Analysis of a Dakar rally truck using a multibody model


 Sophia Bennett
 2 years ago
 Views:
Transcription
1 Analysis of a Dakar rally truck using a multibody model J.A.M. Hopmans DCT Traineeship report Supervisor: Dr. Ir. I.J.M. Besselink Technische Universiteit Eindhoven Department Mechanical Engineering Dynamics and Control Technology Group Eindhoven, January, 2008
2 2
3 Acknowledgements I would like to thank Dr. Ir. I.J.M. Besselink, Eindhoven University of Technology, for his help and solid support. I am grateful to him for giving me the opportunity to work on this project. I extend my sincere thanks to Theo de Ruiter, engineer of DAF, for his enormous effort, time and giving detailed information. I have furthermore to thank Team de Rooy for their time, information and giving the opportunity to validate the model in the future. Apart from my own knowledge, I have relied on two previous reports on this subject: [1] M. Pinxteren, Development of a multibody model of a Dakar Rally truck with independent suspension and [2] G.R. Siau and T.L. Spijkers, Development of a multibody simulation model of the DAF Dakar Rally truck.
4 ii Acknowledgements
5 Abstract Two studies w.r.t. the Dakar rally truck have been completed in the last 2 years ( [1], [2]). These studies have both been finished after the rally. This year the study has been finished before the prototype vehicle has been completed. Therefore, the results obtained can still be taken along in the development of the actual rally truck. A highly simplified multibody model is the basis of this report. The reason for this, is the complexity of the previous models. Models rapidly become too complex which requires too much development time. Factors which influence the vehicle behavior may be difficult to trace. By applying a simplified model it is possible to analyze the influence of factors such as stiffnesses and ride heights and to give a well substantiated recommendation w.r.t. these factors. With the simplified model simulations have been carried out, using the parameters of the 2006, 2007 and the 2008 truck. The largest differences between these trucks can be found in the configuration of the suspension, the mass of the body, the mass of the axles, static axle loads and the stiffnesses. The 2006 and 2008 trucks are equipped with rigid axles. The 2007 truck has been equipped with an independent suspension. Due to the lower unsprung mass and stiffnesses of the 2007 truck a more desirable pitch behavior, when driving over the bump, has been reached. This remained not unnoticed at team de Rooy; the truck was described as a flying carpet. Due to a change in the regulations the mass of the 2008 truck has been increased. The minimum (empty) mass is now 8500 kg. Ultimately a total vehicle mass of 9300 kg is applicable for the loaded vehicle configuration. The influence of this increase can be seen in the pitch behavior. The positive peak becomes lower. By analyzing the simulation time histories of the 2008 truck in detail and describing the characteristic points, a good understanding of e.g. the pitch behavior is reached. The motion of the truck has been visualized. From this analysis a number of requirements for tuning the truck can be derived. Ultimately, the 2008 truck has been improved using these requirements. The following order of steps results from a conversation with Mr. de Ruiter: reducing the stiffness at the front and rear axle, increasing the ride height and raising the damping coefficient. The ride height has been raised to avoid unnecessary contact with the bumpstops. This combination of changes results in e.g. a much better pitch behavior and faster tyre/road contact after taking the bump. To develop the multibody model further it is necessary to get more specific information about the dampers and the fast rebound system. In this model linear dampers are used. The damping coefficient at the front equals Ns/m. The rear has a value of Ns/m (75 % of the front).
6 iv Abstract To validate the model in a simple way, a number of measure methods have been described. With the information obtained the model can be developed further.
7 Samenvatting In de afgelopen 2 jaar zijn er 2 studies met betrekking tot de Dakar rally truck afgerond. Deze studies zijn beide ná de in het betreffende jaar verreden rally voltooid. Dit jaar is de studie afgerond voordat de laatste hand aan het prototype is gelegd. De verkregen resultaten kunnen zodoende nog in de rally truck worden meegenomen. Aan de basis van dit rapport staat een sterk vereenvoudigd multibody model. De reden waarom hiervoor is gekozen moet worden gezocht in de complexiteit van de voorgaande modellen. Modellen worden al snel te complex waardoor de ontwikkelingstijd vaak nodeloos toeneemt. Het traceren van invloedsfactoren met betrekking tot het voertuiggedrag kan dan gecompliceerd worden. Door het toepassen van een vereenvoudigd model kunnen invloedsfactoren, zoals stijfheden en rijhoogten, goed worden geanalyseerd en kan er zodoende een goed beargumenteerd advies worden uitgebracht. Op basis van de 2006, 2007 en de 2008 truck paramaters zijn diverse simulaties uitgevoerd. De grootste verschillen tussen deze trucks zijn terug te vinden in de wielophanging configuraties, de massa van de body, de massa van de assen, statische aslasten en de veerstijfheden. De 2006 en 2008 truck zijn uitgevoerd met starre assen. De 2007 truck heeft een onafhankelijke wielophanging. De 2007 truck laat door de lage onafgeveerde massa en lage veerstijfheden een zeer wenselijk pitch gedrag zien wanneer over de gedefinieerde hobbel wordt gereden. Dit bleef ook niet onopgemerkt bij team de Rooy. De truck werd als een vliegend tapijt omschreven. Door veranderingen in de reglementen van de Dakar rally is de massa van de 2008 truck verhoogd. De minimale ledige massa bedraagt 8500 kg. Uiteindelijk is een totale voertuigmassa van 9300 kg toegepast. De invloed van deze verhoging is direct terug te zien in het pitch gedrag. De positive piek wordt lager. Door de simulatie resultaten van de 2008 truck uit te vergroten en belangrijke punten in deze grafieken te beschrijven is een goed inzicht ontstaan ten aanzien van o.a. het pitch gedrag. De volledige beweging is per punt gevisualiseerd. Uit deze analyse zijn een aantal voorwaarden naar voren gekomen om de wielophanging van de truck te tunen. Met deze voorwaarden is de 2008 truck uiteindelijk verbeterd. Uit conversaties met de heer de Ruiter is de volgende reeks van stappen naar voren gekomen: het verlagen van de veerstijfheid aan voor en achterzijde, het verhogen van de rijhoogte en het verhogen van de dempingscoëfficiënt. De rijhoogte is verhoogd om onnodig contact met de bumpstops te voorkomen. Deze combinatie heeft als resultaat dat het pitch gedrag zichtbaar is verbeterd en dat de 2008 truck eerder band/wegdek contact maakt na het nemen van de hobbel. Om het multibody model verder te ontwikkelen is het aan te bevelen om meer gedetailleerdere informatie omtrent de dempers en het Fast Rebound systeem te vergaren. In dit model zijn lineaire dempers gebruikt met aan de voorzijde een dempingscoëfficiënt van N s/m en aan de achter
8 vi Samenvatting zijde van Ns/m (75 % van de voorzijde). Om het model op een eenvoudige wijze te valideren, zijn een aantal meetmethodes in dit verslag opgenomen. Met behulp van deze informatie kan het model verder worden ontwikkeld.
9
10 viii List of symbols List of symbols Symbol Description Unit a zf Vertical suspension deflection acceleration front [m/s 2 ] a zr Vertical suspension deflection acceleration rear [m/s 2 ] b df Vertical tyre damping front [Ns/m] b dr Vertical tyre damping rear [Ns/m] F df Damper force front axle [N] F dr Damper force rear axle [N] F sf Spring force front axle [N] F sr Spring force rear axle [N] F zf Vertical tyre force at the front [N] F zr Vertical tyre force at the rear [N] h cg Height Centre of Gravity [m] h fa Height centre front axle [m] h ra Height centre rear axle [m] IR Installation Ratio [ ] k sf Spring stiffness front [N/mm] k sr Spring stiffness rear [N/mm] k s Spring stiffness [N/mm] k tf Vertical tyre stiffness front [N/mm] k tr Vertical tyre stiffness rear [N/mm] k w Stiffness at the wheel [N/mm] m urod Mass upper rod [kg] m lrod Mass lower rod [kg] m hub Mass hub [kg] m wheel Mass rim + tyre [kg] m plf Preload front axle [kg] m plr Preload rear axle [kg] m f Mass axle front [kg] m r Mass axle rear [kg] m lf Load front axle [kg] m lr Load rear axle [kg] m unspr Unsprung mass [kg] m body Sprung mass (body) [kg] rh f Ride height front [m] rh r Ride height rear [m] tw Track width [m] v zf Vertical suspension deflection velocity front [m/s] v zr Vertical suspension deflection velocity rear [m/s] wb Wheelbase [m] x cg Longitudinal distance Centre of Gravity w.r.t. front axle [m] z zf Suspension deflection front [m] z zr Suspension deflection rear [m]
11 Contents Acknowledgements Abstract Samenvatting List of symbols i iii v vii 1. Introduction Background/motivation and objective Outline of the report The simulation model General description Vehicle parameters Comparison of the 2006, 2007 and 2008 trucks Driving over a bump Comparison of the 2006, 2007 and 2008 truck Tuning the suspension of the 2008 truck Introduction Spring front/rear configuration Ride height front/rear configuration Damper front/rear configuration Final configuration Conclusions and Recommendations 29 Bibliography 33 Appendix A Vehicle dimensions and parameters 35 Appendix B Output signals from the simulation model 37 Appendix C Simulation results 2008 baseline configuration 39
12 x CONTENTS Appendix D Axle and wheel inertia 45 Appendix E Tuning results 2008 vehicle 47
13 1. Introduction 1.1 Background/motivation and objective Team De Rooy is famous for their long term participation in the prestigious ParisDakar rally. This rally is considered to be one of the toughest rallies in the world. The Dakar Rally (or simply "The Dakar"; formerly known as "The Paris Dakar Rally" and now as "The Lisbon Dakar Rally") is an annual offroad race, organised by the Amaury Sport Organization. The race is open to amateur and professional entries; amateurs typically make up about eighty percent of the participants. The first race was organized in 1978, a year after racer Thierry Sabine got lost in the desert and decided this would be a good location for a regular rally. Despite its name, it is an offroad endurance race rather than a conventional rally the terrain and the competitors traverse is much tougher and the vehicles used are true offroad vehicles rather than the modified sedans used in rallies. Most of the competitive specials are offroad, crossing dunes, mud, camel grass, rocks, among others. The distances of each stage covered vary from several kilometers to several hundred kilometers per day. The three major competitive classes of the Dakar are: motorcycles, automobiles (ranging from buggies to small trucks) and full size Trucks ("T4", "Camions" or "Lorries"). Many vehicle manufacturers exploit the harsh environment the rally offers as a testing ground, and consequently to demonstrate the durability of their vehicles, although most vehicles are heavily modified. After every edition of the Dakar rally team de Rooy evaluates the last rally to see what can be improved. This is done in close cooperation with engineers at DAF. In the last 2 years students had the chance to participate at this stage. In 2006, 2 TU/e students, G.R. Siau and T.L. Spijkers [2], developed a multibody model of the 2006 DAF rally truck which was used to study different rally truck configurations. In 2007 the GINAF rally truck was equipped with independent axles, so a new model had to be made. This model was developed by M. Pinxteren [1]. These 2 studies where completed after the real rally trucks had been built. This time the model has been made before the real rally truck has been finished. This gives the opportunity to give some advice, which are related to the suspension of the truck. Objective: The objective of the assignment is to optimize the suspension of a Dakar rally truck w.r.t. the springs, dampers and ride height when driving over a predefined bump. The optimalization has to be carried out by using a simplified multibody simulation model.
