2. Automobile Engg - IJAERD - Rigid PVC Gearbox Housing.pdf

8/20/2019 2. Automobile Engg - IJAERD - Rigid PVC Gearbox Housing.pdf

1/26

www.tjprc.org [email protected]

RIGID PVC GEARBOX HOUSING FOR AUTOMOBILES

VEDAGYA BAKSH1& SHIVOMENDRA PATEL2

1,2 Rajiv Gandhi Technical University, Medi-Caps Institute of Science and Technology, Indore, India

Abstract

The present invention relates to gearbox housing for automobiles, and more particularly, it relates to a light

weight rigid PVC gearbox housing for automobiles capable of reducing the stress concentration of the gearbox housing.

Keyword: Gear box casing, optimization, rigid poly vinyl chloride (PVC), transmission, automobiles, weightreduction,

power to weight ratio

Received: Nov 15, 2015 ;Accepted: Oct 19, 2016 ; Published: Feb 15, 2016 ; Paper Id.: IJAuERDFEB20162

INTRODUCTION

The gearbox housing is the housing that surrounds the mechanical components of a gear box. It provides

mechanical support for the moving components, a mechanical protection from the outside world for those internal

components, and a fluid -tight container to hold the lubricant that bathes those components.

Traditionally, the gearbox housing is made from cast iron or cast aluminium, using methods of permanent

mould casting or shell moulding. Experimentally, though, composite materials have also been used.

The cast iron is one important material which is used for gearbox housing. The cast iron provides string

housing to the inner component and lasts long, but it is very cumbersome when it comes to welding it to desired

gearbox design. Also spray painting may cause rusting and lead to low life of the gearbox housing.

The cast aluminium is another commonly used material for gearbox housing which is li ghtweight and

can be designed easily. Though, cast aluminium is lightweight and design easy, it is heavier as compared to RIGID

PVC for gearbox housing.

Accordingly, there exists a need to provide a gearbox housing which overcomes above mentioned

drawbacks.

DETAILED DESCRIPTION

• Objects of the Invention

An object of the present invention is to provide a light weight rigid PVC gearbox housing which reduces

the stress concentration which acts on the housing of gearbox.

Another object of the present invention is reduction in body weight which increases the power to weight

ratio. Yet another object of the present invention is reduction in overall cost of the gearbox housing.

Further object of the present invention is to provide an alternative material for gearbox housing.

Or i gi n al Ar t i c l e

International Journal of Automobile Engineering

Research and Development (IJAuERD)

ISSN(P): 2277-4785; ISSN(E): 2278-9413

Vol. 6, Issue 1, Feb 2016, 15-40

© TJPRC Pvt. Ltd.


2/26

16 Vedagya Bakshi& Shivomendra Patel

Impact Factor (JCC): 5.4529 Index Copernicus Value (ICV): 6.1

• Brief Description of the Drawing

Figure 1

Figure 2


3/26

Rigid PVC Gearbox Housing for Automo

www.tjprc.org

biles

Figure 3

Figure 4

17

[email protected]


4/26

18

Impact Factor (JCC): 5.4529

Veda

In

Figure 5

Figure 6

ya Bakshi& Shivomendra Patel

ex Copernicus Value (ICV): 6.1


5/26

Rigid PVC Gearbox Housing for Automobiles 19


Figure 7

Figure 8

Figure 9

Figure 10

Figure 11


6/26



Figure 12

Figure 13

Figure 14

Figure 1 shows a cross-sectional perspective view of gearbox housing, in accordance with the present invention;

Figure 2 shows equation for calculating Centre to Centre distance, in accordance with the present invention;

Figure 3shows equation for design of synchronizers, in accordance with the present invention;

Figure 4, shows equation for selector mechanism, in accordance with the present invention;

Figure 5, shows equation for design of gearbox housing, in accordance with the present invention;

Figure 6 shows values of Eigen Modes analysis generated due to Vibration having minimum value of 2.600E-02

and maximum value of 1.400E+01, in accordance with the present invention;

Figure 7 shows values of Eigen Modes analysis in X -axis generated due to Vibration having minimum value of -

9.911 E+00 and maximum value of 3.051E+00, in accordance with the present invention;

Figure 8 shows values of Eigen Modes analysis in Y -axis generated due to Vibration having minimum value of -

7.495E+00 and maximum value of 5.139E+00, in accordance with the present invention;

