3 Final Year Project Lastly Edited

Final project on
MODELING AND DESIGN OF A SEAT SUSPENSION (ISOLATOR)

OF A TRUCK DRIVER
By: Taddesse Fentie

Advisor:
Ir. Fisseha Meressa June 2007
(M.Sc.Lecture, Mechanical eng’g department)
Modeling and Design of a Chair
Suspension (isolator) of a truck driver
Acknowledgement
Before any thing, I would like to thank my advisor Ir.Fisseha Meressa, lecturer at Mekelle
University Department of Mechanical Engineering for his constructive, my project centered
advice and providing me reference material that are crucial for my project progress and
successful completion.
Secondly, I would like to thank Fanuel who works in Mesfin Industrial Engineering, for his
polite reception and giving me relevant information for my project. And also I want to thank
Mesfin Industrial Engineering PLC
And I want to thank Ato Zeray and Alem Tekle who are foremen in maintenance Department in
Sure Construction Company and spare part workers, mechanics (technicians).
Finally, I would like to thank Mekelle University for last five years giving me educational
services.
Final year project by Taddesse Fentie

Abstract
Seat dynamics is one of the most critical elements affecting truck ride comfort. Good
measurement and evaluation methods for truck seat characteristics are important tools in the
development of better driver environment.
This project mainly focused on the design of passive seat suspension system and the study of
responses of each of ten DOF modeling of the truck. In this project the responses of the driver
seat with different damper has been done using mat lab. The displacement response’s of the seat
that has high damping rate decays out within a short period of time.

List of Symbols
1-D one dimensional
2-D two dimensional
3-D three dimensional
DOF degree of freedom
FBD free body diagram
ISO international organization for standardization
WBH whole body vibration

Tables of content
1. Introduction......................................................................................................................1
1.1 Vibration Theory........................................................................................................1
1.2 Elementary Parts Vibratory System...........................................................................1
1.2.1 Spring Elements..................................................................................................1
1.2.2 Mass or Inertia Element......................................................................................2
1.2.3 Damping Element...............................................................................................2
1.3 Classification of Vibration System............................................................................2
1.3.1 Forced /free vibration.........................................................................................2
1.3.2 Damped and Undamped Vibration.....................................................................3
1.4 Vibration Control.......................................................................................................3
1.5 Suspension System....................................................................................................3
1.5.1 Purpose of Suspension System...........................................................................4
1.5.2 Suspension Springs.............................................................................................4
1.5.3 Dampers: Shock Absorbers.................................................................................5
1.5.4 Types of Suspension System...............................................................................7
2 LITERATURE REVIEW..................................................................................................8
2.1 Human Being Comfort Index of Vibration................................................................8
2.2 Controlling Effects Whole Body Vibration............................................................10
2.3 Geometry of Seat.....................................................................................................10
3. Design of truck driver’s seat..........................................................................................14
3.1 Introduction..............................................................................................................14
3.2 Types of Seat Suspension System............................................................................15
3.3 Passive Seat Suspension System Design for Truck Driver......................................17
3.4 Seat Suspension Spring Design...............................................................................22
3.5 Seat Bolt Design......................................................................................................24
4. Modeling of Truck and Study of Overall Response......................................................26
4.1 Truck Modeling.......................................................................................................26
4.1.1 Modeling Assumptions.....................................................................................27
4.2.1 Rear Wheel Modeling.......................................................................................29
4.2.2 Front Wheel Modeling......................................................................................31
4.2.3 Modeling of the Chassis...................................................................................31
4.2.4 Modeling of the Engine....................................................................................36
4.2.5 Modeling of Cabin............................................................................................38
4.2.6 Modeling of the Driver Seat.............................................................................41
4.2.7 Modeling of Loading Area................................................................................42
4.3 Road Profile.............................................................................................................43
4.4 Mat lab Analysis......................................................................................................45
5. CONCLUSION..............................................................................................................56
6. Recommendation...........................................................................................................57
Bibliography......................................................................................................................58
1. INTRODUCTION
1.1 Vibration Theory

Any motion that repeats itself after an interval of time is called vibration or oscillation. The
theory vibration is based on the study of oscillatory motion bodies and the effects of forces
associated with them.
1.2 Elementary Parts Vibratory System

Mass, spring and dampers are the most fundamental building blocks for the modeling, design and
analysis of vibratory system. A vibratory system in general, includes spring a means for storing
of potential energy (spring elasticity), a means for storing kinetic energy (mass or inertia) ,and a
means by which energy is gradually lost (dampers).
The transfer of potential energy to kinetic energy and kinetic energy to potential energy occur
during oscillation of vibratory system. Some form of energy is dissipated in each cycle of
vibration if the system is damped.
1.2.1 Spring Elements

A linear spring is a type of mechanical link that is generally assumed to have negligible mass and
damping. When relative motion occurs between the two ends of the spring, force is developed in
the spring. This is given as;
F  Kx
where: K is the spring constant or stiffness
x is the deformation or relative of one end with respect to the other
The work done (U) in deforming spring is stored as a strain or potential energy in the spring.
This is given as;

1 2
U Kx
2
1.2.2 Mass or Inertia Element

The mass or inertia element is assume to be a rigid body; it can gain or loss kinetic energy
whenever the velocity of the body changes. Using Newton’s 2nd law of motion the force applied
to the mass is equal to the product of the mass and its acceleration. Work done on the mass is
given by the force multiplied by the displacement parallel to the applied and this energy is stored
in the form of kinetic energy.
1.2.3 Damping Element

The energy involved in mechanical vibration is gradually converted to heat or sound. Due to this
the response, such as the displacement of the system gradually decreases. The mechanism of
dissipating this energy is known as damping. An element that performs this is called damper.
Dampers are assumed to have neither mass nor elasticity, damping force occurs only if there’s a
relative velocity between the two ends of the damper.
Damper can be modeled as
 Viscous damping
 Coulomb or dry friction damping
 Hysteretic damping
1.3 Classification of Vibration System
1.3.1 Forced /free vibration

Free vibration is a system that is set to vibrate after it has been given an initial disturbance.

Forced vibration: if a system is subjected to an external force (usually repeating type of force),
the resulting vibration is known as Forced vibration.
1.3.2 Damped and Undamped Vibration

If there is no energy lost (energy dissipation) in the form of friction or other resistances during
oscillation, the vibration is known as undamped vibration. If there is energy lost (energy
dissipation), on the other hand called damped vibration.
1.4 Vibration Control

1.4.1 Vibration Isolation
Vibration isolation is a procedure by which the undesirable effects of vibration are reduced,
basically it involves the insertion of a resilient member (or isolator) between the vibrating mass
(or equipment or payload) and the source vibration so that a reduction in dynamic response of the
system is achieved under specified condition of excitation.
1.4.2 Shock Isolation

A shock load involves the application of a force for a short period of time (or short duration),
usually for a period of less than one natural time period of the system. Shock isolation can be
defined as a procedure by which the undesirable effects of shock are reduced.
1.5 Suspension System

Suspension system is a system of springs, shock absorbers, and other devices supporting the
upper part of a motor vehicle. Suspension system for vehicles is an integration of various
machine components designed and assembled in such a way that to absorb all the shocks and
vibrations.

Cars need a soft suspension for better comfort, whereas a stiff suspension leads to better
handling. Cars need high ground clearance on rough terrain, whereas a low center of gravity
(CG) height is desired for swift cornering and dynamic stability at high speeds. It is
advantageous to have low damping for low force transmission to vehicle frame, whereas high
damping is desired for fast decay of oscillations
The chassis of automobile is assembled on the axles, with the help of springs.
Obviously this is done to isolate the different parts of machine against shocks.
These shocks cause vehicle to bounce, pitch, roll or sway. No one wants to have a ride which
gives more of roller coaster feelings. Everyone wants the ride to be smooth and comfortable this
is what the suspension does for us. All the machine parts which help in isolating the vehicle
against the road shocks are collectively called a suspension system.
1.5.1 Purpose of Suspension System

The main purpose of the suspension system of a vehicle is to:
1. Maintain correct vehicle ride height
2. Reduce the effect of shock forces
3. Maintain correct wheel alignment
4. Support vehicle weight
5. Keep the tires in contact with the road
6. Control the vehicle’s direction of travel
7. To restrict the vibrations from being transmitted to various components of vehicle.
8. To protect the person sitting inside the vehicle against road shocks.
9. To maintain stability of the vehicle in pitching or rolling when in motion.
10. Attaching the wheels and tires to the vehicle.
11. Maintaining the proper wheel alignment and location as the vehicle traverses bumps
potholes, and uneven road surfaces.
12. Stabilizing the vehicle's attitude during acceleration braking, and cornering, and isolating
the road's roughness from the passenger compartment.

1.5.2 Suspension Springs
The springs are located between the wheels and the vehicle body. After the wheel hits a bump or
pit the spring deflects and is stretched outwards. It is then pulled back due to elasticity thereby
extracting the energy created due to bumps. The amplitude of spring deflection decreases
gradually due to its internal friction and friction of suspension joints until spring comes to rest
The following are some of the types of the suspension spring widely used.
1. Rubber springs: are further classified as: compression spring, compression shear spring,
steel reinforced spring, progressive spring, torsional shear spring, face shears spring.
2. Steel spring: Steel springs are also classified as: leaf spring, coil spring torsional bar and
tapered leaf spring.
3. Plastic spring
4. Air spring
5. Hydraulic springs
1.5.3 Dampers: Shock Absorbers

Shock absorber is device for reducing the effect of a sudden shock by the dissipation of the
shock's energy. On an automobile, springs and shock absorbers are mounted between the wheels
and the frame. When the wheels hit a hole or a raised spot on a road, the springs absorb the
resultant shock by expanding and contracting. To prevent the springs from shaking the frame
excessively, their motion is restrained by shock absorbers, which are also known by the more
descriptive term ‘dampers’. The type of shock absorber found on automobiles is usually a
hydraulic type that has a casing consisting of two tubes, one telescoping into the other. In order
for a spring to expand and contract, it must pull apart and push together the ends of this shock
absorber. But the ends offer so much resistance that the motion of the spring quickly dies out.
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The ends are connected to a piston in an oil-filled chamber in the shock absorbers' inner tube.
The piston can only move if it forces oil past it through valves. This arrangement creates a large
resistance to any motion by the piston and consequently by the ends. On some automobiles a
type of hydraulic suspension is used to function both as a spring and as a shock absorber. It
comprises a sealed spherical container filled with equal volumes of hydraulic fluid and gas under
pressure. Shock absorbers are used on aircraft to ease the impact upon landing. Some machines
are mounted on resilient materials composed, for example. of cork or rubber. The materials act as
shock absorbers, isolating the vibrations of the machine from the surrounding area
Unless a dampening structure is present, a car spring will extend and release the energy it
absorbs from a bump at an uncontrolled rate. The spring will continue to bounce at its natural
frequency until all of the energy originally put into it is used up. A suspension built on springs
alone would make for an extremely bouncy ride and, depending on the terrain, an uncontrollable
car.
Shock absorbers slow down and reduce the magnitude of vibratory motions by turning the
kinetic energy of suspension movement into heat energy that can be dissipated through hydraulic
fluid. To understand how this works, it's best to look inside a shock absorber to see its structure
and function, as shown in the figure below.

