mehanika vo nje - odsek za puteve, eleznice i aerodrome · dinamika vozila u poduºnom ravcup...

Dinamika vozila u poduºnom pravcu

MEHANIKA VO�NJE

Odsek za puteve, ºeleznice i aerodrome

Prof dr Stanko Br£i¢Doc dr Stanko �ori¢Doc dr Anina Glumac

Gra�evinski fakultetUniverzitet u Beogradu

�k. god. 2018/19

S.�ori¢ Mehanika voºnje

Dinamika vozila u poduºnom pravcu

Sadrºaj

1 Dinamika vozila u poduºnom pravcuKretanje vozila u prostoruPravolinijsko kretanje vozila


Dinamika vozila u poduºnom pravcuKretanje vozila u prostoruPravolinijsko kretanje vozila

Sadrºaj




Prostorni i materijalni sistem10. Vehicle Planar Dynamics 585

x

z

Fx

FzMz

Mx

ϕ

ψ

p

r

yFy

My

θq

XY

Z

B G

ψ

FIGURE 10.2. Illustration of a moving vehicle, indicated by its body coordinateframe B in a global coordinate frame G.

The vehicle coordinate frame is called the body frame or vehicle frame, andthe grounded frame is called the global coordinate frame. Analysis of thevehicle motion is equivalent to expressing the position and orientation ofB(Cxyz) in G(OXY Z). Figure 10.2 shows how a moving vehicle is indi-cated by a body frame B in a global frame G.The angle between the x and X axes is the yaw angle ψ and is called

the heading angle. The velocity vector v of the vehicle makes an angle βwith the body x-axis which is called sideslip angle or attitude angle. Thevehicle’s velocity vector v makes an angle β + ψ with the global X-axisthat is called the cruise angle. These angles are shown in the top view of amoving vehicle in Figure 10.3.There are many situations in which we need to number the wheels of a

vehicle. We start numbering from the front left wheel as number 1, and thenthe front right wheel would be number 2. Numbering increases sequentiallyon the right wheels going to the back of the vehicle up to the rear rightwheel. Then, we go to the left of the vehicle and continue numbering thewheels from the rear left toward the front. Each wheel is indicated by aposition vector ri

Bri = xii+ yij + zik (10.6)

expressed in the body coordinate frame B. Numbering of a four wheelvehicle is shown in Figure 10.3.

Example 375 Wheel numbers and their position vectors.Figure 10.4 depicts a six-wheel passenger car. The wheel numbers are

indicated besides each wheel. The front left wheel is wheel number 1, andthe front right wheel is number 2. Moving to the back on the right side,we count the wheels numbered 3 and 4. The back left wheel gets number 5,

G . . . Prostorni (inercijalni) koordinatni sistem OXYZB . . .Materijalni (pokretni) koordinatni sistem Sxyz



Glavni rotacioni stepeni slobode5. Applied Kinematics 231

roll

pitch

yaw

x

z

ϕ

ψ

yθ

FIGURE 5.4. Local roll-pitch-yaw angles.

5.5 F Euler Angles

The rotation about the Z-axis of the global coordinate is called precession,the rotation about the x-axis of the local coordinate is called nutation,and the rotation about the z-axis of the local coordinate is called spin.The precession-nutation-spin rotation angles are also called Euler angles.Rotation matrix based on Euler angles has application in rigid body kine-matics. To find the Euler angles rotation matrix to go from the globalframe G(OXY Z) to the final body frame B(Oxyz), we employ a bodyframe B0(Ox0y0z0) as shown in Figure 5.5 that before the first rotation co-incides with the global frame. Let there be at first a rotation ϕ about thez0-axis. Because Z-axis and z0-axis are coincident, by our theory

B0r = B0

RGGr (5.72)

B0RG = Rz,ϕ =

⎡⎣ cosϕ sinϕ 0− sinϕ cosϕ 00 0 1

⎤⎦ . (5.73)

Next we consider the B0(Ox0y0z0) frame as a new fixed global frame andintroduce a new body frame B00(Ox00y00z00). Before the second rotation, thetwo frames coincide. Then, we execute a θ rotation about x00-axis as shownin Figure 5.6. The transformation between B0(Ox0y0z0) and B00(Ox00y00z00)is

B00r = B00

RB0B0r (5.74)

B00RB0 = Rx,θ =

⎡⎣ 1 0 00 cos θ sin θ0 − sin θ cos θ

⎤⎦ . (5.75)

