Car Transmission Design

Gears Design and Simulation for Gears Design and Simulation for

Car Transmission Car Transmission

Jiting Li, Weidong Guo, Mileta M Tomovic

School of TechnologyMechanical Engineering Technology

Outline

� About the Car Transmission

� Kinematic Design and Simulation of the Transmission

� Strength Design of the Gear Mesh

� Finite Elements Analysis of Gears

Download the file:

http://www.purdue.edu/discoverypark/PLM/SME/car_transmission.zip

About the Car Transmission

The transmission is used to increase the driving torque of the engine.It is connected to the engine through the clutch, and connected with drive shaft through U-joint.

Driving System Illustration [1]

Transmission Graphic Illustration [1]

About the Car Transmission

Kinematic Design and Simulation of the Transmission

-999-3.38:1Reverse

3,3751:14th

2,5381.33:13rd

1,6792.01:12nd

9993.38:11st

RPM at Transmission Output Shaft with Engine at 3375 rpm

RatioGear

Assume that the transmission is required to have five different gear ratios, listed in the following Table.

Input ShaftOutput Shaft

Counter Shaft

Idler Gear Shaft

Z23

Z35

Z27

Z33

Z40

Z18 Z25 Z31

Z20


To meet the requirement, the mechanism of the transmission is designed as follows.


Z18

Z20

Z40

Z25

Gear

33Z33

1831Z31

2027Z27

4035Z35

2523Z23

Tooth numberTooth numberGear

And the tooth number of the gears is listed in the following table.

Note! The transmission mechanism comes from literature [2].


Then the simulation is carried on to validate the kinematic design result. All the gears and shafts are first modeled and assembled incertain solid modeling software, such as Solidworks and Unigraphics, and then imported into the kinematic and dynamic analysis software ADAMS/View to do the kinematic simulations.

First Gear

Z23

Z35

Z40

Z18

38.31823

4035

ZZ

ZZGR

1823

40351 =

×

×==Transmission ratio


First Gear Simulation


Second Gear

Z23

Z35

Z33

Z25

01.22523

3335

ZZ

ZZGR

2523

33352 =

×

×==Transmission ratio



Second Gear Simulation

Third Gear

Transmission ratio 33.13123

2735

ZZ

ZZGR

3123

27353 =

×

×==


Z27

Z31

Z23

Z35

Third Gear Simulation


Fourth Gear

Connecting Shaft

Transmission ratio 1GR 4 =


Fourth Gear Simulation


Reverse

Z23

Z35

Z40

Z18

38.31823

4035

ZZ

ZZGR

1823

4035rev −=

×

×−=−=Transmission ratio

Z20


Reverse Simulation


Strength Design for Gear MeshThere are five pairs of gear mesh to be designed. Here we just take the first pair,

i.e., the gear mesh 1, for an example. The design method for others are same.

Note! All the Equations, Figures and Tables will be used are come from literature [3].

Requirements: Design a 1.522:1 helical-gear reduction for a 18-hp, output shaft of

engine running at 3375rpm. The load is medium shock, providing a reliability of

0.95 at 109 revolutions of the pinion. Using through hardened steel, grade 1 material.

Solutions: Make the a priori decisions as

•Function:18hp, 3375rpm, R=0.95, N=109 cycles, Ko=1.35

•Design factor for unquantifiable exingencies: nd=3

•Tooth system: Φn=20°, helix angle: ψ=20°•Tooth count: NP=23 teeth, NG= 35 teeth

•Quality number: Qv=8, use grade 1 material

•Assume mB≥1.2 in Eq.(14-40), KB=1

Pitch: Select a trial diametral pitch of Pd=8 teeth/in. Thus, dP=23/8=2.875 in and dG=35/8=4.375 in. From Table 14-2,YP=3.334, YG=0.374. From Fig. 14-6, JP=0.345,JG=0.375.

lbf833.233272.2540

1833000

V

H33000W

ft/min272.254012

75π(2.875)33

12

nπdV

t

pp

=×

==

===

From Eqs.(14-28) and (14-27),

404.172.70

272.254072.70

A

VAK

72.70)63.01(5650)B1(5650A

63.0)812(25.0)Q12(25.0B

63.0B

v

3/23/2

v

=

+=

+=

=−+=−+=

=−=−=

Strength Design for Gear Mesh

From Eq.(14-38)

885.0)95.01ln(0759.0658.0)R1ln(0759.0658.0KR =−−=−−=

From Fig.14-14,

874.0)522.1/10(6831.1)522.1/N(6831.1)(Y

862.0)10(6831.1N6831.1)(Y

0323.090323.0

GN

0323.090323.0

PN

===

===

−−

−−

From Fig.14-15,

791.0)522.1/10(466.2)522.1/N(466.2)(Z

773.0)10(466.2N466.2)(Z

056.09056.0

GN

056.09056.0

PN

====

===

−−

−−


From the recommendation after Eq. (14-8), 3p ≤ F≤5p. Try F=1.57 in.

