The document discusses various types of vehicle transmission systems including manual, automatic, semi-automatic, and continuously variable transmission (CVT), emphasizing the efficiency and operational reliability of CVT. It details the design and selection process of a drivetrain using CVT, highlighting calculations for gear ratios and vehicle dynamics, and performance enhancements over traditional manual systems. Additionally, it outlines the design of a gearbox that ensures effective torque transfer and provides material analysis to optimize strength and functionality.

1. 1
CHAPTER 1
INTRODUCTION

2. 2
1. Introduction
The function of vehicle transmission is to adapt traction available from drive unit and
to suit it for vehicle, the surface, the diver, the environment. The transmission has
decisive effect on the fuel consumption, the reliability, road safety, ease of drive.
The main goal when developing a vehicle transmission is to convert the power from the
engine into vehicle traction as efficiently as possible, over a wide range of road speed
this has to be done ensuring a good compromise between a number of speeds, climbing
performance acceleration fuel consumption of vehicle technical and technological
advances has to be taken in to account as do operational reliability and adequate service
life.it is also important to have regard from environmental and social consideration.
1.1. Manual Transmission
In this type of transmission for obtaining different speed, gear changing process was
done manually. This type of transmission need clutch pedal to change the gear ratio.
This type of transmission achieved by using different types of gear box
a) Sliding mesh gearbox
b) Constant mesh gearbox
c) Synchromesh gearbox
Types of Trasmission
Manual Transmission
System
Automatic Transmission
System
Semi-Automatic
Transmission System
Continuously Variable
Transmission

3. 3
1.2. Automatic Transmission
In this type of transmission gear ratio automatically changes as the vehicle move and
there is no need to change manually. This transmission system does not need the clutch
pedal to change the gear ratio. This type of transmission used hydraulic fluid coupling
or torque convertor instead of clutch to change the gear ratio.
1.3. Semi-Automatic Transmission
In this type of transmission gear does not change automatically rather than it uses
electronic sensor, pneumatics and actuator to change the gear ratio. It does not need
clutch pedal.
The type of semi-automatic transmission are
a) Dual clutch transmission
b) Sequential transmission
1.4. Continuously Variable Transmission (CVT)
A continuously variable transmission (CVT) is a transmission which can gradually
shift to any effective gear ratios between a specified ranges.
Different types of CVT are: Pulley based CVT, Toroidal CVT, and Hydrostatic CVT.
Among these, Pulley based CVT is the most affordable and simple one. It has two
pulleys: a) Primary Pulley connected to the Engine b) Secondary Pulley connected to
the Transaxle. These two pulleys are connected by a V belt.
Variation in the pitch radii of these pulleys lead to the change in the gear ratio.

4. 4
CHAPTER 2
CONTINUOUSLY VARIABLE
TRANSMISSION

5. 5
2. CVT (Continuously Variable Transmission)
It works on the principle of continuously varying pulley diameters. In this type of
transmission, infinite gear ratios can be achieved in a fixed range.
The continuously variable transmission (CVT) is a transmission in which the ratio of
the rotational speeds of two shafts, as the input shaft and output shaft of a vehicle or
other machine, can be varied continuously within a given range, providing an infinite
number of possible ratios.
A CVT need not be automatic, nor include zero or reverse output. Such features may
be adapted to CVTs in certain specific applications.
Other mechanical transmissions only allow a few different discrete gear ratios to be
selected, but the continuously variable transmission essentially has an infinite number
of ratios available within a finite range, so it enables the relationship between the speed
of a vehicle, engine, and the driven speed of the wheels to be selected within a
continuous range. This can provide better fuel economy than other transmissions by
enabling the engine to run at its most efficient speeds within a narrow range
About CVT.
2.1. Construction
It consists:
a) Primary or drive pulley: It consists of spring and flyweight assembly. This
pulley is connected to engine shaft.
b) Secondary or driven pulley: It consists of ram and spring assembly. This pulley
is connected to transaxle.
c) And Belt

6. 6
Fig. 2.1: Cut View of CVT component
2.2. Function
The driven pulley consist of two sheaves with the fixed sheaves often
duplicating in function as a break disc. A straight pressure spring pushed
against the movable sheaves and control the belt movement

