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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 absorbed by brake is dissipated in the form 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 brakes must have well anti-fade characteristics i.e. their effectiveness should not decrease with Constant prolonged application. 1.1 PRINCIPLE OF BRAKING 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 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 FUNCTION OF VEHICLE BRAKING 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 Internal expending brake shoe commonly used in automobiles. 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 surface of each brake shoe. The brake shoes are hinged at one end by anchor 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 mechanism to protect it from dust and first . A plate holds whole assembly 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 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 common type of disc brake on modern bikes is the singlepiston 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 caused by surface rough spots that lock together. These spots 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 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 frictionfor 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 will be able to have the same amount of braking force applied 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 asbestos. 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 older vehicles have been fitter with asbestos-free linings. Many other countries also limit the use of asbestos in brakes. 1.6SOLID WORKS SOLIDWORKS Simulation uses the displacement formulation of the finite element method to calculate component displacements, strains, 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 solver. SOLIDWORKS Simulation also offers the 2D simplification 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 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. Product Engineers 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. Users can specify local mesh control at vertices, edges, 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 will be. Since the majority of industrial components are made of metal, most FEA calculations involve metallic components. The analysis of metal 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 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 nonlinear 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 SOLIDWORKS Simulation offers the 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 is automatically generated for sheet metal model and surface bodies 2.
Beam elements are automatically defined for structural members
1.8 MATERIAL DEFINITION All elements are defined by nodes, which have only their location defined. In the case of plate and shell elements there is no indication of thickness. This thickness can be given as element property. Property tables
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for a particular property set 1-D have to input. Different types 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 industries. Etc.,
CHAPTER 2 LITERATURE REVIEW H.P. Kharinar, V.M.Phalle,S.S.Mantha et.al (2010)Presented a paper, 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, acceleration, critical sliding distance, temperature, normal load, 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 Murthy Nathi, T.N Charyulu, K.Gowtham,pSathis Reddyet.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 out to investigate the temperature variation across the disc using 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 analysis is complicated by considering variable thermal conductivity, 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 cost and weight of disc assembly up to some extent. Cost
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of disc is depend on so many factors some factor like transportation cost, material handling cost, different 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 disc rotor is made of grey cast iron and is either solid or ventilated. The ventilated type rotor consists of a wider with cooling fins cast through the middle to ensure good cooling. Some ventilated rotors have spiral fins which creates more air flow and better cooling. BorchateSourabhShivaji, N.S. Hanamapure and Swapnil S. Kulkarni et.al (2014) have reduced the thermal stress disc when it is working at high temperature. In this dissertation work it is proposed to consider any 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 cooling effect of disc rotor So as it will enhance the performance of disc brake. For this dissertation work following proposed methodology is adopted. A suitable Solver would be employed for securing the simulated result. Performed using the finite element analysis. To analyse the thermo elastic phenomenon occurring in the disk brakes , the occupied heat conduction is solved with contact problem. Also, thermo elastic instability (TIE) phenomenon is investigated in the present study, and the influence of
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the materials properties on the thermo elastic behavior is investigated to facilite the conceptual design of the disc brake system. P.K.Zaware, R.J.Patil, P.R.Sonawaneet.al(2014) is to investigate and analyze the temperature distribution of rotor disc during braking operation. The work uses the finite element analysis techniques to predict 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 initial heat flux q0 into the rotor face is directly calculated using the formula. A. Belhcoine, A.R. Abu Baker, M. Bouchetraet.al (2014)Thiswork on Design Modification & Optimization of Disc Brake Rotor deals 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 study. The stress analysis of brake disc and drum is crucial for the performance of wheel and other applications. Many researches are done on Brake disc and drum, their main concern is based on loadingthe brake
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drum/discand their stress analysis. They used theoretical and FEM for 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)
Identifying the critical point of Disc brake and Drum brake wear. Modeling of disc brake and drum brake system to study contact using solid works software to study the stress analysis.