Finally, we consider the B00(Ox00y00z00) frame as a new fixed global frameand consider the final body frame B(Oxyz) to coincide with B00 before thethird rotation. We now execute a ψ rotation about the z00-axis as shown in

roll . . . bo£no ljuljanje (rotacija oko poduºne ose x)

pitch . . . galopiranje (rotacija oko popre£ne ose y)

yaw . . . skretanje (rotacija oko vertikalne ose z)



Generalisane koordinate i sile (pokretni sistem)

10

Vehicle Planar DynamicsIn this chapter we develop a dynamic model for a rigid vehicle in a planarmotion. When the forward, lateral and yaw velocities are important and areenough to examine the behavior of a vehicle, the planar model is applicable.

10.1 Vehicle Coordinate Frame

The equations of motion in vehicle dynamics are usually expressed in a setof vehicle coordinate frame B(Cxyz), attached to the vehicle at the masscenter C, as shown in Figure 10.1. The x-axis is a longitudinal axis passingthrough C and directed forward. The y-axis goes laterally to the left fromthe driver’s viewpoint. The z-axis makes the coordinate system a right-hand triad. When the car is parked on a flat horizontal road, the z-axis isperpendicular to the ground, opposite to the gravitational acceleration g.

xz

y

FxFz

Fy

Mz

My

Mx

ϕθ

ψ

C

pq

r

FIGURE 10.1. Vehicle body coordinate frame B(Cxyz).

To show the vehicle orientation, we use three angles: roll angle ϕ aboutthe x-axis, pitch angle θ about the y-axis, and yaw angle ψ about the z-axis. Because the rate of the orientation angles are important in vehicle



Prostorni i materijalni sistem (kretanje u ravni)586 10. Vehicle Planar Dynamics

X

Y

C

x

y

v

β

d

ψ

BG1

2

3

4r2

r1

r3

r4

Yaw angleSideslipβ

ψCruise angleψ + β

FIGURE 10.3. Top view of a moving vehicle to show the yaw angle ψ betweenthe x and X axes, the sideslip angle β between the velocity vector v and thex-axis, and the crouse angle β + ψ between with the velocity vector v and theX-axis.

and then moving forward on the left side, the only unnumbered wheel is thewheel number 6.If the global position vector of the car’s mass center is given by

Gd =

∙XC

YC

¸(10.7)

and the body position vectors of the wheels are

Br1 =

∙a1w/2

¸(10.8)

Br2 =

∙a1−w/2

¸(10.9)

Br3 =

∙−a2−w/2

¸(10.10)

Br4 =

∙−a3−w/2

¸(10.11)

Br5 =

∙−a3w/2

¸(10.12)



Sile koje deluju na vozilo - reakcije podloge

9

Applied DynamicsDynamics of a rigid vehicle may be considered as the motion of a rigidbody with respect to a fixed global coordinate frame. The principles ofNewton and Euler equations of motion that describe the translational androtational motion of the rigid body are reviewed in this chapter.

9.1 Force and Moment

In Newtonian dynamics, the forces acting on a system of connected rigidbodied can be divided into internal and external forces. Internal forcesare acting between connected bodies, and external forces are acting fromoutside of the system. An external force can be a contact force, such as thetraction force at the tireprint of a driving wheel, or a body force, such asthe gravitational force on the vehicle’s body.

xz

y

Fy2Mz2

My2

Mx2

C

Fz2

Fx2

Fy3

Fz3

Fx3

My3

Mz3Mx3

FIGURE 9.1. The force system of a vehicle is the applied forces and moments atthe tireprints.

External forces and moments are called load, and a set of forces andmoments acting on a rigid body, such as forces and moments on the vehicleshown in Figure 9.1, is called a force system. The resultant or total force Fis the sum of all the external forces acting on a body, and the resultant or



Analiza kretanja drumskih vozila

Analiti£ko i numeri£ko modeliranje drumskih vozila

Motorno vozilo (npr. automobil) je pokretan sistem veomasloºene strukture

Moºe da se posmatra kao skup krutih tela koja su me�usobnopovezana u jednu celinu sa komplikovanim me�usobnimvezama

- razni zglobovi i mehanizmi- elasti£ne opruge- prigu²iva£i

Veliki broj stepeni slobode kretanja

Komplikovane nelinearne veze izme�u pojedinih delova sistema

Veoma sloºen sistem nelinearnih dif. jedna£ina kretanja



Ra£unarska simulacija automobila (SIMPACK)

SIMPACK AUTOMOTIVE

What is SIMPACK?

