Report Group l

8/18/2019 Report Group l

1/17

University of Puerto Rico

Mayaguez Campus

Department of Mechanical Engineering

Design of a Brake Disc

Tania M. Ortiz Menéndez

Liza M. Cardona Gonzalez

Ramón Torres

Objecties


2/17

Design of the rotor component for a disc brake system using load

analysis, stress analysis and fracture analysis system approach.

Descri!tion

A caliper disc brake is the most common type of disc brakes used in

modern cars. It is compound of a piston, a caliper, the brake pads, the rotor

and the hub. The single compound that we will be designing on is the rotor.

The rotor is the compound that receives the force applied by the brake pads

when the brake pedal is pressed and the piston is activated producing the

caliper to close.


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Design Details

We must rst understand what are its function and the parameters that

are important in its use. We need to know all of these things in order to make

a good design. What do braking systems really do The brakes of your car

convert the energy of motion into heat. In other words the brakes in your car

are responsible for converting kinetic energy into thermal energy. !ne

important thing to take into consideration for our design is that small

changes in the speed have a huge impact on the brake temperatures. This is

an inde" that will have to watch very carefully when we take into

consideration our design. There are many forces that can stop our car. An

e"ample of this can be wind or gravity. We need brakes to assist us in the

process of stopping the car. The #rake system is composed of many parts.

The most important are the #rake $edal, The %aster &ylinder, &alipers, The

$ads and the 'otor. We will brie(y analy)e the role of each of these parts and

their role in the process of stopping our car.

We will rst start o* with an analysis of the brake pedal. The purpose is

to harness and multiply the force e"erted by the driver+s foot. or the

analysis of the $edal we assumed an input driver force of -lb a pedal ratio

of /01. We then multiplied the force by the ratio. The resulting force gave us

a value 23 lbf. The brake pedal itself cannot take the car to a complete stop.

The rest of the components are very important. The only modication that


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we can make to the brake pedal is to change the pedal ratio. or our pro4ect

we assumed a pedal ratio of /01.

Another important component in the design of our brake disk is the

master cylinder. The master cylinder is responsible for converting the

amplied force from the brake pedal into hydraulic pressure. It consists of a

cylinder, a piston, break pedal output rod on one side and brake (uid on the

other side of the cylinder. As the pedal assembly output rod pushes on the

piston, the piston moves within the cylinder and pushes against the (uid,

creating hydraulic pressure. We calculate the pressure generated by the

master cylinder by dividing the force created by the pedal divided by its

area. or the master cylinder we used dimension of .5 inches of diameter.

The area of the pedal was2)350.0(∗π . And the force of the pedal as

mentioned before was 23 lbf. The calculation gave us a value of -26.//psi.

This component is also very important. #ut this alone does not stop the car.

There are some things that we can change in our master cylinder in order to

obtain the performance that we want. If we increase its diameter it will

decrease the amount of pressure generated. 7ven the smallest change in

diameter makes a big di*erence in the performance of the cylinder.

!ur third component is the calipers of the car. The caliper is very

similar to a piston with pressuri)ed (uid on one side. The caliper uses

hydraulic force on the input to create mechanical work. The caliper does a

s8uee)ing or clamping force of the brake disk. We calculated this clamping


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force by multiplying the pressure of the cylinder by the area of the cylinder.

This calculation gives us 6955.3 lb. The clamping force of the caliper is very

sensitive to changes in the diameter of the caliper.

The fourth component that we will analy)e is the brake pad. It is a big

misconception that changing brake pad material will magically decrease your

stopping distances. There is actually no relationship between each other. The

brake pads s8uee)e the rotor with the force that is generated by the calipers.

To analy)e the brake pads we needed a friction coe:cient. We took the value

of 45.0= µ .

And last but not least the 'otor. The rotor also assists in the process of

stopping the car, but it does not stop it. The rotor plays ; important roles. It

acts like a frictional interface for the brake pads. It reacts to the output by

absorbing the tor8ue created. or the analysis we assumed a value of ;3//.-


6/17

ftpolar inertial moment? gave us 3312./ 4in . In @I units

the calculation of jTr /=τ gave us a value of 2.56 psi.

or our design material we have two choices a ceramic material and

gray cast iron. We choose gray cast iron as the appropriate material for it

wear resistance and hardness. Also it absorbs and dissipates heat well to

cool the brakes. >'efer to Appendi" #?.


7/17

After the material is selected a fracture analysis can be done. or the

design the fracture analysis was performed for the static and dynamic

aspects. or the static aspect we assumed a value of ;.6 for the stress

concentration factor. The calculated value for the safety factor using the

Internal riction Theory for a brittle material and the ultimate tensile and

compressive strength for the material properties >'efer to Appendi" &? is

2.91. Also a value of ;.; was assumed for the stress concentration factor on

the dynamic aspect. We used the alternating forces e"erted in the disc. The

forces fluctuate from to /25;9-.9 lb. The >amplitude? of /25;9-.9 lb and

the >mean? of ;193//.- lb were corrected with the dynamic stress

concentration factor. With the corrected values the principal stresses were

calculated. The values for B1>amplitude? and B1>mean? were used to

calculate the safety factor with the %odied


8/17

fracture. That is because the force e"erted in the disc is a compressive force.

Thats why the materials used for the manufacturing of brake disc are brittle.

Also for that reason we calculate a big endurance limit.

