MEHANIKA VO NJE - Odsek za puteve, eleznice i aerodrome · Kretanje vozila u ravni Analiza vibracija vozila Diferencijalne jedna£ine kretanja Vremenski odgovor Sile koje deluju na

Kretanje vozila u ravniAnaliza vibracija vozila

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


Sadrºaj

1 Kretanje vozila u ravniDiferencijalne jedna£ine kretanjaVremenski odgovor

2 Analiza vibracija vozilaVe²anje drumskih vozilaRa£unski modeli u analizi vibracijaRa£unski model £etvrtine vozilaRa£unski model polovine vozila



Diferencijalne jedna£ine kretanjaVremenski odgovor

Sadrºaj






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




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




Diferencijalne jedna£ine kretanja vozila u ravni590 10. Vehicle Planar Dynamics

X

Y

C

xy FxFy

v

βvx

vy

d

ψ

BG

FIGURE 10.6. A rigid vehicle in a planar motion.

The velocity vector of the vehicle, expressed in the body frame, is

BvC =

⎡⎣ vxvy0

⎤⎦ (10.29)

where vx is the forward component and vy is the lateral component of v.The rigid body equations of motion in the body coordinate frame are:

BF = BRGGF

= BRG

¡m GaB

¢= m B

GaB

= m BvB +m BGωB × BvB. (10.30)

BM =Gd

dtBL

= BGLB

= BL+ BGωB × BL

= BI BGωB +

BGωB ×

¡BI B

GωB

¢. (10.31)

The force, moment, and kinematic vectors for the rigid vehicle are:

BFC =

⎡⎣ FxFy0

⎤⎦ (10.32)




Kretanje vozila u ravni

Diferencijalne jedna£ine kretanja vozila u ravni

Zakon o kretanju centra mase i promene momenta koli£inekretanja:

m~v = ~FR Jzωz =Mz

Dobija se (u sistemu pokretnih koordinata vozila):

mvx −mωz vy = Fx

mvy +mωz vx = Fy

J ωz =Mz

(1)




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 (za ravno kretanje)

Sile na kontaktu svakog to£ka sa podlogom (u ravni podloge):

- vu£na sila ili sila ko£enja Rx = µxRz

- sila otpora kotrljanju Rr = −µr Rz- bo£na reakcija puta (proporcionalna sa uglom bo£nog klizanja)

Ry = −Cα α

- spreg poravnanja (momenat skretanja) Mz = Ry axα

Ne posmatraju se (direktno) gravitacione sile - kretanje uhorizontalnoj ravni

Obi£no se zanemaruje visinska razlika izme�u podloge (otiskagume), centra to£ka i centra mase vozila (sile u ravni)




Sile na kontaktu to£ka sa podlogom600 10. Vehicle Planar Dynamics

xw

v

α

xβ

δ

xw v xβ

δ

y

yw

y

yw

α

(a) (b)

FIGURE 10.9. Angular orientation of a moving tire along the velocity vector vat a sideslip angle α and a steer angle δ.

10.3.2 Tire Lateral Force

Figure 10.9(a) illustrates a tire, moving along the velocity vector v at asideslip angle α. The tire is steered by the steer angle δ. If the angle betweenthe velocity vector v and the vehicle x-axis is shown by β, then

α = β − δ. (10.104)

The lateral force, generated by a tire, is dependent on sideslip angle α thatis proportional to the sideslip for small α.

Fy = −Cα α

= −Cα (β − δ) (10.105)

Proof. A tire coordinate frame Bw(xw, yw) is attached to the tire at thecenter of tireprint as shown in Figure 10.9(a). The orientation of the tireframe is measured with respect to another coordinate frame, parallel tothe vehicle frame B(x, y). The angle between the x and xw axes is the tiresteer angle δ, measured about the z-axis. The tire is moving along the tirevelocity vector v. The angle between the xw-axis and v is the sideslip angle






Projektovanjem sila na ose koordinatnog sistema to£kaxwywzw se dobija:

Fxw = Rx −Rr cosαFyw = Ry −Rr sinαMzw = Ry axα

(2)

gde je Rx = µxRz, Rr = −µr Rz, Ry = −Cα α




Bo£na sila na kontaktu to£ka sa podlogom600 10. Vehicle Planar Dynamics

xw

v

α

xβ

δ

xw v xβ

δ

y

yw

y

yw

α

(a) (b)

FIGURE 10.9. Angular orientation of a moving tire along the velocity vector vat a sideslip angle α and a steer angle δ.

10.3.2 Tire Lateral Force

Figure 10.9(a) illustrates a tire, moving along the velocity vector v at asideslip angle α. The tire is steered by the steer angle δ. If the angle betweenthe velocity vector v and the vehicle x-axis is shown by β, then

α = β − δ. (10.104)

The lateral force, generated by a tire, is dependent on sideslip angle α thatis proportional to the sideslip for small α.