14 2 1. Introduction 1.2 Outline of the report Chapter 2 discusses the simplified model. This discussion will consist of a general description of the model, as well as a description of the relevant vehicle parameters such as the coordinate system used and the springdamper system. In chapter 3 simulations will be done with the 2006, 2007, as well as the 2008 truck configuration after which a comparison between the three will be made. This chapter also contains a very extensive explanation on the results of the 2008 model. In chapter 4 the final configuration is at issue. This chapter describes a parameter variation for optimizing the vehicle behavior. Following parameters will be discussed here: stiffness, ride height and damping. Finally, conclusions will be drawn and recommendations will be made, which should help team de Rooy setting up their suspension system. Also some remarks will be made concerning the validation of the 2008 model. A number of measure methods are described here.
15 2. The simulation model This chapter discusses the model used for this report. It starts with a general description of the model. Subsequently, the used axis system and vehicle parameters will pass the review. Finally, the implemented fast rebound system will be described. 2.1 General description The model used in this report is simplified w.r.t. the previous SimMechanics models. The models of 2006 and 2007 have been built using the data of the trucks. In these models the suspension has been modeled more in detail. At this stage, this was also required because the kinematic suspension behavior was also studied. The simplified model has been built using the following components: truck chassis, (m body ); dummy chassis, (m 1 and m 2 ); front/rear axle, (m f and m r ); spring/damper combination, (k sf, k sr, b df and b dr ); tyres, (k tf and k tr ). Figure 2.1 is a 2D schematic representation of the 2008 truck: mbody mbody m1 m2 ksf bdf ksr bdr ksf bdf ksr bdr mf mr mf mr ktf ktr ktf ktr WB (a) Rigid axles WB (b) Independent suspension Figure 2.1: 1/2 truck models
16 4 2. The simulation model The sprung mass includes several components such as the loading space (storage for material), cabin and of course the chassis itself. The cabin and loading space are rigidly connected to the chassis. The sprung mass is indicated with m body. In the model, the sprung mass is connected to the ground using a custom joint. The dummy chassis is a simple solution for dealing with rigid axles or an independent suspension. The mass of the dummy chassis, (m 1 and m 2 ), equals zero when we deal with rigid axles. In case of an independent suspension the dummy chassis represents among other things the weight of the differential. In the model, the dummy chassis is rigidly, by a weld, connected to the body. Both axles have one spring (k s ) damper (b d ) combination. This means that four springdamper combinations, for one axle, have been replaced by an equivalent springdamper combination. This simplification may be carried out, because only symmetric obstacles are simulated. Each axle contains bumpstops and rebounds (lower bumpstops). The rally truck has four tyres, this means no double rear is used. This is done because the load of the rear tyres is not as high as with normal trucks and unwanted materials such as stones won t get stuck between the tyres. An illustration in virtual reality can be seen in figure 2.2: Figure 2.2: Rallytruck in virtual reality. 2.2 Vehicle parameters Axis system: The multibody model of the Dakar rally truck is constructed in a three dimensional space. The axis system represents a "righthanded" coordinate system. Definition of the coordinate system: positive xaxis pointing forward; positive yaxis pointing to the left; positive zaxis normal to the plane, pointing upwards. The origin of the axis system is located at the front axle, see figure 2.3 (and 2.4): x= 0 centre front axle; y= 0 plane of symmetry; z= 0 road level.
17 2.2 Vehicle parameters 5 X Z Front axle Rear axle Y Figure 2.3: Used axis system. Vehicle dimensions: Several paramaters are important when developing the model, see figure 2.4. CG CG stroke ksf bdf hcg ksf bdf ksr bdr Z Z X Y hfa X Y xcg tw wb (a) Front view (b) Side view Figure 2.4: Important measures The dimensions can be found in appendix A. Again the location of the coordinate system has been drawn. The concerning figure shows how the spring damper combinations are placed in the model. Vehicle mass: With respect to the mass there is a clear distinction between rigid axles and the independent suspension. This becomes clear when table A.1 in appendix A is examined. The total vehicle mass of the 2008 truck is 800 kg higher than the 2006 truck, because of the new regulations. This can be seen also in the higher sprung mass. Obviously, the load has been increased at the rear tires. This means that the position of the Centre of Gravity (CG) has been moved to the rear. This is, among other things, caused by placing the engine to the rear. This is allowed within a range of 0 to 40 cm w.r.t. last year.
18 6 2. The simulation model One can see, the unsprung mass (m unspr ) of the independent suspension is much lower compared to the rigid axles. The unsprung mass of an independent suspension has been calculated as follows: m unspr = m wheel + m hub (m urod + m lrod ) (2.1) Inertia of the sprung mass: The model consists of multiple bodies with different inertia values. In appendix D a calculation of the axle inertia has been incorporated. An empirical formula has been used for the body inertia I yy : 0.8 (m plf x 2 cg + m plr (wb x cg ) 2 ) (2.2) m plf = m lf m f 2m wheel m plr = m lr m r 2m wheel In figure 2.5 one can see how this formula has been developed. mbody mplf mplr ksf bdf ksr bdr mf mr ktf ktr xcg wb Figure 2.5: Explanation empirical formula. The sprung mass (m body ) has been represented by two point masses (m lf, m lr ) which are located at the front and rear. When the masses are positioned exactly above the axles the following holds for the inertia around the yaxis of CG: m plf x 2 cg + m plr (wb x cg ) 2 (2.3) Because the masses are not exactly above the axles, see dashed circles, formula 2.3 has been multiplied by factor 0.8. From measurements it must become clear if this is a good estimation.
19 2.2 Vehicle parameters 7 Spring and damper characteristics: Here, the spring and damper characteristics are discussed. It must be noted that in contrast to the spring characteristics, the damper characteristics are not fully known. First, the spring characteristics will be covered. Deriving from the spring stiffnesses at the front and rear axles the information from appendix A can be taken. In the case of the 2006 and 2008 trucks the stiffness per wheel (k w ) can be calculated as follows: k w = 2 k scf + k ls Calculation of the stiffness at the wheels of the 2007 truck: (2.4) k w = 2 k scf IR 2 (2.5) For the rear the same calculations can be made. Over the years the stiffness at the wheels has decreased. There was a drop in stiffness between 2006 and 2007 of 22.9 % at the front and 9.1 % at the rear. In paragraph 4.2 the front and rear stiffness will be varied. The solution gives another 48 % drop in stiffness (55 N/mm) at the front. The rear increases a bit (55 N/mm). Concerning the damper characteristics still very little is known. In the model of Siau and Spijkers [2] fixed damping coefficients were used, 7000 Ns/m for the instroke and Ns/m for the outstroke. Between the 2006 and 2007 model new information about the dampers became available in the form of graphs. The base for the 2008 model is a linear damper characteristic for both the in and outstroke with a damping coefficient of Ns/m at the front and Ns/m at the rear (75 % of the front). This is visualized in figure 2.6. Figure 2.6: Damper characteristics for the 2008 truck.
20 8 2. The simulation model Fast Rebound system: The dampers which are mounted on the rally truck are equipped with a so called fast rebound system. The system is frequently used in motorcross. The principle of the system is as follows: strongly reduce the damping force when the wheels have no contact with the ground. Consequence: contact with the ground is reestablished sooner. Below the Fast Rebound system is depicted, see 2.7. Figure 2.7: Fast Rebound system. The indicated massspring combination operates like a valve which normally shuts off an extra orifice. The system works in one direction: the outstroke of the damper. Due to the inertia of the mass, the valve opens the extra orifice after reaching a certain, still unknown, acceleration. By opening the extra orifice the damping fluid can flow easier, which causes a 95 % reduction w.r.t. to the nominal value in damper force. After taking e.g. the bump, the contact with the ground is reestablished sooner, which means that traction and steering manoeuvres are possible faster. The concerning system is implemented in the multibodymodel by taking the acceleration as an input. Schematically this can be represented as follows: Figure 2.8: Fast Rebound system modeled in Simulink. The fast rebound valve is a lookup table. It performs a linear interpolation on input values using the specified table. Because there is no information available w.r.t. the concerning system, the data used for the 2006, 2007 and 2008 model is estimated. For all the models the same data has been used. The fast rebound system will be activated at a acceleration of 20 m/s 2. When the acceleration crosses the established limit, the damper force drops to 5% of the nominal damper force; see figure 2.8 product2. When the acceleration is within the boundaries (< 20 m/s 2 ), the damper force is multiplied by a factor 1.