Fig.1.5.3 Twin-tube shock absorber (source [7])

A shock absorber is basically an oil pump placed between the frame of the car and the wheels.
The upper mount of the shock connects to the frame (i.e., the sprung weight), while the lower
mount connects to the axle, near the wheel (i.e., the unsprung weight). In a twin-tube design, one
of the most common types of shock absorbers, the upper mount is connected to a piston rod,
which in turn is connected to a piston, which in turn sits in a tube filled with hydraulic fluid. The
inner tube is known as the pressure tube, and the outer tube is known as the reserve tube. The
reserve tube stores excess hydraulic fluid.
When the car wheel encounters a bump in the road and causes the spring to coil and uncoil, the
energy of the spring is transferred to the shock absorber through the upper mount, down through
the piston rod and into the piston. Orifices perforate the piston and allow fluid to leak through as
the piston moves up and down in the pressure tube. Because the orifices are relatively tiny, only
a small amount of fluid, under great pressure, passes through. This slows down the piston, which
in turn slows down the spring.
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Shock absorbers work in two cycles: the compression cycle and the extension cycle. The
compression cycle occurs as the piston moves downward, compressing the hydraulic fluid in the
chamber below the piston. The extension cycle occurs as the piston moves toward the top of the
pressure tube, compressing the fluid in the chamber above the piston. A typical car or light truck
will have more resistance during its extension cycle than its compression cycle. With that in
mind, the compression cycle controls the motion of the vehicle's unsprung weight, while
extension controls the heavier, sprung weight.
All modern shock absorbers are velocity-sensitive -- the faster the suspension moves, the more
resistance the shock absorber provides. This enables shocks to adjust to road conditions and to
control all of the unwanted motions that can occur in a moving vehicle, including bounce, sway,
brake dive and acceleration squat.
1.5.4 Types of Suspension System
Front suspensions: of course, must deal with not only the motion of the suspension assembly
caused by road irregularities, but also the steering motion. Front-wheel-drive complicates the
suspension geometry even more, because drive shafts must adjust as wheels change angles
during turns.
Rear suspensions can be much simpler by comparison, since in all but the most sophisticated
rear-wheel-steering set-ups, the track of the rear wheels is a relative constant.
Independent rear suspensions on front-wheel-drive vehicles often use assemblies (McPherson
strut or modified strut) similar to those shown for front suspensions, except that no steering
knuckle is required, and a variety of leading and trailing links are used to maintain wheel
location.

2 LITERATURE REVIEW
2.1 Human Being Comfort Index of Vibration
Whole body vibration is transmitted to the body organ through the supporting parts such as the
feet, buttocks or back. There are various sources of whole body vibration such as standing on a
vibrating platform, floor surface, driving, and construction, manufacturing, and transportation
vehicles. The health effects of whole body vibration on passengers of heavy vehicle versus
workers in a similar environment who were not exposed to whole body vibration have been
compared. Research indicates back disorders are more prevalent and more severe in exposed to
vibration than that of non-exposed passengers.With short term exposure to vibration in the 2 to
20 Hz range at 1 m/sec2, one can feel different symptoms.
 Abdominal pain
 General feeling of discomfort, including headaches
 Chest pain
 Nausea
 Loss of equilibrium (balance)
 Muscle contractions with decreased performance in precise manipulation tasks
 Shortness of breath
 Influence on speech
 Sleeping ,etc
Long-term exposure can cause serious health problems, particularly with the spine:
 disc displacement
 degenerative spinal changes
 lumbar scoliosis
 inter vertebral disc disease
 degenerative disorders of the spine
 herniated discs
 disorders of the gastrointestinal system
 urogenital systems and some other
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Whenever the forcing frequency coincides with one of the natural frequency of the system (in
case of human being the parts (organs) of the body), resonance will occur. The most prominent
feature of resonance is large displacement induce undesirably large strain and stress; can lead to
the failure of the system (in case of human being discomfort and muscle fatigue). Most of the
time, it is difficult to control the excitation frequency; because it is imposed by the functional
requirement of the system or the machine.
There are two types of occupational vibration: segmental and whole body. Segmental vibration is
transmitted through the hands and arms, and is known to cause specific health effects such as
Raynaud’s syndrome. Whole body vibration is transmitted through the body’s supporting
surfaces such as the legs when standing and the back and buttocks when sitting. Along with
musculoskeletal problems, exposure to occupational whole body vibration also presents a health
risk to the psychomotor, physiological, and psychological systems of the body.
Whole Body Vibration Exposure
Industry vehicles
Manufacturing Forklifts
Construction Power shovels, tow motors,
Cranes, wheel loaders, bulldozers,
caterpillars, Earth moving machines
Transportation Buses, helicopters, subway trains,

locomotives and trucks
Agriculture tractors
Table.2.1 (source [12])
When vibrations are attenuated in the body, the vibration energy is absorbed by tissue and
organs. Vibrations lead to both voluntary and involuntary muscle contraction and can cause local
muscle fatigue especially at resonant frequencies. Vertical vibrations in the 5 to 10-Hz range
generally cause resonance in the 'Woracic-abdominal’ system (at 4 to 8 Hz in the spine, at 20 to
30 Hz in the head-neck-shoulder, and at 60 to 90 Hz in the eyeball. There are many studies which
suggest the risk of low-back pain due to the effect of vibration.
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Resonance frequency ranges for different parts of the body (vertical excitation direction)
Parts of body Resonance frequency (Hz)

Head 20
Chest 10-50
Stomach 4-8
Pelvic area(2nd order) 10-12
Spine 10-12
Table.2.2 (Source [8])
2.2 Controlling Effects Whole Body Vibration

The following actions are recommended to reduce the effects of vibration of whole body of
human beings.
1. Reduce the transmission of vibration to the passengers by engineering the equipment or

working on system more effectively.
For example:
 improving vehicle suspension
 altering the position of the seat within the vehicle
 mount equipment on springs or compression pads
 maintain equipment properly (i.e., balance and replace worn parts)
 proper engineering of seating
 use materials that generate less vibration
2. Decrease the amount of vibration to which the passenger is exposed by:

 reducing the speed of travel
 minimizing the exposure period by alternating working tasks where vibration is present
and those where it is negligible
 increasing rest/recovery time between exposures.
3. Modify the seat and control positions to reduce the incidence of forward or sideways leaning
of the trunk, and provide back rest support.
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4. Eliminate awkward postures due to difficulty of seeing displays or reaching control.
5. Where feasible, reduce or isolate passengers from the vibration source. For example:
 in seated tasks, provide a spring or cushion as a vibration isolator
 in standing operations, provide a rubber or vinyl floor mat
 minimize the undulations of the surface over which the vehicle must travel.
2.3 Geometry of Seat
Seat geometry in bus is a restricted seated working posture in which the passengers must interact
with and operate vehicle components. The traveling posture is therefore determined and
influenced by seat characteristics such as surface shape, amount of cushion, seat back and pan
angles, lumbar support, and adjustability as well as the locations of controls (steering wheel and
pedals), field of vision, and available head room.
For the design of the geometry of the passenger's seat, the following geometric parameters are
considered.
Length of the seat,
Height of the seat,
Lumber support,
Seat width, and
Seat pan (back) angle.
The data given below shows the importance of the above geometric parameters.
Factor Estimated importance

1 Vertically-curved lower lumber support 20%
2 Minimum trunk-thigh angle 15%
3 Length of seat 10%
4 Height of seat 5%
5 Open front of seat 5%
6 Tilt of seat 3%
7 Free space for sacrum and elbow
8 Front and top borders rounded or soft
9 Moderate contouring or cushioning porous cloth
1 porous cloth
0
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Table.2.3 (source [9])
A person with a round back feels more comfortable in a seat with a large curvature of the seat
back, while a person with a flat back feels comfortable in a seat with a flatter seat back. It was
found that the distances between the most lordtic points of the lumbar and the most prominent
point in the back (scapular, etc.) were 10-15 mm in the sitting posture (Figure 2.1).
Fig 2.1 sitting posture (source [9])
When the backrest inclination increased, a larger proportion of the body weight was transmitted
to the backrest thereby reducing the stresses on the spine resulting in less disc pressure and less
muscle activity. However, the effect was less pronounced at larger recline angles because the
neck must be flexed to maintain eye position.
A large backrest to seat cushion angle increases the angle of the hips and forces the pelvis to
rotate backwards (suitable hip angles or seat back angle are between 95 to 1200). To preserve the
suggested hip angles, it is necessary to increase the inclination of seat cushion and backrest
simultaneously
To prevent postural overload, 1100 or more of backrest angle, 60 of seat inclination and lumbar
support at L3 level are recommended. These reduce the postural stress, and also reduce the
stresses arising from road shock and vibration. To prevent vibration in the range of 4 to 8 Hz,
soft cushions should be replaced with firm ones, and the seat should be suspended to get a
natural frequency of less than 1.5 Hz. The line of action of pedal-force should pass from the foot
through the hip joint, and the backrest should firmly resist pelvic rotation.
In short
1. Side Support: Side supports would be favorable to the back by keeping the spine in the
appropriate vertical position. Papers written on this issue that proposed a small space
between trunk and side support to allow body movement for fatigue relief. Grandjean et
al. (1973) found that the passenger felt more comfortable when the backrest was gently
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curved (45-cm radius at the lumbar level and 60-cm radius on the upper part of the
backrest).
2. Lumbar Support: Grandjean et al. (1969) found that the highest comfort rating was
obtained when the center lumbar support was 10 to 14 cm above the depressed seat
surface.
3. Inclination of Seat Surface: For the driver's seat, Andersson et al. (1974) recommended a
backrest angle of 1200 and seat surface angle of 140. In the case of passenger seats,
slightly different values (Grandjean et al. 1969) can be applied (i.e., seat surface angle of
210 and 240 with seatback angle of 1220 and 1280).
4. Profile and Shape of the Seat Surface: Grandjean et al. (1973) concluded that a backrest
which is slightly concave in the thoracic region 45 to 55 cm above the depressed seat
surface allows a larger portion of the back muscles to relax. The concavity in the upper
part of the backrest provides a better neck position and therefore reduces the risk of
fatigue in the neck area.
Fig. back angle
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Fig 2.3 seat inclination
A comfortable body posture requires the following angles:
Body part angle