Fy = −Cα α

= −Cα (β − δ) (10.105)

Proof. A tire coordinate frame Bw(xw, yw) is attached to the tire at thecenter of tireprint as shown in Figure 10.9(a). The orientation of the tireframe is measured with respect to another coordinate frame, parallel tothe vehicle frame B(x, y). The angle between the x and xw axes is the tiresteer angle δ, measured about the z-axis. The tire is moving along the tirevelocity vector v. The angle between the xw-axis and v is the sideslip angle





Bo£na sila na kontaktu to£ka sa podlogom

Ravan to£ka zaklapa ugao δ sa osom x vozila - to je ugaookretanja to£ka: δ = ∠(x, xw)

Vektor brzine to£ka zaklapa ugao bo£nog klizanja α sa osomxw to£ka: α = ∠(xw, ~v)

Ugao izme�u ose x vozila i vektora brzine to£ka β je globalniugao bo£nog klizanja: β = ∠(x,~v)

Na slici (a) su prikazani pozitivni uglovi i vaºi relacijaα = β − δBo£na reakcija podloge je proporcionalna sa uglom bo£nogklizanja

Ry = −Cy α = −Cy(β − δ)





Bo£na sila na kontaktu to£ka sa podlogom

U realnom okretanju to£kova vektor brzine je izme�u osa x ixw, kao ²to je dato na slici (b)

Zbog �eksibilnosti gume, vektor brzine zaklapa manji ugao odugla okretanja to£ka

Prema tome, β < δ , odn. pozitivan ugao okretanja to£kageneri²e negativan ugao bo£nog klizanja

Kako je, videti sl. (b), −α+ β = δ, opet vaºi relacija

α = β − δ

Nastajanje ugla bo£nog klizanja generi²e bo£nu reakcijupodloge Ry = −Cα α




Sile na kontaktu sa podlogom to£ka br. 1598 10. Vehicle Planar Dynamics

C

x

y

1

Fx1

Fy1

1δ

xw

yw

Mz1

Fxw

Fyw

BBw

B1

FIGURE 10.8. The force system at the tireprint of tire number 1.

BFy =Xi

Fyi

=Xi

Fyw cos δi +Xi

Fxw sin δi (10.88)

BMz =Xi

Mzi +Xi

xiFyi −Xi

yiFxi . (10.89)

Proof. The coordinate frame of the wheel is a local coordinate called thewheel frame shown by T (xw, yw, zw) or Bw. For simplicity, we ignore thedifference between the tire frame at the center of tireprint and wheel frameat the wheel center. The force system generated at the tireprint in thewheel frame is

BwFw = Fxw ı1 + Fyw j1 (10.90)BwMw = Mzw k1 (10.91)

where

Fxw = Fxw1 − Fr1 cosα (10.92)

Fyw = Fyw1 − Fr1 sinα (10.93)

Mzw = Mzw1(10.94)

The wheel force in the xw-direction, Fxw , is a combination of the longitu-dinal force Fxw1 , defined by (3.96) or (4.59), and the tire roll resistance Fr1defined in (3.64). The wheel force in the yw-direction, Fyw , is a combina-tion of the lateral force Fyw1 defined by (3.130) and (3.153), and the tireroll resistance Fr1 defined in (3.64). The wheel moment in the zw-direction,






Koordinatni sistem gume xtytzt je u sredi²tu otiska gume (napodlozi)

Koordinatni sistem to£ka xwywzw je u sredi²tu mase to£ka

Koordinatni sistemi gume i to£ka su me�usobno paralelni, alije po£etak sistema to£ka izdignut za zt = rd od podloge

Koordinatni sistem vozila Cxyz je u centru mase vozila

Usvaja se da su svi ovi sistemi u istoj ravni - u ravni puta XY






Sile na kontaktu svakog to£ka sa podlogom se izraze ukoordinatnom sistemu to£ka (gume) xwywzw, pa se projektujuna sistem koji je u centru to£ka, ali je paralelan sakoordinatnim sistemom vozila Cxyz

Zatim se redukuju sile sa to£ka na centar mase vozila C

Dobija se, za to£ak broj i (i=1,. . . ,4):

Fx,i = Fxw,i cos δi − Fyw,i sin δiFy,i = Fxw,i sin δi + Fyw,i cos δi

Mz,i =Mzw,i + xiFy,i − yiFx,i(3)

gde je δi = ∠(x, xw,i)




Sile na kontaktu to£kova sa podlogom602 10. Vehicle Planar Dynamics

C

x

y

1 2

34

Fx1 Fx2

Fx3Fx4

Fy1 Fy2

Fy3Fy4

1δ 2δ

Mz1 Mz2

Mz4Mz3

FIGURE 10.10. A front-wheel-steering four-wheel vehicle and the forces in thexy-plane acting at the trireprints.

The vehicle lateral force Fy and momentMz depend on only the front andrear wheels’ lateral forces Fyf and Fyr , which are functions of the wheelssideslip angles αf and αr. They can be approximated by the followingequations:

Fy =

µ−a1vx

Cαf +a2vx

Cαr

¶r − (Cαf + Cαr)β + Cαfδ (10.114)

Mz =

µ−a

21

vxCαf −

a22vx

Cαr

¶r − (a1Cαf − a2Cαr)β + a1Cαfδ (10.115)

where Cαf = CαfL+CαfR and Cαr = CαrL+CαrR are equal to the sideslipcoefficients of the left and right wheels in front and rear, respectively.

