CHAPTER 6.ppt

Dr. B Dayal

STEADY STATE CORNERING

Cornering behaviour equated with handling.handling meant to imply the responsiveness of a vehicle to driver input, or the ease of control. As such, handling is an overall measure of the vehgicle driver combination.Driver and vehicle is a closed loop system.Open loop refers to vehicle response to specific steering inputs, and is more precisely defined as directional response behaviour.

LOW SPEED TURNING

At low speed tyres need not develop lateral forces. Thus they roll with no slip angle.If the rear wheels have no slip angle, the center of turn must lie on the projection of the rear axle.Likewise, the perpendicular from each of the front wheels should pass through the same point called as center of turn.This is called as condition of true rolling.The ideal turning angles on the front wheels can be established by Ackerman Geometry or Ackerman steering.

CONDITION OF TRUE ROLLING

Condition of true rollingtrue rolling occurs only when the direction of motion of the vehicle is perpendicular to the wheel axis, i.e., the wheel is subjected to forward force. on a circular path, true rolling condition occurs when the projected axes of several wheels all moving in different curved path intersect at a single point called the instantaneous centre. whenever a vehicle takes a turn, the front wheel must turn in a definite manner both in relation to eachother and to the axis of the rear wheels so that the lateral slip may be avoided and true rolling for all the wheels is obtained. for this all

Condition of true rollingthe wheels must always rotate about the instantaneous centre. since the rear wheels have a common and fixed axis, it is quite obvious that this common centre o would lie some where on its extensionfrom the figure

cot = (c + x)/b = c/b + x/b = c/b + cot or cot - cot = c/bwhere, = angle of inside lock = angle of outside locka = wheel trackb = wheel base

Condition of true rollingC = DISTANCE BETWEEN PIVOT CENTRESD = LENGTH OF TRACK RODTURNING CIRCLE RADIUSOUTER FRONT WHEEL ROF = b/SIN + (a c)/2INNER FRONT WHEEL RIF = b/SIN - (a c)/2OUTER REAR WHEEL ROR = bCot + (a c)/2INNER REAR WHEEL RIR = b Cot (a c)/2

TURNING RADIUSTURNING RADIUS OF AN AUTOMOBILE VEHICLE IS THE RADIUS OF THE ARC DESCRIBED BY THE CENTRE OF THE TRACK MADE BY THE OUTSIDE FRONT WHEEL OF THE VEHICLE WHEN MAKING ITS SHORTEST TURN.- Society of Automotive Engineer

ROF =[(-b/SIN )2 + c2 + 2bc/TAN ]1/2+(ac)/2

EXAMPLEA motor car has a wheel base of 2.743 m and pivot center of 1.065 m. the front and rear wheel track is 1.217 m. calculate the correct angle of outside lock and turning circle radius of the outer front and inner rear wheel when angle of inside lock is 400.

GRAPHICAL SOLUTION DIAGRAM

GRAPHICAL SOLUTIONCot = PI/HI = (PG +IG)/HI

Cot = QI/HI = (PG IG)/HITherefore, Cot - Cot = 2IG/HI =

2QG/QK = c/b

ANALYTICAL SOLUTION FOR ACKERMANN LINKAGE

ANALYTICAL SOLUTION FOR ACKERMANN LINKAGEIf the slight inclination of the track rod is neglected, the movement of M and N in the direction parallel to the axle beam PQ can be considered as the same, say z. let MN represent the correct steering position and r denotes the cross arm radius.Thensin ( + ) = (y +z)/rAndsin ( ) = (y z)/r\hencesin ( + ) + sin ( ) = 2sin with the help of above equations, the variables and can be calculated for correct steering.

LOW SPEED TURNING

Off-tracking occurs at the rear wheels.Off - tracking distance. = R [1 cos (L / R)]Using the expression for a series expansion of the cosinecos z = 1 z2/2 + z4/4 z6/6 .Then = L2 / 2RFor obvious reasons, off tracking is primarily of concern with long wheel base vehicles such as trucks and buses. For articulated trucks, the geometric equations become more complicated and are known as tractrix equations.

HIGHS SPEED CORNERING

At high speed, the turning equations deffer because lateral acceleration will be present. To counteract the lateral acceleration the tyres must develop lateral forces, and slip angles will be present at each wheel.Slip angle. The angle between direction of heading and direction of travel of a wheel is known as slip angle.The cornering force. The lateral force denoted by Fy, is called as the cornering force when the camber angle is zero.At a given tyre load, the the cornering force grows with slip angle. At low slip angles (50 or less):Fy = CWhere,C = cornering stiffness = slip angleA positive slip produces negative force (to the left) on the tyre.SAE defines cornering stiffness as the negative of the slope, such that C takes on a positive value.

