HCM UNIVERSITY OF TECHNOLOGY

Faculty of Transportation Engineering

Department of Automotive Engineering
---------------------

AUTOMOTIVE DESIGN METHODS

PROJECT
CLUTCH SYSTEM

Group: 6

Class: CC01
Students:
PHAN THANH PHUC 2053345
PHAN HOANG TRUNG KIEN 2052556
NGUYEN HUYNH TUAN KIET 2052562

Lecturer: Assoc. Prof. Dr. HONG DUC THONG

 Table of Contents
1. WORKING CONDITIONS & DESIGN REQUIREMENTS ................... 2
2. PRESENT, ANALYSE PLANS AND CHOOSE PLAN........................... 2
3. GENERAL LAYOUT DESIGN .................................................................. 6
4. TECHNICAL DESIGN ..............................................................................10
5. DESIGN OF TECHNOLOGY ..................................................................21
6. ECONOMIC OF DESIGN ........................................................................21
REFERENCES.................................................................................................21

1
1. Working conditions & design requirements
1.1. Working conditions

 The component is designed to withstand heavy loads, ensuring its durability and
reliability under significant stress. This characteristic is essential for applications
that involve substantial weight or pressure over extended periods.
 It is built to endure dynamic loads, allowing it to perform consistently in
conditions where forces fluctuate. This adaptability makes it suitable for
environments with constant motion or shifting forces.
 The system operates in a state of continuous engagement and disengagement,
highlighting the need for precision and durability in its design. These repetitive
cycles place additional strain on the component, necessitating robust materials and
engineering.
 Friction contributes significantly to wear and tear, especially during prolonged
operation. This issue is exacerbated by the high temperatures generated by the
frictional forces during use.
 The component is subject to damage from environmental factors, including
moisture or oil leaks from adjacent systems. These elements can accelerate wear
and reduce the overall lifespan of the component if not properly managed.

1.2. Design requirements

 Ensure Quick Disengagement and Smooth Engagement:
The system must be capable of rapid disengagement to enable efficient gear
changes without causing delays during operation. Equally important is the smooth
engagement of the clutch, which minimizes jerks and ensures a comfortable
driving experience.
 Use Friction Materials Capable of Maintaining a Stable Coefficient of Friction
Within the Operating Temperature Range:
The friction materials chosen should perform consistently, maintaining a stable

2
 coefficient of friction across varying temperatures. This stability is essential to
ensure reliable performance and prevent unpredictable operation during extreme
or prolonged use.
 Provide Good Heat Absorption and Dissipation Properties:
Components must have excellent heat absorption capabilities to handle the thermal
energy generated by friction during operation. Additionally, efficient heat
dissipation is necessary to prevent overheating, which could compromise the
system's longevity and performance.
 Friction Torque Must Be Sufficient to Transmit the Engine's Torque, but not
excessively high to allow the clutch to act as a safety mechanism in Cases of
Reverse Torque from the Wheels:
The clutch system must transmit the engine's torque effectively to ensure optimal
performance. At the same time, the friction torque should remain within safe limits
to allow the clutch to disengage during reverse torque scenarios, protecting the
drivetrain from potential damage.
 Employ a Reasonable Clutch Actuation Method to Keep the Pedal Force Within
Prescribed Limits:
The clutch actuation mechanism must be designed to minimize the force required
to operate the pedal. This ensures that the system remains user-friendly, reducing
driver fatigue during extended periods of use.
 Order Requirements: The system must adhere to all relevant regulations and
standards applicable to automobile design. Compliance ensures safety, reliability,
and compatibility with current industry practices and legal requirements.
 General Requirements: robustness and Reliability - The system should
demonstrate high durability, withstanding normal and extreme operating
conditions without failure. This requires the use of high-quality materials and
precision manufacturing techniques to ensure long-term reliability.

3
2. Present, analyze plans and choose plan
2.1. Present plans
 Issues that require building the plans:
 Moment transmission method
 Working state
 Method of generating force on the pressure plate
 Friction types
 Clutch actuation types
2.2. Analyze plans
 Moment transmission method

Type Advantages Disadvantages

Friction - Efficiency transmission is equal - Wear out after a period of use.
to 1. - Generate a significant amount of
- Capable of transmitting large heat.
torque.
- Simple design, easy to
manufacture.

