 Open Access
 Total Downloads : 17
 Authors : Neel Shah , Rhythm Prajapati , Harshil Dabhi , Jash Patel
 Paper ID : IJERTV8IS090140
 Volume & Issue : Volume 08, Issue 09 (September 2019)
 Published (First Online): 21092019
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Design of A Hybrid Transmission for Electric and Engine Powered Transmission
Transmission Design for A Parallel Hybrid Vehicle
Neel Shah
Mechanical Engineer Babaria Institute of Technology
Vadodara, Gujarat, India
Harshil Dabhi
Mechanical Engineer
Pandit Deendayal Petroleum University (PDPU) Ahmedabad, Gujarat, India
Rhythm Prajapati
Mechanical Engineer
Pandit Deendayal Petroleum University (PDPU) Ahmedabad, Gujarat, India
Jash Patel Mechanical Engineer Nirma University
Ahmedabad, Gujarat, India
Abstract with the increasing amount of pollution and threat of global warming, the Toyota research team did come up with the idea of the Hybrid vehicles in year 1999. The Battery electric vehicle (BEV) is different than Hybrid electric vehicle (HEV) as they use single source of power generation i.e. electric motor. This project was launched as the Toyota Prius 1st Generation. A general hybrid car have two or more engines. Toyota Prius 1st Generation had one internal combustion engine and two electric motor. Both of this electric motors are AC motorgenerator (MG) with permanent magnet rotor. One of this motor (MG1) is used to start the IC engine and as the alternator to charge the high voltage battery, while other motor (MG2) is used as the primary drive motor and as the alternator to charge the high voltage battery while car being on the regenerative mode. Honda Hybrid cars started with single motor which later changed to dual motor just like as in Toyota Prius.
There are many advantages for using this kind of vehicles. To start with being more environment friendly, providing higher pickup and providing better fuel economy. Though with stated advantages, there are also certain disadvantages as such as less power development, poor handling due high weight of high voltage batteries and higher maintenance. This problems are constantly being worked upon and changes are made to hybrid cars to make it more reliable, economical and safe.
KeywordsTransmission design, ANSYS, Solidworks, Gear calculations, Planetary geartrain, Parallel type hybrid vehicle, 30Ni4Cr1, EN30A.

INTRODUCTION
According to the Wikipedia, the proper definition for the Hybrid vehicle is as follows, A Hybrid vehicle is a vehicle that uses two or more distinct type of power source as such as an IC engine and an electric motor. E.g. in Dieselelectric trains using diesel engines and electricity from overhead liner & Submarines that uses diesel when surfaced and batteries when submerged. Other means to store energy include pressurized fluid in Hydraulic Hybrids [2]
There are total five types of hybrid vehicles. First is a Mild Hybrid vehicle, which unlike BEVs uses IC engine, an electric battery and a motor/generator. It is least electrified type of HEV and uses oversized starter motor which can also
be used as the generator called integrated startergenerator (ISG) or belted alternator starter (BAS). Second is Series Hybrid vehicle, where engine is coupled to a generator that charges the battery which provides power to the wheels through transmission or a direct coupling. Third is Parallel Hybrid vehicle, where there are two parallel paths to power the wheels, an engine path and an electrical path. Transmission couples the generator/motor and engine allowing either or both to power the wheels. This is clearly complex mechanism than the series one. Fourth is Series Parallel Hybrid vehicle, this has both series and parallel energy flow paths. Complexity varies with number of motors used. This one is classified further as complex hybrids, split parallel hybrids and powersplit hybrids. Fifth is Plugin Hybrid vehicles (PHEV), which can be charged by plugs from electricity walls. They can be easily distinguished by larger size of battery packs than other HEVs. This type can be made with any Hybrid configuration. [3]
The topic that is choose by our team is regarding Hybrid vehicles transmission design. In this project, the gear box is designed for a Hybrid car with the max speed, acceleration, fuel economy and distance to achieve max speed that will be generated by motor and motor working with the engine. Alongside this calculations, certain essential simulations are also being done with CAD modelling. During this calculations, the assumptions were made considering a passengers car specifications in general. E.g. gross weight was assumed to be 1900kg, final gear reduction as 3:1 and max speed constraint was taken as 55.56m/s.
In this transmission design, we have considered the parallel HEVs from above discussed types of HEVs we have used a single motor/generator instead of two motors thus to make design compact and one that can be used in practice. We made use of two planetary gear set whose sun gears are connected to each other through shaft that is connected to the motor/generator at the other end.