7. 7
The driving pulley also consist of two sheaves, a fixed & movable. A
pressure spring pushed the movable sheaves away from the belt & a
centrifugal mechanism worked against spring. These centrifugal
mechanism first had to overcome the pretension of the pressure spring to
engage the belt and then goes to higher until it overcome spring in the
driven sheaves and started to shift the belt into higher ratio. The spring in
the driving pulley control the engagement speed. The spring in the driven
pulley controls the shift speed which should coincide with engine’s power
peak for best performance.
2.3. Comparison between Manual Transmission with CVT
Transmission
Speed Diagram for CVT Transmission
Fig. 2.2: Speed diagram for CVT transmission dotted line represent the variation of
speed as vehicle speed increases.

8. 8
Speed Diagram for 4-speed Manual Transmission
Fig. 2.3: Speed diagram for 4 speed Manual transmission. Dotted line show the
variation of speed when vehicle speed increases.
Fig. 2.4: speed diagram of CVT transmission with manual transmission. Curve A & B
shows the variation of speed of CVT and Manual transmission respectively.

9. 9
The fig 2 show the variation of engine speed with vehicle speed. This fig. shows that
once the engagement was don engine speed remains constant this speed is known as
shift speed and it is equal to the peak power rpm.
The fig 2 shows the variation of engine speed with vehicle speed in case of manual 4
speed transmission. Which shows the rise and fall in engine speed as each gear
exchange occurs and the vehicle increases speed.
The fig 3 shows the comparison of variation of engine speed with vehicle speed in CVT
and Manual transmission. Since in CVT transmission engine keep at peak rpm it
increase the performance of engine, fuel economy and utilised the maximum power of
engine.
Also in this case there is no clutch pedal so it increases the driver comfort.
Fig. 2.5: Image of CVT

10. 10
CHAPTER 3
DESIGN AND SELECTION OF DRIVETRAIN

11. 11
3. Design and Selection of Drivetrain
3.1. Objective
The aim was to design a highly efficient transmission system that provides good
acceleration, top speed and is reliable on different terrains. To obtain an infinite range
of gear ratios so as to obtain the highest torque and as well reach the maximum speed
of 60kmph, a CVT along with a self-designed auxiliary reduction gearbox was
incorporated.
3.2. Considerations
 Driver comfort
 To utilize maximum engine power
 Enhance system’s performance
 To obtain maximum Acceleration
 Less weight
 High Power to weight ratio
 Fuel Economy
Keeping the above objectives in mind, CVT (continuously variable transmission)
was chosen over a manual transmission because it gives the infinite number of gear
ratio, keep the engine at peak rpm, better fuel economy, and Driver comfort.
3.3. Selection of CVT
For selecting the CVT first highest and lowest ratio of CVT was calculated.
Engine Specification
BRIGGS AND STRATTON 10HP
Idle Speed (Ni): 1750±100 RPM
Maximum Speed (Ne): 3800 RPM

12. 12
3.3.1. Calculation
 Radius of wheel (Rw)= 11.5in = 0.2921m
 Weight of vehicle = 300Kg
 Efficiency of CVT (ηCVT) = 0.85
 Efficiency of Gearbox (ηG) = 0.95
 Coefficient of Rolling resistance (Kr) = 0.023
Calculation for highest Ratio
 Considering Maximum velocity of vehicle (v) = 60 kmph =16.67 mps
 Overall Highest gear ratio = GH
Velocity = (2*∏*Ne*RW*ηCVT* ηG)/(60*GH)
By substituting all value, GH is calculated as
GH = 5.63
Calculation for Lowest Ratio
Considering maximum Gradiability angle θ = 40 degree
Lowest gear ratio = GL
Fig. 3.1: force distribution on gradient

13. 13
Total resistance forces on the vehicle are
 Drag force
 Force Due to slop (RS)
 Rolling Resistance(RR)
Since the velocity of vehicle is less so the drag force are neglected
Force due to slop is
RS = M*g*sinθ
Therefore, RS = 1891.72 N
Rolling resistance is
RR= Kr*M*g*cosθ
Therefore, RR = 51.85 N
Consider at this gradiability angle our vehicle is in equilibrium condition, then
The total resistance = Tractive force provided by engine
Tractive force (FT) provided by engine is
FT = TW/RW = (Te*GL* ηCVT* ηG)/RW
Torque produced by engine (Te) is
Te = (60*P)/(2*∏*Ne)
Therefore, Te =18.7468 Nm
Lower gear ratio is calculated by equating Tractive force and total resisting force
Therefore, GL = 37.5
Considering reduction ratio of gearbox = 11
Therefore, Highest Ratio of CVT = 0.512
Lowest Ratio of CVT = 3.4

