Detailed Design Report

MECHISMU RACING

Indian School of Mines, Dhanbad

Car no. 26

for

Formula Student India

Wheel Size Selection

FSAE rules set the minimum wheel size to be eight inches in diameter. For the purposes of this design, the smallest applicable wheel size was ten inches. Using a wheel with an inner diameter smaller than ten inches makes packaging of the wheel upright difficult, and greatly increases the loading conditions on the upper and lower control arms.

The consideration categories were based on factors deemed important to the size of the wheel and tire. They were:

1. Mass Moment of Inertia – The rotating mass of different tire sizes varies and affects dynamic performance.

2. Tire Availability – As we are limited to using what is available for purchase, the selection matters.

3. Upright Packaging – Different size wheels dictate how much room there is to place components such as uprights and brake rotors inside of them.

4. Chassis Impact – The tire size ultimately affects the positioning and packaging of the rest of the chassis.

5. Wheel Availability – Less important than tire availability since making a wheel is much easier than making a tire, but this is still a consideration.

6. Cost – The budget is limited for this project and it does play a role in the decision.

7. Mass Effect – This is how the larger mass moment of inertia affects the car with respect to acceleration and braking.

8. Mass Addition – This is the total mass difference of the tire and wheel package.

YAMAHA FZ6-R ENGINE

Stock Engine Specs:

Reasons for Preferring this Engine

Performance:1) The primary ideas behind choosing a four cylinder engine were as followsGreater power developed i.e. Better power to weight ratio.Completely balanced primary forces and use of secondary balancershafts in these engines to balance secondary forces. Thus enginegenerates very less vibrations.Balanced torque output i.e. less torque fluctuation.Single cylinder and double cylinder engine were ruled out due tohigh vibrations, unbalanced forces, torque fluctuations, reliability andless power than counterpart 4 cylinder.FZ6-R Engine-Weight: 65 kgWeight of Car Without Engine: 150 kgPower to Weight Ratio of Car: 66/(150+65) = 0.30697EX500 Engine-Weight: 52 kgType: Double Cylinder 500ccPower: 52hpPower To weight Ratio of Car: 52/(150+52)= 0.257CBR250 Engine-Weight: 38kgType: Single Cylinder 250ccPower: 26 hpPower to weight ratio of Car: 26/(150+38)= 0.1382) The power output of this engine 66 hp @ 9800rpm and 53nm @ 8300rpm of torque was better than considerable options within budget constraints

and performance targets of the car owing to experience of the team.3) The engine is basically designed of good midrange power rather than maximum power and flat torque curve both being suitable for most of the events.Availability and Cost:

The best deal sorted after extensive survey and search of engine had following

Salient features:Cost- Well within the budget constraints. Costing aprox. 1 lac INRincluding taxes and shipping.Ebay list price: $949/-The deal engine was used and clocked 4600 miles. The engine wasremoved from 2009 FZ6-R model with no quoted damages.The package consisted of complete engine and transmission, wiringharness, ECU, injectors, throttle body, starter motor, CDI, sparkplugs, butlacks exhaust manifold and fuel pump.

Calculated Vehicle Performance

Parameters:Stock engine performance of 66hp max. power and 53nm of torque.Driver Weight: 65 kgVehicle Weight: 220 kgTyres: 155/65 R13Primary Reduction: 1.955 (Stock) Gear Ratios (Stock): 1- 2.846 2- 1.947 3- 1.556 4- 1.333 5- 1.190 6- 1.083 Secondary Ratio: 3.4 (Stock: 2.875)Drag Coefficient: .25

Estimated Performance

Top speed: 150.28kmph0-75 m in 4.09 sec (ignoring shift delay)Maximum acceleration (1st gear): 1.5g0-60kmph in 1.65 sec0-100kmph in 3.24 sec

Drag Run (Distance vs Time)

Shiftings:

Gear Shift RPM Top Speed (kmph)1 0-10750 57.22 7550-10750 83.223 8650-10750 104.184 9300-10750 122.115 9650-10750 136.766 9850-10750 150.28

Performance in different gears. (Acceleration vs Speed)

Usage and Planned Modifications:The engine hangs below the motorcycle midframe through relevantmountings. The same mountings will be used to mount engine on thechassis.Engine will be naturally aspirated owing to budget, time, resource andexperience constraints.The intake manifold between the throttle body and restrictor, exhaustmanifold and cooling system were not acquired and will be needed todesign and fabricate to optimize performance.Stock ECU will be replaced with a programmable aftermarket ECU tofacilitate mapping of spark timings and fuel injections for performanceoptimization.Stock injectors will be used although timings will be mapped forperformance.