METHODOLOGY

Selection of the geardrives
A Planetary gearset has way more advantages than that of the conventional gearsets. With multiple kinematic combinations, it also provides reduction in volume which proves to be essential criteria for power to weight calculations. Thus we have used a planetary gearset or to be called epicyclic geartrain instead of the conventional geartrain.
Planetary gear box works on planetary motion principle each stage of the planetary gear box consists of a central Sun Gear meshing with accurately positioned three Planet Gears around it which in turn mesh with the internal teeth of the outer Ring Gear. Normally, the Ring gear is stationary and forms the part of the housing, input is given to the sun gear and output is derived from the three planet gears through a planet carrier. [11]
Working of Planetary gear set: Any one of the three members can be used as the driving or input member. At the same time, another member might be kept from rotating and thus becomes the reaction, held, or stationary member. The third member then becomes the driven or output member. Output direction can be reversed through various combinations.
Advantages of Planetary gearbox:

Compared to conventional gearbox has smaller dimensions.

Easier to sort through the constant rounds of shot.

Greater durability than conventional bikes in gear.

Easy to achieve high transmissions ratio due to the size.
[11]Disadvantages of Planetary gearbox:

More expensive than conventional production of gearboxes.

More complex than conventional transmissions. [11]
Special features of Planetary gearbox:

All shafts are made of special alloy steel and are hardened and tempered.

Good quality bearings for input and output shafts.

High efficiency.

Low noise level.

No oil leakages.

Taper roller bearings on output shafts for bigger models.

Long and trouble free performance. [11]


Calculations

Maximum speed

Kinematics of the Mechanism
Figure 1: Transmission Gearbox for the Parallel Hybrid Car [5]
The figure above represents the schematic diagram of a motorintegrated transmission mechanism. This transmission is made up of 2 planetary geartrains and one motor/generator. In the figure, two boldly marked parallel lines indicates a mesh. The geartrain on left is the input planetary geartrain and one on right is output planetary geartrain. The two Sun gears are connected together by link1 i.e. a common input saft. This link1 here is attached to the rotor of the electric motor. The input ring gear is connected to the output carrier shaft i.e. link3. Link3 is connected further to the final gear reduction unit of the transmission. The engine crankshaft can be coupled to the link1 through by a rotating clutch C1or to the link2 by another rotating clutch C2. The output ring gear can be grounded to the casing by a band clutch B1 and the electric motor can be held stationary by a band clutch B2.
On basis of which clutches are engaged, 5 modes of operation are made possible, namely motor mode (M), power mode (P), engine mode (E), engine/charge mode (EC) and regenerative mode (R). This modes are further classified as shown in the table below into 16 submodes. Note that here one of the charge modes provide CVT capability and X signifies the engaging of that clutch in that mode.
Now applying the fundamental circuit equation for the kinematic analysis of the hybrid transmission, let i and j be the gear pair and k be the carrier and thus forming a fundamental circuit. The equation for the same can be written as,
= Â± ( ) (1)
Where i,, j and k denote angular velocities of link i, j and k respectively and Nij represents the gear ratio between this gears i.e. Nij = Tj / Ti where Tj and Ti represents the number of teeth on respective gears. The sign of this equation is positive or negative depending on whether the gear mesh is internal or external.
The geartrain in fig. 1 contains 4 fundamental circuits: (7, 3, 2), (7, 1, 2), (6, 4, 3) and (6, 1, 3). The circuit equations can be written as:
7 2 = 37 (3 2) (2)
7 2 = 17 (1 2) (3)
6 3 = 46 (4 3) (4)
Table 1: Clutching Condition for Different Modes and Condition of Motor Operation at that Time [5]
Operating Mode
Working clutches
C1
C2
B1
B2
Motor
X
Power
P1
X
X
P2
X
X
P3
X
X
Engine
E1
X
X
E2
X
X
E3
X
X
Engine/Charge
EC1
X
X
EC2
X
X
EC3
X
X
CVT
X
Regenerative
R0
X
R1
X
X
R2
X
X
R3
X
X
6 3 = 16(1 3) (5) From equation (4) and (5),
1 + 414 (1 + 41)3 = 0 (6) Where N31 = N37/N17. Similarly, equation (2) and (3)
gives,
1 + 313 (1 + 31)2 = 0 (7)
The equation (6) and (7) relate the angular velocities of 4 co axial links 1, 2, 3 and 4. With the angular velocities of any 2 links, we can solve equation (6) and equation (7) for the other two.
The motor will be running in the Motor and the Power mode whereas it will be freewheeling in the Engine mode. When the car will be propelled in Engine/Charge mode or Regenerative mode, the motor will act as the generator.