Hydraulic - Can replace the engine flywheel, - The weight of the hydraulic clutch
reducing the vehicle's overall is higher.
weight. - Transmission efficiency is low.
- Reduces dynamic load on the
drivetrain.
- No need for engagement or
disengagement control.

4
  Working state:
- Normally engaged clutch: Primarily used in cars and high-speed
engines.
- Non-normally engaged clutch: Mainly used in tractors and low-speed
engines. The passive part has a large moment of inertia.
 Method of generating force on the pressure plate

Type Advantages Disadvantages

Centrifugal - The engagement and - Not suitable for transmitting large
type disengagement process occurs torque.
automatically. - Maybe slipping and overheating,
- Smooth operation process. with low transmission efficiency
(high load, low speed).
Cylindrical - The pressure force can be easily - The structure is complex.
spring type adjusted. - The size of the clutch will be
- Easy to manufacture. larger.
- Low cost. - The pressure force distribution is
uneven.
Conical - Generates an even pressure force - Clutch shaft area becomes
spring type on the friction surface. cramped, making it difficult to
(centered) - Can shorten the length of the arrange the clutch release bearing."
clutch. - Requires a large clutch
disengagement force.
- When the clutch wears, the
pressure force decreases
significantly.
Diaphragm - Can generate an even pressure - More complex and harder to
spring type on the friction surface." manufacture.
- Simpler clutch structure.

5
 - Clutch disengagement force is
not large, and the pressure force
does not change much.

 Friction types

Type Advantages Disadvantages

Wet type - Durability and wear are lower - Low friction coefficient of 0.1.
compared to dry clutches. - Power transmission area is
- Better heat dissipation. reduced.
- Smoother engagement and
disengagement.
Dry type - High friction coefficient of 0.3. - High temperature.
- Wears faster compared to wet - Operation is not smooth.
clutches. - Requires protection against wear
due to humidity.

 Clutch actuation types

Type Advantages Disadvantages

Mechanical - Simple, easy to manufacture. - The structure depends on the
drive - Easy to disassemble and repair. clutch installation position.
- High reliability. - Requires a large force to be
applied to the pedal.
- Low efficiency.
Hydraulic - Compact structure, simpler - Risk of oil leakage, which can
drive installation compared to the cause the clutch to lose its
mechanical type. engagement and disengagement
capability.

6
 - Easily amplifies the driver's - Requires high precision in the
applied force. hydraulic drive system.

Pneumatic - Suitable for vehicles equipped - Increases the complexity of the

power- with a pneumatic system. system structure.
assisted - Reduces the force required on the -Maintenance, repair, and
pedal. adjustment are more difficult.
- Shortens the pedal travel - Low precision.
distance.

2.3. Choose plan

We choose the dry friction clutch with normally engaged operation using cylinder springs
and a hydraulic pressure spring actuation mechanism.

7
3. General layout design
3.1. General layout dimensions:

1. Flywheel housing
2. Clutch housing
3. Compressed spring
4. Flywheel
5. Gearbox input shaft
6. Damping spring
7. Driven disc
8. Clutch release lever
9. Pressure plate
10. Damping disc with an oil deflector
11. Clutch release bearing

3.2. Preliminary calculation:

- To determine the general layout parameters in mechanical systems or
mechanisms, the following kinematics and dynamics problems need to
be solved:
 Calculation of Sliding Work:
𝐿0 = 𝐿1 + 𝐿2

8
𝐿1 (J) is the sliding work during the first stage of increasing the clutch torque from
0 to the resistance value at shaft Ma, at which point the car begins to start from
rest.

𝜔𝑚 − 𝜔𝑎
𝐿1 = 𝑀𝑎 𝑡1
2

𝐿2 (J) is the sliding work during the second stage when the clutch torque increases
from the resistance value to the engine torque value, at which point slipping no
longer occurs.