Cαf = CαfL + CαfR (10.116)

Cαr = CαrL + CαrR (10.117)

Proof. For the two-wheel vehicle, we use the cot-average (7.3) of the outerand inner steer angles as the only steer angle δ.

cot δ =cot δo + cot δi

2(10.118)






Za sve to£kove se dobija torzor sistema sila za centar masevozila (sistem Cxyz):

Fx =∑i

Fxw,i cos δi −∑i

Fyw,i sin δi

Fy =∑i

Fxw,i sin δi +∑i

Fyw,i cos δi

Mz =∑i

Mzw,i +∑i

xiFy,i −∑i

yiFx,i

(4)





Diferencijalne jedna£ine kretanja vozila u ravni

Prema tome, unose¢i sile (4) u jedna£ine (1), dobija se (usistemu pokretnih koordinata vozila) sistem diferencijalnihjedna£ina ravnog kretanja vozila:

mvx −mωz vy = Fx

mvy +mωz vx = Fy

J ωz =Mz

(5)




Model vozila sa dva to£ka - "Model bicikla"

Pretpostavke modela vozila sa dva to£ka

Zanemaruje se rotacija vozila oko poduºne ose x

Ravan vozila xy je uvek paralelna sa ravni puta XY

Vozilo se aproksimira kao model sa dva to£ka ("model bicikla")

Dva to£ka na jednoj osovini se prikazuju kao jedan zajedni£ki

Telo vozila se prikazuje kao jedan ekvivalentan ²tap

Ista masa, poloºaj teºi²ta i osovina kao i posmatrano vozilo




Model vozila sa dva to£ka10. Vehicle Planar Dynamics 603

O

δ

lR

R1

C

a2

r

a1

vr

Center of rotation

fβfα

vβ

rα

vf

δ

FIGURE 10.11. A two-wheel model for a vehicle moving with no roll.

Furthermore, we define a single sideslip coefficient Cαf and Cαr as (10.116)and (10.117) for the front and rear wheels. The coefficient Cαf and Cαr areequal to the sum of the left and right wheels’ sideslip coefficients.Employing Equations (10.87)-(10.89) and ignoring the aligning moments

Mzi , the applied forces on the two-wheel vehicle are:

Fx = Fx1 cos δ1 + Fx2 cos δ2 − Fy1 sin δ1 − Fy2 sin δ2

= Fxf cos δ + Fxr − Fyf sin δ (10.119)

Fy = Fy1 cos δ1 + Fy2 cos δ2 + Fx1 sin δ1 + Fx2 sin δ2

= Fyf cos δ + Fyr + Fxf sin δ (10.120)

Mz = a1Fyf − a2Fyr (10.121)

The force equations can be approximated by the following equations, if weassume δ small.

Fx ≈ Fxf + Fxr (10.122)

Fy ≈ Fyr + Fyr (10.123)

Mz ≈ a1Fyf − a2Fyr (10.124)

Assume the wheel number i of a rigid vehicle is located at (xi, yi) in the





Rezultuju¢e sile kod modela sa dva to£ka

Upravlja se samo sa prednjim to£kom (okre¢e se za ugao δ)

Rezultuju¢e sile, odn. torzor za centar mase vozila, (4), sudate sa:

Fx = Fx,f cos δ + Fx,r − Fy,f sin δFy = Fx,f sin δ + Fy,f cos δ + Fy,r

Mz = a1Fy,f − a2Fy,r(6)

gde su (Fx,f , Fx,r), kao i (Fy,f , Fy,r) sile u pravcu ose xw i ywprednjeg i zadnjeg to£ka (f, r)






Ako je ugao upravljanja δ relativno mali, onda je cos δ ≈ 1,kao i sin δ ≈ 0

Rezultuju¢e sile, odn. torzor za centar mase vozila, (6), sudate sa:

Fx ≈ Fx,f + Fx,r

Fy ≈ Fy,f + Fy,r

Mz = a1Fy,f − a2Fy,r(7)

Ukupna bo£na sila vozila Fy, kao i momenat Mz, zavise samood bo£nih sila to£kova Fy,f i Fy,r

Bo£ne sile to£kova su funkcije uglova bo£nog klizanja αf i αr






Bo£na sila vozila Fy, kao i momenat Mz mogu da seaproksimiraju na slede¢i na£in:

Fy = (−a1vxCα,f +

a2vxCα,r) r − (Cα,f + Cα,r)β

+ Cα,fδ

Mz = (−a21

vxCα,f −

a22vxCα,r) r − (a1Cα,f − a2Cα,r)β

+ a1Cα,fδ

(8)

gde su Cα,f i Cα,r jednaki zbirovima koe�cijenata bo£nogklizanja to£kova levo i desno na prednjoj i zadnjoj osovini:

Cα,f = C lα,f + Cdα,f kao i Cα,r = C lα,r + Cdα,rS.�ori¢ Mehanika voºnje





Relacije (8) se komplikovano izvode

Parametri Cα,f i Cα,r su koe�cijenti bo£nog klizanja prednjih izadnjih to£kova

Parametar δ je ugao okretanja prednjeg to£ka modela sa dvato£ka:

cot δ =cot δo + cot δi

2

Parametar β je globalni ugao klizanja vozila: β = ∠(x,~v). Zamali ugao δ je

β =vyvx

Parametar r je izvod po vremenu ugla skretanja ψ (ugaonabrzina vozila): r = ψ gde je ψ = ∠(X, x)






