Brake Drum and Disc Brake Final

 

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CHAPTER 1 INTRODUCTION

  In recent years, brake systems have undergone tremendous changes in terms of performance, technology, design and safety.. A brake is device means of which artificial frictional resistance is applied to moving machine member, in order to stop the motion of a machine. In the process of performing this function, the brakes absorb either kinetic energy of the moving member or the potential energy given up by objects  being lowered by hoists, elevators etc. The energy energy absorbed by bra brake ke is dissipated dissipated in the fo form rm of heat. This heat is dissipated in the surrounding atmosphere to stop the vehicle, so the  brake system should have following requirements: 

The brake must be strong enough to stop the vehicle with in minimum distance in an emergency.



The driver must have proper control over the vehicle during  braking and vehicle must not skid.



The brak brakes es mus mustt hav havee wel welll ant anti-f i-fade ade cha charac racter terist istics ics i.e. i.e. their  their  effec ef fecti tive vene ness ss sh shou ould ld no nott de decr crea ease se with with Co Cons nsta tant nt pr prol olon onge ged d application.

1.1 PRINC PRINCIPLE IPLE OF B BRAKIN RAKING G

Braking system is necessary in an automobile for stopping the vehicle. Brakes are applied on the wheels to stop or to slow down the vehicle.

 

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“The kinetic energy due to motion of the vehicle is dissipated in the form of heat energy due to friction friction between moving parts (Wheel or Wheel drum) and stationary parts of vehicle (brake shoes)”. Brakes operate most effectively when they are applied in manner so that wheels do not lock completely but continue to roll without slipping on the surface of road. 1.2 FUNCT FUNCTION ION OF VE VEHICLE HICLE B BRAKI RAKING NG 

To slow down or stop the vehicle in the possible time at the time of need.



To control the speed of vehicle at turns and also at the time of  driving down on hill slope.

1.3 CLASSIFICATION OF BRAKES

On the basis of method of Actuation, 

Foot brake (also called service brake) operated by foot pedal.



Hand brake (also called parking brake) operated by hand.

On the basis of Mode operation, 

Mechanical brakes.



Hydraulic brakes.



Air Brakes.



Vacuum brakes.



Electric brakes.

On the basis of Action on Front or Rear Wheels, 

Front -wheel brakes



Rear – wheel brakes

On the Basis of method of Application of Braking Contact,

 

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

Internally – expending brakes



externally – contracting brakes

Mechanical Brakes

Inte Intern rnal al ex expe pend ndin ing g br brak akee sh shoe oe comm common only ly us used ed in au auto tomo mobi bile les. s. In an automobile, the wheel is fitted on a wheel drum. The brake shoes come in contact with inner surface of this drum to apply brakes. The whole assembly consists of a pair of brake shoes along with brake linings, a retractor spring two anchor pins a cam and a brake drum. Brake linings are fitted on outer  surfacee of each brake shoe. The brake shoes are hinged at one end by anchor  surfac  pins. Other end of brake shoe is operated by a cam to expand it out against  brake drum. A retracting spring brings back shoes in their original position when brakes are not applied . The brake drum closes inside it the whole mech me chani anism sm to pr prot otec ectt it fro from m du dust st an and d firs firstt . A plat platee ho hold ldss wh whol olee assembly assem bly and fits to car axle. It acts as a base to fasten the brake shoes and other operating mechanism. Internal expanding shoe brakes are the generally used braking system in automobiles. In an automobile, the wheel is fitted on a wheel drum. The brake shoes are fitted in contact with inner surface of this drum to apply brakes. The construction of mechanical disk brake is shown in picture. The whole assembly contains of a pair of brake shoes with brake linings, two anchor pins and retractor spring, a cam and a brake drum. Brake linings are attached on outer surface of each brake shoe.

 

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Figure 1.1 Mechanical Brake

The brake shoes are hinged at one end by means of anchor pins. Last end of brake shoe is functioned by a cam to expand it out counter direction to  brake drum. Retracting springs provided are used for bringing the shoes to their original position when brakes are not applied. The brake drum closes inside it the entire mechanism to protect it from dust and sand. A plate holds the total assembly and fits to car axle. It also acts as a base to fasten the brake shoes and other operating mechanism. Hydraulic Brakes

The brakes which are actuated by the hydraulic pressure are called hydraulic brakes. Hydraulic brakes are commonly used in the automobiles. Hydraulic brakes work on the principle of Pascal’s law which states that “Pressure at a point in a fluid is equal in all directions in space”. According to this law when pressure is applied on a fluid it travels equally in all directions so that uniform braking action is applied on all four wheels. A hydraulic braking system transmits brake-pedal force to the wheel brakes through pressurized fluid, converting the fluid pressure into useful work of   braking at the wheels. A simple, single-line hydraulic layout used to operate a drum and disc brake system is illustrated. The brake pedal relays the the driver’s foot effort to the master-cylinder piston, which compresses the brake fluid.

 

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This fluid pressure is equally transmitted throughout the fluid to the front disc-caliper pistons and to the rear wheel-cylinder pistons. As per the regulations a separate mechanical parking brake must be incorporated with at least two wheels. This provision also allows the driver to stop the vehicle in the event of failure of the hydraulic brake system.

Figure 1.2 Hydraulic Brakes Disc brake

Modern motor bikes are fitted with disc brakes instead of conventional drum type brakes. Discbrake consists of a rotating disc and two friction pads which are actuated by hydraulic braking system as described earlier. The friction  pads remain free on each side of disc when brakes are no applied. They rub against disc when brakes are applied to stop the vehicle. These brakes are applied in the same manner as that of hydraulic. But mechanism of stopping vehicle isdifferent than that of drum brakes.