TRANSMISSION

Objective: To transfer the power output of the engine to tyres via torsen differential so as to improve torque biasing in driving wheels during cornering.

The transmission of the vehicle comprises of 3 sub components:

1. Stock Yamaha FZ6-R 6 speed manual gearbox;

2. Torsen differential;

3. Axle.

Rear wheel drive was selected with the engine at the rear of the vehicle for maximum traction during acceleration and better weight distribution during braking.

Torsen differential was opted because it improves the traction in the outer tyre during cornering. It has bias ratio ranging from 3.7- 4:1.

The gear box is connected to the differential by a chain drive. As this was the best possible option we have.

The gear ratio has been set based on for achieving maximum performance on the Auto-Cross event.

The speed of the wheel on the road has been set between 25km/hr - 110 km/hr in 3rd gear so that whole auto-cross event can be covered without a gear change and supplying torque which is approximately under the traction supplied by the road in accelerating condition of 1-g.

The speed of the wheel on road is calculated by-

Speed=(2πN/60)*n1*n2*n3

n1=Primary ratio

n2=Secondary ratio

n3=Final drive ratio(48/13)

25km/hr is considered as the maximum turning speed in a sharp turn.

The chain and sprocket pitch is 0.525 inch (transferring around 600 Nm Torque), it is used in 500cc off-road bikes.

Selection is based on:-

->Performance requirement in autocross event.

->Availability in market.

The differential assembly except for the gears is designed and will be manufactured at college workshop. We choose to use inboard brake on the rear side in order to reduce the unsprung mass of the brake assembly on the upright. Aluminium 6061-T6 was used to reduce the weight of assembly. The differential assembly comprises of 2 differential uprights 1 brake sleeve, drive sleeve, brake flange, 2 bearings and 2 bushings. Brake calliper is mounted on the differential upright and the brake rotor is bolted to brake flange Brake flange is constrained by brake sleeve which is connected with splines to torsen differential housing.

The axles have Tripod joints whose housing is custom designed and manufactured.

Suspension Report

Tires chosen are Hoosier R25B 13in tires because of following reasons :1. Rim diameter being 13in which will allow easy packaging of brake components,

upright and hub.2. Availability of tire data on FSAE TTC.

Basic Decisions

Unequal length a-arm type of suspension design has been chosen, because of the ease with which kinematics of the design can be studied and changed. It allows ease of fabrication, easy calculation of forces and corresponding analysis.

The major setback in the design process is the unavailability of tire data due to lack of funds. However, the best possible option for getting a ballpark value for a number of parameters is to use the available data for another 13in tire, i.e. Michelin FSAE tires.

A MATLAB simulation has been programmed to study the effect of various front and rear slip angles on a vehicle in steady state cornering, for provided roll center heights, wheel base and track widths. It has been used in deciding track widths, calculating jacking force, deciding roll center heights and in general seeing the balance of car for different slip angles.

Deciding camber angle based on pattern given by Michelin dataThe lateral force generated by the outside laden tire in a corner is considered to be the primary factor. In the below shown figures, values of lateral force for different vertical loads is shown, corresponding to negative slip angles from 0 to -5 deg.Following are the curves of lateral vs. normal force for various slip angles, at various camber angles:

Tire pressure = 1.1 MpaAbscissa: Fy, Ordinate: Fz

Camber0

Camber-1

Camber-2

-

Camber-3

Camber-4

Camber-5

Observations:

The lateral force (Fy) for various vertical loads (Fz) is falling on decreasing camber angles beyond 0 deg. The decrease is more pronounced for larger slip angles than for smaller sip angles. The difference in lateral force values for camber 0 and -1 deg is considerably less, than it is for further angles.

Inference:

The necessary force is easily available for all the camber angles. But the value for lateral force is max for camber angle = 0deg and it decreases thereafter.

Result:

It can be seen that the maximum lateral force value occurs at slightly negative camber angle. So, the camber has been chosen to be -1.5 deg. This ensures the camber angle remains at 0 or slightly lesser throughout suspension travel. This negative value is meant to account for the positive camber gain of outside tire during roll and bump, although in the front, it will be tried to be compensated through the negative camber change during steering due to positive castor angle. Provision for camber change will be there.

Caster angle:The caster angle is decided based on following considerations:

1. Desirable mechanical trail so as to provide sufficient feedback to the driver.

2. Will be kept slightly positive to provide self-aligning moment.3. It will be tried to offset the positive camber change due to roll through the

camber change during steering.

