kinematics of machines compiled t.v.govindaraju
TRANSCRIPT
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e-Notes by Dr.T.V.Govindaraju, Principal, Shirdi Sai Engineering College,
Bangalore
4.0 Gears:
Introduction: The slip and creep in the belt or rope drives is a common phenomenon, in thetransmission of motion or power between two shafts. The effect of slip is to reduce thevelocity ratio of the drive. In precision machine, in which a definite velocity ratio is
importance (as in watch mechanism, special purpose machines..etc), the only positive drive
is by means of gears or toothed wheels.
Friction Wheels: Kinematiclly, the motion andpower transmitted by gears is equivalent to that
transmitted by friction wheels or discs in contact
with sufficient friction between them. In order tounderstand motion transmitted by two toothed
wheels, let us consider the two discs placedtogether as shown in the figure 4.1.
When one of the discs is rotated, the other disc will be rotate as long as the tangential forceexerted by the driving disc does not exceed the maximum frictional resistance between the
two discs. But when the tangential force exceeds the frictional resistance, slipping will take
place between the two discs. Thus the friction drive is not positive a drive, beyond certainlimit.
Gears are machine elements that transmit motion by means of successively engaging teeth.
The gear teeth act like small levers. Gears are highly efficient (nearly 95%) due to primarily
rolling contact between the teeth, thus the motion transmitted is considered as positive.
Gears essentially allow positive engagement between teeth so high forces can be transmittedwhile still undergoing essentially rolling contact. Gears do not depend on friction and do best
when friction is minimized.
Some common places that gears can normally be found are:
Printing machinery parts Newspaper Industry Book binding machines
Rotary die cutting
machines
Plastics machinery builders Injection molding machinery
Blow molding machinery Motorcycle Transmissions (streetand race applications)
Heavy earth moving topersonal vehicles
Agricultural equipment Polymer pumps High volume water pumps for municipalities
High volume vacuum
pumps
Turbo boosters for automotive
applications
Marine applications
Boat out drives Special offshore racing drive
systems
Canning and bottling
machinery builders
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Figure 4.1
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Hoists and Cranes Commercial and Military
operations
Military offroad vehicles
Automotive prototype and
reproduction
Low volume automotive
production
Stamping presses
Diesel engine builders Special gear box builders Many different specialmachine tool builders
4.1 Gear Classification: Gears may be classified according to the relative position of the
axes of revolution. The axes may be
1. Gears for connecting parallel shafts,2. Gears for connecting intersecting shafts,
3. Gears for neither parallel nor intersecting shafts.
Gears for connecting parallel shafts
1. Spur gears: Spur gears are the most common type of gears. They have straight
teeth, and are mounted on parallel shafts. Sometimes, many spur gears are used at onceto create very large gear reductions. Each time a gear tooth engages a tooth on the other
gear, the teeth collide, and this impact makes a noise. It also increases the stress on the
gear teeth. To reduce the noise and stress in the gears, most of the gears in your car are
helical.
Spur gears are the most commonly used gear type. They are characterized by teeth, which
are perpendicular to the face of the gear. Spur gears are most commonly available, and are
generally the least expensive.
Limitations: Spur gears generally cannot be used when a direction change between
the two shafts is required.
Advantages: Spur gears are easy to find, inexpensive, and efficient.
2. Parallel helical gears: The teeth on helical gears are cut at an angle to the face of
the gear. When two teeth on a helical gear system engage, the contact starts at one end
of the tooth and gradually spreads as the gears rotate, until the two teeth are in fullengagement.
2
rnal contact
Iinternal contact
Spur gears (Emerson Power Transmission Corp)
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Helical gears
(EmersonPower Transmission Corp) Herringbone gears
(or double-helical gears)
This gradual engagement makes helical gears operate much more smoothly and quietly than
spur gears. For this reason, helical gears are used in almost all car transmission.
Because of the angle of the teeth on helical gears, they create a thrust load on the gear whenthey mesh. Devices that use helical gears have bearings that can support this thrust load.