21 3. Comparison of the 2006, 2007 and 2008 trucks The following three trucks will be covered in this chapter: the 2006, 2007 and 2008 truck. First a description of the dynamic behavior of the 2008 truck when driving over a predefined bump will be given. Then, a description at a glance of the physical differences between the mentioned trucks will be shown. The different settings are subjected also to the bump scenario. With this test it is possible to make a comparison between the dynamic behavior of the 2006, 2007 and 2008 truck. Analyzing the dynamic behavior is done by plotting the simulation results of all trucks. For building and analyzing the multibody models of the Dakar rally truck, the software package Matlab, version (R2006a), was used. Beside Simulink also SimMechanics and the Virtual Reality toolbox are required. 3.1 Driving over a bump The bump is adopted from the report of Pinxteren [2]. A representation of the bump is given in figure m 3 m Figure 3.1: Bump scenario. This scenario occurs very often during a Dakar rally and is also used in the reports concerning the 2006 and 2007 truck ([2], [1]). A DVD compilation of the 2005 ParisDakar rally confirms this situation. When driving over the bump some signals will become available from the model. An overview of these signals is given in appendix B. The initial velocity is put on 140 km/h. No brake, steer or throttle input occur during the simulation. After executing the simulation, the results are processed in eight graphs:
22 10 3. Comparison of the 2006, 2007 and 2008 trucks i) Suspension deflection at the front; ii) Suspension deflection at the rear; iii) Suspension velocity at the front; iv) Suspension velocity at the rear; v) Vertical tyre force at the front; vi) Vertical tyre force at the rear; vii) Centre of Gravity (CG) pitch angle; viii) CG vertical displacement. These graphs are used during the analysis. Appendix C shows an enlarged version of the graph representing the 2008 truck. Using this graph a conversion to the real movement has been made. This is shown in figure C.2, C.3 and C.4. In figure C.1 the most important events which characterize the movement of the truck taking the bump have been indicated by points A till N. Now follows a description of the momentary movement per point. Point A: This is the point where the truck meets the bump. In the time period t < 4 s an initialization of the model takes place. Now can be assumed that the truck drives from a steady state against the bump. This will happen at t = 4 s. Point B: At t = s the front wheels are leaving the top of the bump. The expired time is t = s. This is equal to: S = = 9.09 m v = m/s t = S v = s s (3.1) At this point the vertical tyre force reaches a maximum of F zf 100 kn. In the accompanying drawing, figure C.2 B, the arm CG x has been depicted. This arm in combination with the increased vertical force F fz (F fzmax F fznom ) introduces a moment around the yaxis of the centre of gravity. Mathematical this can be expressed as: I yy ω y = F zf CG x (3.2)
23 3.1 Driving over a bump 11 Thus, F zf causes an negative angular acceleration ω y. Because of ω y the negative angular velocity ω will grow. The consequence w.r.t. the movement of the body will be treated at the following points. At this moment it is sufficient to say that a small negative pitch (front is moving upwards, ) occurs. It is clear that the rear wheels have not yet been in contact with the bump. At the rear, no reaction can be seen w.r.t. the spring/damper combination. At the front, the vertical suspension deflection velocity reaches a maximum of v zf = 2.22 m/s. This results in a maximum damper force of F df = N at the front. The average vertical suspension deflection acceleration in the aforementioned time period equals a zf = 28.5 m/s 2. F sf = F zf F d m axle (g + a zf ) (3.3) At the maximum velocity the acceleration equals a zf already started and has a value of z zf = m. = 0 m/s 2. The vertical suspension deflection Point C: At t = 4.09 s the front wheels lose all contact with the bump. This can be seen by the vertical force F zf = 0N. Furthermore, the suspension deflection velocity decreases, and so does the damper force. A decrease in damper velocity is a result of a negative vertical axle acceleration (at point B it changed of sign). One can see that the suspension deflection at the front still increases, albeit by a small margin (velocity is still positive). This results in a growth of the spring force (F sf ). There is a small increase in negative pitch. The (backwards) rotating body contains kinetic energy which can be expressed as follows: T = 1 2 I yyω 2 (3.4) This is true, because no external forces are applied which will influence the movement of the rotating body. At this point F equals 0 N. As a result from this the angular acceleration ( ω) reduces to 0 rad/s 2 at t = 4.1s. At this point the maximum negative ω has been reached. The rear spring/damper combination is still not used. The centre of gravity will move slightly upwards. Point D: Point D represents t = 4.1 s. At this point a maximum positive suspension deflection z zf 0.14 m occurs. The velocity v zf equals 0 m/s.
24 12 3. Comparison of the 2006, 2007 and 2008 trucks The velocity v zf is influenced by the gravity. Now v zf becomes negative and the spring force at the front: F sf = F d mg (3.5) The negative pitch angle increases with a constant ω (linear part between t = 4.1 s and t = 4.18 s). The rear tyres meets the bump. This can be seen in the graph of F z. A very small increase is visible. Point E: Arriving at point E, t = 4.18 s, the vertical tyre force at the rear F zr reaches a maximum of approximately 86 kn. In the time space that the rear wheels are in contact with the bump, the vertical force grows. In the accompanying drawing it is clearly visible that the concerning force works over an arm X. Following is true: I yy ω y = F zr X (3.6) This will reduce the angular velocity ω of the body and with this the increase of pitch angle (not linear anymore). At the front the suspension deflection and velocity becomes negative (wheels are in the air). At the rear both are positive. Point F: Point F, t = s, indicates that v zf reaches its minimum of 4.5 m/s and v zr its maximum of 2.7 m/s. This is also the point where the rear tyres leave the bump. This can be seen by the drop in F zr. Point G: At G, t = 4.2 s, the pitch enters a minimum of One can also say that the truck is totally in the air (no contact with the ground). Both vertical forces are equal to 0 N. At the front z zf nearly reaches its minimum and v zf becomes less negative. At the rear z zr is still increasing and v zf is decreasing. Point H: At this moment, t = 4.22 s, the front wheels touch the ground. A slight increase in F zf is visible. The next conclusion can be made: the truck has been in the air for t fly = = 0.02 s.
25 3.1 Driving over a bump 13 The pitch angle increases, less negative, because ω of the body is positive. At the rear, the suspension deflection reaches a maximum of 0.16 m. Both the front and rear suspension deflection velocities cross the 0 m/s line. The front suspension deflection is equal to the minimum (rebound) value of 0.2 m. Point I: Now, the front wheels have achieved full contact with the ground. The rear wheels are still in the air, see F zr at t = 4.24 s which is equal to 0 N. The vertical force at the front wheels shows a small peak, which is caused by some bounce effect. Notice that during this bounce the front suspension deflection stays at its minimum. The inertia of the front axle plays a roll. Most of the bounce will be handled by the tyres. The rear suspension deflection decreases. The front velocity stays around 0 m/s. The rear velocity decreases towards a minimum. Here it is not yet visible, but the bounce will have an effect on the shape of the pitch. The pitch angle is still decreasing, caused by a positive ω, but a negative angular acceleration ω is introduced which will decrease the positive ω. The effect is very small because the deflection is at his minimum and the damper dissipates some energy. Point J: Characteristic for this phase is the constant vertical force between t > 4.24 s and t = 4.32 s at the front tyres. In this time period the front suspension deflection increases, and so does the velocity, albeit marginally. This means that part of the bounce at t = 4.24 s is handled by the springs and dampers at the front. Notice that the rear tyres have almost made contact with the ground. At this point the suspension deflection velocity v zr reaches a minimum of 5 m/s. One can see that z zr is almost at the rebound. The tyre force at the rear is much lower than the front. Nevertheless, z zr and v zr reach higher values in comparison to the front. The only difference between the front and rear configuration is the damping coefficient. At the front this coefficient (b df ) is equal to Ns/m at one corner. This value represents two dampers. The coefficient at the rear is a result of the following multiplication: = Ns/m (thus 75% of the front value). This is done because of the difference in static axle loads. The front and rear axles contain both a Fast Rebound System. Due to this system the damper force will drop to 5 % of the nominal value. Front damper force stays higher in comparison to the rear. As a result the rear axle will move faster downwards when flying into the air. The body is oriented totally horizontal. This can been seen by the pitch angle being equal to 0.
26 14 3. Comparison of the 2006, 2007 and 2008 trucks There is still some positive angular velocity, so in the next time period the body will rotate further. This means that the rear is moving upwards and the front is moving downwards. Point K: A small peak at t = 4.34 s in F zr is visible, which indicates that the rear wheels have made contact with the ground. Comparable with the front, the peak illustrates a bounce effect. Rear Suspension deflection and velocity do not react on this bounce. The deflection reaches a minimum (bumpstop). The suspension deflection velocity is equal to 0 m/s. Analyzing the front leads to the following conclusions: z zf increases and v bf is in the vicinity of 0 m/s. This is a response to the vertical tyre force. It is clear to see that the time period between z z = max and z z = min, e.g. for the rear = 0.12 s, is large in comparison to the time period necessary for the total movement, which is t = 5 4 = 1 s. Because there is little information available w.r.t. to the dampers, no extensive analysis has been done on this subject. It is plausible to expect that the time periods will change when the real damping coefficients become available. The pitch angle is still increasing, but at this point the direction coefficient of the line changes to a lower value (less steep). All tyres make contact with the ground, so the following equation applies: M = 0 Fzr X F zf CG x I yy ω y = 0 (3.7) X = wb c gx From the graph of F z it can been seen that F zf will rise and F zr drops. Result: ω decreases because of a more negative ω. This is also the point where a maximum in "Centre of Gravity vertical displacement" is reached. This maximum is equal to 0.19 m. Point L: Till t = 4.5 s the suspension deflection at the rear holds its minimum position (stays at the bumpstops). An increase in F zr will force the suspension to move upwards. The front suspension deflection is almost at his reference position. Pitch angle is still slightly increasing. Point M: At this point the maximum pitch of 1.85 has been reached. The front tyre force is higher in comparison to the rear, but works at a smaller arm (CG x ). However, this results in a anticlockwise moment (M) which decreases the pitch (negative ω). Both v zf and v zr will go to 0 m/s. The suspension deflections will go to their equilibrium position.
27 3.2 Comparison of the 2006, 2007 and 2008 truck 15 Point N: The vehicle reaches a steady state situation again. 3.2 Comparison of the 2006, 2007 and 2008 truck As mentioned before, the simulation results of the different truck settings will be shown in this paragraph. Two wheel suspensions were mentioned, being: rigid axels and independent suspension. The 2006 and 2008 trucks are equipped with rigid axles. The 2007 truck is equipped with an independent suspension. For an overview of all parameters see appendix A. A summation of the differences between the vehicles is represented in table 3.1. Category Description remark Masses [kg] total mass loaded vehicle body mass loaded vehicle front axle load loaded vehicle rear axle load loaded vehicle unsprung mass front incl. wheel unsprung mass rear incl. wheel Dimensions [mm] body CG hor. pos loaded vehicle Inertia [kgm 2 ] I yy loaded vehicle Stiffness [N/mm] stiffness at the front wheel no bump stops stiffness at the rear wheel no bump stops Table 3.1: Differences vehicle parameters. Each vehicle setting has been subjected to the bump scenario. By means of the aforementioned descriptions, paragraph 3.1, a comparison between the different truck configurations can be made. This paragraph will not go into any detail. It only treats the largest differences between the trucks mentioned. The simulation results can be found in figure 3.2. In this case the 2006 truck has been taken as the reference vehicle. The 2008 truck is based on the 2006 truck. The difference between both trucks must be especially ascribed to the prescribed changes such as a higher mass.
28 16 3. Comparison of the 2006, 2007 and 2008 trucks Figure 3.2: Simulation results when driving over the predefined bump with different truck settings.