KPI – A positive value of KPI causes positive camber change during roll. So a slightly positive value will be chosen during iteration on any desired software. The value can’t be very large as that might result in untolerable jacking during steering.

Mechanical trail – The range of desirable values was decided through steering considerations, which could provide just sufficient feedback to driver. Also, the value for trail will be kept sufficiently small so that there might be greater contribution of pneumatic trail in aligning torque. The pneumatic trail decreases with increasing slip angle, thus, giving indications to the driver of breaking grip.

Virtual swing arm length and instantaneous center– The VSAL will be kept large enough to minimize camber change during roll. The height of IC from the ground will be small enough to minimize vertical component of jacking forces.

Top and bottom wishbone lengths – Deciding the top and bottom wishbone lengths and inclination is a compromise in terms of camber change and roll center migration, with dive and roll, as is discussed in detail by Smith(p.54). No geometry can give the best results, thus a compromise will be chosen.

Track width –A min of 40 in track is assumed to be required for packaging all the components (template rule in front and differential in rear) along with giving ample space for a-arms designing.Checking for 40 in track width: Total mass of vehicle = 240 kg Assuming 40:60 weight distribution, weight of either front wheel = 470 N Weight on either rear wheel = 707.2 N RCH front (zf ) = -2.5 in [justification for RC heights given later] RCH rear (zr) = -0.5 in Assuming roll stiffness to be equal in front and rear; and equal to 250 Nm/deg Then the weight transfer on front shouldn’t exceed 470 N, while that on rear shouldn’t exceed 707.2 N

Weight transfer on front = ((la*240)/t f)*((0.4*zf)+(kf*(.254- (0.6*z r)+(0.4*zf))/(kf+kr))) Where, la = acc. to be tested during tilt test = 1.7 g Tf=front track zf=frnt RCH zr=rear RCH kf=front roll stiffness kr=rear roll stiffness Therefore, weight transfer on front= 365.2263 N Similarly, weight transfer on rear = ((la*240)/tr)*((0.6*zr)+(kr*(.254- (0.6*zr)+(0.4*zf))/(kf+kr))) = 435.2697 N So, both the values are well under limits

Now, the narrowest possible track width is desired to be selected because:1.) It lowers yaw moment of inertia contribution by unsprung mass for given

wheelbase.2.) It allows more room to the driver during any corner as he gets more space.3.) It is especially helpful during slalom course; because narrower is the car lesser

the car has to be steered.

Also, it has been decided that the front track will be kept slightly larger than rear. The reason being the curve to be traced by rear has slightly smaller radius than front. So, larger front track will ensure that the car does not fall out of track just due to the rear portion.

Then, various values of track were used in the matlab simulation to get a slightly positive over steering yaw moment, because in the steady state being analyzed by the model, a slight over steer will help the driver to turn sharply for rest of the part of corner. However the value should not be very large. So, the values of track from 40-50in, with rear being narrower were tested and 46in for front and 44 in for rear was worked out. Although other track dimension also gave satisfactory results, these values were chosen as they would allow ample space for designing suspension geometry.Front track = 46 inRear track = 44in

Roll center analysis and justification:Above ground or below ground:Roll center away from ground are known to produce jacking forces. So, to analyze effects of jacking forces, following calculations were done.Jacking force calculations:

Above picture shows an example case where the lateral force on roll center (above the ground at 1in) is reacted by the forces from tire contact patches. It shows a right turn, so F1 represents the force from laden wheel while F2 from the unladen one. The horizontal components of these forces add up to give the lateral force, while the vertical ones result in jacking forces.Now, θ = γ = atan (2RCH/t) [can be seen from the figure]So, if A is the lateral force then,(F1+F2) cos (θ) = AAnd, jacking force = (F1-F2) sin (θ)

Now, using an example case using the data from matlab simulation for front slip angles =2(assuming 0% Ackermann)Rear slip angles = 0.5F1 cos(θ)= 841.1840 N F2 cos( θ)= 366.5166 [for front]Now, θ = atan(2RCH/t) = atan(2*1/46) = 2.489 degF1= 841.978 N F2= 365.861 NJacking force = (841.978-365.861) sin (2.489) N= 20.6765 NAgain, taking the case when RCH has increased to 2in for the same geometryΘ = 4.969F1=844.3573 F2=367.899Jacking force = 41.269 NConsidering ride stiffness = 45000N/mThe deflection in sprung mass due to jacking (RCH=1in) = (20.6765/45000)m=0.4594mmThe deflection in sprung mass due to jacking (when RCH changes from 1in to 2in) = (41.269-20.6765/45000)m=0.4576mmAS we can see, the jacking forces produce practically 0 defection and thus no change in CG height, so in further considerations, jacking forces will not be considered.

