Part 09 Brakes MANF361 Students

MANF 361 Brakes 26 March, 2012

Prof. Samy J. Ebeid ١

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BrakesMANF 361 – Part 09

Part 09 Brakes

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A brake is a device which stops

motion.

Its opposite component is a

clutch.

Most commonly brakes use

friction to convert kinetic energy

into heat.

Part 09 Brakes

Since kinetic energy increases quadratically with velocity (K = mv2/2),

an object traveling at 10 meters per second has 100 times as much

energy as one traveling at 1 meter per second, and consequently the

theoretical braking distance, when braking at the traction limit, is 100

times as long.


Prof. Samy J. Ebeid ٢

٣Part 09 Brakes

Friction brakes on automobiles store braking heat in the drum brake while

braking then dissipate it to the air gradually.

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1. Block Brake

Pressing force

Lever

Block

Rotating wheel

which is required to be stopped

Hinge fixed to frame

Part 09 Brakes


Prof. Samy J. Ebeid ٣

٥Part 09 Brakes

1. Block Brakes

٦

2. Band Brake

Pressing force

Lever

Band(sangle)

Rotating wheel


Hinge fixed to frame Part 09 Brakes


Prof. Samy J. Ebeid ٤

٧Part 09 Brakes

Differential Band Brake

Pt =tangential effort on Brake = T2-T1 T2>T1

T2/T1 = eµθ

µ = coefficient of friction

θ = angle of wrap

٨Part 09 Brakes


µµµµ=0.3

θθθθ=250̊ = 0.7x2π

Pt = T2-T1= tangential effort on brake

T2/T1 = eµθ

µ = coefficient of friction

θ = angle of wrap


Prof. Samy J. Ebeid ٥

٩Part 09 Brakes


Equilibrium of Lever

١٠Part 09 Brakes


In all cases b2 is

kept constant.

Case 1:b1/b2 > eµθ

= e0.3x2πx0.7 = 3

A positive brake and P acts downwards.


Prof. Samy J. Ebeid ٦

١١Part 09 Brakes


Self-locking brake used in hoists (electrically

driven).

Eg: when the current is put off.

The brake is always braked. To open the drive the lever is pushed upwards.

Case 2:b1/b2 = eµθ

P=0

Case 3:b1/b2 < eµθ

P= a –ve value

In all cases b2 is

kept constant.

١٢Part 09 Brakes


There is no braking effect.

There should be a certain ratio between b1 and b2i.e. a certain leverage.

b1/b2 (positive brake) = 2.5 to 3µµµµ=0.3

θθθθ=250 ̊ = 0.7x2π

Thickness of band = Dia. of brake drum/200 = 2 to 4 mm

B = width of band

t = thickness of band


Prof. Samy J. Ebeid ٧

١٣Part 09 Brakes


١٤Part 09 Brakes

Simple Band Brake


Prof. Samy J. Ebeid ٨

١٥

3. Shoe Brake

Pressing force

Tension spring for releasing the brake

Shoe

Fixed frame

Rotating wheel


Part 09 Brakes

١٦Part 09 Brakes

3. Shoe Brake


Prof. Samy J. Ebeid ٩

١٧

3. Shoe Brake

Piston

Oil under pressure

1.

Mechanical

force

generation

Cam

2.

Hydraulic

force

generation

Part 09 Brakes

١٨

4. Disc Brake (Pads)

Friction

liner

Seal

Fixed frame

Rotating disc which is

required to be stopped

Oil under pressure

Part 09 Brakes


Prof. Samy J. Ebeid ١٠

١٩Part 09 Brakes

4. Disc Brake (Pads)

٢٠Part 09 Brakes

Braking Energy

All the Energy of the system

should be absorbed by the brake.

(v1+v2)/2=avg velocity

v * t =distance

P.E.= mass*distance

1. Decrease in KE of weight

2. Decrease in PE of load


Prof. Samy J. Ebeid ١١

٢١Part 09 Brakes

Braking Energy

Er : in kgm

3. Decrease in KE of rotating mass

Brake work in time t (sec) done by tangential force Ft

٢٢Part 09 Brakes

Ex 09-1

Input

Output

1. For the shown cone brake, find an expression for the braking torque in

terms of the externally applied force W acting on the bell crank.


Prof. Samy J. Ebeid ١٢

٢٣Part 09 Brakes

Ex 09-1

Input

Output

2. Draw the pressure

distribution on the contact

surface assuming the uniform

wear case.

3. What is the frictional work

done if the rotating shaft comes

to rest from 300 rpm during 100

revolutions.

٢٤Part 09 Brakes

Ex 09-1


Prof. Samy J. Ebeid ١٣

٢٥Part 09 Brakes

Ex 09-1

UWD case

٢٦Part 09 Brakes

Ex 09-1


Prof. Samy J. Ebeid ١٤

٢٧Part 09 Brakes

Ex 09-1

٢٨Part 09 Brakes

Ex 09-1


Prof. Samy J. Ebeid ١٥

٢٩Part 09 Brakes

Ex 09-1

Documents

Part 09 Brakes MANF361 Students