29 3.2 Comparison of the 2006, 2007 and 2008 truck 17 Per simulation the eight graphs as mentioned at the beginning of this chapter are generated. The results from these graphs are interpreted below. Suspension deflection of front and rear: i) the 2006 and 2008 truck have the same shape at the front. This is selfevident because the masses, spring stiffnesses and damper coefficients are the same. At point I and K there was a interpretation about the bounce effect. This bounce effect occurs at the same time, although it does not influence the suspension deflection. Notice that, at the front, the line of the vertical tyre force is exactly the same; ii) till t = 4.5 s the shape at the rear of the 2006 truck is exactly the same as the 2008 truck. After this time period there is some difference between both trucks. These differences are caused by the fact that the rear tyres of the 2008 truck make contact with the ground much earlier than the 2006 truck. This can be seen in the "vertical tyre force" graph. When the rear tyres of the 2006 truck hit the ground, the 2008 truck already has a small second peak in tyre force; iii) the shape of the 2007 truck is totally different (t > 4.2 s for the front and t > 4.5 s for the rear) in comparison with the other trucks. At the front, the suspension deflection stays longer at its minimum position (not in the rebounds). The vertical tyre force shows a visible peak at the moment that the suspension deflection is on his minimum. This peak does not influence the deflection much (there is no contact with the rebounds anymore). The force drops which allows the suspension to stay on the concerning minimum position. At the rear, the deflection differs a little with the other trucks. This is reflected in the shape of the vertical tyre force. The maximum peak at t = 4.1 s is marginally higher. Deflection velocity of front and rear: i) the same as the suspension deflection, the shapes of the 2006 and 2008 truck are equal. The same interpretation as under i of suspension deflection applies here too; ii) there is only some difference between the 2006 and 2008 truck after t = 4.5 s; The 2008 truck has a much lower peak at the rear. This corresponds with the peak in F zr ; iii) for the 2007 truck, the absolute value of the maximum and minimum peaks at respectively t = 4.1 s and t = 4.2 s are higher. The masses and spring stiffnesses of the 2007 truck are much lower in comparison with the other trucks. Now, the force generated by the springs is lower. Mathematical the following must hold: F s + F d + m axle (a g) = 0. The acceleration is higher, so the deflection velocity. This is also the case at the rear suspension. The peak at t = 4.7 s is a result of the peak in F z (this peak is a bit lower than the one of the 2006 truck). Vertical tyre force of front and rear: i) the lines of the 2006 and 2008 trucks are almost the same for the front suspension. The preload (t < 4 s) differs with 100 kg and the huge peak at t = 4.1 s is a bit lower; ii) the lines between both trucks differ at the rear. In first instance there are some axle load differences. The 2006 truck has a static axle load of 4900 kg at the front and 3600 kg at the rear. For the 2008 truck the following values hold: 5000 kg at the front and 4300 kg at the rear. The difference in preload (700 kg) is clearly visible at the rear (t < 4 s). When considering the time
30 18 3. Comparison of the 2006, 2007 and 2008 trucks instance when the tyres touch the ground again, it can be observed that the 2008 truck touches the ground much earlier than the 2006 truck. When the rear tyres of the 2008 truck touch the ground, both rear axles are at their rebounds. The pitch angle is also the same. The difference in time is an effect of total altitude of the jump. Looking at the "CG vertical displacement" graph it is clear that the 2008 truck does not jump as high as the 2006 truck. The rear wheels will come in contact with the ground sooner; iii) first of all, the preloads of the 2007 truck do not differ much from the 2006 truck (front: 5000 kg and rear: 3500 kg). This explains why the peaks of F zf and F zr at respectively t = 4.1 s and t = 4.2 s are almost the same for both trucks. The bounce force is much higher for the 2007 truck. At the same time the deflection velocity reaches its minimum. The front tyres hit the ground with a much higher velocity. This is the effect of the higher bounce force. The rear gives the same shape as the 2006 truck. CG pitch angle: i) the difference in pitch angles is very large. The front wheels will hit the bump at t = 4 s and as a consequence a negative angle is established. For the 2006 and 2008 trucks there is no difference in pitch till t = 4.3 s. The 2008 truck gives a much lower positive peak than the 2006 truck. Point K gives an interpretation of this phenomenon. The body mass of the 2008 truck is higher than the 2006 truck, because of the new regulations. This increases the mass moment of inertia around the yaxis (I yy ). The influence of adding mass can be seen in figure 3.3. Also the influence of moving the CG to the back has been simulated. Conclusions w.r.t. these 2 subjects are: bringing CG to the back results in a lower negative pitch. There is some improvement in the positive pitch, but this improvement is not as great as compared to adding mass. As baseline vehicle the 2006 truck has been taken. Adding mass gives the same difference in positive pitch as in figure 3.2. The difference in body mass between the 2006 and 2008 truck is equal to = 800 kg. Now it has been proved that the concerning difference is caused by the difference in body mass; ii) for the 2007 truck it is not only the difference in mass ( = 540 kg) but also the balance of forces between front and rear axle plays a role, see also point K. The higher bounce force at the front gives a positive result w.r.t. the shape of the (positive) pitch angle. (a) Influence of CG (b) Influence of mass Figure 3.3: CG and mass influence analysis.
31 3.2 Comparison of the 2006, 2007 and 2008 truck 19 CG vertical displacement: i) the differences in displacement are very small. The 2008 truck has the smallest peak value (lowest jump); ii) the 2007 truck shows the most smooth behavior.
32 20 3. Comparison of the 2006, 2007 and 2008 trucks
33 4. Tuning the suspension of the 2008 truck In this chapter the suspension of the 2008 rally truck has been tuned. First an introduction which covers the requirements will be treated. With these requirements the tuning of the suspension has been completed. This process has been described in the subsequent paragraphs. 4.1 Introduction After analyzing the truck movement a package of requirements can be composed that are related to the tuning of the suspension. The following requirements have been established: i) driving against the bump F z at the front tyres must be kept small. This results in a decrease in negative pitch; ii) the force at the rear wheels must be higher for compensating the negative pitch ( ω). Thereby the front tyres can touch the ground sooner. Disadvantage: a raise in tyre force will reduce the life of the tyres which is not desirable. iii) It is important that, after passing the bump, the vertical tyre forces rapidly return to a stationary value, so that steer and driving forces can be transmitted to the road. iv) the bounce (peak force) must remain small, but a fast rise of the tyre forces at the front and back is again desirable. See also Point I, J and K. v) pitch angle must remain as small as possible and a steady state situation must be guaranteed as soon as possible. Because of this, the altitude difference at the front and rear, w.r.t. the ground, will be kept small. This also has a positive influence on the driver. It can also act as a feedback system, e.g. to know when serving the throttle is allowed or steer operations are permitted. Think of the instability of the vehicle when steer manoeuvres come to soon. vi) axles may not hit the bumpstops constantly. In the next paragraphs the following changes are treated to optimize the behavior of the 2008 truck: i) adjusting the spring stiffnesses with the 2008 truck as the baseline vehicle (paragraph 4.2); ii) after adjusting the springs, the ride height will be changed in order to prevent that the suspension frequently comes into the bumpstops (paragraph 4.3); iii) after implementing the above described changes into the model, the damping is adjusted (paragraph 4.4).
34 22 4. Tuning the suspension of the 2008 truck The order of steps results from a conversation with Mr. de Ruiter. Finally a paragraph has been dedicated to the final configuration. 4.2 Spring front/rear configuration At first, the stiffness has been altered. The baseline vehicle contains the following stiffness configuration: 105/55 N/mm (notation: front stiffness/rear stiffness). Three different coil spring stiffnesses are available: 10, 23 and 35 N/mm. It is assumed that these springs have a linear characteristic. Each axle contains 2 leafsprings which have a stiffness of 35 N/mm. Per corner 2 coil springs and 1 leafspring is mounted. Both coil springs have to be of the same stiffness value for reasons of symmetry. This means that the following configurations are possible: i) = 55 N/mm ii) = 81 N/mm iii) = 105 N/mm These configurations can be used both at the front and rear. This results in 9 possible configurations. This matrix is depicted twice, see table 4.1 and 4.2. Table 4.1 contains the min/max pitch angles in degrees. Table 4.2 contains the fly times of the front/rear axle in seconds. Front r e 552,22 / 0,632,30 / 1,152,38 / 1,85 a 812,14 / 0,652,27 / 0,992,37 / 1,66 r 1052,14 / 0,602,27 / 0,812,37 / 1,48 Table 4.1: Pitch angle results in degrees (min. peak/max. peak) after driving over the bump with above mentioned k s settings in N/mm. Front r e 55 0,13 / 0,13 0,12 / 0,13 0,12 / 0,13 a 81 0,13 / 0,12 0,12 / 0,12 0,12 / 0,12 r 105 0,13 / 0,12 0,12 / 0,12 0,12 / 0,11 Table 4.2: Fly time results in seconds (front/rear) after driving over the bump with above mentioned k s settings in N/mm. From the tables one can see that the 55/105 N/mm configuration scores the best w.r.t. the pitch angle and fly times. All results are plotted after the simulations. An example is given in appendix E, figure
35 4.2 Spring front/rear configuration 23 E.5. From this plot it becomes clear that the best configuration does not satisfy the requirements. This can be seen in the pitch graph. The steady state situation has been reached after a longer time period. Simulations ended up with the following configuration: 55/55 N/mm. Appendix E figure E.1 gives the results obtained with this stiffness configuration. In the same graph the 2008 baseline truck has been plotted. Analyzing the results with the altered configuration w.r.t. the 2008 baseline truck: i) at first, the suspension deflection is almost the same. When the front wheels leave the bump, the new configuration will also end up into the rebounds. This time the suspension remains longer at the minimum value. This is due to the fact that the generated force by the spring is higher in comparison to the baseline 2008 truck. This is illustrated by figure 4.1. The figure contains a reference (stroke = 0 m) and two limit values (stroke = 200 mm and 200 mm). The difference occurs at the front. The static loads (preloads) stay the same, but the stiffness of the new configuration drops. When the suspension is on his minimum, the force stays at a higher value. This means that the bounce at t = 4.22 s is not high enough to compress the suspension; ii) at the front and rear the peak values of F z, generated by the bump, are identical. The fly times of the front and rear axle are also the same (front difference: = 0.01 s longer). After the bounce F zf is higher. At t = 4.38 s the force reaches its static state value. Now, the suspension deflection increases (F zf is high enough to get the axle out of its rebounds). iii) increasing the force at the front tyres to a stationary value after the bounce, has a big influence on the pitch behavior, see point I. This force introduces a negative ω, which decreases the maximum pitch angle to (a) 2008 front, k sf = 105 N/mm (b) 2008 rear, k sr= 55 N/mm (c) 2008 front, k sf = 55 N/mm Figure 4.1: Consequence of lowering the stiffness. In case of lowering the spring stiffness the axles do not reach the bumpstops. There is a big margin between the bumpstop and maximum suspension deflection peak, since the damper dissipates a lot of energy. Because it is possible that higher obstacles can occur during the rally, it may be advisable to raise the ride height. This subject will be covered in the next paragraph.