Keeping RC above the ground ladens the outside wheel instantaneously through geometric weight transfer, while it’s the opposite in case where RC is below ground, where inner wheels are laden first. Then, finally the outer wheels are laden when body roll takes place with time. However this is not expected to create any significant changes in handling or driver feel.

When RC is kept above the ground, as the body rolls during cornering, IC for laden wheel shifts down while that for unladen wheel goes up. The kinematic roll center no longer remains useful as both the tire contribute unequal amount of forces. Since most of the cornering forces are produced by outside wheel, so the roll center height effectively goes down. This causes roll stiffness to decrease progressively during a roll while cornering. While the opposite is true when RC is below ground. This causes roll stiffness to increase with roll as effective RC moves up. Thus, RC is chosen to be below the ground. However a low point is comparatively lesser roll stiffness. Also, although a negative RC can be produced by a positive as well as negative swing arm length, but only in case of positive swing arm length, the IC of laden wheel will rise, leading to increasing height of effective roll center. Thus, producing increasing roll stiffness.

Rear RC will be higher than front RC in an attempt to keep roll-axis parallel to mass inertia axis. This will cause linear roll generation and the roll moment will be equal from front to rear. Also it causes roll under steer, while the opposite causes roll over steer, which is not desirable.

During pitching, roll center heights in front and rear should be kept the same as it gives a stable feedback to the driver. On the other hand, a migrating roll

axis gives confusing feedback. So, roll center migration will be kept to a minimum.Keeping above points in mind, for the decided track and wheelbases, effects of various roll center heights on balance has been observed using the matlab simulation, and RCH = -2.5 in for front and -0.5 in for rear has been chosen.

During cornering, the motion of front and rear roll centers will be tried to be kept similar, within the constraints.

So, using above considerations, suspension geometry has to be decided. The effects of various inputs on various kinematic variables can be studied through following methods:

1. Graphical analysis2. Computational analysis3. Using suitable kinematic software

The 3 rd method was chosen due to ease with which the geometry could be changed to study the corresponding effects. Software used was Suspension Analyzer v2.4.

Anti- considerationThe front and rear heave motions were calculated to be under 1in under braking/acceleration. Anti-geometry is used to decrease squat/lift, whatever is applicable, by dividing the spring force into force through suspension arms. This might be useful when the pitching deflections are large, however, for such small deflections, considering anti geometry will add to complexity along with deadening the feel to driver. It also increases tire compliance, and might be leading to unexpected behavior. Although, due to lesser front ride stiffness, very small anti squat could have been considered, but it would add to difficulty in chassis manufacture with negligible gains. In rear, if only upper points were varied, pro geometry is achieved which is not desirable.

The suspension geometry decided :Front:

Specifications:RC height -2.57 inCamber gain 0.499 deg/deg of rollRC height change (bump) 1.021in/inRC height change (rebound) 1.1 in/inGround clearance (with no ARB) 3.8inTrack 46 in

Rear

Specifications:RC height -1.2 inCamber gain 0.665 deg/deg of rollRC height change (bump) 0.537in/inRC height change (rebound) 0.629 in/inGround clearance 3.8inTrack 44in

Ride and Roll rates calculation

Deciding ride frequencies:The solution for frequencies of a vehicle involves formation of two equations, one for pitching motion, and other for heave motion. However, we will be using an oversimplified case, Assuming (ry

2/ab)=l[ry=radius of gyration about transverse axis through CG, l=wheelbase, a and b= distance of CG from front and rear wheel centerline respect.]. Then, the corresponding solution gives the frequencies at front and rear wheel centerlines itself, separating the front and rear heave motions. Although not exact, this solution gives fair idea of stiffness and other properties, with very simple analysis. On

the other hand, the full solution does not give any further insight into the design, in spite of making the calculations complex.

The rear frequency is kept higher than front due to following reasons:1. A higher rear frequency helps in elimination of pitching motion upon hitting a

bump, by allowing rear end to catch up with the front.2. Also, for a rear weight biased car, a larger rear frequency and ride stiffness are

desirable.

There is no magical number for ride frequencies, however 0.64-1.27 Hz (vertical) are associated with resonance of thorax abdomen resonance, while a frequency greater than 4 Hz is associated with harshness. So, value of frequency selected should be between 1.27 Hz – 4 Hz.

Also, selection of frequency value for a given mass dictates amount of wheel travel available. A lower frequency value implies a softer suspension, and more wheel travel.