 

The most commo common n type of disc brake on modern bikes is the  the  single-

 piston floating calliper. In this article, we will learn all about this type of disc  brake design.

 

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Figure 1.3 Disc Brake 1.4 PRINCIPLE OF FRICTION RELATED OF THE DISC BRAKES

 

Friction is a force that resists the movement of one surface over another.

In some instances it can be desirable; but more often is not desirable. It is caus ca used ed by su surf rfac acee ro roug ugh h sp spot otss th that at lo lock ck toge togeth ther er.. Th Thes esee spot spotss can can be microscopically small which is why even surfaces that seem to be smooth can experience friction.  

Friction is a measure of how hard it is to slide one object over another.

Take a look at the figure below. Both of the blocks are made from the same material, but one is heavier. I think we all know which one will be harder for  the bulldozer to push.  

Different materials have different microscopic structures; for instance, it

is harder to slide rubber against rubber than it is to slide  steel against steel. The type of material determines the  the coefficient of friction, the ratio of the force required to slide the block to the block's weight. (a) Low coefficient of friction for a pair of surface means they can move easily over each other. (b)   High coefficient of friction for pair of surface means they cannot move easily over each other.

1.5TYPES OF DISC BRAKE

 

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SOLID DISC BRAKE

One slightly tarnished, solid disc of metal. Vented discs, on the other  hand more like two discs of metal with ribs in between, allowing air to flow through and provide a cooling effect. These are consequently generally much thicker than solid discs.

Figure 1.4Solid Disc VENTILATED DISC BRAKE

 

Both Bot h wil willl be abl ablee to hav havee the sam samee amo amount unt of bra brakin king g for force ce

applied appli ed to them, but the vented discs are able to shed the heat build-up more quickly than solid discs which leads to a longer period of time before brake fade becomes an issue and more consistent braking accordingly.

Figure 1.5Ventilated Disc

 

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Early brake shoes contained  contained  asbestos. asbestos. When  When working on brake systems of  older cars, care must be taken not to inhale any dust present in the brake assembly. The United States Federal Government began to regulate asbestos  production, and brake manufacturers had to switch to non-asbestos linings. Owners initially complained of poor braking with the replacements; however, technology eventually advanced to compensate. A majority of daily-driven olde olderr ve vehi hicl cles es ha have ve be been en fi fitt tter er wi with th as asbe best stos os-fr -free ee lini lining ngs. s. Many Many othe other  r  countries also limit the use of asbestos in brakes. 1.6SOLID WORKS

SOLIDWORKS Simulation uses the displacement formulation of the finite fin ite ele elemen mentt met method hod to cal calcul culate ate com compon ponent ent dis displa placem cement ents, s, strain strains, s, and stresses under internal and external loads. The geometry under analysis is discretized using tetrahedral (3D), triangular  (2D), and beam elements, and solved by either a direct sparse or iterative solv so lver er.. SO SOLI LIDW DWOR ORKS KS Si Simu mula lati tion on al also so of offe fers rs the the 2D simp simpli lifi fica cati tion on assumption for plane stress, plane strain, extruded, or axisymmetric options. SOLIDWORKS Simulation can use either an h or p adaptive element type,  providing a great advantage to designers and engineers as the adaptive method ensures that the solution has converged.

1.7FINITE ELEMENT ANALYSIS

 

SOLIDWORK SOLIDWORKS S Simul Simulation ation can use eithe eitherr an h or p adaptive adaptive eleme element nt

type, providing a great advantage to designers and engineers, as the adaptive method met hod ens ensure uress tha thatt the sol soluti ution on has con converg verged. ed. Produc Productt Engi Enginee neers rs can review the internal mesh elements with the Mesh Sectioning Tools to check 

 

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the quality of the internal mesh and make adjustments to mesh settings before running the study.  

Use Users rs can spe specif cify y loc local al mes mesh h contro controll at vertic vertices, es, edges, edges, faces, faces,

components, and beams for a more accurate representation of the geometry.  

Integrated with SOLIDWORKS 3D CAD, finite element analysis using

SOLIDWORKS Simulation knows the exact geometry during the meshing  process. And the more accurately the mesh matches the product geometry, the more accurate the analysis results r esults will be.  

Since the majority of industrial components are made of metal, most

FEA FE A ca calc lcul ulat atio ions ns in invo volv lvee me meta tall llic ic co comp mpon onen ents ts.. The The an anal alys ysis is of meta metall components can be carried out by either linear or nonlinear stress analysis. Which analysis approach you use depends upon how far you want to push the design.  

If you want to ensure the geometry remains in the linear elastic range

(that is, once the load is removed, the component returns to its original shape), then linear stress analysis analysis may  may be applied, as long as the rotations and displacements are small relative to the geometry. For such an analysis, factor  of safety is a common design goal.  

