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MIME 4222 Engineering Design II
Page 2
Index
List of Topics
Chapter 1: Brakes
Sl.No.
Topics
Page No.
1
1.1
Introduction on brakes
6
2
1.2
Types of brakes
6
3
1.3
Materials for brake lining
7
4
1.4
Heat to be dissipated during braking
7
5
1.5
Problems on energy conversion
9
6
1.6
Single block brake
11
7
1.7
Pivoted block brake
14
8
1.8
Double block brake
19
9
1.9
Simple band brake
20
Chapter 2: Friction Clutches
Sl.No.
Topics
Page No.
1
2.1
Introduction on friction clutches
27
2
2.2
Friction clutches
28
3
2.3
Material for friction surfaces
29
4
2.4
Considerations in designing a friction clutch
29
5
2.5
Types of friction clutches
29
6
2.6
Design of a plate clutch
31
7
2.7
Multi plate clutches
34
8
2.8
Cone clutches
38
Chapter 3: Wire Ropes
Sl.No.
Topics
Page No.
1
3.1
Introduction on wire ropes
47
2
3.2
Advantages of wire ropes
47
3
3.3
Construction of wire ropes
48
4
3.4
Designation of wire ropes
48
5
3.5
Stresses in wire ropes
48
MIME 4222 Engineering Design II
Page 3
6
3.6
Procedure for designing a wire rope
49
7
3.7
Problems on wire ropes
50
8
3.8
Failure of ropes
54
Chapter 4: Belt and Rope Drives
Sl.No.
Topics
Page No.
1
4.1
Introduction belt drives
62
2
4.3
Velocity ratio of belt drive
62
3
4.4
Slip of belt
63
4
4.5
Creep of belt drive
63
5
4.6
Ratio of driving tensions for belt drives
63
6
4.7
Condition for the transmission of maximum power
65
7
4.8.
VBelt Drives
67
Chapter 5: Ergonomics in Engineering Design
Sl.No.
Topics
Page No.
1
5.1
Introduction on Ergonomics
72
2
5.2
Importance of Ergonomic Principles
72
3
5.3
Important Ergonomic Principles
73
Chapter 6: Product Design
Sl.No.
Topics
Page No.
1
6.1
Introduction to Product Design
77
2
6.2
Strategies in Product Design
78
Chapter 7: Parametric Design
Sl.No.
Topics
Page No.
1
7.1
Definition of Parametric Design
79
2
7.2
Details of Parametric Design
79
MIME 4222 Engineering Design II
Page 4
Chapter I
Mechanical Brakes
Course Outcomes Covered
?
Design wire ropes, sheaves and drum, brakes and clutches.
?
Select and classify appropriate motor components.
?
Perform motor power calculations.
?
Realize the importance of synthesis of components.
?
Apply a complete system design project using calculation, specification and
drawings.
MIME 4222 Engineering Design II
Page 5
Chapter 1
Brakes
1.1
Introduction
A brake is a device used to bring a moving system to rest, to slow its speed, or to
control its speed to a certain value under varying conditions. The function of a brake is to
turn mechanical energy into heat. This heat is dissipated in the surrounding air or water
which is circulated through the passages of the brake drum so that excessive heating if the
brake linings does not take place. The design or capacity of a brake depends upon the
following factors:
?
The unit pressure between the braking surfaces.
?
The coefficient of friction between the braking surfaces.
?
The peripheral velocity of the brake drum.
?
The projected area of the friction surfaces
?
The ability of the brake to dissipate heat equivalent to the energy being absorbed.
The major difference between a brake and a clutch is that a clutch is used to keep
the driving and driven member moving together, whereas the brakes are used to stop a
moving member or to control its speed.
Fig.1.1 A disc brake in action (Source: Wikipedia)
1.2 Types of Brakes
1.
Hydraulic brakes
2.
Electric brakes
3.
Mechanical brakes
MIME 4222 Engineering Design II
Page 6
The hydraulic and electric brakes cannot bring the member to rest and are mostly
used where large amounts of energy are to be transformed while the brake is retarding the
load such as in laboratory dynamometers, highway trucks and electric locomotives. These
brakes are also used for retarding or controlling the speed of a vehicle for downhill travel.
The mechanical brakes according to the direction of acting force, may be divided into the
following two groups:
(a) Radial brakes: In these brakes, the force acting on the brake drum is in radial
direction. The redial brakes may be subdivided into external brakes and internal
brakes.
(b) Axial brakes: In these brakes, the force acting on the brake drum is in axial direction.
The axial brakes may be disc brakes and cone brakes.
1.3
Materials for Brake Lining
The materials used for the brake lining should have the following characteristics:
?
It should have high coefficient of friction with minimum fading which means that the
coefficient of friction should remain constant over the entire surface with change in
temperature.
?
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 and oil.
1.4
Heat to be dissipated during braking
The energy absorbed by the brake and transformed into heat must be dissipated to
the surrounding air in order to avoid excessive temperature rise of the brake lining.
The temperature rise depends upon the mass of the brake drum, the braking time
and the heat dissipation capacity of the brake. The highest permissible temperatures
recommended for different brake lining materials are:
?
For leather, fibre and wood facing = 65 70?C
?
For asbestos and metal surfaces what are slightly lubricated = 90 105?C
?
For automobile brakes with asbestos block lining = 180 225?C
MIME 4222 Engineering Design II
Page 7
Since the energy absorbed (heat generated) and the rate of wear of the brake lining
at a particular speed are dependent on the normal pressure between the braking surfaces,
therefore it is an important factor in the design of brakes.
The permissible normal pressure between the braking surfaces depends upon the
material of the brake lining, the coefficient of friction and the maximum rate at which the
energy is to be absorbed. The energy absorbed or heat generated is given by the equation:
E = Hg = ?RNv = ?pAv (J/s or Watts)
(Equation. 1.1)
Where ? is the coefficient of friction,
p normal pressure between the braking surfaces (N/m2)
A Projected area of the brake drum in m2
V Peripheral velocity of the brake drum in m/sec.
The heat generated may also be obtained considering the amount of kinetic energy or
potential energies which is being absorbed.
In other words, Hg = EK ? EP
where EK is the total kinetic energy absorbed and EP is the total potential energy
absorbed.
Heat dissipated (Hd) = C (t1t2) Ar
(Equation. 1.2)
where C is the heat dissipation factor or coefficient of heat transfer in W/m 2/?C.(t1t2) is the
temperature difference between the exposed radiating surface and the surrounding air in ?C
and Ar is the area of radiating surface in m2.
The expression for heat dissipated is quite approximate and should serve only as an
indication of the capacity of the brake to dissipate heat. The exact performance of the brake
should be determined by the test. It has been found that 10 to 25% of the heat generated is
immediately dissipated to the surrounding air while the remaining heat is absorbed by the
brake drum causing its temperature to rise.
The rise in temperature of the brake drum
?t ?
Hg
mC
(Equation. 1.3)
where ?t is the temperature of the brake drum in ?C
Hg is the heat generated by the brake in J
m is the mass of the brake drum in kg
C is the specific heat for the material of the brake drum in J/kg?C
MIME 4222 Engineering Design II
Page 8
1.5
Problems
1.5.1 A vehicle of mass 1200 kg is moving down the hill at a slope of 1:5 at 72 kmph. It is
to be stopped in a distance of 50 m. If the diameter of the tyre is 600 mm, determine the
average braking torque to be applied to stop the vehicle, neglecting all the frictional energy
except for the brake. If the friction energy is momentarily stored in a 20 kg cast iron brake
drum, what is the average temperature rise of the drum? The specific heat for C.I may be
taken as 520 J/kg ?C. Determine, also the minimum coefficient of friction between the tyres
and the road in order that the wheels do not skid, assuming that the weight is equally
distributed among all the four wheels.
