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ANNA UNIVERSITY SYLLABUS
Semester - VI Mechanical
DESIGN .OF TRANSMISSION SYSTEMS
.1' .. ~
1.
9 hrs.
DESIGN OF TRANSMISSION SYSTEMS FOR FLEXIBLE ELEMENTS
Selection of V belts and pulleys - Selection of Flat belts and pulleys - Wire ropes and pulleys Selection of Transmission chains and Sprockets. Design of pulleys and sprockets.
2.
Gear Terminology - Speed ratios and number of
Dynamic effects ..:..Fatigue strength - Factor of safety - Power rating calculations based on strength
gears - Pressure angle in the normal
and stresses - Estimating the si
3.
9 hrs.
SPUR GEARS AND PARAllel AXIS HELICAL GEARS
is - Tooth stresses and Face width
axis Helical
9 hrs.
BEVEL, WORM
bevel gear: Too
the d~ions
ermino
forces and stresses, equivalent number of teeth.
air of st lght bevel gears.
'ts and dem rits, terminology, thermal capacity, materials, forces and stresses,
the size of the worm gear pair.
~
.
lical: Terminology, helix angles, estimati~g the size of the pair of cross helical gears."
4.
DES'IGN OF GEAR BOXES .
.,
"
9 hrs,
Geometric progression - Standard step ratio - Ray diagram, kinematics layout - Design of sliding
mesh gear box - Constant mesh gear box - Design of multi speed gear box.
5.
DESIGN OF CAM, CLUTCHES AND BRAKES
9 hrs.
Cam design: Types, pressure angle and under cuning - Base circle determination,
surface stresses.
forces and
Design of plate clutches - Axial clutches - Cone clutches - Internal expanding rim c1l1tches-
.Internal and external shoe brakes.
TUTORIALS
30 hrs.
TOTAL: 75 hrs.
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CONTENTS
CHAPTER 1 : FLAT BELTS AND PULLEYS
1.1 -1.43
Introduction - Classification of drives - Types' of belts - Characteristics of belt drives _
Selection of a belt drive - Typ~s of flat belt drives - Belt materials - Velocity ratio of
belt drive - Effect of belt thickness on velocity ratio - Effect of slip on velocity ratio _
Phenomenon of creep - Effect of creep - Law of belting - Geometrical relationships _
Power transmitted by a belt - Tensions in a belt drive - Ratio of tensions for flat belt
drive - Losses in transmission and efficiency - Stresses in the belt - Design of flat belt
pulleys - Design of flat belt drive based on manufacturer's
data - Design of flat belt
drives using basic equations - Review and summary - Review questions - Problems for
practice.
CHAPTER 2: V-BEL TS AND PULLEYS
Introduction
- Construction
2.1 - 2.24
of V-belts - Materials of V-belts - Advantages
and
disadvantages of V-belt drive over flat belt drive - Types of V -belts - Specification of
V-belts - Ratio of driving tensions for V-belt - V-flat drives - Design of sheaves
(or V -grooved pulleys) - Design of V -belt drive based on manufacturer's data - Design
of V -belt drive using basic equations - Review and summary - Review questions Problems for practice.
CHAPTER 3: WIRE ROPES AND PULLEYS
3.1 - 3.20
Introduction - Advantages of wire ropes - Construction of wire ropes - Classification of
wire ropes - Specification of wire ropes - Guidelines for the selection of wire rope Stresses in wire ropes - Design of wire ropes - Failure of ropes - Design of wire rope
sheaves and drums - Review and summary - Review questions - Problems for practice.
CHAPTER 4: CHAIN DRIVES
4.1 - 4.31
Introduction - Advantages and disadvantages of chain drives - Types of chain drives Link chains - Dimensions of a link chain - Classification of link chains - Construction
of link chains - Selection of link chains - Advantages and disadvantages of link chains Transmission (or roller) chains - Construction of roller chains - Specification of a chain
- Geometric relationship of a roller chain and sprocket - Chordal (or polygonal) action -
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Design procedure of roller chain - Design of sprocket wheels - Types of sprockets Silent (or inverted tooth) chain - Construction - Types of silent chains - Advantages and
disadvantages of silent chains - Dimensions of the various parts of the chain - Review
and summary - Review questions - Problems for practice.
5.1 - 5.88
CHAPTER 5: SPUR GEARS
Introduction - Advantages and limitations of gear drive over chain and belt drives Definition of gear - Classification of gears - Spur gears - Gear nomenclature - Law of
gearing - Forms of gear tooth profile - Standard systems of gear tooth - Standard
proportions
of gear systems - Gear materials - Selection of gear material - Gear
manufacturing
- Gear tooth failure - Force analysis on spur gears - Tooth stresses
(Lewis beam strength equation) - Gear blank design - Gear design using Lewis and
Buckingham's
equations - Beam strength of gear tooth - Dynamic effects - Tangential
load on tooth - Dynamic tooth load (Buckingham's
equation for dynamic
load) -
Estimating gear size - Standard module - Fatigue strength of gear tooth (wear tooth
load) - Number of teeth - Face width - Factor of safety - Design procedure - Gear
design based on gear life - Dynamic load - Induced bending stress - Design bending
stress - Design contact stress - Surface compressive stress - Design procedure - Check.
for plastic deformation
- Gear design for variable loading - Design of gears with
reliability factor - Design of internal gears - Design of non-metallic gears - Review and
summary - Review questions - Problems for practice.
6.1 - 6.65
CHAPTER 6 : HELICAL GEARS
Introduction - Advantages - Disadvantages - Types of helical gears - Kinematics and
nomenclature
of helical gears - Virtual or formative number of teeth - Face width of
helical gears - Tooth proportions for helical gears - Basic dimensions of helical and
herringbone gears - Force analysis on helical gears - Design of helical gears - Helical
gear design using Lewis and Buckingham's
equations - Lewis equation
for beam
strength of helical gears - Dynamic load on helical gear tooth - Wear strength of helical
gears - Design procedure - Helical gear design based on gear life - Design formulas for
helical gear design - Design procedure - Herringbone gears - Design of herringbone
gears - Crossed helical or spiral gears - Advantages and limitations of spiral gears Shaft angle - Centre distance - Velocity of sliding between gears - Efficiency - Force
analysis on crossed-helical gears - Review and summary - Review questions - Problems
for practice.
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CHAPTER 7: BEVEL GEARS
7.1 -7.47
Introduction - Types of bevel gears - Bevel gear nomenclature - Virtual or formative
number of teeth - Proportions for bevel gears - Basic dimensions of bevel gears - Force
analysis on bevel gears - Design of bevel gears - Bevel gear design using Lewis and
Buckingham's equations - Beam strength of bevel gears - Dynamic load on bevel gear
tooth - Wear strength of bevel gears - Design procedure - Bevel gear design based on
gear life - Design formulas for bevel gear design - Design procedure - Review and
summary - Review questions - Problems for practice.
CHAPTER 8: WORM GEARS
8.1- 8.38
Introduction - Applications - Advantages and disadvantages - Types of worm gear
drives - Specification of a pair of worm gears - Nomenclature of worm gears - Tooth
proportions of worm gears - Basic dimensions of worm gears - Force analysis on worm
gearing - Efficiency' of worm gearing - Power lost in friction - Self-locking
and
overrunning drives - Design of worm gear drive - Materials for worm and worm wheel
- Failure of worm gearing - Selection of number of starts in the worm - Length of worm
- Face width of the wheel - Thermal rating of worm gearing - Worm and worm gear
design using Lewis-and Buckingham's
equation - Beam strength of worm gear tooth -
Dynamic load on wo\m gear tooth - Wear strength of worm gears - Design procedure-
\
Worm gears design ~sing basic equations - Design formulas for worm gears design Design procedure - Review and summary - Review questions - Problems for practice.
CHAPTER 9: GEAR BOX
9.1 - 9.41
Introduction - Requirements of a speed gear boxes - Methods of changing speed in gear
boxes - Preferred numbers - Step ratio - $tru9tural formula - Kinematic layout - Ray
diagram - Basic rules for optimum gear box -design - Overlapping speed gear box Design of gear box - Design procedure for gear box - Review and summary - Review
questions - Problems for practice.
CHAPTER 10: CLUTCHES
10.1 -10.50
Introduction - Functions of the clutch - Principle of operation of clutch - Classification
of clutches - Friction materials for clutches - Single plate clutch - Design of a single
plate clutch - Multiplate clutch - Design of a multiplate clutch - Service factors - Cone
clutch - Design of a cone clutch - Centrifugal clutch - Design of a centrifugal clutch -
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Internal expanding rim clutches - External contracting rim clutches - Energy dissipation
during clutching (Energy considerations) - Temperature rise - Review and summary Review questions - Problems for practice.
11.1 -11.58
CHAPTER 11 : BRAKES
Introduction - Clutch Vs Brake - Classification of brakes - Brake lining materials Block or shoe brake - Single block or shoe brake - Self-locking and self-energizing
brakes - Double block or double shoe brake - Design procedure for block brake - Band
brake - Simple band brake - Design procedure for band brakes - Differential band brake
- Band and block brake - Internal expanding shoe brake - External contracting shoe
brake - Energy considerations - Temperature rise - Review and summary - Review
questions - Problems for practice.
TWO MARKS Q&A
Q&A.l - Q&A.21
SOLVED ANNA UNIVERSITY QUESTION PAPERS
Q.l - Q.55
SUGGESTED READINGS
INDEX
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Flat Belts and Pulleys
"Live as if you were to die tomorrow;
Learn
(U
if you were to live forever ."
- Mahatma Gandhi
1.1. INTRODUCTION
Whenever power has to be transmitted from one shaft to another shaft, flexible machine
elements such as belts, ropes or chains are frequently used. Pulleys are mounted on the shaft
and a continuous belt or rope is passed over them. In belts and ropes, power is transmitted
due to friction between them and the pulleys. In case of chain
sprocket wheels are
used. When the distance between the shaft is large, then
upon several
smaller distances, gears are used. The amount of power
arc of contact
factors such as velocity of the belt, tensions i
belt,
between the belt and the smaller pUlle.1iilllllt":
1.2. CLASSIFICA
.,
Drives
l
~
Direct drives
Flexible drives
l
1
Chain
Rope
Belt
l
~
~
Gear drive
Cam drive
1.2.1. Types of Belts
Four types of belts used for power transmission are :
I. Flat belts.
2. V-belts,
3. Ribb
1
e d bel ts,
an d
4. Toothed or timing belts.
These four types of belts are shown in Fig.l.l.
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Design of Transmission Systems
1.2
Flat belt
Multiple V-belts
V-belt
Ribbed belts
Toothed or timing belt
Fig. 1.1. Types of belt drives
1.2.2. Characteristics of Belt Drives
The characteristics of different belts are tabulated, as shown in Table 1.1.
Table 1.1. Characteristics 0/ belt drives
,
Flat belts
Characteristics
S.No.
Toothed or
V-belts
timing belts
l.
Maximum velocity ratio
16
12
11
2.
Maximum belt speed (m/s)
35 to 110
25
80
I
3.
Slip
1 to 5%
1 to 5%
Nil
4.
Tension
High
Less
Very less
5.
Shock resistance
Good
Good
Fair
6.
Resistance to wear
Good
Fair
Good
7.
Dressing
Required
Not required
Not required
8.
Initial cost
Less
Less
Moderate
1.2.3. Selection of a Belt Drive
Selection of a belt drive depends upon :
./
Power to be transmitted
./
Speed reduction ratio
./
Speed of driver and driven shafts
./
Centre distance
./
Shaft relationship
./
Positive drive requirement
./
Service conditions
./
Space available
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1.3
-
Flat Belts and Pulleys
1.2.4. Types of Flat Belt Drives
Depending on the requirement, flat belts can be arranged in different ways. The different
types of arrangement and their applications are tabulated, as shown in Table 1.2.
Table 1.2. Types o/flat belt drives alld tl,eir appiicatiolls
Applications
Types of drives
Used
1. Open belt drive:
Sladl side
~
with
shafts
arranged
parallel
and
rotating in same direction.
-Tight side
Fi • 1.2.
2. Open belt drive with one idler pulley:
Used with shafts arranged parallel and when
an open belt drive cannot be used due to
small angle of contact on the smaller pulley.
Idler pulleys (also known as Jockey pulleys)
are provided to obtain high velocity ratio and
when the required belt tension cannot be
obtained by other means.
Fig. 1.3.
3. Open bell drive wuh many idler pulleys:
Used when it is desired to transmit
motion
from one shaft to several parallel shafts.
Fig. 1.4.
)
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Design a/Transmission Systems
1,4
Types of drives
4. Crossed or twisted belt drive:
Applications
Used with shafts arranged parallel
rotating in the opposit~ direction.
and
FI . 1.5.
s.
Used with shafts arranged at right angles and
rotating in one definite direction.
FI • 1.6.
6.
Quart~r twist bett drive will, guide
pul/~y :
Used with shafts arranged at right angles
when the reversible motion is desired.
t
FI • 1.7.
7. St~PfNdor cone pulley driv«:
Used for changing the speed of the driven
shaft while the main or driving shaft runs at
constant speed.
n« t.s.
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I.S
[Iat Belts and Pulleys
----------~--------------I
Applications
Types of drives
8. Fast and loose pulley:
Used when the driven shaft is to be started or
stopped whenever desired without interfering
with the driving shaft.
Fi • 1.9.
9. Compound drive:
Used when several units are to be driven
from one central shaft.
Fi . 1.10.
1.3. BELT MATERIALS
The desirable properties of a belt material are high coefficient of friction, flexibility,
durability and strength. The main materials used for flat belts are:
1. Leather belts: Leather belts are made of animal hides. The best quality leather is
obtained from either sides of the backbone of a steer (bullock). Leathers for belting may be
tanned with oak, or chrome salts. Oak-tanned leather is fairly stiff, whereas chrome-tanned
leather is soft and pliable.
Belts are specified according to the number of layers. e.g., single-ply, double-ply or tripleply belts. Double-ply (or triple-ply) belts are made by cementing two strips (or three strips) of
leather together with hair sides out.
2. Fabric and cotton bells:~ These belts are made by stitching together three or more plies
(or layers) of canvas or cotton duck. The fabric is treated with linseed oil to make it waterproof.These belts are cheap. They are most suitable for farm work, quarry and saw mills.
J. Rubber belts: These belts are made up of plies of fabric impregnated with vulcanised
rubber or synthetic rubber. The main advantage of these belts is that they can be easily made
endless. Saw mills, creameries, chemical plants and paper mills largely use the rubber belts.
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Design of Transmission Systems
J.6
4. Bnlaln ~/ts:
Balata is gum similar to rubber. Balata belts are made in the same manner
as the rubber belts except that balata is substituted for rubber. These belts are acid proof and
water-proof. The balata belts cannot be used at temperature above 40°C because at this
temperature it begins to soften and becomes sticky.
5. Nylon core belts
6. Camel's hair belts.
The commonly used belt materials for various belt types are given in Table 1.3.
Tuble /.3. Commonly used bell materials
Belt types
.
Belt materials
Flat belts
Leather canvas, cotton and rubber
V-belts
Rubberised fabric and rubber
Ropes
Cotton, hemp and manila
1.4. VELOCITY RATIO OF BELT DRIVE
The ratio between the speeds of the driver and the follower or driven is known as velocity
/
ratio.
D and d
=
Diameters of the driver and driven respectively,
N) and N2
=
Speeds of the driver and driven respectively,
and
=
Angular velocities of the driver and driven respectively.
Let
(01
(02
N2
Velocity ratio, -N
(02
=
=
)
(0)
and
D
... (1.1)
d
1.4.1. Effect of Belt Thickness on Velocity Ratio
When the thickness of belt (I) is considered, then velocity ratio is given by
N2
N)
=
D+I
... (1.2)
d+1
1.4.2. Effect of Slip on Velocity Ratio
S~ip is defined as the relative motion between the belt and pulley. The difference between
~he linear sp~ds of the pulley rim and belt is the measure of slip. The reason for slip to occur
that the.re IS a ten~e~cy for ~he belt to carry with it on the underside, between the pulley and
the belt. I.e., the frictional gnp between belt 'and pulley is, insufficient. The presence of slip
red~ces the velocity ratio of the drive.
IS
B
h'
h
.
. y roug ernng t e belt by dressing or by crowning.
avoided.
one of the pulleys
'
the slip can be
• For mere details. refer section 1.17.3.
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1.7
Flal Bells and Pulleys
S, = Percentage slip between the driver and the belt,
Let
S2
S
.
= Percentage slip between the belt and the driven pulley, and
= Total percentage slip = S, + S2
N
.
1'12
:. Velocity ratio,
I
D [
d
=
_ S, + S2 ]
1
.. , (1.3)
100
If thickness of the belt (I) is considered, then
Velocity ratio, ~
= ~:;
[
I -
J%o ]
... (1.4)
1.5. PHENOMENON OF CREEP IN BELTS
When the belt passes from the slack side to the tight side, a certain portion of the belt
extends. And it contracts again when the belt passes from the tight side to slack side. Due to
these changes of length, there is relative motion between the belt and the pulley surfaces. This
relative motion is termed as creep.
The net effect of creep is to reduce the speed of the driven pulley and consequently
power transmitted.
the
1.5.1. Effect of Creep of Belt
crt and cr2 = Stresses in the belt on the tight side and slack side respectively,
Let
E
and
= Young's modulus of the belt material.
VI'e ocity ratio,
.
N
N2,
=
o
x
d
E+Y-C;:
... (1.5)
E+~
I Note lin
practice the combined effect of slip and creep is called simply slip and the combined
effect should not exceed three percent.
1.6. LAW OF BELTING
Law of belting states that the centre line of the belt, as it approaches the pulley, must lie in
a plan~ perpendicular to the axis of that pulley or must lie in the plane of the pulley,
otherWise the belt will run off the pulley.
1.7. GEOMETRICAL RELATIONSHIPS
For open belt drive: An open belt drive is shown in Fig.I.1 I.
Let D and d = D'
iarneters of the larger and smaller pulleys respectively
in metres,
C = Centre distance between the two pulleys in metres,
L
=
Total length of the belt in metres ,
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Design a/Transmission Systems
1.8
2a
as
aL
=
The angle subtended between the straight portions of the belt in degrees,
Wrap angle (or angle of contact / lap) for small pulley in degrees, and
=
Wrap angle for large pulley in degrees.
=
. -1
Sin
-
0- d
I 2C
,~
I,
I
--\----C----Fig. 1.11. Open belt drive
As seen from the Fig.I.II,
sma
also
as
O-d
= --
2C
=
(180-2a)
and
aL
=
(180+2a)
... (1.6)
O-d')
Wrap angle for small pulley, as
=
180 - 2 sirr ' ( 2C
Wrap angle for large pulley, aL
=
180 + 2 sirr ' (O-d)
2C
Length of the belt, * L
=
2C +
and
(¥)
d
(0 + d) + (04-C )2
... (1.7)
For crossed belt drive: A crossed belt drive is shown in Fig.l.12, with notations having
the usual meanings.
. -1
Sin
D+d
--
2C
I
,..--,
I,
1------
C ----
.....
Fig. 1.12. Crossed hell drive
• For derivations of the formulas used in this chapter, the readers are suggested to refer any 'KinematicS of
Machines' book.
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Flat Belts and Pulleys
1.9
As seen from the Fig.I.12,
Sill U
=
(02+cd)
Us
=
uL
and
=
(180 + 2 n)
... (1.8)
Therefore, wrap angles for smaller and larger pulleys are same and is given by
Us
=
=
uL
180 + 2 sirr ' (~ ~d )
Length of the belt, L = 2 C +
1.S. POWER TRANSMITTED
Let
( 21t)
(D + d) +
(0 + d)2
4C
... (1.9)
BY A BELT
P
=
Power transmitted by a belt in watts,
TJ and T2
=
Tensions in the tight and slack sides respectively in newtons, and
v = Linear velocity of the belt in m/s.
I Power
transmitted,
P
(TJ
=
-
T2) v
I
... (1.10)
1.9. TENSIONS IN A BELT DRIVE
1. Tight and slack side tensions (T] and T~ : When a belt is moving round a pulley and
transmitting power, the tension in belt on two sides of pulley will be different. The side of
belt in which tension is higher is the tight side and the other is called slack side.
2. Centrifugal tension (T c) : As the belt moves round the pulley it would experience
a
centrifugal force which has a tendency to separate the belt from the pulley surface. To
maintain contact between pulley and belt, the centrifugal force produce additional tension in
the belt, which is known as the centrifugal tension,
Centrifugal tension is a waste load, because it increases tension without increasing pow.r
capacity.
Let
m = Mass per unit length of the belt in kg/m, and
v
=
Linear velocity of the belt in m/s.
..
I Centrifugal
tension,
Tc
=
mv2
I
...
(1.1 I)
3. Initial tension in belt (To) : The tension of the belt when a belt is fitted to a pair of
stationary pulleys, is termed as the initial tension of tile belt (To).
:.
Initial tension, To
=
=
TJ +T2
2
... [Neglecting centrifugal tension]
TJ+T2+2Tc
2
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... [Considering centrifugal
... (1.12)
tension] ... (1.13)
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Design of Transmission Systems
1.10
INotes I I. Maximum tension to which the belt can be subjected
T = T 1+ T C
Maximum stress x Cross-sectional area of belt
T =
and
due to centrifugal tension,
... (1.14(a))
... (1.14(b))
= cr· b- t
o = Maximum safe stress in N/m2,
where
=
b
Width of belt in metres, and
Thickness of belt in metres.
2. When the centrifugal tension is taken into account, then
Total tension in the tight side, Til = T I + T c
and total tension in the slack side, Ta = T 2 + Tc
3. Effect of centrifugal tension
We know that
0" power transmitted
Power transmitted, P
(Ttl - T/2) V
=
Thus, the centrifugal tension has
110
:
[(T I + Tc) - (T 2 + Tc) ] v = (T I - T 2)
V
effect on the power transmitted.
4. For a belt speed of upto 10 mls the centrifugal tension is negligible. But for belt speed more than
10 mis, the centrifugal tension should be considered without fail.
1.10. RATIO OF DRIVING TENSIONS FOR FLAT BELT DRIVE
Let
TI and T2 = Tensions on tight and slack sides of the belt respectively,
a
= Angle of wrap
f.l
=
Coefficient
11
Tension ratio,
and
INotes I I.
T2
TI-mv2
T2
-mv2
(i.e., angle of contact) of belt with the pulley,
and
of friction between the belt and pulley.
=
ell a
... [Neglecting
centrifugal
tension]
... (1.15)
=
ella
... [ Considering
centrifugal
tension]
... (1.16)
It should be borne in mind that 'a' in the tension ratio equations must be in radians.
2. Condition for the transmission of maximum power: The power transmitted shall be maximum
when the centrifugal tension (Tc) is one third of the maximum belt tension (T).
T ==
and
maximum
velocity. v
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3 Tc
-
... (1.17(a»
(J
\j~
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... (1.17(b))
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Flat Belts and Pulleys
1.11
1.11. BELT SPEEDS
Most efficient power transmission
(Refer PSG data book, page no.7.53).
is obtained f tl
or at belts at speeds of 17.5 to 22.5 m/s
1.12. SPECIFIC WEIGHTS OF BELT MATERIALS
Leather
= 1 x 10-5
Rubber
= 1.4 x 10-5
N/mm3
Balata = 1.11 x 10-5 N/mm3
N/mm3
Canvas
1.22 x 10-5 N/mm3
=
1.13. COEFFICIENT OF FRICTION
The coefficient
of friction
between the belt material and tlie pv IIey sur face
.
.
the belt material,. material . of. the pulley.' surface the belt speed an d th e be It
values of coefficients of friction for design purposes are given in Table 1.4.
d epen d s upon
I' A verage
Sip.
Table 1.4. Mean coefficient of frlctlon, J.I
Pulley material
Belt material
Compressed
Wood
Steel
Cast iron
-
paper
Leather (oak-tanned)
0.33
0.30
0.25
0.25
Leather (chrome-tanned)
0.45
0.40
0.35
0.35
Cotton or fabric
0.25
0.23
0.20
0.20
Rubber
0.35
0.30
0.30
0.30
1.14. CENTRE DISTANCE
(A longer belt will last more tlutn a shorter belt. Why?)
The life of a belt is a function of the centre distance between the driver and driven shafts.
The shorter the belt, the more often it will be subjected to additional bending stresses while
running around the pulleys at a given speed. And also it will be destroyed quickly due 10
fatigue. Hence, a longer belt will last more than a shorter belt.
1.15. LOSSES IN TRANSMISSION
AND EFFICIENCY
The losses in a belt drive are due to :
Slip and creep of the belt on the pulleys (about 3%),
(i)
(ii) Windage or air resistance to the movement of belt and pulleys (usually negligible),
(iii) Bending of the belt over the pulleys (about I%), and
(iv]
Fricti n in the bearings of pulley (about 1%).
1htrcfore the overall efficiency of the drive is about 95 to 96%.
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M·
-
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I.' 2
1.16. STRESSES IN THE BELT
.
The various stresses acting at various portions of the belt are.
1. Stress due
where
maximum working tension, TI (a; :
10
at
=
Tight side tension
Cross-sectional area of the belt
b
=
Width of the belt, and
1 = Thickness of the belt.
2. Stress due
where
10
ab
=
E
=
=
d
3. Stress due
bending of the bell over tile pulley (a,,) :
10
where
P
Young's modulus of the belt material,
and
Diameter of the smaller pulley.
the effect of centrifugal force (uj :
_
ac
E·,
d
2
_ mv =
2
- b· I
Pv
-
Centrifugal force
Cross-sectional area of the belt
=
Density of the belt material in kg/m '.
It. is noted that the stress will be maximum when the belt moves over the smaller pulley.
Therefore the maximum stress in the tight side of the smaller pulley is given by
amax
= at + ab + ac
1.16.1. Permissible Stresses
Leather belts = 2 to 3.45 MPa
Rubber belts = I to ].7 MPa
Fabric belts = Less than ].5 MPa
DESIGN OF FLAT BELT PULLEYS
1.17. INTRODUCTION
In order t~ design a flat belt. drive, we need the diameters of driving and driven
u lie s.
Thus the design of belt pulleys IS to be done first Since the velocitv rati
p
Y
pulley diameters, therefore the pulleys should be s~lected caref:I~~lty ratio depends upon the
1.17.1. Materials Used for Pulleys
The commonly used pulley materials are:
./
Cast iron
./. .
.
Fabricated steel
./
Wood or fibre
./
.
Compressed paper
Cast Iron pulleys are most widely used iIII actual practice.
.
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Flat_Be/ts and Pulleys _
1.13
1.17.2. Types of Pulleys for Flat Belts
Based on the construction
methods, the pulleys are classified as solid pulleys and split
pulleys.
Small pulleys can be made in single casting which is known as solid pulleys. But medium
and larger pulleys are cast in halves, which can be joined at the rim and the hub. This type of
pulleys are known as split pulleys, In the following article, the design of cast iron split pulley
will be discussed.
1.17.3. Design Procedure for Cast Iron Pulleys
The cross-section of a cast iron pulley is shown in Fig.1.13. (Refer PSG data book, page
no. 7.56).
Rim
Fig. 1.13. Cross-section of putley
D
where
b
dl
I
=
=
=
=
Diameter of the pulley,
Thickness of the arm,
Diameter of the hub,
a = Width of the pulley,
t = Thickness of the rim,
d2 = Diameter of the shaft, and
Length of the hub.
1. Dimensions of pulley:
. .
.'
lie (D): Obtain the diameter of the pulley either from velocity
.:v
id tion We know that the centrifugal stress
ratio consideration or centrifugal stress cons: era
.
induced in the rim of the pulley,
(i) DIameter of the pu
0c
where
=
P y2
p = Density of the rim material,
= 7200 k m3 for cast iron, and
7t D N
D being the diameter of pulley and N the
v = Velocity of he rim = 60 '
speed of the pulley.
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Design of Transmission Systems
1.14
Now, select the diameter of the pulley (D) referring to Table 1.5.
Table 1.5. Recommended pulley diameters ill mm (from data book, page
40,45,50,56,63,71.
710,860.900,
80, 90.100,112,125,140,160,180,200.224,250,280.315,
110.
7.54)
355, 400, 450,500,560,630,
1000, 1120, 1250, 1400, 1600, 1800 and 2000.
(ii) Width of tile pulley (a) : If the width of the belt is known, then select the width of the
pulley referring to Tables 1.6(a) and (b).
Table 1.6(0). Pulley width (from data book, page no. 7.54)
Belt width
Pulleys to be wider than the belt width by
Upto 125 mm
13 mm
125 to 250 mm
25mm
250 to 375 mm
38mm
375 to 500 mm
50mm
Table 1.6(b). Recommended series of width of flat pulleys, mm (from data book, page
110.
7.55)
20, 25, 32, 40, 50, 63, 71, 80, 90, 100, 112, 125, 140, 160, 180, 200, 224, 250,
280, 315, 355, 400, 450, 500, 560 and 630.
(iii) Thickness of the pulley rim (t): For C.l. pulleys,
t
D
= 200 + 3 mm, for single belt
D
= 200 + 6 mm, for double belt
where
D
... [From data book, page no. 7.57]
= Diameter of the pulley in 'mm'.
2. Dimensions of arms:
(i) Number of arms (n) :
Number of arms {
4 for diameters upto 450 mm
6 for diameters over 450 mm
... [From data book, page no.7.56]
(ii) Cross-section of arms {b and bI2): The cross-section of the arms is elliptical, with
major axis (b) is equal to twice the minor axis (bt2).
Major.axis of elliPtical}
secnon near the boss
b
= 2.94
_3
fiQ4Dn
-\j ~
~ 2.941"¥f
for single belt, and
for double belt .
... [From data book, page no. 7.56]
Minor axis of elliptical section near the boss
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Flat Belts and Pulleys
1.15
(iii) Arms taper: The arms are tapered from hub to rim.
Taper = 4 mm per 100 mm
(iv) Radius of the cross-section of arms: r
... [From data book, page no.7.56]
=
4'3
b
3. Dimensions of hub:
(i) Diameter of the hub (d 1) :
Diameter of the hub (d1)
d,
or
=
=
(1.7 to 2.0) x Diameter of the shaft (d2)
(1.7 to 2.0) d2
(ii) Length of the hub (/) :
=~a
Minimum length of bore (i.e., length of the hub), I
where
a
Width of pulley.
=
... [From data book, page no. 7.56]
4. Crowning of pulley rim: The face of the pulley rim is crowned, as shown in Fig.l.13,
to keep the belt on the pulley. Otherwise the inaccurate alignment of the pulleys causes the
belt to run off side ways. Thus the crown will force the belt to return to the centre of pulley.
Selection of crown height (II) : Knowing diameter (0) and width (a) of the pulley, select
the crown height (h) referring to Tables 1.7(a) and (b).
Table 1.7(a). Crow" of flat pulleys (40 to 355 mm diameter) (from data book, page no. 7.55)
(crow" is unrelated to the width ill this diameter range)
Diameter D, mm
Crown h, mm
40 to 112
125 and 140
160 and 180
200 and 224
250 and 280
315 and 355
0.3
0.4
0.5
0.6
0.8
1
Table 1.7(b). Crow" offlat pulleys (40 to 2000 mm diameter) (from data book, page no. 7.55)
(crown varies with tire width in this diameter range)
Crown"
Diameter
D,mm
400
450
500
630
800
1000
1250
2000
(in mm) of pulleys of width (in mm)
125 and
140 and
180 and
224 and
280 and
smaller
160
200
250
315
1
1.2
1.2
1.5
1.5
1.5
1.5
1.5
2.5
1.2
1.2
1.5
2
2
2
2
3
1.2
1.2
1.5
1.2
2
2
2.5
2.5
2.5
2.5
3.5
4
I
1
1
1
1
1.2
2
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1.2
1.5
3
3
355
400 and
larger
1.2
1.2
1.5
2
2.5
3
3.5
5
1.2
1.2
1.5
2
2.5
3
4
6
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Design of Transmission Systems
1.16
I Example
I Design
1.1
a cast iron pulley to transmit 20 kW at 300 r.p.m: The diameter
of the pulley is 500 mm and the angle of lap is 180 ~ The pulley has four arms of elliptical
cross-section with major axis twice the minor axis. The coefficient of friction between the
belt and the pulley surface is 0.3. The allowable belt tension is not to exceed 250 N in 10
mm width. The allowable shear stress for the shaft material may be taken as 50 Nlmml.
Given Data:
a = 1800 =
as
= 50
=
20 kW
rad;
n
P
1t
=
20 x 103 W;
4' ,
J.1
=
N = 300 r.p.m.;
0.3;
TI
of the pulley or belt,
v =
=
=
D = 500 mm = 0.5 m ;
2.5 N in 10 mm width of the belt;
N/mm2.
To flnd : Design a cast iron pulley.
© Solution:
Velocity
7t·D·N
=
60
7t x 0.5
60
x
300
=
7.854 m/s
1. Dimensions of pulley :
(i) Diameter
recommended
0/
the pulley (D) is given as 500 mm. Now referring
diameter of the pulley is also 500 mm. Ans. "
(ii) Width of the pulley (a) : In order to find the width of the pulley
Table
1.5, the
let us find the width
of the belt first.
=
Let TI and T2
Tensions
on the tight and slack side
of the belt respectively.
We know that the power transmitted
P
=
(TI - T2) v
20 x 103 = (T, - T2) 7.854
or
T, - T_ = 2546.47
... (i)
TI
and ratio of tensions,
= e~Q
T2
,
=
T2
From equations
1t
or
T, = 2.566 T2
... (ii)
(i) and (ii), we get
TI
INou I Since
eO.3)(
=
4171.68 N
and
T2
=
1625.75 N
the velocity of the belt (or pulley) is less .han )0 mIs, therefore the centrifugal
tension need not to be considered.
b
Let
Since the allowable
width, therefore
=
Width of belt
tension (i.e., maximum
tension)
is 250 N in 10 mm width or 25 N/mm
width of the belt
T,
25
b = -
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=
4171.68
25
= 166.86mm
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1.17
flot Belts and Pulleys
Referring to Table t.t3, the standard width of 4 ply belt is 200 mm.
l1lerefore width of the pulley (a), referring the Table 1.6(a), is given by
= Belt width + 25 mm = 200 + 25 = 225 mm
Then, referring to Table 1.6(b), the standard pulley width is 250 mm. Ans."
(iii) Thickness of the pulley rim (t) :
D
For single belt,
t
= 200 + 3 mm
500
= 200 + 3
[From data book, page no. 7.57]
= 5.5 mm Ans."
2.Dimensions of arms:
(i) Number of arms, n
=4
... [Given]
(ii) Cross-section of arms: Major axis of elliptical section near the boss is given by
b ~ 2.94
where
Wn
for single belt
... [From data book, page no. 7.56]
= Width of the pulley = 250 mm,
D = Diameter of the pulley = 500 mm, and
n = Numberofarms = 4
a
:. Major axis -- 2.94
3
250
x
500
4x4
60
= Major
- 2 axis = 2
=
and Minor axis
=
30
58.34 mm say 60 mm Ans."
mm
A
ns. ~
3
.
(iii) Radius of the cross-sections of arms = 4
x M'ajor axis
3
= 4 x 60
=
45 mm Ans. ~
3. Dimensions of the hub:
(i) Diameter of the /tub : In order to find the diameter of the hub, let us find the diameter
of the shaft first.
Let
d
=
Diameter of the shaft
We know that the torque transmitted by the shaft,
p x 60
20 x 103 x 60
T = 2 1t N = 2 1t x 300
= 636.62 N-m = 636620 N-mm
We also know that the torque transmitted by the shaft (T),
T
= ~
16
xo x
s
d3
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,u
"
r,
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Design of Transmission Systems
1.18
=
636620
Therefore,
r6 x 50 x d3 or d
Diameter of the hub
= 40.17 mm say 45 mm
= 2 x Diameter of the shaft
= 2 x 45 = 90 mm ADs.'"
2
2
= 3"
Length of the hub = 3" x Width of the pulley (a)
(ii)
166.67 mm
=
x 250
ADs.'"
y
4. Crown height of tire pulley (lr) : For 500 mm pulley diameter and 250 mm pulley
width, from Table 1.7(b), the crown height is selected as h = 1.5 mm ADs.
I Example
=
I Design' an overhanging pulley'for
1.2
= 200 r.p.m.; Angle of contact
18 kW; Speed
the following specifications,'
Power
= 0.25;
= 165"; Coefficient of friction
Overhanging length (i.e., the distance of the pulley centre line from tire nearest bearing) =
0.30 m; Belt thickness = 10 mm; Safe sirear stress for shafts = 40 MPa; Safe stress for belt
= 2.5 MPa; Safe stress for rim = 4 MPa; Density of the leather = 1000 kg/m'.
Given Data: P = 18 kW = 18
= 2.88 rad; ~
=
40 N/mm2;
Pleather
=
0.25;
0belt =
L
=
2.5 MPa
OJ
x
103 W;
N = 200 r.p.m.; a
10 rnrn;
111; 1 =
(0)
h
= 2.5 x 106 N/m2; 2.5 N/mm :
n
=
(Jllm
= 165 = 165
40 MPa
=
0
=
rt
x 180
40 x 106 N/m2
4 MPa = 4 x 106 Nzmrn-';
= 1000 kg/m ',
Tofind: Design an overhanging pulley.
©Solution:
I. Dimensions of pulley:
(i) Diameter of the pulley (D) :
Let D = Diameter of the pulley.
0c
= Centrifugal stress or tensile stress in the pulley rim' =
p = Density of the pulley material
We know that centrifugal stress,
0c
4 x 106
Velocity of the pulley is also given by
= p
rim'
and
7200 kg/rn ' for cast iron.
v2
= 7200 x v2 or
v =
23.57
=
0·
v = 23.57 m/s
7tDN
60
= 7txDx200
60
or D = 2.25 m ADS. ~
(i;) Width of tirepulley (a): In order to find the width of the pulley, let us find the width
of the belt first.
Let
T, and T2 = Tensions on the tight and slack sides of the belt re pectively,
b ;; Width of the belt.
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and
.
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FJal Bells and Pulleys
1.19
We know that the power transmitted (P),
P
= (T, - T2) v
18 x 103 = (T, - T2) 23.57
or T, - T2 = 763.68
... (i)
and tension ratio of the belt,
T,
T
= eO.25 x 2.88 = 2.054 or T, = 2.054 T 2
= ella
2
... (ii)
\
From equations (i) and (ii), we get
T)
T2
1487.62 Nand
=
=
725.25 N
Since the velocity of the belt (or pulley) is more than 10 mis, therefore centrifugal tension
must be taken into consideration.
Assuming a leather belt for which the density is given as
1000kg/m-'.
We know that centrifugal tension,
Tc =
In'
v2
where
m = Mass of the belt per metre length
We know that
In
=
Density x Volume
But Area of cross-section
=
of the belt
=
Density x Area x Length
bxt
=
b x 10 = 10 b mm-
=
10 b x 10-6 m2
1000 x (10 b x 10-6) x 1 = 0.01 b kg/m
Then centrifugal tension, Tc = m- v2 = 0.01 b (23.57)2 = 5.55 b N
In
=
and maximum tension in the belt,
T
=
abe)t
=
2.5 x 106 x (lOb x 1O--{) = 2 5 b N
x Area of cross-section
of belt
=
abe)t
x (b x t)
We know that tension on the tight side of the belt (T),
T)
=
T - Tc
or
1487.62 =- 256 - 5.55 b
=
19.45 b
Width of the belt, b = 76.48 mm
Referring to Ta.ble 1. 13, the standard width of the belt = 90 mm
Therefore, width of the pulley (a), referring the Table 1.6(a), is given by
=
Belt width + 13 mm
=
90 + 13
=
103 mm
Then, referring to Table 1.6(b), the standard pulley width is 112 mm.
ADS."
(iii) Thickness of tile pulley rim (t) :
For single belt,
t = 2~0
=
+ 3 mm
2250
200 + 3
=
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... [From data book, page no. 7.57]
14.25 mm
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Design a/Transmission
Systems
1.20
2. Dimensions
0/ arms :
(i) Number of arms (n) :
6 (for diameters over 450 rnm)
7 56]
... [From data book, page no. .
Number of arms, n =
(ii) Cross-section of arms:
.
Major axis of elliptical section near the boss
IS
where
Wn
given by
b
=
2.94
a
=
Width of the pulley = I12 mm,
for single belt
D = Diameter of the pulley = 2250 mm, and
n
Number of arms = 6
=
112 x 2250
4x6
3
..
Major axis = 2.94
and
Minor axis =
(iii)
.
Major axis
2
Radius of cross-sections of arms
-
=
64.38 mm say 65 mm Ans . ...,
65
= 32.5 mm
2
ADS. ~
3
- x Major axis
4
=
3
= '4 x 65 = 48.75 mm
ADS. ~
3. Dimensions of tile II ub :
(i) Diameter of the hub : In order to find the diameter of the hub, let us find the diameter
of the shaft first.
Let
d
=
Diameter of the shaft
We know that the torque transmitted by the shaft,
p x 60
T = 2 1t N
=
18 x 103 x 60
21t X 200
=
859.44 N-m
and bending moment on the shaft due to the tensions of the belt
,
M = (T) + T2 + 2 Tc) L = (1487.62 + 724.25
=
T2 + M2
= ~
= ~
x 5.5 x 90) 0.3
'" (.,' Tc
960.56 N-m
We know that equivalent twisting moment (Te),
T,
+2
= 5.55 b N)
(859.44)2 + (960.56)2
= 1288.92 N-m = 1288.92 x 103 N-mm
We also know that equivalent twisting moment (T ),
e
1288.92 x 103 ==
or
7t
1 6 x as x d3 - ~
16 x 40 x d3
Diameter of the shaft.
.
d = 54 75
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.
mm
say 55 mm,
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Flat Belts and Pulleys
1.21
Diameter of the hub
= 2 x Diameter of the shaft
= 2 x 5S = 110 mm
ADS • ...,
2
(ii) Length of the hub = 3 x Width of the pulley (a)
2
= 3"
x
112
=
74.66 mm
ADS • ...,
4. Crown height of tire pulley (II) : For 2250 mm pulley diameter and 112 mm pulley
width, from Table 1.7(b), the crown height is selected as h = 2 mm ADS • ...,
DESIGN OF FLAT BELT DRIVE
The two different design procedures used are:
Using the manufacturer's data, and
(i)
(ii) Using the basic equations.
1.18. DESIGN OF FLAT BELT DRIVE BASED ON MANUFACTURER'S
DATA
In actual practice, the designer has to select a belt from the manufacturer's catalogue
(which were obtained by their long experience). The required information for the selection /
design of a flat belt are:
(i)
Power to be transmitted,
(ii)
The input and the output speeds, and
(iii)
The centre distance depending upon the availability of space.
The step by step procedure is as follows:
1. Selection of pulley diameters:
Select the pulley diameters and angle of contact (i.e., wrap angie). By using the given belt
speed and assuming number of plies, minimum pulley diameter is chosen. Use Table 1.8 to
choose the diameter of the smaller pulley
Table 1.8. Minimum pulley diameter for the given speed and the number of belting plies, mm
(from data book, page
No. of plies
3
4
5
6
8
==
Maximum
110.
belt speed mls
10
IS
20
90
140
200
250
450
100
160
224
315
500
112
180
250
355
560
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7.52)
25
30
140
200
315
400
630
180
250
355
450
710
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J
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Design a/Transmission Systems
--------------------~~~~==~~--
~I.~n
2. Calculation 0/ design power in kW:
Calculate the design kW by using the relationship given below.
Rated kW x Load correction factor (Kj)
Design kW
=
.. , (1.18)
Arc of contact factor (Ka) x Small pulley factor (Kd)
(i) Load correction/actor (KJ : This factor is used to account for the nature of application
and type ofload. The value of Kscan be selected from Table 1.9.
Table 1.9. Load correction/actor,
Ks (from data book, page no. 7.53)
Ks
Load classification
1.0
1.2
Nonnalload
Stead load-screens, centrifugal pumps, agitators, belt conveyors,
light machine tools, etc.
Intermittent loads - Reciprocating pumps and compressors, heavy
machine tools, heavy duty fans and blowers, etc.
Shock loads - Crushing machinery, hammers, presses, grinders,
rolling mills, etc.
1.3
1.5
(ii) Arc 0/ contact factor (KaJ': The load rating (i.e., rated power capacity) is given for
1800 of contact. So, it has to be corrected for actual arc of contact. A decrease
contact implies additional load.
..
d)
Arc of contact
=
1800
D and d
=
Diameters of larger and smaller puIJeys, and
where
(D ~
_
x 600
in arc of
... [From data book, page no. 7.54]
C = Centre distance.
For the calculated value of arc of contact, the arc of contact factor (K(l) is selected from
the Table 1.10.
Table 1.10. Arc
Arc of
contact
90
Correction
factor
1.68
0
120
130
140
1.33
1.26
1.19
0
0
0
0/
contact factor, Ka (from data book, page
150
0
160
170
180
190
200
210
220
230
240
250
1.08
1.04
1.00
0.97
0.94
0.91
0.88
0.86
0.84
0.82
0
1.13
7.54)
110.
0
0
0
0
0
0
0
0
0
(iii) Small pulley factor (K~ : This factor is used to account for the amount of bending or
flexing of the belt and how this affects the life of the belt. Use Table 1.11 for small pulley
factor.
Table 1.11. Small pulley factor, Kd (from data hook, page no. 7.62)
Small pulley diameter
K"
Upto 100 mm
100-200 mm
200-300 mm
300-400 mm
400-750 mm
Over 750 mm
0.5
0.6
0.7
0.8
0.9
1.0
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Flat Belts and Pulleys
1.23
3. Selection of a belting :
Select a belt referring to Table 1.12.
Table 1.12. Load rating of fabric belts per mm width per ply at 1800 are of contact at
lQ m/s belt speed (from data book, page no. 7.54)
Load rating
Type
HI-SPEED duck belting (light duty)
0.023 kW/mmlply
FORT duck belting (heavy duty)
0.0289 kW/llImlply
4. Load rating correction :
Correct the load rating to the actual speed of the belt by using the relation given below ..
Load rating at V m/s
=
Load rating at 10 m/s x
V
TO
...
[From data book, page no. 7.54]
5. Determination of belt width:
Determine the belt width by using the following relation:
Design power
.
.
Load ratmg x No. of plies
Width of belt =
". (1.19)
Knowing the smaller pulley diameter and velocity of the belt, and consulting Table 1.8,
the number of plies can be found.
The calculated belt width should be rounded off to the standard belt width by consulting
Table 1.13.
Table 1.13. Standard widths of transmisslon belting (from data book, page
4 ply
3 ply
I
mm
25
32
40
44
50
63
76
90
100
mm
25
32
40
44
50
63
76
90
100
112
125
140
152
200
5 ply
mm
76
90
100
112
125
152
180
200
22-l
250
6 ply
R ply
mm
100
112
125
152
180
200
250
mm
200
250
305
355
400
IlO.
7.52)
6. Determilltltioll of pulley width:
3
--
Detcrlllll1e (he pulle
\ idih, b) referring rh Table
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1.6(a) and (b).
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Design a/Transmission
Systems
1.24
7. Calculation of belt length (L) :
Calculate the length of the belt by using the equation given below.
( 27t) (0 + d)
7t)
+ ( "2 (0 + d)
For open belt drive:
L = 2C +
For crossed belt drive:
L
=
2C
dY:
+
(0 4C
+
(0 + d)2
4C
... [From data book, page no. 7.53]
I Example 1.3 lIt
is required to select a flat-belt drive/or a/an running at 360 r.p.m:
which is driven by a 10 kW, 1440 r.p.m: motor. Tile belt drive is open-type and space
available for a centre distance 0/ 2 m approximately. The diameter 0/ a driven pulley is
1000mm.
Given Data: N)
= 1440 r.p.m.; N2 = 360 r.p.m;
C = 2m; D
P = 10 kW
=
10 x 103 W ;
= 1000 mm.
Tofind : Select (or design) a open flat belt drive.
© Solution:
The given arrangement is shown in Fig.l.14.
1. Calculation of pulley diameters:
Driven pulley diameter, D
We know that
=
velocity ratio =
360 r.p.m.
--;--....
1000 mm
o
D
Fan
d
_ Driver pulley speed
N)
1440
- Driven pulley speed - N2 = 360
d
=
4
o
Driver pulley diameter, d = 4
=
1000
-4-
Fig. 1.14.
=
250mm
Consulting Table 1.5, the recommended driver pulley diameter = 250 mm
Ans. ~
2. Calculation of design power in k W :
Design kW
(i)
=
Rated kW x Load correction factor (Ks)
Arc of contact factor (KJ x Small pulley factor (Kd)
Rated kW = 10 kW
... [Given]
(ii) Referring to Table 1.9, load correction factor
(iii) To find arc of contact factor (Ka) :
K, = 1.2 for steady load.
'
)..a
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·1.25
Flat Belts and Pulleys
= 180
0
-
(
1000 - 250 )
0
2000
x 60 = 157.50
Consulting Table 1.10, arc of contact factor for 157.50, Ka::::: 1.08.
(iv) Consulting Table 1.11, small pulley factor, Kd
Design kW
=
lOx 1.2
1.08 x 0.7
=
= 0.7
ADS.~
15.873 kW
3. Selection of belt:
Consulting Table 1.12, HI-SPEED duck belting is selected.
kW/mmJply.
Its capacity
is given as 0.023
4. Load rating correction:
Velocity of the belt, V
=
=
7t
x 0.25 x 1440
= 18.85 mls
60
.
Load rating at V mls = Load rating at 10 mls x
Load rating at 18.85 m/s
= Load rating at 10 mls
V
TO
x (18.85 ItO)
= 0.023 x (18.85 I 10) = 0.04335
kW I mm I ply
5. Determination of belt width :
For 250 mm smaller pulley diameter and velocity of 18.85 mIs, consulting Table 1.8, the
number of plies can be selected as 5.
Width of belt
=
=
Design power
Load rating x No. of plies
15.873
0.04335 x 5 = 73.23 mm
Consulting Table 1.13, the calculated belt width should be rounded off to the standard belt
width.
:. For 5 ply belt, standard belt width = 76 mm
ADS. ~
6. Determination of pulley width :
Consulting Table 1.6(a), the pulley width is given by
Pulley width = Belt width + 13 mrn = 76 + 13 = 89 mm
" Referring Table 1.6(b), the standard pulley width is 90 mm ADS. ~
7. Calculation of length of tile belt (L) :
We know that the length of an open belt,
L
=
2 C + ~ (D + d) + {_D- d)2
.
2
4C
=
2
x
2000 +:!!. (1000 + 250) + 0000 - 2S0}2
2
4 x 2000
= 6033.8 mm
ADS. ~
I
/i
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1.26
Design of Transmission Systems
l
l.tI J Design a belt drive to transmit 20 kW at 720 r.p.m. to an aluminiu",
( Example
rolling machine, the speed ratio being 3. The distance between the pul/eys is 3 m: Diamele,
of rolling machine pulley is 1.2 m.
P == 20 kW
Given Data:
D
== 20 x 103 W ,.
ROlling
N I == 720 r.p.m.;
machine
(driven)
d
Speed ratio == 3 ;
C == 3 m;
D == 1.2 m.
To flnd : Design a belt drive.
© Solution:
Fig. 1.15.
The given arrangement
is shown in Fig.l.I5.
1. Calculation of pulley diameters:
Driven pulley diameter,
We know that
== 1200 mm
D
D
Driver pulley speed
== d == Driven pulley speed
speed ratio
]200
d
3 ==
Consulting
... [Given].
Table 1.5, the recommended
or d==400mm
driver pulley diameter
== 400 mm
ADS."
2. Ca/cularion of design power ill kW:
Rated k W x Load correction
kW == Arc of contact factor (Ka) x Small pulley factor (Kd)
Design
(i)
(ii)
=
Rate kW
From Table 1.9,
20kW
... [Given]
Ks = 1.5 for rolling mills.
=
Arc of contact
(iii)
factor (Ks)
Design
(D-d)
--
x 60°
C
=
180° - (1200-400)
3000
60°
=
1640
x, ~ 1.06.
For 164°, From Table 1.10,
(iv) From Table I.] 1,
180°-
Kd == 0.8
kW ==
20 x 1.5
1.06 x 0.8
== 35.377 kW
3. Selection of belt:
Consulting
Table
1.12,
FORT duck belting is selected. Its capacity
is given as 0.0289
kW/mmlply.
4. Load rating correction:
Velocity
of the belt,
Load rating at V m/s
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-=
::;
1t
d NI
60
==
1t
x 0.4 x 720
60
load rating at ]0 m/s x
==
15.08 m/s
V
10
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1.27
Flat Belts and Pulleys
---w-
= Load rating at 10 mls x (15.08)
Load rating at 15.08 mls
= 0.0289 x
5. Determination of belt width:
(I~.g8) = 0.04358 kW I mm I ply
.
For 400 mm smaller pulley diameter and velocity .of 15.08 mis, consulting Table 1.8, the
number of plies can be selected as 6.
Width of belt
=
Design power
= Load rating x No. of plies
35.377
0.04358 x 6
= 135.29 mm
From Table 1.13, for 6 ply belt, standard belt width = 152 mm Ans. ~
6. Determination of pulley width:
From Table 1.6(a), pulley width
=
Belt width + 25 mm = 152:- 25
=
177 mm
:. Referring Table 1.6(b), the standard pulley width is 180 mm Ans."
7. Calculation of length of the belt (L) :
For open belt, L
=
7t
2'
2C +
(D + d) +
7t
(D - d)2
4C
= 2 x 3000 + 2' (1200+400) +
(1200 - 400)2
_
4 x 3000
= 8~66.6 mm
Ans. ~
1.19. DESIGN OF BELT DRIVES USING BASIC EQUA":IONS
In a belt drive, when the drive is transmitting maximum power without slip (i.e., at the
point of slipping) the tensions in belt are governed by the following equation,
T -my2
1
T2-mY2
where
...!!!!...
= esin ~
T I' T 2' m, y and
(l
=
ella. cosec ~
'" (1.20)
have usual meanings, and
2f3 is the Y-groove angle (= 1800 for flat belts)
In equation (1.20), the term
ella cosec ~
represents the tension ratio at which slip occurs.
The load carrying capacity of a pair of pulleys is determined by the pulley which has the
smaller value of eJ.IIZcosec
fJ. Therefore, while designing a belt drive using basic equations,
first one should find which of the pulley (smaller or larger) governs the design.
Tofind the governing pulley (i.e., pulley governs the design) :
1. If the coefficients of friction are same for both the pulleys (i.e., both the pulleys are of
same material), then the smaller pulley governs the design. Because for open belt drive, the
smaller pulley has the wrap angle smaller than that of the larger pulley and hence the smaller
pulley will have the smaller value of eJ.lQ cosec P than the larger pulley. Therefore smaller
pulley will slip first.
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~
Design a/Transmission
Systems
1.28
2 If the coefficients of friction are different for both the pulleys (i.e., both th~tUlle;s ar~
of different material), then the pulley which has the smaller tension ratio (or sma er va ue 0
eJUlcosec j3 ) governs the design.
~
Use the smaller tension ratio in designing the belt drive.
I Example
1.5
I A leather
belt is used to supply power/rom
motor to a reciprocating air compressor. Tile specifications
a compensator ~tart electric
0/ the
Electric motor
belt drive are as follows:
Air compressor
Power (in k W)
25
-
Speed (in r.p.m.)
1440
360
Pulley diameter (in mm)
250
1200
0/ lap (in radians)
3.8
4.7
0.3
0.25
-Angle
Coefficient
0/ friction
The density of the belt is 1000 kg/mJ. The permissible tension is not to exceed 145 N in
10 mm width. TI,e thickness of tbc belt may be taken as 8 mm. Determine which pulley is
governing the design 0/ the drive and find the necessary width to transmit
power taking into account the centrifugal tension also.
Given Data:
P
= 25
kW
= 25
x 103 W;
d = 250 mm; D = 1200 mm; (11 = 3.8 rad;
P = 1000 kg/rn-; Tl = 145-N in 10 mm width; t =
To find:
(i)
the required
NI ,= 1440 r.p.m.;
N2 = 360 r.p.m.;
(12 = 4.7 rad;
J..LI= 0.3; J.i2 = 0.25;
8 mm.
The pulley which governs the design, and
(ii) Width of the belt.
©Solution:
(i) The pulley which governs the design:
We know that a pulley which has the smaller value of e~a· cosec ~ will govern the design.
Since the coefficients of friction are different for both the pulleys, first we have to evaluate
e~ (since 2~ = 1800 for flat belts) separately.
:. For smaller pulley :
e~l at
= eOJ x 3.8
= 3.127
For larger pulley :
e~2 a2
=
= 3.238
Here
e~1 al
(ii) Width
<
0/ the
e~2 a2.
eO.25 x 4.7
Thus the smaller pulley governs the design. Ans. ~
belt (b) :
Velocity of the belt. v =
1tdNI
60
=
1t x 0.25 x 1440
60
= 18.85 m1s
Let II and T2 be the tensions on the tight and slack sides respectively.
We know that the power transmitted (P),
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"i- T -'fc... '
1.29
Flat Belts and Pulleys
P = (TI - T2) v
25 x 103
(TI - T2) 18.85 or TI - T2
=
=
1326.29
... (i)
We know that cross-sectional area of the belt,
= b x
= b x 8 = 8 b mrn? = 8 b x 1Q-6 m2
t
Mass of the belt per metre length,
m
= Density x Volume = Density x Area x Length
8 b x 10-3 kg/m
=
1000 x 8 b x 1Q-6 x 1
=
m- v2 = 8 b x 10-3 (18.85)2
=
We know that centrifugal tension,
Te
= 2.84 b N
We also know that the tension ratio considering the centrifugal tension,
TI - m· v2
T2-m·y2
=
TI-2.84b
T2 - 2.84 b
or
T)-2.84b
(T)-1326.29)-2.84b
or
'" [e~)
e~) a)
since smaller pulley governs the design]
a) -
= 3.127
. •. [ .:
eJi) a)
=
3.127 ]
... (ii) [.: T2 = TI - 1326.29, from (i) ]
= 3.127
It is given that permissible tension (T) is 145 N in 10 mm width or 14.5 N/mm width.
,
:. TI
= Permissible tension per mm wid~h
r
',1
•
I
x widt~
.
(
= 14.5 b
... (iii)
. .
Substituting the equation (iii) in (ii), we get
14.5.b - 2.84 b
14.5 b - 1326.29 - 2.84 b
On simplification, we get
1l.b(, h
1l.bb b .
= 3.127
= 167.23 mm
b
:. Consulting Table 1.13, standard width of the belt = 180mm Ans."
I Example 1.6 I Find the width of the belt necessary to transmit
7.5 kW to a pulley of
JOOmm diameter, if the pulley makes 1600 r.p.m; and the coefficient of friction between
the belt and the pulley is 0.22. Assume the angle of contact as 2100 and the maximum
tensionin the belt is not to exceed 8 Nlmm width.
GivenData: P = 7.5 kW = 7500 W " d
J.L =
0.22;
C1
= 210 = 210
0
0
x
1:00
=
= 300 mm = 0.3 m ; N - 1600 r.p.m.;
3.6652 rad.
TOjind: Width of the belt (b).
@ Solution: Velocity of the belt, v
=
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7tdN
60
=
7t x 0.3 x 1600
60
=
25.133 rnIs
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Design o/Transmissior1 Systems
1.30
P
We know that,
=
7500 = (T1
TI
Also
T2
=
=
ella
Solving (i) and (ii),
T2) v
(TI -
T2) 25.133 or
-
eO.22
x
=
3.6652
= 539.15 Nand
TI
T2
TI -
2.2396 or
=
=
TI
... (i)
298.415
... (ii)
2.2396 T2
= 240.734 N
T2
Given that, maximum tension is not to exceed 8 N/mm width.
..
=
T1
Maximum tension per mm width x width
= 8b
539.15 = 8 b or b = 67.4 mm
Consulting Table 1.13, standard width of the belt = 76 mm Ans. ~
I Example 1. 71 A flat
belt is required to transmit 35 kW from a pulley of 1.5 m
effective diameter running at 300 r.p.m: The angle of lap is 1650 and JJ = 0.3. Determine,
taking centrifugal tension into account, width of the belt required. It is given that the belt
thickness is 9.5 mm, density of its material is 1.1 Mg/m3 and the related permissible
working stress is 2.5 MPa.
Given Data: P = 35 kW = 35 x 103 W;
1t
a = 1650 = 1650 x 1800 = 2.88 rad; Jl
0'
d = 1.5 m; N = 300 r.p.m. ;
= 0.3;
t
= 9.5
mm ; p = 1.1 Mg/m ' = 1100 kg/m-'
= 2.5 MPa = 2.5 x 106 N/m2.
Tofind:
Width of the belt (b).
1tdN
... Velocity of belt, v = ~
© Solution:
- ~
•
•
~
I
=
1t
x 1.5 x 300
60
= 23.56 m/s.
••
.~ e ,
Let
b = Belt width in mm.
We know that,
p = (T1
T2) v
-
35 x 103 = (TI - T2) 23.56
or TI - T2 = 1485.45
... (i)
T)
T2
Solving
(0 and
= ella
(ii),
=
eO.3 x
2.88 = 2.373 or TI = 2.373 T2
T) = 2568N
Cross-sectional area of the belt
=
bx
t =
and T2 = 1082.19N
9.5'b mrn-
9.5 b x 1~
=
m2
We know that mass of the belt per meter length,
m
Centrifugal tension, Tc
=
Density x Area x Length
=
1100x9.5bx
=
m v2 = 0.01045 b (23.56)2
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l~x
=
1=
p x (b x t) »: I
0.01045b
=
kg/m
5.8 b N
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... (ii)
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Flat Belts and Pulleys
and
1.31
Maximum tension in the belt, T =
bxt)
0" (
= 2.5 x 106 x 9.5 b x lQ-{>= 23.75 b N
We also know that
T=TI+Tc
23.75 b = 2568 + 5.8 b or b = 143 mm
Consulting Table 1.13, standard width of the belt = 152 mm Ans. ~
I Example
1.8
I A belt drive is required
to transmit 12 kW from a motor running at
720 r.p.m: The belt is 12 mm thick and has a mass density of 0.001 gm/mmt. Permissible
stress in the belt not to exceed 2.5 Nlmm2. Diameter of driving pulley is 250 mm whereas
the speed of the driven pulley is 240 r.p.m. The two shafts are 1.25 m apart. Coefficient of
friction is 0.25. Determine the width of the belt.
Given Data:
P = 12 kW
12 x 103 W;
=
p :: 0.001 grn/mrn! = 1000 kg/m'';
d I = 250 mm = 0.25 m; N2 = 240 r.p.m.;
NI = 720 r.p.m.;
= 2.5 N/mm2 = 2.5 x
C = 1.25 m; ~L = 0.25.
0"
t
= 12 mm;
106 N/m2;
Toflnd : Width of the belt (b).
©Solution:
Velocity of belt, v
=
=
N,
0.25
240
= -d
720
2
For an open belt drive,
9.425 m/s
or d2 = 0.75 m
d2-d,
2C
=
0.75 - 0.25
= 0.2
2 x 1.25
a = sirr ' (0.2) = 11.54°
or
INote I Since
a =
=
d2
-
SIl1
x 0.25 x 720
60
d,
N2
Speed ratio = -
We know that
1t
the material is same for both the pulleys, therefore the smaller pulley governs
the design.
Arc of contact for smaller pulley, as = (180° - 2 a) 1;00
= (180° - 2 x 11.54°) x 1;00 = 2.74 radians
We know that
TI = e)lUs
T2
P
Also
12xI03
=
(TI
-
=
eO.25 x 2.74
=
1.983 or TI=1.983T2
T2) v
= (TI-T2)9.425
or TI-T2=
1273.2
... (ii)
T2 = 1272.2 N
Solving (i) and (ii), we get TI = 2545.42 Nand
Let
... (i)
b = Belt width in mm.
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Design a/Transmission Systems
1.32
Cross-section of the belt
= b x t = b x 12 = 12· b mm2 = 12· b x 10"-6m2
We know that mass of the belt per metre length,
m = Density x Area x Length
= p x (b x t) x I
= 1000 x (12 b x 10~) x I = 0.012 b kg/m
Centrifugal tension, Tc = m y2 = 0.012 b (9.425)2 = 1.065 b N
Maximum tension in the belt, T
=
o x (b
=
30b N
T} +Tc
x t) = 2.5 x 106 x (12 b x 10~)
=
30 b = 2545.42 + 1.065 b or b = 87.97 mm
Consulting Table l.13, standard belt width = 90 mm Ans."
We know that
T
, Example 1.9 , Design a rubber belt to drive a dynamo generating 20 kW at 2250
r.p.m: and fitted with a pulley 200 mm diameter. Assume dynamo efficiency to be 85%.
Allowable stressfor belt material = 2.1 N/mml
Density of rubber
=
1000 kg/mJ
Angle of contact for dynamo pulley
=
1650
Coefficient of friction between belt and pulley
=
0.3
= 20 x 103·W ~ N = 2250 r.p.m. ~ d = 200 mm = 0.2 m ~
0.85 ~ o = 2.1 N/mm2 = 2.1 x 106 N I m2 ~ p = 1000 kg 1m3; a = 165 =
Given Data: P = 20 kW
TId
165
= 85%
1t
x
180
=
Tofind:
=
0
2.88 rad;
J.L
= 0.3.
Design a rubber belt (i.e., width and thickness of the belt).
© Solution:
Velocity of belt,
1t
Y
dN
= 60 =
1t
x 0.2 x 2250
60
= 23.6 mls
We know that
... [.: TId is given]
20 x 103
Also
T}
T2
= (T} - T2) 23.6 x 0.85 or T} - T2 = 995
=
... (i)
eJ.1ll= eO.3 x 2.88 = 2.375 or T} = 2.375 T2
Solving (ijand (ii), we get T} = 1719N
and T2
... (ii)
= 724 N.
Let band t be the width and thickness of the belt respectively in mm.
Assume thickness of the belt, t = 10 mm ADS. ~
Cross-sectional area of the belt
= bxt
::: b x 10
= lOb mm-
=
lOb x 10-6 m2
We know that mass of the belt per metre length,
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1.33
Flat Belts and Pulleys
m
= Density
x Area x Length
= pxbxtxI
= 1000 x (10 b x 10-6) x 1 = 0.01 b kg/m
Centrifugal tension, Te
=
m- v2
= 0.01 b (23.6)2 = 5.5696 b N
Maximum tension in the belt, T = c x (b x t) = 2.1 x 106 x (10 b x 10-6)
= 21 bN
We know that
T=T)+Te
21 b = 1719 + 5.5696 b or b = 111.4 mm
Consulting Table 1.13,
standard belt width = 112mm Ans."
( Example 1.10 , A leather belt 125 mm wide and 6 mm thick, transmits power from a
pulley with the angle of lap 1500 and u = 0.3. If the mass of 1 m3 of leather is 1Mg and the
stress in the belt is not to exceed 2.75 MPa, find the maximum power that can be
transmitted and the corresponding speed of the belt.
GivenData:
b=125mm=0.125m;
e = 150
0
t=6mm=6x
1t
= 1500 x 180 = 2.62 rad;
o = 2.75 MPa = 2.75
x
10-3m;
J.L = 0.3;
p = 1 Mg/m3 = 1000 kg/m! ;
106 N/m2.
To find: Maximum power and corresponding speed.
©Solution:
Speed of the belt for maximum power:
T = o xbx
t =
We know that maximum tension in the belt,
2.75 x 106 x 0.125 x 6 x 10-3 = 2062.5 N
and mass of the belt per meter length,
m
= Density x Area x Length = p b . t . I
= 1000 x 0.125 x 6 x 10-3 x 1 = 0.75 kglm
.. Speed of the belt for maximum power,
v=~
=
2062.5
3 x 0.75
= 30.25mls
Ans."
\
Maximum power transmitted:
We know that for maximum power to be transmitted,
T
2062.5
Centrifugal tension, Te = 3 =
3
= 687.5 N
and tension in the tight side of the belt,
Tl = T - Te
_~!tt~. _
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= 2062.5 - 687.5
= 1375 N
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Design of Transmission Systems
1.34
We know that
TI
=
T2
or
TI
T2 = 2.195
Maximum
power transmitted,
P
1.11 lIn
=
2.62
x
1375
_-=
2.195
2.195
626.53 N
= (TI _ T2) v
=
I Example
eOJ
=
ella
(1375 - 626.53) 30.28 = 22.66 kW
Ans. ~
a belt drive, the tension ratio is 2.6. The mass of the belt is J kg/m
length. It runs at a speed of 5 m/s and at this speed, power transmitted is 8 k W. What
should be tile initial tension in the belt and what should be the streng til of the belt?
Given Data:
TI
T =
2.6;
In
=
1 kg/rn ; v
= 5 m/s
; P = 8 kW
=
8 x 103 W.
2
To find : Initial tension (To) and strength of the belt (T).
© Solution:
p
=
(TI - T2) v
8 x 103
=
(T, - T2) 5 or
=
2.6
TI
=
2600 N and T
T
=
111'
We know that
TI
Given that,
T2
On solving (i) and (ii),
Centrifugal
tension,
=
or TI
\ 2
=
Initial tension (T,) : We know that the initial tensi
J
11
(
TI _ T_ = 1600
... (ii)
2.6 T2
_
=
=
... (i)
J
n idcrin
cntrifu
al ten i n,
TI
=
2600
=
1825 N Ans. ~
Strength of the belt (T) :
Strengt.h of the belt = Total ten i n of the tight side
T
I Example
1.12
IA
=
TIT
=
2600 ' 25
2.5 k W of power is transmitted by
(III
=
2625 N Ans. ~
open belt drive. Tlte linear
velocity of the belt is 2.5 m/s. TI,e angle of lap on the smaller pulley is 165 ~ The coefficient
of friction is 0.3. Determine the effect on power transmission in the following cases:
(i)
Initial tension in the belt is increased by 8%,
(ii) Initial tension ill the belt is decreased by 8%,
(iii) Angle of lap is increased by 8% by the use of all idler pulley, for the same speed
and the tension on the tight side, and
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1.35
Flat Bells and Pulleys
(iv) Coefficient of 'friction is increased by 8% by suitable dressing to tile friction surface
of tile belt.
Also state which of the above methods suggested could be more effective?
Given Data:
Tofind:
P
=
a
=
=
2.5 kW
1t
=
1650
v = 2.5 mls ;
2500 W ;
= 2.88
1650 x 1800
.
radians;
~L
= 0.3.
Effect on power transmission.
© Solution:
P = (T I
Power,
2500
T'ens ion ratio,
.
=
T 2) v
!...
(T. - T2) 2.5
T
TI =
=
ellU
... (i)
or T, - T2 = 1000
=
eO.3x2.88
2.37
=
or T,
=
Tl
Solving(i)and(ii),weget
Initial tension,
To
1729.9N
T, +T2
=
and
729.9N
1729.9 + 729.9
2
=
2
=
T2
... (ii)
2.37T2
2
=
1229.9N
-
IO~
(i) When initial tension is increased by 8%:
New increased
initial tension,
To'
T, +T2
or
2
=
108
100 x 1229.9
=
To'
=
=
1328.3 N
1328.3 N or T1 + T2 = 2656.6 N ... (iii)
T, .
As ~ and a remain unchanged,
T
or
ella
is same. So, T,
2
r, .=
Solving (ii) and (iii), we get
Power,
P
.
Increase in power
1868.3N
and
= 2.37
T2 = 78~.3 N
=
(1868.3 - 788.3) 2.5
p.
2:7 -~.5
2.5 V
=
=
T2•
=
P-v
I
2.7 k\y •
0/
0.08 or 8/0
ADS. ~
(ii) When initial tension is decreased by 8%:
New decreased
initial tension,
92
To' = 100 x 1229.9
or
We know that from equation (ii), TI
TI
Solving (ii) and (iv), we get
Then,
Power,
To'
=
=
=
2.37 T2
=
T2
1591.5 Nand
2.S -2.3
2.5
1131.5 N
113 1.5 N or T I + T 2
P = (1591.5 -671.5)
Decrease in power
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=
=
=
=
2.5
=
2263
... (iv)
671.5 N
=
2.3 kW
0.08 or 8°/.,
A
--.
ns. --
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Design of Transmission Systems
1.36
(iii) Wilen angle of lap (a) is increased by 8%:
New angle of lap, a'
and
T1
New tension ratio, T
108
100 x a.
=
=
= ella'
]08
100
=
x 2.88 = 3.1 104 rad
=
eO.3 x 3.1104
2.54 or T I = 2.54 T 2
... (v)
2
i.e., TI
Tension on the tight side remains same as before.
Then from equation (v),
= ] 729.9
T2
=
680.5 N
P
=
(1729.9 - 680.5) 2.5 = 2.624 kW
Power,
. power = [ 2.6242.5- 2.5 ]
... Percentaze
ercen age imcrease in
=
N
0.04 96 or 4 .96 % Ans • ..,
(iv) When coefficient of friction is increased by 8%:
New u'
T
New tension ratio, T I
108
= ] 00
x J.l
=
=
ell's
108
100 x 0.25
=
eO.27
x
2.88
=
=
0.27
2.54 or T I = 2.54 T 2
... (vi)
2
As initial tension is same,
TI +T2
2
Solving (vi) and (vii), we get
-
1229.9 N or TI + T2 = 2459.8
1764.9 Nand
'" (vii)
T2 = 694.9 N
Power, P = (~764.9 - 694.9) 2.5 = 2.675 kW
2.675 - 2.5
Percentage increase in power =
= 0.07 or 70/0 Ans."
2.5
Conclusion: Since the power transmitted by increasing the initial tension by 8% is more,
therefore in order to increase the power transmitted, we shall adopt the method of increasing
the initial tension. Ans."
I Example
1.13
I The
layout of a crossed leather belt drive is shown in Fig. 1.16. The
bell, 6 mm thick, transmits 7.5 kW and operates at a velocity of 13 mls approximately. The
coefficient of friction is 0.3 and the permissible tensile stress for the belt material is
1. 75 N/~.
The density of leather is 0.95 gm/cc. Calculate: (i) the diameter of pulleys;
(ii) the length and width of the belt; and (iii) belt tensions on the tight and loose sides. Take
length of the belt, 1% less to give initial tension.
6 mm = 6 x 10-3 m ;
P = 7.5 kW .= 7.5 x 103 W;
v = 13 mls ;
J.L = 0.3; o = 1.75 Nzrnm? = 1.75 x 106 N/m2;
p
=
0.95
grn/cc
=
950
kg/m '
N 1 = 1000 r.p.m.;
N2 = 500 r.p.m. ;
Given Data:
C
=
t =
1500 mm = 1.5 m.
o
d
.
1---
1500 mm
---1.1
Fig. 1.16.
,.j
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Flat Belts and Pulleys
1.37
@Solution :
(i) Diameter of pulleys:
n
Speed ratio,
d
n
or 0=2d
d
and
velocity, v
=
13
=
1t
d Nl
60
1t
X
... (i)
d x 1000
or d
60
=
Consulting Table 1.5, standard smaller pulley diameter, d
=
Larger pulley diameter, 0
(ii) Length of the belt:
.
L = 2C +
=
=
2 x 250
500 mm
0.248 m
=
250 mm
ADS."
ADS."
'" [From equation (i)]
For a crossed belt, length of the belt is given by
( 1t )
2
2 x 1500 +
(0 + d) +
(0 + d)2
4C
+ 250)2
( 27t) (500 + 250) + (500
(4 x 1500)
=
4271.84 mm
After providing 1% for initial tension,
Belt length
=
=
Power, P =
0.99 x 4271.84
Widthof the belt:
4229 mm
ADS."
(Tl - T2) v
7.5 x 103 = (T, - T2) 13 or Tl - T2 = 576.92
For a crossed belt drive,
a = sirr!
Angle of contact, as
We know that
Solving (i) and (ii), we get
Let
TJ
(D+d)
2C
= sin-J (500+250)
2 x 1500
=
(180° + 2 a) 1;0
=
(180 + 2 x 14.47°) 1;0
T2
= e~
TJ
=
=
eO·3 x 3.646 =
867.56 Nand
=
... (ii)
=
14.470
3.646 radians
2.985 or T J
= 2.985
T 2'" (iii)
T2 = 290.64 N
b = Width of the belt in mm
Cross-section of the belt = b x t = b x 6
We know that mass of the belt per metre length,
m
=
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= 6 b mm? = 6 b x 10-0 m2
Density x Area x Length
=
pxb
x
txI
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J
"
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Design o/Transmission
:
Systems
1.38
lQ-6x 1 = 5.7x 10-3b
= 950x6bx
kg/m
Centrifugal tension, Tc = m- v2 = 5.7 x 10-3 b (13)2 = 0.963 b N
maximum tension in the belt, T
and
... (iv)
=
cs=b>!
=
1.75 x 106 x (6 b x 1Q-6) = 10.5 b N
T=TJ+Tc
We know that
10.5 b
867.56 + 0.963 b or b = 90.96 mm
=
=
Consulting Table 1.13, standard width of the belt
100 mm
ADS. ~
(iii) Belt tensions on the tight and loose sides:
Substituting b = 100 mm in equation (iv), we get
Centrifugal tension, Tc
and
tr-:
,/L
=
0.963 x 100
=
96.3 N
Belt tension on the tight side
=
T I + Tc
=
867.56 + 96.3
=
belt tension on the loose side
=
T2 + Tc
=
290.64 + 96.3 = 386.94 N
963.86 N.
ADS. "
ADS. ~
REVIEW AND SUMMARY
Based on the shape of cross-section of the belts, belts are classified as flat belts, V-belts
and ropes (i.e.. circular belts).
Types offlat belt drives and their applications are tabulated in Table 1.2.
Belt materials: Leather,fabric and colton, rubber, balata and nylon core.
Velocity ratio of belt drive:
(i)
(ii)
(iii)
N2
N,
N2
N/
=
d, + t
d2 + t
=
dJ + I [
S ]
d2 + t 1- 100
NJ
... [when considering the thickness of the belt]
... [when considering the slip of the belt]
a, E+{a;
N2
(iv)
... [when neglecting the thickness of the belt]
=
N2
N/
d,
d2
-
d2 x E + ~
where NJ and N2
d / and d 2
t
'" [when considering the effect of creep of belt]
= Speeds of the driver and driven respectively,
=
Diameters of the driver and driven respectively,
= Thickness of the belt,
S = Total percentage slip,
E
uJ and u2
= Young's modulus of the belt material, and
=
Stresses in the belt on tight and slack sides respectively.
------~
L
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N,
Flat Btlts and Pulleys
./
Velocity ratio of a compound belt drive:
Speed 0/ last dri~en = Product of diameters 0/ drivers
Speed offirst driver
Product of diameters of drivens
./
Length of the belt (L) :
L
= 2 C + !I
L
=
1.39
(D + d) + (D - d)2
'I
4C
... [For open belt drive)
2 C + !I (D + d) + (D + d)2
2
'I
4C
... [For cross belt drive}
2
d and D = Diameters of smaller and larger pulley diameters, and
where
C = Centre distance between the pulleys .
./
Wrap angle or angle of contact of pulleys (a) :
For open belt drive:
and
Wrap angle for smaller pulley, as = 180 - 2 sin ! (D-d)
2C
degrees
Wrap angle for larger pulley, aL = 180 + 2 sin:' (D-d)
2C
degrees
For cross belt drive:
Wrap angle for smaller pulley, as = 180 + 2 . sin ! (D2 ~ d) degrees
and
Wrap anglefor larger pulley, aL
./
Power transmitted,
./
Tension ratio:
./
Centrifugal tension:
where m
./
=
P
TJ
T2
=
=
(IJ - T~ v
=
epa
180 + 2 -sin:' (D2 ~d)
degrees
Tc = mvl
Mass per unit length of bell.
Initial tension (IoJ of the belt is given by
... [Neglecting centrifugal tension)
... [Considering centrifugal tension}
./
Conditionsfor the transmission ofmaximum power are:
(i)
Where
vmax
=
T
=
ff.,
and (ii) T= 3 Tc
Maximum tension in the belt
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=
TI + Tc
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S
In this chapter, the design of flat belt pulleys are presented with step by step proced
{./
./
Design a/Transmission
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Design of flat belt pulleys: To find (i) Dimensions of pulley (includes diameter,
and thickness of the pulley),· (ii) Dimensions of arms (includes number of arms,
. section of arms and arms taper),' and (iii) Dimensions of hub (includes diameter
length of hub).
./
Two different design procedures used for belt drives are: (i) using the manufacture
data,' and (ii) using the basic equations.
,/
Step by step procedure for the design of 'flat belts using the manufacturer's
in this chapter with sufficient example problems.
,/
Flat bell design: To find: (i) Type of bell (i.e., Hi-speed or Fort), (ii) Bell width
number of plies, and (iii) Belt length.
./
Design
•
WIll
ustn« basi
.
'h
equations: Ti e pulley which has the smaller value of tension ra
govern the design of the drive.
G
OSIC
REVIEW QUESTIONS
1.
2.
data is g
Enumerate the various types of belts used for the transmission of pow
What are the materials used in flat belts.
er.
3.
4.
s.
6.
7.
8.
9.
10.
Explain the following terms: (i) velocity ratio (Oo)
I'
Oo.
Wha
' II S JP and (Iii) creep in b It
t is the effect of centrifugal tension on power tr
'.
e s.
Wh t ." be
ansmlsSlon by a flat belt?
. ~ WI
the effect on the limiting ratio of
.
.
Ihetlon between the belt and rim of pull
. d tensIons of a belt if the coefficient of
same?
ey IS oubled while angle of I
.
[Ans . R .
.
ap remams the
What is the purpose of having an idler pull
. b
'. ano of tensIOn will be squared1
L'1Stout the losses in belt d .
ey In elt dnve ?.
,
. fives.
What are the v .
'T
anous s~sses set up in a belt?
What is the ,.ffi
f
.
l'"
ect 0 centre distance and d'
Wh t'
l8ltleter of puJJey
h
a IS crowning of pulleys?
on t e life of a belt?
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Flol Bells and Pulleys
1.41
PROBLEMS FOR PRACTICE
Problems on design of flat facet! pulleys:
1.
Design a cast iron pulley for the following specifications:
Power transmitted = 18 kW; Speed of shaft = 250 rpm; Ratio of tensions = 3: Velocity of
leather belt = 12.5 m/s; Allowable belt tension = 4200 N/m of belt width: Number of
arms with elliptical section = 6; Safe shear stress for shaft = 56 MPa; Safe tensile stress
for Cf. = 14 MPa.
2.
An overhung pulley transmits 3S kW at 240 r.p.m. The belt drive is vertical and the
angle of wrap may be taken as 1800• The distance of the pulley centre line from the
nearest bearing is 350 mm. Jl = 0.25. Determine:
I. Diameter of the pulley,
2. Width of the pulley assuming thickness of belt of 10 mm,
3. Diameter of the shaft, and
4. Size of the arms (six in number).
Take the section of the arm as elliptical. Safe stress for arms = 15 N/mm2; Safe shear
stress for shafts = 50 N/mm2; Safe stress for belt = 2.5 N/ll1m2; Safe stress for rim = 4.5
N/mm2; Density of leather = 1000 kg/rn'.
Problems on design of flilt belt drive using manufacturer's data :
3. It is required to select ~ flat-belt drive to connect two transmission shafts rotating at 800
and 40.0· r.p.rn. respectively.
The centre-to-centre
distance between the shafts is
approximately 3 m and the belt drive is open type. The power transmitted by the belt is
30 kW. Also select preferred pulley diameters and specify the belt.
4.
S.
6.
Design a fabric belt to transmit 15
1200 r.p.m. The diameter of engine
pulleys is 2 nl.
A belt is to transmit 2S kW at 720
Centre distance between the pulleys
kW at 480 r.p.m. from an engine to a line shaft at
pulley is 600 mm and centre distance between the
r.p.m. to a rolling machine with a speed ratio of 3.
is 2.8 m. Design a suitable belt drive if the rolling
machine pulley diameter is 0.9 111.
Design a fabric belt to transmit 7.5 kW at 8 r.p.s. of an engine to a line shah .u 22 r.p.s.
Engine pulley diameter is 550 mill and centre distance is 2 m.
Problems on design of jlllt belt drive using basic equations :
7.
The specifications
of a belt drive from a motor to an exhaust fan are given below :
Motor
ran
n111l
350
1·100
Angle of lap, radians
2.75
4.25
Coefficient of friction
OJ
0.25
Speed.
l(lOO
~5n
20
-
Pulley diameter.
Lp Ill.
POWl:r. kW
AFrB:J.P~
_
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__
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:rss_
..
)
.X
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Design a/Transmission
Systems
1.42
.
. ible tension of 16 N/mm width is used. The density of the
Leather belt Ilavmg perrrnsst
.
. h oultev i
belt is 1000 k~3. The belt is approximately 8 mm thick. Determme w.hlc pu e~ IS
governing the design of the drive and find the necessary width to transmit t~e required
power taking into account the centrifugal tension also. [Ans: Smaller p~lley • 152 mm]
8.
Determine the width of a 97.5 mm thick leather belt required to transmit 15 ~W from a
motor running at 900 r.p.m. T Iie diiameter 0f tlie dri
nvm g pulley of the motor IS 300 mm.
The driven pulley runs at 300 r.p.m. and the distance between the centre of two pulle~s
is 3 metres. The density of the leather is 1000 kg/m''. The maximum allowable str.ess III
the leather is 2.S MPa. The coefficient of friction between the leather and pulley IS OJ.
[Ails:
Assume open belt drive and neglect the sag and slip of the belt.
9.
80 mm]
The following data relate to a flat belt drive: Power transmitted 18 kW; Pulley diameter
1.8 rn; Angle of contact 1750; Speed of pulley 300 r.p.m.; Coefficient of friction between
belt and pulley surface 0.30; Permissible stress for belt 300 N/cm2; Thickness of belt
8 mm; Density of the belt material 950 kg/rn'. Determine the width of belt required
taking centrifugal tension into account.
[Ails: 44 mm]
10. A pulley is driven by a flat belt, the angle oflap being 1200• The belt is 100 mm wide by
6 mm thick and density 1000 kg/m-'. If the coefficient of friction is 0.3 and the maximum
stress in the belt is not to exceed 2 MPa, find the greatest power which the belt can
transmit and the corresponding speed of the belt.
[Ans: 9.67 kW; 25.82 m/s]
11. An open belt drive connects two parallel shafts 1.2 m apart. The driving and the driven
shafts rotate at 350 r.p.m. and 140 r.p.m. respectively and the driven pulley is 400 mm in
diameter. The belt is 5 mm thick and 80 mm wide. Coefficient of friction between belt
and pulley is 0.3 and maximum permissible tension in the belting is 140 N/cm2.
Determine (a) .diarneter of the driving pulley, (b) maximum power that can be
transmitted by the belting, and (c) required belt tension.
[Ans: 160 mm; 462 W; 395.89 N]
12. The power transmitted between two shafts 3.5 metres apart by a cross belt drive round
the two pulleys 600 mm and 300 mm in diameters, is 6 kW. The speed of the driver is
220 r.p.m. The permissible load on the belt is 25 N/mm width of the belt which is 5 mm
thick. ji = 0.35. Deterrnine : (1) necessary length of the belt; (2) width of the belt; and (3)
necessary initial tension in the belt.
[Ans: 8.472 m ~52 mm ; 888 N]
13. Power is transmitted by an open belt drive from a pulley 300 mm diameter running at
200 r.p.m. to a pulley 500 mm diameter. Angle of lap on the small pulley is 1650• The
belt is on the point of slipping when 2.5 kW is being transmitted.
The coefficient
friction is 0025. It is desired to increase the power to be transmitted.
State which of the
followlng two methods suggested could be more effective?
of
(a) Initial tension in the belt
is increased by 10%, and (b) Suitable dressing is given to the friction surface at the belt
to increase the coefficient of friction by 10%.
[Ans: First method is more effective]
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1.43
Flat Belts and Pulleys
14. In an open belt drive, the linear velocity of the belt is 3 mls. The angle of lap on the
smaller pulley is 166°, the coefficient of friction is 0.3 and the power transmitted is
3 kW. Determine the effect of power transmission in the following cases:
(i)
Initial tension in the belt is increased by 10%,
(ii) Initial tension in the belt is decreased by 10%,
(iii) Angle of lap is increased by 10% by the use of an idler pulley, for the same speed and
the tension on tight side, and
(iv) Coefficient of friction is increased by 10% by suitable dressing to the friction surface of
the belt.
[ Ans : (i) % increase in power = 10% ~ (ii) % decrease in power
=
10%;
(iii) % increase in power = 6.23% ~ (iv) % increase in power = 8.85%]
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V -Belts and Pulleys
"The only way to discover the limtts of the possible
is to go beyond them into the impossible. "
- Arthur C Garke
2.1. INTRODUCTION
V -belts are
mixer, grinder,
machinery, and
that V -belts run
used with electric motors to drive blowers, compressors, appliances (like
etc.), machine tools (like lathe, drilling machine,
farm and industrial
so on. V-belts are endless and run in
pul
grooved pulleys
in are called sheaves.
2.2. CONSTRUCTION OF V-BEL T
Fabric and
rubber cover
power is transmitted
t
between the
t and the V-gr
gi~tiol1
ults in
er
groove an
of a sheave
is made
than the belt section
angle. This
ides the wedging action of the
belt in the groove. The exact value of this angle
depends upon the belt section, the sheave
diameter, and the angle of contact. If it is made
too much smaller than the belt, then the force
required to pull the belt out of the grooves will
be excessive. So usually groove angles of 32°
and 38° are used.
Fabric
Cord
Rubber
Fig. 2.1. Cross-section of V-belt
Sufficient clearance must be provided at the bottom of the groove to prevent the belt from
bottoming as it becomes narrower from wear. Multiple V-belts are used in order to increase
the power capacity. As many as 12 or mote belts are commonly used in heavy-duty
applications.
2.2.1. Materials of V-belts
V -belts are made of cotton fabric and cords moulded in rubber and covered with fabric
and rubber, as shown in Fig.2.1.
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2.2
2.3. ADVANTAGES AND DISADVANTAGES
OF V-BElT
DRIVE OVER
FLA T BELT DRIVE
AdvantagD:
_, Power transmitted is more due to wedging action in the grooved pulley.
_, V-belt is more compact, quiet and shock absorbing.
_, The drive is positive because the slip is negligible due to wedge action.
_, Higher velocity ratio (upto 10) can be obtained.
_,
_,
V-belt drive can operate in any position (i.e., horizontal, vertical or inclined).
Multiple V -belts can be used, thus enabling transmission of more power.
Disadvantages :
_,
It cannot be used with large centre distances.
_,
It is not as durable as flat belt (because of high bending stress that is caused due to
higher ratio of belt depth to diameter of pulley than that in case of flat belt drive).
_,
Since the V -belt is subjected to certain amount of creep, therefore it is not suitable
for constant speed application such as synchronous machines and timing devices .
./
It cannot be used for large power .
./
The efficiency of the V -belt is lower than that of the flat belt.
./
The construction of V-grooved pulleys is complicated and costlier compared with
pulleys of the flat belt drive.
2.4. TYPES OF V-BEl TS
According to Bureau 'of Indian standards (IS :2494-] 974), the V -belts are classified as A.
B, C, D and E type (based on the cross-section of V -belts).The various dimensions of on
standard V-belt sections are shown in Table 2.2.
2.4.1. Specification of V-belts
V-belts are designated by its type and nominal inside length. For example, a C2845 belt
has a cross-section of type C and has a nominal inside length of2845 mm.
2.5. RATIO OF DRIVING TENSIONS FOR V-BEL T
TI
=
eJ.Ul!sin
=
Tensions in the tight and slack sides respectively,
P
= ella'
cosec
P
... (2.1)
T2
where
TI and T2
2P =
fl
=
Angle of the groove, and
Coefficient of friction between the beit and sides of the groove.
Total power transmitted
Number of'V-belts = Power transmitted per belt
.,
-
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(2.2)
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r
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2.3
V-Bells and Pulleys
l
•
2.6. V-FLAT DRIVES
In a V-belt drive, if the large grooved pulley is replaced by a flat-faced pulley (and smaller
pulJey remains V-grooved), then the drive is known as V-flat drive. The larger pulley has a
greater contact angle than the smaller pulley, that compensates for the loss of wedging action.
V-flat drives are used in domestic piston pumps, domestic clothes drier and large punch
presses.
I
I .
2.7. DESIGN OF SHEAVES (OR V-GROOVED PULLEYS)
f
1. Materials of V-grooved pulleys:
The commonly
Table 2.1.
used sheave materials and their characteristics
are summarised
In
Table 2.1.
Material of sheaves
1.
Characteristics
and lor applications
Cast iron
It is economical, stable and durable. Also it has
excellent friction characteristics on V-belts.
2. Pressed steel
It is lighter and cheaper but it gives rise to excessive
belt slip, wear and noise.
3.
Formed steel
Primarily used in automotive and agricultural purposes.
4.
Diecast aluminium
Used for special applications.
2. Dimensions of sheaves:
The cross-section of a sheave (i.e., V-grooved pulley) for V-belt drives
Fig.2.2. (Refer data book, page no. 7.70).
IS
~-----------------------------I------------------I
Fig. 2.2. Cross-section of a sheave
where
Ip
=
Pitch width,
I
=
Face width,
=
Edge of pulley to first groove centre,
f
..
',Scanned
,~;;:' \
'.
_--------by CamScanner
'-:;;:
,
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shown
In
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2.4
Design a/Transmission
Systems
= Centre to centre distance of grooves,
dp = Pulley pitch diameter,
e
b
h
= Minimum distance down to pitch line, and
= Minimum depth below the pitch line.
The various dimensions of standard V-grooved pulley are given in Table 2.2.
Table 2.2. Dimensions of standard V-grooved pulley in mm (from data book, page no. 7.70)
No. of
Groove
Groove
Ip
cross-
dp
b
angle (2P)
section
sheave
I
e
h
grooves
in degrees
(n)
A
11
3.3
7S
32,34,38
8.7
IS
10
6
B
14
4.2
125
32,34,38
10.8
19
12.5
9
C
19
5.7
200
34,36,38
14.3
25.5
17
14
0
27
8.1
355
34,36,38
19.9
37
24
14
E
32
9.6
SOO
-
23.4
44.5
29
20
INote I
Face width, I
= (n -
1) e + 2 f
Design Procedure: The two steps involved in designing a V-grooved pulley are:
Select the cross-section of the belt (i.e., type of belt) depending on the power to be
transmitted, by consulting Table 2.3.
I.
Table 2.3. Data on standard V-belt sections (from data book, page no. 7.58)
Recommended
Cross
section
Area, mm2
Usual load
minimum
of drive, kW
pulley pitch
symbol
dia, d, mm
Nominal top
Nominal
Mass per
width, W
thickness, T
metre, m
mm
mm
kg/m
A
80
0.7S - 5
75
13
8
0.106
B
140
2 -15
125
17
11
0.189
C
230
7.5-75
200
22
14
0.343
0
475
22 -ISO
355
32
19
0.596
E
695
30-190
500
38
23
-
2.
For the selected cross-section of the oelt, consulting Table 2.2, select the various
required dimensions of the V-grooved pulley.
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2.5
V-Belrs and Pulleys
, Example 2.1
..,
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I Design
a Vsgrooved pulley of a V-belt drive to transmit 14. 7 k W to a
compressor.
Given Data : P = 14.7 kW
To flnd : Design a V-grooved pulley (i.e., a sheave).
@Solution :
1. Selection of cross-section of belt:
For the given power transmitted (i.e., P
=
14.7 kW), consulting Table 2.3, the belt cross-
section 'C' is selected.
2. Selection of various dimensions of V-grooved pulley :
The cross-section of a V-grooved pulley is shown in Fig.2.2. For the selected belt crosssection C, consulting Table 2.2, the various dimensions of V -grooved pulley are given as
below.
=
Minimum distance down to pitch line, b =
Pulley pitch diameter, dp =
Groove angle, 2J3 =
Minimum depth below pitch line, h =
Centre to centre distance of grooves, e =
Edge of pulley to first groove centre, / =
Member of sheave grooves, n =
We know that face width, I =
=
Pitch width, Ip
19mm
5.7mm
200mm
34°
14.3 mm
25.5 m
17 mm
14
(n - I) e + 2 /
(14-1)25.5+2x
17
=
365.5mm
3. Material selection:
Since the cast iron is economical,
material for V-grooved pulley.
stable and durable, we can choose cast iron as a
DESIGN OF V-BEL T DRIVE
The two different design procedures used are:
(i)
Using the manufacturer's
data, and
(ii) Using the basic equations.
2.8. DESIGN OF V-BELT DRIVE BASED ON MANUFACTURER'S DATA
The design of V -belt is primarily concerned with the selection of belt section, selection of
pulley diameters, determination of number of belts and centre distance required for the given
transmitted power.
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Design cf Transmisston Systems
2.6
Design procedure:
1. Sttection of belt section:
,
Consulting Table 2.3, select the cross-section
of a belt (i.e., type of belt) depending on the
power to be transmitted.
2. Selection o/pulley diameters (d and D) :
Select small pulley diameter (d) from Table 2.3. Then using the speed ratio, calculate the
large pulley diameter (D). These pulley diameters should be rounded off to a standard
diameter by using Table 1.5.
;<Selection
of centre distance (C) :
Select the centre distance, if not given, from Table 2.4.
Table 2.4. Selection of centre distance, C (from data book, page
CIO ratio
2
3
4
5
1.5
1.2
I
0.95
0.9
Cmin = 0.55 (0 + d) + T ~
_getermination
7.61)
I
Speed ratio i = Old
Recommended
110.
C max
=
6
to
9
0.85
2 (0 + d)
of nominal pitch length:
Determine the length of the belt L (which is also known as nominal inside length) by
using the formula,
L
=
2C +
(n)'2
(D-d)2
(0 + d) + - 4 C
For the calculated nominal inside length and belt section, consulting Table 2.5, select the
next standard pitch length.
INote I ./' Tile nominal
pitch length is defined as the circumferential
length of the belt at the pitch
width (i. e., the width at the neutral axis of the belt). The value of the pitch width remains constant for
each type of belt irrespective of the groove angle .
./' For pitch length, add with inside length, 36 mm for A belt, 43 mm for B, 56 mm for C. 79 mm
for 0 and 92 mm for E belt.
5. Selection of various modification factors:
In order to calculate
determ ined.
(i) Length
the design power. the following
correction factor (F J
modification
factors have to be
:
For a given sheave speed. longer belt has more life than a short one. because the shorter
bell is subjected to the action of the load a greater number of times. For this reason. belt
correction factor (Fc).is used.
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2.7
V-Belts and Pulleys
Table 1.5. Nominal Inside length, nominal pitch length and length correction/actor/or
standard
sizes 0/ JI-belts (from data book, page 110. 7.58, 7.59 and 7.60)
(The values for a few cases only given)
Nominll
Correction
Nominal pitch length, mm
factor,
Fe
inside
A
B
C
D
E
A
B
C
D
E
610
645
-
-
-
-
0.80
-
-
-
-
965
1001
1008
-
-
-
0.88
0.83
-
-
-
1905
1941
J948
J9~1
-
-
1.02
0.97
0.87
-
-
2438
2474
-
2494
-
1.08
-
0.92
-
-
2667
2703
2710
2723
-
-
1.10
1.04
0.94
-
-
3048
3084
3091
3104
3J27
-
1.13
1.07
0.97
0.86
-
3251
3287
3294
3307
3330
-
1.14
1.08
0.98
0.87
-
4013
-
4056
4069
4092
-
-
l.J3
1.02
0.92
-
4572
-
4615
4628
465J
-
-
l.J6
1.05
0.94
-
5334
-
5377
5390
5413
5426
-
1.19
1.08
0.96
0.94
-
l.ll
1.00
0.96
-
1.21
1.09
1.05
length,
mm
6045
-
-
6101
6124
6137
._
9093
-
-
9149
9172
9185
-
For the selected belt cross-section, choose length correction factor (Fe) from Table 2.5.
(ii) Correction factor for arc oj contact (Fttl
~
.-
First determine the angle of contact (or arc of contact) of the smaller pulley.
Arc of contact = 1800
-
(D-d)
x 600
C
... [From data book, page no. 7.68]
~
For the calculated arc of contact, select the correction factor from Table 2.6.
~
Arc of contact factor is taken into account because the power transmitted may be
limited by slipping of the belt on the smaller pulley.
Table 2.6. Arc
0/ contact/actor, Fd
(from data book, page no. 7.68)
(The values/or a few cases are given below)
Arc of contact on smaller
pulley (in degrees)
180
171
160
151
139
130
120
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Correction
factor (proportion
of 1800 ratine)
V-V
V-belt on V-pulley
V-Flat
V-belt on flat pulley
1.00
0.98
0.95
0.93
0.89
0.86
0.82
0.75
0.77
0.80
0.82
0.85
0.86
0.82
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Design of Transmission Systems
2.8
(iii) Service factor (F J
:
./
Select the service factor (Fa) consulting Table 2.7.
./
The service factor takes into account the severity of the load transmitted which
depends upon the characteristics of the driving and driven units.
Table 2.7. Service factor for V-betts, Fa (from data book, page no. 7.69)
Driving units type - I
10 hrs.
Service
16 hrs.
Driving units type - ((
Over 16 hrs.
10 hrs.
(per day)
Light duty
Medium duty
Heavy duty
1.0
1.1
I.l
1.2
1.2
1.3
1.4
1.3
Extra heavy duty
16 hrs.
Over 16 hrs.
(per day)
1.2
1.1
1.2
1.3
1.2
1.3
1.3
1.4
1.4
1.4
1.5
1.6
1.5
1.5
1.6
1.8
INote I The details of driving units and driven machines under different duties are available
in the
data book, page no. 7.69.
6. Calculation of maximum power capacity :
Calculate the maximum power capacity (in kW) of a V-belt using the formulas given in
Table 2.8.
Table 2.B. (from data book, page no. 7.62)
Belt cross-section
Maximum value of
'de' in the formula
Formula
symbol
where
mm
=
A
kW
B
kW
=
C
kW
=
D
kW
=
E
kW
(0.45S-O·09 - 1~62 -0.765
x 10-4 S2) S
e
:0:
(0.79s-O·09 - 5~.8 - 1.32 x 10-4 S2) S
e
142.7
(1.4 7 S-O'09 - -d- 2.34 x 10-4 S2) S
e
(3.22S-O·09 - 5~.7
e
- 4.78 x 10-4 S2) S
(4.SSs-O·09 - 9:2 -7.0S x 10-4 S2) S
e
125
175
300 .
425
700
kW = Maximum power in kW at 1800 arc of contact for a belt of average length,
S = Belt speed, mis,
de = Equivalent pitch diameter
dp
= dp x Fb'
= Pitch diameter of the smaller pulley, mm, and
Fb = Small diameter factor to account for variation of arc of contact, from
Table 2.9.
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~V.~-B~e~/u~~an~d~P~u~I~/~~~~
~~
Table 2.9. (from data book, page no. 7.62)
Speed ntio ranle Old
Speed ratio range Old
Small dll. flctor, Fb
1.0 to 1.019
1.00
1.02 to 1.032
1.01
1.033 to 1.055
1.02
1.056 to 1.081
1.03
1.082 to 1.109
1.04
1.11 to 1.142
1.143 to 1.178
1.179 to 1.222
Small dil. flctor, Fb
1.223 to 1.274
1.215 to 1.340
1.08
1.341 to 1.429
1.43 to 1.562
1.10
-
1.09
1.11
1.12
l.OS
to 1.814
1.8tS to 2.948
1.13
1.06
2.949 and over
1.14
I.S63
1.07
7. Determination of number of belts (n!) :
Determine the number of belts (nb) from the relation,
P
nb =
where
x
Fa
... [From data book, page no. 7.70]
kWxFexFd
P
=
Drive power, in kW,
Fa
=
Service factor for V -belts,
kW
Fe
Fd
= Rated power (i.e., rating of a single
= Length correction factor, and
= Correction factor for arc of contact.
V -belt),
8. CaJcuJllIionof actual centre distance:
-'
Calculate the actual centre distance from the relation,
CactuaJ = A +
where
",j A2 -
... [From data book, page no, 7.61]
B
A=~-n[D;dJ
B
:c:
ill~df,
and
L ::: Nominal pitch length of the belt from Table 2.5 (refer step 4).
I ExaMple 2.2 I Design a V-belt drive to the following
Power to be transmitted
=
7.5/rW
specifications :
.
Speed of driving wheel - 1440 r.p.m;
Speed of driven wheel
:=
400 r.p.m.
Diameter of driving wheel =- 300 mm
Centre distance =- 1000"""
Service
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=- 16 hours / day
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Design a/Transmission Systems
2.10
Given Data:
P
= 7.5
d
= 300 111m = OJ m;
kW ;
=
N,
N2 = 400 r.p.m. ;
1440 r.p.m. ;
C
=
1000 mm = 1 m.
Toflnd : Design a V -belt drive.
© Solution:
1. Selectioll of the belt section:
Consulting
Table 2J, for power 7.5 kW, B section is selected.
2. Selection of pulley diameters (If and D) :
Smaller pulley diameter,
d
=
Nt
1440
d = N2 = 400 = 3.6
=
300
... (Given)
mill
Referring Table 1.5, the preferred smaller pulley diameter,
..
Larger pulley diameter,
C~~~)
D
Speed ratio
d
= 315
mm.
D = 3.6 d = 3.6 x 3) 5 = I) 34 mm
Referring Table 1.5, the preferred larger pulley diameter,
D = 1250 mm.
3. Setection of centre distance (C) :
Centre distance,
(0/
C = 1000 mm
... (Given)
4. Determination of nominal pitch lengtt: :
. I' IllSIide Iengt I1, L
Nomina
=... ..,C
+ (
=
2x
1000+(¥)
=
4676.85
2'7t )
(D + d ) + (D4- Cd)2
(1250+3IS)+(I~5~~:d:)2
I11Ill
For this nominal inside length and 0 section. consulting
length is selected as 4996 mm.
Table 2.5. the next standard pitch
S. Selection of various modlflcotion factors :
(i) Length correction factor (F cJ
length correction factor,
:
For B section. referring Table 2.5.
Fe -= 1.18
{ii) Correction factor for arc of contact (F ,J
Arc of contact
tr
:
D-d)
(--C
=
I 80° -
=
1800 _ ( ) 250 - 3 ) 5)
1000
x
60°
x 600 =
-_.-It
_
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2.11
V-Beits and Pulleys
For this arc of contact, consulting Table 2.6, correction factor for arc of contact is selected
as Fd=O.83.
(iii) Service factor (F J:
Consulting
Table 2.7, for light duty 16 hours continuous
service, for driving machines of type.Il, service factor is selected as Fa = 1.3 ..
6. Calculation of maximum power capacity :
Consulting Table 2.8, for B section, power capacity formula is given as
kW = (0.79 S-009 - 5~~8 _ 1.32 x 10-4 S2 ) S
where
7t
x 0.3 ~~ x 1440
Belt speed
de
=
=
=
=
Equivalent pitch diameter = dp' x Fb
F~
de
=
=
=
dp
..
7tdN)
S
60
=
23.75 mls
Pitch diameter of the smaller pulley = d. = 315 mm
Small diameter factor for speed ratio of3.6 = 1.14
315 x 1.14
=
... [From Table 2J]
359.1 mm
But from Table 2.8, maximum value of de in the formula should be 175 mm.
Power, kW
=
(0.79 x 23.75-0.09-
~~.:
-
1.32 x 10-4 x 23.752)
23.75
=
5.445 kW
7. Determination of "umber of belts (n,) :
P x Fa
We know that
=
7.5 x 1.3
5.445 x 1.18 x 0.83
=
1.828:::: 2 belts ADS."
8. Calculation of actual centre distance:
Actual centre distance is given by
Cactual
where
and
=
A + \} A 2 - B
L _ [D+dJ
A = 4
B
Cactual
8
7t
=
634.42
=
{D-d)1
8
=
+ \}
4996
--4
{1250 - 315}2
8
= 634.42
(634.42)1= 1175.92 mm ADS. ~
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- " [ 1250 + 315 ]
8
=
109278
109278
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2.12
Design of Transmisston Systems
I
[!xamele 2.3 A centrifugal pump running at 340 r.p.m: is to be driven by a 100 kW
motor running at 1440 r.p.m. The drive is to work for atleast 20 hours every tillY. Tile
centre distance between the motor shaft anti the pump shaft is 1200 mm. Suggest a suitable
multiple V-belt drivefor this application. Also calculate the actual belt tensions and stress
induced.
Given Data: N2 = 340 r.p.m.; P = 100 kW; N) = 1440 r.p.m.; C = 1200 mm = 1.2 m
Tofind:
(i)
Design a V -belt drive, and
(ii)
Actual belt tensions and stress induced.
@Solution :
1. Selection of tile belt section:
Consulting Table.2.3, for power 100 kW, D section is selected.
2. Selection of pulley diameters (d and D) :
Since diameters of both pulleys are not given, therefore first select the smaller pulley
diameter from Table 2.3.
.. Consulting Table 2.3', for power 100 kW, smaller pulley diameter, d
Speed ratio
.. Larger pulley diameter, D
N)
D
=
355 mm.
1440
= -d = -N2 = -340 = 4235
.
=
4.235 x d
=
4.235 x 355
=
1503.53 mm
Consulting Table 1.5, the preferred larger pulley diameter, 0 = 1600 mm.
3. Selection of centre distance (C) :
Centre distance, C
=
1200 mm
... (Given)
4. Determination of nominal pitch length:
Nominal inside length, L = 2 C
7t
+
'2
(D + d) +
(0 -
d)2
4C
7t
=
2x 1200 +
'2
(1600+355)
+
(1600 - 355)2
4x 1200
= 5793.83 mm
For this nominal inside length and 0 section, consulting Table 2.5, the next standard pitch
length is selected as 6124 mm.
5. Selection of various modification factors:
(i) Length correction factor:
Fe = 1.00.
For 0 section, referring Table 2.5, length correct i '," "actor,
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2. l.i
"-8 !,~
I "lit ."
~ --------
-----'-'-~
(ii) Correction factor for arc of contact (F; :
Arc of contact
= 1800 _ (D ~ d )
=
x 60° = 180° - (
J 600 - 355)
1200
600
117.75°
For 117.75°, consulting Table 2.6, correction factor for arc of contact
For light duty, for over 16 hours continuous
machines of type II, consulting Table 2.7, the service factor, Fa = 1.3.
(iii) Service factor (F,.):
6. Calculation of maximum power capacity (k U1
Fd = 0.81.
service, for driving
:
Consulting Table 2.8. for D section, power capacity formula is given as
kW ~ (3.22
where
S
de
=
5~.7
S"'()09 ~
1t
Belt speed =
= dp
~ 4.78
d NI
=
60
7. Determination
= 355
=
S2 )
S
x 0.355 x 1440
60
1t
=
Fb = Small diameter factor,
Power, kW
10-4
=
26.76 mls
x Fb
dp = Smaller pulley diameter
dt!
x
355 mm
for speed ratio of 4.235, from Table 2.9
=
1.14
x 1.14 = 404.7
( 3.22 x 26.76-0.09
= 21.44 kW
0/ number 0/ belts
506.7
)
404.7 - 4.78x 10-4 x 26.762
26.76
-
(n,) :
P x Fa
We know that
nb
= kW
x
Fe
100
21.44
=
8. Calculation
0/ actual
A
B
Caclual
,
..
X
Fd
1.3
1 X 0.81
=
7.486 ~ 8 belts Ans."
centre distance:
Actual centre distance,
where
X
X
=
Caclual
L
4 -
1t
In_
d)2
= ~8=
A + \] A2 - B
=
[O+d
8
] _ 6124
_ 4 -
(1600-355)2
= -
8
-
=
1t
[1600+355
8
]
= 763.27
193753.125
763.27 + ~ 763.272 - 193753.125
= 1386.83mm Ans. 1.1
~
"'::-.
t
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Design of Transmission Systems
2.14
9. Calculation of belt tensions (TI and T~ :
We know that,
power transmitted per belt =
100 x 103
8
... "(i)
From Table 2.3, mass per metre length, m = 0.596 kglm
groove angle (2(3) = 34°.
From Table 2.2,
Already found that arc of contact for smaller pulley, a. = 117.75
o
1t
x 1800
2.055 radians
=
We know that the tension ratio for V -belts considering centrifugal tension,
Tl - mv'l
= ef.UX I sin
T2-mv2
T, - 0.596 (26.76)2
T2 - 0.596 (26.76)2
=
P
= ef.UX· cosec ~
eOJ x 2.055 x cosec 17°
=
8.237
Tl - 8.237 T2 = - 3088.68
or
T1
Solving (i) and (ii), we get
=
... (ii)
958.45 Nand
T 2 = 491.33 N Ans. ~
10. Calculation of stress 'nduced :
Consulting Table 2.3, cross-sectional area of 0 section
Stress induced
=
=
475 mm2
Maximum tension
Cross-sectional area
= 2.02 N/mm2
=
958.45
475
Ans. ~
2.9. DESIGN OF V-BELT DRIVES USING BASIC EQUATIONS
As discussed in section 1.19, the load carrying capacity of a pair of pulleys is determined
by the pulley which has the smaller value of ef.UX· cosec P . Thus the pulley' which has smaller
tension ratio governs the design.
( Example 2.4
IA
V-belt having a lap of 180 0 has a cross-section
groove angle as 4S~ The density of a belt
is 0.0015
area of 2.5
kglcmJ and maximum
to 400 x 1(J4 Nlml. If J.l = 0.15, find the power that can be transmitted,
mean diameter of 300 mm and runs at 1000 r.p.m:
.....
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stress
if
cml and
is limil~d
the wheel has a
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2.IS
"-Belts and Pulleys
Given Data: a
p
d
=
=
= 180 =
0
0.0015 kg I em!
300
Tofind:
mm
= 0.3
=
=
1800 x 1:00
1t
radians;
=
0.0015 x 1()6 kg 1m3; a
m;N
a = 2.5 em2
= 2.5
400 x 104 N/m2;
"
1
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x 10-4 m2;
J.l = 0.15 ;
= 1000 r.p.m.
Power transmitted (P).
e SO/lIIion :
v
=
1tdN
=
60
T
1t
x 0.3 x 1000
60
= 15.71 m1s
=
e~' cosec p = eO.IS x
=
3.426
TI
=
3.426 T2
Mass per unit length of belt, m
=
Density x Area x Length
We know that,
Tension ratio T:
or
7t x
cosec 22.SO
... (i)
= 0.0015 x 106 x 2.5 x 10-4 x 1
...
and
Centrifugal tension, Tc
maximum tension in the belt, T
= 0.375 kg I m
= mv2 = 0.375(15.71)2 =
=
=
crxa
400 x IQ4x2.5x
92.65N
10-4
= 1000N
We know that the tension in the tight side of the belt,
T
=
TI +Tc
T, = T - Tc = 1000 - 92.55 = 907.5 N
T,
T2 = 3.426
From equation (i),
Power transmitted,
P = (T t
-
907.5
- 3.426
2.5
I Two shafts
264.9 N
T2) v
= (907.5 -264.9)
I Examp/e
=
x
15.71 = 10.1 kW
Ans.1J
whose centres are 1 m apart are connected by a V-belt
drive. The driving pulley is supplied with 100 k Wand has an effective diameter of 300 mfIL
It runs at 1000 r.p.m: while the driven pulley runs at 375 r.p.m: The angle of groove on the
pulleys is 40 ~ The permissible tension in 400 mm2 cross-sectional area of belt is 2.1 MP".
The dellSity of the belt is 1100 kg/ml. Taking JJ = 0.28, estimate the number of belts
nqllued. Also calculate the lengll, required of each belt.
C = l m ; P = 100 k W = 100 x 103 W;
d = 300 mm = 0.3 m ;
N. = 1000 r.p.m.; N2 = 375 r.p.m.; 2P = 40° or p = 20°; a = 400 mm2 = 400 x l~ m2:
Given Data:
(J
= 2.1
MPa
'--.
= 2.1
x
106 N/m2 ; p
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=
1100 kg 1m3;
J.l = 0.28.
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Design o/Transmission Systmr.
2.16
Tofind: Number of belts required and length of each belt.
@Solution:
\1
=
1t
d NI
7t
=
60
x 0.3 x 1000
= 15.71m1s
60
. N2
d
Speed ratio, N I = 0
D
or
=
1000
NI
N2 x d
= 375 x OJ
0.8 m
=
Number of belts required: For an open belt drive,
sin a =
Angle of contact,
e
TI
or
=
0.8-0.3
2x I
= (180° - 2 a)
=
T
Tension ratio, T~
O-d
'""2C:
= 0.25 or a = 14.48°
x 1~O = (180° - 2 x 14.48°) x 1:00
2.636 rad
= e.,.e· cosec Ii = eO.28
= 2.158
x
2.636
x
cosec 20°
= 2.158
... (i)
T2
Mass of the belt per metre length, m = Density x Area x Length
Centrifugal tension, Te
and
Maximum tension in the belt, T
=
1100
=
m y2
= 0.44 kglm
= (0.44)(15.71)2 = IO?'.'~9N
x
400
x
IQ-6 x I
= a· a
= 2.1 x 106 x 400
x IQ-6 = 840 N
We know that tension in the tight side of the belt,
T
=
TI +Te
or
TI
=
T - Te
From equation (i),
T2
.z;
= 2.158 = 2.158 = 338.93 N
= 840 - 108.:59 = 731.41
N
731.41
..
Power transm itted, P
=
(TI - T2) v
= (731.41 - 338.93) 15.71 = 6165.86 W
=
Number of belts
=
Total power transmitted
Power transmitted per belt
100 x 103
6165.86
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=
16.22
=:
17 Ans. ~
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V-Belts and Pulleys
2.17
Length of each belt: We know that length of each belt,
L = 2 C + ~ (0 + d) + (D - d)2
2
2C
=
I Example
2 x 1 +; (0.8 + OJ) + (O.~~ ~J)2
=
3.852 mAns.
"
I Power
is transmitted between two shafts by a V-belt whose mass is 0.9
kglm length. The maximum permissible tension in the belt is limited to 2.2 kN. The angle
0
of lap is 170 and the groove angle 45 ~ If the coefficient of friction between the belt and
pulleys is 0.17,find
(i)
2.6
Velocity of the belt for maximum power, and
(ii) Power transmitted at this velocity.
Given Data:
m = 0.9 kg/m ; T = 2.2 kN = 2200 N ; 2J3 = 45° or J3= 22.5° ;
e = 170° = 170° x
To find:
7t
1800 = 2.967 rad; J.l = 0.17.
(i) Velocity of the belt ("), and (ii) Power transmitted (P).
© Solution:
(i) Velocity of the belt:
Velocity of the belt for maximum power to be transmitted,
fl
. v - -\j];
-
---~
-\j 3 x 0.9
=
28.545 mls Ans."
(ii) Maximum power transmitted:
T
Tension ratio, T~ = e~a· cosec p = eO.)7 x 2.967x cosec 22.5° = 3.736
... (i)
or
Centrifugal tension for the maximum power,
Tc = m v2
=
or For maximum power transmitted, Tc =
=
0.9 (28.545)2
=
733.33 N
Maximum tension
3
and tension in the tight side,
T)
T - Tc
From equation (i),
T)
T2 = 3.736
=
2200
-3= 733.33 N
=
2200 -733.33
=
1466.67
3.736
=
=
1466.67 N
392.58 N
Maximum power transmitted, P = (T) - T2) v
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=
(1466.67 - 392.58) 28.545
=
30.66kW Ans."
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Design of Transmission Systems
2.18
[Example
2.71
A compressor is driven by a 1400 r.p.m. motor by means of a flat belt
of thickness 20 mm and width 250 mm. The motor pulley is 115 mm diameter and the
compressor pulley is 1250 mm. TIle stmft centre distance is 1200 mm anti an idler is used to
make the angle of lap on tile smaller pulley 2100 and Oil larger pulley 280". Coefficient of
friction is 0.25 on tire smaller pulley and 0.3 011 tile larger pulley. The maximum allowable
belt stress is 2.J Nlmml and the belt weighs 1 x I 5 Nrmm', Determine the power of tile
rr
drive.
To eliminate idler pulley, tire smaller pulley is changed with Vsgroove pulley with
groove angle of 14". J.J = 0.1. Larger pulley and other data remaining the same, what shall
be the power transmitted?
Given Data:
N)
=
d = 315 mm = 0.315
as = 210°
=
1 x 104 N/m3
;
Thickness
D = 1250 mm
m;
7t'
= 210°
III = OJ ; o
1400 r.p.rn.;
x 1800 = 3.665 rad ; aL
2.1 N/mm2
=
=
20 mill;
7t
= 280° = 280°
2.1 x 106 N/m2
;
C = 1200 mm
1.25 m
=
=
Width
=
x 1800
250
mm
=
1.2 m
= 0.25
;
1 x 100s Nzrnrn '
=
4.886 rad ; ~s
Weight density of belt
=
213 = 34° or 13= 17°.
©Solution:
J. Power transmitted by the flat belt drive:
Since the friction coefficients for both pulleys are different, therefore first we have to find
the pulley which governs the design. So we have to evaluate ells as and ellL al for the
pulleys.
e!lS as
<
ellL aL.
ellS as
=
elllal
= e°.3x4.886
Therefore
eO.25 x 3.665 =
Weight density
=
1x
104
N/m3
V
=
d N1
60
=
.
or Mass density
Mass of the belt per metre length
Centrifugal
= 4.33
smaller pulley governs the design.
1t
Velocity of belt,
2.5
tension,
Maximum tension,
Tc
x 0.315 x 1400
60
=
=
23.09 m/s
1 x 104
9.81
kg/m '
=
Density x Area x Length
=
I x 104
9.81
=
=
x
(0.25
x
0.02)
x
1
5.096 kg/m
m v- = 5.096(23.09)2
T = Stress x Area
=
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1t
=
=
2717.37N
c x (b x I)
2.1 x 1Of)x (0.25 x 0.02)
= )0500
N
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r;~~~~1
~
~Belts
and Pu eys
__
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---.:.2~19
---...
---------:-
...
Tight side tension, T 1
=
T - Tc
= 10500-2717.37
= 7782.63 N
We also know that
= 7782.63
Slack side tension, T2 =
= 3 113 N
2.5
P = (T 1 - T 2) v
Power transmitted,
=
(7782.63 - 3113) 23.09
=
107.82kW Ans. ~
11. Power transmitted by the V-Flat drive:
Given that, in order to eliminate idler pulley the smaller pulley is changed with V-groovs
pulley and the larger pulley remains flat faced pulley (which is known as V -flat belt drive).
Since smaller pulley is changed to V -grooved pulley without idler, therefore we have to
calculate the new angles of contact.