1 2
𝐿2 = 𝐽𝑎 (𝜔𝑚 − 𝜔𝑎 )2 + 𝑀𝑎 (𝜔𝑚 − 𝜔𝑎 )𝑡2
2 3

- The times t1 and t2 are calculated as follows:

𝑀𝑎
𝑡1 =
𝑘

𝐴
𝑡2 =
√𝑘

𝜔𝑚 : Flywheel speed

𝜔𝑎 :: Clutch shaft speed

𝑀𝑎 : Resistance torque referred to the clutch shaft

𝐽𝑎 : Inertia of the vehicle referred to the clutch shaft

- The coefficient k has the following values:

For passenger cars: k= 50 ~ 150 Nm/s2

For trucks: k= 150 ~ 750 Nm/s2

- The value of A can be calculated using the following formula:

𝐴 = √2𝐽𝑎 (𝜔𝑚 − 𝜔𝑎 )

9
  Calculation of Maximum Friction Torque:
𝑀𝑙 = 𝛽𝑀đ = 𝜇𝑃𝑅𝑡𝑏 𝑝

𝑀𝑙 : Clutch friction torque (Nm)

𝑀𝑑 : Engine torque. For cars taken as the maximum engine torque (Nm)
𝛽: Clutch reserve coefficient can be selected according to the table below.
Passenger car:
Transport, no work with a trailer: 1.3 ~1.5
Transport, with work with a trailer: 1.6 ~ 2.25
Transport, heavy work with a trailer: 2 ~ 3

𝜇 : Coefficient of friction
𝑅𝑡𝑏 : Average friction radius (m)
𝑝: Number of friction surface pairs
𝑃 : Force applied to the friction discs (N)

𝛽𝑀đ
𝑃=
𝜇𝑅𝑡𝑏 𝑝

- Calculation of outer radius(R2) and inner radius(R1) of the friction ring:

2 𝑅23 − 𝑅13
𝑅𝑡𝑏 =
3 𝑅22 − 𝑅12

3.16 𝑀𝑒𝑚𝑎𝑥
𝑅2 = √
2 𝐶

Coefficient C, which is based on empirical experience for different types of vehicles and
usage conditions:

10
 For passenger cars: C=4.7

For transport vehicles under normal conditions: C=3.6

For self-loading trucks and transport vehicles under heavy-duty conditions: C=1.9

𝑅1 = (0.53 ÷ 0.75)𝑅2

0.53 for engines with low rotational speed

0.75 for vehicles with high rotational speed

 Determine the friction coefficient:

- Typically, automotive clutches use friction surfaces made of steel with

Ferodo or copper Ferodo. Considering the decrease in the coefficient of
friction due to rising temperatures and relative slipping, the calculated
coefficient of friction is reduced to 0.25-0.3
- Determine the number of friction surfaces

The number of friction pairs can be chosen based on the existing structure or determined
using the following formula:

𝛽𝑀đ
𝑝= 2
2𝜋𝑞𝜇𝑏𝑅𝑡𝑏

11
b: Width of the friction plate (m) (b = R₂ - R₁)

q: Allowable pressure on the friction surface for the material chosen from the table.

 Mass of the heated components

To account for the operating conditions of the clutch, we use the value of specific sliding
work:

𝐿0
𝑙=
𝐹0

L₀: Sliding work (J)

F₀: Friction surface area (m²), F₀ = (R₂² - R₁²)

l: Specific sliding work (J/m²)

 Sliding work is checked according to the following data:

Transport vehicles with a load up to 50 kN: 150 - 250 kN/m²

Transport vehicles with a load greater than 50 kN: 400 - 600

Passenger vehicles: 1000 – 1200

 In addition to specific slip work, the increase in temperature of heated components

is also calculated, as elevated temperatures significantly increase the wear of these
components.
𝑣𝐿0
𝑇=
𝑐𝑔𝑛
𝑇: Temperature increase of a component (K), not exceeding 8-10 Kelvin
c: Specific heat capacity of cast iron and steel, c ≈ 500 J/kg·K
𝑔𝑛 : Mass of the heated component (kg)
𝑣: A coefficient used to determine the part of the sliding work for heating the
component, which can be calculated as follows:

12
 1
𝑣= for the pressure plate (n is the number of driven disks)
2𝑛
1
𝑣 = for the intermediate driving disks (n is the number of driven disks)
𝑛

The temperature T can be reduced by increasing the component mass or by

ventilating through the clutch windows.

