Module 12
Design of brakes
Version 2 ME , IIT Kharagpur
Lesson 1
Design of shoe brakes Version 2 ME , IIT Kharagpur
Instructional Objectives: After reading the lesson the students should learn:
• Different types of shoe brakes and their operating principles
• Design procedure of different shoe brakes
1. Types of brakes
Brakes are devices that dissipate kinetic energy of the moving parts of a
machine. In mechanical brakes the dissipation is achieved through sliding
friction between a stationary object and a rotating part. Depending upon the
direction of application of braking force, the mechanical brakes are primarily of
three types
• Shoe or block brakes – braking force applied radially
• Band brakes – braking force applied tangentially.
• Disc brake – braking force applied axially.
2. Shoe or block brake In a shoe brake the rotating drum is brought in contact with the shoe by suitable
force. The contacting surface of the shoe is coated with friction material.
Different types of shoe brakes are used, viz., single shoe brake, double shoe
brake, internal expanding brake, external expanding brake. These are
sketched in figure 12.1.1.
drum
shoe
lever
Figure 1(a) Single shoe brake
Version 2 ME , IIT Kharagpur
drum
shoe
lever
Figure 1(b) Double shoe brake
drum
shoe
Figure 1(c): Internal expanding shoe brake
Version 2 ME , IIT Kharagpur
drum
shoe
Figure 1(d): External expanding shoe brake
Figure 12.1.1: Different shoe brakes
Single Shoe brake The force needed to secure contact is supplied by a lever. When a force F is
applied to the shoe (see figure 12.1.2a ) frictional force proportional to the
applied force 'frF Fμ= develops, where 'μ depends of friction material and the
geometry of the shoe. A simplified analysis is done as discussed below.
P F
O
Ffr
Figure 12.1.2a: Free body diagram of a brake shoe
Though the exact nature of the contact pressure distribution is unknown, an
approximation (based on wear considerations) is made as
Version 2 ME , IIT Kharagpur
0( ) cosp pθ θ=
Where the angle is measured from the centerline of the shoe. If Coulomb’s law
of friction is assumed to hold good, then
θμθ cos)( 0pf fr =
Since the net normal force of the drum is F, one has
0
0
( )cos ,Rb p d Fθ
θ
θ θ θ−
=∫
Where R and b are the radius of the brake drum and width of the shoe
respectively.
The total frictional torque is
∫−
=0
0
2)(θ
θ
θθ dRfbT fr
If the total frictional force is assumed to be a concentrated one, then the
equivalent force becomes frTFR
= . A simple calculation yields,
0
0 0
4 sin2 sin 2
μ θμθ θ
′ =+
2θ
Figure 12.1.2(b): Pressure distribution on brake
Version 2 ME , IIT Kharagpur
It may be seen that for very small value of 0 , '.θ μ μ= Even when , 00 30θ =
' 1.0453 .μ μ= Usually if the contact angle is below , the two values of friction
coefficient are taken to be equal.
060
Consider, now single shoe brakes as shown in figures 12.1.3(a) and 3(b).
Suppose a force P is applied at the end of a lever arm with length l. The shoe
placed at a distance x from the hinge experiences a normal force N and a
friction force F, whose direction depends upon the sense of rotation of the
drum. Drawing free body diagram of the lever and taking moment about the
hinge one gets
(a) for clockwise rotation of the brake wheel,
Nx + Fa = Pl
(b) for anticlockwise rotation of the brake wheel,
Nx – Fa = Pl.
Where a is the distance between the hinge and the line of action of F and is
measured positive when F acts below point O as shown in the figure. Using
Coulomb’s law of friction the following results are obtained,
(a) for clockwise rotation PlFx aμμ
=+
,
(b) for anticlockwise rotation PlFx aμμ
=−
,
It may be noted that for anticlockwise rotating brake, if xa
μ > , then the force P
has negative value implying that a force is to applied in the opposite direction to
bring the lever to equilibrium. Without any force the shoe will, in this case, draw
the lever closer to the drum by itself. This kind of brake is known as ‘self-
locking, brake. Two points deserve attention.
(1) If a < 0, the drum brake with clockwise rotation becomes self-energizing and
if friction is large, may be self locking.
(2) If the brake is self locking for one direction, it is never self locking for the
opposite direction. This makes the self locking brakes useful for ‘back stop’s
of the rotors.
Version 2 ME , IIT Kharagpur
Double shoe brake
P
N
lx
Fa
Figure 12.1.3(a): FBD of shoe (CW drum rotation)
P
N
lx
Fa
Figure 12.1.3(b): FBD of shoe (CCW drum rotation)
Since in a single shoe brake normal force introduces transverse loading on the
shaft on which the brake drum is mounted two shoes are often used to provide
braking torque. The opposite forces on two shoes minimize the transverse
loading. The analysis of the double shoe brake is very similar to the single shoe
brake.
External expanding shoe brake An external expanding shoe brake consists of two symmetrically placed shoes
having inner surfaces coated with frictional lining. Each shoe can rotate about
respective fulcrum (say, and ). A schematic diagram with only one shoe is 1O 2O
Version 2 ME , IIT Kharagpur
presented (figure 12.1.4) When the shoes are engaged, non-uniform pressure
develops between the friction lining and the drum. The pressure is assumed to
be proportional to wear which is in turn proportional to the perpendicular
distance from pivoting point (O1N in figure 12.1.4). A simple geometrical
consideration reveals that this distance is proportional to sine of the angle
between the line joining the pivot and the center of the drum and the line joining
the center and the chosen point. This means
0( ) sin , p pθ θ=
where the angle is measured from line OO1 and is limited as 1 2.θ θ θ≤ ≤
Drawing the free body diagram of one of the shoes (left shoe, for example) and
writing the moment equilibrium equation about (say) the following equation is
resulted for clockwise rotation of the drum :
1O
1 ,p fF l M M= −
Where is the force applied at the end of the shoe, and 1F
( )0 2 1 1 21 1( ) sin 2 sin 2 ,2 2pM p bRδ θ θ θ θ⎡ ⎤= − + −⎢ ⎥⎣ ⎦
( )0 1 2 11 (cos ) cos 2 cos 2 ,2 4fM bR R δμ δ θ θ θ θ2
⎡ ⎤= − − −⎢ ⎥⎣ ⎦
where δ is the distance between the center and the pivot (OO1 in figure 12.1.4)
and is the distance from the pivot to the line of action of the force (O1F 1C in
the figure). In a similar manner the force to be applied at the other shoe can be
obtained from the equation
2 .p fF l M M= +
The net braking torque in this case is 2
0 1(cos cos ).T p bR 2μ θ θ= −
Version 2 ME , IIT Kharagpur
C
B
A
F
θN O
O1
Figure 12.1.4: Force distribution in externally expanding brake.
Internal expanding shoe brake Here the brake shoes are engaged with the internal surface of the drum.
The analysis runs in the similar fashion as that of an external shoe brake.
The forces required are
1 ( )p fF M M= + l
and 2 ( )p fF M M l= − ,
respectively.
One of the important member of the expanding shoe brakes is the anchor
pin. The size of the pin is to be properly selected depending upon the face
acting on it during brake engagement.
Version 2 ME , IIT Kharagpur
Module 12
Design of Brakes Version 2 ME , IIT Kharagpur
Lesson 2
Design of Band and Disc Brakes
Version 2 ME , IIT Kharagpur
Instructional Objectives: After reading this lesson the students should learn:
• Different types of band brakes
• Design of band brakes
• Design of disc brakes
• Properties of friction materials
1. Band brakes: The operating principle of this type of brake is the following. A flexible band of
leather or rope or steel with friction lining is wound round a drum. Frictional
torque is generated when tension is applied to the band. It is known (see any
text book on engineering mechanics) that the tensions in the two ends of the
band are unequal because of friction and bear the following relationship:
1
2
,T eT
μ β=
where = tension in the taut side, 1T
= tension in the slack side, 2T
μ = coefficient of kinetic friction and
β = angle of wrap.
If the band is wound around a drum of radius R, then the braking torque is
( ) ( )1 2 1 1brT T T R T e μ β−= − = − R
Depending upon the connection of the band to the lever arm, the member
responsible for application of the tensions, the band brakes are of two types,
(a) Simple band brake: In simple band brake one end of the band is attached to the fulcrum of the
lever arm (see figures 12.2.1(a) and 1(b) ). The required force to be applied to
the lever is
1bP Tl
= for clockwise rotation of the brake drum and
Version 2 ME , IIT Kharagpur
2bP Tl
= for anticlockwise rotation of the brake drum,
where l = length of the lever arm and
b = perpendicular distance from the fulcrum to the point of attachment of
other end of the band.
Figure 12.2.1: Band brakes
(b) Differential band brake: In this type of band brake, two ends of the band are attached to two points on
the lever arm other than fulcrum (see figures 12.2.2(a) and 12.2.2(b)). Drawing
the free body diagram of the lever arm and taking moment about the fulcrum it
is found that
2 1aP T Tl l
= −b , for clockwise rotation of the brake drum and
1 2aP T Tl l
= −b , for anticlockwise rotation of the brake drum.
Hence, P is negative if
b l
T1 T2
P
1(a): Band brake with CW rotating drum
b l
T2 T1
P
1(b): Band brake with CCW rotating drum
Version 2 ME , IIT Kharagpur
1
2
T aeT b
μ β = > for clockwise rotation of the brake drum
and 1
2
T aeT b
μ β = < for counterclockwise rotation of the brake drum. In
these cases the force is to be applied on the lever arm in opposite direction to
maintain equilibrium. The brakes are then self locking.