Ang Ie 0 f contact,
<l
= SID
. -I (O-d)
2C
= SID
. _1(1250-315)
2 x 1200
0
=2293
.
n
and
<lL
n
= (1800 + 2(1) 1800 = (1800 + 2 x 22.930) 1800 = 3.94 rad
Again we have to calculate the tension ratios for both pulleys to find the governing pulley.
TI)
( T2
S
1
and
(T
Wefind
(~:)L
.)
T2
L
_
ell as . cosec
-
= el1QL
<
11 =
eO.) x 2.)4 x
= eO.3
x).94
cosec 17°
= 11.03
= 3.26
(~:)s
Therefore, the larger pulley (i.e., flat faced pulley) governs the design.
Tight side tension, T 1 = 7782.63 N
TI
Slack side tension, T2 = 3.26 =
(Already calculated)
7782.63
3.26 = 2386.6 N
Power transmitted, P = (T1 - T2) v
= (7782.63 - 2386.6) 23.09 = 124.59 kW Ans.
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Design of Transmlsslon SY81ems
2.20
2.10. RIBBED V-BELTS
t2.10.1. Introduction
Ribbed V-belts is a cable cord reinforced
transmission belt with 'V' shaped ribs along the
circumference of the belt, as shown in Fig.2.3. It is
flexible, with high power capacity per unit width and
is designed on completely integrated construction.
The ribbed surface provides maximum area of contact
and reduced face pressure.
Fig. 2.3. Ribbed V-belt
The ribbed V-belt is gaining wide acceptance in the automotive field due to the
advantages it exhibits over conventional V-belts. These belts are particularly good for high
speed and I or high ratio applications which conventional belts are unable to handle. The
smooth, vibration free performance and excellent reverse bend characteristics of this belt
enable the design of single belt, compact drive systems.
2.10.2. Applications
Ribbed V-belts have been selected for use on many applications. These applications range
from iil:: single multi-rib accessory drive concept (SMAC) for passenger cars to water pump,
alten\ator, and air conditioning drives on heavy trucks.
\.
2.10.3. Special Features of Ribbed V-8elts
./
Highly flexible and can be used with smaller pulley diameter resulting in lighter
and compact drive .
./
Higher power rating per unit width compared to conventional belt - 40% higher.
./
Almost eliminates the slippage due to maximum wedge contact on the pulleys .
./
Being thinner, the rear side can be used to drive additional accessories I idlers
without affecting belt life.
./
Gives noise-free power transmission in every application .
./
Eliminates the necessity to use multiple I matched set of belts .
./
Can be used at higher speed more than 40 m/sec .
./
Give lesser wear and longer life to the pulleys due to lesser static tension and belt
slippage.
'"
Reduces down time and hence belt replacement cost.
2.10.4. Construction
Features of Ribbed V-8elts
Ribbed V-belt sections which are designated by the letter "K" and "J" are easily
identifiable, as shown in Fig.2.4, by measuring the distance between the adjacent ribs within
-
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)
q
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~V~-B~p~/~~~a~"~~~P~u~/I~~~'~~
-----------------------------------------------~
a belt. The "J" section belt has a smaller rib than the "K" belt and the overall thickness of the
"J" is less than the "K". Refer Table 2.10.
PK Type
PJ Type
PH Type
W3.56mm
W2.34mm
W1.60mm
(c)
(b)
(a)
(e)
(d)
Fig. 2.4. Construction of ribbed V-betts
The polyester tension cords are embedded into the belt to achieve longitudinal stability.
Then crosswise fibres are incorporated into the polychloroprene substructure for unsurpassed
transverse bending strength, finally the belts are ensured for maximum resistance to
temperature fluctuations and ozone.
Table 2. J O.Dimensions of various ribbed V-belts
Pitch:
P
Height of Rib:
Profile
(mm)
ht
Height of belt:
(mill)
(mm)
H
1.6
1.1
3.0±0.15
J
2.34
1.8
3.9 ± 0.25
K
3.56
2.4
5.5 ± 0.30
L
4.7
4.6
9.0 ± 0.40
M
9.4
9.4
16.0 + 0.60
H
2.11. TIMING (OR SYNCHRONOUS) BELTS
Flat belt and V-belt drives cannot provide a precise speed ratio. because slippage occurs at
the sheaves. But certain applications require an exact output to input speed ratio. In such
situations. timing belts are used. The timing belts have teeth that mesh with bzroovcs in the
sheaves, as shown in Fig.2.5. Most timing belts are reinforced with glass fiber. steel or
aramid.
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Design of Transmission Systems
2.22
Fig, 1,5, Timing belt
REVIEW AND SUMMARY
.,f
The grooved pulleys that V-belts run in are called sheaves .
.,f
Materials
of V-belts : V-belts are made of cotton fabric
and cords moulded
ill rubber
and covered with fabric and rubber .
.,f
Types of V-belts : Based on the cross-section,
as A, B, C, D and E
V-belts are classified
type .
.,f
Specification
of V-belts : C2845 means C is the belt section and 2845 mm is the nominal
inside length.
Tensi
.
enston rano,
T
T2I
=
Number of Vsbelts
=
2;J
ella· cosec p where
Angle of the groove.
=
TOlal power transmilled
Power trans milled per belt
V-flat drive: A V-bett with one flat-faced
pulley and other V-grooved pulley is known as
V-j1at drive.
Materials
of sheaves:
Cast iron, pressed steel, formed
The design of Y-grooved pulleys are presented
Two
different
manufacturer's
design procedures
used
steel and diecast aluminium.
in this chapter.
for
V-belt
drives
are:
(i)
Using
the
data. and (ii) Using the basic equations.
Step by step procedure
for
presented
example problems.
with sufficient
the design
of Vsbelts using
the manufacturer's
V-belt design " To find (i) belt section (A, B, C, D or E), (ii) nominal
data
is
inside length, and
(iii) number of belts.
Design using basic equations
: The pullev which has the smaller
value of lens ion ratio
will govern the design of the drive.
At the end of this chapter, an engineering
brief about the ribbed
V-belts and timing belts
are presented.
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V-Bells and Pulley!
2.23
REVIEW QUESTIONS
1.
What are the advantages and disadvantages
drive?
of V-belt drive when compared with flat
2.
Why slip is less in the case of V-belts when compared with flat belts ?
3.
Why V-belts are preferred than flat belts?
4.
What are the materials used in V-behs ?
S.
How can you specify a V-belt?
6.
What are the materials used for sheaves (or V-grooved pulleys)?
7.
Write an engineering
belt
brief about (a) Ribbed V-belts, and (b) Timing belts.
PROBLEMS FOR PRACTICE
Problems on design of sheaves (or V-grooved pulleys) :
I.
Design a V -grooved pulley of a V-belt drive to transmit 25 kW to an electric generator.
2.
100 kW power is transmitted using a V-belt drive. Design a sheave for the given V-belt
drive.
Problems on design of V-belt drive using manufacturer's data:
3.
60 kW of power at 720 r.p.m. is to be transmitted to a compressor shaft at 300 r.p.m. by
V-belts. Diameter of larger pulley is approximately 1500 mrn. Approximate
distance is 1650 mm. Overload factor is 1.5. Design the V-belt drive.
centre
4.
A 30 kW, 1440 r.p.m. motor is to drive a compressor by means of'V-belts. The diameter
of pulleys are 220 mm and 750 mm. The centre distance between the compressor and
motor is 1400 mm. Design a suitable drive.
S.
Select a V-belt drive for transmitting 1.5 kW from a motor running at 1450 r.p.m. to a
blower at 300 r.p.m. in an air conditioning plant. The centre distance should be atleast
1.5 times the diameter of the larger pulley. Diameter of motor pulley is 300 mm.
6.
Design a V-belt drive to transmit 10 kW at 400 r.p.m. The speed ratio is 3. The distance
between the pulley centres is 600 mm. The drive is for a crusher.
7.
A 10 kW, 720 r.p.m. motor is to drive a mixer at 180 r.p.m. The starting load is heavy
and the service is intermittent. Select suitable drive and determine the main dimensions
of the drive.
8.
A crusher running at a speed of 400 r.p.m. is driven by an electric motor of 10 kW with a
reduction in speed of 3 times. Taking a centre distance of 600 mm, design a V-belt drive.
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Design of Transmission Systems
2.24
Problems on design of V-belt drive using basic equations:
9.
Two shafts whose centres are I m apart are connected by a V-belt drive. The driving
pulley is supplied with 10 kW and has an effective diameter of 300 r.p.m. It runs at 1000
r.p.m. while the driven pulley runs at 375 r.p.m. The angle of groove on the pulley is
40°. The permissible tension in 400 mm? cross-sectional area of belt is 2.1 MPa. The
density of the belt is 1100 kglm3. J.l = 0.3. Estimate the number of belts required.
[Ans: I0 belts]
10. An engine of 25 kW, has a speed of 650 r.p.m. It has to drive an electric generator at
1500 r.p.m. by means of V-belts. Diameter of pulley on the engine is 600 mm and the
centre to centre distance between the shafts is 3 metre. Groove angle of the pulley is 30°
and the coefficient of friction is 0.25. If the maximum safe strength of the belt is 500 N,
determine the number of V-belts required for the drive. Neglect the belt thickness, and
assume the open drive.
[Ans: 3 belts]
II. A belt drive consists of two V -belts in parallel, on grooved pulleys of the same size. The
angle of groove is 30°. The cross-sectional area of each belt is 750 mm2 and J.l = 0.12.
The density of the belt material is 1.2 Mg/rn! and maximum safe stress in the material is
7 MPa. Calculate the power that can be transmitted between pulleys 300 mm diameter
rotating at 1500 r.p.m. Find also the shaft speed in r.p.m. at which the power transmitted
would be a maximum.
[Ans: 171.75 kW ; 2809 r.p.m.]
12. Power is transmitted using a V-bclt drive. 'J he ilfl.:!~:dcd~i1f,le vI" V-groove is 30°. The.
belt is 20 mill deep and maximum width is 20 mm. If the mass of the belt is 0.35 kg per
metre length and maximum allowable stress is 1.4 MPa, determine the maximum power
transmitted when the angle of lap is 140°. J.l = 0.15.
[Ans: 6.53 kW]
Problems on V-flat belt drive:
13. A compressor is driven by a 1500 r.p.m. motor by means of a flat belt of thickness
20 mm and width 250 mm. The motor pulley is 300 mm in diameter and the compressor
pulley is 1440 mill. The shaft centre distance is 1200 mm and an idler is used to make
the angle of lap on the smaller pulley 215° and on larger pulley 275°. Coefficient of
friction is 0.25 on the larger pulley and 0.3 on the smaller pulley. The maximum
allowable belt stress is 2.2 N/mm2 and the belt weighs I x IO-s Nzmrn-'. Determine the
power of the drive.
To eliminate idler pulley, the smaller pulley is changed with V-grooved pulley with
e = 34°. Il = 0.25. Larger pulley and other data remaining the same, what shall be the
power transmitted?
[Ans: 117.2 kW ; 136.78 kW]
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Wire Ropes and Pulleys
..There is a difference between happiness and wisdom.
He that thinks himself the happiest man is really so:
but he that thinks himself the wisest is generally the greatest fool."
- Francis Baro"
3.1. INTRODUCTION
Wire ropes are used whenever large power is to be transmitted over long distances (upto
150 m). The wire ropes are extensively used in elevators, oil
II drilling, mine hoists,
cranes, hauling devices, conveyors, tramways, suspension
s and other material
handling equipments. Wire ropes are preferred than
and man-made fibre
ropes because of its greater strength and reliabil
As compared
advantages:
to
ropes have the following
,/
ighter;;'ittt
erIe
a
in 0
eration even at high working speed .
,/
Less danger for damage due to jerks.
3.2.1. Materials of Wire Ropes
The commonly used materials for wire ropes are \ rought iron, cast steel, plow steel and
alloy steel. For special purposes copper, bronze, aluminium alloys and stainless steel are also
used.
3.3. CONSTRUCTION OF WIRE ROPES
The construction of the \\ ire rope i shown in
Fig.3. l . I·irst separate wires are twisted into strands
and then the strands are t\ isted about a core to form a
rope. The strands arc laid 011 a core made of hemp,
jute, asbestos or a wire of softer steel.
Wire
/
Fig. 3.1. 6 x 7 wire rope
~._.----.
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3.2
Design of Transmtssion Systems
Steel wire ropes are manufactured by special machines. In the process of manufacture the
wire rope is subjected to special heat treatment, which combined with cold drawing, gives
high mechanical properties to the wire.
3.4. CLASSIFICATION
OF WIRE ROPES
Wire ropes are classified depending upon the
direction of twist of the individual wires and that
of strands, relative to each other. They are:
~~
(a)
1. Cross or regular lay ropes: In these ropes,
(b)
the strands are twisted into a rope in the opposite
direction to that of the wires in the strands , as
shown in FigJ.2(a). Such ropes find the greatest
(c)
application.
Fig. 3.2. Lays of steel wire ropes
2. Parallel or lang lay ropes: In these ropes, the direction of twist of the wires in the
strand is the same as that of the strands in the rope, as shown in Fig.3.2(b). These ropes are
more flexible and resist more wear effectively. Such ropes are employed in lifts and in other
hoists with guideways and also as haulage rope.
3. Composite or reverse laid ropes: In these ropes, the wires in two adjacent strands are
twisted in the opposite direction, as shown in FigJ .2(c).
INote I Besides
the above types, the direction of the lay of a rope can be either right-hand or left-
hand. But the right-hand lay ropes are more frequently used.
3.5. SPECIFICATION
OF WIRE ROPES
The wire ropes are specified (or designated) by the number of strands and the number of
wires in each strand. For example, a 6 x 7 rope means a rope made from six strands with
seven wires in each strand. Refer Fig.3.t.
3.6. GUIDELINES FOR THE SELECTION OF WIRE ROPE
The wire rope is selected based on its application. The Table 3.1 shows the standard
designation of wire ropes and their applications.
Table 3.1. Selection of wire rope
Standard designation
Application
6 x 7 rope
Used as haulage and guy rope in mines, tramways and power transmission.
6" 19 rope
Used a
hoisting ropes in rmnes. quarries. cranes. derricks. dredges.
elevators, tramways. well drilling, etc.
6x37rope
Used as an extra flexible hoisting rope in steel mill laddies, cranes. high
speed elevators.
8 x 19 rope
Used as an extra flexible hoisting rope.
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mtd
~~~~~-----~---------------------------------------------.
Pfllleys
Wire RnIW..'
l)
3.7. STRESSES IN WIRE ROPES
The various types of stresses induced in a wire rope are :
1. Dinel slnss dll~ 10 t"~ w~;g"t oft"~ load to M lift~d and w~/ght of the rop« (er;":
Let
.
..
Weight of the load to be lifted,
W,
=
=
A
=
Area of useful cross-section of the rope.
W
Weight of the rope, and
W+W,
Direct dress,
CJd
=
... (ll)
A
2. Bending stress wilen the rope passes over the sheave or drum (ut) :
We know that when a wire rope is wound over the sheave, then the bending stresses are
induced. The bending stress induced is directly proportional to the wire diameter and
inversely proportional to the diameter of the sheave.
.
Bending stress,
where
CJb
=
E, x dw
D
... (3.2)
E, = Modulus of elasticity of the wire rope,
=
0.84 x lOs N/mm2,
=
i
for steel ropes of ordinary construction,
E; E = Modulus of elasticity of the wire material,
dw = Diameter of the wire, and
D = Diameter of the sheave.
J. Stress due to acceleration (ua) :
Due to change in speed, an additional stress is induced. The stress due to acceleration is
given by
... (3.3)
where
a
=
Acceleration of rope and load during hoisting (not at starting or stopping)
V2 -VI
=
I
; (v2 - vI) is the change in speed in 'I' seconds.
4. Stress during starting and stopping (un) :
During starting and stopping, force is required to accelerate the weight of the rope and the
load supported by the rope. This force induces an additional load in the rope and thus an
additional stress in the rope.
(/) When there is no slack in the rope:
... (3.4)
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3.4
Design of Transmission Systems
/iij When there is slack in the rope beta
.
I'
:/1 re starlmg or I
.
h
h
.
considerable impact load on the rope.
s opping, t en t ere will be a
«,
..
= W+W,
[
A
1 + 2'a
1+
S
·h·E
r
]
... (3.5)
ad·j·g
where
as = Acceleration during starting or stopping,
h = Slack during starting, and
j =
Length of the rope.
5. Effective stress:
(i) Effective stress in the rope during normal working,
aen = ad + 0b
... (3.6)
(ii) Effective stress in the rope during starting,
°est
=
ast +
0b
... (3.7)
(iii) Effective stress in the rope during acceleration
of the load ,
,
.,. (3.8)
The greatest of the stresses mentioned above should be less than the ultimate strength of
the wire material divided by factor of safety.
3.8. RECOMMENDED
FACTOR OF SAFETY FOR WIRE ROPES
The recommended factor of safety for wire ropes based on .the ultimate strength are given
in Table 3.2.
Table 3.2. Recommended factor of safety for wire ropes, n' (from data book, page no. 9.1)
Class 1
Classes 2 & 3
Class 4
Fixed guys, jib cranes
3.5
4.0
4.5
Cranes and hoists, in
4.5
5
6
Application
general hoist blocks
3.9. EFFECT OF NUMBER OF BENDS
We know that the rope life is directly proportional to the ratio Dmin / d (where Dmin is the
minimum diameter of the sheave or drum and d is the diameter of the wire rope). But
investigations have shown that at the same ratio Dmin / d rope life is approximately inversely
proportional to the number of bends. Therefore to obtain the same rope life, the effect of the
number of bends should be compensated by an appropriate changes in the ratio Dmin / d.
6
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_lS
Wire Ropes and Pulleys
Ttlb/~ 3.3. (D".;,/d) as a/unction o/tlle number of bends (from data book, page no. 9./)
No. of
I
2
3
4
5
6
7
8
9
10
16
20
23
25
26.5
28
30
31
32
33
bends
Dmi,/d
-
I1
12
13
14
IS
16
34
35
36
37
37.S
38
-
DESIGN OF WIRE ROPES
3.10. DESIGN PROCEDURE FOR A WIRE ROPE
1. Selection 0/ suitable wire rope:
First select the suitable type of wire rope for the given application, from Table 3.1.
2. Calculation
0/ design load:
Calculate the design load by assuming a larger factor of safety, say 15 (or find the design
load by assuming a factor of safety 2 to 2.5 times the factor of safety given in Table 3.2).
..
Design load = Load to be lifted x Assumed factor of safety
3. Selection of wire rope diameter (d):
Select the wire rope diameter (d) from Table 3.4, by taking the design load as the breaking
strength.
Table 3.4. Group 6 x 19 (from data book, page no. 9.5 and 9.6)
Nominal breaking strength of rope, kN
Diameter of rope
mm
Approx. mass
kglm
Tensile strength of wire,
au
= 1600 to
1750 N/mm2
Tensile strength of wire,
au = 1750 to 1900 N/mm2
0.54
0.94
1.25
86
94
134
193
147
211
22
1.84
259
25
2.41
340
284
376
32
38
3.76
533
584
5.43
767
f4
7.38
1047
843
1148
48
8.48
1199
1321
12
16
18
CJu
6
10
0.15
=1100 to 1250 N/mm2
15
40
12
0.4
0.55
16
0.95
96
20
1.45
2.25
150
230
2S
--
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55
CJu
= 1250 to 1400 N/mm2
17
45
60
110
170
260
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I!------------------------------------
o~D~e~si~gn~oJ~T.~r~an~s~m~~~s~iO~n~S~ys~te~m~s~o
4. Calculation of sheave diameter (D) :
Consulting Table 3.5, obtain the diameter of sh
(
•
C:
d
eave or drum). Always larger sh~ave
diameter ISprelerre .
Table 3.5. (from data book
, page
110.
9.1)
* Dtnin
~
Rope
Purpose
(Ratio upto V = 50 m/min)
Class 1
2&3
4
Mining installations
All ropes
100
100
100
Cranes &
6 x 37
15
17
22
Hoists
6 x 19
19
23
27
,.;Ratio for 50 m/min of rope speeds - to be increased by 8% for each 'additional speed of
50 m/min.
5. Selection of the area of useful cross-section of the rope (A):
Consulting Table 3.6, select the area of useful cross-section of the rope.
,
Table 3.6.
Type of construction
Metallic area of r?j)e A, mm2
6x7
6 x 19
0.38 d2
0.4 d2
6 x 37
0.4 d2
6. Calculation of wire diameter (dw)
:
Calculate the diameter of wire using the relation
d
... (3.9)
dw = 1.5-fi
where
i = Number of wires in the rope
= Number of strands x Number of wires in each strand.
7. Selection of weight of rope (W,) :
Obtain the rope weight (Wr) from Table 3.4.
8. Calculation of various loads:
.
bI
Calculate the various loads using the relations gIven e ow.
(i)
Direct load, W d = W + Wr
d
(ii)
Bending load, W b
_
.2:xA
xA=E
0
1 e in the speed of hoisting,
(iii) Acceleration load due to chat g [W +
W a ::: g
a
-
(Jb
r
WrJ
where a
=
v2 - vI
t
(when spee
d of the rope changes
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t":
rrorn VI
to v in t seconds)
2
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~H~~~n~R~~~.~~Qn~d~P~u/~/e~y~~
~
(iv) Starting or stopping load:
(a) When there is no slack in the rope:
Starting load, WS1 = 2· Wd = 2 (W + Wr)
(b) When there is slack in the rope :
Starting load W"
=
CJ"
xA
=
(W + W,) [ I +
1 + 2'a
S
.s .« r ]
(Jd·l·g
9. Calculation of effective loads:
(i) Effective load on the rope during normal working, Wen = W d + W b
(ii) Effective load on the rope during acceleration of the load, Weo = W d + W b + W 0
(iii) Effective load on the rope during starting, West = W b + W st
10. Calculation of working (or actual) factor of safety (FSw):
Working factor of safety,
_ Breaking load from Table 3.4 for the selected rope
FS", Effective load during acceleration (Weo)
... (3.10)
INote I Generally
effective load during starting (W cst) is not used for calculating the working factor
of safety. It is used to calculate the induced stress.
11. Check for safe design:
Compare the calculated working factor of safety (FS",) with the recommended
factor of
safety (n') given in Table 3.2. If the working factor of safety is greater than the recommended
factor of safety (i. e., FSlI'> n'), then the design is safe and satisfactory.
If FSw < n', then the design is not satisfactory. Now choose some other rope with greater
breaking strength or increase the number of ropes.
12. Calculation of number of ropes:
_ Recommended factor of safety
Number of ropes Working factor of safety
n'
... (3.11)
- FS
\tI
I Example 3.1 I Design a wire rope for an elevator in a building 60 metres high and/or
a total load of 20 kN. The speed of the elevator is 4 m/sec and the full speed is reached in
10 seconds.
Given Data:
Height
= 60
m;
W
= 20 kN = 20 x 103 N ; v = 4
m/sec = 240 mlmin;
t=lOsec.
Tofind: Design a wire rope.
e Solution:
1. Selection of suitable wire rope: Given that the wire rope is used for an elevator, i.e..
for hoisting purpose. So lets use 6 x 19 rope (refer Table 3.1).
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3.8
Design o/Transmission Systems
2 Calculatiolt of design load: Assuming a larger facto f f
f
.
.
.
rosa
ety
0
15,
the
design
load
IS
calculated.
= Load to be lifted x Assumed factor of safety
Design load
=
=
20 x 15
300 kN
3. Selectiolt of wire rope diameter (d) : From Table 3.4, taking the design load as the
breaking strength, the wire rope diameter is selected as 2S mrn.
:. d
= 2S mm
for au = 1600 to 1750 N/mm2 and breaking strength
= 340
kN.
4. Calculation of sheave diameter (D): From Table 3.5, for 6 x 19 rope and class 4,
Dmin
d =
.
27 (for velocity upto 50 m/min)
Since the given lifting speed is 240 mlmin (= 4 m/s), therefore Dm;,/d ratio should be
modified. Thus for every additional speed of 50 mlmin, Dm;,/d ratio has to be increased by
8%.
Modified
Dmin
d =
27
(1.08)5-1 = 36.73 say 40.
X
- 5
[ ... 240
50 ....
J
= 40 x d = 40 x 25 = 1000 mm
5. Selection of tire area of useful cross-section 0/ the rope (A): From Table 3.6, for
The sheave diameter,
D
6 x 19 rope,
A
= 0.4 d2 = 0.4 (25)2
6. C.alculationof wire diameter (dw)
Wire diameter,
= 2S0 mm2
:
d
dw == 1.5 {i
i == Number of strands x Number of wires in each strand
where
== 6 x 19 == 114
dw =
25
= 1.56 mm
1.5Tt!4
7.Selection of weight of rope (W,) :
From Table 3.4,
Approximate
mass
=
Weight of rope / m =
and
Weight of rope, Wr ==
2.41 kg 1m
2.41 x 9.81 = 23.6 N/m
23.6 x 60 = 1416 N
B.Calculation of various loads:
W + W r == 20000 + 1416 == 21416 N
(i)
Direct load, W d =
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Wire Ropes and Pulleys
(ii)
3.9
load, W b = Clb x A. ='
Bending
E,·dw
D
x A
5
=
0.84 x 10 x 1.56 x 250 = 32760 N
1000
[Take
,
'jii)
(w:w,)
=
load, Wa
Acceleration
E,
0.84 x lOs N / mm21
=
a
v2 - vI
where
a = Acceleration
of the load =
t
I
= 0.4
Wa =
(iv) Starting load (W Sf)
4-0
--10
m/s2
(2000~.;/416)
=
0.4
873.23 N
:
When there is no slack in the rc.-e, starting load is given by
=
WSf
=
2· Wd = 2 (W + Wr)
2 (20000 + 1416) = 42832 N
9. Calcukn.>n of effective loads on tile rope:
(i)
(ii)
(iii)
Effcci.ve 10,1.' rturing normal working,
Effective k:.ld Juring acceleratio..
Effective load during starting,
Wen
=
Wd + Wb
=
21416+32760
21416 + 32760 + 873.23
=
55049.23 N
=
West
10. Calculation of working factor of safety (FSw)
=
.
,
:I
'('
•
•
'.
..
: ..•
••
0.
•
:
.
".
11. 'Check for' safe
factor'ofsafety
Since
therefore
=
..
."
.
=
•
the design is
32760 ~ 42832
=
75592 N
:
Breaking load from Table 3.4 for the selected ru~
Effective load during acceleration (Wea)
340000
55049.23
'"
=
.
6.176
c
.f".
: .
(,.
t . F(O,m·1:ahle 3.2; for hoists and class ~, .the rt!commend.ed
.'.
~
Q
...
de~ii:it
6.
the working
W d + W b + Wa
=
=
Work ing factor of safety
=
Wea
of the load,
54176N
=
.
factor
of safety
."
is greater
'.
'..
than the 'recommended'
.
factor
safe.
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.
.
of :iafety·
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----------------------------
~D~~~i~gn~o~if~n~ra~m~m~·~~s~w~
~ele
3.~ 1 Fo~ t~~ data of Example 3.1, determine the stress induced in the rope
t Slarling with an initia! slack of rope of 0. 2 m.
due '0
•
.
Given Data: h = 0.2 m.
T'O find: Stress induced in the rope due to starting.
We know that the load dueto starting when there is a slack of 0.2 m on the
@So!ulion:
rope,
1 +. 2·a s ·h·E
ad·/·g
W = 20000 N;
Wr
= 1416 N; E, = 0.84
as
=
=
32760 N;
ad
a = 0.4 m/s2;
= A = 250 = 85.66 N/mm2 = 85.66
x
h = 0.2 m; /
... [From Example 3.1]
= 60 m.
= 24102
106 N/m2 ;
1 + 2 x 0.4 x 0.2 x 0.84 x 1011 ]
85.66 x 106 x 60 x 9.81
WS1 = (20000 + 1416) [ 1 +
..
J
x 105 N/mm2 = 0.84 x 1011N/m2 ;
21416
Wd
Wb
r
N
Effective load on the rope during starting with slack,
West
..
Stress induced
=
Wst
West
=
I Examp/e 3.3 I· Select
A
+
Wb
_
-
= 24102 + 32760 :-: 56862
56862 _
250 - 227.45 N/mm
2
N
Ans. ~
a wire rope for a vertical mine hoist to lift 1500 tons of ore in 8
hours shift from a depth of 900 m: Assume a two-compartment shaft with the hoisting skips
in balance. Use a maximum velocity of 12 m/sec with acceleration and deceleration period
'0/15 sec each and a rest period of 10 sec for discharging and loading the skips. A hoisting
skip weighs approximately 0.6 of the capacity. Take E, = 0.84 x lOS Nlmm2.
= 1500 tons = 1500 x 103 kg = 1500 x 103 x 9.81 =
14715 kN; Depth = 900 m; E, = 0.84 x 105 N/mm2 = 0.84 x 1011 N/m2; v = 12 mls =
Given Data: Weight to be lifted
720 m/min.
Tofind:
Select a wire rope.
@ Solution:
Let us first find the total load to be lifted (W).
v2 - VI
Acceleration,
as
The distance travelled during acceleration
=
=
I
12 - 0
15
21 as (2 = 21
Similarly distance travelled during deceleration period=
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=
=
0.8 m/s2
x 0.8 x (15)2
= 90 m
90 m
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________
--.-------:..:3.11
Total distance truvellcd at full speed = 720 - 2 x 90 = 540 m
and Time required for travelling 540
III at
=
720 Ill/min
540
720
=
075 min
.
=
45 sec
Time required for one trip:
Acceleration period = 15 sec
Deceleration period
=
15 sec
Period of travel with full speed
=
45 sec
Rest period = 10 sec
Total = 85 sec
Number of trips in a shift
Load to be Iifted/tri p
=
=
8 x 60 x 60 ~ 339
85
14715
339
=
43.4 kN
Hoisting skip weight = 0.6 x load = 0.6 x 43.4 = 26 kN
Total load to be lifted by the ropes = Weight of ore + Weight of skip
=
43.4 + 26 = 69.4 kN ~ 70 kN
1. Selection of suitable wire rope: For hoisting purpose, from Table 3.1, 6 x 19 rope is
selected.
2. Design load: Assuming the factor of safety of 15, the design load is calculated.
Design load
=
70 x IS = 1050 kN
From Table 3.4, d
1750 Nzrnm? and breaking strength = 1199 kN.
3. Wire rope diameter (d):
.
4. Sheave diameter (D) : From Table 3.5, we find
50 m/min. But the actual speed is 720 m/min. ti.e.,
Dmin
d
=
48 mm is chosen, for
Dm~
d = 27 for
0'" =
1600 to
class 4, for velocity upto
720
50 ~ 15 times 50 m/min).
Therefore
.
has to be modified.
=
27x(1.08)IS-1
=
79.3 say 80
Sheave diameter, D = 80 x d = 80 x 48 = 3840 mm
5.
From Table 3.6, for 6
A = 0.4 d2
= 0.4
(48)2
x
19 rope, the area of useful cross-section
= 921.6
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of the rope,
rnrn-.
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~D~~~~~n~o~if~n~ra~m~m~U~Si~o
6. WiTt diameter (d..,) :
dw
t. Weight of the rope
=
d
48
_r
I.S V ;
= --;:::.====-
1.5 ~ 6 x 19
= 3mm
(W,): Refer Table 3.4.
= 8.48 x 9.81 = 83.2 N/m
Weight of the rope = 83.2 x 900 = 74880 N = 74.88 kN
Weight of the rope 1m
8. Load calculations:
(i)
Direct load, Wd
=
W
+ Wr = 70 + 74.88 = 144.88 kN
(ii)
Bending load, W b
=
ab
xA
=
0.84 x lOS x 3
3840
x 921.6
(iii)
Acceleration load, Wa
= Er ~ d", x A
= (W+Wr)
g
:. Effective load on the rope during acceleration
=
a
,
=
= 60.48 kN
(70+74.88)
9.81
W ea
.
xO.8
= 11.81 kN
= W·d + Wb + Wa
144.88 + 60.48 + 11.81 = 217.17 kN
9. Working factor of safety (FS..,) :
Working factor of safety
=
Breaking load from Table 3.4 for the selected rope
Effective load during acceleration
1199
= 217.17 = 5.52
10. Check for safe design:
From Table 3.2, the recommended factor of safety (n') = 6
Since FSw < n', the design is not safe. But FSw value is closer to n' value, so the design is
marginally satisfactory.
The safe design (i.e., greater working factor of safety) c~n be achieved either by selecting
the rope with greater breaking strength or by increasing the drum diameter.
11. Redesign of rope: Now select the rope with breaking strength of 1321 kN from Table
3.4. For breaking strength 1321 kN and au = 1750 to 1900 N/mm2, the diameter remains
same as 48 mm. Thus the load and other calculations remain same.
1321
.. Working factor of safety, FSw = 21'i .17 = 6.082
Since the value of working factor of safety. is greater than the recommended factor of
safety,therefore the design is safe.
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----~--------------~213
_!ire
Ropes and Pulleys
W~~~~'~~
3.11. FAILURE OF ROPES
.
The failure of rope is mainly due to fatigue
an d wear while passing around
I . the
fl sheave
. .
.,
it
over
the
sheave
resu
ts
In
uctuabno
The bending and straightening of the rope as I passes
~
.
individ
I'
lide
on
each
other
and
over
the
sheave
stress leading to fatigue failure. The indivi ua wires s I
resulting in gradual wearing of both the rope and the drum.
The amount of wear that occurs depends upon the pressure between the rope and the
sheave and is given by
where
p
=
T
=
2T
'" (3.12)
dxD
Tension in rope,
d = Diameter of rope, and
D
Sheave diameter.
=
3.11.1. Fatigue diagram
The fatigue diagrams for 6 x 19 and 6 x 37 ropes are shown in Fig.3.3 (refer data book,
page no. 9.2). The fatigue diagrams are constructed by using experimental data.
0.006
1
-;
~
\
\
0.004
[\
~
.9:
0.002
o
I~ ~
5
2x10
6 x 37_
r-6
~
s
4x10
x 19
6x105
8x10S
106
Number of bends to failure
Fig. 3.3.
The fatigue diagram is plotted on cartesian co-ordinates, in which the number of bends is
taken as abscissa and a dimensionless quantity (pIau) as ordinate. The figure indicates that
the 6 x 19 rope has long (i.e., infinite) life if the value of ratio p/O'uis less than 0.0012.
I Example
I
3.4
I A 6 x 19 wire rope withfibre
~ore and tensile designation 1420 is used
to raise a load. TI,e nominal diameter of the wire rope and the sheave diameter are 10 and
4S0 mm respectively. Assuming long life on the basis. of fatigue consideration, determine
the maximum load that the wire rope can carry.
_
Given Data: 6 x 19 wire rope; au = 1420 N/mm2;
Tofind:
d
=
10 mm ; D = 450 mm.
Maximum load that the wire rope can carry.
© Solution:
Tensile designation 1420 means au
=
1420 N/mm2.
_'"•
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3.14
Design o/Transmission Systems
We know that the 6
x
19 wire hope has long life when piau = 0.0012 .
. , Contact pressure between the rope and drum, p = 0.0012 x au
= 0.0012
x
1420
= 1.704 N/mm2
2T
We also know that
p
=
where
T
=
Maximum load that the wire rope can carry
T
=
p x;
dD
xD
= 1.704 x 10 x 450 = 3834
N Ans. ~
2
I Example
(i.e.,
O'u
\
3.5 , A 6 x 19 wire rope witll fibre core and tensile designation
= 1750
0;
1750
Nlmm2) is used to raise tile load of 22 kN as shown in Fig.3A. The nominal
diameter of the wire rope is 10 mm and the sheave "as 550 mm pitc" diameter. Determine
the expected life of rope assuming 500 bends per week.
Given Data:
6 x 19 rope;
au
1750 N/mm2;
=
Load
=
22 kN;
d
= 10 mm ;
D = 550 rnm ; Number of bends = 500 I week.
Tofind:
Expected life of rope.
© Solution:
Let
T
= Load acting on the rope
From the Fig.3.4, we can write
2T = 22 kN [ignoring the weight of rope and the acceleration load]
or
T = II kN
We know that the contact pressure between
the rope and the sheave,
2T
P
piau
=
dD
=
2 x 11000
IOx550
=
4 N/mm2
=
=
4
1750
0.00228
Fig. 3.4.
· 3 3 the life of the wire rope is obtained as 2.8 x 105 bends
From the fatigue diagram, F Ig..,.
,
before failure;
2.8 x lOS
= 5bO weeks'
Life =
..
500
2.8
or
= 500
105
x 52
x
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=
10.77 years Ans.~
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3.15
Wire Ropes and Pulleys
DESIGN OF WIRE ROPE SHEAVES AND DRUMS
3.12. DESIGN OF WIRE ROPE SHEAVES
We know that the bending stress induced in the wire rope,
dw
= E,
crb
where
=
D
x
0
Diameter of the sheave
So it is clear that the bending stress induced in the rope is inversely proportional to the
diameter of the sheave. Therefore the sheave diameter should be fairly large in order to
reduce the bending stress in the rope when they bend around the sheaves. Also the larger
diameter sheaves provide better and more economical service.
SheQ11e materials: For light and medium service, the sheaves are made of cast iron, but
for heavy service they are often made of steel castings. They are usually made in the form of
casting or weldment.
a
,.
!
e
b
.1
'I
_j
t
..r:::
r2
Fig. 3.5. Cross-section of standard groove
The cross-section of standard sheave groove for steel wire ropes is shown in Fig.3.5. (refer
data book, page no.9.10). The various proportions of sheave groove are given in Table 3.7.
Table 3.7. Proportions of sheave grooves, mm (from data book, page no. 9.10)
Wire
a
b
c
e
h
I
r
'1
'2
'3
'4
4.8
22
15
5
0.5
12.5
8
4
2.5
2
8
6
8.7
28
20
6
1.0
15
8
5
3
2.5
9
6
13.0
40
30
7
1.0
25
10
8.5
4
3
12
8
19.5
55
40
10
1.5
30
15
12
5
5
17
10
24.0
65
50
10
1.5
37.5
18
14.5
5
5
20
15
34.S
90
70
IS
2.0
55
22
20
7
8
28
20
rope dia.
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3.16
Design o/Transmission Systems
wireTherefore,
rope first. in order to design a sheave for wire rope, one should find the diameter of the
3.13. DESIGN OF WIRE ROPE DRUMS
Whenever more than one wire rope is used, one has to use rope drums to enable the rope
to be wound in several layers. The drum diameter is selected in the same manner as the
selection of diameter of sheaves. Drums for steel wire ropes are made of cast iron, more
rarely of steel castings or weldments.
With a power drive, the drum should always be provided with helical grooves so that the
rope winds up uniformly and is less subject to wear. The radius of the helical grooves should
be selected so as to prevent jamming of the rope. The standard and deep grooved drums for
wire ropes are shown in Fig.3.6. Table 3.8 gives the dimensions of standard and deep grooves
for drums.
(a) Standard groove
(b) Deep groove
Fig. 3.6.
t.able 38
. . D'tmensto. ns of d.rum groovesfor wireropes, mm (from data book,page no. 9.9)
Ropedia, d
..._
'.
Standard
Groove
Deep Groove
S)
C)
'2
S2
C2
4.8
3.5
7
2
5.5
9
4.5
8.7
5
II
3
8.0
13
6.5
13
8
15
4
11.0
19
9.5
27
13.5
19.5
11.5
22
5
15.5
28
15.5
31
8
20.5
36
18.0
39
21
42
12
28.0
50
24.5
CExample
. rope IS. use d for tile elevator in the building. The
Ire
3.6 I A 6 x 19 stee w
if200 m/min. The weight of the elevator cage
.
•
'mum
speed
0
t1J'Pe•
Hlelghtis being lifted with a maxi
•
N. '7'1. tlfting sheaves are of the trac tiIon .J.
.
gers IS 8 k . tne
ts 12 kN and the weight of passen,
k factor of safety as 10.
Design a suitable wire rope sheave. Ta e
. . W
= 12 kN; W"""nge< = 8 kN ;
v
=
200
rn/min
,
elevator
Given Data: 6 x 19 wire rope ;
FS::::: 10.
I
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-
3.17
Wire Ropes and Pulleys
Tofind:
1
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Design a suitable wire rope sheave.
@Solution :
0/ rope diameter (d) : We
Calculation
know that, in order to design a wire rope sheave,
first one should find the diameter of rope.
Total load to be lifted
=
Weight of the elevator cage + Weight of passengers
12 kN + 8 kN
Design load
=
=
Total load to be lifted x Factor of safety
=
20 kN x 10
=
=
20 kN
200 kN
Taking the design load as the breaking strength, consulting Table 3.4, the diameter of the
rope is selected as 25 mm.
:. d
= 25 mm
for au = 1100 to 1250 N/mm2
and breaking strength = 230 kN.
Calculation of sheave diameter (D): Refer Table 3.5.
Dmin I d
For velocity upto 50 m/min,
=
27
As the lifting speed is 200 mlmin, this ratio has to be modified. That is, for every
additional speed of 50 mlmin, Dmin I d ratio has to be increased by 8%.
Dmin
d =
Modified
Then
Diameter of sheave, D
Design
=
27 x (1.08)4 -
I
=
34
= 850 mm
34 x 25
Ans. ~
0/ wire rope sheave:
The cross-section of standard sheave groove for steel wire
rope is shown in Fig.3.5. (refer data book, page no. 9.10).
Knowing the diameter of rope and consulting Table 3.7, the proportions of sheave grooves
are given below.
a
=
65 rnrn ;
h = 37.5 mm;
r2
= 5 mm ;
I Example
3.7
b = 50 mm;
I = 18 mm ;
r3
=
20 mm ;
c = 10 mm ;
r = 14.5 mm ;
r4
=
e
=
1.5 mm ;
15 mm.
I A 6 x J 9 wire rope drive 0/ three 38 mm
diameter ropes is used in a
hoisting equipment. The maximum rope speed is 10 m/s. Calculate the drum diameter and
various dimensions 0/ drum grooves for wire ropes.
Given Data: 6 x 19 wire; n = 3 ; d
To find:
= 38 mm
; v = 10 mls
=
600 m/min.
Drum diameter and various dimensions of drum grooves for wire ropes.
@Solution :
Drum diameter:
From Table 3.5, for 6 x 19 rope and class 4,
=
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27
.. , [For velocity upto 50 m/min]
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l.!!--
Design of Tramm ission Systems
Since the given rope speed is 600 mlmin
ti.e., 6 0 0= 12 times 50 m/min).
5 0
Therefore.
_ / d ratio has to be modified.
D",m
Dmin
d
Drum diameter,
= 27 x (1.08)12 - I
=
62.95 say 65
D = 65 d = 65 x 38 = 2470 mm
Ans. ~
Various dimensions of drum grooves: The standard drum groove for the wire rope is
shown in Fig.3.6(a). (refer data book, page no. 9.9). Knowing the diameter of rope and
consulting Table 3.8, the various dimensions of drum grooves are given below.
rl = 21 mm; SI = 42 mm and C1 = 12 mm Ans."
REVIEW AND SUMMARY
.f
Wire ropes are extensively used in elevators, oil well drilling, mine hoists, cranes.
conveyors, hoisting devices, etc .
.f
Types of wire ropes are: Cross or regular lay ropes, parallel or lang lay ropes, and
composite or reverse laid ropes .
.;
Specification of wire ropes: Wire ropes are designated by the number of strands and
the number of wires in each strand. For example, a 6 x 17 rope means a rope made from
six strands with seventeen wires in each strand.
.f
Selection of wire rope: 6 x 17 rope is suitable as haulage and guy rope, 6 x 19 and 6 x
37 ropes for hoisting applications .
.f
Stresses in wire ropes:
1. Direct stress,
acceleration, and 4. Stress due to starting.
2. Bending stress,
3. Stress due to
The life of the wire ropes are inversely proportional to the number of bends .
.f
In this chapter, the step by step design procedure for wire ropes is presented
with
sufficient example problems.
Wire rope design: To find: 1. Type of construction, 2. Rope diameter, 3. Sheave or
drum diameter, 4. Wire diameter, 5. Various loads and their stresses, 6. Working
factor of safety, and 7. Checkfor safe design .
.f
Thefailure of rope is mainly due to fatigue and wear while passing around the sheave .
.f
The contact pressure between the rope and the sheave is given by
2T
p = dxD
T = Tension in rope,
d = Rope diameter, and D = Sheave diameter.
From the fatigue diagram, for 6 x 19 ropes, the rope has infinite life if the ratio pIau is
where
0/
0/
~
less than 0.0012.
At the end of this chapter. the design of wire rope sheaves and drums are also discussed.
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3.19
~w.~/~re~R~o~p~e~s~an~d~P~u~/~/e~ys~
------------------------------------~-REVIEW QUESTIONS
1.
2.
What are the applications of wire ropes?
What are the advantages of a wire rope when compared with chains and fibre ropes?
3.
How can you specify a wire rope? What do you understand by 6 x 17 construction
in
wire ropes?
4.
Give the application of the following wire ropes:
(a) 6x7rope;
(b) 6x 19 rope,
and
(c) 6x37rope.
5.
Distinguish regular-lay and lang-lay ropes.
6.
Explain the various stresses induced in the wire ropes.
7.
How can you select a wire rope sheave or drum for the given wire rope?
PROBLEMS FOR PRACTICE
Problems on design of wire ropes:
1.
Select a wire rope necessary for a mine hoist carrying a load of 72.5 kN to be lifted from
a depth of 250 metres. A rope speed of 7 mls is to be attained in 10 seconds.
2.
In an office building the elevator rises 400 m with an operating speed of 275 mlmin and
reaches the full speed in 10m. The loaded elevator weighs 22 kN. Design a suitable wire
.
.
v2
(275 / 60)2
rope.
[Hint : Rate of acceleration, a = 2 s = 2 x 10
= 1.05 rn/s]
3.
Select a suitable wire rope to lift 1 tonne of debris from a well of 50 m deep. The weight
of the bucket is 3000 N. The weight is being lifted with a maximum speed of 3 mlsec
and the maximum speed is attained in 1 sec. Determine also the stress induced in the
rope due to starting with an initial slack of 0.2 m.
4.
Select a wire rope for a vertical mine l.oist to lift 12000 kN of ore in each 8-hr shift from
a depth of 800 m. Assume a two compartment shaft with the hoisting skips in balance.
Use a minimum velocity of 12.5 mls with acceleration and deceleration of 15 sec each
and a rest period of 10 sec for discharging and loading the skips. A hoisting skip weighs
approximately
5.
0.6 of its load capacity.
Select a wire rope for the elevator in a building where the total lift is 180 m. The rope
velocity is 4.5 mls and the full speed is to be attained in 12 m. The lifting sheaves are of
the traction type. The weight of the elevator cage is 15 kN and the weight of passengers
is 10 kN.
6.
A workshop crane is lifting a load of 20 kN through a wire rope and a hook. The weight
2
of the hook etc., is 12 kN. The load is to be lifted with an acceleration of 1 m/sec .
Calculate the diameter of the wire rope. The rope diameter may be taken as 30 times the
diameter of the rope. Take a factor of safety of 6 and Young's modulus for the wire rope
0.8 x lOS N/mm2. The ultimate stress may be taken as 1800 Nzmm-'. The cross-sectional
area of the wire rope may be taken as 0.38 times the square of the wire rope diameter.
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1-~.20~------------------------
~D~e~~I~gn~of~~~a~m~m~U~Ji~On~Sy
7. A 6 x 19 wire rope with fibre Core and tensile designation of 1570 is used to raise the
load of20 kN as shown in Fig.3.5. The nominal diameter of the wire rope is 12 mm and
the sheave has 500 mm pitch diameter. Determine the expected life of the rope assuming
506 bends per week.
[Ans: 12.5 years]
8. A 6 x 19 wire rope with fibre core and tensile designation 1600 (i.e., au = 1600 N/mm2)
is used to raise the load. The nominal diameter of the wire rope and the sheave diameter
are 12 mm and 540 mm respectively. Assuming infinite life (i.e., pIau = 0.0012) on the
basis of fatigue consideration, determine the maximum load that the wire rope can carry.
[Ans :
6220.8 N]
Problems on design of wire rope sheaves and drums :
9. A 6 x 19 steel wire rope is used in a hoist to lift the ore in a copper mine. The weight of
the car and cage is 25 kN. The maximum rope speed is 10 m/s. Design a suitable wire
rope sheave.
10. A four 6 x 19 steel wire rope of diameter 28 mm is used in a crane. The maximum rope
speed is 8 mls. Calculate the drum diameter and various dimensions of drum grooves for
wire ropes.
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Chain Drives
"My grandfather once told me that there were two kinds oJpeOple:
those who do the work and those who take the credit. He told me to try to be in the first grOUp;
there was much less competition. "
-Indira
GtuuIIU
4.1. INTRODUCTION
The chain drive is intermediate between belt and gear drives. It has the major advantages
of both belt and gear drives. Chain drives are used for velocity ratios less than 10 with chain
velocities upto 25 mls and power ratings upto 125 kW. Chain drives are popularly used in the
transportation industry such as bicycles, motor cycles and automobile vehicles. They also
find wide applications in agricultural machinery, metal and wood working machines, textile
machinery, building construction and materials handling machinery.
4.2.
ADVANTAGES AND DISADVANTAGES
BELT AND GEAR DRIVES
OF CHAIN DRIVES COMPARED
WITH
Advantages:
./
They can be used for long as well as short centre distances .
./
They are more compact than belt or gear drives .
./
There is no slip between chain and sprocket. So they provide positive drive .
./
One chain can be arranged to drive several sprockets .
./
Higher efficiency (upto 98%) of the drive .
./
They transmit more power than belt drives .
./
Smaller load on the shafts than in belt drives .
./
They can be operated under adverse temperature and atmospheric conditions.
Disadvantages :
./
They require precise alignment of shafts than the belt drives .
./
They require proper maintenance (particularly lubrication) and slack adjustment
compared with belt drives .
Noisy operation .
./
./
They require the take-up devices (to compensate the increase in chain pitch due to
wear of chain joints).
.
More complicated design.
I
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__---~~----------------------------~D~~~~~n~Of[TI~r~a~m~m~~~s~;o~n~s~~s~
~.3. TYPES OF CHAIN DRIVES
The common types of chains are:
1.
Link chains (or welded chains),
2.
Transmission chains (or roller chai
ams,) and
3.
Silent chains (or inverted tooth chains).
All these chains will be discussed , in detail ,ein th
I
10 II'
owing
c.
LINK CHAINS
sections.
.
I
4.4. LINK CHAINS
Link chains, also known as welded load chains, are widely used
./
In low capacity hoisting machines such as hoists, winches and hand operated
cranes as the main lifting appliances .
./
As slings for suspending the load from the hook or other device.
4.5. DIMENSIONS OF A LINK CHAIN
The main dimensions of the liI_lkchain are shown in Fig.4.1.
Fig. 4.1. Main dimensions of a link
c/,ain
They are pitch (t) equal
to the inside length of the link, outside width (B) and diameter (d)
of the chain bar.
4.6. CLASSIFICATION OF LINK CHAINS
1.Depending on the ratio between the pitch and tl,e diameter of the chain bar:
(a) Short link chains:
If t ~ 3 d, then the chains are known as short link chains.
(b) Lollg link chalns : If I> 3 d, then the chains are known as long link chains.
2.Depending on the manufacturing accuracy:
. (a) Pitched chalns : When the permissible deviations from the nominal pitch size is
".'thin i 0.03 d and from the outside width is within i 0.05 d, then the chain is called as
Pitchedchain .
. (6) Calibrated chalns : When the permissible deviations is within i 0.1 d of the nominal
Slze
in pitch and outside width, then the chain is known as calibrated chain.
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Q1ain Drives
4.3
--..
4.7. CONSTRUCTION OF LINK CHAINS
./'
./'
.
.
.
Id d) hains are formed from oval steel hnks an the sequence shown in
Link (or we e c
Fig.4.2 .
Link chains are manufactured from steel 2 and 3.
o
o
(c)
(b)
(8)
Fig. 4.2. Stages ill tue manufacture
(d)
of welded c/,ains
(a) Blanks of cut steel bar, (b) Prebent blanks, (c) Chain assembled prior to welding,
(d) Chain witl, welded links
./'
Links for welded chains are formed by a number of methods. Mostly forge
welding and electric resistance welding are used .
./' In forge welding, a single weld is made in the link. In the electric resistance
method, the link is made from two butt-welded half-links. The welds are made on
the straight sides of the link. The resistance welding method produces more
accurate chains with increased strength .
./' When assembling separate lengths of forged links into chains, the welded ends of
each pair of adjoining links should form one joint to increase the life and strength
of the chain. Chains welded by the electric resistance method can be assembled in
any manner .
./' After manufacturing, the chains are annealed.
4.8. SELECTION
OF LINK CHAINS
In link chains, it is extremely difficult to determine the actual stresses. Because, with
respect to external forces the links of welded chains are statically determinate and with
respect to internal stresses three-fold statically indeterminate. Therefore the stresses are
determined approximately.
The chains are checked for tension. To compensate the statically indeterminate feature of
the chain, somewhat reduced safe stress is taken.
The general formula for selecting link chains in tension is given by
Psafe
where
Pbr
n
Safe load carried by the chain,
Breaking load of the chain, and
=
=
n = Factor of safety from Table 4.1.
Psafe
Pbr
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... (4.1)
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Design a/Transmission
Table 4.1. Data/or selection o/link chains
r-"
~
Systems
.
Chains
Drive
Factor ofsa(eb', n
Welded calibrated and un calibrated
Hand
Power
3
6
2.
Welded calibrated on a pocket sheave
Hand
Power
4.5
3.
Welded uncalibrated on slings
-
6
I.
8
The intensity of wear depends on the following factors: the ratio between the pitch of the
chain and drum, the tension and speed of the chain, the angle of relative turn of the links as
they pass around the sheaves, the environment, etc.
4.9. ADVANTAGES OF LINK CHAINS
./
Good flexibility in all directions.
"
Smaller pulley diameters and drums .
./
Simple design and manufacture.
4.10. DISADVANTAGES
OF LINKS CHAINS
./
Heavy height.
"
Susceptibility to jerks and overloads .
./
./
Sudden failure .
Intensive wear of the links in the joints.
TRANSMISSION
CHAINS AND SPROCKETS
4.11. TRANSMISSION (OR ROLLER) CHAINS
Chain
A roller chain provides a readily available
and efficient method for transmitting power
between parallel shafts. That's why the roller
chains are also called as transmission chains. A
roller chain consists of an endless chain running
Over two sprockets _ driver and driven. A
sprocket is a wheel with teeth of a special
profile. Smaller sprocket is called pinion and
bigger one is called wheel. Typical roller chain
Driven sprocket
Driving sprocket
Fig. 4.3. C/.ain drive
on sprocket is shown in Fig.4.3.
4.12. CONSTRUCTION OF ROLLER CHAINS
.
The
. . h
. Fig 4 4 It consists of alternate hnks made of
.
construction of a roller cham IS sown m
..,
. f k
coupling link whereas
Inner and outer link plates. The outer plates are known as pm m or
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SC~h~a~;n~D~r~N~~=-
---------------------------~
.
Th e 0th er parts of a roller chain are
the inner plates are called roller link.
fi pin,
d i bushing Irwj
roller. Pins are press fitted into the outer plates whereas bus/res ar~ pre.ss Itte into ~heinllt:
plates. The pin and the bush form a swivel joint and the outer !mk IS free to sWivel wi~
respect to the inner link. The chain rollers mounted on the bushings roll over the sprOcket
teeth.
Pitch.
P
Fig. 4.4. Construction of a roller chain
4.13. CHAIN MATERIALS
./'
Link plates are made of cold-rolled, medium-carbon or alloy steels such asC4j.
C50 and 40 Crl .
./'
Pins, bushings and rollers are made of carburizing steels such as CIS, C20, ana
30 Ni4 Crl.
4.14. SPECIFICATION OF A ROLLER CHAIN
Roller chain is specified by three dimensions - pitch, width and diameter.
Pitch: It is the distance fro-n centre to centre of adjacent pins or rivets.
Width: It is the nominal width of the link or the length of the pin.
Diameter: It refers to the actual outside diameter of the roller.
Roller chains are available in single-row or multi-row construction such as simple.\,
duplex o~triplex strands as shown in Fig.4.5. (refer data book, page no. 7.71) .
.
(a) Simplex chain
(b) Duplex chain
(c) liiplex chain
Fig. 4.5.
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Design a/Transmission Systems
".15. GEOMETRIC RELATIONSHIPS OF A ROLLER CHAIN AND SPROCKET:
Fig.4.6 shows a sprocket driving a chain in a counter clockwise direction.
p = Chain pitch,
Let
a
= Pitch angle,
aJ2 = Angle of articulation,
D = Pitch circle diameter of the
sprocket, and
z
Variable
= Number of teeth on the
sprocket.
Pitch circle diameter (D) :
Pitch angle, a
=
360
z
'" (4.2)
From the trigonometry
of the figure,
. a
n/2
sin - = ~
2
D/2
=
or
D
INote I
Fig. 4.6. Engagement of a chain and sprocket
P
p
=
[ .,'a = 3~0 ]
. (180)
z
sin (~)
SIO
'" (4.3)
-
The angle aI2, through which the link swings
as it enters
sprocket, is called the angle of
articulation.
Transmission (or velocity) ratio of chain drive (i) :
Let
N 1 and N2 = Speeds of rotation of driving and driven sprockets respectively,
zi and z2 = .Number of teeth on driving and driven sprockets respectively.
N)
z2
Transmission
ratio, i
=
N-
2
=;- 1
and
... (4.4)
Pitch dituneters of sprockets:
Let
d) and d2 = Diameters of driving and driven sprockets respectively.
Then, the equation (3.2) can be' rewritten as
Diameter of driving sprocket,
and
Diameter of driven sprocket,
d1
=
d2 =
p
sin (1801
... (4.5)
z)
p
... (4.6)
Average velocity of the chain (v) :
.The average velocity of the chain is given by
v
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=
1t·D·N
60
= z·p·N
60
... (4.7)
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~C~h~a'~'n~D~r~w~u~
~
L~lIgth of th« chaill (L) :
The length of the chain is always expressed
.
.
In terms 0
f the number of tanks
(or pitches).
••.
(4.8)
Ip xp
.
L
where
L = Length of the chain in mm, and
.
I = Number of links (or pitches) in the cham.
=
-
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--------------------------
p
N"",ber of lillo in th« chain (I,) :
.
.
. h hai
be determined
The approximate number of links (or pitches) m t e c am can
following formula.
Ip
where
= 2 (~)
+
(ZI ; Z2) + (Z22-;1
r
x
by using the
(!)
...(4.9)
a = Centre distance between axes of driving and driven sprockets in mm
It is known that the calculated number of links (/p) should be rounded to the next digit.
Since the chain consists of alternate pairs of inner and outer link plates, therefore it is always
preferred to have even number of links. When the chain has odd number of links, an
additional link called 'offset' link, is required. But the offset link is weaker than the main
links.
4.16. CENTRE DISTANCE
The centre to centre distance between the axes of the two sprockets corrected to an even
number of links can be determined by using the following relationship.
Centre distance, a
where
e
__ e+..Je2-SM
- 4
=
(Z2 2: 1
z
M =
r'
[from d ata b ook, page no. 7.75] ... (4.10)
xp
a constant.
on ~:ctual Pkractiche~
a small amount of sag is essential for the links to take the best position
sproc et w eel. Therefore in 0 d t
should be decreased b th
r er 0 accommodate the initial sag, centre distance
y e amount fla.
Ila = ~
where
f
[f - (Z;1I-:;
r.
p ] '" 0.5
= Chain sag of driven side in mm
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f ..
1tS
0.5 (0.02 a) .. 0.0 I a
[from data book, page no. 7.75]
0.02 a
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4.8
Design o/Transmission Systems
4.17. CHORDAL (OR POLYGONAL) ACTION
An
.
. important ..factor affecting the operating smoothness of a roller chain driIve, particularly
at. high speeds, IS chord~1 action. The chordal action is illustrated in Fig.4.7(a) and (b). In
Flg.4.7(a), ~oller A has Just seated on the sprocket, and the centreline of the chain is at the
chordal. rad.lUsr c: After the sproc~et rotates through angle (al2), the chain is in the position
shown ID Flg.4.7(b). Here, the cham centreline is at the sprocket pitch radius rc.
I-p-I
Chordal rise, r - rc
~
-~-----~l--, Pitch
circle
---------j--------_.
r
t
r
c
-----L
--1
(b)
(a)
Fig. 4. 7. CI,ordai action of a roller chain
The amount of chain rise and fall (i.e., chordal rise) is given by
tor
= rc-r=r(l-COS~)
=r[l-COSC!O)]
... (4.11)
where
z = Number of teeth in the sprocket.
Thus it is clear that the linear speed of the chain is not uniform but varies from V min to
vlfIIJ%
during every cycle of tooth engagement. This results in a pulsating and jerky motion.
7t
0 N cos (
i)
60
7t
and
Vmax
=
DN
60
... (4.12)
Because of chordal action, a chain drive is analogous to a belt drive running with a prism.
~
In order to reduce the variation in c/,ain speed, the number of teeth on the
sprocket should be increased.
[!Vote) It is better to use an odd number of teeth (such as 17, 19 or 21) for the driving sprocket.
The odd number of teeth of the sprocket, in combination with an even number of chain links, facilitates
moreuniform wear.
4.18. DESIGN PROCEDURE OF ROLLER CHAIN
1. Selection of the transmission ratio (iJ :
Select a preferred transmission ratio from Table 4.2.
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4.9
Chain Drives
Table 4.2. Transmission rallo, I (from dala book, page no. 7.74)
1-2
30
INote I
27
Where space is a problem,
2-3
3-4
4-5
5-7
27 25
25-23
23 - 21
21 - 17
zimin
=
7
3. Determination of number of teeth on the driven sprocket (l; :
Determine the number of teeth on the driven sprocket (z2) by using the transmission
ratio (I) and z r-
Recommended
value of l2:
z2max
=
100 to 120
... [from data book, page no. 7.74]
Now check whether the calculated z2 is less than the recommended z2
z2 is large, the stretched chain may slip off the sprocket for a small pull.
max'
Because, when
4. Selection of standard pilch (P) :
Knowing (or assuming) the initial centre distance (a), detennine the range of chain pitch
by using the relation
a = (30 - 50) P
... (4.13)
From the pitch range obtained, consulting Table 4.4, select a suitable standard pitch.
Table 4.4. (from data book, page no. 7. 74)
No. or teeth on pinion
Pitch,p, mm
sprocket
z.
7
9
s.
15
21
27
35
45
Selection of the chain •
..
9.525
2800
2800
2400
2400
2100
1800
1600
12.7
15.875
2600
2400
2400
2100
1800
1600
1400
2000
1800
1800
1500
1300
1200
,
Select the chain type and hal
..
c am number, by using the selected
Table 4.5. Initially assume simplex or d I
hai
up ex cam.
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1000
-
standard pitch,
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~D~e~~,~g~n~oflTJ~r~a~m~m~~~~
Table 4.5. (from data book, page nos. 7.71, 7.72 and 7. 73.
This table gives some detailsfor afew c/.alns.)
~
Designation
Rolon
ISO No.
No.
Roller
Bearing
Breaking
Mass/metre
diameter,
area, A
load, Q
(average)
d"mm
mm2
newton
m,kg/m
Pitchp,
mm
Width
between
inner plates,
b,mm
088-1
Rl278
12.7
8.51
50
18200
0.70
8.00
088-2
DR1278
12.7
8.51
100
31800
1.32
8.00
088-3
TR1278
12.7
8.51
150
45400
1.95
8.00
lOA-I
R50
15.875
10.16
70
22200
1.01
9.55
IOA-2
DR50
15.875
10.16
140
44400
1.78
9.55
IOA-3
TR50
15.875
10.16
210
66600
3.02
9.55
12A-1
R60
19.05
11.9
105
32000
1.47
11.90
12A-2
DR60
19.05
11.9
210
63600
2.90
11.90
12A-3
TR60
19.05
11.9
315
95400
4.28
11.90
INote I R -
Simplex, DR - Duplex, TR - Triplex
6. Calculation oftotal/oad
on the driving side of the chain (P-d :
Total load on the}
driving side (PT)
Tangential fOrCe}
_
due to power
+
{
transmission (PI)
or
=
PT
Centrifugal
tension (P c)
due to speed
of the chain
Tension
}
+ due to chain
{
sagging (P,,)
... (4.14)
P, + Pc + P s
(i) Tofind tangential force (P,) :
P, -
Where
1020N
v
... (4.15)
N - Transmitted power in kW, and
xpxN]
60 x 1000
Z2 xp x
z]
v
= Chain velocity in rn/s =
(ii) Tofind centrifugal tension (P J
= mil
m = Mass
60 x 1000
:
... (4.16)
P,
Where
or
N2
of chain / metre, from Table 4.5.
(iii) Tofind tension due to sagging (P,) :
p.. = k-
W·
a
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... (4.17)
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I
I
I
-
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Clio;" Driws
where
4.11
ffi'
nt of chain drive,
oe lClent of sag taking into account the arrangeme
from Table 4.6.
k == C
W
== Weight of chain I metre
II:
m- g, and
a - Centre distance in metre.
T.b/~ 4.~ CoqJkklfl/or s.g, k (from dala book, pale no. 7. 76)
Position or chain drive
Coefl"lCieat for sa.
Horizontal
Upto400
More than 400
Vertical
6
4
2
I
k
("J :
7. c.Jcu/atio" of servic« /«Ior
The service factor is used to account for variations in the driving and driven sources for
roller chains.
:.
*s == * r : *2 . *3 . *4 . *s . *6
... (4.) 8)
*1' *2' *J' *4' *s and *6 from Tables 4.7 to 4. )2 correspondingly.
Service factor,
Select the values of
Table 4.7. Loadfactor,
"J (from data book, page no. 7. 76)
kJ
Types of load
Constant load
1.0
Variable load or load with rnild shocks
1.25
Variable load or load with heavy shocks
1.5
Table 4.8. Factor lor distance regulation, k 1(from data book, page no. 7. 76)
kl
Types of distance re2ulation
1.0
1.1
Adjustable supports
Drive using idler sprocket
Fixed centre distance
1.25
Table 4.9. Factor lor centre distance
0/ sprockets, k J (from data book, page no.
k3
Centre distance of sprockets
Ip
zi + z2
Ip
%. +Z2
Ie
> I or
Dp
< 2Sp
1.25
1.0
.. I.S or
Dp
=
(30 to 50)p
~ 2.0 or
Dp "'"
(60 to 80)p
zi +z2
_
7.76)
0.8
.
Length of chain in multiples of pitches (number of links)
== 2 op
z +z
I
2 +
+
2
[(Z2-ZI)/27t]2
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op
[to be corrected to a greater even
be
Dum r
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J
.....
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.~!2
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---------------------Qp
~D~u=.~~~n~o.f~Tr~a~m~m~U~·!~ion~S~~~
= Approximate centre distance in multiples of pitches,
= 001 p.
where
ao = Initially assumed centre distance in mm, and
p
=
Pitch, mm.
Table 4.10. Factor for the position of the sprockets, k4 (from data book, page
Position of the sprockets
Inclination
of the line joining
the centres
110.
7.77)
k..
of the
t
sprockets to the horizontal upto 600
More than 600
1.25
Table 4.11. Lubrication/actor, ks (from datil book, page no. 7.77)
kS
Types of lubrication
Continuous (oil bath or forced lubrication)
0.8
Drop lubrication
1.0
Periodic
1.5
Table 4.12. Rating/actor, k6 (from datil book, page no. 7.77)
Working
k6
schedule (or rating)
Single shift of 8 hours a day
1.0
Double shift of 16 hours a day
1.25
1.5
Continuous running
8. Calculation0/ design load :
Design load - Total load on the driving side of the chain
or
Design load -
x
Service factor
... (4.19)
PT x k,
9. Calculation0/ working factor of safety (FS'; :
Calculate the working factor of safety by using the relation
Factor of safety = Breaking load ~:~:~~
from Table 4.5 -
P';Xk. ... (4.20)
10.Check/or/actor ofsafety:
Compare the working factor of safety with the recommended minimum value of factor of
safetygiven in Table 4.13.
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,
s~
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~C~M~i!n~Dr~~~
~~
Tab/~ 4.13. Factor of safety, n '(from data book, page no. 7. 77)
-
Speed of smaller sprocket, rpm
Pitch
p.lDm
upto 50
200
400
600
800
1000
1200
1600
2000
2400
2800
7.0
7.8
8.55
9.35
10.2
11.0
11.7
13.2
14.8
16.3
18.0
9.525
12.7
15.875
If the working factor of safety (FSw) is greater than the recommended minimum value of
factor of safety (n'), then the design is safe and satisfactory.
If the working factor of safety is not satisfactory, one more chain may be added (i.e.,
simplex to duplex or duplex to triplex) to the existing one or the chain pitch may be
increased.
11. Check/or the bearing stress in the roller:
./
Calculate the bearing stress in the roller using the formula
o
Tangentialload
Bearing area
=
P, x ks
A
=
... (4.21)
Take the bearing area (A) value from Table 4.5 .
./
Now compare the calculated bearing stress (or induced stress) value with the
allowable bearing stress value given in Table 4.14.
For safer design, the induced stress should be less than the allowable bearing stress.
Table 4.14. AUowable bearing stress,
Ia/,
Nlmm2 (from data book, page no. 7.77)
Speed of rotation of smaller sprocket, rpm
Pitch
p, DIDI
<50
200
400
600
800
1000
1200
1600
2000
2400
2800
35
31.5
28.7
26.2
24.2
22.4
21.0
18.5
16.5
15
13.7
9.525
12.7
15.875
12. Calculation of actual length of chain (L) :
./
Calculate the number of links (lp) using the formula
Ip
=
Qp
=
2Qp
"0
p
+
=
[ ZI +2 Z2 ]
(Z2 -;1 )2
+
2
Initial centre distance
pitch
Qp
[from data book, page no. 7.75]
[from equation 4.9]
j
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4.14
Design o/Transmisslon Systems
./
Correct the calculated number of links (/ ) into
./
Now calculate the actual length (L) of chain using the formula
L = / x p
p
[from data book, page no. 7.75]
p
an even number .
1J. Calculation of exact centre distance:
Calculate
. h es) usmg
.
. the exact centre distance corrected to an even numbe r 0f liIn ks ( pitc
the re Ianon
Exact centre distance,
a
xp
where
M
and
=
(Z22~ZI
r
=
[from equation 4.10]
constant
Decrement in centre distance for an initial sag, Sa
Exact centre distance
[from data book, page no. 7.75)
=
=
0.01 a
Q-
0.01 a
=
0.99
Q
14. Calculation of pitch circle diameters (pcd) of sprockets :
Pcd of smaller sprocket, d I
=
Pcd oflarger sprocket, d2
=
p
sin (180 I z)
io
sin (I
I z2)
Smaller sprocket outside diameter,
dOl
Larger sprocket outside diameter, d02
and
where
d,.
=
+ 0.8 d,
... (4.22)
= d2 + 0.8 d,
... (4.23)
dl
= Diameter of roller taken from Table 4.5.
I Example 4.1 I A truck
equipped with a 9.5 kW engine uses a roller chain as theflnal
drive to the rear axle. The driving sprocket runs at 900 r.p.m: and the driven sprocket at
400 r.p.m. with a centre distance of approximately 600 mm. Select the roller chain.
GivenData: N = 9.5 kW; N) = 900 r.p.m.; N2 = 400 r.p.m.; Qo = 600 mm.
TOflnd: Select (i.e., design) the roller chain.
@Solution :
1.Determination of the transmission ratio (i) :
..
..
N)
900
Transmission ratio, , = N2 = 400
(Since the transmission
to COnsultTable 4.2)
'!Etc'
=
2.25
ratio can be calculated from the given data, therefore we need not
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C}taill /)riveS
1. ~electioll O/IIlUf1bero/ti!eth Oilthe driver sprocket (1.J:
From Table 4.3,
%.
= 27 (for ; = 2 to 3) is selected.
1. J)ete,.",;nalioll o/llumber
o/teeth Oilthe driven sprocket (I.,) :
z2 = i x z. = 2.25 x 27
Recommended value,
z2mca -
...
=
1.2
=
60.75 ~ 61
100 to 120
61 is satisfactory .
4. Selection 0/ sIIIIIdardpilch (P) :
...
Centre distance, a = (30- 50)p
600
a
Maximum pitch, Pmax - 30 -30
and
Minimum pitch, Pmin
We know that
=
a
50
600
-50
=
20mm
=
12mm
Any standard pitch between 12 mm and 20 mm can be chosen. But to get a quicker
solution, it is always preferred to take the standard pitch closer to Pmax' Refer Table 4.4.
...
Standard pitch, p
S. Selection
=
15.875 mm is chosen .
0/ the chain:
Assume the chain to be duplex. Consulting Table 4.5, the selected chain number is
10.40-2/ DRSO.
6. Calculation
0/ total load on the driving side 0/ the chain (P r) :
(i) Tangential force (P; :
P, where
t020N
v
N = Transmittedpower
v
=
60 x 1000
(Ii) Centrifugal tension (PJ
Pc
From Table 4.5,
...
m
Pc
=
9.5 kW
Chain velocity in m/s
_ z. xpxNI
PI =
in kW
1020 N
v
-
_ 27 x 15.875 x 900
60 x 1000
6.43 m/s
1020 x 9.5
= 1507 N
6.43
:
-
mvl
1.78 kglm
1.78 (6.43)2 = 73.59 N
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4.16
Design o/Transmission Systems
(iii) Tensio« due to sagging (P J
:
P, = k- w, a
From Table 4.6,
k = 6 (for horizontal)
(iv) Total load (PrJ:
w
= mg = 1.78 x 9.81 = ]7.46
a
=
Initial centre distance
P,
=
=
6 x 17.46 x 0.6
PI + P, + P,
=
J 507
Pr
=
=
N
0.6 m
62.82 N
+ 73.59 + 62.82
=
1643.4 N
7. Calculation of service factor (kJ :
We know that the service factor,
ks
=
k,· k2 . k3 . k4 . k5 . k6
From Table 4.7,
k, = 1.25
From Table 4.8,
k2
From Table 4.9,
k3 =
(',: we have used
From Table 4.10,
k4 =
(for horizontal drive)
From Table 4.11,
= 1
k6 = 1.25
(for drop lubrication)
From Table 4.12,
=
(for load with mild shocks)
(for adjustable supports)
ks
Q
p
=
(30 to 50) p)
(for] 6 hours I day running)
ks = ] .25 x ] x ]
X ] X ]
x 1.25 = ] .5625
8. Calculation of design load:
Design load = Pr x ks = 1643.4 x 1.5625 = 2567.8 N
9. Calculation of working factor of safety (FSwJ :
FSw
=
Breaking load Q from Table 4.5 = 44400
Design load
2567.8 = 17.29
10. Check for factor of safety :
Consulting Table 4.13, for smaJler sprocket speed of 900 r.p.m. and pitch 15.875 mm, the
required minimum factor of safety is 11. Therefore the working factor of safety is greater
than the recommended minimum factor of safety. Thus the design is safe and satisfactory.
11. Check/or the bearing stress in the roller:
We know that
=
PI ~ ks ; where
A
= 140 mrn- from Table 4.5.
= 1507 x ] .5625 = 16.8 N/mm2
140
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~-----------------------------------------------~
. on f.bIe 4.14. for II1\.IIIer .pmc:kd speed of 900 r.p.m. and picdl 1 S.175 nun. lilt
belting .uesl is 22.4 Nlmml. Theftfore !he induced smss is less than lilt
al
Thus till Msip ,. RIf, -'"'
• ...t* bearing.-o·
t1/ I~"gt" of cl"u"
U..CIk_lIIllMf
+~) (
(L) :
%,
=
Number of links, Ip
2 Qp + (
~
p =
I
.'. Actual length of chain,
0/ aact
+
600
- 15.875
pitch
+
2" J2
Qp
Centre distance
<
/
( 27 + 61)
. (61
2
+
=
37.795
- 27) I 2Jt
37.795
J2
::I:
120.36
122 links (rounded off to an even number)
L
=
Ip x P
= 122 x 15.875 = 1936.7S mm
centre dista"c~ (a) :
We know that
Q
= e+
e =
and
M
=
Q
=
y e4
2
-
8M
x
ZI+Z2)
2
Ip - (
P
(27+61)
2
= 122 -
= (61-27)2
(Z2-ZI)2
27t
x
29.28
4
78
_
- 29.28
2n
78 +y 782 - 8
-
x
15.875 = 613.11 mm
= 0.01 a = 0.0 I (613.11) = 6.1311 mm
Exact centre distance = 613.) I - 6.1311 = 606.978 mm
~rement
..
.2
.'
•
I J. ClllcIllIllWII
(%2 - :,)
= 2 (37 <795)
p
1IJIis/adtH7.
in centre distance for an initial sag
14. CIIlcllllllioll
0/ sprocket
diam~ters :
SlIUIIlu sproclcet :
Pcd of smaller sprocket, d I
=
p
sin (180 1 zl)
t 5.875
= sin (180 127) = 136.74111111
and
where
Sprocket outside diameter,
dOl = d I + 0.8 d,
d, = Diameter of roller, from Table 4.5 = 10.16mm
dOl = 136.74 + 0.8 x 10.16 = 144.868111111
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Design ofTrQnsmiss;~n Systems
-
4.18
Larger sprocket:
Pcd of larger sprocket, d 2
=
P
sin (180 / z2)
= 308.38
Sprocket outside diameter, d02
and
15.875
=---sin (180 / 61)
mm
d2 + 0.8 d,
-
- 308.38 + 0.8 x 10.16
= 316.51
mm
[Example 4.21 The transporter of a heat treatment furnace is driven by a 4.5 kW,
1440r.p.m. induction motor through a chain drive with a speed reduction ratio of 2.4. The
transmission is horizontal wlth bath type of lubrication. Rating is continuous with 3 sid/is
per day. Design the complete chain drive.
Given Data: N = 4.5 kW; Nt == 1440 r.p.m.; i = 2.4.
Tofind: Design the chain drive.
@Solution:
1. Transmission ratio,
2. Tojind
ZI:
i = 2.4 (Given)
From Table 4.3,
3. Tojindz2:
z2 -
Recommended z2 max
z,
ixzt
:. N2
N)
=
i
1440
=
2.4 =
600 r.p.m.
= 27 (for i = 2 to 3) is chosen.
= 64.8 ~ 65
= 2.4x27
= 100 to 120. .,
Z2
=
65 is satisfactory.
4.: Standard pitel, (p): Since the centre distance is not given, we have to assume the
initial centre distance, say a = 500 mm.
We know that
a
= (30 - 50) P
...
Pmax
=
a
30
and
Pmin
-
500
a
-50
50
500
-30
- 16.6 mm
=lOmm
From Table 4.4, in between 10 and 16.6 mm, a standard pitch, p = 15.875 mm is chosen.
5. Selection
0/ cl,ain : Assume
the chain to be simplex.
From Table 4.5, the JOA-J / R50 chain number is chosen.
6. Calculation of total load on the driving side (P r) :
PT
- PI + P,
+ P,
1020N
v
(i)
PI -
Where
N - Transmitted power in kW = 4.5 kW
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(Given)
,
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~C~ha~i~n~D~rr~v~~
~~
= Velocity of chain in mls
v
zi x P x Nl
= 60 x 1000 =
P
1020 x 4.5
10.287
=
I
=
27 x 15.875 x 1440
60 x 1000
10.287 mls
446.19 N
= mv2
(ii)
pc
From Table 4.5,
= 1.01 kglm
Pc = 1.01 (10.287)2 = 106.88 N
(iii)
Ps
From Table 4.6,
=
m
= k,w'a
k
6
=
(for horizontal)
= mg = 1.01 x 9.81 = 9.908 N/m and a= 0.5 m
P s = 6 x 9.908 x 0.5 = 29.72 N
w
(iv)
Total load, PT
=
466.19 + 106.88 + 29.72
kl· k2 . k3 . k4 . ks . k6
=
582.79 N
7. Service/actor:
ks
=
From Table 4.7,
kl
= 1.25
From Table 4.8,
~
=
1
(for adjustable supports)
From Table 4.9,
k3
=
1
(since we have used a
From Table 4.10,
k4
=
1
(for horizontal drive)
From Table 4.11,
ks
=
0.8
(for bath type lubrication)
From Table 4.12,
k6
=
1.5
(for continuous running i.e., 3 shifts I day)
(for load with mild shocks)
= 1.25 x 1 x 1 x 1 x 0.8 x 1.5
8. Design load = PT x ks = 582.79 x 1.5 = 874.19N
k,
=
9. Working/actor of safety
FSw
=
=
= (30
to 50) p)
1.5
Breaking load Q from Table 4.5
Design load
22200
874.19
=
25.39
From Table 4.13, for smaller sprocket speed 1440 r.p.m. and pitch 15.875 mm, the
recommended minimum value of factor of safety (n') is 13.2. Since the working factor of
safety is greater than the recommended minimum value of factor of safety, therefore the
design is safe and satisfactory.
f/fI' ..
10. Bearing stress in the roller: From Table 4.5, A = 70 mm2
0'
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=
P, x ks = 446.19 x 1.5 = 9.56 N/mm2
A
70
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-
4.20
Vesign a/Transmission Systems
From Table 4.14, for smaller sprocket speed 1440
.
pm
allowable bearing stress is 18.5 N/mm2 Since the' d r · · . an.d pitch 15.875 mm, the
.
m uce d stress IS I th
bearing stress, the design is safe and satisfactory.
ess an the allowable
11. Actual length of chain (L) :
Number of links,
where
'»
=
a
= ao
2Q +
p
p
p
Ip
(ZI ; Z2)
+ [(z2 - z,) 1 2n ]2
ap
500
= 15.875 = 31.496
2 (31.496) + (27 + 65)
[ (65 - 27) 1 2n
2
+
31.496
=
J2
= 110.153 ~ 112
(rounded off to an even number)
Actual length of chain, L = Ip xp = 112 x 15.875 = 1778 mm
12. Exact centre distance:
a
xp
=
(ZI ; Z2)
[(Z\~ZI)
Ip _
e =
where
and
M
=
a
=
r
66 +
=
C7;
112 _
r
(65;;.27
V 662 4- 8 x 36.57
Decrement in centre distance for an initial sag, /).a
Exact centre distance.
=
x
=
65)
=
66
36.57
15.875
514.92 mm
=
= 0.01 a = 5.149 mm
514.92 - 5.149 = 509.77 mm
=
13. Sprocket diameters:
For smaller sprocket:
and
Pcd
=
p
=
sin (1801 z,)
=
136.74 mm
Sprocket outside diameter, do, = d, + 0.8 d,
d,
From Table 4.5,
do,
= Diameter of roller = 10.16 mm
= 136.74 + 0.8 x 10.16 = 144.87 mm
p
For larger sprocket:
and
15.875
sin (180/27)
Pcd
= sin (180 1 z2) =
Sprocket outside diameter, d02 = d2 + 0.8 d,
=
15.875
sin (180 1 65) = 328.58 mm
328.58 + 0.8
x
10.16
= 336.71 mm
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•
-~~--------------------------------~~
eMi" I!!:!"es
::::;;--'Example 4.3
I A compressor is to be actuated/rom
aID kW electric motor. The ".o~o,
skatis 970 r.p.m: and that o/the compressor is to be JJO r.p.m. The compress~r opertll'''1
in two shifts. The minimum centre distance should be 1000 mm: Design a suuable chili"
drive.
Given Data: N = 10 kW; N. = 970 r.p.m.; N2 = 330 r.p.m.;
Tofind:
ao = 1000 mm.
Design a suitable chain drive.
@Solution :
1. Transmission ratio : i
2. Tofind
1./:
J. Tofind
1.2 :
=
N.
N2
=
From Table 4.3, 1.)
Z2
Recommended
z2max
=
=
i x z.
970
330
=
2.94
=
25 (for i = 2 to 3) is chosen.
=
2.94 x 2S
A:j
74
:. 1.2 = 74 is satisfactory.
100 to 120.
4. Standard pilch (p): We know that a = (30 - 50) P
and
a
Pmax
=
30
P",in
=
50
a
1000
= 30 =
33.33 mm
1000
= So =
20 mm
It can be seen that, in Table 4.4, there is no standard pitch in between 20 mm and
33.33 mm. Therefore we cannot proceed further.
(refer Table 4.9)
a = (60 to 80)p
Now take
..
and
Pmar = 60
a
1000
-60
= 16.66 mm
-a
1000
--80
= 12.5 mm
p"""
=
80
Any standard pitch between 12.5 mm and 16.66 mm can be chosen. Therefore, from Table
4.4, the standard pitch (closer to Pmar) is selected as 15.875 mm.
5. Other design par~ten:
Knowing the standard pitch, the chain selection, total load
calculation, service factor selection, design load calculation, check for factor of safety, check
for bearing stress in the roller, actual length and centre distance calculations, and sprocket
diameters calculations can be proceeded as discussed in the previous examples.
[ EXllmfJle 4.4
IA
bucket elevator is to be driven by a gear motor and a roller chaill
drive. Gear motor power = 7.5 kW: Speed 0/ gear motor = 1400 r.p.m.; Transmission ratio
-10: I; AssulM a minimum centre distance belween sprockets = 550 mill. Sekct a suiJoblt
cltaht.
Glvell Data: N = 7 5 k~;
N = 1400 r.p.n.. ; i = 10 .
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.I•.
-= 550 !nfT'.
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Ttlf",d: Select a suitable chain.
e StJIuJioll :
J frtUISmission ,lI1io, ; = 10
.
•
... (Given)
., Ttlfilld %1: From Table 4.3, for ; = 10 the recommended
be f
h
Ito
'num
r 0 teet on the
. 'on sprocket (zl) is not available. But from Table 43. , for; > 7,can
it
be seen t hat
pIn!
shouldbe lessthan 17.
z,
If we assume Z I as 15, then z2 = lOx 15 = 150 which is greater than the recommended
Z2 /IItl% (,'.,e , 100 to 120) value. Therefore it is not acceptable •
. Nowassume
z,
zi as
Then z2 = ; x zi
= II is acceptable.
11.
= lOx II
=
J 10, which is within the recommended
Z2 max'
Hence
Now proceed the problem as discussed in Examples 4. I and
4.2, to find the other design parameters such as standard pitch, chain type, design load,
workingfactor of safety, induced stress in the roller, actual chain length, exact centre distance
and sprocket diameters.
3. Other design parameters:
DESIGN OF SPROCKET WHEELS
I
4.19. SPROCKETS
It is understood that the operation of a chain drive is largely dependent on the quality of
sprocket wheels. In general sprockets are made of low carbon or medium carbon steels. But
sometimesstainless steel is also used for sprockets.
4.20. TYPES OF SPROCKETS
(a)
(b)
(c)
(d)
Flg.4.8.
F
. d in design practice. They are :
styles·of sprockets are standardize
. . 8( )
.
nsions as shown In Fig.d. a.
~ Style A is a flat plate with no hub exte
,
OUr
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~C~'h~a/~'n~D!!_r~iv~!S:!_
./
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~
4.23
--:-
Style B has a hub extension on one side of the plate (flange),
as shoWn ill
Fig.4.8(b) .
./
Style C has hub extensions on both sides of the flange, as shown in.Fig.4.8(c) .
./
Style D has a detachable hub, as shown in Fig.4.8(d). The style D hub is normally
attached to the flange with bolts.
4.21. SPROCKET DIAMETERS
The five important sprocket diameters are pitch, outside, bottom, caJiper and maximum
hub diameters, as shown in Fig.4.9.
..
'.
Caliper
diameter
Max hub
diameter
I
I~
~
Bottom diameter Pitch diameter __
Outside diameter
14----
....
---i
Fig. 4.9. Roller chain sprocket diameters
The equations for those diameters are :
(i)
Pitch diameter
(ii)
Outside diameter
(iii)
(iv)
(v)
=
p
sin (180/ z)
=
P [ 0.6 cot ( ~)
Bottom diameter
=
Pitch diameter - Roller outside diameter
Caliper diameter
= Pitch diameter x cos
Maximum hub diameter = p [ cot (
'!O) -
]
(:0) _
Roller outside diameter
I ] _ 0.03
4.22. TOOTH FORM
The standard tooth form d
fiI
f
The pri . Idl
.
an pro I es 0 a sprocket wheel are shown in Fig.4.1 O(a) and (b).
ncipa
ImenSlOns of the tooth profile are given in Table 4.] 5.
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4.24
Design a/Transmission Systems
(a) Tooth profile of sprocket
(b) Rim profile for sprocket
Fig. 4.10.
Table 4.15. Proportions of the sprocket wheel (refer Figs.4.10(a) and (b))
Dimension
SoNo.
Notation
Relation
-
1.
Chain pitch
p
2.
Pitch circle diameter
D
D =
P
sin (180/z)
3.
Roller diameter
dr
-
4.
Width between inner plates
bl
-
5.
Transverse pitch
PI
-
6.
Top diameter (outer)
Do
DO = D +0.8dr
7.
Root diameter
Df
Df = D-2 r;
8.
Roller seating radius
r;
(r;}max
= (0.505 d; + 0.069 ~
(r;}m;n
= 0.505 dr
9.
Tooth flank radius
re
(r e )max - 0.008 d; (z2 + 180)
= 0.12 d; (z + 2)
(re )m;n
10.
Roller seating angle
a.
)
0
(lmax
90
= [ 120 -7
(lm;n
90
= [ 140 -7
0
]
0
11.
Tooth height above the pitch polygon
12.
Tooth side radius
13.
Tooth width
.__ 14.
Tooth side relief
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ha
0
]
(ha )max
= 0.625 p - 0.5 d; +
(ha )m;n
= 0.5 (P-dr)
rx
(rx)m;n
p
bj1
bj1
0.93 bl ifp S 12.7 mm
bj1
= 0.95 bl if p > 12.7 mm
ba
ba
0.8p
z
0.1 p to 0.15p
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Chain Drives
4.25
Therefore, knowing the chain pitch (P) and the chain type, we can determine the various
dimensions of the sprocket wheel using the relations given in Table 4.15.
INote I
Instead of using the relations, the various dimensions of the sprocket wheel can be straight
away selected from the data book. So refer data book, page no. 7.79 and 7.80.
I Example
IAssume
the dtda of Example 4.2, and calculate the following
dimensions of the driving sprocket wheel: (i) outer diameter; (ii) roller sealing radius;
(iii) root diluneter; (iv) tooth flank radius; (v) tooth side radius; (vi) tooth width; and
(vii) tooth side relief.
4.5
,Given Data: From the Example 4.2,
pitch, p
=
=
15.875 mm; chain type
Number of teeth on the driving sprocket,
e SolUlion : Pitch circle diameter
simplex and lOA - I/R50;
= 27.
zi
p
of driving sprocket, 0 =
.
SID
(180)
zi
15.875
=
= 136.74 mm
. (180)
SID
27
Roller diameter, d,
From Table 4.5,
and
Width between inner plates, bl
(i) OUler diameter of driving sprocket (DoJ
Do
From Table 4.1 S,
=
10.16mm
= 9.SS
mm
:
=
D + 0.8 d,
=
136.74 + 0.8 x 10.16
=
144.86 mm ADI.'"
{ii) Roller setl/ing radius (rJ :
From Table 4.15,
=
(0.505 d, + 0.069
=
(0.505 x 10.16 + 0.069
-
0.505 d,
=
0.505 (10.16)
5.28 + 5.13
(r,)max
and
(r,)",,"
Roller seating radius, rl
::
2
~)
~10.16
)
=
5.28 mm
= 5.13 mm
= 5.205
mm AD!.-.a
(iii) Root diQIIfd~r (Dr) :
.J
From Table 4.1 S,
0-2r
=
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I
136.74 - 2 (5.205)
= 126.33
mm A ... ..,
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-4.26
Design o/Transmission Systems
(iv) .Toothflank radius (rJ:,
From Table 4.1 5,
0.008 d, (z2 + 180)
(r~)max =
=
0.008 x 10.16 (272 + 180) .: 73.88 mm
0.12 d, (z + 2)
=
0.12 x 10.16 (27 + 2)
73.88 + 35.35
and
Tooth flank radius, r~
(v) Tooth side radius (rJ
From Table 4.15,
=
=
35.35 mm
= 54.6 mm
2
Ans. ~
:
(rX)min
=
p = 15.875 mm Ans. ~
(vi) Tooth width (bpJ:
From Table 4.15,
0.95 b,
0.95
x
if p> 12.7 mm
9.55
=
9.0725 mm Ans. ~
,
(vii) Tooth side relie!(bj :
From Table 4.15,
ba = 0.1 p to 0.] 5 p
= (0.1
or
ba
=
15.875) to (0.15
1.587 + 2.38
x
2
=
x ] 5.875) =
1.687 to 2.38
1.98 mm Ans. ~
DESIGN OF SILENT CHAIN
4.23. SILENT (OR INVERTED-TOOTH) CHAIN
Inverted-tooth chains are also called silent chains because of their relatively quiet
operation. Silent drives are often selected for high-power, high-speed and smooth operation.
Silent chains have inward-pointing teeth that engage the sprocket, as shown in Fig.4.ll.
4.24. CONSTRUCTION
Silent chains consist of toothed link plates that are pin-connected to permit articulation, as
shown in Fig.4. J 1. The link teeth and the corresponding sprocket teeth are usually straightsided.
Various provisions are made to prevent the chain from sliding off the sprockets. They are:
(i) Centerguide chain: It has central guide links that fit central grooves in the wheels,
as shown in Fig.4.] I(b).
(ii)
Side-j1ange silent chain: Fig.4.] I(c) shows a chain with side guide links that
straddle the sprocket face.
(lii,i1
'J
Duplex (or bend back) chain: It has teeth on both sides, as shown in Fig.4.II(d). It
can be used in 'serpentine drives whereby sprockets are driven from both sides of
the chain.
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Chain Drives
--------------------
-'4.27
(hi Centerguidc chain
Ie) Side guide chain
(d)
Duplex (or "bend back") chain
Fig. 4. I I. Inverted-tooth (silent) chains
4.25. TYPES OF SILENT CHAINS
Depending upon the type of joint between links, the silent chains are classified into:
(i) Reynold chain: In Reynold chain, the links are connected by pins resulting in sliding
friction.
(ii) Morse chain: In Morse chain, the rocker pins are used.
4.26. ADVANTAGES AND DISADVANTAGES OF SILENT CHAINS OVER THE
ROLLER CHAINS
Advantages:
./ It can be used for high-speed and high-power applications .
./
./
They operate much smoother and quieter than roller chains .
More reliable due to laminated construction.
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___
4.28__..------------------
!!:_~.~.r~~~~~
eSlgn OJ ransmission Systems
I
Disadvantages:
./'
More heavier .
./'
More complex .
./'
More expensive .
./'
More difficult to manufacture .
./'
Require more careful maintenance.
Due to the above reasons, the silent chains have limited app I'icanons.
.
4.27. DIMENSIONS OF THE VARIOUS PARTS OF THE CHAIN
The dimensions
of the various parts of the chain are usuall y given
"
'.
m terms
The approximate values are given below:
5
I.
Roller diameter,
d,
=
2.
Pin diameter,
dp
=
3.
Chain width (i.e., the distance between the roller link plates) :
b,
8'
16x
8'5
8'
pitch
.
x pitch
I
4.
Thickness of link plates,
tp =
5.
Width between outer plates,
bo
6.
Maximum height of pin link plates (i.e., the outer plates):
=
ho =
7.
x
pitch
b, + 2 tp
0.82
x
pitch
Maximum height of roller link plates (i.e., the inner plates):
hi
8.
f the
the pitch.
ni
x pitch
5
=
0
Length of roller,
=
l, =
0.95
x
pitch
0.9 b, - 0.15
~oteJ As already mentioned, for low and medium speeds the roller chain is recommended. For
highspeedsthe silent chain is preferred.
[EXample 4.6
IA
12. 7 mm pitch silent chain operating under steady load conditions,
transmits 30 kW from an electric motor to a centrifuge. Design the silent chain.
Given Data:
Tofind:
Pitch, p = 12.7 rnm ; Power = 30 kW.
Design the silent chain.
e Solution:
The dimensions of various parts of the silent chains in terms of its pitch are
calculated as below.
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1
/
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~~~~--------------------------------~
£_hain DriveS
dr
I. Roller diameter:
dp
2. Pin diameter:
5
= -8 x
pitch
5
= -8 x
12.7
=
7.9375 mm
5
. h
16 x prtc
=
5
= 16 x
=
12.7
3.9687 mm
3. Chain width (i.e., the distance between the roller link plates) :
4. Thickness of link plates:
i
hi
=
tp
= i1
._
5. Width between outer plates:
bo
ho
7.9375 mm
. h
x Pltc
.1
-8 x . 12.7 = 1.5875 mm
= hi + 2 tp
= 7.9375
6. Maximum height of pin link plates:
=
pitch
+ 2 (1.5875) = 11.1125 mm .
=
0.82 x pitch
=
0.82 x 12.7
7. Maximum height of roller link plates:
hi = 0.95 x pitch
8. Length of roller:
=
l, =
=
= 10.414 mm
0.95 x 12.7 = 12.065 mm
0.9 b, _ 0.15
0.9 x 7.9375 _ 0.15 = 6.9937 mm
REVIEW AND SUMMARY
and
./
Types of chains: 1. Link (or welded) chains. 2. Transmission (or roller) chains.
3. Silent (or inverted tooth) chains.
./
Link chains are widely used in low capacity machines such as hoists. winches and hand
operated cranes, and as slingsfor suspending the loadfrom the hook or other device.
./
The classification, construction, selection, advantages and disadvantages of link chainJ
are discussed in this chapter .
./
Similarly, the construction, specification, geometric relationships of transmission (Le.,
roller) chains are also presented in detail with sufficient illustrations.
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"
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----------------------------
~D~u~~~n~o~if~ff~a~m~m~u~·~
Designprocedure fo~ roller c~ain : 1. Tofin~ number of teeth in the driving and driven
sprockets; 2. Selection of pitch; 3. Selection of chain type; 4. Calculation of PT;
5.Calculation of design load; 6. Check for factor of safety; 7. Checkfor bearing stress
on roller; and 8. Calculation of actual length of chain and exact centre distance.
"
A sprocket is a wheel with teeth of a special profile. The design of sprocket for the
selected chain is also presented
"
The inverted chains are also called silent chains because of their relatively quiet
operation. The construction, types, advantages, disadvantages and dimensions of the
silent chains are discussed at the end of the chapter.
REVIEW QUESTIONS
I.
Discuss the advantages and disadvantages of chain drive.
2.
Write the applications of chain drive.
1 Classify the chains.
4.
Write short notes on link chains.
5.
6.
How can you specify a roller chain?
Write an engineering brief about the chordal action of a chain drive.
7.
In chain drives, the sprocket has odd number of teeth and the chain has even number of
8.
links. Why?
What are the materials used for making chains and sprockets?
9. Write short notes on sprocket wheels.
10. What is a silent chain? In what situations, silent chains are preferred?
II. List out the advantages and disadvantages of silent chains.
12. Explain the construction of silent chains.
PROBLEMS FOR PRACTICE
Problemson design of transmission chains:
. dri t
tuate a compressor from 15 kW electric motor at 1000 r.p.m.,
I . DesiSign a c h am
rive 0 ac
..'
.
t 550 r p in Minimum centre distance IS 550 mm. The cham
the compressor runnmg a
. . .
.
k 8
tensi
b
di t d by shifting the motor on rails. The compressor IS to wor
ension may e a ~us e
hou~day.
.
.
2. Design a roller chain to transmit power from a 7.5 kW motor t:; ~c:roca:mg. p~:~
The pump is to operate continuously 24 hours per day. The spe 0 t e. mo or IS I
. 600
m The motor can be located at a distance not ess
r.p.m. and that of the pump IS
r.p..
than 700 mm.
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~C~h~Q~m~D~r~w~e~s
3.
~
Design a chain drive to run a compressor
r.p.m., the compressor
from a II kW electric motor running at I~
speed being 350 r.p.m. The compresso.r
The centre distance should be approximately
oper~tes 3 shifts a day.
600 mm. The cham tension can be adjusted
by shifting the motor on slides.
4.
A 15 kW squirrel cage motor, ] 250 r.p.m. is driving a centrifugal
The centrifugal
pump at 550 r.p.tn.
pump is located at 700 mm from the motor. Design a chain drive and
draw the arrangement.
5.
Select a suitable chain to transmit 50 kW at 900 r.p.m. of the sprocket pinion. A speed
reduction of 2.5 : I is desired. The driving motor is mounted on an adjustable base. The
load is steady, the drive is horizontal and the service is 16 hours a day.
6.
It is required to design a chain drive with a duplex chain to connect a 15 kW, 1440 r.p.m.
electric motor to a transmission
moderate shocks.
shaft running at 350 r.p.m. The operation
(i)
Specify the number of teeth on the driving and driven sprockets.
(ii)
Select a proper roller chain.
involves
(iii) Calculate the pitch circle diameters of the driving and driven sprockets.
(iv) Determine the number of chain links.
(v)
Calculate the exact centre distance.
During preliminary stages, assume the centre distance as 40 times the pitch of the chain.
Problems on design of sprocket wheels :
7.
Assume the data of problem I, and calculate the following dimensions of the driving
sprocket wheel: (i) outer diameter; (ii) roller seating radius; (iii) root diameter,
(iv) tooth flank radius; (v) tooth side radius; (vi) tooth width; and (vii) tooth side relief.
8.
Design the driving sprocket wheel, for the data of the problem 2.
Problems on design of silent chains :
9.
A centrifuge is to be driven by an electric motor. A 15.875 mm pitch silent chain is used
for this purpose. Design the silent chain.
10. A 19.05 mm pitch silent chain transmits 15 kW from a four-cylinder internal combustion
engine to a vibrating screen. Determine the various dimensions of the chain parts.
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Spur Gears
"Opportunity is missed by most people because it comes
dressed in overalls and looks like 'Work."
- T"om"~ Edison
5.1. INTRODUCTION
Gears are toothed wheels used for transmitting motion
power from one shaft to
another when they are not too far apart and when
locity ratio is desired. In
comparison with belt, chain and friction drive
compact, can operate at
high speeds and can be used wher
e.
Iso gear drives are used when
large power is to be trans, ......
·~
o the gears in position is much less
in lower bearing pressure, less wear 011 the
than in an equivalent frictio
bearing surfac
nd Limitations
of Gear Drive Over Chain and Belt Drives
Advantages
1. Since there is no slip, so exact velocity ratio is obtained.
2. It is capable of transmitting larger power than that of the belt and chain drives.
3. It is more efficient (upto·99%) and effective means of power transmission.
4. It requires less space as compared to belt and rope drives.
5. It can transmit motion at very low velocity, which is not possible with the belt
drives.
limitations
:
1. The manufacture of gears require special tools and equipments.
2. The manufacturing and maintenance costs are comparatively high.
3. The error in cutting teeth may cause vibrations and noise during operation.
5.1.2.·Definition of Gear
.A circular body of cylindrical shape or that of the shape of frustum of a cone and .of
uOifonnsmall width, having teeth of uniform formation, provided on its outer circumferenual
surface, is called a gear or toothed gear or toothed wheel.
-_
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Design of Transmission Syste".,!
5.2
5.2. CLASSIFICATION
OF GEARS
Gears may be classified in different manners as given below:
1. Cla.slflcatlon based on the relative position of their shaft axes:
(il Parallel shafts
Examples: Spur gears, helical gears, rack and pinion, herringbone gears and internal
gears.
(Ii) lnters~ct;ng shafts
Examples: Bevel gears and spiral gears.
(iii) Non-parallel, non-intersecting shafts
Examples: Worm, hypoid and spiral gears.
2. Classification based on the relative motion of the shafts :
(i) Row gears: In this type, the motion of the shafts relative to each other is fixed.
(ii) Planetary and differential gears
3. Classification based on peripheral speed (v) :
(i) Low velocity gears
v <3 m/ s
(ii) Medium velocity gears v = 3 to IS m / s
v > 15 m / s
(iii) High velocity gears
4. Classification based on the position of teeth on the wheel:
(i) Straight gears
(iii) Herringbone gears
(ii) Helical gears
(iv) Curved teeth gears
5. Classification based on the type of gearing:
1--1-
(i) External gearing
(.i.i! Internal ge~ri.ng
(III) Rack and pinion
But from our subject point of view,
gears are broadly classified into four
groups, viz., spur, helical, bevel and
./J;\.I(X-....\
~-'¥'):~-i-'
~_
.......
, '- ..
~
_,,/
.,*
"IIDjITIIIlIIIIIhjU IIIII
worm gears.
0/ Spur gears (sometimes
called
(a) Spur gears
straight spur gears) have teeth parallel
(c) Bevel gears
to the axis of rotation are used to
transmit motion from one shaft to
another parallel shaft, as illustrated in
Fig.5.1 (a).
0/ Helical
gears
have
teeth
inclined to the axis of rotation,
shown
in Fig.5.I.(b).
The
as
double
(b) Helical gears
(d) WOfm geara
helical gears connecting two parallel
shafts
are
k.iown
as
herringbone
gears.
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Fig. 5.1. Typesof gears
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Spur Gears
5.3
./ Bevel gt~rs have tee~h form~d on conical surfaces. They are mostly used for
transmitting monon between mtersectmg shafts. A straight-tooth bevel gear is shown in
Fig.S.l(C).
./ Worm gears consist of a worm and a worm wheel, as shown in Fig.S.l(d). Worm and
worm wheel can be visualised as a screw and nut pair. They are used to transmit motion
between non-parallel non-intersecting shafts.
In this chapter, the design of spur gears will be discussed.
~oltJ
The smaller of two gears in mesh is called pinion and the bigger gear is called whee! or
Itllr.
5.3. SPURGEARS
In spur gears, the teeth are straight and parallel to the axis of the wheel. The gearing so
formed is called spur gearing. They are used to transmit rotary motion between parallel
shafts. This gearing may be internal or external. External gears rotate in opposite directions
while internal gears rotate in the same direction.
5.3.1. Tenninology
Used in Gears (Gear Nomenclature)
The terminology of gear teeth is illustrated in Fig.5.2. The various terms used in the study
of gears have been explained below.
-'
~d8\~\)11'
_ .
,-' .
oeO~\)11'
Clearance
Circle
Oedenaum
CIrcle
Fig. 5.1. Gear IIomenciulure
-
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Design of Transmission Syste1tl3
5.4
Pinion: A pinion is the smallest of two mating gears. The larger is often called the gear
1.
or the wheel.
2. Pitch circle:
It is an imaginary circle which by pure rolling action, would give the same
motion as the actual gear.
Pitch circle diameter: It is the diameter of the pitch circle. The size of the gear is
3.
usually specified by the pitch circle diameter. It is also called as pitch diameter.
4.
Pitch point: It is a common point of contact between two pitch circles.
5.
Pitch sur/ace: It is the surface of the, rolling discs which the meshing gears have
replaced at the pitch circle.
Pitch: Pitch of two mating gears must be same. It is defined as follows:
6.
(a) Circular pitch (Pc) :
It is the distance measured along the circumference of the pitch circle from a point on one
tooth to the corresponding point on the adjacent tooth.
Circular pitch,
I. r, --
where
1t
I
D .
z
... (5.1)
D
=
Diameter of pitch circle, and
z
=
Number of teeth on the wheel.
(b) Diametral pitch (Pd) :
It is the ratio of number of teeth to the pitch circle diameter.
Diametral pitch,
IP
d ~
~
~ ~
I
... (5.2)
(c) Module pitch (m) :
It is the ratio of the pitch circle diameter to the number of teeth.
Module, [ m ~ ~ [
... (5.3)
7.
Addendum circle (or Tip circle) : It is the circle drawn through the top of the teeth and
is concentric with the pitch circle.
8.
Addendum:
tooth.
9.
Dedendum circle (or Root circle): It is the circle drawn through the bottom of the circle.
It is the radial distance of a tooth from the pitch circle to the top of the
10. Dedendum : It is the radial distance of a tooth from the pitch circle to the bottom of the
tooth.
11. Clearance: It is the radial distance from the top of the tooth to the bottom of the tooth,
in a meshing gear. A circle passing through the top of the meshing gear is known as
clearance circle:
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Spur Gears
5.5
12. Total depth: It is the radial distance between th
dd d
e a en urn and the dedendum of a gear.
Total depth = Addendum + Dedendum '
13. Working depth: It is the radial distance from th
dd d
.
en urn circle to the clearance
hi
e 0 mes 109 gears.
14. Tooth thickness: It is the width of the tooth measured al ong th e pitc
. h eire
. Ie.
. I I'
I
eire e. tIS equa to the sum of the addendum ofth
e a
tw
15. Tooth
space: It is the width of space between the two adiacenr t th
d I
. h . I
~
ee measure a ong the
pitc eire e.
16. Backlash: It is the difference between the tooth space and the tooth thickness along the
pitch circle.
Backlash = Tooth space - Tooth thickness
17. Face width: It is the width of the gear tooth measured parallel to its axis.
18. Top land: It is the surface of the top of the tooth.
19. Bottom land: The surface of the bottom of the tooth between the adjacent fillets.
20. Face: Tooth surface between the pitch circle and the top land.
21. Flank: Tooth surface between the pitch circle and the bottom land including fillet.
22. Fillet: It is the curved portion of the tooth flank at the root circle.
23. Pressure angle (or Angle of obliquity) (;) : It is the angle between the common normal
to two gear teeth at the point of contact and the common tangent at the pitch point. The
standard pressure angles are 14 Y:z0 and 200•
24. Path of contact: It is the path traced by the point of contact of two teeth from the
beginning to the end of engagement.
25. Length of path of contact (or Contact length) : It is the length of the common normal
cutoff by the addendum circles of the wheel and pinion.
26. Arc of contact: It is the path traced by a point on the pitch circle from the beginning to
the end of engagement of a given pair of teeth. The arc of contact consists of two parts.
Theyare:
(a) Arc of approach: It is the portion of the path of contact from the beginning of the
engagement to the pitch point.
(b) Arc of recess: It is the portion of the path of contact from the pitch point to the end
of the engagement of a pair of teeth.
27. Velocity ratio: It is the ratio of speed of driving gear to the speed of the driven gear.
I
where
i
= ~
= ~
1
.
". (5.4)
NA and NB = Speeds of driver and driven respectively, and,
= Number of teeth on driver and driven respectively,
ZA and ZB
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Design of Transmission Systel1lJ
5.6
28. Contact ratio: The ratio of the length of arc of contact to the circular pitch is Im.own as
contact ratio. The value gives the number of pairs of teeth in contact.
5.4. LAW OF GEARING (OR CONDITIONS OF CORRECT GEARING)
The law of gearing states that for obtaining a constant velocity ratio, at any instant Of
teeth the common normal at each point of contact should always pass through a pitch point
(fixed point), situated on the line joining the centres of rotation of the pair of mating gears.
The law of gearing states the condition which must be fulfilled by the gear tooth profiles
to maintain a constant angular velocity ratio between two gears. This is the fundamental
condition which must be satisfied while designing the profiles of the teeth of the gear wheels.
5.5. FORMS OF GEAR TOOTH PROFILE
Two curves of any shape that fulfill the law of gearing can be used as the profiles of teeth.
If profile of the teeth of one of the mating gears is arbitrarily chosen and the profile of teeth
of the other gear is determined so as to satisfy the law of gearing, such teeth are known as
conjugate teeth. Gears having conjugate teeth can be successfully used for transmitting
motion but they are difficult to manufacture as special devices are used for this purpose
which are costly. So conjugate teeth are not much common in use.
Therefore the common forms of teeth profiles used in actual practice are:
(i) Involute tooth profile,
and
(ii) Cycloidal tooth profile.
Table 5.1 shows the comparison between the involute and cycloidal tooth profiles.
Table 5.1. Involute Vs Cycloldat toot" profile
Involute Tooth Profile
S.No.
Cycloidal Tooth Profile
1.
Pressure angle remains constant throughout
operation.
the
2.
Variation in centre distance does not affect the
velocity ratio.
Centre distance should not vary.
3.
Interference occurs.
No interference occurs.
4.
Easier to manufacture.
It is difficult to manufacture
and epicycloid are required.
5.
Weaker teeth
Strong teeth and smooth operation
6.
More wear and tear as contact takes place between
convex surfaces.
Less wear and tear as concave flank makes contact
with convex flank.
Pressure angle varies. It is zero at the pitch point
and maximum at the start and end of engagement.
as two ("Irves hypo
5.6. STANDARD SYSTEMS OF GEAR TOOTH
The American Gear Manufacturers Association (AGMA) and the American National
Standards Institute (ANSI) standardised the following four forms of gear teeth depending
upon the pressure angle.
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Spur
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Gears
5.7
I. 14Y2° composite system,
2.
3. 20 full depth involute system, and
.
In actual practice, the 200 involute system'
14~o full depth involute system;
0
5.6.1. Advantages of 14 Va
0
200 stub involute system
id I
.
IS WI e y used.
4.
Involute System
./'
It provides smooth and noiseless operation .
./'
It has stronger tooth.
5.6.2. Advantages of 200 Involute System
./'
It reduces the risk of undercutting .
./'
It has stronger tooth with a higher load carrying capacity .
./'
It has greater length of contact.
5.7. STANDARD PROPORTIONS OF GEAR SYSTEMS
The Table 5.2 shows the standard proportions in module (rn) for the four gear systems.
Table 5.2. Standard proportions of gear systems
Particulars
S.No.
14 YJ ° composite or full
20° full depth
20° stub involute
depth involute system
involute system
system
I.
Addendum
1m
1m
0.8 m
2.
Dedendum
1.25 m
1.25 m
1m
3.
Working depth
2m
2m
1.6m
4.
Minimum total depth
2.25 m
2.25 m
1.8 m
s.
Tooth thickness
1.5708 m
1.5708 m
6.
Minimum bottom clearance
0.25
0.25 m
0.2m
7.
Fillet radius at root
0.4 m
0.4 m
0.4 m
I Examp/~
tin
5.1 lIn
1.5708
Q
III
III
pair of spur gears, the number of teeth on the pinion and the gear
20 and 100 nsp~cliv~/y.
0
The module is 6 mm. The pressure angle is 20 full depth.
Ctdclllale:
the cemr« distance,
(ii) the pilch circle diameters of the pinion and the gear,
(i)
(iii) adtkndulfl
and dedendum,
(iv) tooth thickness and bottom clearance, and
tlte gear ralio.
Given Data: z, = 20; z2 = 100; m = 6 rnrn ; ~
iv)
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= 20° full depth.
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Design o/Transmission
5.8
Systelns
@Solution :
(i) Centre distance (a): The centre to centre distance between two gears is given by
= (01 +02)
C
mZI +mz2
2
=
2
=
m(zi +z2)
2
01 and 02 = Pitch circle diameters of pinion and gear respectively.
where
=
C
6(20+100)
2
-360
-
A
mm
ns. ~
(ii) Pitch circle diameters of the pinion and the gear:
=
02 =
=
m z2 =
01
and
m
(iii) Addendum and dedendum:
and
=
=
6 (20)
zi
6 (100)
120 mm
600 mm
Ans. ~
From Table 5.2, for 20° full depth involute system,
Addendum
=
1m
=
1x 6
Dedendum
=
1.25 m
=
1.25 x 6 = 7.5 mm
= 6mm
Ans. ~
Ans. ~
{iv) Tooth thickness and bottom clearance:
From Table 5.2, for 20° full depth involute system,
Tooth thickness
and
bottom clearance
=
1.5708 m
= 0.25 m
=
1.5708 x 6
= 0.25
x6
=
=
9.4248 mm
1.5 mm
Ans. ~
Ans."
(v) The gear ratio:
100
= 5 Ans."
20
Z2
Gear ratio = -
zi
I Example
5.2
I A pinion
= -
with 25 teeth and rotating at 1200 r.p.m: drives a gear which
rotates at 200 r.p.m: The module is 4 mm. Calculate the centre distance between the gears.
Given Data:
Tofind:
z)
= 25; N) =
1200 r.p.m.;
N2
= 200 r.p.m.;
m
= 4 mm.
Centre distance between the gears (C).
e Solution:
Gear ratio, i
Z2
__ z2
z)
=
_
-
6xz)
N) _ 1200
N2
200
=
=
150
=
6x25
6
Then, centre distance between the gears is given by
C = m (z) + z2) _ 4 (25 + 150)
2
2
= 350 mm
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Ans. ~
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Spllr Gears
5.9
I Example
5.3 ) A pair of Spur
.
gears Wllh a centre dl
speed reduction of 4.5 : 1. The m d. I .
stance of 495 mill is
o u e IS 6 mm. C I I
a cu ate tile "umber (if teeth on the
pinion and the gear.
Given Data:
Tofind:
and
C == 495 rnm : i == 45 .
,
_
.,
m - 6 mm,
Number of teeth on the pin'
d
Ion an the gear (i.e .•
@ Solution:
Gear ratio, i ==
centre distance,
ILfIU/ /111 "
z2
zi
== 4 5
.
or z2 == 4.5
C
zi
and z2)'
zi
... (i)
or 495 = _6_{Z_;_I_+_Z:;:..2)
2
or
... (ii)
Solving equations (i) and (ii), we get
zi
. I Example
== 30 and
z2
== 135
Ans."
I
~.4 A ~oothed wheel lias 112 teeth. Its module is 1.75 mm: Find pitch
diameter, the circular pitch and the diameteral pitch.
~.JvtData:
,,!.f:'d:
z == 112;
m
= 1.75 mm.
(i) Pitch diameter (D), (ii) Circular pitch (Pc)' and (iii) Diameteral pitch (pd)'
©Solution:
(i) Pitch diameter (D) :
(ii) Circular pitch (pc) :
D == m- Z == 1.75 x 112 == 196 mm Ans."
Pc
=
nD
Z
z
==
7t
196
112
x
112
(iii) Diameteral pitch (p tJ: P d = D = 196
=
== 5.497 mm
0.57 tooth/mm
Ans.~
Ans."
5.S. GEAR MATERIALS
In modern industries, a wide variety of gear materials are used. The gear materials are
broadly grouped into two groups viz., metallic and non-metallic materials.
1. Metallic materials:
(a) Steel: ../ So far, the most widely used material in gear manufacture is steel. Almost
all types of steels have been used for this purpose .
../' To combine the property of toughness and tooth hardness, steel gears are heat treated .
../' Steels with BHN < 350 are used in light and medium duty drives. But steels with
BHN > 350 are used in heavy duty drives and also where compactness is required.
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Design of Transmission Systems
5.10
../ The plain carbon steels used for medium duty applications are 50 C 8, 45 C 8, 50 C 4
and 55 C 8. For heavy duty applications, alloy steels 40 Cr I, 30 Ni 4 Cr I and 40 Ni 3 Cr 65
Mo 55 are used. For planetary gear trains, alloy steel 35 Ni I Cr 60 is recommended.
(b) Cast iron: ../ It is used extensively as a gear material because of its low cost, good
machinability, and moderate mechanical properties .
../ Generally, large size gears are made of grey cast iron of Grades FG 200, FG 260
FG 350 .
Or
../ Disadvantage: It has low tensile strength.
(c) Bronze:
../ It is mainly used in worm gear drives because of their ability to
withstand heavy sliding loads .
../ Bronze gears are also used where corrosion and wear are a problem .
../ Disadvantage:
They are costlier .
../ The bronze alloys are either aluminium bronze, manganese bronze, silicon bronze, or
phosphorus bronze.
2.
Non-metallic
materials:
../
The non-metallic materials
compressed paper and synthetic resins like nylon are used for gears .
../ Advantages:
(i) Noiseless operation;
like wood,
(ii) Cheaper in cost;
and
rawhide,
(iii) Damping of
shock and vibration .
../ Disadvantages:
INote'
(i) Low load carrying capacity; and (ii) Low heat conductivity.
The material for pinion should always be better than the material of the mating gear.
Because the teeth of pinion undergo more number of cycles than those of gear and hence quicker wear.
The properties of the various materials used for the gears are given in Table 5.3.
Table 5.3. Gear materials and their properties (from data book, page no. 1.40)
Tensile
Material
Condition
strength (au)'
Yield point
stress (0,),
N/mm2
N/mm2
BUN
I. Grey cast iron
Grade 20
As cast
200
-
179 min
(ii) Grade 25
As cast
250
-
197 min
(iii) Grade 35
As cast
350
-
207 min
(iv) Grade 35
Heat treated
350
300 min
2. Phosphor bronz.e
Sand cast
160
-
Chill cast
240
-
70 min
Centrifugal cast
260
-
90 min
(i)
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5.11
r-Tensile
"bferia'
Condition
strength (au),
Yield point
stress (0,),
N/mm2
N/mm2
BUN
3- CdSlnls
(i) Grade 1 (i.e., CS 65)
-
650
400
(ii) Grade 2 li·e .. CS 71)
-
190 min
710
570
207 min
(iii) Grade 3 (i.e., CS 85)
-
850
710
248 min
(iv) Grade 4 (i.e., CS 105)
-
1050
870
31 i min
(,.) Grade 5 (i_e., CS 125)
-
1250
1020
363 min
Normalised
490
240
137
(ii) C 30
Hardened and tempered
600
300
179
(iii) C 40
"
630
330
217
[iv) C 45
"
700
360
229
750
420
255
4. Fo,,~ SIMs
(.) CtUbtHI stub
(i) C 14, C 15
(v) C 60
Normalised
-
(b) Alloy st~~1s
(i) 40 Cr I
Hardened and tempered
800
540
229 min
(ii) 40 Ni 3
"
900
600
229 min
(iii)40 Ni 2 Cr I Mo 28
"
1175
880
255 min
(iv)40 Cr 2 AI I Mo 18
"
975
700
255 min
5.8.1. Selection of Gear Material
The
../
../
../
../
selection of the gear material depends upon :
Type of service
,/ Peripheral speed
Method of manufacture
,/ Degree of accuracy required
Wear and shock resistance
,/ Cost of the material
Space and weight limitations
../ High loads, impact loads, and
longer life requirements .
../
Safety and other considerations
5.9. GEAR MANUFACTURING
Gears can be manufactured by various processes that can be classified under the following
three topics.
1. Gear milling:'/
Almost any tooth can be milled. However only spur, helical and
straight-bevel gears are usually milled .
../ Surface finish can be held to about 3.2 urn.
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Design of Transmission Syste~
5.12
2. Gear generating:
,( In the generating process, teeth are formed in a series of passes
by a generating tool shaped somewhat like a mating gear. Either hobs or shapers can be used.
Surface finishes as fine as about 1.6 urn can be obtained.
v" Robbing:
The hob is simply a cutting tool which is shaped like a worm. Hobbing can
produce almost any external tooth form except bevel gear teeth. Hobbing closely controls
tooth spacing, lead, and profile.