Static Torque and Power Flow Equations
Consider the coupled planetary gearset as a system then under the steadystate conditions, torque will be exerted on link 1, 2, 3 and 4 about the central axis of the gear set that must be summed up to zero. i.e.
1 + 2 + 3 + 4 = 0 (8)
Where i denotes an external torque that has been exerted on link i. Similarly, the net power flows into the system via the four coaxial links that must sum up to be zero. i.e.
11 + 22 + 33 + 44 = 0 (9)
Eliminating the 3 from equation (8) and (9) and using equation (6) and (7),
411 + (1 + (41 / 1+31)) 2 + 3=0 (10)
Eliminating 4 from equation (8) and (9) and using equations (6) and (7),
(1 + 41)1 + (1 + (41 / 1 + 31)) 2 + 3 = 0 (11) This equations (10) and (11) relate torques exerted on the links 1, 2, 3 and 4 in terms of the gear ratios N31 and N41.
Given torques exerted on any of the two links, we can find those exerted on the other two.

Operating Modes Details
The data for the process are given as follows:
Input Planetary geartrain: T1=20, T7=16 and T3=52. Output Planetary geartrain: T1=20, T6=16 and T4=52. Final reduction ratio=3:0
Tire radius= 0.235m

Motor mode
To start a vehicle from stationary conditions, the band clutch B1 is engaged and all the other clutches are disengaged. Using the output ring member, the electric motor drives the vehicle all by itself through the output planetary geartrain. Meanwhile, the input planetary geartrain spins freely. Thus, the output ring gear 4 = 0. Substituting this value in equation

and (7) gives,
1 = (1 + N41)3 = 3.63 (12)
2 = (1 + (N41 / 1 + N31))3 = 1.723 (13)
By the equation (12), its clear that the speed ratio between the electric motor and the output link is 3.6. As the input planetary geartrain moves freely, we take 2 = 0 and substituting its value in the equation (10) and (11) gives,
4 = 411 = 2.6 (14)
3 = (1+41)1 = 3.6 (15) By the equation (14) and (15), its observed that reaction torque exerted on the link 4 is 2.6 times of the motor torque and the output torque is equal to 3.6 times that of the motor torque. The negative sign in equation (15) is the indication that the torque is being applied on the final reduction unit by the
output shaft.

Power Mode

When the vehicle in climbing a hill or it has the requirement of the maximum acceleration to be achieved, the vehicle goes into the power mode. For this mode to be active, the engine must be in the running condition. There are 3 sub modes available in this mode, P1, P2, P3 & P4.
P1 mode has the rotational clutch C1 and the band clutch B1 engaged while the rest clutches are disengaged. It uses the output ring member, motor and the engine to drive the vehicle through the output planetary geartrain. Meanwhile, the input planetary geartrain moves freely. As both of the motor and engine are connected to the same link which is grounded and have the same speed, the equation (12) and (13) remains valid whereas equation (14) and (15) becomes,
4 = 411 = 2.6( + ) (16)
3 = (1 + 41)1 = 3.6( + ) (17) Hence, from the equation (16) and (17), we can observe that the reaction torque is 2.6 time the combined motor and engine torques and the output torque is equal to the 3.6 times that of the combined motor and engine torques. Moving forward and neglecting the frictional losses, the output power
can be obtained as follows:
= 3.6( + )3 (18)
Here, the negative sign signifies that the power is flowing out of the transmission.
P2 mode has the rotational clutch C2 and a band clutch B1 engaged while the rest of the clutches are disengaged. It uses the output ring member, motor and the engine to drive the vehicle simultaneously at the second reduction. However, the planetary geartrain is no longer freewheeling. The engine transmits the power through entire geartrain whereas the motor transmits the power through the output planetary gear train. Te speed ratios for this are given by equation (12) and

and substituting N31 = N41 = 2.6 in equations (10) and (11), we get,
4 = 411 + (1 + (41 /1 + 31)) 2
= 2.6 + 0.72 (19)
3 = (1+41)1 (1 + (41 / 1 + 31)) 2
= (3.6 + 1.72) (20) Hence, from the equation (19) and (20), we can observe that the motor torque is amplified by 3.6 times and the engine torque is amplified by 1.72 times at the output shaft. Thus, the
overall power output can be obtained as,
= 3.6( + )3 (21) Here, the maximum power occurs at the different point though the formula is same as the P1 mode as the speeds of the electric motor and the engine are related to the output
shaft.
P3 mode has both rotating clutches C1 and C2 in engaged condition whereas the rest of the clutches are in rest. In this condition, the gear set locks up as the rigid body moving the vehicle with 1:1 ratio. The output torque and output power are given as follows:
3 = ( + ) (22)
= ( + )3 (23)