SIMPACK is a general-purpose multi-body

simulation (MBS) software tool which is used

to aid the development of any mechanical or

mechatronic device, ranging from single

components through to complete systems (e.g.

wind turbines, vehicles, and high performance

Formula 1 engines). All SIMPACK products are

100% compatible.

SIMPACK Automotive is an add-on module

tailored to the specific requirements of the

automotive simulation sector.

Applications:

Passenger cars/trucks/busses

Racing (Formula 1)/Formula Student

Land machinery

People movers/motorcycles

Component/complete system analyses

Handling/ride/NVH analyses

SiL, HiL and MiL applications

Stress/durability

DOE and optimization

Highlights:

Unlimited modeling flexibility

Extreme non-linear system behavior

Analyses in the frequency and time domain

Fully parameterized highly detailed

automotive database for all typical

suspension/driveline types available

Scalable detail and complexity

Batch jobs and automatic report generation

Interfaces to third party software

(CDTire, Delft-Tyre, FTire, TMEasy)

Accurate Fast Robust Versatile

Application



Ra£unarska simulacija automobila (veDYNA)




veDYNA - Driver Simulators

DYNAware User MeetingDYNAware User Meeting23rd June 2008




Driver SimulatorsExamples with veDYNA - FordExamples with veDYNA Ford

11All brand names, trademarks and registered trademarks belong to their respective owners • © TESIS GmbH Tokyo , 23 June 2008




Pojednostavljena klasi�kacija kretanja drumskih vozila

Pojednostavljeni slu£ajevi kretanja vozila sa smanjenimstepenima slobode kretanja

Bolja preglednost i razumevanje dominantnog kretanja

Klasi�kacija prema lokalnim osama vozila xyz

Poduºna dinamika vozilaPopre£na dinamika vozilaVertikalna dinamika vozila




Poduºna dinamika drumskih vozila

Vozilo se kre¢e pravolinijskom i/ili krivolinijskom putanjom uvertikalnoj ravni:

- kretanje po pravolinijskoj horizontalnoj deonici puta- kretanje po pravolinijskom usponu ili padu puta- kretanje u vertikalnoj krivini (konkavnoj i konveksnoj)

Usvaja se da se drumsko vozilo kre¢e (globalno posmatrano)translatorno, sa brzinom ~v(t) i ubrzanjem ~a(t)

Uzimaju se u obzir samo sile u ravni kretanja (ostale sezanemaruju)

Najjednostavniji ra£unski modeli kretanja




Popre£na dinamika drumskih vozila

Analiza kretanja vozila u horizontalnoj krivini

Analiza kretanja vozila u serpentinama

Analiza manevrisanja vozila tokom kretanja (preticanje i sl.)

Znatno sloºeniji ra£unski modeli kretanja




Vertikalna dinamika drumskih vozila

Analiza oscilacija vozila tokom kretanja (usled raznihporeme¢aja)

Analiza uticaja oscilacija vozila na udobnost putnika tokomvoºnje

Analiza uticaja oscilacija vozila na kontakt sa putem

Znatno sloºeniji ra£unski modeli kretanja




Specijalizovana dinami£ka analiza kretanja vozila

Analiza ekstremnih situacija tokom kretanja vozila

- nailazak na prepreku ili rupu na putu- prevrtanje vozila u krivini- prevrtanje vozila usled prepreke sa jedne strane vozila

Analiza udara vozila u nepokretnu prepreku

Analiza sudara vozila pri kretanju u istom smeru

Analiza sudara vozila pri kretanju u suprotnim smerovima(£eoni sudar)

Analiza sudara vozila pod (pravim) uglom

Najsloºeniji ra£unski modeli



Sadrºaj






Vozilo se kre¢e po horizontalnom (ili nagnutom) pravcu

Posmatraju se samo poduºne sile (u vertikalnoj ravni)

Vozilo je simetri£no u odnosu na vertikalnu poduºnu ravan

Glavni aspekt razmatranja su otpori kretanju i reakcije puta

Ravnomerno kretanje (~v = const) ili kretanje sa konstantnimubrzanjem ili ko£enjem (~a = const)



Parkiran automobil na ravnom putu

2

Forward Vehicle DynamicsStraight motion of an ideal rigid vehicle is the subject of this chapter.We ignore air friction and examine the load variation under the tires todetermine the vehicle’s limits of acceleration, road grade, and kinematiccapabilities.