36 24 4. Tuning the suspension of the 2008 truck 4.3 Ride height front/rear configuration Increasing the ride height can be seen as moving the rebounds and bumpstops around the reference position. The motivation for increasing the ride height has been described in the previous paragraph. From a conversation with Mr. de Ruiter arose the notion that an increase of the ride height is subjected to a limit. An increase of 10% of the total stroke, which is equal to 400 mm, is possible. This results in 3 different values for the ride height, being: i) 200 mm inward and 200 mm outward stroke; ii) = 220 mm inward and = 180 mm outward stroke; iii) = 240 mm inward and = 160 mm outward stroke; These configurations can be used both at the front and rear. Ultimately, this results in 9 possible configurations. To obtain the best configuration a 3 3 matrix has been implemented into the simulation script (explanation of the used notation: front inward/outward distance). Front r 200/ / /160 e 200/2002,22 / 0,632,22 / 1,062,22 / 1,53 a 220/1802,22 / 0,432,22 / 0,842,22 / 1,29 r 240/1602,22 / 0,292,22 / 0,682,22 / 1,12 Table 4.3: Pitch angle results in degrees (min. peak/max. peak) with k sf above mentioned ride heights in mm. = k sr = 55/55 N/mm and Front r 200/ / /160 e 200/200 0,13 / 0,13 0,13 / 0,13 0,13 / 0,13 a 220/180 0,13 / 0,13 0,13 / 0,13 0,13 / 0,13 r 240/160 0,13 / 0,16 0,13 / 0,17 0,13 / 0,20 Table 4.4: Fly time results in seconds (front/rear) with k sf = k sr = 55/55 N/mm and above mentioned ride heights in mm. Simulation results can be found in appendix E figure E.2. From the tables one can see that the following configuration scores the best w.r.t. the pitch and fly times: rh f = 200/200 mm and rh r = 220/180 mm. Analyzing the results obtained with a ride height of 200/200 mm at the front and 220/180 mm at the rear w.r.t. the stiffness configuration gives the following conclusions: i) the outward deflection at the rear is smaller z k r = 0.18 m. This is obvious because the ride height at the rear has been changed;
37 4.4 Damper front/rear configuration 25 ii) when all tyres have contact with the road, the balance between F zf and F zr is such that the pitch angle reduces much faster to 0. The bounce peak at the rear, t = 4.3 s is lower. The vertical tyre force raises at t = 4.55 s. This is the cause of a much faster steady state situation. The maximum pitch angle value reduces to iii) the fly times for front and rear are equal to the stiffness configuration. This suggests the notion to increase the ride height at the rear. Simulations with a 240/160 mm setting gives better results w.r.t. the pitch angle, but the fly times rises (front t = 0.13 s, rear t = 0.16 s). 4.4 Damper front/rear configuration Finally the damping coefficient is changed. The following four damping coefficients have been used: i) front: = Ns/m and rear = 8100 Ns/m; ii) front: = Ns/m and rear = Ns/m; iii) front: = Ns/m and rear = Ns/m iv) front: = Ns/m and rear = Ns/m. These configurations will be used at the front and rear. Ultimately, this results in 25 possible configurations (including no change in damping coefficient). To obtain the best configuration a 5 5 matrix has been implemented into the simulation script (explanation of the used notation: added or subtracted percentage). Front r 402,08 / 0,372,28 / 0,272,43 / 0,302,38 / 0,402,63 / 0,38 e 201,96 / 0,382,16 / 0,352,31 / 0,362,50 / 0,342,52 / 0,43 a 01,86 / 0,462,05 / 0,432,22 / 0,432,35 / 0,452,45 / 0,49 r +201,77 / 0,661,98 / 0,502,15 / 0,492,27 / 0,512,37 / 0, ,74 / 0,851,93 / 0,542,10 / 0,542,21 / 0,572,31 / 0,59 Table 4.5: Pitch angle results in degrees (min. peak/max. peak) with k sf = k sr = 55/55 N/mm, rh f = 200/200 mm, rh r = 220/180 mm and above mentioned damping settings (%). Front r 40 0,14 / 0,14 0,13 / 0,14 0,13 / 0,14 0,13 / 0,14 0,13 / 0,14 e 20 0,14 / 0,14 0,13 / 0,14 0,13 / 0,14 0,13 / 0,14 0,13 / 0,14 a 0 0,14 / 0,13 0,13 / 0,13 0,13 / 0,13 0,13 / 0,13 0,13 / 0,13 r +20 0,14 / 0,13 0,13 / 0,13 0,13 / 0,13 0,13 / 0,13 0,13 / 0, ,14 / 0,13 0,13 / 0,13 0,13 / 0,15 0,13 / 0,16 0,13 / 0,16 Table 4.6: Fly time results in seconds (front/rear) with k sf = k sr = 55/55 N/mm, rh f = 200/200 mm, rh r = 220/180 mm and above mentioned damping settings (%).
38 26 4. Tuning the suspension of the 2008 truck From the tables one can conclude that the following configuration scores the best w.r.t. the pitch angle and fly times: 20 % at the front and 40 % at the rear. These results can be found in appendix E figure E.6. During the simulations it however became clear that the following setting improves the result: +20 % at both front and rear. The positive pitch peak is almost twice as high in comparison to the first setting. There is no huge difference between the fly times. The line which represents the pitch angle shows a much faster steady state situation. Notice that both front and rear suspension deflection represent a significant difference, they are much smaller. This is desirable for extreme bump situations (safety measure). These are the reasons why the last setting is preferred. Simulation results can be found in appendix E figure E.3. Analyzing the results w.r.t. the ride height configuration: i) the inward deflection is smaller z kf = 0.12 m (7.5 %) and z kr = 0.14 m (12 %); ii) both bounce peaks, at t = 4.24 s and t = 4.34 s are a bit lower. At t = 4.3 s (front) and t = 4.45 s (rear) F zf and F zr become larger. One can see that the pitch behavior has not been influenced much ( = 0.08 ). The bounce peaks are lower and this is a advantage w.r.t. the life time of the tyres (although these tyres are built for high peak forces).
39 4.5 Final configuration Final configuration In this paragraph the total package will be discussed. This package consist of the following settings: i) spring stiffness 55/55 N/mm; ii) ride height front 200/200 mm and rear 220/180 mm; iii) damping coefficient 21600/16200 Ns/m. The baseline configuration and the final configuration are plotted in one graph. The result can be found in appendix E figure E.4. The differences have already been discussed in the previous paragraphs. With the baseline truck and the final configuration a predefined random road simulation (pavé road) of 4152 m length has been executed. The data for simulating this road is obtained from DAF engineers. The random road has been scaled by a factor 5. This factor is an approximation. Scaling have been permitted, because the frequency response of the tyres still shows a 2 slope. Only a shift in amplitude is visible. No further analysis have been done on this subject. Scaling has been done, because desert road conditions had to be simulated instead of Belgium roads. The initial velocity is put on 150 km/h. No brake, steer or throttle input occur during the simulation. The results are listed below. Description Min Max RMS Susp. defl. front [m] Susp. defl. rear [m] Defl. vel. front [m/s] Defl. vel.rear [m/s] Fz tyre front [kn] Fz tyre rear [kn] Pitch angle [ ] CG vertical displ. [m] Table 4.7: Results after driving over a predefined random road with the baseline 2008 truck. Description Min Max RMS Susp. defl. front m Susp. defl. rear m Defl. vel. front m/s Defl. vel.rear m/s Fz tyre front kn Fz tyre rear kn Pitch angle CG vertical displ. m Table 4.8: Results after driving over a predefined random road with the final truck.
40 28 4. Tuning the suspension of the 2008 truck Description Fly time front s Fly time rear s 2008 truck Final configuration Table 4.9: Fly time results after driving over a predefined random road. Analyzing the results w.r.t. the baseline 2008 truck: i) differences in suspension deflection are marginally; ii) for the suspension deflection velocity holds that all the absolute values of the final configuration are lower; iii) the maximum vertical tyre force at the front is, w.r.t. the baseline 2008 truck, 9 % higher for the final configuration. At the rear this force is 11.5 % higher. There exist a marginally difference in RMS values; iv) the minimum pitch peak is smaller for the 2008 truck. This difference equals 0.13 ; v) the maximum pitch peak is smaller for the final configuration. This difference equals 0.24 ; vi) in this case the most important value is represented by RMS (because it is a random road). The final configuration gives a lower RMS value. The difference is 0.09 which is equal to 20 %; vii) at the front, the fly time is lower for the 2008 truck. At the rear, the fly time is lower for the final configuration; viii) differences in CG vertical displacement are marginally. No further analysis has been made w.r.t. the random road. The general conclusion is, that the final configuration performs better.
41 5. Conclusions and Recommendations In the first section of this chapter the conclusions are drawn regarding the multibody model and the behavior of the 2008 truck for both the baseline and final configuration. In the second section recommendations are given w.r.t. the validation of the model. Conclusions The most important conclusions can be drawn from chapter 3 and 4. These are: i) for a good pitch behavior it is desirable that the fly time of the front tyres is as small as possible when driving over the predefined bump. This results in the fast increase in vertical tyre force (F z ) at the front. This introduces a negative pitch acceleration and reduces the positive pitch peak; ii) it is hard to reduce the negative pitch peak. The truck drives with a initial velocity of 140 km/h over the bump. The increase in vertical tyre force is huge and happens in a very short time period; iii) it is desirable that the fly time at the rear is as small as possible. By reducing this time period the balance between the front and rear axle will occur sooner. As a result the truck will arrive earlier in a steady state situation; iv) due to new regulations the mass of the total vehicle has been increased with 800 kg w.r.t. the 2006 truck. This influences the moment of inertia around the yaxis (I yy ) and thus the pitch behavior. In comparison to the 2006 truck the difference equals to a drop of 0.8 in positive peak value; v) over the last 2 years a significant drop in stiffness is visible. Simulations ended up with the following configuration: front and rear 55 N/mm (for the 2006 and the baseline 2008 truck: front 105 N/mm and rear 55 N/mm). Difference in pitch w.r.t. the baseline truck: No huge differences in fly times occur (front: 0.13 s and rear: 0.13 s); vi) the ride height has been changed according to the following configuration: front 200/200 mm, rear 220/180 mm. With this change the space between the bumpstops and the axle increases. It can be viewed as a safety measure. It also results in a drop of 0.2 w.r.t. to the positive pitch peak. The vehicle reaches its steady state much faster. No difference in fly times are observed in comparison to the configuration with a lower stiffness (conclusion point v); vii) increasing the damping coefficients with 20 % has no huge influence on the pitch. The vertical tyre forces are a bit smaller after the bounce peak. The tyre forces are a bit higher when the tyres come in contact with the bump. The inward deflection is lower. The space between the maximum deflection and the bumpstops increases. The fly times stay the same.
