Modelling and Dimensioning
the Rear Suspension of a
Mountain Bike
Brian Frydensberg Damgaard
Gustav Winther Svendsen
Nicolai Christian Bloch
Rasmus Bruus Nielsen
Jonas Morsb¿l
Jesper Kann
Department of Mechanical Engineering
Bachelor project
Aalborg University
Department of Mechanical
Engineering
Pontoppidanstræde 105
9220 Aalborg Ø
Tel (+45) 9940 9300
Fax (+45) 9815 1675
http://www.me.aau.dk/
Title:
Modelling and Dimensioning the Rear
Suspension of a Mountain Bike
Semester theme:
Complex Mechanical Systems
Project period:
The M-sector’s 6th semester
February 2nd 2009
– May 27th 2009
Project Group:
53b
Participants:
Jonas Morsbøl
Gustav Winther Svendsen
Brian Frydensberg Damgaard
Nicolai Christian Bloch
Rasmus Bruus Nielsen
Jesper Kann
Supervisor:
Jens Christian Rauhe
Jonas Morsbøl
Gustav Winther Svendsen
Brian Frydensberg Damgaard
Nicolai Christian Bloch
Rasmus Bruus Nielsen
Copies: 9
Number of Pages: 96
Jesper Kann
iii
Abstract
This report deals with the analysis and design of a mountain bike rear suspension. The suspension is
designed for riding off-road in Danish terrain and a set of simple loading conditions have been created.
The four bar linkage design with Horst link has been chosen as the overall basis for developing
a suspension model. A kinematic model has been developed using the Nikravesh approach with a
Newton-Raphson numerical solving method [1]. This model allows the rear wheel of the bike to be
moved independently of the frame thus allowing a simulation of the rear suspension travel. Based
on the kinematics, a kinetic model has been developed using the Lagrange Multiplier method. A
quick study of rider induced suspension activation has been made. The suspension compression due
to weight of the rider has been determined. Also, various configurations of the four bar linkage have
been examined aiming to prevent suspension activation due to pedalling.
A hydraulic model of a FOX RP32 mountain bike shock absorber has been developed, allowing
a more realistic simulation of the suspension characteristics [2]. The spring and damping forces have
proven to be nonlinear with the damping force also being dependent on the compression level of the
shock absorber. Difficulties in determining part of the shock absorber geometry has led to the proposal
of two laboratory experiments. Using inverse modelling, the accuracy of the damping characteristics
in the model could then be greatly improved.
The kinematic model has been used in combination with the shock absorber model to determine the
equivalent stiffness and damping on the rear wheel from the shock absorber. These calculations have
been used to examine the response of the suspension from vertical drops. The approach used to find
the equivalent stiffness has proven incorrect because of the varying shock absorber spring stiffness.
A vibrational analysis is carried out to determine the response of the bike suspension on uneven
terrain. The results indicate that similar responses can be achieved for different configurations of the
four bar system. This implies that the tendency of the suspension to activate due to pedalling may
possibly be reduced without affecting the suspension performance.
All parts of the model have been programmed in Matlab. Numerical integration has been used
extensively in the model due to the nonlinear nature of the system. The implementation and connection
of the different parts of the model has proven difficult and needs to be developed further.
The upper rear link of the suspension has been dimensioned against static and fatigue failure according to DS419/DS409 [3], [4]. This has been done using beam theory and finite element analysis.
[1] Nikravesh PE. Computer-Aided Analysis of Mechanical Systems. Prentice Hall; 1987.
[2] Fox Racing Shox. Fox Float RP23. http://www.foxracingshox.com/bike/09/shocks/FLOAT; 2009.
[3] Dansk Standard. Norm for aluminiumskonstruktioner. 3rd ed.; 2001.
[4] Dansk Standard. Norm for sikkerhedsbestemmelser for konstruktioner. 2nd ed.; 1998.
v
Preface
This report has been composed by Group 53b on the 6th semester of Design of Mechanical Systems at
Aalborg University. It presents an analysis and design of a mountain bike rear suspension.
In appendix E is a CD containing the following elements:
•
•
•
•
•
Movies of the active bike suspension.
All Matlab programs used in the developed models.
AnyBody configuration files used for simulation.
SolidWorks CAD files for parts and FE analysis.
A copy of this report in PDF format.
Citations are done in accordance with the Vancouver Style in square brackets with references
numbered in order of appearance: e.g. [1] or [2, p. 5] for specific page numbers. The numbers refer to
the bibliography in the back of the report where information about the sources is found.
Figures and equations are numbered according to the chapter in which they are presented—for
instance: figure 1.2 is the 2nd figure in chapter 1.
The following mathematical notation is used throughout the text:
~q
D̄¯
x̂
~ ξ ≡ FD
~ · ξˆ
|FD|
~si
~s j 0i
2
q
Time derivatives:
~q˙ ≡ d~
q¨ ≡ ddt~2q , . . .
dt , ~
h i
~Φ = [Φ1 . . . Φm ]T , i = row index.
∂~Φi
∂~Φ
¯
Partial derivatives with respect to vectors: Φ̄~q ≡ ∂~q ≡ ∂q j
T
Vectors:
Matrices:
Unit vectors:
Vector component ξ:
Global vector from origin to point i:
Local vector from frame j to point i:
(m×k)
~q = [q1 . . . qk ] ,
j = column index.
As proposed by Nikravesh the first, second, and third axes are named x, y, z and ξ, η, ζ for global
and local axes, respectively [1]. Further notation is given where relevant and in the general list of
nomenclature in the back of the report.
Special thanks are addressed to Professor Jon Rasmussen and Ph.D. Student Christian Gammelgaard
for assistance with the AnyBody software and providing qualified assessments of the data obtained.
vii
Contents
1
Introduction
1.1 Hardtail and Full Suspension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Initial Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Analysis of Different Bike Designs
3
2.1 Suspension Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Selection of Suspension Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3
Problem Statement
12
3.1 Chapter Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4
Loading Scenarios for the Bike
4.1 Free Fall . . . . . . . . . . . . .
4.2 Magnitude of the Braking Force
4.3 Pedal Force . . . . . . . . . . .
4.4 Operational Conditions . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
1
2
14
14
15
15
18
5
Modelling and Design Strategy for the Mountain Bike.
6
Elements in the Dynamic Model
6.1 Kinematic Model of the Four Bar Linkage
6.2 Kinetic Analysis . . . . . . . . . . . . . .
6.3 Shock Absorber . . . . . . . . . . . . . .
6.4 Equivalent Stiffness and Damping . . . .
6.5 Response to Frequency Input . . . . . . .
6.6 Modelling the Free Fall . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
21
21
27
36
49
53
61
Results of the Dynamic Analysis
7.1 The Four Bar Linkage . . . .
7.2 The Shock Absorber . . . .
7.3 Response to Frequency Input
7.4 Free Fall . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
66
66
68
70
70
7
8
Strength & Dimensioning
CONTENTS
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
19
72
ix
CONTENTS
8.1
8.2
8.3
9
Dimensioning Based Upon Classical Beam Theory . . . . . . . . . . . . . . . . . . . 72
Weldings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Discussions & Conclusions
94
Nomenclature
97
Bibliography
101
A Additional Calculations to Frequency Response
103
B Solving the Equation of Motion Describing the Free Fall
104
C Sign Convention
105
D Welding
107
E CD
108
F Sketch of the Entire Bike
109
x
CONTENTS
Introduction
1
By observing bicycles, a vast variety in designs is easily noticed. Road bikes usually have a stiff
and light structure, whereas mountain bikes have wide tires, a strong frame, and a front fork with
suspension. So apparently the design depends upon the area of use, and even within mountain bike
design the variety is great.
In recent years the number of different mountain bike designs has increased. This is caused by the
constantly present urge from bikers to overcome even rougher terrains and steeper hills, while keeping
a high level of comfort and safety. As a consequence the full suspension mountain bike has been
brought in to focus.
The optimal performance of the suspension depends on the technical and physical prowess of the
rider, and the range of use the rider expects from the mountain bike. Riders exercising All Mountain
and Down Hill, which include steep drops and high speed, require full suspension to soften the uneven
tracks, allowing them to go faster. Riders exercising Cross Country and Marathon, which include
riding in forests, on hills and on dirt-roads, require light and energy efficient bikes with less travel and
better adjustment of the suspension characteristics.
To summarize, there are two different kinds of demands from mountain bikers. One is satisfied by
a stiff bike with a so called hardtail. The other is dealt with through full suspension mountain bikes.
1.1
Hardtail and Full Suspension
In many years, mountain bikes have had suspended front wheel fixtures as indicated in figure 1.1. By
introducing the full suspension mountain bike, the rear wheel has become suspended, which is the
main feature of a full suspension bike. This concept is illustrated in figure 1.2. The figures 1.1 and
1.2 simply describe the two basic concepts in mountain bike designs, which is why many designs that
deviate a lot from the ones shown, may be found. What makes a bike a full suspension bike is the
Figure 1.1: The regular hardtail mountain bike.
Figure 1.2: Full suspension mountain bike.
system of hinged bars and springs, allowing a displacement of the rear wheel fixture relative to the
main frame, and therefore the ability to level out rough terrain.
CHAPTER 1. INTRODUCTION
1
1.2. INITIAL PROBLEM
1.1.1
Advantages and Disadvantages of The Full Suspension Bike
As mentioned earlier, the full suspension mountain bike provides comfort and safety. Comfort is
achieved through better handling characteristics which enhances the pleasure of mountain biking and
potentially encourages the rider to use his bike more frequently. Safety is provided as the suspension
secures better grip between tire and track which improves braking performance and reduces the risk of
skidding.
When allowing relative movement between rider and rear wheel, increased energy loss when pedalling is a possible side effect. As the rider treads the pedals, the shock absorber suspending the rear
wheel may compress. This reduces the efficiency of the power transmission due to energy dissipation
in the suspension unit.
1.2
Initial Problem
The main issue when designing full suspension mountain bikes is to satisfy the desire for comfort
and safety by absorbing bumps on the road, but at the same time having a complete transformation of
pedalling work to propulsion. This has lead to the initial problem:
How is an easy driven and comfortable full suspension mountain bike designed
for Danish terrain?
The design of the rear wheel fixture is considered as the main characteristic of a full suspension
mountain bike. The focus in this report will therefore mainly be on the rear suspension unit.
2
CHAPTER 1. INTRODUCTION
2
Analysis of Different Bike
Designs
This chapter serves the purpose of clarifying different terms and side effects regarding full suspension
mountain bikes, supported by mechanical considerations. This is followed by a presentation of different suspension designs, concluded by selection of a specific design type for further investigation.
Terms and assertions throughout this chapter are based on material from Titus Cycles [2]. Before this
chapter begins some basic mountain bike terms and components are presented in figure 2.1 and table
2.1. These terms will be used in the rest of this report.
13
11
14
12
4
3
15 16
2
1
8
9
5
10
6
7
Figure 2.1: A typical full suspension mountain bike.
No.
1
2
3
4
5
6
7
Term
Rear dropout
Cassette
Braking calibre
Upper rear bar
Horst link
Chain adjuster
Lower rear bar
No.
8
9
10
11
12
13
14
Term
Near crank pivot
Crank
Chain ring
Rocker arm
Shock absorber
Main Frame
Front fork
No.
15
16
Term
Front shock absorber
Front wheel dropout
Table 2.1
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
3
2.1. SUSPENSION DESIGNS
2.1
Suspension Designs
The development of modern full suspension bikes has given birth to many different rear suspension
designs with different riding characteristics. Most designs can be placed in one of the design groups
which will be presented later in this section. Each group represents a general suspension configuration
with some inherent advantages and disadvantages.
Depending on geometry, the performance of a rear suspension can be affected by pedalling, braking or whether the rider is seated or standing. If pedalling, breaking or standing in upright position
is causing the suspension to be activated so it either resists compression or over-compresses, the suspension design is said to be semi-active or in more severe cases; in-active. On the other hand, if the
suspension system is not, or only slightly, affected by these factors the suspension is called fully-active.
Bio-Pacing and DISC
The terms Bio-Pace and DISC1 are both related to pedalling. Here, pedalling refers to the fact that the
chain exerts a force on the cassette mounted on the rear wheel. Due to the properties of the suspension
linkage this force can result in compression or un-compression of the suspension. When the pedalling
compresses the suspension the design is said to exercise DISC and if the pedalling uncompresses the
suspension the design is Bio-Pacing. Because of the periodic nature of pedalling the force in the chain
is also periodic. This means that if the suspension design tends to DISC the rider will feel that the
saddle is pulled downwards by each pedal stroke. On top of this it is often seen that the mass of the
rider is moving up and down when the pedalling force continuously is moved from one leg to the other.
This up and down movement of the rider will in its self force the suspension to compress periodically
with the same frequencies as the compression due to the DISC. Again it is important to note that
the compression induced by up and down movement of the rider is not a part of the pedalling but a
often seen side effect. If the suspension design Bio-Paces this will reduce the compression made by
the rider’s up and down movement because the Bio-Pacing tries to un-compress the suspension at the
same time as the rider’s moving mass tries to compress the suspension. This means that it is possible
that the rider in some situations can benefit from Bio-Pacing.
To understand the mechanism of how pedalling can affect the compression of the suspension a few
figures can be helpful. On figure 2.2 the suspension linkage is represented by the green bars. To get
a view of the forces that act on the wheel and suspension system, two diagrams are shown in figure
2.3. On the left part of 2.3 the red arrows indicate forces that are applied on the wheel, i.e. gravity on
the wheel from rider and bike, Fvertical , and the chain force due to the pedalling. The green arrows are
reactions to these applied forces. It is seen that Fvertical is balanced by the normal force acting from the
ground. The chain force and the friction force act together on the bike through the suspension linkage
represented by a red arrow on the right part of figure 2.3. Note that ~Fwheel = ~Fchain + ~R f riction . Due
to Newton’s third law, the suspension returns an opposing force, Rbike , on the wheel. On most bike
designs the chain force and Rbike are close to parallel.
As a result of the kinematics of the suspension linkage, it is indeed possible that, when the suspension is activated and the wheel moves through its travel, its travel path has a component parallel to
~Fwheel , which is seen on figure 2.4. When a movement parallel to ~Fwheel is allowed by the kinematics,
the suspension will be affected by the chain force. If the component of the wheels travel path is in the
1 Abbreviation
4
for: Drive train Induced Suspension Compression.
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
2.1. SUSPENSION DESIGNS
Rear
wheel
Shock absorber
mbike
Fvertical
mbike
Fchain
Rbike
Fwheel
Rfriction
Chain
Figure 2.2: Bike with suspended rear wheel. The red
dashed line represents the shock absorber and the green
bars represents the suspension linkage.
Nground
Nground
Figure 2.3: The left part of the figure shows a free body
diagram containing the forces that act on the wheel when
pedalling. The right part shows the forces from the wheel
that act on the suspension linkage.
same direction as ~Fwheel , when the suspension is compressed, the suspension design tends to DISC. On
figure 2.4 this is the case on path no. 1 where the path is shown as a blue dashed curve. Here, when
the wheel moves up it also moves in direction of ~Fwheel . Alternatively, if the component of the wheels
travel path is in the same direction as ~Fwheel when the suspension rebounds, the bike tends to bio-pace.
On figure 2.4 this happens on path no. 2. Two simple examples of suspension designs which generate
DISC or Bio-Pacing, are seen in figure 2.5 and 2.6 respectively.
Possible travel paths
mbike
1
2
Fwheel
Figure 2.4: When the suspension is compressed and the wheel moves through its travel path it is possible that the movement
of the wheel has a component which is parallel to ~Fwheel . The shape of the travel path is determent by the kinematic properties
of the suspension linkage.
Braking
How the suspension is affected by braking can be seen on figures 2.7 and 2.8. When the rear brake
is activated, the wheel creates a force on the braking calibre due to the friction between the braking
surface on the disc and the braking pads in the calibre. The braking force induces a reaction force from
the wheel on its mounting point on the bar. These two forces create a couple, that tends to rotate the
bar on which the braking calibre and wheel are mounted. If the kinematics of the suspension linkage
allows a rotation of the bar, the suspension will be affected by braking. If such rotation of the bar is
not a part of the natural movement of the suspension during its travel, the force couple will only result
in deflection of the bar and not affect the suspension performance.
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
5
2.1. SUSPENSION DESIGNS
Traveling path
Traveling path
Figure 2.5: A full suspension mountain bike which will
tend to DISC.
Figure 2.6: A full suspension mountain bike which will
tend to bio-pace.
Braking calibre
mbike
mbike
Fbrake
Rbrake
Braking disc
Figure 2.7: On this figure the bike is mounted with a disc
brake.
Figure 2.8: The braking force is balanced by an opposing
force from the mounting point of the wheel.
Unsprung Mass
The mass of the wheel, rear drive train, and the bars that the suspension linkage is made of can be
represented by a equivalent mass acting alone on the shock absorber. It means that the equivalent mass
represents the unsprung mass. When the unsprung mass is low the force from the shock absorber can
deliver higher accelerations of the equivalent mass. This can be seen from this equation: ~Fshock =
d ~Punsprung
dt
d~v
= meq unsprung
= meq~aunsprung . This gives a faster rebound when the wheel is leaving the top
dt
of a bump or going to the bottom of a hole and the suspension has to uncompress. A fast rebound is
important to ensure wheel contact to the ground.
Horst Link
The Horst link was originally discovered by Horst Leitner. It is simply a pivot placed in front of and
below the rear wheel dropout. This means that the Horst link is in fact not a link but a joint. By use
of a properly placed Horst link it is possible to reduce both braking influence on the suspension and
Bio-Pacing or DISC. A Horst link can be seen on figure 2.1.
Summary
With this analysis of how the forces interact with the bike design, it should be possible to evaluate
which design is suited for a particular use. The following analysis provides an overview of some more
6
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
2.1. SUSPENSION DESIGNS
or less popular suspension design groups. Based on the advantages and disadvantages presented here,
a specific design will be selected and dealt with in greater depth.
2.1.1
High Pivot
In High Pivot designs, the main pivot point connecting the front and rear part of the bike is placed
above or in line with the front chain rings as seen in figure 2.9.
Figure 2.9: High Pivot suspension system.
Figure 2.10: High Pivot suspension activated.
Advantages and Disadvantages
High Pivot suspensions often exhibit some degree of Bio-Pacing depending on the position of the pivot
point relative to the active chain ring. The design is simply allowing great strength, which makes it
a popular suspension type for down hill bikes. Furthermore, the path of the rear wheel travel can be
set to match that of the front wheel, to maintain a constant wheel base. The High Pivot suspension
is used widely in down hill bike design due to its simplicity which eases manufacturing and calls for
very robust and stiff suspension system. The simple and linear suspension compression may reduce
freedom for fine tuning the suspension characteristics but will in some cases be sufficient.
2.1.2
Unified Rear Triangle
Unified Rear Triangle (URT) designs are characterised by the crank being mounted on the unsprung
rear part of the bike as shown in the figures 2.11 - 2.14. This reduces the influence from chain pull on
the suspension system, but increases the unsprung mass.
Advantages and Disadvantages
Typically, the rider will experience a significant drop in suspension performance when pedalling while
standing. This is due to the relatively large unsprung mass which has a negative effect on the suspension response.
High Pivot/Sweet Spot
In High Pivot URT designs, the pivot point is placed above the chain rings. See figure 2.11. This
reduces the rider’s influence on the suspension, but also causes the system to become semi-active
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
7
2.1. SUSPENSION DESIGNS
when the rider is not seated. Because the crank is positioned far in front of the pivot point, the distance
between the seat and the crank varies considerably when the suspension is activated, as can be seen
in figure 2.12. The High Pivot URT suspensions generally have good resistance towards compression
when the rider is pedalling while standing. Therefore this design is suitable for cross-country and
marathon bikes.
Figure 2.11: URT High Pivot suspension system.
Figure 2.12: URT High Pivot suspension activated.
Low Pivot
The Low Pivot URT design is characterised by having the pivot point placed below the chain rings,
typically near the crank, as it can be seen in figure 2.13.
Figure 2.13: URT Low Pivot suspension system.
Figure 2.14: URT Low Pivot suspension activated.
Compared to the High Pivot URT design, this reduces considerably the variation in distance between
the seat and the crank when riding. The Low Pivot design is more sensitive to suspension activation,
caused by the rider pedalling while standing, compared to most other designs.
URT suspension designs effectively reduce DISC and Bio-Pacing, but tend to become in-active
when the rider is not seated. In cross-country and marathon disciplines this may be beneficial, but it is
unwanted on rough terrain. Depending on the position of the pivot point, the suspension may cause a
considerable variation in seat-to-crank distance, which is considered as a major disadvantage.
2.1.3
Strut Design
The strut design with or without a Horst link has been used for many years. The design was—probably
due to its simplicity—one of the first types of full suspension bikes, and is still used. One of the
8
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
2.1. SUSPENSION DESIGNS
characteristics of the strut design is the use of a rear bar as a shock absorber. The strut design comes
both with and without a Horst link.
The bike design shown in figure 2.15 is the strut design with a Horst link, and like other types
of designs, the strut design has a tendency to either bio-pacing or DISC when it is mounted without
a Horst link, but these effects can be eliminated by including a Horst link in the design. As seen in
Figure 2.15: Strut design with Horst link.
Figure 2.16: Strut design with activated suspension.
figure 2.15, the upper rear bar is substituted with a shock absorber, which must be able to withstand
shear forces. The shock absorber is in this sense acting like a translational joint, as indicated in figure
2.16 where no hinge is indicated between the shock absorber and the green bars. Note that the shock
absorber is hinged on the main frame of the bike.
Advantages and Disadvantages
The shock absorbing bar replaces both a linkage and a part of the frame, which lessen the amount of
material in the design. The reduced amount of material combined with the simplicity of the design
makes it possible to design very light bikes. As mentioned earlier, the design requires a shock absorbing bar, which has an ability to withstand both bending and shear forces. This bending arises when
braking and as a result of this bar’s ability to handle bending, the mountain biker does not experience
any horizontal movement when braking. This requires a special shock absorbing unit, whereas other
mountain bike designs use standard shock absorbers which are hinged in both ends.