14. 14
After extensive market survey COMET 790 model was selected which had almost
equal highest and lowest ratio.
Specification of CVT
Highest ratio = 0.54
Lowest ratio = 3.38
It uses rubber V-belt. The centre distance was selected after the total assembly of
powertrain i.e. engine and gearbox. The belt selected was rubber V-belt with centre
distance 11.5 inches.
After selection of CVT actual gearbox reduction was decided
Gearbox specification
Number of stages = 2
Each stage reduction = 3.333
Overall reduction of gearbox = 11.11
Overall reduction in highest ratio with CVT is
GH = 0.54*11.11 = 6
Overall reduction in lowest ratio with CVT is
GL = 3.38*11.11 = 37.55
3.4. Performance of Vehicle
Maximum velocity of vehicle is
Vmax =
2∗∏∗Ne∗RW∗ηCVT∗ ηG
60∗ GH
Therefore, Vmax = 15.64 mps = 56.3 Kmph

15. 15
Tractive force (FT) provided by engine is
FT = TW/RW = (Te*GL* ηCVT* ηG)/RW
Therefore, FT = 1946.1 N
Maximum gradiability angle had calculated by equating Tractive force and total
resisting force
Therefore, maximum gradiability angle
θmax = 40.1 degree
For straight road
Maximum acceleration =
𝑇𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒−𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑣𝑒ℎ𝑖𝑐𝑙𝑒
amax = 6.26 m/sec2
Torque at wheel is
TW = TE*GR*ɳCVT* ɳG
TW = 568.45 Nm

16. 16
Layout of Drive Train
Fig. 3.2: Layout of Drivetrain

17. 17
CHAPTER 4
GEARBOX

18. 18
4. Gearbox
To provide the sufficient torque to the wheel single speed auxiliary reduction gearbox
was designed which give the constant reduction. During the designing first the ratio of
gearbox was decided.
Overall gear box reduction ratio = 11.11
Number of stage = 2
Reduction on each stage = 3.33
For both stages helical pair of gear was selected because it gives
a) Silent operation: In a helical gear train, the teeth engage a little at a time rather
than the entire face at once. This causes less noisy power transfer in case of
helical gears
b) More strength: For same tooth size (module) and equivalent width, helical
gears can handle more load than spur gears because the helical gear tooth is
effectively larger since it is diagonally positioned.
Fig. 4.1: Layout of Gearbox

19. 19
4.1. Components of gearbox
4.1.1. Gears
After deciding the gear ratio and types of gear design process was commenced. Initially
the material selection was done to get material of the required strength.
Material Selection
Table 4.1: Material Comparison for gear
Material 20MnCr5 EN353 18CrNiMo
Sut 850 MPa 580 MPa 700 MPa
Syt 492 MPa 320 MPa 520 MPa
BHN (after
hardening)
500 600 340
E 268 GPa 190 GPa 230 GPa
Cost 78/Kg 98/Kg 105 /Kg
Availability Available Available Not available
From above material we have selected 20MnCr5 because it has superior Sut and BHN
and after calculation we got the almost equal bending strength and wear strength
which we have not got in any other material.
Calculation
For 1st
Stage
Parameters
Speed =3800/3.38 =1124 rpm
Normal pressure angle (Øn) =20 degree
Helix angle (α) =20 degree
Sut = 850 Mpa
Hardness = 500 BHN
Number of teeth of pinion (Zp) = 15
Number of teeth of gear (Zg) = 50