42 30 5. Conclusions and Recommendations Recommendations In the previous reports by M. Pinxteren [1] and G.R. Siau and T.L. Spijkers [2] it has already been indicated that a validation of the model is required. This is really necessary to obtain a good impression of the validity of the model(s). This model cannot provide a accurate setup without validation. Gathering data of the vehicle behavior in reality can be done by equipping the rally truck with some "simple" sensors and dataacquisition. In this report five important parameters were discussed, being: i) pitch; ii) axle displacement zdirection; iii) axle velocity zdirection; iv) tyre force zdirection; v) moment of inertia around the yaxis of CG (I yy ). The above signals can be measured as follows: Pitch: Making films and/or photographs at the left or right side of the bump (at some distance, because the whole truck and the complete movement has to be analyzed). It is not accurate, but it gives a clear view on what is happening. Figure 5.1 gives an impression of what is meant. Axle displacement: Only in the zdirection. Measuring method: simple potentiometer placed over the spring/damper combination; Axle velocity: Only in the zdirection. No sensor needed. Information can be obtained from measuring the displacement and the used timestep ( t); Tyre force: Only in the zdirection. This is not so easy to measure. Moment of inertia: Only around the yaxis. This can be done by building the setup shown in figure 5.2. The following mathematical expressions hold: M = ka 2 θ + kb 2 θ = kθ(a 2 + b 2 ) (5.1) k(a ω = 2 + b 2 ), ω = 2πf res (5.2) I yy I yy = k(a2 + b 2 ) ω 2 Remark: The result of this measurement is the moment of inertia I yy of the whole vehicle. The inertias and masses of the axles and wheels have been estimated well. Recalculation results in the moment of inertia of the body I yy. It is also possible to drive over a small obstacle without dampers. The stiffness of the (linear) springs is known. Measure the time and count the number of oscillations.
43 31 Bump Camera view (a) Yellow reference line Camera (b) Camera position Figure 5.1: Measurement pitch a b J Rigid connections asin(θ) aθ k θ k bsin(θ) bθ Figure 5.2: Inertia setup. After measuring the above mentioned signals it is possible to validate the multibody model and to improve the accuracy and predictive quality of the multibody model. The values of the snubber (endstops) damping coefficients have to examined, because they are estimated. A small study has already been finished. This study has been done within Matlab. First a road file with a very large (5 m high) bump has been developed. After driving over the bump the truck will be in the air for several seconds. During this time period the axles will hit the rebounds. Now, the intention is such that no oscillation, of the axles, will occur when the axles hit the rebounds. The way in which the Fast Rebound system has been modeled appears not to be correct. This conclusion came at a very late stage. It is important that this point gets extra attention during the next research.
44 32 5. Conclusions and Recommendations
45 Bibliography [1] M. Pinxteren. Development of a multibody model of a dakar rally truck with independent suspension. Technical Report DCT , Eindhoven University of Technology, [2] G.R. Siau and T.L. Spijkers. Development of a multibody simulation model of the daf dakar rally truck. Technical Report DCT , Eindhoven University of Technology, 2006.
46 34 BIBLIOGRAPHY
47 Appendix A Vehicle dimensions and parameters Category Description remark baseline Dimensions [mm] wheelbase track width vehicle length vehicle width vehicle height chassis width width axles Masses [kg] vehicle mass /9300 empty/full total sprung mass /7400 Empty/loaded axle load front Loaded vehicle axle load rear Loaded vehicle unspr. mass front axle unspr. mass rear axle front axle mass No rims & tyres rear axle mass No rims & tyres tyre rim wheel mass Incl. rim & tyre chassis mass 3400 cabin mass 700 motor mass 1500 loading space 400 tank left mass 50/300 Empty/full tank right mass 50/300 Empty/full spare wheel mass wheels Table A.1: Vehicle parameters, part 1
48 36 Appendix A Vehicle dimensions and parameters Category Description remark baseline Stiffness [N/mm] coil springs front coil springs rear leaf springs Not for 2007 stiff. at the front wheel No bump stops stiff. at the rear wheel No bump stops vertical tyre stiff. front vertical tyre stiff. rear Damping [Ns/m] vertical susp. damp. front Instroke/outstroke vertical susp. damp. front Instroke/outstroke vertical tyre damp. front vertical tyre damp. rear Table A.2: Vehicle parameters, part 2
49 Appendix B Output signals from the simulation model Below a summation of all available signals is shown. Signal Description Measure W.r.t. Truck chassis s.chassis.pitch Pitch angle truck chassis deg CGchassis s.chassis.pitchvel Pitch angle velocity truck chassis deg/s CGchassis s.chassis.pos Position chassis in x,y,z dir. CG m Local body/world s.chassis.acc Acceleration chassis in x,y,z dir. CG m/s 2 Local body/world s.chassis.angacc Angular acc. chassis x,y,z dir. CG rad/s2 Local body/world s.vx Velocity chassis xdir. m/s CGchassis s.zcg Height CG m CGchassis Front axle s.front.acc Acceleration front axle x,y,z dir. m/s 2 Local body/world s.front.susp.disp Displacement rear suspension zdir. m CGaxle body s.front.susp.vel Velocity rear suspension zdir. m/s CGaxle body s.front.susp.fspring Spring force rear suspension zdir. N s.front.susp.fdamp Damper force rear suspension zdir. N Rear axle s.rear.acc Acceleration rear axle x,y,z dir. m/s 2 Local body/world s.rear.susp.disp Displacement rear suspension zdir. m CGaxle body s.rear.susp.vel Velocity rear suspension zdir. m/s CGaxle body s.rear.susp.fspring Spring force rear suspension zdir. N s.rear.susp.fdamp Damper force rear suspension zdir. N Tyres s.varinf1() Left front tyre forces x,y,z dir. N s.varinf2() Right front tyre forces x,y,z dir. N s.varinf3() Left rear tyre forces x,y,z dir. N s.varinf4() Right rear tyre forces x,y,z dir. N Table B.1: Signals from models
50 38 Appendix B Output signals from the simulation model
51 Appendix C Simulation results 2008 baseline configuration
52 40 Appendix C Simulation results 2008 baseline configuration Suspension deflection front and rear axle suspension deflection [m] D H L N Suspension deflection velocity front and rear axle deflection velocity [m/s] B F Vertical tyre force front and rear axle vertical tyre force [kn] A C E I J front rear CG pitch angle [ ] CG pitch angle M G CG vertical displacement [m] K CG vertical displacement Time [s] Figure C.1: Enlarged time histories based on the 2008 truck.
53 41 CG x Driving dir. A t < 4[s] t = 4[s] Start bump CG x B Velocity C t = 4.078[s] F n Front: F n = max, max. pos. damp vel. and spring activation, small neg. pitch t = 4.09[s] Front: F n = 0[N], decrease damp vel., more neg. pitch, almost max susp. defl. Rear:In front of bump D X E X t = 4.1[s] F n t = 4.18[s] Front: F n = 0[N], damp vel.=0[m/s], more neg. pitch, max susp. defl. (0.1375[m]) Rear: begin bump Front: F n = 0[N], neg. damp vel., more neg. pitch, neg. susp. defl. Rear: F n =max, almost max pos. damp vel., pos. spring defl. Figure C.2: Movement of the 2008 truck.
54 42 Appendix C Simulation results 2008 baseline configuration F X G F n t = 4.185[s] t = 4.2[s] Front: F n = 0[N], max neg. damp vel., more neg. pitch, neg. spring defl. Rear: decrease F n, max pos. damp vel., pos. spring defl. Front: F n = 0[N], decrease neg. damp vel., max neg. pitch ( 2.38[ ]), neg. spring defl. Rear: F n =0, decrease pos. damp vel., pos. spring defl. H CG x I CG x t = 4.22[s] t = 4.24[s] Front: Increase F n, decrease neg. damp vel., decrease neg. pitch, almost max neg. susp. defl. Rear: F n =0, decrease pos. damp vel., max pos. susp. defl. Front: Small peak F n, damp vel. around 0[m/s], decrease neg. pitch, max neg. susp. defl. Rear: F n =0, increase neg. damp vel., decrease pos. susp. defl. J CG x K X CG x t = 4.32[s] Front: Constant F n, small increase pos. damp vel., increase pos. pitch, small decrease neg. susp. defl. Rear: F n =0, max neg. damp vel., almost max neg. susp. defl. t = 4.34[s] Front: Constant F n, small increase pos. damp vel., increase pos. pitch, small decrease neg. susp. defl. Rear: Peak F n, damp vel. around 0[m/s], max neg. susp. defl., max vert. displ. CG (0.19[m]) Figure C.3: Continue movement of the 2008 truck.
55 43 L CG x M CG x t = 4.5[s] t = 4.56[s] Front: F n almost static value, damp vel. around 0[m/s], almost max pos. pitch, decrease neg. susp. defl. Rear: F n almost static value, damp vel. around 0, begin increase susp. defl. Front: F n almost static value, damp vel. around 0[m/s], max pos. pitch, susp. defl. almost middle position Rear: F n almost static value, damp vel. around 0[m/s], decrease neg. susp. defl. N t = 5[s] End movement (steady state) Figure C.4: Continue movement of the 2008 truck.
56 44 Appendix C Simulation results 2008 baseline configuration
57 Appendix D Axle and wheel inertia The following assumption has been made for calculating the axle inertias: axles are treated as "rods". Calculation inertia rod: I xx = I zz = m 12 l2 I yy = m 12 (l2 + h 2 ) (D.1) (D.2) with, m = mass; l = length; h = height; Next, the calculation of the axle inertia, front and rear, will be treated. Table D.1 shows a summation of the required parameters. Parameter Value [mm] w 310 l 1755 h 350 m f 700 [kg] 600 [kg] m r Table D.1: Axle dimensions. This part requires some extra explanation, because the cross section of the yplane is not rectangular. Nevertheless, in this case it is assumed to be so. The shape is depicted in figure D.1.
58 46 Appendix D Axle and wheel inertia Figure D.1: Cross section yplane axle. Inertia front axle: I yy = (( )2 + ( ) 2 ) 13[kgm 2 ] The inertia tensor then becomes: I yy = [kgm 2 ] (D.3) Inertia rear axle: I yy = (( )2 + ( ) 2 ) 11[kgm 2 ] I yy = [kgm 2 ] (D.4) The inertia I yy of the tyre is given in the "Michelin XZL.tir"file and is equal to 10 [kgm 2 ].
59 Appendix E Tuning results 2008 vehicle
60 48 Appendix E Tuning results 2008 vehicle Figure E.1: Bump simulation results k sf = k sr = 55/55 [N/mm].
61 Figure E.2: Bump simulation results with k sf = k sr = 55/55 [N/mm], rh f = 200/200 [mm] and rh r = 220/180 [mm]. 49
62 50 Appendix E Tuning results 2008 vehicle Figure E.3: Bump simulation results with k sf = k sr = 55/55 [N/mm], rh f = 200/200 [mm], rh r = 220/180 [mm], b df = [Ns/m] and b dr = [Ns/m].
63 Figure E.4: Bump simulation results, final configuration vs 2008 baseline. 51
64 52 Appendix E Tuning results 2008 vehicle Figure E.5: Baseline truck and comparison between different stiffnesses.