One interesting thing about helical gears is that if the angles of the gear teeth are correct,they can be mounted on perpendicular shafts, adjusting the rotation angle by 90 degrees.
Helical gears to have the following differences from spur gears of the same size:
o Tooth strength is greater because the teeth are longer,
o Greater surface contact on the teeth allows a helical gear to carry more
load than a spur gear
o The longer surface of contact reduces the efficiency of a helical gear
relative to a spur gear
Rackand pinion (The rack is like a gear whose axis is atinfinity.):Racks are straight gears that are used to convertrotational motion to translational motion by means of a
gear mesh. (They are in theory a gear with an infinite pitch
diameter). In theory, the torque and angular velocity of the
pinion gear are related to the Force and the velocity of therack by the radius of the pinion gear, as is shown.
Perhaps the most well-known application of a rack is the rack and pinion steering system
used on many cars in the past
Gears for connecting intersecting shafts:Bevel gears are useful when the direction ofa shaft's rotation needs to be changed. They are usually mounted on shafts that are 90
degrees apart, but can be designed to work at other angles as well.
The teeth on bevel gears can be straight, spiral or hypoid. Straight bevel gear teeth actually
have the same problem as straight spur gear teeth, as each tooth engages; it impacts thecorresponding tooth all at once.
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Just like with spur gears, the solution to this problem is to curve the gear teeth. These spiral
teeth engage just like helical teeth: the contact starts at one end of the gear and progressivelyspreads across the whole tooth.
Straight bevel gears Spiral bevelgears
On straight and spiral bevel gears, the shafts must
be perpendicular to each other, but they must also
be in the same plane. The hypoid gear, can engagewith the axes in different planes.
This feature is used in many car differentials. The
ring gear of the differential and the input piniongear are both hypoid. This allows the input pinion
to be mounted lower than the axis of the ring gear.
Figure shows the input pinion engaging the ring
gear of the differential. Since the driveshaft of thecar is connected to the input pinion, this also
lowers the driveshaft. This means that the
driveshaft doesn't pass into the passenger compartment of the car as much, making moreroom for people and cargo.
Neither parallel nor intersecting shafts: Helical gears may be used to mesh two shafts
that are not parallel, although they are still primarily use in parallel shaft applications. A
special application in which helical gears are used is a crossed gear mesh, in which thetwo shafts are perpendicular to each other.
Crossed-helical gears
Worm and worm gear: Worm gears are used when
large gear reductions are needed. It is common forworm gears to have reductions of 20:1, and even up
to 300:1 or greater.
Many worm gears have an interesting
property that no other gear set has: the worm can easily turn
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Hypoid gears(Emerson Power Transmission Corp)
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the gear, but the gear cannot turn the worm. This is because the angle on the worm is so
shallow that when the gear tries to spin it, the friction between the gear and the wormholds the worm in place.
This feature is useful for machines such as conveyor systems, in which the locking
feature can act as a brake for the conveyor when the motor is not turning. One other very
interesting usage of worm gears is in the Torsen differential, which is used on some high-performance cars and trucks.
4.3 Terminology for Spur Gears
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Figure 4-4 Spur Gear
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Terminology:
Addendum: The radial distance between the Pitch Circle and the top of the teeth.
Arc of Action: Is the arc of the Pitch Circle between the beginning and the end of theengagement of a given pair of teeth.
Arc of Approach: Is the arc of the Pitch Circle between the first point of contact of the gearteeth and the Pitch Point.
Arc of Recession: That arc of the Pitch Circle between the Pitch Point and the last point of
contact of the gear teeth.
Backlash: Play between mating teeth.
Base Circle: The circle from which is generated the involute curve upon which the toothprofile is based.
Center Distance: The distance between centers of two gears.
Chordal Addendum: The distance between a chord, passing through the points where the
Pitch Circle crosses the tooth profile, and the tooth top.