2.1 Parked Car on a Level Road

When a car is parked on level pavement, the normal force, Fz, under eachof the front and rear wheels, Fz1 , Fz2 , are

Fz1 =1

2mg

a2l

(2.1)

Fz2 =1

2mg

a1l

(2.2)

where, a1 is the distance of the car’s mass center, C, from the front axle,a2 is the distance of C from the rear axle, and l is the wheel base.

l = a1 + a2 (2.3)

a1a2

2Fz2 2Fz1

x

z

C

mg

FIGURE 2.1. A parked car on level pavement.



Pravolinijsko kretanje drumskih vozila

Parkiran automobil na ravnom putu

Vozilo miruje na horizontalnom ravnom putu

Vozilo je simetri£no u odnosu na poduºnu vertikalnu ravan

Posmatra se kao "prosta greda"

Vertikalne reakcije puta se odre�uju iz uslova ravnoteºe∑Z = 0

∑M = 0

Dobija se

Fz1 =1

2mg

a2`

Fz2 =1

2mg

a1`

(` = a1 + a2)



Parkiran automobil na nagnutom putu46 2. Forward Vehicle Dynamics

2Fx2

2Fz1

xz

C

mg

a2

a1

2Fz2

h

a

φ

FIGURE 2.5. A parked car on inclined pavement.

These equations provide the brake force and reaction forces under the frontand rear tires.

Fz1 =1

2mg

a2lcosφ− 1

2mg

h

lsinφ (2.44)

Fz2 =1

2mg

a1lcosφ+

1

2mg

h

lsinφ (2.45)

Fx2 =1

2mg sinφ (2.46)

Example 46 Increasing the inclination angle.When φ = 0, Equations (2.36) and (2.37) reduce to (2.1) and (2.2). By

increasing the inclination angle, the normal force under the front tires ofa parked car decreases and the normal force and braking force under therear tires increase. The limit for increasing φ is where the weight vectormg goes through the contact point of the rear tire with the ground. Such anangle is called a tilting angle.

Example 47 Maximum inclination angle.The required braking force Fx2 increases by the inclination angle. Be-

cause Fx2 is equal to the friction force between the tire and pavement, itsmaximum depends on the tire and pavement conditions. There is a specificangle φM at which the braking force Fx2 will saturate and cannot increaseany more. At this maximum angle, the braking force is proportional to thenormal force Fz2

Fx2 = μx2Fz2 (2.47)




Parkiran automobil na nagnutom putu

Vozilo miruje na nagnutom putu

Aplicirana je ru£na ko£nica na zadnje to£kove

Reakcije puta se odre�uju iz uslova ravnoteºe∑X = 0

∑Z = 0

∑M = 0

Dobija se

Fz1 =1

2mg

a2`cosφ− 1

2mg

h

`sinφ

Fz2 =1

2mg

a1`cosφ+

1

2mg

h

`sinφ

Fx2 =1

2mg sinφ



Kretanje automobila po horizontalnom putu50 2. Forward Vehicle Dynamics

a1a2

2Fz2 2Fz1

x

z

C

mg2Fx1

ha

2Fx2

FIGURE 2.7. An accelerating car on a level pavement.

Assumingμx1 = μx2 = μx (2.83)

will provide

Fz1 =1

2mg

a2lcosφM −

1

2mg

h

lsinφM (2.84)

Fz2 =1

2mg

a1lcosφM +

1

2mg

h

lsinφM (2.85)

tanφM = μx. (2.86)

2.3 Accelerating Car on a Level Road

When a car is speeding with acceleration a on a level road as shown inFigure 2.7, the vertical forces under the front and rear wheels are

Fz1 =1

2mg

a2l− 12mg

h

l

a

g(2.87)

Fz2 =1

2mg

a1l+1

2mg

h

l

a

g. (2.88)

The first terms, 12mg a2l and12mg a1l , are called static parts, and the second

terms ±12mg hl

ag are called dynamic parts of the normal forces.