~ Shaping: Using a shaper, teeth may be generated with either a pinion cutter or a rack
cutter. They can produce external and internal spur, helical, herringbone, and face gears.
3. Gear molding:
,( Mass production of gears can be achieved by molding.
v" Injection molding produces light weight gears of thermoplastic
material.
v" Die casting is a similar process using molten metal. Zinc, brass, aluminium,
magnesium gears are made by this process.
v" Sinter;ng is used in small, heavy-duty
are mostly used for this process.
gears for instruments
v" Investment casting and shell molding produce medium-duty
rough applications.
and
and pumps. Iron and brass
iron and steel gears for
5.9.1. Gear Finishing
In order to improve accuracy and surface smoothnes . the gears produced must undergo
gear finishing operations. The generally used gear finishing perati n are having, grinding,
lapping and honing.
5.10. GEAR TOOTH FAILURE
The two modes of gear tooth failure are:
I. Tooth breakage (due to static and dynamic loads), and
2. Tooth wear (or surface deterioration)
(a) Abrasion
(b) Pitting, and
(c) Scoring or seizure
1. Toot" bre,dage: The load on any gear tooth is cyclic and therefore fatigue fracture of
tooth may occur at the root. Tooth breakage may also be cau ed by an unexpected heavy load
imposed on the teeth.
1. Tooth wear (or surface deterioration} :
(II) AbrlUion : When some foreign materials such as dirt, rust or metal particles
in between the mating teeth, there will be wear of tooth surfaces.
abrasion wear.
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Thi
deposited
wear is known as
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§J!!!!_Gears
5. I 3
(6) PiJtillg and spalling : Pi~ing is the process during which small pits are formed on the
active surfaces of gear tooth. It IS a surface fatigue failure which occurs when the load on the
gear tooth exceeds the surface endurance strength of the material.
(c) Scorin.g.or seizure: .Sc~ring ~an OCcur under heavy loads and inadequate lubrication.
At this condition, the lubrication 011 film breaks down and metal-to-metal contact occurs.
Hence high temperatures result and the mating spots of the two surfaces weld together. This
phenomenon is known as scoring or seizure.
5.11. FORCE ANALYSIS OF SPUR GEARS
The gears in contact are used to transmit power. It is understood that a tooth of driving
gear pushes the tooth of the driven gear in contact. According to the law of gearing, the'
resultant force F should always act along the pressure line, as shown in Fig.S.3(a). We know
that the pressure line is the tangent line which is tangent to base circles of both pinion and
gear.
.
.'
The resultant force F between mating teeth can be resolved at the pitch point into two
components, as shown in Fig.5.3(b).
(Driving pinion rotates clockwise)
,--="
.~IllP
F
•
(.
F,
F
d,
\._
P
Pitch circle
) (pinion)
•
lZfFr
F
Fr~
~Ft
1>('
/' P
d2)
(
F
\\
Wheel (
\
.
Illg
~;:~rde
•
.~
,
-_./
(b)
(a)
Fig. 5.3..
t F is a useful component.
The tangential componen
t
. ed
1. Tangential component (F,) :
f F the magnitudes of torque and transnutt
ing the value 0 t'
Because it transmits power. U s
power can be detennined.
= F
. .
.
Transmitted load, W t
t
F'
eparating force which IS always
..
d' I mponent
IS a s
ful
.
nt (F ) : The ra ia co
'k
S it is not really a use
2. RadiIIJ com pone
f.
F does no wor.
0
erse
directed towards the centre· of the ~ear'ft~e shaft. The force F, is also called as transv
es bend 109 0
component. This force F , caus
force or bending force.
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Design o/Transmission
Systeltls
5.14
Let
P
M,
=
Power transmitted by gears in watts,
= Torque transmitted by gears in N-m,
Nt and N2 = Speeds of pinion and gear respectively in r.p.m.,
d, and d2 = Pitch circle diameters of pinion and wheel respectively in m, and
~ =
Pressure angle.
The torque transmitted by the gears is given by
M,
=
60 x P
27tN
The tangential component F, acts at the pitch circle radius.
M,
=
F, =
or
From Fig.5.3(b),
d
F, x 2
2· M,
d
Radial component, , Fr
=
F,' tan ~
I
... (5.5)
... (5.6)
Pitch line velocity (v) is given by
v
=
7tdN
60
mls
Then the transmitted power is calculated as
" -P-=-F-,-x-v----,
INote I The tangential
.. , (5.7)
force on pinion and wheel will be same.
5.11.1. Assumptions Made in Force Analysis
The assumptions made for the above force analysis are :
./
Friction losses in the bearings and gears are negligible .
./
The gears mesh at the pitch circles .
./
The gear teeth have standard involute tooth profiles .
./
The shafts for pinion and gear are parallel.
./
The effect of the dynamic forces is neglected .
./
As the point of contact moves, the magnitude of resultant force F changes. This
effect is neglected.
.
I
I
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Spur Gears
5.15
I
[§.rame/~ 5.5 A draft horse walks at a steady pace of 5 miles per hour (lit 2.23 m/s).
What st~ady force must it exert if the power output is exactly 1 horse power (lit 745 W).
Giv~n Data:
Tofmd:
=
v
2.23 m/s ; P = 745 W.
Steady force exerted (F I)'
o Solution:
We know that the transmitted power,
P = F, x v
or
745 = F, x 2.23
Steady force exerted, F,
[§xamele
5.6
I A pinion
= 333.37 N Ans. ~
of 120 mm pitch diameter, running at 900 r.p.m: transmits
5 kW of power to gear whose pitch diameter is 360 mm. For straight tooth, the angle of
pressure is 20 ~ Determine the tangential force, the transverse or bending force on shafts
and torques on driving and driven shafts.
Given Duta:
d1
=
120 mm
=
0.12 m ; N,
=
900 r.p.rn.;
P = 5 kW
=
5 x 103 W;
d2 = 360 mm = 0.36 m ; ~ = 20°.
Tofind:
(i) Tangential and transverse forces (i.e., F, and F r)' and
(ii) Torques on driving and driven shafts (i.e., T, and T2)·
@ Solution:
Gear ratio, i
N2
=
N, _ d2 _ 360
N2 - d , - 120
N,
V
3
900
= 3 = 3 = 300 r.p.m.
1td,N,
The pheripheral speed,
=
=
=
60
1t
=
1td2N2
60
(0.12) 900 = 5 655 m/s
60
.
/:. F. and F) .
(i) Tangential and transverse/Drees u.e., t
f. •
. d p-Fxv
We know that the power transmltte,
- ,
3
_ ~ = 5 x 10 = 884.19 N Ans.~
Tangential force, F, - V
5.655
or
.
onent) force is given by
The transverse (i.e., radial comp
_
19 tan 200 = 321.82 N Ans.~
F = F/ x tan ~ - 884. x
r
'u'r}
{I
'I
Torques on
d . 'ng and driven shafts (i.e., T} and T} :
"VI
..
. '
h ft = Torque transmitted by plDlon
Torque on dnvtng s a
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/
5.16
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Design of Transmission Systems
We know that the torque transmitted by pinion,
60 x P
T) = 27tN)
..
= 53.05 N-m
Torque on driving shaft, T)
=
53.05 N-m
Similarly torque on driven shaft
=
Torque transmitted by gear
Torque transmitted by gear, T2
.
=
60 x 5 x 103
27t X 900
Torque on driven shaft, T2
60 x P
= 2 7t N2
=
Ans. ~
60 x 5 x 103
27t x 300
159.15 N-m
=
=
159.15 N-m
Ans. ~
I Example 5.71 A
train of spur gears is shown in Fig.5A(a). Pinion 1 runs at 1750
r.p.in. hnd transmits 2.5 kW power to idle gear 2. The teeth are cut on the 200 full-depth
system and have a module of 2.5 mm: The number of teeth on gears 1, 2 and 3 are 20, 50
and 30 respectively. Calculate:
(i) the torque on each gear shaft, and
(ii) the components of gear tooth forces.
Also draw a free body diagram of gear 2 showing all the forces which act upon it and
determine the reaction on the idler gear shaft.
y
,
I,
I
-~vt,2 '
b -----
X
Y
I
Fb2 -----,
Fb2
I
I
(a)
(b)
Fig. 5.4.
Given Data: N) = 1750 r.p.m. ; P
z) = 20; z2 = 50; z3 = 30.
@Solution:
= 2.5 kW = 2.5 x 103 W ; ~
=
200;
m
= 2.5 mm ;
The pitch circle diameters of gears I, 2 and 3 are given by
d) = m z) = 2.5 x 20 = 50 mm ;
d2
d3
= .~ z2 = 2.5 x SO
= m z3 = 2.5 x 30
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=
125 mrn ; and
=
75 mm.
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§pur Gears
5.17
All three gears have the same pitch I'
. . .
velocity IS given as
.
me velocity (v) Th
. erefore for pinion 1, th e pitc
. h I'me
_1tdINI
60
VI
_
= 1t(50 x
10-3) 1750
60
=
4.581 m/s
Considering pinion 1 and gear 2 :
Let the tangential force of pinion 1
P = F
on gear 2
F' 12 an d iIS given
.
by,
,x v
= f = 2.5 x
or
.
IS
103
4.581
V
545.67 N
=
ADs."
:. Radial force of pinion 1 on gear 2 is given by
Fr12 = F ,12
x
tan ~
= 545.67 x tan 20° = 198.61 N ADs."
Then the resultant force of pinion 1 on gear 2 is given by
F'
= _!L
F
_ 545.67
cos ~ _ cos 200
12
=
580.69 N
ADs. ~
The torque transmitted by the pinion 1 is given by
TI =
=
60 x P
21t NI
[ .: P = 2 1t6~ T ]
60 x 2.5 x 103
21t x 1750
= 13.64 N-m
ADS. ~
Considering gears 2 and 3: Since gear 2 is an idler, it transmits no torque (power) to its
shaft.
Torque transmitted by the gear 2, T 2
= 0
ADS. ~
Since gear 2 is an idler, whatever torque it receives from pinion 1 is transmitted to gear 3.
Therefore, the tangential component between gears 2 and 3 must be equal to the tangential
component between gears 1 and 2, as shown in Fig.5.4(b). Therefore
and
= 545.67 N
F~2
= F~2
F;2
= F~2 = 198.61 N
F32
=
= 580.69 N
FI2
ADs."
Since the same power is transmitted from pinion 1 to gear 3,
2 1tN I TI
60
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=
2 1tN3 T3
60
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Design of Transmission Systems
5.18
NIT, = N)T)
or
or
by
Torque transmitted
gear 3, T3 = TI (~:)
[ .: i = ~: = ~ ]
= TI (~~)
T3 = 13.64 (~~)
=
20.46 N-m
Ans. ~
Reaction on the idler gear shaft: The shaft reactions in the x and y directions are
R~2 = -(F~2+F;2)and
R~2 = - (F~2 + F;2 ).
Here (-)ve sign is for reaction. Because reaction is always opposite in direction to the
force applied.
R~2
=
-(F'12
+F;2)
=
-(-545.67+
198.61) = 347.06N
... [using sign convention (+) -+ ~(-) t-]
and
R~2 = - (F~2 + F;2 ) = - (198.61 - 545.67) = 347.06 N
... [using sign convention (+)
i ;(-)~]
Therefore the resultant shaft reaction is given by
Rb2
I Example
= ~
(R~2 )2 + (R~2 )2
= ...J
(347.06)2 + (347.06)2
= 490.82 N
An5."
I
5.B A train of spur
gears is shown in Fig.5. 5. Gear J is
the driving gear and transmits 5 k W
power at 720 r.p.m: The number of
leeth on gears J, 2, 3 and 4 are 20,
.~O, 30 and 60 respectively. The
modllie for all gears is <I mm. The
gears Itave a 20° full deptlt invohu«
profile. Calculate lite langOltiaJ and
radial components of the tooth force
between:
(i) Gears I and 2,
and
(ii) Gears 3 and 4.
Given Data : P = 5 kW = 5 x 1()3
Z4 ::
60; ." = 4 rnm ;
Fig. 5.5.
w ~ N.
=
720 r.p.m.;
Zt =
20;
z2 =
50; z).== 30;
q, = 20°.
@ So 'ulion: The pitch circle diameters of the gears 1, 2, 3 and 4 are given by
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SpurGears
5.19
=
mZI
=
4 x 20
=
80mm;
d2 =
mZ2
=
4 x 50
200 mm ;
d3 =
mZ3
=
4 x 30
=
=
d4 =
mZ4
=
4 x 60
=
240 mm,
dl
120 mrn ; and
The pitch line velocity for gear I is given as
1t
60
=
vI
~ x 80 x 10-3 x 720
dIN I
60
=
3.016 mls
=
(i) Considering gears 1 and 2: The tangential force of gear 1 on gear 2 is given by
I
P
=
F 12
=
vI
5xl03
3.016
1657.86 N Ans. ~
=
The radial force of gear 1 on gear 2 is given by
= F;2
F;2
x tan
q, =
1657.86 x tan 200
=
603.41 N Ans. ~
Then the resultant force of gear 1 on gear 2 is given by
F
F;2
= --
cos
12
1657.86
cos 200
q, =
=
1764.26 N Ans. ~
(ii) Considering gears 3 and 4 :
i
=
Speed of gear 2, N2
=
Gear ratio,
We know that,
N,
N2
NI
2.5
-
z2
z,
50
20
720
-2.5
=
=
2.5
288 r.p.m.
Since gears 2 and 3 are coaxial, the speeds are same i.e., N2 = N3·
The pitch line velocity of gear 3 is given by
d3 N3 =
1t
7t
60
v3 =
x 0.12 x 288
60
=
1.809 m/s ... [.: N3
= N2 = 288 r.p.m.]
The tangential force of gear 3 on gear 4 is given by
= f..
F'
34
v3
=
3
5 x 10
1.809
= 2763.10 N Ans. ~
The radial force of gear 3 on gear 4 is given by
F'
34
=
F'
34
x tan
q,
= 2763.1 x tan 200
=
1005.69 N Ans.
l'
Then the resultant force of gear 3 on gear 4 is
F
34 -
F;4
= 2763.1
cos cj>
cos 200
=
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2940.43 N Ans.1?
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Design
of Transmission
5.20
I Example
5.9
I A train
Syste
~
of spur gears with 200 full depth involute teeth Is showII ill
Fig.5.6. Gear 1is the driving gear and transmits 50 kW power at 300 r.p.m: to the gear
train. The number of teeth on gears 1, 2,3 and., are 30, 60, 25 and 50 respectively, White
the module for all gears is 8 mm. Gears 2 and 3 are mounted on the same shaft. Gear 1 ;,
rotating in a clockwise direction when seen from the left side. Calculate: (i) the tangelltiat
. and radial components of tooth forces between gears 1 and 2, and gear 3 and 4; alld
(ii) draw afree body diagram of forces acting on each gear.
I
I
I
I
-----r--I
I
Fig. 5.6.
Given Data: ~ = 20°; P = 50 kW
z3 = 25 ; z4 = 50; m = 8 mm.
Tofind : (i)
= 50 x 103 W' 'I N = 300 •.p.m. ,
r
•
F, and F, of tooth forces between gears 1 and 2, and gears 3 and 4; and
(ii) Free body diagram of forces acting on each gear.
@Solution : The pitch circle diameters of gears 1 2 3 and 4
.
by
_
' ,
are given
d , - m zi = 8 x 30 = 240 mrn .,
=
d3 =
d4 =
d2
m z2
m Z3
m z4
= 8 x 60 = 480 mrn ,.
= 8 x 25 = 200 mm ; and
= 8 x 50 = 400 mm.
(i) Tangential and radial components of tooth forces:
(a) ConSidering gears 1 and 2· We kn
th h .
.
.
ow at t e pitch line velocity of gear 1
VI = 1tdiNI
= 1txO.240x300
'
60
60
= 3.77 mls
We know that th ta
.
e ngential component of tooth force betw
een
gears 1 and 2
F;2 = -P = 50 x 103 _
'
VI
3.77
- 13262.9 N Ans.""tJ
and the radial component of tooth force betw
een gears 1 and 2 ,
L
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Spur Gears
5.21
I
,.
,
,
- F'
-
Then the resultant force,
F
12
12
x
tan ....
't'
= F;2
cos 't'....·
13262.9 x tan 20 = 4827.3 N Ans. ~
=
0
= 13262.9
=
.cos 200
14114.0SN
F34
Fig. 5.7. Free body diagrams
(b) Considering gears 3 and 4 :
We know that,
NI
gear ratio, i
Speed of gear 2, N2
=
N2
NI
300
= 2 = T = 150 r.p.m.
Since gears 2 and 3 are coaxial, therefore speeds are same. i.e., N2
= N3.
The pitch line velocity of gear 3 is given by
1t
v3
=
d3 N3
60
=
1t
x 0.2~g x 150
=
1.571 mls
[.,' N3 = N2
= 150 r.p.m.]
We know that the tangential component of tooth force between gears 3 and 4,
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,
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Design of Transmission Systems
5.22
p
50 x 103
1.571
=
=
=
31830.99 N Ans."
and the radial component of tooth force between gears 3 and 4,
F;4
F34
Then the resultant force,
=
F;4 x tan <p
=
=
11585.53 N
ADS. ~
=
__lL
31830.99 x tan 200
FI
=
cos <p
31830.99
cos 200
=
33873.83 N
(ii) Free body diagrams: The free body diagram of forces acting on gears and the shaft
Be is shown in Fig.5.7.
I Example
5.10
I The layout of a two-
stage gear box is shown in Fig.5.B. The
number of teeth on the gears are as
follows: ZI = 20; Z2 = 100; Zj = 25 ;
z" = 150.
Pinion 1 rotates at 1100 r.p.m. in all
anticlockwise direction when observed
from the left side and transmits 12 k W
power to the gear train. The pressure
angle is 20 ~ Calculate:
(i) torques acting on shafts A, B
and C;
(ii) tangential
and
radial
of tooth forces
components
between gearj' 1 and 2, and
gears 3 and 4,' and
200
100
Fig. 5.8.
(iii) resultant reactions at bearings E and F.
Also draw afree body diagram of the gear tooth forces.
Given Data:
z,
= 20;
z2 = 100; 13 = 25 ;
P = 12 k W
@ Solution:
=
12 x 103 W ; <p
= 150'
Z4
= 20
NI
::
1100 r.p.m. ;
0
.
The module interms of centre distance (a) is given by
m =
2a
(zl+z2)
=
2
(20
ISO
100
=
2.5
['.: a = 150 mm from Fig.5.8]
Pitch circle diamet.ers of gears I, 2, 3 and 4 are gi en by
d , ::
mZI
=
2.5x20
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=
SOmm'
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Spur Gears
5.23
d2 =
mZ2
= 2.5 x 100
= 250 mm ;
=
mZ3
= 2.5 x 25
= 62.5 rnm ; and
d4 =
mZ4
= 2.5 x 150
= 375 mm.
d3
(i) Torquesacting on shafts A, Band C:
Torques acting on shafts A, Band
respectively.
C are equal to torques acti~g on gears I, 2 and 4
Therefore torque acting on shaft A is given by
60 x p
27t NI
-
We know that,
Speed of gear 2, N2 =
Torque on shaft B, (M')B
=
= 104.17 N-m Ans.-c»
100
-- -20 -- 5
Z2
zi
1100
5 = -5- = 220 r.p.m.
NI
60 x P
27t x N2
60 x 12 x J03
=
= 520.87 N-m Ans.-c»
2n x 220
150
Z4
- z -- -25 = 6
COlISidering gears 3 and 4 :
and
60 x 12 x 103
2n x 1100
=
speed on gear 4, N4
220
NJ
= 6 = 6 = 36.66 r.p.m.
[.: N
Torque on shaft C, (M,)
2n
=
= N2 = 220 r.p.m.]
N4
3125.22 N-m
Ans.-c»
(ii) TtulgentiaJ and radial components of tooth forces:
COlUilkr;ng gears J and 1:
Pitch line velocity of gear I,
"I
_ n diN
60
n x 50 x 100J x 1100 = 2.88 m/s
60
=
I
Tangential component of tooth force of gear I on gear 2 is given by
and radial component,
F~2
= ~
F~2
= F~2 x
==
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=
!()3
X
12
2S
tan.
=
=
4166.% N
ADS."C>
4 J 66.96 x tan 20°
1516.65 N ADS. ~
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5.24
Design of Transmission Sys1e1l1s
Then the resultant force of gear 1 on gear 2,
FJ2
=
v3
=
F~2
cos ~
=
4166.96
cos 200
=
4434.38 N ADS. '"
Considering gears 3 and 4 :
Pitch line velocity of gear 3,
=
1t
x 62.5 x 10-3 x 220
60
= 0.72 m/s
Tangential force, F;4
Radial force, F;4
and
Resultant force, F34
=
P
=
v3
12xl03
0.72
= F;4 x tan ~
= 16667.86 N
=
ADS.-tJ
16667.86 x tan 200
=
6066.6 N ADS."
=
F;4
cos ~
16667.86
= cos 20
0
=
17737.56 N ADS. ~
(iii) Resultant reactions at bearings E and F :
The free body diagram of forces acting on gears and the shaft EF is shown in Fig.5.9.
,
F'2
r
lCS$---------
F~
F'2
F34
I •
F12
,
-
Shaft EF
Shaft EF
Fig. 5.9. Free body diagrams
•
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5.25
The forces acting on the shaft EF are shown in Fig.5.1 O.
Fig.5.10.
Tofind reactions (RF)v and (R Fiv : Consider vertical forces.
Sum of upward forces
or
=
Sum of downward forces
(RE)v + (RF)v
=
F~2 + F;4
(RE)v + (RF)"
=
1516.65 + 6066.6
=
7583.25 N
... (i)
Taking moments about bearing E, we get
or
(RF)v x 0.350 - F;4 x 0.250 - F~2 x 0.1
=
0
(RF)v x 0.350 - 6066.6 x 0.250 - 1516.65 x 0.1
=
0
or
(RF)v
=
4766.6 N
Then from equation (i),
(RE)v
=
7583.25 - 4766.6
... (ii)
=
2816.63 N
... (iii)
Tofind (RF)H and (RF)H : Consider horizontaJ forces.
Sum of right hand side forces = Sum of Jeft hand side forces
or
(RE)H + (RF)H
=
F~2 + F~4
(RE)H + (RF)H
=
F~2 + F~4 = 4166.96 + 16667.86
=
20834.82 N
... (iv)
Taking moments about bearing E, we get
(RF)H x 0.350 - F~4 x 0.250 - F~2 x 0.100 = 0
or
(RF)H x 0.350 - 16667.86 x 0.250 - 4166.96 x 0.100
(RF)H
Or
=
=
0
13096.2 N
.,. (v)
Substituting equation (v) in (iv), we get
(RE)H
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=
20834.82 - 13096.2 = 7738.64 N
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Design of Transmission Syste1trs
5.26
Resultant reaction at bearing E :
RE
= ~ [ (RE)H F + [ (RE)y]2
= ~
8235.3 N Ans. ~
=
Resultant reaction at bearing F :
RF
(7738.64)1 + (2816.63)1
= ~ [ (RF)H ]2 + [ (RF)y]2
= ~ (13096.2)1+ (4766.6)1
= 13936.67N Ans. ~
'Example
5.11 , A planetary gear train
consists of three gears A, Band C as shown in
Fig.5.ll. The gear A has 72 internal teeth and
gears Band C have 20 and 32 external teeth
respectively. The gear B meshes with both A and
C and is carried on an arm EF. The sun gear A
rotates in a clockwise direction and transmits
7.5 kW power at 1400 r.p.m. to the gear train.
The module is 5 mm and the pressure angle is
20 ~ Calculate:
I
\
Fig. 5.11.
0) the tangential and radialforces acting on each gear, and
(ii) the torque that the arm EF can deliver to its output shaft.
Also draw afree body diagram offorces acting on each other.
Given Data:
zA
=
Zs
72;
=
Zc
20;
=
32;
P
=
7.5 kW
=
7.5
x
103 W;
NA = 1400 r.p.m.; m = 5 mm; cp = 20°.
.
© Solution:
t:
\
r
The pitch circle diameters of gears A, Band C are given by
dA
= m zA =
5 x 72
=
dB
= m zB =
5 x 20
= ) 00 mm;
de
= mzc
=
5 x 32
- 160 mm.
360 mm ;
and
The length of the arm EF is given by
\
LEF =
dA +dB
2
=
360 + 100
2
=
230 mm
(i) Tangential and radialforces acting on each gear:
Considering gears A and B :
Pitch line velocity of gear A,
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VA
=
=
1t
x
0.360 x 1400
60
=
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26.38 mls
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Spur Gears
5.27
Tangential force of gear A on gear B, F~
p
=
vA
and radial force of gear A on gear B, F~
7.5 x 103
= 284.2 N Ans. ~
26.38
=---
=
F~
=
103.44 N Ans."
x tan
ct>
=
284.2 x tan 200
Then resultant force of gear A on gear B,
i'
F AB
=
F~B
cos
ct>
=
284.2
cos 200 ::-::302.44 N Ans."
"I
j
Considering gears Band C: Since the gear C (also known as ring gear) is fixed, therefore
forces acting on gear B is same as that of gear A, as shown in Fig.5.12.
(;;) Torque thai the arm EF can deliver to its output shaft: The free body diagram of
forces acting on each gear and arm EF are shown in Fig.5.12.
Arm EF
,
,
"
Fee = FAa
Gear A
GearC
Fig. 5./1.
From Fig.5.12, the force acting on the ann EF,
F EF = 2· F AB . cos 20°
= 2 x 302.44 x cos 20°
=
568.4 N
Torque that the ann EF delivered, T = L£F x FEF
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=
0.230 x 568.4
=
130.73 N-m
ADS. ~
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Design of Transmissmn
SYS
.
~
5.28
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5.12. TOOTH STRESSES (BEAM STRENGTH OF GEAR TOOTH - LEWIS EQUATION
The frrs. anal sis of gear tooth stresses was done by Wilfred Lewis in J 892. The for
given by Lewis (also known as Lewis equation) still serves. as th~ basis for gearbending stress analysis. In the Lewis analysis, the gear tooth IS considered as a Canti
beam as shown in Fig.5. 13(a).
5.12.1. Assumptions
made
The Lewis equation is based on the following assumptions :
./'
The effect of radial component F,., which induces
negligible .
compressive
stresses
v: The tangential component F, is uniformly distributed across the full face Width.
./'
The tangential force F, is applied to the tip of a single tooth. In other Words .
, I
assumed that at any time only one pair of teeth is in, contact and takes the
load.
to
./'
Stress concentration in the tooth fillet is negligible .
./'
Forces which are due to tooth sliding friction are negligible.
5.12.2. LeWis Equation
. In Fig.5.13(b), the resultant force F is transferred.
.
to
Intersection of the line of action of resultant tooth I d F Pd°)flt 0 from tip, where 0 is th
/!...............
/.....
, .'
b."
:
.
an the centre of the tooth.
(F,/b)
f
,,
,
,:
.'
,:
•
oa
F
·· .:: .'..
·r----:"
.,., . ..'.: '
f
J
..
I
.'
.'
/(a)
The resultant force F .
JS
Th
Fig. 5.13.
(b)
resolved into tw
0 component
.
e tangential component f
s I. e., FI and F
break the tooth
0 tooth force FI induces a b
. ,..
0/.
eanng st ress
The radial
component F indUces
,.
a compressive stress.
0/
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W
h.ich tends to
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~~~~~------------------
~5~.2~9~
Since the direct compressive stress is very small as compared to the induced bending
therefore the effect of compressive stress on the tooth may be neglected Hence the
streSS'.
'.
.
,
be"d;"g stress IS used as the basISfor desIgn calculations.
From Fig.5.13(b), it is known that the section XX is the section of maximum stress or the
criticalsection.
Let
Mb = Maximum bending moment at the critical section XX
= FI x h,
Ft = Tangential force (or load) acting on the tooth,
h
0b
=
=
Length of the tooth,
Maximum bending stress at section XX,
I = Moment of inertia about the centre line of the tooth
b
=
Face width or width of gear face,
Thickness of the tooth at critical section XX,
1 =
y
Pc
= ~~ ,
= Maximum distance between centre line and the extreme fibre = t 12, and
Circular pitch.
=
At critical section XX,
Maximum bending moment, Mb = F t
X
... (i)
h
But we know that the flexural formula,
Mb
=
I
y
... (ii)
or Maximum bending stress,
where y
= 112;
I
= b13/12;
and Mb
=
F, x h.
Substituting the values of y, I and M, in equation (ii), we have
(F, x h) x (112)
(~~)
or
tangential load, F,
= b .
0b
=
6 Ft x h
... (iii)
btl
(;2h)
." (iv)
MultiplyiQgthe numerator and denominator on the right hand side by PC' we get
or
(6 ~lpJ
F,
=
F,
= Pc' b . 0b . Y
e; -b- "b
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... (5.8)
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Design of Transmission Systems
5.30
y
where
=
(2
6 h p;
=
constant,
known as Lewis/orm/aclor or tooth/orm/actor.
5.8 can be rewritten as
Equation
o
=
b
The beam strength
F,
.
Pc' b . y'
is the maximum
.
which is known as Lewis equation.
value of the tangential
force F, that the tooth can
transmit without failure.
I Beam
strength,
F(
=
7t.
m· b·
0b' ~
[.:
Pc
= 7t.
m]
... (5.9)
The values of the Lewis form factor (y) are given in Table 5.5.
INote I I.
In order to avoid the breakage of gear tooth due to bending, the beam strength should be
more than the effective force between the meshing teeth.
2. In the design of gears, it is required to decide the weaker between pinion and gear. In equation
5.9, the beam strength depends on the product (ob x y) since the m and b are same for pinion as well as
for gear.
(a)
When the same material is used for pinion and gear the pinion is always weaker than the gear.
(b)
When different materials are used, the product (ob
y) decide
the weaker between pinion
and gear. The Lewis form factor y is always less for pinion corn pared with gear.
5.13. GEAR BLANK DESIGN
There are two basic methods of gear design:
(i)
Gear design using Lewis and Buckingham equations (or gear design based on beam
strength), which is recommended
by AGMA (American
Gear Manufacturers
Association);
and
(ii)
Gear design using basic relations (or gear design based on gear life), which is the
conventional method.
The above two methods will be discussed
I.
in the following sections.
GEAR DESIGN USING LEWIS AND BUCKINGHAM
(DeSign of Spur Gear recommended by AGMA)
EQUA'rIONS
5.14. BEAM STRENGTH OF GEAR TOOTH OR LEWIS EQUATION
As discussed in Section 5.12.2, the Lewis equation is given by
F, = p . b . 0h . Y
. The beam .stren~th is the maximum value of the tangential force that the tooth can trl.lnsmil
WIthout bending failure. Replacing F by F o by [].
.
I
S'
b
cr b ' ' we get
,
.>I
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Spur Gears
5.31
=
F, =
Fs =
crb =
[ crb ] =
Fs
where
[ crb ]
::::::
=
Pc =
y* =
b
P»: b . [crb] . y
=
7t.
m· b- [crb]
.y
... [.: Pc
= 7t m]
Tangential load on gear tooth
Strength (or beam strength) of gear tooth,
Induced bending stress,
Allowable static stress, from Table 5.4, or
Ultimate stress (crut
)
3
Face width,
Circular pitch =
7t.
m, and
Lewis form factor based on Pc, from Table 5.5.
The value of Lewis form factor (y) in terms of number of teeth is expressed as follows
(fromdata book, page no.8.50) :
y = 0.124 - (0.684Iz), for 14~0 full depth involute system
=
0.154 - (0.912/z), for 20° full depth involute system
=
0.175 - (0.841/z), for 20° stub system
The following table shows the values of allowable static stresses for the different gear
materials.
Table 5.4. Values of allowable static stresses
Material
Allowable static stress Iob I
(N/mm2)
Cast iron, high grade
56
70
105
Cast steel, untreated
140
Cast steel, heat treated
Alloy steel, heat treated
196
126
224
350
460
Bronze
84
Non-metallic materials (like
Rawhide, Fabroil, Bakelite, etc.)
56
Cast iron, ordinary
Cast iron, medium grade
Forged steel, 'ca'\e hardened
Forged steel, heat treated
Alloy steel, case hardened
• Note that we have two types of form factors, viz., form factor based on circular pitch (y) and form factor
based on diarneteral pitch (Y).
y = Y . In this text we have used form factor 'y'.
1t
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f
,
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Design of Transm iss ion SyStems
----~--
~5.~J2~
TlIb/~ 5.5. Fomt/tlclor
Y III Lewts ~quatlon (y
II:
YIII) (from datil book, page no. 8.53)
!
--
10°FD
20° stub
0.330
0.380
0.454
40
0.336
0.389
0.459
0.339
45
0.340
0.399
0.468
0.295
0.360
50
0.346
0.408
0.474
0.270
0.308
0.377
55
0.352
0.415
0.480
20
0.283
0.320
0.393
60
0.355
0.421
0.484
22
0.292
0.330
0.404
65
0.358
0.425
0.488
24
0.302
0.337
0.411
70
0.360
0.429
0.493
26
0.308
0.344
0.421
75
0.361
0.433
0.496
28
0.314
0.352
0.430
80
0.363
0.436
0.499
30
0.318
0.358
0.437
90
0.366
0.442
0.503
32
0.322
0.3M
0.443
100
0.368
0.446
0.506
33
0.324
0.367
0.445
200
0.378
0.463
0.524
35
0.327
0.373
0.449
Rack
0.390
0.484
0.550
z
14 ~ ° FD
100FD
10° Stub
z
14 Ya ° FD
10
0.176
0.201
0.261
37
12
0.192
0.226
0.289
14
0.236
0.276
16
0.255
18
-
-
5.15. DYNAMIC EFFECTS
When a pair of gears is driven at moderate or high speeds, it is certain that dynamic effects
are present. In order to account for the dynamic effects, a velocity factor Cv (also known as
dynamic factor) is introduced. The velocity factor (cv) developed by Barth are given as
follows:
(i)
For ordinary and commercially cut gears made with form cutters and with v < 10
m/s:
3
c =
3+v
v
(ii)
For accurately hob bed and generated gears with v < 20 mls :
6
c,
(iii)
=
6+v
For precision gears with shaving, grinding and lapping operations and with v> 20
mls:
Cv
=
0.75
1 + v + 0.25
Therefore the dynamic load considering the velocity factor is given as
Fd
=
F,
Cv
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... (5.10)
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Sp~0~e=a~N~
~~
~
5.33
5.16. TANGENTIAL LOAD ON TOOTH (F,)
The tangential load (Fr) to be used in the Lewis e ti .
.
.
qua Ion ISthe load for which the drive is
to be designed. Therefore tangential load considerin th ffi
.
drive (i.e., service factor) is given by
gee
ect of service conditions of the
P
x
Ko
v
where
... (5.]1)
P = Transmitted power in watts,
v
Ko
=
Pitch line velocity in mis, and
= Service or shock factor, from Table 5.6, to account for shock loading.
Table 5.6. Service I Shock factor
Type of load
Ko
Steady
1.0
Light shock
Medium
Heavy
INote'
1.25
shock
1.5
shock
2.0
If the starting and rated torques. data are given in the problem, then service factor can be
calcuJated as below.
.
Starting torque
Service or shock factor, Ko = R ated torque
For example, the starting torque of motor is 140% of the rated torque, then service factor, Ko = 1.4.
5.17. DYNAMIC TOOTH LOAD (Buckingham's
Equation for Dynamic Load)
In addition to the static load due to power transmission, there are dynamic loads between
the meshing teeth. The dynamic loads are due to the following reasons:
Let
./
Inaccuracies of tooth spacing,
./
Irregularities in tooth profiles,
./
Elasticity of parts,
./
Misalignment between bearings,
./
Deflection of teeth under load, and
./
Dynamic unbalance of rotating masses.
Fd = Total dynamic load on the gear tooth,
F(
=
Transmitted load i.e., steady load due to transmitted torque, and
FJ
=
Incremental load due to dynamic action.
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Design of Transmission
SysletnJ
5.34
Then,
= Transmitted load + Incremental load
Dynamic load
=
Fd
F, + FI
The incremental load (FI) depends upon the pitch line velocity, the face width, material of
the gears, the accuracy of cut and the tangential
Buckingham
load given by
load. The incremental
is
21 v (be + F,)
FI =
2 I v + ...J be + F,
Therefore Buckingham dynamic load (also known as Buckingham's
load) is given as
21 v (be + F,)
F, +
_
2 I v + 'J be + F,
equation for dynamic
r
where
e
Constant, known as deformation factor, in N/mm
=
A deformation factor (e) depends upon the tooth form, the material of the gears, and the
expected error in tooth profile. Consulting Tables 5.7(a) and b) the def rmati n value c can
be found.
Table 5.. (b) Expected error (e) In
Table 5.7. (a} Value of c
(fro", data book, page no. 8. 53)
Tooth
Maltrial
form
100111
c
of pinion
and gur
N/mm
14.5°
profiles. ;11 mm
Firil clUJ
artfully
commer'cis]
cut gurl
57.0 c
00
114.tO c
oe
e
depth
MI5
gun
O.OI2~
78 0 c
~I iron
20° full
Prtcllion
o.
I
00
I
7
12~
Ol~
0.017
c
I iron
M
0.019
"
O.020~
4""
0
2
84 0
5.18. ESTIMATING GEAR SIZE (Estimation
We kn \\ thai
modul
In order
agar
l
111
t
III
b
of Module)
on it
m d Ie alue. The
an
av id fai Iurc
f ear
t
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rh,
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~~:.:a~n
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------
~~
5.35
J---
i.e.,
Module, m ~
F(
CV X 7t X
... (5.13)
b x [ ob ] x y
5.18.1. Standard Module
Table 5.8 gives the recommended series of standard module (m).
.-
Table 5.B. Recommended series of standard module in mm (from .I t b k
82"
,
aa a 00 ,page no. . '/
Choice-l
Choice-
I, 1.25, 1.5,2,2.5,3,4,5,
2
Choice
1.125, 1.375, 1.75,2.25,2.75,
6,8, 10, 12, 16,20
3.5,4.5,5.5,
3
3.25,3.75,6.5
7, 9, II, 14, 18
5.19. FATIGUE STRENGTH OF GEAR TOOTH
(Wear Strength of Gear Tooth or Wear Tooth Load)
The failure of the gear tooth is mostly because of wear between two meshing teeth. For
example, pitting is a surface fatigue failure. In order to avoid this type of failure, the
proportions of the gear tooth and surface properties, such as surface hardness, should be
selectedin such a way that the wear strength of the gear tooth is more than the effective load
betweenthe meshing teeth.
The maximum wear load (or wear strength of the gear tooth) mainly depends upon the
radii of curvature of the tooth profiles and surface fatigue limits of the materials. Therefore
themaximum or the limiting wear load of gear tooth is given by
I Fw =
where
d)· b· Q. Kw
I
... (5.14)
Fw
=
Maximum or limiting wear load in newtons,
d)
=
Pitch circle diameter of pinion in mm (use d) irrespective of whether
pinion or gear is designed),
b = Face width of the pinion in mm,
Q
=
Ratio factor
2x i
=
= --i+1
=
2x i
i-I
where
Z
p
=
2zg
, for external gears
Zg +zp
2Z
Zg
g .,
-zp
... (5.15a)
.. , (5.15b)
for internal gears
i = Gear (or velocity) ratio =
z/zp, and
and z g = Number of teeth in pinion and gears respectively.
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Design o/Transmission Systems
5.36
K, = Load stress factor (also known as material combination factor) in
N/mm2, from Table 5.9.
=
(J..+_l)
f;s·sinq,
1.4
Ep
(
Eg
... 5.16)
f es = Surface endurance limit of a gear pair in N/mm2, from
where
Table 5.9,
q, =
Pressure angle, and
Young's moduli of pinion and gear materials respectively.
Ep and Eg =
INote I The surface
endurance limit (f es) for steel may also be obtained by using the relation
f es =
... (5.17)
(2.8 x BHN - 70) N/mm2
Table 5.9. Values of
\
and Kw
Kw for 14 Yz
f es
Material
Pinion
f es
N/mml
Gear
0
Kw for 200 FD
FD
N/mml
N/mm1
Steel
Steel
150 BHN
342
0.206
0.282
250BHN
618
0.673
0.919
400BHN
1030
1.869
2.553
342
0.303
0.414
618
1.0
1.31
445
0.503
0.689
549
1.05
1.42
-
Cast iron
Steel
150 BHN
Cast iron
Steel
250BHN
Bronze
Steel
200BHN
Cast iron
Cast iron
Non-metal
Metal
1.4
5.28. NUMBER OF TEETH
The mmimum number of teeth on pinion to avoid interference is given b
2
zmi" = sin2 ~'
Y
where ~
=
Pressure angle.
ilMlW&e tooth profile with 14 'h 0 or 20° pressure angle is used.
herefolle, aaswne zi as given below:
~lY
./
,
ZI ~
17, for 200 full depth system,
./ z, ~ 32,
and
for 14'120 full depth system.
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. .. (5.18)
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~~--------------------------
~S.~37~
21. FACE WIDTH
5. In the design of gears, it is required to express the face idth i
f
•
Ii
WI
in terms 0 module In
ractice,the optImal range of face width is
.
p
8 m < b < 12 m
f "'1
11'
... (5.19)
However, or imtia ca cu ations of gear design, face width is assumed as ten times of
module (i.e., assume b = 10 m).
5.22. fACTOR OF SAFETY
(i) Factor of safety for bending:
_ Beam strength of gear tooth
Fs
(F . Sknding - Dynamic load on gear tooth - Fd
The recommended factor of safety for bending is between 1.5 to 2.
(;;)Factor of safety for pitting:
_ Wear strength of gear tooth
Fw
(F . S) pnnng
..
- 0 ynarmc. Ioad on gear tooth - Fd
'" (5.20)
... (5.21)
The recommended factor of safety for pitting is 2.
5.23. DESIGN PROCEDURE
The gear is designed on the basis of beam strength using Lewis equation and checked for
dynamicloading and limit wear loading using Buckingham's equation.
1. Selection of material:
If not given, select a suitable pinion and gear materials,
referringTable 5.3.
2. Calculation of z 1 and z 2 :
-/ Ifnot given assume number of teeth on pinion zl ~ 7, say 18.
-/
Then number of teeth on gear,
Z2
= i x zl' where i-gear ratio.
3. C(liculation of tangential load on toot/, (Ft)
:
Calculate the tangential load on tooth using the relation
P
F,
where
= -;
P
=
v
=
x
Ko
Power transmitted in watts,
1tdN.
I
d
Pitch line velocity = 60 ,m m s, an
Ko = Service I Shock factor, from Table 5.6.
4. Calculation of initial dynamic load (FttJ : Calculate the preliminary value of dynamic
loadFd using the relation
Where
.I ..
FI
Fd
= -
c
= Velocity factor. The value of v used in velocity factor formula may
v
. Cv
n't'ally taken from 10 to 15 m/s.
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Design of Transmission Syste11zs
5.38
5. Calculation of beam strength (FJ :
the beam strength (Fs) in terms of module using the relation
Calculate
where
F,
=
1t. m . b . [ crb ] . y
m
=
Module in mm,
b = Face width in mm, initially assume
[ crb]
b = lOx module,
= Design bending stress or allowable
y
static stress, from Table 5.4, and
= Form factor, from Table 5.5.
6. Calculation of module (m) : Since the gear is designed on the basis of beam strength,
therefore
F s ~ F d: So equating Fsand F tP find the module.
Then select the nearest higher standard module value from Table 5.8.
7. Calculation of b, d and v :
b
=
./
Find face width (b):
./
Find pitch circle diameter (d I):
./
Find
m oi
pitc hi' me ve I'ocity ()v :
10m
V
d 1= Z . m
=
l I
1td60
N
8. Recalculation of the beam strength (FJ:
Recalculate
the beam strength
of the gear
tooth using the relation
F,
9.
=
1t. m . b . [ crb] . Y
Calculation of accurate dynamic load (F{J: Calculate
accurately
using Buckingham's
where
In calculating
F +F
e
=
Deformation
by neglecting
the dynamic
=
21 v (be + Ft)
F + ----;::=====-
=
tit
load more
equation as given below.
F
d
the dynamic
21 v + '.} be + F,
factor, from Tables 5.7(a) and (b).
load (F d), the value of tangential
service factor (Ko) i.e., F,
=
load (F,) should be calculated
P/v.
10. Checkfor beam strength (or tooth breakage) :
./
Compare F d and F s'
./
If F d ~ F s' then the gear tooth has adequate
beam strength and it will not fail by
breakage, Thus the design is satisfactory .
./
If Fd > Fs' then the design is not satisfactory.
Now increase the face width, module
or both. Usually, to reduce the dynamic load (F d)' the gear should be carefully cut
(i.e., to reduce the deformation factor (cj), Even for precision gears, F d > Fs' then
increase the face width, till F d < F s:
=
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~~a~n--------------------
~5'23~9
II. Calculatio" of the maximum wear/oad (F.J :
Calculate the maximum or limiting wear load using the relation
r, = dlxbxQxKw
where
Q = R·auo eractor
2i
2 z2
= "+ I
-- z +
I
K,
,an d
z2
= Load stress (or material combination) factor, from Table 5.9.
12. Checkfor wear:
"
Compare the calculated values of dynamic load (Fd) and wear strength (Fw)'
./
If Fd < Fw' then the gear tooth has adequate wear capacity and it will not wear out.
Thus the design is safe and satisfactory.
13. Calculation of basic dimensions of pinion and gear: Calculate all the basic
dimensions of pinion and gear using the relations listed in Table 5.10.
Table 5.10. Basic dimensions of spur gears (from data book, page no. 8.22)
Nomenclature
Notation
Formula
Module
m
2 a / (z, + z2)
Centre distance
a
m (z, + z2) / 2
fo
Height factor
fo = I, for
full depth teeth
f 0 = 0.8, for stub teeth
c
Bottom clearance
0.25 m, for full depth teeth
0.3
h
Tooth depth
In,
2.25
for stub
m,
for full depth teeth
1.9 m, for stub
Pitch circle diameter
d
dl
Tip diameter
da
dal
mz,;
(z,
d2
+ 2/0) m ;
do2 = (z2 +
df
Root diameter
dfl
d
f2
mZ2
(z,
= (z2 -
2/0> m.
2/0)
In
2c;
2/0> m - 2c.
. d
h weaker of the two gears.
is applie to t e
. .
Id b
.
te of tile same material, then tile puuon shou e
2. When both the pillion and II.e gear are mac, .,
mpared with gear therefore the pinion is
'.
.
.
I
less
for
pmlon
co
.'
designed. Since the Lewis factor y IS a ways
[Note
I I. The
Lewis equation
always weaker than the gear.
]
decid
the weaker between pinion
roduct [ob
x y eCI es
3. When different materials are use , t e p . .
d [
]
for the gear are calculated. Tile
0b2 Y2
and gear. That is, the product [obI ] YI fior the pinion. an
element which has lower I ub I y va lue s h uld be designed.
d
h
°
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Design of Transmission Sysle111s
5.40
I Example
I
Given Data:
P = 4? kW;
5.12 Design a spur gear drive required to transmit 451flr at a pinion sPeed
of 800 r.p.m: TI,e velocity ratio is 3.5 : 1. The teeth are 200 f u/l depth involute with 18 teeth
on the pinion. Both the pinion and gear are made of steel with a maximum safe static stress
of 180 Nlmm2. Assume medium shock conditions.
[ 0b ] =
i = 3.5;
N, = 800 r.p.m.;
<I>
=
20°;
Zt
=:
18 .,
180 Nzmm-.'
Tofind:
Design a spur gear.
© Solution:
Since both the pinion and gear are made of the same material, the pinion is
weaker than the gear. So we have to design only pinion.
1. Selection of material: Given that the pinion and gear are made of steel. Assume steel
is hardened to 200 BHN.
2. Calculation of z, and
Z2:
Number of tee.h on pinion, zl = 18
Number of teeth on .gear, z2
... (Given)
= i x zl =
3.5 x 18
=
63
3. Calculation of F, :
Tangential load, F,
P
= -v xK
0
1td1N1
where
V
=
1txmz1xNt
60
=
=
60 x 1000
1t X m x 18 x 800
60 x 1000
=
... [.: d I = m . Z I and 'm' is in mm]
0.754 m
Ko = 1.25, for medium shock conditions, from Table 5.6.
F,
=
3
45 x 10 x 1.25
0.754 m
=
74603
m
4. Calculation of initial Fd:
Initial dynamic load, Fd
F,
= -C
v
where
Cv
=
Velocity factor, assuming v = 12 mls
=
6
6 + v' for accurately hob bed and generated gears with v < 20 m/s
=:
Fd
=
6
6 + 12
:: 0.333
74603
m
1
x 0.333
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=
223809
In
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~::e=~~s----------------
_J~
5.41
5. Calculation of Fs :
Fs =
Beam strength,
b
where
y
7t •
=
=
=
=
m . b . [ 0b ] . y
Face width = 10 x m
... (assume)
Form factor
0.154 - (0.912 1 Z I), for 200 full depth system
0.154-(0.912/18)
= 0.1033
7tXmxl0mxI80xO.l033
F, -
= 584.15m2
6. Calculation of module (m) :
We know that,
6it \mL:.
m2
~84.15
m
If\.
or
• module, m ~ 7.26 mm
From Table 5.8, the nearest higher standard module value under choice 1 is 8 mm.
7. Calculation of b, d and v :
b
=
= 10 x 8 = 80 mm
d , = m- zl = 8 x 18 =
10 x m
'"
Face width (b):
.;
Pitch circle diameter (d1):
.;
Pitch line velocity (v):
1t
v
d1 Nl
60
=
8.· Recalculation of beam strength (F J
Beam strength,
Fs =
=
=
where
F,
=
e
=
P
v
=
45 x 103
6.03
Deformation
= 11860
=
x 144 x 10-3 x 800
60
=
6.03 mls
:
m . b . [ (J b ] • Y
1t.
1t X
8 x 80 x 180 x 0.1033 = 37385.45 N
9. Calculation of accurate dynamic load (Fd)
Dynamic load, Fd
7t
144 mm
:
21 v (be + Ft)
F, + 21 v + '.} be + F,
=
7462.68 N,
factor, from Tables 5.7(a) and (b),
e, for 200 FD, steel and steel, from Table 5.7(a), and
e = 0.038, for module upto 8 and carefully cut gears, from Table 5.7(b).
Then ,
e = 11860 e = 1186<Yx 0.038 = 450.68 N/mm.
21 x 6.03 x 103 (80 x 450.68 + 7462.68)
7462.68 + 21 x 6.03 x 103 +~ 80 x 450.68 + 7462.68
=
50908.19 N
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Design oj Transmission Systems
5.42
10. Check/or beam strength (or tooth breakage) :
Since Fd > Fs' the design is unsatisfactory. That is dynamic load is greater than the beam
strength.
In order to reduce the dynamic load F d' select the precision gears. Therefore, fn>m Tab\~
5.7(b), e = 0.019 for precision gears.
Then, the deformation factor, c
=
11860 x e
= 11860 x 0.01 Q = 225.34
Therefore the dynamic load is recalculated as
Fd
21 x 6.03 x 103 (80 x 225.34 + 7462.68)
= 7462.68 + 21 x 6.03 x 103 + ~O x 225.~4
7467~6~--
+
-v
= 32920.46 N
Now we find F d < Fs' It means, the &~~ft~olh
h~li~g~~u~t~P,~~111strength and it ~ill not
fail by breakage. Therefore ~~f!~~~«.~~·~~fis.faq~q'~
,
11. Calculation of maximum ~eqr 1!J.w.f (f~ :
Wear load,
Fw =
Q
where
~y
= Ratio factor
2i
=
i +- T
2 x 3.5
3.5 + 1
=
= 1.555, and
= Load stress factor.
= 0.919 N/mm2, for steel hardened to 250 BHN,
Fw = 144x80x 1.555 x 0.919 = 16462.6N
from Table 5.9.
12. Check/or wear: Since Fd > Fw, the design is unsatisfactory. That is, dynamic load is
greater than the wear load.
In order to increase the wear load (Fw)' we have to increase the hardness (BHN). So now
for steel hardened to 400 BHN, K, = 2.553 N/mm2, from Table 5.9.
..
Fw
=
144 x 80 x 1.555 x 2.553
= 45733.42 N
Now we find Fw> F d" It means, the gear tooth has adequate wear capacity and it will not
wear out. Therefore the design is satisfactory.
13. Basic dimensions of pinion and gear: Refer Table 5.10 .
m = 8 mm .
../
Module:
./
Number of teeth :
4•
= 18 and
./
Pitch circle diameter:
d1
=
d2 =
./
Centre distance:
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a
144 mm;
m· z2
2
63
and
= 8 x 63 = 504 mm
m(zl+z2)
=
42 =
= 8 (18 + 63)
2
= 324 mm
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•• ,__ -.•• t-,t~
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5.43
./
Face width:
./
Height factor :
./
Bottom clearance:
./
Tip diameter :
b = 80mm
10
= 1, for 200 full depth teeth
d.01
=
=
~zl+2fo)m
d02
=
(~2 + :]. 0)
dfl
=
(zl-2fo)m-2c
(18 - 2 x l) 8 - 2 x 2
4/2
=
=
c
=
0.25 m
0.25 x
=
(l8+2x
=
(63 + 2 x 1) 8
! m
(z2 - 2
8 = 2 mm
f 0) m -
=
= 160mm;and
1)8
=
520 mm.
124 mm;
and
2c
= (63:- 2 x 1) 8 - 2 x 2 = 484
mm
Ij;X4m:f't5:~{J .tt compressor
running at 300 r.p.m: is driven by a 15 kW, 1200 r.p.m.
motor thr(Jllgh ,,14l1z 9ful! dept" spur gears. The centre distance is 375 mm. The motor
pi"lon Is to be of C 30 forged steel hardened and tempered, and tile driven gear is to be of
~ast ir(Jn. Assumlnll mf!4ium slmck condition, design tile gear drive completely.
N~ = 300 r.p.m.;
Given Data:
T(1find:
P = 15 ~W; NI = 1200 r.p.m.; ~ = 14 12 ; a = 375 mm.
1
0
Design the spur gear drive.
@ Solution:
evaluate [crbl
]
",,(i
Since the ",,,(eri41$ (1/ pinion
11~"r'''~ (/iff!!r~nf~fir&t wc? have
• y, and [ crbl ] . Y2 to find out the weaker element,
Gear ratio, i
= Nt =
N2
1200
300
to
= 4
Assume zi = 18.
Z2
i x z,
=
=
4 x 18 = 72
z, = 18, and
Form factor, Y,
=
0.270, from Table 5.5, for
Permissible static stress, [ob]
=
112 Nrrnrn-, from Table 5.4, for forged steel.
y,
=
0.270
112 x -7t-
Forpinion:
[ crbl ]
For gear :
Form factor, Y 2
Permissible static stress, [crb
=
= 9.625
0.360, from Table 5.5, for z2
h = 56 N/mm2,
...
[ ... y 1---]
YI
7t
= 72, and
from Table 5.4, for cast iron.
[ .: Y2
Y2
= -]7t .
We find [ crb2 ] Y2 < [ crbl ] YI i.e., the gear is weaker than the pinion. Therefore, we have
.
eSIgnthe
gear only.
tod
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'J
,n'
0/
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5.44
1. Material selection:
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Design
Transmission
Pinion:
C 30 Forged steel;
Gear:
Syslel
and
Cast iron.
... (given
2. Ca/cuilllion of module (m):
Since the centre distance (a) is given, we need not t
equate F, and Fdto find the module. Here the module can be calculated using the relation
m (z, +z2)
a
375
or
=
2
= m (18 + 72)
2
or
Module, m
=
8.333
From Table 5.8, the nearest higher standard module under choice 1 is 10 mm.
3. Calculation of iJ, d and v :
b
=
10 x m
=
10 x 10
100 mrn
Face width (b):
../
Pitch circle diameter of pinion (d ) . d = m
../
Pitch circle diameter of gear (d ).
,.,
2'
../
Pitch line velocity (v):
l'
= 'It
4. CalCulationof beam strength (FJ
Beam strength, F =
:
'It.
d2 -
d2 N2
60
=
m .b.[
.z, -- 10 x 18 = 180 mm
m . Z2
'It
=
lOx 72 = 720 rnrn
x 720 x 10-3 x 300
60
]
CJh 2'
s
cu/ation of dynamic load (F
Dynamic load, Fd
) :
d
=
x 103
, = fv -_ IS11.31
FI +
= 11.31 m/s
Y2
= 'It x lOx 100 x 56 x (0.3 6)
5. Cal.
where
=
../
'It
=
20160 N
21 v (be + FI)
21 v + \} be + F,
F
= 1326.26
N,
c ;::
Then ,
Deformation factor from T
;:: 7850
c.
'
abIes 5.7(a) and (b)
e, ror 14° FD
'
e
' steel - cast iro
== 0.022, for mOdUle Upto J 0
.n: from Table 5.7(a), and
c ;:: 7850
and preCISion ge
x 0.022 :::: 172.7 N1m
ars, from Table 5.7(b).
m
1326.26 + _21 x 11.31 x 103 (JOO
;::
21 x 11.31 x 103 +
x 172.7 + 1326.26L
19911.85 N
100 x 172.7 + 1326.26-
Vi
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Spur Gears
5.45
6. Checkfor beam strength (or tooth breakage) :
We find Fd < Fs' It means, the gear tooth has adequate beam strength and it wiJI not fail
by breakage. Thus the design is satisfactory. .
7. Calculation of wear load (F.J :
Maximum wear load, Fill
where
=
dI x bx Q
Q
=
Ratio factor = _lj_
i+1
Kill
=
Load stress factor
=
=
X
K,,,
2x4 - I6
4+1
- .,an
d
I N/mm2, for steel (250 BHN)-
cast iron and 14° FD, from Table 5.9.
Fw
=
180 x 100 x 1.6 x I
8. Check forwear:
=
28800 N
We find Fw > Fd" It means, the gear tooth has adequate wear
capacity and it will not wear out. Therefore the design is satisfactory.
9. Basic dimensions of pinion and gear: Refer Table 5.10 .
../ Module: m = 10 mm •
../ Number of teeth: z. = 18, and Z2 = 72.
d1 = 180 mm, and d2
/
Pitch circle diameter:
./
Centre distance:
../
Face width:
../
Height factor, fo = 1, for 14 Y2° FD .
../
Bottom clearance:
../
Tip diameter:
b
= 375
a
=
=
c = 0.25 m
dOl
df
Root diameter:
mm .
100 mm .
d 02
../
= 720 mm •
=
=
I =
«»
0.25 x 9
=
2.25 mm
10) m = (18 +2 x I) 9 = 180 mm; and
(z2 + 2 10) m = (72 + 2 xl) 9 = 666 mm •
(z I - 2 10) m - 2 c = (18 - 2 x I) 9 - 2 x 2.25
(z) + 2
=
139.5 mm;
and
=
(z2-2fo)m-2c
=
(72'-2 x 1)9-2x2.2S
= ~~5mm.
(Example 5.14 , A bronze spur pinion rotating at 600 r.p.m: drives a cast iron gear at a
trow .....
rlllis of 4·: 1. AIIoNIIIIW slatic stress for pinion and gear are 85 and 105
N~
r5/M!Ctively. TM pilliIJ. has 22 stQfHl4l~d20 0full dt!pJj~ ;IW~
uel& The power
Ir.... i*ed is 32 kW. The sill/ace elldluallCe linUJ for Ihe gear pair is 520 NI"".2, modulus
of el_icily of Ihe pillUM ",,*,iaJ is 1.2 x 11J5NI"",,1 aIIII that of gear is 1 x 1fJ,5 NI"""z. If
t~ Sllll'tu., I#WI-- is 25% ,..,e ,... tile IMIIII tlNt-, desig", tile gear drive completely.
Given Dala: NI = 600 rp.m.;
i = 4; [O'bl]
= 85 Nzmm-": [O'b2] = 105 N/mm2 ;
z,
==
E 2- J
22;
q, =
200,
x ]05 N/mm2;
Ko
P
=
=
.
Starting torque
Service factor = Rated torque
32 kW;
fes
=
520 N/mm2;
EI
=]
=
1.2 x 105 N/mm2
.25.
....
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;
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~
25~.4~8
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------~-~.=-~-~-~~~ __~~~
D_e_si~gn~of_Tl_t_an_s_m_~_s_io_n_S~~_~/~ern~s
to ./inti f t}~sigli the geat drive,
SJHtt! the IHlJleflals 0/ plnton and gear are different; first we have to
evaluate [<1bl 1Yi Md ((Jb11 >'2 to find out the weaker element.
@ StJiulitJH!
Given that il
For pinion:
=
22; and
:1== , ':1 == 4 x 22
Y t = 0.330, for zi = 22, from Table 5.5.
Forth factor,
..
88.
===
r 0bi])it
For gear: Form factor,
Yt = ~]
n
= OA40, for %2 = 88 (by interpolation), from Table ~.5.
Y2
[crlJ2]Y:Z
= 8.93 N/mm2
== 8S x 0.330
n
== lOS xQA4
[ .,'
== 14.7N/mm2
n
[.,' Y2= Y2]
n
We find [obI] YI < [ 0621 'vi' i.e., the pinion is weaker than the gear. Therefore we have
to design the pinion only.
1. MateridJ
seuato«
I
Pinion: Bronze, and
... (Given)
. .. (already calculated)
Gear: Cast iron
2.
zi
= 22; and
z2
= 88.
3. Calculation of FI :
p
Tangential load, F/ = v
where
v
Ko
F/
x
Ko
=
=
1txmztxNt
60 x 1000
=
1t x m x 22 x 600
60 x 1000
=
Starting torque
Mean torque
=
_.. (Here module is in rom)
0.691 m
_
- 1.25
= ~~6~/ ~ x 1.25 = 578:4.52
4. Calculation of initial Fd:
Initial dynamic load, Fd =
where
ClI
F/
clI
= Velocity factor, assuming v = 12 m1s
= 6
=
Fd =
6
+ v ' for accurately generated gears with v < 20 m1s
6
6 + 12
=
57874.52
m
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0.333
1
x 0.333
=
173797.37
m
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~~Ge~a~~s ----------~~~------~--------~------------5_.4_7-
::r;--
5. Calculation of
r,:
Beam strength, F,
=
1t.
b
=
Face width == 10 x m
where
m . b · [ ObI]
• YI .
;.. (assume)
YI = 0.154 - (0.9121 ZI)' for 20° full depth system
:::
Fs
0.154-(0.912/22)
= 1tXmx
0;1125
"'=
IOmx85xO.1125
=
300.41",2
6. Calculation of module (m) :
Fs ~ Fd
We know that,
300.41
or
m2
~
173797.37
m
Module, m ~ 8.33
From Table 5.8, the nearest higher standard module under choice 1 is 10 mm,
7. Calculation b, d and v :
./
=
b=10m
Facewidth(b):
./
lOx 10=100mm
Pitch circle diameter (d 1): d 1 = m . Z1 = lOx 22 = 220 mm
./
Pitch line velocity (v):
v =
1t dl
8. Recalculation of beam strength (F J
Beam strength, F, =
1t.
=
1tX
NI
60
=
1t
x 220 x 1O~3 x 600
60
= 6.91 mls
:
m . b . [ 0b]
.Y
lOx 100x85xO.1125
9. Calculation of accurate dynamic load (F J
=
30041.48N
:
21 v (be + F,)
Dynamic load, Fd = F +
,
21 v + \} be + F,
where
F, =
e
=
32 x 103
v 6.91
P
=
4629.96 N,
Deformation factor, from Tables 5.7(a) and (b).
~ 8150e,
from Table 5.7(a), and
e ~ 0.022, for precision cut gears with module upto 10 mm,
from Table 5.7(b).
e ~ 8150 x 0.022 = 179.3 N/mm
Then,
Fd = 4629.96 +
21 x6.91 x 103 (100 x 179.3+4629.96)
21 x 6.91 x 103 + ~ 100 x) 79.3 + 4629.96
= 27166.59 N
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Design of Transmission Systems
5.48
/0. Check for beam strengtl, (or tooth breakage) :.
.
.
We find Fs> Fd" It means, the gear tooth has adequate beam strength and It WIll not fail
by breakage. Therefore the design is satisfactory.
11. Calculation of maximum wear load (Fw) :
Wear load, F",
Q
where
K",
= dl x
bxQx
Kw
2i
2x4
+ 1 = 1.6, and
= Ratio factor = i + I = 4
Load stress factor
=
=
(520)2 x sin 200 [I
1.4
[1Ep + Eg1 ]
/2 . sin,
's 1.4
=
I
1.2 x lOs + I x lOs
•••
F", = 220 x 100 x 1.6 x 1.211
[.:
J
=
1.211 N/mm2
t, and Eg are given]
/ ~S'
42627.2 N
=
12. Check for wear: We find F,.. > F d" It means, the gear t
th has adequate wear
capacity and it will not wear out. Therefore the design is satisfactory.
/3. BlISic
~lISiDlIS
of pinion and gear: Refer Table 5.10 .
./
Module: m
./
Number of teeth:
./
Pitch circle diameters:
= 10 mm
'I = 22;
and
'2 = 88
d, = 220 mm'
and d2
= ", .;;
= 10
88
= 880
mm
10 (22 + 88)
=
2
= 550 mm
dOl =
./
Root diameter :
dill
=
dfJ
=
(z I - 2 /0) m - 2 c
=
(22 - 2 x I) 10 - 2 x 2.5
=
195 mm'
and
d 12 = (z2 - 2 /0) m - 2 c
= (88 - 2
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x
I) 10 - 2
2.5
=
855
BUD.
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~~S-------------------[]Xnmplt 5.1~ I A bakelite pinion driving
~S~.4~9
a cast iron gear. Tile pinion rotating at
700r.p.m. transmits 5 k ~ to a gear. Th~ velocity ratio is 3, the teeth are 200 full depth, and
-d is smooth. Design tire gear drive. Take the allowable static stress for bakelite as
tht I.,..
40NI",nr2·
Given J)ata: N.
= 700
P = 5 kW;
r.p.m.;
; = 3; ~ = 20°;
[ob]
=
40 N/m~2.
To find: Design the gear drive.
Since bakelite is a non-metallic material, bakelite pinion is weaker than the
cast iron gear. Therefore we have to design the pinion only.
@ Solution:
1. Material: Pinion - Bakelite,
2. Assume z.
=
18; and z2
= ;
F,
3. Calculation of F, :
v
where
and Gear--Cast iron.
x z.
P
xK
= -v
=
1t
=
54.
0
d. N.
60
= 0.66
Ko
3 x 18
=
. .. (Given)
1txmxz.xNI
=
60 x 1000
=
1t
X
m x 18 x 700
60 x 1000
m
= 1, for smooth load, from Table 5.6.
F,
=
4. Calculation initial F d:
Fd
=
where
clI
=
5 x 103
7578.81
0.66 m x 1 =
m
F,
clI
V e IOCIity f:actor -- 10.75
+ V + 0 .,25 for non-metallic gears
-_ 0.75 + 0 25 = 0.375, assuming v = 5 m/s
1+5
.
_L_
7578.81
m
x 0.375
= 20210.16
m
I
, 5. Calculation of beam strength :
where
F,
=
1t.
b
=
Face width
m .b .[
(Jb ] . y
.
.., (assume)
= 10m
static stress = 40 N/mm2, and
[ (J b ] = Allowable
- 0 154 - (0 .912/z I) , for 20° full depth system
y = Form factor -.
=
Fs
0.154-(0.912/18)
= 1t
=
X
(Given)
= 0.1033
m x 10m x 40 x 0.1033
129.85 m2
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Design of Transmission SYslenlS
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5.50
6. Ca/cli/ation
0/ module:
We know that, Fs
129.85 m2
or
Module,
Fd:
~
20210.16
m
~
m ~ 5.38
From Table 5.8, the nearest higher standard module under choice-]
is 6 mm.
7. Calculation of b, d and v :
b = 10m
./
Face width:
./
Pitch circle diameter:
./
Pitch line velocity:
= lOx 6
=
d1
=
m- zl
1td1 Nl
v =
60
60 mm
=
6 x 18 = 108 mm
1t
=
x 108 x 10-3 x 700
60
- 3.96 mls
8. Recalculation of beam strengtb :
=
F,
1t·m·b·[crb]·y
= 1t
X
6 x 60 x 40 x 0.1033
=
4673.18 N
9. Calculation of dynamic load: We know that the Buckingham's
equation
for dynamic
load is not applicable to non-metallic gears. Therefore the dynamic load for non-metallic
can be calculated as
=
Fd
F,
Cv
p
F, = -v =
where
c,
5 x 10
= 1262.62 N, and
3.96
=
· . c.
( 0.75 )
V e Iocrty
tactor = ·1 + v + 0.25
=
0.75 )
( 1 + 3.96
+ 0.25 = 0.4012
Fd =
gear
1262.62
0.4012
=
3147.12 N
10. Check/or beam strength (or tooth breakage) :
We find Fs> Fil' therefore the design is satisfactory.
11. Calculation of wear load:
Fw = d I x b x
where
Q
=
Kill
=
Fw =
Q x K,
R . f
2 i
atro actor = i + 1
Load stress factor
=
2x3
---
3+1
=
1.5, and
1.4 N/mm2• for non-metallic,
from Table 5.9.
108 x 60 x 1.5 x l.4 = 13608 N
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j'l
5.51
Spur Gears
, I
12. Checkfor
We find Fw > Ftl' tliererore
C'
I
••
the
design
wear:
13. BaJic dimensions of pin ion anti gear:
Module:
./
Number of teeth :
ZI
.; ,Pitch circle diameter:
./
Centre distance:
./
Face width:
./
Height factor:
./
Bottom clearance:
./
Tip diameter:
Refer Table 5.10 .
b
= 18; and Z2 = 54.
{II = 108 mm ; and
=
d2 = m . z2
=
6 x 54
= 324 mm.
m (z I + z2) = 6 (18 + 54)
a =
2
2
=
216 mm
60 mm
f0= I
c = 0.25
III
= 0.25
x 6
=
1.5 mm
·dal = (zl + 2 f 0) m = (18 + 2 xl) 6 = 120 mm;
da2
dJ1
Root diameter:
df2
I Example
satisfactory.
m = 6 mm
./
./
IS
= (z2 + 2 f 0) m
= (z I - 2 f 0) m
= (z2-2fo)m
5.16 lIt is required to design
II
and
=
(54 + 2 x 6) 6 = 336 mm.
=
(I 8 - 2 x I) 6
=
96 mm ; and
=
(54-2x
=
312mm.
1)6
two stage spur gear reduction unit (Fig. 5.14)
with 200full depth involute teeth. The 'input shaft rotates at 1440 r.p.m. and receives 10 kW
power through a flexible coupdng .. TIre speed of output shaft should be approximately 180
r.p.m. All the gears are made of plain steel 45 C 8 (a"l = 700 Nlmm2). Tire service factor
can be taken as 1.5.
Given Data: $ = 200;
Tofind:
NI
=
,1440 r.p.m.;
•
P = 10 kW; N4
=
180 r.p.m,;
Ko = 1.5.
Design the gear drive.
© Solution:
The layout of
gears is shown in Fig.5.14. For"
ease of manufacturing,
the
pinions I arid 3 are made'
identical, while gears 2 and 4 are
also exactly 'identical.
I
t
In the design of gear train, we
need not to design all the gears. ':
Instead, the pair of gears which
transmits more torque (i. e., either
pair I or pair 2) is designed. So '~
first let us find the pair which
governs the design.
•
Fig. 5.14.
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,
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~~
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----------------------------------~D~e~sl~.gn~O~if~v~~_a_n_s_n_l~_s_i_o_n_S~~_st_~
~.52
Total transmission ratio, if
Speed of first driving gear
Speed of last driven gear
=
==
1440
180
==
8
The speed ratio at each stage (i) is given by
= -{7 = -{i == 2.8284
i x zi = 2.8284 (18) == 50.9]
= z3 = ] 8; and
= Z4 = 51.
i
Assume
zl
=
18...
z2
==
Zl
As stated earlier,
Z2
:::::5 J.
The speeds of the gears are
N. = ]440r.p.m.
N2
and
= N,
N3 = N2
N.
=
xG:)
= 1440GD
= 508.23
r.p.m.
N3 (;~)
508.23 x
=
= 508.23
(!D
=
r.p.rn,
179.37 '" 180 r.p.rn.
. .. (Given)
We know that torque transmitted
is inversely
proportional
to the
speed
[.: T = 60 x P/21tN]. Therefore, in this case, gear 3 transmits more torque titan tile gear 1
because of its lower speed. Helice the second pair consisting of pillion 3 anti gear 4 are to
be designed.
1. Material: Given that the pinion and gear are made of alloy steel. Since both gears are
made of the same material, we have to design the pinion only.
2. Already calculated that z3 = 18 and Z4 = 51.
3. Calculation of F( :
where
P
FI = ~
v
x
Ko
=
=
1t X
m x z) x
N)
60 x 1000
x 18 x 508.23
60 x 1000
IT X Tn
=
=
0.479 m, and
Ko = 1.5.
F(
4. Calculation of initial Fd:
=
... (G' lven )
10 x 103
31315.24
0.479 m x 1.5 =
m
Fd
F
= .:
Cv
where
~
-
.)
= Velocity factor = _]___ =
= 0.375,
3+v
3+5
31315.2-i
83507.3
Fd =
x
=
m
0.375
m
Cv
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assuming
=5
S
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.,
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~~
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----
~5~.5~3~
5. Calculation of beam strength :
F,
=
7t.
m . b . [ 0b ] . Y
=
b = Face width
where
[ ab]
y
F,
=
Allowable
=
Form factor
=
0.154-(0.912/18)
=
tt x m x
6. Calculation of module:
757.22
m2
10 rn,
~
... (assume)
=
ai'
=
static stress
=
7~0
=
233.33 N/mm2,
and
0.154 - (0.912 I zl)' for 20° full depth system
=
0.1033
=
IOmx233.33xO.1033
We know that, F s
~
757.22m2
F d:
83507.3
In
or
m ~ 4.79
Module,
From Table 5.8, the nearest higher standard module under choice-l is 5 mm.
7. Calculation of b, d and v :
=
Face width (b):
-/'
Pitch circle .diameter (d3):
./
Pitch line velocity (v):
b
=
10 x 5. = 50 mm.
d3
=
10 m
./
v
m- z3
7t d3 N3
60
=
=
=
5 x 18
=
90 mm
7t x 90 x 10-3 x 508.23
60
=
Ii
2.395 mls
i' I
8. Recalculation of beam strength (FsJ :
Beam strength,
F s = 7t X m . b . [ a b
=
] .
y
7tx5x50x233.33xO.l033
=
18930.44 N
9. Calculation of accurate dynamic load (F(/J :
21 v (be + F,)
10 x 103
2.395
where
e
e
Fd
=
21 v + \} be + F(
4175.36 N,
=
Deformation
=
11860 e, for steel and steel, from Table 5.7(a), and
=
0.025, for carefully cut gears and module upto 5 mm, from Table 5.7(b).
c = 11860
Then ,
= F, +
load, F d
Dynamic
=
4175.36
x
factor, from Tables 5.7(a) and (b),
0.025 = 296.5 N/mm
21 x 2.395 (50 x 296.5 + 4175.36)
+ 21 x 2.395 +~ 50 x 296.5 + 4175.36
=
9254.76 N
'j
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5 -4
--:?.--.--(0. Chec*/or ben'" strength (or tootl' breakage) :
therefore tl.e design is satisfactOry.
· fi d F ,. F
We nn
o :
S
11. Calcllla/ion 0/ maximum wear load (F",) :
Wear load, f.. ~ d I x b x Q x 1(" 2 i
i+1 ~ 0
Q ~ Ratio factor ~
where
~-
2 . Z4
1<,,, ::: Load streSS factor.
~ 2.553 N/mm2, for 400
BJ:IN
1.478, and
18 + 51
-
steel, from Table 5.9.
f" ~ 90 x 50 x 1.478 x 2.553 ~ 16983
N
J2. Check/or wear: We find F", > F d' therefore t/.e design is satisftlCtOry.
13. Basic dimensions of gears: Refer Table 5.10.
m == 5 mm
j
Module:
./
Face width:
../
Height factor:
,f
Bottom clearance:
b == 50 mm
f 0 ==
For p.inions 1 and 3 :
./
Number of teeth :
1
c ~ 0.25 rn ~ 0.25 x
z. :::·Z3 ==
5
=
1.25 DIm
18.
Pitch circle diameter: d I ~ d 3 ~ 5 x 18 = 90 mm
x
•
al
a3 zl+2!0)'" = (18+2
Tip diameter'
d == d - (
.:
,f
./
Root ,diameter:
mm
1)5=100
d i 1 == d f 3 == (z 1 - 2 f 0) m - 2 c
== (18 - 2 xl) 5 - 2 x 1.25 = 77.5 mm
For gears 2 and 4 :
./
Number of teeth : Zi == Z4 = 51
./
.,f
Pitch
circle
diameter:
•
.
./
TIp
dIameter:
Root diame.tcn:
d 2 == d 4 - 5 x 51
d
da 2 == dda4 -/4
f2
=
=
255 mm
. + 2 xl) 5 - 265 mrn
+ 2 f 0) m = (51
z2 - 2 f 0) m - 2· c
«Z2
_. (51 - 2 xl) 5 - 2 x 1 25 .
- 242.5 mm
II. GEA
R DES.IGN BASED
(GearDesign usin9 Basrc
~NRelations)
GEAR LIFE
5.24.AsDYNAMIC
LOAD (OR DESIGN TORQUE)
di
.
ther
iscussed In
i S ection
.
5.17·
. mg
. . teeth.
e are dynamic loads betw een ,m
addition
to the st aftc ,oad due t 0 power
the-mesh.
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transmission,
~
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~~S
~5~.5~5
--------------~
... Design torque,
where
M/
[M/]
=
I
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=
K . Kd
M/·
=
Pinion torque
(5.22)'
60 x P
2 1t N '
P = Power transmitted,
N
=
Speed of pinion,
K = Load concentration factor, from Table 5.11, to account for uneven
distribution of tooth load along the face width of the tooth, and
Kd
Dynamic
=
load factor, from Table 5.12, to account for profile and
circular pitch errors, and velocity of operation.
Table 5.11. Load concentration foetor, k for cyttnartcut gears (from data book.page no. 8.15)
Bearings close to
gears and
symmetrical
IlIp= bid}
Asymmetrical
Over hung pinion
*Very rigid shaft
**Less rigid shaft
0.2
0.4
I
I
1.05
I
1.04
1.1
1.15
1.22
0.6
0.8
1.0
1.2
1.03
1.08
1.06
1.13
1.16
1.22
1.32
1.45
1.1
1.29
-
1.14
1.18 .
1.23
1.36
-
1.4
1.6
1.19
1.29
-
1.25
1.35
1.45
1.55
ds
=
-
1 = length of the shaft. .
diameter of the shaft;
Table 5.12. Dynamic load factor, Kit (from data book, page
IS Quality
.._
Cylindrical
gear
.._
--
5
6
8
r---.
10
t---
Conical
gear
Pinion
surface
hardness
HB
Spur & Straight
110.
Bevel
8.J 6)
Helical & Spiral Bevel
Pitch line velocity. m/s, upto
1.0
3.0
8.0
12.0
3.0
8.0
12.0
18.0
< 350
-
-
1.2
1.4
-
1
1.1
1.2
> 350
-
-
1.2
1.3
-
I
1.0
1.1
< 350
-
1.25
1.45
-
I
1
1.2
1.3
> 350
-
1.2
1.3
-
I
I
1.1
1.2
< 350
1
1.35
1.55
-
1.1
1.3
1.4
-
> 350
-
1.3
1.4
-
1.1
1.2
1.3
-
< 350
1.1
1.45
--
-
1.2
1.4
-
-
> 350
-
1.4
-
-
1.2
1.3
-
-
s 350
1.2
-
-
-
-
-
-
-
-
5
6
8
10
•
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I'
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- 5-6
).
·gn oifTransmission Systel11s
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---
-
CEO BENDING STRESS, Gb
.
5.25. INDU
.
(.
modified LewIS equa
di stress equatIon i.e.,
Lnducedben mg
t'on for beam strength) is given
I
by
i±I
==
a . m . b· y
°b
where
[M,l
== Form factor, from Table 5.13.
fi
y (from data book, page no. 8.18)
Table 5.13. Form actor,
Y
I Noiil
5.26.
... (5.23)
z
y
z
y
0.308
35
0.452
12
0.330
40
0.465
14
16
0.355
45
0.471
18
0.377
50
0.477
20
0.389
60
0.490
22
0.402
80
0.499
24
0.414
100
0.505
26
0.427
150
0.515
28
0.434
300
0.521
30
0.440
Rack
0.550
The above values are divided by 0.8 for stub teeth.
DESIGN BENDING STRESS [
a" ]
As discussed in Section 5.10, the tooth breakage- is caused by fatigue due to repeated
bending stresses. Therefore permissible bending stress should be determined on the basis of
endurance limit and is given by
=
where
1.4
Kb/
n-
Ka x
x
°_1 ,
0_1,
for gears having one direction of rotation only.
for gears having two directions of rotation
Kb/
=
Life factor in bending, from Table 5.14,
Ka
=
Stress concentration factor for filler, from Table 5.15,
°_1 = Endurance limit stress in bending, from Table 5.16,
n
= Factor of safety, from Tab Ie 5. I 7.
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... (5.24)
and
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5.57
Table 5. U. Life factor for bemllng, Kb/ (from datu book, page no. 8.10)
Material
Surface hardness
Life in number of
HB
cycles, N
s 350
~ 10'
Steel"
~ 25 x
10'
< 25
10'
-
Cast Iron
1
10'
<
> 350
Kb/
x
~ 10' IN
0.7
~IO'/N
-
• If the case hardness HB > 350 and the core hardness HB < 350, then the coefficient
~IO'/N
Kb/
is obtained for HB < 350.
Table 5.15. Stress concentration factor for fillet, Ka (from data book, page no. 8.19)
Material and Heat Treatment
Steel, normalised, surface hardened
1.5
Steel, case hardened (low carbon steels)
1.2
Cast iron
1.2
Table 5.16. Endurance limit, u_I (from data book, page 110.8.19)
a_I' Endurance
Material
limit in reversed bending,
Forged steels
0.25 (au + cry) + 50
Cast steels
0.22 (a" + cry) + 50
Alloy steels
0.35 au + 120
0.45 all
Cast iron
INote I
au -
N/mm2
Ultimate strength, N/mm2,
O'y- Yield stress, N/mm2
Table 5.17. Factor of safety, n (from data book, page no. 8.19)
Material
Mode of
manufacture
Heat treatment
Factor ofsafety, n
Steel,
Cast
No heat treatment
2.5
Cast iron
Cast
Tempered or normalised
2.0
Steel
Cast or Forged
Case hardened
2.0
Forged
Surface hardened
2.5
Normalised
2.0
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Design of Transmission Systems
5.58_------------~~....:.:!-----~~
-
5.27. DESIGN CONTACT STRESS [<1c]
Surface strength is proportional to the surface hardness and it is given by
[ CJ ]
c
== CB x HB x KcI
... (5.25)
== CR x HRC x KcI
C and C
where
=
Coefficients
depending
on the material
and heat treatment,
R
B
from Table 5.18,
HB == Brinell hardness number, and
KcI == Life factor for surface compressive
strength, from Table 5.19.
Table 5.18. Coefficients CB and CR (from data book, pllge 110.8.16)
Surface hardness
Heat treatment
Material
Coefficient
Co or CR
HB S 350
Cs =2.5
any type
Normalised or
Hardened and
Tempered
High strength alloy
Case Hardened
HRC == 55 to 63
CR == 31
Alloy steels
Case Hardened
HRC
55 to 63
CR -28
Carbon and manganese steels
Case Hardened
HRC == 55 to 63
CR =22
Alloy steels, Carbon steels
C40; C45
Hardened and
Tempered
HRC - 40 to 55
CR = 26.5
Alloy steels, Carbon steels
C40;C45
Surface Hardened
HRC - 40 to 55
CR = 23
Cast iron, Grade 20, 25
HB
170 to 200
Cs
2.0
Cast iron, Grade 30, 35
HB
200 to 260
CB
2.3
Carbon steels and alloy steels of
Nickel chromium steels
=
C15; C20; CI5 Mn -'
85' C20
Mn85
Table 5.19. Life factor for surface (contact compressive) strength, K
(from data book ,page 110..
8 17)
cl
Material
Surface Hardness
Life in number of cycles
Life factor
HB
N
KcI
~ 107
Steel
< 107
Steel
1
S 350
~ 25
x
If)7
<25
x
107
~ 107 IN
0.585
> 350
~
107 IN
~
107 IN
Cast iron
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...
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Spur Gears
5.59
5.28. SURFACE COMPRESSIVE STRESS
The gear teeth in contact are nothing b t t
.
.
u wo curved surfaces under pressure. The stress
induced m the surface, known as Hertz contact stre ss, .ISgiven
.
by
o = 0.74 i ± I "'
where
i
= Gear ratio
a
=
=
Nl
= z2
N2
zl'
Eeq [M,}
... (5.26)
Centre distance between pinion and gear,
·b = Face width oftooth,
Eeq
I i± 1 x
\j i b
a
c
and
=
Equivalent Young's modulus, from Table 5.20,
=
2 El . E2
EI + E2 '
... (5.26a)
where EJ and E2 are Young's modulus of pinion
and gear respectively.
2 £1 £2
Table 5.20. Equivalent Young's modulus, E = £1 + £2 (from data book, page no. 8.14)
Wheel
Pinion
Tensile
Young's
Material
Equivalent
EI,
modulus
Material
au'
strength
N/mm2
N/mm2
Steel
2.15
x
lOS
N/mm2
105
2.15x
s 280
1.1 x
105
1.46
x
105
> 280
1.4 x
105
1.7
x
lOS
1.2
105
1.55
x
lOS
7 x 103
1.36
x
105
Nylon
of materials,
N/mm2
105
Bronze
For other combination
modulus E.
2.15 x
Steel
CI
Young's
Young's
modulus E2,
x
usc equation (5.26a).
5.29. CENTRE DISTANCE (a)
..r. ce cOlnpressive stress o should not exceed the design surface
W e k now t h at t h e surra
c
compressive stress [ ac]' i.e., (J'c S [ crc].
i± 1
074
. - a
i±1
ib
x
Eeq
[M,]
s
[ crc ]
By rearranging, we get
Centre distance, a ~ (i + 1)
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.. , (5.27)
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)
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5.60
...::.:::..:......
Design
S"SI'e'lis
_:;_:__~ of Transmisstnn
_::_:::..:..=.:::.:...:::!_J'
where
[ O'c]
=
Design contact stress, from equation (5.25), and
= ~
\jI
=
Ratio of gear width to centre distance, from Table 5.21
Table 5.21. Width to centre distance ratio, f// (from data book, page
110.
8.14)
'" = b I (/
Type of gear transmission
Open type gearing
0.1 to 0.3
Speed reducers (closed type)
(a) High speed 8 to 25 111/s
upto 0.3
(b) Medium speed 3 to 8 mls
upto 0.6
(c) Low speed 1 to 3 m/s
upto 1.0
Gear boxes with sliding gears
I Note I
0.12toO.15
For light and medium duty b s d I
;
For heavy duty b s 1.5 d I
where b - face width, and d I - pinion diameter.
5.30. DESIGN PROCEDURE
1. Calculation of gear ratio (i) :
Calculate the gear ratio (i.e., speed ratio) using the relation
.
N,
z2
-- N2 - zl
1--
If gear ratio is not specified, it may be assumed to be unity. In case of multistage speed
reducers, the sped ratio may be selected from R 20 series.
2. Selection of materials:
Consulting Table 5.3, knowing the gear ratio i, choose the
suitable combination of materials for pinion and wheel.
3. If not given, assume gear life (say 20,000 Ius).
4. Calculation of inittal design torque I M, J
:
Calculate the design torque using the relation
[ Mt]
where
= Mt•
=
K . Kd
.
60 x p
Transmitted torque = 2 7t N '
K = Load concentration factor, from Table 5.11, and
Kd. = Dynamic load factor, from Table 5.12.
,,L
Since datas are inadequate to select the values ofK and K,b initially assume K· Kd::::
(The above details are taken from data book, page no.8.15.)
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I.3.
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G
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5.61
IrS
~
5. Calculation of E~q' f ubi and
I uel :
I
consulting Table 5.20,' calculate the equivalent Young's modulus (Eeq ).
I
consulting Table 5.16, calculate the design bending stress [ CJb
I
Tofilld
].
I ue I : Calculate the design contact stress [ CJc 1 using the relation
[ CJc
1 =
where CB or CR
CB· HB . Kc/ or
[CJc]
= CR . HRC . Kc/
= Coefficient depending on the surface hardness,
from Table 5.18,
HB or HRC
KcI
= Brinell or Rockwell hardness number , and
Life factor for surface strength, from Table 5.19.
=
The above relations are taken from data book, page no.8.16.
Calculate the centre distance between gears based
on surfacecompressive strength using the relation
(i+I)
a~
where
\V
(
Eeq [MIl
i\V
x
[CJc]
[from data book, page no. 8.13, Table 8]
= Qa = Width to centre distance ratio, from Table 5.21.
Unless otherwise stated, take \V = 0.3 for initial calculations.
7. Selection of number of teet" on pinion (z/) and gear (z~ :
(i) Number of teeth on pinion, zi : Assume zi ~ 17, say 18.
(ii) Number of teeth on gear, z2:
=
z2
i x zi
B. Calculation of module (in): Calculate the module using the relation
m
=
2a
(zi + z2)
Using the calculated module value, choose the nearest higher standard module from
TableS.S.
. 9. Revision of centre distance (a) : Using the chosen standard module, revise the centre
distance value (a).
m (zi
a
=
+ z2)
2
10. Clllculation of b, til' v and VIp:
~
Calculate face width (gear width) b:
~
Calculate the pitch diameter of the pinion d, : d,
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b = \V a.
=
.,
I
"
6. Calculation of centre distance (a):
0.74)2
,
(
m- z).
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I
I
)
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Design
of Transmission
Systell1S
5.62
./
Calculate the pitch line velocity v :
./
Calculate the value of 'tip : 'tip
=
b
d
1
•
11. Selection of quality of gears: Knowing the pitch line velocity (v) and consulting
Table 5.22, select a suitable quality of gear.
Table 5.22. Penpheral speed of gear (from data book, page
110.
8.3)
-
Straight bevel
IS quality
Preferred quality
Cylindrical gears
High precision
3&4
4
Above 15
upto 9
Precision
5&6
6
8 -15
upto6
8
1-8
upto 3
. 7,8,&9
Medium
Coarse
10 & 12
upto 1
10, 12
gears
upto 2
12. Revision of.design torque I Mil:
v" Revise K: Using the calculated value of'Vp'
revise the value of load concentration
factor (K) from Table 5.11.
v" Revise Kd:
Using the selected quality of gear and calculated
pitch line velocity,
revise the value of dynamic load factor (Kd) from Table 5.12.
,/ Revise I Mil: Using the revised values of K and Kd, calculate the revised design
torque [Mt]
value. Use [Mt]
= M, . K . Kd.
13. Checkfor hending :
./
Calculate the induced bending stress using the relation
0b
./
=
a~~b?Y
[M/]
Compare the induced bending stress
the value of [ 0b
],
refer step 5. If
0b
0b
[from data book, page no. 8.13]
and the design bending stress [
s [0b l. then
0b ].
For
the design is satisfactory.
14. C~,eckfor wear strengtlt :
./
Calculate the induced contact stress
°c
./
=
i±l
0.74 -;;-
A/i±l
\j ib
],
using the relation
Eeq [Mt]
Compare the induced contact stress
value of [ 0c
0c
refer step 5. If crc s [
0c
[from data book, page no. 8.13]
and the design contact stress [
0c ],
0c ].
then the design is safe and satisfactory·
L
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For the
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5.63
pinion.
./'
If the materials for pinion and gear are different, design the pinion first and check both
pinion and gear.
/
16. C!teck/or gear:
(i) C!teekfor bending:
./
Calculate the induced bending stress using the relation
or
where crbl and
O"b2
YI and'Y2
=
=
bending stress in the pinion and gear respectively,
Form factors of pinion and gear respectively,
./
Calculate the design bending stress for gear
./
Compare the induced bending
If crb2s [(Jh2
(ii)
= Induced
O"b2
],
(Jb2
[(Jb2],
and
t
from Table 5.13.
I
Table 5.16.
consulting
and the design bending stress [
I
(Jb2 ].
,
then the design is satisfactory.
Checkfor wear strength :
Calculate the induced contact stress O"c2 for gear using the equation (5.26). In fact,
./
the induced contact stress will be same for pinion and wheel.
./
Calculate the design contact stress for gear [ (Jc2 ] as discussed
./
Compare the induced
bending
stress
O"c2
i.e.,
O"c2
= (Jc'
I
in step 5.
and the design bending
stress [ (Jc2 ]. If
i
cre2:s; [ (Je2 ], then the design is safe and satisfactory.
I)
,17.
Calculation of basic dimensions
of pinion
anti gear:
Calculate
dImensionsof pinion and gear using the relations listed in Table 5.10.
[§xample 5.17] In a spur gear drive for a stone
el
;te . Tlte pinion is transmitting 30 kW {It 1200 r.p.m.
lOursper d.ay, SIX
',1
•
D'estgn
u(IYS a week and for 3years.
Given Data:
TOjilld·
©SOlul'
lher f
e Ore
.
Pinion and gear materials:
I
all the basic
,,
I(
crusher, tile gears {Ire made of C40
III
Tile gear ratio is 3. Gear is to work 8
t l.te {,lri
rtve.
C 40 steel; P
= 30
kW; N I = 1200 r.p.m.; i
I
= 3.
t
0 .
!
I
eSlgn the spur gear drive.
.
II
SInce the pinion and gear are made of same material
iave to do the design (}f pinion alone.
1011:
We J
(i.e., C 40 steel),
fl
,~r
r
,
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5.64
Design of Transmission
1. Gear ratio:
j
= 3.
SYSletns
... (Given)
2. Material selection:
Pinion and gear are made of C 40 steel.
'" (Given)
Assume surface hardness> 350.
3. Gearlife: Given that the gear i.s to work 8 hours per day, six days a week; and for 3
years. Therefore gear life in terms of hours is given by
Gear life
=
Life in numbers of cycles, N
=
8 x (52 x 6) x 3
449280 x
=
7488 hours = 449280 min
N I = 449280
x 1200
= 53.9 x
107 cycles.
4. Calclilation of initial design torque I MIl:
Design torque, [M,]
K . Kd .
=
MI"
where M,
=
60 x P
, 60 x 30 x 103
= 238.73 N-m
21t NI = 21t x 1200
x,
=
1.3.
=
238.73 x 1.3 == 310.34 N-m
K·
[ M,]
5. Calculatioll of Eeq, I ubI and I uel :
(i) To find Eeq : From Table 5.20, for C 40 steel,
(ii) To find
I ub I:
The design bending stress [ a b
[ab
where
]
=
=
Kb/
n
=
1.4
Eeq
]
=
2.15 x 105 N/mm2
is given by
Kb/
n . Ka
. a_I, assuming rotation in one direction only' .
0.7 for HB > 350 and N ;:::25 x 107, from Table 5.14,
2, for steel tempered, from Table 5.17,
Ka == 1.5, for steel, from Table 5.15, and
a-I ==
But
au ==
Q'
-I
0.35 all + 120, for C 40 alloy steel, from Table 5. ~6.
2
630 N/mm
= 0.35 x 630
from Table 5.3, for C 40 steel.
,
+ 120
== 340.5 N/mm2
Then,
==
1.4 x 0.7
2 x 1.5 x 340.5
==
111.23 N/mm2
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__ -----------------(iii) Tofind
I 0', J: The design contact stress [ 0c ] is given by
= CR' HRC . Kc(
[(Jc]
CR
where
_25~.6~5
26.5, for C 40 steel hardened and tempered, from Table 5.18, and
:c
= 40 to 55, for C 40 steel, from Table 5.18, and
Kcl = 0.585, for HB > 350 and N ~ 25 x 107, from Table 5.19.
HRC
[ CJ,]
26.5 x 55 x 0.585
=
=
852.64 N/mm2
6. Calculationof centre distance (a) :
( 0.74
We know that, a ~ (i + 1)
'V
where
=
[°c ]
Eeq [M/]
i'V
x
... (assumed initially)
0.3
3
( 0.74 )
852.64
a ~ (3 + 1)
..
r
2
x
2.15 x 105 x 310.34 x 103
3 x OJ
~ 152.89 mm or a = 155 mm.
7. Selection of z, and
Z2:
(i) Assume, zl = 17, for 20° full depth system.
(ii)
z2
= i .ZI
3 x 17 = 51.
=
8. Calculationof module (m) :
2(lSS)
_
(17 + 51) - 4.56 mm
=
We know that,
From Table 5.8, the nearest higher standard module, m = 5 mm.
9. Revision of centre distance:
m (zl + z2)
New centre distance, a =
10. Calculation of b, d l'
.f
Face width (b):
.f
P'rtch diameter
.
=
2
V
b ==
\jJ'
a
== OJ
170mm
0/
Pitch line velocity (v):
0/
\V:.E_
P
d
v
=
x 170 = 51 rnm
d I -- m . Z I
f pm
.. IOn (d)'I'
0
= ~85
=
and 'lip:
1t
l
S(l7+Sl}
2
dl NI
60
= 5 x 17
1t
=
= 85 mm
x 85 x 10-3 x 1200 == 5.34 mJs
60
-06
- .,
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c
Design"
. ol"Transmission~
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/ I S~/eclionof qualtty 0 g
locity 5 34 mis, IS quality g
ear:
. 8 ears are selected.
. Table 5.22, for pI•'tchof.line
ve OCI 1M. I:
rom
= 0.6 an d ror
~
12.
F RI!.IS'
. Oil 0'~ design 10rguI! of gear
.;
./
I
Revise
•
. K'
Revise
Kd: From
From Table 5.11, for 'l'p
8, HB > 350
Table 5.12, for IS qua rty ,
K,,= 1.4.
. 1M l : Design torque, [MIl
v"
,.
bearings close to gears, K
Re.lSe
=
=
I'
and
5.3
MI x K x K"
238.73 x 1.03 x 1.4 _
344.24 N-m
/1. Check for hending :
Db:
"" Calculation of illduced hending stress,
crh:::
Where Y = Fonn ,actor -',
c:
v::::
0366
-
x [MIl
(i+l)
a·m·b.y
I'
forz
from Table 5.13.
b
V
We find crb < I crb J
SlIIisfactory.
cr = 170)( 5)(
51I))( 0.366
(3 +
- )( 344.24 x 103 = 86.78 Nlmm2
ti.e.,
86. 78
Nlmm2 <: 111.23 Nlmm2). Therefore Ilze desigll i
14. Clleckfor Wear strengtl,:
"" Calculalion Of indllced COlllaclslress, 0;,:
i+l ~ !i+l
e
cr = 0.74 -;;-"
V ib x Eeq
3+I
170
Qnd Satisfactory.
e
= 0.74 x
::: 765.9 N1mm2
[Mil
3+I
;
J
3x 51 )( 2.15)( 10- )( 344.24 x 10
e
.; We find cr <: ( cr J (i.e., 765.9 Nlmm2 <: 152.64 N1mm2). Therefore Ilze design is safe
J S. .;Ctl/culation
pinion mid gear:
MOdule: Of hasic dilllensions
III = Of
5 mm
V
Face Width:
V
Height factor:
wi'
Bottom clearance:
V
Tooth depth :
.;
Pitch circle di3Jneter:
Refer Table 5. I O.
h :::::51mb)
=
fo
d,
e = 0.25 In = 0.25 )( 5 = 1.25 mm.
===
=
h
tl2
I, for full depth teeth.
=
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=
2.25 )( 5 = 11.25 Ill"I .
tn zi = 5 x 17 = 85 Illlll; and
'" z2 = 5 x 51 = 255 Illlll.
2.25 m
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Spur Gears
~
5.67
Tip diameter:
dOl = (Zl + 2 /0) m = (17 + 2 x I) 5 = 95 mm ; and
do2 = (z2 + 2/ ) m = (51 + 2 x 1) 5
0
~
Root diameter :
dfJ
= (z I - 2 /0)
= 72.5 mm;
m ._ 2 c
265 mm.
(17 - 2 x I) 5 - 2 x 1.25
and
df2 .= (z2-2/0)m-2c
=
=
=
= (51-2x
1)5-2x.1.25'
242.5 mm
I
(LYole In the above problem, both gears are made of same material. Hence design of pinion alone
is sufficient. So, we need not to check for gear.
IA
[Example 5.18
cast steel 24 teeth spur pinion operating at 1150 r.p.m. transmit
3 kW to a cast steel spur wheel. The gear ratio is 2.25. The tooth profile is 200 full depth
involute. Design the gears and check for stresses.
Given Data:
Tofind:
ZI
= 24;
NI
=
I 150 rpm; p = 3 kW; i = 2.25;
~ = 200•
Design the gears and check for stresses.
Since the pinion and gear are made of same material, (i.e., cast steel),
tfierefore we have to do the design of pinion alone. Assuming the gear life, say 20000 hrs, the
given gears are designed in the same manner as solved in Example 5.17.
@) Solution:
I Example
I Design
5.19
a spur gear drive to transmit 22.5 kW at 900 r.p.m. Speed
reduction is 2.5. Materials for pinion anti wheel are C 15 steel and cast iron grade 30
respectively. Take pressure angle of 200 and working life of the gears as 10000 hrs.
Given Data: P = 22.5 kW; N I = 900 r.p.m.; i = 2.5; ~ = 200; N = 10000' hrs.
To find:
Design a spur gear.
@)Solution : Since the materials for pinion and wheel are different, therefo_re we have
to design the pinion first and check both pinion' and wheel.
... (Given)
i = 2.5
1. Gear ratio:
2. Material selection:
Pinion:
CI5 steel, case hardened to 55 RC and core hardness'< 350, and
Wheel:
C.1. grade 30.
3. Gear life:
N
... (Given
10000 hrs
=
Gear life in terms of number of cycles
N
4. Design torque
... (Given)
=
IS
given by
10000 x 60 x 900
=
54 x 107 cycles
I Mil:
[M,]
=
M,·
K . Kd
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__ ----------------_::D:...::e:::.s::.!::ig::....n-o~if-T.-,.a-n-s-.m-is-.s-io_n-..:S~i)'sle",
---..:!.
:
5.68
--
3
=
M,
where
= 238.73
N-m, and
= 1.3
K . Kd
... (assume)
= 238.73 x 1.3 = 310.35 N-m
Design torque, [ M,]
..
60 x P = 60 x 22.5 x 10
21t x 900
21tNI
5. ell/euilltion of E«/ ' { 0"6I and { OJ I :
(i) To jilld E : From Table 5.20, for pinion steel and cast Iron (> 280 N/mm2),
«/52
equivalent Young's modulus,
(ii) Tojint/{
Eeq
= 1.7 x 10 N/mm .
O"bl: The design bending stress [ ab] is given by
[ ab]
1.4 x Kb/
n . Ka
=
x a_I'
..
assuming rotation
s 350)
111
one
di
rrectron only -.
and N ~ ]07, Kb/ = 1.
-/
From Table 5. ]4, for steel (HB
-/
From Table 5.17, for steel case hardened, factor of safety n = 2.
-/
From Table 5.15, for steel case hardened, stress concentration
-/
From Table 5.16, for forged steel, a_I = 0.25 (all + ay) + 50.
= 490 N/mm2
factor, Ka = 1.2.
But from Table 5.3, for CIS,
au
:.
= 0.25 (490 + 240) + 50 = 232.5 N/mm2
a_I
Then,
(iii) To jind
=
I D"cl:
1.4 x 1
2 x 1.2 x 232.5
=
135.625 N/mm2
The design contact stress [ ac ] is given by
[ac]
where
and ay = 240 N/mm2.
CR' HRC . Kc/
=
CR = 22, for C 15 case hardened steel, from Table 5.18,'
HRC = 55 to 63, for C 15 steel, from Table 5.] 8, and
=
Kef
0.585, for HB > 350, N;?: 25 x 107, from Table 5.19.
[ac J = 22 x 63 x 0.585
= 810.81 N/mm2
6. Ca/eulation of centre distance (a) :
We know that,
where
(i+l)
a;?:
b
\jJ
= a
== 0.3
a ~ (2.5 + I)
~
... (assumed
0.74 ) 2
1.7 x 105 x 310.35 x' 103
( 81 0.81
x ---2-.5......:x::....:0:...:.:..:.:3:...::._:...:_:..;:_
135.94 mrn or a -- 136 mm.
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initially)
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~~~
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__ ------------________________________________~
7. Tofind
%/
lind
__
9
Z2 :
I
i
(i) For 20° full depth system, select zl = 18.
..
(ii) z2= i x z, = ~.5 x 18=45
8. Calculation of module (m) :
2 x 136
= 18 + 45 = 4.32 mm
We know that,
I
II
I
I
From Table 5.8, the nearest higher standard module, m = 5 mm.
I
11
9. Revision of centre distance:
New centre distance, a = m (zl + z2) _ 5 (18 + 45) = 157.5 mm
2
2
V'P:
10. Calculation of b, dl, vand
./
Face width (b):
./
Pitch diameter of pinion (dl):
./
Pitch line velocity (v):
./
'Vp
=
b
d
b = 'V . a = 0.3 x 157.5 = 47.25 mm .
47.25
= ~
=
l
v
=
d,
=
1tdl Nl
60
= 5x
Ill' zl
=
1t X
18 = 90 mm,
90 xlO-3 x 900
60
= 4.24
m/s.
0.525.
11. Selection of quality of gear:
From Table 5.22, for v = 4.24 mIs, IS quality 8 gears are selected.
12. Revision of design torque
_
..
I Mt I :
= 0.525,
./
Revise K: From Table 5.] 1, for 'Vp
./
Revise Kd: From Table 5.12, for IS quality 8 and v = 4.24 mIs, Kd = 1.4.
./
Revise
I Mtl:.
[M/]
K ~ 1.03.
= MI' K . Kd
=
238.73 x 1.03 x 1.4 = 344.24 N-m
13. Check for bending:
-/ Calculation of induced ben~iig stress,
O'b :.
(i + I)
(Jb
Where
y
'.
(Jb
'" We find
CJb
= a. m . b . y
[Mt]
=
=
Form factor
0.377, for zi
=
(2.5 + 1) x 344.24 x 10
157.5 x 5 x 47.25 x 0.377
3
=
18, from Table 5.13.
= 85:89 Nzrnm"
< ( OR). Therefore the design is satisfactory .
.....
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----------------------------------_D~e~s./~g~n~O~if~TJ~r~a_n_s_m_i_ss_i_o_n_S~~~
~.'70
14. Check/or wearstrength:
./ Calculation0/ induced contact st_re_s_s,_O"c-:
_--i+1
0.74 --;-
CJc =
../ We find
=
2.5 + I )
0.74 ( 157.5
=
684.76 N/mm2
< [oc]'
CJ
c
i+1
Tb_X_E_eq_[_M_,_]~
(2.5
+ 1 ) x 1.7 x 105 x 344.24 x 103
2.5 x 47.25
Therefore the design is safe and satisfactory.
15. Checkfor wheel:
(i) Calculationof I ublwheel and I uclwheel
Wheel material:
:
CI grade 30.
NI
I
Wheel speed :
Life of wheel =
To find
-------
I ublwheeJ
=
900
2.5
=
360 r.p.m.
10000 hrs = 10,000 x 60 x 360
= 21.6'
x 107 cycles
: The design bending stress for wheel is given by
=
[ CJb ]wheel
1.4 x KbI
n .K
x
CJ_I'
assuming rotation in one direction
only.
c
\]"W ~
./
From Table 5.14, for cast iron wheel,
./
From Table 5.17, for cast iron, n = 2 .
./
From Table 5. I 5, for cast iron, Kcr = 1.2.
../
From Table 5.16, for cast iron, 0_1 = 0.45 aU"
Kb/
~
107
2l.6 x 107
= 0.918 .
But from Table 5.3, for cast iron, au = 290 Nzrnrn-'.
a_I
Then,
To find
I Uc lwheel
C
B
=
0.45 x 290 = 130.5 N/mm2
=
1.4 x 0.918
2 x 1.2
x 130.5 = 69.88 N/mm2
: The design contact stress for wheel' IS given
.
by
=
[ CJc ]wheel
where
=
23
.,
CB· HB . Kc/
for cast iron grade 30, from Table 5.18,
HB = 200 to 260, for cast iron, from Table 5.18, and
Kd
~
=
:.[cr]
c wheel
-\fW =
2.3
X
107
21.6xl07
= 0.879, for cast iron, from Table 5.19.
260 x 0.879 = 525.64 N/mm2
6
\
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Spur Gears
5.71
(ii) Check for bending:
./
Calculation of induced bending stress for wheel O"b2 :
where
O"hl x YI
= O"h2 X Y2
and O"h2
=
O"hl
Induced bending stresses in the pinion and
wheel respectively, and
YI and Y2
From Table 5.13,
= Form factors for pinion and wheel respectively.
Y2
O"hl
85.89 x 0.377
or
../
O"h2
=
=
=
=
0.471, for z2
= 45.
85.89 N/mm2 and YI
O"b2 x
= 0.377
... (already calculated)
0.471
68.75 Nzmm?
We find Gh2 < [ CJb JWhc:e1'
Therefore the design is satisfactory.
(iii) Check for wear strength: Since contact areais same, therefore
= 684.76
CJ
e
wheel
= CJe pinion
Nzrnm-, Here CJe wheel> [ CJe ]wheel' It means, wheel does not have the required
wear resistance. So, in order to decrease the induced contact stress, increase the face width
(b) value 'or in order to increase the design contact stress, increase the surface hardness, say to
340 HB. Increasing the surface hardness will give [ CJe ] = 2.3 x 340 x 0.879 = 687.34
N/mm2, Now we find CJc < [CJc J. SO the design is safe and satisfactory.
16. Calculation of basic dimensions of pin ion and wheel : Refer Table 5.10.
m = 5 mm
./
Module:
./
Face width:
./
Height factor:
./
Bottom clearance:
./
Tooth depth:
./
Pitch circle diameter:
b = 47.25 mm
/0=
1 for full depth teeth,
c = 0.25 m
h = 2.25 m
=
d,
d2
./
Tip diameter:
dal
Root diameter:
x 5 = 1.25 mm •
2.25 x 5 = 11.25 mm .
= m- z, = 5 x 18 = 90 mm; and
= m . z2 = 5 x 45 = 225 mm,
= (z I + 2 f 0) m = (18 + 2 x l) 5 = 100 mm
+ 2 f 0) m
= (45 + 2 xl)
(z2
dfl
=
=
(18 - 2 x l ) 5 - 2 x 1.25
df2
=
=
=
(45 - 2 x 1) 5 - 2 x 1.25
da2
./
= 0.25
______
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; and
5 = 235 mm
(zl-2/0)m-2c
(z2 -
= 77.5
mm; and
2 /0) m - 2 c
~
__
= 212.5
,~,.·T-~_~_.
mm
_
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5.72
~
5.31. CHECK FOR PLASTIC DEFORMATION
It is understood that during starting, braking or sudden stopping, instantaneous
act on the teeth. This instantaneous overloads (or instantaneous
plastic defonnation
maximum torque) may cau
of tooth or crushing of tooth. Therefore
should be done to prevent plastic defonnation
overloa ds
check for plastic deformat;
se
on
of the tooth surface.
(1)Check for bending:
Let
= Induced bending stress due to maximum instantaneous
o b max
torque,
di
(i±l)
Idnuce db en mgstress=ambY[Mt],
0b=
·
I . .
60 x P
M t = N omma pimon torque = 2 1t N '
[Mt ]max
=
=
Maximu.!D instantaneous torque, and
=
Permissible bending stress, from Table 5.23.
=
°b
°bmax
Compare
° b max
0b
2 Mt,
if [M, ]max is not given.
[Mt ]max
Mt
... (5.28)
max value with permissible bending stress
s [ o b ]max
[
0b
] max given by Table.5.23. If
' then the design is satisfactory.
Table 5.23. Permissible bending stress I OJb J max (firom cata
I book, page no. 8.21i
Core Hardness
Material
lOb Imax ' N/mml
No heat treatment
Steel
0.8 cry
HB < 350
Heat treated
0.36
Steel
HB> 350
-
Cast Iron
cU I K (J
0.60u
(ii) Check for wear strength :
Let
°c max
=
°
c
Induced contact stress due to maxi
.
axrmum instantaneous
= Induced contact stress, and
= 0.74 i ± 1 ""
a
= p
I i i±b1
\j
x Eeq [M t ]
..
ermissible contact stress, from Table 5.24.
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torque
'
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5.73
Spur Gea,.,5
Table 5.24. Permissible contact stress
Material
I
cre Inrax (from data book, page
Steel
HB
Steel
HB > 350
42 HRC
Cast Iron
HB s 350
1.8 all
s 350
[ M,
8.21)
I 0e Imax • N/mm2
Surface Hardness
3.1 0y
]ma.x
M
°c max = °c
110.
... (5.29)
,
If 0c max <
[Oc ]maJP
I Example
5.20 , It is desired to determine the proportions of a spur gear drive to
then the design is safe and satisfactory.
transmit 8 kW from a shaft rotating lit 1200 r.p.m. to a low speed shaft, with a reductionof
3: 1. Assume that the teeth are 200 full depth involute, with 24 teeth 011 tll.e pinion. The
pinion is to be of 4i}C 8 normalized steel and gear of 30 C 8 normalized steel. Assume that
the starting torque is 130% of the rated torque.
"Given Data:
P = 8 kW; N, = 1200 r.p.m. ; i = 3 ; ~ = 20° ; = 24 ;
z,
=
Starting torque
1.3 x rated torque.
Tofind : Design a spur gear.
@) Solution:
1. Gear ratio:
i
=3
... (Given)
2. Material selection:
3. Gear life:
Pinion
=
40 C 8 normalized steel; and
Gear
=
30 C 8 normalized steel.
... (Given)
Assume 20,000 hours.
:. N = 20000 x 60 x 1200
4. Design torque
I Md:
=
144 x 107 cycles
[ Mt]
where
K·
x K x Kd
60 x 8 x 103
21t x 1200 = 63.66 N-m, and
M,
=
60 x P
21t N,
x,
=
1.3
=
63.66 x 1.3
[Mt]
5. Calculation of Eeq,
= M,
=
....
= 82.76
(Assume)
N-m
I UbI ami I uel:
(i) Tofind Eeq : From Ta~le 5.20, Eeq
= 2.15
2
x 105 N/mm for steel.
1.4 x Kb/
(ii) To jimll O'b I
:
Design bending stress, [M,]
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Design of Transmission
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,/
=
Kbl
s 350
1, for steel HB
and N ~ 107, from Table 5.14,
,/
n = 2, for .steel normalized, from Table 5.17,
,/
Ka
=
1.5, for steel normalized, from Table 5.15,
,/
a_I
=
0.35 au + 120, for alloy steel, from Table 5.16
=
[ab
372 N/mm2.
=
0.35 x 720 + 120
= 2 x 1.5 x 372 = 173.6 N/mm
I ~/: Design contact
(iii) Tofllld
Where
".
l.4xl
1
stress, [a
c
J=C
['.'
au
=
720 N/rnrn2]
2
x HB x KcI
n
Co = 2.5, for alloy steel nonna/ized, from Table 5.18,
HB
s
350, from Table 5.18, and
1<",
=
I, for steel, HB $' 350 and N ? 107, from Table 5.19.
[ ac
1
== 2.5 x 300 x I == 750 N/mm1
6. Cent,.e distance (a): Assume 4J == 0.3.
a ~ (i + 1) "'\J
(JUi_)
/
V [a
? (3 + I)
x E q [M
1
I
I
1
'JI
2
107.2 or a == 110 mrn
~
7. Given that z,
1
V(~;n
c
:. z2 = i z, = 3
= 24.
24 = 72.
8. MOdule (III): ", == _2 a
_ ~ x 110
z, + zl - 24 + 72
::: 2.29 mill
From Table 5.8, the nearest higher Standard m
9. Revised Cent"edistance: a _ __"'_(._;._·I_~
10. Calculatio"
0"
b d
..,
,/
",.
'"
,
d
,VQIf
f11::::
_.
mill.
2
'11.
P'p.
I 0 mrn
l
.t'
dule,
Face width (h)· h.
- 4J x a == 0.] x 120 :::
Pitch diame''''·
f . .
36 mOl
~ 0 PinIon (d,):
d ==
,
",. ZI == 2.5
24p.Itch line velOCity ( ) .
- 60 mOl .
7[ d N I
V
IjJP
..t. -- 60
~
= d
l
v=6lJ!
=~
=
60
::::0.6
3 77 m/s
II. QlUllity Of Ie.,:
22 t4
From Table 5
.
, Or
v ::::J. 77 mls
IS quality 8
J
r
n
ears are se lecr'ed.
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SY.S"
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Spur Gears
5.75
1M,I:
12. Revised design torque
From Table 5.11, for
\jI
== 0.6, K == 1.03.
From Table 5.12, for IS quality 8 HB
,
[M/]
== M I ·K·K
S;
350 and v == 3.77 mis, Kd== ] .55.
d
63.66 x 1.03 x 1.55 == ]01.63 N-m
13. Check for bending:
==
Induced bending stress,
CJb
==
i+1
a- m . b . y [M/]
y
=
0.414, for
CJb
=
(3 + 1)
.
120 x 2.5 x 36 x 0.4]4 x ]01.63 x ]03
where
..
We find
< [ CJb
CJb
].
.
== 24, from Table 5.13.
Zt
=
90.9 N/mm2
Thus the design is satisfactory.
14. Check for wear strength : Induced contact stress is given by
CJc
We find
CJc
i+]
~i+]
=
0.74 - a
=
1
0.74(31;0 )
=
701.71 N/mm2
--:--b
I
x Eeq [M]I
(::3]6)X2.15X]05XI01.63XI03
< [ CJc ), thus the design is safe anti satisfactory.
15. Check for plastic deformation :
M,
=
Rated torque
=
... (already calculated)
63.66 N-m
Given that starting torque is 130% of rated torque.
..
[M,]max
=
Maximum instantaneous torque
=
1.3 x 63.66
(i) Check for bending:
= 1.3 x M,
= 82.758 N-m
Induced bending stress due to maximum instantaneous torque is
I given by
CJb
max =
CJb
[M/]max
=
M,
82.758
90.9 x 63.66
- 11817NI
2
.
mm
[.:
CJb =
90.9 N/mm2]
From Table 5.23, for steel HB ~ 350, permissible bending stress is given by
[ CJb
Since
]max
=
2
0.8 cry = 0.8 x 540 = 432 N/mm
< [IT)
Vb
mllx'
the design is satisfactory.
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[.:
CJy
= 540 N/mm2]
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Design of Transmission Systems
5.76
(ii) Check for wear strength: Induced contact stress due to maximum instantaneou
torque is given by
S
[ M, ]max
O'c
max
=
O'c X
=
82.758
701.71 x 63.66
M,
=
912.22 N/mm
s 350, permissible
x540 = 1674N/mm2
From Table 5.24, for steel HB
=
[O'c]max
Since
O'c
max
=
3.10'y
<[
O'c
3.1
]max '
2
contact stress is given by
the design is safe and satisfactory against plastic de/ormation
also.
16. Basic dimensions of pinion and gear: Refer Table 5.10 .
m = 2.5 mm
.,/
Module:
.,/
Face width:
.,/
Height factor:
.,/
= 0.25 m = 0.25 x 2.5 = 0.625 mm
Tooth depth:
h = 2.25 m = 2.25 x 2.5 = 5.625 mm
Pitch circle diameter:
dl = m- zi = 2.5 x 24 = 60 mm ;
d2 = m- z2 = 2.5 x 72 = 180 mm.
.,/
.,/
.,/
.,/
b = 36 mm
/0
Bottom clearance:
=1
c
Tip diameter:
Root diameter:
(zi + 2 f 0) m
=
=
(24 + 2 xl)
2.5
= 65 mm
(72 + 2 xl)
2.5
= 185 mm.
do2
=
=
(z2 + 2
dfl
=
(zi - 2 f 0) m - 2 c
(24 - 2 xl)
df2
=
=
dOl
(z2 - 2
f 0) m
2.5 - 2 x 0.625
f 0) m -
and
2c
= (72
- 2
= 53.75 mm
~ and
; and
1) 2.S - 2 x 0.625
= 173.75 mm.
I Example
5.21
I Design
a spur gear drive to transmit 12 kW at 1440 r.p.m: with a
'peed ratio of 3. The starting torque is 50% more than the mean torque. Assume suitable
1If/IJerials.
Given Data:
Tofind:
P
=
12 kW;
NI
=
1440 r.p.m.;
i = 3;
[M, ]max
= 1.5 M,.
Design a spur gear.'
e Solution:
Material selection:
Take same materials for both pinion and gear. Assume case twdened
alloy steel 15 Ni2 Crl Mo 15 material.
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Spur Gears
5.77
Now the problem is very similar to the previous problem. So proceed this problem as
discussed in Example 5.20.
I
I,
Since the starting torque value is given, therefore the design should be checked for plastic
deformation without fail.
I!
I'
I
I'
I,
I
I,
RELATED TOPICS I PROBLEMS
ADDITIONAL
5.32. GEAR DESIGN FOR VARIABLE LOADING
,
In gears, the load (torque, stress or power) generally varies over period of operation. A
gear is generally subjected to higher load in the beginning of a day and then it oper utes at
various levels of lower loads. The number of cycles spent during high loads may be small.
However for designing the gears, the highest load should be considered. Fig.S.IS depicts the
variation of load as function of time.
When the load comprises of a maximum sustained torque MIl acting for (I hours at a mean
speed of N I and small sustained torques Ma, Mt3,
acting for (2' 13, .••..• hours, at
mean speeds ofN2, N3, ..•••.•.. , then the equivalent number of loading cycles is given as
60
where
Neq
= -3
Mtl
MIl
=
t1
L
3
M/i
Ii
X
Ni
... (5.30)
Maximum continuously acting torque.
J
t2
I
I
r
Mt1
f
I
~
I
r
.
Mt3
_t
Mt2
1
~
t3
1
Mt4
+
Time in hrs (or) No. of cycles
Fig. 5.15. Variation of load (torque, stress or power) asfunction of time
5.32.1. DeSign Procedure
While designing the gear for this case,..
. .
./ Use MIl to calculate the maximum induced bending and contact stresses. That IS m
equations (5.23) and (5.26), replace [ M/] by MIl·
./
Use Neq t0 ca lculate the life factors. That is replace N by
factors Kbf andKc/.