Evaluating the effects of post-yield load cycling on the geometry a non-

linear should be carried out. In this case, the impact of strain hardening on the residual stresses and permanent set (deformation) is of most interest. The analysis of non-metallic components (such as, plastic or rubber parts) should be carried out using nonlinear methods, due to their complex load deformation relationship. SOLIDWORKS Simulation uses FEA methods to calculate the displacements and stresses in your product due to operational loads such as:

 

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1 .Forces 2. Pressures 3. Accelerations 4. Temperatures Contact between Components

 

Loads can be imported from thermal, flow, and motion Simulation

studies to perform multiphysics analysis. Mesh Definition

 

SOLID SOLIDWORK WORKS S Simul Simulation ation offers the capability capability to mesh the CAD

geometry in tetrahedral (1st and 2nd order), triangular (1st and 2nd order),  beam, and truss elements. The mesh can consist of one type of elements or  multiple for mixed mesh. Solid elements are naturally suitable for bulky models. Shell elements are naturally suitable for modelling thin parts (such as sheet metals), and beams and trusses are suitable for modelling structural members. Solid Works Mesh Type

  1. Shell mesh mesh is automatically generated for sheet metal model and surface  bodies  2.

Beam elements are automatically defined for structural members

1.8 MATERIAL DEFINITION

 

All ele elemen ments ts are defi defined ned by nod nodes, es, which which have have onl only y the their ir loc locati ation on

defined. In the case of plate and shell elements there is no indication of  thickness. thick ness. This thick thickness ness can be given as element propert property. y. Property tables tables

 

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for a particular particular propert property y set 1-D have to input input.. Diffe Different rent type typess of elements have different properties. Engineering Application

1. Design of aircraft and aerospace structures for minimum weight. 2. Finding the optimal trajectories of space vehicles. 3. Optimal production planning, controlling, and scheduling. 4. Design of optimization pipeline networks for process pr ocess industries. Etc.,

CHAPTER 2 LITERATURE REVIEW H.P. H. P. Kha Kharin rinar, ar, V.M V.M.Ph .Phall alle,S e,S.S. .S.Man Mantha tha et. et.al al (201 (2010) 0)Pre Prese sent nted ed a pa pape per, r,

Comparative Frictional Analysis of Automobile Drum and Disc Brakes In

 

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this paper Todays technological developments in the vehicle technology seek   better control on active safety by addressing the need for better and safer   braking systems. The braking system is integral and vital part of the active safety control of vehicles. The selection of drum brakes for rear wheels and disc brakes for front wheels has been the trend in the earlier vehicles. Better  heat dissipation achieved by disc brakes is responsible for the wide use of  disc braking system in the modern vehicles. The functional dependence of the friction coefficient upon a large variety of parameters including sliding speed, acceleratio accel eration, n, criti critical cal slidi sliding ng dista distance, nce, tempe temperature rature,, norma normall load, humidity, humidity, surface preparation and of course material combination. The variations in the friction coefficient are highly dependent on the frictional materials and The study presents an estimation methodology of the friction coefficient for drumshoe and rotor disc pad interface in the automobiles. The output from deduced equations was compared with the similar obtained from virtual Simulink  models for drum and disc brakes. Guru Gur u Mur Murthy thy Nat Nathi, hi, T.N Cha Charyu ryulu, lu, K.G K.Gowt owtham ham,pS ,pSath athis is Red Reddye dyet.al t.al (2012) The motive of coupled structural and thermal analysis is to study and

evaluate the performance under severe braking conditions and thereby assist in disc rotor design and analysis. A transient thermal analysis has been carried outt to inve ou invest stig igat atee the tem tempe pera rattur uree va vari riat atio ion n ac acro ross ss the disc disc us usiing axisymmetric elements. As a future work, a complicated model of Ventilated disc brake can be taken and there by forced convection is to be considered in the ana analys lysis is is com compli plicat cated ed by consid consideri ering ng variabl variablee the therma rmall con conduct ductivi ivity, ty, variable specific heatand non-uniform deceleration of the vehicle. Chetan T. Jadav, K.R. Gawandeet.al(2014)have worked on optimization of 

disc brakes. Their study showed that by keeping the braking torque constant if we reduce the diameter of disc rotor and increase the friction pad area then we can reduce the the cost and weigh weightt of disc assembly assembly up to some extent extent.. Cost

 

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of disc is depend on so many factors some factor like transportation cost, material mater ial handl handling ing cost, differ different ent kinds of taxes etc. So if we reduce the diameter of disc we can reduce the materials consumption and ultimately we can reduce the cost of disc. A disc brake is a wheel brake which slows rotation of the wheel by the friction caused by pushing brake pads against a  brake disc with a set of callipers. To stop the vehicle, friction material in the form of brake pads is forced mechanically, hydraulically, pneumatically or  electromagnetically against both sides of the disc. Friction causes the disc  brake and attached wheel to slow or stop. Compared to drum brakes, disc  brakes offer better stopping performance, because the disc is more readily cooled and disc brakes recover more quickly from immersion. The brake disc is the disc component of a disc brake against which the brake pads are applied. Generally the di disc sc rotor is made of grey cast iiron ron and is eithe eitherr solid or ventilated. The ventilated type rotor consists of a wider with cooling fins cast through the middl middlee to ensure good cooling. Some ventilat ventilated ed rotors have spiral fins which creates more air flow and better cooling. BorchateSour Borch ateSourabhSh abhShivaji, ivaji, N.S. Hanam Hanamapure apure and Swapn Swapnil il S. Kulka Kulkarni rni et.al (2014) have reduced the thermal stress disc when it is working at high

temp temper erat atur ure. e. In th this is di diss sser erta tati tion on wo work rk it is pr prop opos osed ed to co cons nsid ider er an any y two/three/four wheels vehicle which is used with disc brake now- a-days on the road. For that vehicle it is proposed to carry out Reduction of weight and increase incre ase cooling effect of disc rotor So as it will enhanc enhancee the performan performance ce of  disc brake. brake. For this dis dissertat sertation ion work foll following owing pr proposed oposed met methodolo hodology gy is adopted. adopte d. A suitable suitable Solver woul would d be employed for securing securing the sim simulate ulated d result.. Performe result Performed d using the finit finitee element ana analysis lysis.. To analyse analyse the thermo elas el asti ticc ph phen enom omen enon on oc occu curr rrin ing g in th thee disk disk br brak akes es , the the occu occupi pied ed heat heat conduction conduc tion is solved wit with h contact problem. problem. Also, the thermo rmo elasti elasticc instabil instability ity (TIE) phenomenon is investigated in the present study, and the influence of 

 

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the mat materi erials als prop propert erties ies on the the thermo rmo ela elasti sticc beh behavi avior or is inv invest estiga igated ted to facilite the conceptual design of the disc brake system.