20. 20
σbp and σbg are permissible bending stress for pinion and gear respectively
σbp = σbg = Sut/3 =283.33Mpa
Equivalent number of teeth of pinion is
Zp’ =15/Cos3
α = 18.077
Equivalent number of teeth of gear is
Zg’ =50/Cos3
α = 60.25
Lewis form factor of pinion is
Y’p=0.484-2.865/Z’p=0.3252
Similarly lewis form factor of gear is
Y’g=0.484-2.865/Z’g=0.4363
Since pinion and gear are made of same material hence pinion is always weaker than
gear.
Therefore, design for pinion
Beam strength of pinion is
Fb=σbp*Y’p*b*mn
Where, b is thickness of gear
mn is normal module
Therefore, Fb=829.25* mn²
Wear strength of pinion is
Fw=dp*Q*k*b/Cos³α
Where, dp is pitch circle diameter of pinion
dp=m*15/Cos20= 16.987mn
Q is ratio factor
Q=2*Zg/(ZP+Zg)=1.538
k is surface endurance strength
k=0.16(BHN/100)²=4
After substituting all value wear strength is calculated as
Fw = 894.812 mn²

21. 21
Since beam strength is lesser than wear strength. Hence pinion is weaker in bending.
Therefore, pinion is designed for bending.
Effective force on pinion is calculated as
Feff=Ka*Km*Ft/Kv
Where, Ka is 1
Km is 1
Kv is velocity factor = 3.6/(3.6+0.9693m½
)
Ft is tangential force is calculated as
Ft=P/V
V is pitch line velocity = πDp*Np/60
Module was calculated by equating wear strength with multiplication of equating
effective force and safety factor.
Fw=FOS*Feff
After equating this and solving module was calculated
mn = 2.98 mm
Taking standard value of module, mn = 3 mm
Therefore, face width (b) = 27 mm
Pitch circle diameter of pinion (dp) = 47.89 mm
Pitch circle diameter of gear (dg) = 159.63 mm
Centre distance (C.D) = 103.76
By considering this value dynamic force on the pinion is calculated by using the
formula
Fd =
21V(bcCos²α+Ftmax.)Cosα)
21V+(bcCos²α+Ftmax.)½
After substituting all value Fd was calculated,
Fd = 1500.1 N

22. 22
Effective force is calculated as
Feff=Fd + Ftmax
Therefore, Feff =3989.276 N
Safety factor is calculated as
FOS=Fw/Feff
FOS = 1.87
Therefore, selected module is correct.
Similarly the second stage was calculated
The dimensions of both stages are
For 1st
stage
Module = 3 mm
Pitch circle diameter of pinion = 47.89 mm
Pitch circle diameter of gear = 159.63 mm
For 2nd
stage
Module = 3 mm
Pitch circle diameter of pinion = 47.89 mm
Pitch circle diameter of gear = 159.63 mm

23. 23
Fig. 4.2: Actual pics of Pinion and Gear
Finite Element Analysis
Every component assembled in the gearbox are analysed to validate its calculation as
well as safety. Analysis was done in the software ANSYS 14.0 by applying different
loading condition obtained from calculation.
Analysis of Pinion and Gear
By theoretical calculation we obtained the FOS of pinion and gear. In case of gear the
main objective behind analysing the gears was to make optimization between the
strength and weight. Analysis was done by performing different iterations doing
some hole and other arrangement for weight reduction. The safest combination of
all iterations was then finalized for use in gearbox.
During analysis of pinion and gear we fixed the keyway provided in the gear and
applied tangential, radial and axial force on the gear teeth. The force was considered
to be acting on 3 teeth because 3 teeth were always in mesh condition.

24. 24
First pinion
Boundary condition:
Fixed support: keyway
Force: Ft=2646.263 N, Fr=1024.974, Fa=963.161
The equivalent stress and safety factor as shown in fig and it shows it was safe.
Fig. 4.3: Equivalent Stress
Fig. 4.4: Safety Factor

25. 25
First Gear:
Boundary condition:
Fixed support: keyway
Force: Ft=2646.263 N, Fr=1024.974, Fa=963.161
The equivalent stress and safety factor as shown in fig and it shows it was safe.
Fig. 4.5: Equivalent Stress
Fig. 4.6: Safety Factor

26. 26
2nd
stage pinion and shaft:
For 2nd
stage made the pinion and shaft integrated due to its easy manufacturing,
serviceability and less cost.
Boundary condition:
Fixed support: Meshing teeth (three teeth)
Torque: 211 Nm On gear position
The equivalent stress and safety factor as shown in fig.
Fig. 4.7: Equivalent Stress
Fig. 4.8: Safety Factor