From Eq.(a), Sec. 14-10,

061.18

334.057.1192.1

P

YF192.1K

0535.00535.0

s =

=

=

From Eqs.(14-31), (14-33), (14-35), Cmc=Cpm=Ce=1. From Fig. 14-11, Cma=0.11 for commercial enclosed gear units. From Eq. (14-32),

037.0)57.1(0125.00375.0)875.2(10

57.1F0125.00375.0

d10

FC

P

Pf =+−=+−=


From Eq. (14-30),

147.1)]1(11.0)1(037.0[11)CCCC(C1K emapmpfmcm =++=++=

From Table 14-8, for steel gears, psi2300CP =

°==Ψ

Φ=Φ 17.21

20cos

20tgarctg

cos

tgarctg n

t o

o

in341.117.21cos2

875.2cosrr tPbP =°=Φ=

in040.217.21cos2

375.4cosrr tGbG =°=Φ=

in125.08

1

P

1a ===


[ ] [ ]

581.017.21sin)2

375.4

2

875.2(

040.2)125.02

375.4(341.1)125.0

2

875.2(

sin)rr(r)ar(r)ar(Z

2/1

22

2/1

22

tGP

2/12

bG

2

G

2/12

bP

2

P

=°+−

−++

−+=

Φ+−−++−+=

in393.08P

pn =π

=π

=

143.018

8

2

20sin20cos

1P

P

2

20sin20cosI =

+=

+=

oooo

669.0)581.0(95.0

20cos393.0

Z95.0

cosp

Z95.0

Pm nnN

N ==Φ

==

o


in963.0)862.0(40000

)885.0(1

345.0

)1(147.1)8(061.1)404.1(35.1)833.233(3

YS

KK

J

KKPKKKWn)F(

Nt

RT

P

Bmdsvo

t

dbend

==

=

Pinion tooth bending.

With the above estimates of Ks and Km from the trial dimetral pitch, we check to see if the mesh width F is controlled by bending or wear considerations. Choose the hardness of steel through-hardened HB=350. From Fig. 14-2

Equating Eqs. (14-15) and (14-17), substituting ndWt for Wt and solving

for face with (F)bend necessary to resist bending fatigue, we obtain

kpsi 401280077.3HBSt =+=


psi 41800100912HB223Sc =+=

From Fig. 14-5,

Then

in790.0)143.0(875.2

)1(147.1061.1)404.1(35.1)833.233(3

885.0)1(141800

)773.0(2300

Id

CKKKKWn

KKS

ZC)F(

2

P

fmsvo

t

d

2

RTC

Np

wear

=

=

=


DecisionMake face width 1.2 in. Correct Ks and Km:

046.18

334.02.1192.1

P

YF192.1K

0535.00535.0

s =

=

=

019.0)2.1(0125.00375.0)875.2(10

2.1F0125.00375.0

d10

FC

P

Pf =+−=+−=

129.1)]1(11.0)1(019.0[11)CCCC(C1K emapmpfmcm =++=++=

psi768.10115345.0

)1(129.1

2.1

8)046.1(404.1)35.1(833.233

J

KK

F

PKKKW)(

P

Bmdsvo

t

P

==

=σ


The bending stress induced by Wt in bending, from Eq. (14-15), is

psi768.10115345.0

)1(129.1

2.1

8)046.1(404.1)35.1(833.233

J

KK

F

PKKKW)(

P

Bmdsvo

t

P

==

=σ

The factor of safety in bending of the pinion, from Eq. (14-41), is

852.3768.10115

)]885.0(1[)862.0(40000

)KK(YS

)S( RT

Nt

PF ==σ

=


Decision.Gear tooth bending. Use cast gear blank because of the 4.375-in pitch diameter. Use the same material, heat treatment, and nitriding. The load-induced bending stress is in the rotio of JP /JG. Then

psi507.9306375.0

345.0768.10115

J

J)()(

G

PPG ==σ=σ

245.4507.9306

)]885.0(1[)874.0(40000

)S( GF ==


Pinion tooth wear.The contact stress, given by Eq. (14-16), is

psi593.74914143.0

1

)2.1(875.2

129.1)046.1(404.1)35.1(833.2332300

I

C

Fd

KKKKWC)( f

P

msvo

t

PPC

==

=σ

The factor of safety from Eq. (14-42), is

653.1593.74914

)]885.0(1[1)773.0(141800

)KK(CZS

)S(C

RT

HNC

PH ==σ

=


By our definition of factor of safety, pinion wear is

3n733.2653.1)S( d

22

H P=<==

So make face width larger, F=1.57 in.

The factor of safety (SH)P is

863.1551.66486

)]885.0(1[1)773.0(141800

)KK(CZS

)S(C

RT

HNC

PH ==σ

=

psi551.66486143.0

1

)57.1(875.2

147.1)061.1(404.1)35.1(833.2332300)( PC ==σ

3n471.3863.1)S( d

22

H P=>==


By our definition of factor of safety, both of pinion bending and wear are larger than nd.

Gear tooth wear. The hardness of the gear and pinion are the same. Thus the contact stress on the gear is the same as the pinion,

The wear strength is also the same, SC = 141800 psi. The factor of safety of the gear in wear is

906.1551.66486

)]885.0(1[1)791.0(141800

)S( GH ==

psi551.66486)( GC =σ

By our definition of factor of safety, both of gear bending and wear are larger than nd.

634.3906.1)S( 22

H G==


Rim. Keep MB ≥ 1.2. The whole depth is

in281.08

25.2

P

25.2

P

25.1

P

1dedendumaddendumh

ddd

t ===+=+=

The rim thickness tR is

in337.0)281.0(2.1hmt tBR ==≥

For other pinions and gears in this transmission, we can use the same method and process to calculate their bending and wear stresses.


Finite Elements Analysis of Gears

This will be done in a few days !

Reference

1. http://auto.howstuffworks.com/transmission.htm

2. Richard Stone and Jeffrey K. Ball, Automotive Engineering Fundamentals. SAE International, 2004.

3. Joseph E. Shigley, Charles R. Mischke and Richard G. Budynas, Mechanical Engineering Design (7th edition), Higher Education, 2003.

Acknowledgments

The authors wish to acknowledge the support from the Society for Manufacturing Engineers - Education Foundation,

SME-EF Grant #5004 for “Curriculum Modules in Product

Lifecycle Management.”

Car Transmission Design

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