(iv)
To estimate of stress in the contact condition in disc brake and 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 TEMPERATURE WEAR AND ANALYSIS BY THEORTICALLY
STRESS, DEFORMATION AND WEAR ANALYSIS BY NUMERICAL (FEA)
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 v2 Braking distance, d = 2 μg
(4.1)
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202 = (2 ×0.7 × 9.81)
d= 4.04 m After finding out braking distance we must calculate deceleration, Deceleration, a =
v2 (¿)μg 2d
=
(4.2)
(202 ) (2 ×14.04)
a = 6.867m/s2 Braking force: Already we know that newton’s second law, Braking force, Fb = 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 Rf T = μ × pavg × pad area × Rf 3 3 1 2 Ro −Ri 2 2 =μ × 0.89 × pmax × 2 θ (Ro – Ri ) × 3 ( 2 2 ) Ro −Ri
(4.4)
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360.517 = 0.4 × 0.89 × 0.65 X 106 × 1/2 × 54 × 2/3 × (Ro3 – Ri3) 360.517 =
12.495× 106 × Ro3 – Ri3 3
Ro3 – Ri3 = 8.655 × 10-4 From Reference (7), We know the following condition, Ri/Ro = 0.80 p avg
if p
max
= 0.89 , so
Ri= 0.80 Ro Ro3 – Ri3 = 8.655x 10-4 Substitute ‘Ri’ value in above equation , we get Ro3 – 0.803 Ro3= 8.655 x 10-4 Ro3 (1-0.803) = 8.655 x 10-4 Ro
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8.655× 10− 4 = (1−0.803 )
Ro3 =
8.655× 10− 4 (0.488)
Ro= 0.135 m Ri= 0.80 × Ro = 0.80 × 0.135 Ri = 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 m
Density, ρ = v 1.43
7100 = Area × t π
Area of disc, A = 4 (D2 – d2) π
= 4 (0.2702 – 0.2162) A = 0.02061 m2 1.43
7100 = 0.02061× t 1.43
t = 0.02061× 7100 t = 7.995 × 10-3m t = 4.2× 10-3m Outside Radius
= 0 .135 m
Inner Radius
= 0.108 m
Thickness of disc
= 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−φ mgz q = 2 . 2 . vo / A d ε p
(4.7)
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= 1−0.7 .
6.867 ) 9.81 . (25/0.5) 2 ×0.0206
(1.43 × 9.81×
2
q = 1.356x 106 W/m2 Also consider the temperature distribution among the rotor disc followingFormula can be taken from reference journal, Maximum Temperature, Tmax=
0.527 ×q × √t + Tamb √ ρcpk
(4.8)
0.527 ×1.356 ×106 × √2.45 Tmax = + 30 √ 7100 ×586 ×54
Tmax= 136.208 oC Maximum temperature,Tmax= 136.208 oC Compressive stress can be found by following formula, E
σ = 1−μ × α × ∆T E
125 ×105
∆T = cm = 380× 1.17 = 47.4134 125 ×10 9 σ= × 0.12 × 10-6 × 47.4134 (1−0.27)
= 64.28 Mpa Compressive stress, σ = 64.28 Mpa Deformation can be found by following formula,
(4.9)
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σ Youngs modulus, E = e =
108.6× 106 δd 0.27
108.6× 106 110 × 109 = δd 0.27 δd 108.6× 106 = 0.27 125× 109
Deformation, δd= 0.342 mm CONTACT STRESS FOR ALUMINIUMALLOY Also
consider
the
temperaturedistribution
among
followingFormula can be taken from reference journal, Maximum Temperature, Tmax=
0.527 ×q × √t + Tamb √ ρcpk
0.527 ×1.356 ×106 × √2.45 Tmax = + 30 √ 2700 ×900 ×155
Tmax= 154.929 oC Maximum temperature,Tmax= 154.929 oC Compressive stress can be found by following formula, E
σ = 1−μ × α × ∆T σ=
72 ×10 9 × 0.24 × 10-6 × 47.4134 (1−0.33)
= 63.82 Mpa Compressive stress, σ = 63.82 Mpa
the
rotor
disc
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Deformation can be found by following formula, 122.3× 106 σ Young’s modulus, E = e = δd 0.27 122.3× 106 72 × 10 = δd 0.27 9
δd 122.3× 106 = 0.27 72× 109
Deformation, δd= 0.323 mm CONTACT STRESS FOR CHROME STAINLESS STEEL Also consider the temperature
distribution among the rotor disc