#!!endi$ #

Calc"lations%

$edal

Input driver forceE1lb

'atio 301

lbf lb F 6006100 =∗=


9/17

%aster &ylinder

indp 7.0=

psi A F P p pc 1.1559)350.0(/600 2

/ === π

&alipers

/ pistons

ind c 4=

lb A P C cc f 3.783674))2((1.1559 2

=∗∗=∗= π

$ads

45.0= µ

10r F E lb f 3.3526545.03.78367)( =∗= µ

'otor

ind 4.160 =

ind i 4.8=

)2/(0 R R R ie +=

in Re 2.6)2/()12/2.412/2.8( =+=

lbinT −=∗+= 8.437289)2()2/()12/2.412/2.8(*)3.35265(

44444

0 1.6613)32/()4.8()4.16(()32/()( ind d J i =−=−= π π

J Tr /=τ

psi97.409)1.6613/()2.6)(8.437289( ==τ

%aterial0 Cray &ast Iron

Assumed0 5.2=t K


10/17

psi psicorrected 9.10245.297.409 =×=τ

&nternal 'riction T(eor) *&'T+

81.30

0326.0150000

9.1024

40000

9.10241

150000400001

9.1024

00195.0

9.1024

8.20491050420

0

8.204997.40922

1050420)9.1024(

0

31

3

2

1

3

32

2

1

3

222

3

2222

2

1

=→

=−

−=

==−=

−=→

−=→

=→

−−→

=−+−

=×=−−−+=

−=−=−−+++=

=++=

n

n

psiS psiS whereS S n

psi

I I I

I

I

I

ucut

ucut

xy z xz y yz x yz xz xy z y x

yz xz xy z x z y y x

z y x

σ σ

σ

σ

σ

σ σ

σ σ σ

τ σ τ σ τ σ τ τ τ σ σ σ

τ τ τ σ σ σ σ σ σ

σ σ σ

'ract"re #nal)sis D)namic


11/17

( )

814.09.99@

1@

59.0

06.60766.0

81.2

81.24.1601062.0010462.0

7296.0869.0

995.09.39

016.1409.39

696.516000814.0159.07296.0016.1

)(

1616000400004.04.:

20040

)(

222

95

097.0

995.0

'

'

=

=

=

==

=×==

==

−==

=×==

=×××××=

=

==×==


12/17

psi

psi J

Tr

lbT

psi

psi J

Tr

lbT

psi

psi

psi

I I I

I

I

I

corrected

e!n

!corrected

!plitude



z y x

98.4502.299.204

99.2041.6613

2.69.218644

9.2186442

8.437289

934.9012.297.409

97.4091.6613

2.68.437289

8.437289

93.901

002217.0

94.901

9.18039.813484

0

9.1803934.90122

9.813484)934.901(

0

3

2

1

3

32

2

1

3

222

3

2222

2

1

=×=

=×

==

==

=×=

=×

==

=

−=→

−=→

=→

−−→

=−+−

=×=−−−+=

−=−=−−+++=

=++=

τ

τ

τ

τ

σ

σ

σ

σ σ

σ σ σ



σ σ σ

896.5

40000

98.450

5696

94.901

1

5696400001

98.450

00443.0

98.450

96.90196.203382

0

96.90198.45022

96.203382)98.450(

0

11

3

2

1

3

32

2

1

3

222

3

2222

2

1

=→

+

=

==

+

=

−=→

−=→

=→

−−→

=−+−

=×=−−−+=

−=−=−−+++=

=++=

n

psiS psiS where

S S

n

psi

psi

psi

I I I

I

I

I

eut

ut

e

!

f



z y x

σ σ

σ

σ

σ

σ σ

σ σ σ



σ σ σ

,nd"rance Limit


13/17

( ) ( )

( ) ( )bb

bb

e

bb

ut

!& !& S

!!S

!!S

=→=

=→=

=→=

94.901

10569610

103600109.0

66

33

b

b!b!

b!b!

38.0

)1)(6(log)76.3)(1()10log(log)5696log(

3log56.4)10(log)36000log(

6

3

−=

−+=−→+=

+=→+=

3.227668

267.0

=

−=

!

b

cycles &

&

7

267.0

102.99

3.22766894.901

×=→

= −


14/17

#!!endi$ B

Craph for the material @election


15/17

Craph 'epresenting the Alternating orces

#!!endi$ C


16/17

Subcategory: Ferrous Metal; Gray Cast Iron; Metal

Key Words: Grey Cast Iron, ASTM A 48 Class 40, cast irons

Component Wt. %

C 3.25 - 3.5

Cr 0.05 - 0.45

Cu 0.5 - 0.4

Component Wt. %

Mn 0.5 - 0.!

Mo 0.05 - 0.

"i 0.05 - 0.2

Component Wt. %

# Ma$ 0.2

S Ma$ 0.5

Si .8 - 2.3

Material Notes:Car%on liste& in t'e co()osition a%o*e is t'e total car%on. Can %e oil +uenc' 'ar&ene& ro( 80C toattain a /oc1ell C 50 (ini(u( surace 'ar&ness. ata )ro*i&e& %y t'e (anuacturer, Siltin In&ustries,Inc

Physical Properties Metric English Comments

ensity .5 cc 0.258 l%in6 Ty)ical or Gray Cast Iron

Mechanical Properties

7ar&ness, rinell 83 - 234 83 - 234

7ar&ness, 9noo) 258 258 Con*erte& ro( rinell 'ar&ness.

7ar&ness, /oc1ell ! ! Con*erte& ro( rinell 'ar&ness.

7ar&ness, /oc1ell C 20 20 Con*erte& ro( rinell 'ar&ness.

7ar&ness, :icers 24 24 Con*erte& ro( rinell 'ar&ness.

Tensile Strent', lti(ate Min 2 M#a Min 40000 )si

lti(ate Co()ressi*e Strent' Min 034 M#a Min 50000 )si

Mac'ina%ility 0 < 0 < :ery oo& (ac'ina%ility. "o nu(ericalratin a*aila%le.


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