VARIABLES AFFECTING CORNERING STIFFNESS

Tyre size and type (radial versus bias ply construction)Number of pliesCord angleWheel widthTreadsFor a given tyre:LoadInflation pressure

HIGH SPEED CORNERING

Cornering forces (contd):Cornering coefficientCC = C / FzCornering equations Fy = Fyf + Fyr = MV2 / RWhere,Fyf = lateral cornering force at the front axle.Fyr = lateral (cornering) force at the rear axle.M = mass of the vehicleV = forward velocityR = radius of turnAlso,Fyfb Fyr c = 0ThusFyf = Fyr . c/bSubstituting into equationMV2 / R = Fyr (c/b + 1) = Fyr (b + c)/b = Fyr L/bFyr = Mb/L (V2/R)f = Wf V2 / (CfgR)Andr = Wr V2 / (CrgR)


Cornering equations (contd)It can be seen that: = 57.3 L/R + f r = 57.3 L/R + WfV2 / CfgR WrV2 / CrgR = 57.3 L/R + (Wf/Cf Wr/Cr)V2 / gRWhere, = steer angle at the front wheelCf = cornering stiffness of the front axle tyresCr = cornering stiffness of the rear axle tyresUndersteer gradientThe above equation can be written in the short: = 57.3 L/R + KayWhereK = understeer gradient (deg / g)ay = lateral acceleration


Understeer gradientHere,K = Wf/Cf Wr/Cr There are three possibilities:Neutral steer: K = 0 OrWf / Cf = Wr / CrTherefore,f = rOn a constant radius turn, there is no change in steer angle. Steer angle = ackerman angle = 57.3 L/R.Neutral steer case corresponds to a balance on the vehicle such that the force of the lateral acceleration at the CG causes an identical increase in slip angle at both the front and rear wheels.Understeer: K > 0OrWf / Cf > Wr / CrTherefore,f > rThe steer angle will have to increase with speed in proportion to K timmes the lateral acceleration in gs.


Understeer gradientUndersteer (contd)Thus it increases linearly with the lateral acceleration. In the under steer case, the lateral acceleration causes the front wheel to slip side ways to a greater extent than at the rear wheels. Thus, front wheel require to be steered to a greater angle.Oversteer: K < 0OrWf / Cf < Wr / CrTherefore,f < rThe steer angle will have to decrease with speed. In the this case, the lateral acceleration causes the rear wheel to slip side ways to a greater extent than at the front wheels. Thus, front wheel require to be steered to a lesser angle.


Characteristic speedFor an under steer vehicle,The speed at which the steer angle required to negotiate any turn is twice the ackerman angle. = 57.3 L/R + Kay If = 2 . 57.3 L/RThen2 x 57.3 L/R = 57.3 L/R + Kay OrK ay = 57.3 L/Ray = V2 / R = 57.3Lg/KRV2 = 57.3Lg / KVchar = (57.3Lg / K)0.5Critical speedIn the over steer case, a critical speed will exist above which the vehicle will be unstableVcrit = (-57.3 Lg / K)0.5Long wheel base vehicles have a higher critical speed. An over steer vehicle can be driven at speeds less than critical and becomes directionally unstable above this speed.


Lateral acceleration gainIt is the ratio of lateral acceleration to the steering angleay / = [V2/57.3Lg) / [1 + KV2/57.3 Lg]When K = 0lateral acceleration gain V2When V = Vcrit, lateral acceleration = Yaw velocity gainRatio of yaw velocity to the steering angle. = r/Yaw velocity, r, is the rate of rotation in heading angle.r = 57.3 V / Rr / = (V/L) / [1 + KV2/57.3Lg]Side slip angleAt any point on the vehicle, a side slip angle may be defined as the angle between the longitudinal axis and the local direction of travel.For any spee the side slip angle , at CG will be: = 57.3 c/R r = 57.3 c/R WV2 / (CrgR)


Side slip angle (contd)The speed at which the side slip angle becomes 0 is:V = 0 = (57.3 gc Cr / Wr)0.5Side slip angle is independent of radius of turn.Static Marginthe static margin is defined as the distance the neutral point falls behind the CG, normalised by wheel base.static margin = e / LStatic margin is determined by the point on the vehicle where a side force will produce no steady state yaw velocity (i.e., the neutral steer point). Neutral steer line is the locus of points in the x z plane along which external lateral forces produce no steady state yaw velocity.When the point is behind the CG, the static margin is +ve and the vehicle is under steer.At the CG the margin is 0, and neutral steer.When ahead of CG, -ve, over steer.