Figure 9 shows values of Eigen Modes analysis in Z -axis generated due to Vibration having minimum value of -

9.441E+00 and maximum value of 3.806E+00, in accordance with the present invention;


7/26



Figure 10 shows distribution of Strain Energy Generated in object due to vibration having minimum value of -

8.545E-10 and maximum value of 8.417E -10, in accordance with the present invention;

Figure 11 shows Density of Strain Energy Generated in object due to vibration having minimum value of -

2.600E-11 and maximum value of 3.841E -11, in accordance with the present invention;

Figure 12 shows graph of the variation of Thickness of gearbox housing of tw o materials, in accordance with the

present invention;

Figure 13 shows graph of the variation of Volume of gearbox housing of two materials , in accordance with the

present invention; and

Figure 14 shows graph of the variation of Mass of gearbox housing of two materials, in accordance with the

present invention;

•

Detailed Description of the Invention

The foregoing objects of the present invention are accomplished and the problems and shortcomings associated

with the prior art, techniques and approaches are overcome by the present invention as described below in the preferred

embodiments.

Gearbox housing is used to cover the gear box. Also, to prevent it from external undesired objects and dirt. It is

also used to retain certain amount of gear lubricant inside it, so that the gear train can run smoothly. The proposed design

and analysis is concerned with an alternative material (rigid PVC) which can perform same and has less weight. The

novelty of this proposed invention is that by using the gear box housing of rigid PVC, the weight of the gear box

housing/housing can be reduced as well as the stress concentration which act on the housing of gear box made up of cast

iron/steel/aluminum which are in commercial use today. It can be used for any heavy vehicle and any machines where gear

boxes are in use where there is low temperature of about 60 degree and there is no significant space constraint.

Refereeing now to figures 1 to 5, it shows a cross-sectional perspective view of a gearbox housing (100) and

equations for calculating different units respectively.

Following are the parameters for calculating the design of gearbox housing (100).

Given:

Table 1Power = P

Speed = N

Gear

Ratios=

{ 1

,2

,3

,4

,5

,r }

*5 Forward + 1 Reverse Gears

Design of Gearbox Involves the Following Steps

• Estimating the Centre to Centre distance

•

Calculation of gears and their dimensions


8/26


9/26



a =

24

1 1 33 32

1 1 ,lim

{( )( )( )( )( )( )}( 1)

4( ) { ( )( )( )( )( )( )}

B D H E H

A V H H

H NT L R V W X

Z Z Z Z Z S T k k k k

b d Z Z Z Z Z Z

ε β

β α

µ

µ σ

+

Provided :

1 1( ) st

gear b d = 0.65

1 2( ) st gear b d

= 0.45

1 3( ) st

gear b d = 0.28

1 4( ) st

gear b d = 0.28

1 5( ) st gear b d

= 0.30

1( )reverseb d = 0.65

ak =

0.65, for passenger cars &

0.85, for commercial vehicles

V k =

H k α =

H k β = 1

H Z = 2.25

/ B D Z = 1

Z E

= 0.175 E , For commercial steel =2

189.8 N mm

Z ε = 0.95

Z β = 0.95

Now we have the following result:-

, , , , , NT L R V W X Z Z Z Z Z Z = 1

,lim H σ = 21800 N mm , for commonly used

material of shaft (16MCr5)

H S = 1.2

CALCULATIONS OF GEARS AND ITS DIMENSIONS

• For Permanent Reduction

From empirical data :-


10/26



1d =

1

2

1

a

µ +

1

d is the diameter of pinion on input shaft.

Now,

1 2

2

d d +

= a

1 2d d + = 2a

2d =

12a d −

• For first gear

9 10d d + = 2a

1 µ =

9

10

Z

Z

1 µ =

9

10

d

d

• For second gear

7 8d d + = 2a

2 µ = 7

8

Z

Z

2 µ = 7

8

d

d

• For third gear

5 6d d + = 2a

3

µ =

5

6

Z

Z

3 µ = 5

6

d

d

• For fourth gear

3 4d d + = 2a

4 µ = 3

4

Z

Z

4

µ =

3

4

d

d


11/26



• For reverse gear

DC = 150 mm

θ = 80o

tanθ = DC

DB

DB = tan

DC

θ

AD = AB DB−

In ∆ ADC, By Pythagoras Theorem

2 AC =

2 2 AD DC +

2

9 11

2

d d +

= 2 2 AD DC +

9 11

2

d d +

= 2 2

AD DC +

11d = ( )2 2 92 AD DC d + −

Now,

In ∆ BDC, By Pythagoras Theorem

2 BC =

2 2 BD DC +

2

11 12

2

d d +

= 2 2 BD DC +

12 11

2

d d +

= 2 2

BD DC +

12d = ( )2 2 112 BD DC d + −

• Face width (b)