27. 27
2nd
Stage gear
Boundary condition:
Fixed support: keyway
Force: Ft = 8978.46N, Fr = 3477.62N, Fa = 3267.89N
The equivalent stress and safety factor as shown in fig.
Fig. 4.9: Equivalent Stress
Fig. 4.10: Safety Factor

28. 28
Manufacturing Process
Gear manufacturing process are as follows
a) Material cutting
b) Turning
c) Milling
d) Keyway boring
e) Teeth cutting (hobbing)
f) Heat Treatment
g) Grinding
h) Teeth grinding

29. 29
4.1.2. Shafts
Shaft is used to support the CVT driven pulley, gears and pinion and transmit the
load from CVT to input shaft and further to the gearbox output shaft. Calculations
were done to calculate the dimensions of shaft and check whether the shafts were
safe in bending and could take the torsional loads.
During the designing the shaft first material was selected
Material selection
Table 4.2 Material Comparison of Shaft
Material AISI
4340(EN24)
AISI
4140(EN19)
20MnCr5 EN353
Sut 1258 MPa 1020 MPa 682 MPa 580 MPa
Syt 1012 MPa 655 MPa 375 MPa 320 MPa
Cost 95/Kg 88/Kg 80/Kg 98/Kg
Availability Available Available Available Available
From above material we have selected AISI 4340 normalised because it has superior
bending strength.
Calculation
1st
shaft
Maximum Torque on 1st
shaft
T = 63362 N-mm
Force calculation on shaft
 Tangential force on gear Ft
Ft = 2T/D
Ft = 2646.263 N

30. 30
 Radial force on gear Fr
Fr = Ft tan20/ cos 20
Fr = 1024.974 N
 Axial force on gear Fa
Fa = Ft tan20
Fa = 963.161 N
 Force due to belt tension and pulley on shaft T1 , T2
(T1- T2)Rcvt = T
And T1/T2 = 𝑒μϴ
μ is the Coefficient of friction between belt and pulley = 0.36
Sin ϴ’ = (Difference of diameter of pulley)/(2*centre distance)
Sin ϴ’ = (214.88-35.4)/(2*292.1)
ϴ’ = 17.6020
ϴ. = (180-2ϴ’)* π/180
ϴ. = 2.527
So Calculating T1 and T2
T1 = 988.569 N
T2 = 398.135 N
 Vertical Force Analysis
Fig. 4.11: Vertical loading diagram

31. 31
By applying static force balancing conditions
i.e. ∑ Fy =0 &
∑ MA = 0
Vertical reactions were calculated as
RvA = 2860.746 N
RvB = -1521.328 N
 For the vertical Bending Moment Diagram of shaft 1
At A = 0,
At C = 74379.396 N-mm,
At B = 90759.967 N-mm,
At D = 0
BMD of shaft
Fig. 4.12: Vertical bending moment diagram
The Vertical maximum bending moment act at point B is 90759.967 N-mm

32. 32
Horizontal Force Analysis of shaft
Fig. 4.13: Horizontal loading diagram
By applying static force balancing conditions
i.e. ∑ Fy = 0 &
∑ MA= 0
The horizontal reactions were calculated as
RHA = 1148.903 N
RHB = -364.727 N
 For the Horizontal Bending Moment Diagram
At A = 0,
At B = 16479.865N-mm,
At C = 29871.47 N-mm,
At D = 0
BMD of shaft
Fig. 4.14: Horizontal bending moment diagram

33. 33
The Horizontal maximum bending moment act at point C is 29871.47 N-mm
Resultant bending moment at any point of shaft as
MA
2
=MvA
2
+ MHA
2
MA = 0,
Similarly MB=92244.010 N-mm,
Mc = 80153.597 N-mm,
MD = 0.
So Maximum bending moment act on point B i.e. MB=92244.010 N-mm
 Calculation for actual shear stress and FOS
Since this shaft support to the CVT driven pulley as well as 1st
pinion
and the inner diameter of CVT driven pulley was 19 mm. Therefore, 19
mm shaft diameter was used.
For the 19 mm diameter of shaft maximum shear stress on the shaft was
calculated by using Maximum Shear Stress Principle.
i.e. √𝑀2 + 𝑇2 =
𝜋
16
𝐷3
× 𝜏 𝑝 × 0.75
𝜏 𝑝 = 110.79 𝑀𝑃𝑎
 The allowable shear stress for the material EN24 (normalized) ids
Syt = 1012 MPa and Sut = 1258 MPa
By ASME Analysis Allowable shear stress can be
0.18 Sut or 0.3 Syt
i.e. 226.44 MPa or 303.6 Mpa
So Allowable shear stress is 226.44 MPa
Hence the selected material and dimension was correct.
The safety factor was calculated as
FOS = 226.44/110.79 = 2.04