The important design variables of a band brake are the thickness
and width of the band. Since the band is likely to fail in tension, the following
relationship is to be satisfied for safe operation.
1 TT wts=
where w = width of the band,
t = thickness of the band and
= allowable tensile stress of the band material. The steel bands of the
following dimensions are normally used
Ts
w 25-40 mm 40-60 mm 80 mm 100 mm 140-200
mm
t 3 mm 3-4 mm 4-6 mm 4-7 mm 6-10 mm
Fig.12.2.2(a): Differential Band brake with CW rotation
b l
T1
T2
P
a
Version 2 ME , IIT Kharagpur
b l
T2
T1
P
a
Fig 12.2.2(b): Differential Band brake with CCW rotation
2. Band and block brakes: Sometimes instead of applying continuous friction lining along the band, blocks
of wood or other frictional materials are inserted between the band and the
drum. In this case the tensions within the band at both sides of a block bear
the relation
1
1
1 tan1 tan
TT
μ θμ θ
+=
′ −,
where = tension at the taut side of any block 1T
= tension at the slack side of the same block 1T ′
2θ = angle subtended by each block at center.
If n number of blocks are used then the ratio between the tensions at taut side to slack side becomes
1
2
1 tan1 tan
nTT
μ θμ θ
⎛ ⎞+= ⎜ ⎟−⎝ ⎠
.
The braking torque is ( )1 2brT T T= − R 3. Disc brake:
Version 2 ME , IIT Kharagpur
In this type of brake two friction pads are pressed axially against a rotating
disc to dissipate kinetic energy. The working principle is very similar to friction
clutch. When the pads are new the pressure distribution at pad-disc interface
is uniform, i.e.
constantp = .
If F is the total axial force applied then FpA
= , where A is the area of the pad.
The frictional torque is given by
brakingA
FT rA
dAμ= ∫
where μ = coefficient of kinetic friction and r is the radial distance of an
infinitesimal element of pad. After some time the pad gradually wears
away. The wear becomes uniforms after sufficiently long time, when
constant = (say)pr c=
where dAF p dA cr
= =∫ ∫ . The braking torque is
'brakingAFT pr dA Ac dAr
μμ μ= = =∫∫
It is clear that the total braking torque depends on the geometry of the pad. If
the annular pad is used then 3 3
1 22 2
1 2
23br
R RT FR R
μ⎛ ⎞−
= ⎜ ⎟−⎝ ⎠
1 2
2brR RT Fμ +⎛ ⎞′ = ⎜ ⎟
⎝ ⎠
where 1 and 2R R are the inner and outer radius of the pad.
Version 2 ME , IIT Kharagpur
4. Friction materials and their properties. The most important member in a mechanical brake is the friction material. A
good friction material is required to possess the following properties:
• High and reproducible coefficient of friction.
• Imperviousness to environmental conditions.
• Ability to withstand high temperature (thermal stability)
• High wear resistance.
• Flexibility and conformability to any surface.
Some common friction materials are woven cotton lining, woven asbestos
lining, molded asbestos lining, molded asbestos pad, Sintered metal pads etc.
Review questions and answers: Q.1. A double shoe brake has diameter of brake drum 300mm and contact
angle of each shoe 90 degrees, as shown in figure below. If the coefficient of
friction for the brake lining and the drum is 0.4, find the spring force necessary
to transmit a torque of 30 Nm. Also determine the width of the brake shoe if the
braking pressure on the lining material is not to exceed 0.28 MPa.
S S
250
225
Figure 12.2.3
Ans. The friction force required to produce the given torque is
Version 2 ME , IIT Kharagpur
( )1 230 200
0.150F F N+ = =
The normal forces on the shoes are 11 2, N ,
' 'FN 2Fμ μ
= = where
00
0 0
4 sin' ( ) 0.44.2 sin 2 4
μ θ πμ θθ θ
= =+
= Writing the moment equilibrium equations about
the pivot points of individual shoes (draw correct FBDs and verify)
1 1 10 F 0.718 ,
'
SlSl N x Fa Sxaμ
− + + = ⇒ = =+
and
2 2 20 F 1.1314
'
SlSl N x F a Sx aμ
− + = ⇒ = =−
This yields S = 98.4(N).
Width of the friction lining :
According to the pressure distribution assumed for a shoe brake, the maximum
bearing pressure occurs at the centerline of the shoe. The width of the brake
lining must be selected from the higher values of the normal forces, in this
case . Noting that 2N
/ 42
2 max/ 4
cos ,N Rbp dπ
π
θ θ−
= ∫
Where R = 0.150, the value of b is
calculated to be 5.4 mm or 6 mm (approx.).
6max 20.28 10 , N 1.314 98.4 / 0.44,p X= = ×
Q2. A differential band brake has brake drum of diameter 500mm and the
maximum torque on the drum is 1000 N-m. The brake embraces 2/3rd of the
circumference. If the band brake is lined with asbestos fabric having a
coefficient of friction 0.3, then design the steel band. The permissible stress is
70 MPa in tesnion. The bearing pressure for the brake lining should not
exceed 0.2 MPa.
Version 2 ME , IIT Kharagpur
Ans. The design of belt is to be carried out when the braking torque is
maximum i.e. Tbr = 1000 N-m. According to the principle of band brake
25.01)1( 343.0
11 ×⎟⎟⎠
⎞⎜⎜⎝
⎛−=−=
×−−π
μβ eTReTTbr
Which yield In order to find the pressure on
the band, consider an infinitesimal element. The force balance along the radial
direction yields
1 2 15587 , T T 1587 .T N e μ β−= = = N
N T θ= Δ
Since N p b R θ= Δ so p = TbR
.
The maximum pressure is 1max
TpbR
= .
Hence 6
55870.25 0.2 10
b = 0.112 m (approx.)
Δθ
T
T+ΔT
F
N
=× ×
The thickness t of the band is calculated from the relation
1tS bt T=
Which yields 6
558770 10 0.1117
t =× ×
= 0.0007145 m or 1 mm (approx.).
Version 2 ME , IIT Kharagpur
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
MODULE III
Brakes, Clutches and Flywheel,
Brakes
• A brake is a device by means of which artificial resistance is applied on
to a moving machine member in order to retard or stop the motion of the
member or machine
Types of Brakes
• Different types of brakes are used in different applications
• Based on the working principle used brakes can be classified as
mechanical brakes, hydraulic brakes, electrical (eddy current) magnetic
and electro-magnetic types.
Mechanical Brakes
• Mechanical brakes are invariably based on the frictional resistance principles•
In mechanical brakes artificial resistances created using frictional contact
between the moving member and a stationary member, to retard or stop the
motion of the moving member.
Basic mechanism of braking
The illustration below explains the working of mechanical brakes. An element
dA of the stationary member is shown with the braked body moving past at
velocity v. When the brake is actuated contact is established between the
stationary and moving member and a normal pressure is developed in the
contact region. The elemental normal force dN is equal to the product of contact
pressure p and area of contact dA. As one member is stationary and the other is
in relative motion, a frictional force dF is developed between the members. The
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
magnitude of the frictional force is equal to the co-efficient of friction times the
normal force dN
dA
v dFf =µ.dN=µ.p.dA
dN=p.dA
Figue 3.1.1
The moment of the frictional force relative to the point of motion contributes to the
retardation of motion and braking. The basic mechanism of braking is illustrated
above.
Design and Analysis
To design, select or analyze the performance of these devices knowledge on the
following are required.
• The braking torque
• The actuating force needed
• The energy loss and temperature rise
At this beginning stage attention will be focused mainly on some preliminary
analysis related to these aspects, namely torque, actuating force, energy
absorbed and temperature rise. Torque induced is related to the actuating force,
the geometry of the member and other contact conditions. Most mechanical
brakes that work on the frictional contact basis are classified based on the
geometry.
There are two major classes of brakes, namely drum brakes and disc brakes.
Design and analysis of drum brakes will be considered in detail in following
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
sections, the discussion that follow on disc or plate clutches will from the basis
for design of disc type of brakes.
Drum brakes basically consists of a rotating body called drum whose motion is
braked together with a shoe mounted on a lever which can swing freely about a
fixed hinge H. A lining is attached to the shoe and contacts the braked body. The
actuation force P applied to the shoe gives rise to a normal contact pressure
distributed over the contact area between the lining and the braked body. A
corresponding friction force is developed between the stationary shoe and the
rotating body which manifest as retarding torque about the axis of the braked
body.
Brakes Classification
O
Lining
O
G
O
ωωωRotating body(drum)
ShoeStationary member
Rigid Pivoted
Short Shoe Long shoe
Figure 3.1.2
Various geometric configurations of drum brakes are illustrated above.