Data: m = 1200 kg
H = 50 m
Slope = 1:5
v = 72 kmph = 20 m/sec
d = 600 mm
mb = 20 kg
C= 520 J/kg ?C
Average braking torque to be applied to stop the vehicle
K.E of the vehicle EK =
1 2 1
mv = (1200) (20)2 = 240, 000 Nm
2
2
P.E of the vehicle EP = mgh x slope = 1200 x 9.81x 50x 1/5 = 117,720 Nm
Total energy og the vehicle or the energy to be absorbed by the brake E = EK+ EP
E = 357,720 Nm
Since the vehicle is to be stopped in a distance of 50 m, tangential braking force required
Ft = 357720/50 = 7154 N
Average braking torque to be applied to stop the vehicle TB = Ft r = 7154 x 0.3
= 2146.3 Nm
TB = 2146.3 Nm
Average temperature rise of the drum
Let ?t be average temp. rise of the drum in ?C
The heat absorbed by the brake drum
Hg = Energy absorbed by the brake drum
= 357,720 Nm = 357,720 J
Heat absorbed by the brake drum is also given by Hg =357,720 = mb x Cx ?t
357, 720 = 20 x 520 x ?t
Therefore
?t = 34.4 ?C
Minimum coefficient of friction between tyre and road
Let ? be the minimum coefficient of friction between tyre and road RN be the normal force
between the contact surfaces. This is equal to the weight of the vehicle.
MIME 4222 Engineering Design II
Page 9
RN = m.g = 1200 x 9.81 = 11772 N
Tangential braking force (Ft) = 7154 = ? RN = ? x 11772
? = 0.6
Therefore
1.5.2 A four wheeled automobile has a total mass of 1000 kg. The moment of inertia of
each wheel about a transverse axis through its centre of gravity is 0.5 kgm2. The rolling
radius of the wheel is 0.35 m. The rotating and reciprocating parts of the engine and the
transmission system are equivalent to a moment of inertia of 2.5 kgm2 which rotates at five
times the roadwheel speed. The car is traveling at a speed of 100 kmph on a plane road.
When the brakes are applied, the car decelerates at 0.5 g. There are brakes on all four
wheels. Calculate
(i)
The energy absorbed by each wheel
(ii)
The torque capacity of each brake.
(i)
Kinetic energy of the car
v1= 100 kmph = 27.78 m/sec
v2 = 0
(ii)
K.E =
1
m ??v12 ? v2 2 ?? = 385,802.4 J
2
Kinetic energy of the wheels
? 27.78
?1 ? 1 ?
? 79.37rad / sec
R
?2 =0
0.35
?1
?
Kinetic energy of four wheels = 4 ? I ??12 ? ?2 2 ? ? = 6298.8 J
?2
?
(iii)
Kinetic energy of the engine and transmission system
?1 = 5 (79.37) = 396.83 rad/sec
?1
?
K.E = ? I ??12 ? ?2 2 ? ? = ½ x 2.5 x (396.83)2 = 196837.97 J
?2
?
The energy absorbed by the four brakes consists of K.E of the car, the K.E of the
wheel and the K.E of the engine and the transmission system.
K.E = ¼ (385802.4+629808+196837.97) = 147234.8 J
Braking time t,
?1 ?? 2
t
? 0.5g
t = 5.66 secs
The average velocity during the braking time is
MIME 4222 Engineering Design II
?1 ? ?2
2
or
?1
2
Page 10
?? ?
? ? ? 1 ? t = (79.37/2) 5.66 = 224.6 rad
? 2?
Therefore
Torque, T = E/? = 665.54 Nm
1.5.3 A mass of 2500 kg is lowered at a velocity of 1.5 m/sec as shown in figure below. The
mass of drum is 50 kg and its radius of gyration can be taken as 0.7 m. On applying the
brake, the mass is brought to rest in a distance of 0.5 m. Calculate:
(i)
The energy absorbed by the brake
(ii)
The torque capacity of the brake
Fig.1.2 Load handled by a brake (Source: Design of Machine Elements by Bhandari)
1 2
1
2
mv ? ? 2500 ? ?1.5 ? ? 2812.5 J
2
2
v 1.5
Angular velocity of drum ? = ?
? 2 rad
sec
r 0.75
1
1
2
2
K.E of drum = mk 2? 2 ? ? 50 ? ? 0.7 ? ? ? 2 ? = 49 J
2
2
K.E of mass =
Potential energy of mass = mgh = 2500 x 9.81 x 0.5 = 12262.5 J
Total energy = 2812.5 + 49 + 12262.5 = 15124 J
During braking action, the mass moves through a distance of 0.5 m. If ? is the angle
through which the drum rotates the braking period,
??
? x drum radius = 0.5
Therefore, Torque T =
MIME 4222 Engineering Design II
E
?
=
0.5
? 0.667rad
0.75
15124
? 22686 Nm
0.667
Page 11
1.5.4 Figure shows a 1000 kg mass being lowered by a cable at a uniform rate of 4 m/s
from a drum of 550 mm diameter weighing 2.5 kN and having a 250 mm radius of
gyration. What is the kinetic energy in the system?
Fig.1.3 Load handled by a cable (Courtesy: Fundamentals of Machine Component Design
by Robert Juvinal)
1
2
? 1000 ? ? 4 ? = 8000 Nm
2
?
4
Angular velocity of drum ? ? ?
? 14.54 rad/sec
r
0.275
2.5 ? 103
mass of drum m =
? 255 kg
9.81
1
1
2
2
K.E of drum = ? mk 2? 2 ? ? 255 ? ? 0.25 ? ? ?14.54 ? ? 1686 Nm
2
2
K.E of mass =
Therefore, Total K.E = 8000 + 1686 = 9686 Nm
1.6
Single block or shoe brake
A single block or shoe brake consists of a block or shoe which is pressed
against the rim of a revolving brake wheel drum. The block is made of a softer material than
the rim of a wheel.
This type of a brake is commonly used on railway trains and tram cars. The
friction between the block and the wheel causes a tangential braking force to act on the
wheel, which retard the motion of the wheel.
The block is pressed against the wheel by a force applied to one end of a lever
to which the block is rigidly fixed as shown in figure 1.2. The other end of the lever is pivoted
on a fixed fulcrum O.
MIME 4222 Engineering Design II
Page 12
Fig.1.2 Clockwise rotation of a brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
Let
P = Force applied at the end of the lever
RN = Normal force pressing the brake block on the wheel
r
= Radius of the wheel
2? = Angle of contact surface of the block
? = Coefficient of friction
Ft = Tangential braking force or the frictional force acting at the contact
surface of the block and the wheel
If the angle of contact is less than 60?, then it may be assumed that the normal
pressure between the block and the wheel is uniform. In such cases, tangential braking force
on the wheel
Ft = ? RN and the braking torque TB = FT r = ? RN r
Case1: When the line of action of tangential braking force (Ft) passes through the fulcrum
O of the lever and the brake wheel rotates clockwise as shown in figure 1.2, then for
equilibrium, taking moments about the fulcrum O,
RN x X = P x l
or
RN =
P?l
X
Therefore braking torque TB = ? RN r= ? x
P?l
xr
X
(Equation.1.4)
Please note that when the brake wheel rotates anticlockwise as shown in figure 1.3, then
the braking torque is same, i.e
TB = ? RN r = ? x
MIME 4222 Engineering Design II
P?l
xr
X
(Equation.1.5)
Page 13
Fig.1.3 Anticlockwise rotation of a brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
Case 2: When the line of action of the tangential braking force (Ft) passes through a
distance a below the fulcrum O, and the brake wheel rotates clockwise as shown in figure
3, then for equilibrium, taking moments about the fulcrum O,
RN x = Ft a + P.l
or
RN x = ? RN a + P.l
(Equation.1.6)
Braking torque,
R N ? x?a ? =P.l
P.l
RN =
(x?a)
TB = ?R N r =
?Plr
(x?a)
Fig.1.4 Clockwise rotation of brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
MIME 4222 Engineering Design II
Page 14
Fig.1.5 Anticlockwise rotation of brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
When the brake wheel rotates anticlockwise, then for equilibrium,
RN x + Ft a = Pl
and braking torque,
T B = ? RN r =
? P.l.r
x ? ?a
(Equation.1.7)
In equation 1.6, the moment of frictional force (? RN.a) adds to the moment of force (P.l). In
other words, the frictional force helps to apply the brake. Such types of brakes are said to
be selfenergizing brakes. When the frictional force is great enough to apply the brake with
no external force, then the brake is said to be selflocking brake.