34. 34
Similar Process had done to calculate the dimension of 2nd
and 3rd
shafts.
The specification of all shafts were as follows
Table 4.3 Dimensions of all Shaft
Sr.No. Diameter Maximum
Shear
Stress
Selected
Material
Allowable
Stress
FOS
Shaft 1 19 mm 110.79 MPa EN 24
(Normalized)
226.44 MPa 2.04
Shaft 2 25 mm 122.31 MPa EN 24
(Normalized)
226.44 1.85
Shaft 3 35 mm 116.97 Mpa EN 24
(Normalized)
226.44 1.94
Fig. 4.15: CATIA model of gear and shaft assembly

35. 35
Finite Element Analysis of Shaft
Analysis were done to verify the calculation and check whether the shafts were safe in
bending and could take the torsional loads. Analysis was done by applying the bending
and torsional load in extreme cases and it had proved to be safe for the various loads
that they were projected to in extreme cases.
1st shaft:
Analysis of 1st
shaft was done by supporting the bearing position on the shaft and
applying the torque and force due to tension on the CVT belt at the position of CVT
driven pulley.
Boundary condition:
Fixed support: bearing Position
Force: 1386.7 N
Torque: 63.375 Nm
Fig. 4.16: Equivalent Stress

36. 36
3rd
shaft:
Analysis of 3rd shaft was done by supporting the drive shaft position on the shaft and
applying the force on the key obtained from calculation at the position of gear on
the shaft.
Boundary condition:
Fixed support: Drive shaft position
Force: 17647 N on each key
Fig. 4.17: Equivalent Stress
The shafts were checked after analysis which proved to be safe for the various loads
that they were projected to in extreme cases.

37. 37
Manufacturing Process for Shaft
Manufacturing process of shaft include the different process which were as follows.
a) Material cutting
b) Turning
c) Milling
d) Keyway and spline boring
e) Heat treatment
f) Grinding
4.1.3. Bearings
The function of bearing is to hold the shaft. The bearings were selected to bear the
reaction force i.e. tangential and axial force and it is reliable for the sufficient life.
Calculation
1. For 1st
shaft
The reaction forces on the 1st
shaft
At bearing A FRA= 3082.830 N
Fa=963.161 N
To calculate the equivalent force on the bearing check the ratio of axial and
radial force on the bearing.
e =0.37
Fa
FRA⁄ = 0.312 < e

38. 38
Therefore, X=1, Y=1, Ka = 1.2
The equivalent force Pe was calculated as
Pe=X × Fa × Ka
Pe = 3699.396 N
Consider life of the bearing is 400 Hrs
The life of bearing in terms of number of revolution was calculated as
L10=
400×60×1124.26
106 =26.982 million rev.
The dynamic load (Cr) of the bearing was calculated as
L10= (
𝐶𝑟
𝑃𝑒
)10/3
Therefore, Cr = 9.941 KN
Similarly,
At bearing B FRB= 1564.437 N
Fa=963.161 N
To calculate the equivalent force on the bearing check the ratio of axial and
radial force on the bearing.
e =0.37
Fa
FB⁄ = 0.615 > e
Therefore, X=0.4, Y=1.6, Kb= 0
FAB =
0.5×FB
Y
+ Ra = 1451.911 N
The equivalent force Pe was calculated as
Pe = {FB × Ka Or (X × FB + Y × FAB) × 𝐾𝑎}; whichever was larger.
i.e. Pe = {3538.598 Or 1877.324 N}; whichever was larger.
Therefore, Pe = 3538.598 N