Engine Mode

One of the following modes are used when the demand for power is low and battery state is sufficiently high to handle accessory loads.
E1 Mode is very much similar to the P1 mode but the motor here is on off mode and is freewheeling. Using the link4 as reaction member, equation (12) and (13) counts just right for the speed ratios of the four coaxial links. As the motor is running freely, motor is zero and thus equation (16) and (17) comes out to be:
4 = 411 = 2.6 (24)
3 = (1 + 41)1 = 3.6 (25) Here, you can see that the engine torque is amplified by

times at the output shaft. This corresponds to the first gear of the conventional automatic transmission. The output power is given as:
= 3.63 (26) E2 Mode is very much similar to the P2 mode but the motor here is on off mode and is freewheeling. Using output ring gear as the reaction member while the engine is the only driving member for the vehicle through the entire gear train at the second reduction. The speed ratios for four coaxial links are given by equation (12) and (13) and substituting 1 = 0 into
the equation (19) and (20) gives,
4 = (41 / 1 + 31)2 = 0.72 (27)
3 = (1 + (41 / 1 + 31))2 = 1.72 (28) And the output power is given by,
= 1.723 (29) E3 Mode locks up the planetary geartrain as the single rigid body and transmits the power to the output shaft at 1:1 speed ratio i.e. the direct drive. In this mode, both C1 and C2 clutches are engaged. The output torque and power are given
by,
3 = (30)
= 3 (31)

Engine/Charge Mode
EC1 Mode has a feature where when the motor is working as the generator, the engine also supports by giving power to the vehicle simultaneously. The engine also powers the electrical accessories and also charge the batteries. The kinematic, torque and power relationships is same as that of the P1 mode, only difference is that the applied torque is negative.
EC2 Mode has a feature where when the motor is working as the generator, the engine also supports by giving power to the vehicle simultaneously. The engine also powers the electrical accessories and also charge the batteries. The kinematic, torque and power relationships is same as that of the P2 mode, only difference is that the applied torque is negative.
EC3 Mode has a feature where when the motor is working as the generator, the engine also supports by giving power to the vehicle simultaneously. The engine also powers the electrical accessories and also charge the batteries. The kinematic, torque and power relationships is same as that of the P3 mode, only difference is that the applied torque is negative.

Continuously variable transmission (CVT) Mode

It is used at moderate to high speed or as the E4 Mode. Here, the rotating clutch C2 is engaged and all rest of the clutches are disengaged. The motor is switched into a generator for charging the batteries. The clutches position makes the input planetary gearset to work as a power splitting one input two output device. The part of the power that the engine has produced is directed to the input ring gear to drive the vehicle and the rest of the part is used to drive the input sun gear to drive the generator. In this mode, engine can run at optimum efficiency point while regulating the speed of the vehicle by controlling the speed and load on the generator. Thus, the vehicle runs as a CVT. The speed ratios for the four coaxial links can be given by the equation (6) and (7) and substituting the value 2.6 for N31 in equation (7) gives,
3 = (1 + (1 + 31)2) / 31
= 0.3846 + 1.3846 (32) So, it is possible to run the engine at an optimal operating speed by proper control of the generator speed for and given output shaft speed 3. Say that a car is moving at the speed of 100 Km/hr and output shaft speed is 2775 rpm. Now, if we want engine to move at 2400 rpm, the generator must be running at 1742 rpm. Here, link4 spins freely and the output planetary geartrain carries no load. Substituting value of 4 as
0 in the equations (10) and (11), we get,
1 = (1 / 1 + 31)2 = 0.27 (33)
3 = (31 / 1 + 31)2 = 0.72 (34)
= 0.723 (35)
= 0.271 (36)