Neq
while calculating
life
Now proceed the design, as usual.
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Design of Transmission s,,'le
5;~.7~8
------------------------------~~------~~~~
( E.-cample5.11
I If
spur gear drive is used in a power-operated hoist. The required
service life is to be 95,750 hours. The mean utiliz~tion of the machine ti"'.e is 0.35. 'lire
driv« trallSmits maximum sustained torque Mil durmg 0.15 top; 0.6 Mil dUring 0.5 topa"d
0.1 Mil during 0.35 top where topis the total operating time, constituting 0.25 of the cYcle
time. If the speed of drive is 700 r.p.m., find the equivalent number of cycles at lire
maximum sustained load
N = 95,750
0.25 x cycle time; N 1 = N2
Given Data:
=
top
Tofind:
hours;
= N3 =
Mean utilization
700 r.p.m.
of machine
=
time
0.35
Equivalent number of cycles at the maximum sustained load (Neq)'
© Solution:
We know that when a gear carries varying loads,
60
=
Neq
3
3
~
(i)
Nt It···
Mli
MIl
Mil
=
Maximum sustained (i.e., continuously acting) torque.
Cycle time
=
Required service lift x Mean utilization of machine time
=
95,750 x 0.35
where
=
33512.5 hours
... (Given)
Operating time, lop = 0.25 x Cycle time
0.25 x 33512.5 = 8378.125 hours
=
In this case, equation (i) can be written as
.'
60
3
Neq = M3
3
3
N I . II + Ma . N2 . 12 + M/3 . N3 . 13 ]
[Mil'
II
where
Ma
=
0.6 Mil;
II
=
0.15 lop
12
= 0.5
lop
13 = 0.35 lop
NI = N2
60
..
Neq = -3
MIl
M/3 = 0.2 Mil
... (Given)
=
0.15 x 8378.125
=
0.5 x 8378.125 = 4189.06 hours;
=
0.35 x 8378.125
= N3 =
=
=
1256.72 hours;
2932.34 hours;
and
700 r.p.m.
[M:I x 700 x 1256.72
+ (0.6 M/I)3 x
700 x 4189.06 +
(0.2 Mil?
Neq = 9.177 x 107 cycles
or
I Example
5.23
I In a non-reversible
x 700 x 2932.34 ]
Ans."
type rolling mill drive, a gear is designed to run
20 hours per day, transmitting power in the following manner: (i) 3.5 kW for 4 seconds,
550 times a day, (ii) 1.2 kW for remaining day and (iii) 5.25 kW maximum (i.e., momentary
peak load) all at a constant speed of 50 r.p.m. Take the life of the gears is to be 8 years.
Determine the equivalent number of loading cycles.
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LiJllr Gears
5.79
Gb~n Data: PI - 3.5 kW;
life of gears = 8 years.
P2 = 1.2 kW.
P
'
_
,"cu -
5.25 kW ; NI =
N2
= Nmax = SO r.p.m.
;
Tofind: Equivalent number of Ioa d'IIlg cycles (N )
'
eqr:
SIo ution : We know that f
.
or varyltlg loads ' equrva
. Ient number of loading cycles,
60
e
N
eq
where
Mil
- M3
-
[
II
=
3
Mil'
60 x PI
=
2 7t NI
Ma =
60 x P2
2 7t NI
=
II .
N 1+ M3 . I
.
N ]
12
2
60 x 3.5 x I03
27t x 50
=
668.45 N-m
60 x 1.2 x 103
27t x 50
=
229.18 N-m
2
[Maximum torque]
=
4 seconds,
=
4 x 550 = 2200 sees I day = 602200
= 0 .61 Ilours / d ay
x 60
12
=
20 hours - 0.61 hours
NI
=
N2
Neq
=
60
(668.45)3
=
4174.35 cycles / day
II
=
550 times a day
=
19.39 hours / day
50 r.p.m.
[(668.45)3 x 0.61 x 50 + (229.18)3 x 19.39 x 50]
For a period of 8 years,
Neq
=
(8 x 365) days x 4174.34 cycles / day
=
12.2 x 106 cycles
INote I In order to design the gear drive for the above problems, one has to use the following tips:
./
Use maximum sustained torque,
Mil
= 668.45 N-m, to calculate the maximum induced
bending and contact stresses.
Use Neq = 12.2 x 106 cycles, to calculate life factors.
'"
From the given momentary peak load, P",ax = 5.25 kW, calculate the momentary peak
60 x 5.25 x 103 = 1002.68 N-m. Use this
. momentary peak torque
torque, M, max e =
27t x 50
M
to check for plastic deformation.
I
max
5.33. DESIGN OF GEARS WITH RELIABILITY FACTOR
The reliability
factor (K ) is used to adjust desired reliability levels. The reliability factor
R
is based on a statical analysis of failure data, as shown in Table-5.25.
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Design of Transmission SYstems
5.80 .
Table 5.25. Reliabilityfactor, KR
Reliability
KR
Fewer than one failure in 10
0.90
0.85
Fewer than one failure in 100
0.99
1.00
Fewerthan
0.999
1.25
0.9999
1.50
Requirements
of application
one failure in 1000 .
Fewer than one failure in 10000
where
KR
=
-
-
-
0.7 - 0.15 log (1 - R), for 0.9 ~ R < 0.99
= 0.5 - 0.25 log (1 - R), for 0.99 ~ R < 0.9999
5.33.1. Design Procedure
If the reliability requirements are given in the design problem, then modify the design
bending stress [ crb ] and design contact stress [ crc ] as below.
Design bending stress = [cr
b ]
I KR;
Design contact stress = [cr
c]
I KR.
and
... (5.31)
... (5.32)
Then proceed the design as usual, as discussed in Section 5.30.
I Example
5.24
I Assume
the data of Example 5.17, for the 90% reliability, design the
gear drive.
Given Data:
To find:
Reliability
=
90%
=
•
0.90.
Design the gear drive.
@) Solution:
Refer Example 5.17 .
./
The gear ratio, material selection, gear life, and design torque are same as in
Example 5.17 .
./
Calculation of
I O'b J and I O'c J: From Example 5.17,
[ crb]
=
111.23 N/mm
For the 90% reliability, reliability factor KR
and [ere]
=
Modified design bending stress [ cb]
=
I«e]
=
modified design contact stress,
=
852.64 N/mm2
0.85, from Table 5.25.
-
111.23
0.85
=
852.64 _
0.85 - 1003 N/mm
=
130.86 N/mm2,
and
2
Use this modified [ erb ] and [ ere ] values to calculate centre distance (a), and to
check for bending and wear strength .
./'
Then proceed the design in the same manner as discussed in Example 5.17.
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~5.~8~1
:
~~a~N-----------5.34. DESIGN OF INTERNAL GEARS
The design procedure of internal gears is same as th
t
I
..
e ex ema gears,
modifications 10 the design formulas .
./
i:
10
II .
I' ht
owing, rg
Use negative sign in the induced contact stress formula. Now equation (5.26)
becomes
c = 0 74
C
./
ith
WI
•
L=J.
.... Ii - 1
a \j i b
x Eeq [M I ]
... (5.33)
Similarly use negative sign in the induced bending stress formula also. Now
equation (5.23) becomes
i-I
crb
./
=
... (5.34)
a- m . b . Y [Ml]
Centre distance (a) formula becomes, a
=
(z2-Zt)m
2
... (5.35)
5.35. DESIGN OF NON-METALLIC GEARS
Generally non-metallic pinions are used with cast iron gear, where smooth and noiseless
operation are required. In the non-metallic pinion - cast iron gear pair, pinion is weaker than
thegear. Thus only pinion is designed and check for beam strength is carried out.
5.35.1. Design Procedure for Non-metallic Gears
./
The design procedure for non-metallic gears are as same as that for metallic gears .
./
The design bending stresses are given in Table 5.26.
Table 5.26. Design bending stress/or non-metallic materials (from (lata book, page no. 8.20)
Material
( Gb
I, N/mm2
Textolite
50
Vulcanised fibre gears
36
Plastics
60
Bakelite
56
lli'ot~ ./ The design bending stress [ crb ] for non-metallic materials can also be calculated by
using therelation [crb]
Where
a
u
= au / n.
= Ultimate strength of material, and
n = Factor of safety, 2.5 to 3.
0/
Readerscan refer Example 5.15, to clearly understand the design steps of non-metallic gear.
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~5.82
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----------------------------------D_e=sl~·gn:=o:if=Tl=~=a=ns=m=,=s~si~o~nSYSt --"!"_1tIs
--......
====~----------~R~EV~I~EW~A;NnDisOiUMMiMM,AAlR~Y'------------./
. . 0(the charpter" advantages
At the b egtnntng
presented
./
In spur gears, the teeth are straight and parallel to the axis of the wheel .
./
The terminology of gear tooth (i.e., gear nomenclature)
./
Law of gearing states that for obtaining a constant velocity ratio, at any instant of teeth
the common normal at each point of contact should always pass through a pitch point .
./
Forms of gear tooth profile:
./
Standard systems of gear tooth: (i) 14 VJ ° composite, OJ) 14 Y2°full depth involute
,
(iii) 20 °full depth involute, and (iv) 20 °stub involute system .
./
Gear materials: 1. Metallic materials - steel, cast iron, bronze, etc; and 2. Non-metallic
materials - wood, rawhide, compressed paper, bakelite, etc .
./
Gear manufacturing:
1. Gear milling; 2. Gear generation
shaping; and 3. Gear molding.
./
./
Gear tooth failure:
"
pttting
an d scoring.
limitations and classification
of gears
Q10e
are also discussed
(i) Involute, and OJ) Cycloidal tooth profile .
1. Tooth breakage; and 2. Surface failure
includes
hobbing and
which includes abrasi
lon,
For analysis of gear tooth:
(i)
Tangential component of tooth force,
p
F,
V
(ii) RadiaI component of tooth force, F;
Ij
=
= F, x tan ¢
(iii) Resultant tooth force, F = F, / cos ¢
~ Gear design using Lewis and Buckingham's
equation:
Beam strength of gear tooth (Lewis equation):
where
F,
=
m = module,
b
=
Face width
=
Allowable static stress, and
= 10 x module
,
Y = Lewis form factor.
o accountfior dyna . I ad'
mtc 0 mg, a velocity fi t
.
Buckin ham'.
ac or Cv IS used.
g
s equatIOn for dynamic load (F J :
1;
./
Fd
::
F +
I
--
where c - D.I'.
- eJormation factor.
21 v
(be
+ FJ
21 v + \} be + FI
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Spur Gears
5.83
where
./
2i
i+1
Q
=
Ratio factor
s;
=
Load stressfactor
- --
=
j2
Lr Ep1
sin t/J
es 1.4
=
1 ]
+ Eg
Factor of safety:
(F . shendillg
17 and face width:
8 m < b < 12 m.
./
zJ;?
.;
When both pinion and gear are made of the same material, then pinion should be
designed. When different materials are used, the product [ ab 1 .y decides the weaker
between pinion and gear. The element which has lower [ ab
designed .
.;
1x
y value should
be
2. Calculate
Fs"
1. Equate F, and initial Fd, to get module;
Design procedure:
3. Calculate accurate Fd;- 4. Check for beam strength (or tooth breakage) (F" 5 F..):
5. Calculate Fw;
6. Check for wear (Fd < F,j; and 7. Calculate basic dimensions
of
pinion and gear.
II. Gear design based on gear life:
[Mfl
./
Design torque:
./
Induced bending stress:
./
Design bending stress:
=
M, . K . Kd
i:t1• b .y [Ml ,
a . 111
ab =
[ ab 1
=
1.4 Kbl
n. K
x a_I'
for one direction of rotation
(1
=
./
Induced contact stress:
./
Design contact stress:
./
Centre distance:
x a_I' for two directions of rotation.
i+1
ac = 0.74 --;;[(J'cl
~ /i+1
\j ib
= CB x HB x KcJ
a ~ (i + 1)
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x Eeq
or [(J'cl
0.74)2
( [ (J'c1
=
[Mfl
CR x /-IRe x KcJ
x Eeq.[ M,l
I
IJI
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-t::
D_e_'7s_ign=--_;of:......_1J_ra_'_Js_m
SYstetn,s
5.S4~
~
Design procedure:
J. Select material,' 2. Calculate design torque; 3. Calculate desi
stress [ a" i and [ ac t. 4. Determine centre distance,' 5. Calculate module; 6. Checlc
bending (a"
r'
s [ a" l):
JO,.
7. Check for wear strength (o; S [ ac J); 8. Check for plastic
deformation (if required); and 9, Calculate basic dimensions of pinion and gear .
./' If the material is same for both pinion and gear, only the pinion is designed If th
materials are different, pinion is designed and checks are carried out for both Pinio;
and gear .
./'
At the end cf this chapter, the special topics such as gear design for variable loadin
gear design with reliability factors, design of internal gears, and design of non-melal/~
gears are 'also presented.
REVIEW QUESTIONS
I.
State the advantages and limitations of gear drive over belt and chain drives,
2.
Why is gear drive known as direct and positive drive?
3.
How can you classify gears? What different types of teeth are used in gears?
4.
Make a neat sketch of a gear and indicate the terminology used for it.
5.
Explain the following terms: (i) module, (ii) pressure angle, (iii) circular pitch, and
(iv) pitch point.
6.
State the law of gearing.
7,
Write an engineering brief about various forms of gear tooth profile.
8.
List out the gear materials used.
9.
Write short notes on gear manufacturing.
10. Explain the two modes of gear failures.
11. Deduce the expressions for tangential and radial components of tooth force in a spur
gear.
)2. Deduce the Lewis equation for beam strength.
f
) 3. What are the assumptions made in deriving Lewis equation?
14. Why is a gear tooth subjected to dynamic loading?
15. What is Lewis form factor?
16. Write the Buckingham's equation for dynamic load and expression for limiting wear
load.
17, Why is pinion made stronger than gear?
18. Discuss the design procedure of spur gears recommended
)9. How can you calculate the equivalent number ofloading
to variable loading?
20.
Explain the design procedure of internal gears.
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by AGMA.
cycles when gears are subjected
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~~~~
~5~.8~5~
PROBLEMS FORPRACnCE
Problemon basic dimensions of spur gear:
I. In a pair of spur gears, the number of teeth on the pinion and the gear are 20 and ') 20
respectively ..The mod~le is 4 mm. If the pressure angle is 20° full depth, calculate : (i)
the centre distance, (II) the pitch circle diameters of the pinion and gear, (iii) the
addendum and dedendum, (iv) the tooth thickness, (v) the bottom clearance, and (vi) the
gear ratio.
[Ans: (i) 280 mm; (ii) 80 mm and 480 rnrn; (iii) 4 mm and 5 mm; (iv) 6.2832 mm;
(v) 1 mm; and (vi) 6]
Problemson force analysis of spur gear:
2.
A pinion of 100 mm pitch diameter, running at 1000 r.p.m. transmits 6.25 kW of power
to gear whose pitch diameter is 300 mm. For straight tooth the angle of pressure is 20°.
Determine the tangential force, the transverse or bending force on shafts and torques on
driving and driven shafts.
[Ans: 1193.75 N; 434.43 N; 119.375 N-m; 358.125 N-m]
3. A pinion driving.a gear about 1110 of its speed transmits 20 kW of power at 2000 r.p.m.
The number of teeth for a full depth pinion of addendum equal to module should be such
that undercutting by a rack cutter is eliminated. The pressure angle is 20°. The sum of
teeth on pinion and gear is required to be 200 while module is 5 mm. Calculate:
(i) number of teeth on pinion and gear; (ii) actual gear ratio; (iii) pitch circle diameters of
pinion and gear; (iv) tangential force; (v) radial force; and (vi) torque on driving shaft.
[Ans: (i) 18 and 182; (ii) 10.111; (iii) 90 mm and 910 mm; (iv) 2120 N; (v) 772 N;
(vi) 95.5 N-m]
4.
The pitch circles of a train of spur gears are
shown in Fig.5.16. Gear A receives 3.5 kW
power at 700 r.p.m. through its shaft and rotates
in the clockwise direction. Gear B is the idler
gear while gear C is the driven gear. The number
of teeth on gears A, Band Care 30, 60 and 40
respectively, while the module is 5 mm.
Calculate: (i) the torque on each gear shaft; and
(ii) the components of gear tooth forces; (iii)
Draw a free-body diagram of forces and
determine the reaction on the idler gear shaft.
Assume the 20° involute system for the gears.
[Ans: (i) (M')A
= 47746.48
N-mm; (M,>S
Fig. 5.16.
= 0; (M,)c = 63661.98
N-mm; (ii) 636.62 N
and 231.71 N; (iii) 1228 N]
j
l
I
i.I·
t,
.j
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)
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~5.~86~
5.
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Design of Transmission Syste~
------------------------
__ ~~~~~~~~---
f two-stage gear box is
The la~ouFt.0 5 ~ 7 The number of teeth
shown m Ig..·
. Z = 20;
on gears are as fo]Jows.
1
z2 -- 50',z3
'an
Pinion 1 rotates at 1440 r.p.m. 10
anticlockwise direction when. observed
from the left side and transmits 10 kW
power t 0 the gear train '. The pressure
angle is 20°. Calculate:
(I) the
tangential and radial components of
tooth forces between gears 1 and 2, and
gears 3 and 4; and (ii) the resultant
reactions at bearing E and F. Also draw
a free body diagram of the gear tooth
c
175
200
•
100
Fig. 5.17.
·) F' - 132629 N' Fr
[Ans: CI
12 .
'12
(RE)H = 552.62 N; CRE)y
.1._
-------
forces.
6.
r'
= 20'' and z4 = 50.
= 603.41
= 482.73
N; (RF)H
N; F;4
=
3315.73 N; F;4 = 1206.83 N; (ii)
= 2542.06
5 N]
_
N; (RF)y - 1086.1
A planetary gear train is shown in Fig.5.18. The
sun gear A rotates in a clockwise direction and
transmits 5 kW power at 1440 r.p.m. to the gear
train. The number of teeth on the sungear A, the
planet gear B and the fixed ring gear Care 30,
60 and ISO respectively. The module is 4 mm
and the pressure angle is 20°. Calculate: (i) the
tangential and radial forces acting on each gear;
and (ii) the torque that the arm can deliver to its
output shaft. Also draw a free body diagram of
forces acting on each other.
Fig. 5./8.
[Ans: Ci) F:W = 552.62 N; F~
=
588 N; (ii) 198.95 N-mm]
Problems on spur gear design based on beam strength:
7.
Design the 20° full depth teeth of a pair of gears to transmit a smooth continuous load of
30 kW at 1200 r.p.m. of the 20-tooth pinion. The velocity ratio is 3 and the material is to
be steel. The surface of the teeth may be heat treated as needed for wear.
8.
Design a spur gear to connect an electric motor to a reciprocating
pump both bein.g
mounted on the same bed. Speed of the motor is 1440 r.p.m. Speed reduction desi~;
10 : 1. Motor power 35 kW. The gears are to have 200 pressure angle. Both the pml
and gear are made of steel with a maximum safe static stress of 180 N/mm2.
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Spur Gears
5.87
9.
A pair of 20° full depth involute tooth
r.p.m. of the pinion. The velocity r ti .spur gear is to transmit 30 kW at a speed of 250
a 10 IS4 . 1 The "
.
d
allowable static stress of 100 N/m 2 '.'
piruon ISrna e of cast steel having an
allowable static stress of 55 N/mm2 ;h' whll.e the gear is made of cast iron having
the velocity accounts for the dyn '. Ie service factor can be taken as 1.5. Assume that
arruc oad Design th
.
and suggest suitable surface hardn
f h.
e gears, specify their dimensions
ess or t e gears.
10. Design a gear drive to transmit 20 kW
distance between the shafts is 340
Th 1200 r..p.m. Velocity ratio is 3. The centre
cast steel and cast iron respectiv Immp' e materials for the pinion and the wheel are
. .
"
e y. ressure angle = 20° Th d .
c.
pmion material IS 84 N/mm2 a d tl
.
e esrgn stress lor the
n
ie surface endi
limi f
N/mm2. The modulus of elasticity of the . .
lra~ce. irmt or the gear pair is 60
ear is 1 x 105 NI
2
puuon material IS 2 x 105 Nzmm? and that of
g
mm.
II. A rawhide pinion is to transmit 30 kW at 115
.
is 4' 1 For 200 full de th .
I
0 r.p.m. to a cast Iron gear. Speed reduction
. .
p mvo ute teeth assumi
d'
h
..
.
gear drive.
'
109 me rum sock condition, deSIgn the
12. Design a two stage reverted spur
gear drive
(Fig.5.19)
for the
following data: Input power = 2.5
kW; Input speed = 1440 r.p.m.;
Overall reduction ratio = 9; Centre
distance between shafts = 300 mrn:,
All the gears are of alloy steel.
---~-T-'
L
300mm
2
----'"'-l.
3
Problemson spur gear design based on gear life:
13. Design a spur gear to transmit 2 kW at 1440 r.p.m. Desired speed ratio is 3. Use C45
steel for the gears.
14. Design a spur gear device for transmitting 50 kW from motor running at 1440 r.p.m. to a
machine running 360 r.p.m. Use C45 steel for the gears and specify the heat treatment, if
necessary. Design the gears.
15. Design a spur gear drive to transmit 10 kW at 1440 r.p.m. with speed ratio of 3. The
starting torque is 50% more than the mean torque. Assume suitable materials.
16. A steel pinion meshing with a cast iron gear transmits 7.5 kW. The pinion has a speed of
600 r.p.m. and the speed ratio is 3. Assuming 20° involute tooth profile, find the centre
distance between the shafts, module and exact number of teeth on the wheels. Check for
induced bending stress.
17. A 37.5 kW is transmitted at 450 r.p.m. to a shaft running at approximately 112 r.p.m.
through a spur gear drive. The load is steady and continuous. Design the gear drive and
check the design. Assume the following materials: Pinion - Heat treated cast steel; Gear
- High grade cast iron.
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~
'f_JL.<.:
tf.
.
~.'
"
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----
Design of Transmission Systems .
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.
5.88
.
. 28 kW at 900 r.p.rn. Speed reduction is 2.5. Material
.
.
'ear
drive
to
transmit
18. Design 8 SpUIg ,
I d
t .ron grade 30. Take pressure angle as 20°
for pinion and wheel are CIS stee an cas I
and working life of gears as 10,000 hours.
.
A motor shaft at 1440 r.p.m. is required to transmit 8 kW to.another shaft with a speed
19. reduction of 3 : 1. The starting torque may be taken as 40% hlghe.r than the ~e~n torque.
The gears are made of heat treated steel and the shaft of mild steel. PInion has a
minimum of 20 teeth. Design the gear drive.
Problems 011 gear design for variable loading :
20. The gear drive in a hoisting machine transmits maximum sustained torque M I during 0.2
lop; 0.4 Mil during 0.4 lop and 0.2 MIl during 0.4 lop where lop is the total operating time,
constituting OJ of the cycle time. The mean utilization of the machine time is 0.35. The
required service life is to be 100,000 hours. Find the equivalent number of cycles at the
maximum sustained load, speed being 750 r.p.m.
[Ails: 10.81 x 107 cycles]
21. In a non-reversible type rolling mill drive, a gear is designed to run 20 hours per day,
transmitting power in the following manner: (i) 2000 W: normally (ii) 5420 W for 4
seconds, 550 times a day, (iii) 7900 W maximum (momentary peak load) all at a
constant speed of 45 r.p.m. The life of the gears is to be 7 years. Determine the
equivalent number of cycles at the maximum sustained load. [Ans: 12.5 x 106 cycles]
Problem on gear desigll considering reliability factor:
22. A 3.5 k~ gas turbine running at 1400 r.p.m. will be used to drive the generator at 500
r.p.m. A life of SO hours with 99% reliability is required. Design the spur gear pair.
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·Helical Gears
"Sixty years ago I knew everything; now I know nothing;
education is a progressive discovery of our own ignorance ."
- Durant, Will
lcal gears are simple modification of
or mary spur gears.
e tea gear has tee In the
form of helix around the gear. The use of helical
gears is most common in automobiles, turbines and
high speed applications. A pair of helical gears is
shown diagramatically in Fig.6.1. It can be seen
that the teeth of the two wheels are of opposite
hand. The helixes may be. right handed on one
-
wheel an~~~ed
6.1.~es
on the other.
of Helical
~
Single-helical gear set.
~1.
Ge,J
/
There are three mam reasons why helical gears are preferred than spur gears. They are :
1. JYgise : Helical gears produce less noise than spur gears of. equivalent quality because
the tota\ contact ratio is increased. That is, at any time more than one pair of teeth (upto 10
\
pairs ofteeth) are in engagement. Since more teeth are in contact simultaneously, the load is
transferred gradually and uniformly as successive teeth come into engagement. Thus helical
gears operate more smoothly.
I
I
2. Load carrying capacity: Helical gears have a
greater ~d
carrying capacity than equivalent size
spur gears because the total length of the line of
contact is increased.
In spur gears the line of contact is parallel to the
axis of rotation, as shown in Fig.6.2(a). In this case,
the'total length of contact line is equal to the face
width. In helical gears the line of contact is diagonal
across the face of the tooth, as 'shown in Fig.6.2(b).
In this case, the total length of contact line is,greater
~an the face width. This lowers the unit loading and
Increases load carrying capacity. ,..
a...
_--b--_I
'd_/
T
Single line of
contact
~------------~
(a)
I-
Face width -_
Multiple
contact lines
(b)
Fig. 6.2.
__
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~~
~.2
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----------~D~e~S~ign~o~if_TJ_~_uns--m_I.~U~fu~n~Sy
~
J. Manufacturing:
A limited number of standard cutters are used to cut a wide V8tiel) Of
helicai'gears simply by varying the helix angle.
6.1.~advantage
.
Since the tee
thrust loads. ~
of Helical Gears
.
the ~is of rotation, helical gears are subjected to a\'i.1
are me me 0
.'.
~
~~ ~t
load can be eliminated by usmg Herrmgbone (I.e., doub~
th are i I' d t
I
helical) gears. ~
.
TYPES OF HELICAL GEARS
1. P,
, ~"'Ii
'ei helical gears :
./
They operate on two parallel shafts .
./
The magnitude of the helix angle is the same for the pinion and the gear.
./
They have opposite hand of the helix. i.e., a right hand pinion meshes with a left
hand gear and vice versa.
2. Crossed-helical (or spiral) gears :
_
\
~
,
./
The operate on two non-parallel shafts.
./
They have the same or opposite hand of the ~el~
KINEMATICS AND NOMENCLATURE
OF HELICAL
:~ \_
GEARS
.
The geometry of a helical gear is shown in Fig.6.3: There-are three ways-to view the/teeth
on the rack. They are :
1. Transverse direction: If the teeth are viewed in the direction of motion of the rack (as
shown in Section A-A of Fig.6.3), then the orientation is called as the transverse dirtdion.
When considered in the transverse plane (that is, a plane perpendicular to the axis of the
gear), all helical gear geometry is identical to that for spur gears.
2. Normal direction: If the teeth are viewed in a direction that is aligned with the slant of
the teeth (as shown in Section B-B of Fig.6.3), then the orientation is called as the nornuU
direction.
3. Axial direction: If the teeth are viewed in the direction of the axis of the rack or gear,
then the orientation is called as the axial direction.
Let
p =
Helix angle,
Pt = Transverse circular pitch,
Pn
Po
= Axial pitch,
Pd
=
Diametral pitch,
at and an
=
Transverse and normal pressure angles respectively,
-
Normal circular pitch,
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6.3
m,
and mn -
Transverse and normal modules respectively,
zl and Z2 = Number of teeth on pinion and gear respectively,
d1 and d2 = Pitch circle diameters of pinion and gear respectively,
N 1 and N2 = Speeds of pinion and gear respectively, and
a
Centre to centre distance between pinion and'gear,
=
The various terms used in the study of helical gears have been explained below.
1. Helix angle (or Spiral angle) (P): It
B
is the angle between the tooth axis and the
plane containing the wheel axis. The helix
angle varies from 1S° to 25°.
2. Lead: It is the distance advanced by
each tooth per revolution. measured along
the axis parallel to the axis.
3. Transverse circular pitch
(PJ : The
distance between corresponding points on
adjacent teeth measured
in a plane
perpendicular to the shaft axis is known as
transverse circular pitch .
=
Pt
1t d 1
1t.
ml
= -. zl
.....-, .. :
., p~
=. ~.-
cos ,...'
...(6.1). .
b
4. Normal circular pitch (p,J: The
distance between corresponding points on
adjacent teeth measured
in a plane
perpendicular to helix is known as normal
(a)
Basic rack
circular pitch.
Pn = P I
cos
x
5. Axial pitch
A
I-'
=
n-.. mn,
... (6.2)
(paJ: The distance
(b)
between corresponding points on adjacent
teeth measured in a plane parallel to the
shaft axis is known as axial pitch.
p
a
= - PI
tan
J3
= - Pn
sin
J3
=
1t • mn
. A
stu I-'
•••
3)
(6.
(e)
Fig. 6.3. Helical gear geometry
(a) Gear, (b) Section AA (transverse plane),
(c) Section BB (normal plane).
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Design of Transmission Sy".".,
~
6. Normal diametral pitel. (p,J: The reciprocal. of the normal module is ,knoWnas
normal dlametral pilch.
'" (6.4)
7. Transverse pressure angle
(aJ : The
pressure angle measured in transverse plane i.e.,
along the plane A-A is known as transverse pressure angle, ar
B. Normal pressure angle (a,.) : The pressure angle measured in normal plane i.e., along
the plane B-B, is known as normal pressure angle,
The angles
13, at and an are related
an'
by the equation
an
tanat
tan
=
cos J3
... (6.5)
9. Pitch circle diameters of the pinion and the gear:
./
Pitch circle diameter of the pinion, d t
./
Pitch circle diameter of the gear,
d2
=
J3) Z I
cos J3) z2
(Inn / cos
= (mn
/
10. Centre distance (a) : The centre to centre distance between gears is given by the
relation
... (6.6)
'H, Speed ratio (i) :
... (6.7)
6.4. VIRTUAL OR FORMATIVE NUMBER OF TEETH (Zeq)
In the design of helical gears, an imaginary (virtual) spur gear is considered in normal
plane (i.e., plane A-A). This spur gear is called a 'virtual' or 'formative' spur gear. The
number of teeth on the virtual spur gear in the normal plane is known as virtual or formative
or equivalent number of teeth. It is given by
\
... (6.8)
where
z
=
J3 -
Actual number of teeth on a helical gear, and
lfIelix angle.
This virtual number of teeth value i~.useful during both designing and manufacturing (in
cutter selection) of helical teeth.
6.5. FACE WIDTH OF HELICAL GE~RS (b)
For smooth and continuous operation of the helical gears, it is recommended that the face
width is greater th~ the axial pitch.
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Helical
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i.e.,
b ~
In practice,
b = 10 mn
::=
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1t.
Pa or b ~
.
mn
sm
6.5
... (6.9)
A
p
3 Pn is generally used in the initial stages of the design.
6.6. TOOTH PROPORTIONS FOR HELICAL GEARS
There are no standard proportions for helical gears. The proportions recommended by
American Gear Manufacturer's Association (AGMA) are as follows:
../
Normal Pressure angle (a.n)
-
15° to 250
./
Helix angle (J3)
-
8° to 25°, for helical
- 25° to 40° for herringbone
../
Addendum, maximum
-
../
Dedendum, minimum
- mn
../
Tooth depth
-
2.25 mn
../
Minimum clearance
-
0.2mn
../
Thickness of tooth
-
1.5708 mn
o.s »,
6.7. BASIC DIMENSIONS OF HELICAL AND HERRINGBONE GEARS
All the basic dimensions of helical and herringbone gears are listed in Table 6.1.
Table 6.1. Basic dimensions of heuca! gears (from data book, page no. 8.22)
S.No.
Nomenclature
Notation
Units
I.
Normal module
mn
mm
2.
Transverse module
ml
mm
3.
Centre distance
a
mm
4.
Height factor
10
-
5.
Bottom clearance
c
mm
-
Formula
=
(
ml
=
mn
-cos P
a
=
(.».)
x (Zl.+ z2)
cos p
2
7.
mm
h
Tooth depth
diameter
mm
d
-
-
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by CamScanner
from intsig.com
From : www.EasyEngineering.net
10
0.8, for stub teeth
1;
c
0.25 mn
h
0.3 mn, for stub teeth
2.25 mn
=
Pitch circle diameter or reference
2a
zl +z2
=
10
=
6.
2a ) x cos
zl + z2
J3 s:::-dZ
mn
d1
=
d2
=
1.9 mn,
mn
cos
mn
--x
cos
x zi ; and
p
.
P
for stub teeth
zi.
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f
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Design
6.6
S.No.
8.
Notation
Nomenclature
(Addendum
9.
Units
mm
Tip diameter
mm
(~J} +2/0) m
(c::f3 + 2/0)
dal
-
d a2
=
dfl
=
dj2
=
(C::f3-2/o)mn-2c
P
=
=
8° to 25°, for helical
n;
degrees
Helix angle
(c::
II.
Number of teeth
12.
Virtual number of teeth -
f3
2/0 ) mn-2c;:
+ z2)
Z
'7
-v
PI
Transverse circular pitch
= --
mm
ZI
z2
=-v2 cos3 ~
. and
'7
f3 '
dl
PI
=
'It·m
Pn
=
Plxcosf3='It·mn'
Pa
PI
Pn
'It. mn
= tan f3 = sin f3 = sin ~
I
=1tx-
z]
=
I
14.
Normal circular pitch
Pn
mm
15.
Axial pitch
Pa
mm
I Example
-
2a
=
cos3
13.
p-
25° to 40°, for herringbone
mn (zi
cos
IJId----
mn·
(Dedendum circle diameter)
10.
SYlleltl,
Formula
circle diameter)
Root diameter
Transmission
0/
Pn Icosp
'
I A pair of helical gears consists of a 25 teeth
pinion meshing will, 50
teeth gear. TI,e normal module is 4 mm: Find the required value of the helix angle if tilt
centre distance is exactly 165 mm:
Given Data: zl = 25; z2 = 50; mn = 4 mm; a = 165 mm.
6.1
Tofind: Helix angle (13).
© Solution: We know that the centre distance between gears,
a =
165 Helix angle,
13 -
(...!!!!L)
cos 13
(_i_)
cos 13
COS-I
x (ZI + z2 )
2
x (25 + 50) or cos
2
(0.909) = 24.62° ADS. ~
13 =
0.909
I
, Examp/e 6.2 A pair of parallel helical gears consists of a 20 teeth pinion and the
velocity ratio is 3 : 1. The.helix angle is 150 and the normal module is 5 mm. Ca/culate:
(i) the pitch circle diameters of the pinion and the gear; and
{ii) the centre distance.
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\
\
Helical Downloaded
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Given Data:
Toflnd:
zl
= 20;
i = 3' 't-' r:l = 15° ,m
.
(i) d, and d2;
@So/Iltion:
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6,7
-- 5 mm.
and (ii) centre distance , a.
We know that,
z2
= ;.
n
=
zJ
3 x 20
=
60.
'il d1 and d2: We know that the PI'tC h eire
. Ie diiameters of pimon
..
and gear,
tl~
and
d1
-
(mn / cos B) zJ
d2
=
(mn / cos B] z2 = (5/ cos 15°) 60
(5/ cos 15°) 20 = 103.53 mm
-
=
310.58 mm Ans. ~
(ii) Centre distance (a) : We know that the centre distance ,
a =
(co~'13) x (Z
= 207.05 mm
I Example
6.3
z2 )
I;
= (COs5IS" ) x
eo;
60 )
Ans. ~
I A pair of parallel /Ielical gears consists of an 18 teeth pinion meshing
with a 63 teeth ,gear. The normal module is 3 mm: The helix angle is 230 whlle the normal
pressure angle is 20 ~ Calculate:
(I)
the transverse module;
(ii)
the transverse pressure angle; and
the axial pltch.
Given Data: Z I = 18 ;
(iii)
Tofind:
(i)
m.:
(ii)
z2 = 63 ;
0.1;
@Solutlon :
(I) Transverse module (mt)
mn = 3 mm;
:
We know that the transverse module is identical to module
d
Transverse module,
d,
d2
ml _-=-=Z
zi
Pitch circle diameter of pinion, d 1
Then
= 23°', an = 20°.
and (iii) Po'
for spur gears.
...
A
tJ
ml
Z2
_ (m; / cos B) zi = (3/ cos 23°) 18 = 58.66 mm
_ ~
zi
_ 5~.:6
= 3.26 mm Ans. ~
(U) Transverse pressure angle (at) :
We know that the relation between angles ~, an and 0.1'
tan an
cos ~ - tan 0.1 or tan 0.1 tan 20°
0
or
..
Transverse pressure aug Ie, at
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tan an
cos ~
= 0.395
cos 23
- tarr! (0.395)
= 21.57°
Ans. ~
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6,8
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Design' ofTran~'mI8NI()h6'IIMI
~~--~~=--------------------------------=~~----~.~
.....
(ill) 4U,,/ pitch (p.): We know that
,
Axial pitch, p a
or
Pa =
( E..\'tl"'pI~ 6."
IA
1tx3
= 24.12 mm Ani • ."
. 230
Sin
parallel helical gearset
consists of
(I
19-toolll pinion drlvl", a
a"gle OJ
S14fHHA IMr. Tilt! pinion lias a left-hand helix angle of 20~ a normal pressure
1-1 ~ ,,,,,I II norma! diametral pitch of 0.4 teeth/mm: Find:
(a)
7"'~
IltlTmtll, transverse, and axial circular pitches.
(b) T"~ transverse module and the transverse pressure angle.
(e) 71;« pilch diameters of the two gears.
G;'~1fData :
=1 =
19;
f3 = 200;
57;
=2 =
an =
~.
v Solutiolf: \\ e knov that, normal module, mn
I
nln
=
0.4
=
14!120; Pn = 0.4 teeth/mm.
= Normal
I
diametraJ
pitch (Pn)
2.5 mm.
\ (tl) T11~normal, transverse, and axial circular pitches:
Normal circular pitch:
Pn -
Transverse circular pitch:
Pc-
1t
x mn =
7t
x 2.5 = 7.85 mm
Pn
-
f3
cos
ADS. ~
1t x 2.5
cos 20°
- 8.36 mm Ans.""
",
Axial circular pitch:
Pn
Pa
-
sin
1t
-
f3
sin
f3
x 2.5
sin 200
=
22.96 mm
Ans . ...,
(6) rllt! IraJtSW!l'S~ module and the transverse pressure angle:
",
Transverse module (m ,J-:
nI, =
P, it
_
it
Pn
_.;.;,._
x
cos
... [ .: P, = 11 x m, =
f3
7.85
-
it
x cos 200
=
2.66 mm
Ans."
Transverse pressun angle ( aJ :
_
tan a
l
or
tan a."
COS
_ tan 14.5°
f3 -
cos 200
at - tan-I (0.275)
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= 0.275
= 15.3!JO ADS."
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;';p 1
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Helical Gears
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-
..,
.........
6.9
"S' ..
t
;:a
(c) Tile pilcl. diameters oftllC~ two g'nr~':
d,
2.S
COs ~ x %, ~ cos 200 x ,19
=:
~
&II
~O.S5 mm ; and
2.5
--cos 20Q xS7.,
151.64nun Ans. ..,
I
[Example 6.5 A pair of paralleillel/elli gellrs cOlIs/~·tsof a 1.l/eelll plnlon meshing
willt a 46 teeth gear. Tire helix allcle is 240 mU/llw Iwrlllill pressur« tIIlgle 21 ~ The normal
module is 4 mm. Calculate :
(i) tile transverse
IIlodllle;
(/I) 11111
transverse pres.\'IIre IlIIgle;
(iii) the axial pitch;
(Iv) the pilclt circle (litll"Cler~'of lilt! pinion and the gear;
(v) II,e centre distallee; and
(vi) the addendum and dedendum circle dlameter« (If the pili/Oil,
GivenData:
zl
= 23
;
z2
= 46; 13 = 24°; 0.,,::::: 210;
III"
== 4 111m,
©Solution:
(i)
Transverse module (mJ : We know that
Tn, =
111"
cos
13
4:::::
4,38 mm Ans,,,
= cos 24°
(Ii) Transverse pressure angle (at) :
We know that the relation between angles
13,0."
and o.t'
tan o.n
cos 13
or
Transverse
pressure angle,
0.,
=
tnn 2,)° :::::0.42
cos 24°
= tall-' (0.42) = 22.79°
Ans. ~
(iii) Axial pitch (p a) " We know that,
n -m;
7tx4
Axial pitch, Pa = sin 13 = sin 24°
=
30.89mm Ans."
(iv) Pitch circle diameters OJ·I'tl,epin/on and the gear (i.e, d I anti d ~ "
d,
We know that
=
Inn
cos A.....x
Z,
=
4
cos 24°
X
23 = 100.7 mm ; and
4")40 x 46 -- 201.4 mm Ans."
cos ...
(v) Centre distance
r. ,1 ••
ta)
( cosm) 13
x
-
a b
__!L
We know that
(ZI +2 z2 )=
( cos424
0
) x
(23 +2 46)
= 151.06 mm
ADS."'CI
7
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Design of TrQlUmi&sionsY"t"l1
6.10
'r
~
(111) Addn«dum iDUl dLdnuhon cirde diameJen of lhe pinion (i:e; did and d/ z) :
Addendum circle diameU:r (i.e., tip diameter) of the pinion,
.
do, -
(~IJ +219) m.
dOl -
(~~40
Dedendum circle diameter
djl
f0
where
-
+2 A J ) x 4
f 0 = height factor::: 1]
108.7].mm Ans."
i.e., root diameter) of the pinion,
(c:; Il
-210) m.-2 c
Height factor = 1, and
-
c - Bottom clearance
d"
... [.:
=
(C05~40-2
=
0.25 x mn
x I ) 4 -2 x I
=
0.25 x 4
=
I nun
= 90.71 mm Ans."
6.8. FORCE ANALYSIS ON HEUCAL GEARS
The three dimensional view of the forces acting against a helical gear tooth is shown in
Fig.6.4(a). The point of application of the forces is in the pitch plane and in the centre of the
gear face. The resultant fOJCe(F) whereby one helical gear transmits power to another gear is
resolved into three perpendicular components :
(i)
the tangential component (FJ,
(ii)
the radial component (Fr)' and
(iii)
the axial
O£
thrust force component
(FJ.
c
A
o
J
•
-i-"
.'
~PiIctr
-----*"---+~ x
(b)
Fa
(c)
qtinda
\~.~./
la)
Fig. 6.4. TootIIfoTces tU:ting on II hdicllJ gUT
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GearsFrom : www.EasyEngineering.net
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6.11
6.8.1. Derivations for F" Frand F
a
Let
F -
Resultant or total force
Tangential or t
.'
ransmltted force (or load)
Radial force ,
'
-
FI
Fa M d J3 Fr
tot~J'
t
Axial or thrust force ,
Transmitted torque,
Pitch circle diameter of gear,
Helix angle, and
0,/ and an Transverse and n
I
.
orma pressure angles respectively.
From triangle ABC (refer Fig.6.4(b »,
Fr and
F· sin an
... (i)
BC> - F· cos an
... (ii)
From triangle CBn (refer Fig.6.4(c»,
'mil
Fa -
of~
~
BC
X
sin f3
F· cos an . sin f3
=
'" (iii)
,
:earl
and
FI
~
-
BC x cos f3 = F x cos an . cos f3
... (iv)
Dividing equation (iii) by (iv), we get
Fa
FI
or
f3
F· cos an . cos f3 = tan f3
F . cos an . sin
-
Axial force, Fa -
FIx tan
f3
... (6.10)
Equation (iv) can be written as
F
=
FI I cos an . cos f3
Substituting the above F value in equation (i), we have
F,
F
r
or
Radial force, F,
= cos
- an
=
F,
. cos
f3
x
Sin
an
[:5~]
... (6.11)
We know that the tangential component F, acts at the pitch circle radius. Therefore,
d
F t x-2
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=
'M I
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6. \2
Il
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Design ofTransmi.uiOl'l.S~
"',Yi"' .. css .......
~
or
Tangential force,
F,
l:::
d
'" {6.i2j
60 x P
M I = 21t N
where
[
.: P = 2 it N~ .,
60
j
•
8.8.2. Dlractlon. of Force Components for Helical' Gears
is
I
./ 1~Inlle"t/1I1force direction
tangent to the
pltch circle and lies in the plane' of rotation. It
opposes the motion of the driving gear. For
example, for the gear pair shown in Fig.6.S, the
drivel' turns clockwise, and the tangential force
direction is inward. Tangential force on the driven
gear is equal and opposite to that for the driving
gcar .
1 (driver)
./ Rlld/1I1force dlrectlon is towards the centre
of u gear. Radial forces on the driver and driven
gears are equal and opposite.
Fig. 6.5.
./ Thrust force dlrectlon is given by treating the driving gear as a screw. The foHow~
guidelines can be used to determine the direction of the thrust component:
(i)
For the driving gear, use right hand for RH-helix and left hand for LH-heJix.
(ii) Keep the fingers in the direction of rotation of the gear and the thumb will ind~
the direction of the thrust component for the driving gear.
(iii) The thrust component on the driven gear is equal and opposite to that for the dmrffig
gear.
The above said guidelines can be verified 'by using the illustration shown in Fig.6.5.
li!mnple
6.6
I A pair
of parallel helical gears consists of an 18 teeth pinion meshing
with a 45 teeth gear. A 7.5 kW power at 2000 r.p.m: is supplied to the pinion through itJ
.,·Iut/l. Tile normal module is 6 mm, w!Ii!e the normal pressure angle is 20'! The helix angle
Is 23 '! Determine the tangential, radial and axial components of the resultant toot/t f~
between the meshing teeth.
GlvenData:
z,=18;
z2=4S; P=7.SkW = 7.Sxl()3W;
N1=2000r.p.m.;
mn
=
6 mm ; a; = 20°;
To find:
(i) F/;
(ii) Fr;
J3 = 23°.
and (iii) Fa'
@Solution :
(J) Tllngenli(ll componelll of the resultant tootk force (F J
We know that,
F/
=
:
2· MI
d
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Helical Gears
6.13
J~I
where
M, == Transmitted torque
=
60 x P
21tN
60 x 7.5 x 103
2 1tx 2000 == 35.81,N-m
-
d
=
Pitch circle diameter of pinion
=
...!!!!L
cos
F,
..
-
6 x 10-3
cos 23° x 18 == 0.1173 m
==
2 x 35.81
0.1173
=
f3
x
z,
610.43N Ans. "
(ii) Radial component of tile resultant tooth force (Fr) :
We know that,
F =
[tan an ]
cos f3
F, x
r
610.43 [tan 20° ]
cos 23°
=
=
(iii) Axial component of the resultant tootli force (F
11
We know that,
Fa -
I Example
6.7
I A torque
610.43
oJ :
f3
F, x tan
-
241.37 N Ans. ~
x tan
23°
=
259.11N Ans."
of 250 N-m acts upon tile shaft of a helical gear whose pitch
circle diameter is 300 mm. The gear has 60 teeth and runs at 250 r.p.m: The pressure angle
of teeth in transverse plane is 180 and angle of helix is 28 ~ Calculate (i) Power
transmitted; (ii) Driving force;
(iii) Normal force on gear tooth , and (iii) Force
transmitted to shaft.
Given Data: M, = 250 N-m;
0./
= 300 mm
d,
= 0.3 m;
zl
= 60; N,
== 250 r.p.m.;
= 18° ;. 13 = 28°.
Tofind:
(i) Power (P);
(ii) Ft;
(iii) Fr;
and (iv) F
Q'
@ Solution :,
(i) Power transmitted (P) : We know that,
p
(ii) Driving force
=
21t N Mt
60
t.e, tangentialforce
2 M,
We know that,
F, -
d
2 1t x 2~0 x 250
60
=
6.544 kW Ans."
(Ft):
2 x 250
- 0.3
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=
1666.67N Ans. ~
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,
6.14
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(iii) Normal fora
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Le; radillJforce on gear tooth (PrJ:
We know that.
F,
=
F, [:~
]
The normal pressure angle (an) can be calculated using the relation
tan an -
tan at . cos
f3
= tan 18° x cos
or
an -
tarr ' (0.287) = 16°
Therefore,
Fr
1666.67 [ cos 2SO
tan
-
160]
Fa -
I Example
6.8
tan
f3
1666.67
X
F/
X
= 0.287
= 541.27 N ADS."
(iv) Force transmitted to shaft i.e., axial thrust (FaJ
We know that,
28°
:
Am."
tan 2SO = 886.18 N
I A pair of parallel helical gears is shown
z=30
A
in Fig.6.6. A 7.5 kW power at 900 r.p.m: is supplied to pinion
A through its shaft The normal module is 4 mm and the
normal pressure angle is 18~ The pinion has left hand teeth,
while the gear has right hand teeth. The helix angle is 25 ~
The arrow indicates the direction of rotation when seen from
the left hand side. (i) Determine the components of the
resultant tooth force; and (ii) draw a free body diagram
showing theforces acting on the pinion and the gear.
1.
Pinion A -
Gear 8
RH
Fig. 6.6-
Given Data: P
an = 18°;
=
7.5 kW
= 7.5
x loJ W; NA
= 900
r.p.m.;
mn = 4 mm;
J3 = 25°.
©So/ulion:
(i) Components of the resultant toothforce (i.e; F1' F, and FtJ:
We know that,
Pitch circle diameter of pinion,
Therefore, the tangential component
dA
-
F, =
-
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60 x P
60 x 7.5 x IoJ
= 79.58 N-m
27t x 900
mn
cos t-'f.t
=
x zA
2 (MJA
dA
-
4
cos 250
x
30
=
132.4 mm
2 x 79.58
132.4 x 10-3
1202.11N Au. ~
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·~
\
H~~a~I~G~e~~~
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"
__ -
The radial component,
F _
r
-
Fa -
The axial component,
-
F,
~~
6.15
[ tan an ]
cos (3
1202.11 [tan 18° ]
cos 25°
F t X tan (3
1202.11 x tan 25°
=
430.97 N Ans. ~
= 506.55 N Ans. ~
(ii) Free body diagram of the pinion and the gear:The free body diagram showing the
forces acting on the pinion and the gear is drawn, as shown in Fig.6.7. The/directions of force
components are found as discussed in Section 6.8.2.
I
,
,,
,
'~...... _'-
" .... 1-·--.
'
.... ,
, B
-_-_-.L
~I
"", I
---...._
,
,,
--
Fig. 6.7.
I Example
6.9] In Fig. 6.8, a o. 7~5kW
1800 r.p ttl. In the
electric motor runs at
• ·d· fi m the
•
as VJewe
ro
clockwise direcuon,
is keyed to
positive x-axls: The motor shaft
I
• h aving a norma
an I8-teeth helical pinIon ,
l 01" 300
0
.1" 20
a /,elix ang e 'J
'
pressure angle oJ'
itch of 0.472
and a normal diametral p. is shown in
teethimm. The hand 0/ the helIX
thefigure.
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y
_-
A
_
_.--250mm
17:•• ~
R.
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Design o/TransmissionS)~
~,
(i) Determine the components
0/ the tooth force,
and
(ii) Draw a three dimensional sketch of the motor shaft and pinion, shOwing th~I~
acting on the pinion and the bearing reactions at A and B. The thrust shouldIr
taken out at A.
Given Data: P = 0.745 kW 13 = 30° ; Pn = 0.472 teeth I mm.
© Solution:
745 W;
=
N,
1800 r.p.m.;
zJ
=
18;
a.::: 2fF:
We know that,
Normal diametral pitch, P» -
or
Normal pitch -
The pitch 'Circle diameter of pinion, d, -
1
mn
= 0.472 teeth / mm
1
0.472
= 2.12 mm
2.12
Inn
cos
f3
zi -
X
cos 300
X
18
=
44.06 mm
(i) Components of tooth force:
Transmitted torque, M I
...
-
Tangential component, F, -
60 x P
60 x 745
21tNI = 21t x 1800 = 3.952 N-m
2·M
d
I
-
- 179.41 N
Radial component, Fr - F
I
=
ADS. "
[tan an]
f3
[tan J
cos
179.41
Axial component, Fa - F, X tan
=
2 x 3.952
44.06 x 10-3
20°
cos 30°
f3
= 75.4 N ADs. 1:1
= 179.41 x tan 30°
103.58 N Ans. ~
(ii) Calculation of bearing reactions at A and B: The forces F F and F actingon
" r
a
pinion C is shown in Fig.6.9. The force FI in the + z direction, F, in the _ y direction, and
Fain the - x direction, acting at point C are shown in Fig.6.9.
We assume bearing reactions at A and B as shown.
For
~ Fx
=
0, we get
F~
= Fa =
103.58 N
Taking moments about the z axis, we get
FrX(250+75)-FQx(~I)
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- F~ x250
= 0
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~~h~ca~/~G~e_a_n
~
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~
~~
6.17
or 75.4 x (250 + 75) - 103.58 x
(14.~)
2
- F~ x 250
or
- 0,
FYB = 8a.892 N
Fr
y
IFX -:
F~_~l/F~
/'
l
1-:
I
'/-1
C /
/ 7-~A--rl--:-=:':_ _
z
Mt~z
F,
•
Fa
_j --
x
B
FY
BIB
Fig. 6.9.
For L Fy = 0, we get
F~ -FY
or
A
-F
r
= 0
88.892 - F~ - 75.4 = 0
or
FY
A
- 13.492 N
Now taking moments about the y axis, we get
F~ x 250 - F, x (250 + 175) = 0
or
F~ x 250 - 179.41 x (250 + 75)
=
0
FZB = 233.23 N
or
For L F, = 0, we get
or
FZA _Fz B + FI
=
0
F~ - 233.23 + 179.41
=
0
FZA - 53.82 N
or
Therefore, resultant reaction at bearing A,
RA -
~-------
~ (F~ )2 + (F~ )2 + (F~ )2
- -J (103.58)2 + {13.492)2 + (53.82)2
and resultant reaction at bearing B,
RB -
= 117.5N Ans.1;I
-v (F~ )2 + (F~ )2 + (F~ )2
- -v 0+
(88.892)2 + (233.23)2
- 249.6N Ans."
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6.18
____________________
-:--
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Design
of Transmission
·(\"'1
..:::.....-=:.:.:.:.:.:.:..:::u~}',elJlll
I
'Example 6.10 The pitch circles of a
train of helical gears are shown in Fig. 6. 10.
Gear A, in the figlU~ has 16 teeth, a 200
transverse ang/~ a 150 helix angle, and a
normal module of 3 nun. Gear A drives the
idler gear B on shaft 'b', which has 36 teeth.
The driven gear on shaft 'c' has 28 teeth. If
the driver rotates at 1720 r.p.m. and transmits
5.5 k W,find the radial and thrust load on each
shaft
Fig. 6.10.
Given Data:
NA
= 1720 r.p.m.;
Tofind:
zA
= 16;
at
f3 =
= 20°;
15°;
mn
=3
mm ;
zB
= 36;
Zc = 28;
P = 5.5 k\V.
Radial and thrust load on each shaft.
@) Solution:
The pitch circle diameters of the gears ~ Band C are given by,
dA -
mn
cos 13 x
dB -
--13
cos
de
mn
cos 13 x ze
=
3
zA
mn
3
- cos 150 x 36 = 111.81 mrn ; and
XZB
Normal pressure angle, tan an -
3
= cos 150 x 28 = 86.96 mm
tan at' cos
an -
or
- cos 150 x 16 = 49.69 mrn ;
13 =
tarr ' (0.3515)
tan 20° x cos IS° = 0.3515
= 19.37°
Radial and thrust load on each shaft :
Forces between gears A and B :
60 x P
60 x 5.5 x 1()3
= 30.54 N-m
21t x 1720
21tNA
-
FrAD
-
F'
-
1229[
-
F~
AD
and
2 x (M,)A
Ft
Fa
AD
dA
AD
-
2 x 30.54
- 1229N
49.69 x 1<r3
[tIDcos a.13 ]
tID
1937° ]
. _cos 15°
x tan
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f3
=
447.32 N
= 1229 x tan 15°
=
Ans."
329.31 N
Ans....,
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--
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Helical Gears
Forces between gears B and C . GA
to its shaft. Therefore,
ear B
.
.
-
IS
the idler gear and •
~6.:.!'!_9
----~
coes not pemut my ~
(M/)a == 0
We know that,
betw
velocity ratio
N
een gears A and B,1. :::: ~
N:>D
N
or
::::225 ::::764.44
2.25
28
== Zc
Nc
zB:::
~
36 ::::
Since the same power is transmitted from gear Ate
o gear
,
=
__j
r.p.m,
o.ns
::::982.~5 r.p.m,
• Therefore
(~lt)c :::: 60 x p _ 60 x 5.5 x 103
2 1t Nc - 2 1t x 982.85 ::: 53.4 N-m
F~c
::::
F~c
=
FI
CB
Fr
::: 2 x (M/)c
de:
-
CB -
F'
Be:
2 x 53.+1
::: 86.96 . 10-:
F~B
=
:::
F~c
x tan
329.31 N
=
t!29 ~
[mna.]t3
cos
19.37° ] ~. _
•
= 1229 [ .tan
-cos 150
44,.J2 N
F~c
=
16
764.44
Nc :::: 0.778 :::: 0.778
Torque transmitted
36
::::-
andC,
i ::::~
or
-A
1720
NB == ~
Then velocity ratio between gears B
-8
::::_
p :::
.-\m.. ~
12_9). tun 15°
ADS.~
It can be seen that, since gear B is idler, whatever torque it receives fu."'IID
transmitted to gear C. Thus, the tangential component between gears B and C m~
to the tangential component between gears A and B. Since the tangential ~~:S
equal, the radial and axial components are also be equal.
A ~
be ~~
~.1r'
iL~
I
[Example 6.11
The Fig. 6.11 shows a double reduction hehca! gmrsd. PiD. .. l is tk
driver, and it receives torque of 135.58 Nsm from its shaft in the directio« S*O __ PUt .. J
has a normal diametral pitch of 0.315 teeth/mm (i.e; normal modtde
= 4_:15 = _1..1:5
--.).
14 teeth, and a normal pressure angle of 200 and is cut righl-handf!'d ~ir'II .dil· ~
rI/
30,! The mating gear 2 on shoft 'b' has 36 teeth. Gear 3, ~'hieh is th~ dri..xr for tie SlAvaJ
pair of gears in the train, has a normal dlametral pitch of tu t~"_'" fi.~6' ----
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6.20
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._.___
module •
,
-
De.'18!!.!1..r"Qn6m~
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__,
G
-1.'2 • S mm), J S tellt/I, tlntl tl normat
,__
prelfNure angle of 20 D and INcut left-hUlldtd
wltl' a helix lIngle 0/1 J ~Milling R'lif 1/ 11(181/.1 teeth.
Find tile nllllJnltude 1",,1 dlractll)n III th« force eXINel1 by tl,e beurlnlll lit C IInll D
Oh
"lil,{llb'. A.Nun,e bear/ng C can luke ,)nly rad/fll/l)od whil« bellrlng D Is mounted 10lallt
botl' 'I,d/al l,nlllllfUNt 1o,,,/.
y
--0
---
•
I
76
8
•
37.6-
Fig. 6./1.
(3J ==
(32 ==
30°;
z2
=
135.58 N-m;
mnl = mn2 = 3.175 mm;
zJ = 14; ani = 20° ;
= 36; mn3 = mn4 = 5 mm ; z3 == J 5 ; an3 == 200;
133 = 134 = 15°; z4 = 45.
Given Data : (M')J
Tofind: Magnitude and direction of the force exerted by bearings C and D.
@ Solutio,,: The pitch circle diameters of gears 1,2,3 and 4 are given by
3.175
mnJ
dJ
== cos 131 x zJ == cos 300 x 14 = 51.33 mm;
d2
=
d3 =
mn2
cos 132
mn3
cos 133
xz
xz
2
=
5
= cos] 50 x 15
3
mn4
d 4 = cos
(34
x
3.175
cos 300 x 36
Z4
-
=
]31.98 mm;
=
77.65 mrn ; and
5
cos 15'0 x 45 == 232.94 mm
Forces between gears 1 and 2 :
(M/)I == 135.58 N-m
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... (Given)
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9~::../~G~e~
__~__ --
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~
~~~
~
6.2'1
'd
-
2,(M)
U:::
2 x 13S.58
51.33 x 10-3 ::: 5282.68 N
o_
d1
F~2 and
F'
12
[
tan anI ]
F~2 - F~2 . tan
(M
In -
_
cos (31
[
- 5282.68
fil ::: 528:i:68
F;2 (~2)
=
tan 200 ]
cos 300
= 2220.19 N
-x truf30° ::: 3050 N
C
5282,68
3) =
3 J.9~ x 10-
348.6 N-m
Forces between gears 3 and 4 :
(M ,)3 F~4
-
(M ,)2
d3
34 - F~
F;4
348.6 N-m
2 (3~8.6)
2 (M 1)3 _
F,r
and
=
= F;4
-
77.65 x 10-3
[~Sa;:] =
x tan (33
=
=
8978.75 N-m
8978.75 [:~
8~78.75
X
~~: ]
tan ISo
=
=
3383.28 N
2405.85 N
The forces acting on the gears 2 and 3, and the components of the bearing reactions are
shown in Fig.6.12.
f
~I
;,
"
t:
Fig. 6.12.
.
o
ant and taking moments about bearing C, we get
.'
ConsIdering the forces
Ff> x (75 + 87.5 + 50)'-
III vertical
(d
x 2 ) _1
F~2
(d )
pi
r
75
F 12 (
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+ 87.5) .
F04
3
x -3
2
+ F'34 x 75
= 0
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6.22
Design of Transmulion Sy,
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~
or 212.5 x
Fb - 3050 x ( 1312'98)
- 2220.19 (162.5) - 2405.85 x
( 77.65 )
2
+
3383.28)( 7S
Fb
or
For LV
= 0,
:::0
= 1890.41 N
i.e., considering the equilibrium of vertical forces, we get
Ph -
0
or
F~ - 3383.28 + 2220.19 - 1890.41 -
0
or
pYc -
F~ - F;4 + F;2
-
3053.5 N
Considering the forces in horizontal plane and taking moments about bearing C, we get
- F~ x (75 + 87.5 + 50) + p/12 x (75 + 87.5) + P~4 x 75 -
0
- 212.5 x F~ + 5282.68 x 162.5 + 8978.75 x 75 -
0
or
FXD
or
-
7208.67 N
For E H = 0, i.e., considering the equilibrium of horizontal forces, we get
F~ - F ~4
F~2+ ~
-
or
F~ - 8978.75 - 5282.68 + 7208.67
or
FXc
- 0
-
0
-
7052.76 N
For E F, = 0, i.e., considering the equilibrium of axial forces, we get
...
;:.
F~4-F~2 -F~
or
-
0
2405.85 - 3050 - F~ - 0
F~ - - 644.15 N
or
The negative sign indicates that F~ acts in the opposite direction to the direction shown in
Fig.6.12.
Therefore,
resultant reaction at bearing C -
-v '(F~)2 + (F~ )2 + (F~ )2
- ~ (7052.76)2 + (3053.5)2 + 0
and
resultant reaction at bearing D -
7685.39 N AIJs. ~
-v
(F~ )2 + (Fh )2 + (F~ )2
---
- V,....(-72-08-.6-7~)2-+-(1-8-90-.4-1
)2-=--+
(--644~.15f
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7~80.1 N ~.
~
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Helical Gears
:;;---
6.23
DESIGN OF HELICAL
I.
i
GEARS
I
1
I
I
,
HELICAL GEAR DESIGN USING LEWIS
(Helical Gear Design Recomme n d ed by AND
BUCKINGHAM'S EQUATIONS
AGMA)
I
6.9. BEAM STRENGTH OF HELICAL G
As discussed in Section 6.4 the f
~RS
(Lewis Equation for Helical Gears)
. 1
,ormattve
gear is
.
.
peepend leu ar to the tooth element It .
d
an imagmary spur gear in a plane
•
•
IS un erstood that th b
virtual) gear IS nothing but the beam stre gth
.
e earn strength of formative (or
. .
n
of helical gear Th f
.
equation IS used to determine the beam str
h
.
. ere ore a modified Lewis
F _
engt of a hehcal gear. It is given by
s - 1t. mn· b [ CJb ] • y'
F _ B
t
gth
... (6.13)
s
earn s ren
of helical gears,
where
mn - Normal module ,
b -
Face width
where Pn
[ CJb]
Permissible
-
2Pn to 4 Pn (or) 10 m.,
Normal circular pitch = 1t. mn
=
=
or allowable static stress, from Table 5.4, and
y' _ Tooth form factor or Lewis factor based on virtual or formative
or equivalent number of teeth.
The beam strength (Fs) indicates the maximum value of tangential force that the tooth can
transmit without bending failure.
6.10. DYNAMIC LOAD ON HELICAL GEAR TOOTH (Effective Load on Gear Tooth)
As discussed in previous chapter, in addition to the static load due to power transmission,
there is dynamic load betWeen the meshing teeth. In order to account for dynamic loads, the
following two methods are used.
I. Approximate estimation of dynamic load using the velocity factor, which is used in
.2.
the initial stages of gear design, and
Accurate estimation of dynamic load using the Buckingham's equation, which is used
in the final stages of gear design.
1. Calculation of initial dynamic load (FtJJ :
·,
I
f d amic load can be calculated by using the relation
e
pre
irrunary
va
ue
0
yn
Th
I
~
Fd :::
where
.. , (6.14)
Cv
F, _ Tangential load considering service factor
_ -P x Ko'
v
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I
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~~~.2~4
=
P
where
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--------D_e_s~ign~o~if_n_rmu--m~u~n=on~s~
Power
transmifted
v = Pitch line velocity
Ko -
Service
I Shock
in watts,
=
idN
60
factor,
,in
mis, and
from Table
5.6.
e" ;;:: Velocity factor (c, values are the same as for spur gears)
::;:
3
for
3+\1'
v~
10
mis,
commercially
;;::
6
6+\1'
=
5,;'!
=
10,75
+ V + 0,25, for non-meta II'IC gears,
cut gears
for v = 5 to 20 nils, carefuJIy cut gears
v' for v> 20
mIs,
precision gears
2. Buckingham's equationfor dynllmic load:
Buckingham's
equation, used for accurate estimation of dynamic load, is given by
Dynamic load, Fd
where
F(
v
=
21 v ( c b ' cos2 13 + F,) , cos
F, + 21 v + ~ cb . cos- 13. + F,
=
13
... (6.15)
Tangential load neglecting service factor = ~,
= Pitch line velocity,
b = Face width,
c
=
Deformation or dynamic factor in N/mm,
from Tables 5.7(a) and (b), and
13 = Helix angle.
6.11. WEAR STRENGTH OF HELICAL GEARS (WEAR TOOTH LOAD)
The wear strength equation of the spur gear is modified to suit helical gears. Thus the
modified wear strength equation for helical gears is given' by
Wear load, Fw =
where
d1,b
'Q'~
... (6.16)
cos213
d I = Pitch circle diameter of pinion (use pinion diameter
irrespective of
whether pinion or gear is designed),
b
=
Face width,
Q
=
Ratio factor,
=
2x
j
j
+1
=
2 z2
zl
+z2
!
I
I
, for external gears
__________
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(
....
,,/
&~iJ.. --$.-;,P.~-
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r~
~
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~::i~~I~G~e_~_~
~~
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__ ~
~~
6.25
_ W __2z2
i-I
-
z2 -;;
,
for internal gears
where i == velocl'ty ra ti10 == -z2
zl
-
=
Load stress factor (al kn
.
N/mm2 fr
T
so own as matenal combination factor), in
, om able 5.9 (or)
[f;s
x sin
1.4
an ]
_L ]
E +E
P
f es
where
[.l
== Sm'
U
an -
Ep and Eg
J3
g
ace endurance limit in N/mm2, from Table 5.9,
Normal pressure angle, and
== Young's modulus of pinion and gear respectively.
== Helix angle.
INote I In the design of helical gears, the normal module mn should be selected from standards. For
the recommended series of modules, refer Table 5.8.
IS)
6.12. DESIGN PROCEDURE
Though the design procedure for helical and herringbone gears is exactly the same as for
spur gears, the step by step design procedure for helical gears is given below for ready
reference. In this method, the gear is designed on the basis of beam strength using Lewis
equation and checked for dynamic loading and limit wear loading using Buckingham's
equation.
1. Selection
md
0/ material: If not given, select a suitable pinion and gear materials.
2. Calculation o/Zl and'z2:
-/
If not given assume number of teeth on pinion
-/
Then number of teeth on gear,
3. Calculation
Z2 = ;
0/ tangential load on tooth
x
zl'
ZI ~
17, say 18.
where; - gear ratio.
(Ft) : Calculate the tangential load on tooth
using the relation
P
r, - -; x Ko
.6)
of
where
P -
v -
Power transmitted in watts,
ndN
. mI and
Pitch line velocity = 60 ,an
s,
. / Shock factor, from 'liable 5.6.
Service
.
•
_ .....1". ~ ad (F til : Calculate the preliminary value of dynamiC
4. Calcullltion of initial dyn~
0
load Fd using the relation
Ko -
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Design of TransmisSion SYSletns
6.26
----...::.
e"
where
= Velocity factor.
The value ofv used in velocity factor formula may be initially taken from 10 to 15 mls.
5. Calculation of beam strength (FJ: Calculate the beam strength (Fs) in terms of
normal module using the relation
F,
= 1t·mnb[O'b]Y'
mn =
where
Normal module in mm,
b = Face width in mm, initially assume b = 10 mn,
[ O'b 1
-
Design bending stress or allowable static stress, from Table 5.4, and
y' = Form factor based on virtual number of teeth, from Table 5.5.
6. Calculation of normal module (m,J : Since the gear is designed on the basis of beam
strength, therefore Fs ~ Fd' So equating Fs and Ftb find the normal module.
Then select the nearest higher standard module value from Table 5.8.
7. Calculation of b, d and" :
./
Find face width (b): b = 10 mn
./
Find pitch circle diameter (dt):
./
.
Find pitch line velocity (v):
v
d,
=
m
=~
cos
x zl
p
7td1N1
60
8. Recalculation of the beam strength (F J
tooth using the relation
: Recalculate
the beam strength of the gear
Fof = 7t. mn . b [0' b ] Y'
9. Calculation of accurate dynamic load (FttJ: Calculate
the dynamic load more
accurately using Buckingham's equation as given below.
Fd = F,+Fl
=
Ft +
21 v (cb . cos213
21 v +
+ F,) . cos 13
-.J cb cos213
+ F,
. wh~re
c . = Deformation factor, from Tables 5.7(a) and (b).
Wh1le calculating the dynamic load (Fd), the value of tangential load (F) should be
calculated.by neglecting service factor 0<0).
'
i.e.,
p
F, = -;
l
j
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-
Helical Gears
6,.27
10. Checkfor beam strength (or tooth breaka
•
./ Compare Fd and Fs:
rge) •
./ If F d S FoS' the gear tooth h
b aka
Th th
as adequate beam strength and it will not fail by
re
gee us e design is satisfactory
./ If Fd > FS' then change face width
d .
I d (F ) th
,mo ule or both. Usually, to reduce the dynamic
;a
fi d , e .g~ should be carefuHy cut (i.e., reduce the deformation factor (c».
ven or precision gears, if Fd > Fs then increase the face width till F < F
11. Calculation of the maximum
' ad
'
d
s:
wear ..0 (F.,): Calculate the maximum or limiting
wear load using the relation
d1xbx9xK,.,
cos2
f3
.
2;
I+ 1
Q - RatIO factor
where
~.,
-
= -.-
=
2z2
d
z I + z2' an
Load stress (or material combination) factor, from Table 5.9.
12. Check for wear:
./
Compare the calculated values of dynamic load (Fd) and wear strength (Fw)'
./
If Fd < Fw' the gear tooth has adequate wear capacity and will not wear out. Then
the design is safe and satisfactory.
13. Calculation of basic dimensions ofpinion and gear: Finally, calculate the basic
dimensions of pinion and gear by consulting Table 6.1.
I Example
6.12
I Design
a helical gear to transmit 15 kW at 1400 r.p.m: to the
JoUowing specifications: Speed reduction is 3; Pressure angle is 20~ Helix angle is 15~
The material of both the gears is C45 steel: Allowable static stress 180 N/nun2; Sur/ace
endurance limit is 800 N/nutf1; Young's modulus of materia: = 2 x lOSN/nun2.
Given Data:
[ CJb] =
P
=
Nl
f es = 800 N/mm2;
180 N/nun2;
Tofind:
15 kW;
=
1400 r.p.m.;
El
=~
=
i
= 3;
~
=
20°;
J3
=
15° ;
2
2 x lOS N/mm
•
Design a heJical gear.
@Solution:
1.Material selection:
2. Calculation of
Zl
Pinion and gear - C45 steel.
and
... (Given)
Z2 :
z 1 = 20
./
Assume
./
z2 = i x zi = 3 x 20 = 60
3.. CaicuIaJion of tangential load on tooth (F; :
P
We know that,
FI == -V x
Ko
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6.21
v ... [ .: d. =
-
Ko :
"X
:'-13
X
z. and m_ i. in 'lnln'l
1400 x mn x 20
000
150 ;:: 1.518 mn mil
60x I
XCOI
1.25, assuming medium shock, from Table 5.6
15 x loJ.
5 _ 12353.45
P, ;:: 1.518 mn x 1.2 mn
...
4. CalculllIion of initial dynamk load (F tJJ
We know that,
Fd
=
where
c
=
"
P,
c"
6
6 + v' for v = 5 to 20 mls and carefully cut gears
= 6:
...
:
15
=
0.286, assuming v = 15 mls
12353.45
1
43237.075
x 0.286 -
5. Cakulation
of beam strength
We know that,
PI
=
1t.
(F,) :
mn . b . [
0b ] •
JI
b = Pace width = 10 mn (initially assumed)
where
... (Given)
[ °b ] = 180 N/mm2
y' = 0.154 _ 0.912 , for 20° involute
Zeq
where
-
Zeq
...
y'
Then,
FI -
zi
cos3 ~
= 0.154 1t
X
20
= cos3 15° = 22.192::t: 23
0.912
23
= 0.1143
mn x 10 mn x 180 x 0.1 143
=
646.62 m/
i
6. Calculation of normal module (m,) :
I
We know that,
.1
PI ~ Fd
646.62mi
or
~
43237.075
I
mn ~ 4.058 mm
From Table 5.8, the nearest higher standard normal module is S mm.
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Helical
G_ears
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6,29
7. Calculation of b, d and v :
./
Face width (b): b == 10 m _
n -
./
./
Pitch circle diameter (d ) .
J
Pitch line velocity (v):
•
lOx S
=
d,
SO mm
m
cos f3 x
--!L
N,
v == 7t d,
=
S
Z,
::0
cos IS0 x 20 :: 103,53 mm
== 7t x 103.53
60
8. Recalculation of the beam st l'ength (F,) :
We know that,
FS == 7t. mn ' b . [ <J ]· y'
X
10-3 ~ I4(J(J
= 759 mls
60
b
== 7tx5x50x
180xO.1143
9. Calculation of accurate dunaml
•
'J
C .oad (F,J :
=
16]5878N
.
F d == F + 21 v (cb· cos2 f3 + F,) cos f3
I
21 v + ~ cb . cos2 f3 + FI
We know that,
15 x ]03
7.59
F == ~_
where
I
V
,
-
e, ~3 .
~
=
]976.28 N
c = Deformation factor, from Tables 5.7 (a) and (b).
-
11860 e, for steel and steel, 20° full depth, from Table 5.7(a).
_
e = 0.025, for mn upto 5 and carefully cut gears, from Table 5.7(b). ~ ~ :S~ '
...
Then,
c = 11860 x 0.025
Fd
=
1976.28 +
-
17772 N
= 296.5 N/mm
21 x 7.59 x ]()3(296.5 x 50 x cos2 ]5° + 197628)
21 x 7.59 x ]()3+~ 296.5 x 50 x cos? 15° + 1976.28
10. Check for beam strengtl, (or toot/, breakage) :
We find F, < Fd: So the design is unsatisfactory. In order to reduce Fdo we try with
(b), for precision gears, e = 0.0125.
precision gears. Fro"" T~!~
Then deformation~actor,
c
=
11860 x 0.0125
=
148.25
The newvalue ofFd is given by
21 x 7.59 x ]03 (148.25 x 50 x oos2 15° + ]97628)
Fd = 1976.28 + 21 x 7.59 x 1()3+~ 148.25 x 50 x oos215° + 1976.28
=.
10863.26 N
Now we find Fs > F d' It means the gear tooth has adequate beam strength and will not fail
by breakage. Thus the design Is satis/adory.,
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:;
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load (F,J :
11. Calculation of the maximum wear
We know that,
=
Fw
s, x h x Q x x,
~
f3
COl2
21
Q - Ratio factor :: 1 + J
where
sin
- f ./ 1.4
Kw
::
12. Check for wear:
(800)2
(1..
2~3
3+J
1,5
+ 1. ]
[_'
E.
x
:i+
~
sin 20" [
,
1.4
2 x JO~
h" 'IP
]
1.5635 N/mm2
We find
Pw > Ftl It means the gear tooth ha~ adt€r~
wt;lr
capacity and will not wear out. Thus the design is safe and 6atJsfactOry.
13. Calculation of basic dimensions of pinion and gear:
Refer Table 6, J,
Normal:
./
Normal Module:
mn = 5 mm
./
Number of teeth:
Zl
./
Pitch circle diameter:
= 20; and
d1
Zz = 60.
::
103.53 mm;
::
310.58 mm
and
dz
m;z
= cos
f3 /
Z].
=
5
cm,l~
60
./
Centre distance :
a =
10 -
~
cos J3
(ZI +Z2)
2
5
=
cos J 5°
(20+60)
2
207.05 mm
./
Height factor:
./
Bottom clearance :
c - 0.25 mn
=
0.25 x 5
= L25 mm
./
Tooth depth :
h - 2.25 mn
=
2.25 x 5
=
./
Tip diameter:
do. -
-
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1
(C:~p
11.25 mm
+ 2 f 0 ) m.
(CO:~S. + 2 x , ) 5 = 113.53 DIm; and
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'I
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" (~~
== (
-/
Root diameter:
6.31
+2/0) m.
60
)
~s
1
S;
+
2
x
l
: 5 == 320·58 mm.
.
- (c:~f3 - /0) », 2
2c
- (cO:~50 - 2 x I ) 5 -2 x 1.25= 91.03 mm; and
(c:: /0)
f3 - 2
m. - 2 c
- (cO:~50 -2 x I ) 5 -2 x 1.25 = 298.08 mm,
./
Virtual number of teeth :
-
20
cos! 150
=
22.192 ~ 23;
f3 -
60
cos3 150
=
66 .57 ~ 67
ZI
cos! f3
z2
zv2 = cos!
I Example
6.13
Given Data:
Tofind:
and
I For the above example, calculate the end thrust on the gear.
Refer Example 6.12.
.
'.
..
End thrust on the gear.
@ Solution : We know that the end thrust or axial load on the gear,
p
Fa = FI X tan
J3
=
X tan f3
.
v
[Example
6.14
IA
15 X 103
7.59
X
tan 15°
=
529.54 N Ans."'tJ
compressor running at 360 r.p.m: is driven by a 140 kW, 1440
I'.p.m. motor through a pair of 200 full depth helical gears having helix angle of 25 ~ The
centre distance is appl'oximately 400 mm: The motor pinion is to be forged steel and the
dl'iven gear is to be cast steel. Assume medium shock conditions. Design the gear pair.
Given Data: N2
=
360 r.p.rn;
P
= 40 kW;
Nt = 1440 r.p.m.;
~
= 20°;
J3
= 25°
;
a=400mm.
Tofind: 'Design the ~eJical gear pair,
@ Solution:
Since th~,materials for pinion and gear are different, first we have to
evaluate ,[ (J~1 ] Y~ 'and [(J b2]
y;
to' find out the weaker element.
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6.32
=
Gear ratio, ;
Assume
...
1440
360
=
4
4 x 20
=
zi
Z2
-
; x ZI
zvl
-
cos3
=
20
zi
Virtual number of teeth :
J3 -
80
cos! 250 ~ 27;
and
80
Zv2 = cos3 250 ~ 108
Given that the pinion is to be forged steel and the gear is to be cast steel. ~
consulting Table 5.3, the following steels are selected.
Pinion Gear -
For pinion:
Forged steel, and
Grade 1 i.e., CS 65 cast steel
From Table 5.4,
[ crbl]
-
y; -
Form factor,
-
Y; -
[obi]
For gear: From Table 5.4,
[CJb2]
Y;
Form factor,
[CJb2
We find (crbl)
Y; < [
Cfb1
)
]y;
y;,
112 N/mm2,
for forged steel; and
0.154 - (0;12)
, for 200 full depth
-vI
0.154 - (
0.912)
27
112 x 0.1202
=
= 0.1202
]3.465 N/rrunl
-
]05 N/mm2, for cast steel; and
-
0.154 - (0.912)
, for 200 full depth
-
0.912)
0.154 - ( 108
=
-
105 x 0.1455
z,,2
=
0.1455
15.28 N/~
i.e., the pinion is weaker. Thus, we have to design the
pinion only.
1. Material selection: Pinion - 40 Ni 2 Crl Mo28; and Gear - Grade 1 cast steel
2.
ZI = 20
and
z2
= 80.
. .. (Already caJcutated)
3. Calculation of module: Since the centre distance is given, we need not to equate f,
and Fd to calculate the module. The module can be calculated using the relation
a =
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(;'-li) x(ZI;~)
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-
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mn_)
400 == (-
cos 25°
or
Nonna] module, m == 7 25
n
.
mm
From Table 5.8, the nearest high
er standard no
I
./
Face width (b):
./
Pitch circle diameter (d ). d
b == 10
_
).
_ mn
) - -
cos
Pitch line velocity (v): v ==
We know that,
FS
2
mn - lOx 8 == 80 nun .
NI
60
5. Calculation of beam strength (Fs) :
./
(20 + 80)
rma module is 8 mm.
4. Calculation of b, d and v :
,/
6.33
=
'It.
=
'It X
'It d)
f3
xz
=
'It
8
=-I
cos 250 x 20 =
176.54 mm
x 176.54 X 10-3 x 1440
60
= 13.31 mls
mn . b . [ CJb' ] y'
8 x 80 x 112 x 0.1202
= 27067.76 N
6. Calculation of accurate dynamic load (F,J :
We know that
where
F
'd
=
t
F
= F
,
+ 21 v (cb· cos2 ~ + F,) cos~
21 v + \} cb cos2 ~ + F,
p _ 140 x IcP
v
13.31
= 10518.4N
c = Deformation factor, from·Tables 5.7(a) and (b).
-
11860 e, for steel and steel, 20° full depth, from Table 5.7(a).
e = 0.038 mm, for mn upto 8 and carefully cut gears, from Table 5.7(b).
...
c = 11860 x 0.038
10518.4
Then,
= 450.68 N/mm
+ 21 x 13.31 x
I()3 (450.68 x 80 x cos2 25°
21 x 13.31 x )()3
-
+-v
+ 10518.4) cos 25°
450.68 x 80-x cos2 25° + 10518.4
46865.44, N
7. Check for beam strength (or
breakage): We find Fd> F" So the design is
unsatisfactory. Since the difference between Fd and F, is high, we can increase the face width
I
I
I
10(Jlh
by increasing the normal module from 8 mm to 9 mm.
Then
Face width, b
=
10 mn = 10 x 9 = 90 mm;
mn
9
x z) =
250 x 20
Pitch circle diameter, d 1 = -13
cos
cos
'It
Pitch line velocity,
V
=
d1 NJ
60
_
'It
= 198.61 mm ;
x 198.61 x 10-3 x 1440
60
=
14.97 mls;
I
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6.34
Design of Transmission $y3Ie""
~
p
Tangential load, F/ =
V
140 x 103
14.97
= 9352 N ; and
-
Expected error, e - 0.0205 mm, for mn upto 9 and precision gears
from Table 5.7(b).
:.
Deformation factor, c = 11860 e
-
11860 x 0.0205 = 243.13 Nlmm
Then the modified value of dynamic load is given by
F
d -
-
21 x 14.97 x 103(243.13 x 90 x cos2 25° + 9352) cos 25°
9352 +
21 x 14.97 x 103
243.13 x 90 x cos2 25° + 9352
+-v
34104.29 N
The modified beam strength (Fs) value is given by
F,
=
1t
X
= 34257.64 N
9 x 90 x 112 x 0.1202
Now we find F, > Fd" It means the gear tooth has adequate beam strength and it wiIJ not
fail by breakage. Thus the design is satisfactory.
8. Calculation of limiting wear load (F,.J :
dl xbxQxK"
P
We know that
FlII =
where
Q = Ra tio tactor = i + 1 = ;4 rl-,.1 - 1.6; and
cos-
.
2i
&:
2x4
,
Kw
.
Fw
=
\
=
Load stress factor
=
for steel hardened to 400 BHN, from Table 5.9.
198.61 x 90 x 1.6 x 2.553
cos2 250
= 88892.06 N
2.55) N/mm2,
wear: We find F,., > Fr It means the gear tooth has adequate. wear capacity
and will not wear out. Thus the design is safe and satisfactory.
9. Checkfor
10. Calculation of basic dimensions of pinion and gear: Refer Table 6.1.
-/
Normal'module : mn = 9 mm
./
Face width: b = 90 mm
./
Number of teeth : zJ = 20; and
-/
Pitch circle diameter:
Zl = 80
d I = 198.61 mm;
- d2
_ mn
-:
9
----;i x z2
cos
fJ
I
and
= cos 250
.
x
= 794.43 mm
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I
80
I
i
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,J
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-
Helical Gears
'"
6.35
Centre distance :
a = (~)
cos f3
= _
Height factor:
./
Bottom clearance:
'"
T
d
==
- 2
== 496.52
-
d02
+2 f0
)
mm
20
)
9 == 216.61 mm; and
(3 + 2/0 ) mn
(CO:~50+ 2 I ) 9
(ZJ
) 2
cos f3 - 2 f
X
d
fJ =
mm
mIl
cos 25° + 2 x J
= (::
=
mIl -
0
20
= ( cos 250
d/2
.
(-.:r_f3
_ (
-2 x I
)
=
812.43 mm
c
9-2
-
x 2.25 = 176.11 mm; and
C::
=
(3 - 2/0 ) mn - 2 c
(CO:~50
=
Virtual number of teeth : Zvl
0/
(20+80)
2.25 x 9 == 20.25 mm
cos
aJ
Root diameter:
2
0.25 x 9 == 225
n -
ooth depth: h == 2.25 m _
Tip diameter:
./
(ZJ+Z)
-2
c == 0.25 m _
n -
'"
9_
cos 25°
10= 1
'"
x
=
- 2
27; and
XI)
Zvl
9 - 2 x 2.25
=
771.93 mm
== 108
6.15 , A pair of 20" full depth involute teeth 30 D helical gears having u
velocity ratio of 3. Tile pinion is made of steel with allowable static stress of 100 N/nuttl
[Example
and the gear is made of cast steel will, allowable static stress of 70 N/nuttl. The pinion
transmits 40 kWat 1500 r.p.m. Determine all the basic dimensions of the gear pair.
.
Assume
width of face
0.154 - 0.0912
z'V
To find:
z
is the equivalent number of teeth.
V
Given Data:
P=40kW;
where
as 3 times the normal pitch and tooth form factor as
NJ
q, = 20° te_; f3 = 30
0
=
J500r.p.m.;
;
i = 3;
[CIhJ
]
2
= 100 N/mm ;
[CIh2 ]
b=3p".
Design the helical gear pair.
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2
== 70 N/mm ;
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6.36 Downloaded From : www.EasyEngineering.net
e Solution:
evaluate [ O'bl
] y'l
DesignofT~~
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Since the materials for pinion and gear are diffemJt, f.. ~ ~ II
and [ 0b2 ]
to find out the weaker element. Assume z, = 20.
y;
Then,
=
i x z 1 = 3 x 20
Z2 -
ZI
VirtlUll number of teeth:
zYI
-
cos313
Z2
For pinion :
60
Form factor,
-
20
cos3300
f3 -
zY2 _
cos3
y; -
0.154 _ (0.912),
[Obl]~
_
30.79 ~ JJ, and
60
cos) 300
= 9238 ~ 93
for 2W
z,,1
- 0.154_(°;112)
...
=
fun depIh
=
0.1246
=
0.144
100 x 0.1246
= 12.46 N/mm2
For gear :
Fonn c.ractor, Y2,
[O'b2
We find [(61)
Y; < [ 0'61 )
y;,
]y;
-_
0.154 _ (0:;2)
= 70 x 0.144 = 10.09 NI~
i.e., the gear is weaker. Thus we have to tU:sigII 1MP
only.
1. Material selection: Pinion - Steel, and Gear- Cast steel
2. z 1 = 20; and
z2
= 60.
... (Already cakuJafed)
3. Calculation of tangential load on tooth (FJ :
We know that,
where
P
FI = -v
v
X
Ie
&~
_
=
7t
d2 N2
60
N.
60
= 7t x
...
x 1500 x mn x 20
60 x 1000 x cos 300
[
'.'
(mIJ
cos
P
~)
x 1000
mil
d I = cos
Px
ZI
and mil
. .
IS ID
j; ...
I11'1
'uP"
7t
-
Ko -
...
FI
=
Shock factor
=
=
1.814 mn
1.25, assuming medium shock, from Table 5.6-
40 x 1()l
21566.44
1.814 mn x 1.25 =
mil
4. Calculation of inilial dy1llllllic load :
We know that,
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6
c
·.·
y
==
6+;,
-
6
6
for
'V
== 5 t 20 mI
o
s and carefully cut gears
+ IS = 0.286, assuming v = IS mls.
F d == 27566.44
1
96386.17
x 0.286 = ----0..;..;;..;..
mIl
mil
S. 0Ilcuhrti01l of bmm stre"gth (F j :
Weknow~
F, -
where
7t.
mil • b . [ Gb2 ] •
3 PIt = 3 x
b -
7t
y;
x mII ~ 10m II
·.·
Fs - 7txm"x IOmllx70xO.144
6. Clzkllllllioll of1f017IUJ/ modul~ (m,J:
Weknowth~
Fs
316.67 m2
"
or
Normal module, mil
~
~
~
... (Given)
= 316.67m2
II
Fd
96386.17
mil
6.726
From Table 5.8, the nearest higher standard module is 7 mm.
7. CillCu/atioll ofb, d andv:
-/
Face width (b):
b == 10 mil
-/
Pitch circle diameter (d I) :
-/
Pitch line velocity (v) :
=
lOx 7 = 70 mm
mil
cos p
--A
v
=
_
&. ll«tzIcuJadoll of beam strength (FJ
Fd == F, +
21 v+~
=
7
300 x 20 = 161.66 mm
cos
dt Nt
60
'It
x 161.66 ~OIO-3 x ~500 = 12.7 mls
:
21 v (cb . cos2
We know that,
'It
XZt
f3 + F,) cos f3
cbcoslf3+F,
P _ 40 x 1()3 == 3149.61 N
F, - v 12.7
. factor , from Tables 5.7(a) and (b).
_ J)efonnaDon
c
c.. st I and steel 20° FD, from Table 5.7(8).
_
e -
·.·
11860 e, lor
ee
'
m
upto 7 from Table 5.7(b).
0.017.JDDl, for"
'
c = 11860 x 0.017 == 201.62 N/mm
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6.38
Des ign of Transmission Sy'lenr
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~
Then,
21 x 12.7 x 103 (201.62 x 70 x cos2 30° + 3149.61) COs30
Fd = 3149.61 +
-
21 x 12.7 x 1()3 +
..I
~
v 201.62 x 70 x cos2 30° + 3149=:61'
15038.95 N
10. Check for beam strength (or tooth breakage) : We find F,. > F II- It means the ge
tooth has adequate
satisfactory.
and will not fail by breakage.
beam strength
Thus the des;g"
11. Calculation of the maximum wear load (F~ :
dJxbxQx~
We know that,
Fw -
where
Q - Ratio tactor = i + 1 = 3 + 1 = 1.5; an
cos2
J3
2i
• e.
2x3
d
K,.. - 0.919 N/mm2, for steel hardened to 250 BHN, from Table 5.9.
...
Fw
_
161.66 x 70 x 1.5 x 0.912 = 20799.17 N
cos2300
12. Check for wear: We find Fw> F II- It means the gear tooth has adequate wear capacit
and will not wear out. Thus the design is safe and satisfactory.
13. Calculation of basic dimensions of 'pinion and gear: Refer Table 6.1 .
./
Normal module:
mn = 7 mm
./
Number of teeth:
ZI
./
Pitch circle diameter:
= 20;
and
Z2
= 60
dl = 161.66mm;
and d2=
7
- cos 300 x 60
./
a =
Centre distance :
(_!!l!!_)
cos J3
=
X (ZJ
10 -
cos
A XZ2
tJ
484.97 mm
+ Z2
2
7
20+60)
2
(
cos 30°
./
mn
)
= 323.3 mm
./
Height factor:
./
Bottom clearance:
c - 0.25 mn
=
0.25 x 7 = 1.75 mm
./
Tooth depth :
h = 2.25 mn
=
2.25 x 7
./
Tip diameter:
dOl
-
1
=
15.75 mm
(c:~IJ +210) m.
- (co;~Oo + 2
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XI)
7 = 175.66 mm; and
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HelicalGears
6.39
=
(0::13 10) m.
=
(co:~oo + 2 x I) 7
=
(ZIcos f3 -
=
(co;~Oo -2 x I ) 7 -2 x 1.75 = 144.16 mm; and
d
Q2
~
Root diameter;
d
/1
df2
II.
Virtual number of teeth ;
2f0
)
=
498.97 mm
mn - 2 c
(o~13-2/0)
=
=
~
+-2
m.-2 c
(co:~Oo -2 x I) 7-2 x 1.75 = 467.47 DIm
zvl
=
27; and
zv2
=
108
HELICAL GEAR DESIGN BASED ON GEAR LIFE
(Helical Gear Design Using Basic Equations)
6.13. DESIGN FORMULAS
FOR HELICAL GEAR DESIGN
Almost all the design formulas for helical gear design are same as that for spur gears.
However, the modified design formulas are given here for ready reference.
,
(i) Design torque (or) Design load (M;:
I
,
~M/]
We know that,
M,
where
K x Kd
... (6.17)
=
M,
=
Transmitted torque
X
=
60 x P
2 1t N '
K = Load concentration factor, from Table 5.11, and
Kd
=
Dynamic load factor, from Table 5.12.
(ii) Induced bending stress (u,,) :
,-or
Vb
-
-.
07
i± 1
[Mt]
a· b . mn· y v
NI
where
1
= Gear ratio
= N;
-
... (6.18)
z2
-,
•
zi
a - Centre distance between gears,
b - Face width,
n1n =
Normal modute.
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6.40
Design of n-ansmi3slon
By"'t""
~
=
y"
[ M I]
Form factor based on virtual number of teeth, from Table S.13, 8JJd
= Design torque.
(III) Design bending stress I 0'6 J :
=
[ CJb]
where
Kb/
-
1.4
Kbl
nKa
Kbl
n·K
x a_I' for rotation in one direction
x a_I'
for rotation in both directions
... (6.I9(a))
... (6. J9(b))
0-
Life factor for bending, from Table 5.14,
Ko- - Stress concentration factor for fillet, from Table 5.15,
a_I
-
Endurance limit in reversed bending, from Table 5.16, and
n = Factor of safety, from Table 5.17.
(iv) Induced contact stress (uJ :
«; - 0.7
where
Eeq
-
i±1 ... /i±l
a
\j ib x Eeq [M/]
... (6.20)
Equivalent Young's modulus, from Table 5.20.
2 EJ E2
or
EJ +E2
where El and ~ Young's moduli of pinion and gear respectively.
(v) Design contact stress I UcJ :
... (6.21)
where Co and CR
-
HB HBC Kel
-
Coefficients depending on the surface hardness, from Table 5.18,
Brinell hardness number,
Rockwell hardness number, and
Life factor for surface strength, from Table 5.19.
(vi) Centre distance (a) :
a:2:(i±l)
where
'" -
3 (_0.7)2
[ael
Eeq
x [Mil
... (6.22)
i",
Ratio of gear width to centre distance = b/a
Take '" = 0.3 for initial calculations.
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Helical Gears
6.41
6.14. DESIGN PROCEDURE
Though the design p~ocedure for helical gears is same as for spur gears, .the step by step
design procedure for helical gears is given below for ready reference.
.
1. Calculation
of gear ratio (i): Use 1. = Nt
N2
z2
= _
Zt
2. Select the suitable combinatioll of materials
Table 5.3.
for pinion and wheel,
consulting
3. If not given, assume gear life (say 20,000 Ius).
4. Calculation of initial design torque I Mt
[ M t]
Use
=
Mt
X
i:
K x Kd
Since datas are inadequate to select the values of K and Kd, initially assume K· Kd = 1.3.
5. Calculation of EelJ,
I
Ubi
and
I ucl :
./
Calculate the equivalent Young's modulus, Eeq, consulting Table 5.20 .
./
To find [ob]:
Calculate the design bending stress [ob]
./
To find [ 0c]:
Calculate the design contact stress [ ere] using the equation 6.21.
using the equation 6.19 .
6. Calculation of centre distance (a) :
Calculate the centre distance (a) using the equation 6.22.
7. Selection of number of teeth
./
If not given, assume
./
z2
(Zl
and zJ) :
-
Z I ~ 17.
= i x zl
8. Calculation of normal module (m,) :
Calculate the normal module using the relation
In
n
==
2a
(z) +z2
) x cos
A
I-'
choose the nearest higher
standard normal
Using the calculate d norma I 1nodule value
,',.
,
module from Table 5.8.
•
f. r\ •
9. Revision of centre distance Ia/ .
Using the chosen standard normal module, revise the
centre distance value.
Centre distance, a, ==
{c'ol11s' r:t) x (ZI
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I-'
;Z2)
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_6_.4_2
......,..............,..
............
.........,D_- 0......
\·..;.1[(11 of 1'rat/,vml,Y6JCJll
...
10. Calculatlon of b, ill'
V
b = \Jl x a. It
p,
l;;l
-t
1t
(.\
an ~
Calculate the pitch diameter of pinion (d I): Use d I
1t
Calculate the pitch line velocity (v):
Calculate the value
',~
be noted thut, fuce width b > P fr"
CUll
smooth operation, where Pa = Axial pitch
./'
i~llJ,tjfN
and VIp:
Calculate face width (b):
./
~""'__~fl"I';
Use
of\Vp:
\jlp
Usc v
=
>
u
el,
•
z I x IU n
tl
IJ
111"
cos ~ x z,
el, N,
60
b
ell
11. Select tile suitable quali(v 0/ gear, consulting Table 5.22.
12. Revision 0/ design torque I M,l:
Revise K: Using the calculated value of \Ill" revise the value of load concentration
./'
factor (K), from Table 5.11.
./'
Revise Kd: Using the selected quality of gear and calculated pitch line velocity,
revise the value of dynamic load factor (Kd), from Table 5.12 .
./'
Revise
I MIl:
[ M I].
Use [M I]
13. Check/or
Using the revised values of K and Kd, revise the design torque
=
MI
X
K x Kd
bending:
./'
Calculate the induced bending stress using the equation 6.18 .
./'
Compare the induced bending stress with the design bending stress .
./'
If
14. If
fJb
fJb
s [fJb ], then the design is satisfactory.
>[
fJb ],
then the design is not satisfactory. Then increase the module or face
width, or change the gear material. The procedure is repeated until the design is satisfactory.
i.e.,
Gb
s [fJb ].
15. Check for wear strength :
./
Calculate the induced contact stress using the equation 6.20 .
./'
Compare the induced contact stress with design contact stress .
./'
If
fJ ~ [ fJe],
e
then the design is safe and satisfactory.
. dilinensiOflS
16. Calculation of basic dimensions of tile gear pair: Calculate all the baSIC
of the pinion and gear using the equations listed in Table 6.1.
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~~/~G~e~ar._~_·'_'
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~
-:
~~
6.43
0ot!l
./
./
./
The above listed procedure is for th desi
. .
e esign of pinion .
As discussed in Section 5 31 If Il
(le.fi"II only tile pin!
1'1';"
te malerlals of.tlle pinion and gear are sallie, then
lJ
on. 'J me materials of til I I .
the pin 1011first (III I l 'kfi
e p ~~on «..~-gear are dlfferent, then des/gil
c C tee or botl: pill/Oil (lIldHell,:
The induced bending stress in th
(
-'
e gear ob2) can lit!' determined by using the relation
=
Obi Yvl
where
CJ b I and
cb2
°h2Yv2
=
Yvl and Yv2 =
... (6.23)
Induced bending stresses of pinion and gear respectively, and'
Form factors of pinion and gear respectively based on the
virtual number of teeth.
Since the contact area is same, the induced contact stress is same for both pinion and gear.
i.e., Oct = CJc2'
I Example
'6.16
I For
intermittent duty of
(lII
elevator, two cylindrical gears have to
1'(IIIs»I;112.5 kW lit II pinion speed of 1200 r.p.m: Design tile gear pair for tile following
specifications: Gear ratio 3.5, pressure angle 20 ~ involute fuJI dept", helix angle 15 ~
Gears are expected to work 6 "ours a dayfor 10 years.
Given Data : P = 12.5 kW;
N, = 1200 r.p.m.;
q, =20~
i = 3.5;
FO';
Toflnd : Design the hel ical gear pair.
f3'=
15°.
r
<,
© Solution :
1. Gear ratio:
... (Given)
i = 3.5
2. Selection 01 material : For both pinion and gear, alloy steel 40 Ni 2 Cr 1 Mo 28 can
of same material, we design only tile
be se'I ecte d, consu I'ting T a hie 5.3. AoJ,Since tile gears are
,
pinion.
3. Gear life:
..
Given that gears are to work 6 hours a day for 10 year.
Gear life
=
6 hours / day x 365 days / year x 10 = 21,900 hours
The gear life in terms of number of cycles is given by
N == 21,900 x 1200 x 60 = 157.7 x 107 cycles
4. Calculation l!f initial design torque I M,
W e know that,
J:
[ M /] == M, x K x Kd
3
where
Assume
/I -.1 '
l~)
,'c~-
M/ K· Ko
==
60 x p _ 60 x 12.5 x 10
2 1t N 21t x 1200
1.3
(_PS 6\
=
99.47 N-m
a , \~ )
== 99.47 x 1.3 == 129.31 N-m
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of Transmission
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5. Calculation of
~
~
I O'cl:
Toflnd E eq .' From Table 5.20, for steel pinion and steel gear,
Eeq
~
and
Eeq, 100bl
S
Tofind
I O'bI:
=
r~.
2.15 x lOs N/m~2
14)
We know that the design bending stress,
1.4 Kb/
= n-
[ CJb]
where
= 1.5, for steel hardened, from Table 5.15,
=
~ To find
where
~.
tV
2.5, for steel hardened, from Table 5.17, and
= 0.35 au + 120, for alloy steel, from Table 5.16.
=
1550 N/mm2, from Table 5.3.
= 0.35 x 1550 + 120 = 662.5
0'_1
1.4 x 0.7
x 1.5
= 2.5
[ O'b ]
662.5
x
N/mm2
= 173.133
N/mm2
I C'c J : We know that the design contact stress,
[ O'c] = CR x HRC x Kc/
(
CR = 26.5, for alloy steel hardened, from Table 5.18,
HRC
Kcl
...
{
107, from Table 5.14,
Ka
au
Then,
x
-
0'_1
...
0.7, for HB > 350 and N ~ 25
Kb/
n
But
K, x O'_J , for rotation in one direction
[ O'c ]
= 40 to 55, from Table 5.18, and
= 0.585, for steel HB > 350 and N ~ 25 x 107, from Table 5.19.
=
26.5
x
55
x
0.585
=
852.6 N/mm2
6. Calculation of centre distance (aJ : We know that,
where
\V -
b
-a =03.
... (initially assumed)
3 ( 0.7)2
(3.5 + 1)
852.6
a ~ 117.6 mm say a
7. Assume
Then,
2.}5 x 105 x 129.31 x UP
x
3.5 x 0.3
= 120 mm
zJ = 20.
Z2
= i X Zt =
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3.5
x
20
= 70
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...
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It
~
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G~ea:...,.,~'S=~=~~
Helical
~
~~
6.45
8. Cn/clI/alion of normal module (m,J :
We know that,
a
(z 2+.,.)
m n --
,
2 120
x cos ~ = _x-.:..=-=...
(20 + 70) x cos 15° = 2.576 mm
-2
From Table 5.8, the nearest higher standa d
r normal module is 3 mm.
9. Revision of centre distance:
We know that,
a
(c::.~)
=
10. Calcukuion of b, d I'
'"
Face width (b):
•
V
b
• .
AXIal pitch, Pa
3
cos 150
x
(20 + 70)
~
=
\jJ x a = 0.3 x 139.76 = 41.93:::::42 mm
mn
.
A
Sin tJ
1t
x3
=. Sin 15° = 364" mm We find b> Po'
'"
Pitch diameter of pinion (d,):
'"
Pitch line velocity (v):
'"
Tofind\jJp:
..
d L_.= -cos
--~
n d , N,
11. Selection
= 139.76 mm
and lJIp :
1t X
=
(ZI ;Z2 )
x
=
b
42
\jJp = d, = 62.12
of quality of gear:
=
60
v =
=
f3
7t
X ':"
-I --
cos3150 x 20
x 62.12 x 10-3 x 1200
60
=
62 . 12 mm
= 3.903 mls
0.676
From Table 5.22, for HB > 350 and v upto 8 m/s,
IS quality 8 is selected.
12. Revision of design torque
We know that,
where
[ M/]
K
Kd
...
[M, ]
13. Cileckfor
I Mt}-:
=
MI x K x K"
::::
1.045, for \fIp = 0.676, from Table 5.11, and
= 1.2, for IS quality 8, from Table 5.12.
= 99.47 x 1.045 x 1.2 = 124.74 N-m
bending : We know that the induced bending stress,
0.7(i+
(Jh
where
Yv
=
l)[M,]
a' b . Inn . Yv
= Form factor based (In virtual number of tooth
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6.46'
We find
zvl
-
z.
cos f3
Yv
-
0.402, for z,,)
ab
_ 0.7 (3.5 + 1) 124.74 x 103
139.76 x 42 x 3 x 0.402
ab < [a,,].
Thus
20
= cos! J 5°
3
=
%
22
22, from Table 5. J 3.
=
55.5 N/mm2
the design is satisfactory.
14. Check for wear strength: We know that the induced contact stress,
.... /;+1
.i+l
ae
=
\j ib
0.7 ---;;-
I) -------------------------3.5 + 1
.
42 2.15 UP
JOJ
3.5 +
- 0.7 ( 139.76
We find ac < [ ac
J.
x Eeq [M/]
3.5 x
x
x
x 124,74
645.8 N/mm2
Thus the design is safe and satisfactory.
15. Calculation of basic dimensions of pinion and gear:
~
Normal module: mn = 3 mm
~
Number of teeth:
~
Pitch circle diameter:
ZI = 20;
dI
-
and Z2 = 70
62.U mm;
mn
~
Refer Table 6.1.
R X Z,
cos.....
-
and
=
3
cos 150. x 70
=
217..4 mm
~
Centre distance :
~
Height factor :
~
Bottom clearance :
c -
0.25 mn = 0.25 x 3 = 0.75 mm
~
Tooth depth :
h -
2.25 mn
~
Tip diameter:
a = 139.76 mm
fo -
dOl
-
1
(C~I3+2fo)
=
225 x 3 = 6.75 mm
m.
- (a:~
50 + 2 x I ) 3 = 68.U mm; and
da2
=
(c:13
+ 2 f 0 ) m.
- (CO:~50 + 2 XI)
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3 = ..l23.41-
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Yo
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6.47
./
d /1 -
Root diameter:
cos '(3 -
0
)
n1n - 2 c
(co~~ S° - 2 x I ) 3 - 2 x 0.75
-
(0::
d f2 -
13 - 2 f
Virtual number of teeth :
zvl
22;
~
:\)
"1::"
2 x 0.75
and z
=
v2
[Example 6.17
IA
=
54.6 mm; and
=
209.91 mm
0) »,- 2 c
( cos7015° - 2 x 1 )
./
2f
(~
z2
=
70
cos3 (3
cos! 150 :::::78
helical gear speed up drive is reqllired to drive a centrifugal
compressorrun/ling at 3000 r.p.m. Tile helical gear speed up unit is driven by an electric
motor running at 1000 r.p.m. Tire compressor requires a nominal input power of 12.5 kW.
TIle helix angle of 25 0 may be assumed for the gears. Standard involute profile 200 full
dept" system will he used for the gear teeth. The gear pair is required to last for atleast
10,000 hrs. Design tile gear drive for tilefolio wing gear materia/so
Pillion: Heat treated cast steel.
Gel": High grade cast iron.
Given Data: N I = 3000 r.p.m.;
N2
=
1000 r.p.m.;
P = 12.5 kW;
$ = 20°
13 = 25°;
FD; n = 10,000 hours.
Toflnd : Design the hel ical gear drive.
@) Solution : Since the materials of plnlon and gear are different, we have to design the
pinionfirst and the check for gear.
. :
1. Gear ratio
.
I
N!
=N
2
=
3000
1000
=
3
2. Selection of material :
Pinion
=
Heat treated cast steel - Grade 1 i.e., CS 65
Gear = High grade cast iron - Grade 35 C.1.
J. Gear life: Given that n
cycles is given by
=
10,000 hours. T-hen the gear Iife in terms of number of
N = 10000 x 3000 x 60
4. Calculation of initial design torque I Mt
We know that,
[ M I]
=
MI X K x
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J:
=
180 x 107 cycles
..
~
x,
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Design o/Transmission SySIe""
6.48
where
K· Kd -
..! (initially assumed'
1.3.
= 39.788
[Mt]
Tofind Eeq : From Table 5.20, for steel pinion and C.l. gear (au> 280 N/mm2),
=
Eeq
../
= 51.725 N-m
x 1.3
I OCl :
5. Calculation 0/ Eeq, 1000bl and
../
60 x 12.5 x 1()3
27t x 3000
= 39.788 N-m, and
60 x P
27tN
Tofind
I O"b J:
1.7 x 105 N/mm2
We know that the design bending stress,
].4
-
11
Kb/
K
for rotation in one direction
x a_I'
(J
where
Kb/
-
I, for steel pinion, HB
s 350, N ~
~
-
] .5, for steel hardened,
from Table 5.15;
= 2, for cast steel heat treated, from Table 5.17, and
11
a_I
For CS 65, all
=
v" Tofind
=
400 Nzrnm-', from Table 5.3.
-
0.22 (650 + 400) + 50 = 281 N/mm2
-
].4 x ]
_
2 x 1.5 x 281 - 131.133 N/mm
I ocl : We know
[ae]
where
0.22 (all + ay) + 50, from Table 5.16.
-
650 Nzmm? and ay
Then,
107, from Table 5.14,
2
that the design contact stress,
- CR x HRC x KcI
CR - 22, for cast iron, from Table 5.18,
HRC - 55 to 63, from Table 5.18, and
KcI
[ae]
-
r,
for steel, HB
- 22 x 63 x 1
=
s 350, N ~
107, from Table 5.] 9.
1386 N/mm2
6. Calculation of centre distance (a): We know that,
(i + I)
where
\V -
b
a
=
0.3
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3
(0.7
[ae]
J
2 x
... (initially assumed)
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~::~G~~ __6
~~
6.49
.
..
a
i:.
or
7. As~ume
a
i:.
z. -
54.23 say a
20.
Zl
-
ix
(3
I)
( J~'~6)
2 x
1.7 x lOS x 51.725 x 1<P
3 x 0.3
Then,
8. Calculation
We know that
~
%.I:
60 mm
= 3 x 20 :; 60
ZI
0/ normal module (m,) :
m
n
2a
+ zV
= {z,
x COS
p
2; 60
= (20Y~ 60) x cos 25° = 1.359 mm
From Table 5.8, the
standa
. 2 mm.
. nearest higher
.
car d norma I module IS
9. Revision
0/ centre distance (a) :
a
We know that,
(z,
+ ~)
=
-'!!.!!_ ~
cos J3
-
2
(20 + 60)
cos 25CJ
2
Face width (b):
./
P'itc h eire
. Ie diiameter 0f prnron
. . (d ):
1
./
Pitch line velocity (v):
./
b
To find '¥p: '¥p = d,
a
=
./
y
v
=
= 88.27 mm
'!P:
10. Calculation of b, d I' v and
b = lV
2
=
= 26.48 mm
0.3 x 88.27
1t
d,
d, N,
60
26.48
44.14
=
mn
=--A
cos .....x
1t
= cos225°
Z,
x 20 = 44 14 mm
.
x 44.14 x 10-3 x 3000
60
= 6.93 m/s
= 0.6
11. Selection of qUlIlity of gear: From Table 5.22, for v upto 8 mis, the IS quality 8 gear
is selected.
12. Revision of design torque
We know that,
where
[ M I]
-
I MIl:
M I Y. K x Kd
K ::: 1.03, for 'Vp = 0.6, from Table 5.11, and
Kd:::
1.3, for IS quality 8 and v upto 8 mis, from Table 5.12.
[ M,] ::: 39.788 x 1.03 x 1.3 = 53.275 N-m
13. Check/or bending: We know that the induced bending stress,
0.7 (i
...,=
OL
+ 1) [ M f ]
a' b· mn' Yv
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Design of Transmission SY!te
6 SO
~.~--------------------------------~~------~
where'
Yv
-
Form factor based on virtual number of teeth
20
zi
zvl
Yvl
- cos! J3
~
-
cos
0.43, for
zvl
3
250 = 26.86 ~ 27
= 27, from Table 5.13.
0.7 (3 + 1) 53.275 x 103
ah - 88.27 x 26.48 x 2 x 0.43
...
= 74.21 N/mm2
We find ab < [ ab]. Thus the design is satisfactory.
14. Check for wear strength: We know that the induced contact stress,
Gc
= 0.7
e: -vW
1
)
U8~2~)
-
0.7
-
677.39 N/mm2
x Eeq x[Mtl
(3 ~ ;6~48) x 1.7 x 105 x 53.275 x 1()3
We find ac < [ ac]. Thus the design is safe and satisfactory.
15'. Check for wheel: Wheel material
First we have to calculate [ ab
Given that
]wheel
gear life -
=
Cast iron-Grade 35.
and [a c ]wheel
10,000 hrs
Npinion
Nwheel
Tofind
I Oblwheel
-
3
180 x 107
3
= 60 x 107 cycles
: We know that the design bending stress;:
[ ah ]wheel
where
Kb/
=
-
1.4 x Kb/ (wheel)
K
nx 0"
-\.[W,
Ka
n
a_I
...
Then,
x a_I
for cast iron, from Table 5.14,
107
60 x 107
For C.I Grade 35,
.
= 0.8195
-
1.2, for cast iron, from Table 5.15,
=
2, for cast iron, from Table 5.17,
-
0.45 aU' for cast iron, from Table 5.16.
au - 350 N/mm2, from Table 5.3.
0'_1
[ O'b ]wheel
-
0.45 x 350 = 157.5 N/mm2
-
1.4 x 0.8195
2 x 1.2
x 157.5
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.=
75.29 N/mm2
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~::../~O~ea_~_~
~~
6.51
./Tojlnd
I "clwhtel:
We know that the de~ign contact str
[ °c ]whecl
r:
",
where
eg
ess,
==
en x HB x Kcl
=:
2.3, for C.1. Grade 35, from Table 5.18,
HB = 200 to 260, for CI. Grade 35, from Table 5.18, and
«f
Kc{ =
.for cast iron, from Table 5.19.
'J
t
..
[ crc ]whec:1
(i! Che~k for
=
107
60 x 107
=
2.3 x 260 x 0.742
=
0.742
=
443.62 N/mm2
bending: The induced bending stress for wheel can be calculated using the
reIanon
crbl
where
=
=
Yvl
Yvl
°b2Yv2
0.43, for
zvl
= 27,
from Table 5.13.
60
z2
= cos) (3 = cos) 25° ~ 81
Yv2 = 0.499, for zv2 = 8~, from Table 5.13.
crbl = Induced bending stress for the pinion = 74.21
zv2
74.21 x 0.43
or
0b2
We find
0b2
We find
0c wheel>
0b2
x 0.499
63.95 N/mm2
Thus the design is satisfactory.
< [ CJb ]wheel'
(ii) Check/or wear:
=
=
Since the contact area is same, o c wheel
[crc
N/mm2
= o c pinion = 677.39
lwheel' Thus the design is unsatisfactory.
N/mm2.
In order to inc~
the
design contact stress, increase the hardness to 400 HB.
..
[ crc ]wheel
Now we find
CJ wheel
16. Calculation
v'
c
2.3 x 400 x 0.742 = 682.64 N/mm2
=
< [ 0c ]wheel'
Thus the design is safe and satisfactory.
0/ basic dimensions 0/ pinion
Normal module:
mn = 2 mm
v'
Number of teeth :
v'
Pitch circle diameter:
v' . Centre distance:
and gear: Refer Table 6.1.
Zl
~
= 20;
and
d1
= 44.14
Zl
.>:
= 60.
mm;
and d2
=
2
cos 250 x 60
=
132.4 mm
a = 88.27 mm
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6.52
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/0
../
Height factor:
../
Bottom clearance:
../
Tooth depth:
../
Tip diameter :
h
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=1
c = 0.25 mn
= 0.25
x2
= 0.5
= 2.25 mn = 2.25 x 2 = 4.5 mm
do I =
c:~
f3+ 2 f 0 )
mn
= (CO~~50 + 2 Xl)
d 02
,/
Rootdiameter:
mm
dl1
=
C::
f3+ 2 f 0 )
2 = 48.13 mm;
»,
=
(CO:~50 + 2 Xl)
=
(C:~f3-2fo)
=
2
136.4 mm
mn-2c
= (CO~~50 - 2 Xl) 2 - 2 x 0.5
d12 =
../
Virtual number of teeth :
C::
f3- 2 f
=
= 27;
39.13 mm; and
0) mn - 2 c
(CO:~50 - 2 Xl) 2 - 2 x 0.5 =
zvl
and
and
zv2
=
127.4 mm
~1.
6.15. HERRINGBONE (OR DOUBLE HELICAL) GEARS
Herringbone or double helical gear consists of teeth
having a right and left handed helix cut on tl-e same blank,
as illustrated schematically in Fig.6.13. One of the
disadvantages of the single helical gear is the existence of
axial thrust load. They. are eliminated by the Herringbone
configuration because the thrust force of the right hand is
balanced by that of the left hand helix. Helix angles are
usually greater for Herringbone gears than for single helical
gears because of the absence of the thrust reactions.
Fig. 6.13., Herringbone gear
INote I Though
the terms herringbone and double helical are used interchangeably, when there ~s
no groove in between the gears, is specifically known as herringbone gears (Fig.6.13). When there
groove in between the gears, then it is known as double helical gears.
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IS
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Helical Gears
6.S3
:...---
6.15.1. Design of Herringbone
Gears
The design formulas, the design procedure and other parameters of Herringbone gears are
exactly the same as that for parallel helix gear. Thus in order to design a Herringbone gear,
one can follow the procedure presented in Sections 6.12 and 6.14, without any modifications.
6.16. CROSSED-HELICAL
OR SPIRAL OR SCREW OR SKEW GEARS
A pair of crossed-helical
gears, also known as spiral gears, are shown in Fig.6.14. Spiral
gears are used to connect and transmit motion between two non-parallel and non-intersecting
shafts. As the contact between the mating teeth is always a point, these gears are suitable only
for transmitting a small amount of power.
In order for two helical gears to operate as crossed-helical gears, they must have the same
normal diametral pitch and normal pressure ~n' But the gears need not to have the same helix
angle or be opposite of hand. In most crossed gear applications, the gears have the same hand.
6.16.1. Advantages of Spiral Gears
v"
They provide noiseless operation due to smooth engagement.
v"
They can be used at any angles other than 90°.
v"
They permit a wide range of speed ratios without change of centre distance or gear
size,
6.16.2. Limitations of Spiral Gears
v"-
They transmit relatively small amounts of power because of point contact between
teeth .
../' They have lower efficiency than that of toothed gears because of sliding action.
6.16.3. Shaft Angle (9)
The shaft angle (9) is defined as the angle through which one of the shafts is rotated so
that it is parallel to the other shaft. Referring to the Fig:6.14,
Shaft angle,
e =
[31 + [32' when both gears of same hand
= [31 - [32'
where
[31 and [32
=
when gears are of opposite hand
Helical (or spiral) angles of gear teeth for pinion and
gear respectively.
6.16.4. Centre Distance for a Pair of Spiral ,Gears
The distance between gear centres is given by the shortest distance between the axes of
rotation.
Let
N I and N2
=
Speeds of pinion and gear,
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r
6.54
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Design o/Transmisslon SYlle""
--..:::..
Zl and z2
i-Gear
d 1 and d2
-
Number of teeth on pinion and gear respectively,
-
ratio
=
N)
z2
N2 - z 1 '
Pitch circle diameters of pinion and gear
respectively,
PI)
and Pt2 = Transverse
circular pitch of pinion and
gear respectively,
Pnl and Pn2 = Normal circular pitch of pinion and gear
respectively, and
a
= Least centre distance between the axes of
shaft.
n d,
We know that, PIJ
and P 12 ----
= --
ZJ
1td2
Z2
or
Since the normal pitch is same for both the spiral gears,
therefore
Pn
=
Pnl
=
=
Pt2 x cos 132
Pn2 = PII
cos 131
X
[.,' Pn = PI' cos 13]
Fig. 6.14. Centre distance/or a pail
of spiral gears
or
PII
=
Pn
d
Pn
cos 131 an Pt2 = cos 132
... Oi)
,
Substituting equation (ii) in (i), we get
d
I
=
»,
X
zl
1t x cos 131
Pn x z2
= ___;;_.:.:...._.......::.-
and d
2
1t x cos
.. , (iii)
132
But we know that the centre distance,
a
=
=l[
dl+d2
2
2
pnxz1
1t x cos 131
... (6.21(a»
=
e, zl
21t
[I
cos
i
13, + cos 132
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]
...
[
... i = zl
z2
J ...
(6.24(b»
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~::ca~/_G_e_~_~
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~~
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6.55
1. The centre distance in terms of normal
=
a
d 1 + d2
2
0.5
=
_! [
- 2
d
mo ule can be derived as
mn
m
xz
cos ~ I
=, [co:'I3,
+
+
I
n
cos ~2 x z2
]
... [.: d=
::13 xzJ
co: 13
2. From equation (iii), the speed ratio, i
2
=
]
2
... (6.25)
N I _ z2
N2 - zl
=
d2 cos ~2
dl cos ~I
... (6.26)
This expression is useful while solving problems.