39. 39
Consider life of the bearing is 400 Hrs
The life of bearing in terms of number of revolution was calculated as
*L10=
400×60×1124.26
106
=26.982 million rev.
The dynamic load (Cr) of the bearing was calculated as
L10= (
𝐶𝑟
𝑃𝑒
)10/3
Therefore, Cr = 9.509 KN
From bearing Cat log
Bearing selected was Tapper roller bearing 32004 X/Q
Specifications:
Type: Tapper Roller Bearing
ID: 20 mm
OD: 42 mm
Width: 15 mm
Cr: 24.2 KN
Co: 27.0 KN
Using similar procedure bearings were selected for 2nd
and 3rd
Shaft. Their
specifications were as follows,
Table 4.4 Specification of Bearings
Sr. No. Properties 1st shaft 2nd shaft 3rd shaft
1 Name 32004 X/Q 30205 J2/Q 32007 X/Q
2 Bearing Type Taper Roller Taper Roller Taper Roller
3 ID 20 mm 25 mm 35 mm
4 OD 42 mm 52 mm 62 mm
5 Width 15 mm 16.25 mm 18 mm
6 Cr 24.2 KN 30.8 KN 42.9 KN
7 Co 27.0 KN 33.5 KN 56.0 KN

40. 40
Fig. 4.18: Image of Bearings
4.1.4. Circlip
The function of Circlip is to prevent the axial motion of the gear and bearing on the
shaft. It should be reliable to sustain the axial load. The Circlips were selected for
the standard dimension of shaft.
The Circlips selected and their dimension were as follows,
Table 4.5 Specification of Circlips
Sr. No. Dimension Type Quantity
1 20 External 4
2 25 External 3
3 35 External 2

41. 41
4.1.5. O – Rings and Oil seal
The function of O-rings is to prevent the oil leakages during static and dynamic
condition. The 3 mm diameter of rubber O-ring was selected and for their mounting
a groove was done on the both face of gearbox casing.
The function of oil seal is to prevent the oil leakages during dynamics condition and
reduce the friction losses. The oil seals are selected according to shaft dimensions.
The specification of standard oil seals are
Table 4.6 Specification of Oilseals
Sr. No. Inner
diameter
Outer
diameter
Width Quantity
1 20 35 10 1
2 35 55 10 2
4.1.6. Gearbox Casing
The function of gearbox casing to hold the all components i.e. shaft, gears, bearings,
oil seal, and O-ring. In order to hold the gear and shaft assembly, to have the freedom
to select the material of required strength as well as the one having a sufficient co-
efficient of thermal expansion, the gearbox was designed. A few empirical relations
relating to the thickness of the wall, clearance between the wall and gear, distance
between the two gears on a single shaft were considered. After these considerations,
the casing was designed in CATIA and the analysis was done in ANSYS.

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Material Selection for Gearbox Casing
Table 4.7 Material comparison for Gearbox Casing
Material He20(Al 6061) He9(Al 6063) He30(Al 6082) Al2024
Density 2700 2700 2680 kg/m3
2780
Sut 310 241 295 483
Syt 276 214 240 345
BHN 95 73 89 120
α 23.5*10-6 23.5*10-6
23.1*10-6
22.8*10^-6
Cost 380 /Kg 350 /Kg 370 /Kg 900/Kg
From above material we have selected Al 6061 T6 because it has superior strength
hardness and coefficient of thermal expansion. Also it has less cost and easily
available in market.
Fig. 4.20: CATIA view of casing

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Finite Element Analysis of Gearbox Casing
Analysis of casing was performed by providing fixed support at the mounting points of
the casing and applying bearing loads at their respective positions. A pretension bolt
load was also considered while analysing the casing.
Boundary condition:
Fixed support: All mounting point
Force: All bearing and bolt
The safety factor and equivalent stress on the gearbox casing is as given below
.
Fig. 4.21 Safety Factor of Gearbox Casing

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Fig. 4.22 Equivalent Stress of Gearbox Casing
Manufacturing Process of Gearbox Casing
Manufacturing process of gearbox casing was done on vertical machining centre
(VMC)
Fig. 4.23: Manufacturing of Gearbox Casing on VMC Bed

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Fig. 4.24: Image of Gearbox Casing
Fig. 4.25: Gearbox CATIA model