Ankle 90 to 1100
Knee 110 to 1300
Arms versus vertical line 20 to 400
Hip 100 to 1200
Head-Neck Axis to Trunk Axis 20 to 250
1. Design of bus for passengers seat
1.1. Introduction
Automotive seats need to accommodate a wide range of passengers sizes over relatively long
periods of time and provide isolation from vehicle vibration and shock. To fulfill these
requirements, there have been remarkable advances in automotive seat design during the past
decade incorporating seatback recliners, lumbar support, motorized multi-axes adjustments, and
foam cushions. However, these added features have resulted in increased cost and have been
used in only a limited number of seating environments. Even with the progress that has been
made, however, many passengers continue to experience significant discomfort in automotive
seating, and the factors that contribute to long-term discomfort or improved comfort are still not
clearly understood.
Thus, in spite of abundant research studies in automotive seating, many questions still remain
about what really contributes to seating comfort. As stated by many researches about seat
comfort, one of the most difficult, though apparently simple, problems in ergonomics is the
evaluation of the quality of seating, and perhaps the one dimension which is most difficult is
comfort of seating.
Studies of seating comfort are particularly difficult to conduct due to a large number of
interacting factors. The most difficult challenge in such studies is that of accurately and
consistently measuring the subjective perception of discomfort. Though a researcher called
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Hertzberg (1958) defined comfort as the absence of discomfort, there is no universally accepted
operational definition of discomfort. Furthermore, there is no agreed upon, reliable method for
quantifying the sensation of comfort or discomfort.
Comfort and road handling performance of a passenger car are mainly determined by the
damping characteristic of the shock absorbers. Passive shock absorbers have a fixed damping
characteristic determined by their design. Depending on the road excitation, however, it is
desirable to adjust this characteristic to increase performance. Semi-active and active suspension
systems offer the possibility to vary the damper characteristics along with the road profile e.g. by
changing the restriction of one or two current controlled valves or by changing the viscosity of a
magneto rheological fluid. Semi-active suspensions on the other hand are less complex, more
reliable and commercially available. They do not require an external power source (e.g.
hydraulic pump) and are safer because they can only dissipate energy and therefore cannot
render the system unstable.
One of the most important functions of a seat is its ability to isolate the occupant from road
vibration. This isolation characteristic of the seat can be defined by the transfer function (or
transmissibility) which is the ratio of the output of the seat to the input (via the cabin) as a
function of frequency. When the transmissibility (transfer function) is unity, the seat transfers
cabin floor vibration directly to the occupant. At the natural (or resonance) frequency, the seat
amplifies the input acceleration maximally. Thus, the output acceleration reaches the maximum
at natural frequency.
There are several methods to analyze and evaluate (or design) vehicle seat comfort objectively;
these methods can be subdivided into three categories:
 Vibration analysis
 Pressure analysis
 Methods based on the human body (physiological or orthopedic)
1.2 Types of Seat Suspension System
Similar to the suspension system of a vehicle body, there are four main types of seat suspension
system: passive, semi-active, active, and fully active.
1. Passive seat systems are the most common because they are cheap and effective for most
vibration. Passive systems include springs and passive dampers which reduce the vibration of the
operator’s seat. Passive systems cannot realistically attenuate the entire frequency range of
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whole-body vibration, specifically in the 1 to 7 Hz range. This is due to the amount of travel of
the system necessary to cancel the vibration.
Fig. Model of passive seat system
2. Semi-active seat suspension systems are somewhat common, and they give better results with
damping vibrations than passive systems. The defining trait of semi-active systems is that they
can only dissipate energy and not create energy. Semi-active systems can use springs and active
dampers which generally use electro rheological (ER) or magnetorheological (MR) fluid to
actively damp vibrations. These suspension systems work the following way. A sensor detects the
vehicle’s vibration, and a controller controls the flow and timing of fluid through the active
damper to attenuate the vibration of the seat. This method is slightly more advanced than a
passive seat; however it does not fully attenuate vibrations in the 1 to 7 Hz frequency range.
3. Active seat suspension systems are fairly uncommon due to the cost and power requirements
of the seating system. However, active suspension systems suppress vibrations better than
passive and semi-active suspension systems. Active systems are capable of suppressing
vibrations in the 1 to 7 Hz range, which make them ideal for whole-body vibration cancellation.
These seat suspension systems generally have springs and dampers, but their defining
characteristic is that the active actuators can dissipate energy, as well as create energy. The ability
to dissipate and create energy allows for greater vibration attenuation in the low-frequency range.
4. Fully active seat suspension systems are the most uncommon; however, they perform the best
for attenuating vibrations in the harmful frequency ranges. Fully active systems contain only
active components, and do not include any springs or dampers, which allows them to react faster
and more effectively. Fully active suspension systems cost about the same as active suspension
systems; however, they have a much higher power requirement.
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This is due to the fact that fully active suspension systems are required to use the active actuators
to continuously support the weight of the seat and the operator, while active suspension system
usually have a spring which supports most of the load.
1.3 Passive Seat Suspension System Design for Truck Driver
This particular design is based on the passive seat suspension system as mentioned above that
has a suspension system (isolator) consists of springs and passive damper. In this project this
type of isolator has been selected because since the cost of this isolator is very cheap and its
mechanism is simple as compared to other seat suspension types.
Isolation Maximum Frequency ratio

efficiency transmissibility (r=ω/ ωn)
90% 0.1 3.32
80% 0.2 2.45
70% 0.3 2.08
Table. Isolation efficiency and transmissibility (Source [6])
From the table above selecting an isolator having an isolation efficiency of 80%, the
transmissibility is 0.2and its frequency ratio is 2.45.
Assuming the passengers' seat is exposed to a base excitation (the cabin floor excitation which
itself is exciting sinusoidally due to the sinusoidal profile of the road or unevenness of the road
shape) and has magnitude of 5cm, the base displacement is given by the equation
Y(t)=0.05sin ωt
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where: ω is the forcing frequency
The schematic drawing and modeling of the truck drive seat is given as shown below.
Table 2.4 (source [9])
The following are recommended guidelines for automotive seat design

 A fore-aft adjustment (minimum range of 15 cm) and adjustable backrest angle between
900 and 1200 are essential.
 The seat cushion depth should not be shorter than 44 cm and not exceed 55 cm.
 The seat cushion angle should not be smaller than 100 and not exceed 220.
 The backrest should have a lumbar support. Side supports to seat cushion as well as to the
backrest are advisable to improve the position of the hips and trunk.
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Fig.3.2 modeling of truck driver’s seat
K56 and C56 are the effective or equivalent stiffness and damping coefficient respectively of the
chair isolation system.
The equation the displacement transmissibility for the base excitation is given by the equation
given below.
X 1  (2 r ) 2
Td  
Y (1  r 2 ) 2  (2 r ) 2
ISO-WBV recommended acceleration value for human being in the vertical direction is between
0.4 and 2.0m/s2.For this particular design purpose, taking the mean value 1.2m/s 2, the chair
design is based on this value.
Then the forcing frequency can be calculated as
By assuming the truck travel with a velocity of 30km/h on bumpy road that a bump height
(amplitude) of 0.3m that repeats itself in the interval of 0.75m distance (has wave length of
0.75m), the forcing frequency is given as.
The forcing frequency ω
 30 *1000 
  2    0.75  69.81rad / s
 3600 
For Td =0.2, the frequency ratio r=2.45

21
21

X
 0.2  X  0.2 * 0.05  0.01
Y
69.81
 n   19.028rad / s
2.45
This calculated value is the forcing frequency of the cabin floor. If there is no chair suspension
system (isolator), this frequency will cause the resonance of the chest, pelvic area and spine the
driver. But the seat suspension system gets rid off this risk.
For Td =0.2, from table (4) the frequency ratio r is 2.45.
The natural frequency of the system (the chair component and driver) would be
21.028
 fn   4.535Hz
2.45
But this calculated natural frequency of the system coincides with the natural frequency of the
stomach since the stomach resonates between 4-8Hz (the resonance of stomach occurs when the
forcing frequency is between 4-8Hz).
This problem can be solved by taking an isolation system that has isolation efficiency higher than
80%.Taking an isolator having an isolation efficiency of 90%, Td =0.1 and r=3.32.
69.81
n   21.028rad / s
3.32
21.028
fn   3.346 Hz
3.32
This calculated natural frequency of the system doesn’t coincide with the resonance frequency of
one of the parts of the human body.
The stiffness the suspension system is calculated as follows.

 K  n 2 * m  21.0282 * 90  39376 N / m
Taking to be K=4000N/m
The damping coefficient also given as
c
 
2mn
For displacement transmissibility Td =0.1, the ς is calculated as follows.
22
22

1  (2 r ) 2
Td 
(1  r 2 ) 2  (2 r )2
1  (2 * 0.4 * 3.32)2
Td 
(1  3.322 ) 2  (2 * 0.4 * 3.32) 2
 =0.01
But for effective isolation,  =0.4 is recommended
Then c  2n  m =2*90*21.028*0.4
C=1600Ns/m
The specification the isolator would be K=40000N/m and c=1600Ns/m.
The spring arrangement on the driver seat is as shown below; with 20 numbers of springs that
has 2000N/m each. And the arrangement of the dampers is shown below, four dampers each with
a damping coefficient of 400Ns/m.
23
23

Fig.3.3 shock absorbers and springs arrangement
Fig.3.4 shock absorber
The spring arrangement on the driver seat is as shown below; with 20 numbers of springs that
has 2000N/m each.
Fig.3.5 Shows springs arrangement in parallel along the seat pedestal
24
24

3.4 Seat Suspension Spring Design

The dynamic force amplitude due to the vibration the cabin acted on the base of the seat is
calculated as follows.
1
 1  (2 r )2 2
FT  YK  2 
 (1  r )  (2 r ) 
2 2
1
 1  (2 * 0.4 * 3.32)2 2
FT  0.05 * 40000  2 
 547.44
 (1  3.32 )  (2 * 0.4 * 3.32) 
2 2
The weight the driver and the seat component

W=90*9.81=882.9N
The total force acting on the spring will be

FT=547.44N+882.9N=1430.34N
Since there are 20 numbers of springs in the sit suspension system the force acting on each
spring will be
P=1430.34N/20 = 71.517N
From standard table selecting steel that has the following properties:
G=80KN/mm2
 = 224MPa
Giving factor of safety, F.S = 2
 all = 224/2 = 112Mpa
The spring index is assumed to be 5, that is C=Dm/d=5.
The Wahl’s stress factor K is given by
4C  1 0.615
K 
4C  4 C
4 * 5  1 0.615
K   1.31
4*5  4 5
25
25

Fig.3.4 the coil spring dimension parameters

where:
Dm=the mean diameter of the spring
Do=the outer diameter of the spring
Di=the inner diameter of the spring
The initial torsional stress of the spring

8* P *C
 K
 *d2
26
26

The wire diameter of the spring is given by
N 8 * 71.517 N * 5
112  1.31*
mm 2
d2
d=3.56mm
Taking it to be 4mm, that is d=4mm
Then Dm=. 5*4=20mm

Do=Dm+d=20+4=24mm
Di=Dm-d=18mm
Active number turns (coils) of the spring is give as follows

n=active number of coils
The compression of the spring (δ) is given by
8 * P * C3 * n

Gd
Assuming the static deflection of the springs to be 5mm, that is δ=5mm
8 * 71.517 N * 53 * n
5mm 
80 KN / mm 2 * 4mm
n=13.42
Taking it to be, n=14
For square and ground end, the total number of turns is given by
n'=n+2=14+2=16
Free length of the spring is given by the formula

L=n’*d+ δ+0.15* δ
=16*4+5+0.15*5=67.45mm
Taking it to be L=70mm
Pitch of the spring is given by the formula
freelength
p
n ' 1
27
27

70
p  4.67 mm
16  1
Take it to be p=5mm
3.5 Seat Bolt Design

The bolt which is used to fix the chair and its overall suspension system to the seat rail, its design
is as follows.
The total force acting on the bolt has been already calculated to be 1430.34N.
Taking the number of bolt to be 10, then the force acting on each spring will be
1430.34/10=143.034N.
From standard table selecting steel for bolt material that has a tensile strength of 50MPa and a
factor of safety, F.S of 4.
The allowable tensile strength will be
σall=50/4=12.5MPa
Since the bolt is forced to a tensile force of 143.034N, the design is based on the tensile strength
of the bolt.
143.034
t 
db 2

4
db 2
 12.5 N / mm   2
 143.034 N
4
db =8.675mm
Take it to be 10mm
From standard table for bolts and nuts, its dimension will be M 10  1.5 .
28
28

Fig.3.5 Seat rail and spring base plate assembly
29
29

4. MODELING OF TRUCK AND STUDY OF OVERALL RESPONSE
4.1 Truck Modeling

By using the equivalent values of the stiffness, mass and damping of the system of each part of
the truck, the truck can be modeled to have ten degree of freedom. The ten degrees of freedom
are explained in detail below.
30
30

Fig.4.1.1 Truck modeling
4.1.1 Modeling Assumptions
 The movement of the masses in the horizontal direction is(can be) neglected.
31
31

