BIKE ALL TYPE BRAKE INFO :-

1. • NAME :- Wadiya Viral S.
•ENROLLMENT NO :- 146018319012
•SEMINAR TOPIC :- BRAKES
•GUIDANCE BY :- MR. D.I.SURTI

2. A Y DADABHAI TECHNICAL INSTITUTE
(KOSAMBA)
On Topic : BRAKES
SUB : Design Of Machine Elements PREPARED BY
GUIDE : MR. D.I.SURTI
WADIYA VIRAL S
146018319012

3. This is to certify that Wadiya Viral S. Of mechanical
engineering department Enrollment no. 1460183190012 studies
in 5th semester from A.Y. Dadabhai technical institute,
kosamba has finished the seminar on topic “BRAKES
satisfactorily in subject Design Of Machine Elements.
Signature of Guide: Signature of Principal: Signature of HOD:

4. INTRODUCTION
 A Brake is a device used absorb the energy possessed
by a moving system by means of friction.
 The main pressure of brake is to slow down or
completely stop the motion of a moving system such
as rotating drum, machine or vehicle.
 It is also used to hold the parts of the system in the
position of the rest.
 The energy absorbed by the brake can be either
kinetic or kinetic and potential both.
 In automobile application, the kinetic energy of
moving vehicle is absorbed by the brake.
 The energy absorbed by the brake is converted in to
heat energy and dissipated to surroundings or
absorbed by the additional cooling system provided for
that purpose.

5.  Classification of Brakes :-

6.  Band Brake :-
 A band brake consists of flexible steel strip lined with
friction material which is wrapped and pressed against the
rotating drum.
 The rotating shaft for which the motion is to be controlled,
a brake drum is mounted on the shaft.
 A steel bend is wrapped on the drum covering the required
portion of the circumference of the drum.
 The ends of the steel bend are connected to the lever and
fulcrum in appropriate manner.
 When the band is pressed against the external surface of
drum, the frictional force between drum and band will
induce braking torque on the drum.

7.  Types of Band Brake :-
 There are two types of band brakes.
a) Simple Band Brake
b) Differential Band Brake
a) Simple Band Brake :-
 In this types of brake, one end of the band is attached
to the fulcrum of the lever while the other end of the
band is attached to the lever at a distance from the
fulcrum.
 When the drum rotates in the clockwise direction, the
end of the band connected to the fulcrum O becomes
the tight side and the other side of the band
connected to the lever A becomes slack side.
 When the effort ‘P’ is applied at the end of the lever
at the point C in the downward direction, the band
becomes tight and the brake is applied.

8.  Construction of Simple Band Brake & Differential
Band Brake :-

9. b) Differential Band Brake :-
 In this type of band brake, one end of the band is
band is attached to the lever on the side of the
fulcrum and the other end of the band is attached to
the lever on the other side of the fulcrum.
 In this type of brake when the drum rotates in the
clockwise direction, then the band end on the left
hand side of the fulcrum, O becomes tight side and
the band and on the right hand side of the fulcrum O
becomes slack side.
 When the braking force P is applied at the end lever
at point C
in the downward direction, the band becomes tight
and brake is
applied.

10. 1) Band Brake :-
2) Shoe Brake :-
3) Disc Brake :-

11.  The following characteristics should have
materials used for brake lining :-
 It should have low wear rate.
 It should have high heat resistance.
 It should have high heat dissipation capacity.
 It should have low coefficient of thermal
expansion.
 It should have adequate mechanical strength.
 It should not be affected by moisture.

12. Friction force and braking torque in the
Band brakes :-
 Let,
T₁ = Tension in tight side of the band – N
T₂ = Tension in slack side of the band – N
θ = Angle of lap or wrap of the band on the drum, radius
R = radius of the drum – m
a = Perpendicular distance between line of action of T₁
and
fulcrum O
b = Perpendicular distance between the line of action of
T₂ and
fulcrum O
L = Perpendicular distance between line of action of
effort line
(P) and fulcrum O
Mt = Braking Torque – N.m

13. a) Simple Band Brake :-
 As shown in figure, the simple band
brake behaves like a belt drives.
 The ratio of tensions in the band,
T₁/T₂ = ₑμθ
 Breaking force on drum/friction
force on the drum = ( T₁ - T₂ )
 μ = Coefficient of friction between
drum and band.
 Breaking Torque on the drum, Mt = ( T₁ - T₂ )*R
T₁/T₂ = ₑμθ
 Friction force, F = ( T₁ - T₂ )

14. 1) Assuming clockwise rotation of drum :-
 Considering equilibrium of the lever, Taking moment
about fulcrum.
P * L = T₂ * b ( OR ) P = T₂ * b / L
2) Assuming counter clockwise rotation of the
drum :-
 The arrangement is shown in figure.
 Considering equilibrium of the lever,
taking moment about fulcrum,
P * L = T₁ * a
P = T₁ * a / L

15. b) Differential Band Brake :-
 As shown in figure, the differential
brake also behaves like a flat belt
drive – Here also, T₁/T₂ = ₑμθ
 Friction force, F = ( T₁ - T₂ )
 Friction torque, Mt = ( T₁ - T₂ ) * R
1) Assuming clockwise rotation
of the drum :-
 Considering equilibrium of the lever, taking
moment about fulcrum.
 P * L + T₁ * a = T₂ * b
 P = T₂ * b – T₁ * a / L
 Also, P = T₂ ( b - ₑμθ * a )

16.  EXAMPLE :-
1) In a band brake, used for a winch, the drum for angle of
wrap of 240˚. The drum diameter is 0.75 m. One end of
the band is attached to the lever at the left hand side of
the fulcrum at a distance of 30 mm and the other end of
the band is attached to the lever on the right side shaft
of the fulcrum at a distance of 120 mm. coefficient of
friction is 0.3. If the effort 600 N its applied in the
upward direction on the lever at a distance of 1.2 m
from the fulcrum, find the braking torque available.
 Solution :-
Here, the construction of the differential brake is as
shown in the figure. The drum is assumed to rotate in
anticlockwise direction.
 Data Given :-
a = 30 mm, b = 120 mm, μ = 0.3, θ = 240˚, D = 0.75 m =
750 mm,
P = 600 N, R = 375 N, L = 1.2 m = 1200 mm

17.  T₁ / T₂ = ₑμθ
= ₑ0.3 * ( 240 * π / 180 )
 T₁ / T₂ = 3.51
 Taking moment about fulcrum.
 P * L + T₁ * a = T₂ * b
 600 * 1200 + 3.51 T₂ * 30 = T₂ * 120
 720000 + 105.4 T₂ = 120 T₂
 T₂ = 720000 / 14.59 = 49315 N
 T₁ = 3.51 * 49315 = 173095.65 = 173096 N
 Breaking torque, Mt = ( T₁ - T₂ ) * R
 Mt = ( 173096 – 49315 ) * 375
= 46417875 N.mm
 Mt = 46417.875 N.mm - ( Answer )

18.  References :-
 www. google.co.in
 Theory of machine ( Atul prakashan )
 Design of machine ( R. S. Khurmi )