Drum Brakes are classified based on the shoe geometry. Shoes are classified as
being either short or long. A short shoe is one whose lining dimension in the
direction of motion is so small that contact pressure variation is negligible, i.e. the
pressure is uniform everywhere.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
When the area of contact becomes larger, the contact may no longer be with a
uniform pressure, in which case the shoe is termed as long shoe. The shoes are
either rigid or pivoted, pivoted shoes are also some times known as hinged
shoes. The shoe is termed rigid because the shoes with attached linings are
rigidly connected to the pivoted posts. In a hinged shoe brake - the shoes are
not rigidly fixed but hinged or pivoted to the posts. The hinged shoe is connected
to the actuating post by the hinge, G, which introduces another degree of
freedom
Preliminary Analysis
The figure shows a brake shoe mounted on a lever, hinged at O, having an
actuating force Fa, applied at the end of the lever. On the application of an
actuating force, a normal force Fn is created when the shoe contacts the rotating
drum. And a frictional force Ff of magnitude f.Fn, f being the coefficient of friction,
develops between the shoe and the drum. Moment of this frictional force about
the drum center constitutes the braking torque.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Fa
ro
(a) Brake assembly
Fa a
RyFf cRx o
b
(b) Free-body diagram
Fn
θ
ω
shoe
drum
Ry
Rx
T
Figure 3.1.3
Short Shoe Analysis
For a short shoe we assume that the pressure is uniformly distributed over the
contact area. Consequently the equivalent normal force Fn = p .A, where = p is
the contact pressure and .A is the surface area of the shoe. Consequently the
friction force Ff = f.Fn where f is the co-efficient of friction between the shoe
lining material and the drum material.
The torque on the brake drum is then,
T = f Fn. r = f.p.A.r
A quasi static analysis is used to determine the other parameters of braking.
Applying the equilibrium condition by taking moment about the pivot ‘O’ we can
write
O a n nM F a F b f F c 0= − + =∑
Substituting for Fn and solving for the actuating force, we get,
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Fa = Fn(b+-fc)/a
The reaction forces on the hinge pin (pivot) are found from a summation of
forces,
i.e.
F 0,R fp Ax x a= =
F 0,R p A Fy y a a= = −
Self- energizing
The principle of self energizing and leading and trailing shoes
With the shown direction of the drum rotation (CCW), the moment of the frictional
force f. Fn c adds to the moment of the actuating force, Fa
As a consequence, the required actuation force needed to create a known
contact pressure p is much smaller than that if this effect is not present. This
phenomenon of frictional force aiding the brake actuation is referred to as self-
energization.
Leading and trailing shoe
• For a given direction of rotation the shoe in which self energization is present is
known as the leading shoe
• When the direction of rotation is changed, the moment of frictional force now
will be opposing the actuation force and hence greater magnitude of force is
needed to create the same contact pressure. The shoe on which this is prevailing
is known as a trailing shoe
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Self Locking
At certain critical value of f.c the term (b-fc) becomes zero. i.e no actuation force
need to be applied for braking. This is the condition for self-locking. Self-locking
will not occur unless it is specifically desired.
Short and Long Shoe Analysis
• Foregoing analysis is based on a constant contact pressure p.
• In reality constant or uniform constant pressure may not prevail at all points of
contact on the shoe.
• In such case the following general procedure of analysis can be adopted
General Procedure of Analysis
• Estimate or determine the distribution of pressure on the frictional surfaces
• Find the relation between the maximum pressure and the pressure at any point
• For the given geometry, apply the condition of static equilibrium to find the
actuating force, torque and reactions on support pins etc.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Drum Brakes
Among the various types of devices to be studied, based on their practical use,
the discussion will be limited to “Drum brakes” of the following types which are
mainly used in automotive vehicles and cranes and elevators.
Drum Brake Types:
• Rim types with internal expanding shoes
• Rim types with external contracting shoes
Internal expanding Shoe
The rim type internal expanding shoe is widely used for braking systems in
automotive applications and is generally referred as internal shoe drum brake.
The basic approach applied for its analysis is known as long-rigid shoe brake
analysis. Long –rigid Shoe Analysis
• A schematic sketch of a single shoe located inside a rotating drum with relevant
notations, is shown in the figure below. In this analysis, the pressure at any point
is assumed to be proportional to the vertical distance from the hinge pin, the
vertical distance from the hinge pin, which in this case is proportional to sine of
the angle and thus,
p d sin sin∝ θ ∝ θ
Since the distance d is constant, the normal pressure at any point is just
proportional to sinΘ. Call this constant of proportionality as K
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
>
><
X
Y
RX
dNcos
d dN
N Ndf
sindNcos
df NsinF
FX
FY
RY
d
θ θ
θθf
θ
dθ
Figure 3.1.4
Thus p K sin= θ
It the maximum allowable pressure for the lining material is pmax then the constant
K can be defined as
max
max
ppKsin sin
= =θ θ
max
max
pp s
sinin= θ
θ
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
• The normal force dN is computed as the product of pressure and area and the
frictional force as the product of normal force and frictional coefficient i.e. f dN.•
By integrating these over the shoe length in terms of its angle the braking torque
T, and other brake parameters are computed.
To determine the actuating force F, the moment equilibrium about the pivot point
is applied. For this we need to determine the moment of the normal force MN and
moment of the frictional force about the pivot point. Moment of the normal force
is equal to the normal force times its moment arm about the pivot point. From
the figure it is clear that the moment arm in this case is equal to d sin Θ where d
is the distance between the drum center and pivot center
1 1
2 2
1
2
2max
M p.b.r.d .d sin b.p.r.d.sin .dN
p = b.d.r. sin d
sin
θ θ
θ θ
θ
θ
= θ θ = θ
θ θθ
∫ ∫
∫
θ
( )p b.d.r 1 1maxM (sin 2 sin 2 )N 2 1 2sin 2 4a1
⎡ ⎤= θ − θ − θ − θ⎢ ⎥θ ⎣ ⎦
On similar lines the moment of friction force is computed
( )
( )
1
2
1
2
max
max
M f.p.b.r.d r d sinF
p = f .b.r. sin r d sin d
sin
θ
θ
θ
θ
= θ − θ
θ − θθ
∫
∫ θ
( ) ( )max.
a
f .p b.r d 2 2M r cos cos sin sif 2 1 2sin 2⎡ ⎤= − θ − θ − θ − θ⎢ ⎥θ ⎣ ⎦
n 1
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
The actuating force F is determined by the summation of the moments of normal
and frictional forces about the hinge pin and equating it to zero.
Summing the moment about point O gives
M MN fFc
±=
where,
• MN and Mf are the moment of the normal and frictional forces respectively,
about the shoe pivot point.
The sign depends upon the direction of drum rotation,
(- sign for self energizing and + sign for non self energizing shoe)Where the
lower sigh is for a self energizing shoe and the upper one for a self deenergizing
shoe.
The reaction forces are determined by applying force summation and equilibrium
( ) ( )
2 2
1 1
2 2
1 1
2max max
max max
max.b
max
R dN.cos dF.sinx
b.r.p cos d f b.r.psin d
p p b.r. sin cos d f b.r. sin d
sin sin
p r 1 1 1 1 2 2( ) sin 2 sin 2 f sin sin2 1 2 1 2 1sin 2 2 4 2
θ θ
θ θ
θ θ
θ θ
= θ + θ
= θ θ + θ θ
= θ θ θ + θ θθ θ
⎛ ⎞⎡ ⎤= θ − θ − θ − θ ± θ −⎜ ⎟⎢ ⎥θ ⎣ ⎦⎝ ⎠
∫ ∫
∫ ∫
∫ ∫
θ
The equations can be simplified and put as
p braR (Ax sin a
=θ
∓ fB)
p braR (B fA)y xsin aF= ± −
θ
Where
( )1 2 2A sin sin2 12= θ − θ
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
1 1 1B ( ) sin 2 sin 22 2 42 1 2 1⎡ ⎤⎛ ⎞= θ − θ − θ − θ⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
The braking torque T on the drum by the shoe is of the frictional forces f.dN times
the radius of the drum and resulting equation is,
1
1
1
1
2
max
T f .b.p.r.d .r
pmax = fbr sin dsin
θ
θ
θ
θ
= θ
θ θθ
∫
∫
2fdp r (cos cos )a 1Tsin a
2θ − θ=
θ
T
Double Shoe Brakes Twin Shoe Brakes
Behavior of a single shoe has been discussed at length. Two
such shoes are combined into a complete practical brake unit,
two being used to cover maximum area and to minimize the
unbalanced forces on the drum, shaft and bearings.
• If both the shoes are arranged such that both are leading shoes in which
self energizing are prevailing, then all the other parameters will remain
same and the total braking torque on the drum will be twice the value
obtained in the analysis.
• However in most practical applications the shoes are arranged such that
one will be leading and the other will be trailing for a given direction of
drum rotation
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
• If the direction of drum rotation changes then the leading shoe will
become trailing and vice versa.
• Thus this type of arrangement will be equally effective for either direction
of drum rotation. Further the shoes can be operated upon using a single
cam or hydraulic cylinder thus provide for ease of operation
One leading shoe & one trailing
Two Leading shoe
Figure 3.1.5
However the total braking torque will not be the twice the value of a single
shoe, if the same normal force is applied or created at the point of force
application on both the brake shoes which is the normal practice as they
are actuated using a common cam or hydraulic cylinder.