As a result of the light weight, bikes with strut designs generally require less effort to maneuver.
This, combined with the fact that a strut bike is neutral when pedalling, is particularly advantageous in
disciplines such as cross country where the mountain biker is climbing hills.
2.1.4
Four Bar
The four bar suspension is probably the most common and covers designs in a wide range of uses.
The design is indicated in figures 2.17 and 2.18.
It consists of three bars, one being the triangle placed below the seat called the rocker arm, and the
frame as the fourth bar. The shock absorber is attached between the rocker arm and the frame.
Advantages and Disadvantages
The main advantage of this design is the ability to adjust the amount of DISC and Bio-Pacing, which
depends on the configuration of the bars. The relatively large numbers of bars makes it easy to obtain
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
9
2.1. SUSPENSION DESIGNS
Figure 2.17: The system is not activated.
Figure 2.18: The four bar suspension system activated.
Figure 2.19: The suspension is at its neutral position.
Figure 2.20: The Virtual Pivot Point with activated suspension
specific suspension characteristics, but this also leads to an increased number of pivots, which can
results in a heavier, weaker and more expensive bike.
Virtual Pivot Point
Even though they look different, the Virtual Pivot Point (VPP) is a variant of the four bar with Horst
link. This design is illustrated in the figures 2.19 and 2.20. The difference with (VPP) is, that the Horst
link is located near the crank instead of near the rear wheel dropout, which is the case in the regular
four bar design. Also at first glance, it looks like this design has an extra bar. However, this bar merely
serves as a structural reinforcement of the rear upper link and does not constitute a separate bar in the
linkage system.
The name virtual pivot point refers to the fact, that the rear wheel does not have an absolute center
of rotation, hence the term virtual.
Summary
The suspension characteristics of the four bar design is in many ways superior to those of other designs
and is therefore considered the most popular. It can be designed for many disciplines such as down
hill or cross country, and this flexibility brings some interesting challenges to the design process.
10
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
2.2. SELECTION OF SUSPENSION DESIGN
2.2
Selection of Suspension Design
A variety of suspension designs have been presented ranging from the simple High Pivot single link
to the more complicated four bar system with multiple links. Table 2.2 summarises the strengths and
weaknesses of each suspension type. Based on the analysis of different design types, the four bar
Name
High Pivot
High Pivot
URT
Low Pivot
Strut
Regular
Four Bar
VPP
Pros
+ Simple design
+ Robust
+ Constant wheel base
+ Little DISC/Bio-Pacing
+ Little DISC/Bio-Pacing
+ Simple design
+ Light structure
+ Not affected by braking
+ Little DISC/Bio-Pacing
+ Highly configurable
+ Little DISC/Bio-Pacing
+ Highly configurable
Cons
- Tendency to Bio-Pace
- Fine tuning difficult
- Varying seat-to-crank distance
- Suspension only active when seated
- Sensitive to poor riding style
- Suspension only active when seated
- Tendency to DISC/Bio-Pace
- Non-standard shock absorber needed
- Complicated design
- Complicated design
Table 2.2: Pros and cons of suspension designs presented in section 2.1.
linkage suspension system is chosen, and will be investigated further even though the above analysis
does not necessarily show, that the four bar linkage is well suited for Danish terrain. The reason for
this is, that the four bar linkage provides the greatest number of design parameters when configuring
the design and therefore possibly has the biggest potential in fulfilling as many wishes as possible
regarding suspension characteristics. Consequently the four bar suspension system is chosen as a basis
for the design presented in the following chapters.
CHAPTER 2. ANALYSIS OF DIFFERENT BIKE DESIGNS
11
3
Problem Statement
This chapter outlines the principal areas of work in this report. A list of necessary tasks will be presented, providing an overview of the content in the succeeding chapters.
In accordance with the initial problem and the selected design, the purpose of this report is as follows:
Based on a dynamic analysis, a mountain bike rear suspension system suitable for
cross country riding, must be examined.
As preferences regarding riding characteristics may be very subjective, it has been decided that the
objective of the design process will be to design a fully-active suspension system with a minimum of
energy loss. The project will be focused on developing a design strategy and -model suitable for this
purpose. The project will include the following fields of work:
• Loading inputs on the bike from both the rider and the environment must be determined.
• A mathematical model describing the kinematic properties of the rear wheel suspension must be
made.
• Based on the kinematics of the system, a kinetic model must be established to predict the movement of the suspension system under different loading conditions.
• A further investigation of the system response to different input must be made, leading to a
satisfying configuration of the suspension. Proper damping- and spring coefficients, lengths of
members and other relevant parameters must then be determined.
• One link in the four bar linkage system and one welding joint will be dimensioned. These
structural components are to be dimensioned in accordance with DS 419 and DS 409.
3.0.1
Limitations
The following overall limitations are made for the project:
• Dimensioning of kinematic joints are not conducted.
• The impact of a transverse load on the bike frame is not examined.
• The characteristics of the tires are not included in the dynamic analysis.
12
CHAPTER 3. PROBLEM STATEMENT
3.1. CHAPTER OUTLINE
3.1
Chapter Outline
The above statements lead to following outline of this report:
• Operational conditions
– Determination of loading conditions through the life cycle of the bike.
– Determination of maximum loading conditions.
• Modelling
– System kinematics
– System kinetics
• Dynamic optimisation
– System response to e.g. frequency input into the wheels
– Determination of system configuration and parameter values
• Design and dimensioning
– Classical Beam Theory
– FE analysis
• Discussion
CHAPTER 3. PROBLEM STATEMENT
13
4
Loading Scenarios for the Bike
Before the bike can be designed and its components can be dimensioned, the loading conditions for
the bike must be determined. To obtain a realistic loading scenario, tests and measurements should
be conducted. Since such tests would be too time consuming for this project, a simpler approach is
taken. This chapter presents three different loading scenarios which are assumed to adequately cover
the wide range of possible types of loadings on a full suspension mountain bike.
4.1
Free Fall
When riding a mountain bike in a forest, it is likely to leap off a small hill or bump. A situation where
the rear wheel lands first and absorbs all the momentum from the fall is plausible. Since this project
deals with the design of the rear suspension, such a situation is relevant to analyse. The horizontal velocity and -force components do not need to be taken into account, so the load exerted on the mountain
bike by a vertical fall will be examined. It is presumed that a free fall from one metre is an adequate
design criterion. Figure 4.1 illustrates the exact point just as the rear wheel hits the ground. All the
kinetic energy of the rider and bike will cause the frame to move downwards at an initial velocity
~vin,1 , causing the suspension system to move. As a result the shock absorber compresses at an initial
velocity ~vin,2 . The reason for the different velocities is the gearing due to the linkage system. As the
Vin 2
Vin
Figure 4.1: The free fall inducing an initial velocity downwards on the frame causing the suspension to compress.
geometry changes during compression the gearing will change. So without knowing the nature of the
shock absorber it can already be concluded that it has a nonlinear impact on the suspension system.
The force exerted upon the frame from compression of the shock absorber is to be determined and later
used in the stress analysis.
14
CHAPTER 4. LOADING SCENARIOS FOR THE BIKE
4.2. MAGNITUDE OF THE BRAKING FORCE
4.2
Magnitude of the Braking Force
When braking a force on the braking calibre is created as described in section 2.1. In the following,
the conditions of this force are examined further.
The force on the braking calibre due to braking is called the braking force. The magnitude of the
braking force is determined by the friction force between the ground and the wheel when braking. As
long as slip is not present the friction force is limited by:
Ff riction ≤ µstatic N
where:
µstatic
The coefficient of static friction.
N
The magnitude of normal force.
(4.1)
If the wheel starts to slip the friction force is limited by:
Ff riction = µdynamic N
where:
µdynamic
(4.2)
The coefficient of kinematic friction.
Ff riction is expected to reach its maximum while slip is not present as, in general, µdynamic < µstatic .
The rubber wheels on the mountain bike are used on many different surfaces and, as previously
mentioned, the friction on the surfaces vary with the condition on the surface; e.g. if it is wet, icy,
dusty, or clean. The highest magnitude of the coefficient of static friction is expected to be found when
the wheel is rolling on clean and dry asphalt, and it is assumed not to exceed µstatic = 1 [3].
These considerations regarding the braking force will be relevant for a kinetic analysis, as described in section 6.2.
4.3
Pedal Force
When the rider is pedalling, a force is induced in the chain. This force will obviously cause stresses in
the bars of the suspension linkage. Furthermore, the chain force might also result in either Bio-Pacing
or DISC, depending on the suspension design, as explained in section 2.1.
Pedalling can be characterised by two parameters. One parameter is the force from the rider’s foot
on the pedal. The other parameter is the cadence, i.e. the number of pedal revolutions per minute.
These two parameters are together an expression for the mechanical power output. The physical capabilities of the human body strongly depends on the time provided. In a short amount of time a rider
can deliver a much larger mechanical power output, compared to the mechanical output available over
a longer period of time. Both are interesting when designing the suspension of the mountain bike.
The first event is interesting because here, the force on the pedal reaches its overall maximum due to
pedalling. The other comes into interest in the fatigue analysis. The maximum possible power output
from elite cycle athletes over a few seconds is expected to be 2000 W at a cadence of 100 rpm. The
average mechanical power output from same athletes over a period of hours is expected to be 500 W
at a cadence of 80 rpm [4], [5], [6].
CHAPTER 4. LOADING SCENARIOS FOR THE BIKE
15
4.3. PEDAL FORCE
During one revolution the force on the pedal is not constant as it depends on the angel between
thigh and shinbone. To estimate the pedalling force, the biomechanical simulation software AnyBody is
used. AnyBody was initially developed at Aalborg University. One of the motivations for developing a
biomechanical model of the human body and integrate it in a computer software, is to make it possible
to do ergonomical tests and get an impression of the response from the human body without use of
physical test persons. AnyBody can show the response of the human body in interaction with e.g. hand
tools and different sports equipment [7]. AnyBody is used as a tool to obtain the data for the project
without further investigating the software.
Figure 4.2 shows a model of the human body pedalling on a road bike. This model is a standard
model in AnyBody called BikeModelFullBody. Even though the rider’s position on a road bike is
slightly different from the position on a mountain bike, the required pedal force on a road bike to a
certain cadence and mechanical power output, is expected to be comparable with the force required on
a mountain bike at the same cadence and power output conditions. To minimise this error the human
model is placed in a position on the road bike that is similar to the riding position on a mountain bike.
The setup of the simulation can be found on the CD in appendix E.
r
t
Figure 4.2: Bio-mechanical model of the human body
riding on a bike.
4.3.1
Figure 4.3: The pedalling force can be split into a tangential and a radial force.
Results
Figures 4.4 and 4.5 show the forces exerted by the rider on the left and right pedal respectively for one
revolution. It is important to notice that the force components are based on a local coordinate system
on the pedal arm as shown in figure 4.3. This coordinate system follows the pedal arms as it rotates
during the simulation. The red curve represents the tangential force exerted by the rider perpendicular
to the pedal arm and generates a moment in the crank. The green curve represents the radial force
parallel to the pedal arm, and does not contribute to the crank moment. The force plots in figures 4.4
and 4.5 are based on a simulation of one complete revolution with the simulation starting at θ = 0
16
CHAPTER 4. LOADING SCENARIOS FOR THE BIKE
4.3. PEDAL FORCE
where the pedal arms are vertical. By observing the tangential force on the right pedal in figure 4.5, it
can be seen that the maximum value of 1490 N occurs around θ = π2 . This corresponds to the pedal arm
being in a near horisontal position, enabling the rider to apply his pedalling force perpendicular to the
pedal arm. This leads to a mostly tangential load and thus maximises the achievable crank moment.
The total tangential force can be obtained by summarising the moment contributions from the left
and the right pedal. This is shown in figure 4.6 for a 170 mm pedal arm.
Simulations have been carried out for the expected power output in the two cases of peak physical
activity (maximum) and prolonged physical activity (mean). The results can be seen in table 4.1.
The maximum tangential force of 1490 N, which is obtained by summarising the curves from figure
4.4 or 4.5, corresponds to the rider pressing down on the pedal with a weight of 152 kg. The weight
of the rider is 75 kg, which means that the remaining 77 kg of the total load must be achieved by a
combination of push and pull in both pedals and the handle bar. As these figures apply to short sprints
by top athletes, the results seem humanly achievable and a maximum power of 2000W is considered
realistic. For prolonged physical activity, the power output of 500W results in a load on the right
pedal at θ = π2 of 59 kg which is significantly lower than the weight of the rider. This does not seem
unrealistic compared to activities such as running, where the full body weight is transferred from one
leg to the other.
1200
1600
800
1200
400
0
F[N]
F[N]
800
400
-400
0
-800
-400
-1200
-800
-1600
0
π/3
2π/3
π
θ[rad]
4π/3
5π/3
0
2π
Figure 4.4: Reaction force on the left pedal at maximum power output. The red curve shows the tangential
force on the pedal. The green curve shows the radial
force on the pedal.
π/3
2π/3
π
θ[rad]
4π/3
5π/3
2π
Figure 4.5: Reaction force on the right pedal at maximum power output. The red curve shows the tangential
force on the pedal. The green curve shows the radial
force on the pedal.
CHAPTER 4. LOADING SCENARIOS FOR THE BIKE
17
4.4. OPERATIONAL CONDITIONS
400
Moment [Nm]
300
200
100
0
0
π/3
2π/3
π
θ[rad]
4π/3
5π/3
2π
Power output [W ]
Cadence [rpm]
Angel at max. moment[rad]
Left tangential force [N]
Right tangential force [N]
Resulting moment [Nm]
Max.
2000
100
1.68
761
-1490
379
Mean
500
80
2.09
117
-566
116
Figure 4.6: The tangential force on the left and the right Table 4.1: The table summarises the results from the
pedal can be summarised into a moment in the crank.
AnyBody-simulation. The length of the pedal arm is 170mm
4.4
Operational Conditions
Finally a use case is defined which must be taken into account when dimensioning the bike against
fatigue failure. Based on the data presented above, this is assumed to be:
Pedalling: Pedal cadence is 80 rpm and the speed is 20 km/h in average. Each pedal revolution consists of two pedal strokes. It is assumed that the rider is not pedalling when riding down hill.
The overall pedal stroke frequency is 80 impacts/min in average. For each 10 min of riding the
rider is pedalling at max moment (see table 4.1) at 100 rpm for 10 sec. In average this gives
5/90 impacts/min at max moment. Otherwise the mean moment (see table 4.1) will be present.
Braking: One activation of the brake for each 500 m’s of riding at 20 km/hr in average. This gives
2/3 impacts/min.
These data will be used in chapter 8.3.3, where a link in the suspension system is tested for fatigue
failure.
4.4.1
Concluding Remarks
Altogether, the bike must be able to withstand the following four scenarios:
• The free fall.
• The forces induced by braking.
• The forces induced by pedalling.
• A combination of braking and pedalling.
18
CHAPTER 4. LOADING SCENARIOS FOR THE BIKE
5
Modelling and Design Strategy
for the Mountain Bike.
This chapter describes how the overall mathematical model of the bike is structured. Also a description of how it will be used later is given.
For design purposes a calculation model is developed and implemented in Matlab. An overview of
the calculation model is given in figure 5.1. The calculation model is built up from the functions and
Response to
frequency
input
Legend
Program related to the
dynamic model
Kinetic
Analysis
Function related to
dynamic model
Kinematic
model
Shock
Absorber
Analysis of a
free fall
Equivalent Equivalent
stiffness damping
Figure 5.1: Overall flowchart of the mathematical model of the mountain bike.
programs shown with boxes in the figure. The two terms are defined as:
Function: A function is a piece of code which can be called by any other function or program. A
function requires a predefined number of inputs and gives a number of outputs based on the
given inputs. Thus, a function is a tool accessible from anywhere in the calculation model.
Program: A program is a piece of code that combines different functions to carry out an analysis. A
program does not require any inputs and does not give any outputs.
The entire calculation model is found on the CD in appendix E. A short description of each box in
figure 5.1 is given next.
Kinematic Model: This function solves a set of constraint equations by inputting position, velocity
and acceleration for the rear dropout and crank for a given number of time steps. The outputs are
CHAPTER 5. MODELLING AND DESIGN STRATEGY FOR THE MOUNTAIN BIKE.
19
three matrices containing positions, velocities, and accelerations for any member of the linkage
system.
Kinetic Analysis: This program determines the response of the bike when a rider is placed on the
bike, generating a chain force by pedalling or braking. It returns data for position, velocity,
acceleration, and reaction forces for any member of the suspension system at all time steps.
Shock Absorber: This function solves a hydraulic system and returns stiffness and damping coefficients for the shock absorber as well as the force exerted on the suspension system by the shock
absorber. The inputs are compression and compression velocity of the shock absorber known
from the kinematic analysis.
Equivalent stiffness & damping: Based on the kinematic properties of the linkage system and the
properties of the shock absorber, stiffness and damping coefficients are calculated, replacing
both linkage system and shock absorber.
Response to frequency input: Here the suspension system’s ability to soften out an uneven surface
is tested. Except for reaction forces, this program returns similar outputs to the kinetic analysis.
Analysis of a free fall: This program determines the response to a free fall. Loads are achieved and
used to determine stresses later on.
The exact content of each of the functions and programs in figure 5.1 are described in chapter 6. By
use of this calculation model the design process is carried out through the following steps:
Step 1: The linkage system is designed with regards to, i.e. length of bars, length of Horst link, and
position of shock absorber, to control DISC and Bio-Pacing. This is done by using the kinetic
analysis, where the response to both a chain and a brake force is observed.
Step 2: The coefficients of damping and stiffness for the shock absorber are adjusted so that desired
effects such as effective suspension properties are maximised. This is done by use of Response
to frequency input.
It is expected that a suspension system optimised to withstand DISC and Bio-Pacing, would tend to
have a relatively high stiffness. Whereas a suspension system optimised to smoothen out rough terrain
will tend to have a relatively low stiffness. These two issues are of course contradictory which yields
an iterative design process, where step 1 cannot be carried out without doing step 2, and both analysis
must return acceptable results simultaneously.
The results of the design process are presented in chapter 7 before which the dynamic characteristics will be fully determined.
Step 3: Stresses in the links are calculated to ensure that they are properly dimensioned. Here the
Free Fall analysis is necessary.
The analysis in the third step is subjected to the results from the first two steps and therefore carried
out lastly. The stress analysis is done in chapter 8.
20
CHAPTER 5. MODELLING AND DESIGN STRATEGY FOR THE MOUNTAIN BIKE.
6
Elements in the Dynamic Model
In this chapter each of the boxes in figure 5.1 dealing with dynamics are described in more details,
both how the necessary equations are derived and how the analysis is handled in Matlab. It should be
emphasised that, in general, no results are presented in this chapter. Instead chapter 7 is dedicated to
serve this purpose. A sketch of the entire bike with all named points is given in appendix F. This can
be useful when reading the following chapters.
6.1
Kinematic Model of the Four Bar Linkage
In order to produce a dynamic model of the suspension system a kinematic analysis must be conducted
as the first step. All subsequent analysis are based on, or use, the kinematic model, which is set up in
this section. The approach for performing the kinematic analysis is adopted from Nikravesh [1].
Figure 6.1 gives an overview of the kinematic model of the four bar linkage suspension system. The
C
4
D
1
3
A
2
P3
B
P2
F
O
Figure 6.1: The rigid bodies with global and local coordinate systems.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
21
6.1. KINEMATIC MODEL OF THE FOUR BAR LINKAGE
figure shows the four links in distinct colours, and how these are attached to each other with revolute
joints. Each link is given a number and attached a local coordinate system with coordinate axes ξi and
ηi (i = {1, 2, 3, 4} being the number of the links). Besides, a global coordinate system and some useful
points and angles are shown in the figure.
The motion of the suspension system is examined by three distinct analysis. The position-, velocity-,
and acceleration analysis are described in the following three subsections, respectively.
6.1.1
Position Analysis
For determining the degree of freedom of the suspension system Gruebler’s equation is used:
m = 3(n − 1) − 2 f − h
⇒
m = 3(5 − 1) − 2 · 4 = 4
where:
m
The degree of freedom.
n
The number of links.
f
The number of low-pairs joints.
h
The number of high-pairs joints.
(6.1)
From equation (6.1) it is clear that the suspension system has four degrees of freedom. Three of
those are removed by fixating link 1 to a specific place and rotation in the global coordinate system.
Thus, only one extra information is needed in order for the system to be fully specified.
The relative motion of the four bars are described by formulating kinematic constraint equations.
Five kinematic constraints exist between the origin and the four bars:
• C0: Fixation of link 1 to a specific position and orientation.
• C1: Revolute joint between link 1 and link 2—point A.
• C2: Revolute joint between link 2 and link 3—point B.
• C3: Revolute joint between link 3 and link 4—point C.
• C4: Revolute joint between link 4 and link 1—point D.
The kinematic constraints can be formulated mathematically as:
~~r = ~0
~r1 − C
1
θ1 −Cθ1 = 0
(~r1 + Ā¯ 1~s1 ) − (~r2 + Ā¯ 2~s2 0A ) = ~0
(~r2 + Ā¯ 2~s2 0B ) − (~r3 + Ā¯ 3~s3 0B ) = ~0
(~r3 + Ā¯ 3~s3 0C ) − (~r4 + Ā¯ 4~s4 0C ) = ~0
0A
(~r4 + Ā¯ 4~s4 0D ) − (~r1 + Ā¯ 1~s1 0D ) = ~0
22
(6.2)
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.1. KINEMATIC MODEL OF THE FOUR BAR LINKAGE
where:
~~r
C
1
~~r = [Cx1 ,Cy1 ]T .
The position of point F in global coordinates. C
1
Cθ1
Ā¯
The orientation of point F.
The rotation matrix for frame i.
~si 0 j
The local coordinate from the origin of frame i to point j.
~ri
The global vector from the global origin to the origin of frame i.
i
0j
The rotation matrix Ā¯ i serves the purpose of transforming local coordinates~si into global coordinates.
The rotation matrix is given by:
cos (θi ) − sin (θi )
¯
Āi =
(6.3)
sin (θi )
cos (θi )
The remaining information needed for the system to be fully constrained is given by a driver function
which specifies the position of the rear dropout.
The kinematic and driving constraints can be written in expanded form as:


x1 −Cx1


y1 −Cy1




θ
−C
1
θ
1




~
~
x1 + |FA|ξ cos(θ1 ) − |FA|η sin(θ1 ) − x2




~ ξ sin(θ1 ) − |FA|
~ η cos(θ1 ) − y2
y
+
|
FA|
1




~
~
~
 x2 + cos(θ2 ) |AP2 | + |P2 B| sin(ϕ2 ) − sin(θ2 ) |P2 B| cos(ϕ2 ) − x3 




~
~
~

~Φ(~q) =
y + sin(θ2 ) |AP2 | + |P2 B| sin(ϕ2 ) + cos(θ2 ) |P2 B| cos(ϕ2 ) − y3  = ~0
(6.4)
 2



~ 3 | − |P~3C| cos(ϕ3 ) − sin(θ3 ) −|P~3C| sin(ϕ3 ) − x4 
 x3 + cos(θ3 ) |BP



~ 3 | − |P~3C| cos(ϕ3 ) + cos(θ3 ) −|P~3C| sin(ϕ3 ) − y4 
 y3 + sin(θ3 ) |BP





~ − x1 − cos(θ1 )|OD|
~ ξ − sin(θ1 )|OD|
~ η


x4 + cos(θ4 )|CD|




~ − y1 − sin(θ1 )|OD|
~ ξ + cos(θ1 )|OD|
~ η


y4 + sin(θ4 )|CD|
~ 3 |ξ sin(θ3 ) − cyP3 (t)
y3 + |BP
where:
~Φ
The constraint vector.
~q
The position vector.
cyP3 (t)
The driver function specifying the height of point P3 .
The position vector ~q is given as:


xi
~q =  yi  , i = {1, 2, 3, 4}
θi
(6.5)
The Jacobian matrix is introduced to serve a number of purposes. First of all, in the specific Matlabimplementation, it is needed for solving the system of equations to find the positions, as will be explained in section 6.1.4. Secondly, it is also necessary to determine the velocities and accelerations of
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
23
6.1. KINEMATIC MODEL OF THE FOUR BAR LINKAGE
the system. The Jacobian matrix is defined as:
~
¯ = ∂Φ
D̄¯ = Φ̄
~q
∂~q
where:
D̄¯
¯
Φ̄
(6.6)
The Jacobian matrix.
Also the Jacobian matrix.
~q
The Jacobian matrix for the current system can be calculated to be:











D̄¯ = 









where:
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0 D4,3 −1 0
0
0
0
0
0
0
0
0
1 D5,3
0 −1
0
0
0
0
0
0
0
0
0
0
1
0 D6,6 −1 0
0
0
0
0
0
0
0
0
1 D7,6 0 −1
0
0
0
0
0
0
0
0
0
0
1
0 D8,9 −1 0
0
0
0
0
0
0
0
0
1 D9,9
0 −1
0
−1 0 D10,3 0
0
0
0
0
0
1
0 D10,12
0 −1 D11,3 0
0
0
0
0
0
0
1 D11,12
0
0
0
0
0
0
0
1 D12,9 0
0
0
D4,3
~ ξ − cos(θ1 )|FA|
~ η
= − sin(θ1 )|FA|
D5,3
~ ξ − sin(θ1 )|FA|
~ η
= cos(θ1 )|FA|
D6,6
D9,9
~ 2 | − |P~2 B| cos(ϕ2 )) + cos(θ2 ) |P~2 B| sin(ϕ2 )
= − sin(θ2 )(|AP
~ 2 | − |P~2 B| cos(ϕ2 )) + sin(θ2 ) |P~2 B| sin(ϕ2 )
= cos(θ2 )(|AP
~ 3 | − |P~3C| cos(ϕ3 )) + cos(θ3 ) |P~3C| sin(ϕ3 )
= − sin(θ3 )(|BP
~ 3 | − |P~3C| cos(ϕ3 )) + sin(θ3 ) |P~3C| sin(ϕ3 )
= cos(θ3 )(|BP
D10,3
~ ξ + cos(θ1 )|FD|
~ η
= sin(θ1 )|FD|
D10,12
~
= − sin(θ4 )|CD|
D11,3
~ ξ + sin(θ1 )|FD|
~ η
= − cos(θ1 )|FD|
D11,12
~
= cos(θ4 )|CD|
D12,9
~ 3|
= cos(θ3 )|BP
D7,6
D8,9





