Relacije (8) zavise od tri parametra r, β i δ i mogu da seprikaºu kao linearne funkcije parametara r, β i δ:

Fy = Fy(r, β, δ) = Cr r + Cββ + Cδδ

Mz =Mz(r, β, δ) = Dr r +Dββ +Dδδ(9)

Koe�cijenti bo£ne sile i momenta skretanja su nagibi (izvodi)krivih linija kojima se prikazuje Fy i Mz u funkciji r, β, δ






Koe�cijenti bo£ne sile Fy:

Cr = −a1vxCα,f +

a2vxCα,r

Cβ = −(Cα,f + Cα,r)

Cδ = Cα,f

(10)

Koe�cijenti momenta skretanja Mz:

Dr = −a21vxCα,f −

a22vxCα,r

Dβ = −(a1Cα,f − a2Cα,r)Dδ = a1Cα,f

(11)





Diferencijalne jedna£ine kretanja

Diferencijalne jedna£ine kretanja (5), gde su sile date sa (7),za relativno mali ugao okretanja to£ka, mogu da se prikaºu uobliku:

vx =1

mFx + rvy =

1

m(Fx,f + Fx,r) + rvy (12)

{vyr

}=

[Cβmvx

Crm − vx

DβJzvx

DrJz

]{vyr

}+

{CδmDδIz

}δ (13)

Za vozilo koje se kre¢e sa konstantnom brzinom unapred jevx = 0

Jedna£ina (12) postaje nezavisna, a bo£na brzina vy i ugaonebrzina r su dati sistemom jedna£ina (13)






Ako se posmatra da je ugao okretanja δ ulazna komanda, a dasu bo£na brzina vy i ugaona brzina r odgovor sistema, ondasistem jedna£ina (13) predstavlja linearan kontrolni sistem:

q = Aq + u (14)

Vektor kontrolnih promenljivih i ulazni vektor su dati sa:

q =

{vyr

}u =

{CδmDδIz

}δ (15)






Matrica koe�cijenat kontrolnog sistema je data sa:

A =

[Cβmvx

Crm − vx

DβJzvx

DrJz

](16)

odnosno, u razvijenom obliku, sa

A =

[−Cα,f+Cα,r

mvx

−a1Cα,f+a2Cα,rmvx

− vx−a1 Cα,f−a2 Cα,r

Jzvx−a21Cα,f+a

22Cα,r

Jzvx

](17)




Model vozila sa dva to£ka - ustaljeno okretanje

Diferencijalne jedna£ine kretanja za stacionarno kretanje

Posmatra se model vozila sa dva to£ka i sa upravljanjem naprednjim to£kovima

Posmatra se okretanje pri stacionarnim uslovima (izvodi povremenu su jednaki nuli)

U diferencijalne jedna£ine kretanja

mvx −mωz vy = Fx

mvy +mωz vx = Fy

J ωz =Mz

se unosi: vx = 0, vy = 0, ωz = r = 0






Diferencijalne jedna£ine kretanja postaju

Fx = −mrvyFy = mrvx

Mz = 0

Unose se izrazi (9) za bo£nu silu i momenat skretanja

Fy = Cr r + Cββ + Cδδ

Mz = Dr r +Dββ +Dδδ






Diferencijalne jedna£ine kretanja postaju:

Fx = −mrvyCrr + Cββ + Cδδ = mrvx

Drr +Dββ +Dδδ = 0

(18)

Kod ustaljenog okretanja vozilo se okre¢e po krugu saradijusom R sa brzinom vx i sa ugaonom brzinom ωz = r,tako da je

vx = Rr (19)






Unose¢i (19) u jedna£ine (18), dobija se:

Fx = −mRvxvy

Cββ − (Crvx −mv2x)1

R= mrvx

Dββ +Drvx1

R= −Dδδ

(20)

Prva od jedna£ina odre�uje potrebnu vu£nu silu tako da brzinavx bude konstantna






Druga i tre¢a od jedn. (20) odre�uju ustaljene vrednosti zaizlazne promenljive: za globalni ugao bo£nog klizanja β, kao iza krivinu putanje:

κ =1

R=

r

vx(21)

za konstantan ugao okretanja to£kova δ (ulazni parametar) priokretanju vozila sa konstantnom brzinom vx






Druga i tre¢a od jedn. (20) mogu da se napi²u u matri£nomobliku, unose¢i de�niciju krivine (21):[

Cβ Crvx −mv2xDβ Drvx

]{βκ

}=

{−Cδ−Dδ

}δ (22)

Iz jedna£ina (22) mogu da se odrede β i κ u funkciji δ, pri£emu je vx dato sa (21)





Stacionarno kretanje vozila u krivini

Iz jedna£ina (20) mogu da se odre�uju razli£iti izlazniparametri u zavisnosti od ugla okretanja to£kova δ kaoulaznog podatka

Zavisnost krivina - okretanje ("curvature response"):