followingFormula can be taken from reference journal, Maximum Temperature, Tmax=
0.527 ×q × √t + Tamb √ ρcpk
Tmax =
0.527 ×1.356 ×106 × √ 3.65 + 30 √ 7800 ×460 × 18
Tmax= 209.07 oC Maximum temperature,Tmax= 209.07 oC Compressive stress can be found by following formula, E
σ = 1−μ × α × ∆T 193 ×10 9 σ= × 0.11 × 10-6 × 47.4134 (1−0.28)
= 62.32Mpa
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Compressive stress, σ = 62.32 Mpa Deformation can be found by following formula, σ Youngs modulus, E = e =
139.8× 106 δd 0.27
139.8× 106 193 × 109 = δd 0.27 δd 139.8× 106 = 0.27 193× 109
Deformation, δd= 0.132mm CONTACT STRESS FOR TITANIUM 6AL-4V Also consider the temperature distribution among the rotor disc following formulacan be taken from reference journal, Maximum Temperature, Tmax=
0.527 ×q × √t + Tamb √ ρcpk
Tmax=
0.527 × 4.5 ×106 × √ 2.65 + 30 √ 7850 × 475 ×44.5 × 0.65
Tmax= 128.26oC Maximum temperature,Tmax= 128.26 oC Compressive stress can be found by following formula, E
σ = 1−μ × α × ∆T σ=
2.1 ×10 5 × 1.23 × 10-5 × 47.4134 (1−0.29)
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= 61.78Mpa Compressive stress, σ = 61.78Mpa Deformation can be found by following formula, 102.4 ×106 σ Youngs modulus, E = e = δd 0.27 102.4 ×106 2.1 × 10 = δd 0.27 5
δd 172.4 ×106 = 9 0.27 2.1× 10
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 + Tamb √ ρcpk
Tmax =
0.527 ×1.356 ×106 × √ 3.65 + 30 √ 7750 ×486 × 0.65 ×52
Tmax=122oC Maximum temperature,Tmax=122oC Compressive stress can be found by following formula, E
σ = 1−μ × α × ∆T
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2.1 ×10 5 σ= × 1.25 × 10-5 × 47.4134 (1−0.29)
= 58.21 Mpa Compressive stress, σ = 58.21 Mpa Deformation can be found by following formula, σ Youngs modulus, E = e =
286.2× 106 δd 0.27
286.2× 106 2.1 × 105 = δd 0.27 δd 286.2× 106 = 0.27 2.1× 105
Deformation, δd= 0.20 mm Table 4.3 static structural analysis - disc brake result Compressive stress Theoretical
Materials
(Mpa) Greycast
Analysis
(Mpa)
Deformation Theoretical Analysis (mm)
Maximum temperature
(mm)
(0C)
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
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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. Figure 4.1 shows the bar chart representation of deformation 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
Gray cast iron
Aluminium alloy (1060)
Chromium stainless steel
Copper
Titanium
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 ×
Fn ×S H
Where, v = wear volume Fn = normal force
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H = hardness of the material The sliding distance S can be given as ds=r × dθ 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 n × rω H
Where, r is basic circle radius θ is brake disc angle
V = 0.234µm/sec.
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CHAPTER 5 FINITE ELEMENT ANALYSIS
5.1 MODELLING SOFTWARE
OF
DISC
BRAKE
USING
SOLIDWORKS
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 convective boundary condition are imposed on all boundaries to consider more realistic heat conditions. The initial temperature is T=27 0 C in the study.
Figure 5.1 Isometric view of Disc brake model
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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
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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 The disc brake surface von misses stress is Minimum 60.24 Mpa and Maximum 62.14 Mpa.