(6.7)
In order to determine the positions, the equations in (6.4) are solved for ~q with a Newton-Raphson
solver programmed in Matlab. The specific implementation is described in section 6.1.4.
24
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.1. KINEMATIC MODEL OF THE FOUR BAR LINKAGE
6.1.2
Velocity Analysis
With the Jacobian matrix known, it is possible to find the velocities of the local coordinate systems.
The velocity vector can be found from the velocity equation:
D̄¯ ~q˙ = ~α˙
~q˙ = D̄¯ −1~α˙
where:
~q˙
~α˙
⇒
(6.8)
The time derivative of the position vector, i.e. the velocity vector.
The time derivative of the constraint equations.
The velocity of any other point P on link i in the system can be determined by differentiating its
position vector with respect to time:
˙
~r˙iP = ~r˙i + Ā¯ i~s0P
i
0P
¯
˙
= ~ri + B̄i~s θ̇i
(6.9)
i
where:
6.1.3
B̄¯ i
− sin(θi )
=
cos(θi )
− cos(θi )
− sin(θi )
Acceleration
The accelerations of the coordinate systems can be found from the acceleration equation:
˙
D̄¯ ~q¨ + D̄¯ ~q˙ = ~α¨
⇒
˙
~q¨ = D̄¯ −1 ~α¨ − D̄¯ ~q˙
(6.10)
Thus, with the Jacobian and velocities known the accelerations of the frames are easily found. The
accelerations of any other point P on link i can be determined by differentiating its velocity vector:
˙¯ 0P
~r¨iP = ~r¨i + B̄¯ i~s0P
si θ̇i
i θ̈i + B̄i~
= ~r¨i + B̄¯ i~s0P θ̈i − Ā¯ i~s0P θ̇2
i
i
i
(6.11)
Combining the position-, velocity-, and acceleration analysis gives a complete description of the systems kinematic behaviour. To illustrate the basic motion of the suspension system a simple example is
given next.
Example of the Kinematic Behaviour of the Suspension System
A simple example of the suspension system’s kinematic behaviour is given by using a sinus-function as
driving constraint for the height of the rear wheel. I. e. the latter of the kinematic constraints becomes:
~r3 + Ā¯ 3~s3 0P3 −~cyP3 (t) · ŷ = 0
⇒
~r3 + Ā¯ 3~s3 0P3 − ~rP3 X P3 sin ωP3 t + ψP3 +~hP03 · ŷ = 0
(6.12)
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
25
6.1. KINEMATIC MODEL OF THE FOUR BAR LINKAGE
where:
~cyP3 (t)
= [0, cyP3 (t)]T .
X P3
The amplitude of sinusoidal input at point P3 .
ωP3
The frequency of sinusoidal input at point P3 .
ψP3
The phase of sinusoidal input at point P3 .
P3
h~0
The starting position of point P3 .
The idea of giving a sinusoidal input to the rear wheel is to approximate the situation where a rider
rides over a series of bumps. This aspect will be further investigated later in section 6.5.
An animation of the suspension system subject to the sinusoidal input given in equation (6.12) can
be found on the CD in appendix E under Movies » KinematicSinusInput.
6.1.4
Implementation of the Three Analysis into Matlab
Input driver function
and fixation of link 1
Define constraints:
~Φ
Define jacobian:
D
.. .
~q, ~q, ~q
Position Analysis
based on Newton-Raphson method
Loop over
given time
Velocity
Analysis
Acceleration
Analysis
..
.
~q , ~q , ~q
to all time steps
Figure 6.2: Flowchart of the Matlab implementation of the kinematic model.
The kinematic model has been programmed in Matlab as a function that can be called by any other
function or program. The input data are a driver function for the rear wheel and a fixation of link 1.
The latter will later be used to move the bike’s main frame during a given time span. The outputs of
the function are positions, velocities, and accelerations of the four links in the suspension system. The
three outputs are calculated separately as described above. A flowchart of the function is sketched in
figure 6.2. The function has a built in loop which makes it possible to determine discrete values of
positions, velocities, and accelerations for a given time span. The constraint equations set up in the
position analysis are solved with a Newton-Raphson method. The method requires an initial guess.
For this the results from the previous iteration are used. Since the Newton-Raphson method is used to
26
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.2. KINETIC ANALYSIS
a great extent throughout the project, a brief exposition of the method will be made in the following
subsection.
The Newton-Raphson Method
The Newton-Raphson method is a numerical method used for approximating roots of a real-valued
function. The method converges after a series of iterations to roots of the function from an initial guess.
There are three components needed for the method to work: A real valued function, its derivative and
an initial guess. For the one dimensional case the result of each iteration is calculated as:
xn+1 = xn −
where:
f (xn )
d f (x)
dx |x=xn
f (x)
The function for which a root is sought.
n
The iteration counter.
(6.13)
This can be expanded to hold for a function of an arbitrary number of variables. Hence, for the
problem of positional analysis the Newton-Raphson method can be formulated as:
¯ −1 ~Φ = ~q − D̄¯ −1 ~Φ
~qn+1 = ~qn − Φ̄
(6.14)
n
n
n
~q,n n
This calculation is repeated until the error is less than some predetermined algorithm tolerance. The
error of each iteration is simply calculated as the length of the constraint vector:
q
Error = Φ21 + Φ22 + ... + Φ2m
(6.15)
where:
Φi
The i’th row in ~Φ.
m
The number of rows in ~Φ.
For the position analysis the allowed error (the algorithm tolerance) is set to 1 · 10−9 in SI-units.
6.2
Kinetic Analysis
In the previous section a kinematic analysis of the suspension system was performed. In this section a
kinetic analysis of the suspension system is conducted. That is, the response of the system due to loads
acting on the system is examined, whereas kinematics deals with motion regardless of the forces producing it. Hereby it will be possible to determine the internal reaction forces due to an external load
as well as the resulting accelerations. The analysis is performed by the Lagrange multiplier approach
as proposed by Nikravesh [1].
According to Newtons second law of motion the movement of each link in the suspension system
can be described as:

  

f(x)
m 0 0
ẍ
 0 m 0   ÿ  =  f(y)  ,
i = {1, 2, 3, 4}
(6.16)
0 0 J i θ̈ i
T
i
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
27
6.2. KINETIC ANALYSIS
where:
mi
The mass of link i.
Ji
The mass moment of inertia of link i about the ζi -axis.
fi
The applied force on link i.
Ti
The applied moment on link i.
This can, for the entire suspension system, be shortened into:
M̄¯ ~q = ~g
where:
M̄¯
~g
(6.17)
The diagonal mass matrix diag[mi , mi , Ji ], i = {1, 2, 3, 4}.
The sum of all applied forces. Note that the force vector contains both single forces and force
couples, i.e. moments.
If the applied forces on each link are now divided into external forces and internal reaction forces
(also denoted constraint forces), the Lagrange method states that the equation of motion can be written
as:
M̄¯ ~q = ~gext +~gc
¯ T~λ
= ~gext + Φ̄
⇒
(6.18)
~q
where:
~gext
The external forces.
~gc
¯
Φ̄
The constraint forces.
The Jacobian matrix from the kinematic analysis, section 6.1.1.
~λ
The Lagrange multipliers~λ = [λ1 , λ2 , ..., λn ]T .
~q
Additionally, the acceleration equations (6.10) for a constrained system are:
˙
D̄¯ ~q¨ + D̄¯ ~q˙ = ~α¨ = ~0
(6.19)
Thereby the equations (6.18) and (6.19) can be combined into the fully constrained system of equations:
#
"
# "
ext
¯ T~λ
~
g
M̄¯ ~q¨ − Φ̄
~q
=
⇒
˙
¯ ~q¨
−D̄¯ ~q˙
Φ̄
~q
"
#"
# "
#
ext
¯T
¨
~
g
~
q
M̄¯ Φ̄
~q
(6.20)
=
˙
¯
−~λ
−D̄¯ ~q˙
Φ̄
0̄¯
~q
Given some initial conditions ~q0 and ~q˙0 which satisfies the constraint equations
~Φ(~q0 ) = ~0
and
¯ (~q )~q˙ = ~0
Φ̄
~q 0 0
(6.21)
the system of equations can be solved for ~q¨ and~λ. Hence, the reaction forces and accelerations can be
determined for a given ~gext .
28
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.2. KINETIC ANALYSIS
~ shock
F
~ rear
G
~ brake
F
~ chain
F
~ friction
F
Figure 6.3: The model for analysing whether the suspension system has a tendency to either DISC or Bio-Pace.
6.2.1
Determining the External Force
In order to determine whether the suspension system tends to DISC or Bio-Pace a simple model, as
shown in figure 6.3, is used. The model represents a situation where the main frame is fixed in both
~ rear , and
displacement and rotation and the system is subjected to five forces: ~Fshock , ~Fchain , ~Ff riction , G
~Fbrake . ~Fshock is calculated based on the model of the shock absorber which will be described in section
6.3. ~Fchain , ~Ff riction , and ~FBrake are examined in the following two subsections. It should be noted that
~ rear is the reaction force at the rear dropout, due to the mass of rider and bicycle.
G
The Force Induced by Pedalling
~chain
F
~pedal
F
~pedal
F
F~friction
Figure 6.4: The forces acting on the suspension system during pedalling.
As described in section 2.1 the force induced by pedalling has two contributions; a force from the
a
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
b
29
6.2. KINETIC ANALYSIS
chain and a friction force from the wheels contact with the ground. The two forces are calculated as:
~
F
chain rcassette
~Ff riction =
êground
(6.22)
rwheel
~Fchain = Mrider êchain
(6.23)
rchain
where:
rcassette
The radius of the cassette depending on the current gear. See table 2.1 no. 2 for bike terms.
rwheel
The radius of the wheel.
Mrider
The moment provided in the crank by pedalling obtained from AnyBody. See chapter 4.
rchain
The radius of the chain ring, depending on the current gear. See table 2.1 no. 10 for bike terms.
êchain
A unit vector in direction of the chain (with positive x-component).
êground
A unit vector in direction of the ground (with positive x-component).
Hence, the force induced by pedalling can be calculated as:
~Fpedal = ~Ff riction + ~Fchain
~
Fchain rcassette
Mrider
=
êground +
êchain
rwheel
rchain
for a given pedalling moment and gear of the bike.
(6.24)
The Force Induced when Braking
b
a
h
y
~G
~friction
F
x
~ rear
N
~ front
N
Figure 6.5: A very simple model of the bike used for determining the force induced when braking.
Activating the rear brake will, as described in section 2.1, affect the suspension system with a
force ~Fbrake acting at point R. The magnitude of the force induced by braking is determined from a
very simple model of the bike, as shown in figure 6.5. Summation of forces on the free body gives:
~ + ~Nrear + ~N f ront + ~Ff riction = m~a
Σ~F = G
mgŷ + Nrear ŷ + N f ront ŷ − Ff riction x̂ = −max̂
30
⇒
(6.25)
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.2. KINETIC ANALYSIS
where:
~
G
The gravity force on the body.
~Nrear
The normal force on the rear wheel.
~N f ront
The normal force on the front wheel.
~Ff riction
The friction force between rear wheel and ground.
m
The mass of the bike.
Taking moment about the center of mass gives the following equation:
ΣMCM = bN f ront − aNrear − hFf riction = 0
(6.26)
Combining equations (6.25) and (6.26) and solving for Ff riction gives:
Ff riction =
where:
µs
bmg
+h
b+a
µs
(6.27)
The coefficient of static friction originating from Ff riction = µs Nrear .
The friction force causes a moment on the wheel around the rear dropout which must be opposed by
the corresponding moment contribution from the braking force. Hence, the force induced by braking
is found to be:
~
~Fbrake = rwheel |Ff riction |
|P~3 R|
rwheel bmg
=
|P~3 R| b+a
+
h
µs
where:
|P~3 R|
The distance from P3 to R.
g
The gravitational acceleration.
(6.28)
The total external force ~gext can now be constructed by assembling the three distinct forces including each of their moment contributions. The external forces are assembled into the resulting force
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
31
6.2. KINETIC ANALYSIS
vector:

~gext
where:










=









gx1
gy1
gθ1
gx2
gy2
gθ2
gx3
gy3
gθ3
gx4
gy4
gθ4


0
0
0
0
0
0
 
 
 
 
 
 
 
 
 
 
=
~ rear |x
|~Fchain |x + |~Fbrake |x + |G
 
 
~ rear |y
|~F
| + |~F
|y + |G
 
  P3 ~chain y R brake
~ rear
~
  ~s3 _ Fchain +~s3 _ Fbrake +~sP33 _ G
 
 
|~Fshock |x
 
 
|~Fshock |y
P4 ~
~s _ Fshock





















(6.29)
4
_
The planar cross product operator. ~a _~b ≡ [~aT , 0]T × [~bT , 0]T · [0, 0, 1]T ∀ ~a,~b ∈ R2 .
~sXi
The global vector from the origin of frame i to point X.
The response of the suspension system in the given scenario can now be found by substituting the
external force vector into equation (6.20) and solving for ~q¨ and ~λ. The two vectors are to be interpreted as:
T
~q¨ = ẍ1 , ÿ1 , θ̈1 , ẍ2 , ÿ2 , θ̈2 , ẍ3 , ÿ3 , θ̈3 , ẍ4 , ÿ4 , θ̈4
(6.30)
~λ = [RFX , RFY , τF , RAX , RAY , RBX , RBY , RCX , RCY , RDX , RDY ]T
(6.31)
and
where:
6.2.2
Ri j
The j-component of the reaction force in point i.
τi
The reaction moment in point i.
Solving the System of Constrained Equations by Use of Matlab
The kinetic analysis is programmed in Matlab as a program that utilises several results and functions
¨ ~λ, ~q,
˙ and ~q for a specified
from other parts of the project. The results of the program are values of ~q,
time span.
A flowchart of the program is shown in figure 6.6. Before any analysis can be performed, an initial
bike configuration needs to be set up. I.e. the variables needed to describe the bike, that is to be
subjected to an external force, are defined. The rest of the analysis is performed in a loop which loops
over some predefined time steps. Generally seen, the loop can be divided into three parts, as illustrated
in the flowchart. The three parts are described here:
32
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.2. KINETIC ANALYSIS
Part 1: In the first part, the four external forces are determined and the system of constrained equations is set up, cf. equations (6.20). It should be noted that all external forces are time dependent.
~ rear is calculated based on a calculation of the
~Fpedal and ~Fbrake are calculated as described above. G
~ rear is considered to be:
centre of mass. The size of G
~ rear | = (1 − |P~3 rCM |x )mg
|G
where:
|P~3 rCM |x
(6.32)
The x-component of the distance from the rear dropout to the centre of mass.
~Fshock is determined based on the model of the shock absorber, which will be described in section
6.3. The system of equations is set up based on the bike variables from the previous iteration. I.e.
for the first iteration the system of equations relies on the initial bikes positions but for the n’th iteration the system of equations relies on the (n − 1)’th iterations results. This is indicated with an arrow
¨ ~λ, ~q,
˙ and ~q.
surrounding the loop which returns ~q,
The velocity used for each iteration is approximated as:
~q˙n = ~q¨n−1 ∆t + ~q˙n−1
where:
∆t
(6.33)
The time step.
This is a rather crude assumption. However, for small time steps the results are believed to be reasonable.
Part 2: In the second part of the loop, the system of equations is solved. This is done with a NewtonRaphson solver as explained in section 6.1.4. However, as earlier stated this solver requires a derivative
of the function for which roots are searched. In order to ensure the flexibility of the system this
derivative is not determined algebraically (as for the kinematic model) but approximated numerically.
The derivative is approximated as:
Ψ [~q¨T ,~λT ]T +~ε j − Ψ [~q¨T ,~λT ]T
i
i
ψi, j =
(6.34)
|~ε j |
where:
Ψi
The i’th row of the left side of equation 6.20.
~ε j
A vector of same height as Ψ with 1 · 10−10 in the j’th row and zeros in the remaining rows.
ψi, j
The i, j’th element of the approximated derivative of Ψ.
The correctness of this approximation relies on perturbation theory which is beyond the scope of
this text. However, it is worth noticing that the use of an approximated derivative does not falsify, or
approximate the solution of the equations further. It only slows down the iterative process.
Part 3: The results from part 2 is a number of reaction forces and accelerations of the links. In part 3
the kinematic model is used to update all positions and velocities with respect to the new accelerations.
This is done by first converting the acceleration of link 3 into an acceleration of the rear dropout, ÿP3 ,
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
33
6.2. KINETIC ANALYSIS
by use of equation (6.11). This acceleration of the rear dropout is then passed to the kinematic model
where the driver function is specified as:
1
cyP3 (t) = ÿP3 ∆t 2 + ẏP3 ∆t + yP3
2
yP3
ċ (t) = ÿP3 ∆t + ẏP3
yP3
c̈
⇒
(6.35)
⇒
(t) = ÿP3
When the loop is finished reaction forces, accelerations, velocities, and positions are determined for
the entire time span. Hence, the analysis is complete.
34
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.2. KINETIC ANALYSIS
Setup initial
bike
.
..
~q, ~λ, ~q, ~q
Loop over
specified time
Determine
~Fbrake(t)
Determine
D and D
Shock
Absorber
Determine
~ rear(t)
G
Determine
~Fpedal(t)
.
Setup ~gext
..
M ΦqT
Φq 0
~gext
~q
=
~
~
-λ
-Dq
. .
Setup:
~Fshock(t)
Approximate
derivate
..
Solve for ~q and ~λ
by Newton-Raphson method
Kinematic
Model
.
..
~q, ~λ, ~q, ~q
to all time steps
Figure 6.6: Flowchart of the Kinetic Analysis program.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
35
6.3. SHOCK ABSORBER
6.3
Shock Absorber
This section describes the shock absorber model. A brief description of the shock absorber will be
given, explaining central components and functionality. This will create the basis for modelling the
fluid circuits in the shock absorber. Finally, the characteristics of the spring and damper model will be
presented and discussed.
6.3.1
How the Shock Absorber Works
Ai
rs
l
e
e
ve
Ai
rs
pr
i
ngc
hambe
r
s
Ne
e
dl
epor
t
Vi
s
c
ousdampe
rc
hambe
r
s
Ni
t
r
oge
nc
hambe
r
I
FP
Figure 6.7: The shock absorber in a semi-compressed. IFP is an abbreviation for Internal Floating Piston.
A typical mountain bike shock absorber consists of an air spring and a viscous damper and can
be loaded axially leading to compression or extension. The shock absorber, which the model will be
based on, is a Fox Float RP23 [8]. This is a very popular suspension unit which is widely used for full
suspension mountain bikes.
This particular shock absorber consists of five chambers containing air, nitrogen and oil. Figure
6.7 shows the shock in a semi-compressed state. It is important to note that the damper unit is mounted
inside the air spring causing their motion to be closely related.
The Spring
Figure 6.8 shows a principal drawing of the cross section of the shock absorber. The spring consists
of two air chambers (1) and (2) separated by a piston. Neutral position is found when the shock
absorber is fully extended and thus allowing compression only. As the shock absorber compresses the
air pressure in the positive air chamber (1) rises. This creates a spring effect where the air pressure in
(1) balances the external compressive force F at some level of compression.
36
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.3. SHOCK ABSORBER
F
F
1
3
2
7
5
4
6
Figure 6.8: Principal drawing of a shock absorber.
To control the stiffness of the spring both chambers are filled with compressed air. The spring
stiffness can be increased by increasing the air pressure in the positive chamber. Since the air in the
positive air chamber is compressed, it will generate a force Fpreload = p1,preload · A1 on the piston even
when the shock absorber is fully extended. The second air chamber (2) is called the negative air
chamber.
The compressive force F must exceed this preloaded spring force to produce any compression. By
adding pressure in the negative air chamber on the other side of the piston, the Fpreload force can be
fully or partially cancelled out.
Only the pressure in the positive air chamber can be adjusted externally. A special valve between
the positive and negative air chamber ensures, that the ratio between the pressures in the two chambers
remains as intended when the positive air pressure is changed. It is possible to change the pressure of
the two champers, but since the ratio between the two pressures remains constant, a pressure change
only affects when the spring is active. This is discussed further and more detailed in section 6.3.3.
The Damper
The damper consists of three chambers separated by a damping piston and an internal floating piston
(IFP) as shown in figure 6.7. The oil chambers (3) and (4) in figure 6.8 are separated by the damper
piston (7) which contains a number of pressure dependent orifices allowing the oil to flow from one
chamber to the other. The damping force comes from the viscous friction caused by the oil flows
through the orifices. The amount of friction depends on the flow and thus varies with the velocity of
the damper piston.
The orifices in the piston allowing oil to flow between the chambers, are often referred to as ports.
Figure 6.9 shows a frequently used piston design, which not fully corresponds with the piston in figure
6.7 but uses the same principle. Note that top and bottom are similar on the piston. A top plate is
holding a stack of shims covering a set of holes in the piston marked as bump ports in figure 6.10.
Between the bump ports are set of rebound ports restricted by a shim stack on the other end of the
piston.
The flow through the rebound ports are bypassing the shim stack restricting the bump ports and
vice versa. As the pressure difference between the two sides of the piston increases, the oil in the ports
press on the shims. Dependent on which chamber contains the larger pressure, one of the shim stacks
will deflect allowing the oil to flow through the ports and the piston to the other chamber. During
compression of the shock absorber, the bump shim stack will deflect while the rebound shim remains
closed. Extension of the shock absorber will lead to deflection of the rebound shim stack. As the orifice
area depends on the deflection of the shims, the flow through the piston can be adjusted individually
for each flow direction.
Besides the shim ports, the damper also has an adjustable unidirectional valve called a needle port
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
37
6.3. SHOCK ABSORBER
Shims
Top plate
Rebound ports
Figure 6.9: The piston with a stack of shims [9].
Bump ports
Figure 6.10: The piston with ports. The bump ports are
blocked by the shims when the flow is running through
the rebound ports [9].
which is active during compression. It can be seen as the small needle in the middle of the damping
piston mounted with a spring just behind it (figure 6.7). The pretension of the spring behind the needle
can be adjusted by the rider. Depending on the amount of pretension, the needle port will only open
when a sufficient pressure difference called the crack pressure has been reached. The preloading of
the spring is called lock-out and affects the damping at low speeds when the pressure difference in the
damper chambers is small.
The piston rod reduces the surface area of the damper piston in chamber 3 compared to chamber 4. Therefore the volume changes in the chambers ∆V3 (∆x1 ) < ∆V4 (∆x1 ) for a given piston displacement ∆x1 . Chamber (5) is separated from chamber(4) by the IFP (6) and contains nitrogen
which compresses easily compared to the oil. If the oil is considered incompressible, the displacement caused by the piston rod will be equalled by the compression of the nitrogen chamber such
that ∆V3 (∆x1 ) + ∆V5 (∆x2 ) = ∆V4 (∆x1 ) for a given damper piston displacement ∆x1 and resulting IFP
displacement ∆x2 .
The difference in piston area between chamber (3) and (4) creates a resulting force on the piston
when it is at rest or travelling at low speeds:
v piston ' 0 ⇒ p3 ' p4 ⇒ p3 A3 < p4 A4
Fdamp,static = p4 A4 − p3 A3
where:
⇒
(6.36)
v piston
The damper piston travel speed.
A3
Damper piston area in chamber 3.
A4
Damper piston area in chamber 4.
Fdamp,static The static damper force.
The nitrogen pressure p5 increases when the shock absorber is compressed whereas the static pressures in chamber (4) and (5) are equal. This causes Fdamp,static to vary with travel thus creating a spring
effect in the damper. If the nitrogen is treated as an ideal gas with isothermal compression, the static
38
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.3. SHOCK ABSORBER
spring force can be calculated using equation 6.36:
∆p3 = ∆p4 = ∆p5 =
Fdamp,static =
where:
6.3.2
p5,iniV5,ini
p5,iniV5,ini
=
V5
V5,ini + ∆V5
⇒
p5,iniV5,ini
(A4 − A3 )
V5,ini − (A4 − A3 ) x1
p5,ini
The nitrogen pressure at zero travel.
V5,ini
The nitrogen chamber volume at zero travel.
x1
The piston travel starting at x1 = 0.
(6.37)
Modelling the Shock Absorber
F
x1 ,v1
Qtop
l3
l2
V,p
2 2
A3
8
V,p
3 3
2
l4
13
A1
l1
l5
11
5
6
Q2
Qbottom
A4
8
V,p
4 4
A5
V,p
1 1
12
Q1
1
A2
3
10
x2,v2
7
V,p
5 5
4
9
Figure 6.11: A schematic drawing of the fluid system in the shock absorber. The left part is the spring and the right side is
the damper. Note that there is a spring component build into the damper unit.
Figure 6.11 illustrates how the spring and the damper are connected. The spring is represented by
the left part and the damper is represented by the right part of the figure. For every circled number
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
39
6.3. SHOCK ABSORBER
in the figure 6.11, an equation will be set up. The equations are listed below and will be described
afterwards.