• This is because the effective contact pressure (force) on the trailing shoe
will not be the same, as the moment of the friction force opposes the
normal force, there by reducing its actual value as in most applications
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
the same normal force is applied or created at the point of force
application on the brake shoe as noted above
• Consequently we may write the actual or effective pressure prevailing on
a trailing shoe
F.a'p p .a a (M M )n f
⎡ ⎤= ⎢ ⎥
+⎢ ⎥⎣ ⎦
Resulting equation for the braking torque
p2 aT f .w.r . (cos - cos )(p pB 1 2sin a= θ θ
θ')a a+
Some pictorial illustrations of the automotive drum brakes are presented
below
Figure 3.1.6
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Figure 3.1.7
Oblong Cam Actuator
Leading shoe
Rotating Drum
Pivot point(Fixed axis)
The anatomy of the single leading shoe drum Brake
Trailing shoe
Animation
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Figure 3.1.9
Figure 3.1.10
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
External Contracting Shoe
• The same analysis can be extended to a drum brake with external contracting
type of shoes, typically used in elevators and cranes.
A schematic sketch of as single shoe located external to the rotating drum is with
all relevant notations is shown in the figure below.
Figure 3.1.11
• Corresponding contact geometry is shown in the figure
• The resulting equations for moment of normal and frictional force as well as the
actuating force and braking torque are same as seen earlier.
• For convenience they are reproduced here again
( )2fbp r cos cosa 1Tsin a
2θ − θ=
θ
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
M MN fFc
±=
( )p bra 1 1aM (sin 2 sin 2 )N 2 1 2sin 2 4a1
⎡ ⎤= θ − θ − θ − θ⎢ ⎥θ ⎣ ⎦
( )a
a
fp br d 2 2M r cos cos (sin sinf 1 2 2sin 2⎡ ⎤= θ − θ − θ −⎢ ⎥θ ⎣ ⎦
)1θ
TWIN SHOE BRAKES
As noted earlier for the internal expanding shoes, for the double shoe brake the
braking torque for one leading and one trailing shoe acted upon a common cam
or actuating force the torque equation developed earlier can be applied.
i.e p2 aT f .w.r . (cos - cos )(p pB 1 2sin a
= θ θθ
')a a+
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Brake with a pivoted long shoe
When the shoe is rigidly fixed to the lever, the tendency of the
frictional force (f.Fn) is to unseat the block with respect to the
lever. This is eliminated in the case of pivoted or hinged shoe
brake and it also provides some additional advantages.
Long Hinged Shoe
In a hinged shoe brake - the shoes are not rigidly fixed but
hinged or pivoted to the posts. The hinged shoe is connected to
the actuating post by the hinge, G, which introduces another
degree of freedom - so the shoe tends to assume an optimum
position in which the pressure distribution over it is less peaked
than in a rigid shoe.
Bo
H
FF
G
OH=a
OG=b
X
BG
φ
y
x
Figure 3.1.12
As wear proceeds the extra degree of freedom allows the linings
to conform more closely to the drum than would be the case to
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
rigid shoes. This permits the linings to act more effectively and
also reduces the need for wear adjustment.
The extra expense of providing another hinge is thus justified on
the grounds of more uniform lining wear and consequently a
longer life. This is the main advantage of the pivoted shoe brake
This is possible only if the shoe is in equilibrium.
For equilibrium of the shoe moments of the forces
about the hinge pin should balance -
i.e ΣMG=T+Fxby-Fybx=0 where bx = b.cos θG
by = b.sinθG
This needs that the resultant moment due to the frictional force
(and due to the normal force) about the pivot point should be
zero, so that no rotation of the shoe will occur about the pivot
point. To facilitate this location of the pivot is to be selected
carefully.
The actuating force P is applied to the post HG so the shoe itself
is subject to actual and ideal contacts only - the (ideal) at pin G
and the actual as distributed contact with the drum.
The location is in such a way that the moment of frictional force
(and the normal force) about the pivot is zero. i.e the actual
distributed contact leads to the ideal (concentrated)contact at the
hinge or pivot.
i.e the actual distributed contact leads to the ideal contact at the
hinge or pivot Further it is desirable to minimize the effect of pin
reaction for which the shoe pivot and post pivot points are made
con current.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Let us now look how this can be met, satisfying the conditions set above and
consequently the derive the equations relating the location of the pivot from the
center of the drum
A schematic sketch of a single shoe is shown in the figure
dN cosθrcosθ
h
θ
fdN
cosfdN θ
θ
(hcosθ−r)
Rx
yR
fdN sin θ
dNr
Force acting on shoe
An element of friction lining located at an angle Θ and
subtending to a small angle dθ is shown in figure. The area if
the element is ( )r.d .bθ , where b is the width of the friction lining
parallel to the axis of the brake drum. If the intensity of pressure
at the element is p, the normal reaction dN on the element is
given by
dN (rd b)p= θ (a)
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Distribution of pressure
If the shoe is long then the pressure will not be uniform
We need to determine the distribution of pressure along the
lining; the pressure distribution should be conducive for
maintaining a uniform wear
Since the brake drum is made of a hard material like cast iron or
steel, the wear occurs on the friction lining, which is attached to
the shoe. As shown in fig the lining need to retain the cylindrical
shape of the brake drum when wear occurs. After the radial
wear has take place, a point such as X’ moves to X in order to
maintain contact on the lining with the brake drum. In figure δx is
the wear in the X direction and δr is the wear in the radial
direction. If it is assumed that the shoe is constrained to move
towards the brake drum to compensate to wear, δx should be
constant because it need to be same for all points. Therefore,
rxcosδ
δ =θ
= constant (b)
θ
Y
X
x
δx
x'θ
δx
wear
wear of friction lining
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
The radial wear δr is proportional to the work done by the
frictional force. The work done by the frictional force depends
upon the frictional force ( fdN ) and the rubbing velocity. Since the
rubbing velocity is constant for all points on friction lining,
r fdNδ ∝
Or ( )r frd bpδ ∝ θ
Therefore (c) r pδ ∝
From the expression (b) and (c)
1p cons tan t or p C cos
cos= =
θθ
C
(d)
Where C1 is the constant of proportionality. The pressure is
maximum when . 0θ =
Substituting,
max 1p = (e)
From Eqs (d) and (e),
maxp p cos= θ
Substituting this value in Eq. (a
maxdN (rd b)p cos= θ θ (f)
The forces acting on the element of the friction lining are shown
in figure. The distance h of the pivot is selected in such a manner
that the moment of frictional force about it is zero.
Therefore, fM 0==
dMf=f.dN moment arm
moment arm in this case = (h cos r)θ −
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
fM fdN(h cos r)θ
−θ
0= θ − =∫
f maxM fp rd cos (h r cos )θ
−θ
= θ θ −∫ θ
Substituting dN from Eq. (f),
( )
( )
2
0
0 0
0
0
h cos r cos d 0
1 cos 2or h d r cos d 02
1 sin 22or h R sin 0
2
4R sin h2 sin 2
θ
θ θ
θ
θ
θ − θ θ =
+ θ⎡ ⎤ θ − θ θ =⎢ ⎥⎣ ⎦
⎡ ⎤φ + φ⎢ ⎥− φ =⎢ ⎥
⎢ ⎥⎣ ⎦
θ=
θ + θ
∫
∫ ∫
The elemental torque of frictional force fdN about the axis of
brake drum is fdNR . Therefore
BT 2 fdNθ
−θ
= r∫
Substituting the value of dN from Eq.(f)
2B max
2B max
T 2fr bp cos d
T 2fr bp sin
θ
−θ
= θ θ
= θ
∫
The reaction can be determined by considering two components XR
( ) ( )dN cos and fdN sinθ θ .
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Due to symmetry, the other two vertical components of the force
balances
i.e
fdN sin 0
dN sin 0
θ =
θ =
∫∫
Therefore,
x
2max
max
x max
R dN cos
rbp cos d
2 sin 2 2rbp4
1or R rbp (2 sin 2 )2
θ
−θθ
−θ
= θ
= θ θ
θ + θ⎡ ⎤= ⎢ ⎥⎣ ⎦
= θ + θ
∫
∫
Note that is also = xR nF
The reaction yR can be determined by considering two
components ( ) ( )dN sin and fdN cosθ θ
Due to symmetry,
=0 dN sin∫ θ
Therefore,
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
y
2max
y max
R fdN cos
frbp cos d
1or R frbp (2 sin 2 )2
θ
−θθ
−θ
= θ
= θ θ
= θ + θ
∫
∫
As noted earlier,
2B maT 2fr bp si= θx n
Rewriting it,
B max
n
2 sin 4r sinT frbp .2 2 sin
= fF h
θ + θ θ=
θ + θ
DOUBLE BRAKE SHOE
A double block brake with two symmetrical and pivoted shoes is show in
figure.
If the same magnitude of actuating forces are acted upon the posts, then
B n1 n2 nT f .(F F ).h 2f .F .h= + =
Θ
Q
P P
Pivoted double block brake
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Pivoted shoe brakes are mainly used in hoists and cranes. Their
applications are limited because of the physical problem in
locating pivot so close to the drum surface.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Energy Consideration
It has been noted that the most common brakes employ friction to transform
the braked system's mechanical energy, irreversibly into heat which is then
transferred to the surrounding environment -
• Kinetic energy is absorbed during slippage of either a clutch or brake, and this
energy appears as heat.
• If the heat generated is faster than it is dissipated, then the temperature rises.
Thorough design of a brake therefore requires a detailed transient thermal
analysis of the interplay between heat generated by friction, heat transferred
through the lining and the surrounding metalwork to the environment, and the
instantaneous temperature of the surface of the drum as well as the lining. For
a given size of brake there is a limit to the mechanical power that can be
transformed into heat and dissipated without lthe temperatures reaching
damaging levels. Temperature of the lining is more critical and the brake size is
characterized by lining contact area, A.