P.K.Zawar P.K. Zaware, e, R.J. R.J.Patil Patil,, P.R. P.R.Sonawa Sonawaneet.a neet.al(2014 l(2014))  is to inve invest stig igat atee an and d

analyze the temperature distribution of rotor disc during braking operation. Thee wo Th work rk uses uses th thee fi fini nite te el elem emen entt an anal alys ysis is tech techni niqu ques es to pr pred edic ictt the the temperature distribution on the full and ventilated brake disc and to identify the critical temperature of the rotor by holding account certain parameter such as the materials used, the geometric design of the disc and the mode of   braking. The analysis also gives us, the heat flux distribution for the two discs.. The iiniti discs nitial al heat heat fl flux ux q0 into the rotor face is directly calculated using the formula.

A. Bel Belhco hcoine ine,, A.R A.R.. Abu Bak Baker, er, M. Bou Bouche chetra traet.a et.all (20 (2014) 14)Thi Thiswo swork rk on

Design Des ign Modi Modific ficati ation on & Opt Optimi imizat zation ion of Dis Discc Brake Brake Rotor deal dealss with with the study of discbrake rotor by modeling & analysis of different shapesof slots of  different vehicles disc brake rotor with same outer diameter &innermounting  position of holes on wheel hub. Therefore, it gives optimal stress, deformation & weight of the modified disc brake rotor & also good heat dissipation.

2.1RESEARCH GAP

The term contact stress is crucial in many area of tribology. Brake disc and brake drum is a component which has to be evaluated fortribological st stud udy. y. Th Thee stre stress ss an anal alys ysis is of br brak akee disc disc an and d dr drum um is cruc cruciial fo forr the the  performance of wheel and other applications. Many researches are done on Brak Br akee disc disc an and d dru drum, m, th thei eirr ma main in co conc ncer ern n is base based d on load loadin ingt gthe he br brak akee

 

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drum drum/d /dis isca cand nd th thei eirr stre stress ss an anal alys ysis is.. Th They ey us used ed theo theoret retic ical al an and d FEM FEM fo for  r  comparing various parameters. This work is based on finite element analysis using solidworks software for contact stress analysis of brake disc and drum to evaluated wear. The results obtained from finite element analysis software is compared with theoriticalformulae .

CHAPTER 3 METHODOLOGY 3.1GENERAL OBJECTIVES 

Surface Wear analysis of disc brake and drum brake

 

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3.2 THE SPECIFIC OBJECTIVES OF THIS THESIS WORK 

(i)

Force analysis. (ii)

Stress analysis.

(iii) (ii i)

Ide Identi ntifyi fying ng the the cri critic tical al point point o off Dis Discc bra brake ke an and d Dru Drum m bra brake ke we wear. ar. Modeling of disc brake and drum brake system to study contact using solid works software to study the stress analysis.

(iv)) (iv

To es estim timate ate o off str stress ess iin n the con contac tactt con condit dition ion iin n dis discc brake brake an and d drum drum  brake

by finite element method using SOLID WORKS finite

element software. (v)

Comparing the solid works software results with theoritical formulation.

3.3

METHODOLOGY

OF

WORK

DONE

THISPROJECT

STUDY LITERATURE REVIEW

PROBLEM IDENTIFICATION

RELATED

TO

  

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STRESS, DEFORMATION, MAXIMUM

STRESS, DEFORMATION AND WEAR ANALYSIS

TEMPERATURE WEAR  AND ANALYSIS BY

BY NUMERICAL (FEA)

THEORTICALLY

COMPARING SOLID WORKS SOFTWARE RESULTS WITH THEORETICAL RESULTS

END

CHAPTER 4

DESIGN AND ANALYSIS OF DISC BRAKE

The caliper disc brake and drum brake is operated hydraulically. Two  pads are pressed against opposite side of the disc brakes to provide a braking

 

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torque. In this chapter to design the dimension of disc brakes and drum brake such as outside and inside diameter, thickness,etc. and also determine the analysis parameter such maximum temperature, compressive stress, heat flux and deformation etc.