Proof. The vehicle is considered as a rigid body that moves along a hor-izontal road. The force at the tireprint of each tire may be decomposedto a normal and a longitudinal force. The equations of motion for the ac-celerating car come from Newton’s equation in x-direction and two static




Kretanje automobila po horizontalnom putu

Vozilo se kre¢e po horizontalnom putu sa ubrzanjem ~a = a~ı

Zakon o kretanju centra mase i Zakon o promeni momentakoli£ine kretanja:

m~a = ~FR JS ε =M(S)R

Vozilo se kre¢e translatorno (kao mat. ta£ka), pa je ε = 0,

odnosno, M(S)R = 0

Predpostavlja se pogon na sva £etiri to£ka 2Fx1 i 2Fx2





Zakon o kretanju centra mase (zanemaren otpor vazduha):

ma = 2Fx1 + 2Fx2 (1)

0 = 2Fz1 + 2Fz2 −mg (2)

Zakon o promeni momenta koli£ine kretanja (za sredi²te mase):

0 = −2Fz1 a1 − (2Fx1 + 2Fx2)h+ 2Fz2 a2 (3)

Tri jedna£ine, 4 nepoznate reakcije puta Fx1, Fx2, Fz1 i Fz2

(nepoznato je i ubrzanje a, ali se posmatra kao poznatparametar)





Unose¢i jedn. (1) u (3) elimini²u se Fx1 i Fx2:

0 = 2Fz1 a1 +mah− 2Fz2 a2 (4)

Iz jedn. (2) se elimini²e, npr. Fz1:

2Fz1 = mg − 2Fz2

pa se dobija, unose¢i u (4),

0 = mg a1 − 2Fz2 a1 +mah− 2Fz2 a2





Dobija se re²enje za vertikalne reakcije puta:

Fz2 =1

2

a1`mg +

1

2

h

`mg

a

g

Fz1 =1

2

a2`mg − 1

2

h

`mg

a

g

kao i za zbir horizontalnih reakcija puta:

Fx1 + Fx2 =1

2ma =

1

2mg

a

g





Stati£ke komponente vertikalnih reakcija puta:

Fz2,st =1

2

a1`mg Fz1,st =

1

2

a2`mg

Dinami£ke komponente vertikalnih reakcija puta:

Fz2,din =1

2

h

`mg

a

gFz1,din = −1

2

h

`mg

a

g

Dinami£ki deo vertikalnih reakcija zavisi od ubrzanja vozila,kao i od vertikalnog poloºaja centra mase vozila





Tokom voºnje sa ubrzanim kretanjem (a > 0) - ubrzavanje,prednji to£kovi su rastere¢eni (za dinami£ki deo), a zadnjito£kovi su vi²e optere¢eni u odnosu na stati£ki deo

Tokom voºnje sa usporenim kretanjem (a < 0) - ko£enje,prednji to£kovi su vi²e optere¢eni (za dinami£ki deo), a zadnjito£kovi su rastere¢eni u odnosu na stati£ki deo

Za vozilo sa prednjom vu£om je Fx2 = 0, dok su komponentevertikalnih reakcija puta iste (kao i za pogon na sva 4 to£ka)

Za vozilo sa zadnjom vu£om je Fx1 = 0, dok su komponentevertikalnih reakcija puta iste (kao i za pogon na sva 4 to£ka)




Maksimalno ubrzanje na horizontalnom putu

Ubrzanje sredi²ta mase vozila je:

a =1

m(2Fx1 + 2Fx2)

Maksimalno ubrzanje vozila je proporcionalno sa "vu£nim"silama, odn. sa horizontanim reakcijama podloge

Maksimalno ubrzanje je proporcionalno sa koe�cijentomprianjanja izme�u guma i puta (jer je Fx = µx Fz)

Predpostavlja se da su koe�cijenti prianjanja prednjih i zadnihto£kova me�usobno isti i da se maksimum ostvari u isto vreme

Horizontalne komponente reakcija podloge su, tada:

Fx1 = ±µx Fz1 Fx2 = ±µx Fz2





Iz uslova ravnoteºe vertikalnih sila je 2Fz1 + 2Fz2 = mg,

pa se dobija maksimalno ubrzanje vozila

a = ± 1

mµx(2Fz1 + 2Fz2) = ±

1

mµxmg

Prema tome, maksimalno ubrzanje vozila je dato sa

a = ±µx g

Maksimalno ubrzanje (usporenje) vozila direktno zavisi odkoe�cijenta prianjanja µx (za asfaltne i betonske kolovoze jeµx ≈ 0.8÷ 0.9)





Sa stanovi²ta udobnosti voºnje, ubrzanje (usporenje) vozila upoduºnom pravcu se ograni£ava

Ograni£enje ubrzanja vozila u poduºnom pravcu:

Ubrzanje vozila Udobnost voºnjeu [m/s2] u faktoru od g

do 2.65 0.27 g udobna voºnjado 3.45 0.35 g neudobna voºnjado 4.25 0.43 g posebni uslovi