Let

module = m = 8mm

Helix Angle = β = 35o

Now, from K.MAHADEVAN design datebook, page number 213, equation 12.23(b)

According to AGMA, the minimum face width,

minb =

(1.15)

tan

mπ

β

&Equation 12.23 (c)


12/26



And the maximum value of face width;

maxb =

20

tan

m

β

Now, from data book page number 214, eq. 12.24 (a)

Lewis equation for helical or herringbone gears

t F =

d V n

w

C bYm

C

σ

nm = cosm β

d σ = 30 MPa ,……table-12.22, Page-241

wC = 1.15 ,……table-12.22, Page-241

V C =

6.1

6.1 v+ = 0.378 ,……eq-12.25,

Page-241

Y = yπ

y = 0.148 ,……table-12.21, Page-232

From data book page number 214, equation number 12.26 (a)

Now, According to Buckingham, the inertia force;

iF =

2

3

2

3

( cos )cos

cos

t

t

k V Cb F

k V Cb F

β β

β

+

+ +

3k = 6.60

, Dynamic Load Factor C = 786.5 ,……table-12.12,

Page-236

The Dynamic Load,

d F =

t iF F +

Now from equation 12.26 (b) of Databook

The dynamic strength of gear is given by following formula;

sF = d nbYmσ

Condition for safe working:

sF ≥

d F

Now from data book page number 214;

. , No of Teeth Z = d m


13/26



,Circular Pitch p = d

Z

π

, d Diametrical Pitch p = 1

m

, r Dedendum Circle Diameter d = 2( ) fn cn nd t t m− +

,n

Tooth Factor for Standard Tooth t = cos

f t

β =

1

cos β

,cnTooth Clearance Factor t

=

cos

ct

β

= 0.2

cos β

, o Addendum Circle Diameter d = 2r d h+

h = (2 ) f c

t t m+ = 2.2m

DESIGN OF SYNCHRONIZERS

In automobile, a synchronizer is a part of synchromesh manual transmission that allows the smooth engagement

of gears. Synchronizers serve to let shafts and gears engage with each other smoothly after their speeds have been

synchronized.

• Design of Cones:

We know, the friction torque to be transmitted;from design data book page number 259, equation number 13.10

(d)

1T = 2sin

f a mF D µ

α =

2

2 m

bpDπ

µ

p = 0.07

f µ = 0.12

q = m D

b = 4.5 8to

From equation 13.10(h),page number 260

b = 32

49

n

f

T

pπµ

m D = 7b

Now we know,

m

D = 1 2

2

D D+

For Second Gear


14/26



Gear Ratio = 2 µ

1 µ =

1

2

N

N

2 N = 1

1

N

µ = 9 N

Now we know that;

8

7

N

N

= 2 µ

8

2

N

µ =

7 N

7T = ( ) 77 2t

d F

7P =

7 72

60

N T π

Now

( )2nd m gear

D = 3 7

32

10 f

P kq

pnπ µ

2cb = ( )

2

nd m gear

D

q

We already observe that;

( ) ( )2 22

c co id d −

= ( )2 sincb α

( ) ( )2 2c co id d − = ( )22 sincb α

Since;

( )2c od ≠

7d

( )2nd m gear D < ( )2c od < 7d

Taking average value

( )2c od = ( ) 72

2

nd m gear D d +

( )2c id = ( )

( )72

22 sin2

nd m gear

c

D d b α

+

−

For Third Gear

Gear Ratio = 3 µ


15/26



1 µ =

1

2

N

N

2

N = 1

1

N

µ =

6

N

Now we know that;

6

5

N

N =

3 µ

6

3

N

µ =

5 N

5T = ( ) 55 2t

d F

5P =

5 52

60

N T π

Now

( )3rd m gear

D = 3 5

32

10 f

P kq

pnπ µ

3cb =

( )3rd m gear

D

q


( ) ( )3 32

c co id d −

= ( )3 sincb α

( ) ( )3 3c co id d − = ( )32 sincb α

Since;