Keeping above points in mind, following values are decided:Front ride frequency = 2.5 HzRear ride frequency = 2.65 Hz

Calculations:Front Front ride frequency desired = 2.5 HzSo, front ride rate desired (from single wheel) = 4(π)2mf ff

2

= 4(π)2 (48)(2.5)2 =11831.52 N/m = Kf

Tire rate for given load = 140143 N/m Wheel rate = tire rate*ride rate/(tire rate-ride rate) = 140143*11831.45/(140143-11831.45) N/m =12922.49 N/m Wheel rate = spring rate (IR)2

For spring rate = 19267.42 N/mIR = 0.8189MR = 1/IR = 1.221Roll rate = ((π)(tf

2)Kf)/360 = 140.87 Nm/deg [tf=46in]

Wheel travel due to static weight = 48*9.81/(12922.49) = 36.438 mmWhich is greater than minimum wheel travel required min rebound (=1in)Sag developed in the spring = 29.84 mmFree length of spring = 285mm

RearRear ride frequency desired = 2.65 HzSo, rear ride rate desired = 4(π)2mr fr

2 =Kr

= 4(π)2(72)(2.65)2 N/m = 19949.84 N/m Tire rate for given load = 141490.38 N/m

Wheel rate = tire rate*ride rate/(tire rate-ride rate) = 140143*19949.84/(140143-19949.84) = 23213.96 N/m Wheel rate = spring rate (IR)2 For spring rate = 21733 N/mSo, IR = 1.0408 MR = 1/IR = 0.9607Roll rate = ((π)(tr

2)Kr)/360 = 242.53 Nm/deg [tr=44in]

Wheel travel due to static weight = 72*9.81/(23213.96) = 30.4265 mmWhich is greater than minimum wheel travel required in rebound (=1in)Sag developed in the spring = 31.67 mmFree length of spring = 320mm

Front roll rate is appreciably less so ARB has to be used in the front, else significant over steer will occur during cornering due to more weight transfer in rear. Also, ARB will be a very effective tool during testing of the car.

Roll gradient calculations:Assuming unsprung mass = 35kg, sprung mass = 240kg-35kg = 205kgMoment generated = ma*(CG to RC) = 205*9.81*12.221*0.0254 N/m (a=1g, CG to RC = 12.221in) = 624.256 NmRoll gradient without ARB = Moment generated/(Kf+Kr) = 624.256/(140.87+242.53) deg/g = 1.628 deg/g

Front roll stiffness will be increased by ARB upto 65Nm/deg (tunable).

ARB calculations:ARB Motion ratio in starting condition (M) =3.2 (approx.)

T=GJ (θM )/L,

G = shear modulus of MS = 76 GPa

J = (3.14 d4)/32 m4

θ = 1 deg = 0.0174444 rad

M = 3.2

L = 350 mm

T = 45 Nm

Putting the values in the formula we get:

Or, d = 2r =14 mm (approx.)

New roll gradient = 624.256/(45+140.87+242.53)

= 1.46 deg/g

Check:

Assuming max twist = 5 deg

The max shear stress = Gθr/L

= 132 MPa

Which gives FOS = 180/132 =1.36

Dimensions of pushrod, ARB and rocker arm (FRONT)

Pushrod: For a wheel travel of 1.5 in, compression in spring = 285-223.774 mm = 64.35 mm Spring force = kx = 19267.42*0.061226 N = 1289.85 NBalancing moments about the chassis pivot, Force in pushrod = 1460.004 N

ARB: For max ARB twist = 5 deg,

Torque generated = 70.3125 Nm

Force in ARB arm = 70.3125/ (1.998*0.0254)N

= 1385.5 N

It has been assumed that this force remains in plane with rocker arm plane.

The shape of rocker arm has been optimized considering the hypothetical case of maximum roll +heave, where the rocker arm will be experiencing the maximum forces due to the spring and the arb-arm.

The corresponding force in pushrod = 3007.33 N

Dimensions of Pushrod

Inner Diameter of pushrod = 15 mm

Outer diameter of pushrod= 18mm

Dimensions of Arb

Diameter of arb arm = 14 mm

[The diameters have been checked for buckling]

Dimensions of pushrod, ARB and rocker arm (REAR)

ADJUSTMENT OF ANGLES:

For the front suspension, emphasis was put on independence and decoupling, since adjustment is needed in the suspension for factors such as camber, caster and toe. The method of adjustment is typically contained within the connection between various components, i.e. Heim joints on the ends of control arms to adjust camber. The important factor to remember here though is that depending on how the adjustment method is designed, it may end up changing more than one suspension parameter when adjusted. In Figure 7, a control arm design that uses Heim joints on the ends of the control arms to adjust both caster and camber.