Chordal Thickness: The thickness of the tooth measured along a chord passing through the
points where the Pitch Circle crosses the tooth profile.
Circular Pitch: Millimeter of Pitch Circle circumference per tooth.
Circular Thickness: The thickness of the tooth measured along an arc following the Pitch
Circle
Clearance: The distance between the top of a tooth and the bottom of the space into which it
fits on the meshing gear.
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Contact Ratio: The ratio of the length of the Arc of Action to the Circular Pitch.
Dedendum: The radial distance between the bottom of the tooth to pitch circle.
Diametral Pitch: Teeth per mm of diameter.
Face: The working surface of a gear tooth, located between the pitch diameter and the top ofthe tooth.
Face Width: The width of the tooth measured parallel to the gear axis.
Flank: The working surface of a gear tooth, located between the pitch diameter and the
bottom of the teeth
Gear: The larger of two meshed gears. If both gears are the same size, they are both called
"gears".
Land: The top surface of the tooth.
Line of Action: That line along which the point of contact between gear teeth travels,
between the first point of contact and the last.
Module: Millimeter of Pitch Diameter to Teeth.
Pinion: The smaller of two meshed gears.
Pitch Circle: The circle, the radius of which is equal to the distance from the center of the
gear to the pitch point.
Diametral pitch: Teeth per millimeter of pitch diameter.
Pitch Point: The point of tangency of the pitch circles of two meshing gears, where the Lineof Centers crosses the pitch circles.
Pressure Angle: Angle between the Line of Action and a line perpendicular to the Line of
Centers.
Profile Shift: An increase in the Outer Diameter and Root Diameter of a gear, introduced tolower the practical tooth number or acheive a non-standard Center Distance.
Ratio: Ratio of the numbers of teeth on mating gears.
Root Circle: The circle that passes through the bottom of the tooth spaces.
Root Diameter: The diameter of the Root Circle.
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Working Depth: The depth to which a tooth extends into the space between teeth on the
mating gear.
4.2 Gear-Tooth Action
4.2.1 Fundamental Law of Gear-Tooth
Action
Figure 5.2 shows two mating gear teeth, in
which
Tooth profile 1 drives tooth profile 2
by acting at the instantaneous contact point
K.
N1N2 is the common normal of the two
profiles.
N1 is the foot of the perpendicular from
O1 toN1N2
N2 is the foot of the perpendicular from
O2 toN1N2.
Although the two profiles have differentvelocities V1 and V2 at pointK, their velocities
along N1N2 are equal in both magnitude and
direction. Otherwise the two tooth profileswould separate from each other. Therefore, we
have
( )1.4222111
NONO =
or
( )2.411
22
2
1
NO
NO=
We notice that the intersection of the tangency N1N2 and the line of centerO1O2 is point
P, and from the similar triangles,
( )3.42211PNOPNO =
Thus, the relationship between the angular velocities of the driving gear to the driven gear, or
velocity ratio, of a pair of mating teeth is
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Figure 5-2 Two gearing tooth profiles
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( )4.41
2
2
1
PO
PO=
Point P is very important to the velocity ratio, and it is called the pitch point. Pitch pointdivides the line between the line of centers and its position decides the velocity ratio of the
two teeth. The above expression is the fundamental law of gear-tooth action.
From the equations 4.2 and 4.4, we can write,
( )5.411
22
1
2
2
1
NO
NO
PO
PO==
which determines the ratio of the radii of the two base circles. The radii of the base circles isgiven by:
( )6.4coscos222111
PONOandPONO ==
Also the centre distance between the base circles:
( )7.4coscoscos
221122112121
NONONONOPOPOOO
+=+=+=
where is the pressure angle or the angle of obliquity. It is the angle which the common
normal to the base circles make with the common tangent to the pitch circles.