Sκ =κ

δ=

1

Rδ=

CδDβ − CβDδ

vx(DrCβ − CrDβ +mvxDβ(23)

Zavisnost ugao bo£nog klizanja - okretanje ("sideslipresponse")

Sβ =β

δ=

Dδ(Cr −mvx)−DrCδDrCβ − CrDβ +mvxDβ

(24)






Zavisnost ugaona brzina - okretanje ("yaw rate response"):

Sr =r

δ=κ

δvx = Sκ vx =

CδDβ − CβDδ

(DrCβ − CrDβ +mvxDβ(25)

Zavisnost bo£no ubrzanje - okretanje ("lateral accelerationresponse")

Sa =v2x/R

δ=κ

δv2x = Sκ v

2x =

(CδDβ − CβDδ)vxDrCβ − CrDβ +mvxDβ

(26)






Zavisnost krivina - okretanje ukazuje kako se menja radijuskrivine sa promenom ugla upravljanja (okretanja) to£ka

Zavisnost Sκ data sa (23) moºe da se prikaºe kao:

Sκ =κ

δ=

1/R

δ=

1

`

1

1 +Kv2x(27)

gde je

K =m

`2(a2Cα,f

− a1Cα,r

) (28)

i zove se faktor stabilnosti





Stacionarno kretanje vozila u krivini - faktor stabilnosti

Znak faktora stabilnosti K odre�uje promenu zavisnosti Sκ sabrzinom vx - da li se Sκ pove¢ava ili smanjuje sa vx

Za K > 0a2Cα,f

>a1Cα,r

parametar Sκ je opadaju¢a funkcija sa vx

Krivina putanje κ = 1R se smanjuje za konstantan ugao

okretanja to£ka δ

Smanjenje krivine κ zna£i da se radijus putanje R priustaljenom okretanju pove¢ava sa pove¢anjem brzine vx

Pozitivan faktor stabilnosti K je poºeljan i vozilo sa K > 0 jestabilno. To je podupravljivost ("understeer")






Za K < 0a2Cα,f

<a1Cα,r

parametar Sκ je rastu¢a funkcija sa vx

Krivina putanje κ = 1R se pove¢ava za konstantan ugao

okretanja to£ka δ

Pove¢anje krivine κ zna£i da se radijus putanje R priustaljenom okretanju smanjuje sa pove¢anjem brzine vx

Negativan faktor stabilnosti K je nepoºeljan i vozilo sa K < 0je nestabilno. To je nadupravljivost ("oversteer")

Treba da se smanji ugao okretanja to£ka δ ako se pove¢avabrzina vx da bi zadrºali isti radijus krivine






Za K = 0a2Cα,f

=a1Cα,r

parametar Sκ ne zavisi od vx

Krivina putanje κ = 1R ostaje konstantna za konstantan ugao

okretanja to£ka δ

Konstantna krivina κ zna£i da se radijus putanje R priustaljenom okretanju ne menja sa promenom(pove¢anjem/smanjenjem) brzine vx

Nulti faktor stabilnosti K zna£i da je vozilo sa K = 0 nagranici stabilnosti. To je neutralna upravljivost ("neutralsteer")

Kada se vozi sa K = 0 ne mora da se menja ugao okretanjavozila ako se pove¢ava ili smanjuje brzina kretanja u krivini





Stacionarno kretanje vozila u krivini - upravljivost vozila

Upravljivost vozila je mera podudarnosti stvarnog pravcakretanja vozila sa pravcem de�nisanim uglom upravlja£kihto£kova za ugao bo£nog klizanja α = 0, u slu£aju stacionarnogobrtanja (konstantna brzina, ugaona brzina i radijus kivine)

U odnosu na ugao okretanja upravlja£kih to£kova δ, vozilo seu odnosu na slu£aj α = 0 kre¢e:

po krivini odgovaraju¢eg polupre£nika (neutralna upravljivost,K=0)po krivini ve¢eg polupre£nika, odn. po "blaºoj krivini"(podupravljivost, K<0)po krivini manjeg polupre£nika, odn. po "o²trijoj krivini"(nadupravljivost, K>0)




Faktor stabilnosti K - promena sa brzinom vx10. Vehicle Planar Dynamics 629

[ ]/xv m s

Sκ

K 0>

K 0<

K 0=

FIGURE 10.19. Comarison of the curvature response Sκ for a car withK = 1.602× 10−3, K = −2.21× 10−4, and K = 0.

the speed vx is equal to the following critical value

vc =

r− 1K

(10.309)

thenSκ →∞ (10.310)

and any decrease in steering angle cannot keep the path. When vx = vc, thecurvature κ is not a function of steering angle δ, and any radius of rotationis possible for a constant δ. The critical speed makes the system unstable.Controlling an oversteer vehicle gets harder by vx → vc and becomes un-controllable when vx = vc.The critical speed of an oversteer car with the characteristics

Cαf = 57296N/ rad (10.311)

Cαr = 52712N/ rad (10.312)

m = 1400 kg ≈ 95.9 slug (10.313)

a1 = 125 cm ≈ 4.1 ft (10.314)

a2 = 130 cm ≈ 4.26 ft (10.315)

is

vc =

r− 1K= 67.33m/ s (10.316)

because

K =m

l2

µa2Cαf

− a1Cαr

¶= −2.2059× 10−4. (10.317)