Development of a multibody model of a Dakar Rally truck with independent suspension
Development of a multibody model of a Dakar Rally truck with independent suspension M.Pinxteren DCT 2007.043 Internal Traineeship report Supervisor: Dr. Ir. I.J.M. Besselink Technische Universiteit Eindhoven
More informationSuspension systems and components
Suspension systems and components 2 of 42 Objectives To provide good ride and handling performance vertical compliance providing chassis isolation ensuring that the wheels follow the road profile very
More informationParameter identification of a linear single track vehicle model
Parameter identification of a linear single track vehicle model Edouard Davin D&C 2011.004 Traineeship report Coach: dr. Ir. I.J.M. Besselink Supervisors: prof. dr. H. Nijmeijer Eindhoven University of
More informationResponse to Harmonic Excitation Part 2: Damped Systems
Response to Harmonic Excitation Part 2: Damped Systems Part 1 covered the response of a single degree of freedom system to harmonic excitation without considering the effects of damping. However, almost
More informationEngineering Feasibility Study: Vehicle Shock Absorption System
Engineering Feasibility Study: Vehicle Shock Absorption System Neil R. Kennedy AME40463 Senior Design February 28, 2008 1 Abstract The purpose of this study is to explore the possibilities for the springs
More informationDevelopment of a multibody simulation model of the DAF Dakar rally truck
Development of a multibody simulation model of the DAF Dakar rally truck G.R. Siau T.L. Spijkers Report No. DCT 6.9 Internal Traineeship Supervisors: Dr. Ir. I.J.M. Besselink Prof. Dr. H. Nijmeijer Eindhoven
More informationEquivalent Spring Stiffness
Module 7 : Free Undamped Vibration of Single Degree of Freedom Systems; Determination of Natural Frequency ; Equivalent Inertia and Stiffness; Energy Method; Phase Plane Representation. Lecture 13 : Equivalent
More informationFree Fall: Observing and Analyzing the Free Fall Motion of a Bouncing PingPong Ball and Calculating the Free Fall Acceleration (Teacher s Guide)
Free Fall: Observing and Analyzing the Free Fall Motion of a Bouncing PingPong Ball and Calculating the Free Fall Acceleration (Teacher s Guide) 2012 WARD S Science v.11/12 OVERVIEW Students will measure
More informationPhysics 2A, Sec B00: Mechanics  Winter 2011 Instructor: B. Grinstein Final Exam
Physics 2A, Sec B00: Mechanics  Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry
More informationDetermination of antipitch geometry acceleration [1/3]
Determination of antipitch geometry acceleration [1/3] 1 of 39 Similar to antisquat Opposite direction of D Alembert s forces. Front wheel forces and effective pivot locations Determination of antipitch
More informationChapter 11 Equilibrium
11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of
More informationThe Effects of Wheelbase and Track on Vehicle Dynamics. Automotive vehicles move by delivering rotational forces from the engine to
The Effects of Wheelbase and Track on Vehicle Dynamics Automotive vehicles move by delivering rotational forces from the engine to wheels. The wheels push in the opposite direction of the motion of the
More informationPHYSICS 111 HOMEWORK SOLUTION #9. April 5, 2013
PHYSICS 111 HOMEWORK SOLUTION #9 April 5, 2013 0.1 A potter s wheel moves uniformly from rest to an angular speed of 0.16 rev/s in 33 s. Find its angular acceleration in radians per second per second.
More informationTech Tip: Steering Geometry
Designing Steering Geometry When you re designing steering kinematics, the goal is to orient the tire to the road in the optimal orientation. But, how do you know the optimal orientation? Tire data, of
More informationUnit 4 Practice Test: Rotational Motion
Unit 4 Practice Test: Rotational Motion Multiple Guess Identify the letter of the choice that best completes the statement or answers the question. 1. How would an angle in radians be converted to an angle
More informationChapter 3.8 & 6 Solutions
Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled
More informationA Road Crash Reconstruction Technique
A Road Crash Reconstruction Technique Mukherjee S, nonmember Chawla A 1, member Lalaram Patel, nonmember Abstract The purpose of reconstruction is to identify the critical factors involved in a road
More informationLongitudinal and lateral dynamics
Longitudinal and lateral dynamics Lecturer dr. Arunas Tautkus Kaunas University of technology Powering the Future With Zero Emission and Human Powered Vehicles Terrassa 2011 1 Content of lecture Basic
More informationResearch on Vehicle Dynamics Simulation for Driving Simulator Fang Tang a, Yanding Wei b, Xiaojun Zhou c, Zhuhui Luo d, Mingxiang Xie e, Peixin Li f
Advanced Materials Research Vols. 308310 (2011) pp 19461950 Online available since 2011/Aug/16 at www.scientific.net (2011) Trans Tech Publications, Switzerland doi:10.4028/www.scientific.net/amr.308310.1946
More informationRotation: Moment of Inertia and Torque
Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn
More information1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition. t + ") # x (t) = A! n. t + ") # v(0) = A!
1.1 Using Figure 1.6, verify that equation (1.1) satisfies the initial velocity condition. Solution: Following the lead given in Example 1.1., write down the general expression of the velocity by differentiating
More informationState Newton's second law of motion for a particle, defining carefully each term used.
5 Question 1. [Marks 28] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding
More informationPHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013
PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be
More informationSUSPENSION AND STEERING OVERVIEW
SUSPENSION SUSPENSION AND STEERING OVERVIEW The S40/V50 has a wide track and a long wheelbase for its relative size and weight. This gives the car stable and predictable driving characteristics. It also
More informationPHY121 #8 Midterm I 3.06.2013
PHY11 #8 Midterm I 3.06.013 AP Physics Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension
More informationSOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS  VELOCITY AND ACCELERATION DIAGRAMS
SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS  VELOCITY AND ACCELERATION DIAGRAMS This work covers elements of the syllabus for the Engineering Council exams C105 Mechanical and Structural Engineering
More informationHW Set VI page 1 of 9 PHYSICS 1401 (1) homework solutions
HW Set VI page 1 of 9 1030 A 10 g bullet moving directly upward at 1000 m/s strikes and passes through the center of mass of a 5.0 kg block initially at rest (Fig. 1033 ). The bullet emerges from the
More informationMidterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m
Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of
More informationSTATIC STRUCTURAL ANALYSIS OF SUSPENSION ARM USING FINITE ELEMENT METHOD
STATIC STRUCTURAL ANALYSIS OF SUSPENSION ARM USING FINITE ELEMENT METHOD Jagwinder Singh 1, Siddhartha Saha 2 1 Student, Mechanical Engineering, BBSBEC, Punjab, India 2 Assistant Professor, Mechanical
More informationPHYS 211 FINAL FALL 2004 Form A
1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each
More informationPrepared by Graham from SuperPro August 2011
Understanding di Steering and Wheel Alignment Angles Prepared by Graham from SuperPro August 2011 Remember: Tyre Wear Patterns Tell The Technician A Story Provide Vital Information For Determining Final
More informationPractice Exam Three Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Practice Exam Three Solutions Problem 1a) (5 points) Collisions and Center of Mass Reference Frame In the lab frame,
More informationTennessee State University
Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an Fgrade. Other instructions will be given in the Hall. MULTIPLE CHOICE.
More informationINTERACTION BETWEEN MOVING VEHICLES AND RAILWAY TRACK AT HIGH SPEED
INTERACTION BETWEEN MOVING VEHICLES AND RAILWAY TRACK AT HIGH SPEED Prof.Dr.Ir. C. Esveld Professor of Railway Engineering TU Delft, The Netherlands Dr.Ir. A.W.M. Kok Associate Professor of Railway Engineering
More informationPHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?
1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always
More informationPhysics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives
Physics 9e/Cutnell correlated to the College Board AP Physics 1 Course Objectives Big Idea 1: Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring
More informationAn Introduction to Race Car Engineering Table of Contents Page Chapter 1 Some Basic Truths... 1/1 to 1/4
An Introduction to Race Car Engineering Table of Contents Page Chapter 1 Some Basic Truths... 1/1 to 1/4 Basic concepts and truths leading to the successful study and understanding of vehicle dynamics
More informationPHY231 Section 1, Form B March 22, 2012
1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate
More informationVehicle Chassis Analysis: Load Cases & Boundary Conditions For Stress Analysis
Vehicle Chassis Analysis: Load Cases & Boundary Conditions For Stress Analysis Ashutosh Dubey and Vivek Dwivedi ABSTRACT The current work contains the load cases & boundary conditions for the stress analysis
More informationSolving Simultaneous Equations and Matrices
Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering
More information2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration.
2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration. Dynamics looks at the cause of acceleration: an unbalanced force. Isaac Newton was
More informationState Newton's second law of motion for a particle, defining carefully each term used.
5 Question 1. [Marks 20] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding
More informationPhysics 201 Homework 8
Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 Nm is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kgm 2. What is the
More informationUnit  6 Vibrations of Two Degree of Freedom Systems
Unit  6 Vibrations of Two Degree of Freedom Systems Dr. T. Jagadish. Professor for Post Graduation, Department of Mechanical Engineering, Bangalore Institute of Technology, Bangalore Introduction A two
More informationSlide 10.1. Basic system Models
Slide 10.1 Basic system Models Objectives: Devise Models from basic building blocks of mechanical, electrical, fluid and thermal systems Recognize analogies between mechanical, electrical, fluid and thermal
More informationDesigning a Hydraulic Actuator for the Tyre Measurement Tower
Designing a Hydraulic Actuator for the Tyre Measurement Tower H.N.L. de Wispelaere DCT 26.26 Bachelor End Project Supervisors: dr. ir. A.J.C. Schmeitz dr. ir. W.J.A.M. Post Technische Universiteit Eindhoven
More informationMotorcycle accident reconstruction in VL Motion
Motorcycle accident reconstruction in VL Motion Presenter Nicola Cofelice ESR 14 AGENDA 1 Motorcycle accident reconstruction  Overview 2 Research progress 3 What s next? copyright LMS International 
More informationA ball, attached to a cord of length 1.20 m, is set in motion so that it is swinging backwards and forwards like a pendulum.