 The movement of the masses in the vertical direction is only under consideration.i.e the
model can be considered as one dimensional.
 The masses are concentrated in the centre of gravity.
 The rear wheel, front wheel and the chair can be assumed to don’t have an angular
displacement with respect to their centre gravity. That is, J1=J2=J6=0.
 The installation (mounting) of the spring-damper system is in the line of gravity, that is
the spring damper systems are always in vertical direction (horizontal movement is
neglected).
 Initially the spring-damper system are deflected by the masses carried by them
System parameters description
 ai is distances from the center of gravity of the chassis to the corresponding spring
damper system.
 bi is distances from the center of gravity of the engine to the corresponding spring
damper system.
 ci is distances from the center of gravity of the cabin to the corresponding spring damper
system.
 m1 is the mass of the rear wheel in Kg.
 m2 is the mass of the front wheel in Kg.
 m3 is the mass of the chassis in Kg.
 m4 is the mass of the engine in Kg.
 m5 is the mass of the cabin in Kg.
 m6 is the mass of the driver seat in Kg.
 m7 is the mass of the loading area and the load in Kg.
 J3 is the moment inertia of the chassis in Kg.m2
 J4 is the moment inertia of the engine in Kg.m2
 J5 is the moment inertia of the cabin in Kg.m2
 J7 is the moment inertia of the loading area in Kg.m2
 Ki is the stiffness coefficient Ns/m.
 Ci is the damping coefficient in Ns/m.
32
32

33
33

Fig. 4.1.1 Spring-damper system location
4.2.1 Rear Wheel Modeling
The rear wheel can be modeled as having stiffness and damping properties (actually it has these
properties because the tire is made from rubber and it is known that rubber has stiffness and
damping
Assuming the rear wheel travels or oscillates in the vertical direction only (i.e. in the y-axis), the
coordinate Y1 (t) is used to describe the linear displacement of the rear wheel.
For convenience, assigning the damping and stiffness properties of the tire as C13 and K13
respectively, the equation of motion derived as follows.
34
34

Fig.4.2.1 the modeling and the FBD of the rear wheel
Where:
(Fo)C10 is the reaction force of the tire acted on the rear wheel due to the damping property of the
tire.
(Fo)K10 is the reaction force of the tire acted on the rear wheel due to the stiffness property the
tire.
(Frear sus)C13 is the reaction force of the rear suspension system acted on the rear wheel due to
the shock absorber or the damper designated by C13.
(Frear sus)K13 is the reaction force of the rear suspension system acted on the rear wheel due to
the stiffness property of the spring (leaf spring) designated by K13.
The values the above expressions are given below.
 
(Fo)C10= C10 (Y1 (t )  u1 (t ))

(Fo)K10=K10 (Y1 (t)-u1 (t))
(Frear sus)K13= K13 (Y3 (t)-Y1 (t) +a13sinф3 (t))
  
(Frear sus)C13= C13 ( Y3 (t )  Y 1 (t )  a13 cos 3 (t ) 3 (t ) )

Applying Newton’s 2nd on the FBD of the rear wheel, the equation of the rear wheel can be
written as follows.
   
m1 Y 1 (t )  k13 (Y3 (t )  Y1 (t )  a13 sin 3 (t ))  k10Y1 ((t )  u1 (t ))  C13 (Y3 (t )  Y 1 (t )  a13 cos 3 (t ) 3 (t ))
 
C10 (Y 1 (t )  u1 (t ))  m1 * g  K13 * st
But m1 * g = K13 * st
After considering the above assumption and simplifying, the equation motion of the rear wheel
will be:
  
m1 Y 1 (t ) + k10Y1 (t ) - k10 u1 (t ) - k13 a13 sin 3 (t ) - k13Y3 (t ) + k13Y1 (t ) + C10 Y 1 (t ) - C10 u1 (t ) -
  
C13 a13 cos 3 (t ) 3 (t ) - C13 Y3 (t ) + C13 Y 1 (t ) =0
35
35

For very small ф3, sin ф3 (t) = ф3 (t) and cos ф3 (t) =1
   
m1 Y 1 (t )  k10Y1 (t )  k10 u1 (t )  k13 a133 (t )  k13Y3 (t )  k13Y1 (t )  C10 (Y 1 (t )  C10 u 1 (t )  C13 a13 3 (t )
 
C13 Y3 (t )  C13 Y 1 (t )  0
4.2.2 Front Wheel Modeling
The front wheel can be modeled as in the fashion as the rear wheel has modeled. The coordinate
Y2 (t) is used to designate the linear displacement the front wheel in the vertical direction.
36
36

Fig.4.2.2 Front wheel modeling and its FBD
Applying Newton’s 2nd law on the FBD of the front wheel (the above figure), the equation of
motion of the front wheel can be written as shown below.
   
m2 Y 2 (t )  k20Y2 (t )  k20u 2 (t )  k 23 a233 (t )  k 23Y3 (t )  k 23Y2 (t )  C20 (Y 2 (t )  C20 u 2 (t )  C23 a23 3 (t )
 
C23 Y3 (t )  C23 Y 2 (t )  0
4.2.3 Modeling of the Chassis
The actual modeling of the chassis is complex but by take into account many assumptions, we
can model the chassis as having two DOF. The assumptions are given be as follows.
1. Neglecting the stiffness and damping property of the chassis components
2. The chassis can be assumed as a rigid body having a lumped mass at centre of gravity
and a mass moment of inertia (J3) about the axis through the centre of gravity it and
perpendicular to the axis along its length.
3. The stiffness and damping properties are only due to the suspension system.
4. For simplicity the chassis has to have two DOF.That is Y 3 (t) and ф3 (t) describes the
linear and angular displacement of the centre gravity of the chassis respectively.
Assigning the damping and stiffness of the suspension system by K 13, C13 and K23, C23 for the rear
and front suspension system respectively, the modeling and FBD of the chassis is shown below.
37
37

Fig.4.2.3.1 Modeling of chassis
Fig.4.2.3.2 FBD of the

chassis
Where:
 (Fcabin)35 is the
resultant force
acting on the chassis due to the combined effect of the damping and stiffness of the cabin
isolation system designated by 35.
 (Fcabin)53 is the resultant force acting on the chassis due to the combined effect of the
damping and stiffness of the cabin isolation system designated by 35.
 (Fengine)34 is the resultant force acting on the chassis due to the reaction force of the
isolation system of the engine(engine mounting) designated by 34.
 (Fengine)43 is the resultant force acting on the chassis due to the reaction force of the
isolation system of the engine(engine mounting) designated by 43.
 (Fl.area)37 is the reaction force the isolator of the loading area that is exerted on the
chassis assigned by number 37.
The expressions (values) the above forces are given below:
(Fcabin)35=(Fcabin)C35+(Fcabin)K35
Where: (Fcabin)C35 and (Fcabin)K35 is the reaction force of the isolator of the cabin assigned
by 35 due to the damping and stiffness respectively.
   
C35 [Y 5 (t )  c35 cos 5 (t )  5 (t )  (Y3 (t )  a35 cos 3 (t )  3 (t ))]  K 35 [Y5 (t )  c35 sin 5 (t )
(Fcabin)35= (Y3 (t )  a35 sin 3 (t )]
38
38

(Fcabin)53 =
   
C53 [Y 5 (t )  c53 cos 5 (t )  5 (t )  (Y3 (t )  a53 cos 3 (t )  3 (t ))]  K53 [Y5 (t )  c53 sin 5 (t )
(Y3 (t )  a53 sin 3 (t )]
   
C34 [Y 4 (t )  b34 cos 4 (t )  4 (t )  (Y3 (t )  a34 cos 3 (t )  3 (t ))]  K 34 [Y4 (t )  b34 sin 5 (t )
(Fengine)34= (Y3 (t )  a34 sin 3 (t )]
   
C43 [Y 4 (t )  b43 cos 4 (t )  4 (t )  (Y3 (t )  a43 cos 3 (t )  3 (t ))]  K 43 [Y4 (t )  b43 sin 4 (t )
(Fengine)43= (Y3 (t )  a43 sin 3 (t )]
(Fl.area)37=(Fl.area)C37+(Fl.area)K37
(Fl.area)37=
 
d37 [a y1 cos 7 (t )  7 (t )  (Y3 (t )]  a37 cos 3 (t )  3 (t ))  K 37 [ ay1sin 7 (t )  (Y3 (t )  a37 sin 3 (t ))]
1. Applying Newton’s 2nd on the chassis for linear displacement of it, the equation of motion
can be derived as follows.
  
( m3  m7 ) Y 3 (t )  m7 eY 1 cos 7 (t )  7 (t )  m7 aY 1 cos 3 (t )  3 (t )  k53 (Y5 (t )  c53 sin 5 (t ))  K 35{(Y5 (t )
 c35 sin 5 (t ))  (Y3 (t )  a53 sin 3 (t ))}  [ K 34 {Y4 (t )  b34 sin 4 (t )}  K 43{Y 4 (t )  b43 sin 4 (t )]  {K13
{Y3 (t )  Y1 (t )  a13 sin 3 (t ))  K 23 (Y3 (t )  Y2 (t )  a23 sin 3 (t ))}  {K37Y3 (t )  K 37 a37 sin 3 (t ) 
    
k37 eY 1 sin 7 (t )}  {C35 (Y 5 (t )  c35 cos 5 (t )  5 (t ))  C53 (Y 5 (t )  c53 cos 5 (t )  5 (t ))}  {C34 (Y 4 (t )
    
b34 cos 5 (t )  5 (t ))  C43 (Y 4 (t )  b34 cos 5 (t )  5 (t ))}  C13 (Y 3 (t )  Y1 (t )  a13 cos 3 (t )  3 (t )) 
    
C23 (Y 3 (t )  Y1 (t )  a23 cos 3 (t )  3 (t ))}  {C37 Y 3 (t )  C37 a37 cos 3 (t )  3 (t )  C37 eY 1 cos 3 (t )  3 (t )}
 m7 * g  m3 * g  K 37 * st  ( K13  K 23 ) * st
For very small (infinitesimal values of) values of ф3, ф4, ф5 and ф7 the respective sine value will
be ф3 (t), ф4 (t), ф5 (t) and ф7 (t) respectively and their cosine value is approximately 1.That is,
sin ф3 (t)= ф3 (t), sin ф4 (t)= ф4 (t), sin ф5 (t)= ф5 (t), sin ф7 (t)= ф7 (t),cos ф3 (t)=1, cos ф3 (t)=1, cos
ф4 (t)=1, cos ф5 (t)=1, cos ф7 (t)=1.
And
 
( 3 (t ))2  0 , ( 7 (t )) 2  0
39
39

m7 * g  K37 * st and m3 * g  ( K13  K23 ) * st
Applying the above assumption and simplifying the above equation, the equation of motion of
the chassis is given as below.
  