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4. Technical design
4.1. Driven plate:
- Area clutch disc:

𝐴𝜇 = 𝜋(𝐷 2 − 𝑑 2 )𝑧𝜇

- Strength of driven plate:

 Rivet

Material: aluminum or bronze.

Diameter: 4 – 6 mm.

Strength: stamping stress and shear stress.

Distribution : arrange multi-row circle. ( 2-row )

𝑀𝑒𝑚𝑎𝑥 𝑟1
𝐹1 = force affecting on rivet
2(𝑟12 +𝑟22 )

𝑀𝑒𝑚𝑎𝑥𝑟2
𝐹2 = force affecting on rivet
2(𝑟12 +𝑟22 )

In that, 𝑟1 𝑎𝑛𝑑 𝑟2 is a radius of circle distribution of rivet on passvie plate

4𝐹1 4𝐹2
𝜎𝑐1 = ; 𝜎𝑐2 = – shear stress with limiting 𝜎𝑐 ≤ 40𝑀𝑃𝑎
𝑛1 𝜋𝑑2 𝑛2 𝜋𝑑2

4𝐹1 4𝐹2
𝜎𝑐𝑑1 = ; 𝜎𝑐𝑑2 = - stamping stress with limiting 𝜎𝑐𝑑 ≤ 25𝑀𝑃𝑎
𝑛1 𝑙𝑑 𝑛2 𝑙𝑑

Where: d – diameter of rivet

𝑛1 – number of rivet that was arranged in the circle having radius 𝑟1

𝑛2 – number of rivet that was arranged in the circle having radius 𝑟2

The rivet link the passive plate with spline and are manufactured by steel with the
diameter 6 – 10 mm. The calculation of this rivet is the same above equation, but the
stress limitation is :

14
 𝜎𝑐 ≤ 30 𝑀𝑃𝑎

𝜎𝑐𝑑 ≤ 80 𝑀𝑃𝑎

 Passive hub:

- The hub usually is designed with a suitable parameter to reduce

unstable, the length of hub is chosen equal to diameter of spline in
clutch shaft: 𝑙 = 𝐷

If the clutch work in hardness condition so: 𝑙 = 1.4𝐷

- The strength of hub’s spline is calculated according to shear and

stamping stress.
8𝑀𝑒𝑚𝑎𝑥
𝜎𝑐𝑑 = ; 𝜎𝑐𝑑 ≤ 20 𝑀𝑁/𝑚2
𝑧1 𝑧2 𝑙(𝐷 2 −𝑑2 )

4𝑀𝑒𝑚𝑎𝑥
𝜎𝑐 = ;𝜎𝑐 ≤ 10 𝑀𝑁/𝑚2
𝑧1 𝑧2 𝑙𝑏(𝐷+𝑑)

Where: b – the length of spline

𝑧1 – number of hub

𝑧2 – number of spline

D – the outer diameter of spline

D – the inner diameter of spline

 Clutch shaft:

Clutch sually is primary shaft of transmission, in the end of this shaft have a shaft gear.

Material: steel using for gear.

Calculating strength of shaft according to total torsion and bending stress:

15
 √𝑀𝑢2 + 𝑀𝑥2
𝜎= ≤ 50 − 70 𝑀𝑃𝑎
0.1𝑑 3

Where :

𝑀𝑢2 - bending moment at a dangerous cross – sectional area

𝑀𝑥2 - torsion moment at a dangerous cross – sectional area

d - diameter of shaft

 Pressure plate
𝑊𝜇 4𝑊𝜇
Specified Sliding work: 𝜔𝜇 = =
𝐴𝜇 𝜋(𝐷 −𝑑2 )𝑧𝜇
2

Where:

D – outer diameter of friction plate (m)

D – inner diameter of friction plate (m)

𝑧𝜇 – coefficient of pairs of friction surfaces

Sliding work: 𝑊𝜇 = ∫ 𝑀𝑐 (𝑡 )(𝜔𝑑 − 𝜔𝑎 )𝑑𝑡

Where:

𝑀𝑐 (𝑡 ) – friction moment of clutch, changing from 0 to maximum value when engaging