( )3c od ≠

5d

( )3rd m gear

D < ( )3c od < 5d


( )3c od = ( ) 53

2

rd m gear D d +

( )3c id = ( )

( )53

32 sin2

rd m gear

c

D d b α

+

−

For Fourth Gear

Gear Ratio = 4 µ

1 µ = 1

2

N

N


16/26



2 N = 1

1

N

µ = 4 N

Now we know that;

4

3

N N

= 4 µ

4

4

N

µ =

3 N

3T = ( ) 33 2t

d F

3P =

3 32

60

N T π

Now

( )4thm gear

D = 3 3

32

10 f

P kq

pnπ µ

4cb =

( )4thm gear

D

q


( ) ( )4 4

2

c co id d −

= ( )4 sincb α

( ) ( )4 4c co id d − = ( )42 sincb α

Since;

( )4c od ≠

3d

( )4thm gear

D < ( )4c od < 3d


( )4c od = ( ) 34

2

thm gear D d +

( )4c id = ( ) ( )

34

42 sin2

thm gear

c

D d b α

+−

• Design of Lockers

Between 2nd and 3rd Gear

As we know that diameter of 3rd gear is smaller than that of 2nd gear and the diameter of annular ring of

slider is to be constant.

Therefore, we need to design the cotter corresponding to that of 3rd

gear. Therefore torque loses to be neglected

and;

3rd Torque on gear = 5T


17/26



Now,

Shifting Force is given by

sF =

3

2 sin

( ) rd

r

s m Gear

T

d

α

µ

Now, from reference, AT by Gisbert Lechner and Herald Naunheimer

page number 252,equation 9.18

Z T = ( )2

2

cFd Cot

β

105o < β < 125o

Now we know

Z T ≤ r T

Neglecting the loses,

Let us assume

Z T = r T

cd = ( ){ }2

2

r T

F Cot β

cd = ( ){ }

2tan

2r

T

F

β

Let ZB be the number of teeth’s on synchronizers

Therefore, pitch of synchronizers

From databook page number 205,equation 12.8, Bachi's formula for beam strength of tooth

m = 532

0.88s

T

k kz

Where

k = s

s

b

m

Strength Coefficient = 2k =

( )24.90 M m for bronze

6 ≤ k ≤ 20

k = 7

Therefore

,s

Circular Pitch P = s

s

d

z

π

( ), d s Diametrical Pitch P = s

s

z

d

( ) , o s Addendum Circle Diameter of Synchronizers d = (2 )r f cd t t + +


18/26



( ) , r s Dedendum Circle Diameter of Synchronizers d = 2( )

f c st t m+

f t = 1

ct = 0.2

Here addendum of synchronizer is the dedendum for synchronizer ring,

s z =

R z

DESIGN OF SELECTOR MECHANISM

Let the angle of contact of selecting fork over synchronizer ring be:

f θ = 180o

Let the selecting fork arms have a T-cross section.

•

Selector Mechanism

In manual transmissions, an interlock mechanism prevents the engagement of more than one gear at any one

time and a decent mechanism holds the gear in the, in detention, in the selected position.

σ = 55 MPa

sF =

(1.5 )

f

f l t

σ

l = 360

cd θπ

t =

(1.5)

f

sl F

σ

Here,

l = Length of Fork

sF = Force of Shift

• Design of shifting lever :

Let L be the affective length of the lever and P be the manual force applied at the handle.

Let B and t be the height and thickness of handle near the boss assuming that the lever is extended to the centre

of shaft (for strength of lever).

We get,

. Max Bending moment on the lever = PL

Section Modulus near the boss = 2

6

tB

As we know,

M = b Z σ


19/26



PL = 2

6

btB σ

bσ = Permissible Bending Stresses In Material

Also, for empirical data

B = 5t

Also,

1 B =

2

B

Where

P

= 70 100 N − for Continuous Shifting

= 200 250 N − for Intermittent Shifting

= 350 400 N − for Instantaneous Shifting

The diameter d2 is subjected to combined bending and twisting.