6.16.6. Velocity of Sliding between Gears (v )
s
(vs) acts at the pitch point tangentially along the tooth surface is given
The sliding velocity
by
Vs,.
=
[vi + v; - 2
VI •
V2
x
cos
e ]112
... (6.27(a»
I
where
and
VI
v2
= Pitch line velocity of pinion and gear respectively, and
e =
Shaft angle.
When the shafts are at right angles (i.e., e = 90°), then the sliding velocity will be
V
6.16.5. Efficiency
s
=
VI
of Spiral Gears
We know that the efficiency
of spiral gears,
Work output
11 = Work input
and the maximum
efficiency
11max -
d
A
an
_ cos (9
+ cj) + cos «(31 - (32 -
- cos (9 - cj) + cos (~-
cj)
cos (9 + CP) + 1
cos (9 - cj) + 1
Shaft angle,
4> -
Angle of friction,
... (6.29)
and
= Helical (or spiral) angles of gears.
A_
P2
).
. I
s in terms of coefficient of friction (J.l
[Note I The efficiency of spira gear
.
11
... (6.28)
J31 + cj)
of spiral gears,
9 -
where
PI
... (6.27(b»
cos (32
=
1 - tan 132 x J.l
l+tanJ3lxJ.l
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.
15 given
b
y
... (6.30)
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6.56
Design of Transmission SYllebrJ
6.16.6. Force Analysis on Crossed-Helical
Gears
All the expressions derived for parallel helical gears in Section 6.8, is also applicableto
crossed-helical gears. Thus the forces on the crossed-helical gears are
2M
./
Transmitted force:
./
Radial
./
Axial thrust load: Fa
01'
F,
=
T
separating force: F,.
F,
=
tan all
cos \lI
X [
J
= F, X tan f3
where notations have usual meanings.
( Example 6.18
1/11
a spiral gear drive connecting two shafts, tile approximate centre
distance is 400 mill and the speed ratio = 3. Tile angle between the two shafts is 500andthe
normal pltcl: is 18 111m.Tile spiral angle for tile driving and driven wheels are equal: Find:
(1) Number of teeth
Oil
each wheel,
(2) Exact centre distance, and
(3) Efficiency of the drive, iffriction lingle
Given Data:
a
© Solution :
= 6~
= 400 mm = 0.4 m; i = ziz2
(1) Number of teeth
Given that the spiral angle
l3,
Oil
=
e
3;
= 50°; P n' = 18 mm . 't''" = 6°.
each wheel:
for the driving is equal to the spiral angle
l32 for the driven
wheel. Therefore,
or
e
=
l3, + l32
l31
=
f32 = 2"
e
=
25°
Centre distance between two shafts 'a' is given by,
a = Pn zl
[
27t
400
or
z,
and
z2
=
18 x
z)
I +
j
cos l31
cos
[
27t
1+3 ]
cos 25°
J_
f32 -
Pn
z)
I+i ]
[
27t
cos
l3)
•••
A =A2]
[ •.• 1-')
P
= 12.64 z)
400
= 12.64 = 31.64 or 32 ADS.
= i- zi = 3 x 32 = 96 ADS. "
"
(2) Exact centre distance:
Exact centre distance,
al
=
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)
[ cos
l3) +
i
cos
l32
]
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't
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~----___
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.
"
--
~6~.5!_
== 404.5 mm
(3) Efjlt/eney oJ the d,lvI! :
:<'j} (~ ~) cos f3 t
11
CUs (~,
""
JlI -Itll
e
~
C{)s
-
-
cos 132
ADS. ~
== COs (f3) + $)
cos (~_~)
[.:
PI == I3V
(25(' ~~
(25(' - 6(,) == 0.907
90.7%
ADS. '"t>
I
Example 6.19 ) A pair {)jNrpira/
'
'. . 2
gears c(Jnnects two sh "ts"
,
rauo IS and driver ha« 25 t th .r
0.1' inclined at 80 ~ The velocity
.'
ee oJ normal pitc« or J 2
the centre distance between the .vhaftv,
'J
mm and spiral angle of 30 ~ Find
Given Data :
e = 8(10.
'
", ,=(iJ2
@ Solution:
~
(J) J
Shaft angle,
== 2;
= ZJ
= 25
zi
'0'
P n -- 12 mm;
A
....
,
=
30°.
e f3, =
If zJ
=
25, then Zz = i
Z
zJ =
2 /. 25
We know that ,
dJ
=
and
d2
=
= 50.
Pn
xZ1
Pn~
1t
Centre distance between shafts. a =
f32
cos
d , + d2
2
= 203.5 mm
[Example
6.20
IA
12 x 25
= 110 mm
it cos 30°
=
1t cos P,
12 x 50
= 297 mm
1t x cos 50°
=
=
1] 0 + 297
2
ADS."
spiral wheel reduction gear, of ratio 3 to 2, is to be used on a
mac/line, with the angle between the shafts 80'! The approximate centre distance between
the shafts is 125 mm. The normal pilch of the teeth is 10 mm and wheel diameters are
equal. Find the number of teeth on each wheel, pitch circle diameter and spiral angles.
Find the efficiency of the drive
Given Data :
if the friction
Zz
3
i = - = 2';
z,
e
angle is 5 ~
Q
== 80
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;
a
= )25
mm;
Pn
= ]0
mm ;
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6.58
Design of TransmissionSY.fl
----------------------------------------------~~------
Tofind:
__ .~
e~
1. Number of teeth on each wheel (zi and z2),
2. Pitch circle diameters (d , and d2),
3. Spiral angles (PI and P2), and
4. Efficiency of the drive (11)·
(PI & P:J :
@ Solution: Spiral angles
d2
We know that,
-
dl
z2
Given that,
cos
zi
cos
WI
=
zi
z2
-
-
PI
P2
3
(l)2
2 '
3
cos
e
= PI + P2 =
80°
and d , = d2
PI
2 cos (80° - PI)
2 (cos 80° . cos PI + sin 80° sin PI) = 3 cos PI
or
131 = 53.4° ADS."
tan PI = ~:~~~: or
132
and
80° - 53.4°
=
=
ADS."
26.6°
Number of teeth on each wheel (ZI and z,J :
a
Centre distance,
125
=
r, zl
21t
_
5.34
=
z2
i z)
i
zi
]
131 + cos 132
cos
lOx zi
21t
125 and
[1
[1
cos 53.4°
or
zi =
+
(~) ]
cos 26.6°
23.4 say 24
3
= 2: x 24 =
ADS."
36 Ans. ~
Pitch circle diameters (d I and d~ :
Circular pitch,
Then,
. Efficiency of tile drive:
PCl
-
=
dl
=
d2
-
11
=
Pnzi
7t
=
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cos
131
Pn
cos PI
=
10 x 24
1t cos 53.4°
d;
=
cos
(132 +~)
cos PI
cos
(13) - ~)
cos
128 mm
=
128mm
Ans.1:)'
ADS. "
132
cos (26.6° + 5°) cos 53.4°
cos (53.4° - 5°) cos 26.6°
= 0 855 =
85.5%, Ans. ...,
.
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e-
-101 (;eJlTS_
~
~-l-e-6.-2-1"')A
»:
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6.59 .
angled dr~ve on Ii machine is to be made by two spiral wheels.
fIdls are of equal diameters with a normal pitch of 10 mm and centre distance is
ttl~.;moJely 150 mm: If the speed ratio is 2.5 to 1, find (a) spiral angles of
~b)
",,--
=:
0/
teeth on each wheel, (c) the exact centre distance,
~cy
if the coefficient
~iven
Data:
e
e Solution:
of friction between the teeth is 0.105.
d1
= 90°;
i
and (d) transmISSion
001
= 2
=
(J)
=
d2;
Pn
2.S;
J.1
a = 150 mrn ;
= 10 rnm ;
0.10S.
=
(a) Spiral angles of teeth :
d2
z2 cos (31
We know that,
d1 -
Z1 cos (32
(32 == 90° - (31
(31 + (32 = 90° ; or
cos (31
1 == 2.5 cos (90° - (31)
...
- 2.S cos (31
cos 90° cos (31 + sin 90° sin (31
sin (31
cos (31
or
tan (31
=
2.5
and (32
=
21.8° Ans.1:I
or
== 2.5
(31
68.2°
=
or
IJ.\
'7o.Tumber
of teet /;I on
tUI
J'Y
each wheel:
1
Pnzl
[
a = ~
We know that,
ISO
zl
or
z2
:::::
10 Z
__l
21t
a
E.n ~
>"
_
21t
154.2
:::::
45 Ans.
~~ [~+
1
[--A
.
1
coS ......1
OUD
go]
1 _ + _ 2.~
- 68 2°
cos 2 .
cos
.
Ans.1:I
-
d'stance:
I
[
17.5 say 18
_25x18=
i x z1 - .
and
(c) Exact centre
~
J
i
+ -cos (32
-1-
~
coS
]
-e
~]
c082\.8°
132
,ADS. 1:1
.'
.fjiciency:
. (d) TransnllsSIOn eJJ
~.::::
11 ::::: 1 + tall 131 X J.l
1 go x 0.10S
~
1 + tan 68.2 ~
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1
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6.60
Design of Transmis.fion S)l.f~
~
...
REVIEW AND SUMMARY
,
./'
A helical gear has teeth in the form of helix around the gear .
./'
Advalltage.f of hetica! gears: Less noisy, greater load carrying capacity, and
manufacturingfeasibiltty .
./'
Disadvantages of helical gears : They are subjected to axial thrust loads .
./'
Types: 1. Parallel helical gears,' and 2. Crossed-helical (or spiral) gears .
./'
The helical gear nomenclature and its kinematics are presented in the beginning ofthiJ
chapter .
./'
Helical gear nomenclature:
(i)
Transverse circular pitch:
-
z,
(ii) Normal circular pitch :
Pn =
(iii) Axial pitch:
PI
-.!!.E_
Pa = tan B = sin B
(iv) Normal diametral pitch:
Pd -
(v) Relation between fJ, at and an:
./'
xd,
PI = 7r' ml
tan an
PI
7
J.
n1
n
=
X
cos
-
P
-
Pn
cos
f3
= «<»,
-
7r·mn
sin
f3
-7r
Pn
tan a
.
l
cos
f3
An imaginary spur gear considered in the normal plane is called as the virtual or
formative spur gear. The number of teeth on the virtual spur gear in the normal plane is
known as virtual or formative or equivalent number of teeth. It is given by
z
eq
=
z
cost
p
Tooth proportions and basic dimensions of helical and herringbone gears are tabu/aled
in Table 6.1.
Force Qllalysis on helical gears:
Three components of resultant force on the gear tooth are :
(i)
-
2MI
d
Tangential component:
(ii) Radial component:
Fr
(iii) Axial or thrust force component:
v"
=
F [tan an ]
I
cos P
Fa = F, X tan
p
Two methods of deSigning a helical gear: 1. Helical gear design using Lewis
Buckingham
equations; and 2. Helical gear design based on gear life.
s
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and
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~~e~~=~================~
~~
_
I
The step by step procedure for the above
illustrative problems.
sa
id
6.61
two methods are preseme d with
wi sufficient
I Lewis beam strength equation for helical gears :
Beam strength,
I
F
= tr x m
S
n
X
b
X
[
uhi xy
,
Buckingham's equations for helical gears :
(i)
Fd = F + 21 v (cb .cost f3 + Fl) cos
I
21 v + cb . cos2 f3 + F,
Dynamic. Ioa 'd.,
trr
(ii) rr ear toot
-v
h I 0ad,
F
__ d J
x
W
A
fJ
bx Qx~
cos2 P
.; .Herringbone or double helical gear consists of teeth having a right and left' handed
helix cut on the same blank. Herringbone gears are used to overcome the difficulty of
end thrust in the single helical gear.
Crossed-helical or Spiral gears:
.; For connecting non-parallel and non-intersecting shafts, crossed helical gears are used
.;
Crossed helical gears transmit relatively small amounts of power because of point
contact between teeth
.f
Kinematics of spiral gears :
{l)
Shaft angle,
0
PI + P2' when both gears of same hand
=
= pI - P»- when gears are of opposite hand
(ii) Centre distance (a):
(iii) Gear ratio:
/
a
I
=
~;
=
NJ
N2
(iv) Velocity of sliding between gears.
+ c~2P2 ] =0.5
[c~JpJ
zl
= zl
.
=
=
V.f
m.[
2
c~JpJ +
CO:
P2 ]
d2 cos P2
d, cos PJ
tv} + v1 -2 v xv, x cos Ojll2
I
2
1_
(v) Efficiency of spiral gears:
17 =
cos (0 + ¢) + cos (PI - Pl - tPJ
cos (0- ¢) + cos (Pl - PI + ;)
l-tanPl
=
and
17max =
xp
.. .fin terms of angle offriction ;]
... [in terms of coefficient offriction pj
J + Ian PI x u
cos (0+ ;J + J
cos (0- ¢) + J
where the notations have usual meanings.
liP) MOM
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.i"~
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Design o/Transmission 8\1
6.62
~~teln3
~
REVIEW QUESTIONS
\
1.
What 'are the advantages of helical gears over spur gears?
2.
What is the major disadvantage
of single helical gear?
How can you overcome that
difficulty ?
3.
Differentiate the following terms with respect to helical gears:
(i) Transverse circular pitch; (ii) Normal circular pitch; and (iii) Axial pitch.
4.
What is virtual number of teeth in helical gears?
s.
Differentiate helical and herringbone gears.
6.
What are the components of resultant forces acting on a gear tooth of a helical gear?
Deduce the expressions of it.
7.
Write the expressions for beam strength, dynamic load, and limiting wear load for
helical gears and explain the various terms used in it.
8.
What is a spiral gear? What hands of helix are used?
9.
Why is the crossed helical gear drive not used for power transmission?
10. Deduce the expression for centre distance of spiral gears.
PROBLEMS FOR PRACTICE
Problems on kinematics of parallel helical gears :
I.
A pair of helical gear consists of a 20 teeth pinion meshing with a 70 teeth gear. The
normal module is 3 mm. Find the required value of the helix angle if the centre distance
is exactly 150 mm.
[Ans : 25.84°]
2.
A pair of parallel helical gears consists of a 25 teeth pinion and the velocity ratio is
3.5 : I. The helix angle is 20° and the normal module is 5 mm. Calculate: (1) the pitch
circle diameters of the pinion and the gear; and (ii) the centre distance.
[Ails: (i) 133 mm and 465.58 mm; (ii) 299.3 mm]
3.
A pair of parallel helical gears consists of an 20 teeth pinion meshing with 60 teeth ge~.
The normal module is 4 mm. The helix angle is 21 ° while the normal pressure angle IS
18°. Calculate : (i) the transverse module; (ii) the transverse pressure angle; and (iii) the
axial pitch.
[Ans: (i) 4.28 mm; (ii) '19.19°; (iii) 35.06 mm]
4.
A pair of parallel helical . gears consists of a 20 teeth pinion meshing with a 40 teeth gear.
. 3
The helix angle is 25° and the normal pressure angle is 20°. The normal module IS
mm. Calculate: (i) the transverse module; (ii) the transverse pressure angle; (iii) the
axial pitch; (iv) the pitch circle diameters of the pinion and the gear; (v) the centre
distance; and (vi) the addendum and dedendum circle' diameters of the pinion.
[Ans: (i) 3.31 mm; (ii)
21.88°; (iii) 22.3
n
mm; (iv) 66.2.mm and 132.4 mm; (v) 99.3 Jlu ;
(vi) 72.2 mm; 58.7 mm]
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tl.
~
5.
/icalG:,_e_o-rs---__
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---
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A stock helical. gear has a n
------6.63
ormal pr
transverse module of 4.25 rn
m, and hesc;ure angle of 141~o a h I'
transverse, the normal , and th e axial
.
.as 18 teeth .: FI'nd" (a) th·,e IX angle of 450' ~and a
Cd)the transverse pressure angle.
Circular pitches; (c) the n~~c~
dl~
(b) the
[Ans:(a) 76.5 rnm; (b) 13.35 mm: 9
dlametra) pitch; and
6. A parallel.helical
gearset Uses a 1'7-t.44
oot mhrn;'. 13.35 mm; (c) 0 .333 teeth! mm' (d) 20 O'.Fl
right-hand helix angle of 300
Pinion driving a 34-tooth
'
~
pitch 0.2 teeth/mm. Find' Ci),tha normal pressure angle of 200 an~ear. The pini~ has a
.
.
e normal tr
,a
normal diametraJ
transverse diarnetral pitch and th e transverse
,ansverse,
and axial circul at . prtc·
' hes:
. ,{ n..-J the
of each gear.
[H'.
pressure angle; and (iii) the itch
.
tnt . Transverse dia
.
.
p
tameter
[Ans: (I) 15.71mm; 18.14mm' 31 41m"
metral pitch = I/transverse module]
' .
m (11) 0 I 73mm' 22 79° (iii)
.,.;
111 98, I Smm; I963mm]
A para..IIe I-s h a ft gearset consists 0 fl'an 8-tooth helical
..
. .
pmion driving a 32-tooth gear.
Th e puuon has a left-hand helix angle of 250
normal module of 3 mm. Find' (a) th
' a normal pressure angIe of 2W, ana 2\
.
e normal transverse and . I'
I
the transverse module and the transv
'
,axta
circu ar pitches; (b)
the two gears.
.
erse pressure angle; and (c) the pitch diameters of
[Ans. (a) 9.42 mm; 10.4 rnrn; 22.3 mm; (b) 3.31 rnrn;. 2L8SC',
d'
7.
(c) 59.58 mrn; 105.92 mm]
Problems on force analysis of parallel helical gears:
8. A pair of parallel helical gears consists of an 22 teeth pinion meshing with a 66 teeth
gear. A 10 kW power at 2200 r.p.m. is supplied to the pinion through its. shaft. The
normal module is 5 mm, while the normal pressure angle is 20°. The helix angle is 25
Determine the tangential, radial and axial components of the resultant tooth force
between the meshing teeth.
[Ans: 715.24 N; 287.14 N; 33352 ~1
C
•
9.
A helical gear of 60° helix angle is required to transmit 10 kW of power at 500 r.p.m,
The pressure angle measured in a transverse plane is 20°. Calculate (a) driving force; (b)
radial force acting as bending force on the shaft; and (c) the axial force on the shaft.
The gear meshes with a pinion rotating at a speed of 914.5 r.p.rn. and sum of number of
teeth on pinion and gear is 99. The module of gears in a plane normal to the tooth is 4
mm. Calculate the angle of pressure in a plane normal to the tooth.
[Ans: (a) 12993 N; (b) 473 N; (e) 750 Nand 115°1
RH
z=2D
10. A pair of parallel helical gears is shown in Fig.6.15 '. A 5 kW
.'
A th ough Its shaft
power at 720 r.p.m. is supplied to pmron
r
..
. 5 rnrn and the normal pressure angle IS
Th e norma I mo d u Ie IS.
d t th while the gear has left200. The pinion has nght-han . 300
ee ,The arrow .indIcates
.
the
hand teeth. The helix angle IS
from the right-hand side.
direction of rotation when seen th fi
Also draw a free
ts fthe too
orce.
Determine the componen
0
. h
'nion and the gear.
body diagram showing the forces on t e pi
[Ans: 11486., N' 482 .73 N; 663.14 N]
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A
t
LH
Fig. 6./5.
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6.64
.
.
"--
. 5h own In
. Fig .6 -16_ Pinion A 'ISni!!l.e W-Jl~
I I' I gear .box 15
1 The layout of a double- ie rca
li d to it through ;1:5 shan D .J. ~ llllJnle
I .
d 10 kW power at 720 r.p.m. IS supp ie . • ~ 50' z ~ 2{)' and _ ~ [,
""
gear
an
. I
are' Z = 20, -B
~
C
O
f teeth on different helica gears
. A
.
f helical .o:-e.arsA and IR :bf fut:h
. 200 For the pair I(} ...,11::=>
_
_
normal pressure angle for all gears IS "
F the pair C and I), the MIX an e''k> '!5 2::'"
angle is 30°, and the normal module IS, :~~ ~~t handed helical teerh, ,,~ • pirilD .
and the normal module is 5 mm. Pml~n and B are mounted on shaft no.] i:n ;;ocih > 't'
has a left handed helix. The beanngs
I
. 2
.
B an take hoJ;b "",,,,I... ,~ :a;
I
k
adial load while bearing
2
c
that bearing B) can on y ta e r
"
f bearinz reactions on
. t Iie magn itude and direction
~
thrust load. Determine
D
RH o
shpn.:J
RH
Fig. 6./6.
[Ails:
FBI
r
= 5044.58
FJ~2 = ) 559.9)
N ; FYBI
N; F~2
= 2'64
.) .);- N
= 592.13
N;
, = 479456 N:
FJ:B_~
FBl
== 557 L23 N: FE:! = sa:«
Problems on helical gear.design using Lewis and Buckingham's
~x
equation:
12. A pair of helical gears are used to transmit 5 kW at 720 r. p. m. of the pinion. Gears are ""
C45 steel. The speed reduction ratio is 2. Number of teeth on piniOn is 10. ~,. •
pressure angle is 20°. Normal module is 5 rnrn. Helix angle is 300. Design the gear drive
13. Design a helical gear to transmit 7.5 kW at 1400 r.p.m, to the fullowing specificsnons:
Speed reduction 3. Pressure angle 20·. Helix angle 10•. Design surface compressive
stress x 900
2.05
105 N/mm'.
N/mm2. Design bending stress 210 N/mm'. Modulus of elasticity ofmaiem!;
14. A pair of helical gears is to transmit 40 kW at 1800 r.p.m. The speed reduction n:quin><i
is 4 and helix angle is 15'. Design the drive for COntinuous service with moderate sMct
loads. Pressure angle is 20·. Assume the suitable materials for the drive.
IS. Design a helical gear drive to transmit 5 kW at 1440 r.p.m. Speed ratio is 2.5. SeI«'
suitable material for the gears.
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~~~~~~~----------------~
.'
6.65
A helical cast steel with 30° helix an I .
}6.
th determi
gels to transm't 36 kW
bas 24 tee
etermme the necessary modul
.
I.
at 1500 r.p.m. If the gear
depth teeth. The static stress for cast st I~' pitch diameter and face width for 20° full
. h F' d al
ee IS 55 N/mm2 F
.
.
nonnal prtc . in
so the end thrust on the
..
ace Width IS three times the
. fh l'
gear.
17. Design a parr 0 e ical gears to transmit 5 kW
..
transmission ratio is 5. Take minimum
b at a PIllion speed of 1440 r.p.m. Required
gears is C 45 steel. Sketch the arrangem:~
er of teeth as 22. The material of both the
frOblems.on "eli~al gear design based
011
gear life:
18. A pair of helical gears subjected to moderate shock loadi'
. .
r.p.m. of the pinion. The speed redu ti
"
ng IS to transmit 30 kW at 1500
c Ion ratio IS4 and the helix
I' 20° Th
.
is continuous and the teeth are 200 FD'
h
ang e IS
. e service
hours, design the gear drive.
~ t e normal plane. For the gear life of 10,000
19. A pair of helical gears with 23° helix angle is t tr
. 2 k
. .
'.
.
0 ansrnit
.5 W at 1000 r.p.m. of the
pmron. The velocity ratio IS 4 : 1. The pinion is t b f
d
I
.
.
o e orge stee and the driven gear IS
to be cast steel. The gears are of 20° full depth involute fiorm d th "
.
h
. th
d .
. an
e pirnon ISto .ave 24
tee th . D esign e gear rive.
20. For the data of problem 19, if the gear is to work 8 hours per day, six days a week and
for 3 years, design the gear drive.
Problems on herringbone gear design :
21. A .power of ~O kW at 1440 r.p.m. is to be transmitted with a gear ratio of 4. Design a
suitable herrmgbone gear drive.
22. Design a herringbone drive, for a 2 kW steam turbine running at 30,000 r.p.m., to reduce
the speed to 2500 r.p.m. The helix angle is 30°. Take the gear life as 10,000 hours.
Problems Oil kinematics of crossed-helical (spiral) gears:
23. A pair of spiral gears is required to connect two shafts 175 mm apart, the shaft: angle
being 70°. The velocity ratio is to be l.5 to 1, the faster wheel having 80 teeth and a
pitch circle diameter of 100 mm. Find the spiral angles for each wheel.
[Ans: 54.65°; 15.35°]
24. Two shafts, inclined at an angle of 65° and with a least distance between them of 175 '
mm are to be connected by spiral gears of normal pitch 15 mm to give a reduction ratio
3: 1. Find suitable diameters and number of teeth. Determine also the maximum
efficiency of the spiral gear, if the friction angle is 7°.
[Ails: 88.5 mm; 245.7 mm; 15; 45; 85.5%]
25. A drive on a machine tool is to be made by two spiral gear wheels, the spirals of which
are of the same hand and has normal pitch of 12.5 mm. The wheels are of equal diameter
and the centre distance between the axes of the shafts is approximately 134 mm. The
angle between the shafts is 80° and the speed ratio 1.25'. Determine: 1. The spiral angle
of each wheel; 2. the number of teeth on each wheel; 3. the efficiency of the drive; if the
friction angle is 6°; and 4. the maximum efficiency.
.
[Ans: 1. 32.46~; 47.54°;
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2. 30; 24; 3. 83%;
4. 83.8%]
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Bevel Gears
"A good scientist is a person with original ideas. A good engineer is a perUJriwh&
makes a design that works with as few original ideas as possible
There are no prima donnas in engineering. .-
- Freeman DyJO"
7.1. INTRODUCTION
(Bevel
gears are used to transmit power
~o
intersecting shafts. Bevel gears are
commonly used in automotive differentials. The
gears are formed by cutting teeth along the
elements of frustum of a cone. That is, the pitch
surface in the bevel gears are truncated cone,
one of which rolls over the other, as shown in
Fig.7.!. When teeth formed on the. cones are
straight, the gears are known as slrlliglzt bev!!j
and when inclined, they are known as spiral or
helical beveL
Fig. 7.1. Bevel gear
Bevel gears are mounted on intersecting shafts at any desired angJe, although 90° shaft
angle is most common. Bevel gears are not interchangeable. Because they are designed and
manufactured in pairs.
The bevel gear teeth can be cast, miJIed, or generated. But the generated teeth is more
accurate than cast and milled teeth.
_.
-
••2. TYPES OF BEVEL GEARS
--'
e teet on the bevel gears are parallel to
the lines generating the pitch cones, then they are
called straight bevel gears. As shown in Fig.7 .2, the
teeth are straight, radial to the point of intersection
of the shaft axes and vary in cross-section
throughout their length. Usually, they are used to
connect- shafts at right angles which run at low
speeds.
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Fig. 7.2. Stratght bevel gellr
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~-~~~=-------~-------------__
_JD~e~s~ign~O~if~Tl~ran~s~m~~~s~ro~n
iral sevel Gears
2.SP~
Jrhell the teeth of a bevel gear are inclined at an
I to the face of the bevel, they are known
.
a"g,e
. .
as spITal
earS ' as shown In Flg.7.3. To reduce th e norse,
.
)eve Ig
beII'cal teeth is.,used on these bevel gears . TIley are
sllloother in action and quieter than straight tooth bevels
as there is gradual load application and low impact
stresses.
However, an axial thrust exists in the spiral bevel
geMS, so it requires stronger bearings and supporting
assemblies.These gears are used for the drive to the
differentialof an automobile,
Fig. 7.3. Spiral bevel gear
3.Zerol Bevel Gears
Spiral bevel gears with curved teeth but with
a zero degree spiral angle are known as zerol
bevel gears. Refer Fig. 7.4. Their tooth action and
the end thrust are the same as that of straight
bevel gears. Zero I bevel gears are quicker in
actionthan the straight bevel type as the teeth are
curved.
4.~oid
Fig. 7.4. Zerol bevel gear
gears :
Hypoid gears are similar in appearance to
spiral-bevel gears. They differ from spiral gears
inthat the axis of pinion is offset from the axis
of the gear, as shown in Fig.7.S. The other
difference is that their pitch surfaces are
hyperboloids rather than cones.
In general, hypoid gears are most desirable
for those applications involving large speed
reduction ratios. They operate more smoothly
andquietly than spiral bevel gears.
Fig. 7.5. Hypoid gears
7.2.2. Classification Based on tch Angle
. 1. Crown gear: A bevel
ar having a pitch angle of 900 and a plane for its pitch surface
IS
known as a crown gear.
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Bevel Gears
2. Internal bevel gear: When the pitch angle of a bevel gear exceeds 90°, it is called
internal bevel gear. Because of the manufacturing difficulties, the internal bevel gears ~
rarely used.
3. Mitre gears: When two meshing bevel gears have a shaft angle of 90° and bave the
same number of teeth, they are called mitre gears. In other words, mitre gears have a speed
ratio of 1. Each of the two gears has a 45° pitch angle.
~ .3. BEVEL GEAR NOMENCLATURE
The geometry of a be~1 gear set is shown in Fig. 7.6. The various terms used in the study
of b~~el gears have been.explained
below.
1. Pilcll cone: It is the cone containing the pitch elements of the teeth.
2. Con~e:
It is the point where the axes of two mating ~ars
other words, it is the apex of the pitch cone.
in~sect_eacl!_ other. In
3. Pitch angle (or semi-cone angle) (8) : It is the angle made by the pitch line of a gear
with the gear axis.
Cone centre
..
,
\
I
,/,
\,
Back
cone
""I'
'\JY
I
\
/
6\~~
! /
\ I
,.
,
.vO~0~
~~~
.
Fig. 7.6. Bevel gear nomenclature
4. Cone distance (or pitcll cone radius) (R) : It is the length of the pitch cone
Mathematically,
QI~Dleot.
cone distance (R) is given by
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)
•
1
I
1
I
,,
Transmission Systems
~---------------------------_Q~~~~~~~~
Design of
R == Pitch radius _ (d 112)
sino
- == (d/2)
sin 0,
sin O2
R
or
d, and d-,
where
z 1 and
°
1
=1
=
>
.;
J.;
111,x zi == 111,x z2
2 sin 1
2 . sin 02
°
[ .: m, = dlz]
... (7.1)
Pitch circle diamete rs 0f oi
punon and gear respectively
== Number of teeth f . .
'
.
0 pimon and gear respectively,
=
and 0) == PItch angles of pnuon
. . and gear respectively, and
111, =
Transverse module
Cone distance can also be given by
_ [(dl)22
(d2)2]~
+"2
R -
For right angle gears, R = 0.5
~
where
j
=~ + =~
111,~
1
= 2" ~
=
... (7.2)
d~ + d;
0.5 m, x
z, ...J ;2 + I
... (7.3)
'i
I,
i•
i
= Gear ratio = -I
~ t:-2'
!
angle (oa) : It is the angle subtended by the addendum of the tooth at the
5. Addendum
cone centre. Mathematically,
e
Addendum angle,
j
~
= tarr ' (
addendum)
cone
distance
=
tan- 1
\
(ha)R
... (7.4)
6. Dedendum angle (Od) : It is the angle subtended by the dedendum of the tooth at the
cone centre. Mathematically,
Dedendum angle,
'-',
ad =
tarr ' ( cone
ded~~d=
IS
ce )
=
tarr '
(hRf)
. .. (7 .)S"
7. Tip (orfa.cM.~~/e : It is the angle subtended bv the face of the tooth at the cone centre.
Mathematically,~
./
Tip angle
8. Root angle:
Mathematican@_)
=
Pitch angle + Addendum angle
= 0 + ea
...
(7.6)
It is the angle subtended by the root of the tooth at the cone centre.
.
Root angle
r
,I
11
~
=
Pitch angle - Dedendum angle
= 8 - e/
... (7.7)
9. Back (or normal) cone: It is an imaginary cone, perpendicular to the pitch cone at the
end of the tooth.
10. Back cone disttmce (or back cone radius) : It is the length ofthe back cone.
r
:!
!
1
11. Backing (B) : It is the distanc~ of the pitch point from the back of the boss, parallel to
the axis of the gear. "
12. MOUII/ing height : It is the distance of the back of the boss from the cone centre.
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~B~ev~e=I~G~e~a~~~
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~
13. Pitch diameter: It is the diameter of the largest pitch circle.
14. Outside or addendum cone diameter: It is the maximum diameter of the teeth of the
gear. It is equal to the diameter of the blank from which the gear can be cut. Mathematically,
Outside diameter -
ha -
where
+ 2 x ha x cos
Pitch diameter
Addendum, and
o -
Pitch angle.
,
I
15. Inside or dedendum cone diameter : It is given by
Inside diameter
" where
hi
'" (7.8)
<)
=
Pitch diameter - 2 x hi x cos
=
Dedendum
°
... (7.9)
Angle relationsfor different bevel gears:
e
1. Acute angle bevel gears: If the shaft angle < 90°, then the bevel gears are knownas
acute angle bevel gears. The pitch angles are given by
tan 01 and
where
tan
°
2
e
°
I and
02
-
sin
e
(z2 / ZI) + cos
sin
(zI / z2)
e
+ cos
e
e
- Shaft angles =
=
... (7.10)
°+
1
Pitch angles of pinion and gear respectively.
2. Right angle bevel gears: If shaft angle
e = 90°, then the bevel gears
angle bevel gears. The pitch angles are given by
tan
°
2
02' and
=, ~
zi
= i ; and
°
1=
90° -
are known as right
°
2
...
(7.11)
3. Obtuse (Ingle bevel gears: If shaft angle 8 > 90°, then the bevel gears are known as
obtuse angle bevel gears. The pitch angles are given by
sin (1800
-
8)
and
7.4.
VIRTUAL OR FORMATIVE OR EQIJIVALENT NUMBER OF TEETH FOR
BEVEL GEARS
In order to simplify the design calculation and analysis, bevel gears are replaced by
equivalent spur gears. An imaginary spur gear considered in a plane perpendicular to the
toroth at the larger end, is known as vinuai orformative or equivalent spur:gear. The virtual
spur gear has pitch circle radius equal to the pitch cone radius 'R' of the bevel gear.
...
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__
~Design 0/11
."
umber of teeth z on thi
.
ransnwuJlon Systems
The n
v
IS ImagInary
.
b if
.
spur
gear
.
I
equivalent num er 0 teet h. It IS given by
IS ca Jed virtual or /orlllllilve or
z
z
cos 0
where z = Actual number of teeth on th b
e evel gear.
~
The virtual number of teeth is Used f
I .
~
or se ectIngthe cun
v
-
-
--
bevelgears.
... (7.13)
ers and
.
In all
.
desJgn calculations of
7.5. PROPORTIONS FOR BEVEL GEARS
The proportions for the bevel gears are given below:
I. Addendum,
ha - I ml
2. Dedendum,
hf
-
].2111,
3.
Clearance,
4.
Working depth, hw - 2111,
c - 0.2112,
5. Thickness of tooth
where
- 1.5708 m
l
111, =
7.6. BASIC DIMENSIONS
Transverse module.
OF BEVEL GEARS
The basic dimensions of straight bevel gears are listed in Table 7.1.
Table 7.1. Basic dimensions of bevel gears (from data book, page no. 8.38)
S.No.
Nomenclature
Units
m(
mm
I.
Transverse
2.
Mean module
mm
mm
3.
Normal module
mn
mm
4.
Cone distance
R
mm
Reference or pitch diameter
d
mm
'- 5.
6.
module
Symbol
mm
Tip (or face) diameter
Face width
Formula
b sin 8
m(=mm+-zInm =
dol
do2
b
mm
b
:::s
mt-
b sin 0
z
m,(zl
=
+2cos81)
nil (Z2 + 2 cos 82)
0.3 R or 10m" whichever is smaller
2R
8.
Number
of teeth on crown
zClV
-
m
l
wheel
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BB~eevvEe~ll(GIEe~a~rs~
---------------------------------~
Nomenclature
S.No.
Minimum number of teeth
avoid
to
pinion
on
9.
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Symbol
Units
Zu
-
Formula
2 X cos °
-
Zu
. ?~
sm-
,
; (X
I. for ~
l
gears)
I
undercutting
I
20° usually
~
degrees
~
Pm
degrees
P2
tan 02
10.
Pressure angle
11.
Mean spiral angle
12.
Reference or pitch angle
0
degrees
13.
Addendum angle
°a
degrees
14.
Dedendum angle
8f
degrees
15.
Height factor
fo
-
16.
Clearance
c
-
17.
Tip angle
°a
degrees
oa -
18.
Root angle
Of
degrees
8
19.
Tooth height
Ii
mrn
h
20.
Working depth
hw
mm
hw
21.
Addendum
ha
mm
ha = mr
22.
Dedendum
hi
mm
hi = 1.1236m1
23.
Virtual number of teeth
-v
-
I Example
7.1
I A pair
30 to 35°;
t , 01
35° preferred
90°
= tan 8a2 =
tan 0al
°2
mt
=I»
R
-
mtUO+c)
= tan °/2 =
tan 8/1
R
-
1
/0
J
= 0.2
c
I
0 + Sa
/ - 0 - S/
= Iza + hi
...
-\"
2 mr
=
-=-. -
.
. =
= cos 0 . -\' 1m tn
I
18 for 6. =20"
of straight bevel gears consists of a 30 teeth pinion ~
with a 48 teeth gear. Tile gears are mounted on shafts, which are inJers«ting
angle. Tile module at the large end of the tooth is 4 nun. Calculate:
(i)
I
111
tile pitch circle diameters of tile pinion and tire gear;
(ii) the pitch angles for tile pillion and gear; and
(iii) tile cone distance.
Give" Data:
To find:
©Solution:
zl = 30;
(i) d) and d2;
.
z2
I
= ~
z2 = 48' 'I 111 = 4 rnrn :,
(ii) 0) and
°
2
;
e = 900
.
and (iii) R.
48
= 30 = 1.6
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I
n:-
. .IS
...
', .....
, . .•. (her
,.j..
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;<.
HI.
1',."
J
!
.
,!
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From : www.EasyEngineering.net
".bh'iQ'W'
;1
.' .. ,_j
.
-
i·
.-
Iii) Cone distance (R) :
~'e kno\\ thar
R = 0.5 m t
=
0.5 x 4
\}
-oJ
z~ + ~
302 + 482
=
113.21 mm
Ans."
1.7. FORCE ANALYSIS ON BEVEL GEARS
In force analysis of bevel gears, it is assumed that the resultant tooth force between two
~g
gears -"15 concentrated at the midpoint along the face width of the tooth. The forces
anmg at the centre of the tooth are shown in Fig.7.7.
The components of the resultant force are :
L Tangential or useful component (Ft), and
_.
~~.'
J'l.JXU
cuing
c.
force
(F)s:
It
I'S
resolved into two components. They are
(i) Axial force (Fa), and
(ii) Radial force (Fr)'
:· ¥
·, .' 1
7
,
,,
:, ,,/ Fs
"
11
_
(b)
(a)
Fig. ...7. 7. Gear tooth forces
tf
j
J
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-,B~~~\~~·/~.!~t~1~~·~~
--------------------------------__
~
7
(i) COnlpfmenll' t'lille tooth force on the pin 1011:
Tojill(1 14,: The tnngential force can be determined
2 M,
M,
F~-:.:,
dI
(1\1
'" (1.14)
"",
M, ::.: Transmitted torque
where
P
==
using the familiar relationship
=
60 x P
21tN '
Power transmitted,
N = Speed of the gear,
d1tll,
1'",
= Average diameter of the pinion, at midpoint along the face width
=
(111,- b .::n Il ), and
=
ZI'
=
(d21 _
=
Mean radius of the pinion at midpoint along the face width
ilia,
b
ZI
S~l
0, )
To find F!i: The analysis is similar to that of spur gears and the separating force can be
determined using the relation
Separating force, Fs = F, x tan
where
=
(l
Tofind F;
(lilt!
... (i)
(l
Pressure angle
Fa : The separating force is further resolved into radial and axial forces,as
shown in Fig.7.7(b).
From the geometry of the Fig. 7.7(b), we can write
Radial force, F,
= F, x cos
IS
.,. (iiJ)
axial force, Fa = Fs x sin IS
and
.., (ii)
Substituting equation (i) in the above equations, we get
Radial force, F, = F,' tan
axial force, Fa
and
=
F,' tan
(l .
cos IS
(7.15)
(l •
sin IS
(7.16)
The above derived expressions are used to determine the components
the pinion.
of the tooth forceon
(ii) Components of the tooth force on the gear:
From the Fig. 7.7, the following conclusions can be made for the right angle bevel gears:
./
The radial component on the gear is equal to the axial component on the pini()l1,
but in opposite dlrecuon.
:.
(Fr)gcar
= - ( Fa )plnlon
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~----------------------------
~D~~~ign~~of~TJ~~~a~ns~n~u~ss~io~n~~
./ Similarly, the
axial component
on the gear is equal to th e radilaI component on the
.
'"
pinion, but m opposite direction,
.
.. (F a )gear = - (F r )pinion
§i!J The three
forces FI' Fr' and
Fa are perpendicular to each other and can be used to
detennine the bearing loads by using the methods of statics.
I A pair
[§xample 7.2
of straight bevel gears has a velocity ratio of 2: 1. Tile pitch
circle diameter of the pinion is
80
mm at the large end of the tooth.
A 5 kW power
is
suppliedto the pinion, which rotates at 800 r.p.m: TI,e face width is 40 mm and the
pressure angle is 20 ~ Calculate the tangential, radial and axial components of the
resultallt
toothforce acting on the pillion.
GivenDtua : i=2;
d1=80mm;
To find: FI' Frand
Fa on pinion.
@) Solution:
(°1 and
°
2),
N1=800r.p.m.;
P=5kW;
b=40mm;a=200.
In order to calculate the force components, first let us find pitch angles
and the mean radius of pinion.
Weknow that,
tan 02 -
i
=
I
e = 90°.
2, for shaft angle,
or
02 -
tarr ' (2)
=
63.43°
and
01 -
90° - 02
=
90° - 63.43°
= 26.57°
. of the puuon
. , at rruidnoi
t long the face width is given by,
TIie mean radius
porn a
d1
b . sin 81
r; = "2"
2
80
- 2
40 x sin 26.57°
2
=
31.054 mrn
(i) Tangential component (Ft) :
We know that,
r, =
2Mt
-d
=-:-1
lav
III
3
Where
..
60 x p _ 60 x 5 x 10
M, = 27tN1' 27t x 800
F, =
59.68
31.054
= 59.68 N-m
= 1921.91 N ADS. ~
x 10-3
(Ii) Radial component (F,) :
We know that,
I
°1
F' ::; F, x tan a x cos
r = 1921.91 x tall 20° x cos 26.57°
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= 625.64
N
ADS. ""SJ
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v:
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i
Bevel Gears
7
~~~~----------------------------------~
(iii) Axial component (FoJ :
Fa - F,
We know that,
tan a x sin 0,
X
= 1921.91 x tan 20° x sin 26.57°
I Example
I For the
7.3
= 312.89 N
Ans."
data of above example, calculate the tangential, radilll and
axial components of the resultant tooth force acting on the gear wheel
Given Data: Refer Example 7.2;
To find:
N2
Nl
800
i =2 =
=
400 r.p.m.
F" Frand Fa on gear.
@)Solutio_,,: 0,
= 26.57°
and O2 = 63.43° from Example 7.2.
(i) Tangential component (FJ : We know that the tangential force on the pinion and gear
are equal in magnitude and opposite in direction. Therefore,
=
(Fr)gear
(F,)pinion
INote I The above result can be verified
where (M r)gear =
rm2
60 x P
2n N2
=
=
1921.91
N
Ans."
as below.
60 x 5 x 103
=
2n x 400
119.366 N-m
= Mean radius of the gear at the midpoint
b· sin 02
2
d2
2
==
] 60
2
[':d2=ixdl=2x80=I60mm]
40 x sin 63.43°
2
=
62.11 mm
) 19.366
.-
(Fr)gear == 62.11 x 10-3 == 1921.85 N, which is the same value.
(ii) Radial component (F,) :
We know that ,
(F)r gear = Fr
=
INote I It can be seen that,
(F)
r
X
tan a. x cos O2
1921.91 x tan 20° x cos 63.43 = 312.88
gear
=
N Ans. ~
(F ) __
a ptruon •
(iii) Axial component (FJ :
We know that
,
(F )
a gear = FI x tan a. x sin 0
INote I It can be seen that,
2
::: 192] .9] x tan 20° x sin 63.43 = 625.64 N
(F )
a gear
=
ADS.
-e
(F) _ _
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r Pinion .
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Design of
Transmission
Systems
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~-
~
7.4) A pair of bevel gears is to be used to transmit 10 kW from a pinion
~20
r.p.m.. ttl a gear mo.u~ted .on a shaft which intersects the pinion shaft q.t an
tatill
rO
700. ,Assunuftg that th~ pinion IS to have an outside pitch diameter of 180 mm, a
!if
angle of 20 ~ a face width of 45 mm, and the gear shaft is to rotate at 140 r.p.m;
presslll'~ (i) the pitch angle for the gears; (ii) tile forces'on the pinion and gear; and
dtltrtnlne ue produced about the shaft axis.
"iI the torq
(iJll
Data. P == 10 kW; N} = 420 r.p.m; e = 70°; d} = 180 mm; a. = 20°; b = 45 mm;
Given
•
N2::: 140r.p.m.
d8
(ii) F; F, and Fa on pinion and gear; and (iii) (M,)gear .
Tofind: (i) 8} an 2;
.
N}
420 _ 3 _ z2
i
==
N2
==
140 - zl
@ Soilltion :
IfIIIlt
(i) Pitch angles for the gears (i.e., 81.and 8.) :
sin B
We know that,
...
sin 70°
:::' 0.281
3 + cos 70°
Ans . ...,
tan~1(0.281) =..15.70
or
e
We also know that,
or
.
== &1 + &2
.
e - 8\
&2 ==
_
rml
-
Let us first
d1
b . sin 01
- -2 -
2
180
-
-2
. d
the pinion,
Torque transffiltte on
. lion .
(a) Forces on the p';.
(i)
Tangentilll force:
(ii) Radial force:
F rl
31
Ans· ~
dius of the pinion at
f d the mean ra 1
10
45 x sin 15.7: ::::83.91 mm
-
2
'. 60 x' 10' x 1Q_3
- §9 x !::::
-:21t x 420
M11 - 21t N1
j.
.
22~7
~.36. n--i __ 2709.62 N·.
3
:::: _!l. ::::83.91 X 10-
M
F ,.
I :::
0
and the geqr :
(ii) Forces on rile pinIOn
.
We know that,
.
=
. _ 700' _ 15.7° ::::54.3
n..... '.,
midpoint.
Z2
..
e
==
70°
;
-Zl
(•
J'
rna I
A
ns,
~
.
.'
x cos 8.
.N
.
- 15 70 == 949.43
I
200 x cos .
::: 2709.62 x tan
.
:::: FIx
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tan
0.
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Bevel
Gears
(iii) Axial force :
Fal
-
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Ftl
x tan ex x sin
8,
2709.62 x tan 200 x sin 15.7° = 266.87 N
~."
4 ....,
(b) Forces on the gear:
(i)
Tangential force :
Fa -
(ii)
Radial force:
Fr2
(iii) Axial force:
Fa2
Ftl = 2709.62 N
ADS. ~
-
Fa x tan ex x cos 82
-
2709.62 x tan 20° x cos 54.3°
- Fa x tan ex x sin
=
°
=
575.5 N Ani.."
2
2709.62 x tan 20° x sin 54.3° = 800.89 N kos. ~
{iii} Torque produced about the shaft axis (i.e., torque on the gear shaft}:
60 x p
Ma = ix Mtl
We know that,
-
, Exanlple 7.5
I A pair
[.: Ma= 21tN2
3 x 227.36
=
NJ (60 x ')
21tN
=j.<M,!
=N
2
J
682.08 N-m Ans."
of bevel gears transmitting
15 kWat
600 r.p.m: .as ~ho"'nin
Fig.7.8(a). The pressure angle is 200. Determine the components of the resuitantgearJooIlJ
force and draw a free body diagram of forces acting on tire pinion and the gear.
2751.93 N
600 rpm
(a)
(b) Free body diagram of
torces
Fig. 7.8.
Given Datil: P
=
IS kW;
N}
d2 = 320 mm;
To find:
I. Components
=
600 r.p.m. ;
a = 20°' ,
d,
=
200 mrn ;
b = 50 mm.
of the resultant gear tooth force, and
2. Draw a free body diagram acting on the pinion and gear.
NJ
i = -
@ Solution:
We know that,
tan
°
N2 -
2 =
z2
- -
ZI
d2
320
- d} - 200 -- I .6
- -
i = 1.6, for shaft angle
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e = 90°.
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Design if n
o 'ansmission Systems
or
Then,
:::::90° - 580
M
Torque transmitted on pinion
,
Mean radius of the pin'
_
II
ron,
60 x P
2;N- :::::~
-_
:::::
ml
:::::32°
rl
21t X 600
I
d
h . sin 0 ]
2 - ..1
2
_10
x Si03~]
2
1. (i) Force conlponents acti
Ing on the pinion:
Tangential force:
./
Radial force :
FtJ _ _Mil::::
'mJ
Frl
./
..
.
(Ii) :orce componen~
./
Radial force :
I
2751.93 x tan 20° x cos 320 == 849.42 N
-
F'I x tan ex. x sin 0
-
2751.93 x tan 200 x sin 320 == 530.78 N
Ans. "
I
acting on the gear:
Tangential force:
./
86.75 x 10-3 == 2751.93 N Aus."
-
Axial force :
Fal
== 86.75 rnrn
238.73
Ft) x tan a. x cos 0
-
Ft2::::
_
Fr2
F'l
F/2
N-m
-l
~ [~
./
== 238.73
Ans, "
:::: 2751.93 N
x tan ex. x cos O2
-
2751.93 x tan 200 x cos 580 = 530.78 N An~."
-
2751.93 x tan 200 x sin 580 == 849.42 N An
Axial force :
2. Free body diagram of the forces:
s. "
,
p' The free body diagram of forces acting on the pinion and the gear are drawn, as shown in
Ig.7.8(b).
7. 6) A he vel pinion shown in Fig. 7. 9 rotates at 600 r.p.m. in the direction
~o,.,n and transmits 3.75 k W. The mounting distances, the location 0/ all bearings, and
[§xamele
1. e average pitch
radii of the pinion and gear are shown in Fig.7.9. For simplicity, the
.teetIi /Iave been replaced by the pitcll cones. Bearings A and C shotdd take the radial and
thrllSt10(ltis, while bearings Band D can only take radial loads. Find the bearing forces (or
reacti
°ns at the bearings) on the gears/loft.
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~B~ev~e~IG~e~a~~~'------------------------------~
y
160 _-~
1----
7S
-,----"0
65
.
+J0;-::~-~'"~
)(
92
J__
1_---
225
---
Fig. 7.9.
Given Data : N, = 600 r.p.m;
Fig. 7.J(J. Forces aCliJIC
P = 3.75 kW;
= 75 mm; b = 35 111m;
d,
011 gCI"
shoft
=, = 15; :1::
d2 =225 mm.
Tofind: Bearing forces on the gearshaft.
© Solution
N,
': Gear ratio, i
z2
d-,
z, -
= N2
- -- -
-
d,
-
...
-
J
tan 82
=
i, for right angle bevel gears
or
82
=
tarr ' (i)
=
tarr ' (3)
Then,
81
=
90° - 82
=
90° - 71.56°
We know that,
Then, mean radius of the pinion,
1'",
=
=
Torque transmitted to the pinion, M'l
Forces acting
../
../
../
011
=
71.56°
=
18.43°
dl
b· sin 81
')
2
7S
35 x sin 18.43°
2
2
60 x P
= 2rrN,
=
=
60 x 3.75 x 103
2rr x 600
31.97 mm
=
59.68 N-m
tile pinion :
M'l
=
59.68
= IS66.75N
31.97 x 10-3
Tangential force:
F'I
Radial force :
Frl = F'I x tan a x cos 81
Axial force:
Fal
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=
rm,
=
=
F,[ x tan a x sin 81
=
1866.75 x tan 20° x sin 18.430
1866.75 x tan 20° x cos 18.43°
=
644.59 N
=
218.8 N
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,.'hO ~
V
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_JD~e~s~i~~o'[_~Til~·a·
. Systems
I1srmSSlOn
-----------------------
~cti"g 011 tile gear:
forces,
.'
I TangentIal force:
I
Fa - Fit:::: 1866.75 N
Radialforce:
I Axial force:
Pr2
-
Fat::::
F a2
-
F/') :::: 644.59 N
218.8N
corees acting on the gearshnft : The forces acting on the gearsna1 ft :IS shown In
.. Flg.7.1Q.
I",
Forces ill the xy plane:
ForL x == 0, i.e., considering
equilibrium
FXo + F r2
=
of forces, we get
pXC
.. ,
(i)
Taking moments of forces about bearing D, we get
F'~ (92 + 65) - F r2 x (1'1/1
)pinion
=
F02 x (1'/11 )gear
-
[.,'
(I'm )gear = i
X
0
(rm )pinion = 3
or F~ (92 + 65) - 218.8 - 8 x (31.97 + 65) - 644.59 x 95.91
=
X
31.97
==
95.91 mm]
0
-
F~
or
== 528.91 N Ans. ~
Now from equation (i), we have
8.91 N or F:Co== 310.11 N Ans. ~
F~ + 218. 8 - 52
Forcesin the xzplane:
~
For ~y
For I.z = 0, i.e., considering
0 FY
-==
F '" == 644.59 N
-
'C
ADS.
-e
az
equilibrium of forces, we get
... (ii)
F~ + F~ == Fa
Taking moments of forces about bearing D, we get
FZ x (92 + 65) - Fax 95.91
c
1866.75 x 95.91 == 1140.38 N Ans. ~
or
FG - - (92 + 65)
Now from equation (ii), we have
1866.75 or F~ == 726.37 N Ans.-«J
FZ + 1140.38 ==
D
Resultant bearing force:
Resultant bearing force at bearing C, Fe ==
(528.9If + (644.59f + (1140.38)2
:::: 1412.69 N Ans . ...,
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Bevel Gears
= '.} (F~)2 + (F~)2 + (F~)2
Fo
- -v (310.11)2 + 0 + (726.37)2
I Example 7. 71 For
=
789.8 N ADS. ~
the data of Example 7.6, determine the bearing forrn
pinion shaft ie., reactions at the bearing A and B. Assume that the bearing A
radial as well as thrust load, while the bearing B can only take radial load
011
dt
C'tUr ,.
Given Data: Refer Example 7.6.
Tofind:
Bearing forces on the pinion shaft (FA an Fa)
@Solution:
Refer Fig.7.9.
The forces acting on the pinion shaft is shown in Fig.7.11.
~
z
Y
~I
!
I \fP
<,~
. I.'".>
a
l.S,;.
F
x
./'.:,...--"<,
0""""'-;:
.'"
."'.'"
FI1
•
",.
",.
->
'>t'"
.',
",.',
15.->· <, ,..)
FX
~
.
F~
81
F~
t~"
'Fz
A
<,
P
I'
Fr1
Fig. 7.11.
Forces in the xy plane: For L Y = 0, i.e., considering equilibrium of forces, we get
FYA
=
Frl + FYa
... (.)I
Taking moments offorces about bearing B, we get
Fal
or
x 31.97 -
Frl
x [(160 - ~5.91) + 75 ] +.f~ x 75 = 0
218.8 x 31.97 - 644.59 x 139.09 + F~ x 75 = 0
or
F~
-
1102.15 N
ADS. ~
New from equation (i), we have
pY
B
For LX = 0,
FXA
Forces in the xz plane: For I:z =
=
1102.15 - 644.59
= 457.56
N
ADS.-CJ
= Fal = 218.8 N ADS."
0 .
'.
..
, I.e., ~onsldenng equilibrium of forces, we get
A
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Y1,18------·-----------=~--~--------iD~~~i~~O'[-Tl~~~.~.~~
ransmiSSIon Systems
F~ - F 11 + FZB
... (ii)
'og moments of forces about bearing B
'f ale 1
, we get
·F/lx[(160-95.91)+751-F~
1866.75 x 139.09 ::::
or
F~
or
NOW
x75 :::: 0
=
F~ x 7S
3461.95 N ADs. ~
from equation (ii), we have
pZ
B
=
pz
A -
r
F,t
=
3461.95 -1866.75
=
1595.2 N Ans. ~
Resllitant bearing force:
1,
~
Resultant bearing force at bearing A, FA
= \} (218.8)2
+ (1102,15)2 + (3461.95}2
= 3639.74 N Ans."
and
resultant bearing force at bearing B, FB
= \)
=
I Example
7.8
I A differential
(457.56}2 + (1595.2}2
1659.52 N
ADs."
planetary gear train is shown in Fig.7.12(a). The input
sllaft receives 7.5 kW power at 300 r.p.m. The pitch circle diameters of bevel gears A, B, C
Qlld D at the midpoint along the face width are 200, 100, 200 and 400 mm respectively. The
pitchcircle diameters of spur gears E and Fare 200 and 300 mm respectively. The gears
rotateat constant speed. Determine the various forces acting Oil various gears and the
torqueon each of the two output shafts.
D
Power at
N1 rpm
Output
shaft 1
I-------I-t---_.__
\
Output shaft-2
(b)
(a) Differential planetary gear system
Fig. 7.12.
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i
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Bevel Gears
P = 7.5 kW; N) = 300 r.p.m ; (dm)A = 200 mm', (dm)n·;
200 mm ; (dm)D = 400 rnm ; dE = 200 mm ; dF = 300 mm. ;
Given Data:
(dm)c
=
Tofind :
100 rnrn.
(i) Tangential forces acting on various gears, and
(ii) Torque on each of the two output shafts.
@) Solution : The torque acting on the input shaft is given by
(M),
60 x p
60 x 7.5 x 103
.
21t N 1 21t X 300
-
1 -
(i) Tangential forces acting
Oil
=
238.73 N-m
various gears :
Between gears A and B :
2 x 238.73
200 x 10-3 = 2387.32 N
=
ADS. ~
Between gears C and D: We know that, for constant speed,
or
(F')AB x (r m)9
=
(F,)co x (r m )c
2387.82 x (0.1
-2- 00)
=
(F,)co
(F,)eo
=
1193.91 N
x (0.200)
-2-
-.
ADS. ~
Between spur gears E and F :
or
= (F')EF x (r m )E
(F,)cb x (rl1l)o
We know that,
1193.91x(02.4)
=
(F')EF
(0.2)
"2
x
(ii) Torque on each of the two output shafts: The torque on the output shaft 1, referring
Fig.7.12(b), is given by
(M,»)
=
(F')BA x (rm)A + (F')CD x
~ 2387.32 x
(Oi2)
tr; )D
+ 1193.91
x
(Oi4) ~ 477.5 N-m
Ans.'"
The torque on the output shaft 2 is given by
(M')2 = (F')EF x (rm)F
=
2387.82 x
(°
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3
. ) = 358.17 N-m
2
Ans."
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,
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I
I.
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I
DESIGN OF BEVEL GEARS
BEVEL GEAR DESIGN USING LewiS AND BUCKINGHAM'S
(Helical Gear Design Recommended by AGMA)
EQUATlON
.
7.8. BEAM STRENGTH OF BEVEL GEARS (OR LEWIS EQUATION)
Since the tooth thickness varies along its length, a modified Lew IS equation \,1' t"~m
strength is used for bevel g~ars. It is given by
Beam strength,
F,
where
In, = Transverse module,
=
b
~~
The factor
(R;b)
1tXIn,xbx[ab]xy'
= 10 In, or OJ R,
'" ~.n
=
Face ~"i~th
=
Permissible or allowable static stress from Table 5.4.
y'
=
Lewis form factor based on virtual number of teeth, and
R
=
Cone distance =
(R-b)
~
0.5' j
~t
z~ +
~ ....
z;
~
may be called as be~ljactor.
7.9. DYNAMIC LOAD ON BEVEL GEAR TOOTH (Effective Load on Gear Tooth)
As discussed in Section 6.10, in order to account for dynamic loads, Ole following
methods are used.
1. Calculation of initial dynamic load (FD):
tw,
Approximate value of dynamic load, using
the velocity factor, which is used in the initial stages of design, is given by the relation
F,
where
... ( .1 S)
F, = Tangential load considering service factor
C
v
Ko
= Velocity factor, and
=
=
Shock / service factor, from Table 5.6.
3.{;
3.5 +
=
, for commercially cut gears and v:<;5 m/s
v
5.{;,
5.6+
v
for generated teeth.
. I d
2. Buckingham's equation for dynamIC oa :
accurate estimation of dynamic load, is given by
Buckinllhsm's equation. used for
0
----_ ...
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~B~ev~e~/~G~ea~r!s
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y
---------------------------------~
21 v (be + F()
Dynamic load, Fd
= F( +
21 v+~
'" (7.19)
bc+F(
p
where
F( = Tangential load neglecting service factor = -; ,
v
=
b
= Face width, and
Pitch line velocity,
c = Deformation or dynamic factor, from Tables 5.7(a) and (b).
7.10. WEAR STRENGTH OF BEVEL GEARS (Wear Tooth Load)
The modified wear strength equation to suit bevel gears is given by
Wear load, FlY =
where
0.75 d , x b x Q' x KlJI
°
cos
ell
=
Pitch circle diameter of the big end of the pinion,
b
=
=
Face width,
Q'
=
=
... (7.20)
1
Ratio factor, based on virtual number of teeth
2x
zvl
zv2
+ zv2
2x
, for external gears
zv2
, for internal gears
zv2 -zvl
K,~ = Load stress factor, from Table 5.9 (or)
= [
f es
2
••
X Sill
a
1.4
f es =
where
=
Eg =
ex.
Ep and
°
1 =
INote I In the wear load formula,
Surface endurance limit, from Table 5.9,
Pressure angle, and
Young's modulus of pinion and gear respectively.
Pitch cone angle of the pinion.
use dl and 8., irrespective of whether pinion or gear is designed.
7.11. DESIGN PROCEDURE
The design procedure for bevel gears are the same as for spur gears.
1. Select the materials.
2. Calculate zl and
z2'
If not given, assume zl ~ 17.
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\,
1.22
___
Design o/Transmission Syste~s
3. Calculate
the pitch angles (i • e ., 0 1 and ~) and th .
•
%,,2)using the following relations.
e Virtual number of teeth (i.e., z"l and
Pitch angles:
.f
tan O2 == i and 0 == 900
1
.f
zl
z,,1 ==
-
02, ~lor right
. angle bevel gears
Z
cos 01
an d zv2 == cos2 O
2
4. Calculate the tangential load on to oth using
. the relation F
I
= -PV
X Tl _
""0
5. Calculate the preliminary value of dvnarni
ynamic load usmg the relation Fd
=
F
_j .
6. Calculate the beam strength Fs in te rms 0f transverse module using the relation
e"
F
"
s
==
1t X
b' x [ a
mt x
Initially assume b = 10 mt
b ]
x y' x (
R; b)
•
7. Calculate the transverse module m, by equating F, and Fd:
8. Calculate the values of b, d 1 and v using the following relations:
.f
Face width:
.f
Pitch circle diameter:
.f
Pitch line velocity :
b
= 10 ni,
v
=
9. Recalculate the beam strength using the relation
r,
= 1t
x m, x
bx [
O"b ]
x y' x
(R ~ b)
10. Calculate the dynamic load more accurately using Buckingham's equation,
21 v (be + Ft)
Fd = Ft + 21 v +
..J be + F
t
11. Check for beam strength (or tooth breakage). If Fd ~ FoJ' the gear tooth has adequate
beam strength and will not fail by breakage. Thus the design is satisfactory.
.
.
0.75xd1xbxQ'xl<w
12. Calculate the maximum wear load using the relation Fw =
cos Ul
!:!
13. Check for wear strength. If Fd < Fw' the gear tooth has adequate wear capacity and
will not wear out. Thus the design is safe and satisfactory.
14. Calculate the basic dimensions of pinion and gear using the Table 7.1.
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Bevel Gears
7.23
\ Example
7.9
I Design
a pair of bevel gears
Ir"II.\'",11
If)
J(J
kW
a pln;(.m SPetd ..
(1/
1440 r.p.m. Required transmission rutlo ls 4. Malerlallilr
gears ts 15 Ni ler J
.
Tile tootlt profiles of tile gears lire (~l2() cmnpo,\'lIe/i)rm.
III J
mo
SIS Of
ltd.
(J
Given Data : P> 10 kW~ NI r.:: 14401'.1',111, ; i ..-: 4;
To find,'
Design the pair of bevel gears.
© Solution : Since the same material
1. Muterlalfor
3. Calculation
15 N i 2 Cr I Mo 15
geurs:
2. Calculatlon of
und
ZI
ofpltch
Pitch angles:
Z2.'
(I
ASSlIme
To find
ZI
We know that,
and
21/'
.
(.0(/
=112 :
~
zvl
= COS 01
zI/2
::;
_
then
cos 02
~ x K
4
;;
x
20
= 80
of teeth "
i= 6
or 02 = tarr ' (4) = 7~,96°
"
20
14.040
=
20 61 :::::21, and
HO
c:
329.7() :::::330.
Oil
I(J(JIII (PI) :
0
1td,N,
=
60
v ==
= i ,.< Z I
cos 75.96°
4. Calculation (~ltll,.,gelltillllo{/d
F, == v
=
z2
- °2
_._;;...__-
L:OS
Z2
where
= 20,
tan 02
()I ~,)
Z,
We know that,
'" (Given)
IIg les and virtllfllllll",ber
T1Cn,
I
..!
is used for both pinion and gear, the pinion ii
we have 10 destgn only pin/OIl.
weaker than the gear. Therefore,
../
= 20°.
CJ.
_1tNI
(In/XZI]
60
1000·"
[.. d
. .
=
I
lln
x ZI andm , isin'n '!
m
I
1440 (1'11,..< 20) = 1.508 m m/s
60
1000
'
1.25, assuming medium shock, from Table 5.6,
1t y,
==
Ko
c:
10 x 103
..
F[
:=
1.508
alculation of lnltlal
5. C
_
x 1.25 -
R2R9.32
.
111,
111[
dYIl(lmic lotu! (f~/) :
P,
We knoW that,
Fd == -Cv
where
cv ~
5.6
,f~r generated teeth
S,6+{;
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~-----------~----------------
__ ~D~e~SI~'gn~Of~T~~a~n~sm~l~ss~;~on~~~s~r
5.6
5.6 + ,\(5-
-
- 0.714, assuming
8289.32)
(
.'
1
xO.714-
n1,
Fs b
where
1t X
11599.23
117,
mix bx lab]
x x(R;b)
j/
10 m,
=
= 450
[ CYb ]
mls
(F.J :
~ Calculation
of beam strength
We know that,
v == 5
... (initially assumed)
N/mm2, for alloy steel, from Table 5.4.
y' - Form factor based on virtual number of teeth
-
0.154 -
0.912
zvl
, for 20° full depth
0.912
21
=
0.1106
R - Cone distance
=
0.5
- 0.154-
- 0.5 x
Fs
=
202 + 802
117,~
1t X 111, X
nl,
10
n7,
-0 z~ + z~
41.23
=
117,
41.23177,-10117,)
41.23111,
x 450 x 0.1106 x (
- 1184.38111;
!. Calculation
We know that,
oftransversemodule (mJ :
Fs
~
1184.38 m~ ~
In, ~
Or
Fd
11599.23
111,
2.14
From Table 5.8, the nearest higher standard transverse module is 3 rnm,
S.Calculation 0/ b, d, and v :
b
-/
Face width:
-/
Pitch circle diameter:
dI
=
10m,
= m, x ZI
1t
-/
Pitch line velocity:'
=
v::::
=
10 x 3
::::
3 x 20
d) NI
60::::
1t
30 mrn
= 60 rnm
x 60 x 10-3 x 1440 ::::4.52
60
rn/3
9. Recalculation of beam strength : We know that,
Fs
=
1184.38
m; ::::1184.38
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X
32
=
10659 N
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"
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2212.4 + 21 "4.52,,
1()3
8866.S8N
1/, Ch6ck for b6am IlnmBIII (or tooth breakage): We find Fa >
tooth hal adequate beam strength and wHJ not fail by breakage ... ~;u,LIil!
IlIIlIfllCllJry.
/1, Calculallon of mflX/mum wear load (F.,) :
We know that, Fw
where
0.75 x d I x b x Q' x Kw
cos 81
2 x zv2
2 x 330
21 + 330
Q'
Ratio factor
Kw
2.5S3 N/mm2, for steel gears hardened to 400
F
0.7S x 60 x 30 x 1.88 x 2.553 = 66 1'1
cos J 4.040
79
w
Zvt
+ zv2
Ia
11, CI16ckfor wear: Since Fw < Fd, the design is unsatisfactow.
I
Now we bave to ~ncrease~~ FAAs
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Transmission Systems
F, Fd -
P
v
-
lOx 103
== 1326.26 N
7.54
1326.26 + 2121 x 7.54 x 1Q3 (50 x 148.25 + 1326.26)
x 7.54 x 103 + -v 50 x 148.25 + 1326.26
10059.9N
Wefind F s > F d' so the design is safe against beam strength.
r,
_ 0.75 x 100 x 50 x 1.88 x 2.553
cos 14.040
= 18552.89
N
NoW we find Fw > F,t· It means the gear tooth has adequate wear capacity and will
wear out. Thus the design is safe against wearfailure also.
not
14.Calculation of basic dimensions of pinion and gear: Refer Table 7.1.
mt = 5 mm
.; Transverse module:
.;
Number of teeth :
Zl
./
Pitch circle diameter:
{II
./
Cone distance:
=
=
and
20;
= 80
Z2
100 mm ; and d2 = 111( x z2 = 5 x 80 = 400 mm
R == 0.5
In/ ~
z~ + z;
= 0.5 x 5
-J 20
1
+ 802
= 206.15 mm
./
Face width:
./
Pitch angles:
./
Tip diameter:
b
=
do, -
da2
Height factor:
./
Clearance:
./
Addendum angle:
=
tn, (z,
Dedendum angle :
= 109.7 mm ; and
= m, (z2 + 2 cos 02)
= 5 (80 + 2 x cos 75.96°) = 402.43 mm,
10
= 1
c
= 0.2
mt x fo
tan 9al
eal
=
tan 9a2 ==
R
0.02425
==
eal
== 1.4°
m/Uo +c)
tan ell ==
taD
912 ==
R
=
5 (1 + O.~ == 0.0291
206.15
0
9
or
= 75.96°
5 (20 + 2 x cos l4.04°)
.2 xL;:::
./
50 mm
+ 2 cos 0,)
;::: 206.15
or
=
lOx 5
01 = 14.04°; and ~
=
./
10m /
fl
== 9/2 == .67
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Bevel Gears
7
~~~~-------------------------------------~
,/
0a2
./
e
°
2+
=
8 Jl = 0, -
Root angle:
Sa2
=
Virtual number of teeth :
Zl'l
= 21;
15.44°; and
75.96° + 1.4° = 77.36°
efl =
0
14.04° - 1.67 = 12.37°; and
= 75.96° - 1.67 = 74.29°
0
8/2 = 82 - S/2
,/
=
0a' = 0, + a, = 14.04° + 1.4°
Tip angle:
and
zv2 =
330
f
\ EXClmpie 7.10 , A pair of 200 full depth involute teeth bevel gears connect two s/uifts
ClI right angles havtng a velocity ratio 3.2:1. The gear is made of cast steel with IlJt
allowable static stress as 72 Nlmm2 and the pinion is made of steel having a static stress oj
100 N/IJIllr2. TILe pinion transmits 40 kW (It 840 r.p.m. Find the module, face width IIIId
pitch diameter from the statui point of beam strength and check the design from the sttmd
point of wear.
e = 90°; i = 3.2;
Give" Data: CL= 20° ;
P
=
40 kW;
Tofind:
NI
[ob2]
=
72 N/mm2;
[obi] = 100 N/mm2~
= 840 r.p.m.
Module, face width and pitch diameter of the gears.
© Solution
: Since the materials of pinion and gear are different, we have tc evaluate
[obi] y, and [ob2])'2
Assume .:,
= 20,
to find out the weaker element.
then
22
=
i
x
z,
=
3.2 x 20
=
64.
The pitch angles are given by
Then,
tan 82
=
i
= 3.2
81
=
90° - 82
or 82 = tan-I (3.2)
=
=
72.64°
90° - 72.64° = 17.36°
The virtual number of teeth on gears are given by
and
2,,1
=
2v2
=
ZI
cos
z
cos
=
8,
20
~ 21
cos 17.36°
=
82
64
cos 72.64°
~ 215
Then form factors based on virtual number of teeth are given by
y'l
=
0.154 - 0.912
=
0.154 _ 0'29112 = 0.1106
=
0.154 _ 0~~2
zvl
and
y;
=
0.l54 - 0~:2
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=
0.1497
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8llltllllJruJn of tangential/oad
FI
=
v
=
Ko =
Ft =
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:
p
V ~
1[
Ko
dl Nt
60
1.25, assuming medium ~~'J.l~ll~·_~~:~
40 x 1()3
56841
0.879 ml x 1.25 = m,
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m, ~ 6.94 mm
b
./
=
= m, x Z 1 = 7 x 20
Pitch circle diameter:
d1
:PitCh line velocity :
v2 -
r,
=
VI
237.62 x
= 140
1t d1 N1
=
60
-
m;
Calculation of accurate dynamic load (F J
:
21 V (bc + F/)
F = F +--~==d
I
2 I V + be + Ft
-v
p
F/
= V
=
40 x 1()3
6.16
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= 6493.5 N
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De.\'
237.62 x m2 = 23762
x 142
pt.
F t _ - __ 40 x
v
Transmission
r::
46573.52 N
103
12.315
=
3248N
c ~ , 11860 x e = II 860 x 0 ,025 ~ 296.5 N/l11m
r,
3248+
21 x 12.315 x 103(I40x 296.5+32~
21 x 12.3) 5 x 103 \) 140 x 296,5 + 3248= 47969.4 N
=
We find F, >. F d' Now the design
. tile tooth failure.
. is safe and satisfacto ry llgmllst
12. CalculatIOn of wear load (Fw) :
0.75 x d1 x b x Q' x K IV
We know that,
FlI' -
where
Q' - Ratio factor
COS
0,
2 x zv2
1<.,,,
- 0.919
=
z,,1
± zv2
=
2 x 215
21 + 215
=
1.822, and
for steel gears hardened to 250 BHN , from
N/mm2,
Table 5.9.
0.75 x 280
Fw =
x
140 x 1.822 x 0.919
= 51578.25N
cos 17.360
13.Check for wear: We find Fw> Ft/. It means the gear tooth has adequate wear capacity
andwill not wear out. Thus the design is safe against wear failure also.
14.Basic dimensions ofpillion and gear,' Refer Table 7.1 .
./
Module:
./
Face width:
./ 'Pitch diameter:
m, -
14 mm
lOx 14 == 140mm
b - lOxn7{ ::
14 x 20 == 280 mm; and
dl - m, x Zl ==
= 896 mm.
. d - m, x z2 == 14 x 64
2
r
II. BEVEL GEAR DESIGN BASED ON GEAR LIFE
(Bevel Gear Design using BasiC Equations)
7.12.DESIGN FORMUL~S'FO~
BEVEL GEAR DESIGN
... (7.21)
(i) Design torque (or Design load) [Mil:
[M{J '::' M,xKxKd
Where
60 x P
21t N '
_ Transmitted tOFque ==
. f: t from Table 7.2, and
:: Load concentration ac or,
K
." 1 ad factor from Table 5.12.
Kd -
Dyna.n1JC
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0
"'
"
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R-Vfl+l
[MIl
(R-o.s b)2· b- m,· Yv
R
=
Cone distance,
N. =
i = Gear ratio = -
N2
z2
z. '
b - Face width,
ml
-
Transverse module, and
Form factor based on virtual number of ~_ .....
Dea.lgn bending stress I Db J :
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Design __
,~ Transmissi
oJ
.
:.==~I~on~Sys~te~1tI6~
(vJ Design cOlltact stress {
CdC
Vi
ere
h
[ <5c]
CB
-
x
Uc
J:
HB
K
x
c/ or [O"c]
= C
Coefficients
denendi
x HRC x Kcl
.
epending on the Rsurfa
••• (7.25)
_ Brinell hardness number
ce hardness. from Table 5.18.
B an HBR _
HBC
K
-
Rockwell
hardn ess number
.
Life factor for surface strenzth
. and'
(vi) COile di>'tallce (RJ :
0
• trom Table 5.19.
cI
-
... (7.26)
'V
where
y
Take
= Ratio of cone distance to face" idth = R
.
'4')' -_. 3 for initial
calculations.
b
. Ilowing
(vii) Transverse
module
I'
(m ".
1 • Th e transverse mod I
to
cone distance
equation
ue
R
=
0.5 m (\j
-
r
can be found by using the
(m.)
,- ')..,)
•• , \ 1.-
.. , t":.2S)
R
or
n1t
=
0.5
I
_2 -r_2
-1
-2
-J=~
-l-
=~
7.13. DESIGN PROCEDURE
1.
Calculate gear ratio and the pitch angles.
5'
SUI
e com matlOn or mal"TID!S or pinion
2.
Select the
3.
.,.
'tab}
bi
-
.
fi
ann wheel,
consulrin" Table
-
If not given, assume gear lift! (say 20.000 hrs).
Calculate the initial design torque
4.
I M,l-
Use [:vi,] = :-1, " J.:. ... J.:.J
Initially assume K· Kd::= 1.3.
s.
Calculation of El!q' {dbl and I O"cl:
../ Calculate the equivalent Young's m.,dulus. E",. consulting Table 5.20.
../
../ 'To find [ (Jc 1: calculate
Ctdculote tire colle ~
6.
suess [ "b ] using the equation 7.23.
the design contaCt suess [ "c] l)5ingthe equation 725.
To find [ (Jb 1: Calculate the design bending
(ll) usiJtg the equation 7:/fJ.
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R
b -
'i'y
b sin 51
Calculate average module (mav):
maY
Calculate average pcd (d lav)
d laY = maY •
:
Calculate pitch line velocity (v) :
Calculate 'l'y: Use
'l'y = -d
=
v
ml
zi
-
ZI
=
b
lav
1. Select the suitable quality of gear , consulting Table 5.22.
12; tRevision of design torque I Mt J :
~
Revise K, using '1'.)1 and Table 5.11.
Revise Kcbusing Table 5.12.
1i.ftI1~'fI:K;
[M/], using the revised values ofK and K • Use [M ]
d
t
~~l~q.lcWlatethe induced bending stress using the equation 1..22
;omll8l'ethe induced bending stress with the design ~~If!P.l!g~~§i
], then the design is satisfactory.
design is not satisfactory.
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•
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of Transmwion
"
c"lell/II/I011 of basic ci,nlensions 01" h
. .
d
.
'J I e gear pai .
of the pll\lOn an gear using the equ a tiIons listed
.
r : Calculate all th bas"
.
in T bl
e
ic dimensions
~
a e7_1.
16.
,
Tuc above listed procedure
is for tlte d CSI"n
. at' pi .
=:
mien.
As disc\lssed earlier. if the materials of th "
.'
1t' t IIt! materials of the pi . e plmon and eea
the pillIon.
:: r are same. then desien onlv
.
inIon and !!ear a d'ffi
.
first and check lor both pinion and ae
re I erent, then design the pirnon
I
~ ar.
Th~ induced bending stress in the ::>zear t C; h'J.) can be determined by using (he relation
O"bl
where
=
)'\'1
ab2 . J\~
a b I and a b2 = Induced bending stresses of pinion and gear respectively, and
=
y", and .",'2
Form factors of pinion and gecr respectively based on the
\ irtual number of teeth .
Since the contact area is same. the induced contact stress is some for both pinion and
.;
gear. i. e.. acl = O'c'2'
[!:xlIlJlp/e 7.11 ] Design
a cast iron bevel gear drive for a pillar drilling nuichlne 10
Irlllls/llit 1875 JV at 800 r.p.tn.
TO
a spindle at .IOUr.p.m. The gear is
per week/or 3 yelLrs. Pressure angle is 200.
Given Data: P = 1875 \V: Nl = 800 r.p.m.:
}..2
= 400 r.p.m.:
u.
10
work/or.:flJ Iuutrs
= :u=_
To [lntl : Design a bevel gear drive.
.
.'
d
e same \\ e hav e
©So/Iltioll:
Since the materials ofplnlOIl an gearar:)
.
to
desien onlv the
.~.
pinion.
I
\. Gear ratio :
Pitch angles :
-
NI
N,
800
=
400
""I
=
For right angle bex el gears. mil O2
==:=
"
== tarr ' (2) == 6 J.-U
,-'0
s::
U,
or
51 -
and
2. Material for
3 . Gear life in
Gear life in
4. Calculation
We know that.
"00
6""
'"'\
~o
--0
== _(J.:-
90° - 5~ == '"I - .).-t...
.
G d "'",near tn:~lte,l
Cast Iron.
J1l e .r ,
pillio/l ant! gear :
~
,-.,bl.,>""
== 350N'JllJl12. trom .. ~-'all
I _
\"lr:-":';
\ cars) ==
(52 wecj-> r-:
.
= (-lO hrs ! \\cek) x
hours
== 29.Q52'- llf c~~ks
_ 6'1-l0 x 800 " 60
cycles, N - I J .
. .' I I" ton torqllt! [ll I •
of mu!« {eS1z,
_ '[\'1 I x K x Kef (.M/1
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_9 ~
~\J.'~'
9
=
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107
29.952
X
107
1.2, for C.I, from Table 5.15.
0.45
CJu'
350 N/mm2, for C.l., from Table 5.3
0'_1
CJb]
=
0.45 x 350 = 157.5 N/mm2
_
1.4 x 0.8852
_
2
2 x 1.2
x 157.5 - 81.33 N/mm
-
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of Transmission
=
'Vy
R/ b -- 3,Initially
. "
assumed.
3rr[~~0.~72~~J~2~~~C-----(3 _ 0.5) 498.08
..
x 1.4 x 10.1
X ~
7.905 x loJ
'
~
or
7. Assume
R ZI
50.2
51 mm
== 20;
Then
z2 :::;
i
x Z I :::;
2
x
20
=
ZI
Virtualnumber of teeth :
-
cos 01
Z2
O2 =
cos
8. CalcIllation of trails verse module (mJ :
R
20
cos 26.57°
~ 23 " and
40
cos 63.430 ~ 90
51
=
We know that,
40
0.5 ~ 202 + 402
= 2.28 rnrn
From Table 5.8, the nearest higher standard transverse module is 2.5 mrn
9. Revision of cone distance (RJ :
We know that, .
R = 0
10. Calculation of b,
.;
Iflav'
Face width (b):
b
r
.5 mt\J'l z
-
d lay'
R
=-
V
Average module
2
Average
3
b sin 8
pcd of pinion (d1av):
1
1
/1 I -
==
(l71av ) : may
X
18.63 x sin 26.57°
== 2.5 20
zl
2.083 mm
dlay ==
v ==
Pitch line velocity (v) :
55.9 mrn
== 18.63 mm
1(
./
=
\,/.
55.9
==
==
.;
0.5 x 2.5 - r202 + 402
and If'Y :
\lfy
.;
=
+ z2
2
== 2.083 x 20 == 41.66
may
x ZI
d lay
X
60
N 1 ==
1t
r1
x 41.66 x 10-3 x 800
60
1.745 m/s
b
./
To find '?y: \jJy == -d
lav
11 L'C'!.
. ~ quality 6 bevel gear
.
IS
_ ~.61 == 0.447
- 41.66
d from Table 5.22.
assume ,
12. Revision of design torque I Mt J :
We know that,
[Mt]
== Mt)( K)( Kd
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...
x 33.24 x 103
18.63 x 2.5 x 0.408
=
e find
Thus the design is not satisfactory.
0'6> [ 66]'
TriIIl2:
again, we get
Imm2
100.75
0
R = 0.5
b
m
a.
m, x ~ z~ + z;
R
67.08
= -'l'y = = m -
v»
2.5 x 20 = SO mm
x Nt
- 7t X SOx 6010-
60
b
d11n'
22.36 x sin 26.570
20
.;;..;_;,;;....;;....~=.;;;.;;;;.;.;;..;_
%1 =
1t X d11n'
= -
22.36 mm
= 3-
%1
I
v =
=
b sin 01
d1tw = min' x
\11
3
= 0.5 x 3 x \} ~02 + 4~
22.36
=-
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SO
3
x 800
= 0447
.
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Now~. find
(16
.oJIi
< [ (1b]'
0.72
::::
Transmission
Thus the design is satist;
'Jactory.
J' Check/or wear strength:
(Jc ~
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(R - 0.5 b)
We know that th .
e Induced contact str
• I
[
+ 15£
V (;2
ess,
J1
ix b
x Eeq [M,']
[..J (2 + 1)3
2
.
2
0.72
(67.08 - 0.5 x 22.36)
2 x 22.36
1
x 1.4 x .lOS x 33.24 x loJ ] 1:
:::; 439.33 N/mm2
Wefind
O'c; < [ <Jc ].
Thus the design is satisfactory.
15. Calclliation of basic dimensions
I
Transverse rnodu le :
I
Number of teeth :
I
Pitch circle diameter:
Zl
of pinion and gear: Refer Table 7.1.
3 mm
nIt
=
=
20;
Cone distance :
.;
Face width :
.;
Pitch angtes :
.;
x
d1 -
117t
d
mt x z2
-
2
.;
and Z2 = 40.
=
zl
3 x 20
67.08 mm
b -
22.36 mm
d
1
!
60 mm ; and
== 3 x 40 == 120 mm.
R -
°
=
and 82 == 63.43°
_
26.57°;
-
m, (ZI + 2 cos 81)
=
3 (20 + 2 cos 26.57°)
a
Tip .diameter :
_ 65.37 mm ; and
-
m, (Z2
+2cosb)
0
== 3 (40+2
cos 63.43
)
2
_ 122.68 mID
.;
Height factor:
.;
Clearance:
.;
Addendum angle '
10 -
1
c - 0.2
tan
eal
or
.;
Dedendum angle:
eal
tan
3 (1 +O·Y
== - 67.08
ef 1
or
./
Tip angle :
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./
Root angle:
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o fl
- 01 -
efI
= 26.57° - 3.07° = 23.So I
0/2
=
O2-
e/2
=
Virtual number of teeth: Zvl = 23;
./
and
tUld
63.43° - 3.07° = 60.36°
zv2
= 90.
IIn""
IExample
7.12 , Design a straight bevel gear drive between two shafts at right
to each other. Speed of the pinion shaft is 360 r.p.m: and 'lie speed of the gear wl'te/ s/,
is 120 r.p.m: Pin;oll is of steel and wheel of cast iron. Eac" gear is expected to HlO'k~
hours I day for 10 years. The drive transmits 9.37 kW.
e = 90
Given Data:
Tofind:
= 360 r.p.m.; N2 =
Nl
0;
120 r.p.m.;
P = 9.37 kW.
Design the bevel gear drive.
© Solution:
Since the materials of pinion and gear are different, we have to designIlle
pinion first and check the gear.
1.
= NI
i
Gear ratio:
=
N2
Pitch angles: tan 82
=
0
=
360
120
3
i = 3 or 82 = tarr ' (3)
=
71.560
Then,
2.
Pinion - C45 Steel, crll = 700 Nzrnm? and cry = 360 N/m2
Material selection:
Gear - CI grade 35, all = 350 Nzrnrn-,
3.
..
Gear life in hours
=
from Table 5.3.
(2 hours/day) x ( 365 days / year x ] 0 years)
=
7300 hours
Gear life in cycles, N = 7300 x 360 x 60 = 15.768 x 107 cycles
4. Calculation of initial design torque
[M/]
We know that,
=
I Md :
Mj x Kx K;
60 x p
60 x 9.37 x 103
=
2 1t x 360
= 248.6 N-m, and
21tNJ
where
K . Kd
=
1.3, initially assumed.
[ M/]
=
248.6 x 1.3
=
323.28 N-m
5. Calculation of E eq I O"b J and I DC J :
r
'"
Tofind E~q :
Eeq
=
1.7
X
105 Nzrnrn-', from Table 5.20.
Tofind I UbI J: We know that the design bending stress for pinion,
1.4
[ O'bl]
=
Kbl
n- K
x 0'_1 ,
for rotation in one direction
a
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elmo
[ Gel] = CR· lIRC x ~
CR
=
HRC -
23 from Table S.I8,
40 to 55, tiom Table S.I8, aod
Kcl [ ael]
_3 x 50 x I
-
=
l150Nfmml
, CtIl~tdatioll of cone distance (R) :
ekno that, R >
\!1) ~
{l
T
1
'Vy = R b = 3 initially assumed
3-[---0.n--~]2~1-7--'~~~~
R ~ 3 ~ 32 + I
(3 _ 05) lISO
x
> 99.36
R = lOOmm.
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~WiJjFOJ~.tlRD
toTlple
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IMtl :
[M/]
= M/xKx~
K = 1.1, from Table 7.2, and
Kd
= 13S, from Table S.I2.
[ M/] = 248.6 x 1.1 x 13S = 3"., ...
o-v_,..a
.'0..' bending ofpinion:
We know that the indl~U)(l:DdB:
R ,,;2+ I [Mil
(R-O.S b)2 x b x m, xY;I
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2
= 612.33 N/mm
O'cl < [ act ]. Thus the design Is sotl'sftlc~tomvJM'J1mJ
"",.__ '•.,. geat' (1.&, wheel): Gear /II(IIerlal :
c
I~'""",
....
.,....n"I'''' ... to calculate [O'b2] and [C1 2]'
_ N··
_PlnlQ!L ==
, N 3
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·_6.~
107
5.256 x 107
6
-\j. N:
=
2.3 x 260 x 0.758 = 453.284 N/mm2
-
CI
ta x ;)1,,2
Y"l -
0.402, for z"l
Y,,2
~
0.520, for
-
CIb2
=
77.6 N/mm2
CIb2
=
Z,,2 =
22,
from Table 5.13, and
190, from Table 5.13.
x 0.520
!J"~!fl.e~:"forwearing0/ gear: Since the contact area is same,
CIc2
O"a>
[ O"c2
J.
=
O"cl
=
612.33 N/mm2
It means the gear does not have adequate
the wear strength of. the gear, surface hardness may
= 2.3 x 360
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X 0.758
= 627.62 N/mm2•
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virtual number oftettlt ofbtv., plr refe" to 7
are the varfou. fore .. lOt Ina on a bev., liar 7
WftlUWW
.1I.'.JlIH
Write the exprellion. for beam .trenath, dynamic lOad, and 11
par. and explain the varlou. term. u.ed in It,
12. Write the .tep by .tlp procedure for the dc.iall of beyel &ea,.,
PROBl
M8 FOR PRACTIC
P'01J161111
on kln6mlllle, ol/JflVtl/lltl(lfN:
1.
A 20° .traight-tooth
beve! pinion haYing J 4 teeth and a tran.Ye....
drive. a 32-tooth gear, The two shafts are at right angle. and In the
the cone di.tance; (ii) the pitch allgle.; (iii) the pitch diameter; and (Iv)
(An,: (i) 69,8S mm; (ii) 23,63°; 66.37°; (iii)'6 mm; J28
2,
A pair of bevel gears consists of a 24-tceth pinion meshina with a
gear. are mounted on shafts, which arc intersecting at right anal e.,
Jarge end of the tooth i. 3,S mm. Calculate: 0) ihe pitch diametert of
gear; (if) the pitch angle. for the pinion and the gear, and (Hi) the con.
P,oblellll
3.
[AnI: (i) 84 mm ; 2 J 0 mm; (ii) 21,8° ; 68,20 ;
on lorce analya/N (if bl!vl!/lIl!(lrS :
A pair of straight bevel gears h8M a speed ratio of 3, A 7,S kW oower .._
pinion, which rotate. at 1200 r.p.m. 1he pitch circle diameter oid»
the large end of the tooth, The face width is 4S mm and the Dr...
Calculate the tangential, radial and axial components of the re.ultant
on the pinion,
[AnI: J391.86N;
5~V""",+I... data of previous problem, calculate the tangential, radial and
JIIiJ.II,nClU
tooth force actina on the gear wheel.
[A",:
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13Y,J..JOU .......
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a pair of bevel
are given in Fig.7.14. The
delivers 5 kW pcwer at 500
to the output shaft. The
A and B are mounted on
.n.. t.......
+ shaft in such a way that
B can take radial as
as thrust load, while the
A can only take the radial
Determine the reactions at
bearings.
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~(11ImenS14C)DS of
5kW
8
500 r.p.m.
Fig. 7.14.
: F~ = 145.25 N; F~ = 380.84 N; F~ =341.4 N; F~
= 316N; F~ =
planetary gear train is shown in Fig.7.l2(a). The input shaft ..<>ft~"'''';~::-~
at 500 r.p.m. The pitch circle diameters of bevel gears 'A,
along the face width are 250, 125, 250 and 500 mm resiJeCtiVellvslJAi'e.btl
diameters of spur gears E and F are 250 and350 mm teslbeC1tive
~llS1tant
speed. Draw a free
...,..,.,IoLI~
body diagram of forCeS ac~uUt,·.on:i.ygt.tOl
torque on each of the rwo output shafts·
: F:W = 1527.89 N; F~D
= 763.95 N; lIfu,
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(iiJ Gears are of different materials:
13. A pair of straight bevel gears is to transmit 3.5 kW at 1440 r.p.m.
machine shaft that has to run at 200 r.p.m. The pinion is of cast steel
cast iron. Design the drive.
14. A pair of 20° full depth involute bevel gears connect shafts at ri~
velocity ratio 3 : 1. Gear is made of cast steel having allowable static C!1'..,~nft'tti'
and the pinion is of steel with allowable static stress of 100 N/mm2
transmits 35 kW at 750 r.p.m. Design the drive.
15. A pair of 20° full depth involute teeth bevel gears connect two shafts _......~
having a velocity ratio of 3. The pinion is made of steel with allowable
2
100 N/mm and the gear is made of cast steel with allowable static ln~~'.~',~[JlJHJ
The pinion transmits 40 kW at 750 r.p.m. Determine the module, .ua"'V'<lIr')I,O
pitch circle diameters of the gears. Assume width as one-third the ........
~"!'!,I'{WJ
and tooth form factors as 0.154 - (0.912 I zv) where z" is the virtual
Rroblems on bevel gear design, based on gear life:
(1) Gears are of same material :
16. Design a bevel gear drive to transmit 7.5 kWat 1440 r.p.m. Gear- ratio
gear are made of C45 steel; Life of gears 10000 hrs.
IV.. Design a bevel gear drive to transmit a power of 9 kW at 20 r
ratio is to be 3. Material to be used is C20.
CJ
u
= 500 N/mm2;
expected gear life as 10000 hours.
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Worm Gears
"/"1m II"'y ignore you, then lhey laugh ot you,
IIIe" they fi,g h I you, then you win. "
- Mahatma Gandhi
8.1. GUCTION
(Rte. worm gears ~re used to
tri\l\SRllt
.
. . power between t wo nonmtersectmg, non-parallel shafts, The
Worm lhreAd.
angle ~etween the nou-lntersccting
sha~s IS usually, but not necessarily,
a n~ht angle. As discussed in tho
previous chapters, crossed hellcal
and hypoid gears are also used to
connect
non-intersecting
nonparallel shafts. But crossed-helicnl
and bypoid gears are suitable only
for low speed ratios (upto 8) and low
power ratings. Whereas worm gears
can b
df
.
hi g h as
e use
300 :.'01.j.r hlgh speed ratios as
.
GIIU' WhGlll
~
Fig. 8.1. Worm "lid Wor", gears
The worm
.r drive consists of. worm nt d II worm wheel, as shown in Fig.S.I. If a tooth
of a helical gear makes complete revolUlio s on the pitch cylinder, the resulting gear is
known as a worm: The maling Sear is called w!!!!!!J!."'tr or """", Hlh"!!: The worm in worm
and worm ge;;;: drIve is same as screw In scr~Wand nut pair.
8.1.1.
Qatlons
Worm
gear drives arc wide
used as
1\
speed reducer in materials handling equipmen~
machine tools and automobiles.
8.~VANTAGES
AND' ISADVANTAGESOFWORMGEARDRIVE
.
. 08.
. . r be
used for .peed ratios as high 8S 300: I.
8.~.eAdv~ntages
of Worm
Drives
l'he worm gear drIves can
.
./
The operation
th nnd silent .
IS slllOO
-,,'"
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1
Design of
~.l.
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~
The worm gear drives
Transmission ~
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~
are compact compared with equivalent spur or helical ~
for the same speed reduction.
It means that the motion C8nnot
'
.
~
from worm wheel to the worm. Th IS property of Irreversible.
~
The worm gear drives
are irreversible.
transmitted
advantageous in load hoisting applications like cranes and -lifts-.,)
~dvantages
~_
ofWonn Gear Drives
IS
_/
'"
The efficiency is low compared with other types of gear drives.
'"
They are costly.
'"
Since sliding occurs, the amount of heat generation is quite high.
'"
The power transmitting capacity of worm gear drive is low (upto 100 kW).
rreas
~types
Of WORM GEAR DRIVES
of worms in use are :
1. Single-envelaping worm drive: If the worm addendum forms a cylindrical surface,and
if the worm-gear teeth are curved, then the drive is called singte-enveloplng.
-
2. D_ouble-enveloping worm drive: If the worm addendum has an hourglass form,and if
the worm-gear teeth are also curved, then the drive is called double-enveloping.
\ Note \ Worm with one thread is called single threaded worm~~~~~ing
two, three or four
threads cut on it is called double,triple or quadruplethreaded worm ~
8.4. SPECIFICATION OF A PAIR OF WORM GEARS
A pair of worm gears can be specified and designated by four parameters as
.. , (8.1)
(zl I z21 q I mx)
where
Zl
= Number of starts on the worm,
z2 =
Number of teeth on the worm wheel,
q - Diametral quotient or diameter factor
=
d }/mx'
d 1 = Pitch circle diameter of the worm, and
mx
=
Axial module.
For example, a R5/2517/5 worm drive means, a right hand worm of starts 5, mesheswith a
worm wheel of 25 teeth and of diameter quotient 7, and with module 5 mm.
8.5. NOMENCLATURE OF WORM GEARS
.
The geometry of a worm gear set is shown in Fig.8.2. The various terms used in the stUdY
of worm gears have been explained below.
_,
.,
l
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B.3
(P.J: It is the distance between two consecutive teeth,
1. {",GIl ,.._~ .,. liaftJr pile'
s.k~ the a.'ris of the
~
~"l(U\
worm. In other words, it is the distance between a point on a
throtd and a. CQm!SpOnding point on the adjacent thread measured parallel to the
Px =
~tically~
,
It)( nix
axis,
..•
"Ix - Axial module of the
(B.2)
worm.
l. L811 1.) : It is the distance travelled by a thread when one complete revolution is given
ro ~ worm, Mathematically,
Lead L
=
=
Px X!,
1t X nix
x Z,
'"
(B.3)
- Axial pitch x Number of threads in the worm
Tberefore, fur single start worm, lead is equal to axial pitch; for double start worm
thresd, lead is twice the axial pitch and so on.
It is also clear that the axial pitch of the worm will be equal to the circular pitch of
the worm gear i.e., Px
= Pc =
1t
X
nix'
Pitch
ciImeter (d,)
Pitch cylinder
~
I
Fig. 8.2 Nommcla/UTe 0/ a worm gellr set .
.
,
. th e an gle between the tangent to the pitch helix and the plane of
J.. UtuI tIIIgl~ (r) : It IS
~on.
.
To find
When one thread of a wonn IS
developed, it becomes the hypotenuse of ,a
triangle, as shown in Fig.B.3. The base of this
triangle is equal to the circumference of the
Worm., While the altitude is equal to the lead of
r:
thewonn.
--
Fig. 8.3, Dn~/opme"t o/wonn tltru4
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!8.~4
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Design o/Transmission .(\.
------------------~------~~
, ,
From the geometry of the Fig.8.3,
But we know that, diameter factor, q = -d, . Therefore su bsti
sntuting.
d J -- q . mx 10
. the
111,\'
above equation, we get
1t X
tan y =
Lead angle, y ;
1t
m.\. x
(q.
ZI
ZI
--
111,)
... (8.4(a))
q
... (8.4(b))
rarr ' ( ~ )
4. Tooth pressure angle: It is measured in a plane contain ing the axis of the worm and is
equal to one-half the thread profile angle.
5. Helix angle (P) : It is the angle between the tangent to the thread' helix on the pitch
cylinder and the axis of the worm. The worm helix angle is the complement of wonn lead
angle, i.e.,
J3
=
90° - 'Y
6. Normal pitch: It is the distance measured along the normal to the threads between two
corresponding points on two adjacent threads of the worm. Mathematically,
Normal pitch, Pn
=
and normal diametral pitch, Pdn
=
J.
X
cos y
." (8.5)
1t
d1 • sin y
... (8.6)
Pn
The reciprocal of the normal pitch is known as normal module (m.).
..
2.
Px
"'"
Normal lead,
=
~ =
- Pcin
dl • sin 'Y
zl
d2 . cOS'Y
= -----z2
= PII
." (8.7)
1t
.. ' (8.8)
L x cos 'Y
7. Velocity or gear ratio (i) :
Gear ratio, i = Number of teeth on the worm gear
Number of starts on the Worm
.. , (8.9)
8.6. TOOTH PROPORTIONS OF WORM GEARS
The various proportions of the worm and worrn gears are tabulated in Table 8~I.
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~---------------Table 8.1. Tootll pro1'0rtlo
-
./
nso/w arm gears
particulars
Symbol
Unit
ha
rnm
hal-m
h
f
mm
h_tt = (2.2 cos y
c
mm
c
da
mm
dal
~
~
Addendum
u- Dcdendum
Clcnt1lI1Ce
.u- outside diameter
~
~
~
8.5
.,
Worm
Worm gear
x
ha2
1) m x
mx(2 cosy
I)
hp - mx (I + 0.2 cos 1)
0.2 mx cos y
d I + 2 hal
mx(q + 2)
4-
da2 -d2+2ha2
mx (z2 + 4 cos y z)
l-- Root dinmeter
df
5.
mm
"It
dl 2 hj1
=
d /2
nlx(q +2
4.4 cosy)
= d2 -
2 hfl
2 0.4 cosy)
mx(z2
L--
8.6.1. BasiCDimensions of Worm Gears
Thebasic dimensions of worm and worm gear s are tabulated .In Table 8.2.
Table 8.2. Busic dimensions of worm gea rs ('I;'
- book ,pa ge no.
,. rom data
Symbol
Nomenclature
S.No.
Unit
Worm
.
8 43~
Worm gear
I.
Centre distance
a
mm
a = 0.5 mx(q + z2)
2.
Axial module
mx
mm
mx = 2 a I (q + z2)
3.
Number of teeth on
z2
-
q
-
z2 = i xZI
wheel
4.
Diameter factor
q
= d1/mx
..'(Diameter quotient)
l-
10= 1
10
Height factor
~
c = 0.2 mx to 0.3 mx
u, Bottom clearance
c
mm
u,
Pitch diameter
mm
dl =q x mx
d2 =z2 x mx
d
da
mm
dal = dl+ 2/0 . mx
da2 = (z2 + 2 fo) mx
Tip diameter (outside
,diameter)
8.
r-~~ot
diameter
[Example' 8.1
I A pair
.
~
:,
dl
mm
d11 = d 1- 2j 0 ·m«:2 -c
dj2=(Z2-2f
oJ
mx-2.c
I
.:
of worm gears is designate_das 2154/11015. Calculate: (i) the
~.tre ~ista"ce; (il) the speed reduction; (iii) the dimensions of the worm; and (Iv) the
IntenSion
.1'
•
S OJ the worm wheel .
. Given Data: (2/54/10/5) = (Zl/z2Iqlmx);
TOfind:
(i)
Q,
@So
•
W IUllon:
e know that,
.'
zl
= 2;
(ii) i ~ (iii) d-, dol' djl' Px;
z2 = 54 ~ q = 10 ; mx = 5 mm.
and
(i) Centre distance (a) :
Q ==
0.5 mx (q + Z2)
== 0.5)( 5 (10 + 54)
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il'".
"
a2
2
(iv) d2, d , d/ •
= 160 mm
Ans.1;I
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Design o/'Iransmiasloll
8.6
~
(ii) Speed reduction (i.e; gear ratio) (i) :
gear ratio,
We know that,
(iii) Dimensions
0/ the
./
Pitch diameter:
./
Tip diameter:
=
dl - q x mx
dOl -
Root diameter:
=
27
ADS. '"CI
d I + 2 . to' mx
50 + 2 x 1 x 5
SO mm
where
=
=
d/I
=
50 - 2 x 1 x 5 - 2 x 1.25
»,
-
7t • mx
=
7t X
5
f 0:= l.
2 .c
mx -
Bottom clearance
Axial pitch :
height factor,
60 mm
=
c
..
=
10 x 5
= d I - 2 to'
dfl
where
./
T
worm:
=
./
54
Z2
ZI =
i =
=
0.25 mx
=
=
0.25 x 5 = 1.25 mm
37.5 mm
mm
15.71
(iv) Dimensions of worm wheel:
./
Pitch diameter:
./
Tip diameter:
d2
-
Z2
(54
+2
Root diameter:
d/2
where c = 0.25 mx
= 0.25 x 5 =
.
=
IA
xl)
where
f0= 1
5 = 280 mm
2 to) mx - 2 c
= (z2 -
df2
, Example 8.2
54 x 5 = 270 mm
do2 = (z2 + 2 to) mx
-
./
=
x mx
1.25 mm
(54 - 2 x 1) 5 - 2 x 1.25
=
257.5 mm
triple-threaded worm, having a lead of 70 mm meshes with a gtll1
having 42 teeth and a normal circular pitch of 22 mm. Find the centre distonCt bt/Wtt1f
the shafts if they are 900 apart
Given Data: zl = 3; L = 70 mm; Z2 = 42; Pn = 22 mm; 9 = 90°.
To find:
Centre distance between the shafts (a).
© Solution:
We know that,
Lead L
Axial pitch, Px
and
= Px x Z I =
=
Axial module, mx =
L
70
= '3
zl
L
1t
x zl
=
1t X
mx x Z J
= 23.33 mm
70
7t X
3
=
7.43 mm
The normal pitch is given by,
Pn = Px
or
cos y
=
X
r,
P
x
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-
cosy
22
23.33
= 0.943
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,
~~~-=:.::~~~~~-----------ars
'
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T
d
1
Lea ang e, y
~
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cos-1 (0.943)
8.7
19;440
=:
.'
cue know that,
y .= tarr+ (~)
aut "
>;
«-
or
3
tan 19.44
tan Y
=:
8.5
Therefore, the centre distance is given by, .
a = 0.5 mx (q + Z2)
=
AlternateMethod:
0.5 x 7.43 (8.5 + 42) = 187.6 mm
a ::: d 1 + d 2
Centre distance,
2
Normal module,
mn
:::
Pn
7t
dl
and
a
centre distance,
7t
=
=
d1 x sin Y
:::
1t
Z2
7t
22 x 3
x sin 19.44°
7t
22 x 42
= 311.9 mm
X cos 19.44°
-
x sin Y
Pn
d2 cosy
zl
», x zl
:::
d2 =
Therefore,
Ans. "
We know that,
xZ2
x cos Y
=
= 63.12
mm
63.12 + 31.1.9
= 187.5 mm Ans. "
2
CEXmnDle8.3] A triple-thread worm has a pitch diameter of 125 mm. The hob for
cutting the worm wheel has a normal dlametral pitch equal to O.OBlmm. Find the pitch
diameterof the worm wheel if the reduction is 12 : 1.
GivenData : zl = 3; dl = 125 mm ; Pdn = 0.08/ mm; i= 12.
TOflnd: Pitch diameter of the worm wheel (d2)·
@SOlution: We know that, i =
"
z2
=
The normal diametra)'pitch
i X zl
z2 /
=
zl
12 x 3
=
36
is given by,
... (i)
P dn = d 1 • sin y
From eq
.
uatton (i), we can write
sin y =
Or
P
Lead angle, y -
p;'
Zl
X
3
dl = 0.08 x 125
=
0.3
sin-l 0.3 = 17.46°
t.
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8.8
Now again from equation (i), we have
36
Z,
d2
I
E.:r:ample 8.4
I A triple
=
x ~os y = 0.08 x cos 17.46° = 471.73 mm
Pdn
threaded worm has a pitch diameter of 100
=«
cutting the worm
has a pitch perpendicular to the teeth of 8 modu/a
diameter of the worm wheel if the reduction is 15: 1.
Given Data : zj=3;
d, = IOOmm;
I1ln=8mm;
i=15.
nun.
Alas. "'t
Tire 1106/"
rllld tire /HtQ
Toflnd: Pitch diameter of the worm wheel (d2).
We know that,
©Solution:
We also know that, normal module,
From equation {i), we can write
or
NO\\1
=
Y
•
sin y
,
d2 . cos Y
'" (i)
ZI
x mn
d
,
3x 8
= 100 = 0.24
y = sirr ! (0.24)
13.88°
=
again from equation (i), we have
__z2
d
2
I
d1
=
I1ln
Sill
l
= i x z, = 15x3=45
z2
8.5
Example
I A double
x
I1ln
45 x 8
__
COS Y
=
cos 13.88°
370.84 mm Ans."
threaded worm has a lead of 65 mm: The gear has 411etJII
and is cut witll a hob of module 8 mm perpendicular to the teeth. Find the pitch dillltldtlS
of the worm and gear, and the centre distance of the shafts.
Given Data: zl = 2; L = 65 mm; z2 = 42; I1ln = 8 mm.
To find:
(i) d, and d2;
© Solution:
and (ii) centre distance, a.
We know that,
Lead L
Axial pitch, Pr
=
L
z,
= -
We also know that,
or
X Z
-
I
65
_?
=
32.5 mm
Pn
The normal module,
or
Pr
Normal pitch, Pn
normal pitch Pn
-
=
... [From equation (S.7)J
x mn = 1[ x 8
P:r X cos y
1[
Pn • 25.13
cosy = - =
'»,
32.5
= 25.13 nun
= 0.773
.
Pitch angle, y = cos ! (0.773) = 39.350
(i) Pitch diameters of the worm and gear :
zJ x mn
2x8
We' know that,
d) smy
sin 39.35 = 25.24mm
z2 x mn.
42 x 8
d2 and
cosy = cos 39.35 = 434.51 mm
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ADS.~
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,
I
, , bltef ANAL YS1S ON,WORM
~1,f
\
I
11
I
&1.1.
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GEARING
,
,Components of Resultant Forces, N89~ecting'FriCtion
.
.' ;
'.
..
. u-_..Aearlier, in the force analysis of worm gearing, it is assumed thai the wonn is
AS1SC~.
I
.
d
dri"ingmember, while the worm wheel is the driven member. Fig.8.4 illustrates the three
the nentsof the gear tooth force acting on a worm and worm wheel.
C(lIl1PO F F 1 and F,al = Tangential, radial and axial components of the worm
Let
t1' r
'.
respectively, and
Fa. F r2 and F a2 = Tangential. radial and axial components of the worm wheel
respectively .
._
,~
/
/
(b)
:It
/
\
·•~
F,
I
I
B •••••• -.-.- •• ------.-.-
••
(e)
.. G
(d)
"
()
\
.'
Fig·,
I
\
I
I
For the shaft angle of 90°
can be mad~:'
J
, .~ is'o,.
C
.
---z
••
••
•
I
::,.~;:;;
...
(8)
f_..
.,
'_Ing
",0"'" g
:
ii Force analYS
f F'i .s.4(a), the fo
'" tnetrY 0 g
"owing cooclUS
d frOOl the geo
an
'-
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ionS
Design oI'Transrn,'ss'
",8Downloaded
10
From : www.EasyEngineering.net
Downloaded From : www.EasyEngineering.net ,~~ ~
~.~------------------------------~----~~~~~~
<
.~
./
The forces on a worm wheel are equal in magnitude to that of worqt, but
in direction.
-'
. , : .-, o~Pas~
./
Therefore,
Fa -
-Fa)
Fr2 -
-Frl
= -
Fa2
and
F'l
•
'" (8:91)
Derivations of F" F, and Fa: The resultant force F acting on the tooth of a Worm'
resolved into three components FI' F rand Fa' as shown in Fig.8.4(b).
IS
a - Normal pressure angle, and
. Let
y -
Lead angle.
Resolving the normal reaction F in the plane ABGH, by referring Fig.8.4(c), we get
FN
and
=
F cos a
'" (i)
Fr = F sin a
... (")
IJ
Resolving the component FN in the plane ADHE, by referring Fig.8.4(d), we get
and
FI = FN· sin 'Y
... (iii)
Fa
... (iv)
=
FN· cos y
Substituting equation (i) in (iii), we get
IF
I
=
F x cos a . sin y
I
... (8.l0)
Now substituting equation (i) in (iv), we get
Fa
or
From equation (ii),
I Fa
I Fr
=
F
=
F I / tan y
=
F x sin a = Fax tan a
x
cos a . cos y
I
(8.11)
I
(8.12)
Thus the equations (S.10), (8.11) and (S.12) are used to determine the components oftbe
resultant tooth force, neglecting friction.
8.7.2. Components
of Resultant Forces, ConSidering
Since the relative motion between the
frictional force is significant in worm gears.
WOIll).
Friction
and the worm wheel is pure rolling,the
The resultant friction force, F f = J.l x F
where J.l = coefficient of friction.
Fig.S.4( e) depicts the two components of the frictional force and their directions. 'fhe twO
components are: J.l F cos y and J.l F sin y.
" .
Considering 'frictional forces, the equatioris'(Bvl 0), (S: 1'1) and (S.12) become,
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-,
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FII
rI
FaJ
JIIIi
Fx
COs Q • sin
-
F (COs
IX.
Y+ Ji F
•
.)(
SmY+1I
COsy
F COs Q •
,. • COs y)
COsY-Ji·F
. Y
_ F
. SIQ
(cos a . COs~·
- F· sin IX
Y - " • sin y)
-
F,I
~
-
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[.The tangential component (F
.
t')
.-JreJaIIOD
F'I
•.. (8.13)
•.. (8.14)
(8 )5)
..•.
itted torque using the
can be detenn.lned from the Iransm·
'" (8.16)
M,
=
Transmitted torque = 60 x P
2
P
=
Power transmitted,
NI
=
Speed of the worm, and
d)
=
Pitch diameter of the
2.1be axial force (F.,) in terms oftangeotiaJ force
I
---;:=- _
F
=
F
tl x
01
_:_(F..::.I)~'
1t N)
,
worm..
fro=m:..:equatJ~:ODS (8.13) and (8.14), is given as
(cosa·cosy-J.l·siny)
.-
+".
3. Similarlv.jhe radi
(cos a· sm r
cos 7)
given as
y, the radial force in terms of tan gen tiaI forte (F,).1iom equatioos
I'F 1
die 4. The directi
00 of
-
F
II
... (8.17)
(8.14) '"'" (&'15), is
sin a
x (cos a . sin 7 + " . cos 7)
... (&.18)
rotation of the worm wheel can be found by CQIISideriDg the wonn as saew """
worm wheel as nut.
[!'XiiInf1e 8.6 IA pair
of worm and worm whul is ~
lIS
1IS2I1fJ14.
t~O kif' power at 720 r.p.m. is supplied to the worm shtift. TIle coe.IfideJII D/.frit:tiot' is
of~ lind the pressure angle is 20 ~ Calculate the tllllgeJdial, tIJCiII/ IllUi mdittI ~
. resultant gear tooth fora oding on the worm and the worm wh«L
p~
(2152110/4): (z/z/qlm);
ZI = 2; ~ = 52; q = 10; m" = 4 mm ;
~::"~ata:
, Nt
= 720 r.p.m;
J..L
== 0.04; a == 20°.
~TO/illd : Components of the gear tooth force·
vSoIUl."IOn:
...
We know that,
.Lead angle,
-I
Y == tan
-I
Y = tan
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(z)q
(1.)
10
_!
== 1131°
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I
8.12
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Pitch diameter of worm,
Transmitted torque, M, 1. Force components
'" [.. III
on the worm:
F/I
-
10-3
T. = 240x x132.63
= 6631.45 N
ADs. "
(cos a . cos Y - Jl . sin y)
- f 11 x (cos a . SID
. Y + J.1 • cos Y) ... [From equation (8 .I~
(ii) Axial force:
(iii) Radial force:
F r1
"
-
{cos 20° x cos 11.31 ° - 0.04 x sin 1l.3!.j
6631.45 x (cos 10° x sin 11.31 ° + 0.04 x cos11.310)
-
27105.74 N Ans. ~
-
F 11 x ( cos a . sin
.
'Y + Jl . cos 'Y) ... [From equation (8.18)]
sma
.
-
sin 20°
6631.45 x (cos 20° x sin 11.31° +0.04 x cos 11.31°)
-
10147.45 N Ans. ~
2. Force components on the worm wheel: We know that,
(i)
Tangential force: Fa - Fal = 27105.74 N
Ans . ...,
(ii)
Axial force:
Fa2
Ans. ~
(iii)
Radial force:
Fr2 -
I Example
8.7
-
Fit = 6631.45 N
Fr}
I A double threaded
=
10147.45 N
Ans • ...,
worm drive has an axial pitch of 25 mm and apitd
circle diameter of 70 mm. The torque on the worm gear shaft is 1400 N-m. Thepitch ~
diameter of the worm gear is 250 mm and the tooth pressure angle is. 25~ FiJfJ·
-flrcto~
1. Tangential force on the worm gear; 2. Torque on the worm shaft; 3. Separat/lfg 0 (tIi
the worm; and 4. Velocity ratio. Take the coefficient of friction benveen the worliith
and gear teeth as 0.04.
~ 1400 ~-J1I;
Given Data: z} = 2;
Px = 25 mm;
d 1 = 70 nun; (M/)worm gear
d2 = 250 mm; mm; a = 25°; J.1 = 0.04.
Tofind:
(i) Ft2; (ii) Mil;
(iii) Frl
;
and (iv) i.
_ or • 11ft]
© Solution:
'
60 x p
60 x 10 x 103
. .t~df~
21t N. 21t x 720
- 132.63 N-rn
2M,
(i) Tangential force:
~
d. - "'x x q = 4 x 10 = 40 mm
(i)
Axial module,
mx
=
x
P
1t
Diametral quotient, q
=
dl
mx
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=
=
25
=
7.96 mm
[':Px-"
1t
70 = '8.8
7.96
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'[bell,
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1Ii1Torqueon t/le
I~
Lead
.
angle, Downloaded From : www.EasyEngineering.net
~l~;.
We mow that,
tan-I (ZI)~
'Y :::;::
Wor", sh .,;
•
q
::=:
tan-I
Fa:::;:: F
We alsoknow that,
LCUI
F11:::::
12.810
{I.
~eparatlng fo,ce
Weknow that,
2X
~I
d
I
orrn shaft, M
~ie'
(. ., 'ad,Q! fi
orce)
- 89
II
-
_
-
2546.48N
.
.13N-rn A
the Hlor",(F,.
Oil
1'1 -
(iv) Gear ,atio (il :
v
FQI
'-I)
x tan ex
- 11200 x tan 250
normal module , mn
lIS
•
,J.
F
We know that,
12.810
16
254648
_ 2 xM
.
-::;::--1!_
70 x 10-3
Torque on the w
or
"":}
::=:
Fn - F
aJ :::;:: FIJ / tan
- a» tan y '" 11200)( :.~ 11200 N
or
,Ulj
(.1.)
8.8
== P i n == P
x
_x
n
== 5222.65 N
ADS. "
cos_y
n
:::: 25 x cos 12.810
== 7.76 rnrn
7r
We also know that ,
d2 . cos y
normal module , m. n
z2
Or
z2
d2 x cos y
::::
mn
=
250 x cos 12.810
7.76
:::: 31.4 ~ 32
'.
G
'.
ear ratio,
z2
I
== -;-
="232
= 16 Ans. ~
8.8JA 2-tooth right-hand worm ;ansmits 0.746 kW at 1200 r.p.m:
j~1e
/0
a
'Po,,..,, Hlo,,,, gea" as shown in Fig.8.5. The gear has an axial pitch of 13.5 mm: TIle
." as a'pUch diameter
.
lricHof 50 mm. The normal pressure angle is 14-'h'! Coefficient0/
·IfIII :::::0.03.
(i)
Fi.
/,.;~ IIId the lead anete the lead the normal pitch and the centre distance.
(~
~.
~'
,
,
IlJ.d(he force components on the worm and worm gear.
Q The gea, is supported by bearings A and B, as shown in Fig.8.S. Findthe/orees
. exerted by ti,e bearings againj't the worm-gear shaft, and the outputlorIJU~.
GlVenData: zl = 2; P = 0.746 kW; NJ = 1200 r.p.m.; z2 = 30; Px = 13.5 mm ;
(iii
d,
::=: SO
- nun·
rt
= 14 ~o;
#l = 0.03.
.
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x~~
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----------------_D_e_~=gn~of_n_rmu~m~u~~fun&
~
~.14
y
'"
Worm pitch
cylinder
),,/-
I "
I'"
,/'"
x
Fig. 8.5.
(i) L, P» and a,
Tofind :
(ii) Force components on the worm and worm gear, and
(iii) Forces exerted by the bearings and the output torque.
@Solution:
= pjn = 13.51n = 4.3 mm
Axial module, mx
(i) Lead angle, lead, normal pilch and centre distance:
Leadangle:
. where
=
q
'Y = tan-I (Il.!35)
=
Lead:
Nonnalpitch:
L = PxxzI
= 13.5x2 = 27mm Ans.1I
Pn = Pxx cosy
Centre distance:
9.750 ADS."'tJ
13.5x cos 9.750
=
13.3 mm Ans....
a = 0.5 mx (q + z2,)
-
O.S?<4.3 (11.635 + 30)
- 89.52 mm ADS."'tJ
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----
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~
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8.15
componen~on the won« and
(ill
Wor", gea
force co",ponents on the worm:
.,
:
:1 force
I Tangential component, F II
_
2
X
Mil
d
l
M
where
60 x P
60
2-1[ N == - x 0.7~6 x 1()3
I
27t x 1200 _ 2 x 5.94
tl
-
F
.'
I
tl
50 x 10-3
== Ftl x
Fa I
Axial component,
-
= 5.94 N-m
== 237.46 N Ans.-tJ
(cos
(COS'
cos Y Il sin y)_
-IX • Sin Y + Il . cos Y)
IX •
_ 237 .46 x (cos 14.5°° x cos. 975°
. 9.75°)
0.03 x Sin
(cos 14.5 . sin 9.75° + 0.03 x cos 9.75°)
I
1164.57 N Ans.
Radial force, Fr 1 _ F II x
'"tI
sm IX
(
cos
= ,237.46
IX .