INPUT PARAMETER  Brake disc

Mass of vehicle = 150 kg  Mass of disc

= 2.483 kg

Initial velocity

= 20 m/sec

Radius of wheel = 350 mm Drum brake

Mass of vehicle = 150 kg  Mass of disc

= 4.23 kg

Initial velocity

= 20 m/sec

Radius of wheel = 420 mm

4.1 DETERMINATION OF DISC BRAKING TORQUE

2

v Braking distance, d = 2 μg

(4.1)

 

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 =

2

 

20

( 2 × 0.7 × 9.81)

d= 4.04 m After finding out braking distance we must calculate deceleration, Deceleration, a =

 v

2

2d

( ¿ ) μg

(4.2)

2

   

=

  ( 20 ) ( 2 × 14.04 )

a = 6.867m/s2

Braking force:

Already we know that newton’s second law, Braking force, F b = ma

(4.3)

  = 150×6.867 Fb = 1030.05N Braking Torque:

Braking torque, T= force × radius of wheel   = 1030.05 × 0.35 T=360.517 N-m

Torque equation from reference (9), Torque = μ P R f f 

(4.4)

T = μ × pavg × pad area × R f f  3

3

i   =μ × 0.89 × pmax × 1  θ (R o2 – R i2) × 2 ( R − R ) 2 3  R − R o i o

2

2

 

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360.517 = 0.4 × 0.89 × 0.65 X 106 × 1/2 × 54 × 2/3 × (R o3 – R i3) 360.517 =

12.495 × 10 3

6

  × R o3 – R i3

R o3 – R i3 = 8.655 × 10-4 From Reference (7), We know the following condition, Ri/Ro = 0.80  p avg

if 

 p max

 = 0.89 , so

R i= 0.80 R o R o3 – R i3 = 8.655x 10-4

Substitute ‘R i’ value in above equation , we get R o3 – 0.803 R o3

=

8.655 x 10-4

R o3 (1-0.803) = 8.655 x 10-4 3

R o   =

3

R o  =

−4 8.655 × 10 3

( 1−0.80 ) −4 8.655 × 10

( 0.488 )

R o 0.135 m =

R i= 0.80 × R o  

= 0.80 × 0.135

R i = 0.108 m

Outside radius = 0.135 m

 

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Inner radius = 0.108 m Thickness of Disc

Mass of the disc = 2.48kg Density, ρ = m v

7100 =

1.43

 Area  Ar ea× × t 

Area of disc, A =

 

π  4

=

 (D2 – d2)

π  4

 (0.2702 – 0.2162)

  A = 0.02061 m2 7100 =

t=

1.43 0.02061 × t 

1.43 0.02061 × 7100

t = 7.995 × 10-3m t = 4.2× 10-3m  

Outside Radius

 

Inner Radius

 

Thickness of disc

=

0 .135 m

=

0.108 m

= 4.4 mm

(4.5)

 

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Determination Inner diameter Outer diameter Thickness

Brake disc 270mm 216mm 3.8mm

TABLE 4.1 Design of disc brake and drum brake

Determination Braking distance Braking force Braking torque Heat flux

Brake disc 4.04 m 1050N 360.517 N-m 4.5x106

TABLE 4.2 Determination of braking torque and braking force

4.2 DETERMINATION OF CONTACT STRESS

From following Motion equation braking time should be calculated, that follows S= ut+ ½ g t2

(4.6) 1

14.04 =13.88 + 2 × 9.8 l t2 t = 3.65 s CONTACT STRESS FOR GREY CAST IRON

From tribology journal we can take this heat flux formula, Heat flux of the disc, 1

q=

−φ mgz 2

.

2

. v o / A d ε p

(4.7)

 

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  = 1−0.7 .

( 1.43 × 9.81 ×

=

9.81

)

 . (25/0.5)

2 × 0.0206

2

q

 6.867

1.356x 106 W/m2

Also consider the temperature distribution among followingFormula can be taken from reference journal,

the

rotor

disc

Maximum Temperature, Tmax=

0.527 × q × √ t  t 

 ρcpk  √  ρcpk 

 + Tamb

(4.8)

× √ 2.45 2.45  + 30 √ 7100 × 586 × 54 6

Tmax =

0.527 × 1.356 × 10

o

T

max

= 136.208 C

Maximum temperature,Tmax= 136.208 oC Compressive stress can be found by following formula, σ=

 E  × α × ∆T 1− μ

 E  = ∆T = cm

 

125 × 10

(4.9) 5

380 × 1.17

= 47.4134

σ=

125 × 10

9

( 1− 0.27 )

 × 0.12 × 10-6 × 47.4134

  = 64.28 Mpa Compressive stress, σ

=

64.28 Mpa

Deformation can be found by following formula,

 

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6

σ  Youngs modulus, E =  = e

108.6 × 10

δd 0.27

6

108.6 × 10

110 × 109 =

δd 0.27

6

δd 0.27

 =

108.6 × 10 9

125 × 10

Deformation, δd 0.342 mm =

CONTACT STRESS FOR ALUMINIUMALLOY

Also

consider

the

temperat ratured redistribution

among

followingFormula can be taken from reference journal, Maximum Temperature, Tmax=

0.527 × q × √ t  t 

 ρcpk  √  ρcpk 

 + Tamb 6

Tmax =

0.527 × 1.356 × 10

× √ 2.45 2.45

2700 × 900 × 155 √ 2700

 + 30

Tmax= 154.929 oC o

Maximum temperature,Tmax= 154.929 C  Compressive stress can be found by following formula, σ=

σ=

 E  × α × ∆T 1− μ 72 × 10

9

( 1− 0.33)

 × 0.24 × 10-6 × 47.4134

  = 63.82 Mpa Compressive stress, σ

=

63.82 Mpa

the

ro rottor

disc

 

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Deformation can be found by following formula, 6