SUSPENSION EFFECT ON CORNERING

Oileys definition for understeer / oversteer vehicle.When the front axle is more compliant than the rear (understeer vehicle), a lateral disturbance produces more side slip at the front axle, hence the vehicle turns awayfrom the disturbance. If the rear axle exhibits more cornering compliance (oversteer), the rear of the vehicledrifts out and it turns into the disturbance. The lateral acceleration acting at the CG adds to the disturbance force further increasing the turning response and instability.


Roll moment distribution.Understeer / oversteer of a vehicle depends upon the balance of roll moments distributed on the front and rear axles. More roll moment on the front axle contributes to underrsteer, whereas more roll moment on the rear axle contributes to oversteer.K = 0.5 Kss2Where, K = roll stiffness of the suspensionKs = vertical rate of each of the left and right springs.s = lateral seperation between the springs

Roll center. The point at which the lateral forces are transferred from the axles to sprung mass.Roll center can also be thought of as the point on the body at which lateral forces application will produce no roll angle.It is the point around which the axle rolls when subjected to a pure roll moment.Fzo Fzi = 2Fyhr/t + 2K/t = 2Fz


Roll moment distribution.Where,Fzo = load on the outside wheelFzi = load on the inside wheel Fy = lateral force = Fyo + Fyi hr = roll center height = roll angle of the bodyLateral load transfer arises from two mechanisms:2Fy hr / t = load transfer due to cornering force2K / t = lateral load transfer due to vehicle roll.M = [W h1 sin + h1 cos . W V2 / Rg ]cos For small angle sin = cos = 1 and cos = 1Then,M = Wh1[ + V2/Rg]But M = Mf + Mr = (Kf + Kr)Thus = [Wh1 V2/Rg] / [Kf + Kr Wh1]Roll rate of the vehicle = d/day = Wh1 / [Kf + Kr Wh1]


Roll moment distribution.For front and rear axles momentsMf = {Kf[Wh1V2/Rg)] / [Kf + Kr Wh1]} + Wf hf V2 / Rg = Fzf tfMr = {Kr[Wh1V2/Rg)] / [Kf + Kr Wh1]} + Wr hr V2 / Rg = Fzr trWhere,Fzf = Fzfo Wf/2 = - (Fzfi Wf/2)Fzr = Fzro Wr/2 = - (Fzri Wr/2)


Roll moment distribution on vehicles tends to be biased towards the front wheels due to a number of factors:Relative to load, the front spring rate is usually slightly lower than that at the rear (for flat ride), which produces a bias towards higher roll stiffness at the rear. However, independent front suspensions used on virtually all cars enhance front roll stiffness because of the effectively greater spread on the front suspension springs.Designers usually strive for higher front roll stiffness to ensure understeer in the limit of cornering.Stabiliser bars are often used on the front axle to obtain higher front roll stiffness.If stabilizer bars are needed to reduce body lean, they may be installed on the front or on the front and rear. Caution should be used when adding a stabilizer bar only to the rear because of the potential to induce unwanted oversteer.


Fy = C Where,Fy = Lateral force developed von the axleC = cornering stiffness of two tyres, each at one half the axle load. = slip angleThe cornering stiffness of each tyre can be represented by a second or higher order polynomial and the lateral force developed by either can be written as:Fy = C = (a Fz b Fz2)WhereFy = lateral force of one tyreC = cornering stiffness of one tyre = first coefficient in the cornering stiffness polynomialb = second coefficient in the cornering stiffness polynomialFz = load on one tyre (assumed equal on both tyres in previous analysis)


For a vehicle cornering, the lateral force of both tyres Fy is given by:Fy = (a Fzo b Fzo2 + a Fzi b Fzi2 ) C Now let the load change on each wheel be given by FzFzo = Fz + FzFzi = Fz FzThenFy = [a(Fz + Fz) b (Fz + Fz)2 + a Fz Fz) - b (Fz Fz)2]This equation reduces to:Fy = [2aFz 2b Fz2 2bFz2]C = 2aFz 2bFz2OrFy = [C 2bFz2]Now = 57.3 L/R f rFor the two tyres on the front we can write:Fyf = [Cf 2bFz2] f = WfV2 / RgAnd on the rear Fyr = [Cr 2bFzr2] r = WrV2 / RgThus, = 57.3L/R + [(WfV2 / Rg) / (Cf 2bFzf2)] - [(WrV2 / Rg) / (Cr 2bFzr2)]


