied would be T = WtdP/2 = (40.4)(1.732/2) = 35 lbf.in


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Bevel Gear - Force Analysis

Bevel Gears I

Figure 14 : Forces acting on a bevel gear tooth


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Bevel Gears II

Transmitted load is given by

Wt =Trav,

where T is the torque and rav is the pitch radius at the midpoint ofthe tooth.Radial force is given by

Wr = Wt tanϕ cosγ

Axial force is given by

Wa = Wt tanϕ sinγ


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Bevel Gears III

The bevel pinion shown in the figure rotates at 600 rev/min in thedirection shown and transmits 5 hp to the gear. the mountingdistances, the locations of all bearings, and the average pitch radii ofthe pinion and gear are shown in teh figure. For simplicity, the teethhave been replaced by pitch cones. Bearings A and C should take thethrust loads. Find the bearing forces on the gear shaft.


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Bevel Gears IV


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Bevel Gears V

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Free body diagram

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design of spur and helical gears - qltien.com

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