chapter 3.ppt

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Prof (Dr) B Prof (Dr) B Dayal Dayal

Transcript of chapter 3.ppt

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Prof (Dr) B DayalProf (Dr) B Dayal

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Prof (Dr) B Prof (Dr) B DayalDayal

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BASIC EQUATIONSBASIC EQUATIONS

THE GENERAL EQUATIONTHE GENERAL EQUATION

M . aM . axx = - WD = - WDxx / g = - F / g = - Fxfxf – F – Fxrxr – D – DAA – W sin – W sin θθ

Where,Where, W = vehicle weightW = vehicle weight

DDxx = -a = -axx = linear deceleration = linear deceleration

FFxfxf = front axle braking force = front axle braking force

FFxrxr = rear axle braking force = rear axle braking force

DDAA = Aerodynamic drag = Aerodynamic drag

θθ = gradient = gradient

CONSTANT DECELERATIONCONSTANT DECELERATION

DDxx = F = Fxtxt / M = - dV/dt / M = - dV/dt

WhereWhere FFxtxt = the total of all longitudinal deceleration forces on = the total of all longitudinal deceleration forces on

the vehiclethe vehicle

V = forward velocityV = forward velocity

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BASIC EQUATIONSBASIC EQUATIONS

∫∫VfVfV0V0 dV = -F dV = -Fxtxt / M(∫ / M(∫tsts

00 dt dt

VV00 - V - Vff = F = Fxtxt . T . Tss / M / M

Where,Where, ttss = time for velocity change = time for velocity change

VVoo22 – V – Vff

22 = 2F = 2Fxtxt . X / M . X / M

Where X = distance travelled during the deceleration.Where X = distance travelled during the deceleration.

If vehicle comes to full stop – VIf vehicle comes to full stop – Vff = 0 = 0

OrOr SD = VSD = V0022/ (2F/ (2Fxtxt/M) = V/M) = V00

22 / 2D / 2Dxx

The time to stop:The time to stop:

ttss = V = V00 / (F / (Fxtxt / M) = V / M) = V00 / D / Dxx

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BASIC EQUATIONSBASIC EQUATIONS

Deceleration with wind resistance.Deceleration with wind resistance.

DDAA = = ρρ . A . V . A . V22 . C . Cdd / 2 / 2

OrOr DDAA = CV = CV22

ThusThus ∑F∑Fxx = F = Fbb + C V + C V22

WhereWhere FFbb = total brake force of front and rear wheels = total brake force of front and rear wheels

C = aerodynamic drag factorC = aerodynamic drag factor

Therefore,Therefore, ∫∫SDSD00 dx = M∫ dx = M∫00

V0V0 [VdV / (F [VdV / (Fbb + CV + CV22)])]

SD = (M / 2C) . Ln[(FSD = (M / 2C) . Ln[(Fbb + CV + CV0022) / F) / Fbb]]

Energy / powerEnergy / power

The energy absorbed is the KE of motionThe energy absorbed is the KE of motion

Energy = M(VEnergy = M(V0022 – V – Vff

22) / 2) / 2

power absorption = M Vpower absorption = M V0022 / 2t / 2tss

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BRAKING FORCESBRAKING FORCESThe brakes are the primary source of force available for The brakes are the primary source of force available for

deceleration. Other sources for deceleration are:deceleration. Other sources for deceleration are:Rolling resistanceRolling resistance

Always opposes vehicle motion. Hence, adds to the braking.Always opposes vehicle motion. Hence, adds to the braking.

RRxfxf + R + Rxrxr = f = frr (W (Wff + W + Wrr) = f) = frrWW

ffrr = specific rolling resistance or rolling resistance coefficient = specific rolling resistance or rolling resistance coefficient

Aerodynamic dragAerodynamic drag

DDAA = = ρρ . A . V . A . V22 . C . Cdd / 2 / 2

GradeGrade RRgg = W sin = W sin θθ for small angles: R for small angles: Rgg = W = Wθθ

Driveline drag. Arises from bearing and gear friction in the Driveline drag. Arises from bearing and gear friction in the transmission and differential and engine braking. Engine braking is transmission and differential and engine braking. Engine braking is equivalent to the “motoring” torque. On a manual transmission with equivalent to the “motoring” torque. On a manual transmission with clutch engaged, , the engine braking is multiplied by the gear ratio. In clutch engaged, , the engine braking is multiplied by the gear ratio. In automatic transmission, engine drag does not contribute significantly.automatic transmission, engine drag does not contribute significantly.

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DRUM BRAKES AND DISC BRAKESDRUM BRAKES AND DISC BRAKES

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This is a typical brake system showing all typical components. These are known as service brakes, base brakes, or foundation brakes.