122.3 × 10

σ 

Young’s modulus, E = e  =

δd 0.27

6

72 × 109 =

122.3 × 10

δd

0.27

6

δd 0.27

 =

122.3 × 10 72 × 10

9

Deformation, δd 0.323 mm =

CONTACT STRESS FOR CHROME STAINLESS STEEL

Also Al so cons consid ider er th thee tem tempe pera ratu ture re

di dist stri ribu buti tion on am amon ong g the roto rotorr dis discc

followingFormula can be taken from reference journal, Maximum Temperature, Tmax=

t  0.527 × q × √ t   ρcpk  √  ρcpk 

 + Tamb

× √ 3.65 3.65  + 30 √ 7800 × 460 × 18 6

Tmax =

0.527 × 1.356 × 10

o

Tmax= 209.07 C Maximum temperature,Tmax= 209.07 oC  Compressive stress can be found by following formula, σ=

σ=

 E  × α × ∆T 1− μ 193 × 10

9

( 1− 0.28)

 × 0.11 × 10-6 × 47.4134

  = 62.32Mpa

 

26

Compressive stress, σ

62.32 Mpa

=

Deformation can be found by following formula, 6

139.8 × 10

σ 

Youngs modulus, E = e  =

δd 0.27 6

139.8 × 10

193 × 109 =

δd 0.27

6

δd 0.27

 =

139.8 × 10 9

193 × 10

Deformation, δd 0.132mm =

CONTACT STRESS FOR TITANIUM 6AL-4V

Also consider the temperature distribution among tthe he rotor disc follow following ing formulacan be taken from reference journal, Maximum Temperature, Tmax=

0.527 × q × √ t  t 

 ρcpk  √  ρcpk 

 + Tamb

× √ 2.65 2.65 Tmax= √ 7850 × 475 × 44.5 × 0.65  + 30 6

0.527 × 4.5 × 10

Tmax= 128.26oC Maximum temperature,Tmax= 128.26 oC  Compressive stress can be found by following formula, σ=

 E  × α × ∆T 1− μ

σ=

2.1 × 10 1 0.29

5

( −

)

 × 1.23 × 10-5 × 47.4134

 

27

  = 61.78Mpa Compressive stress, σ

=

61.78Mpa

Deformation can be found by following formula, 6

σ   = e

Youngs modulus, E =

102.4 × 10

δd 0.27

6

102.4 × 10

2.1 × 105 =

δd 0.27

6

δd 0.27

 =

172.4 × 10 2.1 × 10

9

Deformation, δd 0.22 mm =

CONTACT STRESS FOR COPPER ALLOY

Also consider the temperaturedistribution among the rotor disc following formula can be taken from reference journal, Maximum Temperature, Tmax=

0.527 × q × √ t  t 

 ρcpk  √  ρcpk 

 + Tamb

× √ 3.65 3.65  + 30 7750 × 486 × 0.65 × 52 √ 7750 6

Tmax =

0.527 × 1.356 × 10

Tmax=122oC Maximum temperature,Tmax=122oC  Compressive stress can be found by following formula,  E

σ = 1− μ  × α × ∆T

 

28

σ=

2.1 × 10

5

( 1− 0.29)

 × 1.25 × 10-5 × 47.4134

  = 58.21 Mpa Compressive stress, σ

=

58.21 Mpa

Deformation can be found by following formula, 6

Youngs modulus, E =

σ   = e

286.2 × 10

δd 0.27

6

286.2 × 10

2.1 × 105 =

δd 0.27

6

δd

286.2 × 10

0.27  =

5

2.1 × 10

Deformation, δd 0.20 mm =

Table 4.3 static structural analysis - disc brake result Compressive stress Theoretical

Materials

(Mpa)

Analysis

(Mpa)

Deformation Theoretical (mm)

Analysis

Maximum temperature

(mm)

(0C)

Greycast

64.28

62.14

0.34

0.337

136.208

63.82

61.39

0.32

0.3108

154.929

stainless

62.32

61.75

0.13

0.1108

209.07

steel Ti-6Al-4v copper

61.78 58.21

61.56 60.89

0.22 0.20

0.2079 0.1893

128.26 138.42

iron Aluminium

alloy

Chrome

Table 4.3 static structural analysis - disc brake result

 

29

 

In table 4.3 the modeled disc brake is analysed in Finite element

software solid works v 2016 and their results are tabulated. The simulation is done with grey cast iron, aluminium alloy, chrome stainless steel, titanium alloy and copper alloy.  

Figur Figuree 4.1 shows the bar chart repres representat entation ion of deform deformation ation in

modeled disc brake with respective to various materials. DISC BRAKE DEFORMATION (mm) 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

ay ca cas t i ro ron G rra

Al um um in ini um um al all oy oy (1060)

Chromium stainless steel

Copper

T itani um

Figure 4.1 Comparison of disc brake deformation 4.3 WEAR ANALYSIS OF DISC BRAKE

Wear Analysis of Grey cast iron Brake disc as given below, Archards wear law equation, V = k ×

 F n  × S  H 

Where, v = wear volume Fn = normal force

 

30

 

H = hardness of the material

The sliding distance S can be given as ds =r × d θ V = K × F n × rθ  H 

The angular displacement θ can be expressed as: θ =ωt  V =

 K × F n ×rωt   H 

The rate of wear is Q= K × F  × rω  H  n

Where, r is basic circle radius θ is brake disc angle

V = 0.234µm/sec.

 

31

CHAPTER 5 FINITE ELEMENT ANALYSIS

5.1

MODELLING

OF

DISC

BRAKE

USING

SOLIDWORKS

SOFTWARE

The inner radius, outer radius and thickness of disc are as 0.108m, 0.135m, and 0.08m, respectively. The hydraulic pressure is applied to the boundary along radius of the piston side pad and the immobility condition in the axial direction is applied to the boundary along the radius of the finger side one. The heat finite element model of disc brakes with boundary condition are shown.. The convectiv shown convectivee bounda boundary ry condi condition tion are imposed imposed on all boundaries boundaries to 0

consider more realistic realistic heat conditions. The initial temperature is T=27 C in the study.