Negativan faktor stabilnosti K<0 - "Understeer"




Pozitivan faktor stabilnosti K>0 - "Oversteer"





Stacionarno kretanje vozila u krivini - kriti£na brzina

Ako je K < 0, kako je

Sκ =1

`

1

1 +Kv2x

onda se Sκ pove¢ava sa pove¢anjem brzine vx

Ugao okretanja to£ka treba da se smanji da bi sa zadrºalaputanja sa konstantnim radijusom

Kada je 1 +Kv2x = 0 , odn. kada je

vx = vcr =

√− 1

K

onda Sκ →∞ i u pitanju je kriti£na brzinaS.�ori¢ Mehanika voºnje





Kada je vx = vcr nikakvo smanjenje ugla okretanja to£ka nemoºe da zadrºi vozilo na putanji

Krivina tada nije funkcija ugla okretanja to£ka i bilo kojiradijus R je mogu¢ za konstantan ugao okretanja δ

Kriti£na brzina zna£i da je sistem nestabilan

Kontrola vozila sa K < 0 ("oversteer") postaje teºa zavx → vcr, a za vx = vcr vozilo postaje nekontrolabilno






Primer numeri£kih podataka:

Cα,f = 57.3 [kN/rad]

Cα,r = 52.7 [kN/rad]

m = 1400 [kg]

a1 = 1.25 [m] a2 = 1.30 [m]

Dobija se K = −2.2059× 10−4, pa je kriti£na brzina

vcr =

√− 1

K= 67.3 [m/s] = 242.4 [km/h]




Sadrºaj








Da bi se analiziralo pona²anje vozila tokom vremena i kako ¢eda se reaguje u zavisnosti od promene upravljanja δ = δ(t),mora da se re²i spregnut sistem jedna£ina kretanja:

vx =1

mFx + rvy (29)

{vyr

}=

[Cβmvx

Crm − vx

DβJzvx

DrJz

]{vyr

}+

{CδmDδIz

}δ(t) (30)

Re²avanjem jedna£ina se dolazi do vremenskog odgovora vozila

vx = vx(t) vy = vy(t) r = r(t)






Ako se pretpostavi voºnja sa konstantnom brzinom vx, onda jejedna£ina (29) jednostavnija:

Fx = −mr vy (31)

Jedna£ine (30) su nezavisne od jedna£ine (31) i mogu da sepi²u u obliku

q = Aq + u (32)

gde je:

q =

{vyr

}u =

{CδmDδIz

}δ(t) (33)

dok je matrica A data sa (17)



Ve²anje drumskih vozilaRa£unski modeli u analizi vibracijaRa£unski model £etvrtine vozilaRa£unski model polovine vozila

Sadrºaj






Ve²anje drumskih vozila

Osnovni ciljevi ve²anja vozila

Preno²enje gravitacionog optere¢enja

Obezbe�ivanje korektnog poloºaja to£kova

Kontrola pravca kretanja vozila

Obezbe�ivanje kontakta to£kova sa drumom

Smanjenje efekta udarnih sila tokom voºnje





Osnovni elementi sistema ve²anja vozila

Elementi za vo�enje

- poluge- vezni ²tapovi- lisnate opruge

Elementi koji proizvode sile

- spiralne opruge i druge opruge (vazdu²ne, lisnate)- viskozni prigu²iva£i- "anti-roll" osovine (stabilizatori protiv okretanja)

Gume




Sistemi ve²anja kod vozila




Sistemi ve²anja kod vozila




Razne opruge kao sistemi ve²anja




Elasti£ne opruge i sile veze





Sila u elasti£noj opruzi

Osnovne karakteristike elasti£ne opruge:

- duºina opruge u nenapregnutom stanju- koe�cijent krutosti opruge k(sila koja izaziva jedini£no pomeranje opruge u odnosu nanenapregnuto stanje)

Sila u linearno elasti£noj opruzi

Fop = k · x

gde je x pomeranje u odnosu na nenapregnuto stanje

Sila u opruzi je restituciona sila: teºi da vrati oprugu unenapregnuto stanje





Sila u elasti£noj opruzi

Ako je opruga nelinearno elasti£na onda je

Fop = f(x)

gde je f(x) neka funkcija pomeranja x

Po prestanku delovanja aktivne sile, opruga se vra¢a unenapregnuto stanje (elasti£na opruga)




Viskozni prigu²iva£ (sa jednom cevi)





Sila u viskoznom prigu²iva£u

Viskozni prigu²iva£ proizvodi silu otpora zavisnu od brzine

Fd = Fd(v)

gde je v brzina prigu²iva£a

Osnovna karakteristika viskoznog prigu²iva£a je funkcija Fd(v)

Smer sile viskoznog prigu²iva£a je suprotan od brzine

Obi£no se posmatra linearan prigu²iva£:

Fd = c v




Elasti£na opruga i viskozni prigu²iva£




Sadrºaj






Model bicikla u analizi vibracija 13. Vehicle Vibrations 853

a1a2

x

C

m

y2 y1

θ

m1m2

FIGURE 13.9. A bicycle vibrating model of a vehicle.