The problem here is that it adjusts both at the same time, while also putting stress on the control arm members during adjustment. This is a coupled design since multiple functional requirements are controlled by one design parameter.

CAMBER ADJUSTMENT:

Adjustment for camber is controlled by a one piece offset plate. Two of these are incorporated into each upper control arm pickup. By changing the location of the hole in the plate, the distance the upper control arm pivots from the chassis can be adjusted, in turn adjusting camber. Although this has a slight effect on roll center location and migration.

Load Calculations In order to determine appropriate dimensions and materials for components to be used, we first needed to determine the forces they would be loaded under. Milliken's Race Car Vehicle Dynamics book contains both equations and methods to aid in calculating these loads. With this information, a MathCAD file was created allowing easy reference for forces in the uprights, control arms, and pick-ups under braking, acceleration and cornering. The scenarios used were 2g lateral and g longitudinal braking along with a "worst case scenario" 3g bump where it was assumed that the entire vehicle was lifted

by one corner. This provided the information necessary to perform FEA of all suspension components. Stresses and buckling loads could then be calculated.

WHEEL ASSEMBLY

CALCULATIONS FOR UPRIGHT

Assuming

Mass (m) = 300kgWheelbase (W) = 1575.42mm

Front track width ( ) =1168.86mm

Rear track width ( ) = 1118.04mmCentre of mass height=279.4mm

Worst Case Assumed For Design:

For front:Braking=gLateral Acceleration=2gVertical Acceleration=3g

For Rear:Braking=0Lateral Acceleration=2gVertical Acceleration=3g

FOR FORNT UPRIGHT:

LONGITUDNAL WEIGHT TRANSFER:

FOR REAR:

LATERAL WEIGHT TRANSFER:

=Normal on outer front tyre

=Normal on outer rear tyre

FOR FRONT:

Similarly,

FOR REAR:

=1763.93N

FORCES ON TYRE:

ON FRONT TYRE:

Similarly,

ON REAR TYRE:

FORCES THROUGH BEARING ON UPRIGHT:

ON FRONT TYRE:

Similarly,

ON REAR TYRE:

THE UPRIGHT DESIGN:

The uprights were the most complex machined components on the car. While not ideal from a machining standpoint, this allowed the uprights to be lightweight while still having the strength necessary to take the loading on the front suspension. The inputs of suspension pickup points were taken directly from suspension geometry. After this, while considering the packaging inside the wheel selected, a basic design which connected all the suspension points were connected to the wheel using bolts. Aluminium 6061-T6 was selected as the material for the upright due to its higher strength/ density ratio. Now, simulation of the loading conditions was done using SOLIDWORKS. Mass was reduced by making circular holes in order to reduce mass and promote good loading path.

Problem Definition1. Need StatementThe brake system generates the necessary force to slow the car, both for racing andfor emergency situations. The system must be able to able to easily dissipate heatand handle the energy dissipated by braking without compromising the safety orperformance of the car. A powerful and properly balanced brake system will allow thedriver to slow to necessary cornering speeds in shorter distances, and better use thefull capacity of the braking system. Functional RequirementsBelow is a brief summary of the rules that will most affect the brake system design.T7.1.3 – The brake system must be capable of locking all four (4) wheelsduring the brake test specified below.T7.1.8 – The brake pedal must be designed to withstand a force of 2000 Nwithout any failure of the brake system or pedal box. This may be tested bypressing the pedal with the maximum force that can be exerted by any officialwhen seated normally.Brake TestT7.2.1 – The brake system will be dynamically tested and must demonstratethe capability of locking all four (4) wheels and stopping the vehicle in astraight line at the end of an acceleration run specified by the brakeinspectors.In complying with these rules, the brake system is need to be designed to stop thecar at a rate of up to 1.2g’s, with a FoS of 2 in all components being manufactured.Despite efforts to lower the FoS, the inability to accurately model heat transfer withconfidence and the importance of a reliable brake system forces the brake system isbe designed with extra redundancy to ensure driver safety.

Selection of Master Cylinder and calipers.

Various master cylinders of Wilwood,tilton ,Brembo and readily available maruti and Honda master cylinders were taken into consideration. Final decision was taken on following basis:

1. Keeping braking force not too high nor too low for driver as well as keeping bias ratio close to 1.

2. Due to weight cost considerations, wilwood master cylinders having bore diameter of 5/8 inches were selected.

Calculations:

Mass of car = 300 kg

Height of center of gravity from ground =10.42 inches=264.77mm.