Regenerative Mode
When the brakes are applied, this motor works as the generator and meanwhile 4 regenerative modes are possible.
R0 Mode where only clutch B1 is engaged and rest are disengaged the way the kinematic energy of the whole vehicle is directed through the output planetary gearset for charging of those batteries. During this, the power flows in opposite direction of the motor mode. Maximum recovery of the kinematic energy is achieved when no brakes are applied on the vehicle.
R1 Mode has clutch C1 and B1 engaged and rest of the clutches disengaged. Part of the kinetic energy of whole vehicle is directed to output planetary gearset for charging of the batteries while the rest of the kinematic energy is directed to the engine through the entire gearset. During this, the power flows in opposite direction of the P1 mode. Thus, both engine and generator are responsible for generating the braking efforts on the vehicle.
R2 Mode has clutch C2 and B1 engaged and rest of the clutches disengaged. Part of the kinetic energy of whole vehicle is directed to output planetary gearset for charging of the batteries while the rest of the kinematic energy is directed to the engine through the entire gearset. During this, the power flows in opposite direction of the P2 mode. Thus, again, both engine and generator are responsible for generating the braking efforts on the vehicle.
R3 Mode has clutch C1 and C2 engaged and the rest of the clutches are disengaged. In this setup, the gearset locks up together and forms a rigid body. The power flows in the opposite direction to that of P3 mode. Both engine and the generator will provide the braking efforts to the vehicle at 1:1 ratio.


Shifting between the modes
We know by knowledge of above mentioned details that shifting from M to R0 mode can be achieved very easily by merely changing the motor into a generator and viceversa also hold true. Similarly, shifting amongst P1, E1, EC1 and R1 modes; P2, E2, EC2 and R2 modes; and P3, E3, EC3 and R3 requires only changing of motor to freewheeling state or into a generator. Hence, we will consider M and R0 modes as the motor mode, P1, E1, EC1 and R1 modes as the first gear, P2, E2, EC2 and R2 modes as the second gear, P3, E3, EC3 and R3 as the direct drive. So rearranging the table X on basis of above mentioned classification, we get table X.
It is also possible for a parallel hybrid to have wo different driving schedules i.e. Performance schedule and economy schedule. Where first option provides good performance characteristics on expense of less fuel efficiency, the other option provides better mileage feature. In both cases, transmission starts with motor mode and then shifts to first, second, direct and then overdrive or CVT mode.
CVT mode can be applied to moderate to high speeds for charging of the batteries. When the batteries are charged enough, the transmission is further shifted to overdrive to increase the fuel economy. In each gear, the transmission can be shifted from the power mode to the engine mode or engine/charge mode when power demand is low and to regenerative mode when the brakes are applied. Below is the example for a typical performance schedule, economy schedule and engine charge schedule respectively: [5]
M P1 P2 P3 or E3 CVT M E1 E2 E3 CVT
EC1 EC2 EC3 CVT
Table 2: Results for Max. Speed and Max. Acceleration for different Schedules and their respective Modes
Operating schedule
Power Modes
Max. Speed (m/s)
Max.
Acceleration (m/s2)
Performance schedule
M
14.804
7.548
P1
11.386
13.129
P2
16.121
10.083
P3
45.914
2.904
CVT
47.476
0.243
Economy schedule
M
14.804
7.5485
E1
11.386
5.1467
E2
23.803
2.0431
E3
40.994
0.7677
CVT
47.288
0.24285
Table 3: Clutching Condition for Different Modes re arranged and Condition of Motor Operation at that Time [5]
Operating Mode
Working clutches
Motor mode
C1
C2
B1
B2
Motor
M
X
Motor
R0
X
Generator
First Gear
P1
X
X
Motor
E1
X
X
Free wheeling
EC1
X
X
Generator
R1
X
X
Generator
Second Gear
P2
X
X
Motor
E2
X
X
Free wheeling
EC2
X
X
Generator
R2
X
X
Generator
Direct Drive
P3
X
X
Motor
E3
X
X
Free wheeling
EC3
X
X
Generator
R3
X
X
Generator
CVT
CVT
X
generator


Number of teeth
The criteria while choosing the number of teeth was to make overall reduction from Sun gear to Carrier gear of 3.6. Thus, by Trial and error method, we took the value of teeth on Sun gear as 20 and the value of teeth on planet gear 16, we get the value of teeth on annular ring gear as 52 by the formula stated below.
From the basic diagrams of the planetary gear train, we can say that,
dr = 2 (dp + ds) (37)
But m = d / t, Thus,
tr = 2 (tp + ts) (38)
No: of teeth
Sun gear
20
Planet gear
16
Annular ring gear
52
No: of teeth
Sun gear
20
Planet gear
16
Annular ring gear
52
Table 4: Assumed & Derived Total Number of Teeth of Planetary Gearset