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CHAPTER 5
DRIVE SHAFT

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5. Drive Shaft
To transmit the power from the gearbox to the wheels, constant velocity (CV) joints
were decided to be used because of their ability to transmit torque with negligible
losses.
There are mainly two type of CV joint
5.1. Tripod joint
It is used at the inboard end of driveshaft, it enables power transmission even
in case of angle shifting. It has needle bearing or barrel-shaped rollers mounted on a
three legged spider or three-pointed yoke, instead of balls bearings. These fit into a
cup with three matching grooves, attached to the differential. The rollers are mounted
at 120 degrees to one another and slide back and forth in tracks in an outer “tulip”
housing. This three-legged spider with tripod has only limited operating angles, but is
able to plunge in and out with a longer distance as the suspension moves. A typical
Tripod joint has up to 50 mm of plunge travel, and 26 degrees of angular articulation.
Fig. 5.1: Components of Tripod type CV joint

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5.2. Articulating Joint
It is used outboard side of drive shaft, it enables power transmission even in
case of angle shifting. It has six balls mounted in cage at sixty degree each. It gives
the only angular articulation of 40 degree.
Fig. 5.2: Components of articulating type CV joint
Fig. 5.3: Articulating type CV Joint Working
The Drive shaft that was chosen gave a plunge of 30mm on the gearbox side and an
articulating joint that gave a travel of 40 degrees on the wheel side.

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CHAPTER 6
COST ANALYSIS

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6. Cost Analysis
For doing any project cost is the very important parameter, during the project cost of
various component is also considered.
The cost summery of different component is as given below
Table 6.1 Cost Summery
Sr. No. Component Cost Supplier
1 CVT & Belt 27000 Quality drive system(california)
2 Gear box casing 33500 Adinath materials and tool room
engineering(Bhosari)
3 Gears 9000 Precision gears(Bhosari)
4 Shaft 3500
5 Bearing 2200 Mehta enterprises (Bhosari)
6 Fasteners 145 S G traders(Bhosari)
7 ‘O’ ring 100 Marina (Bhosari)
8 Gearbox oil
(20W40 Shell)
285 Radhesham auto care(Bhosari)
9 Circlips 25 S G traders(Bhosari)
10 Shaft with C V joint
( Maruti Suzuki zen)
6000 Radhesham auto care(Bhosari)
Total cost 81755

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CHAPTER 7
FAILURE MODE, EFFECT AND ANALYSIS

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7. Failure Mode, Effect and Analysis (FMEA)
It is a process that identifies all the possible types of failures that could happen to a
product and potential consequences of those failures.
An FMEA is a rigorous, step by step process, to figure out everything that could go
wrong and what can be done to keep those things from happening. It identify the actions
which could eliminate or reduce the chance of potential failure.
The FMEA of the project was as follows
Fig. 7.1: FMEA Chart 1

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Fig. 7.2: FMEA Chart 2

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CHAPTER 8
CONCLUSION

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Conclusion
The main objective of the Drivetrain was to transmit the power from engine to wheel
as maximum as possible, provide good acceleration, top speed and is reliable on
different terrains.
Thus Drivetrain was designed and manufacture in such a way that it provides good
acceleration, top speed and is reliable on different terrains. To obtain an infinite range
of gear ratios so as to obtain the highest torque and as well reach the maximum speed,
a CVT along with a self-designed auxiliary reduction gearbox was incorporated. Also
driver comfort and fuel economy were include by using CVT.
All the self-design part was analyse in ANSYS. Various project plans like DVP,
DFMEA, Gantt Chart were used.
All result were verified by Design Validation Plan (DVP) and sufficient testing of
vehicle was done.
Thus we can conclude that the Drivetrain designed for ATV satisfies all the
requirements, the same was verified by testing.
.

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CHAPTER 9
REFERENCES

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References
 “Automotive Transmissions” , Harald Naunheimer; Springer: 1999;
 “Fundamentals of Vehicle Dynamics”, Thomas. D. Gillespie; Society of
Automotive Engineers. Inc; 2002;
 “Clutch Tuning Handbook”, Olav Aaen’s; 2007;
 “Baja SAE India Rule book” , SAE India; 2015;
 “A Kinematic Analysis and Design of a Continuously Variable Transmission”,
Christopher Ryan Willis; 2006;
 “Quality drive system”,2009;

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CHEPTER 10
APPENDIX

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Fig. 10.1: exploded View of Powertrain assembly
Fig. 10.2: Image of Drivetrain assembly

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Fig. 10.3: Rear View of vehicle which shows the assembly of Drivetrain