(gde je g = 9.81m/s2 ubrzanje zemljine teºe)





Sa stanovi²ta udobnosti voºnje, ubrzanje (usporenje) vozila upopre£nom pravcu se tako�e ograni£ava

Ograni£enje ubrzanja vozila u popre£nom pravcu:

Ubrzanje vozila Udobnost voºnjeu [m/s2] u faktoru od g

do 2.50 0.25 g udobna voºnjado 3.00 0.31 g neudobna voºnjado 3.50 0.36 g posebni uslovi

(gde je g = 9.81m/s2 ubrzanje zemljine teºe)





Pogon na zadnjoj osovini:

a

g=

µx

1− µx h`

a1`

ia

g≤ a2

h

Pogon na prednjoj osovini:

a

g=

µx

1 + µxh`

(1− a1`)




Maksimalno usporavanje na horizontalnom putu

Ko£nice na prednjoj osovini:

a

g= − µx

1− µx h`

(1− a1`)

Ko£nice na zadnjoj osovini:

a

g= − µx

1 + µxh`

a1`



Maksimalno usporavanje na horizontalnom putu




Trenutna promena ubrzanja - "trzaj"

Vektor poloºaja materijalne ta£ke . . .~r = ~r(t)

Vektor brzine materijalne ta£ke . . .~v = ~̇r(t)

Vektor ubrzanja materijalne ta£ke . . .~a = ~̇v(t) = ~̈r(t)

Vektor promene ubrzanja materijalne ta£ke - trzaj ("jerk")

~t = ~̇a(t) = ~̈v(t) =...~r (t)

Najve¢a dozvoljena vrednost trzaja, sa stanovi²ta udobnostivoºnje, je

|~t|max = 2.50 [m/s3]



Kretanje automobila po nagnutom putu56 2. Forward Vehicle Dynamics

2Fx2

2Fz1

xz

C

mg

a2

a1

2Fz2

h

a

2Fx1

φ

FIGURE 2.9. An accelerating car on inclined pavement.

The dynamic parts, ±12mg hl

ag , depend on acceleration a and height h of

mass center C and remain unchanged, while the static parts are influencedby the slope angle φ and height h of mass center.

Proof. The Newton’s equation in x-direction and two static equilibriumequations, must be examined to find the equation of motion and groundreaction forces. X

Fx = ma (2.123)XFz = 0 (2.124)XMy = 0. (2.125)

Expanding these equations produces three equations for four unknownsFx1 , Fx2 , Fz1 , Fz2 .

2Fx1 + 2Fx2 −mg sinφ = ma (2.126)

2Fz1 + 2Fz2 −mg cosφ = 0 (2.127)

2Fz1a1 − 2Fz2a2 + 2 (Fx1 + Fx2)h = 0 (2.128)

It is possible to eliminate (Fx1 + Fx2) between the first and third equations,and solve for the normal forces Fz1 , Fz2 .

Fz1 = (Fz1)st + (Fz1)dyn

=1

2mg

µa2lcosφ− h

lsinφ

¶− 12ma

h

l(2.129)




Kretanje automobila po nagnutom putu

Zakon o kretanju centra mase (zanemaren otpor vazduha):

ma = 2Fx1 + 2Fx2 −mg sinφ (5)

0 = 2Fz1 + 2Fz2 −mg cosφ (6)

Zakon o promeni momenta koli£ine kretanja (za sredi²te mase):

0 = −2Fz1 a1 − (2Fx1 + 2Fx2)h+ 2Fz2 a2 (7)

Tri jedna£ine, 4 nepoznate reakcije puta Fx1, Fx2, Fz1 i Fz2

(nepoznato je i ubrzanje a, ali se posmatra kao poznatparametar)





Zbir tangencijalnih reakcija se iz (5) unosi u (7)

Jedna od normalnih reakcija se iz (6) unosi u (7)

Re²enje za normalne reakcije i za zbir tangencijalnih reakcija je

Fz1 =1

2mg(

a2`cosφ− h

`sinφ)− 1

2ma

h

`

Fz2 =1

2mg(

a1`cosφ+

h

`sinφ) +

1

2ma

h

`

Fx1 + Fx2 =1

2ma+

1

2mg sinφ





Ubrzanje centra mase (pri kretanju uzbrdo):

a =1

m(2Fx1 + 2Fx2) −g sinφ

Ubrzanje centra mase (pri kretanju nizbrdo):

a =1

m(2Fx1 + 2Fx2) +g sinφ

U slu£aju pogona samo na zadnje ili samo na prednje to£koveje Fx1 = 0 ili Fx2 = 0