Note:
?
The condition for the brake to be selflocking is x ? ?a .
?
The brake should be selfenergizing and not be selflocking.
?
In order to avoid selflocking and to prevent the brake from grabbing, x should be
greater than ?a.
?
When a single block or shoe brake is applied to a rolling wheel, an additional load
is thrown on the shaft bearings due to heavy normal force (R N) and produces
bending of the shaft. In order to overcome the drawback, a double block or shoe
brake is used.
1.7
Pivoted Block or Shoe Brake
When the angle of contact is less than 60?, then it may be assumed that the normal
pressure between the block and the wheel is uniform. But when the angle of contact is
greater than 60?, then the unit pressure normal to the surface of contact is less at the ends
than at the centre. In such cases, the block or shoe is pivoted to the lever as shown in figure
instead of being rigidly attached to the lever. This gives uniform wear of the brake lining in
MIME 4222 Engineering Design II
Page 15
the direction of the applied force. The braking torque for a pivoted block or shoe brake (when
2?>60?) is given by
TB ? Ft ? r ? ? ‘ RN .r
where ? ‘ = Equivalent coefficient of friction =
4? sin ?
and ? is the actual coefficient of
2? ? sin 2?
friction.
Fig.1.6 Pivoted Brake Shoe
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
Problem 1: A single block brake is shown in figure below. The diameter of the drum is 250
mm and the angle of contact is 90?. If the operating force of 700 N is applied at the end of a
lever and the coefficient of friction is 0.35, determine the torque that may be transmitted by
the block brake.
Data: d = 250 mm
2? = 90?
P = 700 N
? = 0.35
Since the angle of contact is greater than 60?, equivalent coefficient of friction,
?’ ?
4? sin ?
= 0.385
2? ? sin 2?
Fig.1.7 Pivoted Brake Shoe
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
MIME 4222 Engineering Design II
Page 16
Let RN be the normal force pressing the block and the brake drum and
Ft be the tangential braking force = ? RN
Taking moments about the fulcrum O,
700 (250+200) + Ft x 50 = RN x 50 = RN x 200 =
520 Ft – 50 Ft = 700 x 450
(or)
Ft
?
‘
x 200 =
Ft
? 200 ? 520Ft
0.385
Ft = 670 N
Torque transmitted by block brake TB,
TB ? Ft ? r ? 670 ?125 ? 83,750 N ? mm ? 83.75N ? m
Problem 2: Figure shows a brake shoe
attached to a drum by a lever AB which is
pivoted at a fixed point A’ and rigidly fixed
to the shoe. The radius of the drum is 160
mm. The coefficient of friction of the brake
lining is 0.3. If the drum rotates clockwise,
find the braking torque due to horizontal
force of 600 N applied at B.
Fig.1.8 Pivoted Brake Shoe
(Source: Machine Design by R.S.Khurmi
and J.K.Gupta)
Taking moments about A, and equating clockwise and anticlockwise moments,
RN ? 350 ? Ft ? 200 ? 160 ? ? 600 ? 400 ? 350 ?
Ft
? ??? ? ?? Ft ? 600 ? ???
0.3
Ft ? 372.8 N
Therefore,
Braking Torque TB = Ft x r = 372.8 x 0.16 = 59.65 Nm
Problem 3: Figure shows a single block brake. The brake drum diameter is 400 mm and
rotates at a speed of 150 rpm. The friction material permits a maximum pressure of 0.5 MPa
and µ=0.25. Face width of the block is 50 mm. If the brake is applied for 10 seconds at full
capacity to bring the shaft to stop, determine (i) Effort (ii) Maximum Torque (iii) Heat
generated.
MIME 4222 Engineering Design II
Page 17
Since 2? ? 90º,
?’ ?
4?sin?
4 ? 0.25 ? sin 45?
?
? 0.275
2?+sin2? 90?? ? ? sin 90?
180
Taking moments about fulcrum point O, F x 500 = RN X 250
F?
250
R N = 0.5 RN
500
Ft
RN
(Dimensions are in mm)
Fig.1.9 Single Brake Shoe
(Source: Machine Design by J.B.K.Doss and J.L.Srinivasa Murthy)
(i)
Effort
pressure p b ?
RN
Ab
RN
RN
?
b ? 2 ? r ? sin ? 50 x 2 x 200 x sin 45?
R N ? 7071N
0.5 ?
F?
250
R N = 0.5 RN = 3535.5 N
500
(ii) Maximum Torque
Ft = ? ‘ RN = 0.275 x 7071 = 1944.5 N
TB = Ft x r = 1944.5 x 200 = 388,900 Nmm
TB = 388.9 Nm
(iii) Heat Generated
v=
? d N ? x 0.4 x 150
=
= 3.14 m/sec
60
60
Hg = µpAv = µ RN v = 0.275 x 7071 x 3.14 = 6105.8 J/sec
Therefore, Heat generated in 10 seconds = 6105.8 x 10 = 61058 J = 61.058 kJ
MIME 4222 Engineering Design II
Page 18
Problem 4: A single block brake with a torque capacity of 15 Nm is shown in figure below
and the maximum pressure on the brake linings is 1 N/mm2. The width of the block is equal
to its length. Calculate:
(i)
Actuating force
(ii)
Dimensions of the block
(iii)
Resultant hinge pin reaction
(iv)
Rate of heat generated if the brake drum rotates at 50 rpm.
F2
Ft
F1
RN
(Dimensions are in mm)
Fig.1.10 Single Brake Shoe
(Source: Machine Design by J.B.K.Doss and J.L.Srinivasa Murthy)
Taking moments about point O and equating clockwise and anticlockwise reactions,
P x 650 + Ft x 60 = RN x 200
On solving, RN = 3.57 P
15 x 103 = 0.3 x RN x 150
Torque T = Ft x r
Therefore, RN = 333.33 N
(i) Actuating force:
RN = 3.57 P and hence,
P = 93.33 N
(ii) Dimensions of the block:
Let l and b represent the length and width of the brake shoe.
braking pressure pb ?
1 ?
RN
Ab
333.33
l2
MIME 4222 Engineering Design II
Ab = l ? b ? l 2
(since l = b)
l = b = 18.26 mm
Page 19
(iii)
Resultant hingepin reaction:
Considering vertical forces at hinge O, ? Fv ? 0
? RN ? F2 ? P ? 0
F2 ? RN ? P ? 333.33 ? 93.33 ? 240 N
Considering horizontal forces at hinge O, ? Fh ? 0
Ft = F1 = 99.99 N = 100 N
Therefore resultant hingepin reaction, R =
F12 ? F2 2 ? (100)2 ? (240)2 ? 260 N
R = 260 N
(iv)
Heat Generated:
? (? x 0.3 x 50) ?
H g = ?R N v = 0.3 x 333.33 x ?
?? = 78.54 J/sec
60
?
Therefore,
1.8
Hg = 78.54 J/sec
Double block or shoe brake
When a single block brake is applied to a rolling wheel an additional load is thrown
on the shaft bearings due to normal force (RN). This produces bending of the shaft. In order
to overcome this drawback, a double block or shoe brake is used. It consists of two brake
blocks applied at opposite ends of a diameter of the wheel which eliminates or reduces the
unbalanced force on the shaft.
This type of brake is used in electric cranes and the braking action is doubled by the
use of two blocks and two blocks may be operated practically by the same force which
operates one. In case of double block or shoe brake, the braking torque is T B = (Ft1+Ft2) r
where Ft1 and Ft2 are the braking forces on the two blocks.
Fig.1.11 Double block or shoe brake
(Source: Machine Design by R.S.Khurmi & J.K.Gupta)
MIME 4222 Engineering Design II
Page 20
Problem 5: A double shoe brake as shown in figure below is capable of absorbing a torque
of 1400 Nm. The diameter of the brake drum is 350 mm and the angle of contact for each
shoe is 100?. If the coefficient of friction between the brake and the lining is 0.4, find:
1.