Radius of tire = 0.533/2m=0.266 m.

Diameter of piston of TVS Fiero F2 caliper= 20.42 mm

Radius of front disc = 0.11m

During braking:-

ΣT1 = 0

2N2l – mg(0.4l) + mah = 0

N2 = 0.3mg – mah/2l …………………(1)

ΣT2 = 0

2N1l – mg(0.4l) – mah = 0N1 = 0.2mg + mah/2l ………………….(2)

How to limit value of maximum deceleration:

ΣFy = 0

2(N1 + N2) = mgN1 = N2 = mg/2 …………………..(3)

Also, 2(μ1N1 + μ2N2) =ma

Assuming μ1 = μ2 = μ

2 μ(N1 + N2) = ma

For front calipers:

F*r-μtNfR = I*a/r

For pure rolling

a = R*α

α=a/R = (1.2 * 9.81)/0.266

Iα=mR2a/R ( In worst case assuming whole mass as peripheral)

=mRa

=6 x 0.266 x 1.2 x 9.81

=17.22 N.m.

μtNR= 1.2 x 860.63 x 0.266

= 251.82N.m

=μ(2*p*A)x2

Now,

reff= radius of disc – (width of pad/2)

= 11.00cm – (2.5/2)cm

= 9.75cm

On disc, caliper pads are present on both sides, so friction forces are acting from both sides.

Therefore, Friction force on disc

=2(μNd)

=2(μ x 2pA)

Fondisc = 4*μ*p*A

Therefore, 4μpA x r = Iα + μtNR

p = (17.22 +251.82)/4μAreff

= 263.04/( μ x 4 x 3.404 x 10-04 x 9.75 10-2)

= (269.04 x 106)/( μ x 132.756)

= (2.02 x106)/ μ N/m2

μ front = 0.4

Pfront = 5.066 x 106 N/m2

= 734.76 psi

Area of master cylinder = 1.9808 x 10-4 m2 (Bore dia= 5/8 inches)

Therefore, Force required

= p x Am.c

Ffront =752.8 N

For rear:

Total piston area = 2.4 sq inches

= 15.48 x 10-4 m2

Friction force = 2(μNd) [Since pads are on both sides]

= 2 μpA

F*r - 2μtNR = Iα

Here, Iα taken to be approx. double as that of single tyre on front tyre

2μpAr = 2μtNR + Iα

= 357.55 +30

= 387.55 N.m.

P = 388/2μpAr = 2.506 x 106 N/m2F = 496.038 N

Therefore, total force on combined master cylinders=(752.8+496.0)N=1248.8N

Taking mechanical advantage as 3.5, total maximum force required by driver to put on pedal=1248.8/3.5=356.8N

Bias Ratio=752.8/496.0=1.52.

STEERING DESIGN REPORT1. STEERING TYPE

Pitman arm type

Rack and pinion typeOut of the two, RACK AND PINION is selected as it has following advantages:i) Simple and light weight

ii) Less number of components

iii) Quick Response

iv) Low cost

v) No Slop or slack associated as with steering box pitman arm type system.

2. PARAMETERS IMPORTEDFront track width = 46 inRear track width = 44 inWheelbase = 62 inCamber= -1.5 degKPI = 4.55 degCG height = 11 inCaster angle= 4.55 degScrub radius = 0.49 in

3. MAXIMUM STEER ANGLEMinimum turn radius= 4.5 mMinimum width of track = 3.5 mTrack width of car = 46 inWheelbase= 62 inTaking minimum radius of the car as 3.5 m, and distance between rear axle and cg considering 60-40 weight distribution on rear and front axle respectively = b=

Thus, radius about cg =

Now,

Also,

Thus turning angle at outer wheel at hairpin turn is approx. and that at inner

wheel is approx. .But after calculating the maximum steerable angle in Solidworks, it was found out

that after wheel rotation of , the wheel hits the lower wishbone. Therefore, it was decided to use a steering geometry as close to parallel Ackerman as possible to get maximum lateral force due to weight transfer on outside tyres considering that the obtained steerable angle is less than the required value.

4. STEERING RATIOConsidering that driver does not need to move his hands considerably during the entire course and maximum turning angle is 24 deg, steering ratio was decided as 4:1. This way driver does not need to move the steering wheel more than approx. 96 deg for the worst case, thus providing efficient turning situation during Autocross.