Vozilo sa pogonom na zadnje to£kove:

Vu£ne (tangencijalne) sile . . .Fx1 = 0 i Fx2 6= 0Normalne reakcije puta Fz1 i Fz2 . . . iste kao i za 4WDZa isto ubrzanje kao i za 4WD vu£na sila Fx2 mora da budeve¢a nego za slu£aj pogona 4WD

Vozilo sa pogonom na prednje to£kove:

Jedina razlika je . . .Fx1 6= 0 i Fx2 = 0

Normalne reakcije su iste bez obzira da li je pogon na prednje,zadnje ili na sve to£kove (za umereno ubrzanje i prav put)

Prednosti i mane pogona se javljaju u manevrisanjima, naklizavom putu i kada su potrebna maksimalna ubrzanja




Maksimalno ubrzanje na nagnutom putu - prianjanje

Ubrzanje sredi²ta mase vozila (za kretanje uzbrdo) je:

a =1

m(2Fx1 + 2Fx2)− g sinφ

Maksimalno ubrzanje vozila je proporcionalno sa "vu£nim"silama, odn. sa horizontanim reakcijama podloge

Maksimalno ubrzanje je proporcionalno sa koe�cijentomprianjanja izme�u guma i puta (jer je Fx = µx Fz)

Predpostavlja se da su koe�cijenti prianjanja prednjih i zadnihto£kova me�usobno isti i da se maksimum ostvari u isto vreme

Horizontalne komponente reakcija podloge su, tada:

Fx1 = ±µx Fz1 Fx2 = ±µx Fz2





Iz uslova ravnoteºe sila upravno na put je

2Fz1 + 2Fz2 = mg cosφ,

pa se dobija maksimalno ubrzanje vozila

a = ± 1

mµx(2Fz1 + 2Fz2)− g sinφ = ±µx g cosφ− g sinφ

Prema tome, maksimalno ubrzanje vozila je dato sa

amax

g= ±µx cosφ− sinφ





Za ubrzano kretanje uzbrdo . . . a > 0, φ > 0

Za usporeno kretanje nizbrdo . . . a < 0, φ < 0

Koe�cijent prianjanja treba da bude

µx ≥ | tanφ|

Za suv i dobar kolovoz (beton, asfalt) je µx ≈ 0.8÷ 0.9

Uglovi nagiba puta sa stanovi²ta prianjanja

- za µx = 0.8 . . .φ ≈ 390

- za µx = 0.9 . . .φ ≈ 420




Maksimalno ubrzanje na nagnutom putu - geometrija

Normalne komponente reakcija puta:

Fz1 =1

2mg(

a2`cosφ− h

`sinφ)− 1

2ma

h

`

Fz2 =1

2mg(

a1`cosφ+

h

`sinφ) +

1

2ma

h

`

Reakcije veza moraju da budu pritisci na podlogu:

Fz1 ≥ 0 Fz2 ≥ 0

Iz relacije Fz1 ≥ 0 se dobija

a

g≤ a2

hcosφ− sinφ





Iz relacije Fz2 ≥ 0 se dobija

a

g≥ −a1

hcosφ− sinφ

Zajedno, oba uslova mogu da se prikaºu kao

−a1h

cosφ ≤ a

g+ sinφ ≤ a2

hcosφ

U grani£nom slu£aju za a→ 0 se dobija:

−a1h≤ tanφ ≤ a2

h





Za realne prose£ne podatke o putni£kom automobilu:

- osovinski razmak . . . ` = 2.6 [m]- rastojanja osovina do teºi²ta . . . a1 ≈ a2 = 0.5 ` = 1.3 [m]- visina teºi²ta . . .h ≈ 0.56 [m]

Geometrijski uslov nagiba puta

tanφ =a1h

= 2.321 ⇒ φ ≈ 670

Realni (dozvoljeni) poduºni nagibi puteva su neuporedivo manji




Maksimalni dozvoljeni poduºni nagibi puteva

Niveleta puta je prostorna linija - osovina kolovoza

Maksimalni mogu¢i poduºni nagibi nivelete puta zavise odsnage motora vozila i uslova prianjanja

Za prose£na motorna vozila (automobile) je oko 30%(φ ≈ 170)

Za prose£na teretna vozila je oko 15% (φ ≈ 8.50)

Maksimalan dozvoljeni poduºni nagib nivelete zavisi odpredvi�ene brzine kretanja vozila i od vrste puta

Red veli£ine dozvoljenih poduºnih nagiba puta je 5÷ 12%



Automobil sa prikolicom po nagnutom putu

Zglobna veza izme�u automobila i prikolice (sa jednom osovinom)Zanemaruje se otpora vazduha60 2. Forward Vehicle Dynamics

2Fx2

2Fz1

xz

C

mg

a2

a1

2Fz2

h

a

2Fx1

φ

Ct

mt g

b2

b1

b3

2Fz3

FIGURE 2.11. A car moving on an inclined road and pulling a trailer.