.
sin Y + Il' cos 1)
sin 14.5°
x .'
(cos 14.5° x sin 9.75° + 0.03 x cos 9.75'
== 307.23 N Ans. ~
Forcecomponents on the worm gear:
{
Tangential force, F t2 = - FaJ
{
Axial force, "Fai =', -
F/I
Radial force 'r,.F.., = - PrI
(iii) Force exerted by the bearing
= 1164.57 N ADS.
= '237.46 N ADS.
= 307.~3·N ,ADS.
~
~
~
A a~d B 'against the worm gear:
The force~ acting on the worm ge~,and its shaft~are shown in Fig.8.6.
For L Fx::: 0, i.e., consideringthe. ~uilibrj1.!m offorces in x direction, we get
..
,
.,."
'
.
.-
FxB' = F~ == 237A6 N Ans.'"
.
lakin
.'
g moments .about
. F~ x (37.5
,
the z axis, we nave
+ 6~.5) _ ~~~?<,37~5- Fa2 x 62.5 - 0
Or F~ x (3.7~5+ 62.5) -'307.23 x 37.5 - 237.46 ~ 62.5
..
: 'pY
,_
,
/
= 0
== 263.62 N ADs.-CJ
B .
\
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De8ign ofT'OIII~
8.16
Fig. 8.6.
For I Fy = 0, i.e., considering the equilibrium of forces in y direction, we get
FYA + pYB - F
Jt
T,I.
or
= 0
F~ + 263.62 - 307.23 = 0
F~
= 43.61 N ADs. ~
Taking moments about the y axis, we have
- F~ x (37.S + 62.5) + Fax 37.S or
- F~ x 100 + 1164.5 x 37.5
...
0
0
=
F~ = 436.686 ~ Au. ~
For I F, = 0, i.e., considering the equilibrium of forces in z directio~ we gef
FZA + FZB
or
-
Fa = 0
F~ + 436.686 - I I 64.S7
:.
F~
Resultant bearing force at bearing A, FA
=
0
== 727.88N
=
-v
(F~ )2 + (F~ )2
,
= " (43.61~
= 729.18
and resultant bearin, force at ~g
B, FB
=
-v
= "
Au. ~
+ (727.~a)2
N ADI. ~ .
(F~ )2 + (F~ )2 + (F~ >"
__
(237.46)2 + (263.62)2 + (43~.6861.
- 562.65 N Au. ..
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()e_ar~
.~
__
.
--------
", torque: Taking moments about~, Vf,e,get
ofIIP
\
T - fax ()~.5 x Ltr3
T-
J8~.1~7
0
=::
11~.57 x :62.5 x 1~0-3:::: 0
OQij>ut touque, T - 7
'~.78SN-Ill Ans. '"tJ
..
'
EfFtC'~NCYOFWO~M GEAR~NG
....,.
fth
.',The efficiencr
0
. .
gear dri
.Jllv~ '~s
Ig1'V~nby
e worm
.
Power Qutput
POw~r input
=
n
(
f'I
-
t
=
x VI
anyx(
(cos
(cos
where
~l
and
V2
=
'(
f/2 x ~2
n-
(co,s a . cos 1 - JJ. • sin 1}
,
, cos ~ . sin 'Y + J.1. • cos y)
... (8.19(a»
a - JJ. . tan y)
a + Ii' coty)
... (S.19(b»
fitch line velocity of worm and worm gear respectively.
The coefficient of fr!cHop (J.1) in worm gear drives depends upon the sliding velocity, i.e.,
rubbing speed.
Sliding velocity,
Vs
-
The variation
of the coefficient
,
Fig.8.?
,
\
Q.
e
of friction with respect to rubbing velocity is shown in
0.12
I 1 1
LL
.s
'L\
II
:::L
..:
c 0.09
.
... (8.20)
sin y
~
CD
15
"."
"
Steel worm
I
j
<,
...... ._
I~
-............
~
o 0.06
16
s
'" Cast Iron worm wheel
BronzeiWOhn.Wheel
r-- k::'
r-..
L
r-- c
e:s
00."
,
1;5
1.0
'.'.
-
13
it O.O.~
,
.r
,
..
2.5
2.0
3.0 3.5
Sliding velocity. mls
F/~. ~.7.
,.~.1.P,Q'WerLost
in Friction
It is give~ ~y~~Tebltio~
,
~er
,/
~
'1,10=-.
;-,
,
lost in friction -
(l-J]) x
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pJ
... (8.21)
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a.is
Design of TrrDU'_i ..'S.
"'''''''otJ, IOI'l.t\.'
':I
~:
,-
I
8.8.2. Self-locking and Overrunning Drives
Self-locking drive: The drive is calied self-locking if the worm gear cannot dri
worm, It is possible only when friction force exceeds the driving force.
i
Ve the
In other words, if p ~cos a .tan y, then the drive is self-locking.
Overrunning drive: If the worm gear can act as driver, then it is known as an
overrunning or backdriving gearset
The theoretical criterion for an overrunning wonn drive is
J.l < cos a . tan y
I Example
B.9
I A pair
of worm gears is designated
as 1/52/10/8.
The worm transmits BOOW power at 1000 r.p.m: and-the normal pressure angle Is 20~
Determine the coefficient of friction and the efficiency of the worm gears. Also fllld IlIl
power lost in friction.
Z I = 1, z2 = 52;
Given Data: (I / 52 / 10 / 8) : (z I / z2 / q / mJ;
p = 800 W; Nt = 1000 r.p.m.; a = 20°.
To find:
(i)' J.l ; (ii) 11 ; and
©Solution:
=
10; mx = 8 mm
(iii) Power lost in friction.
Lead angle, y
We know that,
or
q
- tarr ! (~)
= tarr ! (
I~)
= 5.71°
q - dl / mx
d1 = qxmx
=
10 x 8 = 80 mm
(i) Coefficient of friction of tile worm gears:
Velocity of sliding,
Vs
__ n d , Nl
= ~
cos y
1t X
=
60 x cos 'Y
80 x 10-3 x 1000
60xcos5.710
= 4.21 mls
. From Fig.8.7, for Vs = 4.21 mls and bronze worm wheel , the coefficient of friction is
Interpreted as 0.027. Ans."
.
(ii) Efficiency of tile Worm gears :
11 = (cos a - J.L tan y)
(cos a + I..L' coty)
=
(cos 20° - 0.027 x tan 5.71°) _
5% Ans.1J.
(cos 20° + 0.027 x cot 5.710) -77.4
(iii) Power lost in friction:
Power lost in friction = (1 - 11)x P
= (1 -
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0.7745) x 800 = 180.36 W A.ns.
"
11
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.
;
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8.19
~~ae4fS-
A double threaded wo"", drivr is IWfIlilq lor 1'0_ ""-rissio.
"rfll having their axes at right angln to ~ OlItu. 1lre _
..... U %.
tfIO slra IScenJn distance is apProximatdy 200 I11III. If rile axial pitd 01tile _
~ ttttJr. fh~ angle is 23 ~find 1. lead; 2 pbelt clrcIe dUrntetus of ""''''' IIIfII ""'''''
~~J' filii'
IIId /eo I if the worm; and 4. efficiency 01 the ~
if tile IWffIdat of/rit:IiInt
,j. "tli.~ang e 0
tI;j.
":r'"
0
Data..
zl
= 2' '
'1 = 230; Jl = O.OS.
GiVt"
Tofidill
e
'
1. L' , 2. d 1 and d2;
n
eSDIuJio :
3.(3;
P,
n" -Pilch diameters.
•
4. 11.
30 x 2
=
j
60 mm ADs."
!
We know that,
Z[ •
,U,
Zl
tan Y =
=
mx
where
= (d['
nI.>
Axial module -- Px /
=
d,
= __
Zl
q
Z
Weknow that,
and
=
L = Px x z[
Lead (L):
:1
9 = 90"; a = 14 Y, 0; a = 200 mm ; p. = 30 nun;
1
xm
tan
centre distance, a
y
.T
m.
d[
1C
__
=
30 /K
2 x 9.55 = 45 mm
°
- tan23
=
9.~) mm
ADS. ~
-
=
45 +d2
200 =
2
or
2
d
(iii) Helix angle (/1) :
=
13 =
355 mm Ans. ~
0 _ 230
90° - 'Y = 90
cos a _ fl' tan 1)
=
.
~
6'" AIls.
=!
+ fl ' cot y)
(cos a 0 005 x tan 23°) = 87.2.h Aas. ~
145 - .
~O)
- {cos 14'50 + 0.05 x cot 23
rile ~
(cos .
. e w"dltu or
.
r-r&anw--t-e-8:""":.l::-:l-"Fo,' the above pro b/~nr, dderllUn
(iv) EffICiency (11) :
n
'I
If'"
is
sdf-
""'king.
10 J.1 = 0.05.
E ample 8. ,
Refer x
If-locking,
di 'on for se
We kIlow that the con IU
a' tan Y
0411
._
0
> cos
23 = ,
. a o~,
J.l.
14,5° x tan
u: lockiJtg. It IS
In this case cos a . tan Y = COS the drive .s, not se1_,-t1
We find ~ < cos a . tan 'Y. Thus
~
~ Solution:
nning or
-
backdriVinggearset.
-
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··..tiM'·,·W&t-1'J!'·tj·
dt N
eft"
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Design
8.20
o/Trans",. .
lSSIO"~
DESIGN OF WORM ~R
L-
]
DRlyE
--------~----~
'~
8.9. MATERIALS FOR WORM AND WORM WH~EL
•
1
The following guidelines may be used while selecting the materials for Worm and ""~
wheel. Refer Table 8.3.
Table 8.3. Selection of Mal~r411
Mat~rial
Condition
S.No.
Worm
WormWhed
5
1.
Light loads and low speed
Steel
2.
Medium service conditions
Case
. haFdened
steel of
Cast iron
1
Phosphor bronze
~,
3.
High
speeds,
heavy
loads
shock
with
.Hardened
molybdenum
I
I
BHN250
Phosphorbronze(~
steel or chrome vanadium
conditions
steel
8.10. FAILUREOF WORMGEARING
The different worm gear tooth failure are :
1. Seizure:
./
Since significant sliding occurs between the teeth 'ofthe worm wheel and thread of
the worm, the possibility of seizure is very high in worm gear drive.
./
The seizure has greater probability to occur in the zone where oil has squeezed
out.
2. Pitting and rupture:
./
· softer
The worm wheel wears off more than worm. Because the worm whee1 IS
than wonn .
./
Only the worm wheel is affected by the pitting phenomenon.
8.11. SEI,ECT'PNOF NU'QI;~ OF ~T~TS IN THE WORM (Z1):
Table 8.4 shows the approximate efficiencies for the number of starts in the worm·
Table 8.4. Approximat« efficiencies In worm g~ar drive (from data book, page no. 8.46)
Number of starts in the worm <2;.)
Efficiency, 11
1
0.7 to 0.75
1.
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2
0.7 ,t,o 0.82
3 and 4
0.8 to 0.92
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~
~
Jfj.
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OF WORM (OR lEAD), L
8.21
Table 8.5. Length of the ff1fJnnfll
f"
" v v70m
Number
or"-rts <ZJ>
Jor2
dtzItl boo
t. Pllf~ 110. 8.41
.
L~(JJ
301"4
-
~of.ora
+O.06z )m
J
z
L ~ (12.S + 0.09 zJ> m
z
,13.
fACE WIDTH OF THE WHEEL (b)
.
Table 8.6. Face width of whee/., b (fi7om -'_._ boo
'"""
%1
b
J
0.75 dl
..
k •pag~ no I.48)
dd (.aL dialBeUr' oldie wiled)
de2 ~ dol + 2· mz
2or3
0.75 dJ
de2 ~ do2 + 1.5 mz
4
O.67dJ
de2 ~ do2 + m;c
8.14.EFFICIENCY
In Section 8.8, we have discussed the efficiency of wonn gears without considering the
klsses in the worm gearing. In actual design practice, losses in the worm gear drive should be
considered.The main losses in the worm gear are
-/ due to friction in sliding (i.e., gearing loss), and
-/
due to the churning and splashing of lubricating oil.
The efficiency of the wonn gearing, considering only the gearing losses
11 where
is given by
tan a
tan (y + p)
.,. (822(a»
p _ Angle of friction = tan-I (J1), and
Coefficient of friction, from graph 8.7.
.'
unt all the losses, is given by
The efficiency of the wonn gearing, taking
into aero
.
J.1 -
T)
tan 'f
= (0.95 - 0.96) tan (y + p)
8.15 TH
GEARING
.
ERMAL RATING OF WORM
heatIn.the Worm gearing, when the
worm gear ~
.. , (822(b»
.
I
'derable amount of
contInUOUS y, CODSa
•
•
lost (i.e., power loss) in frictJOO must
.IS generated. This heat generated due to
driv and lubricating oil.
be dlSs'
.'
fthe
e
lpated in order to avoid over heaf1Jlg 0
The
. f the drive is given by
COndition to avoid overbeatiDg 0..
to the atmOSPhere (H~
Heat generated 0Ig) := Heat dJSSlpated
pSi.
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_gAP
.~. Li
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.
'"
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~8.=22~
V
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Design o/Transrnu .
--------------------~~~~
where
Hg
=
(1 -11) x Input power,
Hd
=
K, x A x (to - t a)'
K, - Heat transfer coefficient of housing walls (W/m20C),
A -
Effective surface area of the housing (m2),
10 =
Temperature of lubricating oil
t
. a
. t 1 - 11)x Input power
eel, and
=
Temperature of the atmospheric air (OC).
=
K, x A x (to - tal
I
... (813)
\Note' The power transmitting capacity based on thermal considerations is given by
P
=
'K,xA(to-tJ
(I-TIl"
...(114)
, Example 8.12 , A worm gear box ;"ith an effective surface area of 0.25 ",z.Afad
mounted on the worm shaft to circulate air over the surface of the fins. The coeffldtnJ~
heal transfer can be taken as 25 Wlm2 °C The permissible tempera/we rise of 1M
lubricating oil above the atmospheric temperature is 45 cr:: The coefficient of jridio. i
0.035 and the normal pressure angle is 20'! The pair of worm gears is desig1ulItd IS:
1/40/ 10/4. Calculate the power transmitting capacity based on thermal considofdiDtU.
Given Data: A = 0.25 m2; K, = 25 W/m20C;
zl = 1 ; z2 = 40; q = 10; mx = 4 mm.
to - fa
= 45°C; J.l = 0.035; ex ==20°;
Tofind: Power transmitting capacity based on thermal considerations.
@Solution:
The power transmitting capacity based on thermal considerations
P
=
is given by
K,A (to-ta)
(1-11)
So first let us find the efficiency of the drive.
Lead angle, y
=
tarr l (~)
=
tan-I (
1~)
= 5.7'°
11 = cos a - J.l • tan 1
cos a + f.l • cot 'Y
cos 200 - 0.035 x tan 5.71 °
= cos
Now,
20° + 0.035 x cot 5.710 = 72.5~1o
Power P = 25 x 0.25 x 45°
(I - O.n59)
= I026'Y
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ADs. ~
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c~
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)~
8.2.3
I
and Wonn Gear Design Us'
I~
and Wonn Gear Design Reco~ng
('P-
,
,.~
LeWis and Bucki
mended by AGMA)
;,
~AII sTRENGTH OF WORM GEAR TOOT
:I ~ ~
equation of beam strength is given by
F, -
1t X
m); -
Axial module ,
b -
ngham's Equation
H(ORlEWI
S EQUATION)
m); x b x [ CI ] x y'
b
... (8.25)
Face width, from Table 8.6,
[ crb] -
Permissible static stress, from Table 8.9, and'
y' - Form factor for worm wheel, from Table 8.7.
..
TUk &. 7. Form factor for worm wheel (fj10m data b00k, page no 8.' 52)
~angle(a)
!
y'=-
--
14~0
200
25°
30°
0.1
0.125
0.15
0.17
Y
1t
117.DYNAMICLOAD ON WORM GEAR TOOTH (EFFECTtVE LOAD ON GEAR TOOTH)
Ia wormgear drive, dynamic load is not so severe due to the sliding action between the
gear.
Thus the value of dynamic load, using the velocity factor, which can be used in the initial
tmn and worm
iii fmaI stages of the design, is given by the relation
Fd
FI
Ft
... (8.26)
= C
v
_ Tangential load considering .servicefactor,
_ E x~
v
. factor from Table 5.6, and
Shock / Service
'
Ko Cv
_ Velocity factor.
-
-
6
6 +v
where
l1l
'¥EAR STRENGTH
lbe li_:..c·
QIll{UDg
•
.
I'ty
_ p'tch hne ve oct
v -
t
O"!AGS
~
OTH LOAD FOR WORM ~EAR)
(WEAR TO
Of WORM GP"I'
. . en by
I ad for wear (F....
) IS gIV
or maximum
0
F.. · = d2 x b x
. mI
of tl]e worm gear 10
s.
.. , (8.27)
l'
I<"
..}i( ..
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IUf
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8.24
where
d2
=
Pitch diameter of wonn wheel,
b
:=
Pace width, and
Kw -
Wear factor, depends on the materials of worm and wocrn
from Table 8.8.
Table 8.8. Wearfactor /o~ worm gun
~
(x.J UTomdata book, Page no~8.54)
.
Worm wheel
Worm
-
Wear (actor, Nlm.2
Material
S.No.
.
J4~O
W
L
Hardened steel
ChiJIedbronze
0.63
0.88
2.
Hardened steel
Bronze
0.42
0.56
3.
Steel, 250 HB
Bronze
0252
0.35
4.
High te5t CJ
Bronze
0.56
0.805
S.
Steel, 250 HB
Laminated phenolic
038
0.448
-
8.19. DESIGN PROCEDURE
J • Selection of the materials : Select materials for worm and worm wheel.
2. Calculation ofzJ and Z2:
./
Depending upon the efficiency requirement, select the number of starts (zl) in the
worm, -referring Table 8.4 .
./
Then, z2 = i x zi
3. CakuJation of diameter factor (q) and lead angle (,; :
./
Diameter factor, q
./
Lead angle, y
d I / m;r If not given, assume q = II .
=
(zi / q)
= tan-I
4. CalcuJaJlon of tangential load (FJ acting on the wheel in terms.ofaxial modMJt :
Tangential load, FJ
=
5. CalcuJaJion of dynamic load (F J
J
Dynamic load, Fd d:: F ,
Cv
6. CalcuJaJion of beam strength
Beam strength, F, =
Ko
P/v x
1t X
7. Calculation ofaxioJ module
:
assuming initial pitch line velocity.
{ffin terms of axial module :
mx x b x [CJb] x y'
(mJ :
Calculate axial module by equating F, and Fd:
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~uv
~~s __
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~-
Colculati(Jnof b, d 2 and v.
8·
BtC
'S.2S
alculation of beam strength:
Us P' _
e
- 'It X
. ,eetJku/lldon of dynamic load: Usin t~
III. x b x [ a b 1 x y'
IO·ul te the dynamic load, F d = F / c • g_ e calculated pitch Ii
.
~(i
a
I
V'
ne velocity of the wh· I
Checkfor beam strength: If Fd ~ F th
. . ee ,
11.
, . . 8' e gear tooth has ade '
'
'dDOlfai! by breakage. Thus the design IS satisfactory.
quate beam strength rnd
9
WI
·
12. Calcul'.1'
atlOn oJ maximum
Wear load (F.J :
Fw = d2 x b x ~
13.Check for wear strengtb•• : If F d < F W ,the gear
t tl1 has. adequate wear capacity and
,00
\Villnotwear out. Thus the design IS safe and satisfactory.
14.Check for efficiency:
If 11actual >
tbanthe desired, increase lead angle (y).
l'Jdesircd ,
then the design is satisfactory. If it is less
'
15. Calculate the powe~ loss and the area required to dissipate the heat.
16. Calculate the' basic dimensions of the worm and worm wheel using the Table 8.2.
[Example 8.13 I A hardened steel worm rotates at 1440 r.p.m: and transmits 12 kW to
.phosphorbronze gear. The speed of the worm wheel should be 60 :t 3% r.p.m.Designthe
wormgear drive if an efficiency of atleast 82% is desired.
GivenData: Nt = 1440 r.p.m.; P = 12 kW; N2 = 60 ± 3% r.p.m.; 1ldcslred = 82%.
Tofind: Design the worm gear drive.
@ Solution:
Gear ratio required,
i
_ 146400 ± 30/0
Worm wheel -
For 11 = 85%,
zl ~
.f
Then , z2 = i x
Zt =
3. ~alculation olq and
3, from Table 8.4 .
24 x 3 == 72.
(assumed)
r!
~
.f
... [Gi~~n]
Phosphor bronze
Z2:
.f
.f
24 '± 0.72
Worm - Hardened steel, and
1. Material selection :
2. Selection olz} and
=
Diameter factor :
== 11
q ==
"'x
r~
~I(~)
Lead angle :
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::::tan
-1
(1-)
::::15.25°
11
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:i
Ii
I
'i
''''W:-'
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Design o/TrOllS/llissf
~
'
~:,~
,', (',,/('ul{Itltm tfl'lt,
oj'
/11I(!flll,f
"'x. /
S·
p
'rl\"~~l\tlftll(l~1(I, II', -
\I
7e
II -
KO
~
60
cI N
1000
X
m
_ 1r X 72 x
x 60
9
226 m mls
F
60 x 1000
0, -
x
Ko ,.... 1,25. ftRHUming medium shock, from TabJe 5.6.
..
\
12 )(
I", "'"" 0,226
1(J3
n1
66371.68
mx
)( 1,25,:;;
x
Dynomlc lend, Pd
F
..L
F
ell
6
where
where v = 5 rrils is assumed.
6+v'
F
...
6
6+5
:;:=
66371.68
rd
1
;;:;1
mx
0.545
1
x 0.545
-
be
121681.4
mx
6. e(I/cli/(lII(m (if belli" .\·lrIJIII/I/, (F,,) In terms of axial module:
Bellm strength, F" ;;:; 1t x mx x b x [ (Jb ] x y'
h
where
(ob]
r;;;
0.75 d,. from Table 8.6,
I:l
0.75xqmx
= 80 N/mm2,
=
0.75 x Ilmx
=
8.25m,l'
c
from Table 8.9, and
y' ;:::;O. J 25, assuming a = 20°, from Table 8.7.
,
,,
7, Ctl/CII/III/on
We know thnt,
or
01'
F" ;;:
0/ (Lt/t,1 module
2
1t
x mx x 8.25 mx x 80 x 0.125 = 259.18 m,l'
(mx) :
F,f ~ Fd
2
259.18 mx
~
12168 ],4
mx ~ 7.77 mm
From Tobie 5.8, the nearest higher 8tand~d axial pitch is 8 mm.
,1,#
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db
i .
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8.27
!f!r
...,.;#,~
••1
Calc",a"
.~
~/ono/b, d2 and v:
8, I face width (b): b= 8.25 mx = 8.25 x 8 = 66 mm
I Pitch diameter of worm wheel (d2):
d2 =
x mx = 72 x 8 = 576 mm
v = 0.226 mx =-0.226 x 8 = 1.808 mls
I Pitch line velocity of worm wheel (v):
leu/at/on of beam strength (FJ :
'
Beam strength, F, = 259.18
9, Rtca
10.
z2
m;
= .2~18.(8)2
= 16587.52 N
'ltcalcu/ation of dynamic load (FdJ :
Checkfor beam
= -
!/
I
c"
6
6
cv =
6+v
Ft
=
66371.68
mx
Fd
=
8296.46
0.768
where
..
,
Ft
Dynamic load, Fd
= 6 + 1.808
=
= 0.768, and
66371.68
8
= 8296.46 N
= 10802.68 N
Fd < FS" It means that the gear tooth has adequate
beam strength and will not fail by breakage. Thus the design.is s.atisfactory.
12, Calculationof maximum wear load (F.J :
II.
strength : We fmd
~ar
load, F11'
where
,
"
=
d2 x b x K,
~v
= 0.56 N/mm2, from
F 11'
=
Table 8.8.
576 x 66 x 0.56 = 21288.96 N
13. Check/or wear: We find F d < FW"It means that the gear tooth has adequate wear
C4pacity and will not wear out. Thus the design is safe and satisfactory.
14.Check/or efficlency : We know that,
tany
TJactuaI
Where
= 0.95 tan (1 + p)
p = Friction angle =
= tan-I (0.03) = 1.7
0
..
-
",!;
TJ -.
find that the actUlll efflciency
••• ['.'
tan 15250
0 95 x
')
tan (15.250 +1.70
Is greater
[.: J.1 = tan p]
tarr! J.1
J.1 = 0.03, assumed]
.'
'I
=
.
0.8498 or 84.98%
than the desired efjlcJency. Thus the design
rtfletDry.
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"
.~
D eSlgn of Trafl81f1Lt'K/on
8.28
~
15. Calculation of basic dimensions of worm and worm gears: Refer Table 8.2.
./
Axial module:
./
Number of starts:
./
Number of teeth on worm wheel:
./
Face width of worm wheel:
b = 66 mm
./
Length of worm :
;?:
(12.5 + 0.09 z2) mX' from Table 8.5.
;?:
(12.5 + 0.09 x 72) 8 = 151.84 mm
~'-'.
mx = 8 mm
./
Centre distance :
./
Height factor:
./
BottQID'clearance :
./
Pitch diameter:
/0
=
Zl
L
3
Zl = 72
a = 0.5 mx (q
=
= 332 nun
+ z2) = 0.5 x 8 (J I + 72)
1
c = 0.25 mx =
s=»,
d) =
=
0.25 x 8 = 2 mm
11 x 8 = 88 mm;
and
d2 = z2 x mx = 72 x 8 = 576 mm
./
./
dOl = d 1+ 2 /0 . mx = 88 + 2 x 1 x 8
=
da2 = (z2+-2/0)mx
=
Tip diameter:
Root diameter:
dfl
=
(72+2x
1)8
104mm;aoi
592mm
=
d 1-_ 2 /0 . mx _ 2 . c
=
88 _ 2 x 1 x 8 --.2 x 2 = 68 mm ; and
df2 = (z2 _ 2 f 0) mx _ 2 . c
(72 _ 2 x I) 8"'- 2 x 2 .= 556 mm.
_
Example 8.14 I For the data of above example, determine the required coolingattlf
the ovenall heat transfer coefficient for the housing can be assumed as 15 WI"r C(' tUJ4 tit
temperature rise of the lubricant is restricted to 50 't:
I
=
Given Data: Refer Example 8.13; K,
Tofind:
10 W 1m2 °C;
At
=
to =!» = 5OOC.
Required cooling area (A).
@ Solution:
We know that in 'order to avoid overheating,
Heat generated (i. e., power loss) = Heat emitted into the atmosphere
(J _ 11)x Input power -
(l-O.8498)xI2xl03
...
-
K, A (to - to)
IOxAx50
Required cooling area, A - 3.605 ml ADs.
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8~29
"IJ GEARS
~~~
,
,
I
~.
DESIGN USING BASIC EQUATI
ONS
DESIGN FORMULAS FOR WORM GEARS DESIG
1 Design torque (or design load) on GearWh
IJO..
,
[ M/]
•
= MI
eel (M,)
K x Kd
... (8.28)
M, -
where
X
N
wheel torque = 60 x p
2nN2
Transmitted
K - Load concentration factor ,
- 1, when load is almost constant
Kd - Dynamic load factor
- 1, for v2 < -3 m1s
- 1.2, for
8.20.2.Induced Bending
V2
> 3 m1s
Stress in the Wheel (ab)
1.9 [M/]
... (829)
mx3 x q x Z2 xYv
where
n1x
q
= Axial module,
=
z2 -
Y
v
-
.8.20.3.Design Bending
Diameter factor (or diameter quotient),
Number of teeth on wonn wheel, and
Form factor based on virtual number of teeth, from Table 5.13 .
Stress for the Worm Wheel [ ab]
I
Designbending stresses [ab] for various wonn wheel materials are tabulated in Table 8.9.
Table 8.9. Design bending stress {uJ, N~
r--Wheel
lIlaterial
--Bronze
Method of
au
casting of wheel
N/mml
Sand
>390
Chill
>390
Sand
<390
Chill
<390
Centrifugally cast
<390
(from dalD book,pagt! 110.8.45)
60
GIldc25
~
30
-
250
20
27.5
:
40
350
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~
~
~
!
Design beading stress. N/mm2
Rotation in both
Rotation in one
directions
direction only
64
78
90
110
35
50
40
55
47
Cast iron
1
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.!
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8.30
Design ofT'ans",;.~
~--------------------~~~
.
8.20.4. Induced Contact Stress in the Wonn Wheel (GJ
"
[ (~ I q) + IJ
540
efe
= (z2Iq)
[M,]
2
a
10
'--/t..})
,
~
a = Centre distance
8.20.5. Design Contact Stress of the Wonn Wheel [ Gc]
Design contact stresses [
efe ]
for various combination
are ~\.._,
of materials
~n
Table 8.10.
Tabl« 8.10. D6ign contaa stress
I ur.J, NhttJn2
(from dIlJa book, JNlK~ no.1.4S)
Slidiag velocity,
Material
v.., .Js
Worm
Wheel
0.25
0.5
1
2
3
Steel
Cast iron
170-140
120
100
70
-
Bronze
190
18S
176
168
159
•
-
U;
8.20.6. Centre Distance (a)
a = [(z21 q) + 1 ]
3
[
540
(~I q) [ef
]
e
J[Mtl
-
.'!-
10
--- OJJ)
8.21. DESIGN PROCEDURE
1.
Select the suitable combination of materials for worm and worm wbeel,
Tables 8.3 and 8.9.
2.
Calculate the initial design torque [ M,]. Use [M,
K· Kd = 1.
3.
Selection
0/Z1
1 = M, x K x Kd•
coomDn!
InitiallyJ55II!It
and Z2:
'"
Select the number of starts of worm depending on the efficiency ~
consulting Table 8.4 .
./
Then,
z2 = ;
x Zt.
4. Selection 0/10'61 and I O'el:
Select the design bending-stress and design contact stress of the wonn wheel frOID T~
.8.9 and 8.10 respectively. To select [efe], ~1y
take v,r= 3 mls.
S. Calculate the centre distance (a) using the equation 8.31. Choose
factor, q = 11. q can vary from 8 to 13.
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initiallY ~
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r
--=~====~~~~--
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f~ ~~s
..,,,.tt the axial module (m.J using the rei ti
calc",'"
a 10n m
_!~
8.3]
=2a I( +
6. 'gb r standard axial module from Table 5 8
x
q zx)· Then, choose the
bl e,
.
. .
e centre dIStance (a) using the relation a = 0 5
1. BtvlS
• mx (q + z2)
a1ld vs :
1.1 ulate d, v,
~. CUle
I Pitch diameter (d) :
qx mx and d2 =z2 x
r' .
r
»,
_ 7td2N2
I Pitch line velocity (v) :
and v2 -
I Lead angle (y) :
y = tan-i
,
v
Sliding velocity :
s
60
(~)
= ~
cos y
9. Recalculatetile design contact stress [ucl for the actual
VS'
using Table 8.10.
10. Revise1(, Kd and [M,l for the actual velocity of the worm wheel (v2)'
11. Checkfor bending:
,
Calculate the induced bending stress using the equation 8.29 .
./ Compare the induced bending stress with the design bending stress .
./ If
<Jb
s [O"b ],
then the design is satisfactory.
12.If crb > [ O"b], then the design is not satisfactory. Then increase the axial module.
13. Check/or wear strength :
.; Calculate the induced contact stress using the equation 8.30 .
.; Compare .the induced contact stress with the design contact stress.
If <Jc s [O'c
"
],
then the design is safe and satisfactory.
14. Check/or efficiency:
"
If l'Jcalculated
~ TJdesired '
then the design is satisfactory.
-/' Otherwise increase the lead angle y.
15. Calculate the power loss and the area required to dissipate the heat.
r
16.. Calculate all tile basic dimensions of the worm and worm wheel using the equations
lSted In Table 8.2.
l7r.~e
8.15 ] A steel worm running at 240 r.p.m: receives 1.5 kW from i~ shaft.
dttt speedredUction is 10 ' 1 Design the drive so as to have an efficiency 0/ 80%. Also
","lie the cool'
".
d. if tI,e temrperature rise is restricted to 4S DC. Take
area requtrea,
eat transfer coefficient as 10 Wlml DC
O)IeraJIh
tng
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"
&1n
-
! r :
, F I
xeisr i' Wi.!", 1 '.
F imt'nuz tFrom
to', m
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6'g ., r
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7
8.32
Design of Trans",. .
Given Data: Nt = 240 r.p.m.;
K,= 10 W/m2 DC.
~'.
i = 10 ; lldcsired = 80%;
P = 1.5 kW;
t _I
o
o::t 4sec
I
Tofmd:
1. Design the wonn gear drive, and
2. The cooling area required (A).
N2 = Ntl i
@ Solution:
= 240/10
1. Selection of material: Wonn
Wheel
=
24 r.p.m.
- Steel
.. , (Give.j
- Bronze' (sand cast), selected from Tables 8.3andl.~.
2. Calculation of initial design wheel torque I Mt 1:
-
We know that,
[ M,] = M, x K x Kd
60 x P
where
M, = Wheel torque = 21t N2 -
K· Kd
-
596.83 N-m
=
1, assumed initially.
:. Design wheel torque, [ M,] - 596.83 x I
60 x 1.5 x I ()3
2x x 24
596.83 N-m
=
3. Selection of'l.1and'l.1 :
./'
Forn = 80%,
./'
Then
z2
zl
= 3 or 4, from
zl
= 3 is selected .
= i x zl = 10 x 3 = 30.
4. Selection ofl CTbl and
./'
Table 8.4. Here
I CTcl:
For bronze wheel, au < 390 N/mm2, [ab]
= 50 N/mm2 is selected, for rotatJOO
•
one direction, from Table 8.9 .
./' 'From Table 8.10, [ac]
= 159 N/mm2
is selected, assuming
Vs -=
3 mls.
5. Ca/cuJotion of centre distance (a) :
We know that,
where
...
3
[
Q'
=
[(zfq) + 1 ]
q
=
11, initially chosen.
Q
=
[(30/11)
+ 1]
540
(zfq) [ (Jc]
3 [
540,]
] 2 [M,]
(30/11) 159
10
2
x
596.83 x 1~
10
= 168.6 mm
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•
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:
8.33
i
.
f~
(;4lCJllllli"" 0/ oxiDl module (mx) :
6.
2 a _ 2 x .168.6
(q + z2) - (11 + 30) = 8.22 mm
m =
x
f(Olll 'fable5.8,
,I
the nearest higher standard axial mo du Ie .IS 10 mm
BeVision o/centre distance (a) :
1.
a = 0.5 m,x (q + z2) = 0.5 x 10 (11 + 30)
:H
'I
::I
.
II
= 205 mm
I
,,
I
8. Calculation of d, v, rand
vs:
Pitch diameters:
,
= 11 x
= 30 x
dl
-
d2
=
q x mx
z2 x mx
_
1t dIN 1 _ 1t
VI
Pitch line velocity :
,
il
60
1t
v2
=
-
d2 ij2 _
60
-'
1t
Y = tarr! (~)
.; Lead angle:
X
10 = 110 mm ; and
10
=
!I
;
300 mm.
110 x 10-3 x 240
60
== 1.382 m/s ; and
x 300 x 10-3 x 24 == 0.377 m/s .
60
= tarr" (
:1)
= 15.25'
",
II
)
/
1.382
cos'Y
cos
15.250 ==. 1 432 mIs
VI
Sliding velocity:
v
Vs
-
9. Recalculation of design contact stress
For
Vs =
1.432
mis,
10. RevisuJII of
"
For
v2
[CJ ]
c
~
I O'c J :
172 N/mm2, from Table 8.10.
I u,J,:
<3
mis, Kd
== 1
:. [M ] = M,)( K. Kd = 596.8~)( I x I = 596,83 N-m
,
11. Check for bending: We know that the induced bending IstresS,
1.9 [M/]
mx3
x q x Z2 xYv
Form factor based on virtual number of teeth, from Table 5.13.
where
Yv
-
30
cos3 'Y == -;s3 15
Z
Zv
==
Y. '"
==
Then,
We find
CJb,
06 <
.250 ~
34
0.452, for z. = 34, from Table 8.7.
_1.9 x ,596.83 x ~~
== 7.6 N/mm2
(10)3 x 11 x 30 x 0:452
0
I06]' thus the deSign is saJis/adory agolllSt bending.
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I Tip diameter:
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dal Downloaded
::::
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dQ].
8
::::
I Root diameter:
::::
d'l
::::
::::
d'2
::::
RevIew AND SUMMARY
parallel "ear
shafts.drives are USed to transmit PO"'e,.b
I
1U.
I
I
I
worm
o·
'h..
•
e,."een two nOn-tnlerseClingnon.
"
Types: I. Single-enveloPing;
and 2. DoUhle-enVeloPingWortndrive.
SfwJ:lflcfllion: A pair of Worm gears can he spec!fied hy zJ / z} / q / tn.-
I chapter.
The worm gear nomenclQture
and its lcinenzatics are presented in the beginning of this
8.1 and 8.2. and basic dimensions of Wortn and worm gears ore labulaled in
I Tables
Tooth proportions
,I
Force analysis on Worm gearing .-
1. Components of the result(11ltforces, neglectlngfrlctlon :
F; =
F x cos a x sin
Fr = F x sin a
Fa
=
F
X
cos a
=
X
r
Fax Ian a
r =
cos
2. Components of the resultant forces, consldenng
.
F;I
=
F;IIan
. fiiction·
r
.
2x~
d,
sin a
= ~I
Frl
x (cos a . sin r+ p .cos JI)
(cos a .cos r-If .sin 12
= ~I
Fal
" Fol'ces on lhe
WOr",
(cosa . sin
. F
"'heel:
F
X
= -Fal'
r+ p . cos r)
; and Fa2 = _ F,J
=_F
r2
rl
r
t2
• 12
cos Q_ p .tan
" I!JJklency: '1 = tan y x (cos
(cos aa..sin
cosyy·sm ~"' = cos a + II .col r
+ IIII .cos
" PoWerIOSlIn friction = (I _ 1J) x P
.£'
•
" !f "~cos a . tan y, then the drive
. IS. se!f-loo.mg.
" If
, Ii ~ cos a . Ian y,
. ' then the drIVe
. l" oVl!rrllJlnmg.
•
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Hl~
_~.36
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------------------_D_e_S~ign~o~if_TJ_ra_n~~m~u~~~
. - ~
./
Materitll: Usually, worm is of steel and worm wheel may be brass or cast iron.......____
./
Two method" of designing a worm and worm gear: 1. Worm gear de.fign UN/nK
l,f'ltll
and Buckingham's equations; and 2. Worm gear design based on basic equal/on.',
The step by step procedure for the above said two methods are presented wllh luflklttt
./
illustrative problems.
Lewis beam strength for worm gears:
F,
./
Dynamic load on worm gear: Fd
/
./
Wear tooth load on worm gear:
./
Thermal rating 0/ worm gearing: The condition to avoid overheating of the "'ire k
given by (l - 7]) x Input power = K/ x A x (t 0 - t oJ
= F/
FlY
= 1! X mx x b x [ O"bi x r'
Cv
= d2 x b x Kw
Worm fallure : 1. Seizure, and 2. Pitting and rupture.
REVIEW QUESTIONS
1.
Under what situation worm gear drives are used?
2.
What are the advantages and disadvantages
3.
What is irreversibility in worm gear drives?
4.
What are single-enveloping
5.
How can you speci fy a pair of worni gears?
6.
Define the following terms used in worm gearing:
angle; and (d) Helix angle.
7.
What are the various forces acting on a worm and worm gear? Deduce the expressions
of them.
8.
Differentiate self-locking and overrunning drives.
9.
Write the expressions for beam strength, dynamic load, and limiting wear load for worm
and worm gears and explain the various terms used in it.
of worm gear drives?
and double-enveloping
worm drives?
(a) Axial pitch;
(b) Lead;
(c) Lead
10. Write an engineering brief about the thermal calculations in worm gearing.
11. Why is the efficiency of a worm gear drive comparatively
low?
PROBLEMS FOR PRACTICE
Problems all klnematlcs of worm gears :
1.
A pair of worm gears is designated as : 1 /30/
10/8.
Calculate:
worm;
(i) the centre distance; (ii) the speed reduction'
(iii) the dimensions oftbe
and (iv) the dimensions of the worm wheel.
'
[An.t'" (i) 160 mm; (ii) 30; (iii) d1 = 80 nun' , dal
240 mm; do2 = 256 mm; dl2 = 220 mm]
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= 96
mm : d
,
II
= 62 mm ':
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(iv)tiz:::
-~-=~~~~~~~ ~~
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~~s
~w.
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831
fi
"'''readed worm, having a lead of 75 rnm m h.
,.
triple-up
. h f 24'
es es with a €It-.. L .. '
J.
I circular pitc or
mm. Find the centre di
...
gear Havmg 45 tt.tth and
! ,
stalll:e betw~ the shafl.. if Ihey are
90 . _threaded wonn has a pitch diameter of 115
..
[AIlS: 22(J mm J
J. J. trlrhas a pitch perpendicular to the teeth of)2 :' J~ ~ob for ~ing the worm
wbeegear if the reduction is 12 : 1.
m U es. Find the pitch diamettr c;f
onn
w ble-threaded worm has a lead of 60 mm The j . h
[Alii: 454.36 mm]
A dou
'
gear as 30 teeth and is cut 'Ih
4. b of module 8.5 mm perpendicular to the teeth. Find the itch dis
'.
WI.
a
gear, and the centre distance of the shafts.
P
JamL1ers of the worm
p:~
'.
:d
.
[Ans: 37.3 mm ; 286A8 mm ; 161,89 mm)
onforce analysis of worm gears:
!TObie
d
hi'
d .
5 A pair of worm an worm w ee . IS . eSlgnated as 3 / 60 / )0 I 6. The worm is
ttfS
. tranSmitting5 kW power at 1440 r.p.m. to the worm wheel. The coefficient of friction is
O.land the normal pressure angJe is 20°. Determine the components of the gt:ar tooth
force acting on the worm and the worm wheel.
[Ans: F/I = Fa2 = 1105.24 N; Frl = Fn = 1('33.35 N; FaJ = Fa = 2632.55 NJ
6. A worm drive transmits 15 kW at 2000 r.p.m. to a machine carriage at 75 r.p.m. The
worm is triple threaded and has 65 mm pitch diameter. The worm gear has 90 teeth of
6 mmmodule. The tooth form is to be 200 full depth involute. The coefficient of friction
is0.1 O. Calculate: 1. tangential force acting on the worm; 2. axial thrust and separating
force on worm; and 3. efficiency of the worm drive.
[Ans: 1. 2203 N; 2. 7953 N; 2895 N; 3. 70.2%J
7. A 5 kW power at 720 r.p.m. is supplied to the worm shaft, as shown in Fig.8.8. The
worm gear drive is designated as 2 / 40 / 10 / 5. The worm has right hand threads and
pressureangle is 200• The worm wheel is mounted between two bearings A and B and it
can be assumed that the bearing B takes the complete thrust load. Determine the
reactionsat the two bearings.
z
[
Flo. 8.8.
. Ans: F~ = 5548.5N ; F~
= 135.83N;
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b'
s
. y =
Fa = 2653N , Fa
5548.5N;
F
z
a
=
42858N]
.
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8.38
Design ofTraru~
~
8.
1 kW power at 720 r.p.m. is supplied to the worm shaft. The number of
starts for
has 30 Ieetb~
Calculate the efficiency of the Worn. .~
[AIls : 90.12%; 98.1~
of worm are four with a 50 mm pitch circle diameter. The worm wheel
. a 5 mm module. The normal pressure is 200•
drive and the power lost in friction.
Problems on worm gear drive design:
9.
Design a worm gear drive to transmit 8 kW at 720 r.p.m. The desired velocity rabo ~
36 : 1.
10. Design a worm drive for a speed reducer to transmit IS kW at 1440 r.p.m, of the "tllI!
shaft. The desired wheel speed is 60 r.p.m. Select suitable worm and wheel materials.
11. Design a worm gear drive to transmit 10 kW at 1440 r.p.m. with a gear ratio of 12 ~
steel worm and cast iron wheel.
12. A hardened steel worm rotates at 1440 r.p.m. and transmits 12 kW to a pbospborbrooze
gear with a gear ratio of 15. Calculate the centre distance and module required for ~
drive.
13. Design a worm gear drive to transmit 15 kW from a worm at 1440 r.p.m. to the worm
wheel. The speed of the worm wheel should be 40 ± 2% r.p.m.
14. Design a worm and worm wheel drive to transmit 50 kW from an electric motor nmning
at 1440 r.p.m. to a rolling mill required to run at 100 r.p.m. Selecting case hardened
alloy steel for the worm and centrifugally cast phosphor bronze for the wheel, design all
the details of the geared set.
If the complete heat generated is dissipated by the housing, what should be the housing
area required? The overall heat transfer coefficient for the housing can be assumed as 15
W/m2 °C and the temperature rise of the lubricant is restricted to 50°C.
15. Design a worm gear drive to transmit a power of 22.5 kW. The worm speed is 1440
r.p.m. and the speed of the wheel is 60 r.p.m. The drive should have a minimUJP
efficiency of 80% and above. Select suitable materials for the worm and wheel and
decide upon the dimensions of the drive.
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Gear Box
"There is sOmethin
g Wore than a difJi
If you try t
' 'cully." is inertia
.
pe , 'cultie. yrw, uecay
..I
...
o escao diffi
",
,,1. INTR
UII
ODUCTION
.
.
- He,bllt UwOIl
,11 knOW that machme
tools like lathe, milling
hi
"e
.
mac mes, etc reoui
.
. die speeds. Because a machme tool is adoptabi c:
."
qUire a wide range of
spiO
..
e lor cuttmg diff
. g different properties usmg varying grades of cutti
I
erent types of metals
haV1D
• •
109 too s on work- .
.
J'ameters. Thus the provision of variable spindle speed .
~Ieces of different
ul
•
he vari
s IS necessary m order to
~jfferent reqUirements. T e varIOUS methods used for obtainin d'ffi
m~t
•
Co II
.
g I erent speeds of machme
1001 spmdleare as 10 OWS.
(i) By using a gear box mechanism,
(ii) By using a cone pulley arrangement,
(iii) By using a variable speed electric motor, and
(iv) By hydraulic operation.
Among these methods, the gear box method is very popularly used. In this chapter, we
shall discussthe design of gear boxes, in detail, in the following sections.
9.2.REQUIREMENTSOF A SPEED GEAR BOXES
A speedgear box should have the following requirements:
.;
It should provide the designed series of spindle speeds .
.; It should transmit the required amount of power to the spindle.
.; I
.
f the transmission.
t should provide smooth silent operatIOn a
.; It should have simple construction.
.
th t it is easier to
.;
Id be easily accessible so a
Mechanism of speed gear boxes shou
carry out preventive maintenance.
8.3
.
ES ARE IN GEOMETRIC
. THe SPEEDS IN MACHINE TOOL GEAR BOX
.
PROGRE
. (A P) geometric
'l'L
SSION. WHY?
.
. progresSiOn .' :
(L P.).
-ne S
ed in artttuneUc
.
rogresSIOn .
PrOD.
J>eeds in gear boxes can be arrang
d logarithmic P
~essio
.'
. (H P), an
n (G.P.), harmonic
progressIOn
.
l
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~
~9.2
------------
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~D~es_:;g~n~O_if_1i~r~Q~ns~m~is_s_iO_"-!SY&lelft.&
---.:..
However, when the speeds are arrange
d i G P it has the following advantages Over the
an .',
other progressions.
1 The speed loss is minimum.
.
.
d Avai lable speed
.t.e., Speed loss = Desired optImum spee 2.
3.
The number of gears to be employed is minimum.
f . die speeds at each step.
G.P. provides a more even range 0 span
4.
5.
The layout is comparatively very compact.
led'
Productivity of a machining operation, i.e., surface area of the meta remov
6.
time, is constant in the whole speed range.
G ..P mac hime tool spindle speeds can be selected .easily from preferred numbers.
Because preferred numbers are in geometric progression.
In unit
9.4. METHODS FOR CHANGING SPEED IN GEAR BOXES
The two important methods widely used are :
1.
Sliding mesh gear box, and
2.
Constant mesh gear box.
9.4.1. Sliding Mesh Gear Box
It is the oldest and simplest form of gear box. Sliding type gear boxes are quite commonly
used in general purpose machine tools. In order to mesh gears on the main shaft with
appropriate gears on the spindle shaft for obtaining different speeds, they are moved to the
right or the left. It derives its name from the fact that the meshing of the gears take place by
sliding of gears on each other.
9.4.2. Constant
Mesh Gear Box
It derives its name from the fact that all the gears whether of the countershaft or the main
shaft are in constant mesh with each other. It is also known as a silent or quite gear box. It
gives a quieter operation and makes gear changing easier by employing helical gears for the
constant mesh. In order to connect the required gear wheel by means of teeth on the side of
the gear wheel, a separate sliding member is employed.
9.5. PREFERRED NUMBERS
Preferred numbers are the conventionally rounded off values derived from geometric
series. There are five basic series, denoted as R 5, RIO, R 20, R 40 and R 80 series. The
symbol 'R' is used as a tribute to French engineer Charles Renard, who introduced the
p~efer:red num~rs first. Preferred numbers assist the designer in avoiding the selection of
sizes m an arbitrary manner.
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9.3
ae
~
'.
'eshas I·t: own step ratio
ch ser I
68. fable 9.1.
'vell 111
t.e.,
• /
gl
.
senes factor Th'
.
e serres factor for various
series are
.
•
liable 9.1. Step ratio sertes tacto« (from data book , p age no. 7.19
. '/11
Basic series
Step ratio <cI»)
Vw
=
1.58
RIO
l~
=
1.26
R20
2~ 10
=
1.12
R40
4~
=
1.06
R80
8~
=
1.03
R5
The series of preferred numbers is obtained by multiplying a step ratio with the first
number to get the second number. The third number is obtained by multiplying a step ratio
wilh the second number. Similarly the procedure is continued until the series is completed.
Table 9.2 shows the basic series of preferred numbers.
Table 9.2. Basic series of preferred numbers (from data book, page no. 7.20)
Preferred numbers
Basic series
R 5 (~= 1.6)
1.00, 1.60, 2.50, 4.00, 6.30, 10.00
R 10 (~= 1.25)
1.00,1.25,1.60,2.00,2.50,3.15,4.00,5.00,6.30,8.00,10.00
R 20 (~= 1.12)
1.00, 1.12, 1.25, 1.40, 1.60, 1.80, 2.00, 2.24, 2.50, 2.80, 3.15,
3.55,4.00,4.50,5.00,5.60,6.30,
R 40 (~ = 1.06)
7.10, 8.00, 9.00, 10.00
1.00,1.06,1.12,1.18,1.25,1.32,1.40,1.50,1.60,1.70,1.80,
1.90, 2.00, 2.12, 2.24, 2.36, 2.50, 2.65, 2.80, 3.00, 3.15, 3.35,
3.55,3.75,4.00,4.25,4.50,4.75,5.00,5.30,5.60,
6.00, 6.30,
6.70, 7.10, 7.50, 8.00, 8.50, 9.00, 9.50,10.00.
I EXample: Machine tool spindle speeds under R 20 series is given by 100, 112, 125, 140,
60, 180 and 200 r.p.m.
9,6,STEP RATIO (OR SERIES RATIO OR PROGRESSION RATIO) (cI»)
the
:hen t~e spindle
speeds are arranged in geometric progression, then the ratio between
If 0 adjacent Speeds is known as step ratio or progression ratio. It is denoted by ~.
N N
.
. th
I, b N), ...... , N are the spindle speeds arranged in geometric progression,
en
n
N2
~
If' , .
n
IS
the n
=
N3
N2
=
N4
N3
Nn
= ......... Nn - 1
= constant = ~
umber of steps of speed, then
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Design O/T"QIU",isSiol1
9.4
~
= 4»n-1 or / -Nmax =
N
_!!
NJ
I Nou I Permissible
,l.n-I
'f'
L_N~m=m
deviation
=
/1
~
". (9.1)
± 10 (4)> - 1)%.
I Example
9.1 } Find the progression ratio for a 12 speed gear box having 'Ptttls
between 100 and 355 r.p.m. Also find the spindle speeds.
Given Data: n = 12; Nmin = 100 r.p.m.; Nnrax = 355 r.p.m.
To flnd : I. Progression ratio (<1», and 2. Spindle speeds.
@ Solution:
1. Progression ratio (f)): We know that,
Nmax
=
<t>n-
=
4»12- 1 or
«:
355
100
or
1
<t> =
(3.55)1111
=
1.122 Ans. ~
2. Spindle speeds: Since the calculated <t> (= 1.12) is a standard step ratio for R 20 series.
Therefore the spindle speeds from R 20 series are
100,112,125,140,160,180,200,224,250,280,3]5
and 355 r.p.m. Ans. ~
I Example
9.2 , Select the spindle speeds for the following data: 12 speeds, between50
and 600 r.p.m:
Given Data: n = 12;
Tofind:
Nmin
= 50 r.p.m.;
Nmax
= 600 r.p.m.
Spindle speeds.
@ Solution:
We know that
600
50
or
=
q,n-1
=
q,12 -1
or
q, =
1.253
We find the calculated <P is a standard step ratio for R 10 series. So from R 10 series, the
spindle speeds are 50, 63, 80, 100, 125, 160, 200, 250, 315, 400, 500 and 630 r.p.m. ADS. "CJ
It can be seen that the calculated Nmax
630 r.p.rn., which is greater than the required
maximum speed. Therefore we have to check whether the deviation is within the pennissib1e
range or not.
Permissible deviation
Then,
Actual deviation
=
=
± 10 (<I> - 1) %
- ± 10 (1.253 - 1) %
50
- (630 _..:600) x 600
= ± 2.53%
=
2.5
Since the actual deviation is less than the permissible deviation, therefore we can accept
the deviation. Ans • ..,
I Note I If the actual deviation is more than the permissible deviation, then non-standard speedsmay
be used.
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9.S
Select spillllle speeds, for Y lp('ec/ gear box b tw
It 9.3
9' N
pa/(I· . n - , "lin
·nn'l'._
_
G~I/"
= 80
N max
r.p.m.;
, e een
=)
80
285 r .p.m.
and J 185 r.p.m.
d
d' spindle spee s.
foft" .
.
~ So/pilon.
Nmax
. We know that,
=
1285
80
~n-I
= ~9 - I
or ~ = 1.415
or
fi d ~::: 1.4 I 5 is not a standard ratio. So let us find out whether multiples of standard
We n 1 06 come close to ).4) 5.
'0
or .
(3n 112
.
Wecanwrite,
1.12 x (1.12 x 1.12)
So ~ :: 1.12 satisfie.s the requirement.
. . 2 speeds , are giveu by
sklpprng
=
1.405
Therefore,
'" (skip 2 speeds)
the spindle speeds from R 20 series,
80, 112, 160, 224, 315, 450, 630, 900 and 1250 r.p.m. Ans."
Alternate solution: For the above problem, we can also write
= 1.418
1.06 x (1.06 x 1.06 x 1.06 x 1.06 x 1.06)
So $ = 1.06 also satisfies the requirement.
sbpping
5 speeds, are given by
80,112,160,224,315,450,630,900
... (skip 5 speeds)
Therefore the spindle speeds from R 40 series,
and 1250 r.p.m.
Ans."
IIcan be noted that both R 20 and R 40 series gives the same spindle speeds.
[Example 9.4
GivenData'
Tofind:
•
I Select the spindle
Nmm.
= 50
rpm'
. .
.,
speeds, 50 - 800 r.p.m., 12 speeds.
Nmax
=
800 r.p.m. " n = 12.
Spindle speeds.
© 110
o I
Nmox
tuion : We know that N.
= ~n-I
m1l1
800
50
Or
We find
stand'
J. 'I' -.
I 2866 IS
.
=
110t
A.12 - I
'I'
or ~
=
1.2866
a s t an d ar d ratio . So let us find out whether multiples
of
Ward ratio 1. 12 or 1.06 come close to 1.2 86 .
e can write,
1.12 x 1.12 = 1.2544- and 1.24 x 1.12 x ]. 12 = 1.4
Then I
1 06 1 06 x 1 06 x 1.06 = 1.34
, .06 x 1.06 x 1.06 x 1.06 = 1.26 and 1.06 x.
x.·
.
Soboth
.'
t It means 1.286 IS not a
staoda
d ~tandard ratios 1.12 and 1.06 not satisfy the reqUlrem.en .
r raho· Th
. dl
dare obtamed as
50
. erefore the non-standard spm e spee s
, 50 x 1 28
3
2
6 rhus the '. 6, 50 x 1.286 , 50 x 1.268 ,
758 226.2, i'90.8, 374,481,
81.5and79~p~ndle speeds are 50, 64.3, 82.7, 106.3, 136.7,]
.,
. r·P·m. Ans."
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Design Of7"~
9.6
1
~iSlin
9.7. STRUCTURAL FORMULA.
= Number of speeds available at the spindle
Let
n
,
Stage numbers in the gear box, and
PI'P2'P3'
.. · .. · Characteristic of the stage.
XI'
X2, X3,
......
~\,
=
Then, the structural formula is given as
n = PI (Xl)
P2 (X2)
2nd stage
1sl stage
x, =
where
1,
X2
= PI;
X3
= Pl'
rd
3 stage
P2;
X4
41h
= Pl'
stage
'"
(~.
,
P2' P3
9.7.1. Preferred Structural Formulas
The Table 9.3 shows the preferred structural formulas for the different
boxes.
sPteds of
Table 9.3. Preferred structural formulas
'.
S.No.
l.
2.
Preferred
Number of speeds
6 speeds
8 speeds
structural formula
(i)
3(1)
(ii)
2 (I) 3 (2)
(i)
2 (I) 2'(~
(ii)
4(1)
2(3)
2 (4)
2(4)
3,
9 speeds
(i)
3(1)3(3)
4,
12 speeds
(i)
3(1)2(3)2(6)
(ii)
2 (1) 3 (2) 2 (6)
(iii)
2(1)2(2)3(4)
(i)
3(1)3(3)2(5)
(ii)
4(1) 2(4) 2(6)
5.
14 speeds
6.
15 speeds
(i)
3(1)3(3)2(6)
7.
16 speeds
(i)
4(1) 2(4) 2(8)
(ii)
2 (1) 4 (2) 2 (8)
(iii)
2(1)2(2)4(4)
(i)
3(1) 3(3) 2(9)
(ii)
3(1)2(3)3(6)
(iii)
2 (1) 3 (2) 3 (6)
8.
18 speeds
-
-
----
9.8. KINEMATIC LAYOUT (OR KINEMATIC ARRANGEMENT)
Th ki
.
. Fig 9.1.
e mernatic arrangement of a multi-speed gear box is shown In
.
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~
r
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[ rfb
pulley
l
[
Shan 1 (Driving shaft)
J
J
Shaft 2 (I I
'
n ermedaate shaft)
Shaft 3 (Driven shaft)
Pulley
Fig. 9.1. Kinematic arrangement of a 9 speed gear box
From the Fig.9.1, it is clear that the kinematic layout shows the arrangement of gears in a
gear box.The kinematic layout provides the following informations required for gear box
design.
,
The number of speeds available at the spindle, i.e., at the driven shaft.
,
The number of stages used to achieve the required spindle speeds.
,
The number of simple gear trains required to obtain the required spindle speeds
and their arrangement.
,
The overall working principle of the gear box.
.; The information required for structural formula and ray diagram.
.
. d fr
driving shaft to driven shaft
fflustra/ion: In Fig.9.1, the power is transmltte
om d h . g I'S obtained using
drro
•
I
box spee c angm
ugh a intermediate shaft. In this conventlOna gear,
d [J om driving shaft to
Ilid'
h the number of spee s r
Iflg gear mechanism. It can be seen t at
.
h ft is 3 Then the number
intermediate
shaft is 3 and that from intermediate shaft to dnven sa·
ofspindlespeeds is equal to 3 x 3 == 9.
Th
.
, e structural formu la for the kinematiC arran
~venby
n
where
=
PI -
gement of gear
box shown in Fig.9.1, is
PI (XI) .P2 (X2)
3 eeds available)
1 there are sp
3 (i.e., in stage ,
eeds available)
.
2 there are 3 sp
3 (i.e"lOstage
,
1; and X2::::: PI :::::3
I
,
, Stru
P2
=
XI
=
z
etural formula
~re
== 3 (1) . 3 (3).
'
n
= Number ofspee
ds available at
_ PI . P2 == 3 x 3
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the driven shaft
:::::9
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y DIAGRAM (OR SPEED DIAGRAM)
9.1. RA
The ray diagram is a graphical representation
of
the drive arrangement in general form. In ot~er words,
the ray diagram is a graphical representatIon of the
Ns
t-----t-~~L__j
structural formula, as shown in Fig.9.2.
N1
It provides the following data on the drive:
./
Ns
r-_____:::~....::::!""4E-~~-.J.
The number of stages (a stage is a set
of gear trains arranged on two
- -
N4
consecutive shafts).
.....
N
I
./
The number of speeds in each stage.
./
Ns
I
~ 1----
~+--~
en
en
~
The order of kinematic arrangement of
~
N,
the stages .
Fig. 9.2. Ray diagram 0/9 speed gear box
./
The specific values of all the transmission
ratios in the drive .
./
The total number of speeds available at the spindle.
9.9.1. Procedure
./
In this diagram, shafts are shown by vertical equidistant and parallel lines .
./
The speeds are plotted vertical on a logarithmic scale with log
./
Transmission
engaged at definite
speeds of the driving
$ as
a unit.
and driven shafts are
shown on the diagram by rays connecting the points on the shaft lines representing
these speeds .
./
Fig.9.2 shows the ray diagram for a 9 speed gear box, having the structural
formula, z = 3 (I)· 3 (3).
9.10. BASIC RULES FOR OPTIMUM GEAR BOX DESIGN
The basic rules to be followed while designing the gear boxes are as follows:
I. The transmission ratio (I) in a gear box is limited by
1
4 s
In other words,
i
i".;n
=
ilNQX
=
s
2. Refer Fig.9.3.
-.,
N;npul
Nmax
N;npul
I
Ninput
>- -4 and
s2
... (9.3a)
Nmin
.
Fig. 9.3.
2. For stable operation, the speed ratio at any stage should not be greater than 8.
In other words,
Nmax
--
.. , (9.3b
~ 8
N":'1;"
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9.9
t
,,90.
Ii
II StBg
es e]{C
ept in the first stage, Nmax
.
~
N.
mput
>N
min'
'.
th of matlOg gears In a given stage must be the sam e.
oftee
e lor same module in
e
J fb s~ gear set.
~, 'dJllg
s sit.
number of teeth on smallest gear in drives should b
th
il1UllUJ11
e greater an or
The t1l
I" al to 17.
eqU .'
01 difference between the number of teeth of adjacent gears must be 4
il1UllU
The rn
. .
ibl
'.
.
D,
I ould be of mrmmum POSSI e size, Both radial as well as axial dimensions
boX s 1
1, Gear
small as possible.
hou1d be as
~
Tile minimum anti maximum speed of a six speed gear box are to he
I~B
J.
J1l
~
I~a"d500 r.p.
~
f"
I1Il'
Construct tile kinematic arrangement and the ray diagram of the gear
.:::: 6' N . == 160 r.p.m.; Nmax == 500 r.p.m.
, nun
truction of the kinematic arrangement and the ray diagram.
enData.
n
•
Toftnd: CODS
@Solution :
Selection of spindle speeds :
_N"lax
Weknow that,
500 ==
160
= 1.256
",6 - I
or
'I'
~ == 1.256
.
find out whether multiples of stsn
Wecan write,
So
dsrd
is not a standard ratio. So let us
1.12 or 1.06 come close to 1.256.
llIio
~
~n- I
Nmin
or
Wefind ~
==
1.12 x 1. f2
= 1.254
•
~ = 1.12 satisfies the reqlllrement.
. dl
Therefore the spm e sp
... (skip one speed)
eeds from R 20 series,
,
ppmg one speed, are given by
160,200, 250, 315, 400 and 500 r.p.rn.
I&'.rmula ::::3 (1)
2 (3).
f d structura 10
sl
2nd stage
truetura/formula: For 6 speeds, the pre erre
1 stage
S
laUd'
'I lagram'
•
PfOced
.
idi tant lines to
. caI eqUI
IS
d ew 3 verU
{ S.
. .
atic layout, r
Inee there are 3 shafts In kinem
.
to represent
'd'stant
[ines
represent shafts.
6 horizontal equ1 I . Fig.9.4(a).
{ S'
d draW
shown lfi
d
lnee there are 6 spindle spee s,
.ZO ntallines, as
In the secon
he hort
0 stages.
d (3)
{ s))eeds.Then mark the speeds on t
that there are tw"1 ble in a stage an
F
it is clear
ds aval a
rom the structural formula, I
mber of spee d
sta
ts the nu
spee s.
ge, i.e., in 2 (3), 2 represen
those tWO
UTe :
represent"
thp ctpn,
or intervals
betWeen
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o:.!~ __
_2-. J 0
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-------------
./
D_es.:...::jg~n.:...o'2._ifTrQ'" . ~
tnlssio
.
B
After 3 steps above, locate the secon d point
and B are the two output speeds.
./
11 Sy!/
.
Locate the first point A on the lowest spee d I. e., at 160 rp.m
at
3
~
" on the I
These tw aSI8~afJ,
15 r.p.m,
()POintl A
To locate the input speed point from the preceding shaft (i
.
.e., shaft
following requirements should be met. That IS,
2),
I 1
>
- -4 and
nmax
-
ninput
t1t
nmax
-~2
ninpllt
Locate point C at the input speed of250 r.p.m.
Nmin
At point C, we get
N
=
input
]160
1
250 ~ 4 and
Nmax
Ninpl!t
Thus the requirements are satisfied. Note that the above conditions are met for other inp.r
speeds.
'3 (1)
'2 (3)
250
.,...
.,...
.200
/
I
IV
==
s:
IV
==
s:
CJ)
CJ)
.".
I
IV
==
s:
CJ)
160
A
A
(a)
(b)
Fig. 9.4. Ray diagram for 6 speed gear box
./
.~
d available III
In the first stage, i.e., in 3(1), 3 represents the number of spee s
ed is 81.(.
represents the step between those speeds. The lowestspehOve~
·
h
i
I
w hIC IS a ready located. Now locate points D and E on t he ZOd shaft, at sta~J ~
.
d . the ftfS
C'·
stage and (I)
, m a smgle step interval. For these three output spee s
10
input should be from shaft 1.
.
./ I
. . the ratiO
nput speed can be located anywhere on shaft 1 satlsfymg
In this case, the point F is located at 500 r.p.m.
./
'
uirettlc#
reQ
t A~~~
peeds a . ~~
I~ t~e second stage, we find input speed at C gives twO output seds. 'fhal \~ ,
SimIlarly, input speeds at D and E should give two output sped aW lioeS
.
. t
point D, draw lines parallel to CA and CB. Then from pom
E
'
r
to CA and CB, as shown in Fig.9.4(b).
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9.11
-------P1
=3
----
-----.
__
...
--
--
----.
--------------
. Input shaft - 1
Shaft-2
--0 Output shaft - 3
Fig. 9.5. Kinematic arrangement/or 6 speedgearbox
[§xample 9.6 , A nine speed gear box, used as a "ead stock gear box of a turret lathe,
is to providea speed range of 180 r.p.m. to 1800 r.p.m: Using standard step ratio, draw the
spttd diagram, and the kinematic
layout. Also find and fIX the number of teeth on all
gtars.
GivenData:
n
= 9;
=
Nmin
Tofind: Construction
Nmax
180 r.p.m.;
of speed diagram
=
1800 r.p.m.
and kinematic layout.
@Solution:
Selectionof ~pin{lle speeds:
We know that ,
Nmax
Nmin
1800
Or
-
cpn-l
::::
cp9 - 1 or cp
18Q
.Wefind ~
== 1.333 is not a standard
== 1J33
h ther multiples of standard
ratio. So let us find out w e
d 1 12)( 1 12)( 1.12 := 1.405
1 2544 an
.
.
e can write,
1.12 x 1.12::::
.
..' (skip 4 speeds)
l'h
1 338
.
en 1.06 x (1.06 x 1.06 x 1.06 x 1.0.Q)::::·
indle speeds from R 40 series,
So ~ :::::.
.
Therefore the sp
s(ip.
1.06 satIsfies the reqUirement.
ratiO
1.12 Or 1.06 come close to 1.333.
W
Ping 4 s
1&
.
peeds, are given by
d 1800 f.p·m.
0, 236, 31 S, 425, 560, 750, 1000, 1320 an
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9.12
:3 (3)
3 (1)
....-----,..------.11,
,
,,
,,
G
Structural formula: For 9 speeds) the prdtn~
structural formula =, 3 (I) 3 (3).
1800
Speed diagram: The speed diagram is dra ~:,
.
shown in Fig.9.6. The procedure is the sarn~ a:.
discussed in the previous problem.
.,_----+-~':._____,. 1320
Stage 2:
Nmin
Ninput
180
= 560
1
= 0.32> 4'
,
,
.t: I----.t:
IV
s:
en
Nmax
N
ttl
236
-+----
Ninpul
and
1000
= 560
s:
en
= 1.78 < 2
.. Ratio requirements
are satisfied.
Fig. 9.6. Speed diagram for 9 speed gear box
Kinematic arrangement:
shown in Fig.9.7.
The kinematic arrangement
Z3
for 9 speed gear box is drawn, as
= 29
------------.---
Shaft-1
Zg
= 53
Shaft - 2
Ze
= 48
Z,0
= 30
Z,2
= 63
Fig. 9.7. Kinematic arrangement for 9 speed gear box
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------
~9~.13~
Calculat~on of number of teeth:
NI, N2, N3,
......
Number of teeth ofth
Spe d
e gears 1, 2, 3, ... 12 respectively and
e s of the gears 1 2 3
•
, , , ...... , 12 respectively.
_
NI2
Second stage:
First pair: First consider the ray th t .
a glves the maxi
.
diaoraI1l, we find that the speed is reduc d f
mum speed reduction. From the speed
e
.
e rom 560 r
.
may assume that this speed reduction is achi
d
.~.m. to 180 r.p.m. (Refer Flg.9.6). We
1
ieve by uSing the gears 11 and 12
We know t iat,
zmin ~
17. Therefore assume Z - 20 (dri ) .
II fiver
zll
_ NI2
20
180
z12
Nll or zI2 == 560 ;
or
z12
62.22 ~ 63
-
Second pair: Now co.nsider the ray that gives the minimum speed reduction from
560 r.p.m. to 425 r.p.m. This can be achieved by using the gears 7 and 8.
Ng
z7
Zg
N7
-
425
560
=
or
z7
= 0.76
... (i) .
Zg
We know that the centre distance between the shafts are fixed and same. Therefore, the
sum of number of teeth of mating gears should be equal. So we can write
z7
On solving equations
+ Zg
= zil
... (ii)
+ z12 = 20 + 63 = 83
(i) and (ii), we get
Zg = 47.16 ~ 48 and
z7
= 83 -48
=
35
Thi dON
.
ider the ray that gives the speed increase from 560 r.p.m. to 1000
ow conSl
r.p.m. This can be achieved by using the gears 9 and 10.
lr
I. I
pazr:
N
_!Q
z9
~
Also,
z9
On solving equations
-
+ zlO
N9
Z 11
-
First stage:
+Z
12
=
=
h
t e m
can be achieved by gears 5 and 6.
o
Zs
Zs
z6
-
z9
= 1.786
... (iii)
zlO
560
.
axunum
•
Ptrst patr° : Consider
..
or
spee
'... (iv)
20 + 63 == 83
(iii) and (iv), we get
29.79 ~ 30 and
zlO
Assume
1000
= -
=
83 _ 30
=
53
z9
. f
1320 r.p.m. to 560 r.p.m. This
d reductIOn rom
20 (driver)
N6
- Ns
J.6Q_ == 0.4242
== 1320
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Design a/Transmission Syste1n3
9.14
Z5
or
0.4242
z6 =
_
20
- 0.4242
=
47.14 ~ 48.
be
id the speed reduction from 1320 r.p.m. to 750 r.p.m. This can
onsi er
achieved by gears 1 and 2.
750
N2
ZI
... (v)
=or Zt = 0.57 z2
=
..
1320
Nt
z2
.
C
Second pair:
and
zl
+ z2 = z5 + z6 = 20 + 48 = 68
On solving equations (v) and (vi), we get
Z2 = 43.3 ~ 44 and z,
=
... (vi)
68 - 44 = 24
Third pair: Finally consider the speed reduction from 1320 r.p.m. to 1000 r.p.m. This can
be achieved by gears 3 and 4.
Z3
..
=
z4
and
z3
=
+ z4
N4
N3
=
z5
+ z6
1000
1320
=
or z3
20 + 48
=
=
'" (vii)
0.76 z4
... (viii)
68
On solving equations (vii) and (viii), we get
z4
I Example
9.7
=
38.64 ~ 39 and z3
=
I For the data of the above problem,
68 - 39
=
29
calculate the percentage deviation
of the obtainable speeds from the calculated ones.
Given Data: Refer Example 9.6.
Tofind:
Percentage deviation of the speeds.
@ SolllJion :
It is understood from the kinematic arrangement (Fig.9.7) that the
combinations of gears I and 2 (in the first stage), gears 7 and 8, gears 9 and 10, and gears 11
and 12 (in the second stage) are to give three output speeds. Then, the combinations of gears
3 and 4 (in the first stage), gears 7 and 8, gears 9 and 10, and gears 11 and 12 (in the second
stage) provide the next three output speeds. Similarly when gears 5 and 6 are engaged, we
obtain three more speeds. Thus we can achieve totally 9 output speeds.
Calculatlon of output speeds: Let N, and No:: Input and output speeds of the gears.
From the ray diagram (Fig.9.6),
No, = Nxt
Z,
z2
z,
z7
x-
Zg
z9
N02 = Nx, %2 xTO
input speed N,
=
1320 r.p.m.
=
24
35
1320 x 44 x 48
=
525 r.p.m.
=
S3
1320 x 24
x
30
44
=
1272 r.p.m.
z,
zll
x= 1320
N03 = NxI z2
zl2
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x
24
20
44 x
63
= 228.57 r.p.m.
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9.15
==
Nos
Nos
N09
==
z3
zJJ
N,x - xz4
z,2
:::
N06
N07
Z3
z9
N,x - xZ4 zlO
=
==
1320 x ~
35
39 x 48
]320 x
¥
]320
~
9
x~
30
Zs z7
NIx - xz6
Zs
-
=
Zs
z9
NIx- xz6
zlO
- ]320
-
Zs zJJ
N1x- xz6
zl2
-
1320
== 3]].6 r.p.m.
~
35
48
20
53
x 48 x
x 48 x 30
1320 x 20 20
48 x 63
Calculated
speed
== 401 r.p.m.
== 971.66 r.p.m.
==
,
speed
1734 r.p.m.
20
x 39 x -
Calculationof % deviation:
rs.NO. I
==
63
=
Obtainable
== 715.7 r.p.m.
174.6 r.p.m.
exo deviation = Nb-N
0 I
N
(Nobt' r.p.m.)
(Ncal' r.p.m.)
1.
174.6
180
2.
228.57
236
3.'S
3.
311.6
315
LOS
4,
401
425
5.65
5.
525
560
6.25
----.:.
6,
715.7
750
4.57
_7.
971.66
1000
2.83
t--!.
1272
1320
3.64
9,
1
-
I
1734
/~Ol1X1UnPle
9.81 A gear
cal
x 100
1
3
I
1800
cal
3.66
box is to he designed to provide 12 output speeds ranging/rom
Ifti.to 2000 r.p.m. The input speed of motor
is J600 r.p.m: Choosing a standard speed
, COllstruCt the
r£
.
speed
diagram and the kinematic arrangement.
vtve« Data ; n == 12' N . = 160 rpm . N
= 2000 r.p.m, ; N;1rplll
/".
r.
@ofilld:
'mrn'
.
.,
=
1600 r.p.m.
max
ConstructlOn
. of the speed diagram and the kmematIc
.'
arrangement .
SOlutio" •
Select
•
10" 01'
•
..,spindle
speeds:
\Vek
now that ,
Nmax
N
_
~n-I
min
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Design a/Transmission
9.16
s
~
2000
160
or
= cpl2 - 1 or cp = 1.258
We can write,
1.12 x 1.12 = 1.254
.
'" (skip one speed)
So cp = 1.12 satisfies the requirement. Therefore the spindle speeds from R 20 series
skipping one speed, are given by
,
160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600 and 2000 r.p.m.
Structural/ormula:
For 12 speeds, the preferred structural formula
= 3 (1)
2 (3)
2 (6)
1st stage 2nd stage 3rd stage
Speed diagram (or Ray diagram) :
Procedure:
../ Since there are 4 shafts, draw 4 vertical equidistant lines to represent shafts .
../
Since there are 12 spindle speeds, draw 12 horizontal equidistant lines .
../
From the structural formula, it is clear that there are three stages. In the third stage,
i.e., in 2 (6), 2 represents the number of speeds available in that stage and (6)
represents the steps or intervals between these two speeds .
../
Locate the first point A on the lowest speed i.e., at 160 r.p.m., on the last shaft. After
6 steps above, locate the second point B at 630 r.p.m. These are the two output
speeds .
../
Locate the input speed at any point on the preceding shaft (i.e., shaft 2), meeting the
ratio requirements. We find, the input speed 400 r.p m. at point C satisfies the ratio
requirements .
./
In the second stage, there are two speeds. Lowest speed is at C, which is already
located. Now locate point D on the 3rd shaft, above point C, in a three step interval.
For these two output speeds in the second stage, the input should be from shaft 2. We
find, the input speed 630 r.p.m. at point E on shaft 2 satisfies the ratio requirements .
../
In the first stage, there are three speeds. Lowest speed is at E, which is already
located. Now locate points F and G on the shaft 2, above point E, in a single step
interval.
./
Input speed can be located anywhere on shaft 1 meeting the ratio requirements. But in
this problem, given that, input speed is at 1600 r.p.m.
,./' In stage 2, we find input speed at E gives two output speeds at C and D. Similarly,
input speeds at F and G, should give' two output speeds. This can be achieved by
drawing lines parallel to EC and ED, from points F and G, as shown in Fig.9.S.
,./' Now for stage 3, to get the output speeds to all the input speeds in shaft 3, draw lines
parallel to CA and CB. Thus we have located all the input and the output speeds. The
completed ray diagram is shown in Fig.9.8.
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2.-
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-
Njnpu1
160
1
= 400 = 0.4 > -
N;nput
-
= 400 = 1.57 < 2.
-
= 630
Nmin
Stagt 3 :
Nmax
Nm;n
Stogt 2:
N;npu'
4
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9.17
and
630
400
3 (1)
H
1
= 0.63> 4"
Nmax
N;npu,
-
and
800
630
= 1.27 < 2.
Nmin
Siage 1:
Nin;UI
=
630
1600
400
1
= 0.39 > 4"
Nmax
N;npu'
. 315
and
1000
= 1600
250
.
N
;::
nI
;::
s:
.c
III
.
C')
,
III
Stage 1
IJ)
Stage 2
c
III
.c
IJ)
C
III
StagB 3
IJ)
160
= 0.625 < 2.
:. Ratio requirements
Kinematic arrangement:
shownin Fig.9.9.
200
.c
are satisfied.
Fig. 9.S. Ray diagram/or 12speed gear box
The kinematic arrangement for 12 speed gear box is drawn, as
-------_ .. _-_---_ .. _ ..
Shaft-1
p, = 3
Shaft-2
Shaft - 3
Shaft - 4
-o
-------- ------
£.
xa", Ie 9.9
g'"g fj
'Gil'
j ~4
l
Fig. 9.9. Kinematic arrangementfor
12 speed gear box
tis with the output speeds
A machine tool gear box
d diaoratnS for 2 x 2 x 3, 3 x
. t have 12 spee ,
IS 0
the spee"
r-p.m: to 2800 r.p.nt. Draw
h' h is better anti why.
Qlld"
h
mes
w IC
)(3 schemes. Among these sc e
rOIn 63
2x2
,
?
.
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ueslgn
~~
~.18
oj Transmission SYSI
--------------------------------------~~~------~~
=
12', N
= 63
'
e~v
~
r.p.m.;
Nmax = 2800 r.p.m.
ml n
Given Data: n
Tofind: Construction of speed diagrams for various schemes.
© Solution:
Selection of spindle speeds:
Nmax
We know that,
= ~n-I
2800
63 = ~ 12 -
or
We can try,
or ~
1
=
1.4 12
... (skip 5 speeds).
1.06 x (1.06 x 1.06 x 1.06 x 1.06 x 1.06) = 1.418
So ~ = 1.06 satisfies the requirement.
skipping 5 speeds, are given by
63,90,125,180,250,355,500,710,1000,
Therefore
the spindle speeds from R 40 series,
1400,2000 and 2800 r.p.m.
Speed diagrams for various schemes:
(i) 2 x 2 x 3 scheme: Structural formula = 2 (1) 2 (2) 3 (4). Refer Fig.9.1 O.
Nm;n
N
Stage 3 :
=
63
1
250 = 0.252> 4 ~ and
=
1000
= 4> 2
250
input
Nmax
N;npUI
In this stage, it is not possible to satisfy the ratio requirement
case.
2(1)
Stage 2:
Nm;n
Nrnpul
Nmax
N;npul
=
250
500
=
1
0.5 >4
=
500
500
=
. So treat as an exceptional
2(2)
3(4)
-------.--------~
:
2800
:
I--------+--------~----~~
2000
:
j--------t--------~--~~:~1400
and
710
1 < 2.
500
Ratio requirements are satisfied.
Stage 1 :
Nmrn
N;npUf
=
500
1000
355
2.50
=
Nmax
NrnpUf
1
O.S > "4
and
180
710
= -1000
125
..,
,
=
0.71<2.
Ratio requirements are satisfied.
'i
fA
r-----
c::-t----
•
~
s:
90
c::
::
"
--------~----~~~~~~~
Stage ,
~
Stage 2
rJl
Stage J
s:
(fJ
A
Fig. 9.10. Ray diagram for 2 x 2 x 3 scheme
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'I·
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9.19
_.fBO~
Structural formula = 3 (I) 2 (3) 2 (6). Reter Fig.9.11.
~A'''l
2 x2...scheme:
~
(ii) 3 x 1'1 .
• _.!!!!!!-
3(1)
63
== 0.252 >
1'1",ax
2(6)
== 250
Stllg' 3 . 1'1input
.-:--1'1input
2(3)
1
4 ; and
500
== -250
== 2~2
N",in
Stage 2:
I).
Ninput
250
-
== 500
1
- . and
== O .5 > 4'
e\
N",ax
Ninput
710
-
== 500
1.42 < 2
500
1
'
== -1400 = 0.36> -4 ; and ~
«rii
==
N",in
Stagel:
N
input
Nmax
-
....
N
1---Stage 1
~
('I')
'+--«rii Stage
~
~
......
-+----'
2
'i
~
Stage 3
90
!
(/)
~------_63
A
1000 == 0.71 < 2
1400
==
Ninput
Fig. 9.11. Ray diagram/or 3 x 2 x 2 scheme
3 (1)
4 (3)
.. Ratio requirements are satisfactory.
(iii) 3 x4 Scheme:
Refer Fig.9.12.
Structuralformula == 3 (1) 4 (3)
Nmin
Stage 2: --
63
==
1
4.
710 <
and
Ninput
Nmax
Ninput
1400
==
== 1.97 < 2.
710
In this stage, it is not possible to satisfy
the ratio requirements.
exceptionalcase.
Slage 1:
Nmin
==
Ninput
So treat
71 0
2000
= 0.35>
Nmax
N
tnpu:
as an
I
4
and
1400
= 2000
=
L_ __ ----~------~63
A
Fig. 9.12. Ray diagramfor 3 x 4 scheme
0.7 < 2.
., Ratio requirements are satisfied.
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~
9.20
Design o/Trans",.
.
ISS 1011
IS
:)ISte"",
(iv) 4 xJ Scheme:
Structural formula
=
4 (I) 3 (4).
Rt:fer Fig.9.13.
4 (1)
Nmin
Stage 2:
=
NinpU1
63
500
1
- 0.125 < 4'
Nmax
3 (4)
H
and
1000
- 500
Ninpul
- 2~2
In this stage, it is not possible to satisfy
the ratio requirements.
So treat as an
exceptional case.
Nmin
Stage 1:
-
Ninpul
500
--
1400
1
- 0.36>4'
Nmax
=
Ninpul
:. Ratio requirements
1400
1400
=
and
1<2.
are satisfied.
Fig. 9./3. Ray diagram/or 4 x 3 scheme
Conclusion:
Out of the four schemes, 3 x 2 x 2 scheme is better than other schemes.
Because, only 3 x 2 x 2 scheme satisfies the ratio requirements. i.e.,
Nmax
Ninput
in all stages.
Nmin
< 2;
N
input
It means the speed reduction
compact units.
, Example 9.10
I Sketcb
1
Nmax
> 4- ; and -N
min
s
8,
in all stages is minimum which results in
the speed diagram and the kinematic layout/or an 18 speed
gear box for the following data:
Motor speed = 1440 r.p.m; ,. Minimum output speed = 16 r.p.m: ,. Maximum output
speed = 800 r.p.m: ,. Arrangement = 2 x 3 x 3.
List the speeds of all the shafts when the output speed is 16r.p.m:
Given Data:
n = ] 8 ; Ninput = ] 440 r.p.m. ;
Nmin = 16 r.p.m. ;
Nmax =
800 r.p.m. ;
2 x 3 x 3.
To find: Construction of the speed diagram and the kinematic layout.
@Solution:
Selection of spindle speeds :
We know that,
Nmax
_ cpn -
I
Nmin
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800
16 -
9.21
~18-1
Or
~::::
1.258
or
1.12 x 1.12 - 1.254
cafl write,
we
1 12 satisfies the requirement. Therefore
.
the e-«
'" (skip one speed)
e sPindle speeds
. from R 20 series,
.1\::;
•
b
SO
'l1gIf'one speed, are given y
~1Ppl
25 31.5,40, 50, 63, 80, 100, 125, 160,200,250, 315 400 500
'.
.'
,
, 630 and 800 r.p.m.
16, 20,urUl-'fiormula:
Given that, 2 x 3 x 3 (l.e',Pr
P2 .P )
Ilct
3
St1 (OIo\\' that,
We
where
Structural formula
X::: 1;
I
X2
= PI = 2;
=
PI (Xl) P2 (X ) P3 (X )
2
X3 = PI . P2
Structural formula
==
=
3
2 x 3 ::::6
2 (1) 3 (2) 3 (6)
..ttd diagram: The speed diagram is drawn, as shown in Fig.9.14, using the procedure
51 _All' Example 9.8.
~iSCUSscu JO
2
(1)
3 (2)
3 (6)
~----------Ir-----------r--------~800
,
,,'
,
,,'
,
630
500
R:·l.;·-'!,,;·:·_--·- .... 80
Dr\..·-·"
.......
". B 63
I" ..
'\.
,,"..
'.
". ".
..
..
'\. ••••• \
.,....
I
l
1..-----
-------~-
~-I-
l....-------.2!
¢:
en
50
•••• ".
40
"
•••••• ~
31.5
'\.
25
••~
~20
_--1------
~
". ".
"\
16
~
-t-
A
18 speed gear boX
914. Ray diagram/or
rig. .
6
N .
L
_ m~
- 18
c·
Stage 3 :
i
_ . an
:::
0.2 < 4 '
d
Ninput
N
-
ma.!.
N input
~
:::
:::3.125
80
:>
2
b treated
So it can e
as an
exceptional
t satisfied.
., I .
.
ts are no
n this stage, ratio reqUlremen
1,
Case,
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_9_.2_2
~D~~~~~n~o~if~mu~~
I,sS;o" 8
Nmin
Siage 2:
Nit;IPUI
Nmax
=
=
250
125
=
=
-
125
500
= 0.25 s 4" ; and
N;nput
Nm;n
Siage 1:
NinpU1
Nmax
:)I"tellt,
80 .
1
125 = 0.64 > 4" ; and
2
1
160
= 500 = 0.32 <2
N;npul
., Ratio requirements are satisfied in stages 2 and 1.
·
Kinematic layout: The kinematic layout for the 18 speed gear box is d
9 15
.
F Ig..
rawn, as shoUt.. .
"II
....f.--I4..-.
---
-
--
--
--
--
--
--
--
1
-
III
Shaft - 1
Fig. 9.15. Kinematic arrangement for 18 speed gear box
Speeds of all the shafts when'the output speed is 16 r.p.m.:
From the speed diagram (Fig.9.14), speeds of all the shafts when the output speed is 16
r.p.m., are given by
./
Speed of shaft 3
= 80 r.p.m.,
./
Speed of shaft 2
=
./
Speed of shaft 1
= 500 r.p.m.
9.11. OVERLAPPING
125 r.p.m., and
SPEED GEAR BOX
th ear bOX
If the engagement of two different sets of gears provide the same speed, then e g best
is known as overlapping speed gear box. The concept of overlapping of speeds can be
explained by Example 9.11.
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9.23
11 A 14 speed gear box is reqUired to turn' "
./lrrfl'_lt__9,
Draw the speed diagram and the kine IS. Olltputspeeds in t"e rang.
$!;
"tOO
r.p.~
N _ == 250"'0aIle
pl#'> '/I: n - 14'.' N min. = 125 r.p.m.,
•
r,P.m.arrange",ent
jl,,:.I111'_
trUctlon of the speed diagram and the kinernat'
~
11'''''",
Cons
Ii" ..,,:
f,r
IC
8lTangement.
~SD/~of spindle speeds :
"J,M'tlO1l
N max
== ~n-l
)f'~"
kOowthat
N min
We
2500
125
== ~14-1
Or ~ ==
1.259
ite
1.12 x 1.12 = 1.254
'" (skip one speed)
1
'"
_
1.
We caJl_ W0 satisfies the requirement.
Therefore the spindle speeds from R 20 series,
So If'
ed are given by
It
i
one~:
;50, 315, 400, 500, 630, 800, 1000, 1250, 1600, 2000 and 2500 r.p.m.
125,160, . 'mula: For 14 speeds, the preferred structural fonnula
StrJl(/urtdfo~
= 3 (I) 3 (3) 2 (5)
jiflIin8
Speed diagram: The speed diagram is drawn, as shown in Fig.9.16.
min
N
S1Dgt3 : N;npul
-
Nmax
N;npul
125
315
3 (1)
'-- _ _':"_~-'
1
04>
-4' . and
.
I
400
- 315
3 (3)
2 (5)
I __
-_-~--:::--:-:-:-1 2500
L--------+-----~~--~-----~~~2000
. ~'" , _.- ..... .....
l_-------+--~-~··~·~--~'~~·~·~··11600
- 1.27 < 2.
~t2:
Nmin
N;npul
N
~ mat
N;"PUI
'e]..
-
315
800
-
0.39>
-
1250
800
-
l.56 < 2
==
800
1600
N.
~
N;lIput
N~
~
'npUt
1
4' ,. and
1
05>
-4'- . and
.
-
1250
==
1600
=
0.78 < 2.
for 14 speeds
ddiagram
.
gspee
• fa}6 Over/appm
Fig.
7.
•
. ents ar e satisfied.
.
requlrem
.. RatiO
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~.
~
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Design o/Transmi .
'"
----------------------------~~~-==~~~
9~2~4~
~
In Fig.9.16, the ticked ( ......
) four speeds are overlapping speeds.
th ..... 9 16 we find that 400 r.p.m. is achieved through two different rout
From Ie rig. . th'at two different sets of gears are used to achieve the same Srll>"~s,1FCa
and IHJB. t means
r""'U.ln til
.
th
d
overlaps
at
speeds
500
630
and
800
r.p.m.
also.
e
same manner e spee
,
Kinematic arrangement: The kinematic arrangement for J 4 speed gear box is draWn
,as
shown in Fig.9.17.
Shaft - 1
-----------------------_.
r-
.-
l-
Shaft - 2
I- -1-1--
~
t- rf-
Shaft - 3
--------~
'-'-
Spindle (Shaft - 4)
Fig. 9.17. Kinematic layout/or 14 speed gear box
DESIGN OF GEAR BOX
I
9.12. DESIGN PROCEDURE FOR GEAR BOX
1. Selection of spindle speeds:
,/
Determine the progression ratio (<I» using the relation
<l>n-I
,/
Nmax
= --
Nmin
For the calculated <1>, select the standard spindle speeds using the seriesof
preferred numbers, from Table 9.2.
2. Construct the ray diagram, as discussed earlier.
3. Construct the kinematic arrangement for the given gear box, as discussed earlier.
4. Calculation of number of teeth on all gears: Calculate the number of teeth of all the
gears engaged in all stages of the gear box.
5. Select the suitable material, consulting Tables 9.4 and 9.5.
6. Calculation of module:
,/
Calculate the torque for the gear which has the lowest speed using the relation,
T = P x 60
... (9.4)
21t N
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001
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rt
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I ulate the tangential force on the.
.
9.25
I Ca c
gear In term
s of module using th J'
(referfig.9.18), T
2T
e re atJon
F, = -;: = z x m
'" (9.5)
J
[.: T=F,xr and r=\'!!.
I Now calculate the module using the relation
m where
'4'm -
-v F, / '4'm x M
Gear
... (9.6)
Fig. 9.18.
Ratio between the face width and modue=1 h = 10
m
M -
'
Material constant, from Table 9.4.
Table 9.4. Materia~constant (M)
Material
Material constant (M)
C45
30
15 Ni 2 Cr I Mo 15
80
40 Ni 2 Cr I Mo 28
100
7. Calculationof centre distance in all stages:
!!age by usingthe relation
a --
zx and zy
where
=
(Zx2+Zy)
Calculate the centre distance in each
m
... (9.7)
Number of teeth on the gear pair in engagement in each stage.
l Calculationof face width:
b
=
lOx m
Calculationof distance between the bearings ie., length of shafts :
Cal
.
h f [lowing assumptions, (refer
f' culatethe distance between the bearings by using teo
9.
Ig.9.19) :
-I
.
Give 10 mm clearance between the gear an
-I
d the bearing on both sides.
of gears as 20 mm.
.
~ th e pairs gear group
u as 4h and tor re
.
OU S
Take the distance between the adjacent gr p
t/
Take the total length for two pairs gear gro p
.
as Ib, as shown in Fig.9.19.
t/
.
Assume the width of the bearings as 25 mm.
'. Djsta
nee between the bearings is given
b
Y
L ::::25 + 10 + 4b (or 7 h) + 20
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+ 7h (or 4h)
10 + 25
." (9.8)
+
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---r
•
20 mm
.1.
Shalt-a
7b
Fig. 9.19.
10. Design of shafts :
(i) Design of_sp~ndle i.e., outpu~ shaft:
Design the ou~put shaft for maximum bend"
moment by considering the shaft as Simply supported on bearmgs (refer Fig.9.20).
tng
L
r
Shaft - 3
Gear
Fig. 9.20.
./
Calculate the maximum bending moment due to normal load (Fn) using
the
relation
... (9.9)
M=
"
where
./
Fn
= Normal load on gear = --
F,
cos a.
where
T
=
-v M.2 + T2
."
Torque on the spindle
=
2
(9.10)
P x 60
N
1t
./
I
Calculate the equivalent torque using the relation
Teq =
low
Calculate the diameter of the spindle using the relation
I
=
d,
where
[t
[16
x Teq
1t [
't]
13
J
[ ..
. ['t]
_ 16 Teq
-
1t
d/
]
.. ' (9.11)
J = Permissible shear stress, from Table 9.5.
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I
I,
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9.27
Table 9.5. Permissible sl, ear stress I TI,N
'Imml
Shaft material
S.r-Io.
~
C14 (as supplied)
\.
2.
3.
25
C45 (case hardened)
30
Low carbon alloy steel (case hardened)
40
40 Ni 2 Cr 1 Mo 28 (hardened and tempered)
55
4.
L--
(ii)
l 't 1. N/mml
Determine
Desig~ of otheTr shaft OS 2: d s3
. th relation
[ 1:
-.
uSing e
~-p-'e-9.-1-~-"'l1Design
the diameter of the input and intermediate shafts
1
... (9.12)
a 12 speed gear box for an all geared headstock of a lathe.
. ..... and minimum speeds are 600 r.p.m: and 25 r.p.m: respectively. Tile drive isfrom
f,{axJnI..,'·
",electric nWtor giving 2.25 kW at ]440 r.p.m:
Given Data: n ~ 12; Nmox
Ninpul
~
600 r.p.m. ; Nmin = 25 r.p.m.; p = 2.25 kWi
= 1440 r.p.m.
Tofind: Design the 12 speed gear box.
@Solution :
1. Selection of spindle speeds :
-
Nmax
Weknow that
<\>n-
1
<\>12-1
or
or
We can write,
- Nmin
=
c\> ~
600
2s
1.335
1.06 x (1.06 x·' 1
..' (skip 4 speeds).
06)
06 x 1 06)( 1.
:=:
1.338
fr
.es
R 40 sert ,
indle speeds om
.
ent. Therefore the sp
So ~ = 1.06 satisfies the requlrem
~. Plngfour speeds, are given as
450 and 600 r.p- ro .
250 335,
9 21
25, 33.5 45 60 80 106 140, 190,'
h .~t11 in Fig. . .
, ,
,
'
'
d as S OvY
2.
. constructe ,
lay diagram: The ray diagram IS
l
'
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Design. (JITrammbsiol1
9.28
~
3 (1) 2 (3) 2 (6)
Strllctural/orlHula:
Stale J:
Nm1n
N;1ffJII1
=
2~
80
=
1
0.31 > 4 and
=
140
80
2 (3)
3 (1)
2 (8)
eoo
4SO
H
.......'.'
•
335
'
250
Nmax
N;1ffJII1
Stage 2:
Nmin
N;npul
140
=
1.75 < 2
=
80
140
=
Nmax
N;npul
190
=
0.57
106
80
60
1
>4 and
.....
190
140
1i
Stage 1
Nm;n
N1nput
Nmax
N;nput
=
=
1i
.c
=
IV
Stage 2
CIJ
.c
CIJ
Stage 3
25
A
= 1.36 < 2
Stage 1:
,
I
.c
CIJ
45
M
('oj
I
Fig. 9.21. Ray diagram/or
12 speed gear box
140
1
450 = 0,311 > 4' and
250 = 0.56 < 2
450
.. Ratio requirements are satisfied.
3. Kinematic arrangement: The kinematic arrangement for the given 12 speed gear box
is constructed, as shown in Fig.9.22.
10mm
7b
20mm
r---------------------- ,,,_,,
Shaft - ..
Z10 ::
35
Fig. 9.22. Kinematic arrangement for 12 speed gear box
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~
-r-:':
y lation of number
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9.29
of teeth on all gear. . Th
calCu
.
h
l' •
e numb
f
below, followmg t e procedure used in E
er 0 teeth on all gears are
IlJ,ted as
xample 9.6.
J.
cJCJ
Slalt) :
if' Consider the ray that gives maxim
";,sf po '.
'Urn
reduction I
Ii
8
f'
The correspondmg gears are 13 and 14 on shaft 4
.e., rom 0 r.p.m, to
15 r p.rn.
7 Th
.
u:We jaJO w that ' zmin ~ 1.
erefore assume Z ::: 20 (d .
13
rIver).
zJ3
_
NI4
-
zI4
20
25
z14:::
80 ;
or -
NI3
:.
z14:::
64
Second pair: Consider the other ray that gives speed increase from 80
1
nding gears are 11 and 12.
r.p.m. to 40 r.p.m,
The correspo
140
80
NI2
zll
-
=-=-
zl2
Nil
orz
-175.
II -
zl2
... (i)
We alsoknow that the sum of number of teeth of mating gears should be equal.
"
=
+ zl2
zJJ
z13
+ zl4
:::
= 84
20 + 64
... (ii)
II
On solving equations (i) and (ii), we get
= 30.5 ~ 31 and zlJ ::: 84 - 31 = 53
zl2
Stage 2:
Fisf pair : Consider the ray that gives maximum reduction from 140 r.p.m. to 80 r.p.m.
11Jecorresponding
gears are 9 and 1'0. Assume z9 = 20 (driver).
NIO
z9
20
= N9
Z10
or
ZJO
=
80
_
140;
:.
zlO -
35
Second pair : Consider the other ray that gives speed increase from 140 r.p.m, to
190r·p.m. The corresponding gears are 7 and 8.
'.
and
Ns
190
N7 = 140 or
z7
=
Zs
Z
+ Zs
7
- ""9 + Z 10
-
~
=
z7
20 + 35
=.
=
1357
... (iii)
Zs
... (iv)
55
On SOlvingequations (iii) and (iv), we get
Zs
=
23.3 ~ 24 .an
Sta,e 1 .
Fit. •
d
- 55 -24
Z7 -
== 31
. fr 450 r.p.m. to J 40 r.p.m.
reductJon om
'IleCorre
: Consider the ray that gives maxImum (d'
r)
sponding gears are 5 and 6. Assume Zs == 20 nve·
l't
'St Pai,
'.
Zs
-
z6
N6
= -
Ns
20
or -
140
-'
== 450'
.
..
z6
== ~4.28
= 65.
z6
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9.30
Design olTransm/ssl
~
S~colldpair: Consider the ray that gives speed reduction from 450 r.p.m. to 190
The corresponding gears are 3 and 4.
r,p·llI.
z3
%4
..
190
450 or
N4
:a::
N3 =
13
= 0.422.%4
'" (\I)
and
'" (vi)
On solving the equations (v) and (vi), we get
%4
="
59.77
!ItS
60 and
.%3 ==
85 - 60 == 25
Thirdpair: Consider the ray that gives speed reduction from 450 r.p.m. to 250 r.p.m. The
corresponding gears are 1 and 2.
%1
N2
250
z2 = Nl = 450
:.
%1 == 0.555 Z2
or
'" (vii)
and
..• (viii)
On solving the equations (vii) and (viii), we get
z2 = \ 54.66 ~ 55 and zl == 85 - 55 == 30
5. Material selection:
C 45
6. Calculation of module:
./
To find torque: In this case, the lowest speed 25 r.p.m. is obtained by meshing
gears 13 and 14. Therefore torque at 25 r.p.m. is given by
T 14
./
T
'14
= r =
2x
T14
Z14
xm
2 x 859.44 x 103
64xm
We know that,
where
or
2.25 x 103 x 60
2 1t x 25
= 859.44 N-m
Tofind tangentialforce on gear 14 :
F
./
P x 60
= 2 1t N =
=
26857.5
m
module, m = ~ F, I \JIm
\JIm
... [where module (m) is in mm]
X
M
= hIm = 10, and
M
= Material constant = 30, for C45, from Table 9.4.
m
=
module, m
=
(26857.5 1m)
10 x 30
= ~
89.525/ m
or m2 == 89.525 I!"
4.47 mm
From Table 5.8, the nearest higher standard module is 5 mm.
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9.3)
.
~
1.
cme
1II(11lo0/
n centre dIStance:
I Centre distance in stage 1,
at
= (~
;Z2)
In
= (30+55)
2
x 5 = 212.5 mm
a2 = (Z7; Zs)
.; Centre distance in stage 2,
= (31 +
2
.; Centre distance in stage 3,
aJ
=
=
(Zll;
In
24) x
Z12)
5 = 137.5 mm
In
(53 + 31) 5
2
2
= 10mm
I, Calculationof face width: b = \V x m = lOx 5 = 50 mm
9, Calculationof length of shaft (i.e., distance between the bearings) :
Lengthof shaft,
L = 25 + 10 + 7 b + 20 + 4 b + 20 + 4 b + 10 + 25
= I 10 + (15 x 50) = 860 mm
10, Designof shafts :
~)Designo/spindle i.e., output shaft:
,I Tofind maximum bending moment (M) :
Fn x L
where
M
=
Fn
=
Normal load on gear
=
(26857.5/5)
cos 20°
F,
=
cos a.
=
(26857.5 / m)
cos a.
= 5716.23 N
M =
5716.23 x 860
4
= 12.29 x 105 N-mm
Tor",d the equivalent torque (Tu) :
T
= -VI M2 + r214 = ~
eq
=
0/
IS b
4
.. Maximum bending moment,
0/
= 110 +
D~~r
(12.29 x 105)2 + (859.44 x 103)2
1.5 x 106 N-mm
of the spindle is given by
I
= [16
d
x req
7t(t]
s
J 3 where
t = 30
N/mm2, from Table 9.5.
I
0/
R
_ [16
x
-
7t x 30
1.5 x 1()6J3
oUnded off value of the diameter, using
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= 63.38 mm
R 40 series is 67 mm.
,
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iO'
~i
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Design of T,ansrn iss
9.32
(;i) Design of other shafts:
(a) Diameter ofshaft 1:
Input speed = 450 r.p.m.
p x 60
2.25 x 103 x 60
=
Torque =
27t x 450
=
47.746 N-m
We know that,
d;l
47.746 x 103
=
0.2·
ds1
=
19.96 mm :::::20 mm (R 40 series)
or
30
(b) Diameter of shaft 2 :
Minimum speed = 140 r.p.m.
Torque
p x 60
2.25 x 103 x 60
= 153.47 N-m
= 27tN
=
27t x 140
T = 0.2
We know that,
d;
['t]
d;2
153.47 x 103
=
0.2 x
ds2
=
29.46 mm :::::30 mm (R 40 series)
or
x 30
(c) Diameter of shaft 3 :
Minimum speed = 80 r.p.m.
Torque
=
P x 60
2.25 x 103 x 60
=
= 268.57 N-m
27tN
27t x 80
268.57 x 103 = 0.2 x
..
ds3
or
I Example
9.13
I Design
= 35.5
d;3
x 30
mm
a gear drive to give 18 speeds for a spindle of a milling
machine. The drive is from an electric motor of 3. 75 k W at 1440 r.p.m: Maximum and
minimum speeds of the spindle are to be around 650 and 35 r.p.m. respectively.
Given Data:
n = 18; P
Nmin =
Tofind:
=
3.75 kW; Nmotor
=
1440 r.p.m.;
Nmax
= 650 r.p.m.;
35 r.p.m.
Design the gear box.
@Solution:
1. Selection of spindle speeds :
We know that,
or
4>n-l
4>18-1
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=
=
Nmax
Nmin
650
3S
or 4> = 1.1875
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r
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~-:'-O-6X-(l-'-06--XI.-0-6)-910"
1.191
-
9.33
.
AI:::; 1.06satIsfiesthe requirem
SO
•
ent, Therefore the
. ., 2 speeds, are given
by
~PPI~g
.
35. , 42.5, 50, 60, 71, 85, 100, 118, 140 1
d 6705 r.p.m.
.0
, Bay diagram.
i'
•
Th
e ray
diagram is
35.5
Nm~
N;npul
Nmin
Stage 2:
Ninput
1
0,253 > 4 and
280
- 140
I
2 (1)
3 (2)
3~
•
~
140
236
""-~H
_ ""'
....
.' "
"
'.: __.
...
...
f
E
." ,
,.."
,
,....
''(.,'/
'f' ,.....
','
280
"
- 236
\
236
- 475
-
280
....
236
....
,: ........
,"
,
..
140
'II
" ',',",:,
100
" "
' \ " ''-' '' 85
, "
\ ,' ,'
,,"
N;npul
'
...
. , ' , ..
\
Nm;n
.:
and
- 1.186
Stage 1:
~475
,__
, ,. ....
··e
_"_
'• .'(.-7
,."
..
D ~':
Nmax
NinpUI
'
400
_------:"...' ........335
_.........
F '..f-:,~...
.._
_
..,,:"
.' ,
.'
G ~.-
1
......,560
670
,,1'
I"
2
0.59>4
, 475, 560
,:====---t---_::+=~="'
. . l'
-
-
'
, as shown in Fig.9.23.
-
=
, 335 400
constructed
~
= -140
-
s)
s from R 40 series,
'
SJfuctllra}lormula: 2 (I) 3 (2) 3 (6)
---Stille 3'• N
inpJlI
." (skip 2 speed
' 70, 200, 236 280
.
.
Nm;n
.
SPIndle speed
\
~'
,' ,
71
, '. ' 60
'. '
.... 50
\
1
0.497> 4 and
\\
42.5
~ __S_ta_ge_1~I--S_m~ge_2--~~S=ta~ge~3~\~ 35.5
A
-
.
280
475
_ 0.59 < 2
Fig. 9.23. Ray diagram/or 18 speed gear box
'.
J.
Ratio re qUlrements
.
are satisfied.
.
,~draWn
. KlIIelllat·tc arrangement: The kinematic arrangement for the given 18 speed gear box
, as shown in Fig.9.24.
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9.34
1
T
58
79
Fig. 9.24. Kinematic arrangement/or 18 speed gear box
4. Calculation of number of teeth on all gears:
Stage 3:
First pair: Consider the ray that gives maximum reduction from 140 r.p.m. to 35.5r.p.m.
The corresponding gears are 15 and 16.
Assume
ZIS = 20 (driver)
20
35.5
--- 140
Second pair: Consider the other ray that gives speed reduction from 140 r.p.m. to
100 r.p.m. The corresponding gears are 11 and 12.
and
ZlI
zll
zI2
=
NI2
Nll
+ zI2
-
zI5 +zI6
100
= 140 or
=
= 0.714 z12
zll
20 + 79
=
99
... (i)
... (ii)
On solving the equations (i) and (ii), we get
zI2 =
57.75 ~ 58 and
Ztt
= 99
- 58 -= 41
Third pair: Consider the other ray that gives speed increase from 140 r.p.m. to 280r.p.ro.
The corresponding gears are 13 and 14.
NI4
280
- NI3 = 140 or
and
zl3
+ zI4
-
zI5
+ zI6
=
z13
20 + 79
=2
.. , (iii)
xZl4
= 99
On solving the equations (iii) and (iv), we get
zI4 =
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33 and
zI3 =
99 - 33
= 66
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•. , (iv)
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Bo1
i
siage 1'r: ' Consider
. the ray that giv
tpal '
es rna .
firS p00ding gears are 9 and 10.
Xlrnum reducti
"'~
. on from 236
II'.
e
Z
"'0
r·P·m. to 140
A~~'"
9
~
(driver)
r·p.rn.
z9
zION
-
N
20
-lQ
or -
_ 140
--2'
.
, C id tl
h
10
36'"
Z
- 33
lid paIr,' onsi er ue ot er ray tha .
10 .71 ~ 34
secO
di
at gives s
peed reduction from 236 r.p.rn. to 200
r.p.nt. The correspon 109 gears are 5 and 6 •
.'
9
Zs
_ N6
z6
Ns
z +z
d
.
an
.
5
Z
200
== 236 or
Zs
== 0.847
Z
6
6 -
...
(v)
Z9+ZIO==20+34=54
On solvmg the equatIOns (v) and (v i), we get
... (v i)
z6 = 29.24 ~ 30 and
,
.
Zs -- 54 - 30 = 24
Third paIT: Consider the other ray that giv es th e speed mcreas
.
f
2
rIm.Thecorresponding gears are 7 and 8.
e rom 36 r.p.m. to 280
Ng
z7
Z7
and
Zg
-
N7
+ Zg
-
z9
280
=
236 or
+ zlO
=
z7 =
1.186
...
z8
( VlIii)
... (viii)
20 + 34 = 54
On solvingthe equations (vii) and (viii), we get
Zg
=
24.69 ~ 25 and
z7
== 54 - 25 == 29
Slage 1"
First pair,' Consider the ray that gives maximum speed reduction from 475 r.p.m. to
.J~r.p.m.The corresponding
Assume
gears are 3 and 4.
z3 20 (driver)
Z3
St
20 _ ~ . :.
or - - 475 '
N3
z4
•
,.~f411d Pill': Consider
N4
_ -
z4
•
the other ray that gIVes spee
zl
z2
and
Z1
vn sol'
Vlog
40.25 ~ 41
d reduction from 475 r.p.m. to 280
.Thee orresponding gears are 1 and 2.
I'L
Z4 ==
+ z2
... (ix)
N2 _ ~
:::: -
- 475
NJ
:::: z3
or
::::0.59z2
Z
..' (x)
1
.
:::: 20 + 41 ::::61
+ Z4
the equations (ix) and (x), we get
d
z2 :::: 38.37 ~ 39 an
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ZI ~
61 - 39 ~ 22
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Design o/Trans",is
-~~
.
------------------------------------~~~--~~1~
~.36
~
'.
5. Material selection. Take
40 N i 2 Cr I Mo 28 (hardened and tempered)
6. Calculalion of module:
./
_
16 has the lowest speed of .;,5.5 r.p.m .
Tofind torque: T he gear
p x 60
=
T 16
./
21t N
3.75 x 10J x 60 _
2
35 5
- 1008.7) N-m
1t x
.
To find tangential force on gear 16 :
T
F1
16
./
=
where
b
\11
Tm
= -m
r
=
2 x 1008.73 x 103
79 x m
= ~
m
We know that, module,
=
=
= 10 and
'
=
(25537.43 1m)
10 x 100
=
module, m
or
25537.43
m
FI I 'Vm x M
M = Material constant
m
=
100, for 40 Ni 2 Cr 1 Mo 28, from Table 9.4.
= ~
25.5371m
or m2
= 25.537/ m
2.945 mm
From Table 5.8, the nearest higher standard module is 3 mm •
./
Calculation of centre distance:
./
Centre distance in stage 1, al
./
Centre distance in stage 2,
./
Centre distance in stage 3, a3
8. Calculation offace width:
=
a2 =
b
=
(
ZI
+Z2)
2
(Z5 +Z6)
2
(ZII
m =
m
=
(22+39)
2
(24 + 30)
2
3
=
3
= 81
91.5 mm
mm
+2 Z12) m -_ (41 +2 58) ~ _- 148.5mm
= '-V x m = lOx 3 = 30 mm
9. Calculation of length of shaft (i:e., distance between the bearings) :
Length of shaft, L
=
25 + 10 + 4 b + 20 + 7 b + 20 + 7 b + 10 + 25
=
110 + 18 x 30 = 650 mm
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=
110 + 18b
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r
i
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60.1
I~
•
II O/sha/lS :
l~·pes". II o/sp;ndle Le., output shaft .
~ peS"
•
I M8J{i[llu[ll bending moment, M _ ~ x L
4
where F.
= Normal load on gear
lcos a
=
9058.79 x 650
4
gSS 37.43 I "!l
cos a
_ (25537.43/3}
cos 200
= 9058.79 N
M
...
,
=
=
=
1.472
x
106 N-mm
Equivalent torque,
-
~ (1.472 x 106)2 + (1008.73 x 103)2
-
1.784 x 106 N-mm
Diameter of the spindle is given by
,
ds
= [
J
1
161t x[ ~Teq
1 3 where [t
1= 55 N/mm', from Table 9.5.
1
_16
_
6
x 55x 10 J3 = 54.87
x 1t
1.784
mm '" 56 mm (R 40 series)
[
(u) Design 0/ other shafts:
(a) DiamJ!terojshaft
1:
Input speed = 47S r.p.ro. 3
p x 6~ _ ~ 75 x 10 x @ = 75.39 N-m
Torque :::: 21t N 21t x 475
T
::=
0.2 x
d; x [
't ]
Weknow that ,
75.39 x 103 -
ds1
or
0.2 x d~l x S5
19
:::: 18.99 mm ~
Ill
(R 40 series)
IIl
(b)Diameter 0/ shaft 2 :
Minimum speed
Torque
Or
da
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:::::
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Q~~
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·
~
D_e_Sl=gn---=of:..__Tr....::.:a..:..:ns~m;ss;O
-------
~.38
~
(c) Diameter of shaft 3 :
Minimum speed = 140 r.p.m.
3.75 X 103 x 60
p x 60
=
21t x 140
= 255.78 N-m
Torque - 21tN
255.78 x 103
d s3
or
=
0.2 x
d;3
= 28.54 rnm
x SS
=
30
mm
(R 40 series)
REVIEW AND SUMMARY
../
Gear boxes are used to obtain different spindle speeds in most of the machine tools.
../
The speeds in machine tool gear boxes are in geometric progression .
../
The concept of preferred numbers and their significance are presented in the beginning
../
of this chapter .
Step ratio (or series ratio or progression ratio) (¢) is given by
t/Jn -
N
=
I
max
N.
where n
= Number
of spindle speeds required
mm
../
Structural formula : n = PI (XI) 'P2 (X2/ 'P3 (X3)
where XI
= J;
X2 = PI" X3 = PI 'P2
../
The kinematic layout shows the arrangement of gears in a gear box .
../
Ray or speed diagram is the graphical representation of the structural formula .
../
Ray diagram serves to determine the specific values of all the transmission ratios and
speeds of all the shafts in the drive .
../
The basic rules to befollowed while designing the gear boxes are also discussed.
../ If the engagement of two different sets of gears provide the same, then the gear box is
known as overlapping speed gear box.
../
The step by step procedure for gear box design is presented at the end of this chapter.
REVIEW QUESTIONS
I.
What situations demand use of gear boxes ?
2.
What are the requirements of a speed gear box?
3.
The speeds in machine tool gear boxes are in geometric progression. Why?
4.
What are preferred numbers?
5.
Explain briefly the term 'progression ratio'.
6.
Write an engineering brief on :
(i) Kinematic layout of gear box; and (ii) Ray diagram.
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:t
.
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he significance
of structuralDownloaded
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W Ite t
9.3
a.
r
tiate ray diagram and structural d'
1. Oitferen
lagram.
6.
t the basic rules to be followed for
.
9·
~ist 011
Optrmum gear box d .
. the step by step procedure to design th
eSlgn.
W(Ite
e gear box.
/0.
PROBLEMS. FOR PRACTICE
blelflS
construction of ray diagram and kine-,. .
0"
.
layout:
f 1
.
".utlC
~ . need gear box IS to provIde a speed rang
A SIX Srd h k"
e 0 00 rpm t 100
/.
d diagram an t e rnemanc layout of the gear b
. . . 0
0 r.p.m. Draw the
spec
.
_'
ox.
minimum and maximum speed of 6 Speed ge bo
d d'
2.
. truct the spee
ragrarn and the kinematic arranar ox are to be 500 and 1600 r.p.m.
Cons
'.
gement of the gear box.
'ne speed gear box IS required to give output spe d
.
, AnI
.
e s rangmg from 100
6
J.
The input power IS 4 kW at 1000 rpm D
h
r.p.m. to 00
r.p·m.
f
. . . raw t e structural diauram and th
inematic
arrangement
0
gears.
Also
calculate
the
0-"
e
k
f
percentage
deviation
of
th
obtainablespeeds rom t h e calculated ones.
e
The
w the kinematic
4.. Dra
arrangement and the speed diagram of th h d t k
.'
e ea soc gear box ofa
tuITetlathe having arrangement for 9 spmdle speeds ranging from 50 r.p.m. to 1500
r.p.m.Calculate the number of teeth on each gear if the minimum number of teeth on a
gear is 23.
5. A machine tool gear box is to have 9 speeds. The gear box is driven by an electric motor
whoseshaft rotational speed is 1400 r.p.m. The gear box is connected to the motor by a
belt drive. The maximum and minimum speeds required at the gear box output are 1000
r.p.m.and 200 r.p.m. Suitable speed reduction can also be provided in the belt drive.
Whatis the step ratio and what are the values of 9 speeds? Sketch the arrangement.
Obtainthe number of teeth on each gear and also the actual output speeds.
6. A 12 speed gear box is to provide
a minimum speed of 30 r.p.m. with a step ratio of
l.I2. Using standard step ratios, find the number of teeth on all gears.
7. Draw the speed diagram of a 12 speed gear box to give speeds in the ran~e of 63 to 2800
f·p.m. Consider any 4 possible different alternates and indicate the best wIth reasons.
8. A mach'me tool gear box is to provide
. 14 spm. dl e speeds ranging from 20 to 400 r.p.m.
Draw the kinematic
'
arrangement and the ray d'lagram.
9. AI'
of 100 r.p.m. to 560 r.p.m. Sketch
6 speed gear box is to furnish speeds m the range
.
.
the k'
V
t dard progresSIOnrano.
lOematic layout and speed diagram. se san.
.
d
10. A
'.
.
of a milling mach me. Maxlm~m an
.~ear box IS to grve 18 speeds for a spmdle
800 rpm respectively. Fmd the
nlIn
d 16 to
. . .
Imum speeds of the spindle,. are to be aroun
d
the structural diagram and the
:~ ratios which will give the desired speeds and raw
Inemat'
ICarrangement of the drive.
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--------------------------_D--eS_ig~n~Of~Tr_a~nr~rn=/~ss~ns
Q~~
!!.40
~
Problems on design of gear box:
.
.
.
.
ed
box with a step ratio of 1.25. The Input IS from a motor ru .
11. D esign a SIX spe gear
f
. 3k
nnlo
.,
m speed is 112 r.p.m. power 0 motor IS
W.
g
at 1440 r.p.m. TIte mmimu
.
12. A six speed gear box is required to provide output spe;d s ~~the ~~~ge of 125 to 400
5
r.p.m., with a step ratio of 1.25 and transmit ~ power 0
at
r.p.m. Draw the
speed diagram and kinematic diagram. Determ.lne the number of teeth, ~odule and face
width of all gears, assuming the suitable materials for the gears. Determine the length of
the gear box along the axis of the gear shaft.
13. Design a headstock gear box of a lathe having nine spindle speeds ranging from 30 to
1000 r.p.m. The power of the machine is 4.5 kW and the speed of the motor is 1440
r.p.m. Minimum number of teeth on the gear is to be 25. Sketch the layout of the gear
box and calculate the number of teeth on the gears.
14. Design a 9 speed gear box to give output speeds between 280 and 1800 r.p.m. The input
power is 5.5 kW at 1400 r.p.m. Draw the kinematic layout diagram and the speed
diagram. Determine the number of teeth on all gears and the length of all the shafts.
15. Design a nine speed gear box with a minimum speed of 200 r.p.m. and speed ratio 1.5.
The input is from a motor of2 kW at 1400 r.p.m.
16. Design a 12 speed gear box for an all geared headstock of a lathe. Maximum and '
minimum speeds are 900 r.p.m. and 23 r.p.m. respectively. The drive is from an electric
motor giving 2.2 kW at 1440 r.p.m.
17. The spindle of a pillar drill is to run at 12 different speeds in the range of 70 r.p.m. and
325 r.p.m. Design a three stage gear box with a standard step ratio. The gear box
receives 4 kW from an electric motor running at 330 r.p.m. Sketch the layout of the gear
box, indicating the number of teeth on each gear. Also sketch the speed diagram.