The capacity of a clutch or brake is therefore limited by two factors:
1. The characteristics of the material and,
2. The ability of the brake to dissipate heat.
Heat Generated In Braking
During deceleration, the system is subjected to an essentially constant torque
T exerted by the brake, and in the usual situation this constancy implies constant
deceleration too.
Application of the work or energy principle to the system enables the torque
exerted by the brake and the work done by the brake, U, to be calculated from:-
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
U = ∆E = T ∆θ (2)
Where ∆E is the loss of system total energy which is absorbed by the brake
during deceleration, transformed into heat, and eventually dissipated.
The elementary equations of constant rotational deceleration apply, thus when
the brake drum is brought to rest from an initial speed ωo :-
Deceleration = ωo2/ 2 (1) ∆θ = ωm.∆t ; ωm = ωo/2 =
∆θ/∆t where ωm is the mean drum speed over the deceleration period.
The mean rate of power transformation by the brake over the braking period is
:-
Pm = U / ∆t = T ωm ( 3 ) which forms a basis for the selection or
the design of the necessary brake dimensions.
The rise in temperature in the lining material is also important as rate of wear is
also a function of the temperature. Further for any lining material, the maximum
allowable temperature is also another performance criteria.
Temperature Rise
The temperature rise of the brake assembly can be approximated by the classic
expression,,
ETC.m
∆ =
Where is temperature ∆T is rise in temperature in oC,’C’ is the specific heat of
the brake drum material – (500J/Kg for steel or Cast Iron) and m is the mass
(kg) of the brake parts dissipating the heat into the surroundings.
Though the equation appears to be simple, there are so many variables involved
that it would be most unlikely that such an analysis would even approximate
experimental results.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
On the other hand the temperature-rise equations can be used to explain what
happens when a clutch or brake is operated frequently. For this reason such
analysis are most useful, for repetitive cycling, in pin pointing those design
parameters that have the greatest effect on performance.
An object heated to a temperature T1 cools to an ambient temperature Ta
according to the exponential relation
Time-temperature relation
(AU / WC)tT T (T T )ei a 1 a−− = −
Where T1 = instantaneous temperature at time t, ˚C
A= heat transfer area, m2
U= Heat Transfer coefficient, W/(m2.s. ˚C)
T1 = Initial temperature, ˚C
Ta = Ambient temperature, ˚C
C - Specific heat
t - time of operation, s
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
T
T
t t t
C
B
T
A
T
A
a
B C
1
T2
Time t
Figure 3.1.16
Figure shows an application of Eq. (a). At time tA a clutching or braking
operation causes the temperature to rise to T1 at A. Though the rise occurs in a
finite time interval, it is assumed to occur instantaneously. The temperature then
drops along the decay line ABC unless interrupted by another braking operation.
If a second operation occurs at time tB, the temperature will rise along the dashed
line to T2 and then begin an exponential drop as before. About 5 -10 % of the
heat generated at the sliding interface of a friction brake must be transferred
through the lining to the surrounding environment without allowing the lining to
reach excessive temperatures, since high temperatures lead to hot spots and
distortion, to fade (the fall-off in friction coefficient) or, worse, to degradation and
charring of the lining which often incorporates organic constituents
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
In order to determine the brake dimensions the energy need to be absorbed
during critical braking conditions is to be estimated.
Energy to be Absorbed
If t is the time of brake application and ωm the mean or average angular velocity
then the energy to be absorbed in braking E
E = T. ωm .t = Ek+ Ep+ Ei
where Ek is the kinetic energy of the rotating system
Ep is the potential energy of the moving system
Ei is the inertial energy of the system
Energy to be absorbed
( )
k
2
2 22 1
1E2 g1 mv21 v v2 g
ω=
=
ω= −
( )
p
2i
2 22 1
E mgh h
1E I21 I2
= = ω
= ω
= ω − ω
Frictional Material
A brake or clutch friction material should have the following characteristics to a
degree, which is dependent upon the severity of the service.
• A high and uniform coefficient of friction.
• Imperviousness to environmental conditions, such as moisture.
• The ability to withstand high temperatures together with good thermal
conductivity.
• Good resiliency.
• High resistance to wear, scoring, and galling.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Linings
The choice of lining material for a given application is based upon criteria such as
the expected coefficient of friction; fade resistance, wear resistance, ease of
attachment, rigidity or formability, cost, abrasive tendencies on drum, etc. The
lining is sacrificial - it is worn away. The necessary thickness of the lining is
therefore dictated by the volume of material lost - this in turn is the product of the
total energy dissipated by the lining throughout its life, and the specific wear rate
Rw (volume sacrificed per unit energy dissipated) which is a material property
and strongly temperature dependent. The characteristics of a typical moulded
asbestos lining material is illustrated in the figure below. The coefficient of
friction, which may be taken as 0.39 for design purposes, is not much affected by
pressure or by velocity - which should not exceed 18 m/s. The maximum
allowable temperature is 400°C. However at this temperature the wear is very
high. From a lower wear or higher life point, the maximum temperature should
not exceed about 200 oC
Temperature ( C ) 4000
0
0.5
o
Figure 3.1.17
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Linings traditionally were made from asbestos fibers bound in an organic matrix,
however the health risks posed by asbestos have led to the decline of its use.
Non-asbestos linings generally consist of three components - metal fibers for
strength, modifiers to improve heat conduction, and a phenolic matrix to bind
everything together.
Brake Design Section
The braked system is first examined to find out the required brake capacity that is
the torque and average power developed over the braking period. - The brake
is then either selected from a commercially available range or designed from
scratch ff a drum brake has to be designed for a particular system (rather than
chosen from an available range) then the salient brake dimensions may be
estimated from the necessary lining area, A, together with a drum diameter- to-
lining width ratio somewhere between 3:1 and 10:1, and an angular extent of 100
°C say for each of the two shoes.
Worked out Example 1
An improved lining material is being tried on an existing passenger car drum
brake shown in Figure. Quality tests on the material indicated permissible
pressure of 1.0 MPa and friction co-efficient of 0.32. Determine what maximum
actuating force can be applied for a lining width of 40 mm and the corresponding
braking torque that could be developed. While cruising on level road at 100
kmph, if it is to decelerated at 0.5g and brought to rest, how much energy is
absorbed and what is the expected stopping distance?
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
While cruising on level road at 100 kmph, if it is to decelerated at 0.5g and
brought to rest, how much energy is absorbed and what is the expected stopping
distance?
300
300
0
R=125
0
100.9
50 50300 0
5
AUTOMOTIVE DOUBLE SHOE BRAKE
05
30
F F
120 120Pin Pin
0
300
86.6
Figure 3.1.18
Analysis based on leading shoe
a
2 2
P 1 MPa f 0.32 a 187.5 mm 0b 40 mm 90 d = 100 +86.1max
= = =
= θ =
= 99.99 100mm0 05 120 r =125 mm1 2
≈
θ = θ =
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
( )
( )
a 2 1n 2 1
a6 -3 3
n
p rd 1M (sin 2 sin 2 )
sin 2 4
10 *40*10 *125 *10 * 0.1 115 1 = * sin 240 sin10
1 2 180 41
=40*125*0.1 1.003- 1.034
M 631.459 N.m
b
−
θ − θ= − θ − θ
θ
π− ° − °
−
=
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥⎣ ⎦
( )f .b.rpm d 2 2M r(cos - cos ) - sin - sinf 1 2 2 1sin 2a-3 -3 6 2 20.32*40*10 *125*10 *10 0.125(cos5 - cos120 ) - 0.04(sin 120 - sin 5 )
M 224.85 N - mf
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥⎣ ⎦
= θ θ θ θθ
= ° °
=
° °
F*a= N fM M−
n fM M 631.459 224.85F 2174.3Na 0.187 Max. actutating force
− −= = =
( )2 a a
B 1 2
a n f
3 2 6
B
B
p FT fbr cos cos ) 1
sin M M
406.609T 0.32 * 40 *10 * (0.125) *10 (Cos5 Cos120) 1856.36
T 441.329 N-m
−
⎡ ⎤= θ − θ +⎢ ⎥α +⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦=
Running at 100 kmph =100*5/8
= 27.7 m/s
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
U= 27.7 m/s
Deceleration =0.5*9.8=4.9
2 2
2
2
V U 2aS
0 (27.7) 2* ( 4.9) *S
27.7S 78.29 m2* 4.9
− =
− = −
= =
E T. .tav1 27.7 78.24 441.329 .2 0.125 27.7
138206 138.2KJ
= ω
⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
==
Worked out example 2
A spring set, hydraulically released double shoe drum brake, schematically
shown at Fig 2 is to be designed to have a torque capacity of 600 N.m under
almost continuous duty when the brake drum is rotating at 400 rpm in either
direction. Assume that the brake lining is to be molded asbestos having a
friction coefficient of 0.3 and permissible pressure of 0.8 MPa. The width of the
brake shoe is to be third of drum diameter and the remaining proportion's are as
shown in figure.
0
1 413
0 6 0 69330
=
=
= =
a . D
b D
. Dd .cos
D
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Double block brake
90
30ο
ο1.4 D
0.6 D
Figure 3.1.19
Determine the required brake drum diameter, width of the lining and the spring
force required to be set.