𝜔𝑑 – angular motion of active plate

𝜔𝑎 – angular motion of passive plate

𝛾𝑊𝜇
Temperature: ∆𝑡 =
𝑐 𝑚𝑑

Where : 𝛾 – portion of heat transfer for pressrue plate

With: 1-disc clutch 𝛾 = 0.5

16
 2-disc clutch 𝛾 = 0.25 for pressure plate and 𝛾 = 0.5 for
intermediate disc

𝑚𝑑 – mass of plate (kg)

𝑊𝜇 – sliding work of clutch (J)

c – specified heat of material (steel and cast iron: 𝑐 = 481.5 𝐽/𝑘𝑔℃ )

 Spring
- Torsion damper of clutch disc:

𝑀𝑔 = 𝑀𝑙𝑥 + 𝑀𝑚𝑠 = 𝑃𝑙𝑥𝑔 𝑅𝑙𝑥 𝑧𝑙𝑥 + 𝑃𝑚𝑠 𝑅𝑚𝑠 𝑧𝑚𝑠

Where: 𝑀𝑙𝑥 − The moment of spring forces

𝑀𝑚𝑠 − The moment of friction (𝑀𝑚𝑠 = 0.25𝑀𝑚𝑎𝑥 )

𝑃𝑙𝑥𝑔 − The force exerted by a single damper spring.

𝑅𝑙𝑥 − The radial distance where the spring force acts.

𝑧𝑙𝑥 − Number of damper springs.

𝑃𝑚𝑠 − The force exerted on a single friction ring.

𝑅𝑚𝑠 − The radial distance where the frictional force acts.

𝑧𝑚𝑠 − number of friciton ring

 In the calculation process, the maximum value of the damping torque is usually
taken according to the traction torque, calculated from the active wheels attributed
to the clutch shaft

𝐺𝑏𝑥 𝜑
𝑀𝑔 ≤ 𝑀max = 𝑟
𝑖0 𝑖1 𝑖𝑓1 𝑏𝑥

Where: 𝐺𝑏𝑥 − the weight acting on the driven wheels

17
 𝜑 − traction coefficient (usually equal to 0.8)

𝑟𝑏𝑥 − working radius of wheels

𝑖0 − Final drive ratio

𝑖1 − Gear ratio for 1st

𝑖𝑓1 − Gear ratio of secondary gearbox at low gear ratio level

Arrangement of spring: distributing in the circle with the range of the radius 80 – 120 mm
Types:
Number of spring: 6 -10 springs

 Release fork

Material: cast iron and steel

The durability test is calculated according to bending stress with any cross-sectional by x
axis

18
 𝐹Σ′ 𝑎𝑥
𝜎𝑢 =
𝑧𝑑𝑚 𝑐𝑊𝑢

Where:

x, a, c – the parameter according figure

𝑧𝑑𝑚 - number of release fork

𝑊𝑢 – The momnet of anti-bending at a calculated cross-sectional area

 Release bearing
- The release bearing is a crucial component that connects and
disconnects the engine from the transmission in a vehicle.

Structure: Inside the release bearing, there are components such as ball bearings and
thrust washers which can be made of sintered metal.

- Function: These components facilitate smooth movement of parts,

reducing friction and wear.
- Lubrication: To ensure smooth operation, the components of the release
bearing require regular lubrication. Common lubrication methods
include: Direct grease injection into the bearing, grease injection
through a flexible tube into the grease reservoir, using of felt oil seals
and bearings without grease fittings are often housed in a steel shell and
sealed with a grease reservoir.
 Clutch body
- The clutch body is bolted to the flywheel, providing a solid base for
other clutch components.
- The clutch body acts as a support for the pressure plate springs, helping
to apply pressure between the friction disc and the flywheel.

19
 - It serves as a bearing for the release levers, facilitating the engagement
and disengagement of the clutch.
- It has ventilation holes to help cool the internal components of the
clutch.
- The clutch body is typically made of low-carbon steel, although cast
steel or cast iron may be used in some cases.
 Clutch housing
- Structure:

The clutch housing and transmission case are integrated into a single unit.

The clutch housing is a separate component.

The shape, size of the housing, and ventilation openings vary depending on the
structure of clutch.

The clutch housing is aligned with the engine via dowels on the flywheel housing
and with the transmission via the flange on the primary shaft bearing cover.