18. A gear box is to be designed for the following specifications;
Power to be transmitted
-
Number of speeds
15 kW
18
Minimum speed
-
16 r.p.m.
Step ratio
-
1.25
Motor speed
-
1400 r.p.m .
Arrangement
scheme
2 •x 3 x 3
Sketch the speed diagram and the kinematic arrangement.
19. Design the layout of a gear box for a milling machine having an output of speeds
ranging from 180 to 2000 r.p.m. Power is supplied to the gear box by 6 kW induction
motor at 1440 r.p.m. Choose standard step ratio and construct the speed diagram. Decide
upon the various reduction ratios and number of teeth on each gear wheel. Sketch the
arrangement of the gear box.
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I.
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Clutches
"One Ues ore t ,L'
0'
a t,,/IIgsfro",
/L_
TIe
valley.
only s",al/ things from the peat ..
I~f.
INTRODUCTION
.
'he clutch
l
.
.
- G.K. Cllmenoll
a mechanical device which .
.
IS used to conn t
.
,,~J at the operator s Will. The use of clutch
. .
ec or disconnect the SOurce f
,\1
h .
. .
es IS mostly f d'
0
.i jII~JlI'''Iobile clutc IS a transmiSSion device whiICh IS
. used to
oun In automobile . In an
,nlltl the engine to the rest of the system.
engage and disengage the power
IS
For me above said reason,
the clutch is locat e d im be tween the
.
.
_. When the clutch IS engaged to the engin
.
engme and the transmission
~).
.
e, power IS transm'tt d
JL~aged, the power IS not transmitted to the rest f hie
to the wheels. If it is
J1JIIIIing and hence the vehicle stops. Therefore for co 0 , t thesyste~ even though engine is
1
.
.
.
up mg
e engme smooth I t th
i Q1IISlI1isslon
during starnng from rest and gear shifting , c Iu tCh IS
. used.
y 0 e power
lD.2. FUNCTIONS OF THE CLUTCH
Theimportant functions of a clutch are :
"
"
To connect and disconnect
the shafts at will.
T~ start or stop a machine (or a rotating element) without starting and stoppin the
prime mover.
g
"
To maintain constant
speed, torque and power .
.f
To reduce shocks transmitted
.f
For automatic
disconnect,
between machine shafts .
quick start and stop, gradual starts, and non-reversing
and over running functions.
'J. PRINCIPLES OF OPERATION OF THE CLUTCH
The clutch works on the principle of friction. When two friction surfaces are brought in
~lact with each other and pressed they are united due to the friction between them. The
~ on between the two surfaces d~pends upon the area of the surfa~es, pres~u~e applied
~~ ~d coefficient of friction of the surface materials. The principle of fnctlOn cl~tch
·~;)lOOdwith the help of Fig. I0.1. The two surfaces can be separated and brought .mto
Wh
. .
b and the other as driven
en reqUired. One surface is considered as dnvJng mem er
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J1.Q~~
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------------------_D_~~ig.~n~of~T~~
.
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""33iO
.;
.
drivi
member is kept rotating.
When the driven
member is brought.
member. Th e
vmg
.
.
th
.
ber it also starts rotating. When the driven member is se
I
to e dri vmg mem ,
. .
..
h'
parated
vmg member I·it does not revolve ThIS ISthe principle be ind the operation 0f the
dri .
B
C
Bee
A
Disc
Disc
Driving disc
(a) Before engagement
Driven disc
(b) After engagement
Fig. 10.1.
10.4. CLASSIFICATION
OF CLUTCHES
The clutches are classified in two ways:
1. Based on the engage1lU!ntor actuation method used* :
(a) Mechanical,
tb) Pneumatic,
(c) Hydraulic,
(d) Electrical, and
(e) Automatic.
2. Based on the basic operating principle used** :
(a) Positive contact clutches
./ Square jaw
./
Spiral jaw
./
Toothed
(c) Overrunning clutches
./
Roller
Sprag
./
Wrap-spring
(e) Fluid coupling
./
(b) Frictional clutches
./
Axial
./
Radial
./
Cone
(d) Magnetic clutches
./
Magnetic particles
./
Hysteresis
./
Eddy current
Dry fluid
Hydraulic
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r~
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01 Mel/lods :
,co"tr
.
Mechanical control IS achiev db'
I
e y ltnka '
cams or we d ges. Th e actuatinb f
ges at by b
.
'.
e orce can 1.._
ails Or t II
electrtC motor, air cylmder, or hYd .
ue suppli.....
0 trs workinabv
rauhc ra
~ rnanuall
e
er
lectrical
control
of
friction
or
t
h
Ill.
"1
Or
by
solenoid
I E
oot cl h
•
electrically and releasing it by spr] f utc es ofteh invol
Ing orce
ves engagingth
I Pneumatic or hydraulic Control'
.
e clutch
'nflatable tubes or bladders.
IS accomplished b
I
y actuating P'iston nd
.
I f
s a
I AutomatIc contro 0 clutches implie th
s at they react to ed
•• Coupling Methods:
pr eterrnmed conditions.
I
I positive-contact
clutches have interloc k"Ing engaglh•
mechanical junction.
I
.
Frictional clutches are used most frequently T
into firm frictional contact.
I
J.
g SUnates to
form a rigid
.
. wo opposingsurfacesate forced
Overrunning
are used when two members are t0 run freeIy relati
. clutches
..
anve to
do
each other
one direction but are to lock in the other.
I Magnetic clutches use magnetic forces to couple the rotating membersor to
provide the actuating force for a friction clutch.
I Fluid couplings make use of a hydraulic oil or a quantity of heat-treatedsteel shot
But from the subject point of view, the five important types of frictionclutchesare :
1.
2.
3.
4.
In
Disc or plate clutches,
(a) Single plate clutches,
(b) Multi-plate clutches
Cone clutches,
Centrifugal clutches,
Internal expanding rim clutches, and
External contracting rim clutches.
.'
l t hts Thecentrifugal.internal
known as axial frictIon cue .
~
The disc and cone clutches are
d
dial/riction clutches.
. . I tches are calle as ra
expanding rim and external contractmg nm c u
IM:::l
5.
10.5.FRICTION MATERIALS
FOR CLUTCHES
d Friction Material
0.5.1. Required Qualities of a Goo
.
ualities :
wtng
the follo
q
A good friction material should have
.'
d tivity.
. t offnctlon.
. ood heat con uc
-/ A high and uniform coefficlen
together Withg
. h temperatures,
0/ The ability to withstand hlg
1
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Dull" o!T1'aJUmLJllo" e..,
10.4
u~/f""
----.:::..
~ High resistance to wear. scoring, and galling.
~ Resistance against envlronmenral conditions, such
85
moisture, salt water or fun .
IJ.
~ Adequate mechanical and thermal strengths.
10.5.2. Commonly u•• d Friction Materlall are
I.
Wood,
2.
Cork,
J.
Leather;
4.
S.
Asbestos based friction materials, and
Powdered metal (or sintered metal) friction materials.
¥'
Wood, cork and leather are used only for light loads and low speeds. But in
practice, for high speeds and heavy loads, asbestos based materials and powdered
metals are widely used.
10.1.3. Asbestos baled Friction Materlall
The two types of asbestos friction materials used are :
(i) Woven type, and
(ii) Moulded or composite type.
(I) Wov~n type: Woven type friction material is obtained by spinning threads from
asbestos fibres, woven around brass, copper or zinc wires. Then it is impregnated with a
bonding material.
(1/) Moulded or composite type: In mould or composite type, the asbestos fibres in their
natural state are mixed with a bonding material and then moulded in dies under pressure at
high temperature.
10.5.3.1. The bonding materials used are
-/'
Asphaltic bases with additions of natural gums and oils,
-/'
Vegetable gums,
-/'
Rubber, and
-/'
Synthetic resins soluble in alcohol or oil.
10.5.3.2. Difference between woven and moulded asbestos materials
./
Woven material is flexible, while moulded asbestos is rigid.
-/'
Woven material has higher coefficient of friction than the moulded asbestos
material.
-/'
Woven materials wear at a faster rate than the moulded asbestos materials.
-/'
Woven materials are costlier than the moulded asbestos materials.
The main drawback of both the woven and moulded asbestos materials is their inability to
withstand high temperature.
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,/~red
(or .Intered)
r Apo"tAI"IIddeer.
metal
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frlctl
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.
on material.
6"
e materials are obtatned from rn
f~'. I fheSlIurgy in the form of 0.25 to 6 0 metallic POwder by th
metS
. rn thick
e proee
steel discs on one or both 8ides
straps. These
55 of POwder
onto
·
straps can be
twO types of sintered metal frictioh
.
pressed
h
. d
d' .
matenals u
If (i) eBronze-base
,an
(II) Iron-based.
sed are:
Advantages ~f sintered matal friction
1~!·I.l. friction materials)
materials(overaSbestosbased
ey have high and un iform coefficient of f " .
I Th
r rlchon.
They can withstand high temperatures.
I
I They are lighter, cheaper and compact in construction.
I They are unaffected
by environmerttal conditions.
;
Table 10 . 1. shows the values of coefficient of friction for diffierentcom b'matlons.
Table /0.1. Coefficient ollrlction
Coefficient of fridlon
Contacting surfaces
Wet
Dry
0.1-0.2
OJ -0.6
Moulded asbestos - cast iron
0.08-0.12
0.2-0.5
Bronze-base sintered metal - cast iron
0.05 - 0.1
0.1-0.4
Bronze-base sintered metal - steel
0.05 -0.1
O.I-OJ
Wovenasbestos - cast iron
10.6.SINGLE PLATE CLUTCH
Fly wheel _---...
Clutch spring
Engine
shaft -_
clUtCh shaft
Fig. /0.2. Single plate clutch
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_ Design olTI'QfJ31hL
'0.6
•
331
0"
.
S~,fle
r.his type of clutch is mosdy used In ~otor vehicles, It.co~sists of one clutch
~
shaft, dutch spring, pressure plate, friction lining and bearing. The flYWheel. Plate, CI~
the eHgine crankshaft and rotate! with it. The pressure plate is bolted to the fl~hrnounted
~
dutch springs ..The fHttiCU1 I infngs are on both sides of the clutch plate. Fig.I 0.2eelthrOU&h
artltHgement of single plate clutch.
shows the
OpetallOJl: When the clutch is engaged, the clutch plate is gripped between the
and the pressure plates. Due to friction, the clutch plate and shaft revolves. Whe thflY\vheel
pedal is pressed. the pressure plate moves back against the force of the springs, an~ theclutch
plate becomes free between the flywheel and the pressure plate.
e clutch
Thus the flywheel remains rotating as long as the engine is running and the clutch
speed reduces slowly and finaJJy it stops rotating.
shaft
10.8.1. DesIgn of a SIngle Plat. Clutch
(Torque transmitted by the Single Plate Clutch)
Consider two friction surfaces held together by an axial thrust W, as shown in Fig.IO.3(a).
Single disc
OIp1ale
p
Friction surface
<a)
(b)
Fig. /0.3. Forces on a single disc or plate clutch
Let
T
p
=
=
rl -
r2
=
Torque transmitted by the clutch,
Intensity of axial pressure acting on contact surfaces,
External radius offriction surface,
Internal radius of friction surface, and
J.l = Coefficient of friction.
Consider an elementary ring of radius r and thickness dr, as shown in Fig.l O.3(b).
Area of the elemental ring
Normal or axial force on the ring,
oW
= 21tr· dr
=
Pressure x Area
=
p x 2nr' dr
and the frictional force on the ring acting tangentially at radius r is given by
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·
'"p )(27t/' . d.r
' Fr.'ctiona' torque actmg on the ring , r,.:::: F
"
,.)(r
T ::: ,.
,.
,.p)( 2nr, dr)( r
'go of friction clutch is done based on a ::::2n", p r2 dr
tbe des'
ny one of the fOlio '
(i) When there is a uniform pressure, and
Wingassumptions:
(ii) When there is a uniform wear.
I
1.l
ConslU'e
ring uniform pressure:
~ Area 0f the friction
surface, A == n (1'2J
_ 1'
2)
2
,~nn intensity of pressure (P) is given by
VnlJo
W
W
p-------
A -n(r:-r;)
... (10.1)
Totalfrictional torque actin~ ~n the friction surface or on the clutch is obtained by
mtegT8 ting the equation of the ffictlOnal torque on the elementary ring within the limits from
o
'2 to '"
T
=
J
27t Il P r2 dr
=
27t 11P [ ~ ] :
=
2" 11P [ r: ; r~]
r2
Substituting the val ue of p from equation (10.1),
W
T = 21t #l x 1t (I'; _
I'~) [I': 3-ri]
or
where
... (10.2)
R - Mean radius offriction surface
2
[
3 3]
rJ
... (10.3)
-1'2
R = 2
2
3
r I - r2
"
"ty of pressure varies
"
r
uniform
wear,
the
rntens•
. (Ii) Considering uniform wear: Fo
m~ersel '
&'.
Y WIththe distance. ThereJore,
c
= constant::=
=
P2' 1'2 =
C or p == I'
C]
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108
Design a/Transmissions
.
~
where PI and P2 are intensities of pressure at radii rl and r2 respectively.
We know that normal or axial force on the elementary ring,
flW
[ ••.p
= P 21t r dr = 21t (p . r) dr = 21t C dr
C
~;l
:. Total force acting on the friction surface,
'1
W
=
f 21t C dr
= 21t C
[r
(I =
21tC (rl - r2)
'2
'2
Ie
or
=
W
'" (l0.4)
21t (rJ - r2)
We know that the frictional torque acting on the elementary ring,
T,
= 21tfl pr2 dr = 21tfl
=
C
r2 . dr
x -;: x
[ ': p
21t fl . C . r . dr
:. Total frictional torque on the friction surface,
T =
c
= -r ]
f
2x,,·
e .r . dr
= 2x"
e[ ~
r
l
'2
'2
T = 1t fl C [r~ - r; ]
Substituting the value ofC from equation (l0.4),
T
I
=
7t
II
r:
W
X
21t (rl - r2)
X
(r2
1
_ r2 )
2
T = ~ x,,· W (rl + r2) = ". W . R
I
... (10.5)
R = Mean radius of the friction surface
where
or
... (10.6)
INote I
I. In general, totalfrictional torque acting on thefriction sur/ace or on the clutch is given by
T
where n
=
=
Number
n·J.l·W·R
0/pairs
2. For single disc plate clutch, n
offriction or contact sur/aces.
= 2.
Since both the sides a/the disc are in contact.
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C/~fC"es
10.9
. .a the problem, if the word 'new clutch' i
.
'Iesolvmb
S gIVen, then considi
J If'"
Bul if Ihe word old clutch is given, then II I.
er as an uniform prt!$sur~
, ble""
s an uniform
'
prO hen uniform wear should be considered
w~arprob/~m. If nothing ;s
f
fPted. I
•
l Intensll)!
S
J
uf pressure
rfi
'ctionsu ace.
thefr I
roue pressure
'{he {IV e
b
o:
=
r a»
10,1•
.'
IS
,
maximum at the inner radius (r_j
]I
•
a"d mlmmum at the outer radius (r·1 or
1/
(Pd on the friction
or contact suna
'J'
.
.
ce IS given by
Total {orce on/ticlion surface
Cross-sectional area 0/'frtctton s.r.
UIjQce =
frlr1 W
_ }
J
r1
1
...(10.7)
MULTIPLATE OR MULTIPLE DISC CLUTCH
Friction rings (spIined)
Driven aha1!
Engine shaft
Fig. 10.4. Multiplate disc clutch
A multiplate clutch is used when large amount of torque is to be transmitted. In a
~ultiplateclutch, the number of frictional linings and the metal. plates are increased which
1DjCl'easeS the capacity of the clutch to transmit torque, as shown in Fig. 10.4. The multiplate
Cutchw orkS In
. the same way as the single plate clutch by operating
. the clutch pedal. They
are extensive I'y used In motor cars, machine
. tools, etc.
10.7.1 D
. eSignof a Multiplate Clutch
let (Torquetransmitted on Multiplate Clutch)
nl = Number of discs on the driving shaft, and
"2 =
.. Nurnbe
Number of discs on the driven shaft.
r of pair of contact surfaces,
n
= "1+"2-1
l
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'J
--------------------~
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10.10
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---------------------D--es-l=gn~of~Tr-a-n-s-m~~~s~;n~ns
~
Then, total frictional torque acting on the clutch is given by
IT
= n'
J.l • W . R ]
'" (10.8)
R = Mean radius of the friction surfaces
where
R
=
R
=
~ [r1-rn
3
[For uniform pressure]
rl-r2J
rl
+r2
[For uniform wear]
2
10.8. SERVICE FACTORS
We know that the design of disc clutch requires the calculation of (I) the dimensions of
the disc, (2) the number of discs, and (3) the clamping or engaging force . .It may be noted
that a clutch should have a torque capacity substantially greater than the nominal torque
requirement so that the load can be accelerated without excessive slip. So while designing,
one has to consider service factor which includes driver and driven dynamic characteristic
factors, wear factor and frequency of operation factor. Refer Tables 10.2 to 10.5.
Design torque
[T)
=
Nominal torque x Service factor
=
T x ks
... (10.9)
[from Tables 10.2 to 10.5]
where
Table 10.2. Driver dynamic characteristic factor, k J (from data book, page no. 7.9IJ)
Factor
Type of drh'ing system
k)
For direct coupling
For belt transmissions
or when ~ar box is used
Electric motor directly connected
to mains
0.5
0.33
Machines with low starting torque
characteristics, turbines, etc.
0.33
Table 10.3. Driven dynamic characteristic/actor,
-
k2 (from data book, page no. 7.91)
Type of driven system
kl
Metal cutting machine tools, wood working machine tools, reciprocating
compressors, centrifugal pumps, etc.
.
1.25
Metal cutting machine tools with reverse motion (planing),
turbo
1.6
compressors, cranes, heavy machinery, piston pumps, heavy drills, etc.
Forging machines, presses,
paper mach'
. wire drawing machines ,mery,
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etc.
2.5
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-
-
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Tablt 10.4. Wearfaclo~ k
t
160
240
~#',
j
vro", dlltll b
DOt, page
110.
400
7.91)
0.43
Igements in En
~ofeng
~r sbift
~
I
1
\
0
.
~
a oolc,page no. 7.91)
r.;:;l The service factor
~
I t h
.
IS
8
0.2
16
32
0.55
48
0.75
0.9
used mostly an those probl
96
240
480
1.2
1.8
2.00
hi
.
ems w ich require the calculation of
~ions of the c u c .
~lemson Single Plate Clutch
r:
~le
10.1] An automoti~~ sing~e
cltach consistso/two pairso/contacting
If/acts.The inner and outer radii of 'friction plate are 120 mm and 250 mm respectively.
Ik cot/flCkntof friction is 0.25 and the total axial force is 15 kN. Calculatethepower
_mitting capacity of the clutch plate at 500 r.p.m: using
(i) Uniform wear theory, and
(ii) Uniform pressure theory.
GivenData:
n = 2;
'I = 250 mm
~=
0.25 ;
e Solution:
W
=
15 kN
=
=
0.25 m;
15 x 103 N ;
•
IL.
(i) Using uniform wear t"eory .
'2 =
120 mm
=
0.12 m;
N = 500 r.p.m.
. .
Torque transmitted on clutch IS given by
('I + '2)
T -
n·J.1·W
2
{Q.25 + O.l~
_ 2 x 0.25 x 15 x 10
2
3
.
.
'. Power transmitted
~':I •
2
:.p~
1387.5N-m
21t x 500 x 1387J = 72.65 kW Ans."",
21t NT == 60
.
=
60
lutch is given by
itted on c
• Torque transm
", USl'lg uniform pressure theory •
T -
=
n f1 W
3 ]
,31 -'2-
'3 [ ;~
-r~
.
0.25 _ 0.12 ] ~ 1444.6N-m
x ~ x [ 025 - (0.12i
_ 2 x 0.25 x 15 x 103 3
( .
4 k'W Ans. ~
" .... '>l 500~
:::::75.6
21tNT
== ~
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10.12
olT,an.r
Daig,.
IExturtJ!!.r
/0.21
A sillrlr piau frlctla"
cllllch, ",lIh bollt .,..
'"Utio"
I~
~
~ff«tiw,Is 11$«1 to trtJlISlfIitpower
al1440 r.p.m. It has out~r and Inn~,,1UIU ~ ~
60 ".". rap«tivdy.
TII~ maximum int~nsity of pressure is Ilmit~d to 10 )( 1()4ao".". '"
CH.ffkU1It of friction is 0.1, d~t~rmin~ :
/V/tttl. 1/,..
(I) Total pnssun
exerted on th~ plal~, and
(Ii) Power transmitt~d.
Given Data:
N = 1440 r.p.m.;
n = 2;
'2
=
60 mm
=
60 x
10-3 m
;
Pmax
'1 =
=
80 mm == 80
10 x 1()4N/m2...
,
)( 1()-3
,. ==
0.)
Ill;
e Solution:
(i) Total pressure exerted on the plate : Since the intensity f .
(P) is maximum at the inner radius ('2)' therefore for uniform wear
0 PTessurt
= C or C =
Pmax"2
10-3
lOx l()4x60x
=
6000N/m
We know that the axial thrust,
W = 2lt C ('1 - '2)
= 2lt x 6000 (80 x 10-3 - 60 x 10-3) = 754 N
:. Axial thrust or total pressure exerted on the plate
=
754 N Ans. ~
{il) Power transmitted:
Torque transmitted, T
('1 + '2)
= n·J.1·W,
2
3
= 2 x 0.3 x 754 (80 x 10- ;
Power transmitted,
I Example
P
=
2lt NT
60
=
3
60 x 10-
2lt x 1400 x 31.67
60
)
= 31.67 N-m
= 4.643 kW ADS.
~
I
10.3 A single plate clutch transmits 25 kW at 900 r.p.m. The mtJXiIruII
pressure intensity between the plates is 85 kNlntZ. The ratio of radii is 1.25. Both the ~
of the plate are effective and the coefficient of friction is 0.25. Determine (i) the bfM1
diameter of the plate, and (ii) the axial force to engage the clutch. Assumt theory~
uniform wear.
Given Data:
P = 25 kW = 25 x 103 W ;
N = 900 r.p.m.: ,,/'2:::: 1.25;
n
e Solution:
W
=
2;
Pmax
=
85 kN/m2
=
85 x 103 N/m2;
~ == 0.25.
(i) The inner diameter of the plate:
.
e know that the power transmitted, P =
21tNT
60
25 x 1f)3 = 21t x 900 x T
60
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t
"
of
•
e an
'~ce th
t
T = 265.26 N-rn
tensity of pressure is maximurn at th .
e anne
"
,
-
.
d the aX
ial thrust transmitted
W == 21t C (I'.
,JJ
to the frictional
-1'2)
.
r radius ( )
Pmax . 1'2 = C or C - 85
51
t
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x
1()3
'2 ,
'2
N/rnrn
..c.
SUI
lace,
= 2n x 85 x 103
'2
(1.25 r
== 1.335 x lOs (1'2)2
1be me
_
)
2'2
'" [.,' 'I == 1.25 '2]
an radius for uniform wear is given by
R _
r. +
L25'2 + '2
2 - == 1.1251'2
1'2
_
2
T - n·IJ.·W·R
265.26 - 2 x 0.25
Torque transmitted,
x J .335 x
lOs (1'2)2
x
I.1251'2
= 75.104 x 100,~
or
1'2 -
and
1'1
0.1523 m or 152.3 mm
- 1.25
1'2 =
1.25 x ] 52.3
==
190.375 mm
ADS. ~
(Ii) Theaxial force to engage the clutch:
W
= 27tC
=
IExample10.4
(I'I - 1'2)
= 1.335 x lOs ('2)2
== 1.335 x lOs (0.1523)2
3096.57N Ans."
, Outer and inner diameters of a single plate clutch, having ferodo
fridion linings on both sides are 300 mm and 200 mm respectively. The coefficient of
frrtionbetween the contacting surfaces is 0.35. Assuming uniform rate of wear, determine
~ -imum power tllat can be transmitted by the clutch at 1500 r.p.m: when the
trxUnum pressure is not to exceed 15 x 1fJ4 Nlm2. Also, determine the axial thrust to be
~U~~e¥rinponpn~unpm~
3
Given Data:
d , = 300 mm = 300 x 10-3 m; d2 = 200 mm = 200 x 10- m;
.
= 15 X )04 N/m2,
IJ. = 0.35;
N = ] 500 r.p.m.: Pmax
@ Sol'
.
.
. m at the inner radius ('2)'
~
utlOn: Since the intensity of pressure (p) IS maxunu
fore for 'c
unhOrm wear
0-3 == 15000 N/m
Prnax.·r2 = C or C = 15 x l()4x ]OOX ]
A>:iaJ thru
st exerted by the springs is given by
3 100 x 10-3)
150X
10-(
W = 27t C (I'I - 1'2) == 27t x 1500 O x
ro,
qUe
trans
= 4712.39 N Ans. 1)
.
llutted for uniform wear,
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_1_O._14
D_a~ign~Ozif~TJ~rQ~~~-.
.
.. ,ISS IOn
S
~
T = n- J.I • W (";
'2 )
150 x 10-3 + 100 x 10-3)
= 2 x 0.35 x 4712.39 x (
2
= 412.33 N-rn
:. Maximum power transmitted,
21t N T
p
I Example
10.5
21t x 1500 x 412.33
60
=
I Determine
60
=
=
64.77 kW Ans. ~
the maximum, minimum and average pressure in
,
a Pl(lIe
clutch when the axial force is 5000 N. The outer and inner. diameters of the fricr
sUrfaces are 200 mm and 100 mm respectively. Assume uniform wear.
rail
Given Data: W
5000 N ;
=
dl
=
200 mm or rl = 100 mm
=
100 x IO-3m.
,
d2 = 100 mm or r2 = 50 mm = 50.x 10-3 m.
@ Solution:
(i) Maximum pressure:
Since the intensity of pressure is maximum at the inner radius (r2)' therefore
=
Pmax x r2
=
C or C
50 x 10-3 Pmax.
Axial force exerted on the contact surface (W) is given by
W = 21t C (rl - r2)
5000 = 21t x 50 x 10-3 Pmax (100 x 10-3 - 50 x 10-3) = 0.0157 Pmax
Pmax
(ii) Minimum
31.83 x 1()4N/m1 Ans. ~
=
pressure:
Since the intensity of pressure is minimum at the outerradius
(r1), therefore
Pmin
x rl
=
C or C
=
100 x 10-3 Pmin
Axial force exerted on the contact surface (W) is given by
W = 21tC (rl - r2) Pmin
5000
or
=
Pmin -
21tx 100x 10-3x(100x
10-3-50x
10-3)
=
0.0314Pmin
15.92 x 1()4N/m1 Ans."
(iii) Average pressure:
Pay
=
Total normal force on contact surfaces
Cross-sectional area of contact surfaces
W
= 1t [r~ - r~ ]
5000
= 1t [(100 x 10-3)2 - (50 x 10-3)2]
= 21.22 x 104 N/lIll
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Ans. ~
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1#
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JO.6 A car engine
has Downloaded
ils rale~ From : www.EasyEngineering.net
I
u Oil/PilI
0.15
.~
00
1ll_/II. Tile clulcll used is 0" •
Of /2 Itw. "'L
P:.
, J
J Y·
':I .Wngle 'l
.
'-M1IS /Ire is nolto exceed 85 kN/nt2.~."
p a/~lyp~Itav;
11IaxJ"'11111 IO'f/II~
e exte" ,.1
"1 G •
illter"al diameter; Determine I"
.
'II", diollle/~' ..,
~llv~Ilil/aen.,
"'-.~.d,lte
.
e d1Illens.·
'?t '''~ In .
rled by Ihe spnngs. COefficient of.."·'
lOllS Of lite fo"
~/on plll/e is
)J {ilrr/1
I,,~
""0
..J":~press.
''O'r
/' r!J,ct
,JA!J~
~e
r' palo.
P
=
Pmax
=
•
6i¢"
d,
fJ So
=
12 k W == 12
x
::: 0.1
fl'lcllon
I ()3
W.
'lCIIIJIIplllle and
• •
T
85 kNlm2 == 85 x 103 N'
::: 100 N-m;
1m2• ,
1.25 d2
r,
~ - I2
or
d2 -
or -
• 5
Iplion : (i) Dimensions of the friction plate· A
•
Given that the torque developed,
== 1 25 .
r2
"
n == 2.
Il == OJ
ssume uniformwear
T == 100 N-m
.
s
service factor, Ie == Ie + Ie + 1- + k:
where
From Table 10.2,
k, -
From Table 10.3,
k2
From Table 10.4,
k3 -
0.32
(assuming 1000 r.p.m.)
From Table 10.5,
k4 -
0.9
(assuming 48 engagements/shift)
ks
=
0.33 + 1.25 + 0.32 + 0.9
=
100 x 2.8
s
Design torque, [T]
0.33
'2
113
4
(for machines with low start'
109 torquecharacteristics)
= 1.25 (assume)
=
=
2.8
280 N-m
Sincethe intensity of pressure is maximum at the inner radius (r2)' therefore
=
Pmax. x r2
C or C = 85 x 103 x r2 N/m
and the axial force exerted by the springs,
= 2nC (r, - r2)
= 2n x 85 x 103 r2 (1.25 r2 -r2)
W
torque transmitted, [T]
280
Or
and
,
Design torque, [T] == T· k
.'
=
=
n- Jl . W
=
1.335 x 105 r; N
(~+~)
2
2 x 0.3 x 1.335 x 105 (r2)2 (
1.25 r2 + r2) = 90124 x 104 (r2J3
2'
.
r2 = 0.1459 m or 145.9 mm ADS.1:l
r,
-
1.25
rz
= 1.25 x 0.1459
0.1823 m or 182.3 mm
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'L
Ine
ADS. l'
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Design o/Transmission S
10.16
~
(ii) Axial force exerted on the springs:
W = 21tC ('1 - '2)
= 1.335 x lOs
x
= 1.335 x
lOs ('2)2
(0.1459)2
= 2841.79 N Ans. ~
I Example 10.7 I A friction
clutch is used to rotate a machine from a shaft rotatingat Q
uniform speed of 250 r.p.m: The disc type clutch has botll of its sides effective, 'he
coeffICient of friction being 0.3. The outer and inner diameters of the friction plate are200
mm and 120 mm respectively. Assuming uniform wear of the clutch, the intensityOf
pressure is not to exceed 100 kN/mI. 1/ the moment of inertia of the rotating parts Of 'he
machine is 6.5 kg-m1, determine the time to attain the full speed by the machine and'he
energy lost in slipping of the clutch.
What will be the intensity of pressure, if the condition of uniform pressure of the clutch
is considered? Also determine tke ratio of power transmitted with uniform wear to 'hill
with uniform pressure.
N = 250 r.p.m.;
Given Data:
Pmax
n = 2;
J.1 = 0.3 ;
= 100 mm or 0.1 m; d2 = 120 mm or '2 = 60 mm = 60 x 10-3m:,
= 100 kN/m2 = 100 x 103 N/m2; I = 6.5 kg-m-
@Solution:
(i) The time to attain thefull speed by the machine (witll uniform wear) :
Since the intensity of pressure (P) is maximum at the inner radius '2' therefore
Pmax
Axial thrust exerted,
"2
= C or C = 100 x 103 x 60 x 10-3 = 6000 N/m
W
=
27tC ('1 - '2)
W = 27t x 6000 (100 x 10-3 - 60 x 10-3) = 1507.96N
Torque transmitted, T
=
n- J! . W (r. ; r
- 2
x
0.3
x
1507.96
2
)
x (
100
x
10-]2+ 60
x
10-]) = 72.38 N·m
We know that the Power transmitted P = 2nNT = 27t x 250 x 72.38 = 1895W
'60
60
Also
T = I a, where a isangular acceleration.
72.38 = 6.5 x a or a = 11.135 rad/secWe also know that,
(J)
a
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= I =
11.135
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11.135
..
t
fltLlS
he full speed
oe energy lost
(P) I/J
A
IS
21t X 250
60xt
•••
11.135
==
attained by the.
In slipping
d
0
r I::: 2JS
of the I
C utch:
ogle turned by the driving shaft, 91
.~
.35 seconds
.
== (1)1 ::: ~
60
== 61.52 rad
by the driven shaft, 92
I
==
(1)0 1 + 2"
== 0
.' Energy lost in friction
sec Ana
machine in 2
r»
all angle turned
10.17
XI:::
~
60
X
2.35
a. 12
+ -21 ·X II. 135 X (2.35)2 .::: 30.75 rad
-
72.38 x (61.52 - 30.75) ::: 2226 N-m Ans.~
(iii) Intensity of pressure, if the condition is unifior.m pressure :
Intensity of pressure
,
p
2W 2
=
1t
(r I - r2)
=
1507.96
1t [(100 x 10-3)2-(60 x W-3)2]
=
75000 N/m2 or 75 kN/m2 Ans. ~
(iv)Ratio of power transmitted with uniform wear to that withuniform pressure:
Power transmitted with uniform wear == 1895 W [.: already calculated, refer (i)]
Torquetransmitted with uniform pressure == n-
[r: -r~]
2
W . 3' r: - r~
J.L •
== 2 x OJ x 1507.96xj'
T
Pow
er transmitted with uniform pressure
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_(0.06t]
(0.1)2-(0.06)2
== 73.89 N-m
.
IS
. n by
give
2nNl
p == - 60
2n x 25~
~Wer tr~ .
po~ansmitted
with unifonn wear_
er transmitted with unifonn pressure
2 [{g.lf
:::
~
:::
t9J4
::: 1934 W
1895 ::: 0.98 ADS. ~
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10.18
~
Design o/Transmiss;
I Example
~
10.8
I Determine the time required to accelerate a counter shaft 0/ tottJt;"
mass 500 kg and radius of gyration 200 mm to t~e fuJI speed of 250 r.p.1II./ro", r
through a sillgle dutch of internal and exlernal radII 125 mm and 200 mm, laking
'!It
and the sprillg force as 600 N. Assume Ihat only one side of clutch Is ejJectivt!.
/l a.r O.J
Given Data:
k
=
=
0.125 m;
m = 500 kg;
'2
=
125 mm
200 mm
0.2 m;
'1
J.l = 0.3 ;
W = 600 N ;
=
n
=
N
200 mm
=
=
=
250 r.p.rn. ;
0.2 m;
I.
@ Solution: Moment of inertia of shaft is given by
I = mk2 = 500 x (0.2)2 = 20 kg-m-
T = I· a
Torque transmitted is given by,
nJ.lW
('. +'2)
= la
2
125
I x 0.3 x 600 (0.2 +2°. )
or
aa
Angular acceleration,
We know that,
=
20 xa
= 1.4625 rad/sec-
Angular speed
=
Angular acceleration x time
(J)
=
a· t or t
(J)
= -
a
1.
t = (27tN)
60
= (27t x 250)
xa
1
x 1.4625
60
= 17.9seconds Ans."
Problems on Multiplate Clutch
I Example
10.9 , A multiplate clutch has three pairs of contact sutfaces. The older and
inner radii of the contact sutftices are 100 mm and 50 mm respectively. The axial spring
force is limited to 1kN. Assuming uniform wear, find the power transmitted al1500 r.p.IIL
Take II = 0.35.
Given Data:
n = 3;
W = I kN
e Solution:
'1
=
=
100 mm
1000 N;
=
N
0.1 m;
=
'2
= 50 mm =
1500 r.p.m.;
50 x 10-3 m ;
J.1= 0.35.
For uniform wear, torque transmitted is given by
T = n·J.1·W
('1 + '2)
2
- 3 x 0.35 x 1000 ( 0.1 00 + 0.50)
2
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= 78.75 N-m
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power transmitted,
1500 x 78.75
~7t X
-
•
't
60
10.10 A muJtlplate disc c/~
-
:: 12.37 kW
Alii. ~
,L
en trans",.
• ,,1 o//riction f~r the friction sU1:facesis o. tis 55 Ie", of POHle, III 1&
c,e/ftcJe kN/"r. The Internal radius is 80
1. Axial /nleIUJ#o' III
00 '.p.1II.
...Jill 60
mIll and is 0. 7 .
"J
P'esslUe is t
f}P"~_ltLr 0/plates needed to transmit the refJu. d
· tlllleslite exte,nal
no. to
,1J1P'., lre, torque.
raulllS.F",d
;t 'h't" Datil:
P = 55 k W = 55 x 1()3 W; NG
p
=) 60 kN/m2
_ 16
- 1800 r.p.m.;
== O.J "
max
0 X J()3 N/m2 ,•
,..
,,,I'
II
1'2
=
80 mm = 80 x 10-3 m .
Il"0Pmd: Number ofpJates
1'2 == 0.71',
,
needed to transmit the re qUI'red torque.
@Solulion:
1'2 = 0.71',
or
r,
"
"
1'2
,
or -; == 0.7
80 x 10-3
0.7
=0.JJ43m
1'2
= 0.7
'2
or - == 0 7
=
Assuminguniform wear, axial force exerted is given by
W = 27tC (I', We know that the maximum
..
intensity of pressure (Pmax) is at the inner radius ('2)'
Pmax . r2 = C or C
Then,
1'2)
= 160 x
1()3 x 80 x 10-3
=
12800 N/m
W = 21tC (I', - r2)
= 21t x 12800 (0.1143 - 0.08)
=
2758.57 N
Torquetransmitted by a single friction surface is given by
(I', + 1'2)
T=f.L·W·
2
_ 0.1 x 2758.57 x
{O.J J 43 + 0.081
2
.:
Torque required per surface, T = 26.8 N-m
The total
I d as given below.
torque required can be calcu ate
2nNl
power, P == - 60
2n x 1800 xl
55 x 103 == -
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60
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1~~
~~
_!.0.20
or
__
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----------------------_D--~~ig~n_O~if_n_r_a_m_m~U~~~funs
~
T -
T~tal torque required,
291.78 N-m
Total torque required
Number of friction surfaces required = Torque required per surface
=
Total number of lates
=
2i~~;8
=
'" (10.10)
10.887 ~ 11
Number of airs of contact surface + 1
'" (10.11)
= 11 + I = 12 surfaces
Hence, there wiD be 12 total plates, in which driving and driven shafts having Jix
plates each. Ans. ~
I Example
10.11
I A multi-disc
clutch has three discs on the driving shaft and two 0"
the driven shaft is to be designed for a machine tool, driven by an electric motor of 22 kW
running at 1440 r.p.m: The inside diameter of the contact surface is 130 nun. The
maximum pressure between the surfaces is limited to 0.1 Nlmm2. Design the dutch. Take
J.l = 0.3 ; n I = 3 ; n2 = 2.
r2
Given Data: P = 22 kW = 22 x ]03 W;
= 65 mm ;Pmax = 0.1 N/mm2 = 0.1 x 106 N/m2•
N
=
1440 r.p.m.;
d2 = 130 rnrn or
To find: Design the clutch (i.e., determine the outside diameter of disc, total number of
discs, and clamping force).
© Solution: Assume uniform wear.
I. Outside ~iameter of disc (d J: We know that the torque transmitted,
T
=
Design torque, [T]
where Service factor , ks
=
P x 60
21tN
T x ks
=
k, +~+~
=
0.5
From Table 10.2,
From Table 10.3,
From Table 10.4',
From Table 10.5,
k.
~
= 1.25
=
22 x 103 x 60
2 x 1t x 1440 = 145.89 N-m
+k4
(for electric motor)
(for machine tools)
k3 = 0.38
k4
=
Ks
=
(for 1440 r.p.m.)
075
(
.
.
assummg 32 engagements / shift)
0.5 + 1.25 + 0.38 + 0.75 = 2.88
Design torq~e, [T] = 145.89 x 2.88 = 420.16 N-m
We know that maximum intensity of pressure
.
'.
. r _
(Pmax) ISat the Inner radius (,~. Therefore,
max
.
P
2 C or C = 0.1 x 106 x 65 x 10-3 = 6500 N/m
For umform wear, axial force exerted'
.
IS given by
W = 21tC(', - '2)
= 21tx 6500 (,
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65
•-
3
x 10-
)
=
40840.7 ('. _ 0.065)
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.. , (i)
r~
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that number of pairs of contact suJ'.
'.Iaces ,
o~
WeId1
n - n 1 + n2 - 1 == 3 +
~
.11
2- J
"forque
transmitted, [TJ
=
n·".
W (~
;3) ..
==
4
420. J 6 = 4 x 0.3 x 40840 7 (
. r - 0.065)
420.16 = 24504.42 (1'2 _ 4 225
,
.
x J 0-3)
= 0.14618 m or 146.18 mm
(I'
_!
+00
65)
2 .
T,
~router diameter of the disc, d, = 292.37 mm Ans•."'CI
"
l TolOI number of disc:
Total number of disc -
Number of pairs of contact surface + I = n + 1
4 + 1 = S ADS. "
J. C/(JIIIjJ;ng force (or axial force exerted) :
substitutingthe value of
T, in equation
(0, we get
Axialforce exerted, W = 40840.7 (0.14618 - 0.065)
=
3315.45 N
ADS. ~
[§xample 10.12 , A multiplate clutch consisting of 6 plates, each plate 0/ external
_Ier 150 mm and internal diameter 100 nun, is to transmit 7.5 kWat 900 r.p.m.
Assuming jJ = 0.1, determine the pressure on each qJective pair 0/ sUrfacesin contact.
Given Data : Number of plate, np = 6; d, = 150 mm or
r, = 75 mm =
P = 7.5kW
@Solution:
75 x I cr3 m;
= 7.5
x
d2 = 100 mm or
JoJ W; N
900 r.p.m.; I.L
=
As given, number of plates, np
:. Number of pair of surfaces in contact,
We know that the power transmitted,
=
'2
= 50 mm = 50 x 10-3m ;
=
0.1.
6
n = np - I = 6 - J = 5
21tNT
P = 60
21t x 900 x T
7.5 x J03 =
60
Total 1hctlon
'.
. d, T = 79.57 N-rn
torque to be rransmitte
Smceno assumption is given, we assume UnJifCrrn wear.
+ '2)
T J!' W x
2
(75 x 10-3 + 50 x JO-~
Or
,
(I',
79.57
-
O.J x W x
W
-
1273J N
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2
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10.22
Design o/Trarumuslon SYltt""
----.;:_
Total axial force or pressure on contact surfaces = W = ]273 I N
]273] - 2~"£
But n = 5, so pressure on each effective surface =
5
- "'""' N ADJ. ""
, Example 10.13 , A muJtiplate clutch has three discs on the driving shaft and two
0"
the driven shaft The outside diameter o/the contact sutfaces is 240 nun and ilUitk
diameter is 120 mm. Assume uniform wear coefficiem offriction as 0.3, find the I1IIVcintIUlr
axial intensity of pressure between the discs/or transmitting 25 kW at 1575 r.p.m.
Given Data:
n, = 3;
n2 = 2;
r,
d, = 240mmor
= J20mm=O.12m;
d2 = 120 mm or r2 = 60 mm = 60 x 10-3 m;
e Solution:
= 0.3;
P = 25 kW = 25 x ] ()3 W;
Number of pairs of contact surfaces,
n = n, + n2 -] = 3 + 2 - 1 = 4
J.L
Power transmitted, P
25
x
=
N = J775 r.p.m.
21tNT
60
103 = 21t (1575) T
60
T = 151.6 N-m
For unifonn wear, torque transmitted is given by
T =
151.6 =
or
w
=
11 •
f.I •
w
r' ; J
'2
4xO.1 x Wx (0.12;0.06)
1404 N
The axial force exerted (W) can also be given by
W =
or
or
21tC (r, - r2)
W = 21t x Pmax X r (r
2 l
]404
=
Pmax =
-
r2)
21t x Pmax x 0.06 (0. ]2 - 0.16)
62.07 x I()3 N/m2
=
62 kN/m2
Ans. "
10.8. CONE CLUTCH
In a cone clutch, contact surfaces are in th h
.
e s ape of cones
h
. .
o cones A and B are 10 contact when the cl t h .
' as SOwn rn Fig.l 0.5(a). The
tight with the help of springs. Thus torque]
u c . IS engaged. The contact is complete and
fro .
.' e IS transmitted d .
ictron cones from driving shaft to driven sh ft F
. urmg engagement of clutch through
pulled back by means of a lever system co a . ~r disengaging, the clutch of the cone B is
friction surface in these cone clutches The lo pressmg the springs. There is only one pair of
.
s ope angle a
.
vanes from 80 to ]50.
tw
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~
~-----~::::~~----------------
JI~O~.2~3
Clutch pedal
Driven shaft
SprIngs
Driving (conical surfaces)
'----
Driven
Fig. 10.5. (a) Cone elutel.
10.9. .Design of a Cone Clutch (Torque transmitted on the cone clutch)
1
consider a pair of friction surface as shown in Fig.l O.S(b). Fig.l O.S(c) shows a small
elementalring of radius r and thickness dr.
Cone surface
dr
(d)
(c)
(b)
Fig. 10.5.
Let
P
=
n
Normal intensity of pressure on friction surface,
.
c.
angle of the cone or the angle of the friction
Semi-angie of cone or lace
surface with the axis of the clutch,
a
=
r)
= Outer radius of friction surface,
r2
=
Inner radius of friction surface,
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...:.1~O~.2~4
D__u~~~n_o~if~Tro~mus·
-~S}'.Jt
rl
R = Mean radius of the friction surface =
fl
=
~
+ r2
2
'
Coefficient of friction between contact surfaces, and
b = Face width of clutch plate.
Let x is length of ring of friction surface. Then from the geometry of Fig. IO.5(d),
dr
=
sin
X·
or x = dr- cosec a
Ct.
27tr . x = 27tr· dr . cosec a
Area of the ring, A =
The design of conical clutch is done based on anyone of the following a.ssumptions.
(i) When there is a uniform pressure, and
(ii) When there is a uniform wear.
(i) Considering uniform pressure:
Normal load on the ring, 'OWn = Normal pressure x Area of ring
'OWn = Pn
27tr· dr - cosec a
X
Axial load acting on the ring, 'OW = Horizontal component of 'OWn
= 'OWnX sin a
=
Pn
27tr . dr . cosec a x sin a
=
27tPn . r . dr
X
Total axial load taken by the clutch,
f'. 27tr·dr·P
W=
n
'2
=
W =
or
Pn
=
[ r2 ] r.
27tPn 2" r2
»», [2r.
[y2 _y2 ]
= 27tPn
•
2
2
-r2 2 ]
W
2
7t[r 1
-
... (10.12)
2
r2 ]
The frictional force on the ring acting tangentially at radius r,
F, = fl' 'OWn = u- Pn
X
27tr· cosec a· dr
., Frictional torque acting on the ring,
T, = F, x r = 21tfl' Pn • cosec a· r2 dr
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102S
'1
..
'
ts1 frictional torque
~'
T
-
f
27tJ.l . P
"2
to
~ 2ltll' p.'
.
" . COsec
a.
coseca[ t
-_ 2 7t1l.p.
,.. "COsec a
r
[,11-'21]2
. ring the value of Pn from equation (10.12)
~~w
2
r . dr
3
t~~
[r~~ ri
1 x cosec a
T - ~ J.1 w . cosec a
[r~ - r} ]
T = 2ltll x
It
[ r! ~ ri ]
... (10.13)
rl-'2
or
[
or
where
J..l • W . R cosec a
TR
=
2
3'
[rI3-r23]rl
rl2 _
=
I
Mean radius of the friction plate
(ii) Consideringuniform wear: Let us assume p is the intensity of pressure
C
Weknow that
P . r = C = constant or p = -;
at
radius r.
The normal load acting on the ring = 'OWn
= Normal pressure x Area of ring
'OWn = p,' 21tr' dr- cosec a
'OW == 'OWn' sin a
_ Pr' 21tr . dr . cosec a .sin a
AXialload acting on the ring,
== p, 21U . dr
== C x 2rcr . dr == 21t C . dr
r
C
[.,' p,
= -;]
:. Totalaxial load transmitted to the clutch,
f 2ltC . dr
'1
W =
'I == 2rcC (rl- r2)
= 2lt C [ r 1'2
.. , (10.14)
w _
or
C ==
2tt ('1 - '2)
The fro .
• • iven by
Ictaonaiforce acting on the f1ng IS gr'
dr . cosec a
F = 11. IiW. = 11' p.: 2lt
r
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.1~O=.2~6
D_u~~~n~o~if~Tr~a~~~m·
.
-
la!la"
S
~
and frictional torque acting on the ring is given by
T
=
F r x r = J.1' Pr . 2nr . dr . cosec a x r
=
J.1x - x 2nr2 • dr . cosec a
=
21tJ.1 . C cosec a x r . dr
C
r
:. Total frictional torque acting on the clutch,
rl
T =
f 21tJ.1· C .
cosec a .r . dr
r2
= 21tJ.1· C . cosec
[ r~-2 r; ]
a
Substituting the value ofC from equation (10.14), we get
W
T
I
= 21tJ.1· 27t (rl
T = 1"
INote)
x cosec a
R =
rl + r2
2
=
[r12 -rl]
2
'1;'2 ]
cosec a [
J.1. W . R . cosec a
or
where
w·
_ r2)
'" (10.15)
I
Mean radius of the friction plate
Axial force required at the engagement of clutch is given by
or
WI!
WIt (sin a + J.i . cos a)
We
WIt (I + J.i . COl a)
... (10.16)
and Axial force required at the disengagement is given b)
Wei
01"
Wd
=0
WIt (sin a - II . cos a)
=
WIt (I - II cot a)
... (10.17)
10.9.2. The Semi<one angle is normally taken as 12.5°. Why?
We know that the torque capacity
is inversely proportional
to sin a. The value of a
should be as small as possible. But when a is less than the angle of static friction (+), the
has a tendency to grab, resulting in self-engagement. The self-engagement is not
desirable because the clutch should engage or disengage only at the operator's will.
clutch
To avoid self-engagement
and to facilitate disengagemen~
a >
Angle of static friction (.)
a >
tarr '
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(J.1)
... (': ~. tan.)
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J{!
. .
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From : www.EasyEngineering.net
J/'
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etlicient of friction as 0.2 w
the cO
, e get
tan-1 (0.2)
a >
f~iflg
•
I
•
~hesemi-cone angle
IS
Or a::> I I 30
as 12
taken
0
•
fore, .'
.5 .
1f!ete
/0.14 Determine the axial fi
Ie
orce re .
t.t:d, 25 k1J' ofpower at 750 r-p.m: Average n ~'lIred to ellgage a
:Jf,IIg Ie 100 and coefficient of friction 025.fi ctloll diameter of th
COileclutch
('(Jllt an,
. .
e COileis 400
Data:
P == 25 k W = 25 x 103 W .
mm,
t6iVtll
d == 400 mrn
or
.
power transmitted,
.
@ solullon :
200
r:::
p:::
N::: 750
'
mm ::: 02
_2nNT'
_
r·P·m.;
.
m ; a:::: 100;
J.L ::::
60
0.25.
25 x 103 == 2n x 750 x T
60
-
T = 3 18.3 I N-m
or
Nonnal load acting on friction surface can be obtained by
T
W
or
==
J.!' Wn·
==
L
==
11r
n
r
318.31
0.25 x 0.2 = 6366.2 N
The axialforce required to engage the cone clutch is given by
+ J.! cos a) = 6366.2 (sin 10° + 0.25 cos 10°)
2672.85 N Ans. -e
We == W n (sin a
==
IExample10.15
, The following data relate to a cone clutch : Minimum and maximum
'llface contact radii are 125 mm and 150 mm respectively; Semi-cone angle = 200;
,Ylo~able normal pressure is 14 x 1(11Nlm2; p = 0.25. Find (a) The axial load, and
~)The powertransmitted, if the speed is 700 r.p.m.
GivenData : r} == ISO mm = 0.15 m ;
r2 == 125 mm = 0.125 m;
a:: 20°;
J.!
Pn = 14 x 104 N/m2;
0.25;
==
N
= 700 r.p.m,
@So/ution:(a) The axial load transmitted to the clutch:
W _
(6) Po
To
rrPn(rl
2
2)
x 14x 104[(0.150)2-(0.125)2]
==
-r2
3023.78 N Ans.
1t
"'61
wer transmitted:
rque, T == -2
3
II
r-
3
3 ]
'1 -'2
W
[
,2 _,2
1
2
.
cosec a.
115\3] cosec 200
{9 150)3 - (0
_ .!:.;:.L-
304.73 N-m
::::
== ~ x 0.25 x 3023.78 [ (0:150)2 - (0.125)2
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_
D~u~~~n~O~if~ff~Q~~~m~;.
'
1_O_.2_8
...$,o" S
Power transmitted,
P
=
=
27t NT
60
=
27t x 700
x
~
304.73
60
22.34 kW Ans. ~
•.
I A torque of 350 N-m is transmitted through a cone c/Ulchh .
.
mean duuneter of 300 mm and a semi-cone angle of 15~ The maximum normal P'us""
[ Example 10.16
~
the mean radius is 150 kNI",z. The coefficient of friction is 0.3. Calculate the width lit lit
oJ 'ltf
contact surface. Also find the axial force to engage the clutch.
Given Data:
= 350 N-m;
T
Pn = ISO kN/m2
a = 15°;
@ Solution:
D = 300 mm or R = ISO mm = 0.15
= ISO x
t()3 N/m2;
m .,
f.1 = OJ.
(i) Width of the contact surface:
Torque transmitted,
or
T
=
J.l' W n . R
350
=
Wn
=
0.3xWnxO.15
7778 N
Let ' b' be the width of the contact surface.
We know that the normal load acting on the friction surface,
W n = Intensity of cross-sectional pressure x Area of frictional surface
Wn=Pnx(2nRxb)
or
7778
Face width, b
=
=
ISO x 103 x 2n x 0.15 x b
0.055 m or 55 mm Ans. ~
(ii) Axial force to engage the clutch:
We = Wn (sin a + J.l cos a) = 7778 (sin 15° + 0.3 cos 15°)
I Example
10.17
I A leather faced
= 4.267 kN
Ans."
conical friction clutch has a cone angle of 30 ~ The
intensity of pressure between the contact surface is not to exceed 6 x 1(J4Nlm2 and the
breadth of the conical surface is not to be greater than 113 of the mean radius if p = 0.20
and the clutch transmits 37 kW at 2000 r.p.m: Find the dimensions of contact surface.
Given Data:
a
Pn = 6 x 104 N/m2
= 30°;
R
b = 3
,
;
J.l = 0.2 ;
P
- 37 kW = 37 x 103 W;
N
= 2000
r.p.m.
R
Tofind: Dimensions of contact surface ('1 and '2)'
P
=
21tN T
60
37 x 103
=
21t x 2000 x T
60
@ Solution :Power transmitted,
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Fig. 10.6.
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.
T == 176.66 N-m
Assuming servIce factor, ks == 2.5,
of
Design torque, [T1 == 176 .6 6X25. - 441 6
. 5 N-m
srTlitted is also given by, T == 2
,(1Ue tran
1tJ.l P" . R2 . b
rOI'f
== 27t x 0 2
441.65
. x6x l04xR2(E)_
R
Fsccw
3 - 25132.74 R3
R
m or 259.98 mm
h == - _ ~25998
3 3 -
idth is given by,
== 0.25998
== 0.08666 m or 86.66 mm
From Fig.10.6, we find that
==
or
r
-'2
== b- sin a == 0.08666 x sin 150
r
-'2
== 0.02242
Mean radius, R ==
and
sin a
m
... (i)
r +'2
2
= 0.25998 m
r +'2 == 0.5199m
or
... (ii)
Solvingequations (i) and (ii), we get
Outerradius of contact surface, r = 0.2711 m or 271.1 mm and
Inner radius of contact surface, '2 = 0.2487 m or 248.7 mADS. "
[Example10.18 , A cone clutch with a semi-cone angle of 150 transmits 10 kW
01
The normal pressure between the surfaces in contact is not to exceed
1600 iNlm2.The widtll of tile friction surfaces is Iial/ of the mean diameter.Assume p =
Us.Determine:
i80 r.p.m.
(i) The outer and inner diameters of the plate, and
. (ii) The axial force 10 engage tile clutcll.
GIVe" Data: a == 150'
P == 10 k W = lOx J 03 W; N
,
D
=
P" == 100 kNlm2
(ii) Axial force to engage the clutch (We)'
SOllllio
~1Ja1 to th
'
(i) Outer and inner diameters of the plate (d, and d2), and
lO/l"d.,
e
b - - 2
2
100 x 103 N/m;
= 600 r.p.m. ;
= R' Jl = 0.25.
":
(i) Outer and inner diameters o/Ihe p
.'
h t face width b is
late' It IS gIVen t a
.
e lIlean radius of the friction surface R.
..
~ekn
Ow that ,
h -
R
. s problem]
[Refer prevlOU
sin a
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Design a/Trans"" ... ~:
.~~
_!OJO
-----------------_;:::.--=::.....::...:...::.:..:::
'1 -'2
So,
sin
\
'1 + '2
\
2
I
=
a
=
'1 -'2
'1 =
or
to
sin a
Mean radius, R =
and
"Slon
'1 - '2
b =
or
~;
1.296 '2
21tNT
60
P -
We know that the power transmitted,
21t x 600 x T
60
10 x 103 =
T = 159N-m
or
Assuming service factor k, = 2.5,
=
Design torque, [T]
=
T x ks
159 x 2.5
=
397.5 N-m
Assume uniform wear.
Since intensity of pressure is maximum at the inner radius, so
Pmax
X
C
'2 =
Torque transmitted, T
= ~W
. cosec
(l
[71 ; 72 ]
But W = 21t C (r - '2) = 21t Pmax x '2 (" - '2)
T
Then,
or
or
1 cosec
(l
[71 ; 72 ]
=
J..L
397.5
=
0.25 x 1t x 100 x t03 x cosec 15° x '2 [ (1.296 '2)1 - r~]
397.5
=
2.062 x 105
=
=
0.1244 m or 124.4 mm
=
=
1.296'2
Inner diameter
and
72 (71 - 72)
[T]
'2
or
= ~ [2n Pm ax X
r
Outer diameter
1tPmax . cosec a . '2 (', 2 -, 2
2)
,~
248.8 mm Ans. '"
=
322.6 mm
1.296 x 0.1244
=
0.1613 m or 161.3 mm
Ans."
(ii) Axial force required:
W
=
=
21tC ('I - '2)
=
21tPmax'2 ('1 - '2)
21t x 100 x 103 x 0.0917 (0.1188 - 0.0917)
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= 1561 N
ADS.
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-e
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s
The semi-cone
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angle
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Ie'.
a Cone clutch .
~
",ean diameter of 80 1111n. CO'I"I: •
IS 12.$0 Il d
/laJ'ea
J
e.,JIClento/fi .
II the COlltllCt
jpS
'I! required to proauce slipping of th ,I
I'lction is 0.32, DJ
/orqll<
e c,ulch /01' an'
. "'/rat is tile
.~i1IJJIII .1. is used to connect an electric 1110101' • 'h
OXlal/ol'ce0/200 N ?
r e clutell
I
WII. a stalio
tth u..1 to attain the fU,'/ speed and the enel'gy losl '" . naryflY",heel, ",IIat' ,L
etle
.1' •
•
uUl'lngsli" .
IS tne
:.til. !Ie
dthe
moment
oJ
Inertia
of
the
flywheel
is
0.
~
k
.,JPIIIK
r
Motol'SfJeet'.
,.- ",. all
•., g_"';'
r
IS
10.19
-,
,r
,lAf,p'
". ell /) ta:
GiV
a
12 50.
-
a -
.
D - 80
,
-
U
mm or R::::40
mm :::: 40 x 10-3 .
Jl - 0.32;
W = 200 N; N:::: 900
m,
. .
.
r.p.m.; I:::: 0.4 kg-m2•
·
(i)
MInimum
torque
reqUired
to
produce
slippin
fth
10flIId •
g 0 e clutch
(ii) Time needed to attain the full speed, and
'
(iii) Energy lost during slipping.
@Solulion: (i) Minimum
torque required to Pl'oduce slipping:
Weknow that the torque required to produce slipping,
Jl . W . R cosec ex
T or
0.32 x 200 x 40 x 10-3 x cosec 15°
T -
11.828N-m Ans."
(ii) Timeneeded to attain the full speed:
We know that,
or
[where a = Angular acceleration 1
T
-
I a
11.828
-
0.4 x ex
29.57 rad/s2
a -
.. Angular speed of flywheel,
co
= a·
I -
or
I
co _
I
=~ -
(27rN)
60
1
a
21t x 900).
_j_ = 31.187
60
x 29.57
S
(
ADS.
-e
(iii) Ellergylost during slipping:
Angle turned by driving shaft, 9, - co I ==
21t x 900 x 3.187
60
== 300.4 rad
Angle t
urned Ly driven shaft, 92 Energy lost in friction
_
coo' I
I
7 x (3. I 871
1 2 _ 0 + - x 29.5
- al 2
+2
150.2 rad
e ) ::::'11.828
T (e, - 2
_ 1777 N-IP
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(300.4 _ 150.2)
AdS. ~
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~
Design of r,.al7SlI1' ,
10.32
13Sl0'J
S
~
10.10. CENTRIFUGAL
CLUTCH
Centrifugal clutch is being increasingly used in automobiles and machines. Ob .
.
carri the sPider
.
VloUsl,
works on the principle of centrifugal
force. T h e drinvmg s h a ft carnes
h ~,n
r
.
II
h
'.
,s oes
springs while the driven shaft IS connected to the pu ey, as s own In Flg.IO.7.
~
r
Pulley
Brake lining
oriven
'\
Spring
~
J-------------
--
VI/~
~
-----J
,
\_
~
Driven shaft
-Cover
,\\,I:~'1
Fig. 10.7. Centrifugal clutch
The shoes are mounted radially and springs keep them away from inner rim of the pulley,
Shoes have some mass. As the speed of the driving shaft rises, the centrifugal forceonthe
shoes increases causing them to move radially outward within the guides provided.
When the centrifugal force is less than the spring force, brake lining cannot makeany
contact with the pulley rim. But when the centrifugal force is equal to the spring force,the
shoe is just floating. When the centrifugal force just exceeds the spring force, the shoemoves
outward and comes into contact with the driven member and presses against it to transmitthe
torque.
10.10.1. Design of a Centrifugal
Clutch
1. Mass of the shoes:
Let
n = Number of shoes,
=
R =
r =
N =
m
(0
(01
Jl
Mass of each shoe,
Inside radius of the pulley rim,
Distance of centre of gravity of the shoes from the centre of the spider,
Running speed of the pulley,
= Angular speed of the pulley =
2:ON,
= Angular speed at which the engagement begins to take place, and
= Coefficient of friction between the shoe and rim.
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'1
I
torque transmitted ' T = nXI1(F
= I1(Fc- Fs) R
."d rollli frictional
-F)s R = F
.,
I
C
Icll
torque transmitted' expressIon, the mass ofthn- . R ... (1018)
fJOlll !be tO
.
ona
~~.
e
shoes
(m)
can
be
fr
llll
Fe
Fig. 10.8.
1.Size of the shoes:I ::: Contact length of the shoes,
Let
b ::: Width of the shoes,
R '" Contact radius of the shoes.
e '" Angle subtended by the shoes at the centre of thespider.and
p '" Intensity of pressure exerted on the shoe.
,., (10.19)
e
I - R·
From Fig.l 0.8 ,
•• rea of contact of the shoe. A '"
Net force acting on the shoe '"
/. b
Fe - Fs '" A xp lne'" /. b' P
r:=IS
From tho expression, the width of the shoes (b) can be oblll. h d . •11l ricJiDn.1S kg tJIId
~/e
1
I
h eOC hlJ1l g a tJIIISS OJ
"'u,U. 0.20 A centrifugal clldch hUSfour s oes
of thef
.urflJ£t 411
is
:~...... ;,ntre of gravity at a radiUS of 60"''''' The dwn;tt;:OO r.PO""/)eIe,.",;nethl
10
"" _.. e cllllch is to trans",it 6 kif' pOHIer ttl II speed of",. 11 Is gf{1llUJlity ;ncrt/lS
fro
leh lit
eed of rot...
J~
_'<II
ust be exerted bv each spring. I/ the sp
, rake p'" 0. , •
• /11 w6
J
•• ",it torque ·
at speed will the clutch begins to trll'~
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_~~
Design ofT'a" .........
----------------------------------~=-~~J~~i~~
~
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_!_O.34
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'}}':Ie
n
Give" Data:'
Toji"d:
_ 5kg·
m -
= 4·
r = 60mm
,
=
60x IO-3m·
R = 75 rnm = 75 x 10-3 m;
D = ISO mrn or
p = 6 kW = 6 x 103 W ; N = 1000 r.p.m.;
.) Force exerted by each spring, Fs,
,
;
J.l == 0.35
and
(I
(ii) Speed at which transmission begin, N.
@ Solutio,,: (i) Force exerted by each spring:
p
=
6 x 103
=
We know that,
T =
or
2ltN T
60
2lt
X
1000 x T
60
57.29N-m
Total frictional torque transmitted can also be given by
T
=
n· (Fe - Fs) . J.L. R
57.29
=
4 x (Fe - Fs) x 0.35 x 75 x 10-3
. Radial force required at each shoe,
Fe - Fs
=
Centrifugal force, Fe =
=
545.67 N
m (1)2 r
=
5x (
2lt x 1000)2
60
x 60 x 10-3
3289.9 N
Then the force exerted by the spring is obtained by
Fe - Fs =
Fs
or
=
545.67
Fe-545.67
= 3289.9-545.67
(ii) Speed at which transmission begin: Transmission
= 2744.1N Aas.~
will begin when centrifugalforc
is equal to spring force.
When m (2:~
i.e.,
J
r = Fs = 2744.2
5 x 4 x lt2 x N2 x 60 x 10-3
3600
or
I Example
= 2744.2
N -
10.21
I A centrifugal
913.31 r.p.m,
Ans. ~
clutch is to transmit 15 kW at 900 r.p.m. ThtshotS~
four in number. The speed at which the engagement begins is .yJh of the rU/lnings~
lie
The inside radius of the pulley rim is 150 mm and the centre of gravity of the shoe.
120 mm from the centre 0/ the spider. The shoes are lined with Ferrodo f01
wit,,"
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· tOf/riction may be taken as 0.25 D
.diclt"
'1.
b '" db'~ the silo
eter",' lire: J t.t
10 3
.A/.jJ'
:f angle su tenae
.5
cr' hotS, ".
es at the c• lIss of th
;tS
the slloes IS 0.1 Nlmml.
"'re Oft/'e'
e shoes, a d
¢fill
sP.der is 600
n 2. Si7,e 'if
t
Data:
P - 15 kW == 15 x 103
and the prus
Give"
W; N
ure
R = 150 mrn == 0.15 m ;
:::: 900 r,p.rn.; n::: 4.
0"
_
e
roJi"d:
r :::: 12
= 60°;
0 rnrn ::: 0.12 rn.'
p :::: 0.1 N/rnrn2 _
'.l'= 0.25 ;
- 0.1 x I06N/m2
(i) Mass of the shoes (rn), and
(ii) Size of the shoes (b),
@solution: (i) Mass of the shoes (m) :'
Angular speed,
(0
= 27tN
==
27t x 900
60
60
== 94.26
rad/s
Given that the speed at which the engagement b egms
. (co ). :y,th f
I IS 4 - 0 runningspeed ( )
3
3
ro,
(01
== 4" (0 == 4" x 94.26 == 70.7 rad/s
power transmitted,
P
=
15 x 103
=
T =
or
27tNT
60
2x 900 x T
60
159N-m
Centrifugalforce on each shoes,
Fe
=
m
(02
r
=
m (94.26)2 0.12 = 1066 m N
Springforce on each shoe, i.e., the centrifugal force at the engagementspeed (I)"
Fs
=
m «(O1}2 r
=
m (70.7)2 0.12
=
600 m N
Frictionalforce acting tangentially on each shoe,
F
=
Jl(Fe-F )
s
= 0.25 (1066m-600m)
= 116.5mN
We knowthat the torque transmitted,
T 159 _
Or
m -
n·F·B
4 x 116.5 m x 0.15 = 70 m
2.27 kg Ans. ~
1! rad
600 :::: 60° x 180 - 3
. f hoe
ContactradiUS 0 S
db the shoe x
Angle subtende Y
'It
15 :::::0.1571 m
'It
(U)Site of the shoes:
e-
Contact length of shoe
i.e "
-
I -
_
e. R ::::3' x o.
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------------------------D_u_,~gn~Of~TJ~ra~~..~
II'O~.~36~
-
I.f& 10"
We
Fe - Fs =
know that,
b
I Example
=
"""
~Ie~
l- b· P
i
6
1066 m - 600 m = {).1571 x b x 0.1 x 10
466m = 466 }(2.27 = 0.1571 x b
or
S
I
x
0.1
x
106
0.0673 m or 67.3 mm Ans. ~
I
10.22 A centrifugal friction clutch has a driving member c .
•
onsUUn
spider carrying four shoes which are kept from contact wIth the clutch case by , 0/ Q
flat springs until increase of centrifugal force overcomes the resistance of the! ~Q1tt Of
the power is transmitted byfriction between the shoes and the case.
rp"ng, lIIfd
Determine the necessary mass of each shoe if 22.5 kW is to be transmitted
with engagement beginning at 75% of the running speed The inside d!n_-* III 750,.p.""
.
-,ac,e, Of th
IS 300 mm and the radial resistance of the centre of gravity of each shoe from th
e tit".,
is 125 mm. Assume IJ = 0.25.
e ShaftIltir
Given Data:
n = 4;
N
= 750 r.p.m. ;
D
= 300 mm or
r
Tofind:
P
22.5 kW
=
=
22.5 x 103 W ,.
N I = 75% N or COl
= 75% co;
R = 150 mm = 0.15 m ,
.
= 125 mm = 0.125 m ; J.l = 0.25
Mass of each shoe (m) :
@Solution :
Given that,
(1)
(1)1
Power transmitted , P
22.5 x 103
= 27tN
=
60
=
27t x 750
60
= 78.54 rad/s
0.75 co = 0.75 x 78.54
= 58.91 rad/s
= 27tN T
60
= 27t x 750 x T
60
or
286.48 N-m
Centrifugal force acting on each shoe ,
T
=
Fe =
m co2r
=
m (78.54)2
x
0.125
771.06 m N
Spring force on each shoe i e th
.
given by
, . ., e centnfugal force at the engagement speed
=
Fs = m (co~) r = m (58.91)2 x 0.125 ==
433.79 m N
Frictional-force on each shoe
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-
J.L (Fc - Fs)
-
0.25 (771.06 m - 433.79 m) == 84.317111
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.
(I), 15
r
.
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'lfl":,
~I
tsl torq
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10.37
.
by
ue transm itte d' IS given
8~II~eto
T
=
n- F . R
286.48 = 4 (84.317 m) 0.15
m
=
5.66 kg Ans. ~
or
NAL EXPANDING RIM CLUTCHES
INff~
.
.
implies, the Internal expanding rim clutch tra
.
'(5 name
hown i F'
nsmus
torque due to the
AS I
. t rnal rim, as sown
m Ig.I0.9.
'00 of 10 e
!~f1.
~Sl
ents of internal expanding rim clutches are:
Three ele'" .
. .
I theroating friction surface,
I_
the means of transmitting the torque to and from the surfaces, and
I the actuating mechanism.
From Fig.l0.9, it is unders.tood th~t the enga~ement or disengagement of external and
. al rimsis achieved by usmg a suitable actuatmg mechanism. The outer diameter of the
mtem
I
I'
di
.
.temal rimis slightly lesser t ian t re inner
rarneter of the external rim. As the internal
rim
~tes,it expands. The actuating force is controlled by a suitable actuating mechanism.
Because of this expansion of internal rim, it is engaged with external rim.
Forthe disengagement of the two rims, the actuating force is applied on the internal rim in
~opposite
direction. As the internal rim contracts, it automatically disengages from external
mn.
I acting rim clutch
Fig. J 0.9. An internal expanding centrifuga
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Design o/Transm; .
------------------~-.:!.....:...:...::.:.:
_!O.38
10 11 1 Types of Internal expan
~
$$10"
,
S
.
ding rim clutches
. ..
.
mechanism,
Depending upon the operatmg
.
I
the mterna expanding
rim clutche
S
are
classified as
(i) Expanding-ring clutch,
(ii) Centrifugal clutch,
(iii) Magnetic clutch, and
(iv) Hydraulic and pneumatic clutches.
(i) Expanding-ring clutclles :
./ They are mostly used in textile machi~~ry, excavators,
the clutch may be located within the driving pulley .
and machine tools where
./
They transmit high torque even at low speeds .
./'
They require both positive engagement and more release force.
{li) Centrifugal chaches :
./
They are used mostly for automatic operation .
./
If no spring is used, the torque transmitted is proportional to the square of the
speed. This is particularly useful for el cctric-motor during starting. Because
without producing shock, the driven machine comes to the driving member.
./
Springs can also. be used to prevent engagement
been reached, but some shock may occur.
until a certain motor speedhas
(iii) Magnetic clutches:
./
They are mostly used in automatic and remote-control
systems .
./
They are also useful in drives subject to complex load cycles .
./'
They use magnetic forces to couple the rotating
actuating force for a friction clutch.
members
or to providethe
(iv) Hydraulic and pneumatic clutches:
./
They are also useful in drives having complex
machinery, or in robots .
loading cycles and in automat~
./'
In these clutches, the fluid flow can be controlled remotely using solenoid valves.
INo'~lln the arrangement shown in Fig.lO.9, if the external rim is fixed then the devicefunctions
as a internal expanding rim brake. For force analysis and other details, refer Chapter II.
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CJ~/&
1Z·
~~R
NAL CONTRACTING RIM CLUTCHE
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S
ction, arrangement and working f
jD· onstfll
d"
0 external
.(he c
. terna) expan 109 rim clutches exc
contracting .
P'
the In
.'
ept that th
run clutch
, 'Ist'to.
external rim instead of expanding int
. e actuating fore'
. es are
SIII1I(ltJ'llctiOg.
ernal nm.
e ISprovIded by
~eco
I contractmg clutch that is engag d b
!J1 e~tdern.~as shown in Fig.I 0.1 O.
e y expanding the flexible tube .
sse arr,
wIth
ll1pre
ca
sists of three elements: the mating fricf
I
IsO COO
c.
rona surface th
It a
and from the surfaces, and the actuating
hani
' e means of transmitt'
rqUe to
mec enism
mg
~w
.
Classification of Actuating Mechanisms
.
.
Theactuating or operatmg mechanisms are classified as :
10.12.1.
(i) solenoids,
(ii) Levers, linkages, or toggle devices,
(iii) Linkages with spring loading, and
(iv) Hydraulic and pneumatic devices.
10.12.2. Working
From Fig.10.10, it is understood that the
externalrim is rigidly bolted with the outer
casing. Whenever the compressed air is fed into
theflexibletube, the external rim contracts (i.e.,
theflexible tube expands internally because it
cannot expand externally). This contraction of
external rim provides the required clamping
force. Thus the driving and driven members are
engaged. In order to disengage these members,
~e pressureof the compressed air is regulated by
asuitable control mechanism.
~
asan ext
Fig. 10.10. An external contracting clutch
In the arrangement shown in Fig.1 0.1 0, if the internal rim is fixed,thenthe devicefunctions
. d h d tails referChapter 11.
emal contracting rim brake. For force analYSISan ot er e ,
10,13. ENERGY DISSIPATION DURING CLUTCHING (ENERGYCONSIDER_A~IONS)
\Vh
k done (against frictional forces
n
OpPo .e one solid body slides over another, most of the wor C nsequently the temperature
SlOg th
.
t the interface. 0
h
oftl._
e motion) will be liberated as heat a
even destroy the clute .
rubbin
. .
d
temperature
may
E
g surface will increase. This mcrease
I es of surface temperature
us
ven und
. h .nstantaneo va u
(about 10 er moderate and slow speeds, hig 1
f h two surfaces.
OQoC)
I ontact 0 t e
are reached at points of actua c
'lit
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II~O~.4~O
_;:
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----------------------------~D~e~s~~nofn
rans ..., ~;
""SS'
/'
/o"S
10.13.1. Effect of temperature rise in clutches aJ.e·
./
./
:Yll~
Most premature clutch failures can be resulted due to
eXees.
temperature.
Stve s
off·
In case of metallic clutch plates, high temperature existing at th
~
may cause the individual plates to be welded together.
e rUbbing int
err~
./
In non-metallic or semi-metallic
excessive wear.
clutch plates, the high, tern
perature
'
can ~St
./
Distortion of the shape of the plates .
./
Surface cracks in solid metallic plates caused by thermal stresses.
./
In case of lubricated clutches, oxidation of oil resulting in the f
deposits on the working surfaces and in grooves.
Onnation Of
10.13.2. Derivation of Energy Eq uation
The operation of a simple clutch is depicted as a mathematical model of
a
system in Fig. I0.11.
Shaft 1
.
twO-fOe'
nij
Shaft 2
-6_f_:ngin~-i-L·H-~-·F~~~+6
.
W1
11
~
12
(1)2
Clutch
Fig. 10.1 I. Mathematical model of a clutch
Let
IJ and
wJ
= Moment of inertia and angular velocity of the driving memoo
(engine) respectively,
12 and
and
002 = Moment of inertia and angular velocity of the driven member(103li)
respectively.
We know that when the clutch plates are brought in contact some slippage occursbetw~
them before the speeds of driving and driven shafts become equal. The kinetic energywill~
absorbed in work by friction during the clutching operation. The design of the clutchsyst~m
.
t muchn~
should be such that the energy thus absorbed must dissipate fast so that there IS no .' "
.
fri ti n IImng, tti
m temperature. The uncontrolled rise in temperature may cause damage to nc 10
desired to calculate the total energy which will cause the temperature rise.
oue 10
It is assumed that the two shafts are rigid and that the clutch torque is constant.
clutching operation, the shaft I is decelerated
d28
II . dt2 = - T
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and its equation of motion will be
[~l
.: T:= I a::: I . dl2
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,J)
T'
,
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~.
011 the
as
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other hand, elutching will induce
acceleration in sha ft 2 and its
d2
isgi~.o
10.41
.
equatIon of motion
9
12 . dt2
=
T
re e :::: Angle displacement in time 'I'
weh
.
.n.e
angular velocity of shaft 1 is determined bY .mtegratin th
1J'
deI
_T
g e equation (i).
2
or
dt2
=
de,
dl
=
,
-1-
-T
I,
'"
(ii)
'" [From (i)]
.t + C
de,
But at
C =
I:::: 0,
del
-dt
=
dl = co,
-T
-
+ co I
.I
II
... (iii)
Similarly for shaft 2,
de2
... (iv)
=
dt
Thedifference in velocities (i.e., relative velocity) is given by
del
de
dt
=
dl - dl
= (~~
=
de2
x 1+
01, - 002-
T
001) - (~
('III ~ :: )
x I+ ~ )
I
..,(10.20)
. Tillierequired for complete operation {tJ: The clutching operation is completed at the
Instant in w hiich the two angular velocities
~t
Then
COl
and
become equal.
. operation
.
I, = Time required for the entire
0)
= 0 when
(1),
= (1)2 • Hence equation (10.20) can be written as
I, .12 (ro, - ro~
t
I
(1)2
... (10.21)
-
T (I, + 12)
The abo
. ed £. th gagement operation is directly
PrOpo.
ve equation shows that the time reqUlr tor e eo
rtlonal to the velocity difference and inversely proportional to the torque.
,-
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IIO~.~42~--------------------------------~D~~~~n~~Q
~
11.tltli!!iO
l ment (8)· By integrating the equation (10.20) on
II ~'Ftl
Angular uup ace
.
.
"
, e can ti
.
e d urm. g clutching operation over a time mterva\ of t).
Ind 11.._
displacement
"It ~
..I'
e
=
,')
I
=
de
o
T (I + I'
I
(00)-002)dl-~
~
JI
1)'12
0
o
"d,
or
Substituting the value of
e
I) .
=
we get
I),
12 (00) - (02)2
T(I)+12)
or
'" (10.22)
Energy dissipated during clutching operation (E) :
The work done by torque T or the energy dissipated during clutching operation' .
E
or
E
IS glven~
.
=
T·e
=
I) .12
1
2 x
(00) - (02)2
(I) + '2)
'" (\0,23)
The above equation indicates that energy dissipated is independent of clutch torquebm
directly proportional to square of the difference of the angular velocities of drivinganddriven
shafts at the start of clutching.
10.14. TEMPERATURE
RISE
The temperature rise of the clutch can be calculated by using the relation
I aT
where
= {;
I
.1T = Temperature
.. , (\0.24)
rise in "C,
E = Energy dissipated by the clutch,
C = Specific heat of clutch material,
and
m = Mass of the clutch.
h
hen a clute
The above temperature-rise equation can be used to explain what happens W
vera!
.
'
.
ffi
t due to se
b
k
or ra e rs operated. Yet the actual situation may be very much d) eren
variables being involved.
'~l~,
entl
The clutch heated to a temperature T I cools to an ambient temperature T a expOn
as shown in Fig.l 0.12.
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I
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T
··
l~'~t
·
..
T,
A
. ....
'
AT
1
.
2Te",perature-tlmerelationshipfor
a
--
l.
c utch aJi~r ill h
second clutchmg operation after time I wh
C tc Ing 0P~'1lI1011.I
.
B en I~"'p~ralur~ rtJ••
rott1t [11I~ S"OtffJ
(""I 0.1.
f,&'
TIme (I)
.
.
.
f I
'
IUIC~d 10 T"
T
this figure, time 0 c utchmg (I) is ve
'
J
'
t
Hrf
A
In
.'
ry small so the tem
.
. ntaneous I.e., at time lA' temperature rises f
T
perature nse is shoUhO>.
instB
rom D. to T .
..."
rature
drops
along
the
curve
ABC
unless
interne
t
d
b
,Instantaneously.
Then
the
pe
tern
"
P e y another clut h'
.
second operation occurs at time I B' then the temperat
'1"
c tng operation. If
a
ure WI nse to T2 as shown in dotted
line.
1D,15.ENERGY DISSIPATION
ONLY IS CONSIDERED
WHEN MOMENT OF INERTIA OF DRIVEN MACHINE
In the preceding section, the energy dissipated from the clutch is calculated by considering
ilie moment of inertias of both driving and driven machines (i.e., I, and '2)' The problem
recomesrelatively simpler if moment of inertia of driving side is not considered. For this
condition,
the following relations can be used.
12 . (1),
Time required
Energy dissipated
for completing
clutching operation,
I, =
. cute
I hi109, E
in the form of heat during
= -2I I2
I
and the angle turned through
I
.,
haft during
by the d nvmg s
II'
... (10.25)
T
2
(1)
.. , (1016)
I
2
(I) I
e = 2 12 . T
." (10.27)
d ;ves a ",achin~IhrouK", a
r;;. at 350 r.p."" r
tis
L!:a"'ple 10.23 An electric motor rotatmg
h is engaged illakes J stcO_n
p/tue l
. Wilen the clute
If 'nerlia 0/ drIven
(. C UlchWhose both sides are effectIve.
The ",olllenl OJ I
Also
,0, th
d of ",olor.
d its power.
e driven mae/line to attain the spee
" d bv Ihe ",olor an
"och'
prouuee ;,
COIc lilt is 4.5 kg-m2. Calculate the torque
llIQ1e
the energy dissipated by the clutell.
21t 1350) _ 36 65 radls; II ::::J sec ;
_~
==~60
- .
Gille" D
ata , N I - 350 r.p.m.; WI - 60
Where
tati
no ticns have usual mean rngs.
I
12 -
4.5 kg-m2.
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~----------------~--~~
~44
Tojind:
De.rigno/r,UIU'
...,
'"
d by the motor (1'),
(i) Torque Produce
(ii) power developed (P), and
'ssl'pated by the clutch (E).
(iii) Energy d I
@Solulion:
(i) Torque produced by II,e molor ('T) :.
.
.
uired for completmg the clutching Operation (
We know that the time req
COns~.
only 12),
... [using eqUation(I
12 • 0>1
or
Torque, T
Power developed (P) :
Power, P
We know that
=
=
II
0.2~)1,
I
4.5 x 36.65
3
=
54.98 N-m Ans. ~
=
Tx
I
(iJ)
0>
= 54.98 x 36.65 = 2014.94 W Ans."
(iii) Energy dissipated by the clutch (E) :
We know that
E.
1
= '2
=
'21
2
... [from equation(10.26)]
120>.
x 4.5 x (36.65)2 =·3022.25 J Ans.,.
I
I Example
Ilf
10.24
in example 10.23, the moment ~/ inertia of electric motor is U
kg-m2, calculate the same quantities, assuming that driving tr'flChine startsfrom mt II'm
clutch is engaged and takes 1.B sec to anain motor speed. Also compare Ihe t~
dissipated in two cases.
Given Data:
II =
1.4 kg-m2; I. = 1.8 sec
@Solution :
(i) Torque produced by the motor (7) : We know that the time required for completi",
the clutching operation (considering both J I and 12),
I = I •. 12 (co I - CO2)
.I
T (II + 12)
Given that the driven shaft starts from rest. i.e.,
•••
0>2 =
[from equation(IO.211]
O.
From example 10.23, .12 = 4.5 kg-m- ; 0>1 = 36.65 rad/s
Substituting these values in 'II' equation, we get
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_
1.4 x 4.5 (36.65
T (1.4 + 4.5)
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1.8
Torque,
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21.74 N-m Ans. ~
T -
rI
r developed (P) :
~) for;t h t
power,
Wekf1
Q1
p
owt a
= T x (co. - co2)
=
21.74 x (36.65 - 0) = 796.83 W
ADs.~
~ dissipated (EJJ: The energy dissipated is given b
~~~
y
E2 ~
!
x
2
_! x
2
I.· 12 (co. - co2)2
(I. + 12)
•..
[from equation (10.23)J
1.4 x 4.5 (36.65 - 0)2
(1.4 + 4.5)
= 717.14 J Au • ...,
(M Comparison:
E.
~
3022.25
= 717.14
= 4.214 Ans . ...,
It shoWSthat the moment of inertia of the driving machine heJps to reduce the energy
dissipation
and hence rise in temperature.
REVIEW AND SUMMARY
I
, Atthe beginning of this chapter, functions and principle of operation of the clutch are
presented
, The threeimportant types offriction clutches i.e., disc or plate clutches, cone clutches
and centrifugal clutches are discussed, in detail.
Single PlateClutch :
{ Torquetransmitted on the single plate clutch is given by
where
T
f.i
=
f.i.W·R
= Coefficient offriction,
W = Axial thrust exerted by the springs,
R = Mean radius offriction surface.
_! [ r/-r/]
- 3 r/-rl
r, + r2
[Considering uniformpressure)
[Considering uniform wear}
2
I ...J,'uso'jriction surface, and
r, - Externa raa. 'J •
r 2 = Internal radius offriction sur/ace.
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I
J
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Design o/r,aru",#&
~1~O_'4_6
------------------------------------~
Multiplat~ Clutch:
, ,
./
itted on the multiple plate clutch IS given by
Toi'qtletransmt
T = n- J.l . W ,R
where n = Number of pair of contact surfaces == n +
n I = Number of discs on the driving shaft. ~
"2 - J
n1 = Number of discs on the driven shaft,
./
Axial force to engage the clutch. W = 2nC (rI - rJ)
./
Average pressure on thefriction surface is given by
Total force on/riction swface
p
= Cross-sectional area offriction surface
W
=
n(r/ -ri)
(IV
,
Number offriction surfaces required
=
Total torque required
Torque required per surface and
Total number of plates = Number of pairs of contact surface + J
Cone Clutch:
./
Torque transmitted on the cone clutch is given by
T = J.l' W . R cosec a
where R = 2
3
=
./
[rl-rl]
r/-ri
[Considering uniform pressure]
('J; '1)
[Considering uniform wear]
Axial force required at the engagement of clutch is given by
We = Wn (sin a + J.l cos a)
and axialforce required at the disengagement is given by
Wd = Wn (sin a- p. cos a)
Centrifugal Clutch:
Tofind mass of the shoe:
(i)
T
=
where F
T == Totalfrictional torque transmitted.
=
=
=
Number of shoes.
Centrifugal force on each shoe.
Spring force exerted by each spring on theshoe,
= Inside radius of the pulley rim. and
p
=
.
CoeffiCient of friction between the shoe and rlln.
~
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I
I
I
I
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· 1 ~:-.-~~----__
!
II
fO
ft
10,41
"d size of the shoes:
--------
_
Fe - Fs - I· b .p
where
1·= Contact length of the h
s oes
b - Wi'dh
I t. of the shoes, and
'
p = Intensity of
,
pressure exerted on the shoe
'[he internal expandmg and external contracting rim clutc
.
, c/VJ[lltr.
.
hes are also presented in this
dissipation during clutching (considerinae}@
I dI'-~
7:
~
I, Ti1lltrequired for complete
"
operation,
'I
=
1(1/+1,)
, Angulardisplacement during clutching operation, 0
, I Energydissipated during clutching operation,
E =
=
1x
2
1x
I
I
,I
fIJi
11 .12((8/ -
2
((J)/ - (l)J)2
T (II + I:;)
1/ .12 ((1)1 - (J)J)]
T(II + I:;)
Eatrgy dissipation during clutching (considering I] only) :
. t/
, , limerequiredfior comp I'eting cIutc hitng operouon,
'I
I
~'
'
=
I] . (1)/
T
1
I ' Angulardisplacement during clutching operation, 0
1 ,
:/
1/ . I]
=
~
'2 I] . T
Energydissipated during clutching operation, E = ~ I] ~
where
1/ and I] = Mass moments of inertia of driving and
driven machines respectively,
(1)/ and (I)] = Angular velocities of driving and driven
machines respectively, and
T = Torque produced.
(
Tttnptrature rise is given by
, I
LJT =-
E
em
di innted by the clutch,
E = Energy ISSIr'0/, and
C = specific heat 0/ clutch mate" ,
m
=
Mass
0/ the clutch.
I
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Design a/T"ans",. .
10.48
!»Ie~
~
REVIEW QUESTIONS
I.
What are the functions of a clutch?
2.
Enumerate the various types of clutches.
3.
Write an engineering brief about the friction materials used in clutches.
4.
List out the required qualities of a good friction material.
5.
Explain with a neat sketch the working of a single plate clutch. Derive an eq
.
h
the torque transmitted by the single plate clute .
6.
.
uatlon fI
~
Describe with a neat sketch a multiplate clutch and explain how the number of J
P ales art
calculated in the multiplate clutch.
7.
Which of the two assumptions - uniform intensity of pressure or unifonn rate of w
would you make use of in describing clutch and why?
ear,
8.
Show that the torque transmitted in case of a cone clutch is given by,
T
9.
=
2
-3 Il W . cosec a
[rI3-r23J
2
2
rl
=r:
with usual notations.
Explain with a neat sketch the working of a centrifugal clutch. Deduce an expressi~n f~
the total torque transmitted.
10. Why service factor must be considered while designing a clutch?
II. The semi-cone angle in cone clutches is normally taken as J 2.5°. Why?
J 2. Contrast internal expanding and external contracting
rim clutches.
13. List out the actuating mechanisms used in external contracting
rim clutches.
14. While designing a clutch, energy dissipation during clutching should be considered.
Why?
15. What are the effects of temperature rise in clutches?
16. Deduce an expression for energy dissipated during clutching operation.
17. Write short notes on 'temperature
rise'.
PROBLEMS FOR PRACTICE
Problems on single pkue clutch:
J.
A single plate clutch, with both sides effective,
has outer and inner diameters
300:
the:~ ;
and 200. mm respectively. The maximum intensity of pressure at any point in
surface IS not. to exceed 0.1 N/mm2. If the coefficient of friction is OJ. dete~9kWl
power transmitted by a clutch at a speed 2500 r.p.m.
[AIlS: 6).
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!
!
,#. ale P
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clutch, both sides
From : www.EasyEngineering.net 1049
effective Downloaded
d
.
, an of 300
.
A S,tJD diatTleter transmits power such that the
. mrn Outside di
t ;~side JOO x 1()4 N/m2. Assuming uniform nlllJurnurn pressure ameter and 100 mm
I e%~if1e the maximum power that can be tran:e~ and COetlicien;:;he. Pl~te is not to
1
tfe~
of pressure at the outer radius of the I mltted at 1500 r,p.m AI frlctron of 0.35,
/
;~tenSJtY
.
p ate. fAns : 3438
. so determine the
. Ie plate clutch, effective on both'd'
. 3 kW; 33.3 x 1()4N
A slflg
.
h
Sl es, IS req .
1m2]
').,. r p.m. Determme t e outer and inner radii of . . Ulred to transmit 25
f 3~ ~ is 0.255, the ratio of radii is 1.25 and th frlctl~nal surface if the coeffi ~W at
(rICtlO
determi
he axl
e maxlmu
rcrentof
""!rnm2 Also etermine t e axial thrust to be
m pressure is not t
I 0 II"
.
provided b'
0 exceed
juniform wear.
rA .y springs. Assume the theory
lOt.
ns : 120 mm . 96
'ngle plate clutch with both sides effective t
'.
,mm;
1447 N1
I A 51.
, ransmJSSlon is 10 kW
. rilue developed IS 120 N-m. Axial pressure is not t
. The maximum
0 exceed 100 kNl 2 0
.
I
f frJ'ctionplate IS 1.3 times the inner diameter' De.'
m. uter drameter
0
•
,ennlne the d'
.
d
axial
force
exerted
by
the
springs.
Assume
unitConn
d,mens,ons
of the plate
an
"ear an Il = 0.25.
I
to.,
".
I
lOt
fAns: 268.98 mm; 206.9 mm; 2.017 kW]
frObI- onmultiplate clutch:
5. A multiple disc clutch has five plates having four pairs of active friction surfaces. If the
intensity of pressure is not to exceed 0.127 N/mm2, find the power transmitted at
500i.p.m. The outer and inner radii of friction surfaces are 125 mm and 75 mm
respectively.Assume uniform wear and Il = OJ.
fAns: 18.8 kWJ
I
~. A multiplate clutch has six plates each of outer and inner diameters of 150 mm and 100
I
mmrespectively. Assuming Jl = 0.3 and uniform intensity of pressure of30 x J()4 N/m2,
determinethe power transmitted at 1200 r.p.m.
fAns: 46.88 kW]
11. A mUltipledisc clutch has 5 plates having four active frictional surfaces. Determine the
axialintensity of pressure between the discs transmitting 37.3 kW at 500 r.p.m. The
I .
...
.
75
d 125 mm respectively. Assu.ne
lOnerand outer radii of friction surface are
mm an
2
unl'~onn
d
rAns: 25.3 x J ()4 N/m ]
,
It
wear an 1.1 = 0.3.
t·
r.
.
.
2 kW at 3000 r.p.m. The plates
A multJplate friction clutch is required to transmit 89.5
."
The coefficient of
,I are I
d they run In or .
a temately of steel and phosphor bronze an
. 0 8 times the radius of the
I &icti .
fro ti surface IS •
on IS 0.08. The internal radius of the IC on
JI\.f N/m2. If the maximum
e~- I
. I' 't d to 20 x U"
be f
d:'''''na surface. The axial pressure IS nm e
250 mm d.:termine the num r 0
laJneterof the frictional surfaces is not to exceed'
fAns: JOJ
! Plates reqUired.
I
I'
0" COile
dutch:
.01. The
seOli-cone angle !s
e bearing surface rs
~COni '.
. 90 kW at J500 r.p
200 cal fr,ction clutch is used to transmit th mean diameter of thti d the dimensions
2
3iS and the COefficient of frictic!~ is 0.2. If e eed 0.25 N/mm , '"
QlIn
• not to exc
Ofthe
~d the intensity of pressure IS. load required.
785 mm ; 5045 NJ
COnIcal bearing surface and the ax.al
fAns: J 96.5 mm; J •
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10.50
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Design o/TI'QlUI1J'
"'8;011 S
~I~
)O. A leather faced conical clutch has a cone angle of 30°. If the intensity f
between the contact surfaces is limited to 0.35 N/mm2 and the breadth 0 Ptes
surface is not to exceed one-third of the mean radius, find the dimensio of the Coo~t
.
ns Ofth ~
surfaces to transmit 22.5 kW at 2000 r.p.m. Assume Uniform rate of w e ~
coefficient of friction as 0.15.
[Ans: 103.27 Inrn~9 8fId take
.
di
, 4.73
II. Power is transmitted by a cone clutch of mean rameter 200 mrn, face width
111m]
clutch angle of 15° (semi-cone angie). The coefficient of friction == 0.25
Illlll, 8IId
2
pressure on the cone is not to exceed 20 x ) ()4 N/m • Calculate the In!!!':' e n0llna!
.
h
. I Co
•
OCJ\lmulll
transmitted at 750 r.p.m. Also determme t e axia rorce required to hold
~er
Assume uniform pressure.
[Ans: 17.15 kW~e CIIitch.
12. A cone clutch with cone angle 20° is to transmit 7.5 kW at 750 r.p.m. Th .78 kN]
intensity of pressure between the contact faces is not to exceed 0.12 Nt e ~onnal
th
coefficient of friction is 0.2. If face width is 1/5 of mean diameter, find (i)~~ . ~
dimensions of the clutch, and (ii) Axial force required while running.
e main
[Ans: b = 46.8 mm ; r, = 125 mm ; r2 = 109 rnm . 139
ri
,
S N]
Problems on centrifugal clutch:
13. A centrifugal clutch has four shoes with an inside diameter of the drum 300 mm and
radial distance of the centre of gravity of each shoe from the shaft is 120 mm . Assummg
~
J.L = 0.25, determine the necessary mass of the each shoe if 29.42 kW poweris to be
transmitted at 800 r.p.m. with engagement beginning at 75% of the running speed.
[Ans : 1.8 kg]
14. A centrifugal clutch has four shoes which slide radially in a spider keyed to thedriving
shaft and make contact with the internal cylindrical surface of a rim keyed to thedriven
shaft. When the clutch is at rest, each shoe is pulled against a stop by a springso as to
leave a radial clearance of 5 mm between the shoe and the rim. The pull exertedby the
spring is then 500 N. The mass centre of the shoe is 160 mm from the axis of theclutch.
If the internal diameter of the rim is 400 mm, the mass of each shoe is 8 kg, the stiffness
of each spring is 50 N/mm and the J.L is 0.3; then find the power transmitted by theclutch
at 500 r.p.m.
[Ans: 36.1 kW]
Problems on energy dissipation in clutches :
15. An electric motor rotating at 300 r.p.m. drives a machine through a plate whoseboth
sides are effective. When the clutch is engaged it takes 2.6 sec for the driven machi~eto
attain the speed of motor. The moment of inertia of driven machine reduced to dn7n
shaft is 4.78 kg-m-, Calculate the torque produced by the motor and its power.~;
calculate the energy dissipated from the clutch. [Ans: 57.2 N-m; 1797.22 W ; 2334. late
16. If in the previous problem, the moment of inertia of electric motor is 1.2 kg-ro2, calc~ine
the same quantities with same properties of clutch lining, assumin~ that drive~~c.AlS(1
starts from rest when clutch is engaged and takes 1.5 sec to attain motor s~
compare the energy dissipated in two cases.
.
4951
. E /~:: .
fAns: 20.05 N-m; 0.63 kW ; 427 J, ,
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r
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..An eye for an eye
nJ
o yends up malring
Brakes
the whole ""'~Id blind. .,
INTRODUCTION
- Mullbltil Ga1UIJU
.•
~lte is a mechamcal device by means of which mot'
f
.
Br~
Ion 0 a body IS retarded f I.'
Of to bring it to rest, by applying artificial frictional'
. or S owing
down
.
h b ak
reststance. In this process f
lating the motIOn, t ere
absorbs either kinetic
.
0
regu
.'
f'
energy or potential energy In
omobiles, the kinetic energy 0 the moving vehicle is absorbed by th brak I hoi .
aUt
• I
I
.
e
e. n oists and
1 ators the potentia
energy re eased by the obiects
during th e brakimg period
" IS absorbed
elev
,
'J
11.1.
by the brake. ~he energy absorbed
to the surroundmgs.
by the brake is converted into heat energy and dissipated
The capacity of any brake depends upon the unit pressure between the braking surface, the
coefficientof friction between them, velocity of brake drum, heat dissipation capacity of the
brake, etc.
A dynamometer is a brake incorporating a device to measure the frictional resistance
applied. This is used for measuring
power developed by the machine.
the driving force or torque transmitted and hence the
It may work on the principle of absorption or transmission.
11.1.1.Clutch Vs Brake
,
The functional difference between a clutch and a brake is that a clutch connects two .
movingmembers of a machine whereas a brake connects a moving member to a stationary
member.That is, if any one of the moving member of a clutch is fixed, then the device
becomesa brake.
11.2.CLASSIFICATION OF BRAKES
A classification scheme for brake is presented in Fig.II.I.
F
.
h main types of mechanical brakes:
rom Our subject point of view, the followlOg are t e
I.
Block or shoe brake
,
(i) Single block brake,
and (ii) Double block brake
2. . Band brake
,
(i) Simple band brake,
. I b d brake
and (ii) Differentia
an
_j
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_!1.2
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---------
D_e_S....;::ign=---O-=-if_1i~,.,=ans=-m.;.:_l.S:::.::.·
sian
~
8)1
$;L_
Simple
Band
Differential
~
External
Actuation
~
0
L-------'
Short shoe
Mechanical
L..--_In_temal
__
Pneumatic &
Hydraulic
....
Leading &
Trailing shoe
Electrical
Two leading
shoes
Drum
-0~
~
~
~
Automatic
DuoseNO
Full disk
Caliper disk
Electrically on
Magnetic
Electrically off
Fig. 11.1. Classification of brakes
3.
Band and blocs, brake,
4.
Internal expanding shce brake, and
S.
External contracting shoe :~-...ke.
The mechanical brakes, according
following two groups :
.t
.tion of active force, may be divided into the
(a) RadiIII b,akes : In radial brakes, the force acts radially on the drum.
Examples: Band brakes, block brase., ....ld internal expanding brakes.
(b) Axial b,aIu!s : In axial brakes, the force acts axially on the drum.
Examples : Cone brakes and disc brakes.
11.3. BRAKE LINING MATERIALS
The required qualities of a good brake lining material are :
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~
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prpkes ~.
~
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11.3
ffi .
A high and uniform coe rcient of friction.
I h ability to withstand high temperatures t th
.
.
Ire
'ty
, oge er with high heat dissipation
capac I .
I Adequate mechanical and thermal strengths.
I }-lighresistance to wear.
sistance against environmental conditions such'
.
I Re
.
,
as moisture and OIl.
Types of Brake Linings
11.3.1.
f I' .
basic types 0 nungs are:
Three
1.Organiclinings: Organic linings are generally compounded of six basic ingredients.
(i)
Asbestos: For heat resistance and high coefficient offriction.
(ii)
Friction modifier:
To give desired friction qualities. Example: The oil of
cashew nut shells.
(iii) Fillers: To control noise. Example: Rubber chips
(iv) Curing agents: To produce the desired chemical reactions during manufacture.
(v)
For improved overall braking performance. Example:
Powdered lead, brass chips, and aluminium powders.
Other materials:
(vi) Binders: For holding the ingredients together. Example: Phenolic resins.
2. Semi-metallic linings: These linings substitute iron, steel, and graphite for part of the
asbestos and organic components of an organic lining.
J. Metalliclinings: For further details, refer Chapter 10, Section 10.5.
The Table 11.1 shows the materials commonly used for brake lining and their properties,
Table 11.1. Properties of brake lining materillis
,..._
Material
t--._
WOOdon metal
Metal on metal
leather on metal
Asbe
stos on metal in oil
POWdered
C.
metal lining on
·I.IIJOil
J!
Allowable
pressure (P1JUlX> MPa
Max. Temp, -c
0.25
0.48
6S
0.25
1.4
315
0.17
65
0.35
0.40
0.34
0.15
2.8
260
260
114
'. BLOC
I . K ~R SHOE BRAKE
d
inst the rim surface of a
h s are presse agal
Otake d ')'pe of brake, one or more blocks Of s oe . bbe compound, cast iron or of any
rum. Th e blocks are made of wo od,as bestos In ru r
L
Il thiS
h,
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Design
11!1~'4~
!
O/Tri11ls",..
~
------------------------------~~~~~~jIOn~~
Tol'e""
j
j
, I of the block or shoe is softer than that of th
h
itabl metal The materia
&-.
ed
ot er SUI
e
',th'
by the application of a rorce through suitabl I'\JfrI. 'h.
blocks are pressed agamst e ram
e levtra~~
brake hanger,
Or
11.5. SINGLE BLOCK OR SHOE BRAKE
. .
,
h brake is shown in Fig. 11.2, The friction between the bl
A single block or s oe
.
.
Ock :I" ....
he' tarding of the drum, This type of brake IS cOmmonlyu~
..."'the
brake drum causes t ere,"""
on rail
trains and tram cars.
~
b
F
Drum
Fig. 11.2. Clockwise rotation of brake drum
The block is pressed against the drum by a force (F) applied on one end of a lever. The
other end of the lever is pivoted on a fixed fulcrum 0,
Let
r
RN
F
=
=
=
Radius of drum,
Normal reaction of the block,
Force applied at lever end,
J.1 = Coefficient of friction,
J.1 RN - Frictional force, and
TB
=
Braking torque.
11.6.1. When the Rotation of the Drum Is Clockwise
Fig. I1.2, shows the clockwise rotation of brake drum.
Braking torque on the drum is given by
TB
.. , (i)
= J.1 RN· r
Taking moments about pivot 0,
F . I + J.1 RN . C
F ' 1- RN . a + J.1 RN . C
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RN· a
= 0
-
.. , (ii)
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F~~
RN :: F·
or
.
-L:
Q-
... (iii)
J.lC
'wting RN value from equation (iv) in
.
SiJbstJ .
equation (i), We have
T8-1l'-
'" (iv)
F./.,.
-
Q-Ilc
'" (JJ.I)
1.51
If-lOCking Brake
....,uation (iii) the value of F is zero when Q :: If I
III ....,
r- c. t means that
h h
ly the brake is zero, then the brake is applied auto t' II
w en t. e force required
10app
ma lea y Wllell thefiWct.· alfi
. ..lrlCielll enough to QPPIythe brake whit no exle,nalfi'
'lOll Of'Ce
ISIllJr
orce, '"ell 'lie b,.aje Is said to be
stI/.JocAbrg bf'{lkL
2. Self-energizlng Brake
From equation (ii), it is observed that the moment of applied force (F .1) and the moment
of the frictional force (Jl RN . c) about 0 are in the same direction. Thus frictional force
~~) helps in applying brake. This type of brake is known as a seJ/-ellergised 1J1'tIke.
'1.5.2.When the Rotation of the Drum Is Antlclockwfse
Consider the anticlock wise rotat ion of brake drum as shown in Fig.llJ.
Normal reacti
and
n f the block, RN
Brak ing t rque,
Ta
=
=
F· ,
a+J.&c
a, F·'· r
... (11.2)
a+J.&c
•
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Design ofTran
11.6
.
.
Sy!/
~
INote I
PIIq
I. Pivoted Block or Shoe Bl'tlke (21J> 40? .
.
d th t th nonnal reaction and the frlctional force act at the mid .
It IS assume
a
e
-POintof
Generally angle of contact (29) is small. When the value of angle of Contact is
the ,
(.I. e.• w hen 29 > 40°) • the equivalent coefficient of friction. (u') is used in torque equat'IOn·IllOrethan
4J.1 sm9
.
Equivalent coefflcient of friction:
u' = 29 + sin 29
where
fl
29
=
=
Actual coefficient of friction, and
Angle of contact.
4(J<
~-
1 Dimensions ol"the block: The dimensions of the block are determined by the rei .
~
RN = p x b x w
.
where
RN
=
Normal reaction
0f
"·01.4)
t h e bl oc,k
p
= Permissible pressure between the block and the brake drum,
b
=
w
= Width of the block.
Breadth of the block (i.e., shoe), and
3. Rate of heat generated during the brakin~ a~tion : The rate of heat generated duringthebra!;"
period is equal to the rate of work done by the frictional force.
Illg
:. Rate of heat generated = Frictional force x Average velocity = fl' RN x \I
I Example
J 1.1
I A single block brake is shown
'"
(11
.S)
in Fig. 11.4. The diameterof the drll1ft
is J 80 mm and the angle of contact is 60 ~ If the operating force of 400 N is applitdat IItt
end of a lever and the coefficient of friction between the drum and the lining is 0.30,
determine
(i)
the torque that may be transmitted by the block brake,
(ii)
the rate of heat generated during the braking action,' when the initialbrake spttd
is 300 r.p.m, and
(iii)
the dimensions of the block if the intensity of pressure between the block
brake drum is 1 Nlmm2. Tile breadth of the block is twice its width.
200mm
250mm
,
I
,
I
o
'160
400N
mm
Fig. 11.4.
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QIId
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RAIIJ :
d
=
~~'
600 == 60
a ~
Z"
" [IJI :
"
J 80 mm or ,. ::: 90
JI.7
~e~
-L
o
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)( 10-,1 rn
7t
x J 800
= 3 rad;
N::: 300
-
J! ::: 0.30;
by th bl
(i) Torque transmitted
; P ~ 400
r.p,If)' h
=:
2 It'.
e ock shoe
(ii) Rate of heat generated dun b .'
rmg raJung,
(iii) Dimensions of the block.
and
o solution:
a:
~IIIC
Since the augle of contact .
,
ient of friction is given by
IS greater than 400, therefore
Il' =
4J.! s~n 9
ftJn'
== 4 x 0.30 x sin 300
29 + SID 29
7t
0.313
::
3" + sin 60°
(0 Torquetransmitted hy the hlock shoe (1) :
Let
RN = Nonnal reaction or force on the block, and
Il' RN = Braking force.
Taking moments about the fulcrum 0, we get
400 (250 + 200) + u' RN x 60
=
400 x 450 + 0.313 x RN x 60
= 200
RN x 200
RN
RN = 993.3 N
or
Braking force == 11' RN
Torque transmitted
... (i)
0.3 J3 x 993.3
=
=
by the block brake, Ta
Ta = 27.98 N-m
(ii) Rate of
and
heat generated during hraking :
Initial velocity of the drum,
VI
Final velocity of the drum,
V2
Average velocity of the drum, v
We know that
,
Rate of heat generate
_
7t toN
=
0
=
vj +
2
V2
= 310.9 N
Il'RN· r = 310.9
0.09
ADS. ~
o.!~
x 300 = 2.827 mls
=
7t
=
2.827 + 0
2
x
x
=
1.414 mls
d - Frictional force x Average velocity
== ",RNXV
=
OJ x 993.3 x 1.414
= 421.36 N-m/s or W Aas."'SJ
(iii) Dilllensions
Let
\II
hand
b
Wet
e know that'
,
of the hrake shoe:
respectively.
h b d h d width of the brake shoe
[From equation (I 1.4)J
e rea t an
bxw
., .
RN == P x
_ 2 w2
xw993.3 == 1 x 2 w
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1_
~urVI~nt
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Design ofTrans_"
)) 8
~~,~
----------------------~~~~~"'~slon
• ~.
Width, W = 22.285 mm and
Breadth, b = 2 W = 2 (22.285) == 44.57 nlln
or
I Example
11.2
I The diameter
Ails
of the brake drum of a single block shotv" ill Fi . ~
1 m: It sustains 240 N-m of torque at 400 r.p.m. TI,e coefficient o"fi'
. 'g./ I
•
....,
rICIIO,,·
Determine the required force 10 be applied when the rotation of the drum is ,.
1$
(b) counter clockwise, and the angle of contact (i). 35 ~ and (ii) 100 ~ Gi~e aJ CloCkIV
n Ihal
mm, b = 150 nun and c = 25 mm. Also find the new values of 'c' for self loc . a::::
brake.
Ie'"g Of
a
b
a
= 800 mm
b= 150mm
c= 25mm
r= 0.5 m
Fig. J 1.5. Clockwise rotation of drum
Given Data:
d
=
)
or r
= 0.5 m;
240 N-m " N == 400 r.p.rn ;
11 == 0.32; a == 800 mm == 0.8 m; b == 150 mm == 0 15 .
_
© S I, •
.
m, C - 25 mm = 0025
0 'utton : Braking torque is given by
. m.
mm
TB ==
!o.o!
TB ==
11 RN· r
240 == 0.32 x RN x 0.50
RN == 1500N
(i) When angle of contact, 2 (J == 35 0:
(a) Rotation of drum clockwise:
Taking moments about 0, we get
or
F . a == 11 RN . C + RN . b
F x 0.8 == 0.32 x 1500 x 0.025 + 1500 x 0.15
F == 296.25 N 'Ans ~
(b) Rotation of drum counter clock . .
•
:.
Taki
mg moments about O,we get
or
Wise.
RN x b == ·F· a + 11" RN . C
1500 x 0 15 == F
.F _
x 0.80 + 0.32 x 1500 x 0.025
- 266.25 N Ans. ~
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.. , (i)
I
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IIIe 01'
'c' for self-locking of the b k
'J.
ra e :
Ie)fre~
king the externally appl ied force F
I'
If~IOC
'
d
Th
must be
~(l
forse
tion of the
~ise rota
c'f
rum.
us equation (i) c be
zero. This is
.
RN . b == 0 + R an Written as
POssIble for COUnter
. I! N' c
Or bc == b == QJ1
- I!c
I!
"
angle of contact, 2 (}
';1 fhi"
0.32 == 0.469 m
== 469 mm
= 1000.
Ans. "
.
(II! th angle of contact is more than 400 the c
s'nce e
' rei ore the coeffi .
I
u' = 4 J.l sin 8
IClentof friction
28 + sin 28
4 x 0.32 x sin 500
1t
100 x 180 + sin ]00"
=
Then,the braking torque is given by, T B
240 -
Il' RN r
x RN x 0.50
= 1337 N
RN
or
0.359
= 0.359
(a) Rotation of drum clockwise:
Takingmoments about fulcrum 8, we get
F . a = RN· b + Il' RN . c
F x 0.80 = 1337 x 0.15 + 0.359 x 1337 x 0.025
F = 265.7 N ADS. ~
(6)Rolation of drum counter clockwise:
Takingmoments about fulcrum 0, we get
F . a = RN· b - Il' RN c
.., (ii)
F x 0.80 = 1337 x 0.15 - 0.359 x 1337 x 0.025
"
F = 235.7 N
ADS. ~
F(c)New Val Ill! of 'c' fOT self-locking of the brak e :
.
Or self I ki
be
This is possIble for counter
tb:k\y'
OC lng, the externally applied force F must
.zero.
ISerotation of the drum. Thus equation (ii) can be Wiltten as
o =
b
Or
C
''icrio" 0
It
RN' b - Il' RN C
.j
.
_...QJl
= J.1'
- -
== OA17 m ::: 417 Dun
0359
.'
ADS. l'
.
.1'
116. For a coe/flclellt OJ
The short shoe block brake shown In Fig. b' orbed What ac/ua/jllg
is 10 be a >S
tree it'e ' ~frICtional power of 14. 924 k W at 650 r.p.""
91lired , C
". '6 ,
. an the brake be self-loe"zn" .
......._
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~
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!IQ___-----------------------------------D-e-s~ig.~n-O~if~TJ~ra~nsrn·
01H ..
ISS1.
_!1I.10
~
a
d
a = 1.0 m
r = 0.375m
d = 0.375 m
c = 0.05 m
Fig. 11.6.
GivenData:
r
=
I-l
=
rn ; a
0.375
Toflnd:
0.3;
=
P
=
14.924
].0 m ;
c
=
=
14.924
m;
d
kW
0.05
=
3
x 10
0.375
W ; N = 650 r.p.m;
m.
(i) Actuating force (F) required, and
(ii) Checking the brake for self-locking.
@) Solution:
Let
RN
Jl RN
(i) Required actuating force:
= Normal force pressing the block against the wheel,
=
Frictional force, and
28 = Angle of contact of the block with the wheel.
Taking moments about fulcrum 0, we have
= F· a + I-l RN . c
= F· a
RN . d
or
RN (d - Jl c)
or
RN
=
F'a
d-I-lc
Braking torque is given by,
TB
=
I-l RN r
TB
=
0.3 x F x 1.0 x 0.375
0.375 - 0.3 x 0.05
TB
=
0.3125 F
Power transmitted is given by, P
=
14.924 x 103
=
or
..
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TB
=
=
Jl'
F'a
d-I-lc
." (iJ
·r
.. ' (iiJ
27t N TB
60
27t x 650 x TB
60
219.25 N-m
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.. (iii)
'
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~!
te~
~
(f
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11.11
. ns (ii) and (iii),
frOtfl
we get
2 I 9 . 25 --
eqtJ8Uo
F
OJ125 F
701.6 N Ans."
=
king condition: In equation (i) if d
fel/'/Oc
. . 0 f self-Iockmg
. distance
'
(til'J1tIS: .IS the condlllon
.
d ='
of
ase
'
~rO' .
IS C
I~th
J..l
OJ 75 m
c = OJO x 0.05
s f.l . c' th en
=
force F will be
.
negative or
0.0015 m
aJld •.I is IIotless than u c. Thus the brake cannot be self-locking.
ppe.
Ans. "tI
pOlJS
BLOCK OR DOUBLE SHOE BRAKE
LE
If,6,If on lone
block is u ed for braking, then there will be SIide thrust on the bean
y.
f h
Tloisdrawback can be removed by providing two blocks on th
.
ng 0 w eel
~aft.
,
TI'
I d bl
.
e two SIdes of the drum
~slt0' !f1 in flg.)t. 7,.' us a so ou es the braking torque . Th e d ou ble shoes on the drum'
~_
~A
the unbalanced force on the shaft, The blocks or shoes are held on the d rum~m~m
ofspring
LeI force.
S ~ Spring force required t
et the blocks on the drum,
r ::: Radiu
RNI and
RNl
fl
RNI
and fl RN I
~
f drum,
N rmal reaction and the braking f ret on the left hand side shoe, and
=
N ron a I react i
lhe drum i r tatin ' in the
I
I'
and the brak ing f ret on the right hand side shoe.
kwise dire tion. Taking moments about the fulcrum 0"
." (i)
·b
=
RNI'
i)
[)o
ta
blL silO( brfd.
Fig. II.·
M
I runl
\"e get
l
", (ii)
"I'
ill
1'1
I
ment
::::: R 2' a
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~
Design ojTransmisSio
11.12
.
.
~
ke the braking torque IS given by
In case of double block or s hoe b_r_a
__ , __ -----:::-~:_::=--:-~-;;;_-:-:::_
T
= (u RN 1 + J.l RN2) r - J.l r (RN 1 +
I
.
8
.
The values 0 f RN1 an d R N2 can be obtained from equations
substituted in equation (11.6) to get T B'
Let
=
b
g
.....
.. (11.6)
(I) and (II). This l'Ilay
be
Width of brake shoe
Then, projected bearing area of one shoe is given by
IA
=
2rb sin
e
I
'" (11.7)
Bearing pressure on the lining material is given by
Ip
". (11.8)
RN = Maximum normal ioad on any shoe.
where
INote I
I
= ~
1. Total energy to be absorbed by a brake is given by
ET
=
Change in K.E. of load + Change in P .E. of load
rotating parts
I
= 2"
m
2
(VI -
2
v2)
+
w
x
X
+
-4-
Change in K.E. of all other
)
2" I w2
'" (11.9)
2. Brakillg torque in terms of total energy absorbed by a brake is given by
TB =
where
NI
=
60 x
E-r
7txNlxt
". (11.10)
Initial speed of brake drum, and
= Time of application of brake.
I Example
11.4 J TI.e
block
brake
shown ill Fig. 11.8, is set by a spring tlun
produces a force S Oil each arch equal to
3500 N. Tke wheel diameter is 350 mm and
the angle of contact for each block is J 20 ~
Take coefficient of friction as 0.35,
determine (i) the maximum torque that IIIe
brake is capable of absorbing, and (ii) the
width of the brake shoes, if the bearing
pressure on the lining material is not to
exceed 0.3 Nlmml.
Fig. 11.8.
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~
s
it"" p
= 3500 N;
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d = 350 mm
11.13
Or
.1L -
r
== 17 S
mm ::::0.175 m .
G 200:::: 120 x 180 - 2.094 rad , J.l:::: 0.35 ;
p:::: OJ N/mm2.
'
8~ 1
1
(i) Torque absorbed by the brake
"d:
' and
fD ftI
(ii) Width of the brake shoe .
. n: Since angle of contact
"olu/IO'
@oJ'
~cient 0
f friction
.'
IS
given
b
(28) is
y
,~
,therefore th
'
=
II'
4g sin 0
20 + sin 20
r
(0 forq
greater than 400
.
e equivalent
::::! x 0.35 x sin 60°
2.094 + sin 120-;- == 0.409
ue absorbed by tile brake (T sJ :
Consider
the left hand side brake shoe:
raking moments about 02' we get
S (25 + 20) - RN2 x 20 - F2 (17.5 - 4)
.
45 S -
RN2 x 20 - F., x 135
.
- [0.!~9 -13.5 ] Fl
[.:
F2
== Jl' RN2;
F2
RN2
== Jl' ==
F2
0.409 ]
45 x 3500 = 35.4 F2
F2
or
= 4449.2 N
Consider
right hand side brake shoe :
Takingmoments about fulcrum
°
we get
1,
S (20 + 25) -
45 S
FI -
Braking torque T B is given by, T B
+ Fl (17.5 -4)
= FJ [ 0.!~9 + 13.5]
45 x 3500 Or
RNI x 20
=
R .
[ ..• F 1 = II'
r Nl >
62.39 F)
2524.1 N
(F) + F2) r
_ (2524.1 + 4449.2) x 0.175
_ 1220.32 N-m Ans. ~
(ii) JJlit/1
I.e
t
Wekn
h of the brake shoe (b) :
. mm
'b
_ Width of the brake shoes m
.
Ow that projected bearing area for one shoe,
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Design ojTransmissi
I]~I.II~4
--------------------~~~--~~
A == 2 r b sin 9
... [From equation (I 1
== 2 x 175 x b x sin 600
=
303.11 b mm2
.7
" Nonnal force on the right hand side of the shoe,
=
RNI
FI
Il'
=
2524.1
0.409
=617J.39N
and normal force on the left hand side of the shoe,
F2
RN2
=
Il'
4449.2 ._
== 0.409
-
10878.24 N
Since the maximum normal force is on the left hand side of the shoe, therefore
.
.
i:
design the shoe for RN2 i.e., the maximum normal rorce.
We
h
aVe to
We know that the bearing pressure on the lining material,
RN2
P
=
0.3 =
or
breadth of the block shoe, b
=
... [From equation (J 1.8)J
A
10878.24
303.116
119.63 mm Ans."
.I
Example J 1.5 , The layout of a double block
brake is shown in Fig.J1.9. The brake is rated at25(J
N-m at 650 r.p.m. TI,e drum diameter is 250· mm.