Therefore by GUEST’S FORMULA:

3

216

bd

π σ = 2 2

b t M M +

3

216

bd

π σ = 2 2( ) ( )Pl PL+

3

216

bd

π σ = 2 2P l L+

2d =

13

2 216

b

Pl L

πσ

+

Or by RANKINE formula

2d = ( )

1

32 232

b

P l l L

πσ

+ +

Now for dimension of boss,

We know,

PL = 2

2 22

t

d t l t σ

+

Now,

2

2

l

t = ( )1.25 Standard ratios

Let

2t = ( )3 mm General considerations

Therefore,


20/26



2l = 5.75 mm

P = ( )25.625

t sbd t

Lσ +

sbd = 2

5.625t

PL t σ

−

DESIGN OF SHAFTS

• Design of Main Shaft:

As we know, size of gear synchronizers increase gradually, consider the load as UVL.

We know,Deflection in simply supported beam subjected to UVL is,

δ =

( )

4 2 2 47 10 3360

o x

l l x xlEI

ω − +

For UVL,

maxδ =

4

0.00652 o sl

EI

ω

Here

oω = Identity of load

As we know for standard gearbox max. Permissible deflection of shaft

maxδ = 0.0003

s L

0.0003 = 3

0.00652 o sl EI

ω

3

0.4601

s

EI

l =

oω

Net load acting on shaft

nω = o s

lω

nω =

2

0.4601

s

EI

l

A B R R+ = nω

Taking moments about A

B R = 2

3 n

ω

A R = 1

3 nω

( ) Bc

BM = 2

3

B s R l


21/26



( ) Bc

BM = 2 1. .3 3

n slω

( ) Bc

BM = 2

9

n slω

Considering A

( ) Ac

BM = 2

3 A n s

R lω

Maximum bending moment of main shaft

( )b m M = 2

9 n s

lω

Since first gear is subjected to maximum torque i.e T9.

Therefore using GUEST’S THEORY

eT = 2 2

9b M T +

3916 ms

T d π

= 2

2

9

2

9 s nl T ω

+

Let

s

ms

l

d = 10

2

6

916

msT d

π

=

( )22 2

9

410

81 n msd T ω +

2

32 2

256 s msd π τ = 2 2 29

40081

n msd T ω +

Thus diameter of the main shaft is determined and hence length of the shaft is also determined.

• Design of counter shaft:

Approx. Volume of gears

V = 2

4 nad b

π

Let the density of material be ρ ,

g Mass of Gears m = V ρ

( )g nm = 2

4 na

d bπ

ρ

Weight of Gears = ( )g nw = ( ) .g nm g

AC BC R R+ = 1 2 3 4 5 6 g g g g g gW W W W W W + + + + +

Taking moment about Ac

BC s R l =

1 2 3 4 5 63 6 9 12 15

2 2 2 2 2 2

b b b b b bw w b w b w b w b w b

+ + + + + + + + + +


22/26



BC s R l =

1 2 3 4 5 6

7 13 19 25 31

2 2 2 2 2 2

b b b b b bw w w w w w

+ + + + +

BC s R l = [ ]1 2 3 4 5 67 13 19 25 31

2

bw w w w w w

+ + + + +

B R = [ ]1 2 3 4 5 67 13 19 25 31

2 s

bw w w w w w

l

+ + + + +

AC R = ( )g Bnw R∑ −

BM will be max either at 3 or at 4.

( )3

BM = ( ) ( ) ( )2 13 6 62

A

bw b w b R b

+ − +

( )4 BM =

( ) ( )( )5 63 6 6 2 BC b

w b w b R b

+ − +

The greater of ( )3

BM or ( )4

BM is equal to equivalent bending moment, ( )b c M Considering torque maximum due to first gear only

ceT = 2 2

9bc M T +

csd =

1

3

16

ce

s

T π

τ

• Design of idler shaft:

Approx. Volume of gears

igV = 2

4 na

d bπ

Let the density of material be ρ

,ig

Mass of Ge s mar = igV ρ

igm = 2

4 na

d bπ

ρ

Weight of Gears =

ig

w =

igm g

igw = Ai Bi R R+

Taking moment about A

B s R l =

2

sig

lw

2

igw

= A

R = B

R

Also, we know

isl = 2

2

bb +


23/26



bi M =

4

ig

is

wl

Torque is considered to be maximum

ieT = 2 29ib M T +

isd =

1

3

16

ie

s

T π

τ

DESIGN OF BEARINGS

Design of bearings is purposed for selection of bearings of standard sizes.