- Materials:
One-piece types are typically made of cast iron.
Composite types are made of cast iron or have a stamped steel lower half.
 Clutch balancing
- The accuracy of clutch balancing depends on several factors, including:
Rotational speed of the engine, clearances between components and
clutch size.
- Friction discs are often balanced together with the wheel hub, and this
practice is more common in passenger cars.
- A common balancing method involves:
- Drilling: By drilling small holes in specific locations on the disc, it is
possible to reduce mass in the required areas to achieve dynamic
balance.

20
  Control mechanism
- Structure and type of clutch:

The quality of clutch mechanism is evaluated by many parameters: the work done by the
driver to disengage the clutch 𝑊𝑐 , the applied maximum force on the clutch pedal 𝑄𝑏đ𝑚𝑎𝑥
and the highest distance of clutch pedal 𝑆𝑏đ𝑚𝑎𝑥 .

With sedan and SUV

𝑊𝑐 ≤ 23 𝐽 ; 𝑄𝑏đ 𝑚𝑎𝑥 ≤ 150 𝑁; 𝑆𝑏đ 𝑚𝑎𝑥 ≤ 140 − 160 𝑚𝑚.

With truck, lorry and passenger vehicle

𝑊𝑐 ≤ 30 𝐽 ; 𝑄𝑏đ 𝑚𝑎𝑥 ≤ 250 𝑁 – no power assist; 𝑄𝑏đ 𝑚𝑎𝑥 ≤ 150 𝑁 - with power assist

𝑆𝑏đ 𝑚𝑎𝑥 ≤ 180 𝑚𝑚

- Many types of clutch mechanism: mechanical mechanism for sedan and

truck, hydraulic mechanism for almost sedan and passenger vehicle.
- To reduce the force applying on the clutch pedal, the system are
arranged the assist power systems for clutch pedal such as mechanical,
hydraulic, pneumatic and vacuum mechanism. Nowadays, the sedans
used hydraulic clutch with vacuum assisting power system, while truck
used pneumatic assisting power system.
- Calculating the clutch mechanism
- Calculating the transmission ratio from clutch pedal to pressure plate:

𝑆𝑙𝑣
𝑖𝑑đ = = 𝑖𝑏đ 𝑖𝑑𝑡 𝑖𝑐𝑚 𝑖𝑑𝑚
∆

Where: 𝑆𝑙𝑣 – the working distance of clutch pedal

∆ - the distance of pressure plate

𝑖𝑏đ – the ratio of clutch pedal

𝑖𝑐𝑚 – the ratio of the opening fork (1.5 – 2.3)

21
 𝑖𝑑𝑚 – the ratio of the release forks (3.8 – 5.5)

𝑖𝑑𝑡 – the ratio of the driving part being from behind the clutch pedal to the
opening fork ( approximately 1)

According to hydraulic clutch:

2
𝑥𝑐 𝑑𝑐ℎ
𝑖𝑑𝑡 = = 2
𝑥𝑐ℎ 𝑑𝑐

𝑥𝑐 – the replacement of piston

𝑑𝑐 – the diameter of the main cylinder’s piston

𝑥𝑐ℎ - the replacement of actuated piston

𝑑𝑐ℎ - the diameter of actuated cylinder’s piston

- The total of clutch pedal:

′
𝑆𝑏đ = 𝑆𝑙𝑣 + 𝑆𝑡𝑑 = ∆𝑖𝑑đ + 𝛿𝑖𝑑đ

Where: 𝛿 – the clearance

′
𝑖𝑑đ - the ratio of actuating part from clutch pedal to opening fork

𝑆𝑡𝑑 – the freedom distance

𝑆𝑙𝑣 = 20 − 25% 𝑆𝑏đ

- The applying maximum force on clutch pedal:

𝐹Σ′ 𝐹𝑙𝑥
𝑄𝑏đ = +
𝑖𝑑đ 𝜂 𝑖𝑙𝑥 𝜂𝑙𝑥

Where: 𝜂 – the efficiency of total mechanism

𝐹𝑙𝑥 – the force of return spring

22
 𝑖𝑙𝑥 𝑎𝑛𝑑 𝜂𝑙𝑥 – the ratio and efficiency of actuating part from return spring to
clutch pedal, respectively.