Selection of material
Gears are the important element of a mechanical system, which are used for variation of speed and power, failure of even a single tooth of a gear will make the machine to stop. Hence our aim is to strengthen the gear which is a key element
Step 1:
= 2 / (39)
= [tan / cos] (40)
= tan (41)
of gear box. Thus, we need to select appropriate gear material by considering its strength, cost, hardenability and machinability. The material properties and costing of pinion and gear material were studied, and standard gear materials were identified from PSG Design Data Book. The material sorting is done on the basis of availability, cost and strength of the material.
From the below mentioned materials and their properties, the underlined material i.e. 30Ni4Cr1 or as per the British Standards, EN30A, has very high tensile and yield strength. Which is very useful for beam strength criterion. It also has very high BHN value which helps in wear strength criterion. Even from above mentioned advantages, the cost is also very economical and thus is used as our gear material.
Table 5: Different Types of Materials Available in the Market and their Properties [12]
Indian Standards
British Standards
Su (N/
)
Sy (N/
)
Brinell Hardness (BHN)
Cost (Rs/Kg
)
40Ni2Cr1 Mo28
EN24
950
600
27
80
C45
EN8
70
460
229
80
C55MN75
EN9
780
360
255
67
35Ni1Cr6
0
EN111
850
540
248
80
30Ni4Cr1
EN30A
1550
1330
444
67
40Cr1
EN18
850
540
248
–

Shaft design
Table 6: Given or Assumed Data used for the Calculations of the Shaft Design
In this starting step, as we want to determine the shaft diameter, we have to calculate two values (0.30)Syt and (0.18)Sut on the basis of maximum shear stress theory and select the lower most value amongst the both. The selected value is multiplied by 0.75 which results into the max. This value is the permissible maximum shear stress
Step 2:
Considering horizontal components of forces acting on the shaft,
Figure 2: Horizontal Forces Equilibrium
Taking equilibrium of forces and then taking the moment
@ A, we will get the values for Ra and Rb. Now, similarly we will consider all of the vertical components of forces acting on the shaft and similarly we ill take equilibrium of the forces and take moment @ A, we will get the values for
Data obtained from the fundamental force equation of the helical gear:[]
For planetray gearset 1
For planetary gearset 2
Pt
5331.31 N
Pt
7649.2 N
Pr
2092.83 N
Pr
3002.75 N
Pa
2153.99 N
Pa
3090.5 N
Material for the shaft
45C8
Sut
600 N/mm2
Syt
380 N/mm2
Kb
2
Kt
1.5
Data obtained from the fundamental force equation of the helical gear:[]
For planetray gearset 1
For planetary gearset 2
Pt
5331.31 N
Pt
7649.2 N
Pr
2092.83 N
Pr
3002.75 N
Pa
2153.99 N
Pa
3090.5 N
Material for the shaft
45C8
Sut
600 N/mm2
Syt
380 N/mm2
Kb
2
Kt
1.5
Figure 3: Vertical Forces Equilibrium
Step 3:
Here, Pt, Pr and Pa are the different type of the forces that are acting on the respective planetary gearset while being in operation. Pr is the Radial component of the force, Pt is the Tangential component of the force and the Pa is the thrust component of the force. [13]
In this step, we will find the maximum bending moment for given scenario. As we can see from the figure, bending moment occurs at point C and point E. Among this points, the maximum value for the bending moment occurs at point E
Step 4:
Based on the ASME code made on theory of shear stress failure, the formula for the shaft diameter for different Mt can be given as follows: [13]
D3 = (16 /.) {(kb.Mb)2+(kt.Mt)2} (42)
Table 7: Results for the Shaft Calculations
= ( + ) (49)
max
81 N/mm2
(Mb)c
394376.82 Nmm
(Mb)e max
913087.94 Nmm
d @ 111.8 Nm
48.67
d @ 230 Nm
48.89
d @ 330 Nm
50
max
81 N/mm2
(Mb)c
394376.82 Nmm
(Mb)e max
913087.94 Nmm
d @ 111.8 Nm
48.67
d @ 230 Nm
48.89
d @ 330 Nm
50
= {21(2 + )} / {21 + (2 +)}
(50)
Where Pd is the Dynamic load.
For Wear Strength, effective load can be given as,


Safety considerations

Beam & Wear strength of gears
We assumed value of module as 2.5, 3, 3.5 and 4 and out of this values, values other than 4 came unsafe so by trial and error method, the value of the module for the planetary gear set was assumed 4mm.

Beam Strength
In order to determine the beam strength of the helical gear, it is considered to be equivalent to a formative spur gear
i.e. an imaginary spur gear in a plane that is perpendicular to the tooth element. [14] The formula for finding maximum value of tangential force that can be transmitted by the gear without undergoing the bending failure i.e. beam strength is,
S = mbY (43)
Here, Y is the Lewis form factor which is dependent on the virtual number of teeth.