0
a 1.4D1b D30.6Dd 0.693D
cos30
=
=
= =
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
( )
( ) ( )( ) ( )
1
1
6
3N
aJ
2 2a 1 2 2
6
3
1 D sin(210 ) sin 300.8*10 * D * * 0.6028D.3 2 4 4
M 92373.3D *1.035
P brfM r a cos sin d
1
d P brf r cos cos sin sin2
D D D 0.6928 0.8*10 * * * 0.3 1.225 0.8663 2 2 2
12500D
θ
θ
1
π ° − °⎡ ⎤= −⎢ ⎥⎣ ⎦
=
= − θ θ θ
⎡ ⎤= θ − θ − θ −⎢ ⎥⎣ ⎦θ
⎡ ⎤= −⎢ ⎥⎣ ⎦
=
∫
a
' N Fa a
N F
a
a
P 'for the trailing shoe
M MP P
M M
95606.36 12500 P95606.36 12500
0.7687P
⎛ ⎞−= ⎜ ⎟+⎝ ⎠
−⎛ ⎞= ⎜ ⎟+⎝ ⎠=
Torque due to trailing and leading shoe=
( ) ( )
2 2
1 1
a
a
2
a aa
2
a a 1 2a
fdN.r
P sin f .r .rd .b
sin
fbr P sin d P ' sin dsin
fbr . P P ' . cos cossin
θ θ
θ θ
τ =
θ= θ
θ
⎡ ⎤⎢ ⎥= θ θ +
θθ θ
⎢ ⎥⎣ ⎦
= + θ −θ
∫
∫
∫ ∫
θ
( )
total
2
6
3
D D0.3* *3 4 1.7687 *0.8*10 cos15 cos1.51
43324 DT 600 Nm
Therefore D 240.14mm
⎡ ⎤= °⎣ ⎦
==
=
− °
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Actuating force due to spring
2N FM M 83106 * 0.240F 3423.22N
1.4 * D 1.4−
= = =
Actuating force =3423.22 N
DLiving width3
80.04 mm
=
=
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
CLUTCH
Clutch Introduction
A Clutch is ia machine member used to connect the driving shaft to a driven
shaft, so that the driven shaft may be started or stopped at will, without stopping
the driving shaft. A clutch thus provides an interruptible connection between two
rotating shafts
Clutches allow a high inertia load to be stated with a small power.
A popularly known application of clutch is in automotive vehicles where it is used
to connect the engine and the gear box. Here the clutch enables to crank and
start the engine disengaging the transmission Disengage the transmission and
change the gear to alter the torque on the wheels. Clutches are also used
extensively in production machinery of all types
Mechanical Model
Two inertia’s and traveling at the respective angular velocities ωI and I1 2 1
and ω2, and one of which may be zero, are to be brought to the same speed
by engaging. Slippage occurs because the two elements are running at different
speeds and energy is dissipated during actuation, resulting in temperature rise.
ω1
Ι1 Ι1
ω2
Clutch or brake
Dynamic Representation of Clutch or Brake
Figure 3.2.1
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Animated
Figure 3.2.2
To design analyze the performance of these devices, a knowledge on the
following are required.
1. The torque transmitted
2. The actuating force.
3. The energy loss
4. The temperature rise
FRICTION CLUTCHES
As in brakes a wide range of clutches are in use wherein they vary in their are in
use their working principle as well the method of actuation and application of
normal forces. The discussion here will be limited to mechanical type friction
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
clutches or more specifically to the plate or disc clutches also known as axial
clutches
Frictional Contact axial or Disc Clutches
An axial clutch is one in which the mating frictional members are moved in a
direction parallel to the shaft. A typical clutch is illustrated in the figure below. It
consist of a driving disc connected to the drive shaft and a driven disc
co9nnected to the driven shaft. A friction plate is attached to one of the
members. Actuating spring keeps both the members in contact and
power/motion is transmitted from one member to the other. When the power of
motion is to be interrupted the driven disc is moved axially creating a gap
between the members as shown in the figure.
Figure 3.2.3
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Flywheel
Clutch plate
Pressure plate
Clutch cover
Diaphragmspring
to transmission
Throw outBearing
Animated
Figure 3.2.4
METHOD OF ANALYSIS
The torque that can be transmitted by a clutch is a function of its geometry and
the magnitude of the actuating force applied as well the condition of contact
prevailing between the members. The applied force can keep the members
together with a uniform pressure all over its contact area and the consequent
analysis is based on uniform pressure condition
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Uniform Pressure and wear
However as the time progresses some wear takes place between the contacting
members and this may alter or vary the contact pressure appropriately and
uniform pressure condition may no longer prevail. Hence the analysis here is
based on uniform wear condition
Elementary Analysis
Assuming uniform pressure and considering an elemental area dA
dA = 2Π.r dr
The normal force on this elemental area is
dN 2 .r.dr.p= π
The frictional force dF on this area is therefore
dF f .2 .r.dr.p= π
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
<
< >do
Fr
di
dr
lining
A single-Surface Axial Disk Clutch
Figure 3.2.5
Now the torque that can be transmitted by this elemental are is equal to the
frictional force times the moment arm about the axis that is the radius ‘r’
i.e. T = dF. r = f.dN. r = f.p.A.r
= f.p.2.π.r. dr .r
The total torque that could be transmitted is obtained by integrating this equation
between the limits of inner radius ri to the outer radius ro
ro 22 3T 2 pfr dr pf (r r )o i3ri
= π = π −∫ 3
Integrating the normal force between the same limits we get the actuating force
that need to be applied to transmit this torque.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
( )
a
2 2a o i
roF 2 prdr
riF r r
= π∫
= π − .p
Equation 1 and 2 can be combined together to give equation for the torque
3 3o i
a 2 2o i
(r r )2T fF .3 (r r )
−=
−
Uniform Wear Condition
According to some established theories the wear in a mechanical system is
proportional to the ‘PV’ factor where P refers the contact pressure and V the
sliding velocity. Based on this for the case of a plate clutch we can state
The constant-wear rate Rw is assumed to be proportional to the
product of pressure p and velocity V.
Rw= pV= constant
And the velocity at any point on the face of the clutch is V r.= ω
Combining these equation, assuming a constant angular velocity ω
pr = constant = K
The largest pressure pmax must then occur at the smallest radius ri ,
max iK p r=
Hence pressure at any point in the contact region
imax
rp p
r=
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
In the previous equations substituting this value for the pressure term p and
integrating between the limits as done earlier we get the equation for the torque
transmitted and the actuating force to be applied.
I.e The axial force Fa is found by substituting imax
rp p
r= for p.
and integrating equation dN 2 prdr= π
r ro o riF 2 prdr 2 p rdr 2 p r (r r )max max i o irr ri i
⎛ ⎞= π = π = π −⎜ ⎟∫ ∫ ⎜ ⎟
⎝ ⎠
Similarly the Torque
ro 2 2T f 2 p r rdr f p r (r rmax i max i o iri
= π = π −∫ )
Substituting the values of actuating force Fa
The equation can be given as
(r r )o iT fF .a 2
+=
Single plate dry Clutch – Automotive application
The clutch used in automotive applications is generally a single plate dry clutch.
In this type the clutch plate is interposed between the flywheel surface of the
engine and pressure plate.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Flywheel
Frictionplanes
Clutch plate(driven disk)
Pressure platePressure spring
HousingReleasebearing
Enginecrankshaft
To release
Totransmission
Figure 3.2.6
Single Clutch and Multiple Disk Clutch
Basically, the clutch needs three parts. These are the engine flywheel, a friction
disc called the clutch plate and a pressure plate. When the engine is running
and the flywheel is rotating, the pressure plate also rotates as the pressure plate
is attached to the flywheel. The friction disc is located between the two. When
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
the driver has pushed down the clutch pedal the clutch is released. This action
forces the pressure plate to move away from the friction disc. There are now air
gaps between the flywheel and the friction disc, and between the friction disc and
the pressure plate. No power can be transmitted through the clutch.
Operation Of Clutch
When the driver releases the clutch pedal, power can flow through the clutch.
Springs in the clutch force the pressure plate against the friction disc. This action
clamps the friction disk tightly between the flywheel and the pressure plate. Now,
the pressure plate and friction disc rotate with the flywheel.
As both side surfaces of the clutch plate is used for transmitting the torque, a
term ‘N’ is added to include the number of surfaces used for transmitting the
torque
By rearranging the terms the equations can be modified and a more general form
of the equation can be written as
T N.f .F .Ra m=
T is the torque (Nm).
N is the number of frictional discs in contact.
f is the coefficient of friction
Fa is the actuating force (N).
Rm is the mean or equivalent radius (m).
Note that N = n1 + n2 -1
Where n1= number of driving discs
n2 = number of driven discs
Values of the actuating force F and the mean radius for the two conditions of
analysis are summarized and shown in the table
mr
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Clutch Construction
Two basic types of clutch are the coil-spring clutch and the diaphragm-spring
clutch. The difference between them is in the type of spring used. The coil
spring clutch shown in left Fig 3.2.6 uses coil springs as pressure springs (only
two pressure spring is shown). The clutch shown in right figure 3.2.6 uses a
diaphragm spring.
Figure 3.2.6
The coil-spring clutch has a series of coil springs set in a circle.