Assuming coefficient of friction to be 0.3 and for
conditions of service a pv value of J 000 (kPa) m/s
may be assumed Determine:
(a) Spring force'S'
and
Spring force
T
required to set the brake,
(b) Width of shoes.
Which shoe will have greater rate of wear and
what will be the ratio of rates of wear of the two
shoes?
Fig. 11.9.
Given Data: TB = 250 N-rn; N = 650 r.p.m; d = 250 rnm or r = 125 rnm ;
1.1 = 0.3 ; pv
=
1000; 20
= 110
0
= 110 x ~
180
=
1 92
.
@ ~olution : Since angle of contact (20) is greater than 400 therefore the equivaJenl
coefficient of friction is given by
,
1.1' =
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4 fJ. sin 0
29 + sin 20
4 x 0.3 x sin 550
- 1.92 + sin 1100
= 0.344
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lit
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e (S) required: Considerrig"t"
,
.1fore
and side br.
~spri:Ort1ents,about fulcrum 0" we get
like sltoe,'
,=
S (160 + 160)
f~jP8
(RN1 x 160) + FI (125 -60)
160 + 65 ]
. 320 S = F, [ 0.344
'" [..
F, = 0.604 S
• RNI -
II,silfer
left hand side brake shoe:
CD. moments about fulcrum 02' we get
rakJ08
S (160 + 160) = (RN2 x 160) - F2 (125 - 60)
= F2
320 S
=
F2
or
160
]
[ 0.344
- 65
0.8 S
Brakingtorque is given by
TB = (F) + F2) r
250 - (0.604 S + 0.8 S) x 0; 125
Spring force, S - 1424,S N Ans, ~
or
(ii) Widtlt of the brake shoe (b) "
llt
h
Width of the brake shoes in mm
=
Weknow that the projected bearing area for one shoe,
A
e
... [From equation (11. 7)]
=
2 r h sin
=
2 x 125 x h x sin 55°
==
204.79 b mm2
.. Normal force on the right hand side of the shoe,
RN J
F)
0.604 S _ 0.604 x 1424.5 == 2501.16N
= Jl' =
Il'
0.344
andnoonal force on the left hand side of the shoe,
F2
R
S~ce
~ess
We J,_
N2
R
lire.
P\QOW
= -,
N2
th
>
RNJ,
.
therefore
Jl
0.8 S
== --;-
==
Il
0.8 x 1424.5
0.344
'JJ bused
RN2 WI
e
'I
JO
•
==
3312.79N
,
lating the maximum bearmg
ca cu
.
1
I"
materia ,
at the bearing pressure on the JfiJfig
.
_ 1..6.~ x 106 N/m2
RN2
1312.?2..:= .!.§.E Nlmml - .b
P - A = 204.79 b
b
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"itF ]
I
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•_!_~ __
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-----------
D_eS_;;gn~~Of~Tr_a_n_.Jm..:..:..:=u,::.J;01J
S
_I1.16
~
0.25 )( 650 := 8.S I mls
60
60
rubbing
velocity,
V ==
and
1.376)( 108 NI
:= 16 ·18)( 106 x 8.51 :=
b
m-s
pv
b
3
= 1000 (kPa) mls := 1000)( 10 N/m-s
pv
Given that,
",376 x 108
1000 x loJ :=
b
1t )(
7tdN
== -
:...;.---
or Width of block shoe, b := 137.66 mm Ans. ~
...
.
W k
that the wear of block shoe depends upon the friction force
(II;) Wear rauo :
e now
.
F,
Wear ratio
F2
:=
F,
=
F2
As F2 > F
,
1
0.604 S = 0.755
0.8 S
Ans. ~
left hand side shoe will have greater wear.
Ans.·~
11.6.1. Design Procedure for Block Brake
1.
Calculate the total energy absorbed by the brake.
1
1
E-r = '2 m v2 + '2
2.
0)2
\
+ W· x
,
Calculate the torque capacity (or the braking.torque) by using the relation
60
where
E.r
Ts
=
Nl
:=
Initial speed of brake drum, and
:=
Time of application of brake.
t
3.
I
7tXN1xt
Calculate the initial braking power by using the relation
27tNITB
p:::
60
4.
Select (or assume) the brake drum diameter.
S.
Select the suitable brake drum and block shoe materials.
For the chosen materials,
consulting Table 11.1, the coefficient of friction is obtained.
6.
Consulting Table 11.2, calculate the induced bearing pressure (P).
-
Table 11.1. Limiting values of pv (from data book, page no. 7.130)
pv (mPa) (mls)
Operating conditions
Continuous service, poor heat dissipation
1.05
Intermittent service, poor heat dissipation
2.1
Continuous service, good heat dissipation as in oil bath
3.0
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-
-
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calculate the projected area of th
1.
I
rosily calcu ate the breadth ad'
S· I
r
I"',]
£.~~
f th
d
OtoOjectearea
& 11.6
0
e shoe b
Y Usin
n Width of til
h
e s oe, A ::: B
. .
II 17
g the rei .
e shOeb.
~
atlon, A - R.N
P
-
readth}'
lJ51f1g tit
the c
)( Width
e relation
apacity
and the tnQ;" d'
Determine
r the following data :
~ktfio.
''''e''sion,
0'
brake sheave IS mounted on th "
".J • doll"" 11_
'[)t
•
e "ru", sh Iii
t vwvci
yes downwards with a velocue ol'}}
a.,L rhe hO' .
,.J ",0
•v .., • 5 "'Ps (4'
1St tvUh it,
~ dru'" is 1.25 m. The hoist must be sto e I.,e.,1.15 IIIIs). The ./olld tvti,," 45 iN
f'II1J 0/ the drum may be neglected.
'PP d W,thi" a distance .;,tc/r dill/ltfte,o/tlrt
oJ
~elbl
G;,tlf Data : Load = 45 kN;
3.25
IN. The
1.15 rn/ .
ity d
in di
s ; D == 1.25 m; x::: 3 2S
Tofind: C apaci an main Imensions of db'
m.
,
a ou Ie block brake.
v:::
k'
.
Uletic
@Solutlon:
1. Calculationof the total energy absorbed by the brake:
The varioussources of energy to be absorbed are :
1
1
=
2
v
=
Velocity at the time of applying the brake, and
V2
=
Initial and final velocities of the load.
(b)
Potential energy
=
Weight x Vertical distance = W x x
(c)
Kinetic energy of rotation
(a) Kinetic energy of translation
where
VI
and
= ~
2 m (v~ - v~)
mvl =
2
I00
1
1
,.:;'_ = -2 m
Total energy, A-rr
(2
VI -
v2 ) + W . x + 2- I ro2
2
Neglecting
the kinetic energy of the drum,
1
= 2
Er
Initial velocity of loa d ,
final velocity of load,
and
'.
==
VI
m
(2
_ v2 ) + W . x
VI
2
1.15 m/s
V2 == 0
1
~
(1.152 - ()2) +
== -2 x 9.81
,.:;'_
A-rJ'
.. ' [Given]
(45000 x 3.25)
;\1lS. -S'
== 149.283 kN-Pl
.
2. C
~~:
.
r torque cap
Q/cU/ationof braking torque. (0
d' given by
Btakin
.
g torque in terms of energy
absorbe
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IS
.
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where
T8
:=
NJ
':=
t
:=
xNJ
xl
1t
Initial speed of brake drum, and
Time of application of brake.
x
Given that, the distance travelled by the load,
=
3.25 m
1
x =
'2
3.25
=
'2
Time of application of brake, t
=
5.652 sec
(v J + v2) t
But we know that,
1
(1.15 + 0) x t
or
or
Initial speed of brake
Braking torque, T 8
=
drum, N I =
60 x 149.283 x 103
n x 17.57 x 5.652
60 x vI
[.: v, ~~
nO
=
60 x 1.15
n x 1.25
=
28.71 kN-m
=
1
17.57 r.p.m.
Ans.~
3. Calculation of initial brakingpower:
We know that,
Braking power, P =
=
2n Nt Ts
60
2n x 17.57 x 28.71 x 103
60
= 52.82
kW Ans.~
4. Selection of brake drum diameter: Assume a brake drum diameter = 1.Sm
.
I
5. Selection of brake drum and block shoe material: A cast iron brake drumand
sintered metal block shoe may be chosen, from Table 11.1.
.From Table 11.1, safe value of coefficient of friction, f..l
= 0.15.
6. Selection of induced bearing pressure: From Table 11.2, for continuous service,poor
heat condition,
or
But from Table 11.1,
=
pv
s 1.05 (MPa) mls
p
Pmax
1.05
s ill s 0.913 MPa
=. 2 8 MP a
Therefore let us use, bearing pressure p
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1.05 (MPa) mls is selected.
pv
=
2.5 MPa.
'[
,
;
'
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../,r,iOll 0/ 'profected area or Iltl1l It
~ekfJO
""'--.
"J
~ S. From
oe: : www.EasyEngineering.net
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()IC/II'"
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RN
w that induced bearing pressure
A
, P ;::,
RN =:: N
O~/fu~
A _
' and
- Projected
Ii"d RAI: Assume equal friction fore
area of the ~h
fo1''' ,.
e on each sh
~ oe.
~
<>e.
Braking torque
=::
F
28.71 x 103 == F
Friction force, F
or
R
Normal reaction,
==
D
"2
x2
l.5
2
x
x
2
19140 N
==
N
x
E :::
191~
JJ
0.15
. d area of the shoe, A == ~R
Therefore, projecte
=:
127.6kN
127.6 x 1()3
2.5 x 106
I. Calculation of hread/I, and widllt of lite shoe' A
. b
==
P
~twice its width (w).
.
Projected area of the shoe, A
0.051
or
Width,
and
Breadth,
.f!!ij]
The above problem
=
Breadth
=
2 w2
x
ssummg
Width
==
m2
== 0.051
readth(h) of the block ~
2 HI X
HI
= 2 w2
W = O.J 5968 m or 159.68 mm AnJ."
b = 2w
== 2 x 159.68
=
J19J7 mm AIIJ."
'can also be solved by selecting the initial brake drum
diamettr and
~ldthfrom the Table 11.3, and checking it for permissible induced bearing pressure. If the dtsign is
00/ safe, then increase brake drum diameter
and width, and again check it for permissible inductd
~g pressure. Continue the same procedure till the safe design has been reached.
But in th
.
.'
d be . g pressure ;s cbostn and tbtn
1.....
e above problem, first the safe permissible. mduce
arm
lIf~edrum d'
.
lameter and wIdth are calculated.
'll. eiNNO'
BRAKE
'
. .
T:
In band b
d ' belt is
JiaJria]f rake system, a flexible ban or
.
Y
Wra d
A force IS
applied
ppe around a brake drum..
1he d at One end of the band through lever mechanIsm.
It 1'Ilrn rota
. an! When
-efo".
tes and the band remains statJonuoJ'
d
~
Ce IS a
l'
I
the ban
Pres
pp Jed at the free end of the ever,
th
k.. Sed a .
ThuS e
''I,'liOI)
gamst the surface of the drum.
L. retard
F' 11 10.
c..er
s Or stops the motion. Refer Jg... t side,
TJ :::: Tension in the band on tJgh
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T
I
Fit. /1.16.
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J
t9
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Design o/Transm' .
11.20
vY,J1ellJ.t
~.
T2
=
Tension in the band on slack side,
1.1
Coefficient of friction between the band and the drum,
e
=
=
r
=
Radius of the drum.
Angle of lap, and
Then, we know that,
···(11.11)
~
and the brake torque on the drum,
~--------------I
T B = (T 1 - T 2) x r
'" (11.12)
Types of band brakes:
1. Simple band brakes, and
2. Differential band brakes.
11.7.1. Simple Band Brakes
Fig.II.II
shows a simple band brake. The band or rope is wrapped round the cylindrical
drum. When a force F is applied to the lever at B, the lever turns about the fulcrum pin 0 and
tightens the band on the drum and hence the brakes are applied. The friction betweenthe
band and the drum provides the braking force.
a
a
B
B
"-
lever
Lever
Slack side
Tight side
(a) Clockwise rotation of drum
(b) Anticlockwise rotation of drum
Fig. 11.11. Simple band brake
Let
F = Force applied on the lever,
r = Radius of the drum,
t =
Thickness of the band,
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:::: Effective radius of the band Downloaded
:::::r + ! From : www.EasyEngineering.net
~
2'
a :::: Length of lever ::::OB, and
b :::: Distance between fulcrum 0 and .
. of the drum, takmg
. morne POint
,,,vise rotanon
t b A:::: OA .
IOC~yY
n S a Out0
for C
F .a = T .b
[R Ii'
, We gel
1
F :::: T1.- b
or
e er Flg.l1.11(a)1
a
braking torque of the drum is given by
r = [TI - 2L]r [N .
e~e
eglectlng the thickness of band]
- T, [ I - e~9]r = r , ~ (I -;!o) r
Ta
j1IeIl
···(11.13)
=
(T, - T2)
F anticlockwise rotation of the drum, taking moments about 0
or
F· a
or
F
=
T2· b
= T2·;
.we get
b
In thiscase, tight and slack sides of the band will be interchanged.
IN.tel
'" (J 1.14)
If the pennissib Ie tensile stress (0") for the material of the band is given, then maximum
!elISion in the band is given by
where
T,
=
cr· w . t
w
=
Width of the band, and
t = Thickness of the band.
11.7.2.Design
Procedure for Band Brakes
l. Calculate the hraking torque required from the data given.
2. If not given, select the suitable diameter (D) of the brake drum, conSUlting
Table11.3.
r-- Table 11.3. Dimensions of brake drum (from dala book,page no. 7.98)
~er
of the motor, kW
7.36
11.04
14.72
25.76
44.16
73.6
110.4
184
Brake drum diameter, mm
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160
200
250
320
400
500
630
800
Brake drum width, mm
50
65
80
100
125
160
200
250
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Design o/Transm'
11.22
-----------------------...:::_~.....::..:.~
.
~
3.
Determine the tight and slack side tensions.
4.
Calculate the thickness (t) of band. Take thickness of band as 0.005 x O.
'arneter
brake drum.
5.
.
S
ISSlon
of
Calculate the band width (w) based on the induced tensile stress (cr)
following relations:
I .
Induced tensile stress, crt
where
=
Tl
w
x t
'" (iUS)
T1
=
Tight side tension in the band,
w
=
Width of the band, and
t
=
Thickness of the band = 0.005 D.
Permissible tensile stress for steel band, [crt]
T
Using w ~ $ [crt],
...
Use the
=
50 to 80 N/mm2
width of band is obtained.
.,.
6. Check for bearing pressure: Calculate the maximum bearing pressure betweenband
and drum using the relation
Pmax
=
r
=
where
... (11.16)
w·r
Radius of the drum
Now compare the calculated bearing pressure p with the safe permissible bearing pressure
[P] obtained from Table 11.4. Ifp < [p], then the design is safe and satisfactory.
7. Calculate theforce to be applied at the end of the lever.
Table 11.4. Safe bearing pressure in band brakes (from data book, page no. 7.98)
Materials of the rubbing surfaces
Type of brake
Steel band on C.I.
or steel drum
Asbestos brake
band on steel or
C.I. drum
Rolled, press
formed and shaped
Wood on (.I.
friction material
drum
Holding
1.5
0.6
0.8
Lowering
r--------.,.
1.0
0.3
0.4
I Example
11. 7
IA
---
on metal drum
simple band brake is operated by a lever of lengt
0.6
0.4
h 500 "",. /o1lg·
if the
5/8
0
The brake drum has a diameter of 500 mm and the brake band embraces h ot~e'
hile t e
.
circumference. One end of the band is attached to the fulcrum of the lever w. d to the e"J
is attached to a pin on the lever 100 mm from the fulcrum. If the effort app/,e .tnJ1lt ~J
of the lever is 2000 N and the coefficien: of friction is 0.25, then design the SI
brake.
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11.23
r
~I
fof'"
00 ml11
0.5 m; d= 500 rnm or
5
:::: 0.1 m ; F = 2000
r ::;250 m m == 0.25 m .
'
(l::::
6fV'''P
b~ 1
=
a ~ 50;
00 mm
~'(l,'
~(I:
•
i: peslgn
Il
N;
== 0.25.
the simple band brake.
.
.
.
.
Considermg anticlockwise rotatIon of drum th
filIP/ion:
. Fig 1 I II(b).
' e arrangement of simple
~Jv
• hown JO
I.
.
5
brake IS S
~
e =
Angle of contact,
8 x 21t
=
3.927 rad
T)
= ella = eO.25 x 3.927 == 2.669
Tension ratio is given by, T2
T,
2.669 T2
=
... (i)
or
. moments about the fulcrum 0, we get
rakJIlg
F· a = T2· b
2000 x 0.5
=
T2 x 0.1
or
T2
=
10000 N
From equation (i),
TJ
=
2.669 T2
-
T2) r
= 2.669 x 10000 = 26690 N
1. Brakingtorque:
Braking torque, TB = (TJ
=
(26690 - 10000) x 0.25 = "4172.5N-m
2. Brakedrum diameter:
d
500 mm
=
Tight side tension, T, = 26690 N
J. Tightand slack side tensions:
and
... (Given)
T2 = 10000 N
Slack side tension,
l Thicknessof hand:
t
." (already calculated)
=
0.005 x Brake drum diameter
=
0.005 x 500 = 2.5 mm
5, Band width (w) :
lnd.uced tensile stress
(J
=
W
'I
(J
=
26690
twx
26690
w x 2.5
Or
T)
xt
2.5
< 50 N/mm2
-
~ [at]
[ '.' [ o, ] = 50 N/mm2 is assumed]
Width of band, w = 213.52 mm ::= 215 rom
,
.
ck /0,. he •.
. .
re between band and drum IS
>'
anng pressure:'
Maximum beanng pressu
~ 6, eire
Yen b
.
.
I
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------------------------_D_e_S~ign~O~if_n_r_a_ns~m~l=Ss~wns
IIII~.2~4
~
.;
26690
_
.
250 - 0.496 N/mm
215 x
==
Pmax
w·r
[]
2
:..: 1 5 N/mm2
From Table 11.4, for steel band on stee I d rum, P
.
Since PIltllX< (p I, therefore the design is safe and satisfactory.
F == 2000 N
7. Force to be applied at the end of the lever,
... (Given)
a
, Example 1J.81 A band brake acts
on the ~ of circumference of a drum of
450 mm diameter which is keyed to the
shaft
The band brake provides a
braking torque of 225 N-m. One end of
the band is attached to a fulcrum pin of
the lever and the other end to a pin 100
mm from the fulcrum, as shown in
Fig. 1J.12. If the operating force is
applied at 500 mmfrom the fulcrum and
coefficient of friction is 0.25. Find the
operating force when the drum rotates
in the (a) anticlockwise direction, and
(ii) clockwise direction.
Given Data: d = 450 mm or r
b = OA
=
To find:
100 rnm
=
a = 500mm
b
Fig. 11.12.
=
=
225 mm
0.225 m ; T B = 225 N-m ;
a = OB = 500 mm = 0.5 rn ;
0.1 m;
= 100mm
J.L
=
0.25.
The. operating force when the drum rotates in the (i) anticlockwise, and
(ii) clockwise direction.
@ Solution:
Let F == Operation force
(i) Operating force when drum rotates in anticlockwise direction:
Angle of wrap,
e
=
%th of circumference
3
= 4
x 3600
Tension ratio is given by,
TI
T2
or
TI - 3.248 T2
=
ella
=
= 270 = 270
eO.25 x 4.713
0
=
0
x
180
1t
== 4.713 rad
3.248
.. , (i)
We know that the braking torque (Ts),
=
225 = (T) - T2) x 0.225
Ts
or
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= 1000 N
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.. , (ii)
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11.25
~
.
...
uations (I) and (11), we get
solving eq
T
= 444.84 Nand
2
.
faking m
TI
= 1444.84 N
oments about fulcrum 0, we get
F· a = T2 x b
.
F x 0.5 = 444.84 x 0.100
F = 88.97 N Ans. ~
ii) Operatingforce when drum rotates in clockwise direction:
~
. c Ioc kwi
When the drum rotates In
wise d'irection, the tensions TI and T2 will interchange
their positions.
Takingmoments about 0, we get
F
x a
=
TI x b
b
F _" T x1 a
or
= 1444.84
0.100
x 0.500
= 288.97 N Ans. ~
[_Example 11.9
IA
simple
band brake
shoHinin Fig.lJ.13, is applied to a shaft
carryinga flywheel of mass 250 kg and of
radiusof gyration 300 mm. The shaft speed is
200 r.p.m: The drum diameter is 200 mm and
the coefficient of friction
is 0.25. The
dimensions'a' and 'I' are 100 mm and 280
mm respectively and angle p = 135 ~
Determine:
(i) The brake torque when a force
J 20 N is applied at the lever end,
1= 280 mm
T,
a
of
Fig. 11.13.
(ii)
The number of turns of the flywheel before it comes to rest, and
(iii)
The time taken by the flywheel to come to rest.
GivenData:
m
D==200mm;
© Solution:
=
Jl=0.25;
250 kg;
k
= r = 100 mm
= 300 mm = OJ m; N = 200 r.p.m. ;
a= 100mm = 0.1 m;
1= 280mm
=
0.28m;
p
=
135°.
(i) Brake torque applied at the lever end:
Angle of contact,
e
=
e
= 225 x 180 = 3.927 rad
360
0
P = 360
0
0
-
-
135
= 225°
It
J
J!
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I
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11.26
Design of Trans miss'
Ion
T)
Tension ratio is given by,
= eJ.19 =
eO.25
=
x 3.927
SYsle'ns
2.67
T2
r,
or
= 2.67 T2
". (i)
Taking moments about the fulcrum 0, we get
F·/
T)· a
=
120 x 0.280 = T) x 0.100
or
T) = 336 N
From equation (i),
T)
=
2.67 T2
336
=
2.67 x T2
or
T2 = 125.84 N
Braking torque is given by,
T8
= (T) - T2) r
=
(336 - 125.84) x 0.10
= 21.01 N-m Ans. "
(ii) Number of turns of flywheel before it comes to rest (n):
We know that kinetic energy of flywheel,
K.E
= -21 I (1)2 = -21 mk2 (1)2
=
1
2"
x 250 x (0.30)
2 [
21t x 200 ] 2
60 .
= 4934.80 N-m
This kinetic energy is used to overcome the work done due to the braking torque (TB)'
K.E of flywheel
4934.8
n
or
=
TB x
=
21.0 I x 21t n
=
37.38 revolutions
(I)
=
T 8 x 21t n
[n
= numberof turns]
Ans. ~
(iii) Time taken by the flywheel to come to rest:
Time taken
n
37.38
=
N· -
=
0.1868 minutes
=
11.208 sAns.
200
"
11.7.3. Differential Band Brake
.. ed to the
In a differential band brake, as shown in Fig.ll.14, the ends of the band are JOUl bandto
lever DOB at points D and A. Point D is the fulcrum. It may be noted that for the
tighten, the length Ol) must be greater than the length OA.
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11.27
a
B
A
~----4-------~
b
F
T,
T1
'
·······(:r········
.
~
.~
.
8
8
(b) Anticlockwise rotation of drum
(a) ClockWise rotation of drum
Fig. 11.14. Differential band brake
(i) Downwardforce on lever for clockwise rotation
0/ drum:
This type of arrangement is shown in Fig.ll.14(a). Taking moments about 0, we get
F'a
- TI·b
= T2·c
Thus T2 . C > T 1 • b
Thus c > b for the system to work satisfactorily.
If ~
T
= T~ , the
external applied force F = 0, which is the condition/or self-locking.
(ii) Downwardforce
on lever for anticlockwise rotation
0/ drum:
Taking moments about 0, we get
F· a
=
TI· b - T2 . C
Thus TI· b > T2·
Condition/or self-locking:
C
TI C
or T2 > b
If TI b = T2 c, then external applied force F = 0
=
c
b
Upwardforce on lever for anticiockwise rotation of drum:
.
This type of arrangement is shown in Fig.l1.14(b). Taking moments about 0, we get
(iii)
F .a
=
TI· b - T2 . C
Thus T 1 b > T2
Or
TI
T2
>
C
-b
C
!
I
I
I
'n
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I ].28
Design of Transmi
.
S
S$I011
~
(iv) Upwardforce on lever for clockwise rotation of drum:
Taking moment about 0, we get
F .a
T2· e - TJ
=
•
b
T)
e
Thus
b
T2· e > T)· b or
> T2
Condition for self-locking:
If
T)
T2
= be , then F = O.
In this case, e must be Jess than b for proper braking.
I
I
Example J1.J 0' Design a differential band brake for a crane lifting a load
•
I(
oJ 50
through a rope wound round a barrel of 550 mm diameter, as shown in Fig.ll.l
S
brake drum to be keyed to the same shaft is to be 650 mm in diameter and the an I. . 1
of the brake band over the dr~m is 240'! Operating arms of the brake are 45 ",,,,K e of l
.
•
•
••
and 2
mm, as shown In Flg.J1.J6. Operating lever IS 1.5 m long. Take u= 0.25.
.
T
~
550mm
j_
777 "_""'~~--J
-
----
I
650 mm
j_
777
Fig. 11.15.
Fig. 11.16.
Given Data: Load = 50 kN; Barrel diameter = 550 mrn;
Brake drum diameter = 650 rnrn;
To find:
e = 240
0
= 240 x
0
.2!_ -
180
4.188 rad "
Design a differential band brake.
@Solution :
1. Calculation of braking torque:
Braking torque, TB = Load x Barrel radius
- 50000
2. Brake drum diameter:
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D -
x
(0.;50)
~ 13750 N-m
650 mrn
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II
r
= 0.25.
~
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~~~------------------------------~---~~~
~
ltion ofT} and T]:
Ca1ell a
J
T,
T2
Tension ratio,
=
=
el!9
eO.25 x 4.188
== 2.849
"
T, = 2.849 T2
... (i)
or
Braking torque, TB = (T I
We 1,"0"' that,
-
T2)
x r
r.J'
13750 = (T, - T ) x (0.~5)
2
'" (ii)
= 42307.69
or
... (ii)
solvingequations (i) and (ii), we get
Tl
4. Thickness of band:
t
= 65189 Nand
= 0.005
x
0
1'2 = 22881.39
N
= 0.005.x 650 = 3.25 mm
5. Calculation of band wid til (w) :
T,
Induced tensile stress, crt =
w x t
65189
crt
65189
w x 3.25
or
Band width,
W
=
s [crt]
w x 3.25
s 50 N/mm2
[.: [ crt]
=
50 N/mm2 is assumed]
= 401.16 mm ~ 405 mm
6. Check for bearing pressure:
Maximum bearing pressure between band and drum is
given by
Pmax
=
65189
= -__;_;_--~
w -r
650)
405 x ( T
From Table 11.1, for steel band on steel drum, [p]
==
= 0.495 N/mm2
1.5 N/mm2
Since Pmax < (p ), therefore the design is safe and satisfactory.
7.Calculation of the force 10 be applied at the end of the lever:
Referr'109 to Flg.ll.16,
.
taking moments about 0 ,we ge t
F x 1500 + T, x 45
Or
F
= T2 x 210
x 1500 + 65189 x 45 == 22881.39 x 2 IO
Or
F
= 1247.72 N
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ADS.
-e
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Design o!Transmi:J ~.
11.30
~.
[ Example 11.11 I A differential. band brake has a force of 220 N applied III
~t
II ~daJ as shown in Fig.ll.17. The coefficient of friction between the band a"d th
Ill! oJ
'.40. Angle of lap = 180 ~ What is the maximum torque the brake may SUStai" I'. e d,u",it
JO'CQ
clockwise rotation ?
u~
T2
F
= 220N
0
a
b
120mm
60mm
c
240mm
Fig. 11.17.
Given Data :
F = 220 N ,.
II
r-
= 0.40 ,.
e
= 1800 = 1800 x _1L
1800 =
7t
rad;
= 180 mm or r = 90 mm = 90 x 10-3 m; a = 60 mm = 60 x 10-3 m ,.
b = 120 mm = 0.12 m; c = 240 mm = 0.24 m.
d
To find:
Maximum braking torque.
@Solution:
Tension ratio is given by,
Tl
=
3.513
." (i)
or
Taking moments about 0, we get
F· c
=
T2· b - TI . a
220 x 0.240
=
T2 x 0.120 - TJ x 0.06
52.8
=
0.120 x T2 - 0.06 x (3.513)T2
. T -3513T1l
[From equation (I), 1-'
or
T2 = - 581.63 Nand
Braking torque is given by,
TB
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T1) x r
TI
2043.25 N
= (2043.25 _ 581.63) x 0.09
=
(T2
=
131.55N-m Ans. ~
-
= -
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11.31
B,(J
kes
. C lockwi
. T and T are reversed. In that case
ation of the d rum IS
oc wise, then the tension
~
If the ro t
~
.
.
1
. ibrturo ro
_
tile eqllJI
F . c - T I . a - T2 • b c
2
ay be obtained by taking moments about O.
am Ie
11.12
A differential
0
.
band brake is operated by a lever as shown in
The brake drum has a diameter of 500 mm and the maximum torque on the
S
fig.1J.1 . 1'1-"" Iflhe coefficient of friction between the brake lining and drum is 0.3,Ji.d
isJk
ll
dr '" r atingforce.
the ope
J)~
p
500mm
B
Fig. 11.18.
Given Data:
TB = 1 kN-m
=
b = OA
e = 240
0
=
=
r = 250 mrn = 0.25 m ;
d = 500 rnm or
1 x 103 N-m; J.1 = 0.3 ; a = 08 = 500 mm = 0.5 m ;
80 mm
=
80 x 10-3 m ; c
2400 x 1;0
=
= 00 =
100 mm
=
0.1 m ;
4.188 rad.
Tofind: The operating force P.
© Solution
: Tension ratio is given by,
T)
=
ells
=
e0.3x4,)88
= 3.5136
T2
.. , (i)
T B ::: (T) - T2) r
3 ::: (T) - T2) x 0.25
... (ii)
or
Braking torque is given by,
1 x 10
Or
T) - T2
=
,
"
4000
SOlVing equations (i) and (ii), we get
-
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11_!1~3~~
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------
D_es_i=g_n__:of::.._TJ_ra_
~
~
T2
T, :: 5593.6 Nand
=
1593.6 N
Taking moments about the fulcrum 0, we get
3
p x 0.500 :: T, x 0.100 - T2 x 80 x 10P
or
x
Operating force,
I Example
11.13
0.5 = 5593.6
x
0.100 - 1593.6
x
80
x
10-3
P = 863.34 N ADs. ~
I A differential
band brake used for a winch, is wound round a dr
of o. 75 m diameter , keved
to the shaft. The two ends of the• band are att{lched 10 the P
'J
on the opposite sides of the fulcrum of the brake lever at distances of 25 nun and J 00"
from the fulcrum.
~he angle of lap on the drum is 240 ~ Ti,e coefficient offriction.is 0.25..Find the lorq
which can be applied by the brake when a force of 500 N IS applied to the lev
(II) upwards, (b) downwards, at a distance of 1 m from the fulcrum. Consider ho
directions of rotation.
Given Data:
d = 0.75 m or r = 0.375 m ; c = 25 mm = 25 x Hr-3 m·,
b
= 100 mm = O. 1m;
F = 500 N;
To find:
e
a = 1 m and
=
2400
1t
=
2400 x 180
= 4.188 rad; Jl = 0.25 ;
b > c.
1. Torque applied when force F acts in the upward direction:
(a) Brake drum rotating anticlockwise
direction, and
(b) Brake drum rotating clockwise direction.
2. Torque applied when force F acts in the downward
(a) Brake drum rotating anticlockwise
direction,
direction:
and
(b) Brake drum rotating clockwise direction.
©Solution:
(1) Torque applied when force F acts in the upward direction:
(a) Brake drum rotating anticlockwise direction:
F
o
A
~----:.r------..J
B
b
a
Fi1!. J.1_ 1s. (a'/.I b > C«rF uswar dS, drum antictockwise
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~
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--------
wn
AS sho
t{1 Fig.11.18(a),
~1~1.3~3
the ends of band are att h d
.
T
ac e to pomts 0 and A.
Tension ratio, TI2
=
ello =
TI
=
2.849 T2
eX.0.25 4 188' = 2:849
... (i)
or
lakin. g moments about fulcrum 0, (Fig "II 18(a» , we get
T2·b
=
T2xO.IOO
= Fx1.0+Tlx25xI0-3
4 T2
or
F·a+TI·c
=
20000 + T I
=
4950.4 Nand
... (ii)
solving equations (i) and (ii), we get
TI
T2
=
1737.6 N
Braking torque T B applied by the brake,
TB
=
(T1-T2)r
=
1204.8 N-m Ans. ~
=
(4950.4-1737.6)xO.375
(b)Brake drum rotating in clockwise direction:
This case is shown in Fig.ll.18(b).
.
T·ension ratio,
T
TI
= ell e =
[From (i)]
28. 49 (same)
2
Taking moments about fulcrum 0, we get
TI.b
TI x 0.10
or
=
=
F·a+T2·c
500 x 1.0+T2 x 0.025
... (iii)
4 T I = 20000 + T2
F
a
o
_.-J.I
Fig. /l.J8(b)
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11~1~.3~
----------------_D_e_S~ign~o~if_Tl_r_an~s~m~~~s;on~.
-
~
Solving equations (i) and (iii), we get
=
T,
17376 Nand
T2
=
6099N
Braking torque T8 applied by the brake,
T8 = (T,-T2)r
=
= 4228.88 N-m
(17376-6099)
x 0.375
ADS. 1:1
2. Torque applied when force F acts in the downward direction:
(a) Brake drum rotating in antic/ockwise direction:
This case is shown in F ig.ll.l8( c). Again tension ratio is given by
TI
T2
2.849
=
... [same as in equation(i)]
Taking moments about fulcrum 0, we get
TJ
=
• C
F· a + T2 . b
TI x 0.025 = 500 x 1.0 + T2 x 0.10
T I = 20000 + 4 T2
or
'" (iv)
Solving equations (i) and (iv), we get
TI = - 4950.4 N and T2 = - 1737.6 N
Both the tensions are negative, so the band will be loosened on the drum. In order to apply
brakes, the direction of force F should be reversed.
..
Braking torque, T8 = (TI - T2) r = (4950.4 - 1737.6) x 0.375
=
1204.8N-m Ans."
·1
Aa
b
F
T1
(c) b > c, F downwards, drum anticlockwise
(d) b > c, F downwards, drum clockwise
Fig.II.JB.
(b) Brake drum rotating in clockwise direction:
This case is shown in Fig.I1.18( d).
. TI
T·ension ratio,
T
= 2.849
. .. [same as equation (i)l
2
Taking moments about the fulcrum 0, we get
F . a + Tl . b - T2·
C
500 x 1.0 + Tl x 0.10 = T2 x 0.025
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11.35
... (v)
or
. geq uations (i) and (v), we get
~v~
T
1
band will be loosened
""
apply brakes.
, tlOo to
dire'
8
Braking torque,
= -17376N
and
T2=-6099N
as T I and T 2 are negative. Thus force F must act in upward
=
TB
(17376 - 6099) x 0.375
=- 4228.88 N-m
Ans. ~
SAND AND BLOCK BRAKE
11. . VIOU
, sly this arrangement
Ob
,
wn 'In FI'g ' 11 . 19, The band
".
is a combination
of both the band and the block brakes, as
•
lmed with a number of wooden blocks , each of which is In
with the rim of the brake drum. When the brake is applied, the blocks are pressed
shD
con~ac: the
drum.
The
IS
advantage
of using
wooden
blocks
agaJO~ientof friction and they can be easi ly and economically
coeffi
T = Tension in the band on tight side,
Let
is that they provide
higher
replaced after being worn out.
n
To
= Tension
in the band on slack side,
T 1 = Tension
in band between the first and second block,
T2 = Tension in band between second and third block,
T3 =
Tension
n
= Number
Jl
=
28
in band between third and fourth blocks and so on,
of wooden blocks,
Coefficient
= Angle
of friction between the block and the drum,
subtended
RN = Normal
by each block at the drum centre,
and
reaction on the block,
(a)
Fig. 11.19. Band and block brake
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',I
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~
J J .36
Design of Transmission S
~'
~
Consider one of the blocks (say first block) as shown in Fig.ll.19(b).
This is in equilibrium under the action of the following forces.
I. Tension in the band on tight side, T,
2. Tension in the band between first and second block, T I
3. Normal reaction of the drum RN of the block, and
4. The frictional force, J..l RN·
Resolving the forces radially, we get
(T I + To) sin
e
=
RN
'" (i)
J..l RN
'" (ii)
Resolving the forces tangentially, we get
(T I - To) cos
e =
Dividing equation (i) by (ii), we get
u tan B
1 + Jl tan
I-Jltane
e
=
T.-To
=
T. = 1 + Jl tan
To
I - J..l tan
or
e
e
Similarly, it can be proved for each of the blocks that
T2
T.
=
1 + Jl tan e and T3
1 - J..l tan e
T2
=
1 + J..L
tan e
1 - J.l tan e
l+J..Ltane
= ---
Therefore,
1 - J.l tan
e
So the ratio of tensions for all 'n' blocks is given by
r,
T.
T2
r,
T3
To = T0 x:r I xT2
X .•••••
xT
n-I
=
[1
e] n
e
.. , (11.17)
[Neglecting
the thickness of the belt]
+ J..l tan
1 - J.l tan
Braking torque on the drum is given by
TB = (T I - T2) r
T B = (T I - T 2) ( d ~ 21)
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[Considering
.
the thickness
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[the belt]
0
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u
11.14 , A band brake'
r£ii,mple
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11.37
tned with 10
angle of 18 0 at the centre of the br k
wooden blocks each if
.
an
h
l I' 03
a e drum. If th
which subtends
e coefficient of friction b t
blocks and t -e W lee IS • 6, find the ratio b t
th
b
k
.
.
e
Ween
th
e ween
the
I
, band w ten te ra e IS In action
e greatest and th I
.
tl,e
.
e east tensIOns in
Given Data:
n = 10; 26::: 180
Or 6 == 90.
, J.l == 0.36.
~
IS
°
(T J
Tofind:
Ratio between greatest and th I·
e east tensions in the brake _Q
. : Wke now that the tension rati ~ b
~
© SolutIOn
10 lor and and block brake
T n = [ 1 + IJ. tan 6 n - [ 1 + QJ6 tan 9° ] 10
,
To
1 - ,.H
. an 6·
- 1_ 0.36 tan 90
= (1.209)10
J
Tn
To
11.15
[Example
= 3.13
I In
Ans. ~
the band anti block brake shown in Fig.l1.20, the band is lined
witlt 12 blocks each of which subtends an angle of 15° at the centre of the rotating drum.
Tirethickness of the block is 75 mm and the diameter of the drum is 850 mm: If, when the
brake is ill action, the greatest and least tensions in the brake trap are T] and T')7 show that
oJ
T] _ [1 + J,l tan 7.5
T) 1- utan 7.50
12
I
. tl. coefficient of friction for the blocks. Find the least force required at 'C'
W tere f.1IS
te
W 240
m: The coefficient of friction between the band
for the blocks to absorb 225 k at
r.p.
F
and blocks is 0.40.
O.5m
Given Data:
= 12 ,.
n
26 = 15° or 6 - 7.5° ;
T1
= 75 rnrn
t
= 75xlo-3m;
12
= 850 mrn
d
11
= 0.85 m ;
p
= 225 kW
= 225 x 103 W ;
N = 240 r.p.m ;
Il
Fig. 11.20.
= 0.4 ;
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--------------_D_e_sl=g_n_o~if_Tl_ra_n_s_m~~~s~funs
....:.]~I~.3~8~
~
-
= c
OB = b
OA
OC
=
=
150mm
=
30 mm
a = 500 mm
0.15 m;
=
lo-3m;
= 30x
0.5 m
=
© Solution: (i) Ratio between the greatest and least tensions:
Since OA > OB, so force at C must act downwards. Also the drum rotates clockwis
. havi
. T2 (I east tension) ande,thS
t h e b and attached to A will experience slack side
avmg tension
.
band attached to B will be tight side having tension T, (greatest·tensJOn).
For derivation, refer article 11.8.
T, = [1 + J.1 tan 7.5 ]
T2
1 - J.1 tan 7.5
So,
Ans. ~
12
(ii) Least force required at C: Effective diameter of the drum,
D
d + 2t
T,
T2
=
[1
T,
T2
=
[1 + 0.4 tan 7.5] 12
1 - 0.4 tan 7.5
=
3.5449
T,
=
3.5449 T2
Power absorbed, P
=
(T, - T2) x
7tDN
60
225 x 103
=
(T, - T2) x
7t x 1 x 240
60
T,-T2
=
17905
We know that the tension ratio,
or
or
0.85 + 2 x 0.075
=
=
+ J..l tan 8
] - J..l tan e
J
= ]
m
n
= [
1.05266] 12
0.9473
... (i)
.. , (ii)
Solving equations (i) and (ii), we get
T,
=
24940.7 N and T2
Let F
=
Least force required at C
=
7035.70 N
Taking moments about fulcrum 0, we get
,
=
T2·b-T1.c
F x 0.5
=
7035.7 x 0.15 -24940.7
F
=
614.3 N Ans. ~
F'a
1
x 0.03
J
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r ~------------------------------~
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11.39
E~a'"
Ie 11.16
A band. and b~oCk brake having 12 blocks, each of which subtends
0
ngle OJ.r J 6 at the centre, IS applied to a rotating drum of diameter 600 mm. The blocks
an a
m thick. The drum and the flywheel mounted on the same shaft have a mass of
are 75 mand
00
bi
d
have a com me
ra
if gyrattOn
. of
dl
IUS 0
600 mm. The two ends of the band are
18 k;d to pins on the opposite sides of the brake fulcrum at distances of 40 mm and 150
attach
If a force of 250 N
the fulcrum.
",m [rom
nd'
fulcr"m,' fi
is
applied at a distance of 900 mm from the
.
(i) The maximum
braking torque,
(ii) Tire angular retardation of the drum, and
":' The time taken by the system to be stationary from the rated speed of 300 r.p.m.
(Ill,
-tticient of'J friction
between the blocks and tire drum as 0.3.
Take coe)}
.
I
Given D a taa .:
1'1
=
10;
20
=
d
=
16° or 0 = 8°'
600 mrn = 0.6m
D
or r = 0.3
=
I
I'
a
"I
~O-+A-T2------:(
III ;
75 mm
= 0.075
m;
m = 1800 kg ;
= 0.6
k = 600 mill
h
=
=
D
:::
A
:::
4
.1
=
4
In'
Drum
mIn
III ;
150 mtn
::
Fig. /1.11.
m :
F' -= 20N;
I:;;;:
B::: 900
rnrn ::
0.9
rn
he braking torque will be maximum
.
king torque TB' T
© Solution:
(i) Maximw" bra .
.
di .
are satisfied.
wh n the C IIL\ in c n III n
D
a
b>c
.
OA>O,
I. e.
drum rotates an
tic!ockwise, and
b
In thi
t
Brake
15 upwards.
. T and the end of the
lied
force
ac
.'
ht
under
tenSion
I
The ap P
h d to A IS tJg
1
. Fig . 11. 2 .
the end )f the b an.danae e
hown 10
Sl n T 2 as
B i
la k under ten
Ten i n rat i i gi en by
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::; 2.752
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~----
Design of TrnnsmissirJn SYSletns
11.40
.i;
". 0)
or
Taking moments about 0, we get
= T2·b
250 x 0.9 + T, x 0.04 = T 2 x 0.15
225 + 0.04 T, = 0.15 T 2
F·a+T,·c
or
'" (ii)
. d (..) we get
Solving equations (I) an II,
36 Nand
T2 = 56
T,
=
15511 N
. torque T B is given by
Maximum braking
T B = (T, - T 2)
(
= (15511 - 5636) [
We know that,
06+2
x
2
thickness of the band]
0.075 ]
Ans."
3703 N-m
(ii) Angular retardation
[Considering
d ~ 21)
0/ the drum (a) :
TB = I a =
3703 = 1800
mk2 a
(0.6)2
x
x
a
a = 5.71 rad/s! Ans."
or
(iii) Time taken by the system to come to rest from tile speed of 300 r.p.m.:
Initial angular speed,
Final angular speed,
We know t Iiat,
(1)
=
(1)0
2n60N
=
2n 60300
= 31.4 radls
O.
I.' = r·\o - a
UJ
o
or
=
UJ
I
[a is negative due to retardation]
= 31.4 - 5.71 I
1 =
5.5 sAns.
~
11.9. INTERNAL EXPANDING SHOE BRAKE
11.9.1. Introduction
Ino~:
As the name implies this type of brake is provided internally on the brake drum.
days band brakes were used in automobiles, which were exposed to din and water. The" ally
diIssipalion
'.
. was a Iso poor. These days, band brakes have been rep Iace d by Intern
capacity
1 in
expandmg
'. shoe brakes havmg atleast one self-energizing shoe per wheel. Th'IS resu [5
tremendous friction, giving great braking power without excessive use of pedal pressure.
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11.41
~
Jt'( -------
v-
orking principle
11.9.2. 'II ) 5ho'\l5 an internal shoe automobile brake. It consists of two semi-circular
I 22(a
. d itl
~..
I
.
fig. I . d 5, ",11ich are line WIt 1 a I nctiona material such a fcrrodo. When brakes are
.,eS51 all . -tores which pushes the shoes outx ards to press the brake lining against the rim
~
arn ,0 a
,pplied. C A soon as the brakes are off, the shoes are pushed inside by the spring.
, drum. s
• ~lf1he
oted that for the anticlockwise direction, the left side shoe i known as primary
I•
a)' be n
.
It!l1. , e while the right hand shoe ISknown as trailing or secondary shoe.
dmg s,1O
Dr len
Cam operating level
/
._.y,)
J-.-.
Fulcrum
(b) Forces on the brake
(a) Internal expanding brake
Fig. 11.22.
11.9.3.
Determination
of Pressure
and Brake Torque
Considerthe forces on the brake when the drum rotates in anticlockwise
snown
in Fig.II.22(b).
leI
PI = Maximum intensity of normal pressure,
PN
=
r =
b
direction,
as
Normal pressure,
Internal radius of the drum,
= Width of brake lining,
TB
::::
Braking torque,
FI
=:
Force exerted by the cam
F2
=:
Force exerted by the cam on the trailing or secondary shoe,
RN
::::
Normal force,
011
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the leading or primary shoe,
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Design of Transmission S
11.42
~
F
= ~.}'rictional
force,
~J.l = Coefficient of friction between shoe and drum,
MN = Moment of normal force, and
MF
=
Moment of frictional force.
Consider a small. . element AB of brake lining subtending an angle 09 at the centr e of tL_
drum. Join 01to ..It is assumed that the pressure distribution on the shoe is nearly ~~. "Ie
However, the. shoe wears out more at the free end. The rate of wear of the shoe linin v ."
directly as the perpendicular distance from 01 to B i.e., ole.
ganes
°
From the geometry of the Fig.11.22(b)~
-
~.
I
1
and Normal
pressure at B,
.
.
.
=
.- ',PN cc sin 9 or PN
PI
sin 9
:. Normal force acting on the element,
oRN . = Normal pressure
..
=
x
PN x (b· r . 89)
=
. ,:
Area'of the dement
PI
sin":'9 b . r . 09
...~
Friction force on the element.
8F = J.l' oRN
=
J.l PI sin 9 . b . r . 89
°,
Braking torque due to the element about
~ oTB
=
of· r
=
J.l PI sinB . b . r 09 . r
=
J.l PI sin 9 . b r2 09
Total braking torque for whole shoe abo~t 0,
"..
.
92
T B . = J.l PI b r2
,., ,
..1.
J
J.lPI b r2 [- cos 9]
9I
~ .flPI br2 (cos 91
-
. .,_I
a
2
91
cos (2)
Moment of nm·M'al force 8R of th
~
Nee
8MN ='loRN x ole
=
=
sin 9 . d9
I
ement about the fulcrum 0
t .
I'
8RN (001 sin 9)
= PI
sin 9 (b . r . 09) (001 sin 9)
= PI
sin2 9 (b . r . 09) 00
:. Total moment ofth
1
e normal force about the f I
u crum 01 ,
~
>
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.c->:
V
11.43
92
J
MN -- P , . b· r- 00,
sin29 . d9
1
92
::=
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J 2 (I - cos 29) . de
p, . b . r . 00,
[.: sin2 9 =
9,
=
t
PI' b- r- 001 [ 6 _ sin/6]
t
(I - cos 29)]
2
9
9,
::=
=
21 p,
t
. b . r . 00, [sin
92
2292 - 8, + sin 229, ]
-
[(62 - 61) +
PI . b . r .0°1
t
(sin 261 - sin 26
)J
2
'" (11.18)
Moment of frictional force of about the fulcrum 0"
8MF = 8F x Be = of (r-OO,
. cos 9)
substitutingthe value of of, we get
8MF
=
J.lp,'
=
J.l Pi . b . r [ r sin 8 -
b· r· sin 8 (r - 00, cos 8) 88
00, sin 29]
2
. 88
[.: 2 sin 8 cos 8 = sin 28]
Total moment of frictional force,
2
MF
=
J.l p, . b . r
9
J
[
r sin 9 -
00, sin 28]
2
d9
91
= J.l PI . b . r
2
J9
[ - r cos 9 + -4- cos 28 9
00,
1
= Jl.PI· b . r [ r (cos 61 - cos 62) + O~I (cos 262 - cos 261)
ForIe di
a mg shoe, take moments about 0"
F
1
x I ::: MN - MF or
F1 =
J ... (11.19)
.
MN-MF
I
... (11.20)
and for tr T
at 109 shoe, take moments about 02'
F2 x I = MN + MF
COlli/it'
IOn for
IfM :::.M
F
F,
or
F2
=
MN+MF
I
... (11.21)
Self-Locking:
ki
then the brake becomes self-Ioc mg.
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) ).44
Design o/Transmission S
~
[Examele 11.17 I Fig. 11.23 shows the arrangement 0/ two brake shoes ..,hich act
the internal surface of a cylindrical brake drum. The braking forces FI and F2 are a ~
•
rpPlttd
as shown and each shoe pivots on its fulcrum 01 and 02" The width of the brake linin .
35 mm and the intensity of pressure at any point A is -I x lOS sin (J Nlml, ..,here ~ ~
measured as shown from either pivot. TIle coefficient of friction is 0.40. Determine thI
braking torque and the magnitude of the forces FI and F2"
e
I
.~
30'
:1
ir
Fig. 11.23.
Given Data:
To find:
b
=
=
35 mm
0.035 m; PN
=
4 x ] 05 sin
e N/m2;
Il
=
0.4.
I. Braking torque (TB)' and
2. Magnitude of the forces FI and F2.
© Solution:
Intensity of normal pressure is given as,
PN
4 x 105 sin 8 N/m2
=
Then, maximum intensity of pressure (i. e., 8 = 90°),
=
PI
4 x 105 N/m2
(i) Braking Torque (To) :
Distance offorce FI from fulcrum 01'
I
=
200 mm
=
0.2 m
Distance of force F2 from fulcrum 02'
I
=
200 mm = 0.2 m
We know that the braking torque T B '
Ts
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=
IlPI ·h·r2(cos81-cos82)
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11.45
= 0.40 x 4 x 105 x 0.035 x (0.15)2 (cos 25°;_ cos 125°)
= 186.46 N-m
. ce there
are two shoes, so total braking torque
T B = 2 x 186.46 = 372.92 N-m
Sl~
(ii) !Jag'"
'
ADS. ~
'tudes of the forces F1 and F] :
and
=
92
= 125° = 125° x
e geometry of Fig.ll.23,
Fromth
00
25°
25° x 1~o
9,
=::
=
0.436 rad
= 2.]8 rad
I~O
we get
I
O,B
100
= cos 25° = cos 250
= 11034
. mm
Thetotal moment of the normal forces about the fulcrum 0, is given by
MN == ~
== ~
=
p,' b· r· 00,
[(82 -91)
+ ~ (sin 281-Sin282)]
x 4 x lOs x 0.035 x 0.] 5 x 0.11034 [ (2.18 - 0.436) + ~ (sin 50° - sin 250°) ]
300.86 N-m
Momentof friction force Mr about 01 is given by
M, " ~p,. b . r [ r (cos 6, - cos 62) + O~, (cos 262 - cos 26,) ]
:: 0.4 x 4 x 105 x 0.035 x 0.15 [ 0.15 (cos 25°--<:os125°) +
0.11034
4
(cos 250°--<;os 50°)
]
.
:: 163.65 N-m
For leading shoe : Taking moments about the fulcrum 01' we get
FI
I = MN-Mr
x
= 300.86 + 163.65
F1xO.200
Or
F, = 686.05 N Ans. ~
For trailing shoe: Taking moments about the fulcrum
F2
°
2, we
get
I = MN+Mr
F2 x 0.200
=
300.86 + ) 63.65
Or
F2
= 2322.55 N
Ans. ~
J
J
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...:
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lSSIOn S .
------------------------D_e_SI=·gn~Of~T~~an~sm·.
1~1~.4~6~
I Example
11.18
I An
.
.
""
~
automotive type Internal expanding
double-shoe sir
o~"
Fig. 11.11 is 300 mm in diameter and is actuated by a mechanism
'~~h that exerts thesa~~ ill
F on each shoe. The shoes are identical and have a fiace WI", of 32 mm; The .. JO,Cf
moulded asbestos having a coefficient of friction of 0.31 and a pressure lilllitat:lIn"" ;,•
kPa
(i)
~n~J~
Determine the actuating force F, and
{ii) Find the braking capacity (i.e., torque absorbing capacity of the brake).
I
•
27'
~
126'
Fig. /1.24.
Given Data:
Pmax
Tofind:
= 0 .3 m; b = 32 mm;
= 1000 kPa = 1000 x 103 N/m2
d - 300 mm
Jl ~ 0.32 ;
(i) Actuating force F, and
(ii) Braking capacity (T B)'
@S~I u tiron : ,1/
1:;1
Actuating force (F) : The right hand shoe is self-energizing. So the
force F IS found on the basis that the maximum pressure will occur on this shoe. For the right
hand shoe,
00;
e
=
900;
92 = 1260 = 1260 x 180
7t 0
and
sin
e
=
=
122.65 mm
=. 2
199 raad ,:
From the geometry of Fig . 11 .24 ,we get
00 I
=
V 1122 + 502
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,
"
l
j
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"
J 1..47
pktS
rn of the normal forces about the fulcrum 0'
tsl 010 ent
I
p~
I
.
file to
1
... 'PI
i'
.~I~:::2
,
<
r
. b . r : 001
_
IS
.
given by
1
(92
-
91) +
3
(03)
1000 x 103 x 32 x 10- x
2 (sin 2 91 -
sin 2 9 ) ]
2
[
x 0.12265 (2.199-0) + ~ (sin 0° - s'in (2x J 260»]
2
I ,;: 787.27 N-m
I
ffriction force MF about 01 is given by
Moment
0
MF ,;: ~. PI . b . r
[ r (cos 9
°°1
cos 92) + 4 (cos 2 82 - cos 2 (1)
1 -
0.32 x 1000 x 103 x 32 x 10-3 x (
z:
T
[ (0.3)
[cos
°i3)
]
x
0° - cos 126° ] + 0 . 12265
4
[ cos (2
x
126°) - cos 0°] J"
::: 304.174 N-m
Then theactuating force is given by
F=
MN - MF
I
. (Ii) Braking
787.27 - 304.174
= (0.112 _ 0.100)
capacity (T JJ
= 2278.75 N Ans.
1)
:
Weknowthat the braking torque (T B) (i.e., torque applied by right hand side shoe),
T B = Jl PI' b . i2 (cos 9 I
-
cos 92)
= 0.32 x 1000 x 103 x 32 x
IO-J x
(Oi3
J
(cos 0° - cos 126°)
= 365.83 N-m
~e torque contributed by the left hand shoe cannot be obtained until we know its
.
.
.
and M operatlllg
pressure. For identical shoes, It
can be seen from th e ~xpres~lor.s
~f M N
F
that both are proportional to P For left hand shoe, the maximum mtensity of
press
.
I.
I
ure IS taken as p ,
III
aXlmum
I
I.
'/
M'N
=
M'F
=
and
1 POr the I
I
I
j
787.27 PI'
PI
304.174 PI'
PI
781.27 PI'
= 1000 x 103
304.174 PI'
= 1000 x
== 7.813 x 10-4 PI'
==
3.042
x
10- 4 PI'
)03
eft-hand shoe ,
F
==
MN '+M' F
I
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11.48
....~ 8. -.
r
..
44 .
11
w
t
Ilpl'
10 "PI'
0.112 .
<1
lin
I
'
== 445.04
kPa
rque applied by left hand shoe i gi en by
To'
The braking
=
J..l' PI' . b . r (cos 91
=
0.32
=
162.8 N-m
-
C
)
'
10-) x
445.04
2
(03)2
apacity is the total torque .
.. Total torque, T
=
TB
T'
= 365.83 + 162.8
=
528.64 N-m
Ans. ~
11.10. EXTERNAL CONTRACTING SHOE BRAKES
As discussed earlier, in block (or shoe) brakes, the shoes are brought in COntact a
pressed on external brake drum surface. Thus the block brakes are often called e.\1ernaJl,
contracting or closing brakes. The construction and working principles are same as thal:r
the external contracting clutches (refer Section I0.12).Such brakes are commonly used in a
lift or elevator.
11.10.1. Construction and Working Principle
An externally contracting brake
use in a hoist is shown in Fig.II.25.
Since it is used in hoisting device,
the friction torque remains applied
upon the drum all the time, except
when the drum is required to move.
The spring is kept compressed
bet-veen retainer and left side brake
shoe lever, Thus the spring forces
shoe on left lever to press against the
drum. The displacement
of arm
causes the pivot to move towards
right. Then the link turns upwards
about pivot and pulling rod moves
toward left,
Levers
Spring
Pi
ivot
I
Retainer
Brake
Brake shoe lever
. g brake
F·Ig. 11.25. All externally cIOJIII
t~(
Because of this the .: rht J d I
.
. . shoe on .
'
IIg
ian ever IS pulled towards left thus pressmg Its
veOil
d rum. Thus both the shoes'
d
.
'.
. narY,e
lif
h'
.
pi esse against the drum and hence it remalllS statiO
th e I tor orst IS loaded.
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When
the machine
is requirtd
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to op
I) .49
erate movin
to flow through the electric Illoto·
g up or down an I '
\ve d
.
I coupled witl
h
,e
ectnc current is
aliaell floWS through M which bouses on e Iectromag 1 t e drum. At the same time
'
the
t
Cllrf 'olTlagnet in M and pulls down the lower Ie
net. .The current flow energizes the
IceO
c.'
I
ver resu It In .
e
he pivot, rorcing t te two levers apart
d
.
g Into rotation of upp I
bollt t
. .
an causlOg s '
er ever
a
the brake drum from friction torque a d
pring to compress further. This
e1eases
.'
I
n motor can rotate .
r I auah the CirCUIt, t le magnet releases the
II'
It. As Soon as the current is
'Lit t If
;::,
,PLI
mg rod and'
,
c I and shoe levers closer. Thus It creates fro ti
spnng force brings the left and
right I
IC Ion torque on the drum.
11. .2.Determination of Braking Torque
10
Consider the forces on the brake
the
drum
rotates
in
VI hel1
anticlockwise direction, as shown in
Fig.II.26.
The moments
of the
frictional
and normal forces
about the hinge
pin are the same as for the internal
expanding shoe.
Therefore moment
MN
of normal force,
=
~
and moment of frictional
PI . b . r- 0°1
Fig. 11.26. Forces on the brake
[
(92
-
91) +
t
(sin 281
The actuating
=
sin 2 92) ]
force,
00,
MF
-
M PI' b . r [ r (cos 8, - cos 82) + 4 (cos 2 82 - cos 2 8 I)
]
force (F) is given by
MN-MF
F
where the notations
=
I
have usual meanings.
11.11. ENERGY CONSIDERATIONS
. e to conv'ert t(inetic and
k is to absorb energy I. "
.
We know that the basic function of a bra e ld 1'0 dissipate the resulting heat WIthout
Pt.
A brake shou a s
o entlal energy into friction heat.
devel OPlllg
. very high temperatures.
11
Absorbed are
.11.1. Sou rces of Energy to be
1
(i)
Kinetic energy
0
f tran
sJation:
KE
- mv
:::=
2
2
,
__j
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jl..:I::.5~O~
--
D_es__,;;gn:::::....._.....::of:..._T~_a:.:...:n=sm~issio
-
~
.
Kinetic energy of rotation:
(ii)
KE
=
1
-2 I (1)2
(iii) Potential or gravitational energy (as in an elevator being lowered or an aut
olllob',
descending a hill) :
Ie
P.E.
.. Total energy absorbed,
ET
= W x x = Weight x Vertical distance
=
t
+
mv2
t
I (1)2 + W . x
'" (11.22)
This energy is equated to the work done by the brake.
The work done by the brake = T x
Er =
where
e
Txe
T
=
Braking torque in N-m, and
e
= Angle through with the brake drum rotates during the braking
period in radians.
=
21t (N, + N2)
where N) and N2
60
x
I
2
= Initial and final speeds; and
I
= Time of application of brake.
11.11.2. Heat Generated in Brakes
The heat is generated in the brake during braking operation due to rubbing betweenthe
block (or band) and the drum. In fact this heat is nothing but the work done against the
friction.
..
Heat generated,
= Friction force x Rubbing velocity
= F x v = (u RN) x v
Hg
Hg = f.l·p·A,v
where
[ .: p= ~
] .. , (11.23)
f.l = Coefficient of friction,
p = Average pressure
RN =
A
RN
= A '
Normal reaction,
= Projected area, and
" = Peripheral velocity of drum.
11.11.3. Heat Dissipated in Brakes
.
"
. . . iveo bY
The heat generated due to friction should be dissipated. The rate of dissipatIOn IS g
)
4
_
.
" (11,2
C· A ·.1T = C x A (Is -10)
.
a, -
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~P~-----~--~--~~----------------
V
~1~1.5~1
Hd - Rate of heat dissipatl'o
~ltere
C
= Overall
A
heat tra
W/m2°C
. W
n In
11
atts
.
,
ns er coefficIent in W1m2 "C
for III = 400e
'
-
29.5
=
Exposed heat dissipating surface area in m2,
Is = Average
temperature
of heat-dissipating surfaces in 0C, and
Ambient temperature
Ia -
in "C.
Exal1l Ie 11.19 Two externally closing shoes 120 mm wide and subtending angles of
at tne drum centre are applied on drum whose diameter is 450 mm and width
90° eac,'
• .
I
}50 m""
Determine tile heat dissipated by tile brake. Tile radiation heat transfer coefficient
~given by
Co;:Co(1
+ a LJt), where Co = 2.9
TIre temperature difference
' Data' . b sh oe
Gwen
C=Co(l +allt);
=
between tile drum surface and surrounding is 150 K.
120 mm ;
D
=
450 mm;
Co=2.9W/m2K;
Tofind: Heat dissipated
© Solution:
Wlm2 K and a= 0.076 per K
bdrum = 150mm ;
a=0.076perK;
~I = 150K.
by the brake (Hd)'
We know that the heat dissipated by the brake,
Hd
C x A x ~t
=
0076 x 150) = 35.96 W/m2 K
ISO K C = Co (1 + a Ill) = 2.9 (I + .
,
.'
n the brake drum is both on the outer and
The available surface area (A) for radIatIon 0
For st
=
inner surfaces.
Take thickness of drum,
I = 0.005 D
=
..
A =
outer surface
area of drum
I
nner surface area of drum
=
=
and Area of the shoes (for total
=
0 005 x 450 = 2.25 mm
f }
.
}
{Inner surface}
{ Area 0
Outer surface
+
f drum
- the shoes
f drum
area 0
{
area 0
W'dth = 7t D x bdrum
0 ter circumference x I
u
Width == (7tD - t) x bdrum
I
circumference
x
nner
o :
for semi-circle)
di angle 180 I.e.,
subten JOg
1!. D
x
bshoe
2
7t
3
)(
45 - 2.25 x 10A
= 7t
(0.45) (0.15)
+
1t
0 15)_ - (0.45) (0.12)
.
2
(0.
0.338 m2
x 150 == 1824.4
l'hen, heat dissipated,
Hd ::: 35.96 x 0.338
2 W Ans.
"'CJ
=
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Design of Transmiss;o
11.52
~
~
I Example
E
te problem, if tile shaft of the drl4",
the above xamp
d.
will be over heated or not when b
transmitting 45 kW., check ",IIether the rum
ralce sl.~
apply brake torque on the drum.
Given Data: From Example 11.19, Heat dissipated, Hd = 1824.42 W = 1.824 kW
11.20
© Sit'
o U IOn :
mce
th
.
l.r
shaft of the drum is transmitting
45 kW, therefore th
.
h
c
e hea
.
d
b
di
d
i
45
kW
But
from
our
calculatIOns
we
ave
round
that
th
requIre
to e isstpate
IS
.
e he t
diIsslpa
. t e d' IS on Iy. 1 824 kW . Therefore the drum will be overheated and reqUires
at
coolmg.
S·
I For
e
ADS. ~
11.12. TEMPERATURE RISE
The temperature
rise of the brake can be calculated
by using the relation
E
III = Cxm
III =
where
rise in °C,
Temperature
=
E
'" (l1.25)
Energy dissipated
by the brake in J,
C = Specific heat of brake material
=
m
INote I
in J/kg DC, and
Mass of the brake in kg.
Since the concepts of energy considerations
and temperature
rise are same for both
clutches and brakes, readers can see Chapter 10, Section 10.14 for more details.
I Example 11.21 I Aflywheel of mass 950 kg and radius of gyration 300 mm is rotating
at 400 r.p.rn. It is brought to rest by means of a brake. The mass of the brake drum
assembly is 4 kg. The specific heat of cast iron brake is 460 J/kg 'C. Assuming tha: the
total heat generated is absorbed by the brake drum only, calculate the temperature rise.
Given Data:
mr
=
Temperature
111m
=
N 1 = 400 r.p.m. ;
0.3 m ;
C = 460 l/kg °C.
m = 4 kg;
Tofind:
k, = 300
950 kg;
rise (~t).
© Solution : We know that the total energy to be absorbed by the brake,
Er
1
2
= 0 + 2" I (w~
- CD; )
=
I
2
2" mv2 + 2" I (WI
- W2 )
I
=
I
Initial angular velocity,
and
?
2" I(Wi -CD2)
(1)1
Final angular velocity, CD2
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=
=
2
+W xx
+0
2
7t
NI
60
: 2 7t (400)
=
41.88 radfs
60
0
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11.53
I
= - I «(1)2
Er
Total energy,
2
I -
2
(1)2)
=
2I
m,
k; «(J)~ -
(I)~ )
1
2
=
rllen t Ile
E,.
~t -
tet11perature nse,
x 950 x (0.3)2 (41.882
Cxm
74980.69
=-_.:....;;,..:..::..:.
460 x 4
=
-
0)
=
40.7SoC
74980.69 J
Ans. ~
j
I
REVIEW AND SUMMARY
,
,
!
Brake is a mechanical
I
device by means of which motion of a body is retarded for
1 I slowing down or to brmg
It
to rest, by applying artificial frictional resistance.
Thernain types of mecltanical
1
brakes are block or shoe brakes, band brakes, band and
I blockbrakes, internal expandmg shoe brakes, and external contracting brakes.
I
i
Threetypes of brake linings used are organic, semi-metallic and metallic linings.
I Braking torque in Block or shoe brake is given by
I
TB
/ ,I
/:!..F·l·r
= ~
[when the rotation of drum is clockrviseJ
,a/
a - f.1 c
[when the rotation of drum is anliclockwiseJ
/:!..F·l·r
TB = a + uc
(b)
where T8 - Braking torque,
r = Radius of drum,
F = Force applie.} at lever end,
f1 .- Coefficient offriction, and
a, c & I
=
Dimensions
of lever.
t
••
! I Equivalent coefficient
f.1'
where
I
2() =
offTictlOn
use w
{iiI
4/:!. sin 0
20 + sin 2()
Angle of contact
l uble sltoe brake is given by
Braking torque in Double block or c 0
T8 = f.1 r (RN/ + RNJ)
where
r
RNJ & RN2
.(
=
0
d hen 2B> 40 ;s given by
I)
/
=
=
Radius of drum,
. d . ht hand side shoes respectively.
Normal reaction on the left an rig
In Band brake system,
Tj
Tension ratio,
r;
Braking torque, TB
Where
TJ and T2
e
=
eJ.lB,and
= (T - TJ) r
J.
.
= Tenszon in
::::Angle
r ::::Ra
di
d slack sides respectively,
.h
the band on
Ilg I
an
0/ lap. and
--------~~~-IUS
Of
'J
the drum.
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-= 1.54
Design of Transl1Jissio
r~~------------------------------------------------------~
../ Force applied on the lever in Simple band brake is given by
(i)
F =
T,!!.a
[For clockwise rotation of the drum}
(iO
F =
T,!!.
·a
[For anticlockwise rotation of the drum}
For notations a, b, T, and T2, refer Fig. II. II.
Tension ratio in Band and block brake is given by
where
Tn
To
Tn
To
= [
=
=
I + J.l tan ()] n
I - f..J tan ()
Tension in the band on tight side (maximum tension),
Tension in the band on slack side (minimum tension),
2 () - Angle sub tended by each block at the drum centre, and
n = Number of wooden blocks.
Internal expanding shoe brake (Refer Fig. 11.22) :
Moment of normal force, MN ~ ~ .Pi . b . r . DO, [ (0, - 0,) + ~ (sin 20,- sin 20J)
Moment of frictional force, M F
=
f..J . Pt . b . r x
[ r (cos 0, - cos 0,) + O~J
(cos 2 02 - cos 20,)
MN-MF
I
Actuating force on leading (or left hand) shoe, F,
and Actuatingforce
= ---
on trailing (or right hand) shoe, F2
MN+MF
I
=
where the notations have usual meanings .
./'
Energy considerations:
'12 mv?
(i) Total energy absorbed by brake:
Er
(ii) Heat generated in brakes:
f..J x RN xv
Hg
=
=
+
'I2
loJ + W'x
f..J' p .A . V
=
i
(iii) Heat dissipated in brakes:
Hd
=
C x A x L1t
=
CxA x
«- te)
where the notations have usual meanings.
.
Temperature rtse : L1t
=
C
1
E
xm
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1
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I 1.55
REVIEW QUESTIONS
.do
ntiate a brake and a clutch.
DllIere
I· Differentiate brakes and dynamometers.
2·
r
cnOl
nerate the qualities
of the friction material used in brakes
.
3· What are the various types of brakes?
4.
5.
6.
I
What is a self actuating
or self energizing
brake?
When a brake becomes
a self-
locking?
What is the difference
in simple
and differential
band brake?
Which
would
be
preferred?
7.
Explain the arrangement
to
8.
done in band and block brake system. In what way it is superior
band brake?
Show that, in a band and block brake, the ratio of maximum and minimum
tensions
in
the brake strap is
=
I +J.Ltan9]n
[ 1-J.Ltan9
9. Describe with the help of a neat sketch the principles of operation
expanding shoe. Also deduce the expression for the braking torque.
of an internal
10. Describe with the help of a neat sketch the principles
of an external
of operation
contracting sh e.
II. Write short note
(ii) Temperature
on (i) Heat generated and heat dissipated in brakes, and
rise.
PROBLEMS
FOR PRACTICE
Problems on single block brake:
I. A single blo k brake is arranged as shown in Fig.II.2. In Fig.II.2, a = 200 rnm;
b == 250 mm; and c = 50 mm. The diameter of the drum is 250 mm and the angJe of
Contact i 90°. If the operating force of 700 N is applied at the end of a lever and the
coefficient f fricti n between the drum and the lining is 0.35, determine the torque that
rna be tran mined by the bl ck brake.
[Ans: 83.75 N-m]
'.
A ingle-hi
k brake with a torque capacity of 15 N-m as shown in Fig.II.2. In Fig.II.2
a == 200 rnrn: b = 4
rnrn ; and
= 60 mm. The diameter of the drum is 300 mm and
the angle f
nta t i 900. The coeffi ient f friction is 0.3 and the maximum pressure
n the brake lining i I /rnrn-. The" idth fthe block is equal to its length. CaJculate :
i)
the actuating f r e;
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11.56
Design a/Transmission
s.
----------------------------------------------~~------~
(ii) the dimensions of the block; and
(iii) the rate of heat generated, if the brake drum rotates at 50 r.p.m.
[Ans: (i) 93.33 N; (ii) 18.26 mm; 18.26 mm;
(iii) 39.27 WJ
Problems on double shoe brake:
3. The double shoe brake shown in Fig.1 1.7 must provide a braking torque of J 400 N~
In Fig.II. 7, a = 200 mm ; b = 450 rnrn ; and c = 80 mm. The drum diameter is 3;'
mm. Assuming coefficient of friction 0.4 for the brake lining and p = OJ Nlmm20
determine (i) spring force'S' required to set the brake, and (b) width of shoes. Which
shoe will have greater rate of wear and what will be the ratio of the rates of wear of th
oe
two s hoes? Take angle of contact for each block = 100 .
,[Ans: 1372.5 N; ] 55.5 mm; Left hand shoe; 0.533]
4.
5.
A double
shoe brake, as shown in Fig.II.7, is capable of absorbing a torque of
1400 N-m. In Fig.I1.7, a = 200 mm; b = 450 mm and c = 80 mm. The diameter of the
brake drum is 350 mm and the angle of contact of each shoe is 100°. If the coefficient of
friction between the brake drum and lining is 0.4; find (I) the spring force necessary to
set the brake; and (2) the width of the brake shoes, if the bearing pressure on the lining
material is not to exceed 0.3 Nzmm-'.
[Ans: 3587 N; 142.2 rnm]
Determine the capacity and the main dimensions of a double block brake for the
following data: The load weighs 60 kN and moves downward with hoist drum speed of
22 r.p.m. The hoist drum diameter is 1.2 m. The load must be stopped within a distance
of3.2 m.
Problems on simple hank brake :
6.
7.
A simple band brake is operated by a lever of length 500 mm. The brake drum has a
diameter of 500 mm and the brake band embraces 5/8 of the circumference. One end of
the band is attached to the fulcrum of the lever while the other end is attached to a pin on
the lever] 00 mm from the fulcrum. If the effort applied to the end of the lever is 2 kN
and the coefficient of friction is 0.25, find the maximum braking torque on the drum.
[Ans: 4.2 kN-m]
The simple band brake system, as shown in Fig. I 1.13, is applied to a shaft carrying a
flywheel of mass 400 kg. The radius of gyration of the flywheel is 450 mm and runs at
300 r.p.m. The coefficient of friction is 0.2 and the brake drum diameter is 240 mm. The
dimensions 'a' and '/' are 120 mm and 300 mm respectively and J3 = 150°. Determine:
(i) The torque applied due to a hand load of IOO'N,
(ii) The number of turns of the wheel before it is brought to rest, and
(iii) The time required to bring it to rest, from the moment of the application of the
brake.
[Ans: 32.4 N-m; ] 97 turns; 39.3 sec]
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11.57
0" di/ferentlfli hand brake:
1 ~~t",s. h the rope supports a load Wand is W
d
p'
"llnc,
k
oun round a b
I
In II
. I band bra e acts on a drum 800
d'
arre 450 mm diameter. A
1
cr. rentla
mm lameter wh' h .
l
'dille
1
the barrel. The two ends of the bands
rc IS keyed to the same
ft as
are attached to .
Shll I rum of the brake lever and at distances
f? 5
prns on opposite sides of
the fu C Ie of lap of the brake band is 2500 and tOI... mrn ~nd 100 mm from the fulcrum.
l e ang
.
ie coefficient of fricti .
f1
'mum load W which can be supported b h
ncnon IS 0.25. What is
" maxI
.
y t e brake wh
f
.
IJle . d to the lever at a distance
of 3000 rnm fi
h
en a orce of 750 N IS
apphe
rom t e fulcrum?
.
[Ails: 309 kN when drum rotates CW]
'a'erential band brake acting on the :xth of the c·
c
..
Ircumlerence of a drum of 450 mm
~. dlam
. eter
. attached to a
. , is to provide a braking torque of 225 N-m . 0ne en d 0 f t h e band IS
. 100 mrn from the fulcrum '. of the lever and the other end to ano th er pin
. 25 mm firom
pm
i a Iso actmg.
.
If t h e
the fulcrum on .'the other Side of It where the operating force
. IS
operatingforce IS applied at 500 mm from the fulcrum and the coefficient of friction is
0.25, find the two values of the operating force corresponding to two directions of
rotationof the drum.
[Ails: 16.6 N for CW; 266.6 N for CCW]
A dIll'
I
, ~ A differential band brake,
as shown in Fig.l1.14, has an angle of contact of 225°. The
,I. bandhas a compressed
woven lining and bears against a cast iron drum of 350 mm
I diameter. The brake is to sustain a torque of 350 N-m and the coefficient of friction
betweenthe band and the drum is 0.3. The dimensions 'a', 'b' and 'c' are 500 mm, 35
mm and 150 mm respectively.
Find: (1) The necessary force (P) for the clockwise and
anticlockwise rotation of the drum; and (2) The value of 'OA' for the brake to be selflocking,when the drum rotates clockwise.
[Ans: (1) 64 N for CW; 804 N for CCW; (2) 1]4 mm]
Problems
or.band and block brakes:
II.Ina band and block brake, the band is lined with] 4 blocks, each of which subtends an
angleof 200 at the drum centre. One end of the band is attached to the fulcrum .of the
. 150 m from the fulcrum. Find the force required at
brake Iever and the other to a pin
m
.
f 4 kN
TI
h f I n to give a torque 0
-m. re
~e end of the lever I metre long from t e_ u crut
[Ans : 1712 N]
diameterof the brake drum is 1 metre and J..L _ 0.25,
I)
h f which subtends an angle of 15° at the
i -. A band and block brake having 14 blocks, e~c o.
Th drum and the flywheel on
cent .
f 900
ffectlve dIameter.
e
re ISapplied to a drum a
111me
.
di
fgyration 0£500 rnm. The two
the
d h
ombtned ra IUS0
same shaft weigh 20 kN an
ave a c
.
id
f the brake lever at distance of
end f
.
PPoslte SI es 0
s 0 the band are attached to pll1S 011 0
f 200 N .s applied at a distance of 750
30 mm and 120 mm, from the tu
. Icrum. If a force. 0
I angular retardatIOn
. 0 f t he
. (b)
nlln ~.
.
braktng torque,
f360
10m the fulcrum,
find (a) maxlInum
f
the rated speed 0
r.p.rn.
drum'
t orne to rest rom
2 8 04
]
. and (c) time taken bv the system 0 c
.2390 N-m; 4.69 rad/s;.
sec
lake Jl:::: 0.25.
[Ans.
I
..
j
I
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11.58
Design of Trc .nsmt .
-----------------------____:~-=-..:...:....:::..:.::..!ISSIOn S
fulcrum centres = 75 mm; Distance of fulcrum centres and that of cam~ a .
.
fb
.
XIS both
the drum centre = 115 mrn; Distance of line 0 f action 0 raking f~ce from the
fro'll
= 100 mrn: Distance between the points where the cam acts on the tWo brak
cam a)(i
,
,
e sho.
S
mm. Each shoe subtends an angle of 90° at the drum centre. If the brak~ fore .es '" 30
and the coefficient of friction is 0.3, find the braking torque on the drum·~A e IS SOON
.' ssu
the reactions between the brake shoes and the drum passes through the poi k ~e that
n." blse .
the contact angle. Also assume that forces exerted by the cam ends on the tw
Ctlng
0 S",oes a
equa I.
[Ans: 555 N re
14. An automobile type internal-expanding
double-shoe shown in Fig.l1.24 is 250 -~,l
diameter and is actuated by a mechanism that exerts the same force F on each sh I11Ill In
shoes are identical and have a face width of 40 mm. The coefficient of frictl.o
The
".
n IS 032
The angle
can be assumed to be zero. In Flg.II.24, OA = 100.9 mrn; OB:::: 866 ' .
0,°2 = 100 mm and 92 = 120°. The maximum intensity of normal pressure is li " III Ill;
Irnltedto
I Nzrnrn-. Calculate:
?e.
e
(i)
the actuating
(ii)
the torque-absorbing
Problems
011
capacity of the brake.
[Ans: (i) 2089.88 N; (ii) 435 Nm]
energy considerations:
15. Two externally
drum centre
radiation
a
force F; and
closing shoes 100 mm wide and subtending
are applied on drum whose diameter
heat transfer coefficient
= 0.076
per K;
surrounding
is 160 K. Calculate
is 500 mm and width 150 mm.The
is given by C = Co (1
and the temperature
If the shaft of the drum is transmitting
+ a·.1t)
difference
the heat dissipated
angles of 90° each at the
between
where Co = 2.9 W/m2 K;
the drum surface and
by the brake.
60 kW, check whether
the drum will be over
heated when brake shoes apply brake torque on the drum.
[Ans: 2445 W ; Drum will be over heated]
Problems on temperature rise:
16. A flywheel
of mass 1000 kg and radius of gyration
350 mm is rotating at 500 r.p.m.1tis
brought
The
to rest by means of a brake. The mass of the brake drum assembly ·Skg
IS
.
specific
heat
generated
of cast-iron
is absorbed
brake
drum
is 460 J/kg
by the brake drum only, calculate
DC. Assuming
that the total heat
the temperature
rise.
[Ans: 73°C]
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TWO MARKS Q & A ..£du
ca tion is not a product:
mark, degree, job, money .: ,
In
h
t at order- it .
.
,
IS
a process, a never-ending one."
- Bel Kaufman
CHAPTER - 1 : FLAT BELT AND PULLEYS
t
Nanle the four types of belts used for transmissiOlI of power.
2. V -belts ,
\. Flat belts'
3. Ribbed belts;
and
4. Toothed
or timing belts.
2. Differentiate between open drive and cross drive of a belt drive.
I Open belt drive:
sed with
hafts arranged parallel and rotating in same direction.
I Cross belt drive:
ed with hafts arranged parallel and rotating in opposite direction.
3. Why are idler pulleys used in (I belt drive?
I Idler pulley are pr vided t obtain high velocity ratio.
I Man.' idler pulley
e eral parallel
are u ed when it is desired to transmit motion from one shaft to
haft.
/. "/len do ) ou use step/Jed pilI/e) drive?
h dri
h ft
ieppcd
r
n pu IIe drive i used f r changing the speed of t e riven saw
haft run
th drivin
S. Wire" do., ou
fa I and
U~'C
at
tant
dri c i
puucy
.n
. d \ .,It1
ppcd x h nevcr d sire
.1
. I
6. Wlwl are II,e mat riots
I. Lather.
1I
.
hi e shaft is to be started or
ed when the dri en or mac in
.
('.
't11the driving shaft.
Ut uucrtennu
WI
I [or bellllri,'e ?
WiC( J'
_. I-abri . and
11
pe d.
fast and loose pulley drive ?
II
I
1\
I ile
Rubber, 4.
tt n,
B I
ta
and 5. Nylon.
aa ,
bell drive.
d .
.
d the follower or riven.
. , f the driver an
It i th rati
et« ecn vel It~·
I '1 ratio?
. rJjrreclS II,e ve OCI.v
8, R helher lire 11,; kness of a belt OJ'
.
.r
Define velo il)' rauo oJ
y
ut
It I'
(I
i I .
n
1
=
9. De'ine
J
"f
,,'p 0
I
_
1-
""here
(
a h II. rVlral is lire effe
el)
d
It.
•
th
ali
2
r tsnv
=
111
u
1-10
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,
I 'Iy ralio of a bell drive .
li on .'e OCI
of SiP
he belt and pulley.
n bcn",een t
I
II
I
Belt thickness.
::::SI,po
.
J
where
.
fthebeltdnve.
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'
,
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Q~&=A~.2~
~
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Sle".,.
--------------_D_e_s~~~n_O~if_T<_ra_n_s_m~~~Sl~'o~nsy
--.::
....:::
".)
i ,
t.ji
10. Why the phenomenon of creep results in a belt drive?
The phenomenon of sudden contraction and expansion of belt when it passes fromslack
'.,_
i:
side to tight side is called as creep.
11. State the law of belting?
Law of belting states that the centre line of the belt as it approaches
, ' hiethe pulley must Ii 'In
I
a plane' perpendicular to the axis of that pulley or must re 111 t e pane of the pulley,
ot herwise the belt will run off the pulley,
~at~
.11 ~
·O~
I" ((''Ctl ~
J~aJ"e•
12. What is meant by ply in aflat belt?
Belts are specified according to the number of layers, e.g., single-ply, double-ply or
triple-ply.
13. What are the factors on which the coefficient of friction bet wee" the belt and pulley
If'hat
V·
I
'fOe \c
I
(i) ,
(ii)
depend?
1. Material of belt;
2. Material of pulley;
(iii)
4. Speed of belt.
3. Slip of belt; and
14. In an open belt drive, which side of the belt is tight, whether upper side or lower side?
J-Vhy?
./ In an open belt drive, the lower side of the belt is tight.
(iY)
)1. A 10
The
./' Because the driving pulley pulls the belt from lower side and delivers it to the upper
side. So it is obvious that the lower side of the belt is tight.
The
wh:
fati
15. What is centrifugal effect on belts?
./' In operation, as the belt passes over the pulley the centrifugal effect due to its self
weight tends to lift the belt from the pulley surface. This reduces the normal reaction
13. WJ
.f
and hence the frictional resistance .
./' The centrifugal force produces an additional tension in the belt.
16. Centrifugal tension in a belt drive
the power transmitted.
[Ans : will not effect]
17. Centrifugal tension has no effect on the power transmitted. Justify the statement.
When centrifugal tension is taken into account,
then total tension in the tight side, T'I
and
= T I + Te
total tension in the slack side, T 12 = T 2
We know that,
power transmitted,
+ Te
P = (Ttl - T 12) v
=
[(T I + Tc) - (T2 + Te)1 V
=
(TI - T2) v
Thus, the centrifugal tension has no effect on the power transmitted.
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14. (
25.
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6 ,rhot is initial tension in belt drives?
Q&A.3
/ . The tension of the be1t when a b I .
stationary,
. t
IS
erme
.
d as i
e t IS fitted
as mitial tension ofth
to a pair of pull
e belt
eys when th
.
00wertransmitted by a belt drive"
e system IS
19. ('
'l'~,
Without Co '
hen the tight Slue tenSion is equ I
nSldering the ' "
W
a to
initial tension'
,
--__
tim 1i
' ISmaxim11m
es t te centrifugaltension.
].OIl/hat
"j
will be the effect on the Il'm't'
[Ans . th ]
I Ing ratio 01'
'
'
ree
riction
between
the
belt
and
rim
,/,
'J tenSIOnsof a belt if th
.I'r. .
fi ?
oJ pulley is doubt d'
e coeJJ,clentof
same.
[Ans .
e .whlle angle of lap remains
. The ratro of tension will be squared]
21. Whatare the losses in belt drives ?
The losses in a belt drive are due to :
(i)
Slip and creep of the belt on the pulleys,
(ii) Windage
or air resistance
(iii) Bending
of the belt over the pulleys,
(iv) Friction in the bearings
to the movement
o
fbi
d
e t an pulleys,
and
of pulley.
22. A longer belt will last more than a shorter belt. Wily?
The life of a belt is a function of the centre distance between the driver and driven shafts.
The shorter the belt, the more often it will be subjected to additional bending stresses
while running around the pulleys at a given speed, and quicker it will be destroyed due to
fatigue. Hence, a longer belt will last more than a shorter belt.
23. Whatis wipping ? How it can he avoided in hell drives? .
/ If t he centre diistance b e twee n two pulleys are too long then the belt.' begins to vibrate
di ti
f motion of belt. ThIS phenomenon IS
III a direction
perpendicular to the irec Ion 0
y
•
0
called as wipping.
v"
Wipping can be avoided
. Iidl er s pulleys
by usrng
0
•
the lendencyof the belt to slip,
24. Crowning is done on a pulley
[Ans : reduce]
10 -----
of a pulley is crowned?.
25. Wlun is crowning of pulley?
(or) wily Ilteface
. htl
v"
The pulley rims are tapered
as crowning.
o
v"
26 L'
.
The crowning
ISIII,e
slig
d
This slight convexity is known
towards the e ges.
y
.
on a pulley nrn
I belt in centre
tends to keep t re
.
It drives,
0
.
while in motion.
used In be
J
different types of pu IIe"'S
(0)I Solid pulleys,
and
. 0) S lit pulleys.
(11
p
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_9&A.4
~
27. What are the materials used for pulleys ?
./ Fabricated steel,
./ Cast iron,
./ Wood or fibre.
v' Compressed paper,
28. In designing a helt drive, if hoth driving and driven pulleys are of same nul/erial,the
_____
pulley governs the design.
[Ans : smalIer~
CHAPTER - 2: V-BEL T5
1.
What type of belt would you prefer
if the
centre distance between the pulleys is small?
V-belt.
2.
State reasons for V-belt drive being preferred to flat belt drive ?
V-belt drive is preferred to flat belt drive due to the following advantages:
./ Power transmitted is more due to wedging action in the grooved pulley .
./' Higher velocity ratio (upto 10) can be obtained .
./' V-belt drive is more compact, quiet and shock absorbing .
./' The drive positive because the slip is negligible.
3.
Why slip is less in the case of V-helts when compared with flat belts?
The slip is less due to the wedging action in the grooved pulley.
4.
The included angle for
_____
degrees.
V-belt groove
is usually
5.
The included angle for the V-belt is usually
6.
What are the different cross sections of a V-belt ?
between
and
[Ans : 32° ; 38°]
_
A, B, C, D and E type V-belts.
7. How can you specify (or designate) V-belts ?
V-belts are designated by its type and nominal inside length.
8.
How will you determine the number of belts required in the design of V-belt drives?
Nurn ber 0fV - beIts
9.
=
Total power transmitted
.
Power transmitted per belt
Wluu is a 'Vsflat' drives?
In V-belt dr~ve, if the large grooved pulley is replaced by a flat faced pulley (and smaller
pulley remams V-grooved), then the drive is known as V -flat drive.
10. What are the materials used for Vrgrooved pulleys ?
Cast iron, pressed steel, formed steel and die cast aluminium.
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/l&A
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, ~~------------------
~Q~&~A.~5
CHAPT~R - 3: WIRE ROPES AND PUllEYS
1I'/ltn
doyOUprefer a wire rope drive ,
.
ropes are preferred whenever large po
.
Wire 50 rn).
wer IS to be transmitted over long distances
(Upto1
j.
1f'rittanyfour wire rope applications.
J. ,
J.
. I
're ropes are extensive y used in elevato
WJ'
rs, mine hoists, cranes, conveyors, and
d
suspension
bri ges.
alhatare the advantages
"j.
of wire ropes when comp
d . 'It
•
are WII chams andfibre ropes?
The advantages of WIre ropes are :
I Lighter weight and high strength to weight ratio.
I More reliable in operation.
I Silent operation even at high working speeds.
I Less danger for damage due to jerks.
I. Distinguishregular-lay and lang-lay ropes.
(0 Regular lay ropes: In these ropes, the strands are twisted into a rope in the opposite
directionto that of the wires in the strands.
(iOLang lay ropes: In these ropes, the direction of twist of the wires in the strand is the
sameas that of the strands in the rope.
Howare wire ropes designated?
S.
Wireropes are designated
wiresin each strand.
(or specified) by the number of strands and the number of
6, Whatdoyou understand hy 6 x 19 construction in wire ropes?
A 6 x 19 wire rope means a rope is made from 6 strands with 19 wires in each strand.
7. Give the
(a) 6
applications of the following wire ropes: (a) 6 x 7 rope ; (b) 6 x 19 rope.
x 7 rope:
Used as haulage and guy rope in mines, tramways and power
transmission.
(h) 6 x 19 rope: Used as hoisting ropes in mines, quarries, cranes, derricks, elevators,
8
•
well drilling, etc.
, list Out the various stresses induced in the wire ropes.
-I
Direct stress due to the weight of the load to be lifted and weight of the rope,
-I
-I
Bending stress when the rope passes over the sheave,
Stress d ue to acceleration
.
d
an
v
'
Stress d .
.
unng starting stoppmg.
I
f
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Q&A.6
Design a/TransmisSion
s
:~
~
9. Ropel'fail mainly by
and
[Ans: fatigue.
.
.
' wear}
J O. "While designing a wire rope drive, always larger sheave diameter IS pre/erred."
Justify the statement.
. ,
We know that the bending stress induced in the wire rope,
Cjb
= Er
dw
x
where
D
dJII = Wire diameter, and
D = Diameter of the sheave.
So it is clear that the bending stress induced in the rope is inversely proportional to h
diameter of the sheave. Therefore the sheave diameter should be fairly large in orde:
reduce the bending stress in the rope when they bend around the sheaves.
t:
CHAPTER - 4:
CHAIN DRIVES
When do you prefer a chain drive to a belt or rope drive ?
1.
Chain drives are preferred for velocity ratio less than 10, chain velocities upto 2S m1s
,
and for power ratings upto 125 kW.
2.
The limiting speed of a chain drive is
_
3.
Give any three applications of elwin drives.
Chain drives are widely used in transportation
and wood working machines.
4.
[Ans : 2S m1~]
industry, agricultural machinery, metal
What are the different types of chains?
I.
Link (or welded load) chains,
2. Transmission (or roller) chains, and
3.
5.
Silent (or inverted tooth) chains.
What are the applications of link (or hoisting) chains?
Link (or hoisting) chains are widely used
6.
v"
in low capacity machines such as hoists, winches and hand operated cranes as t'
main lifting appliances, and
v"
as slings for suspending the load from hook or other device.
How call you specify a link chain?
A link chain is specified by the pitch, outside width, and diameter of the chain bar.
7. Distinguish short link and long link chains.
(i)
Short link chains:
If pitch
known as short link chains.
s
3 x diameter
of the chain bar, then the chains
(ii) Long link chains: If pitch < 3 x diameter of the chain bar, then the chains
known as IOIlt! lin!. chains,
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r-:
Q&A.7
l,forks Q & A
rIIOM~----~------------------------------------------~~~
, \II
s. pepe"ding onandtile manufacturing
\,..-------;'
dOW
9.
accuracy, the link chains are classified as
does a ftoi~·ting-ihain differ from
P'
l.
[Ans: Pitched; Calibrated]
.
a ro Ier cham?
.'
s, also known as link cams
I I-loistll1g
h '
, " challl
in hOlstmg
machines.
are used t suspen d and lor lift the loads
,0
chains
d f
' .
I Roller chains, also known as trans rmssion
' ,
between parallel shafts using sprock t
' are use or transmlttmg power
,
e s.
_____
,and ----[Ans: Pitch, width; diameter of roller]
10. A roller cftain is specified by -----
11. What is chordal action in chain drives?
.
or s lOSea 0 a contmuOus
When cha in passes over a sprocket, it moves as a series of ch d ' t d f
.
arc as tn the case of a belt drive. It results in varying speed of the chain drive. This
phenomenon IS known as chordal action,
/2. What is the effect of chordal action in chain drives? How canyou reduce that effect?
.; Chordal action results in a pulsating and jerk motion of a chain.
.; In order to reduce the variation in chain speed, the number of teeth on the sprocket
,
, I
should be increased.
13.In chain drives, the sprocket has odd number of teeth and the chain has (!liennumber
0/
/inks.
Reason:
Why?
To facilitate more uniform wear, i.e., the wear will be evenly distributed and
thus total wear wi II be lower.
14. What are the materials used for making chains?
.; Link I
de of medium-carbon or alloy steels such as C 45, C 50 and
m
pates
are rna
de of carburising steels such as C 15, C 20, 30 NI. 4
40 Cr ].
.; Pins, bushings and rollers are ma
Cr ] and 15 Ni 2 Cr 1 Mo 15.
h . ?
15. What is an offset link in a roUer c aIH·
. . Irk
called 'offset' link, is
f linkS, an addltlona tn,
. .
When the chain has odd number 0
. links That's why It ISpreferred to
provided. But the offset link is weaker than the maIO I
!1~
1
16
.
.
.
h'
are pre/erred?
tions SIlent carns
.
What is a silent chain ? In ",hat sJluO
'
. b use of their relatively quiet
'1 nt chams eca
have even number of links.
•
" Inverted tooth chains are cailed 51 e
operation.
.
eed and smooth operation.
.
h' h_pOwer,hlgh-SP ,
" Silent drives are preferred for Ig
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----
D-eS_;ign~...::of~n_r_a_.:.n:.:.sm~;ss;onS
-
~
17. What are the possible waysby wh!IC h a chain drive mayfail ?
The four basic modes of chain failure are:
(i) Near;
(ii) Fatigue;
(iii) Impact; and (iv) Galling.
18. What is 'back sliding' in chain drives?
The wear of the chain results in the elongation of the chain. In other words, the pit h
length is increased. This makes the chain 'to ride out' on the sprocket teeth, resultinginc
faulty engagement. This is known as 'back sliding' of chain.
a
19. Roller chains are designedtofail not in tension but by wear. Explain.
The chain links may not fail by tensile loading. But the rollers and pins mostly willfail
by wear. Therefore roller chains are designed based on wear.
20. Whatyou mean by galling of roller chains? .
Galling is a stick-slip phenomenon between the pin and the bushing. When the loadis
heavy and the speed is high, the high spots (i.e., joints) of the contacting surfacesare
welded together. This phenomenon of welding is called as galling of roller chains.
CHAPTER - 5: SPUR GEARS
1.
What are the advantages of toot/led gears over the other types
systems ?
0/ transmission
Advantages of toothed gears are :
0/ Since there is no slip, so exact velocity ratio is obtained.
0/ It is capable of transmitting larger power.
0/ It is more efficiency and effective means of power transmission.
2. Medium velocity gears have a pheripheral velocity range
0/
mls.
[Ans: 3 to 15 rols]
3.
The size of the gears in metric system are usually specified by their
-[Ans: Module]
4. Back lash for spur gears depends on which twofactors.
l , Module;
and 2. Pitch line velocity.
5. Define transmission ratio with reference to spur gears ?
It is the ratio of speed of driving gear to the speed of the driven gear.
6. Define module.
It is the ratio of the pitch circle diameter to the number of teeth.
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:~
Q&A.9
pair of spur gears, the module is 6
;~t,al
pilch:
.
mm: Deter"'ine the circular pitch and the
f,
(i)
Pc
Circular pitch,
(ii) Diametra
8. Statethe
= 7t
X
m
== rr x 6 == 18.85 mm
Ans."
- 2!_
I
I
I . h
pitc , Pd - Pc == m == 6' == 0.166 mm Ans."
-r': of correct gearing (or) state the law o/gearing.
obtaining a constant I"
.
The laW of gearing states I that for
I
ve ocny ratio, at any Instant of
teeth the comm?n ~o.r~a at eac 1 of contact should always pass through a pitch point,
situated on the line joirung the centres of rotation of the pair of mating gears.
IAGMA' standsfor what ?
9.
The American Gear Manufacturers
Association.
10.If the centre distance of the mating gears having involute teeth is increased then the
pressure angle is
.
[Ans: increased; because centre distance is inversely proportional
to the cosine of pressure angle.]
11.Whatare the conlmon forms of gear tooth profile?
I. Involute tooth profile,
and
2. Cycloidal tooth profile.
12, W/,atare the standard interchangeable tooth profiles?
14 ~
0
composite
(ii) 14 ~
0
full depth involute system;
(i)
system,
(iii) 200 full depth involute system, and
(iv) 20° stub involute system.
full deplh.
13, Thebest gear system to resist wear is
.
[Ans: 20° involute]
[Ans: 200 involute system]
14. Thecommonly used gear looth profile is
d.
th 10 lake heavy loa'S.
15.
system ahs a strong too
[Ans: 200 stub involute system]
16 D
'
.r:
e 10 spur gears.
.
ld give the same
, efine pitch circle With reJerenc
Il'ng action, wou
.
h' h by pure ro I
· circle is an imagmary
.
rcle w IC
PItch
CI
motion as the actual gear.
17 .
. JII"atare the materials used for g
ear manu/aclurl
'ng?
.
.
.
and bronze.
nthetic resms.
I, Metallic gears: Steel, cast Iron,
d paper and sy
se
2
whide, compres
. Non-metallic gears: Wood, ro
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Design of Transmission SlJ
Q&A.IO
.l'stetn,s
~
18. When do you use non-metallic gears?
Non-metallic gears are wherever noiseless operation is required. They are cheaper in
Cost
and provide damping for shock and vibration.
19. What are the non-metallics used as gears?
Wood, rawhide, compressed paper and synthetic resins like nylon.
20. Why is pinion made harder than gear?
Because the teeth of pinion undergo more number of cycles than those of gear and hence
quicker wear.
21. List out the various methods of manufacturing a gear.
I. Gear milling.
2. Gear generating:
3. Gear molding:
(i) Gear hobbing;
(ii) Gear shaping.
(i) Injection molding; (ii) Die casting; (iii) Investment casting.
22. What are the main types of gear tooth failure?
1. Tooth breakage (due to static and dynamic loads).
2. Tooth wear (or surface deterioration) : (a) Abrasion;
(b) Pitting; and (c) Scoringor
seizure.
23. What are the assumptions made in deriving Lewis equation ?
1. The effect of radial component, which induces compressive stresses, is negligible.
2. The tangential component is uniformly distributed across the full face width.
3. The tangential force is applied to the tip of a single tooth.
4. Stress concentration in the tooth fillet is negligible.
24. The Lewis beam strength equation for spur gears is
----[Ans: F, = 1t x m x b x [<Jb] xy]
25. What is Lewis (or tooth) form factor ?
The Lewis equation is given by, Fs =
where [6 x ~2x Pc ]
= y,
c.
c.
L ewrs. rorm
ractor
a
.
1t
x m x b x [ crb] x [
(2
6 x h xPc
]
known as Lewis form factor.
di
.
d pare
imensionless quantity. Since the dimensions t, h an c
.
d
.
I
. alwayS
Increase proportionate y when the gear is enlarged, the form factor value IS
constant.
IS
26. Why is a gear tooth subjeclet/lo dynt,mic loading?
In a gear tooth, dynamic loads are due to the following reasons:
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,{ar*J
Q~---------
r=:
j
----
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I
I In:JCcuracies of tooth spacing..t'
I Elasticity 0 f parts,
I
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.t'
---~-Irrcgul '.
. antics in tooth profile
__Q&AJ_!_
Misalignment bctwec be . '
I Deflection of teeth undcr load ,
v'
D
n ~nngs
ki
h
'
ynamic
ubI
.
B
write the uc Ing ams equation fi dna
ance of rota"
1.
or ynumic load a
. 109 mas ses.
load
lid aprasion for li"'iting W~ar
I
I Dynarnicload,
Fd == F, + _21 "(~C+Ff)_
I Limiting wear load,
F
W
21 v+Vb.c+F
== d x b Q
I
x x K
f
\II
)
WIlenboth the pillion and the gear are
_,
made of II,e s
.
should be designed
ame malenal, then
D
(I..
----
19. Why are the load correction factor and d na .
[Ans: pinion]
design ?
.v mu: loadfactor are considered in gear
./ Load correction factor is considered to aceo t c
. .
.
s:
idth
f
J
un
lor
uneven
d,stnbutlon
of tooth load
a Iong tJie race WI
0 t te tooth,
./ Dynamic
, d crrcu
ci Iar puc
. h errors, and
.. load factor is considered to account for profile an
velocity of operation.
30. Whal are the conditions required for interchangeahilityin toothgears?
Pressure angle and module should be same.
31. Whal are the effects of increasing and decreasing the pressure angle in gear design?
./ Increasing the pressure angle will increase the beam and surface strengths of tooth.
But gear becomes noisy .
./ Decreasing the pressure angle will increase the minimum number of teeth required on
the pinion to avoid interference / undercutting.
32. The difference between the tooth space and the tooth thickness measured along the
., h . I •
[Ans: Back lash]
pile ctrcte
IS
•
CHAPTER - 6: HELICAL GEARS
1. Who do you prefer helical gears than spur gears?
v'
v'
Helical gears produce less noise than spur gears. .
. than eqUivalentspur gears.
Helical gears have a greater load capacity
2. Where do we use helical gears?
d'
He real gears are commonly use
3. fl;1."
"nal is the major disadvantage
V
0'(
'J
sinole helical gear.
b
.
loads.
aXIal thrust
. I (. herringbone)
SIngle helical gears are subJecte
. double hellca i.e.,
,
d by uSing
.
b elimmate
ThIs axial thrust loads can e
gears.
difficulty?
v'
Jl1
.
nd high speed applications.
tomobiles, turbllles, a
thai
au
.
., HoW can you Ol'ercome
.
.
d
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Q&A.12
4.
.DeSignOfTranSmiS~
In helical gears, the right hand helixes on one gear will mesh
the other gear.
5.
q
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[A
heli"._ ..
-c ..011
ns : Left hand]
What hands of helix are used in parallel helical gears ?
In parallel helical gears, opposite hands of helix are used. i.e., a right hand pinion meshes
with a left hand gear and vice versa.
6. Define axial pitch of a helical gear.
The distance between corresponding points on adjacent teeth measured in the plane
parallel to the shaft axis is known as axial pitch.
7.
The helix angle for single helical gears ranges from
[Ans : 15°to 250]
8.
What helix angle is adopted for a double helical gear ?
45°
9.
What is virtual (orformative) number of teeth in helical gears?
The number of teeth on the virtual spur gear in the normal plane is known as virtual
number of teeth (zeq ).
Z
Zeq
-
cos! J3
where
Z
Actual number of teeth on a helical gear, and
-
J3 -
Helix angle.
10. A pair of helical gears consists of a 20 teeth pinion meshing with a 70 teeth gear. The
normal module is 3 mm. Find the required value of the helix angle if the centre
distance is exactly 150 mm.
@Solution
:
Centre distance, a -
150
or
or
Helix angle,
11. What are the components
(i)
-
( =, )
cos
x
J3
( cos3 J3 )
x
(Z
I
+ z2 )
eo
Z
+
2
70)
J3 - 25.840 Ans. "
of resultant forces acting on a gear tooth of a helical gear?
Tangential component:
Ft =
2 x Mt
d
(ii) Radial component:
F, ~ FIx [ t::s~ ]
(iii) Axial or thrust force component:
Fa - F t X tan ~
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~~#~~Q&~A~
~Q~&A~.~J3
(tt'o
•
",rile the expressions for beam strength, dynamic load, and limiting wear load for
J]' helicalgears.
I
I
I
Beam strength :
Fs
=
1t x Inn X
X [ CJb ] X
Y
F + 21 v (cb . cos2 J3 + F,) cos J3
Dynamic load:
,
Wear load:
b
Fw
=
21 v +
-v cb . cos
d1xbxQxK
cos- J3
2
J3 + F,
'"
IJ. Differentiate double helical and herringbone gears.
./ When there is groove in between the gears, then the gears are specifically known as
double helical gears .
./ When there is no groove
in between the gears, then the gears is known
as
herringbone gears.
J4. Whatare spiral (or skew) gears? What hands of helix are used?
./ A pair of crossed-helical
./ In most applications,
gears are known as spiral gears .
the spiral gears have the same hand.
15.Wheredo we use skew (or spiral) gears?
Skew gears are used to connect and transmit motion between two non-parallel and nonintersecting shafts.
16.Whyis the crossed helical gear drive mostly not usedfor power transmission?
As the contact between the mating teeth of crossed helical gears is always a point, these
gears are suitable only for transmitting a small amount of power. That's why mostly
these gears are not used for power transmission.
17.Whatis the condition and expression for maximum efficiency in spiral gears?
(i) Condition for maximum efficiency:
(ii)
Maximum efficiency:
where
9
llmax
= 9; $
Spiral angle, J3
cos (9 + <1» + 1
+1
= cos (9 -~)
= shaft angle, and ~ = angle of friction.
CHAPTER - 7: BEVEL GEARS
I. Underwhat situation, bevelgears are used?
Bevelgears are used to transmit power between two intersecting shafts.
2. Differentiate a straight bevel gear and a spiral bevel gear.
.
When t h
'
. ht the gears are known as straight bevel and
eet formed on the cones are straig ,
Wheninclined, they are known as spiral or helical bevel.
,
I
I
lScanned by CamScanner
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--------
~Q&A. 14
S'
----::t.!!
etnss
=<
D_ej_'/..:::·g:_n_o...:.if_r;_rU_1I_1.\_,m--....:..:ission
.
vel ears over straight bevel gears?
What
are
the
advantages
of
spiral
be
g
.
3.
. and quieter than straight bevel gears
. bevel gears are srnoo ther in action
.
Spiral
What is zerol bevel gear?
5.
'.
h b t with a zero degree spiral angle
Spiral bevel gear with curved teet
u
bevel gear.
_____
gearl' are use«i ln automobile differential unit.
6.
What is a crown gear ?
4.
known
IS
as Zerol
[Ans: HYPOid)
·
A b eve I gear Iravmg
a piit c h angle of 90° and a plane for its pitch surface
.
is know n as a
crown gear.
7.
What is tile difference setween
OIl
angular gear lind
miter gear?
II
./ When the bevel gears connect two shafts whose axes intersect at an angle other than a
right angle, then they are known as angular bevel gear.f .
./ When equal bevel gears (having equal teeth and equal pitch angles) connect two
shafts whose axes intersect at right angle, then they are known a miter gears.
8.
Equal bevel gears when they connect two shafts at right angles are known
OJ
[AII~' : miter gears)
9.
What is meant by pitch angle ill bevel gears? Explain how tile pitch angle of a bevel
gear determines whether tile gear is an external or internal bel el gear .
./ Pitch angle (or semi-cone angle) is the angl
gear aXIs.
mad
by rh pit h line of a gear wilh the
./ If sum of pitch. angles is less than 0°. Ihell III
gears. Those gears having urn of pit h anal . III rrc
gears.
bevel
than
rear are external bevel
0, they
ar internal bevel
10. For bevel gears, define 'back cone distance ..
Back cone distance i the len til f the ba k
perpendicular t the pitch c nc at the end I rh
11. What are the forces acting
I. Tangential
force,
2. Axial force
and
011
, )Ill:.
I
0," k
IlC
an rmaginar
cone,
)111.
a bevel gear ?
3. Radial force.
12.
If the axial force componellt
pinion is'
011
.
13. If tile radial force componellt
pinion is
Oil
the
I:enr
.
IS
625 N, then lite radial component on Jlre
[Ails:
_ ~ . but in oppo it dire lion]
'lte'~
gear ts 1200
fAn
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"hell
: J _0
,
'he axial component on lIre
: but in opp ite dire Ii 0]
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}.larks Q & A
14. Write the expressions for bea
bevelgears,
III strength , drynanrlc.
Q&A.IS
10 d.
a , and lim't'
./ Beam strength :
I
Fs
./ Dynamic load:
==
7t X In
t
xb
x
[
(
0b]Xy'X
mg weal' load for
B.;b)
F + _ 21 v (be + F )
t
t
21 v + \[be + F-
Fd ==
t
./ Wear load:
0.75 x d x b
F w ==
I
X
Q' x~
cos 01
CHAPTER - 8: WOR
1. Under what situation, worm gears are used?
M GEARS
,
The worm gears are used to tran
.
ft
df
.
srmt power betwee
.
sha s, an or high speed ratios as high as 300 : 1. n two non-intersecting, non-parallel
2. Where do we use worm gears?
Worm gears are used as a speed reducer in
teri I
.
.
and automobiles.
rna ena s handling equipment, machine tools
3. What is irreversibility in worm gears?
The worm gear drives are irreversible. It means that the motion cannot be transmitted
fr~m.worm ~he.el to .the worm. This property of irreversible is advantageous in load
hoisting applIcatIOns like cranes and lifts.
4. What are single-enveloping
and double-enveloping worm drives?
q _ Diameter factor:::: d)/mx'
mx ==
6
Axial module.
. Define normal pitch of a worm gear.
?
the nonnal to
It IS the distance measured alo g
.
.
d of the wonn·
POmtson two adjacent threa s
.
n
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di
h threads between rwo correspon rug
t e
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Design of Transmiss'
IOns
Q&A.16
.
~
h
hni arrelationship between the normal lead (L.I
7. For worm gear, give t e tee ntc
IV" IOlJd (
.
-14
and lead angle (r).
LN = L x cosy
B.
A patr'.1"OJ worm gears
e Solution: (2/54/
... Gear ra ti10,
9.
t
•
is
as 2/
"I-Ie'''',·gnated·
'"
54 / 10/ 5. Find the gear ratio.
10/5) : (zl / z2 / q / m)
z2
54
= Z1
- = -2 = 27 Ans. ~
What is the velocity ratio range of worm gear drive ?
Velocity ratio ranges from 10 : 1 to 300 : 1.
10. The direction of rotation of the worm wheel can be found by considering the WO'IfIIlJ
_____
and worm wheel ~.
[Ans : screw; nut]
11. Differentiate self-locking and overrunning worm drives.
,/ The drive is called self-locking,
if f.l ~ cos
(l •
tan y.
./ The drive is called overrunning, if f.l < cos y . tan y.
is the material widely used in the manufacture of worm wheeL
12.
[Ans : Phosphor bronze]
13. Why phosphor bronze is widely used for worm gears?
Phosphor bronze have high antifriction properties to resist seizure. Because in worm gear
drive, the failure due to seizure is more.
14. List out the main types of failure in worm gear drive.
1. Seizure;
2. Pitting and rupture.
15. In worm gear drive, only the wheel is designed. Why?
Since always the strength of the worm is greater than the worm wheel, therefore onlythe
worm wheel is designed.
16. For transmitting
Why?
large power, worm reductions gears are not generally prefemd.
In worm drive, meshing occurs with sliding action. Since sliding occurs, the amountof
heat generation and power loss are quite high.
17. Why is dynamic loading rarely considered in worm gear drives?
In worm gear drive, dynamic load is not so severe due to the sliding action betweenthe
worm and worm gear.
18. What are the various losses in the worm gear ?
./
Losses due to friction is sliding (i.e., gearing loss), and
./
Losses due to the churning and splashing of lubricating
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oil.
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~j~~~-----------~
I"
-~
,.,0'''' gearing
~Q~&~A~.1~7
+:
heat
is an important design requirement. Why?
/9.
the worm gear drives produce much h t U I
eeause.
..
ea . ness proper heat removal
S 'd d the drive may eventually fall by seizure
proVI e ,
.
Jl'haJ situations
J.
Gear
=:
IS
CHAPTER - 9: GEAR BOX
use of gear boxes ?
boxes are required wherever the variable spindle
Jl'rite any two requirements
ds i
spee s IS necessary.
of a speed gear box.
1. I Gear box should provide the designed series of spindle speeds.
I Gear box should transmit the required amount of power to the spindle.
].
1J'/Iy G.P. series is selected for arranging the speeds in gear box?
I The speed loss is minimum, if G.P. is used.
I The number of gears to be employed is minimum, if G.P. is used.
I G.P. provides a more even range of spindle speeds at each step.
I The layout is comparatively
very compact, if G.P. is used.
4. List any two methods used for changing speeds in gear boxes.
I. Sliding mesh gear box, and
2. Constant mesh gear box.
5. What are preferred numbers?
Preferred numbers are the conventionally rounded off values derived from geometric
series.There are five basic series, denoted as R 5, RIO, R 20, R 40 and R 80 series.
~ What is step ratio? (or) Define progression ratio.
Whenthe spindle speeds are arranged in geometric progression, then the ratio between
thetwo adjacent speeds is known as step ratio or progression ratio.
7, What is kinematic arrangement
as applied to gear boxes?
The kinematic layout shows the arrangement of gears in a gear box. It also provides
infonnations like number of speeds available at each spindle and the number of stages
used.
S,
What does the ray-diagram
of gear box indicates?
The ray diagram is a graphical representation of the drive arrangement in general form. It
9
serves to determine the specific values of all the transmission ratios and speeds of all the
shafts .
In the drive
•
, Slale any three basic rules to be followed while designing a gear box.
l. The transmission ratio (i) in a gear box IS
. I'irnrte
. db y J;.;4 -< i -< 2 •
.'
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o/Transmiss;o
~
------z
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stage should not be great
.
seed ratio of any
,
er than 8
2. For stable operation, the p
i.e., Nmax / Nmin ~ 8.
N
>N .
tage N
~ input
min
J In all stages except in the first s
, max
.
t
·/,ieve 12 speetll'from a gear box?
10. What are the possible arrangements
0 (IL
The possible arrangements are:
and (iii) 2 x 2 x 3 scheme.
..)
2
3
x
2
scheme
;
(i) 3 x 2 x 2 scheme; ( II
x
d
b
11. List out the possible arrangements
(i) 4 x 2 x 2 scheme;
(ii) 2 x 4 x
achieve 16spee gear ox.
I
0
2
he'
an
sc em ,
d ( ...) 2
111
12. WI,at is a speed reducer?
x
2
x
4
h
sc erne
.
.
.
h
.sm with a constant speed ratio, to reduce the a
Speed reducer IS a gear mec am
.
h ft
ngular
speed of output shaft as compared with that of input sa.
CHAPTER - 10: CLUTCHES
1.
What is the function 0/ a clutch?
The clutch is a mechanical device which is used to connect or disconnect the SOurce of
power at the operator's will.
2.
Give examples for axial and radialfriction clutches .
./' Axial friction clutches:
Disc and cone clutches .
./' Radial friction clutches:
Centrifugal, internal expanding rim and external contracting
rim clutches.
3.
Whut are the properties required of the material used as afriction surface?
The properties required of the friction materials are:
./' A high and uniform coefficient of friction .
./' Geed resiliency .
./' The ability to withstand high temperatures,
./' High resistance to wear, scoring.and
together with good heat conductivity.
galling.
4. Name few commonly used friction materials.
Wood, cork, leather, asbestos
materials.
5.
based friction
materials,
and powdered
metal friction '
Clutches are usually designed on the basic of uniform wear. Why?
In clutches, the value of normal pressure,
axial load for the given clutch is limited by the
assumptio:::
rate of wear that c~ be tolerated in the brake linings. Moreover, the
un~form wear rate gives a lower calculated clutch capacity than the assumpuo
uniform pressure. Hence clutches are usually designed
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on the basis of uniform
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wear·
r:
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~
.
pistl"g"
~
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_
ish between dry and wet operation of clutches.
.
hen a clutch operates m the absence of a lubricant, then that clutch is known as dry
I ~utch. In dry clutch the torque capacity is high but the heat dissipating capacity is
iowI When a clut~h ~perates '~et.' (i:e., with lubrication), then torque capacity is low but
the heat disslpatmg capacity IShigh.
Jrhy a service factor is used/or calculating tile design capacity of a clutch ?
7. _ In order to start a load from rest an~ accelerate it, a clutch should have torque capacity
b tantially greater than the nom mal torque requirement so that the load can be
SU s
.h
.
I' S
.
accelerated Wit ~ut. excessive Sip. 0 service factor is used to account for driver and
driven source variation and frequency of operation.
8.
In cone clutches semi-cone angle should begreater than 12 ~ Why ?
We know that the torque capacity is inversely proportional to sin a. The value of a
should be as small as possible. But when 'a' is less than the angle of static friction (4)>),
the clutch has a tendency to grab, resulting in self-engagement. The self-engagement is
not desirable because the clutch should engage or disengage only at the operator's will.
To avoid self-engagement and to facilitate disengagement,
a > Angle of static friction.
a > tarr ' {u)
Taking the coefficient of friction as 0.2, we get
a > tarr-' (0.2) or a> 11.3°
Therefore, the semi-cone angle should be greater than 12°.
9. What is the axial force
clutch?
required at the engagement and disengagement
0/
cone
W = W n (1 + Jl cot a)
10.Whatis the difference between cone and centrifugal clutches?
Coneclutch works on the principle of friction alone. But centrifugal clutch uses principle
ofcentrifugal force in addition with it.
1I.Compare disc clutches and cone clutches .
I
./' In disc clutches , friction lined flat plates are used.
./' In cone clutches friction lined frustum of cone is used.
'
2. ListOutthe elements of internal expanding rim clutc/Jes.
1. The mating friction surface,
2. The means of transmitting the torque, and
3. The actuating mechanism.
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Q&A.20
Design afTrans
-----------------------
~. .
""SSIOn
S
Y3/el1J.t
13. Wi,y heat dissipation is necessary in clutches?
When clutch engages, most of the work done (against friction forces
motion) will be liberated as heat at the interface. Consequently the tern oPPosin&
th
bbi
.
.
d
t
t
d
perature
ru mg surface will increase. This mcrease empera ure may estroy the I
of tL e
di . . .
c utch S lie
tssrpation IS necessary in clutches.
. 0 heat
14. Give the relation to find temperature rise in clutches.
Temperature rise, dT
where
E
=
=
E
Cxm
Energy dissipated by the clutch,
C = Specific heat of clutch material, and
m
=
Mass of the clutch.
CHAPTER - 11 : BRAKES
1.
Wtuu is thefunction of a brake?
Brake is a mechanical device by means of which motion of a body is retarded for I .
down or to bring.it to rest, by applying artificial frictional resistance.
S OWmg
2. Differentiate a brake and a clute".
A clutch connects two moving members of a machine, whereas a brake connectsa
moving member to a stationary member.
3. Differentiate a brake and a dynamometer.
A dynamometer
applied.
4.
is a brake incorporating a device to measure the frictional resistance
Give examples for radial and axial brakes.
Radial brakes: Band brakes, block brakes, and internal expanding rim.
Axial brakes: Cone brakes and disc brakes.
5.
What are the types of brake linings?
I. Organic linings,
6.
2. Semi-metallic
linings, and 3. Metallic linings.
What is a self-locking brake?
When the frictional force is sufficient enough to apply the brake with no externalforce,
then the brake is said to be self-locking
7.
brake.
What you meant by self-energizing brake?
fi' . nal force
When the moment of applied force (F . l) and the moment of the r~ctJo brake,
(11 . RN . c) are in the same direction, then frictional force helps in applymg the
This type of brake is known as a self-energizing
brake.
___
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--------------
~---
,",0 blocks diametrically
s.
If on
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opposite to each other are used in a block brake. Why ?
Iy one block is used for braking, then there will be side thrust on the bearing of
.
.
.
wheel shaft. ThiS. drawback c~n be removed by providing two blocks on the two sides of
the drum diametrically opposite.
Writean expression for the ratio between the tight and slack sides of a band and block
9.
brake.
T1
T2
10. The brake commonly
=
(1 +
J.l tan e
1- J.l tan
e
)n
used in automobiles is __
--
[Ans : internal expanding brake]
[Ans : disc brake]
11.Thebrake widely used in motor cycles is_---
[Ans: block brake)'
12.Thebrake used in railway coaches is_---
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