The spring force necessary to set the brake
2.
Width of the brake shoes if the bearing pressure on the lining material is
not to exceed 0.3 N/mm2.
Dimensions are in mm
Fig.1.12 Double Shoe Brake (Source: Machine Design by R.S.Khurmi & J.K.Gupta)
Data: TB = 1400 Nm
d = 350 mm
r = 175 mm
2? = 100? = 100 x ?/180 =1.75 rad
1.
? = 0.4
pb = 0.3 N/mm2
Spring force necessary to set the brake
Let S = spring force necessary to set the brake
RN1 and F1 be the normal reaction and the braking force on the right hand side sjoe
RN2 and F2 be the corresponding values on the left hand side
Since the contact angle is greater than 60?, the equivalent coefficient of friction,
?’ ?
4? sin ?
= 0.45
2? ? sin 2?
Taking moments about O2, and equating clockwise and anticlockwise moments,
S ? 450 ? RN 1 ? ??? ? Ft1 ?175 ? 40 ?
?
Ft1
? ????? Ft1 ???? ? ????? Ft1
0.45
Ft1=0.776 S
S ? 450 ? Ft 2 ?175 ? 40 ? ? RN 2 ? ??? ??
Ft2
? ??? ? ??????Ft2
0.45
Ft2 = 1.45 S
MIME 4222 Engineering Design II
(Equation 1)
(Equation 2)
Page 21
From equations (1) and (2), Torque capacity of the brake (TB),
1400 x 103 = (Ft1+Ft2).r = (0.776 S+1.454 S) 175 = 390.25 S
S = 3587 N
Spring Force
2. Width of the brake shoes
Let b = width of the brake shoes in mm
The projected bearing area for one shoe,
Ab = b (2rsin?) = b (2x175x sin 50?) = (268 b) mm2
Normal force on the right hand side of the shoe, RN 1 ?
Normal force on the left hand side of the shoe, RN 2 ?
Ft1
?
Ft 2
?
‘
‘
?
?
0.776 ? S
? 6186 N
0.45
1.454 ? S
? 11,590 N
0.45
The maximum normal force is on the left hand side of the shoe. Therefore, designing the
shoe for maximum normal force , i.e RN2,
Pb = 0.3 = RN2/Ab = 11590/268 b
therefore, b = 144 mm
Width of shoe
Problem 6: A double shoe brake is shown in figure below. The drum rotates at 200 rpm
when the applied force is 1000 N and coefficient of friction is 0.25. Determine (i) braking
torque and (ii) amount of heat generated due to braking action.
Dimensions are in mm
Fig.1.15 Double Shoe Brake (Source: Machine Design by J.B.K.Doss and J.L.Srinivasa
Murthy)
Taking moments about point A and equating clockwise and anticlockwise reactions,
1000 x 100 = P x 50
P = 2000 N
Solving for the right hand side, taking moments about fulcrum O 1 and equating clockwise
and anticlockwise moments,
MIME 4222 Engineering Design II
Page 22
P x 450 = Ft1 x 50 + RN1 x 200
2000 x 450 = 0.25 x 50 x RN1 + 200 RN1 = 212.5 RN1
Therefore
RN1 = 4235 N and Ft1 = 1058.8 N
Solving for the left hand side, taking moments about fulcrum O 2 and equating clockwise
and anticlockwise moments,
P x 450 = RN2 x 200
and
(i)
RN2 = 4500 N
Ft2 = 1125 N
Braking Torque:
TB = (Ft1 + Ft2) r = (1058.8 + 1125) x 0.5 = 546 Nm
(ii)
Amount of heat generated: (Hg)
??
? dN
60
?
? ? 0.5 ? 200
60
? 5.2 m/sec
Hg = (µ RN1 + µRN2) x ? = (1058.8+ 1125) x 5.2 = 11355.76 J = 11.35 kJ
Problem 7: A twoshoe external drum brake shown in figure below has shoes 80 mm wide
that contact 90° of drum surface. For a coefficient of friction of 0.20 and an allowable contact
pressure of 400 kN per square meter of projected area, estimate the maximum lever force
F that can be used, and the resulting braking torque.
A
Ft2
RN2
RN1
Ft1
O2
O1
All dimensions are in mm
Fig.1.13 Double Shoe Brake (Source: Fundamentals of Machine Component Design by
Robert. C.Juvinall and Kurt. M. Mashrek)
MIME 4222 Engineering Design II
Page 23
Taking moments about point A and equating clockwise and anticlockwise moments,
??=
F x 400 = S x 100
??? =
??
4
4?? sin ??
4 × 0.2 × sin 45
=
= 0.22
2?? + sin 2?? 90° × ?? + sin 90°
180°
Taking moments about O1 and equating clockwise and anticlockwise moments,
S x 700 = Ft1 (25080) + RN1 x 300; substitute Ft1 = ??? RN1
RN1 = 2.07 S
Ft1 = 0.456 S
(1)
Taking moments about O2 and equating clockwise and anticlockwise moments,
S x 700 + Ft2 (25080) = RN2 x 300
RN2 = 2.66 S
Ft2 = 0.586 S
Ab = b (2rsin?) = 80(2 x 250 x sin 45) = 28,284 mm2
??
???? = ????2
??
??
??2
0.4 = 28,284
(2)
Since RN2 is greater,
RN2 = 11,316 N and Ft2 = 2489 N and from equation 2,
S = 4247 N and
??=
4247
= 1061.8 ??
4
TB = (Ft1 + Ft2) x r = 1106 Nm
1.9 Improving Wear Performance of Brakes
Actual wear in a given application depends on a combination of many variables. Friction
materials are relatively softer and weaker than the metallic materials used for discs and
drums. Wear is often characterized as adhesion. As the surface of the friction material
rubs over the high spots of the metal, plastic deformation occurs at the surface and
particles are sheared off, breaking the bond between particles or dislodging filler
materials from the polymer bonding agents. This process accelerates when surface
temperatures rise as the brake absorbs the energy required to stop the rotating system.
The thermal behavior of the system is critical to good life. If temperatures rise above
400°F (200°C), wear rate increases significantly and the coefficient of friction
decreases, leading to poorer braking performance called fade. It is difficult to predict
the life of a given brake system analytically and testing under real operating conditions
is recommended for new designs. The following lists the general principles for
improving wear performance:
?
Specify friction materials that have relatively low adhesion when in contact with the
disc or drum material.
MIME 4222 Engineering Design II
Page 24
?
Specify friction materials that have high bonding strength between constituent
particles.
?
Provide high hardness on the surface of the disc or drum by heat treatment.
?
Keep the pressure between the friction material and the material of the disc or drum
as low as practical.
?
Maintain the surface temperature at the interface between the friction material and
the material of the disc or drum as low as practical by promoting heat transfer away
from the system by conduction, convection and radiation. Forced airflow or cooling
with water is often applied in critical situations.
?
Provide a smooth surface finish on the discs and drums.
?
Provide lubricants such as oil or graphite at the friction interface.
?
Exclude abrasive contaminants from the friction interface.
?
Minimize slipping between the clutch or brake elements by promoting lockup of the
engaging elements.
References:
Text Books:
1. Fundamentals of Machine Component Design, Third Edition by Robert. C.Juvinall
and Kurt. M. Mashrek, John Wiley & Sons Publications, 2003.
2. Machine Design by R.S.Khurmi and J.K.Gupta, Fourth Edition, S.Chand
Publications, 2005.
Reference Books:
1. Shigleys Mechanical Engineering Design, Eighth Edition by Budynas?Nisbett,
McGraw Hill Publications, 2006.
2. Mechanical Design by Peter.R.N.Childs, Second Edition, Elsevier Butterworth
Heinmann Publications, 2004.
3. Design of Transmission Systems by Dr.P.Kannaiah, Scitech Publications (India)
Pvt.Ltd, 2007.
MIME 4222 Engineering Design II
Page 25
Problems for Practice
1.