- If applying force on clutch pedal larger than limited value : 𝑄𝑏đ > [𝑄𝑏đ ]
 We need to design the applying power system for the clutch system
- The force of cylinder on the applying power system – 𝐹𝑡𝑙

𝐹𝑡𝑙 = (𝑄𝑏đ − [𝑄𝑏đ ])𝑖𝑡𝑙 𝜂𝑡𝑙

Where: 𝑖𝑡𝑙 𝑎𝑛𝑑 𝜂𝑡𝑙 – the ratio and efficiency of actuating part from the clutch pedal to the
position of the cylinder in the assisting power system, respectively.

- The diameter of the cylinder in the assisting power system:

𝐹𝑡𝑙
𝐷 = 2√
𝜋𝑝𝑖

Where: 𝑝𝑖 = 0.65 − 0.75 𝑀𝑃𝑎 with the pneumatic assisting power system

𝑝𝑖 = 0.05 𝑀𝑃𝑎 with the vacuum assisting power system

 The diameter of the cylinder in the assisting power system depend on the position of this
on mechanism system
- Calculation sequence
- Mechanical clutch
1. Choosing the applying force on the clutch pedal is suitable for Vietnamese regulation and
ensuring light control
𝑄𝑏đ ≤ [𝑄𝑏đ ]
2. The formula of the transmission ratio:
𝐹Σ′
𝑖𝑑đ =
𝑄𝑏đ 𝜂𝑑đ
With 𝜂𝑑đ = 0.5 − 0.8
The ratio of actuating part is distributed depending on specified mechanism:

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 𝑎2 𝑏2 𝑐2
𝑖𝑑đ =
𝑎1 𝑏1 𝑐1

3. Testing the distance of clutch pedal is suitable for 𝑆𝑏đ ≤ [𝑆𝑏đ ]:

𝑏2 𝑐2
𝑆𝑏đ = ∆𝑖𝑑đ + 𝛿
𝑏1 𝑐1
Where: 𝛿 = 2 − 4 𝑚𝑚

∆ - took in the position when the clutch is completely disengaged.

∆= 1.5 − 2.0 with 1 – disc clutch

∆= 2.0 − 3.0 with 2 – disc clutch

- Hydraulic clutch:

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The calculated process of Hydraulic clutch is same with mechanical clutch, however the
ratio of mechanism is add the ratio of hydraulic part.

𝑎2 𝑏2 𝑐2 𝑑22
𝑖𝑑đ =
𝑎1 𝑏1 𝑐1 𝑑12
- The distance of the clutch pedal:
𝑏2 𝑐2 𝑑22
𝑆𝑏đ = ∆𝑖𝑑đ + 𝛿
𝑏1 𝑐1 𝑑12
- The clutch with the assisting power system

According to The clutch with the assisting power system, the diameter of cylinder need
determine.

- The reaction force from the clutch pedal is 𝑄𝑛𝑙 , so the assisting force is
calculated:
𝑄𝑡𝑙 = 𝑄𝑏đ − 𝑄𝑛𝑙
- Ensuring the standard, we need 𝑄𝑛𝑙 = [𝑄𝑏đ ], however reducing the reaction
force for the driver, we need choose 𝑄𝑛𝑙 < [𝑄𝑏đ ]

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 - The force at the cylinder of the assisting power system is calculated to the
arrangement of the cylinders:
𝑐2
𝑄𝑥𝑙 = (𝑄𝑏đ − 𝑄𝑛𝑙 ) 𝜂
𝑐1
With 𝜂 : the efficiency of actuating part from the clutch pedal to the position of actuating
cylinder
The diameter of cylinder assiting power:

𝑄𝑥𝑙
𝑑𝑡𝑙 = 2√
𝜋𝑝𝑖

With 𝑝𝑖 : the pressure the pneumatic mechanism

5. Design of technology
6. Economy of design

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REFERENCES

1. Hồ Tấn chuẩn, Nguyễn Đức Phú, Trần Văn Tế, Nguyễn Tất Tiến: Kết cấu và
tính toán động cơ đốt trong tập III, (1996).
2. Nguyen Huu Can, Phan Dinh Kien. “Thiết kế và tính toán ô tô máy kéo”

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