1
: Fdamp
= p3 · A3 − p4 · A4
→ Force equilibrium on piston.



2
: Qtop
= A3 · v1
→ The flow into/out of chamber (3).




3
: Qtop
= Q1 + Q2
→ Flow through node .
3



4
: Qbottom
= Q1 + Q2 q
→ Flow through node .
4



ρ

5
: Q1
= Cd,1 · Ad,1 · 2 (p4 − p3 )
→ Flow through needle port when p4 > p3 ,




Q1
=0
→
Flow
through
needle
port
when
p
<
p
,

4
3
q


ρ


6
: Q2
= Cd,2 · Ad,2 · 2 (p4 − p3 )
→ Flow through shims when p4 > p3 ,


q

ρ
→ Flow through shims when p4 < p3 ,
Q2
= Cd,3 · Ad,3 · 2 (p3 − p4 )
(6.38)

7
: p4 · A4
= p5 · A5
→ Force equilibrium on IFP.



8
: V4 +V3
= A4 · l4 + A3 · l3 ⇒



A4 (l4 −
x1 + x2 ) + A3 (l3 + x1 ) = A4 l4 + A3 l3 → Conservation of volume of oil, since the




fluid is assumed to be incompressible.


9
: p5 ·V5
= nRT ⇒ p5 · A5 (l5 − x2 ) = nRT
→ Pressure in nitrogenchamber (an ideal gas). 



10 : Qbottom
= v1 · A4 − v2 · A5
→ The flow in to/out of chamber(4)




11 : Ad,2
= f (∆p)
→ The area of the orifice (shims) in eq. .
6



12 : Ad,1
= g(∆p)
→ The area of the orifice (needle) in eq. .
5


13 : Fairspring
= p2 · A2 − p1 · A1
→ Force equilibrium on air piston.
Modelling the Damper
The damper can be modelled using equations 1 to 12 from equations (6.38). Equation 1 and 7
are simple applications of Newton’s third law where the mass of the system and friction between the
pistons and the chamber walls have been neglected. Equations 2 to 4 and 10 are flow equilibriums
for various parts of the system for which the oil is considered incompressible and of negligible mass
[10, p.7]:
Q = A·v
where:
Q
Flow.
A
The flow area.
v
Velocity of the fluid.
(6.39)
Equations 5 and 6 are applications of the orifice equation with the assumption that the flow is turbulent [10, p.19]:
r
ρ
· (p1 − p2 )
(6.40)
Q = Cd · Ad ·
2
where:
40
Q
Cd
The flow through the orifice.
The discharge coefficient, based on the area of vena contracta, the area of the physical orifice
and an empirical factor.
Ad
The area of the orifice.
ρ
The density of the fluid.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.3. SHOCK ABSORBER
p1
The higher pressure.
p2
The lower pressure.
It should be noted that even though the system actually contains two shim orifices only one of them
can be active at a time. Therefore, they can be modelled as shown in equation .
6
Equation 5 is an application of the Ideal Gas Law based on the assumption that the nitrogen
exhibits ideal gas behaviour in an isothermal compression/expansion process.
When these equations are established, it is possible to solve the problem numerically. Given the
travel velocity and level of compression of the shock, the 12 equations with 12 unknowns can be used
to find the damping force and additionally determine the various pressures and flows in the damper
system.
From the kinematic model described in chapter (6.1), it is possible to find the relative position and
velocity of the shock mounting points. These data can then be used to find the damping force for any
given state of the rear suspension.
Modelling the Spring
Equation 13 is used for modelling the spring. Assuming that the air exhibits ideal gas behaviour in
an isothermal compression/expansion process, the Ideal Gas Law can be used to determine the spring
force:
Fairspring = A2 · p2 − A1 · p1
n2 RT
n1 RT
= A2 ·
− A1
V2
V1
n2 RT
n1 RT
−
=
l2 + x1 l1 − x1
n2
n1
= RT
−
l2 + x1 l1 − x1
where:
Fairspring
Applied force on the piston.
R
The ideal gas constant.
T
Temperature of the gas assumed to be T = 298 K.
n
Amount of gas molecules determined from the initial volume and pressure.
V1
Volume of chamber (1): V1 = A1 (l1 − x1 ).
V2
Volume of chamber (2): V2 = A2 (l2 + x1 ).
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
(6.41)
41
6.3. SHOCK ABSORBER
The spring stiffness k can be obtained by differentiating equation (6.41) with respect to travel x:
−k · x1 = Fairspring
d
d
(−k · x1 ) =
Fairspring
dx1
dx1
n2
n1
k = RT
+
(l2 + x1 )2 (l1 − x1 )2
⇒
(6.42)
⇒
(6.43)
Using equations (6.41) and (6.43), the spring force and stiffness can now be found for any given level
of shock compression.
Determination of Flow Through Shims
The shims deflect when there is a pressure difference between the two chambers. This means that the
orifice area is dependent on the fluid pressure on the shims. This is expressed in equation (6.38) by 6
and :
11
r
ρ
Q2 = Cd,2 · Ad,2 ·
(p4 − p3 )
6
2
Ad,2 = f (∆p)
11
Before it is possible to determine the flow through this orifice the function f (∆p) must be found. The
following description applies to f (∆p) in 6 for p4 > p3.
To find f (∆p), a finite element analysis is performed to determine how much the shims deflect
under loading. When this deflection is known, it is possible to determine the area of the orifice at any
given pressure difference between the two sides of the shims. Since the material of the shims is linear
elastic, the deflection is almost linear with the load. Therefore, it is possible to find a particular level
of deflection by interpolating the data set from the FE simulations (further information about the FE
simulation is given in section 8.3.3). The data set from the FE simulation is a set of force-deflection
pairs. The results are presented later in table 8.1 in section 8.3.3. Before the pressure differences can
be translated into orifice areas, it is necessary to study how the shims deflect. This can be seen in
figure 6.12. When pressure from the oil in port (C) presses on the plate (B), a force is applied on the
edge of the shim (A) causing it to deflect. This way, the plate (B) is pushed back and an orifice opens
between the piston and the plate. The area of this orifice is shown in figure 6.13 and depends on the
plate displacement hA and the cross section of the port (C). hA can be found by determining the shim
deflection at point of contact between plate and shim. Then the total orifice area Ad,2 can be calculated
from the displacement hA and the total circumference of all the bump ports:
Ad,2 = hA (F)Cports
where:
42
(6.44)
hA (F)
Displacement function found by interpolating the shim deflection data from the FE analysis.
Cports
The total circumference of all the bump ports.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.3. SHOCK ABSORBER
A
C
B
hA
Orifice area
Port
Figure 6.12: In this position the shims have been deflected which makes the orifice open.
Figure 6.13: The hatched area shows the orifice area
Ad,2 .
The size of the ports and the fluid pressure in the ports are known. Therefore it is possible to find
the force that each port exerts on plate (B). The force from each port is multiplied by the number of
ports which gives the total force on the plate above the ports and therefore also on the shim. Equation
(6.44) can then be rewritten as:
f (∆p) = Ad,2 = hA (F)Cports = hA ∆p A ports Cports
(6.45)
where:
A ports
Total cross sectional area of the bump ports.
Since hA (F) is linear, the function f (∆p) can also be written as:
f (∆p) = α1 · ∆p
where:
α1
with α1 =
hA (F)Cports
A ports
(6.46)
Slope of the function.
The same principle can be used to determine f (∆p) for p4 < p3 and g(∆p) which is the area of the
orifice in the needle port. Here the linear relationships f (∆p) = α2 · ∆p and g(∆p) = α3 · ∆p are also
valid. The expressions are more obvious since the deflections depend on regular springs. The forces
on the springs can easily be calculated when the pressures and pressure areas are known. For g(∆p)
there is a slight variation because of the fact that an initial preload in the spring has to be exceeded
before the orifice opens.
Suggestions for Refining the Model
The compression of the nitrogen and air chambers have been modelled as an isothermal process. This
may be accurate for the static compression of the shock absorber where the heat is allowed to dissipate
through the chamber walls. As the shock absorber is used on a mountain bike, compression will
typically happen quickly and only during shorts periods of time. Because of that, a more accurate
shock absorber force could possibly be obtained by modelling the compression as an adiabatic process.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
43
6.3. SHOCK ABSORBER
This could be done using the following property: PiVi γ = Pf V f γ where P and V are the pressures and
the volumes at initial and final conditions respectively and γ = 57 for a diatomic gas.
The orifice equations used in 5 and 6 of equation 6.38 assume turbulent flow through the orifices
in the damper. This may be accurate for most of the travel velocity interval, but may not apply to
small velocities. The following equation can be used to approximate the discharge coefficient Cd for
Reynolds numbers Re < 10:
√
Cd = δ Re
where:
δ
An empirically found coefficient.
Re
The Reynolds number of the flow.
(6.47)
Finally, the orifice geometries have been difficult to determine accurately. These data could be found
by making detailed measurements the shock absorber, but since the main purpose of the model is to
find an accurate shock absorber response, an alternative approach is given in the following section.
6.3.3
Experimental Determination of Damping Characteristics
Since the parts of the shock absorbers geometry are determined by use of dimensionless drawings,
it is likely that the characteristics of the mathematical model are different from the characteristics of
an actual physical shock absorbing unit. It may be possible to reduce the deviation by correcting the
model with experimental data from measurements on the physical shock absorber.
Because the damping varies nonlinearly with velocity, the damping coefficient cannot simply be
determined by measuring the settling time of a constant mass with a known initial velocity and displacement. The nonlinear behaviour is caused by the variable orifices in the system. A way to obtain
a more accurate shock model could be by obtaining a set of orifice areas with respect to pressure
difference. Also, the pressure in the nitrogen chamber needs to be determined.
One way of determining the orifice area functions is to first determine the damping characteristics
of the shock absorber. Then inverse modelling can be applied to find the area functions. The damping
characteristics can be determined by measuring the damping force at various damper positions and
velocities. This equates to a system with two independent variables (position and velocity) and one
dependent variable (damping force).
Measurements carried out at different velocities and fixed travel distance is expected to be impractical. Therefore the following assumption is made to simplify the experiment: As mentioned in
section 6.3.3, the spring component of the damper part, constituted by the nitrogen chamber, has linear
stiffness. Therefore, the damping force can be seen as independent of travel except for a linear spring
contribution which can be subtracted when the spring constant is known.
The damping characteristics are to be determined from two different experiments. In both experiments the air sleeve (see figure 6.7) for the air spring should be removed while conducting the
experiment to eliminate any influence from the air spring. In principle this means that the left part of
figure 6.11 is removed.
The first experiment is to determine the pressure in the nitrogen chamber, and the second serves
the purpose of determining the functions for the orifice areas.
44
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.3. SHOCK ABSORBER
Determining the Nitrogen Pressure
First, the static force of the shock absorber will be measured throughout its travel. This is to be done
in a laboratory setup as shown in fig. 6.14, where the force of the damper is determined from the
normal strain in Element A by a strain gauge positioned away from the ends. Measurements of force
and travel are to be taken for zero velocity at various levels of compression by varying the distance
between points B and C. These data will be used to determine the spring constant and confirm the
linear relationship between the force and travel. The nitrogen pressure in the shock model can then be
adjusted to fit the measurements.
Element A
Damper
B
C
Strain
gauge
C0
Travel
Figure 6.14: Experimental setup for static damper force measurement.
Determining the Area Functions
The actual damping characteristics are to be measured using the setup shown in fig. 6.15. One end of
the damper is mounted to a fixed circular rod, the other end is mounted to the periphery of a rotating
disc. Both mounting points are revolute joints allowing in-plane rotation.
θ, ω
B
Strain
gauge
Max travel
Figure 6.15: Experimental setup for determining the area functions.
The diameter of the disc is equal to the maximum travel of the shock absorber. This way, the
damper will go through its full travel when the disc is rotated. During one revolution, the damper
will be subjected to different compressions and compression velocities even if the disc is rotating at
constant velocity.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
45
6.3. SHOCK ABSORBER
Throughout the experiment, the angular positions and velocities of the rotating disc are to be
measured along with the damping force.
The damping force is determined from the bending strain in the rod measured by a strain gauge
mounted as shown in section view A-A in figure 6.15. The damper velocity is determined from the
angular velocity and position of the disc. In practise, the best measurement approach may be to use a
servo/step motor and then calculate the angular velocities from time and position records. It may be
necessary to vary the rotational velocity of the disc to get good measurements throughout the relevant
velocity interval.
Finally, a velocity-damping force curve can be plotted, where the contribution from the nitrogen
spring is subtracted, as this is known from the previous experiment.
Now the mathematical shock model can be fitted to this velocity-damping force curve by adjusting
the orifice area functions at zero travel. The term zero travel refers to the shock absorber in its uncompressed state where there is no contribution from the nitrogen spring. This may be a difficult task,
but it is expected that the needle port orifice is closed at low velocities, when the system is in lock-out
mode as described in section 6.3.1. Consequently, it may be possible to adjust area function to fit the
first part of the velocity-damping force curve before the needle port opens. Then, the needle port area
function can be adjusted by fitting the model at higher velocities.
So basically the results of this second experiment are more precise area functions. This can be
done by assuming that the orifice at the shim stack to become effective at low compression velocities,
and the needle port orifice at relatively high compression velocities.
General Considerations of Setups
When designing the test setups, general care should be taken in dimensioning the individual components so as to minimise unwanted deflection. Also, the strain gauges should be positioned in accordance with Saint-Venant’s principle to prevent any local stress concentrations from affecting the
measurements. Care should be taken not to overload the damper by spinning the disc in Experiment 2
too fast. The mounting points of the shock absorber should be checked for adequate rigidity to prevent
any bending in the damper components, bending on the shaft driving the disc, and critical shearing
near the joints.
Implementing the Shock Absorber into Matlab
The shock absorber is programmed in Matlab as a function that can be called by any other other
program. A flowchart of the function is shown in figure 6.16. The outputs of the function are the shock
absorber’s damping and spring forces along with the spring stiffness.
The inputs of the function are the compression and compression rate of the shock absorber. From
the compression the air spring force and stiffness can readily be calculated from equations (6.43) and
(6.41).
With the velocity also known, the system of equations for the damper can be solved. These are
solved using the Newton-Raphson method as explained in section 6.1.4. It should be noted that the
technique of approximating the derivative, as explained in section 6.2.2, is also used here. When the
system of equations are solved, the force from the damper Fdamper is determined. Furthermore, the
damping coefficient is determined as a divided difference between two damper forces with a small
46
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.3. SHOCK ABSORBER
Input shock absorber
travel and velocity
Setup system of
equations
(
k = RT
)
n2
n1
+
(l2+x1)2
(l1-x1)2
Solve equations
by Newton-Raphson method
Determine
cshock
Fshock = Fdamper + Fspring
Output
kshock, cshock, Fshock
Figure 6.16: Flowchart of the "Shock Absorber" function in Matlab.
velocity difference between them.
Finally, the forces from the spring and the damper are added to give the total force, exerted on the
suspension system from the shock absorber.
Characteristics
For convenience some results from the shock absorber are presented next, since their interpretations
are closely related to the terms used in this section. Using the model developed above, it is possible to
determine the spring and damping characteristics of the shock absorber. Figures 6.17 and 6.18 show
the spring and damping characteristics of the shock absorber model.
Solving the equations using a numerical method has shown difficulties at small velocities. Especially the lower part of the negative velocities lead to singularities when using the Newton-Raphson
method. As the damper force needs to be defined for the full velocity interval, a polynomial fit has
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
47
6.3. SHOCK ABSORBER
-500
-1000
Spring force [N]
-1500
-2000
-2500
-3000
-3500
-4000
-4500
-5000
0
0.01
0.02
0.03
0.04
Shock absorber travel [m]
0.05
Figure 6.17: The air spring force as a function of shock
absorber compression.
Figure 6.18: The damper force as a function of velocity
(positive velocity lead to compression). The three lines
correspond to different compression levels.
been made of the original damper model. Here, the velocities leading to singularities have been interpolated from the neighbouring force values. The blue curve in figure 6.18 shows the polynomial fit
which may be inaccurate in the velocity interval marked .
1 The red curve indicates the expected size
of the damping force for this interval.
Interval 2 represents the damping force under compression of the shock absorber with the needle
port closed. The difference in deflection properties between the bump and rebound shims is evident
when comparing the slope in this interval with the slope at negative travel velocities. Interval 3 shows
the gradual opening of the needle port thus leading to a noticeable change in slope. At the end of the
interval the slope changes again because the needle port reaches its maximum orifice area. Figure 6.19
shows plot of the damping force and the orifice areas Ad1 and Ad2 of the needle port and the bump port
shim stack respectively. The effect of the needle port opening around 0.75s is clearly visible on the
damping force and shim stack orifice area graphs.
Three graphs are shown in figure 6.18. Every graph represents the damping characteristics for
different compression levels. It is clear from this figure, that the slopes are identical, which means that
the damping coefficient is the same for every instance of compression and independent of the position
of the shock absorber. The graphs are vertically displaced due to the spring effect from the nitrogen
chamber explained previously. .
It is seen that the slope of the curves in figure 6.18 are almost constant for negative velocities and
velocities larger than 1 ms . This means that the damping coefficient is almost constant for these parts of
the velocity interval whereas it is significantly larger at low compression velocities.
The correlation between the crack pressure of the needle port and the change in magnitude of the
damping coefficient shows that it is possible to specifically control the low speed damping. This can be
used to reduce unwanted suspension activation caused by the rider while preserving bump absorbtion.
Figure 6.17 shows that the spring constant for the air spring remains almost constant until the
shock absorber is compressed approximately 0.025 m, which is half of the total travel. This causes
the suspension to be very responsive on small bumps while still preventing the shock absorber from
bottoming out on larger bumps. The pressure in the air chambers can be adjusted to utilise the full travel
48
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.4. EQUIVALENT STIFFNESS AND DAMPING
Figure 6.19: F(v) is the force from the damper w.r.t. velocity. Ad1 is the area of the orifice of the adjustable valve w.r.t.
velocity. Ad2 is the area of the orifice of the shims w.r.t. velocity.
of the shock absorber dependent on the type of terrain and rider weight. A higher pressure yields a
greater spring stiffness thus enabling the suspension to support a heavier rider or absorb larger bumps
without bottoming out. Ideally, the air spring should be adjusted such that it provides just enough
stiffness to prevent the shock absorber from exceeding its travel at the largest bump experienced.
6.4
Equivalent Stiffness and Damping
In this section a short description for calculating equivalent damping and stiffness coefficients are
presented. The equivalent stiffness and damping are used for moving the shock absorbers properties
to other points of the rear suspension on a model basis. This is convenient when modelling inputs to
the rear wheel.
In the previous section stiffness and damping properties of the shock absorber are determined. But
since the suspension of the bike is a combination of the properties of the shock absorber and movement of the four bar linkage, it is relevant to consider what influence the shock absorber has on the
rear wheel.
In principle this means that instead of having the shock absorber S placed as it actually is, another
shock absorber Seq is imagined attached to the rear wheel. The idea is now to determine stiffness and
damping coefficients of Seq , leading Seq to have the same influence on the suspension system as S and
the linkage. This is shown in figure 6.20.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
49
6.4. EQUIVALENT STIFFNESS AND DAMPING
S
Seq
Figure 6.20: The figure shows what is in principle done when determining Seq . Note that no revolute joints are present,
when using the Seq as it represents both S and the four bar linkage.
6.4.1
Equivalent Stiffness
The stiffness of the shock absorber only depends on its position. The basic principle is that the work
done by the equivalent stiffness must equal the work done by all springs in the system. For the mountain bike this can be expressed as:
1
1
2
2
keq xwheel
= kshock xshock
2
2
where:
keq
The equivalent stiffness.
xwheel
The displacement of the rear wheel.
kshock
The stiffness of the shock absorber.
xshock
The displacement/compression of the shock absorber.
(6.48)
The equivalent stiffness can therefore be written as:
keq = kshock
xshock
xwheel
2
In this formula the link system can be interpreted as a gearing, where
(6.49)
xshock
xwheel
2
is the gearing.
Correction
The method above is found not to be valid as equation 6.49 only takes the nonlinearity from the linkage
into account. When calculating keq using equation 6.49 at a certain position, χw, f inal , it is assumed that
50
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.4. EQUIVALENT STIFFNESS AND DAMPING
kshock and keq are constant at χw, f inal , which is considered correct while assuming small time steps. But
they are also assumed to be constant at any previous position, which is definitely not correct.
Therefore the equivalent stiffness should have been calculated by solving the following equation:
Z χw, f inal
keq (xw ) xw dxw =
0
where:
Z χs, f inal
ks (xs ) xs dxs
0
(6.50)
χw, f inal
The displacement of the rear wheel where the equivalent stiffness is to be calculated.
χs, f inal
Compression of the shock absorber due to χw, f inal .
xw
Deflection of the rear wheel.
xs
Compression of the shock absorber due to xw .
keq (xw )
The equivalent stiffness being a function of xw .
ks (xs )
Stiffness of shock absorber being a function of xs .
Inasmuch as the expression for ks (xs ) is not known in its explicit form, it is necessary to determine
keq (xw ) by a numerical method. This is done using Riemann sums. Figure 6.21 illustrates the principle
of determining the equivalent stiffness with the work theorem using Riemann sums. Keep in mind that
the gearing is non-constant during displacement of xwheel .
keq x
kshock x
xwheel
xw;i-1 xw;n
xshock
xs;i
Figure 6.21: Illustrating the principle of determining the equivalent stiffness with Riemann sums.
From the principle of figure 6.21 the work theorem can be formulated:
n−1 ∑
i=1
n
2
keq,i (xw,i − xw,i−1 )2 + keq,n (xw,n − xw,n−1 )2 = ∑ ks,i xs,i − xs,i−1
n
∑ ks,i xs,i − xs,i−1
keq,n =
⇒
i=1
i=1
2
n−1
− ∑ keq,i (xw,i − xw,i−1 )2
i=1
xw,n − xw,n−1
2
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
(6.51)
51
6.4. EQUIVALENT STIFFNESS AND DAMPING
n−1
The sum
∑ keq,i (xw,i − xw,i−1 )2 is illustrated in figure 6.21 as the red areas.
i=1
n
shown as the green area. The grey area illustrates the sum
∑ ks,i
keq,n (xw,n − xw,n−1 )2 is
2
xs,i − xs,i−1 .
i=1
6.4.2
Equivalent Damping
For the equivalent damping, Rayleigh’s dissipation function is used known from Rao [11, p. 503].
Similar to the equivalent stiffness, the damping equivalents can be determined by:
1
1
2
2
ceq ẋwheel
= cshock ẋshock
2
2
ẋshock 2
ceq = cshock
ẋwheel
where:
ceq
The equivalent damping.
ẋwheel
The velocity of the rear wheel.
cshock
The damping coefficient of the shock absorber.
ẋshock
The compression velocity of the shock absorber.
⇒
(6.52)
(6.53)
Since the damping is not only dependent on velocity but displacement as well, equation (6.53) is
able to be carried out at various positions.
Note that the correction done to the equivalent stiffness does not necessarily apply to the equivalent
damping. Since the nature of the work done by a spring is reversible, the correction was done. But the
work done by a damper is irreversible and therefore the method used in equation 6.53 is valid. This
can also be seen by the Rayleigh dissipation theorem. It determines the power done by a damper at a
certain velocity and can thus be said to have no memory.
6.4.3
Implementing the Equivalences into Matlab
The equivalent stiffness and equivalent damping are implemented into Matlab as two separate functions.
The Equivalent Stiffness
A flowchart of the equivalent stiffness function is shown in figure 6.22. The inputs of the function
are simply the current positions of the bike. From these positions the function uses the shock absorber
function to get kshock . This is used in equation (6.49) to calculate keq which is the output of the function.
The correction of the calculation of equivalent stiffness described above has not been implemented
into Matlab.
52
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.5. RESPONSE TO FREQUENCY INPUT
Input positions
and velocities
Input positions
Calculate
Input
positions
(q,q)
v
~~
shock
and velocities
Shock
InputAbsorber
positions
.
2
( )
Calculate
Shock
Absorber
vshock(q,q)
~~
.
wheel
xshock
xwheel
2
.
.
keq
keq
= kshock
xshock
( )
( )
ceq
keq
Figure 6.22: Flowchart of the equivalent stiffness function
in Matlab.
2
ceq Shock
= cshock
Absorber
x
ceq
= cshock
wheel
.
.
xshock
keq Shock
= kshock
Absorber
x
xshock
xwheel
2
( )
Figure 6.23: Flowchart of the equivalent damping function in Matlab.
keq
The Equivalent Damping
The structure of the equivalent damping function is sketched in figure 6.23. The inputs of the function
are positions and velocities of all local frames on the bike. From these the absolute velocity of the
shock absorber, vshock , is calculated. This is now used as input to the shock absorber function which
returns a damping coefficient, cshock for the shock absorber. cshock is then used in equation (6.53) to
calculate the equivalent damping, ceq , which is the output of the function.
6.5
Response to Frequency Input
In the previous sections tools for designing the suspensions system were presented. In this section the
performance of the suspension system is studied. This is done through an analysis where the impact,
of oscillations of front and rear wheel, on the rider and frame is calculated.
The entire motivation for designing a full suspension mountain bike is to soften uneven tracks. If
it is assumed that the rider does not move relative to the bike frame, the rider and bike can be modelled as one unit with one centre of mass. By representing an uneven track with a series of cosine
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
53
6.5. RESPONSE TO FREQUENCY INPUT
functions, the magnitude of oscillation of the centre of mass can be observed. The goal is to calculate
the displacement of the center of mass, compare it to the displacement of the wheels, where a possible
difference must be ascribed to the suspension system. This is illustrated in figure 6.24.
Oscillations of frame and rider
y
x
Oscillations of track
Figure 6.24: By oscillating the front and rear wheel, the rider will also be oscillating.
For calculation purposes the bike is considered as a Mass-Spring-Damper-system and is illustrated
in figure 6.25.
l1
l2
x
c1
k1
Rear base
y1
y2
k2
c2
Front base
Figure 6.25: The oscillating situation interpreted on model basis.
The box represents the sprung mass. The sprung mass is considered the mass of rider and frame.
The model uses stiffness and damping coefficients, k1 and c1 , related to the rear suspension, which
were calculated in section 6.4. Figure 6.25 indicates that only vertical and rotational displacements are
54
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.5. RESPONSE TO FREQUENCY INPUT
possible. Horizontal displacement is simply not an option because of the way the inputs are applied.
6.5.1
Assumptions
The calculation is based on the following assumptions:
1. Rotation of the sprung mass θ is small, hence sin(θ) ∼
= θ.
2. The distance l2 in figure 6.25 is constant.
3. Damping and stiffness of the front fork are assumed to be k f ront = 200 N/m and c f ront = 700 kg/s.
4. Only damping and spring properties of the shock absorber and four bar linkage are considered,
hence the tires are considered as stiff elements.
5. The oscillations are forced both upwards and downwards. In other words: The tires will not
loose grip with the ground.
6. While solving the equations, stiffness and damping coefficients are treated as constant.
7. The system starts from rest (x = θ = ẋ = θ̇ = 0).
6.5.2
Equations of Motion
While deriving the equation of motion, the freebody diagram in figure 6.26 is helpful. Note that the
displacements, at the instant of time which figure 6.26 represents, at the rear and front end of the
sprung mass, x − l1 sin(θ) and x + l2 sin(θ) respectively, are both mutually larger than y1 and y2 .
l1
l2
θ
θ
θl1
Fk1
Fc1
Fk1
Fc1
x
y1
y2
Rear base
Fk2
Fc2
Fk2
Fc2
Front base
Figure 6.26: Free body diagram of the sprung mass.
By investigating the force and moment equilibrium on the sprung mass the following equations are
derived.
Mbike ẍ = −k1 (x − y1 − θl1 ) − c1 (ẋ − y˙1 − θ̇l1 ) − k2 (x − y2 + θl2 ) − c2 (ẋ − y˙2 + θ̇l2 )
Jbike θ̈ = k1 (x − y1 − θl1 ) + c1 (ẋ − y˙1 − θ̇l1 ) l1 − k2 (x − y2 + θl2 ) + c2 (ẋ − y˙2 + θ̇l2 ) l2
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
(6.54)
(6.55)
55
6.5. RESPONSE TO FREQUENCY INPUT
where:
Mbike
The sprung mass.
Jbike
The mass moment of inertia of the sprung mass.
The rest of the symbols are seen in figure 6.26. By rearranging both (6.54) and (6.55), the following relation is achieved:
Mbike
0
ẍ
c1 + c2
c 2 l2 − c 1 l1
ẋ
+
...
0
Jbike
θ̈
c2 l2 − c1 l1 c1 l12 + c2 l22
θ̇
k1 + k2
k 2 l2 − k 1 l1
x
F
...+
=
(6.56)
k2 l2 − k1 l1 k1 l12 + k2 l22
θ
M
where
F
M
=
k1
k2
−k1 l1 k2 l2
y1
y2
+
c1
c2
−c1 l1 c2 l2
y˙1
y˙2
(6.57)
By this rearrangement, it is seen that moving the wheels corresponds to applying force and moment
to the system, given by the right side of (6.56). Also it is noted that the system has both static and
velocity coupling. This is the case as the stiffness and damping matrices are not diagonal as stated in
Rao [11, p. 402].
Since the equation of motion is derived, the next steps are defining the input functions y1 (t) and
y2 (t) and solving the equation.
6.5.3
Definition of Input
The purpose of this analysis is to investigate the suspension systems capability to soften out an uneven
surface, which is why the input functions are defined as a combination of cosine functions. Thus:
where:
y1 (t) = Y11 cos(ω1t + φ1 ) +Y12 cos(ω2t + φ2 )
(6.58)
y2 (t) = Y21 cos(ω2t + φ2 ) +Y22 cos(ω1t + φ1 )
(6.59)
y1 (t)
the input function for the rear wheel.
y2 (t)
The input function for the front wheel.
Y jk
Are all amplitudes.
ωj
Are angular velocities.
φj
Are phases.
These functions are used to represent an uneven track as they allow arbitrary combinations of amplitudes, frequencies and phases.
Insertion of (6.58) and (6.59) in (6.57) yields:
56
F = k1 y1 (t) + k2 y2 (t) + c1 ẏ1 (t) + c2 ẏ2 (t)
(6.60)
M = −k1 l1 y1 (t) + k2 l2 y2 (t) − c1 l1 ẏ1 (t) + c2 l2 ẏ2 (t)
(6.61)
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.5. RESPONSE TO FREQUENCY INPUT
Using complex notation equations (6.60) and (6.61) can be written as:
F = (Y11 (k1 + c1 ω1 i) +Y22 (k2 + c2 ω1 i)) ei(ω1t+φ1 ) . . .
|
{z
}
F1
. . . + (Y12 (k1 + c1 ω2 i) +Y21 (k2 + c2 ω2 i)) ei(ω2t+φ2 )
{z
}
|
(6.62)
F2
M = (Y11 l1 (−k1 − c1 ω1 i) +Y22 l2 (k2 + c2 ω1 i)) ei(ω1t+φ1 ) . . .
|
{z
}
M1
. . . + (Y12 l1 (−k1 1c1 ω2 i) +Y21 l2 (k2 + c2 ω2 i)) ei(ω2t+φ2 )
|
{z
}
(6.63)
M2
This turns out to be a convenient way of writing the input.
The solution to the coupled second order linear differential equations (6.56) subjected to the excitations given by (6.62) and (6.63), is then given by a the sum of a particular solution and a complementary solution to equation (6.56), from the homogeneous case of (6.56) when the right side equals
~0.
6.5.4
Particular Solution
The method for finding a particular solution in the case of forced vibration, is guessing functions for
x(t) and θ(t), similar to the input functions. The following guesses are suggested:
x p (t) = X̃1 eiω1t + X̃2 eiω2t
iω1 t
θ p (t) = Θ̃1 e
where:
(6.64)
iω2 t
+ Θ̃2 e
(6.65)
x p (t)
The function translating the sprung mass related to the particular solution.
X̃ j
Are amplitudes for x p (t).
θ p (t)
The function rotating the sprung mass related to the particular solution.
Θ̃ j
Are amplitudes for θ p (t).
Since the first and second order derivatives are needed to describe the motion of the system they
are presented here. The first order derivatives are:
ẋ p (t) = iω1 X̃1 eiω1t + iω2 X̃2 eiω2t
iω1 t
(6.66)
iω2 t
(6.67)
ẍ p (t) = −ω21 X̃1 eiω1t − ω22 X̃2 eiω2t
(6.68)
θ̇ p (t) = iω1 Θ̃1 e
+ iω2 Θ̃2 e
The second order derivatives are:
θ̈ p (t) =
−ω21 Θ̃1 eiω1t
− ω22 Θ̃2 eiω2t
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
(6.69)
57
6.5. RESPONSE TO FREQUENCY INPUT
By inserting the guesses (6.64) - (6.69) on the left hand side in the equation of motion (6.56) along
with (6.62), (6.63) on the right hand side in the equation of motion, the following can be obtained:
(. . .)eiω1t + (. . .)eiω2t = F1 eiω1t+φ1 + F2 eiω2t+φ2
(. . .)e
iω1 t
+ (. . .)e
iω2 t
= M1 e
iω1 t+φ1
+ M2 e
(6.70)
iω2 t+φ2
(6.71)
The brackets (. . . ) represents some large expression. As an attempt to match the expressions in the
brackets to F1 eiφ1 , F2 eiφ2 , M1 eiφ1 and M2 eiφ2 , which is a necessity for the equal signs in (6.70) and
(6.71) to be satisfied, the following separations are made:
(. . .)eiω1t = F1 eiω1t+φ1
(. . .)e
iω2 t
iω2 t+φ2
= F2 e
(. . .)eiω1t = M1 eiω1t+φ1
(. . .)e
iω2 t
= M2 e
iω2 t+φ2
⇔
(. . .) = F1 eiφ1
(6.72)
⇔
(. . .) = F2 e
iφ2
(6.73)
⇔
(. . .) = M1 eiφ1
(6.74)
⇔
iφ2
(6.75)
(. . .) = M2 e
This set of equations can be rewritten to the form:
Z11 X̃1 + Z12 Θ̃1 = F1 eiφ1
(6.76)
iφ2
(6.77)
Z33 X̃2 + Z34 Θ̃2 = F2 e
Z21 X̃1 + Z22 Θ̃1 = M1 eiφ1
(6.78)
iφ2
(6.79)
Z43 X̃2 + Z44 Θ̃2 = M2 e
where:
Z jk
Are impedances, the actual expression for Z jk can be seen in appendix A.
The term impedance for Z jk is relevant to use as they can be interpreted as frequency depended resistances. Hence large frequencies result in small amplitudes.
If the impedance equations are written using matrix notation it is solved:

Z11 Z12 0
0
X̃1
 Z21 Z22 0


0   Θ̃1

 0
0 Z33 Z34   X̃2
Θ̃2
0
0 Z43 Z44


F1 eiφ1
  M1 eiφ1 

=
  F2 eiφ2 
M2 eiφ2


(6.80)
If the impedance matrix is denoted by Z̄¯ , the amplitudes are achieved by:
58
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.5. RESPONSE TO FREQUENCY INPUT



X̃1
 Θ̃1  ¯ −1 



 X̃2  = Z̄ 
Θ̃2

F1 eiφ1
M1 eiφ1 

F2 eiφ2 
M2 eiφ2
(6.81)
The particular solution ~p p is therefore found and given by (6.64), (6.65) and (6.81).
6.5.5
Complementary Solution
As mentioned earlier the solution to the homogeneous equation of motion along with the particular
solution gives the full solution.
By inverting the mass matrix and multiplying it in the homogeneous equation of motion the accelerations can be expressed as:
ẍ
θ̈
=
1
Mbike Jbike
+
Jbike (−c1 − c2 )
Jbike (c1 l1 − c2 l2 )
Mbike (c1 l1 − c2 l2 ) Mbike (−c1 l12 − c2 l22 )
Jbike (−k1 − k2 )
Jbike (k1 l1 − k2 l2 )
Mbike (k1 l1 − k2 l2 ) Mbike (−k1 l12 − k2 l22 )
ẋ
θ̇
x
θ
(6.82)
(6.83)
T
By introducing the vector ~p = xc , θc , ẋc , θ̇c the above equation can be expanded to:
 
ẋc
 θ̇c  

 
 ẍc  = 

θ̈c

0
0
0
0
1
0
0
1
−k1 −k2
Mbike
k1 l1 −k2 l2
Jbike
k1 l1 −k2 l2
Mbike
−k1 l12 −k2 l22
Jbike
−c1 −c2
Mbike
c1 l1 −c2 l2
Jbike
c1 l1 −c2 l2
Mbike
−c1 l12 −c2 l22
Jbike


xc

  θc 
 = B̄¯ ~pc

 ẋc 
θ̇c
(6.84)
Now, by using the eigenvalue method, the differential equation ~p˙ = B̄¯~p given in (6.84) can be solved.
The complementary solution is then given by:


xc
 θc 

~pc (t) = 
 ẋc  =
θ̇c
where:
4
∑ w j ~v j eλ t
j
(6.85)
j=1
wj
Are arbitrary constants.
λj
Are the eigenvalues in matrix B, equation (6.84).
~v j
Are the eigenvectors corresponding to λ j .
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
59
6.5. RESPONSE TO FREQUENCY INPUT
6.5.6
Full Solution
T
By introducing a vector similar to ~pc related to the particular solution ~p p = x p , θ p , ẋ p , θ̇ p , the entire
solution ~p(t) to (6.56) is:
~p(t) = ~pc (t) + p~ p (t)
(6.86)
due to the linearity of (6.56).
The constants w1 . . . w4 in ~pc can be determined by the use of initial conditions. Since it is assumed
that the bike starts from rest, the initial conditions must be:
~p(0) = ~0
(6.87)
So by finding the derivatives ẋ p (t), θ̇ p (t) and evaluating ~p(t) = ~0 at t = 0, that is eλ j ·0 = eiω j ·0 = 1, the
constants are determined as:





 
w1
X̃1 + X̃2
 w2  
Θ̃1 + Θ̃2
 
~p(0) = v~1 v~2 v~3 v~4 
 w3  +  i(ω1 X̃1 + ω2 X̃2 )
w4
i(ω1 Θ̃1 + ω2 Θ̃2 )



w1
X̃1 + X̃2


w2 
Θ̃1 + Θ̃2
 = − v~1 v~2 v~3 v~4 −1 

 i(ω1 X̃1 + ω2 X̃2 ) 
w3 
w4
i(ω1 Θ̃1 + ω2 Θ̃2 )