Wear Strength
In order to determine the wear strength of the helical gear, it is considered to be equivalent to a formative spur gear
i.e. an imaginary spur gear in a plane that is perpendicular to the tooth element. [14] The formula for finding the wear strength of the helical gear is given by,
= / cos2 (44)
Where Q is ratio factor, K is the K factor and is the pressure angle.
= 2 / ( + ) (45)
= / cos (46)
= 0.16[ / 100]2 (47)

Factor of Safety

Factor of Safety can be defined as the ratio of the structures strength to that of the actually applied load. Higher the value of factor of safety, safer is the design in the working conditions. But as the economy is also a factor to keep in mind, usually the Factor of Safety lies between 1.5 and 4. The power can be calculated from the following formula: [14]
= 2 / 60106 (48) For Beam Strength, effective load can be given as,
Peff = CsPt/Cv (51)
Factor of Safety can be determined by the following formula,
= () / (52)
The programming was based on python. The code for helical gear design is given in the Appendix I. The program will calculate the value of constant such as lewis form factor, sum of errors between two meshing teeth etc. automatically from input parameters, i.e. we do not need to refer to the standard tables in order to find constant values. We have developed an algorithm in such a way that it will automatically extrapolate the standard value of module and will ascertain the other dimensions based on that value. In addition to determining dimensions, it is also able to determine whether design is safe for beam and wear strength. Dynamic condition between two meshing teeth are also considered. The results are mentioned in the Table (9).
Table 8: Given & Assumed Data used for the Beam & Wear Strength Calculations
Data that is taken from above calculations or that is assumed are mentioned below:
Zs
20
Zp
16
P
100 KW
N
3000 rpm
Sut
1550 N/mm2
BHN
450
22Â°
Assume:
B
60
mn
4mm
Cs
1.25
20Â°
Table 9: Results of the Beam & Wear Strength
Calculations
Sb
41002.67 N
Sw
17932.767 N
Peff (considering Dynamic loads)
15842.18 N
Peff (not considering Dynamic loads)
15742.33 N
FoS of Beam Strength
2.589


Modelling of the Transmission
The transmission model was made with the results of various calculations done till now. The model was prepared using various platforms as such as Dassault systems SolidWorks 2017 & 2018 and NX 10.
Due to the time constraints, we were unable to model the friction clutch plate but in the fig the detailed location of the friction clutch plate is shown.
Figure 4: Solidworks Model of the Transmission Gearbox (Covered)
Figure 6: SolidWorks Model of the Transmission Gearbox (Uncovered)

Safety Analysis
For the analysis part, we used ANSYS software. This software is very famous and performs very complex analysis and thus is able to give very accurate results in comparison to that of the results from other similar software. Thus, its very common software that is used in various industries and engineering companies.
ANSYS develops and markets finite element analysis software used to simulate engineering problems. ANSYS is used to determine how a product will function with different specifications, without building test products or conducting crash tests. For example, ANSYS software simulate how a bridge will hold up after years of traffic.
Basically, ANSYS users break down larger structures into small components by a process called meshing that are each modelled and tested individually [2]. A user may start by defining the dimensions of an object and then adding weight, pressure, temperature and other physical properties. Finally, the ANSYS software simulates and analyses movement, fatigue, fractures, fluid flow, temperature distribution, electromagnetic efficiency and other effects over time.
For the analysis, the 3Dimensional solid model of gear set is imported in ANSYS as .igs file and analysis is performed by finite element program ANSYS Workbench

Firstly, the 3D solid model is assembly of two mating gear as shown in figures below. The number of elements and nodes generated are 30996 and 142192 respectively. The FE model of gear is shown in figure 8 showing this mesh formed on the gears.
The load is applied in the form of moment. The moment of 330 Nm applied on the faces of the pinion. The frictionless support is applied on the bore of pinion and gear. Frictionless support places a normal constraint on an entire surface. Translational displacement is allowed in all directions except in and out of the supported plane.
The ANSYS version used for the analysis is R15.0 and the input parameters used for it is as follows:
Material EN30A Moment 330 N/mm Module 4mm Thickness 60mm Teeth on pinion 16 Teeth on gear 20
Environmental Temperature 32 C Helix angle 22
Normal pressure angle 20
624.6 MPa and minimum at tip. The material has the strength of 1550 MPa and thus its clear that,
<
= / = 2.48 (52)
Thus,
% = 100 = 0.011% (53)
Figure 7: Meshing Results for the Prepared Helical Gears in the ANSYS