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Typical disc brake assembly. Disc brakes are used on the front of most vehicles built since the early 1970s and on the rear wheels of many vehicles.

Courtesy of Wagner Division, Cooper Industries Inc.

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BRAKE FACTORBRAKE FACTOR

The brake factor is a mechanical advantage that can be utilised in The brake factor is a mechanical advantage that can be utilised in drum brakes to minimize the actuation efforts required.drum brakes to minimize the actuation efforts required.

∑∑MMpp = e P = e Paa + n + n μμ N NAA – m N – m NAA = 0 = 0

Where,Where,e = perpendicular distance from e = perpendicular distance from actuation force to pivotactuation force to pivotNNAA = normal force between lining = normal force between lining

and brake drum and brake drumPPaa = actuation force utilised = actuation force utilised

n = perpendicular distance from n = perpendicular distance from lining friction force to pivotlining friction force to pivotm = perpendicular distance from m = perpendicular distance from the normal force to the pivotthe normal force to the pivot

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FORCES ACTING ON THE SHOES FORCES ACTING ON THE SHOES OF A SIMPLE DRUM BRAKEOF A SIMPLE DRUM BRAKE

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BRAKE FACTORBRAKE FACTOR

The friction force developed by each brake shoe is:The friction force developed by each brake shoe is:

FFAA = = μμ N NAA andand F FBB = = μμ N NBB

therefore,therefore, FFAA / P / Paa = = μμe / (m – e / (m – μμn)n)

AndAnd FFBB / P / Paa = = μμe / (m + e / (m + μμn)n)

In the “leading” shoe, the moment produced by the friction force on In the “leading” shoe, the moment produced by the friction force on the shoe acts to rotate it against the drum and increase the the shoe acts to rotate it against the drum and increase the friction force developed. Ths “self-servo” action yields a friction force developed. Ths “self-servo” action yields a mechanical advantage characterised as the brake factor. mechanical advantage characterised as the brake factor.

If If μμ is too large is too large m = m = μμn n and brake factor = ∞, the brkae will and brake factor = ∞, the brkae will lock.lock.

In trailing shoe the friction force acts to reduce the application force. In trailing shoe the friction force acts to reduce the application force. The brake factor is too low.The brake factor is too low.

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INERTIA DYNAMOMETER TORQUE INERTIA DYNAMOMETER TORQUE MEASUREMENTSMEASUREMENTS

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BRAKE PROPORTIONINGBRAKE PROPORTIONING

Brake proportioning is adjustment of brake torque output at front and Brake proportioning is adjustment of brake torque output at front and rear wheels in accordance with the peak traction forces available.rear wheels in accordance with the peak traction forces available.

The aim of brake proportioning is to bring both axles up to the lock up The aim of brake proportioning is to bring both axles up to the lock up point simultaneously.point simultaneously.

the first order determnants of peak traction force on an axle are:the first order determnants of peak traction force on an axle are:The instantaneous loadThe instantaneous loadThe peak coefficient of friction.The peak coefficient of friction.

During braking, a dynamic load transfer from the rear to the front axle During braking, a dynamic load transfer from the rear to the front axle occurs such that the load on an axle is the static plusthe dynamic l;oad occurs such that the load on an axle is the static plusthe dynamic l;oad transfer contributions. transfer contributions.

Thus, for a deceleration DThus, for a deceleration Dxx::

WWff = cW / L + hW D = cW / L + hW Dxx / Lg = W / Lg = Wfsfs + W + Wdd

AndAnd WWrr = bW / L - hW D = bW / L - hW Dxx / Lg = W / Lg = Wrsrs - W - Wdd

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BRAKE PROPORTIONINGBRAKE PROPORTIONING

Where,Where, WWfsfs = front axle static load = front axle static load

WWrsrs = rear axle static load = rear axle static load

WWd d = hW D= hW Dxx / Lg = Dynamic load transfer / Lg = Dynamic load transfer

Then on each axle the maximum brake forceThen on each axle the maximum brake force

FFxmfxmf = = μμWWff = = μμpp (W (Wfsfs + hW D + hW Dxx / Lg) / Lg) (1)(1)

AndAnd FFxmrxmr = = μμ W Wrr = = μμpp (W (Wrsrs - hW D - hW Dxx / Lg) / Lg) (2)(2)

Where Where μμpp = peak coefficient of friction = peak coefficient of friction

DDxx = (F = (Fxmfxmf + F + Fxrxr) / M) / M

And for rearAnd for rear DDxx = (F = (Fxmrxmr + F + Fxfxf) / M) / M

Substituting in equation (1) and (2)Substituting in equation (1) and (2)