MECHANICS: SIMPLE HARMONIC MOTION QUESTIONS THE PENDULUM (2014;2) A pendulum is set up, as shown in the diagram. The length of the cord attached to the bob is 1.55 m. The bob has a mass of 1.80 kg. The
More informationNo Brain Too Small PHYSICS. 2 kg
MECHANICS: ANGULAR MECHANICS QUESTIONS ROTATIONAL MOTION (2014;1) Universal gravitational constant = 6.67 10 11 N m 2 kg 2 (a) The radius of the Sun is 6.96 10 8 m. The equator of the Sun rotates at a
More informationTyre Awareness Training
Tyre Awareness Training Tyre Law Tread depth Tyres on cars, light vans (not exceeding 3,500kg gross weight) and light trailers must have a tread depth of at least 1.6mm across the central threequarters
More informationPRODUCT SPECIFICATION
DATE: February 2012 DESCRIPTION: Old Man Emu Suspension System. APPLICATION: Jeep Wrangler JK 2 & 4 Door 4 Lift Click for 2Door App Guide Click for 4Door App Guide PART NO: OMEJK4 RETAIL: $1522.62 AVAILABILITY
More informationRACING BUGGY The Racing Buggy represents Fornasari s attitude to innovation and creativity. Introduced in 2010, it is a car which combines the Fornasa
RACING BUGGY RACING BUGGY The Racing Buggy represents Fornasari s attitude to innovation and creativity. Introduced in 2010, it is a car which combines the Fornasari 600 offroad racing features with a
More informationA Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion
A Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion Objective In the experiment you will determine the cart acceleration, a, and the friction force, f, experimentally for
More informationDynamic Analysis of the Dortmund University Campus Sky Train
Dynamic Analysis of the Dortmund University Campus Sky Train Reinhold Meisinger Mechanical Engineering Department Nuremberg University of Applied Sciences Kesslerplatz 12, 90121 Nuremberg, Germany Abstract
More informationAP Physics C. Oscillations/SHM Review Packet
AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete
More informationCenter of Gravity. We touched on this briefly in chapter 7! x 2
Center of Gravity We touched on this briefly in chapter 7! x 1 x 2 cm m 1 m 2 This was for what is known as discrete objects. Discrete refers to the fact that the two objects separated and individual.
More informationLecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is
Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.49.6, 10.110.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of
More information3600 s 1 h. 24 h 1 day. 1 day
Week 7 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationCHAPTER 15 FORCE, MASS AND ACCELERATION
CHAPTER 5 FORCE, MASS AND ACCELERATION EXERCISE 83, Page 9. A car initially at rest accelerates uniformly to a speed of 55 km/h in 4 s. Determine the accelerating force required if the mass of the car
More informationAckerman? AntiAckerman? Or Parallel Steering?
www.racingcartechnology.com.au/steering Ackerman4.doc Here we tackle the tough questions : "Ackerman? Or not? Does it matter?". Dale Thompson from Racing Car Technology looks for some answers. What's
More informationSoil Dynamics Prof. Deepankar Choudhury Department of Civil Engineering Indian Institute of Technology, Bombay
Soil Dynamics Prof. Deepankar Choudhury Department of Civil Engineering Indian Institute of Technology, Bombay Module  2 Vibration Theory Lecture  8 Forced Vibrations, Dynamic Magnification Factor Let
More informationAn Introduction to Using Simulink. Exercises
An Introduction to Using Simulink Exercises Eric Peasley, Department of Engineering Science, University of Oxford version 4.1, 2013 PART 1 Exercise 1 (Cannon Ball) This exercise is designed to introduce
More informationwww.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity
More informationMSR offroad tuning guide.
MSR offroad tuning guide. Mugen Seiki Racing spends a lot of time developing the kit setup and it's always a great place to start and a good place to go back to if you get lost. For updates and the latest
More informationManufacturing Equipment Modeling
QUESTION 1 For a linear axis actuated by an electric motor complete the following: a. Derive a differential equation for the linear axis velocity assuming viscous friction acts on the DC motor shaft, leadscrew,
More informationTorque and Rotary Motion
Torque and Rotary Motion Name Partner Introduction Motion in a circle is a straightforward extension of linear motion. According to the textbook, all you have to do is replace displacement, velocity,
More informationPhysics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension
Physics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension Conceptual Questions 1) Suppose that an object travels from one point in space to another. Make
More informationModeling Mechanical Systems
chp3 1 Modeling Mechanical Systems Dr. Nhut Ho ME584 chp3 2 Agenda Idealized Modeling Elements Modeling Method and Examples Lagrange s Equation Case study: Feasibility Study of a Mobile Robot Design Matlab
More informationGear Trains. Introduction:
Gear Trains Introduction: Sometimes, two or more gears are made to mesh with each other to transmit power from one shaft to another. Such a combination is called gear train or train of toothed wheels.
More informationQuarter Midget Chassis Glossary of Terms
Quarter Midget Chassis Glossary of Terms Ackerman Steering: Alignment Bars: Axle Lead: Baseline Setup: Bicycling: Birdcage Timing: As the front wheels turn through the corner the left front turns a shaper
More informationThe dynamic equation for the angular motion of the wheel is R w F t R w F w ]/ J w
Chapter 4 Vehicle Dynamics 4.. Introduction In order to design a controller, a good representative model of the system is needed. A vehicle mathematical model, which is appropriate for both acceleration
More informationPavel Krulich, Jan Čapek
MultiBody Simulation of Influence of Bogie Interconnection on Vehicletrack Interaction, VÚKV a.s. Bucharova 1314/8 158 00 Praha 5 krulich@vukv.cz, capek@vukv.cz www.vukv.cz 1 Content Introduction Principal
More information1 of 10 11/23/2009 6:37 PM
hapter 14 Homework Due: 9:00am on Thursday November 19 2009 Note: To understand how points are awarded read your instructor's Grading Policy. [Return to Standard Assignment View] Good Vibes: Introduction
More informationChapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc.
Chapter 10 Rotational Motion Angular Quantities Units of Chapter 10 Vector Nature of Angular Quantities Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems
More informationF1 Fuel Tank Surging; Model Validation
F1 Fuel Tank Surging; Model Validation Luca Bottazzi and Giorgio Rossetti Ferrari F1 team, Maranello, Italy SYNOPSIS A Formula One (F1) car can carry more than 80 kg of fuel in its tank. This has a big
More information2After completing this chapter you should be able to
After completing this chapter you should be able to solve problems involving motion in a straight line with constant acceleration model an object moving vertically under gravity understand distance time
More informationCORRECTION OF DYNAMIC WHEEL FORCES MEASURED ON ROAD SIMULATORS
Pages 1 to 35 CORRECTION OF DYNAMIC WHEEL FORCES MEASURED ON ROAD SIMULATORS Bohdan T. Kulakowski and Zhijie Wang Pennsylvania Transportation Institute The Pennsylvania State University University Park,
More informationProblem Set 1. Ans: a = 1.74 m/s 2, t = 4.80 s
Problem Set 1 1.1 A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine her constant acceleration. How long does it take her to
More informationLecture 07: Work and Kinetic Energy. Physics 2210 Fall Semester 2014
Lecture 07: Work and Kinetic Energy Physics 2210 Fall Semester 2014 Announcements Schedule next few weeks: 9/08 Unit 3 9/10 Unit 4 9/15 Unit 5 (guest lecturer) 9/17 Unit 6 (guest lecturer) 9/22 Unit 7,
More informationF N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26
Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250N force is directed horizontally as shown to push a 29kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,
More informationDYNAMIC RESPONSE OF VEHICLETRACK COUPLING SYSTEM WITH AN INSULATED RAIL JOINT
11 th International Conference on Vibration Problems Z. Dimitrovová et al. (eds.) Lisbon, Portugal, 912 September 2013 DYNAMIC RESPONSE OF VEHICLETRACK COUPLING SYSTEM WITH AN INSULATED RAIL JOINT Ilaria
More informationState of Stress at Point
State of Stress at Point Einstein Notation The basic idea of Einstein notation is that a covector and a vector can form a scalar: This is typically written as an explicit sum: According to this convention,
More informationPhysics 125 Practice Exam #3 Chapters 67 Professor Siegel
Physics 125 Practice Exam #3 Chapters 67 Professor Siegel Name: Lab Day: 1. A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the
More informationKINEMATICS OF PARTICLES RELATIVE MOTION WITH RESPECT TO TRANSLATING AXES
KINEMTICS OF PRTICLES RELTIVE MOTION WITH RESPECT TO TRNSLTING XES In the previous articles, we have described particle motion using coordinates with respect to fixed reference axes. The displacements,
More informationChassis* design. *pronounced: chas e singular chas e z plural
Automotive design Chassis* design *pronounced: chas e singular chas e z plural Introduction Loads due to normal running conditions: Vehicle transverse on uneven ground. Manoeuver performed by driver. Five
More informationAP1 Oscillations. 1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false?
1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The
More informationACTIVE SAFETY OF TRUCKS AND ROAD TRAINS WITH WIDE BASE SINGLE TYRES INSTEAD OF TWIN TYRES
ACTIVE SAFETY OF TRUCKS AND ROAD TRAINS WITH WIDE BASE SINGLE TYRES INSTEAD OF TWIN TYRES Dr.Ing. KlausPeter Glaeser Federal Highway Research Institute, BASt Germany Paper No. 497 1 ABSTRACT The development
More informationAddis Ababa University Addis Ababa Institute of Technology (AAiT)
Addis Ababa University Addis Ababa Institute of Technology (AAiT) School of Mechanical & Industrial Engineering Railway Engineering Stream Effect of Track Stiffness Variation on the Dynamic Response of
More informationPhysics 2101, First Exam, Fall 2007
Physics 2101, First Exam, Fall 2007 September 4, 2007 Please turn OFF your cell phone and MP3 player! Write down your name and section number in the scantron form. Make sure to mark your answers in the
More informationFluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  20 Conservation Equations in Fluid Flow Part VIII Good morning. I welcome you all
More informationSolution Derivations for Capa #11
Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform
More informationF f v 1 = c100(10 3 ) m h da 1h 3600 s b =
14 11. The 2Mg car has a velocity of v 1 = 100km>h when the v 1 100 km/h driver sees an obstacle in front of the car. It takes 0.75 s for him to react and lock the brakes, causing the car to skid. If
More information4.2 Free Body Diagrams
CE297FA09Ch4 Page 1 Friday, September 18, 2009 12:11 AM Chapter 4: Equilibrium of Rigid Bodies A (rigid) body is said to in equilibrium if the vector sum of ALL forces and all their moments taken about
More informationTriumph TR3 & 4 Suspension Geometry by Larry Young, Forever Young Racing
Triumph TR3 & 4 Suspension Geometry by Larry Young, Forever Young Racing Above is a schematic of the Triumph TR3/4 suspension system taken from the Service Instruction Manual. The figure below is a simplified
More informationMechanical Principles
Unit 4: Mechanical Principles Unit code: F/601/1450 QCF level: 5 Credit value: 15 OUTCOME 4 POWER TRANSMISSION TUTORIAL 2 BALANCING 4. Dynamics of rotating systems Single and multilink mechanisms: slider
More informationStatics problem solving strategies, hints and tricks
Statics problem solving strategies, hints and tricks Contents 1 Solving a problem in 7 steps 3 1.1 To read.............................................. 3 1.2 To draw..............................................
More informationSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s.
More informationMagnetic / Gravity Loading Analysis
Magnetic / Gravity Loading Analysis 2 ELEMENTS JUL 7 2006 ELEMENTS MAT NUM 2:5:0 MAT NUM POR Design JUL 7 2006 2:5:0 L2 L L q Assumed Location of Gap Encoder(s) ELEMENTS MAT NUM JUL 7 2006 2:5:0 Materials:
More informationAP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s
AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s Answer the multiple choice questions (2 Points Each) on this sheet with capital
More information