5.3 WEAR ANALYSIS OF DISC BRAKE USING SOLID WORKS DESIGN Stress Formula as given below, σ=
−Pmax z2 1+ √ b2
Where, Pmax= maximum pressure z = module b = contact width
(5.1)
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√
Contact Width, b = 2 2 F lπ
P(
( 1−μ1) (1−μ 2) + ) E1 E2 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 from Solid works value is 82.24Mpa 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 . Pm ax ¿ – { }] b
(5.5)
z σ y =−Pmax ¿ – 2{ }] b
(5.6)
σ z=
−Pmax z 2 (5.7) √ 1+ 2 b
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Vonmises stress = 30.26 mpa Archards wear law equation, V =k ×
Fn ×S H
Where, v = wear volume Fn = normal force H = hardness of the material The sliding distance S can be given as ds=r × dθ 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 n × rω H
Where, r is basic circle radius θ is Brake disc Angle
V = 0.023 µm/sec.
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CHAPTER 6 RESULT AND DISCUSSION
A series of comparisons have been made between the theoretical 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 for the Materials are given in 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 discussions it is evident that the von misses and temperature rise of the Grey cast iron for disc brake design. Theoretical von Design
misses stress, (Mpa)
Grey Cast Iron Disc Brake
61.12
Simulated
Error
stress(Mpa)
percentage
62.14
0.16
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 OF DISC BRAKE The material for the disk brake is selected mainly based on the wear. In the table wear for selection material are given. Material
Gray cast iron Aluminum alloy Chrome stainless steel Copper alloy Titanium 6al-4v
wear For disc brake (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
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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 cast iron
Aluminium alloy (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
Ri 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 Ro, Ri, t Where R0 = outer radius of Disc plate Ri= inner radius ofDiscplate t = Thickness of the Disc plate The radii of clutch plate 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 Case
1 2 3 4 5 6 7 8 9
Ro
Ri
T
mm
mm
mm
125 125 125 140 140 140 160 160 160
115 110 105 115 110 105 115 105 115
3.8 4.6 5.2 3.8 4.6 5.2 3.8 4.6 5.2
Contact pressure N/mm2
Surface wear µm/sec
61.12 64.45 48.26 46.52 49.12 45.34 29.12 28.42 26.23
0.02019 0.01154 0.01985 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
Case
Ro
Ri
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
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
0.016483
0.018087
0.014147
Delta
0.005173
0.016275
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
A
0.024
B
C
Mean of Means
0.022 0.020 0.018 0.016 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
inner radius
thickness
Mean of SN ratios
44 42 40 38 36 34 32 125
140
160
105
110
115
3.8
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 a discplate with its film thickness can be increased 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, deformation, maximum temperature and wear of the disc and drum.In addition to verify the virtual simulation were performed and compared with theoretical calculations.
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BIBLIOGRAPHY
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4. BorchateSourabhShivaji,Prof. N.S. , N.S. Hanamapure and Swapnil S. Kulkarni “Design , analysis and performance Optimization of disc brake”, International Journals of Advanced Engineering Research and Stdies/III/III,page 25-27,(January 2014). 5. P.K.Zaware, R.J.Patil, P.R.Sonawane,
Design Modification
&
optimisation of Disc Brake Rotor,International Journals of research in Engineering & Advanced Technology, Vol.02 Issue : 03, (July 2014). 6. N. Balasubramanyam , prof. Smt.G. Prasnthi, “Design and Analysis of Disc Brake Rotor for a Two Wheeler ’’, International Journals of Mechanical And Industrial Technology , Vol. 01 Issue: 01, Page 07-12 (March 2014). 7. Guru Murthy Nathi ,T N Charayulu, K. Gowtham, P Sathis Reddy, “coupled Structural and Thermal analysis of Disc Brake, International Journals of Research in Engineering and technology, Vol.01, Issue : 04, (December 2012). 8. A. Belhcoine, A.R. Abu Baker, M. Bouchetra, “ Numerical Modeling of Disc Brake System in Frictional Contact’’, Tribology in Industry, Vol. 36, No.01, Page 49- 66 , (2014). 9. Shigley’s “Mechanical Engineering Design, Eighth Edition’’, Budynas – Nisbett. 10.Joseph E. Shigly, Charles R. Mischake, “ Standard Handbook of Machiine Design ’’. 11.
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