Since C >> 2bFz2Then1/ (C 2bFz2) = 1/[C (1 2bFz2 / C )] = 1/(1 + 2bFz2 / C) / C The equation for can be written as: = 57.3L/R + [(Wf /Cf - Wr / Cr) + (Wf 2bFzf2 /Cf2 Wr 2bFzr2 /Cr2)] V2 / RgFirst term in the bracket is simply the understeer gradient arising from the nominal cornering stiffness of the tyres, Ktyres, as was developed earlier. The second term represents the understeer gradient arising from lateral load transfer on the tyres, i.e.,Kllt = Wf 2bFzf2 /Cf2 Wr 2bFzr2 /Cr2Since all the variables in the above equation are positive, the contribution from front axle is always understeer; that from the rear axle is always negative, meaning it is oversteer.


Camber change:The inclination of a wheel outward from the body is known as the camber angle. The camber on a wheel will produce a lateral force known as camber thrust.Camber angle produces much lesser force than slip angle.Camber thrust is additive to the cornering force from slip angle, thus affecting understeer.The total camber angle during cornering will be:g = b + Where, g = camber angle with respect to the ground b = camber angle of the wheel with respect to the body = roll angle of the vehicle.The camber angle arising from the suspension is a function of the roll angle, because the jounce on the inside wheel and the rebound on the outside wheel relate directly to roll angle.


Camber change:Fy = C + CThus, = Fy / C C / Cf = Wf . ay /C (C/C) . (f / ) . ( / ay)ayand r = Wr . ay /C (C/C) . (r / ) . ( / ay)ay Substituting these in the turning equation = 57.3L / R + [(Wf/Cf Wr/Cr) + (Cf f / Cf Cr r / Cr ) . /ay] . V2/RgKcamber = (Cf f / Cf Cr r / Cr ) . /ayRoll steerRoll steer is defined as the steering motion of the front or rear wheels with respect to the sprung mass that is due to the rolling motion of the sprung mass.Let be the roll steer coefficient on an axleThe understeer gradient contribution from roll steer = Kroll steer = (f r) . /ayA +ve roll steer coefficient causes the wheels to steer to the right in a


Roll steerRight hand roll.Positive roll steer on the front axle steers out and is understeer.Conversely positive roll steer on the rear axle is oversteer.On solid axles, roll steer coefficient is equal to the inclination angle of the trailing arms.Lataral force compliance steer.With a forward yaw center on a rear axle, the compliance allows the axle to steer towards the outside of the turn, thus causing oversteer. Conversely, a rearward yaw center results in understeer.On a front axle, just opposite is true a rearward yaw center is oversteer and a forward yaw center is understeer.Lateral force compliance steer coefficient A = c / FyWhere,c = steer angleFy = lateral forceOn the front axlecf = Af . Fyf = Af . Wf . ay


Lataral force compliance steer. (contd)Klfcs = Af . Wf - Ar . Wr Aligning torque.The aligning torque is the source of understeer effect.Kat = WP / L . [(Cf + Cr) / (Cf . Cr)]Because C values are positive, the aligning torque effect is positive (understeer) and cannot ever be negative (oversteer).

EFFECT OF TRACTIVE FORCES ON CORNERING

WfV2 / (Rg) = Fyf cos (f + ) + Fxf sin (f + )Wr V2 / (Rg) = Fyr cos r + Fxr sin rWhere,Wf , Wr = load on the front and rear axlesV = forward speedR = radius of turnFyf , Fyr = cornering forces on front and rear axles.Fxf , Fxr = tractive forces on the front and rear axlesf , r = slip angles at front and rear wheels.Solving for f and r and assuming small angles, i.e., cos = 1 and sin = , and substituting into equation = 57.3 L / R + f r = [57.3 L/R / (1 + Fxf/Cf)] + [(WfV2/CfRg)/(1 + Fxf/Cf )] + [(WrV2/CrRg)/(1 + Fxr/Cr )]Since Fxf / Cfand Fxr / Cr are much less than one1 / (1 + Fxf/Cf) = 1 - Fxf/CfThen = [57.3 L/R / (1 + Fxf/Cf)] + [(Wf/Cf Wr/Cr) (WfFxf/Cf2 WrFxr/Cr2)] . V2/Rg