2Fx1 + 2Fx2 − Fxt −mg sinφ = ma (2.152)

2Fz1 + 2Fz2 − Fzt −mg cosφ = 0 (2.153)

2Fz1a1 − 2Fz2a2 + 2 (Fx1 + Fx2)h

−Fxt (h− h1) + Fzt (b1 + a2) = 0 (2.154)

If the value of traction forces Fx1and Fx2 are given, then these are six equa-tions for six unknowns: a, Fxt , Fzt , Fz1 , Fz2 , Fz3 . Solving these equationsprovide the following solutions:

a =2

m+mt(Fx1 + Fx2)− g sinφ (2.155)

Fxt =2mt

m+mt(Fx1 + Fx2) (2.156)

Fzt =h1 − h2b2 − b3

2mt

m+mt(Fx1 + Fx2) +

b3b2 − b3

mt g cosφ (2.157)

Fz1 =b32l

µ2a2 − b1b2 − b3

mt +a2b3m

¶g cosφ

+

∙2a2 − b1b2 − b3

(h1 − h2)mt − h1mt − hm

¸Fx1 + Fx2l (m+mt)

(2.158)



Automobil sa prikolicom po nagnutom putu2. Forward Vehicle Dynamics 61

Fzt2Fx2

2Fz1

xz

C

mg

2Fz2

h

a

2Fx1

φ

Ct

mt g

Fxt

FztFxt

φ

2Fz3

h1

h1h2

a1

a2

b1

b2

b3

FIGURE 2.12. Free-body-diagram of a car and the trailer when moving on anuphill road.

Fz2 =b32l

µa1 − a2 + b1

b2 − b3mt +

a1b3m

¶g cosφ

+

∙a1 − a2 + b1

b2 − b3(h1 − h2)mt + h1mt + hm

¸Fx1 + Fx2l (m+mt)

(2.159)

Fz3 =1

2

b2b2 − b3

mt g cosφ+h1 − h2b2 − b3

mt

m+mt(Fx1 + Fx2) (2.160)

l = a1 + a2. (2.161)

However, if the value of acceleration a is known, then unknowns are: Fx1+Fx2 , Fxt , Fzt , Fz1 , Fz2 , Fz3 .

Fx1 + Fx2 =1

2(m+mt) (a+ g sinφ) (2.162)

Fxt = mt (a+ g sinφ) (2.163)

Fzt =h1 − h2b2 − b3

mt (a+ g sinφ) +b3

b2 − b3mt g cosφ (2.164)





Prikolica je ukupne mase mt i ima samo jednu osovinu

Vr²i se dekompozicija automobila i prikolice

U zglobu izme�u automobila i prikolice (kuka) se javljaunutra²nja sila veze, sa komponentama Fxt i Fzt

Jedna£ine kretanja za prikolicu

mt a = Fxt −mt g sinφ

0 = 2Fz3 + Fzt −mt g cosφ

0 = 2Fz3b3 − Fztb2 − Fxt(h2 − h1)(8)





Jedna£ine kretanja za automobil (pogon na sva 4 to£ka)

ma = 2Fx1 + 2Fx2 − Fxt −mg sinφ

0 = 2Fz1 + 2Fz2 − Fzt −mg cosφ

0 = 2Fz2a2 − 2Fz1a1 − (2Fx1 + 2Fx2)h

+ Fzt(b1 + a2)− Fxt(h− h1)

(9)

Ako se smatra da su "vu£ne sile" Fx1 i Fx2 poznate, ondasistem (8) i (9) sadrºi 6 jedna£ina po nepoznatim veli£inama:

- ubrzanje vozila i prikolice . . . a- normalne reakcije puta . . .Fz1, Fz2 i Fz3

- sile veze izme�u automobila i prikolice . . .Fxt, Fzt


mehanika vo nje - odsek za puteve, eleznice i aerodrome · dinamika vozila u poduºnom ravcup...

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