At high rotational speeds, problems can arise with multi coil spring clutches
owing to the effects of centrifugal forces both on the spring themselves and the
lever of the release mechanism.
These problems are obviated when diaphragm type springs are used, and a
number of other advantages are also experienced
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Clutch or Driven Plate
More complex arrangements are used on the driven or clutch plate to facilitate
smooth function of the clutch
The friction disc, more generally known as the clutch plate, is shown partly cut
away in Fig. It consists of a hub and a plate, with facings attached to the plate.
Figure 3.2.7
First to ensure that the drive is taken up progressively, the centre plate, on which
the friction facings are mounted, consists of a series of cushion springs which is
crimped radially so that as the clamping force is applied to the facings the
crimping is progressively squeezed flat, enabling gradual transfer of the force
On the release of the clamping force, the plate springs back to its original
position crimped (wavy) state
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
This plate is also slotted so that the heat generated does not cause distortion that
would be liable to occur if it were a plain plate. This plate is of course thin to keep
rotational inertia to a minimum.
Plate to hub Connection
Secondly the plate and its hub are entirely separate components, the drive being
transmitted from one to the other through coil springs interposed between them.
These springs are carried within rectangular holes or slots in the hub and plate
and arranged with their axes aligned appropriately for transmitting the drive.
These dampening springs are heavy coil springs set in a circle around the hub.
The hub is driven through these springs. They help to smooth out the torsional
vibration (the power pulses from the engine) so that the power flow to the
transmission is smooth.
In a simple design all the springs may be identical, but in more sophisticated
designs the are arranged in pairs located diametrically opposite, each pair having
a different rate and different end clearances so that their role is progressive
providing increasing spring rate to cater to wider torsional damping
The clutch plate is assembled on a splined shaft that carries the rotary motion to
the transmission. This shaft is called the clutch shaft, or transmission input shaft.
This shaft is connected to the gear box or forms a part of the gear box.
Friction Facings or Pads
It is the friction pads or facings which actually transmit the power from the fly
wheel to hub in the clutch plate and from there to the out put shaft. There are
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
grooves in both sides of the friction-disc facings. These grooves prevent the
facings from sticking to the flywheel face and pressure plate when the clutch is
disengaged. The grooves break any vacuum that might form and cause the
facings to stick to the flywheel or pressure plate. The facings on many friction
discs are made of cotton and asbestos fibers woven or molded together and
impregnated with resins or other binding agents. In many friction discs, copper
wires are woven or pressed into the facings to give them added strength.
However, asbestos is being replaced with other materials in many clutches.
Some friction discs have ceramic-metallic facings.
Such discs are widely used in multiple plate clutches
The minimize the wear problems, all the plates will be enclosed in a covered
chamber and immersed in an oil medium
Such clutches are called wet clutches
Multiple Plate Clutches
Figure 3.2.8
The properties of the frictional lining are important factors in the design of the
clutches
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Typical characteristics of some widely used friction linings are given in the table
Table Properties of common clutch/ Brake lining materials
Friction MaterialAgainst Steel or Cl
Dynamic Coefficient of Friction
Maximum Pressure Maximum Temprerature
Molded
Woven
Sintered metal
Cast iron of hard steel
0.25-0.45
0.25-0.45
0.15-0.45
0.15-0.25
0.06-0.09
0.08-0.10
0.05-0.08
0.03-0.06
1030-2070
345-690
1030-2070
690-720
204-260
204-260
232-677
260
dry in oilKPa oC
Table 3.2.1
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Energy considerations
Kinetic energy is absorbed during slippage of a clutch and this energy appears
as heat.
The clutch or brake operation is completed at the instance in which the two
angular velocities θ1 and θ2 become equal. Let the time required for the entire
operation be t1, then,
( )( )
I I1 2 1 2t1 T I I1 2
ω −ω=
+
This is derived by writing the equations of motion involving inertia
i.e
1 11 11
2 22 22
TI T tI
TI T tI
• • •θ = − θ = − + ω
•• •θ = − θ = − + ω
1 21 2
1 21 2
1 2
T Tt t1 2 I I
I IT t
I I
• • • ⎛ ⎞θ = θ − θ = − + ω − + ω⎜ ⎟
⎝ ⎠⎛ ⎞+
= ω −ω − ⎜ ⎟⎝ ⎠
from which ( )( )
1 2 1 2
1 2
I It
T I Iω −ω
=+
as at the instance of completion of clutching
operation ω1-ω2 = 0
Assuming the torque to be constant, the rate of energy dissipation during the
operation is then,
1 21 2
1 2
I IU T T T .t
I I⎡ ⎤⎛ ⎞+•= θ = ω − ω −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
The total energy dissipated during the clutching operation or braking cycle is
obtained by integrating the above equation from t=0 to t = t1 . The result can be
summed up as,
( )( )
( )( )
1 2 1 2
1 2
1 21 2
1 2
1 1 1 21 2
1 2
2I IE
2 I I
I IU T T T .t
I I
t t I IE udt T T t dt
I I0 02
I I1 2 1 2E2 I I1 2
ω −ω=
+
⎡ ⎤⎛ ⎞+•= θ = ω −ω −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞+= = ω −ω −∫ ∫ ⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
ω − ω=
+
Thus the energy absorbed during clutch slip is a function of the magnitude of the
inertia and the angular velocities only. This energy compared to the brake
energy may be negligible. Heat dissipation and temperature rise are governed
by the same equations presented during brakes. To contain the temperature rise
when very frequent clutching operations, wet clutches rather than dry clutches
are often use
WORKED OUT EXAMPLE 1
Design an automotive plate clutch to transmit a torque of 550 N-m. The
coefficient of friction is 0.25 and the permissible intensity of pressure is 0.5
N/mm2. Due to space limitations, the outer diameter of the friction disc is fixed as
250 mm.
Using the uniform wear theory, calculate:
The inner diameter of the friction disc
the spring force required to keep the clutch in engaged position
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Solution: As noted the friction disc of the automotive clutch is fixed between the
flywheel on one side and the pressure plate on the other. The friction lining is
provided on both sides of the friction disc.
Therefore two pairs of contacting surfaces-one between the fly wheel and the
friction disc and the other between the friction disc and the pressure plate.
Therefore, the torque transmitted by one pair of contacting surfaces is (550/2) or
275 N-m
( ) ( )( ) ( )( ) ( )3 2
i i
2 2M p r r rt a i o if
275 10 0.25 0.5 r 125 r
= πµ −
× = π − 2
From the Eqr ( )2 28r 125 r 5602254i i− =
Rearranging the terms, we have
The above equation is solved by the trial and error method. It is a cubic equation,
with following three roots:
(i) ri = 87.08 mm
(ii) ri = 56.145 mm
(iii) ri =-143.23 mm
Mathematically, all the three answer are correct. The inner radius cannot be
negative. As a design engineer, one should select the inner radius as 87.08 mm,
which results in a minimum area of friction lining compared with 56.145. For
minimum cost of friction lining.
ri=87 mm
Actuating force needed can be determined using the equation
( ) ( )aF 2 p r r r 2 (0.5)(87) 125 87 10390.28Na i o i= π − = π − =
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
WORKED OUT EXAMPLE 2
A multiple-disc wet clutch is to be designed for transmitting a torque of 85 N.m.
Space restriction limit the outside disk diameter to 100 mm. Design values for the
molded friction material and steel disks to be used are f=0.06(wet) and pmax
=1400 kPa. Determine appropriate values for the disc inside diameter, the total
number of discs, and the clamping force.
Solution
Known: A multiple – disc with outside disc diameter, d0≤ 100 mm,
dynamic friction coefficient, f=0.06(wet)
and maximum disc allowable pressure, pmax =1400 kPa,
To transmits a torque, T= 85 N.m
Find: Determine the disc inside diameter di, the total number of disks N, and the
clamping force Fa.
Decisions and Assumptions
Use the largest allowable outside disc diameter, do=100 mm (ro =50 mm).
Select ri =29 mm (based on the optimum d/D ratio of 0.577)
The coefficient of friction f is a constant.
The wear rate is uniform at the interface.
The torque load is shared equally among the disc.
Design Analysis:
Using design equation for torque under constant wear gives
( )2 2N T / p r f r r 6.69max i o i⎡ ⎤= π − =⎢ ⎥⎣ ⎦
Since N must be an even integer, use N= 8. It is evident that this requires a total
of 4+5, or nine discs, remembering that the outer disks have friction surfaces on
one side only. 3. With no other changes, this will give a clutch that is over
designed by a factor of 8/6.69= 1.19. Possible alternatives include (a) accepting
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
the 19 percent over design, (b) increasing ri, (c) decreasing ro, and (d) leaving
both radii unchanged and reducing both pmax and F by a factor of 1.19
4. With the choice of alternative d, the clamping force is computed to be just
sufficient to produce the desired torque:
r r 0.050 0.029o iT Ff N 85 N.m F(0.06) m 8,2 2
F 4483 N
+⎛ ⎞ +⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠=
Rounding up the calculated value of F, we
Find that the final proposed answers are (a) inside diameter= 58 mm, (b)
clamping force= 4500 N and (C) a total of nine discs.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Flywheel
A flywheel is an inertial energy-storage device. It absorbs mechanical
energy and serves as a reservoir, storing energy during the period
when the supply of energy is more than the requirement and releases
it during the period when the requirement of energy is more than the
supply.