A single bloc
Page 2
Index
List of Topics
Chapter 1: Brakes
Sl.No.
Topics
Page No.
1
1.1
Introduction on brakes
6
2
1.2
Types of brakes
6
3
1.3
Materials for brake lining
7
4
1.4
Heat to be dissipated during braking
7
5
1.5
Problems on energy conversion
9
6
1.6
Single block brake
11
7
1.7
Pivoted block brake
14
8
1.8
Double block brake
19
9
1.9
Simple band brake
20
Chapter 2: Friction Clutches
Sl.No.
Topics
Page No.
1
2.1
Introduction on friction clutches
27
2
2.2
Friction clutches
28
3
2.3
Material for friction surfaces
29
4
2.4
Considerations in designing a friction clutch
29
5
2.5
Types of friction clutches
29
6
2.6
Design of a plate clutch
31
7
2.7
Multi plate clutches
34
8
2.8
Cone clutches
38
Chapter 3: Wire Ropes
Sl.No.
Topics
Page No.
1
3.1
Introduction on wire ropes
47
2
3.2
Advantages of wire ropes
47
3
3.3
Construction of wire ropes
48
4
3.4
Designation of wire ropes
48
5
3.5
Stresses in wire ropes
48
MIME 4222 Engineering Design II
Page 3
6
3.6
Procedure for designing a wire rope
49
7
3.7
Problems on wire ropes
50
8
3.8
Failure of ropes
54
Chapter 4: Belt and Rope Drives
Sl.No.
Topics
Page No.
1
4.1
Introduction belt drives
62
2
4.3
Velocity ratio of belt drive
62
3
4.4
Slip of belt
63
4
4.5
Creep of belt drive
63
5
4.6
Ratio of driving tensions for belt drives
63
6
4.7
Condition for the transmission of maximum power
65
7
4.8.
VBelt Drives
67
Chapter 5: Ergonomics in Engineering Design
Sl.No.
Topics
Page No.
1
5.1
Introduction on Ergonomics
72
2
5.2
Importance of Ergonomic Principles
72
3
5.3
Important Ergonomic Principles
73
Chapter 6: Product Design
Sl.No.
Topics
Page No.
1
6.1
Introduction to Product Design
77
2
6.2
Strategies in Product Design
78
Chapter 7: Parametric Design
Sl.No.
Topics
Page No.
1
7.1
Definition of Parametric Design
79
2
7.2
Details of Parametric Design
79
MIME 4222 Engineering Design II
Page 4
Chapter I
Mechanical Brakes
Course Outcomes Covered
?
Design wire ropes, sheaves and drum, brakes and clutches.
?
Select and classify appropriate motor components.
?
Perform motor power calculations.
?
Realize the importance of synthesis of components.
?
Apply a complete system design project using calculation, specification and
drawings.
MIME 4222 Engineering Design II
Page 5
Chapter 1
Brakes
1.1
Introduction
A brake is a device used to bring a moving system to rest, to slow its speed, or to
control its speed to a certain value under varying conditions. The function of a brake is to
turn mechanical energy into heat. This heat is dissipated in the surrounding air or water
which is circulated through the passages of the brake drum so that excessive heating if the
brake linings does not take place. The design or capacity of a brake depends upon the
following factors:
?
The unit pressure between the braking surfaces.
?
The coefficient of friction between the braking surfaces.
?
The peripheral velocity of the brake drum.
?
The projected area of the friction surfaces
?
The ability of the brake to dissipate heat equivalent to the energy being absorbed.
The major difference between a brake and a clutch is that a clutch is used to keep
the driving and driven member moving together, whereas the brakes are used to stop a
moving member or to control its speed.
Fig.1.1 A disc brake in action (Source: Wikipedia)
1.2 Types of Brakes
1.
Hydraulic brakes
2.
Electric brakes
3.
Mechanical brakes
MIME 4222 Engineering Design II
Page 6
The hydraulic and electric brakes cannot bring the member to rest and are mostly
used where large amounts of energy are to be transformed while the brake is retarding the
load such as in laboratory dynamometers, highway trucks and electric locomotives. These
brakes are also used for retarding or controlling the speed of a vehicle for downhill travel.
The mechanical brakes according to the direction of acting force, may be divided into the
following two groups:
(a) Radial brakes: In these brakes, the force acting on the brake drum is in radial
direction. The redial brakes may be subdivided into external brakes and internal
brakes.
(b) Axial brakes: In these brakes, the force acting on the brake drum is in axial direction.
The axial brakes may be disc brakes and cone brakes.
1.3
Materials for Brake Lining
The materials used for the brake lining should have the following characteristics:
?
It should have high coefficient of friction with minimum fading which means that the
coefficient of friction should remain constant over the entire surface with change in
temperature.
?
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 and oil.
1.4
Heat to be dissipated during braking
The energy absorbed by the brake and transformed into heat must be dissipated to
the surrounding air in order to avoid excessive temperature rise of the brake lining.
The temperature rise depends upon the mass of the brake drum, the braking time
and the heat dissipation capacity of the brake. The highest permissible temperatures
recommended for different brake lining materials are:
?
For leather, fibre and wood facing = 65 70?C
?
For asbestos and metal surfaces what are slightly lubricated = 90 105?C
?
For automobile brakes with asbestos block lining = 180 225?C
MIME 4222 Engineering Design II
Page 7
Since the energy absorbed (heat generated) and the rate of wear of the brake lining
at a particular speed are dependent on the normal pressure between the braking surfaces,
therefore it is an important factor in the design of brakes.
The permissible normal pressure between the braking surfaces depends upon the
material of the brake lining, the coefficient of friction and the maximum rate at which the
energy is to be absorbed. The energy absorbed or heat generated is given by the equation:
E = Hg = ?RNv = ?pAv (J/s or Watts)
(Equation. 1.1)
Where ? is the coefficient of friction,
p normal pressure between the braking surfaces (N/m2)
A Projected area of the brake drum in m2
V Peripheral velocity of the brake drum in m/sec.
The heat generated may also be obtained considering the amount of kinetic energy or
potential energies which is being absorbed.
In other words, Hg = EK ? EP
where EK is the total kinetic energy absorbed and EP is the total potential energy
absorbed.
Heat dissipated (Hd) = C (t1t2) Ar
(Equation. 1.2)
where C is the heat dissipation factor or coefficient of heat transfer in W/m 2/?C.(t1t2) is the
temperature difference between the exposed radiating surface and the surrounding air in ?C
and Ar is the area of radiating surface in m2.
The expression for heat dissipated is quite approximate and should serve only as an
indication of the capacity of the brake to dissipate heat. The exact performance of the brake
should be determined by the test. It has been found that 10 to 25% of the heat generated is
immediately dissipated to the surrounding air while the remaining heat is absorbed by the
brake drum causing its temperature to rise.
The rise in temperature of the brake drum
?t ?
Hg
mC
(Equation. 1.3)
where ?t is the temperature of the brake drum in ?C
Hg is the heat generated by the brake in J
m is the mass of the brake drum in kg
C is the specific heat for the material of the brake drum in J/kg?C
MIME 4222 Engineering Design II
Page 8
1.5
Problems
1.5.1 A vehicle of mass 1200 kg is moving down the hill at a slope of 1:5 at 72 kmph. It is
to be stopped in a distance of 50 m. If the diameter of the tyre is 600 mm, determine the
average braking torque to be applied to stop the vehicle, neglecting all the frictional energy
except for the brake. If the friction energy is momentarily stored in a 20 kg cast iron brake
drum, what is the average temperature rise of the drum? The specific heat for C.I may be
taken as 520 J/kg ?C. Determine, also the minimum coefficient of friction between the tyres
and the road in order that the wheels do not skid, assuming that the weight is equally
distributed among all the four wheels.