Searching for the first-unit expression of u1 and u2 provides the followingmode shapes.

u1 =

∙1

−3.1729× 10−3¸

(13.200)

u2 =

∙1

0.157 58

¸(13.201)

Therefore, the free vibrations of the quarter car is

x =nXi=1

ui (Ai sinωit+Bi cosωit) i = 1, 2 (13.202)∙xsxu

¸=

∙1

−3.1729× 10−3¸(A1 sin 8.8671t+B1 cos 8.8671t)

+

∙1

0.157 58

¸(A2 sin 55.269t+B2 cos 55.269t) (13.203)

13.4 Bicycle Car and Body Pitch Mode

Quarter car model is excellent to examine and optimize the body bouncemode of vibrations. However, we may expand the vibrating model of avehicle to include pitch and other modes of vibrations as well. Figure 13.9illustrates a bicycle vibrating model of a vehicle. This model includes thebody bounce x, body pitch θ, wheels hop x1 and x2 and independent roadexcitations y1 and y2.The equations of motion for the bicycle vibrating model of a vehicle are




Model bicikla u analizi vibracija 13. Vehicle Vibrations 855

a1a2

x

y2 y1

m1m2

Cm, θIy

kt2 kt1

k2 k1 c1c2 x1x2

FIGURE 13.10. Bicycle model for a vehicle vibrations.

the Lagrange method. The kinetic and potential energies of the system are

K =1

2mx2 +

1

2m1x

21 +

1

2m2x

22 +

1

2Iz θ

2(13.208)

V =1

2kt1 (x1 − y1)

2+1

2kt2 (x2 − y2)

2

+1

2k1 (x− x1 − a1θ)

2 +1

2k2 (x− x2 + a2θ) (13.209)

and the dissipation function is

D =1

2c1

³x− x1 − a1θ

´2+1

2c2

³x− x2 + a2θ

´. (13.210)

Applying Lagrange method

d

dt

µ∂K

∂qr

¶− ∂K

∂qr+

∂D

∂qr+

∂V

∂qr= fr r = 1, 2, · · · 4 (13.211)

provides the following equations of motion (13.204)-(13.207). These set ofequations may be rearranged in a matrix form

[m] x+ [c] x+ [k]x = F (13.212)

where,

x =

⎡⎢⎢⎣xθx1x2

⎤⎥⎥⎦ (13.213)

[m] =

⎡⎢⎢⎣m 0 0 00 Iz 0 00 0 m1 00 0 0 m2

⎤⎥⎥⎦ (13.214)




Model polovine vozila u analizi vibracija13. Vehicle Vibrations 859

y2m2 y1

m1

Cm, Ix

b1b2

x

ϕ

FIGURE 13.11. A half car vibrating model of a vehicle.

This model includes the body bounce x, body roll ϕ, wheels hop x1 and x2and independent road excitations y1 and y2.The equations of motion for the half car vibrating model of a vehicle are

as follow.

mx+ c (x− x1 + b1ϕ) + c (x− x2 − b2ϕ)

+k (x− x1 + b1ϕ) + k (x− x2 − b2ϕ) = 0 (13.234)

Ixϕ+ b1c (x− x1 + b1ϕ)− b2c (x− x2 − b2ϕ)

+b1k (x− x1 + b1ϕ)− b2k (x− x2 − b2ϕ) + kRϕ = 0 (13.235)

m1x1 − c (x− x1 + b1ϕ) + kt (x1 − y1)

−k (x− x1 + b1ϕ) = 0 (13.236)

m2x2 − c (x− x2 − b2ϕ) + kt (x2 − y2)

−k (x− x2 − b2ϕ) = 0 (13.237)

The half car model may be different for the front half and rear half due todifferent suspensions and mass distribution. Furthermore, different antirollbars with different torsional stiffness may be used in the front and rearhalves.Proof. Figure 13.12 shows a better vibrating model of the system. Thebody of the vehicle is assumed to be a rigid bar. This bar has a mass m,which is the front or rear half of the total body mass, and a longitudinalmass moment of inertia Ix, which is half of the total body mass momentof inertia. The left and right wheels have a mass m1 and m2 respectively,although they are usually equal. The tires stiffness are indicated by kt.Damping of tires are much smaller than the damping of shock absorbersso, we may ignore the tire damping for simpler calculation. The suspensionof the car has stiffness k and damping c for the left and right wheels. It




Model polovine vozila u analizi vibracija860 13. Vehicle Vibrations

y2

m2

y1

m1

b1b2

x

m, Ix

x2

ϕC

kt

k c

kt

kc x1kR

FIGURE 13.12. Half car model for a vehicle vibrations.

is common to make the suspension of the left and right wheels mirror.So, their stiffness and damping are equal. However, the half car model hasdifferent k, c, and kt for front or rear.The vehicle may also have an antiroll bar with a torsional stiffness kR in

front and or rear. Using a simple model, the antiroll bar provides a torqueMR proportional to the roll angle ϕ.