Figure 5.1 Isometric view of Disc brake model

 

32

5.2 ANALYSIS RESULTS OF DISC BRAKE

Material properties of Grey cast iron Young’s modulus, E = 1.25 × 109 N/m2 Density, ρ =7200 kg/m3 Specific heat, c = 510J/kgK  Thermal conductivity, k =45 w/mK  Thermal co-efficient of expansion, α = 1.2 × 10-5/ oC Poisson ratio, μ = 0.27 ANSYS analysis CASE 1: Deformation in grey cast iron disc. CASE 2: Stress in grey cast iron disc.

CASE 1: Deformation in grey cast iron disc.

Total deformation of Grey cast iron brake disc for applied force of 1030.4 N has been shown in figure 5.2,

Figure 5.2 Deformation in grey cast iron disc

 

33

CASE 2: Stress in grey cast iron disc.

The von misses distribution for disc brake is shown in figure 5.3,

Figure 5.3 Stress in grey cast iron disc

Thee disc Th disc brak brakee su surf rfac acee vo von n mi miss sses es stre stress ss is Mi Mini nimu mum m 60 60.2 .24 4 Mpa Mpa an and d Maximum 62.14 Mpa.

5.3 WEAR ANALYSIS OF DISC BRAKE USING SOLID WORKS DESIGN

Stress Formula as given below, σ = − Pmax 2  z √ 1+ 2 b

Where, Pmax= maximum pressure  

z = module

 

b = contact width

(5.1)

34  

 

 

√

Contact Width, b = 2

2 F 

lπ 

 P (

( 1− μ ) ( 1− μ ) ) + 2

1

 E 2

 E 1

(

 1

+

 1

d1 d 2

) (5.2)

Section Module, z = (1/12) (2.5t(6t)3 – 1.5t(4t)3) / (6t/2)

(5.3)

Where, t = thickness The Compressive Stress Stress from Solid works value is 82. 82.24Mpa 24Mpa Then, the contact pressure is 72.86N/mm2 The contact pressure formula is, 2 Fn

 

P0 = πbl

(5.4)

Where, Fn= normal force  

b = contact width

 

l = Contact length of the Brake disc

From equation (5.4), Braking Normal Force, Fn = 981.2 N.

The Von- Misses Stressare given by equations 5.5, 5.6, 5.7,

 z σ  x =−2. v .Pmax ¿  – { }] b

(5.5)

 z σ  y =− Pmax ¿ – 2{ }] b

(5.6)

σ  z=

− Pmax 2

√ 1 +  z 2  (5.7) b

 

35

Vonmises stress = 30.26 mpa Archards wear law equation, V = k ×

 F n  × S  H 

Where, v = wear volume Fn = normal force  

H = hardness of the material

The sliding distance S can be given as ds =r × d θ  K × F  × rθ V =

n

 H 

The angular displacement θ can be expressed as: θ =ωt  V =

 K × F n ×rωt   H 

The rate of wear is Q= K × F n × rω  H 

Where, r is basic circle radius θ is Brake disc Angle

V = 0.023 µm/sec .

 

36

CHAPTER 6 RESULT AND DISCUSSION

A se seri ries es of co comp mpar aris ison onss ha have ve be been en made made betw betwee een n the the theo theore reti tica call  predictions and analysis results to examine the validity of the present theoretical model. These comparisons include; quantitative comparisons of  von misses stresses and wear prediction calculated from the contact stress results of both theoretically as well as analysis results. 6.1 COMPARISION OF THEORETICAL AND FEA ANALYSIS OF ANALYSIS OF DISC BRAKE

Von mises stress values values for the Material Materialss are given in Table Table 6.1. And it is found that Grey cast iron has allowable stress value compared to above set of stress values. From the above discuss discussions ions it is evide evident nt that the von misses and temperature rise of the Grey cast iron for disc brake design. Theoretical von Design

misses stress, (Mpa)

Simulated

Error

stress(Mpa)

percentage

62.14

0.16

Grey Cast Iron Disc Brake

61.12

Table 6.1 Compressive stress analysis of disc brake

Table 6.2shows the Deformation result of the disc brake and it shows the variation of the theoretical and analytical disc rotor, percentage reduction in material of the disc. It is the found that the grey cast iron disc brake is well suitable for the Bajaj pulsar 150cc bike. Design

Theoretical

Ansys

Error

 

37

Grey Cast Iron Disc Brake

Deformation,

Deformation,

mm

mm

0.322

0.337

percentage

0.04

Table 6.2 Deformation analysis of disc brake 6.3 WEAR WEAR OF D DISC ISC B BRAKE RAKE

The material for the disk brake is selected mainly based on the wear. In the table wear for selection material are given. Material

wear For disc brake

Gray cast iron Aluminum alloy Chrome stainless steel Copper alloy Titanium 6al-4v

(μm/sec) 0.023 0.024 0.0242 0.032

0.041 Table 6.4 Comparison of material based wear of disc brake

In the below graph material vs wear represented

 

38

Disc brake Wear ( μm/sec) 0.05 0.04 0.04 0.03 0.03 0.02 0.02 0.01 0.01 0 Gray cas Gray castt iro iron n

Aluminiu Alumin ium m all allo oy (1060)