Data: m = 1200 kg
H = 50 m
Slope = 1:5
v = 72 kmph = 20 m/sec
d = 600 mm
mb = 20 kg
C= 520 J/kg ?C
Average braking torque to be applied to stop the vehicle
K.E of the vehicle EK =
1 2 1
mv = (1200) (20)2 = 240, 000 Nm
2
2
P.E of the vehicle EP = mgh x slope = 1200 x 9.81x 50x 1/5 = 117,720 Nm
Total energy og the vehicle or the energy to be absorbed by the brake E = EK+ EP
E = 357,720 Nm
Since the vehicle is to be stopped in a distance of 50 m, tangential braking force required
Ft = 357720/50 = 7154 N
Average braking torque to be applied to stop the vehicle TB = Ft r = 7154 x 0.3
= 2146.3 Nm
TB = 2146.3 Nm
Average temperature rise of the drum
Let ?t be average temp. rise of the drum in ?C
The heat absorbed by the brake drum
Hg = Energy absorbed by the brake drum
= 357,720 Nm = 357,720 J
Heat absorbed by the brake drum is also given by Hg =357,720 = mb x Cx ?t
357, 720 = 20 x 520 x ?t
Therefore
?t = 34.4 ?C
Minimum coefficient of friction between tyre and road
Let ? be the minimum coefficient of friction between tyre and road RN be the normal force
between the contact surfaces. This is equal to the weight of the vehicle.
MIME 4222 Engineering Design II
Page 9
RN = m.g = 1200 x 9.81 = 11772 N
Tangential braking force (Ft) = 7154 = ? RN = ? x 11772
? = 0.6
Therefore
1.5.2 A four wheeled automobile has a total mass of 1000 kg. The moment of inertia of
each wheel about a transverse axis through its centre of gravity is 0.5 kgm2. The rolling
radius of the wheel is 0.35 m. The rotating and reciprocating parts of the engine and the
transmission system are equivalent to a moment of inertia of 2.5 kgm2 which rotates at five
times the roadwheel speed. The car is traveling at a speed of 100 kmph on a plane road.
When the brakes are applied, the car decelerates at 0.5 g. There are brakes on all four
wheels. Calculate
(i)
The energy absorbed by each wheel
(ii)
The torque capacity of each brake.
(i)
Kinetic energy of the car
v1= 100 kmph = 27.78 m/sec
v2 = 0
(ii)
K.E =
1
m ??v12 ? v2 2 ?? = 385,802.4 J
2
Kinetic energy of the wheels
? 27.78
?1 ? 1 ?
? 79.37rad / sec
R
?2 =0
0.35
?1
?
Kinetic energy of four wheels = 4 ? I ??12 ? ?2 2 ? ? = 6298.8 J
?2
?
(iii)
Kinetic energy of the engine and transmission system
?1 = 5 (79.37) = 396.83 rad/sec
?1
?
K.E = ? I ??12 ? ?2 2 ? ? = ½ x 2.5 x (396.83)2 = 196837.97 J
?2
?
The energy absorbed by the four brakes consists of K.E of the car, the K.E of the
wheel and the K.E of the engine and the transmission system.
K.E = ¼ (385802.4+629808+196837.97) = 147234.8 J
Braking time t,
?1 ?? 2
t
? 0.5g
t = 5.66 secs
The average velocity during the braking time is
MIME 4222 Engineering Design II
?1 ? ?2
2
or
?1
2
Page 10
?? ?
? ? ? 1 ? t = (79.37/2) 5.66 = 224.6 rad
? 2?
Therefore
Torque, T = E/? = 665.54 Nm
1.5.3 A mass of 2500 kg is lowered at a velocity of 1.5 m/sec as shown in figure below. The
mass of drum is 50 kg and its radius of gyration can be taken as 0.7 m. On applying the
brake, the mass is brought to rest in a distance of 0.5 m. Calculate:
(i)
The energy absorbed by the brake
(ii)
The torque capacity of the brake
Fig.1.2 Load handled by a brake (Source: Design of Machine Elements by Bhandari)
1 2
1
2
mv ? ? 2500 ? ?1.5 ? ? 2812.5 J
2
2
v 1.5
Angular velocity of drum ? = ?
? 2 rad
sec
r 0.75
1
1
2
2
K.E of drum = mk 2? 2 ? ? 50 ? ? 0.7 ? ? ? 2 ? = 49 J
2
2
K.E of mass =
Potential energy of mass = mgh = 2500 x 9.81 x 0.5 = 12262.5 J
Total energy = 2812.5 + 49 + 12262.5 = 15124 J
During braking action, the mass moves through a distance of 0.5 m. If ? is the angle
through which the drum rotates the braking period,
??
? x drum radius = 0.5
Therefore, Torque T =
MIME 4222 Engineering Design II
E
?
=
0.5
? 0.667rad
0.75
15124
? 22686 Nm
0.667
Page 11
1.5.4 Figure shows a 1000 kg mass being lowered by a cable at a uniform rate of 4 m/s
from a drum of 550 mm diameter weighing 2.5 kN and having a 250 mm radius of
gyration. What is the kinetic energy in the system?
Fig.1.3 Load handled by a cable (Courtesy: Fundamentals of Machine Component Design
by Robert Juvinal)
1
2
? 1000 ? ? 4 ? = 8000 Nm
2
?
4
Angular velocity of drum ? ? ?
? 14.54 rad/sec
r
0.275
2.5 ? 103
mass of drum m =
? 255 kg
9.81
1
1
2
2
K.E of drum = ? mk 2? 2 ? ? 255 ? ? 0.25 ? ? ?14.54 ? ? 1686 Nm
2
2
K.E of mass =
Therefore, Total K.E = 8000 + 1686 = 9686 Nm
1.6
Single block or shoe brake
A single block or shoe brake consists of a block or shoe which is pressed
against the rim of a revolving brake wheel drum. The block is made of a softer material than
the rim of a wheel.
This type of a brake is commonly used on railway trains and tram cars. The
friction between the block and the wheel causes a tangential braking force to act on the
wheel, which retard the motion of the wheel.
The block is pressed against the wheel by a force applied to one end of a lever
to which the block is rigidly fixed as shown in figure 1.2. The other end of the lever is pivoted
on a fixed fulcrum O.
MIME 4222 Engineering Design II
Page 12
Fig.1.2 Clockwise rotation of a brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
Let
P = Force applied at the end of the lever
RN = Normal force pressing the brake block on the wheel
r
= Radius of the wheel
2? = Angle of contact surface of the block
? = Coefficient of friction
Ft = Tangential braking force or the frictional force acting at the contact
surface of the block and the wheel
If the angle of contact is less than 60?, then it may be assumed that the normal
pressure between the block and the wheel is uniform. In such cases, tangential braking force
on the wheel
Ft = ? RN and the braking torque TB = FT r = ? RN r
Case1: When the line of action of tangential braking force (Ft) passes through the fulcrum
O of the lever and the brake wheel rotates clockwise as shown in figure 1.2, then for
equilibrium, taking moments about the fulcrum O,
RN x X = P x l
or
RN =
P?l
X
Therefore braking torque TB = ? RN r= ? x
P?l
xr
X
(Equation.1.4)
Please note that when the brake wheel rotates anticlockwise as shown in figure 1.3, then
the braking torque is same, i.e
TB = ? RN r = ? x
MIME 4222 Engineering Design II
P?l
xr
X
(Equation.1.5)
Page 13
Fig.1.3 Anticlockwise rotation of a brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
Case 2: When the line of action of the tangential braking force (Ft) passes through a
distance a below the fulcrum O, and the brake wheel rotates clockwise as shown in figure
3, then for equilibrium, taking moments about the fulcrum O,
RN x = Ft a + P.l
or
RN x = ? RN a + P.l
(Equation.1.6)
Braking torque,
R N ? x?a ? =P.l
P.l
RN =
(x?a)
TB = ?R N r =
?Plr
(x?a)
Fig.1.4 Clockwise rotation of brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
MIME 4222 Engineering Design II
Page 14
Fig.1.5 Anticlockwise rotation of brake wheel
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
When the brake wheel rotates anticlockwise, then for equilibrium,
RN x + Ft a = Pl
and braking torque,
T B = ? RN r =
? P.l.r
x ? ?a
(Equation.1.7)
In equation 1.6, the moment of frictional force (? RN.a) adds to the moment of force (P.l). In
other words, the frictional force helps to apply the brake. Such types of brakes are said to
be selfenergizing brakes. When the frictional force is great enough to apply the brake with
no external force, then the brake is said to be selflocking brake.