Since there is no axial motion on shaft thus radial roller bearing is design.

From data book page number 373, equation number 16.7(b).

The equivalent load P for radial roller bearings

P = r a XVF YF +

From data book page number 384, table number 16.5

Considering rotating inner ring

V = 1.0

Considering, Maximum Permissible Force r F in gearbox, is due to first gear.

From databook page number 206, equation 12.8(b)

1r F =

1tan

t F α

And from databook page number 211, equation 12.21

aF = F tant β

Now,

Bearing number is selected from databook page number 394; table number 16.13(a).

On the basis of diameter of shaft

aF = F tant β

• For Main Shaft:

Series NU22,

5

msd

•

For counter shaft:

Series NU22,

5

csd

• For Idler Shaft:

Series NU22,

5

5

id

From the table corresponding value of basic load factors,


24/26



Static Co and dynamic C are selected.

Now,

e = 1.5tanα = 0.545

X =

0.4 Y = 0.4cot α = 1.098 Now,

Life of roller bearings is given by following formula,

nl = ( )

10

3c p

DESIGN OF HOUSING

The gear housing is the housing that surrounds the mechanical components of a gear box. It provides

mechanical support for the moving components, a mechanical protection from outside world for those

internal components, and a fluid-tight container to hold the lubricant that bathes those components.

Maximum vertical measure of gearbox element is measured by considering all the following terms

H = (radius of countershaft permanent reduction gear) + (centre to centre distance) +

(radius of synchronizer ring) + (total height of selecting fork cross section) +

(diameter of boss of shifting shaft).

H = ( )

( )2 1.52 2

o ssb

d d a t t d + + + + +

Now, since it is clear that vertical measure of gearbox elements is dominant on horizontal thus, inner

diameter of the housing can be

i D = H Clearances+

Now, thickness of housing is given by t,

From databook page number 113 equation numbers 8.10

t = 1 i

PC D

σ

Let

1C = 0.54

mid and cid are determined as the outer diameter of the bearing outer case.Therefore selecting the value of D from databook page number 394, table number 16.13(a).

Now

mod = 3mid mm+

cod = 3

cid mm+

The number of studs required

I = 0.015 4id +

The core diameter of studs

c

D = maxi

t

P

d nσ


25/26



Nominal diameter of the studs

N D =

0.84

cd

From databook page number 433, equation number 21.2(a)

,g

Thickness of gasket t = 0.25 1.5mm to mm = 1.5 mm

CONCLUSIONS

This invention has various advantages. Some of them are as follows :

• Reduced weight

• Simplified design

• Increased power to weight ratio of the vehicle

• Alternate material for manufacturing of gear box casing.

The foregoing descriptions of specific embodiments of the present invention have been presented for purposes of

illustration and description. They are not intended to be exhaustive or to limit the present invention to the precise forms

disclosed, and obviously many modifications and variations are possible in light of the above teaching.

The embodiments were chosen and described in order to best explain the principles of the present invention and

its practical application, to thereby enable others skilled in the art to best utilize the present invention and various

embodiments with various modifications as are suited to the particular use contemplated. It is understood that various

omission and substitutions of equivalents are contemplated as circumstance may

Suggest or render expedient, but such are intended to cover the application or implementation without departing

from the spirit or scope of the present invention.

REFERENCES

1. The refences of the article were cited from the following :

2.

A Textbook of Machine Design by R.S. Khurmi& J.K. Gupta ,S.Chand Publication

3.

Design Data Handbook for Mechanical Engineers in SI and Metric Units by K. Mahadevan & K. Balaveera Reddy.

4.

Automotive Trasnmission, Gisbert Lechner and Herald Naunheimer.

5. MASTER'S THESIS IN THE INTERNATIONAL MASTER PROGRAMME IN APPLIED MECHANICS

6. “Synchronization Processes and Synchronizer Mechanisms in Manual Transmissions” , ANA PASTOR

7.

BEDMAR, Department of Applied Mechanics,CHALMERS UNIVERSITY OF TECHNOLOGY, Göteborg, Sweden.

8.

Machine Design,Mubeen and Mubeen.


26/26

Date post:	07-Aug-2018
Category:	Documents
Upload:	anonymous-s8ydmr
View:	224 times
Download:	0 times

2. Automobile Engg - IJAERD - Rigid PVC Gearbox Housing.pdf

Documents