5. WEIGHT DISTRIBUTION DURING CORNERINGMass of car with driver = 300 kgConsidering moment equilibrium about inner wheel and three forces namely cornering (centripetal) force, weight of car and reaction on tires to be acting on the car:

Thus force on inner tire will be outer tire load subtracted from total load:

Considering 60-40 weight transfer ratio on rear and front wheels respectively:

For critical condition so that inner wheels lift off +the ground:

Velocity at this much value of acceleration is:

Considering weight distribution, vehicle is safe only up to 27.8 kmph at turning radius of 3 m. But as lateral acceleration of 2g is never achieved by tires, this much speed is not attained at hairpin bends, the design is completely safe.

6. CASTER AND SCRUBCaster was optimized at 4.55 deg using Susprog as this value provides optimum camber change and body roll during roll. Moreover, adequate value of self-centering effect is obtained, as calculated below, without putting too much pressure on steering wheel.Scrub was taken as 1.24 cm because it imposes a further moment of 10 N-m on kingpin, as calculated below, with coefficient of friction of tires taken as 1 when wheel starts slipping due to longitudinal acceleration of the vehicle. This length cannot be more than this for it will require greater steering force and this much value is required for adequate road feedback for the driver. Moreover this value was also obtained from iterations using Susprog, at required kingpin inclination.

7. STEERING TORQUE ABOUT KINGPINDue to vertical force:

Due to lateral force:

Due to tractive force, taking rolling resistance coefficient as 0.03:

8. MOMENT DUE TO LONGITUDINAL SLIP OF TIRESTaking coefficient of friction as 1, moment due to tractive force is:

Thus total torque at kingpin during cornering =

9. STEERING RACKThe rack length was determined using bump steer geometry and the vertical position was determined as per the mounting constraints on chassis. Thus it was

decided to keep the steering rack from the ground and rack length was

found out to be at the aforesaid height.Further, the movement of steering rack was analyzed using OptimumK and it was found that the rack moves 1.111 in (28.2 mm), for the full rotation of wheels.

10.PINION GEARFace width = 25 mmModule = 1.5Pressure angle = 20 degNo of teeth = 23Involute teeth profileMaterial = Tempered high carbon steel

Diameter = (Thus actual rack displacement obtained with this pinion configuration is 28.9 mm

for of pinion rotation)

11.STEERING ARMSteering arm geometry was first found out by iterating in Solidworks at desired Ackerman configuration and then was optimized using OptimumK to get a

decreasing Ackerman progression at maximum possible steerable angle. Finally

it was decided to have the steering arm 85.6 mm long at an angle of from the lateral vertical plane.

12.STEERING EFFORTSteering wheel radius = 5 in

Thus the force exerted by the driver with each hand is 4.25 kgf which is within

attainable limits.

13.TORQUE AT GEARSTorque transmitted by gears =

Lateral force on rack =

Force on tie rod =

DESIGN OF CHASSIS

Objective: To make chassis stiff enough so that it does not affect the suspension performance during cornering, hard braking and acceleration and at the same time should have minimum weight. Driver ergonomics and safety were foremost priorities in the design.

The chassis is triangulated node to node, especially at the mounting points to provide optimum strength and rigidity. The weight of the chassis for 2014 car is 37 kg, taking AISI 4130 round tubes as the material to be used as ratings for AISI 4130 steel as 3 and AISI 1020 steel as 2.04 were inferred on comparison of ultimate tensile strength, yield strength and density.

Design procedures followedPreliminary sketches of chassis design were prepared on paper and then modelled in Solidworks 2013, in accordance with the rules specified in the rule book. A PVC 1:1 prototype was then fabricated to ensure proper driver ergonomics, positioning of mounts and brackets and auxiliary equipment.

FEA OF CHASSISTest for torsional rigidity:The chassis model was constrained at the rear and a moment was applied by loading front suspension pickup points by 1000 N on either end. The rigidity was then calculated based on the obtained twist and was found to be 2285.9 Nm/Degree. This stiffness value is approximately more than four times the sum of front and rear roll stiffness value of the suspension which ensures that the chassis is stiff enough not to affect the suspension performance.

Test for frontal impact:

For frontal impact case using deformation energy and considering maximum speed of vehicle at the impact, u1= 14m/sec and crash time t of the order of 100ms. F=13650 N when this force was applied in frontal impact condition the maximum stress obtained was 29.893 MPa which was well under limits and the deflection was 15.8 mm which is under acceptable limits.

Side impact test

A force of 7kN was applied on the side impact structure connecting the front and the main hoop. The deformation of cockpit was to be kept at minimum and the maximum deflection obtained was 2.96 mm which does not deform the cockpit appreciably and is safe from the point of view of driver’s protection.