Note:
?
The condition for the brake to be selflocking is x ? ?a .
?
The brake should be selfenergizing and not be selflocking.
?
In order to avoid selflocking and to prevent the brake from grabbing, x should be
greater than ?a.
?
When a single block or shoe brake is applied to a rolling wheel, an additional load
is thrown on the shaft bearings due to heavy normal force (R N) and produces
bending of the shaft. In order to overcome the drawback, a double block or shoe
brake is used.
1.7
Pivoted Block or Shoe Brake
When the angle of contact is less than 60?, then it may be assumed that the normal
pressure between the block and the wheel is uniform. But when the angle of contact is
greater than 60?, then the unit pressure normal to the surface of contact is less at the ends
than at the centre. In such cases, the block or shoe is pivoted to the lever as shown in figure
instead of being rigidly attached to the lever. This gives uniform wear of the brake lining in
MIME 4222 Engineering Design II
Page 15
the direction of the applied force. The braking torque for a pivoted block or shoe brake (when
2?>60?) is given by
TB ? Ft ? r ? ? ‘ RN .r
where ? ‘ = Equivalent coefficient of friction =
4? sin ?
and ? is the actual coefficient of
2? ? sin 2?
friction.
Fig.1.6 Pivoted Brake Shoe
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
Problem 1: A single block brake is shown in figure below. The diameter of the drum is 250
mm and the angle of contact is 90?. If the operating force of 700 N is applied at the end of a
lever and the coefficient of friction is 0.35, determine the torque that may be transmitted by
the block brake.
Data: d = 250 mm
2? = 90?
P = 700 N
? = 0.35
Since the angle of contact is greater than 60?, equivalent coefficient of friction,
?’ ?
4? sin ?
= 0.385
2? ? sin 2?
Fig.1.7 Pivoted Brake Shoe
(Source: Machine Design by R.S.Khurmi and J.K.Gupta)
MIME 4222 Engineering Design II
Page 16
Let RN be the normal force pressing the block and the brake drum and
Ft be the tangential braking force = ? RN
Taking moments about the fulcrum O,
700 (250+200) + Ft x 50 = RN x 50 = RN x 200 =
520 Ft – 50 Ft = 700 x 450
(or)
Ft
?
‘
x 200 =
Ft
? 200 ? 520Ft
0.385
Ft = 670 N
Torque transmitted by block brake TB,
TB ? Ft ? r ? 670 ?125 ? 83,750 N ? mm ? 83.75N ? m
Problem 2: Figure shows a brake shoe
attached to a drum by a lever AB which is
pivoted at a fixed point A’ and rigidly fixed
to the shoe. The radius of the drum is 160
mm. The coefficient of friction of the brake
lining is 0.3. If the drum rotates clockwise,
find the braking torque due to horizontal
force of 600 N applied at B.
Fig.1.8 Pivoted Brake Shoe
(Source: Machine Design by R.S.Khurmi
and J.K.Gupta)
Taking moments about A, and equating clockwise and anticlockwise moments,
RN ? 350 ? Ft ? 200 ? 160 ? ? 600 ? 400 ? 350 ?
Ft
? ??? ? ?? Ft ? 600 ? ???
0.3
Ft ? 372.8 N
Therefore,
Braking Torque TB = Ft x r = 372.8 x 0.16 = 59.65 Nm
Problem 3: Figure shows a single block brake. The brake drum diameter is 400 mm and
rotates at a speed of 150 rpm. The friction material permits a maximum pressure of 0.5 MPa
and µ=0.25. Face width of the block is 50 mm. If the brake is applied for 10 seconds at full
capacity to bring the shaft to stop, determine (i) Effort (ii) Maximum Torque (iii) Heat
generated.
MIME 4222 Engineering Design II
Page 17
Since 2? ? 90º,
?’ ?
4?sin?
4 ? 0.25 ? sin 45?
?
? 0.275
2?+sin2? 90?? ? ? sin 90?
180
Taking moments about fulcrum point O, F x 500 = RN X 250
F?
250
R N = 0.5 RN
500
Ft
RN
(Dimensions are in mm)
Fig.1.9 Single Brake Shoe
(Source: Machine Design by J.B.K.Doss and J.L.Srinivasa Murthy)
(i)
Effort
pressure p b ?
RN
Ab
RN
RN
?
b ? 2 ? r ? sin ? 50 x 2 x 200 x sin 45?
R N ? 7071N
0.5 ?
F?
250
R N = 0.5 RN = 3535.5 N
500
(ii) Maximum Torque
Ft = ? ‘ RN = 0.275 x 7071 = 1944.5 N
TB = Ft x r = 1944.5 x 200 = 388,900 Nmm
TB = 388.9 Nm
(iii) Heat Generated
v=
? d N ? x 0.4 x 150
=
= 3.14 m/sec
60
60
Hg = µpAv = µ RN v = 0.275 x 7071 x 3.14 = 6105.8 J/sec
Therefore, Heat generated in 10 seconds = 6105.8 x 10 = 61058 J = 61.058 kJ
MIME 4222 Engineering Design II
Page 18
Problem 4: A single block brake with a torque capacity of 15 Nm is shown in figure below
and the maximum pressure on the brake linings is 1 N/mm2. The width of the block is equal
to its length. Calculate:
(i)
Actuating force
(ii)
Dimensions of the block
(iii)
Resultant hinge pin reaction
(iv)
Rate of heat generated if the brake drum rotates at 50 rpm.
F2
Ft
F1
RN
(Dimensions are in mm)
Fig.1.10 Single Brake Shoe
(Source: Machine Design by J.B.K.Doss and J.L.Srinivasa Murthy)
Taking moments about point O and equating clockwise and anticlockwise reactions,
P x 650 + Ft x 60 = RN x 200
On solving, RN = 3.57 P
15 x 103 = 0.3 x RN x 150
Torque T = Ft x r
Therefore, RN = 333.33 N
(i) Actuating force:
RN = 3.57 P and hence,
P = 93.33 N
(ii) Dimensions of the block:
Let l and b represent the length and width of the brake shoe.
braking pressure pb ?
1 ?
RN
Ab
333.33
l2
MIME 4222 Engineering Design II
Ab = l ? b ? l 2
(since l = b)
l = b = 18.26 mm
Page 19
(iii)
Resultant hingepin reaction:
Considering vertical forces at hinge O, ? Fv ? 0
? RN ? F2 ? P ? 0
F2 ? RN ? P ? 333.33 ? 93.33 ? 240 N
Considering horizontal forces at hinge O, ? Fh ? 0
Ft = F1 = 99.99 N = 100 N
Therefore resultant hingepin reaction, R =
F12 ? F2 2 ? (100)2 ? (240)2 ? 260 N
R = 260 N
(iv)
Heat Generated:
? (? x 0.3 x 50) ?
H g = ?R N v = 0.3 x 333.33 x ?
?? = 78.54 J/sec
60
?
Therefore,
1.8
Hg = 78.54 J/sec
Double block or shoe brake
When a single block brake is applied to a rolling wheel an additional load is thrown
on the shaft bearings due to normal force (RN). This produces bending of the shaft. In order
to overcome this drawback, a double block or shoe brake is used. It consists of two brake
blocks applied at opposite ends of a diameter of the wheel which eliminates or reduces the
unbalanced force on the shaft.
This type of brake is used in electric cranes and the braking action is doubled by the
use of two blocks and two blocks may be operated practically by the same force which
operates one. In case of double block or shoe brake, the braking torque is T B = (Ft1+Ft2) r
where Ft1 and Ft2 are the braking forces on the two blocks.
Fig.1.11 Double block or shoe brake
(Source: Machine Design by R.S.Khurmi & J.K.Gupta)
MIME 4222 Engineering Design II
Page 20
Problem 5: A double shoe brake as shown in figure below is capable of absorbing a torque
of 1400 Nm. The diameter of the brake drum is 350 mm and the angle of contact for each
shoe is 100?. If the coefficient of friction between the brake and the lining is 0.4, find:
1.