( m7  m3 ) Y3 (t )  m7 eY 1 7 (t )  m7 eY 1 3 (t )  K35Y5 (t )  K 35Y3 (t )  K 53Y5 (t )  K 53Y3 (t )  K 43Y4 (t )
 
 K 43Y 3(t )  K34Y4 (t )  K 34Y3 (t )  K13Y1 (t )  K 23Y2 (t )  K 23Y3 (t )  K13Y3 (t )  C43 Y 3 (t )  C53 Y 3 (t )
       
C35 Y 5 (t )  C35 Y 3 (t )  C53 Y 5 (t )  C34 Y 4 (t )  C43 Y 4 (t )  C34 Y 3 (t )  C13 Y 3 (t )  C13 Y 1 (t ) 
      
C23 Y 2 (t )  C23 Y 3 (t )  C53 a53 3 (t )  C53 c53 5 (t )  C35 a35 3 (t )  C35 c35 5 (t )  C43 a43 3 (t ) 
    
C43b43 4 (t )  C34 a34 3 (t )  C34 b34 4 (t )  C13 a13 3 (t )  C23 a23 3 (t )  K13 a133 (t )  K 23 a233 (t )
 K 35 c355 (t )  K 35 a353 (t )  K 43b434 (t )  K 43 a433 (t )  K 34 b344 (t )  K 34 a343 (t )  K 53 a533 (t )
 K 53 c535 (t )  0
2. Applying Newton’s 2nd law for the rotational motion the chassis, the angular displacement
equation motion of the chassis is written as below. That is the equation of motion of the
chassis can be found taking moment equation about the centre gravity of the chassis.
40
40

1 1  1 1
( m7 aY21  m7 aY21 cos 23 (t )  J 3 )  3 (t )  { m7 aY 1eY 1 cos(3 (t )  7 (t ))  m7 aY 1eY 1
2 2 2 2
 
cos(3 (t )  7 (t ))} 7 (t )  m7 Y 3 (t )aY 1 cos 3 (t )  k13 a132 cos 3 (t ) sin 3 (t )  k13 a13 cos 3 (t )Y3 (t )
 k13 a13 cos 3 (t )Y1 (t )  k23 a232 cos 3 (t )sin 3 (t )  k23 a23 cos 3 (t )Y3 (t )  k23 a23 cos 3 (t )Y2 (t )
k34 a34 2 cos 3 (t ) sin 3 (t )  k34 a34 cos 3 (t )Y4 (t )  k34 a34 cos 3 (t )Y3 (t )  k 43 a43 cos 3 (t )Y4 (t ) 
k43 a43 cos 3 (t )Y3 (t )  k43 a432 cos 3 (t )sin 3 (t )  k35 a35 2 cos 3 (t ) sin 3 (t )  k35 a35 cos 3 (t )Y5 (t )
 k35 a35 cos 3 (t )Y3 (t )  k53 a532 cos 3 (t ) sin 3 (t )  k53 a53 cos 3 (t )Y5 (t )  k53 a53 cos 3 (t )Y3 (t )
K 37 aY21 cos 3 (t ) sin 3 (t )  K 37 a37 2 cos 3 (t )sin 3 (t )  k35 a35 cos 3 (t )c35 sin 5 (t ) 
k53 a53 cos 3 (t )c53 sin 5 (t )  2 K 37 sin 3 (t )a37 cos 3 (t )  K 37 sin 7 (t )e37 aY 1 cos 3 (t ) 
K 37 sin 7 (t )e37 a37 cos 3 (t )  K 37 eY 1 sin 7 (t ) aY 1 cos 3 (t )  K 37 eY 1 sin 7 (t ) a37 cos 3 (t ) 
1 
k34 a34 cos 3 (t )b34 sin 4 (t )  k43 a43 cos 3 (t )b43 sin 4 (t )  m7 aY21 ( 3 (t )) 2 sin 23 (t ) 
2
1  1 
m7 aY 1eY 1 ( 7 (t )) 2 sin 3 (t )  m7 aY 1eY 1 ( 7 (t )) 2 sin( 3 (t )  7 (t ))  2C37 aY 1 cos(3 (t )) 2 a37 
2 2
  
cos 3 (t )C37 cos 7 (t )  7 (t )a37 e37  cos 3 (t )C37 cos 7 (t )  7 (t ) a37 aY 1  cos 3 (t )C53 a53 c53 cos 5 (t )  5 (t )
 
 cos 3 (t )C43 a43 b43 cos 4 (t )  4 (t )  cos 3 (t )C34 a34 b43 cos 4 (t )  4 (t )  cos 3 (t )C37 eY 1 cos 7
    
(t )  7 (t ) a37  cos 3 (t )C34 a34 Y3 (t )  cos 3 (t )C35 a35 Y3 (t )  cos 3 (t )C53 a53 Y3 (t )  cos 3 (t )C23 a23 Y3 (t )
   
 cos 3 (t )C13 a13 Y1 (t )  cos 3 (t )C35 a35 Y5 (t )  cos 3 (t )C43 a43 Y4 (t )  cos 3 (t )C34 a34 Y4 (t ) 
   
cos 3 (t )C53 a53 Y5 (t )  cos 3 (t )C23 a23 Y2 (t )  cos 3 (t )C13 a13 Y3 (t )  C35 a35 2 cos(3 (t )) 2  3 (t ) 
  
C43 a432 cos(3 (t )) 2  3 (t )  C53 a532 cos(3 (t )) 2  3 (t )  C37 a37 2 cos(3 (t )) 2  3 (t )  C37 aY 12
   
cos(3 (t )) 2  3 (t )  C23 a232 cos(3 (t )) 2  3 (t )  C13 a132 cos(3 (t )) 2  3 (t )  C34 a34 2 cos(3 (t )) 2  3 (t )
 
 cos 3 (t )C35 a35 c35 cos 5 (t )  5 (t )  cos 5 (t )C37 eY 1 cos 7 (t )  7 (t )aY 1  m7 * gaY 1 cos 3 (t )  0
Assuming for very small angular motion of the chassis, cosф3 (t) =1,
Cosф4 (t) =1, cosф5 (t) =1, cosф7 (t) =1, sinф3 (t) = ф3 (t), sinф4 (t) = ф4(t),
41
41

   
Sinф5 (t) = ф5(t), sinф7 (t) = ф7(t), ( 7 (t ))  0 , ( 3 (t ))  0 ’ ( 4 (t ))  0 , ( 5 (t ))  0

2 2 2 2
Taking into account the above expression, the simplified angular equation motion of the chassis
will be written as follows.
  
( m7 aY21  J 3 )  3 (t )  { m7 aY 1eY 1 } 7 (t )  m7 Y 3 (t ) aY 1  k13 a1323 (t )  k13 a13Y3 (t )  k13 a13Y1 (t ) 
k23 a2323 (t )  k23 a23 (t )Y3 (t )  k23 a23 (t )Y2 (t )  k34 a34 23 (t )  k34 a34Y4 (t )  k34 a34 (t )Y3 (t )  k 43 a43Y4 (t )
 k43 a43Y3 (t )  k43 a4323 (t )  k35 a35 23 (t )  k35 a35Y5 (t )  k35 a35Y3 (t )  k53 a5323 (t )  k53 a53Y5 (t ) 
k53 a53Y3 (t )  K 37 aY213 (t )  K 37 a37 23 (t )  k35 a35 c355 (t )  k53 a53 c535 (t )  2 K 37 (t )a37 
K 377 (t )e37 aY 1  K 377 (t )e37 a37  K 37 eY 17 (t )aY 1  K 37 eY 17 (t )a37  k34 a34 b344 (t )  k 43 a43b434 (t )
    
2C37 aY 1 a37  C37  7 (t )a37 e37  C37  7 (t )a37 aY 1  C53 a53 c53  5 (t )  C43 a43b43  4 (t )  C34 a34 b43  4 (t )
     
C37 eY 1  7 (t )a37  C34 a34 Y3 (t )  C35 a35 Y3 (t )  C53 a53 Y3 (t )  C23 a23 Y3 (t )  (t )C13 a13 Y1 (t ) 
      
C35 a35 Y5 (t )  C43 a43 Y4 (t )  C34 a34 Y4 (t )  C53 a53 Y5 (t )  C23 a23 Y2 (t )  C13 a13 Y3 (t )  C35 a35 2  3 (t ) 
     
C43 a432  3 (t )  C53 a532  3 (t )  C37 a37 2  3 (t )  C37 aY 12  3 (t )  C23 a232  3 (t )  C13 a132  3 (t ) 
  
C34 a34 2  3 (t )  C35 a35 c35  5 (t )  C37 eY 1  7 (t )aY 1  m7 * gaY 1  0
4.2.4 Modeling of the Engine
The engine can be modeled as a rigid body having a lumped mass at its center of gravity and
mass moment of inertia, J4 about the axis which pass through its centre gravity and perpendicular
to the axis its length. That is, the engine is assumed to have two DOF as shown in the figure
below
.
42
42

Fig.4.2.4.1 modeling of the engine
Fig.4.2.4.2 FBD of the engine
Where:
   
(Fengine)C34= C34 {Y 4 (t )  b34 cos 4 (t ) 4 (t )  (Y 3 (t}  a34 cos 3 (t )  3 (t ))}

(Fengine)K34= K34 {Y4 (t )  b34 sin 4 (t )  (Y3 (t}  a34 sin 3 (t ))}
   
(Fengine)C43= C43 {Y 4 (t )  b43 cos 4 (t ) 4 (t )  (Y 3 (t}  a43 cos 3 (t )  3 (t ))}

(Fengine)K43= K43 {Y4 (t )  b43 sin 4 (t )  (Y3 (t}  a43 sin 3 (t ))}
43
43

1) Applying Newton’s 2nd for the linear displacement of the engine, the following equation
can be derived.

m4 Y 4 (t )  -{(Fengine)C34+(Fengine)K34}-{(Fengine)C43+(Fengine)K43}-m *g+(K +K )Δst
4 34 43

m4 Y 4 (t )  K34Y4 (t )  K 34 b34 sin 4 (t )  K 34Y3 (t )  K 34 a34 sin 3 (t )  K 43Y4 (t )  K 43b43 sin 4 (t ) 
   
K 43Y3 (t )  K 43 a43 sin 3 (t )  C34 Y 4 (t )  C34 b34 cos 4 (t ) 4 (t )  C34 Y 3 (t )  C34 a34 cos 3 (t )  3 (t )) 
   
C43 Y 4 (t )  C43b43 cos 4 (t ) 4 (t )  C43 Y 3 (t )  C43 a43 cos 3 (t )  3 (t ))  0
Considering the assumptions: for very small angular rotations of the cabin and the chassis, sin ф 3
(t) = ф3 (t), cosф4 (t) =1 and m4*g= (K34+K43) Δst.The linear equation of motion of the engine can
be written as follows.

m4 Y 4 (t )  K34Y4 (t )  K 34b344 (t )  K 34Y3 (t )  K 34 a343 (t )  K 43Y4 (t )  K 43b434 (t )  K 43Y3 (t ) 
     
K 43 a433 (t )  C34 Y 4 (t )  C34 b34 4 (t )  C34 Y 3 (t )  C34 a34  3 (t ))  C43 Y 4 (t )  C43b43 4 (t ) 
 
C43 Y 3 (t )  C43 a43  3 (t ))  0
3. Applying Newton’s 2nd law for the angular displacement of the engine about its centre
gravity, its equation of motion can be derived as follows, i.e. taking moment about the
C.G of the engine.
44
44


4 (t ) J 4  K34 b34 cos 4 (t )Y4 (t )  K 34 b34 2 cos 4 (t ) sin 4 (t )  K 34 b34 cos 4 (t )Y3 (t )  K 34b34 cos 4 (t )
a34 sin 3 (t )  K 43b43 cos 4 (t )Y4 (t )  K 43b432 cos 4 (t )sin 4 (t )  K 43b43 cos 4 (t )Y3 (t )  K 43b43 cos 4 (t )
  
a43 sin 3 (t )  cos 4 (t )C34 b34 Y 4 (t )  C34 b34 2 cos 4 ((t )) 2 4 (t )  cos 4 (t )C34 b34 Y 3 (t )  cos 4 (t )C34 b34
   
a34 cos 3 (t )  3 (t )  cos 4 (t )C43b43 Y 4 (t )  C43b432 cos 4 ((t )) 2 4 (t )  cos 4 (t )C43b43 Y 3 (t ) 

cos 4 (t )C43b43 a34 cos 3 (t )  3 (t )  0
For infinitesimal angular displacement of the chassis and engine:
cos 4 (t )  1,cos 3 (t )  1,sin 4 (t )  4 (t ),sin 3 (t )  3 (t )
Then the above equation can be simplified as:

4 (t ) J 4  K 34 b34Y4 (t )  K 34 b34 24 (t )  K 34b34Y3 (t )  K 34b34 a343 (t )  K 43b43 (t )Y4 (t )  K 43b4324 (t ) 
   
K 43 b43Y3 (t )  K 43b43 a433 (t )  C34 b34 Y 4 (t )  C34 b34 2 4 (t )  C34 b34 Y 3 (t )  C34 b34 a34  3 (t ) 
   
C43b43 Y 4 (t )  C43b432 4 (t )  C43b43 Y 3 (t )  C43b43 a34  3 (t )  0
4.2.5 Modeling of Cabin
For simplicity the cabin can be modeled as a rigid body having a lumped mass at its center
gravity and mass moment inertia J5 about an axis through it center of gravity and perpendicular
to the axis along its length.
45
45

Fig.4.2.5 cabin modeling and

FBD of cabin
1. Applying
Newton’s 2nd on the cabin for
its linear displacement, the equation of motion can be derived as follows.

m5 Y 5 (t )  m5 g  K 35Y5 (t )  K 35 c35 s in5 (t )  K 35Y3 (t )  K 35 a35 s in3 (t )  K 53Y5 (t )  K 53c53 s in5 (t )
  
 K 53Y3 (t )  K 53 a53 s in3 (t )  K 56Y5 (t )  K 56 c56 s in5 (t )  C35 Y5 (t )  C35c35 cos 5 (t )  5 (t )  C35 Y3 (t )
     
C35 a35 cos 3 (t )  3 (t )  C53 Y5 (t )  C53 c53 cos 5 (t )  5 (t )  C53 Y3 (t )  C53 a53 cos 3 (t )  3 (t )  C56 Y6 (t )
 
C56 Y5 (t )  C56 c56 cos 5 (t )  5 (t )  0
For infinitesimal angular displacement of the chassis and cabin, the simplified equation can be
written as follows.

m5 Y 5 (t )  K35Y5 (t )  K35 c355 (t )  K 35Y3 (t )  K 35 a353 (t )  K 53Y5 (t )  K 53c535 (t )  K 53Y3 (t ) 
    
K53 a533 (t )  K56Y5 (t )  K 56 c565 (t )  C35 Y5 (t )  C35 c35  5 (t )  C35 Y3 (t )  C35 a35  3 (t )  C53 Y5 (t )
     
C53 c53  5 (t )  C53 Y3 (t )  C53 a53  3 (t )  C56 Y6 (t )  C56 Y5 (t )  C56 c56 (t )  5 (t )  0
2 Applying Newton’s 2nd law for the angular displacement of the cabin about its centre
gravity, its equation of motion can be derived as follows, i.e. taking moment about the C.G
of the cabin.
46
46


5 (t ) J 5  K 35 c35 s in5 (t )Y5 (t )  K 35 c352 cos 5 (t )s in5 (t )  K 35 c35 s in5 (t )Y3 (t )  K 35 c35 cos 5 (t )a35
s in3 (t )  K 53 c53 s in5 (t )Y5 (t )  K 53 c532 cos 5 (t )s in5 (t )  K 53 c53 s in5 (t )Y3 (t )  K 53c53 cos 5 (t )a53

s in3 (t )  K56 c56 cos 5 (t )Y6 (t )  K 56 c56 cos 5 (t )Y5 (t )  K 56 c56 2 cos 5 (t )s in5 (t )  cos 5 (t )C35 c35 Y5 (t )
  
C35 c35 2 cos(5 (t )) 2  5 (t )  cos 5 (t )C35 c35 Y3 (t )  cos 5 (t )C35 c53 a35 cos 3 (t )  3 (t )  cos 5 (t )C35 c53
   
Y5 (t )  C53 c532 cos(5 (t )) 2  5 (t )  cos 5 (t )C53 c53 Y3 (t )  cos 5 (t )C53 c53 a53 cos 3 (t )  3 (t ) 
  
cos 5 (t )C56 c56 Y6 (t )  cos 5 (t )C56 c56 Y5 (t )  C56 c56 2 cos(5 (t )) 2  5 (t )  0
For infinitesimal angular displacement of the chassis and cabin and after simplification, the
equation of motion of the cabin will be written as shown below.

5 (t ) J 5  K 35 c355 (t )Y5 (t )  K 35 c35 25 (t )  K 35 c35 (t )Y3 (t )  K 35 c35 a353 (t )  K 53c53Y5 (t )  K 53c5325 (t )

 K 53c53Y3 (t )  K 53 c53 a533 (t )  K 56 c56Y6 (t )  K 56 c56Y5 (t )  K 56 c56 25 (t )  C35 c35 Y5 (t )  C35c35 2 (t )) 2
      
 5 (t )  C35 c35 Y3 (t )  C35 c53 a35 (t )  3 (t )  C35 c53 Y5 (t )  C53 c532  5 (t )  C53c53 Y3 (t )  C53 c53 a53  3 (t )
  
C56 c56 Y6 (t )  C56 c56 Y5 (t )  C56 c56 2  5 (t )  0
47
47

4.2.6 Modeling of the Driver Seat
The modeling the chair is as shown below.
Fig.4.2.6 Seat modeling and FBD of the model
Where:
(Fchair)K56 is the force exerted on the cabin due to the reaction force of the spring designated
by K56.
(Fchair)C56 is the force exerted on the cabin due to the reaction force of the damper or shock
absorber assigned by C56.
  
m6 Y6 (t )  m6 g  K 56Y6 (t )  K 56Y5 (t )  K 56 c56Y5 (t ) s in5 (t )  C56 Y6 (t )  C56 Y5 (t ) 

C56 c56 cos 5 (t )  5 (t )  0
For infinitesimal angular displacement the cabin, sin ф5(t)= ф5(t) and cos ф5(t)=1.
After simplification, the equation of motion of the driver’s seat will be:
48
48

   
m6 Y6 (t )  K56Y6 (t )  K56Y5 (t )  K56 c56Y5 (t )5 (t )  C56 Y6 (t )  C56 Y5 (t )  C56 c56  5 (t )  0
49
49

4.2.7 Modeling of Loading Area
The modeling of the loading area is modeled as a rigid body having a lumped mass at its center
of gravity and having mass moment inertia J7 about an axis through its centre of gravity and
perpendicular to the axis along its length. This component is assumed to be pivoted at the back of
the truck and has a rotation motion about the pivot.
.
Fig.4.2.7 Modeling of loading area and its FBD
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50

1 1  
{ m7 aY 1eY 1 cos(3 (t )  7 (t ))  m7 aY 1eY 1 cos(3 (t )  7 (t ))} 3 (t )  m7 Y3 (t )eY 1 cos 7 (t ) 
2 2
1 1  
{ m7 eY 12  m7 eY 12 cos 27 (t )  J 7 }7 (t )  C37 cos 7 (t )aY 1 cos 3 (t )  3 (t )e37  C37 cos 7 (t )aY 1
2 2
  
cos 3 (t )  3 (t )eY 1  C37 cos 7 (t )a37 cos 3 (t )  3 (t )e37  C37 cos 7 (t )a37 cos 3 (t )  3 (t )eY 1 
  
C37 cos(7 (t )) 2  7 (t )e37 2  C37 cos(7 (t )) 2  7 (t )eY 12  2C37 cos(7 (t )) 2  7 (t )eY 1e37  m7 geY 1 cos 7 (t )
1  1 
 K 37 s in7 (t )e37 2 cos 7 (t )  m7 aY 1 ( 3 (t )) 2 eY 1 s in(7 (t )  3 (t ))  m7 aY 1 ( 3 (t )) 2 eY 1 s in(7 (t )  3 (t ))
2 2
 K 37 a37 s in3 (t )e37 cos 7 (t )  K 37 aY 1 s in3 (t )e37 cos 7 (t )  K 37 aY 1 s in3 (t )eY 1 cos 7 (t )  2 K 37 eY 1
1 
s in3 (t )e37 cos 7 (t )  m7 eY 12 ( 7 (t )) 2 s in 27 (t )  K 37 a37 s in3 (t )eY 1 cos 7 (t )  K 37 eY 12 s in(7 (t )
2
cos 7 (t )  0
For infinitesimal angular displacement of the chassis and loading area:
 
( 3 (t ))2  0 , ( 7 (t )) 2  0 ,sin ф (t)= ф (t) sin ф (t)= ф (t ) , cos ф (t)=1 and cos ф (t)=1
3 3 7 7 7 3
Taking into account the above assumption, the equation of motion of loading area is given as
shown below.
     
m7 aY 1eY 1  3 (t )  m7 Y3 (t )eY 1  {m7 eY 12  J 7 }7 (t )  C37 aY 1  3 (t )e37  C37 aY 1  3 (t )eY 1  C37 a37  3 (t )e37
   
C37 a37  3 (t )eY 1  C37  7 (t )e37 2  C37  7 (t )eY 12  2C37  7 (t )eY 1e37  m7 geY 1  K 37 e37 2  K 37 a37 e373 (t )
 K 37 aY 13 (t )e37  K 37 aY 13 (t )eY 1  2 K37 eY 13 (t )e37  K37 a373 (t )eY 1  K 37 eY 12 (7 (t )  0
4.3 Road Profile
The road profile varies from asphalt road to bumpy road (off road).This variation of road profile
induces different vibration related problems to the moving vehicle on this roads and causes
structural , vehicle component and leaf spring breakage when the car travels with a higher on
such roads.
I observed Ayenalem road that some what bumpy, the measurement I have taken shows that road
has uneven road profile due improper the founding stone. This improper arranged stone induces
acceleration and vibration to the passenger. The height of the bumps which I have taken varies
51
51

from 2cm to 10cm (some big stone).But some road like Lima limo has higher bump height (0.05-
0.3m).
For this assuming the road profile varies sinusoid ally, with amplitude of 0.25m with a wave
length of 0.75m and the truck travels with a horizontal speed of 30km/h.
Then the forcing frequency is given as
 30 *1000 
  2    0.75  69.81rad / s
 3600 
Implies
u  0.25sin 69.81t
From the ten equation of motion of the parts of the truck, the mass matrix, stiffness matrix and
damping coefficient matrix are written as shown below. The parameters m1, m2 and the like has
been defined already. Their estimated numerical value is given below in the table.
System parameter for component masses and overall dimension is taken from Mesfin Industrial
Engineering specification document for truck purchase and from internet Eurotrakker as shown
below. The damping coefficient and the stiffness is calculated (guessed from), using   0.35
for damping and 0.01m static deflection for the stiffness.
.
Eurotech
EuroTrakker
Dimensions 750E42HT 380E42W MP180E27W 750E44HTE 4500/48
(in mm) 380E37H 6X4 6x6 4x4 6x4 Tractor 6x4
(OR) Truck Freight Freight Tractor
6x4 Tractor Carrier Carrier
Tipper
OL - Overall 8175 6805 8495 7862 6844 6720
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52