Total deformation
As its shown in the figure below, the deformation occurs maximum at the tip of the tooth. And minimum at the base of the tooth. The red section marked on the tip of the tooth signifies the maximum deformation on the tooth i.e. 0.0335 mm. the maximum deformation of the gear is negligible and thus there is no problem of gear deformation for given material and environment.
Figure 8: ANSYS Results for the Total Deformation of the Gear

Equivalent Stress
Engineering stress is the applied load divided by the original crosssectional area of a material and is also known as nominal stress. The value of Stress determines the fatigue failure for the gearset. As its shown in the figure above, the value of the Stress is maximum at the base of the tooth i.e.
Figure 9: ANSYS Results for the Equivalent Stress

Equivalent elastic Strain

The elastic Strain is pretty much selfexplanatory as we all are familiar with the following formula:
= / ; (54)
Where E signifies Youngs Modulus. From the formula the fact is clear that the Stress and Strain are directly proportional to each other. Thus, Strain will also be maximum at the base of the tooth i.e. 0.003 m/m which is under control. The results from the ANSYS are shown in the figure below.
Figure 10: ANSYS Results for Equivalent Elastic Strain


RESULTS
As we have discussed in the sections before, there will be two schedules on which the hybrid car would be driven. Those modes are Performance schedule and the economy schedule. The above mentioned graph is for the Performance schedule. This graph shows the maximum speed and maximum acceleration for the vehicle at different modes in Performance schedule. This mode will give more performance characteristics while reducing the fuel efficiency figure. The Xaxis defines various modes and Yaxis defines values for maximum velocity and maximum acceleration in km/h and m/s2 respectively.
The following graph has similar detailing on X and Y axis and units. The graph below is for economy schedule. This schedule wont give high performance as in Performance schedule but the fuel consumption efficiency would be better than the Performance schedule. Thats the reason why the acceleration is less than that in the Performance schedule.
Figure 11: Graphical Representation of the Results for Performance Schedule
Figure 12: Graphical Representation of the Results for Economy Schedule
The following graph is the combination of both of the above graphs. Thus, it shows performance as well as the economy schedule and their maximum acceleration and maximum velocity at each modes of operation.
Figure 13: Combined Graphical Representation of the Results

SUMMARY
There are 2 planetary gearset and 4 clutches used in this transmission for which feasible number of teeth as well as shaft is designed as per the requirements. This design is feasible even in terms of safety considerations, for this the beam and wear strength was calculated theoretically as well as with the help of the software, namely Python and ANSYS. The values of the factor of safety for Python and theoretical calculations were nearly same whereas the value of the same in theoretical calculations and ANSYS had error of mere 0.011%.
This design can be used for RWD as well as FWD vehicles. In case of the RWD vehicles, transmission is inline to the final reduction drive whereas in FWD vehicles, the transmission is coaxial to the final differential unit.
This transmission design has 15 in total modes of operation under 2 schedules. This 2 schedules are used according to the demand of the situation. The name of this schedules explains the output itself, Performance schedule & Economic schedule.
This transmission design has many advantages & features. Some of which are compactness due to the integrated motor design, high efficiency, high reliability & availability of different modes of operation.
REFERENCES

https://en.wikipedia.org/wiki/Hybrid_vehicle.

MIT Electric Vehicle Team, Electric Powertrains, April 2008.

https://www.conserveenergyfuture.com/advantagesand disadvantagesofhybridcars.php.

L. W. Tsai and G. Schultz, A motorintegrated hybrid transmission University of California Energy Institute, U. S.A, October 2002.

http://www.motioncontrolguide.com/learn/faqs/gearboxesand gears/planetarygearbox/whatarethedisadvantagesofaplanetary gearbox/.

G. Schultz, L. W. Tsai, N. Higuchi and I. C. Tong, Development of a novel parallel hybrid transmission SAE International, Michigan, March 2001.

Dr. Enrico Galvagno, Epicyclic geartrain dynamics including mesh efficiency Levrotto & Bella, Torino (Italy), January 2010.

L. W. Tsai, G. Schultz and N. Higuchi, A novel parallel hybrid transmission ASME, Japan & U. S. A, June 2001.

L. W. Tsai, Enumeration of the kinematic structures according to the functions CRC Press, U. S. A, 2001.

http://www.motioncontrolguide.com/learn/faqs/gearboxesand gears/planetarygearbox/whatarethedisadvantagesofaplanetary garbox/.

PSG Data Hand Book.

V. B. Bhandari, Design of machine elements 3rd Edition, Chapter 9, 330344, Mcgraw Hill Publications, India, 2014.

V. B. Bhandari, Design of machine elements 3rd Edition, Chapter 18, 694703, Mcgraw Hill Publications, India, 2014.