The spring force necessary to set the brake
2.
Width of the brake shoes if the bearing pressure on the lining material is
not to exceed 0.3 N/mm2.
Dimensions are in mm
Fig.1.12 Double Shoe Brake (Source: Machine Design by R.S.Khurmi & J.K.Gupta)
Data: TB = 1400 Nm
d = 350 mm
r = 175 mm
2? = 100? = 100 x ?/180 =1.75 rad
1.
? = 0.4
pb = 0.3 N/mm2
Spring force necessary to set the brake
Let S = spring force necessary to set the brake
RN1 and F1 be the normal reaction and the braking force on the right hand side sjoe
RN2 and F2 be the corresponding values on the left hand side
Since the contact angle is greater than 60?, the equivalent coefficient of friction,
?’ ?
4? sin ?
= 0.45
2? ? sin 2?
Taking moments about O2, and equating clockwise and anticlockwise moments,
S ? 450 ? RN 1 ? ??? ? Ft1 ?175 ? 40 ?
?
Ft1
? ????? Ft1 ???? ? ????? Ft1
0.45
Ft1=0.776 S
S ? 450 ? Ft 2 ?175 ? 40 ? ? RN 2 ? ??? ??
Ft2
? ??? ? ??????Ft2
0.45
Ft2 = 1.45 S
MIME 4222 Engineering Design II
(Equation 1)
(Equation 2)
Page 21
From equations (1) and (2), Torque capacity of the brake (TB),
1400 x 103 = (Ft1+Ft2).r = (0.776 S+1.454 S) 175 = 390.25 S
S = 3587 N
Spring Force
2. Width of the brake shoes
Let b = width of the brake shoes in mm
The projected bearing area for one shoe,
Ab = b (2rsin?) = b (2x175x sin 50?) = (268 b) mm2
Normal force on the right hand side of the shoe, RN 1 ?
Normal force on the left hand side of the shoe, RN 2 ?
Ft1
?
Ft 2
?
‘
‘
?
?
0.776 ? S
? 6186 N
0.45
1.454 ? S
? 11,590 N
0.45
The maximum normal force is on the left hand side of the shoe. Therefore, designing the
shoe for maximum normal force , i.e RN2,
Pb = 0.3 = RN2/Ab = 11590/268 b
therefore, b = 144 mm
Width of shoe
Problem 6: A double shoe brake is shown in figure below. The drum rotates at 200 rpm
when the applied force is 1000 N and coefficient of friction is 0.25. Determine (i) braking
torque and (ii) amount of heat generated due to braking action.
Dimensions are in mm
Fig.1.15 Double Shoe Brake (Source: Machine Design by J.B.K.Doss and J.L.Srinivasa
Murthy)
Taking moments about point A and equating clockwise and anticlockwise reactions,
1000 x 100 = P x 50
P = 2000 N
Solving for the right hand side, taking moments about fulcrum O 1 and equating clockwise
and anticlockwise moments,
MIME 4222 Engineering Design II
Page 22
P x 450 = Ft1 x 50 + RN1 x 200
2000 x 450 = 0.25 x 50 x RN1 + 200 RN1 = 212.5 RN1
Therefore
RN1 = 4235 N and Ft1 = 1058.8 N
Solving for the left hand side, taking moments about fulcrum O 2 and equating clockwise
and anticlockwise moments,
P x 450 = RN2 x 200
and
(i)
RN2 = 4500 N
Ft2 = 1125 N
Braking Torque:
TB = (Ft1 + Ft2) r = (1058.8 + 1125) x 0.5 = 546 Nm
(ii)
Amount of heat generated: (Hg)
??
? dN
60
?
? ? 0.5 ? 200
60
? 5.2 m/sec
Hg = (µ RN1 + µRN2) x ? = (1058.8+ 1125) x 5.2 = 11355.76 J = 11.35 kJ
Problem 7: A twoshoe external drum brake shown in figure below has shoes 80 mm wide
that contact 90° of drum surface. For a coefficient of friction of 0.20 and an allowable contact
pressure of 400 kN per square meter of projected area, estimate the maximum lever force
F that can be used, and the resulting braking torque.
A
Ft2
RN2
RN1
Ft1
O2
O1
All dimensions are in mm
Fig.1.13 Double Shoe Brake (Source: Fundamentals of Machine Component Design by
Robert. C.Juvinall and Kurt. M. Mashrek)
MIME 4222 Engineering Design II
Page 23
Taking moments about point A and equating clockwise and anticlockwise moments,
??=
F x 400 = S x 100
??? =
??
4
4?? sin ??
4 × 0.2 × sin 45
=
= 0.22
2?? + sin 2?? 90° × ?? + sin 90°
180°
Taking moments about O1 and equating clockwise and anticlockwise moments,
S x 700 = Ft1 (25080) + RN1 x 300; substitute Ft1 = ??? RN1
RN1 = 2.07 S
Ft1 = 0.456 S
(1)
Taking moments about O2 and equating clockwise and anticlockwise moments,
S x 700 + Ft2 (25080) = RN2 x 300
RN2 = 2.66 S
Ft2 = 0.586 S
Ab = b (2rsin?) = 80(2 x 250 x sin 45) = 28,284 mm2
??
???? = ????2
??
??
??2
0.4 = 28,284
(2)
Since RN2 is greater,
RN2 = 11,316 N and Ft2 = 2489 N and from equation 2,
S = 4247 N and
??=
4247
= 1061.8 ??
4
TB = (Ft1 + Ft2) x r = 1106 Nm
1.9 Improving Wear Performance of Brakes
Actual wear in a given application depends on a combination of many variables. Friction
materials are relatively softer and weaker than the metallic materials used for discs and
drums. Wear is often characterized as adhesion. As the surface of the friction material
rubs over the high spots of the metal, plastic deformation occurs at the surface and
particles are sheared off, breaking the bond between particles or dislodging filler
materials from the polymer bonding agents. This process accelerates when surface
temperatures rise as the brake absorbs the energy required to stop the rotating system.
The thermal behavior of the system is critical to good life. If temperatures rise above
400°F (200°C), wear rate increases significantly and the coefficient of friction
decreases, leading to poorer braking performance called fade. It is difficult to predict
the life of a given brake system analytically and testing under real operating conditions
is recommended for new designs. The following lists the general principles for
improving wear performance:
?
Specify friction materials that have relatively low adhesion when in contact with the
disc or drum material.
MIME 4222 Engineering Design II
Page 24
?
Specify friction materials that have high bonding strength between constituent
particles.
?
Provide high hardness on the surface of the disc or drum by heat treatment.
?
Keep the pressure between the friction material and the material of the disc or drum
as low as practical.
?
Maintain the surface temperature at the interface between the friction material and
the material of the disc or drum as low as practical by promoting heat transfer away
from the system by conduction, convection and radiation. Forced airflow or cooling
with water is often applied in critical situations.
?
Provide a smooth surface finish on the discs and drums.
?
Provide lubricants such as oil or graphite at the friction interface.
?
Exclude abrasive contaminants from the friction interface.
?
Minimize slipping between the clutch or brake elements by promoting lockup of the
engaging elements.
References:
Text Books:
1. Fundamentals of Machine Component Design, Third Edition by Robert. C.Juvinall
and Kurt. M. Mashrek, John Wiley & Sons Publications, 2003.
2. Machine Design by R.S.Khurmi and J.K.Gupta, Fourth Edition, S.Chand
Publications, 2005.
Reference Books:
1. Shigleys Mechanical Engineering Design, Eighth Edition by Budynas?Nisbett,
McGraw Hill Publications, 2006.
2. Mechanical Design by Peter.R.N.Childs, Second Edition, Elsevier Butterworth
Heinmann Publications, 2004.
3. Design of Transmission Systems by Dr.P.Kannaiah, Scitech Publications (India)
Pvt.Ltd, 2007.
MIME 4222 Engineering Design II
Page 25
Problems for Practice
1.
A single bloc
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