EFFECT OF TRACTIVE FORCES ON CORNERING

= [57.3 L/R / (1 + Fxf/Cf)] + [(Wf/Cf Wr/Cr) (WfFxf/Cf2 WrFxr/Cr2)] . V2/RgThis is the final turning equation for the case where tractive forces are taken into account. The three terms on the right side are:Term 1; this is the ackerman steer angle altered by the reactive force on the front axleTerm 2: this is the understeer gradientTerm 3: this term represents the effect of tractive forces on the understeer behaviour of the vehicle.

SUMMERY OF UNDERSTEER EFFECTS

Understeer componentssourceKtyres = Wf/Cf Wr/CrTyre cornering stiffnessKcamber = (Cf f/Cf - Cr r/Cr) . /ayCamber thrustKroll steer = (f r) d/dayRoll steerKlfcs = Af Wf Ar WrLateral force compliance steerKat = Wp/L[(Cf + Cr) / Cf . CrAligning torqueKllt = (Wf/Cf) . (2bFzf2/Cf) - (Wr/Cr) . (2bFzr2/Cr) Lateral load transferKstrg = -Wf(rv + p) / KssSteering system

EXPERIMENTAL MEASUREMENTS OF UNDERSTEER GRADIENT

CONSTANT RADIUS METHOD

CONSTANT SPEED METHOD

EXAMPLE PROBLEM 1

A car has a weight of 1901 lb front axle and 1552 lb on the rear with a wheel base of 100.6 inches. The tyres have the following cornering stiffness values:

Determine the following cornering properties for the vehicle;Ackerman steer angle for 500, 200, 100 and 50 ft turn radiusUndersteer gradientCharacteristic speedLateral acceleration gain at 60 mphYaw velocity gain at 60 mphSide slip angle at the CG on an 800 ft radius turn at 60 mphStatic margin

loadCornering stiffnessCornering coefficient225670.2684501210.2696751710.2539002250.25011252570.22813503000.222

EXAMPLE PROBLEM 2

A passenger car has an equal arm (parallel) independent front suspension and a conventional solid rear axle with leaf spring suspension. The front suspension has a roll stiffness Kf of 1500 in-lb/deg. The leaf springs have a rate of 115 lb/in and a lateral seperation of 40 inches.What is the rear suspension roll stiffness.If the sprung mass is 2750 lb at a CG height 8 inches above the roll axis, what is the roll rate?Assuming a camber stiffness that is 10% of the cornering stiffness, estimate the under steer gradient due to camber effects.The rear leaf springs have an effective trailing arm angle of -70 (the negative sign means that the pivot of the arms is below the wheel center), what is the under steer gradient due to rear roll steer?

GEOMETRY OF A TURNING VEHICLE

TYRE CORNERING FORCE PROPERTIES

VARIABLES AFFECTING TYRE CORNERING STIFFNESS

CORNERING OF A BICYCLE MODEL

CHANGE OF STEERING ANGLE WITH SPEED

YAW VELOCITY GAIN AS A FUNCTION OF SPEED

SIDE SLIP ANGLE ON A LOW SPEED TURN

SIDE SLIP ANGLE ON A HIGH SPEED TURN

NEUTRAL STEER LINE ON A VEHICLE

OILEYS DEFINITIONS FOR UNDER STEER / OVER STEER

LATERAL FORCE VERTICAL LOAD CHARACTERISTICS OF TYRES

FORCE ANALYSIS OF A SIMPLE VEHICLE IN CORNERING

FORCE ANALYSIS FOR ROLL OF A VEHICLE

LATERAL FORCE CAUSED BY CAMBER OF A TYRE

CAMBER CHANGE IN CORNERING OF A VEHICLE

ROLL STEER WITH A SOLID AXLE

INFLUENCE OF REAR AXLE TRAILING ARM ANGLE ON UNDER STEER

STEER DUE TO LATERAL COMPLIAQNCE IN THE SUSPENSION

CORNERING MODEL WITH TRACTIVE FORCES

EXAMPLE MEASUREMENT OF UNDER STEER GRADIENT BY CONSTANT RADIUS METHOD

EXAMPLE MEASUREMENT OF UNDER STEER GRADIENT BY CONSTANT SPEED METHOD

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CHAPTER 6.ppt