FFxmfxmf = [ = [μμpp (W (Wfsfs + h F + h Fxrxr / L)] / [1 – / L)] / [1 – μμpp h/L] h/L]

FFxmrxmr = [ = [μμpp (W (Wrsrs - h F - h Fxfxf / L)] / [1 + / L)] / [1 + μμpp h/L] h/L]

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MAXIMUM BRAKE FORCES AS A MAXIMUM BRAKE FORCES AS A FUNCTION OF DECELERATIONFUNCTION OF DECELERATION

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MAXIMUM BRAKING FORCES ON THE MAXIMUM BRAKING FORCES ON THE FRONT AND REAR AXLESFRONT AND REAR AXLES

““brake proportioning” brake proportioning” describes the relationship describes the relationship between the front and rear between the front and rear brake forces determined brake forces determined by the pressure applied to by the pressure applied to each brake and the gain of each brake and the gain of each. It is represented by a each. It is represented by a line on the graph starting line on the graph starting at the origin and extending at the origin and extending upward to the right. A upward to the right. A fixed or constant fixed or constant proportioning is a straight proportioning is a straight line. line.

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BRAKING EFFICIENCYBRAKING EFFICIENCY

The braking efficiency of a vehicle is defined as the braking force The braking efficiency of a vehicle is defined as the braking force produced as a percentage of the total weight of the vehicle.produced as a percentage of the total weight of the vehicle.

braking efficiency, braking efficiency, ηη = braking force x 100 / weight of the vehicle = braking force x 100 / weight of the vehicle

Since Since braking force = frictional force and normal load = weight of braking force = frictional force and normal load = weight of the vehiclethe vehicle

Coefficient of friction = frictional force / normal loadCoefficient of friction = frictional force / normal load

= braking force / weight of the vehicle= braking force / weight of the vehicle

Therefore,Therefore, = = ηη

HenceHence μμ = = ηη

The brake efficiency can be derived from the kinetic energy possessed The brake efficiency can be derived from the kinetic energy possessed by the vehicle and the work done in bringing the vehicle to a stand by the vehicle and the work done in bringing the vehicle to a stand still.still.

LetLet F = braking forceF = braking force

μμ = coefficient of friction = coefficient of friction

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BRAKING EFFICIENCYBRAKING EFFICIENCYW = vehicle weightW = vehicle weight

U = initial vehicle speedU = initial vehicle speed

M = vehicle massM = vehicle mass

S = stopping distanceS = stopping distance

ηη = brake efficiency = brake efficiency

Then equating work and kinetic energyThen equating work and kinetic energy

FS = 1.2 MU2 = WU2 / 2gFS = 1.2 MU2 = WU2 / 2g

OrOr F / W = F / W = μμ = = ηη

Thus stopping distance S = UThus stopping distance S = U22 / 2g / 2gηη

Therefore, Braking efficiency Therefore, Braking efficiency ηη = (U = (U22 / 2gS) . 100 / 2gS) . 100Braking efficiency for actual brake system is defined as the ratio Braking efficiency for actual brake system is defined as the ratio

of actual deceleration achieved to the “best” performance possible of actual deceleration achieved to the “best” performance possible on the given road surface.on the given road surface.

ƞƞbb = D = Dactact / / μμpp

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BRAKING EFFICIENCYBRAKING EFFICIENCY

Braking efficiency is determined by calculating the brake forces, Braking efficiency is determined by calculating the brake forces, deceleration, axle loads, and braking coefficient on each axle as a deceleration, axle loads, and braking coefficient on each axle as a function of application pressure.function of application pressure.

Braking coefficient.Braking coefficient. The braking coefficient is defined as the ratio of The braking coefficient is defined as the ratio of brake force to load on a wheel or axle.brake force to load on a wheel or axle.

Example problem:Example problem:

Determine the braking efficiency of a vehicle if the brakes bring the Determine the braking efficiency of a vehicle if the brakes bring the vehicle to rest from 60 km/h in a distance of 15 m.vehicle to rest from 60 km/h in a distance of 15 m.

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EFFICIENCY PLOT FOR A TRACTOR EFFICIENCY PLOT FOR A TRACTOR - SEMITRAILOR- SEMITRAILOR

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OPTIMAL PEDAL FORCE GAINOPTIMAL PEDAL FORCE GAIN

Braking effortDisplacement of pedalNational highway traffic safety administration sponsored a research to determine the ergonomic properties for the brake pedal that would give the most effective control.The research identified an optimum range for pedal force gain – the relationship between pedal force and deceleration.

Influential design variables:The positioning of pedal

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