Flywheels-Function need and Operation
The main function of a fly wheel is to smoothen out variations in the
speed of a shaft caused by torque fluctuations. If the source of the
driving torque or load torque is fluctuating in nature, then a flywheel is
usually called for. Many machines have load patterns that cause the
torque time function to vary over the cycle. Internal combustion
engines with one or two cylinders are a typical example. Piston
compressors, punch presses, rock crushers etc. are the other systems
that have fly wheel.
Flywheel absorbs mechanical energy by increasing its angular
velocity and delivers the stored energy by decreasing its velocity
T
T
T A B C D θ
ω
2
1
m
A B C D θ
max
ωmin
1 CYCLE
Figure 3.3.1
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Design Approach
There are two stages to the design of a flywheel.
First, the amount of energy required for the desired degree of
smoothening must be found and the (mass) moment of inertia needed
to absorb that energy determined.
Then flywheel geometry must be defined that caters the required
moment of inertia in a reasonably sized package and is safe against
failure at the designed speeds of operation.
Design Parameters
Flywheel inertia (size) needed directly depends upon the acceptable
changes in the speed.
Speed fluctuation
The change in the shaft speed during a cycle is called the speed
fluctuation and is equal to ωmax- ωmin
Fl max min= ω − ω
We can normalize this to a dimensionless ratio by dividing it by the
average or nominal shaft speed (ωave) .
max minCfω − ω
=ω
Where ωavg is nominal angular velocity
Co-efficient of speed fluctuation
The above ratio is termed as coefficient of speed fluctuation Cf and it is defined as
max minCfω − ω
=ω
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Where ω is nominal angular velocity, and ωave the average or mean
shaft speed desired. This coefficient is a design parameter to be
chosen by the designer.
The smaller this chosen value, the larger the flywheel have to be and
more the cost and weight to be added to the system. However the
smaller this value more smoother the operation of the device
It is typically set to a value between 0.01 to 0.05 for precision
machinery and as high as 0.20 for applications like crusher
hammering machinery.
Design Equation
The kinetic energy Ek in a rotating system
= ( )1 2I2
ω
Hence the change in kinetic energy of a system can be given as,
1 2 2E IK m max min2⎛ ⎞= ω − ω⎜ ⎟⎝ ⎠
K 2E E E1= −
( )max minavg 2
ω + ωω =
( )( )1E I 2 CK s avg f avg22E E C I2 1 f
EkIs 2Cf avg
= ω ω
− = ω
=ω
Thus the mass moment of inertia Im needed in the entire rotating
system in order to obtain selected coefficient of speed fluctuation is
determined using the relation
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
( )( )1E I 2 CK s avg f avg2EkIs 2Cf avg
= ω ω
=ω
The above equation can be used to obtain appropriate flywheel inertia
Im corresponding to the known energy change Ek for a specific value
coefficient of speed fluctuation Cf,
Torque Variation and Energy
The required change in kinetic energy Ek is obtained from the known
torque time relation or curve by integrating it for one cycle.
( )@ max
T T d El avg @ min
K
θ ω− θ =∫
θ ω
Computing the kinetic energy Ek needed is illustrated in the following example
Torque Time Relation without Flywheel
A typical torque time relation for example of a mechanical punching
press without a fly wheel in shown in the figure.
In the absence of fly wheel surplus or positive enregy is avalible
initially and intermedialty and enery absorbtion or negative energy
during punching and stripping operations. A large magitidue of speed
fluctuation can be noted. To smoothen out the speed fluctuation fly
wheel is to be added and the fly wheel energy needed is computed as
illustrated below
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Torque
Area+20 073
Area+15 388
BA
Area-26 105
Area-9 202
7 020
0
34 200
C
D A
rms
Average
Shaft angle time t
3600
-34 200
θ
ωmaxωmin
Figure 3.3.2
Accumulation of Energy pulses under a Torque- Time curve
From Area= Accumulated sum =E Min & maxE
A to B
B to C
C to D
D to A
+20 073
-26 105
+15 388
-9 202
+20 073
-6 032
+9 356
+154
ωmin@B
ωmax@C
Total Energy= E E@ωmin@ωmin-=(-6 032)-(+20 073)= 26 105 Nmm2
Figure 3.3.3
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Torque Time Relation with Flywheel
7020
8730
0
Average
Time t
C =0.05f
Torque
360Shaft angleθ
Figure 3.3.4
Geometry of Flywheel
The geometry of a flywheel may be as simple as a cylindrical disc of
solid material, or may be of spoked construction like conventional
wheels with a hub and rim connected by spokes or arms Small fly
wheels are solid discs of hollow circular cross section. As the energy
requirements and size of the flywheel increases the geometry
changes to disc of central hub and peripheral rim connected by webs
and to hollow wheels with multiple arms.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
d
D
b
d
b
D0 Do
Figure 3.3.5
D
b
d
a
D
0
Arm Type Flywheel
Figure 3.3.6
The latter arrangement is a more efficient of material especially for
large flywheels, as it concentrates the bulk of its mass in the rim which
is at the largest radius. Mass at largest radius contributes much more
since the mass moment of inertia is proportional to mr2
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
For a solid disc geometry with inside radius ri and out side radius ro,
the mass moment of inertia I is
m2 2I mk (r rm o2= = + 2 )i
The mass of a hollow circular disc of constant thickness t is
( )W 2 2m ro ig gγ
= = π − r t
Combing the two equations we can write
( )4 4I r rm o i2 gtπ γ
= −
Where is material’s weight density γ
The equation is better solved by geometric proportions i.e by
assuming inside to out side radius ratio and radius to thickness ratio.
Stresses in Flywheel
Flywheel being a rotating disc, centrifugal stresses acts upon its
distributed mass and attempts to pull it apart. Its effect is similar to
those caused by an internally pressurized cylinder
2 2 2t i o
2 22 2 2 2i o
r i o 2
3 v 1 3vr r rg 8 3 v
r r3 v r r rg 8 r
γ + +⎛ ⎞⎛σ = ω + −⎜ ⎟⎜ +⎝ ⎠⎝⎛ ⎞γ +⎛ ⎞σ = ω + − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
2 ⎞⎟⎠
γ = material weight density, ω= angular velocity in rad/sec. ν= Poisson’s ratio, is the
radius to a point of interest, ri and ro are inside and outside radii of the solid disc
flywheel.
Analogous to a thick cylinder under internal pressure the tangential
and radial stress in a solid disc flywheel as a function of its radius r is
given by:
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
σ
σ
t
r
Tang. stress
Radial stress
Radius
Radius
The point of most interest is the inside radius where the stress is a
maximum. What causes failure in a flywheel is typically the tangential
stress at that point from where fracture originated and upon fracture
fragments can explode resulting extremely dangerous consequences,
Since the forces causing the stresses are a function of the rotational
speed also, instead of checking for stresses, the maximum speed at
which the stresses reach the critical value can be determined and safe
operating speed can be calculated or specified based on a safety
factor. Generally some means to preclude its operation beyond this
speed is desirable, for example like a governor.
Consequently
F.O.S (N) = Nos yield
ω=
ω
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
WORKED OUT EXAMPLE 1
A 2.2 kw, 960 rpm motor powers the cam driven ram of a press through a gearing of 6:1
ratio. The rated capacity of the press is 20 kN and has a stroke of 200 mm. Assuming
that the cam driven ram is capable of delivering the rated load at a constant velocity
during the last 15% of a constant velocity stroke. Design a suitable flywheel that can
maintain a coefficient of Speed fluctuation of 0.02. Assume that the maximum diameter
of the flywheel is not to exceed 0.6m.
Work done by the press=
3U 20*10 *0.2*0.15 600Nm
==
Energy absorbed= work done= 600 Nm
Mean torque on the shaft:
32.2*10 21.88Nm9602* *60
=π
Energy supplied= work don per cycle
2 * 21.88* 6825 Nm
Thus the mechanical efficiency of the system is =600 0.727 72%825
= π=
η = = =
There fore the fluctuation in energy is =
E Energy absorbed - Energy suppliedk =
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
( )
( )
600 825*0.075 21.88*6* *0.15538.125Nm
E 538.125kI2 2960C 0.02 2 *f avg 60
2 2.6622 kg m
− π
= =⎛ ⎞ω π⎜ ⎟⎝ ⎠
=
( )
( )
r 2 2I . r r .to i2 griAssuming 0.8ro
78500 4 42.6622 * 0.30 0.24 t2 9.86
59.805t
π= −
=
π= −
=
2 .6622 t 0.044559.805
or 45 mm
∴ = =
r 3 1 32 2 2 2r r rt i og 8 378500 3 0.3 1.92 2 2. 0.24 0.3 *0.24t 9.81 8 3.3
29600.543* 2 *t 602 55667N / m
0.556MPa
⎛ ⎞+ γ + γ⎛ ⎞σ = ω + −⎜ ⎟⎜ ⎟+ γ⎝ ⎠⎝ ⎠+⎛ ⎞⎛σ = ω + −⎜ ⎟⎜
⎝ ⎠⎝
⎛ ⎞σ = π⎜ ⎟⎝ ⎠
==
2 ⎞⎟⎠
)
( )( )( )(
or if 150 MPat6 2150 *10 7961.4 0.4125 0.0376 0.090 0.0331
2 0.548216544 rad / sec
σ =
= ω
= ω
ω =
16544yieldNOS 32 164.65
ω= =
ω π=
Top Related