 = ~0

⇒
(6.88)
(6.89)
Both the transient and steady state solutions are known and the problem is in principle solved. But
from section 6.4 clearly the damping and stiffness coefficients are definitely not constants, which was
an assumption while deriving transient and steady state response. Therefor the solution to the problem,
equation (6.86) is only used in small time steps.
6.5.7
Implementing the Frequency Analysis into Matlab
The analysis is implemented into Matlab as a program as sketched in figure 6.27. The results of the
program are the positions of the bike and the movement of the centre of mass to all the specified
time steps. Before the analysis can be performed, an initial bike is set up and the force and moment
subjecting the system are defined.
The rest of the analysis is performed in a loop over a specified time span. In the loop the equivalent
stiffness, equivalent damping, and the centre of mass are first calculated from the current position of
the bike.
From these both the partial- and the complementary solutions are calculated, as described in this
section. These are now added and the relevant constants are found. Thus, the full solution is now
known.
60
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.6. MODELLING THE FREE FALL
Define: F(t)
M(t)
Setup initial
bike
~q
Loop over
specified time
span
Equivalent
stiffness
Equivalent
damping
Determine partial
solution
Calculate
Centre of Mass
Determine
complementary
solution
Add solutions to get
full solution
Kinematic
Model
~q, x, θ
for all time steps
Figure 6.27: Flowchart of the program "Response to Frequency Input".
The full solution gives a movement of the centre of mass. This is now converted into a movement
of the crank which is used as input to the kinematic model. By running a kinematic analysis the new
positions of all local frames on the bike are now known. The positions are sent back to the beginning
of the loop so the next run through the loop will use this as the initial bike configuration. Hence, when
the loop is finished, coordinates of the bike are known to all time steps. From this the behaviour and
movement of the bike can be described.
6.6
Modelling the Free Fall
In chapter 4 four types of loading scenarios were established. To obtain a design criterion that provides the greatest force from the shock absorber on the suspension system, a free fall from one meter
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
61
6.6. MODELLING THE FREE FALL
was presented as a loading scenario. This section deals with modeling the free fall scenario. This will
serve the purpose of determining the greatest force of the shock absorber on the four bar linkage.
As mentioned in section 4.1 there is a non-constant gearing between the rear wheel and the shock
absorber. The response of the system is only relevant to examine up until the point, where the shock
absorber is fully compressed. What happens afterwards due to the free fall is of no interest in this
analysis. To model the response of the suspension system, up until the shock absorber is fully compressed, it is useful to ascribe the properties and behaviour of the system to one point. This simplifies
the model and makes it easier to derive the equation of motion. Figure 6.28 shows how the properties
of the suspension system are ascribed to the point P3 , where the rear wheel is mounted. keq , ceq , and
m refer to the equivalent quantities for the spring, damper, and mass, when they are ascribed to point
P3 . These are determined in section 6.4. Figure 6.28 illustrates the moment just as the rear wheel
ceq
keq
meq P3
x
Figure 6.28: Model of the suspension properties ascribed to point P3 .
hits the ground. In reality, the rear wheel would stand still relative to the ground, while the frame and
linkage system would move. For calculation and programming purposes, the situation is modelled as
if the kinetic energy is provided instantaneously to the rear wheel, with initial velocity in the positive
x-direction. See figure 6.28. The initial velocity of the equivalent mass m can be calculated using
conservation of mechanical energy:
⇒
∆Ekin = ∆E pot
1
m∆v2 = mg∆h
2
p
∆v = 2g∆h
p
ẋ0 = 2gh f
where:
⇒
⇒
(6.90)
ẋ0
The initial velocity of the equivalent mass.
hf
The height of the free fall, 1 m.
Figure 6.29 shows the free body diagram and the corresponding kinetic diagram for the free fall scenario just as the wheel hits the ground. For calculation purposes the gravity force Fg is drawn to act
in the reverse direction. This is done because the gravity force in reality will cause the suspension
system to compress. To ease the implementation of this effect into the kinematic analysis in Matlab,
the gravity force is calculated inversely. In this way the gravity force will affect the rear wheel to move
62
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.6. MODELLING THE FREE FALL
Fd
Fs
ma
Fg
Figure 6.29: To the left a free body diagram and to the right the corresponding kinetic diagram.
upwards and hence the suspension system to collapse.
6.6.1
Equation of motion without gravity force
If the gravity force is disregarded the free body diagram can be combined with the kinetic diagram to
give:
−Fs − Fd = mẍ
⇒
−keq x − ceq ẋ = mẍ
⇒
mẍ + ceq ẋ + keq x = 0
(6.91)
Equation (6.91) is the equation of motion of the system in figure 6.28, where gravity is not taken
into account. The equation of motion is a homogenous second order linear differential equation. The
behaviour of the system described by equation (6.91) is thus only depend on displacement. To solve
this equation a guess is proposed:
x(t) = eRt
(6.92)
By inserting the guess in (6.91) the constant R can be found:
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
63
6.6. MODELLING THE FREE FALL
mR2 eRt + ceq ReRt + keq eRt = 0
⇒
2
⇒
mR + ceq R + keq = 0
ceq
keq
R2 +
R+
=0
m
m
2
ceq
ceq 2 keq
+
−
=0
R+
2m
m
2m
2
ceq 2
ceq
keq
R+
=
−
2m
2m
m
s ceq 2 keq ceq
−
−
R1,2 = ±
2m
m
2m
⇒
⇒
⇒
(6.93)
Since the equivalent values are non-constant but vary with the displacement, there are three possible
scenarios for the value of R. R can obtain to distinct values, R can be a complex number or R can be a
repeated root with repetition of two. The solutions to equation (6.91) for the three incidents of R can
be found in appendix B.
6.6.2
Equation of motion with gravity force
If the gravity force is taken into account a particular solution to the equation of motion can be found.
This can be formulated as:
mẍ + ceq ẋ + keq x = mg
(6.94)
A guess of a solution is a constant B. Inserting this into (6.94) gives:
m · 0 + ceq · 0 + keq B = mg
mg
B=
keq
⇒
(6.95)
So if the gravity force is taken into account, the full solution can be found by super positioning the
homogeneous solution xH , found in either (B.1), (B.8), or (B.11) in appendix B, with the particular
solution xP . That is:
x(t) = xH + xP = xH +
mg
keq
(6.96)
Since the gearing in the suspension system is not constant, the behaviour of the system is non-linear,
when ascribing the properties of the suspension to the point P3 . To solve the equation of motion for
the free fall scenario, the math software Matlab is used. An overview of how this is done, is given in
the following section.
64
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
6.6. MODELLING THE FREE FALL
6.6.3
Solving the equation of motion using Matlab
Figure 6.30 shows a flowchart of how the equation of motion is solved using Matlab. Furthermore the
force in the shock absorber is calculated. Before being able to solve the equation of motion, a set of
Initialising
variables
Loop over all
time steps
x(ti)
Kinematic
Model
Shockforce
Equivalent
stiffness
Equivalent
damping
Solving
equation of
motion
~Fshock
~x(t)
Figure 6.30: Flowchart of how the equation of motion is solved in Matlab.
variables need to be initialised. These are quantities like gravitational acceleration, mass of bike and
rider, initial positions of links etc.
The kinematic model is implemented to give positions, velocities and accelerations for given points
in the suspension system. These values are passed on to calculate the equivalent stiffness and damping
coefficients. Then these values are put into the equation of motion to give the response x(ti ) for the
given time step. These results are saved in the vector ~x(t) to give the full response over the given
time loop. As mentioned earlier only the response up until the point of full compression of the shock
absorber is of interest.
Simultaneously the kinematic model is used for calculating the force in the shock absorber for the
given response of the suspension system.
CHAPTER 6. ELEMENTS IN THE DYNAMIC MODEL
65
7
Results of the Dynamic Analysis
In this chapter the objective is to present how the mathematical model is employed to design the mountain bike. This is based on the strategy described in chapter 5 and the calculation models in chapter 6.
As suggested in the problem statement, chapter 3, and the strategy for settling the bike design presented in chapter 5 an optimisation process is necessary to determine an energy efficient suspension
design. But because of problems with the mathematical calculation model, the actual design process
has not been possible. The difficulties have mainly been related to numerical solving, where division by zero and singular matrices appear. Additionally the calculation model has proven itself very
sensitive to changes in external forcing and changes in internal geometry.
Consequently this chapter only demonstrates simple changes in design and their impact on the
dynamic responses.
7.1
The Four Bar Linkage
The four bar linkage is designed by the use of the kinetic analysis, which is described in step 1 in
chapter 5.
The scenario is as follows: A rider is placed on the bike and starts pedalling. This scenario is
carried out at two different configurations of the linkage system, denoted by A and B, shown in figure
7.1 and 7.2. For the two configurations it is investigated, what happens when the length of the bar
between the Horst link and rear dropout is changed. For both configurations the loads on figure 7.3
y
y
0.9
0.8
0.8
0.7
Normalforce
Position
Position
0.7
0.6
0.9
0.5
0.4
0.6
Normalforce
0.5
0.4
0.3
0.3
Chainforce
Chainforce
0.2
-0.5
0.2
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
Position
Figure 7.1: Configuration A.
0.3
0.4
x
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
Position
0.3
0.4
x
Figure 7.2: Configuration B.
are applied to the rear dropout. The normal force is a result of the rider taking a seat on the bike. This
force is applied linearly during 0.05 s. After 0.27 s the rider starts pedalling, which results in a chain
force as shown in figure 7.3. The chain force is a result of the mean moment obtained from AnyBody.
By running configuration A and B, both subjected to the loads in figure 7.3, through the kinetic
analysis the plots in figure 7.4 and 7.5 are obtained.
66
CHAPTER 7. RESULTS OF THE DYNAMIC ANALYSIS
Wheel position [m]
0.32
0.31
0.3
0.29
7.1. THE FOUR BAR LINKAGE
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
1
1.2
1.4
1.6
1.8
2
1
1.2
1.4
1.6
1.8
2
3000
2000
1000
Normal force [N]
Chain force [N]
Time [sek]
0
0
0.2
0.4
0.6
0.8
0
0.2
0.4
0.6
0.8
600
400
200
0
Time [sek]
Figure 7.3: The loads applied on the rear dropout. Note that the absolute value of the chain force is plotted, since the sign is
dependent upon the position of the linkage system.
The plots clearly show that the response depends on the configuration. With this loading scenario
configuration A, figure 7.1, gives DISC when a chain force is applied, whereas configuration B BioPaces when a chain force is applied. One chain cycle indicates that the chain force obtains values
from zero to maximum to zero. From figure 7.4 and 7.5 it is seen that during one cycle of chain force
configuration A gives approximately 2 cm displacement of the rear wheel and configuration B gives
approximately 0.4 cm displacement. Since there is no preference for neither DISC nor Bio-Pacing
figure 7.4 and 7.5 indicate that configuration B might be the preferable one, since this gives the most
neutral response.
Animations of the response to a normal and chain force for configuration A and B can be found on
the CD in appendix E under Movies » ReponseA and ResponseB.
CHAPTER 7. RESULTS OF THE DYNAMIC ANALYSIS
67
Wheel position [m]
7.2. THE SHOCK ABSORBER
0.32
0.31
0.3
0.29
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time [sek]
Wheel position
[m] [N]
Chain force
ChainNormal
force [N]
force [N]
3000
Figure 7.4: Response to a pulsating chain force using configuration A.
2000
0.295
1000
0.29
7.2
0.285
0
0
0.28
0.2
0.4
0.6
0.8
0.275
600 0
0.2
0.4
0.6
0.8
1.2
1.4
1.6
1.8
2
1
1.2
1.4
1.6
1.8
2
Time [sek]
400
3000
200 Figure
2000
0
10000
1
7.5: Response to a pulsating chain force using configuration B.
0.2
0.4
The Shock00 Absorber
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
1.2
1.4
1.6
1.8
2
Time [sek]
0.6
0.8
1
Normal force [N]
Since the problems600with the mathematical model originates from implementation of the hydraulic
model, the effect of400changing characteristics of the shock absorber are not shown. Instead an interpretation of the current shock absorber is given. By plotting the equivalent damping as a function of
200
position and velocity
of the rear wheel, figure 7.6 is obtained.
From figure 7.6 it0 is seen that the shock absorber along with the linkage system perform relatively
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
high equivalent damping at low compression velocities
and low equivalent damping at high compresTime [sek]
sion velocities. The waves in the regions away from zero velocity are only present due to inaccuracy
from polynomial fitting. The curve should be more planar.
Pedalling is assumed to give a relatively slow response compared to riding over a bump. Bumps
on the road are in general considered to generate higher velocities of the rear wheel than pedalling.
From this the equivalent damping can be interpreted to have the following consequences on dynamic behaviour of the bike. The suspension system performs big resistance to compress at low
velocities, which is positive when pedalling as it prevents energy loss due to DISC and Bio-Pacing.
On the contrary the low resistance to compress at higher velocities increase the ability to absorb the
sudden impacts from the track.
Furthermore it is seen that the equivalent damping is higher during compression (wheel speed > 0)
than during rebound (wheel speed < 0). This is also seen as positive effect since this provides the bike
with a fast rebound which gives better grip.
Regarding the stiffness of the suspension system, its characteristics are shown in figure 7.7. As it
has not been possible to fully integrate it with the rest of the dynamic analysis, no further comments
are given to it, apart from the fact that it qualitatively develops as it is expected by an air spring.
68
CHAPTER 7. RESULTS OF THE DYNAMIC ANALYSIS
7.2. THE SHOCK ABSORBER
Figure 7.6: A plot of the equivalent damping as a function of rear wheel position and velocity.
4
7
x 10
Equivalent stiffness [N/m]
6
5
4
3
2
1
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
Wheel travel [m]
Figure 7.7: The stiffness of the suspension.
CHAPTER 7. RESULTS OF THE DYNAMIC ANALYSIS
69
7.3. RESPONSE TO FREQUENCY INPUT
7.3
Response to Frequency Input
When testing the response to a frequency input it has been necessary to remove the spring model derived in section 6.3 and apply a constant stiffness instead. The equivalent stiffness which was supposed
to be applied returns singularities when used in the frequency analysis. This might by caused by the
error described in section 6.4.
By applying the same frequency input to both rear and front wheel the plots in figure 7.8 and 7.9
are obtained. The plots are seen to be similar, and the amplification for A and B are determined to be
0.35
0.35
0.3
0.3
0.25
Rear wheel
Center of Mass
Displacement [m]
Displacement [m]
0.25
0.2
0.15
0.1
Rear wheel
Center of Mass
0.2
0.15
0.1
0.05
0.05
0
0
-0.05
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
-0.05
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time [sek]
Time [sek]
Figure 7.8: Displacement of the dropouts and of centre
of mass for configuration A.
Figure 7.9: Displacement of the dropouts and of centre
of mass for configuration B.
0.586 and 0.596, which means that configuration A possibly damps an uneven track slightly better than
configuration B. Concurrently it should be mentioned that the differencess are sufficiently small to be
caused by numerical inaccuracies.
The fact that the two configurations have almost the same response to a frequency input, is an
interesting result. It implies that by changing the amount of DISC or Bio-Pacing, the ability to damp
unevenness of the terrain is not affected remarkably. This might permit a less advanced optimisation
process since the two analysis can be optimised more or less independently.
Animations of the response to a frequency input for configuration A and B can be found on the CD
in appendix E under Movies » FrequencyA and FrequencyB.
7.4
Free Fall
This analysis has shown itself very sensitive, as it results in large compressions of the shock absorber,
which the mathematical model can not handle. Therefore it is only possible to model the free fall from
a height of 0.4 m. Hereby it is possible to obtain the displacement of the crank as shown in figure 7.10.
If the analysis is executed at initial height larger than 0.4m, the compression of the shock absorber
is larger than the maximum 50mm, which corresponds to a crank displacement of 100mm, and the
solver returns singularities.
From the free fall analysis it is obtained that the bars in the suspension system must be able to
withstand a shock force like the one shown in figure 7.11. However, since this result does not corre70
CHAPTER 7. RESULTS OF THE DYNAMIC ANALYSIS
7.4. FREE FALL
0.02
Displacement of crank [m]
0
-0.02
-0.04
-0.06
-0.08
-0.1
-0.12
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Time [sec]
Figure 7.10: A possible movement of the crank. This plot is obtained by scaling the equivalent stiffness and damping
5000
4500
4000
Shock force [N]
3500
3000
2500
2000
1500
1000
500
0
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Time [sec]
Figure 7.11: The force the shock absorber delivers during a free fall
spond to the requirements determined in chapter 4, where the free fall was said to be performed from
an initial height of 1m, therefore a different approach of estimating the forces exerted on the four bar
linkage is presented in chapter 8.
Animations of the response to a free fall for configuration A can be found on the CD in appendix
E under Movies » FreeFall1 and FreeFall2. FreeFall1 demonstrates a free fall from 0.4m. FreeFall2
demonstrates a free fall from 1m where the analysis fails.
CHAPTER 7. RESULTS OF THE DYNAMIC ANALYSIS
71
Strength & Dimensioning
8
This chapter describes how one of the bars of the suspension linkage can be designed and dimensioned
to withstand the loading conditions described in the previous chapters. The chapter consists of two
parts. In the first part the dimensioning of the bar is based upon classical beam theory. In the second
part the stresses in the same bar is calculated through FE modelling, and these results are compared
with the results from the classical beam theory. The scope for this chapter is not to do a thorough
dimensioning of the selected bar. As a consequence this chapter does not take loads from the ground
into consideration when dealing with fatigue. Instead the purpose is to demonstrate how a mechanical
part of a dynamic system can be isolated and then designed to carry loads that will run through it
when the system is used.
8.1
Dimensioning Based Upon Classical Beam Theory
This section is based on J. M. Gere [12], Danish standards for aluminium constructions DS419 [13],
and safety of structures DS409 [14]. The focus is concentrated on the bar, on which the wheel and
brake are mounted – link 3, cf. figure 8.1. This is because this bar is subjected to both normal force,
shear force, and bending moment. Torsion in the bar due to transverse loading on the wheel or due to
asymmetrical loading from the braking calibre on the bar is not taken into account.
8.1.1
Forces on Link 3
One of the purposes with the kinetic analysis of the suspension linkage in chapter 6 section 6.2 was to
determine the reaction forces in the joints, and use these as force inputs on the bars. Because of the
problems described in chapter 7, the kinetic analysis has not been able to produce valid reaction forces.
Instead the forces on link 3 have been estimated through a static approach. The consequence of this is
that the bar is designed to be strong enough to withstand the loads in case of static equilibrium. When
it comes to the bars ability to withstand the braking force and the force due to pedalling, the static
approach is considered not fare from the truth, if the suspension linkage is designed to be fully-active
as described in chapter 2 section 2.1. It is also assumed that the loading from the braking force and the
force from pedalling is not significant affected by the position of the wheel – i.e. whether the wheel is
in its uncompressed position or it is compressed due to the weight of the rider and bike.
Because link 3 is supported by a hinged bar in point B the reaction force on link 3 in point B has a
certain direction, α. α is found on figure 8.1 as the angel of the connection line between A and B and
global horizontal. In point C link 3 is supported by the rocker who is hinged to the mainframe in point
D and the shock absorber in point E. In case of static equilibrium of link 3 the air spring in the shock
absorber and the force in point E create a force in point C so that the sum of forces and moments on
link 3 are neutralised and the static equilibrium is reached. This assumption leads to the constraints
72
CHAPTER 8. STRENGTH & DIMENSIONING
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
~breake
F
C
~
F
P3
E
D
β
N
y
α
R
x
α
B
Figure 8.2: Because link 3 is supported by a hinged bar
in point B the reaction force on link 3 in point B has a
certain direction, α. In point C link 3 is supported by the
rocker which delivers a force so that the sum of forces
and moments on link 3 are neutralised. The x-y-frame is
the global frame.
A
P3
Figure 8.1: Link 2, 3, and 4 of the suspension linkage.
The bar on which the wheel and brake is mounted – link
3 – is selected for further examination.
on figure 8.2 and the free body diagram on figure 8.3. I.e. a hinge constraint in C and a rolling hinge
constraint in B. Here the forces are related to a local frame with its origin in point P3 with the ξ-axis
coinciding with the neutral axis of a prismatic beam which is connecting point P3 and point C. Note
that ~FP3 is the force from the wheel hub on link 3 – i.e. the sum of the chain force, the friction force
between the ground and tire due to the chain force, and the normal force from the ground on the wheel.
~breake
F
η
~breake
F
~
F
R’
~
F
P3
~rBR
~ C,η η̂
R
ξ
~rBP3
~rBC
~B
R
~ R’
M
R’
~ C,ξ ξ̂
R
α
~ R’
M
Figure 8.3: The forces on link 3 is refereed to a local
frame with its origin in P3 .
~
F
R’
Figure 8.4: The brake force acts on the beam through a
bracket mounted in point R’.
Σ~F = ~FP3 + ~Fbrake + ~RB + ~RC = ~0
~ B =~rBP3 _ ~FP3 +~rBR _ ~Fbrake +~rBC _ ~RC = ~0
ΣM
where:
R’
(8.1)
(8.2)
~FP3
The sum of chain force, friction force due to the chain force, and the normal force from the
ground on the wheel.
~Fbrake
The force from the braking disc on the braking calibre due to braking.
~RB
The reaction force from link 2 on link 3.
CHAPTER 8. STRENGTH & DIMENSIONING
73
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
~RC
The reaction force from link 4 on link 3.
~rXY
The vector from point X to point Y.
Equation 8.1 can be expanded to scalar equations:
FP3 ,ξ + Fbrake,ξ + RB,ξ + RC,ξ = 0
(8.3)
FP3 ,η + Fbrake,η + RB,η + RC,η = 0
(8.4)
As explained above ~RB must be parallel to the connection line between A and B. This means that:
rAB,η
RB,η
RB,ξ = rAB,ξ = tan(π − (α − β)) ⇒ RB,η = tan(π − (α − β))RB,ξ where β is found on figure 8.2 as the
angel from the global to the local frame and ~rAB is a vector from A to B. This can be substituted into
equation 8.4:
FP3 ,η + Fbrake,η + tan(π − (α − β))RB,ξ + RC,η = 0
(8.5)
Equation 8.2 can be rearranged as:
rBC,ξ RC,η − rBC,η RC,ξ = − ~rBP3 _ ~FP3 +~rBR _ ~Fbrake
(8.6)
Equation 8.3, 8.5, and 8.6 con now be arranged in a matrix equation:

RB,ξ + RC,ξ = − FP3 ,ξ + Fbrake,ξ 

tan(π − (α − β))RB,ξ + RC,η = − FP3 ,η + Fbrake,η
⇒ (8.7)

−rBC,η RC,ξ + rBC,ξ RC,η = − ~rBP3 _ ~FP3 +~rBR _ ~Fbrake 

 


− FP3 ,ξ + Fbrake,ξ 1
1
0
RB,ξ


 tan(π − (α − β))
0
1   RC,ξ  =  − FP3 ,η + Fbrake,η
⇒ (8.8)

0
−rBC,η rBC,ξ
RC,η
− ~rBP3 _ ~FP3 +~rBR _ ~Fbrake


 
−1 
− FP3 ,ξ + Fbrake,ξ RB,ξ
1
1
0


 RC,ξ  =  tan(π − (α − β))
0
1   − FP3 ,η + Fbrake,η
(8.9)

~
~
0
−rBC,η rBC,ξ
RC,η
− ~rBP3 _ FP3 +~rBR _ Fbrake
By use of matrix equation 8.9 the reaction forces on link 3 in point B and C can be calculated. Now all
external forces on link 3 are known and thus it is possible to determine the internal forces in link 3.
8.1.2
Internal Forces
The prismatic beam between point P3 and point C is assumed to follow the classical beam theory and
will therefore be dimensioned by use of this. This beam consists of two sections. One section between
point P3 and the point where the braking calibre bracket is mounted and another section between the
braking calibre bracket and point C. Section I can be seen on figure 8.5 where the internal forces ~N
74
CHAPTER 8. STRENGTH & DIMENSIONING
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
~
F
R’
η
η
~
F
P3
~
F
P3
~ M
~
V
~
N
~rP3B
ξ
~ξ
~ M
~
V
~
N
ξ
~rP3B
~B
R
Figure 8.5: Section I: Between P3 and R’ the forces ~N
~ equilibrate ~FP3 and
and ~V and the bending moment M
~RB .
~ R’
M
~rP3R’
~ξ
~B
R
Figure 8.6: Section II: Between R’ and C the forces ~N
~ equilibrate ~FP3 , ~RB ,
and ~V and the bending moment M
~
~
0
0
−FR , and −MR .
~ equilibrate ~FP3 and ~RB . Through Newton’s II law the internal forces
and ~V and the bending moment M
and the bending moment in this section are found to be:
NI (ξ)
= − ~FP3 + ~RB
(8.10)
VI (ξ)
NI (ξ)
~
~
MI (ξ) = − ~rP3 B _ RB + ξ _
(8.11)
VI (ξ)
Equation 8.11 is found by taking the moment about point P3 . In the section between the braking
calibre bracket and point C the braking force must be taking into account. Figure 8.4 shows how ~Fbrake
~ R0 which are acting from the beam on the braking calibre bracket through the
is balanced by ~FR0 and M
~ R0
point R’. Using Newton’s III law this means that the braking calibre bracket acts with −~FR0 and −M
on the beam through point R’. How this affects section II is seen on figure 8.6. The internal forces and
the bending moment in this section is then found to be:
NII (ξ)
= − ~FP3 + ~RB + (−FR0 )
(8.12)
VII (ξ)
N
(ξ)
II
~
~ R0 | + ξ _
MII (ξ) = − ~rP3 B _ ~RB +~rP3 R0 _ −~FR0 + −|M
(8.13)
VII (ξ)
Remark that
8.1.3
~
d M(ξ)
dξ
= −V (ξ) due to the sign convention of ~N, ~V , and M used at figure 8.5 and 8.6.
Internal Forces at Different Loading Conditions
In chapter 4 section 4.4 the loading conditions for the bike are summarised. On figure 8.7 equation
x
~
8.10, 8.11, 8.12, and 8.13 are plotted where the maximum pedal moment is applied. Here ξ =
0
and N and V are equal to the normal force and shear force. Note that equation 8.10 and 8.11 are used
when 0 ≤ x < rP3 R0 ,ξ and equation 8.12 and 8.13 is used when rP3 R0 ,ξ ≤ x ≤ rP3C,ξ . Also note that the
suspension linkage consists of a set of two link 3’s – i.e. one on each side of the wheel. Consequently
the loads from the wheel hub are distributed into two bars while the braking calibre is place on only
one of the bars. Here the bar with the braking calibre mounted on is dimensioned. Figure 8.8 shows the
CHAPTER 8. STRENGTH & DIMENSIONING
75
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
same plot, but instead of pedal moment the brake is activated. A combination of maximum moment
and braking is found on figure 8.9.
600
Normal Force
Shear Force
Bending Moment
200
100
200
0
[N] - [Nm]
[N] - [Nm]
Normal Force
Shear Force
Bending Moment
400
-100
0
-200
-400
-200
-600
-300
-800
-400
0
0.05
0.1
0.15
0.2
x[m]
0.25
0.3
0.35
Figure 8.7: Forces and bending moment in the beam
when pedalling at maximum moment.
0
0.05
0.1
0.15
0.2
x[m]
0.25
0.3
0.35
Figure 8.8: Forces and bending moment in the beam
when the brake is activated.
As explained in chapter 7 section 6.6 the mathematical model can not handle a 1 m free fall. Instead the force from the ground on the wheel during the impact is estimated through a few rough
assumptions. During the fall the mechanical energy is assumed to be conserved. By this assumption
the impact velocity can be determined exactly as in 6.90 by:
vf =
where:
i
Initial.
f
Final.
v
Velocity.
h
Hight.
p
2ghi
(8.14)
During the impact the mechanical energy is again assumed to be conserved by converting kinetic
energy into potential energy in the spring. This spring can be imagined as a equivalent spring acting
directly on the wheel. It is also assumed that the equivalent spring is linear and is able to absorb all the
kinetic energy during the wheel travel.
∆Ekin + ∆Espring = ∆Emec = 0
1 1 2
m v f − v2i + k t 2f − ti2 = 0
2
2
1 2 1 2
− mvi + kt f = 0
2
2
mv2
k = 2i
tf
76
⇒
⇒
⇒
(8.15)
CHAPTER 8. STRENGTH & DIMENSIONING
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
where:
k
The spring stiffness.
t
The wheel travel.
By inserting equation 8.14 into equation 8.15 the spring stiffness can be calculated. Note that the
final velocity in equation 8.15 is equal to the initial velocity in equation 8.14.
2
√
m 2ghi
2mghi
k=
=
(8.16)
2
tf
t 2f
In the end of the impact the equivalent spring is compressed to maximum wheel travel. Then the force
from the ground on the wheel is:
Ff ree f all =
where:
m
= 85 kg.
g
= 9.82 sm2 .
h
= 1 m.
t
= 0.1 m.
2mgh
2mgh
t=
= 16694N
2
t
t
(8.17)
This force from the equivalent spring is expected to be much larger than what is found under a real 1 m
fall on a bike. In reality when a rider makes a 1 m fall on a bike, the rider is expected to be standing
in the pedals. Then the force from the equivalent spring is limited by the strength of the riders legs.
When the force from the equivalent spring closes the strength of the riders legs they will begin to collapse. Because the legs are the primary connection between the bike and the rest of the body, which
represents a large amount of the entire mass, the collapse of the legs can be seen as an increment of t
in equation 8.17. Even though the above force is larger than expected it will be used later on to proof
the ultimate strength of the beam. The internal forces an the bending moment in the beam at the end
of the impact after the free fall are plotted in figure 8.10. Here it is important to notice that the bending
moment has opposite sign compared to the bending moment under the other loading conditions. This
means that under the impact after the free fall the beam experience compression in the part under the
ξ-axis from both the normal force and the bending moment.
8.1.4
Ultimate strength of the beam
To ensure a satisfying safety against yielding the following criterion must be satisfied:
σ0d ≤ fy,d
fy,k
γ f σ0k ≤
γm
where:
σ0
The Von Mises stress.
fy
The yield strength.
CHAPTER 8. STRENGTH & DIMENSIONING
⇒
(8.18)
77
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
Normal Force
Shear Force
Bending Moment
600
400
0
200
-2000
[N] - [Nm]
0
[N] - [Nm]
Normal Force
Shear Force
Bending Moment
2000
-200
-4000
-6000
-400
-8000
-600
-10000
-800
-1000
0
0.05
0.1
0.15
0.2
x[m]
0.25
0.3
0.35
Figure 8.9: Forces and bending moment in the beam
when pedalling at maximum moment and the brake is
activated at the same time.
γ
The safety factor.
d
Corrected.
k
Characteristic.
-12000
0
0.05
0.1
0.15
0.2
x[m]
0.25
0.3
0.35
Figure 8.10: Forces and bending moment in the beam
at the end of the impact after a 1 m fall landing on the
back wheel alone.
According to DS419 and DS409 a normal degree of safety is reached when [14] [13]:
1.3σ0k ≤
fy,k
1.17
(8.19)
This criterion must be satisfied in all the length of the beam and in any point in the cross-sectional
area. Because the beam has to carry bending around the ζ-axis (the axis normal to ξ and η) and torsion
when the wheel of the bike is loaded transversely (transverse load is not taken further into account
here) a closed hollow cross-section like the one at figure 8.11 is selected. As inputs to Von Mises yield
criterion normal stresses are calculated by use of:
N(ξ)
A
M(ξ)h
σξ,M = −
I
σξ = σN ξ, N + σM ξ, M
σξ,N =
where:
78
σ
The normal stress.
A
The cross-sectional area.
h
The distance from the neutral axis to where the stress is to be found.
I
The moment of inertia.
(8.20)
(8.21)
(8.22)
CHAPTER 8. STRENGTH & DIMENSIONING
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
When calculating the stresses from the shearing forces from the previous section, they are found to
be very small. The shearing forces are caused to that neglected in the further calculations. Therefore
the stresses in the beam are only calculated in the top or the bottom of the cross-section where the
shear stresses are always zero. These points are marked at figure 8.11 as a and b.
20
η
c
ξ
ζ
16
26
a
P3
C
b
15
welding
Figure 8.11: The dimensions of the cross-section are in
mm.
Figure 8.12: The beam is welded in three areas.
In the beginning of this chapter it is stated that the Danish standard for aluminium constructions
is used to insure a satisfying safety of the bike. It is chosen to use the 7005 aluminium alloy as
construction material for the beam. This has been chosen because it is strong compared to other
aluminium alloys, extrudeable, and is suitable for welding. The yielding strength for this alloy is
290 MPa.
The strength of aluminium decreases due to welding – so called HAZ. As indicated on figure 8.12
the beam is welded right after point P3 along the braking calibre bracket and right before point C.
Around these weldings fy,d has to be multiplied by 0.8 if the weldings are made by use of a TIGwelder, cf. [13].
On figure 8.13 the corrected yielding criterion is tested throughout the length of the beam, when
the maximum moment is applied, cf. 8.7. On figure 8.14 a combination of maximum moment and
braking is applied. In both scenarios the bending moment and normal force creates compression in
point a on figure 8.11. The bending moment due to the 1 m free fall is as earlier mentioned negative
which creates compression in point b on figure 8.11. Therefore the yielding criterion has been tested
in this point on figure 8.15. In these three scenarios, it is obvious that the yielding criterion is fulfilled.
8.1.5
Fatigue strength of the beam
In chapter 4 section 4.4 the riding conditions for the bike are stated. A satisfying safety against fatigue
failure under these conditions is reached by using Palmgren-Miner’s formula:
ni
∑ n f at,i ≤ 1.0
where:
n
The number of loads.
n f at
The number of loads at the actual stress range leading to fatigue failure.
(8.23)
The relation between the number of loads and the stress range is determined by the Wöhler curve
which can be found in figure 8.18 and is in DS419 expressed as [13]:
CHAPTER 8. STRENGTH & DIMENSIONING
79
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
300
300
Von Mises Stress
Stress Limit
Von Mises Stress
Stress Limit
250
Stress [MPa]
Stress [MPa]
250
200
150
200
150
100
100
50
50
0
0
0
0
0.05
0.1
0.15
0.2
x [m]
0.25
0.3
Figure 8.13: Corrected Von Mises stresses when the
maximum moment is applied.
300
0.1
Von Mises Stress
Stress Limit
0.25
0.3
0.35
Fatigue in Upper Rear Bar
Fatigue Limit
Partial Fatigue
1
150
0.8
0.6
100
0.4
50
0.2
0
0
0.05
0.1
0.15
0.2
x [m]
0.25
0.3
0
0
0.35
Figure 8.15: Corrected Von Mises stress’s when the
back wheel hits the ground after an 1 m fall.
0.05
0.1
log(a)−m·log(σv )
n = 10
n=
0.15
0.2
x [m]
0.25
0.3
0.35
Figure 8.16: Fatigue in the beam after approximately
375 hours of riding.
log(n) = log(a) − m · log(σv )
⇒
log(a)
= 10
10
−m·log(σv )
aσ−m
v
⇒
(8.24)
a
The material constant found in DS419.
m
The material constant found in DS419.
σv
The stress range.
This can be inserted into equation 8.23 cf. figure 8.18:
aσd, f at,i −m
σv,d,i m
∑ aσv,d,i −m = ∑ σd, f at,i ≤ 1.0
80
0.2
x [m]
1.2
200
where:
0.15
Figure 8.14: Corrected Von Mises stresses when pedalling at maximum moment and the brake is activated at
the same time.
250
Stress [MPa]
0.05
0.35
(8.25)
CHAPTER 8. STRENGTH & DIMENSIONING
8.1. DIMENSIONING BASED UPON CLASSICAL BEAM THEORY
Fatigue in Upper Rear Bar
Fatigue Limit
1.2
Partial Fatigue
1
0.8
σfat
0.6
0.4
σfat,i
0.2
σv,i
0
0
0.05
0.1
0.15
0.2
x [m]
0.25
0.3
0.35
Figure 8.17: Fatigue in the beam after approximately
2,400 hours of riding when the brake is not contributing
to the fatigue.
ni
nfat,i
n
Figure 8.18: A sketch of a Wöhler curve.
Under the actual loading conditions equation 8.25 can be written as:
∑
σv,d,i
σd, f at,i