Chromium stainless steel

Copper

Figure 6.1Comparison of wear in disc brake

CHAPTER 7

Titanium

 

39

TAGUCHI METHOD Case 1 2 3

Ro 125 140 160

 

R i 115 110 105

T 3.8 4.6 5.2

TABLE 7.1 Parameters for taguchi method 7.1 PARAMETERS USED

 The parameters taken for the Taguchi method were R o,o, R i, t  

Where R 0 = outer radius of Disc pl plate ate

R i= inner radius ofDiscplate t = Thickness of the Disc plate The radii of clutch plat platee were taken from PSG design data book pg.no.7.20, By using the concept of l9 orthogonal array TABLE 7.2 L9 Orthogonal array R i

T

mm

mm

mm

1

125

115

2 3 4 5 6 7 8 9

1 12 25 5 140 140 140 160 160 160

1 11 00 5 115 110 105 115 105 115

Case

Ro  

Contact pressure N/mm2

Surface wear µm/sec

3.8

61.12

0.02019

4 5..6 2 3.8 4.6 5.2 3.8 4.6 5.2

6 44 8..4 25 6 46.52 49.12 45.34 29.12 28.42 26.23

00..0011195845 0.01366 0.0024 0.02664 0.02041 0.02148 0.0206

TABLE 7.2 L9 Orthogonal array

 

40

Here the contact pressure and surface wear are calculated theoretically

Figure 7.1 Stress plot of disc brake for case 1

 

41

Figure 7.2 Deformation plot of disc brake for case 1

Figure 7.3 Stress plot of disc brake for case 2

 

42

Figure 7.4 Deformation plot of disc brake for case 2

Figure 7.5 Stress plot of disc brake for case 3

 

43

Figure 7.6 Deformation plot of disc brake for case 3

Figure 7.7 Stress plot of disc brake for case 3

 

44

Figure 7.8 Deformation plot of disc brake for case 4

Figure 7.9 Stress plot of disc brake for case 5

 

45

Figure 7.10 Deformation plot of disc brake for case 5

Figure 7.11 Stress plot of disc brake for case 6

 

46

Figure 7.12 Deformation plot of disc brake for case 6

Figure 7.13 Stress plot of disc brake for case 7

Figure 7.14 Deformation plot of disc brake for case 7

 

47

Figure 7.15 Stress plot of disc brake for case 8

 

48

Figure 7.16 Deformation plot of disc brake for case 8

Figure 7.18 Stress plot of disc brake for case 9

Figure 7.19 Deformation plot of disc brake for case 9

von misses stress, deformation of clutch plate were given below ; TABLE 7.3 L9 Orthogonal array (ANSYS)

 

49

R i

T

Contact pressure N/mm2

Surface wear µm/sec

Mm

Mm

Mm

1

125

115

3.8

64.40

0.02019

2

125

110

4.6

69.42

0.01154

3

125

105

5.2

71.11

0.01985

4

140

115

3.8

50.26

0.01366

5

140

110

4.6

53.26

0.0024

6

140

105

5.2

56.02

0.02664

7

160

115

3.8

32.17

0.02041

8

160

105

4.6

28.15

0.02148

9

160

115

5.2

23.26

0.0206

Case

Ro  

TAGUCHI ANALYSIS

Surface wear versus outer radius, Inner radius, Thickness

LEVEL

OUTER  RADIUS

INNER  RADIUS

THICKNESS

1

33.90

45.58

34.39

2

36.70

32.77

36.28

3

39.23

35.00

40.11

Delta

5.33

12.81

5.72

Rank

3

1

2

Smaller is better  Table 7.4 Response Table for Signal to Noise Ratios

 

50

LEVEL

OUTER  RADIUS

INNER  RADIUS

THICKNESS

1

0.020190

0.006970

0.020150

2

0.015017

0.023245

0.015975

3 Delta

0.016483 0.005173

0.018087 0.016275

0.014147 0.006003

Rank

3

1

2

Table 7.5 Response Table for Mean

The detailed summary of taguchi analysis is given below Main Effects Plot for Means Data Means B

A

C

0.024

0.022

  s 0.020   n   a   e    M0.018    f   o   n   a 0.016   e    M 0.014

0.012

0.010 125

140

160

105

110

112

3.8

4.6

5.2

 

51

Figure 7.19Main effect plot for means

Main Effects Plot for SN ratios Data Means outer radius

46

in ner radiu s

thickn es s

44

  s   o 42    i    t   a   r    N40    S    f   o   n 38   a   e    M 36

34

32 125

140

160

105

110

115

3.8 3.8

4.6 4.6

5.2

Signal-to-noise: Smaller is better 

Figure 7.6 Main effect for SN ratios

From the results, the evaluation of wear rates under various width and radius , the contact pressure and wear rate decreases with increase in inner radius wear occur mainly on the plate surface when it is subjected to enormous load when the thickness of plate is low. It is concluded that the wear rate can be reduced by having having a discpl discplate ate with its its film thi thickness ckness can be inc increased reased and also by increasing the inner radius.

 

52

CHAPTER 8 CONCLUSION

In this project, the disc brake for the Bajaj pulsar 150cc has been designed and it is analysed using theoretical and finite element method. Grey cast iron disc brake is suggested for this bike based on the Von misses stress, defo de form rmat atio ion, n, ma maxi ximu mum m temp temper erat atur uree an and d we wear ar of the the disc disc an and d dr drum um.I .In n addition to verify the virtual simulation were performed and compared with theoretical calculations.

 

53

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