Length
OW - Overall
2500 2500 2500 2500 2500 2550
Width
OH - Overall
3134 2998 3194 3102 3054 3530
Height
FOH - Front
1440 1440 1440 1440 1440 1380
Overhang
ROH - Rear
1855 785 1850 1780 785 1480
Overhang
WB - Wheelbase 4190 3890 5200 4500 3890 3860
BBC - Bumper to
1885 2980 1855 1855 1940 2120-2590
back of cab
CA - Cab to Rear
3775 3640 2660 4085 1855 3005
Axle/Unit
TR - Turning
8325 7975 8450 8375 7400 14200
Radius
Mass Data (in kg)

GVM -
Manufacturer's
38000 33000 38000 18000 33000 26000
Gross Vehicle
Mass
GCM -
Manufacturer's
Gross 85000 75000 60000 36000 76000 72000
Combination
Mass
GA -
Manufacturer's 8000 7500 8000 7500 75000 7500
Front Axle Mass
GA/GAU -
Manufacturer's 30000 26000 30000 11500 26000 21000
Rear Axle Mass
53
53

V - Permissable
Maximum 25500 25500 25700 16500 25500 25500
Vehicle Mass
AF - Permissable
Maximum Front 7500 7500 7700 7500 7500 7500
Axle Mass
AR - Permissable
18000 18000 18000 9000 18000 18000
Rear Axle Mass
UF - Unladen
4540 4775 4955 4655 4800 4860
Front Axle Mass
UR - Unladen
4990 4805 5285 2295 4600 3520
Rear Axle Mass
UT - Total
9530 9580 10240 6950 9400 8380
Unladen Mass
DT - Permissable
Max Drawing 65000 56000 56000 34000 56000 56000
Vehicle Mass
4.4 Mat lab Analysis
M-file for mat lab

function[tad]=fun(z,u,t)
m1=1200;m2=1000;m3=6010;m4=700;m5=900;m6=90;m7=23590;
J3=100367;J4=560;J5=80000;J7=33000;
k10=235440;k20=196200;k13=718108;k23=393054;k34=54936;k43=82404;
k35=105948;k53=70632;k56=40000;k37=4628358;
d10=20186.88;d20=16822.4;d13=29236.34;d23=58472.68;d34=11775.68*.6;
d43=11775.68*.4;d35=15177.6*.6;d53=15177.6*.4;d37=396840.416;d56=1600;
a13=1.865;a23=1.955;a34=1.255;a43=2.555;a35=2.248;a53=4.050;a37=2.150;
az1=4.050;az2=4.050;b34=0.7;b43=0.6;c35=0.6;c53=1.225;c56=0.01;e37=3.100;
ez1=3.100;ez2=5.000; %%%% numeric values for simulation
m=[m1 0 0 0 0 0 0 0 0 0;
0 m2 0 0 0 0 0 0 0 0;
54
54

0 0 m3+m7 m7*az1 0 0 0 0 0 0;
0 0 m7*az1 m7*az1*az1 0 0 0 0 0 0; %%%% mass matrix
0 0 0 0 m4 0 0 0 0 0;
0 0 0 0 0 J4 0 0 0 0;
0 0 0 0 0 0 m5 0 0 0;
0 0 0 0 0 0 0 J5 0 0;
0 0 0 0 0 0 0 0 m6 0;
0 0 -m7*ez1 -m7*az1*ez1 0 0 0 0 0 m7*ez1*ez1+J7];
c=[d13+d10 0 -d13 -d13*a13 0 0 0 0 0 0;
0 d20+d23 -d23 d23*a23 0 0 0 0 0 0;
-d13 -d23 d43+d34+d35+d53+d23+d13 -d35*a35-d53*a53-d43*a43-d34*a34+d13*a13-
d23*a23 -d34-d43 d43*b43-d34*b34 -d35-d53 d53*c53-d35*c35 0 0;
-d13*a13 d23*a23 -d43*a43-d34*a34-d35*a35-d53*a53-d23*a23+d13*a13
2*d37*az1*a37+d37*a37*a37+d37*az1*az1+d35*a35*a35+d43*a43*a43+d53*a53*a53+d23*a
23*a23+d13*a13*a13+d34*a34*a34 d43*a43+d34*a34 d43*a43*b43+d34*a34*b34
d35*a35+d53*a53 -d53*a53*c53+d35*a35*c35 0 -d37*e37*e37-d37*e37*az1-d37*ez1*a37-
d37*ez1*az1;
0 0 -d34-d43 d34*a34+d43*a43 d34+d43 d34*b34-d43*b43 0 0 0 0;
0 0 d43*b43-d34*b34 -d43*b43*a43+d34*b34*a34 d34*b34-d43*b43
d34*b34*b34+d43*b43*b43 0 0 0 0; %%%%%% damping coefficient matrix
0 0 -d53-d35 d35*a35+d53*a53 0 0 d53+d35+d56 d35*c35+d53*c53 d56 0;
0 0 -d35*c35+d53*c53 d35*c35*a35-d53*c53*a53 0 0 d35*c35-d53*c53-d56*c56
d35*c35*c35+d53*c53*c53+d56*c56*c56 d56*c56 0;
0 0 0 0 0 0 -d56 d56*c56 d56 0;
0 0 0 -d37*az1*e37-d37*az1*ez1-d37*a37*ez1-d37*a37*e37 0 0 0 0 0
d37*e37*e37+d37*ez1*ez1+2*d37*e37*ez1];
k=[k10+k13 0 -k13 -k13*a13 0 0 0 0 0 0;
0 k20+k23 -k23 k23*a23 0 0 0 0 0 0;
-k13 -k23 k35+k43+k34+k23+k13+k53 k13*a13-k23*a23-k35*a35-k43*a43-k34*a34-k53*a53
-k43-k34 k43*b43-k34*b34 -k35-k53 -k35*c35+k53*c53 0 0;
-k13*a13 k23*a23 k13*a13-k23*a23-k34*a34-k43*a43-k35*a35-k53*a53
k13*a13*a13+k23*a23*a23+k34*a34*a34+k35*a35*a35+k53*a53*a53+k37*az1*az1+k37*a3
7*a37+2*k37*az1*a37+k43*a43*a43 k34*a34+k43*a43 k34*b34*a34-k43*a43*b43
k35*a35+k53*a53 k35*a35*c35-k53*a53*c53 0 -k37*e37*az1-k37*e37*a37-k37*az1*ez1-
k37*ez1*a37;
0 0 -k43-k34 k34*a34+k43*a43 k34+k43 k34*b34-k43*b43 0 0 0 0;
55
55

0 0 -k34*b34+k43*b43 k34*b34*a34-k43*b43*a43 k34*b34-k43*b43
k43*b43*b43+k34*b34*b34 0 0 0 0; %%%%%%%%%%%%% stiffness matrix
0 0 -k35-k53 k35*a35+k53*a53 0 0 k35+k53+k56 k35*c35-k53*c53-k56*c56 -k56 0;
0 0 -k35*c35+k53*c53 k35*c35*a35-k53*c53*a53 0 0 k35*c35-k53*c53-k56*c56
k35*c35*c35+k53*c53*c53+k56*c56*c56 k56*c56 0;
0 0 0 0 0 0 -k56 k56*c56 k56 0;
0 0 0 -k37*a37*e37-k37*az1*e37-k37*az1*ez1-k37*ez1*a37 0 0 0 0 0
k37*37*e37+2*k37*e37*ez1+k37*ez1*ez1];
mc=m\c;mk=m\k;
A=[zeros(10) eye(10);-mk mc ]';
tad=A*[z(1) z(2) z(3) z(4) z(5) z(6) z(7) z(8) z(9) z(10) z(11) z(12) z(13) z(14) z(15) z(16) z(17)
z(18) z(19) z(20)]'+u;
end
4.4.1 Output of Mat lab

x=
Columns 1 through 8
0.3000 0.2953 0.2854 0.2733 0.2605 0.2475 0.2347 0.2222

0.2500 0.2429 0.2268 0.2067 0.1855 0.1649 0.1459 0.1288
0.0800 0.0876 0.1036 0.1224 0.1411 0.1581 0.1728 0.1849
0.0001 -0.0027 -0.0090 -0.0189 -0.0308 -0.0433 -0.0554 -0.0665
0.0300 0.0302 0.0308 0.0319 0.0331 0.0346 0.0362 0.0379
0.0001 0.0002 0.0003 0.0003 0.0004 0.0004 0.0004 0.0003
0.0500 0.0503 0.0512 0.0525 0.0542 0.0560 0.0581 0.0602
0.0001 -0.0001 -0.0008 -0.0018 -0.0032 -0.0048 -0.0068 -0.0090
0.0400 0.0399 0.0396 0.0391 0.0384 0.0375 0.0362 0.0348
0.0001 -0.0015 -0.0044 -0.0055 -0.0037 0.0011 0.0086 0.0185
0 0.0209 0.0311 0.0366 0.0399 0.0422 0.0440 0.0455
0 0.0190 0.0295 0.0348 0.0369 0.0372 0.0366 0.0356
0 0.0165 0.0412 0.0685 0.0958 0.1221 0.1468 0.1698
0 0.0125 0.0428 0.0787 0.1151 0.1503 0.1832 0.2135
0 0.0042 0.0113 0.0207 0.0318 0.0442 0.0574 0.0713
0 -0.0001 -0.0005 -0.0010 -0.0014 -0.0017 -0.0019 -0.0020
0 0.0058 0.0133 0.0218 0.0309 0.0403 0.0500 0.0598
0 -0.0008 -0.0034 -0.0069 -0.0110 -0.0154 -0.0200 -0.0248
56
56

0 0.0040 0.0082 0.0124 0.0168 0.0214 0.0260 0.0308
0 -0.0012 -0.0013 -0.0006 0.0008 0.0026 0.0044 0.0063
Columns 9 through 16
0.2099 0.1979 0.1862 0.1749 0.1639 0.1533 0.1430 0.1331

0.1137 0.1007 0.0895 0.0800 0.0720 0.0653 0.0598 0.0551
0.1945 0.2015 0.2062 0.2088 0.2095 0.2086 0.2062 0.2025
-0.0762 -0.0843 -0.0908 -0.0956 -0.0987 -0.1002 -0.1002 -0.0987
0.0396 0.0412 0.0429 0.0445 0.0461 0.0476 0.0490 0.0504
0.0002 0.0000 -0.0002 -0.0004 -0.0006 -0.0009 -0.0011 -0.0014
0.0625 0.0648 0.0672 0.0697 0.0721 0.0746 0.0770 0.0795
-0.0114 -0.0142 -0.0172 -0.0205 -0.0240 -0.0278 -0.0319 -0.0363
0.0330 0.0310 0.0288 0.0262 0.0235 0.0204 0.0171 0.0136
0.0305 0.0440 0.0587 0.0741 0.0900 0.1059 0.1216 0.1368
0.0467 0.0477 0.0485 0.0491 0.0496 0.0500 0.0502 0.0504
0.0344 0.0333 0.0324 0.0317 0.0311 0.0307 0.0305 0.0305
0.1911 0.2107 0.2288 0.2454 0.2607 0.2745 0.2872 0.2986
0.2412 0.2662 0.2887 0.3089 0.3268 0.3427 0.3566 0.3688
0.0854 0.0997 0.1139 0.1281 0.1420 0.1557 0.1690 0.1819
-0.0020 -0.0020 -0.0019 -0.0017 -0.0016 -0.0014 -0.0012 -0.0010
0.0698 0.0797 0.0897 0.0996 0.1094 0.1192 0.1288 0.1383
-0.0295 -0.0344 -0.0393 -0.0443 -0.0493 -0.0543 -0.0594 -0.0646
0.0356 0.0404 0.0453 0.0501 0.0550 0.0597 0.0644 0.0689
0.0080 0.0095 0.0109 0.0121 0.0132 0.0140 0.0147 0.0153
0.1235 0.1142 0.1053 0.0967 0.0885 0.0806 0.0730 0.0657