m
σv,d,braking mbraking
σv,d,mean mmean
σv,d,max mmax
=
+
+
σd, f at,mean
σd, f at,max
σd, f at,braking
mbraking
mmean 
mmax 
 γm γ f σv,k,mean 
= 
−mmean 
a
 γm γ f σv,k,max 
+
−mmax 
nmean
where:
a
nmax
mean
The mean moment.
max
The maximum moment.
 γm γ f σv,k,braking 
+
−mbraking 
≤ 1.0
(8.26)
a
nbraking
The values of a and m depend on whether the fatigue criterion is tested close to a stress concentration or on a smooth surface. In figure 8.16 the fatigue criterion is tested throughout the beam after 1.8
million pedal strokes at mean moment, 1,250 pedal strokes at maximum moment, and 15,000 activations of the brake, cf. section 4.4. Here γm = 1.43 and γ f = 1.3 which ensures a normal safety when the
inaccuracy of the stress range is about 30 % according to DS419 and DS409. This gives about 375 h
of riding time or almost 10, 000 km when riding at a average speed at 20 km/h. The welding along
the braking calibre bracket is placed on the cross-sectional area around point c in figure 8.11. This is
discussed further in the next section. Consequently the stress range used in equation 8.26 is calculated
at point c but still without including shear stresses because of the neglectable shear forces. It can be
seen that the fatigue is much more pronounced around the weldings than throughout the pure beam
(cf. figure 8.16). This means that fatigue in the beam will occur first right after the welding between
the braking calibre bracket and the beam.
CHAPTER 8. STRENGTH & DIMENSIONING
81
8.2. WELDINGS
8.1.6
Summary
The ultimate strength of the beam is proven to be more than satisfying. Even under the heavy load
from free fall the yield criterion is not close to be broken (cf. figure 8.15). When it comes to fatigue
the lifetime is about 375 h which is expected to be a relatively short time. If riding two hours each
week the bike is tired up in less than four years – and here the incessant loading from the tails in the
forest such as roots and stones are not taken into account. Compared to manufactures of full suspension
mountain bikes, the outer dimensions of the cross-sectional area of the discussed beam are smaller than
the dimensions presented in this chapter. If the baking force’s contribution to the fatigue is excluded
then the fatigue in the beam is as illustrated on figure 8.17. But here the number of pedal strokes is
11.5 million at mean pedal moment and the number of strokes at maximum moment is set equivalent
according to section 4.4. This indicates that the way the braking calibre bracket is mounted on the bar
is not favourable. A redesign of the connection between the braking calibre and the beam can properly
reduce the dimensions of the beam but according the intro of this chapter this is beyond the scope.
8.2
Weldings
In the following the strength of the welding joint between the break bracket and the upper rear bar is
demonstrated. This is only demonstrated for the weld throat1 since the weld root 2 has been treated in
section 8.1. The strength is demonstrated against both static and fatigue failure.
The rear braking calibre is attached to the frame by a bracket welded on to the top of the upper rear bar.
The bracket is load bearing and positioned on a loaded member, so the strength of the joint needs to be
properly demonstrated. This will be done in accordance with DS 419 [13]. As can be seen from figure
h1
h2
~F
~F
h3
Weld throat
l
y
z
Vertical
part of
wall
y
x
x
z
Figure 8.19: Schematic drawing of welding.
8.19, the bracket is welded on to the upper half of an oval-shaped tube. The force ~F has non-zero x
1 From
2 From
82
Danish: ’sømsnit’
Danish: ’tåsnit’
CHAPTER 8. STRENGTH & DIMENSIONING
8.2. WELDINGS
and y components and creates bending around all three axes.
As can be seen from the section view in figure 8.19 (right), the bottom width of the bracket has
been adjusted such that the weldings are positioned directly above the vertical part of the tube wall.
This configuration offers the best transfer of bending loads around the x and z axes from the bracket to
the tube.
Even though the welding may look like a butt weld3 , it will be treated as an fillet weld4 as a
conservative estimate.
The direction of the weld throat is indicated on figure 8.19.
8.2.1
Stresses In the Weld Throat
The weld throat will be dimensioned against static failure. The load originates from the braking calibre
and the highest possible braking force will be used.
Figure 8.20 shows the weld throat section with the braking force transferred to the centroid of the
welding and the resulting moments added. The subscripts on the moment Mm,F indicate the axis m of
the moment caused by the force ~F.
The stresses from each force and moment component have been indicated on the figure where the
subscripts in σm,n and τm,n indicate the tensor component m and the source n. A comment should be
made about the moment about the z-axis, Mz,F . One could be tempted to believe that Mz,F acts in the
opposite direction than in figure 8.20. This is not the case since h3 is clearly larger than h1 and because
of the angle of ~F, the moment will act about the z-axis as illustrated in figure 8.20.
l
τ
τ
yx,Fx
yx,MyF
a
σ
y,MzF
σ
y,Fy
σ
2
h4
y,MxF
MzF
~F
y
M zF
MyF
x
~F
MxF
x
y
MxF
z
MyF
z
σ
y,MzF
σ
1
y,MxF
A
τ
σ
y,Fy
τ
yx,Fx
yx,MyF
B
Figure 8.20: Schematic drawing of weld throat section.
The normal stress σy,Fy from the normal force Fy is assumed evenly distributed on the full weld
throat area:
3 From
4 From
Danish: ’stumpsøm’
Danish: ’kantsøm’
CHAPTER 8. STRENGTH & DIMENSIONING
83
8.2. WELDINGS
σy,Fy =
where:
Fy
2al
(8.27)
σy,Fy
The normal stress on the weld throat from the force ~F.
Fy
The y component of force ~F.
a
The weld throat thickness.
l
The length of weld.
The following sign convention is used when calculation the stresses according to figure 8.20. If the
stress component acts in the positive y-direction, the sign is positive and vice versa. The normal stress
σy,Mx,F and σy,Mz,F from the moments around the x and z axes are found using the flexure formula
σy,Mx,F (z) = −
where:
Mz,F · x
Mx,F · z
and σy,Mz,F (x) =
Ix
Iz
σy,Mx,F
The normal stress on the weld throat from the moment Mx,F .
z
The z coordinate on the weld throat section.
Ix
The moment of inertia of the weld throat around the x axis.
σy,Mz,F
The normal stress on the weld throat from the moment Mz,F .
x
The x coordinate on the weld throat section.
Iz
The moment of inertia of the weld throat around the z axis.
(8.28)
The shear stress τyx,Fx from the force component Fx is found using the shear formula:
τyx,Fx (x) =
where:
Fx Qz (x)
2 Iz h4
(8.29)
τyx,Fx
The shear stress on the weld throat from the force component Fx .
x
The x coordinate on the weld throat section.
Qz
The first moment of the weld throat area above x with z as the neutral axis.
The length between the centroid of the weld throat and the outer edge of the weld throat in the
z direction. See figure 8.20.
h4
The full expressions for Qz (x) and Im can be found in appendix D. Finally, the shear stress τyx,My,F
from the moment My,F in-plane with the weld throat can be calculated using the following expression
[15]:
τyx
i ,Mw =
Mw h
i
k
∑ hj
2
(8.30)
A
j
j=1
84
CHAPTER 8. STRENGTH & DIMENSIONING
8.2. WELDINGS
where:
τyx
i ,M
The shear stress on the weld throat on the ith weld from the moment Mw .
Mw
The moment at the centroid of the weld.
h
i
The perpendicular distance between the centroid and the outer edge of the ith weld throat.
k
The number of welds carrying the load.
h
j
The perpendicular distance between the centroid and the outer edge of the jth weld throat.
A
j
The area of the jth weld throat.
In this particular situation with the joint consisting of two welds symmetric about the x and z axes,
the stresses τyx
1 ,My,F and τyx
2 ,My,F on figure 8.20 can be calculated as:
τyx
1 ,My,F = −τyx
2 ,My,F
My,F h4 − a2
My,F
=
=
a
a 2
2 h4 − 2 l a 2 h4 − 2 l a
(8.31)
As can be seen from figure 8.20, the direction of the moments Mx,F and Mz,F both lead to negative
normal stresses in the second half of weld .
1 Since σy,Fy is negative, the largest total normal stress
must be found in weld 1 at point A shown in figure 8.20.
Weld 1 also experiences the largest shear stress, because τyx,Fx and τyx
1 ,My,F have the same direction. Since τyx
1 ,My,F is evenly distributed on the throat area and τyx,Fx recedes towards the ends of the
welding, the maximum shear stress is found at the centre of the weld at line .
B
8.2.2
Static Failure Criterion
In accordance with DS 419, the strength of the weld throat must be calculated as [13]:
fuwd =
where:
fuw
= 182 MPa
γm,weld
fuwd
The corrected ultimate tensile strength of the weld material.
fuw
The tabulated ultimate strength set to fuw = 260 MPa for the chosen weld material.
γm,weld
The material correction factor set to γm,weld = 1.43 for correction of tensile strength.
(8.32)
The standard stipulates that a correction factor γ f must be applied to the loading condition. This is
done by multiplying the original braking force with the value γ f = 1.3 found in 8.1.4, thus correct~ Weld class I5 has been selected for this particular
ing both the braking force ~F and the moment M.
welding.
Then the following criterion from DS 419 is used to demonstrate the strength of the welding against
static failure at each point on the weld throat [13]:
q
σy 2 + 3 τyx 2 + τyz 2 ≤ c0 fuwd
5 From
(8.33)
Danish: ’sømklasse’
CHAPTER 8. STRENGTH & DIMENSIONING
85
8.2. WELDINGS
where:
c0
The strength reduction factor set to c0 = 1.0 because weld class I is used.
fuwd
The tensile strength of weld material as determined in equation 8.32.
Calculations show that the left hand side of criterion 8.33 attains its largest value at point A shown in
figure 8.20. Here the failure criterion in 8.33 yields:
q
σy,Fy + σy,Mx,F + σy,Mz,F
2
+ 3 τyx,Fx + τyx,My,F
2
≤ c0 fuwd
30 MPa ≤ 182 MPa
(8.34)
Since this inequality holds true for all areas of the welding, the strength of the joint against static
failure has been properly demonstrated.
8.2.3
Fatigue Failure Criterion
According to DS 419, the endurance limit σi, f at,d,weld of the weld throat can be determined as [13]:
σi, f at,d,weld =
where:
σi, f at,weld
γm,weld
(8.35)
σi, f at,d,weld The corrected endurance limit for stress components σi and τyi .
i
The stress direction.
σi, f at,weld The endurance limit found in DS 419.
γm,weld
The material correction factor for correction of endurance limit.
Here, the value of σi, f at,weld is determined according to the stress direction i = x or i = y, notch
specification6 , weld class, and number of cycles n f ,weld . Both σx, f at,weld and σy, f at,weld are needed,
but only σy, f at,weld can be determined with the correct notch specification. Since DS 419 does not
provide a notch specification for fillet weldings with shear loading parallel to the weld, an alternative
notch specification number of 21 is chosen as recommended by Ole Ø. Mouritsen [15]. This notch
specification applies to fillet weldings loaded perpendicular to the weld, but since the weld throat lies
at a 45 degree angle to load, it will experience shear stresses. Therefore, it is assumed that the loading
situation for notch specification 21 is somewhat similar to the actual load on the weld and thus may be
a reasonable estimate. With this assumption, the endurance limits are now found to be:
σx, f at,d,weld = σy, f at,d,weld =
where:
The endurance limit for notch specification 21, weld class I and n
86
(8.36)
≈ 10.000 cycles set to
f ,weld
σy, f at,weld σ
y, f at,weld = 45 MPa. n f ,weld is the number of times the rider applies the brake.
γm,weld
6 Form
σy, f at,weld
= 31 MPa
γm,weld
The material correction factor set to γm,weld = 1.43 for correction of endurance limit.
Danish: ’kærvanvisning’
CHAPTER 8. STRENGTH & DIMENSIONING
8.2. WELDINGS
Since the welding is only loaded when the break is applied and γ f = 1.3 as in the static case, the
stress ranges σy,v,weld and τy,v,weld in the weld are equal to the maximum static stresses and can be
calculated as:
where:
σy,v,weld = σy,Fy + σy,Mx,F + σy,Mz,F
(8.37)
τyx,v,weld = τyx,Fx + τyx,My,F
(8.38)
σy,v,weld
The normal stress range on the weld throat in the y direction.
τyx,v,weld
The shear stress range on the weld throat in the x direction.
Finally, the fatigue failure criterion for the weld throat requires that:
σy,v,weld
m
σy, f at,d,weld
where:
m
m
τyx,v,weld
+ 1.3
≤ 1.0
σx, f at,d,weld
(8.39)
The slope of the fatigue curve for notch specification 21 and weld class I set to m = 3.2.
Calculations show that the left hand side of criterion (8.39) attains its largest value at point a shown
in figure 8.20. Here the failure criterion in (8.33) yields:
σy,v,weld
σy, f at,d,weld
m
m
τyx,v,weld
+ 1.3
≤ 1.0
σx, f at,d,weld
0.35 ≤ 1.0
(8.40)
Since this inequality holds true for all areas of the welding, the strength of the joint against fatigue
failure has been properly demonstrated.
8.2.4
General Consideration
The static loading condition is based on the largest possible braking force. This may not actually be
the largest possible load on the bracket, which could alternatively occur during a crash. To find a
better static loading condition, more attention should be given to determining the actual conditions
and risks during use. Furthermore, the estimated notch specification for the shear stresses in the weld
throat should be examined further. This could be done by comparing the stresses caused by the actual
loading condition to those assumed in the notch specification using FE analysis. Alternatively, the
strength of the weld against fatigue failure could be demonstrated using a FE analysis.
The number of loading cycles used in the fatigue failure criterion accounts for heavy braking
only. Equation (8.40) shows that the strength of the welding far exceeds the required fatigue strength.
Still, a damage contribution from light braking should be added and the proper fatigue strength be
demonstrated using Palmgren-Miner’s linear damage theorem as describe in DS 419 [13].
CHAPTER 8. STRENGTH & DIMENSIONING
87
8.3. FINITE ELEMENT METHOD
8.3
Finite Element Method
This section deals with the use of the Finite Element Method (FEM). Through this chapter the application of FEM on the upper rear bar and a shim in the shock absorber will be described. The focus will
mainly be the description of: Constraints, material, choice of mesh, applied forces, and interpretation
of results. The results from the FE analysis will be compared with the results found through BernoulliEuler beam theory. As an introduction a brief description of the fundamentals of FE simulation is
given. All stresses presented in this chapter are von Mises stresses.
8.3.1
FEM Outline
When dealing with structures of complex geometry the Bernoulli-Euler beam theory often becomes
inadequate. By dividing the structure into small elements limited by nodes, a continuum can be discretised as seen in figure 8.21. From the displacement of each element it is possible to compute the
stresses and strains for the complete structure.
P
Figure 8.21: A continuum divided into discrete elements [16, p.13].
Since the material used for the construction is assumed to be linear elastic there is a linear relation
between the forces on the nodes and the displacements of the nodes. This can for each element be
formulated as [16, p.23]:
~p = k̄¯ · d~
where:
~p
The local force vector containing the forces acting on each node.
d~
k̄¯
The local displacement vector containing the displacements of each node in the element.
(8.41)
The stiffness matrix for the element.
In the use of FEM the stiffness matrix for each element is generated. When the stiffness matrices
for each element are found, it can be expanded into the global stiffness matrix describing the entire
structure. Since the relation between the displacement and the force is linear, the global stiffness matrix can be constructed for the entire continuum by superposing the extended stiffness matrices from
88
CHAPTER 8. STRENGTH & DIMENSIONING
8.3. FINITE ELEMENT METHOD
each element. When the global stiffness matrix is constructed a global connection between the applied
force and the displacement of the continuum can be established. This can be expressed analogeously
to equation 8.41, just using global notation [16, p.37]:
~P = K̄¯ · ~D
(8.42)
The obstacle to overcome before constructing the global stiffness matrix is to construct the stiffness
matrices for the discrete elements. The stiffness matrix for a single element is based on [16, p.25]:
Wp = Welast
where:
Wp
The work done by the external forces acting on the element.
Welast
The elastic energy stored in the element.
(8.43)
The work done by the force on the element is the work from a spring force, i.e.
n
1
Wp = ∑ ~pi · d~i
i=1 2
where:
n
(8.44)
The number of nodes in the element.
The elastic energy stored in the element is given by (in tensor notation):
Z
Welast =
V
where:
σi j
The stress tensor for the element.
εi j
The strain tensor for the element.
V
The volume of the element.
1
σi j εi j dV
2
(8.45)
Substituting equations (8.44) and (8.45) into equation (8.43) gives an expression in ~p and d~ (the right
side of the equation is given in tensor notation):
~p · d~ =
Z
σi j εi j dV
(8.46)
V
However, it is not possible to isolate the stiffness matrix directly from this equation. If the displacement
functions are assumed linear and the change of volume negligible, applying Hooke’s law to equation
(8.46) can convert it into the form of equation 8.41. Hereby the stiffness matrix can be identified.
When all stiffness matrices are constructed and have been converted into global notation, one of
three components in equation 8.42 are known. Thus, if another component is specified by the user
the last component can be found. Typically one specifies the applied forces and the displacements are
CHAPTER 8. STRENGTH & DIMENSIONING
89
8.3. FINITE ELEMENT METHOD
determined. The strains are easily found from the displacements using Cauchys definition of strain
(using tensor notation):
εi j = ui, j + u j,i
(8.47)
From this the stresses can be found by the use of Hooke’s law.
8.3.2
Upper Rear Bar
To verify the calculations in section 8.1, the upper rear bar is modelled in SolidWorks and simulated
in COSMOSWorks.
The upper rear bar is restrained by two hinges as shown in figure 8.22. A force from the wheel and
a force from the braking calibre is applied to the bar. The forces on the bar are applied as in section
8.1 and the loading comes from mean performance, cf. table 4.1, together with braking. The restraints
and the forces on the bar can also be seen in figure 8.24.
~FBrake
~FChain
Figure 8.22: The restrains applied in the FE model.
The mesh used is a three dimensional solid mesh of high quality in COSMOSWorks, with an element size of 2.96 mm. According the COSMOSWorks documentation high quality solid mesh generates 10-node parabolic tetrahedral solid elements, whereas a draft quality solid mesh generates 4-node
linear solid elements. More tangible the 10-node tetrahedral element is a pyramid with a node at each
corner and a node at the midpoint of each edge. The advantages of parabolic elements are that for
the same number of elements they represent a curved surface to a greater extent and generate better
mathematical approximations than the linear elements.
The material used is aluminium 7005 which has to be configured manually. The yield strength is
Sy = 290 MPa and the modulus of elasticity is E = 72 GPa [17].
90
CHAPTER 8. STRENGTH & DIMENSIONING
8.3. FINITE ELEMENT METHOD
Verification of FEM Through Bernoulli-Euler Beam Theory
At first the upper rear bar is modelled without the braking calibre and braking force. It is discovered,
that there is a clear difference between the stresses found through the FE analysis and the manual calculations. In figure 8.23 the stresses on the surface of the upper rear bar found through FE analysis are
plotted. It is assumed that the maximum stresses occur on the surface. The topmost line indicates the
stresses found by the manual calculations. It is seen that the stresses found in the manual calculations
are twice as big as the stresses from the FE analysis. The red line shows the stresses if only half of the
loads are applied. Here there is a clear resemblance. This entails that there is some kind of error in
either one of the analysis. This error has not been found.
Figure 8.23: The results from the manual calculations (seen as the blue line), the manual calculations divided by two (seen
as the red line) and the stresses on the surface found in the FE analysis.
Because the focus in this chapter solely is on the upper rear bar, the braking calibre is not fully
analysed in the FE model. The braking calibre is drawn as a solid box which ensures that the braking
force is applied at the correct distance and angle from the upper rear bar.
The largest stresses occur two places; right above the welding that holds the braking calibre to
the bar and at the joint on the Horst link. The characteristic stresses are approximately 110 MPa at the
braking calibre and 198 MPa at the Horst link. The stresses are indicates in figure 8.25. For comparison
the stresses from the FE analysis found in different cross sections are plotted as corrected stresses (i.e.
the stresses found in the FE analysis is multiplied with γ f before they are plotted). The plot is seen in
figure 8.26.
The difference between the results from the manual calculations and the FE analysis is obvious in
figure 8.26. By dividing the stresses found through manual calculations by two, the line will instead
be located close to the line illustrating the stresses found through the FE analysis. Another difference
in the curves is seen at the location of the braking calibre. Here the area moment of inertia of the beam
increases locally which decrease the stresses locally.
The difference between the two plots is so big that they barely can be compared. However, the
tendency is the same. The location where the largest state of stress occurs, found in the FE analysis, is seen as the topmost point in figure 8.26. The stress in this point is an expression of a stress
concentration factor caused by the abrupt ending of the braking calibre.
CHAPTER 8. STRENGTH & DIMENSIONING
91
8.3. FINITE ELEMENT METHOD
110 Mpa
198 Mpa
Figure 8.24: The applied forces and restrains on the
upper rear bar.
Figure 8.25: A closer look at the most stressed areas
on the upper rear bar. The greatest stresses occur at the
Horst link and right above the braking calibre.
150
Stress [MPa]
100
50
0
0
0.05
0.1
0.15
0.2
x [m]
0.25
0.3
0.35
0.4
Figure 8.26: The dots mark the maximum stresses in the cross section from the FE analysis and the line marks the stresses
found by manual calculations.
8.3.3
Deflection of Shim in the Shock Absorber
As described in section 6.3, the deflection of the shims must be determined. For this purpose the
FEM is applied to a shim. The shim is constructed in AISI 1020 steel which has the yield strength
Sy = 351.6 MPa and the modulus of elasticity E = 200 GPa.
The inner surface of the shim is fixed in rotation as well as in translation in the (x, y)-plane, cf.
figure 8.27. The spring force grows linearly with the displacement in the z-direction. The plate is able
to move freely in vertical direction as seen in figure 8.28. The loads of 2 N, 5 N, 7 N, and 10 N are
applied perpendicularly to the plate below the shim one after another. The deflections from the FEM
simulation are listed in table 8.1 where the deflection is measured at the periphery of the plate. The
deflection corresponding to a load of 10 N can be seen in figure 8.29.
92
CHAPTER 8. STRENGTH & DIMENSIONING
8.3. FINITE ELEMENT METHOD
Inner surface
Spring force
z
x
Applied force
y
Applied force
Figure 8.27: The forces working on the plate and on the shim.
Figure 8.28: The displacement of the plate which is
used for calculation of orifice area.
Figure 8.29: Deflection of the shim when applying 10 N
to the plate below.
Concluding Remarks
It has through application of the FEM been possible to find the deflection needed for calculation of an
orifice area in the shock absorber model. These results have not been verified through any other calculations and the correctness of these results are therefore unknown. Since the geometry and restrains
are very simple the results are considered reasonable.
Force [N]
2
5
7
10
Deflection [mm]
3.66 · 10−2
9.15 · 10−2
12.8 · 10−2
18.3 · 10−2
Table 8.1: The deflection of the plate.
CHAPTER 8. STRENGTH & DIMENSIONING
93
Discussions & Conclusions
9
The chapter discusses the results of the preceding chapters and whether the Problem Statement has
been fulfilled. Thus this text has its foundation in the Problem Statement in chapter 3
Loading Scenarios for the Bike
The following loading scenarios have been examined:
• Free fall
• Braking Force
• Pedalforce
• Operational Conditions
During the study of the free fall it is assumed that the entire mass of the bike and rider is fixed to the
main frame of the bike as a rigid body. It is also assumed that all the energy from the fall is absorbed
by the rear suspension. Both assumptions contribute to a large load on the rear wheel and is properly
far beyond the actual load. As broached in chapter 8 8.1.3 a part of the kinetic energy from the fall
is absorbed by rider’s body even though the bike is fully suspended. This will soften out the impact
which reduces the load on the suspension system. When landing on the rear wheel after a fall the force
from the ground on the wheel will instantaneously cause the bike to rotate and the front wheel will
soon after hit the ground. The rotation and the fact that the front wheel hits the ground, relieves the
rear wheel from absorbing all the kinetic energy through it.
The braking force is determined in that case where the magnitude of the friction force is equal to the
product of the coefficient of static friction and the normal force on the wheel. I.e. when applying the
maximum amount of braking which does not cause the wheel to slip. The coefficient of static friction
is chosen to correspond dry friction between rubber and asphalt. This is a conservative simplification
of the real range of environments that a mountain bike is used in. In many of these environments the
friction is lower than friction between rubber and asphalt. But it is also possible, that the friction is
higher in some of the environments. I.e. when riding on hard frozen soil or gravel it might is possible
that the friction exceeds friction the one on asphalt. The assumption, that the braking force is largest
the instant before the wheel blocks, may also be wrong. This is due to the fact that the change in
rotational inertia, caused by blocking a spinning wheel, may cause an even higher braking force.
The pedal force is determined by use of the biomechanical simulation software AnyBody. The
pedal forces obtained by AnyBody does not seem to be unrealistic and is expected to be reachable
for a athletic 75 kg person. But whether the mean crank moment is instantaneously reachable over a
longer period of time is uncertain.
The operational conditions for the bike is based on assumptions about how a ride in Danish terrain
proceeds. The loads from the ground during such a ride are not taken into account.
94
CHAPTER 9. DISCUSSIONS & CONCLUSIONS
Common to all the loads on the bike, including those above, is that they are difficult to predict
without doing actual experiments. Relevant experiments could include strain gauge measurements to
determine strains in different places of the bike and under different riding conditions. Relevant experiments also include acceleration measurements i.e. on both the bike and the rider during the impact
after a free fall. When designing a mountain bike such experiments are expected to be necessary.
Mathematical Model Describing the Dynamics of the Full Suspension Mountain Bike
The kinematic model has been developed using a method by Nikravesh [1]. This method has proven
itself successful. As it can be seen on figure 5.1 in chapter 5 the kinematic model has been successfully
used when determining kinetic response, frequency response, equivalent spring stiffness and damping,
and used in analysis of free fall. Note that the problems described in chapter 7—i.e. when analysing
the free fall—are not expected to be caused by the kinematic model. These problems are believed to
be linked to solving the equations in the shock absorber model. They are also expected to be caused
by errors in combining the nonlinear behaviour of both the kinematic model and the shock absorber
model.
As explained in chapter 7 the kinetic model has been difficult to control and has turned out to be
very sensitive when applying external forces on the four bar linkage such as chain force or braking
force. Some of these problems are expected to be caused by errors in the program code.
The shock absorber is a complex system of a combined air/nitrogen spring and a viscous damper.
The compression of both the air spring and the nitrogen string are assumed to be isothermal. But because of the fast compressions when the shock absorber is used in practise, an assumption of adiabatic
compression may be more accurate.
Both the kinematic model, kinetic model, and the model of the shock absorber use the numerical
Newton-Raphson solver to handle their nonlinear behaviour. To understand the problems with the two
last mentioned models, a solver using an adaptable time step might have been useful. An adaptable
time step can provide a deeper insight in the behaviour of the system when it gets unstable and accelerations grow rapidly. Of course the time step in the Newton-Raphson solver can be decreased but this
increases the simulation time.
If a complete model of the mountain bike’s dynamics was desired, this could be done by including
the front suspension in the kinetic analysis. This corresponds to modelling the entire bike by the
Nikravesh approach. By reconsidering the model’s types of inputs, it would be able to replace both the
frequency and free fall analyses, and the equivalent stiffness and damping would also be unnecessary.
An option would also be to involve tires in the analysis, initially just modelled as linear springs and
viscous dampers.
This report demonstrates how a mathematical model, describing a mechanical system where the
dynamical aspects are the main issues, can be established. When designing a full suspended mountain
bike this model can be used. To get reliable results from the model, valid inputs must be determined
through experimental data. Especially the loadings on the bike are difficult to estimate without experimental data.
CHAPTER 9. DISCUSSIONS & CONCLUSIONS
95
Investigation of the System Response
Due to the time consuming obstacles during the implementation of the mathematical models and the
interaction between their components, the available project time did not leave room for optimisation
on the design of the suspension linkage—i.e. length of members, positions of the kinematic joints, and
placement of the shock absorber. However, the foundation for such optimisation has been made.
Dimensioning
Link 3 in the four bar linkage is dimensioned according to DS419 and DS409. By Danish law it is
required to follow Danish Standards (DS) when a failure in a construction has the potential to be life
threatening or entail great economical costs for society. Even though a bike is not imposed by such
law, DS is still applied since it is believed to give satisfying safety without unnecessary strength. The
lifespan of the examined bar is short, despite the outer dimensions, which are larger than what is seen
on other bikes. As explained in section 8.1.6 this is due to the construction of the braking calibre
bracket on the bar.
The FE model of the examined bar cannot be verified by classical beam theory. In section 8.3 it
can be seen that the stresses obtained by beam theory are about twice as large as the stresses obtained
by the FE model. The reason for this has not been found.
96
CHAPTER 9. DISCUSSIONS & CONCLUSIONS
Nomenclature
Latin Letters
A
Ā¯
Area.
i
Cd
Rotation matrix for frame i.
Discharge coefficient, based on the area of vena contracta, the area of the
physical orifice and an empirical factor.
ceq
The equivalent damping.
cshock
D̄¯
The damping coefficient of the shock absorber.
d
êchain
Corrected, when used as subscript.
Local displacement vector containing the displacements of each node in the
element.
Unity vector in direction of the chain (with positive x-component).
êground
Unity vector in direction of the ground (with positive x-component).
Fairspring
Applied force on the piston.
~Fbrake
Force from the braking disc on the braking calibre due to braking.
d~
Jacobian matrix.
Fdamp,static Static damper force.
~Ff riction
Friction force between rear wheel and ground.
fi
Applied force on link i.
fuw
The tabulated ultimate strength.
fuwd
Corrected ultimate tensile strength of the weld material.
fy
The yield strength.
~
G
Gravity force.
g
The gravitational acceleration.
CHAPTER 9. DISCUSSIONS & CONCLUSIONS
97
98
~g
The sum of all applied forces. Note that the force vector contains both single
forces and force couples, i.e. moments.
~gc
The constraint forces.
~gext
The external forces.
h
Height.
I
Moment of inertia.
Ix
Moment of inertia around the x axis.
Iz
Moment of inertia around the z axis.
Jbike
The mass moment of inertia of the sprung mass.
Ji
The mass moment of inertia of link i about the ζi -axis.
k
Spring stiffness.
k
k̄¯
Characteristic, when used as subscript.
keq
The equivalent stiffness.
kshock
The stiffness of the shock absorber.
l
Length.
M
M̄¯
Moment.
Mbike
The sprung mass.
m
Mass.
N
Magnitude of normal force.
~N f ront
The normal force on the front wheel.
~Nrear
The normal force on the rear wheel.
p
Pressure.
~p
The local force vector containing the forces acting on each node.
Q
Flow.
~q
~q˙
Position vector.
R
The ideal gas constant.
Stiffness matrix for the element.
Diagonal mass matrix diag[mi , mi , Ji ], i = {1, 2, 3, 4}.
Time derivative of the position vector, i.e. the velocity vector.
CHAPTER 9. DISCUSSIONS & CONCLUSIONS
Re
The Reynolds number of the flow.
Ri j
The j-component of the reaction force in point i.
r
Radius.
~ri
Global vector from the global origin to the origin of frame i.
~rXY
Vector from point X to point Y.
0j
~si
Local vector from the origin of frame i to point j.
~sXi
Global vector from the origin of frame i to point X.
T
Temperature.
Ti
Moment at point i.
V
Volume.
v
Velocity.
~v j
Eigenvectors corresponding to λ j .
Welast
Elastic energy.
Wp
Work done by external forces.
X̃ j
Amplitudes for x p (t).
ẋ0
Initial velocity of the equivalent mass.
xshock
Displacement/compression of the shock absorber.
ẋshock
Compression velocity of the shock absorber.
xwheel
Displacement of the rear wheel.
ẋwheel
Velocity of the rear wheel.
Y jk
Amplitudes.
y1 (t)
Input function for the rear wheel.
y2 (t)
Input function for the front wheel.
Z jk
Impedances, the actual expression for Z jk can be seen in appendix A.
CHAPTER 9. DISCUSSIONS & CONCLUSIONS
99
Greek Letters
~α˙
Time derivative of the constraint equations.
γ
Safety factor.
γm,weld
Material correction factor for correction of endurance limit.
εi j
Strain tensor for the element.
Θ̃ j
Amplitudes for θ p (t).
θ p (t)
Function rotating the sprung mass related to the particular solution.
λj
~λ
Eigenvalues.
µdynamic
Coefficient of kinematic friction.
µstatic
Coefficient of static friction.
ρ
Density of the fluid.
σ
Normal stress.
σ0
Von Mises stress.
σv
Stress range.
σi, f at
Endurance limit found in DS 419.
σi, f at,d
Corrected endurance limit for stress components σi and τyi .
σi j
Stress tensor.
τ
Reaction moment.
τyx
Shear stress.
~Φ
¯
Φ̄
Constraint vector.
Jacobian matrix.
φj
Phases.
Ψ
The left side of equation 6.20.
ψ
Approximated derivative of Ψ.
ωj
Angular velocities.
~q
Lagrange multipliers~λ = [λ1 , λ2 , ..., λn ]T .
Other Symbols
_
100
The planar cross product operator.
[0, 0, 1]T ∀ ~a,~b ∈ R2 .
~a _ ~b ≡
[~aT , 0]T × [~bT , 0]T ·
CHAPTER 9. DISCUSSIONS & CONCLUSIONS
Bibliography
[1] Nikravesh PE. Computer-Aided Analysis of Mechanical Systems. Prentice Hall; 1987.
[2] Titus Cycles. Full Suspension Unveiled; 2006.
[3] Strandgaard E, Jespersgaard P, Østergaard OG. Databog fysik kemi. F & K Forlaget; 2001.
[4] Ball, Burrows, Sargeant. Human power output during repeated sprint cycle exercise: the influence of thermal stress. Eur J Appl Physiol Occup Physiol. 1999 Mar;(79 (4)):360–6.
[5] http://en.wikipedia.org/wiki/Human-powered_transport; Visited 18-05-2009.
[6] http://en.wikipedia.org/wiki/Petawatt#cite_note-3; Visited on 18-05-2009.
[7] http://www.anybody.aau.dk; Visited 18-05-2009.
[8] Fox Racing Shox. Fox Float RP23 shock. http://www.foxracingshox.com/bike/09/shocks/FLOAT;
Visited 18-05-2009.
[9] Black Art Designs. http://www.blackartdesigns.com/BAD-How it works.html; Visited 18-052009.
[10] Hansen MR, Andersen TO. Fluid Power Circuits, System Design and Analysis. Compendium
from Aalborg University; 2004.
[11] Rao SS. Mechanical Vibrations. 4th ed. Pearson Prentice Hall; 2004.
[12] Gere JM. Mechanics of Materials. 6th ed. Thomson Canada Limited; 2006. ISBN: 0495073075.
[13] Dansk Standard. Norm for aluminiumskonstruktioner. 3rd ed.; 2001.
[14] Dansk Standard. Norm for sikkerhedsbestemmelser for konstruktioner. 2nd ed.; 1998.
[15] Mouritsen OØ. Svejste konstruktioner (kursus); 2009.
[16] Mouritsen OØ. Introduktion til Elementmetoden. 2nd ed. Institut for Maskinteknik, AAU; 2005.
[17] MatWeb Material Property Data.
http://www.matweb.com/search/DataSheet.aspx?MatGUID=34c308934f7a4be589a80ecbee94406e;
Visited 18-05-2009.
BIBLIOGRAPHY
101
.
APPENDIX
102
BIBLIOGRAPHY
Additional Calculations to
Frequency Response
A
The impedances introduced in section 6.5 are presented here:
Z11 = −Mbike ω21 + (c1 + c2 )iω1 + (k1 + k2 )
Z12 = (c2 l2 − c1 l1 )iω1 + (k2 l2 − k1 l1 )
Z21 = (c2 l2 − c1 l1 )iω1 + (k2 l2 − k1 l1 )
Z22 = −Jbike ω1 + (c1 l1 + c2 l2 )iw1 + (k1 l12 + k2 l22 )
Z33 = −Mbike ω22 + (c1 + c2 )iω2 + (k1 + k2 )
Z34 = (c2 l2 − c1 l1 )iω2 + (k2 l2 − k1 l1 )
Z43 = (c2 l2 − c1 l1 )iω2 + (k2 l2 − k1 l1 )
Z44 = −Jbike ω22 + (c1 l1 + c2 l2 )iw2 + (k1 l12 + k2 l22 )
APPENDIX A. ADDITIONAL CALCULATIONS TO FREQUENCY RESPONSE
103
B
Solving the Equation of Motion
Describing the Free Fall
Here the three different instances of the roots R to the characteristic equation are presented.
x(t) = C1 eR1t +C2 eR2t
(B.1)
The
p constants C1 and C2 are determined be means of initial conditions. The initial velocity is ẋ0 =
2gh f cf. equation (6.90). For t = 0 the initial position is x0 = 0, thus:
x(0) = C1 eR1 ·0 +C2 eR2 ·0 = 0
p
⇒
(B.2)
ẋ(0) = C1 R1 eR1 ·0 +C2 R2 eR2 ·0 = 2gh f
0
1 1
C1
= p
(B.3)
R1 R2
C2
2gh f
−1 0
C1
1 1
p
(B.4)
=
C2
R1 R2
2gh f
For the case of repeated roots the solution is:
x(t) = C1 eR1t +C2teR1t
ẋ(t) = C1 R1 eR1t +C2 eR1t +C2tR1 eR1t
x(0) = C1 = 0
p
ẋ(0) = C1 R1 +C2 = 2gh f
p
C2 = 2gh f
⇒
(B.5)
⇒
(B.6)
(B.7)
i.e. the actual solution for repeated roots is:
x(t) = C2teR1t =
p
2gh f teR1t
(B.8)
The solution for complex roots given the initial conditions is:
x(t) = eat (C1 cos(bt) +C2 sin(bt))
ẋ(t) = eat ((aC1 + bC2 )cos(bt) + (aC2 − bC1 )sin(bt))
x(0) = C1 = 0p
ẋ(0) = bC2 = 2gh f
where:
a
The real part.
b
The imaginary part.
⇒
(B.9)
(B.10)
Thus the actual solution for complex roots is:
p
2gh f at
e sin(bt)
x(t) = C2 e sin(bt) =
b
at
104
(B.11)
APPENDIX B. SOLVING THE EQUATION OF MOTION DESCRIBING THE FREE FALL
r
r
V z +ΔV z
r
Vz
r
Δx
C
x
O
r
r
My +ΔMy
r
M
Sign Convention
y
z
The following sign convention is used when determining the internal forces and bending moments in
a beam section:
• Internal forces and moments are listed in positive direction on section faces with a positive
normal vector.
• As a consequence the internal forces and moments are listed in negative direction on section
faces with a negative normal vector.
This sign convention leads to minor deviations from [12]:
Relationship Between My and Vy
y
r
r
Vy +ΔVy
r
Vy
r
Δx
O
r
Mz
x
r
r
Mz +ΔMz
Figure C.1
Moment equilibrium:
∑ M~ O = −M~ z +
~ z + ∆M
~ z + ∆~x × V
~y + ∆V
~y
M
~ z + ∆~x × V
~y + ∆~x × ∆V
~y
= ∆M
= ~0
Now, let ∆~x → 0:
~ z + d~x × V
~y = ~0
dM
~ z + dxx̂ ×Vy ŷ = ~0
dM
~ z + dxVy ẑ = ~0
dM
~z
dM
= −Vy ẑ
dx
APPENDIX C. SIGN CONVENTION
⇒
⇒
⇒
(C.1)
105
Relationship Between My and Vz
r
r
V z +ΔV z
r
Vz
r
Δx
O
r
My
x
r
r
My +ΔMy
z
Figure C.2
Moment equilibrium:
∑ M~ O = −M~ y +
~ y + ∆M
~ y + ∆~x × V
~z + ∆V
~z
M
~ y + ∆~x × V
~z + ∆~x × ∆V
~z
= ∆M
= ~0
Now, let ∆~x → 0:
y~ y + d~x × V~z = ~0 ⇒
dM
~ y + dxx̂ ×Vz ẑ =r~0 ⇒ r
dM
r
V~y0+⇒
ΔVy
~ y + dxVz (−ŷ) =
V
dM
dM
r ~ y = Vz ŷ
Δ xdx
r
Mz
106
O
(C.2)
x
r
r
Mz +ΔMz
APPENDIX C. SIGN CONVENTION
D
Welding
Here the first moment of area and moment of inertia for the welding are presented.
The first moment of area about the z-axis is:
l
Qz = 2a
−x
2
l
Qz = 2a
+x
2
l
2
−x
+x
2
!
l
2
+x
−x
2
!
,
l
≥x≥0
2
,
0≥x≥−
(D.1)
l
2
(D.2)
The moment of inertia about the x- and z-axis are:
l (2h4 )3 l (2(h4 − a))3
−
12
12
2h4 l 3 2 (h4 − a) l 3
Iz =
−
12
12
Ix =
APPENDIX D. WELDING
(D.3)
(D.4)
107
CD
108
E
APPENDIX E. CD
Sketch of the Entire Bike
APPENDIX F. SKETCH OF THE ENTIRE BIKE
F
109
10
4x
2,0 Thru
4x
2,0 Thru
A
45°
19
A
Ø 25
0,2 A
Ø 21,5
0,3 A
8,75 0,05
Aalborg University
Unless otherwise specified
1. Tolerance on: x is 0.2
x.x is 0.1
Title:
M. sector
Gr.: 53b
6th Semester
Scale: 3:1
Shock piston
6
Matr.: EN AW - 6012
Drawn By:
53b
Drw. no.:
1
PROJECTION
2. Angular Tolerance: 0.5
DIMENSION: MM
0,02 A
6 H7
Inner edges: 0.4 mm fillet
Outer edges: 0.1 mm champfer
1,5
A
8,25 0,05
2,5
1,0
5,0
SECTION A-A
File name: shockpiston
Date:
25/05-2009
Sheet:
1/1
BOM Table
ITEM NO.
PART NAME
DESCRIPTION
QTY.
1
shockpistonrod
1
2
shockpiston
1
3
propedalglider
1
4
propedalshim
1
5
reboundglider
1
6
damperspring
2
7
pistonneedle
1
8
pistonneedlepin
1
9
needlspring
1
10
shockpiston_endcap
1
1
7
9
8
6
4
Aalborg University
Title:
Dimensions in MM 3rd Angle Projection
3
2
M. sector
6th semester
Assembly
Filename: exploded view
5
6
Gr.: 53b
10
Matr.: EN AW - 6012
Date: 25/05-2009
Drawn by:
53b
Drw. no.:
2