MR = −kRϕ (13.238)

However, a better model of the antiroll bar effect is

MR = −kRµϕ− x1 − x2

w

¶. (13.239)

To find the equations of motion for the half car vibrating model, we usethe Lagrange method. The kinetic and potential energies of the system are

K =1

2mx2 +

1

2m1x

21 +

1

2m2x

22 +

1

2Ixϕ

2 (13.240)

V =1

2kt (x1 − y1)

2 +1

2kt (x2 − y2)

2 +1

2kRϕ

2

+1

2k (x− x1 + b1ϕ)

2+1

2k (x− x2 − b2ϕ) (13.241)


D =1

2c (x− x1 + b1ϕ)

2+1

2c (x− x2 − b2ϕ) . (13.242)

Applying the Lagrange method

d

dt

µ∂K

∂qr

¶− ∂K

∂qr+

∂D

∂qr+

∂V

∂qr= fr r = 1, 2, · · · 4 (13.243)




Model kompletnog vozila u analizi vibracija

13

Vehicle VibrationsVehicles are multiple-DOF systems as the one that is shown in Figure 13.1.The vibration behavior of a vehicle, which is called ride or ride comfort, ishighly dependent on the natural frequencies and mode shapes of the vehicle.In this chapter, we review and examine the applied methods of determiningthe equations of motion, natural frequencies, and mode shapes of differentmodels of vehicles.

x

ϕθ

FIGURE 13.1. A full car vibrating model of a vehicle.

13.1 Lagrange Method and Dissipation Function

Lagrange equation,

d

dt

µ∂K

∂qr

¶− ∂K

∂qr= Fr r = 1, 2, · · ·n (13.1)

or,d

dt

µ∂L∂qr

¶− ∂L

∂qr= Qr r = 1, 2, · · ·n (13.2)

as introduced in Equations (9.243) and (9.298), can both be applied tofind the equations of motion for a vibrating system. However, for smalland linear vibrations, we may use a simpler and more practical Lagrange




Model kompletnog vozila u analizi vibracija

13. Vehicle Vibrations 867

θ

ϕ

x

y2m4ktf

kf

y1

x1 y3

cr

y4

x4

m, Ix, Iy

m2

ktfktr

ktr

kfkr

cf

cf

kRf

m1

x2 kRr

a1

b2a2b1

FIGURE 13.14. Full car model for a vehicle vibrations.

However, a better model of the antiroll bar reaction is

MR = −kRfµϕ− x1 − x2

wf

¶− kRf

µϕ− x4 − x3

wr

¶. (13.280)

Most cars only have an antiroll bar in front. For these cars, the moment ofthe antiroll bar simplifies to

MR = −kRµϕ− x1 − x2

w

¶(13.281)

if we use

wf ≡ w = b1 + b2 (13.282)

kRf ≡ kR. (13.283)

To find the equations of motion for the full car vibrating model, we usethe Lagrange method. The kinetic and potential energies of the system are

K =1

2mx2 +

1

2Ixϕ

2 +1

2Iy θ

2

+1

2mf

¡x21 + x22

¢+1

2mr

¡x23 + x24

¢(13.284)




Sadrºaj






Model £etvrtine vozila u analizi vibracija












Model £etvrtine vozila u analizi vibracija - zadatak
















Sadrºaj






Model bicikla u analizi vibracija - zadatak13. Vehicle Vibrations 855

a1a2

x

y2 y1

m1m2

Cm, θIy

kt2 kt1

k2 k1 c1c2 x1x2

FIGURE 13.10. Bicycle model for a vehicle vibrations.

the Lagrange method. The kinetic and potential energies of the system are

K =1

2mx2 +

1

2m1x

21 +

1

2m2x

22 +

1

2Iz θ

2(13.208)

V =1

2kt1 (x1 − y1)

2+1

2kt2 (x2 − y2)

2

+1

2k1 (x− x1 − a1θ)

2 +1

2k2 (x− x2 + a2θ) (13.209)


D =1

2c1

³x− x1 − a1θ

´2+1

2c2

³x− x2 + a2θ

´. (13.210)

Applying Lagrange method

d

dt

µ∂K

∂qr

¶− ∂K

∂qr+

∂D

∂qr+

∂V

∂qr= fr r = 1, 2, · · · 4 (13.211)

provides the following equations of motion (13.204)-(13.207). These set ofequations may be rearranged in a matrix form

[m] x+ [c] x+ [k]x = F (13.212)

where,

x =

⎡⎢⎢⎣xθx1x2

⎤⎥⎥⎦ (13.213)

[m] =

⎡⎢⎢⎣m 0 0 00 Iz 0 00 0 m1 00 0 0 m2

⎤⎥⎥⎦ (13.214)








Model polovine vozila u analizi vibracija




Model polovine vozila - podaci o masi




Prikazivanje puta u analizi kretanja vozila




Model puta sa paralelnim trakama




Model deterministi£kih prepreka na putu




Merenje vibracija pri voºnji preko prepreka




Statisti£ki model pro�la puta




Detaljno merenje pro�la puta


Documents

MEHANIKA VO NJE - Odsek za puteve, eleznice i aerodrome · Kretanje vozila u ravni Analiza vibracija vozila Diferencijalne jedna£ine kretanja Vremenski odgovor Sile koje deluju na