0.0513 0.0482 0.0456 0.0435 0.0418 0.0404 0.0392 0.0383
0.1976 0.1918 0.1852 0.1778 0.1698 0.1612 0.1522 0.1427
-0.0958 -0.0915 -0.0860 -0.0791 -0.0711 -0.0619 -0.0517 -0.0404
0.0516 0.0527 0.0538 0.0547 0.0556 0.0563 0.0569 0.0575
-0.0017 -0.0019 -0.0022 -0.0025 -0.0027 -0.0029 -0.0031 -0.0033
0.0819 0.0842 0.0865 0.0887 0.0909 0.0930 0.0949 0.0968
-0.0409 -0.0457 -0.0507 -0.0560 -0.0615 -0.0672 -0.0730 -0.0790
0.0098 0.0057 0.0015 -0.0030 -0.0077 -0.0126 -0.0177 -0.0230
57
57

0.1512 0.1649 0.1774 0.1888 0.1989 0.2076 0.2150 0.2208
0.0504 0.0503 0.0502 0.0499 0.0496 0.0493 0.0488 0.0483
0.0306 0.0307 0.0310 0.0313 0.0316 0.0320 0.0324 0.0328
0.3090 0.3183 0.3267 0.3341 0.3406 0.3464 0.3513 0.3556
0.3793 0.3883 0.3960 0.4024 0.4076 0.4117 0.4149 0.4172
0.1944 0.2064 0.2179 0.2289 0.2394 0.2494 0.2588 0.2677
-0.0009 -0.0007 -0.0006 -0.0005 -0.0005 -0.0004 -0.0004 -0.0005
0.1477 0.1568 0.1657 0.1743 0.1826 0.1906 0.1982 0.2054
-0.0699 -0.0752 -0.0805 -0.0859 -0.0914 -0.0969 -0.1024 -0.1080
0.0733 0.0775 0.0816 0.0854 0.0890 0.0923 0.0953 0.0980
0.0156 0.0159 0.0160 0.0159 0.0158 0.0155 0.0151 0.0147
0.0588 0.0521 0.0458 0.0397 0.0340 0.0285 0.0233 0.0183

0.0375 0.0369 0.0363 0.0359 0.0355 0.0351 0.0348 0.0345
0.1330 0.1229 0.1126 0.1022 0.0916 0.0808 0.0700 0.0592
-0.0281 -0.0150 -0.0009 0.0139 0.0295 0.0457 0.0626 0.0800
0.0579 0.0583 0.0585 0.0586 0.0587 0.0587 0.0585 0.0583
-0.0035 -0.0036 -0.0037 -0.0038 -0.0039 -0.0039 -0.0039 -0.0039
0.0986 0.1002 0.1017 0.1032 0.1044 0.1056 0.1067 0.1076
-0.0851 -0.0913 -0.0977 -0.1040 -0.1105 -0.1170 -0.1234 -0.1299
-0.0285 -0.0341 -0.0398 -0.0457 -0.0517 -0.0578 -0.0639 -0.0701
0.2252 0.2281 0.2295 0.2294 0.2279 0.2249 0.2206 0.2150
0.0478 0.0473 0.0466 0.0460 0.0453 0.0446 0.0439 0.0432
0.0332 0.0335 0.0339 0.0342 0.0345 0.0348 0.0350 0.0352
0.3591 0.3620 0.3643 0.3660 0.3671 0.3677 0.3678 0.3674
0.4186 0.4192 0.4192 0.4185 0.4172 0.4154 0.4131 0.4103
0.2760 0.2838 0.2910 0.2977 0.3038 0.3094 0.3145 0.3190
-0.0005 -0.0006 -0.0007 -0.0008 -0.0010 -0.0011 -0.0012 -0.0014
0.2123 0.2186 0.2245 0.2300 0.2349 0.2393 0.2432 0.2465
-0.1136 -0.1193 -0.1250 -0.1307 -0.1364 -0.1421 -0.1478 -0.1535
0.1003 0.1023 0.1040 0.1052 0.1060 0.1064 0.1064 0.1060
0.0141 0.0134 0.0127 0.0119 0.0111 0.0101 0.0092 0.0082
58
58

0.0136 0.0091 0.0049 0.0009 -0.0029 -0.0064 -0.0098 -0.0129
0.0342 0.0339 0.0336 0.0333 0.0330 0.0327 0.0323 0.0320
0.0483 0.0374 0.0265 0.0157 0.0049 -0.0058 -0.0165 -0.0270
0.0978 0.1162 0.1349 0.1538 0.1731 0.1925 0.2120 0.2316
0.0581 0.0577 0.0572 0.0567 0.0561 0.0555 0.0547 0.0540
-0.0038 -0.0038 -0.0037 -0.0035 -0.0034 -0.0032 -0.0030 -0.0028
0.1084 0.1091 0.1096 0.1101 0.1104 0.1106 0.1107 0.1107
-0.1363 -0.1426 -0.1489 -0.1550 -0.1610 -0.1669 -0.1725 -0.1780
-0.0764 -0.0827 -0.0889 -0.0952 -0.1014 -0.1076 -0.1137 -0.1197
0.2082 0.2001 0.1909 0.1806 0.1694 0.1572 0.1442 0.1304
0.0424 0.0416 0.0408 0.0400 0.0392 0.0383 0.0375 0.0366
0.0354 0.0355 0.0356 0.0356 0.0356 0.0355 0.0355 0.0353
0.3666 0.3653 0.3636 0.3615 0.3591 0.3562 0.3530 0.3495
0.4071 0.4035 0.3995 0.3952 0.3906 0.3856 0.3804 0.3750
0.3230 0.3265 0.3295 0.3320 0.3340 0.3355 0.3366 0.3372
-0.0016 -0.0017 -0.0019 -0.0020 -0.0022 -0.0023 -0.0024 -0.0025
0.2493 0.2515 0.2531 0.2541 0.2546 0.2544 0.2537 0.2524
-0.1592 -0.1648 -0.1705 -0.1761 -0.1816 -0.1871 -0.1925 -0.1979
0.1050 0.1037 0.1018 0.0995 0.0967 0.0934 0.0897 0.0854
0.0072 0.0061 0.0050 0.0039 0.0028 0.0017 0.0005 -0.0006
Column 41
-0.0159
0.0316
-0.0375
0.2512
0.0531
-0.0026
0.1106
-0.1832
-0.1257
0.1159
0.0358
0.0352
0.3456
0.3693
0.3373
59
59

-0.0026
0.2505
-0.2032
0.0808
-0.0017
Command for plotting
>> t=0:0.01:40;
>> xo=[0.3 0.25 0.08 0.000125 0.03 0.000120 0.05 0.000121 0.04 0.000122 0 0 0 0 0 0 0 0 0 0]';
>> u=zeros(20,length(t));
>> x=vtb9_3('tadtrial',u,t,xo);
VTB9_3 has been grandfathered. Please use VTB1_3 in the future.
plot(t,x(1,:));plot(t,x(2,:));plot(t,x(3,:));plot(t,x(4,:));plot(t,x(5,:));
plot(t,x(6,:));plot(t,x(7,:));plot(t,x(8,:));plot(t,x(9,:));plot(t,x(10,:));
4.4.2 Mat lab Output Graphs
Fig .4.4.2.1 time response for the rear wheel
60
60

Fig. 4.4.2.2. linear displacement response of the front wheel
61
61

Fig.4.4.2.3 linear displacement response of the chassis
Fig 4.4.2.4 angular displacement response for the chassis
62
62

Fig 4.4.2.5 linear displacement response for the engine
Fig 4.4.2.6 linear displacement response of engine
63
63

Fig 4.4.2.7 linear displacement response for the cabin
64
64

Fig 4.4.2.8. angular displacement response of the cabin
Fig 4.4.2.9. Angular displacement response of the loading area
65
65

Fig.4.4.2.10 Response of the driver seat assembly
Fig .4.4.11 the combined responses the ten DOF truck model
66
66

5. CONCLUSION
I. From the mat lab graph it has been found that by increasing the damping coefficient of
the driver the displacement response decays within a short period of time.
II. From Ayenalem road profile observation, it is easy to say that even small pieces of stone
can induce a vibration to the car. Uneven roads have an up and down (bump) shape with
different distance of travel. So it is difficult to assume the bump repeats itself after a fixed
distance of travel .for example, it is difficult to say the bump repeats itself after 1m or 2m
etc.
67
67

6. RECOMMENDATION
I. To obtain a much accurate design, for the isolation system of the driver’s seat;
accelerometers and seismometer should be used to record the actual displacement and
acceleration transmitted to the seat during the truck travels on rough road(or at different
road profile).
II. To increase the accuracy of the modeling of the truck, it is recommended to 3-D
modeling.
68
68

Bibliography
1. Robert, F Jr .Steidel, .An Introduction to mechanical vibration, 3rd edition, USA
2. Bensen ,H .Tongue. Principle of Vibration, New York, 1996
3. Singiresu S RAO.MECHANICAL VIBRATION, 3rd edit, Addison-Wesley, USA, 1995
4. www.truck.html
5. Reference materials from my advisor
6. www.Howstuffwork.com
Literature Review
7. Test Methods for Ride Comfort Evaluation of Truck Seats

A Master’s Thesis by Johan Lindén
July 2003
8. REVIEW OF SELECTED LITERATURE RELATED TO SEATING DISCOMFORT

Nahm Sik Lee
Lawrence W. Schneider
Leda L. Ricci
April 1990
9. Field Study to Evaluate Driver Fatigue on Air-Inflated Truck Seat Cushions

by Christopher Matthew Boggs
July 2004
10. Suspension System Optimization to Reduce Whole Body Vibration Exposure on an
Articulated Dump Truck
By JC.kirstein
Thesis:Msc Ing(Meg)
Sept 2005
11. WHOLE BODY VIBRATION: Occupational Health Clinics for Ontario

69
69

workers Inc.
70
70

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Final project on

MODELING AND DESIGN OF A SEAT SUSPENSION (ISOLATOR)

By: Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

1.1 Vibration Theory

1.2 Elementary Parts Vibratory System

1.2.1 Spring Elements

where: K is the spring constant or stiffness

x is the deformation or relative of one end with respect to the other

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

1.2.2 Mass or Inertia Element

1.2.3 Damping Element

Damper can be modeled as

 Coulomb or dry friction damping

1.3 Classification of Vibration System

1.3.1 Forced /free vibration

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

1.3.2 Damped and Undamped Vibration

1.4 Vibration Control

1.4.2 Shock Isolation

1.5 Suspension System

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

1.5.1 Purpose of Suspension System

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

1.5.2 Suspension Springs

1.5.3 Dampers: Shock Absorbers

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Fig.1.5.3 Twin-tube shock absorber (source [7])

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

1.5.4 Types of Suspension System

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Whole Body Vibration Exposure

Transportation Buses, helicopters, subway trains,

Table.2.1 (source [12])

Final year project by Taddesse Fentie

Final year project by Taddesse Fentie

Parts of body Resonance frequency (Hz)

Table.2.2 (Source [8])

2.2 Controlling Effects Whole Body Vibration

1. Reduce the transmission of vibration to the passengers by engineering the equipment or

2. Decrease the amount of vibration to which the passenger is exposed by:

Final year project by Taddesse Fentie