4.2.2 Constant Velocity Ratio
For a constant velocity ratio, the position ofP should remain unchanged. In this case, themotion transmission between two gears is equivalent to the motion transmission between
two imagined slip-less cylinders with radius R1 and R2 or diameterD1 and D2. We can get
two circles whose centers are at O1 and O2, and through pitch pointP. These two circles aretermed pitch circles. The velocity ratio is equal to the inverse ratio of the diameters of pitch
circles. This is the fundamental law of gear-tooth action.
The fundamental law of gear-tooth action may now also be stated as follow (for gears with
fixed center distance)
A common normal (the line of action) to the tooth profiles at their point of contact must, in
all positions of the contacting teeth, pass through a fixed point on the line-of-centers called
the pitch point
Any two curves or profiles engaging each other and satisfying the law of gearing areconjugate curves, and the relative rotation speed of the gears will be constant(constant
velocity ratio).
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4.2.3 Conjugate Profiles
To obtain the expected velocity ratio of two tooth profiles, the normal line of their profilesmust pass through the corresponding pitch point, which is decided by the velocity ratio. The
two profiles which satisfy this requirement are called conjugate profiles. Sometimes, we
simply termed the tooth profiles which satisfy the fundamental law of gear-tooth action the
conjugate profiles.
Although many tooth shapes are possible for which a mating tooth could be designed to
satisfy the fundamental law, only two are in general use: the cycloidaland involute profiles.
The involute has important advantages; it is easy to manufacture and the center distancebetween a pair of involute gears can be varied without changing the velocity ratio. Thus
close tolerances between shaft locations are not required when using the involute profile. The
most commonly used conjugate tooth curve is the involute curve. (Erdman & Sandor).
conjugate action : It is essential for correctly meshing gears, the size of the teeth ( themodule ) must be the same for both the gears.
Another requirement - the shape of teeth necessary for the speed ratio to remain constant
during an increment of rotation; this behavior of the contacting surfaces (ie. the teeth flanks)
is known as conjugate action.
4.3 Involute Curve
The following examples are involute spur gears. We use the word involute because the
contour of gear teeth curves inward. Gears have many terminologies, parameters and
principles. One of the important concepts is the velocity ratio, which is the ratio of the rotaryvelocity of the driver gear to that of the driven gears.
4.1 Generation of the Involute Curve
The curve most commonly used for gear-tooth
profiles is the involute of a circle. This involute
curve is the path traced by a point on a line as
the line rolls without slipping on the
circumference of a circle. It may also be defined
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Figure 4.3 Involute curve
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as a path traced by the end of a string, which is originally wrapped on a circle when the
string is unwrapped from the circle. The circle from which the involute is derived is calledthe base circle.
4.2 Properties of Involute Curves
1. The line rolls without slipping on the circle.2. For any instant, the instantaneous center of the motion of the line is its point of
tangent with the circle.
Note: We have not defined the term instantaneous center previously. The instantaneouscenter orinstant center is defined in two ways.
1. When two bodies have planar relative motion, the instant center is a point onone body about which the other rotates at the instant considered.
2. When two bodies have planar relative motion, the instant center is the point at
which the bodies are relatively at rest at the instant considered.
3. The normal at any point of an involute is tangent to the base circle. Because of theproperty (2) of the involute curve, the motion of the point that is tracing the involute is
perpendicular to the line at any instant, and hence the curve traced will also be
perpendicular to the line at any instant.
There is no involute curve within the base circle.
Cycloidal profile:
Epicycliodal Profile:
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Hypocycliodal Profile:
The involute profile of gears has important advantages;
It is easy to manufacture and the center distance between a pair of involute gears canbe varied without changing the velocity ratio. Thus close tolerances between shaftlocations are not required. The most commonly used conjugate tooth curve is the
involute curve. (Erdman & Sandor).
2. In involute gears, the pressure angle, remains constant between the point of tooth
engagement and disengagement. It is necessary for smooth running and less wear of gears.
But in cycloidal gears, the pressure angle is maximum at the beginning of engagement,
reduces to zero at pitch point, starts increasing and again becomes maximum at the end of
engagement. This results in less smooth running of gears.
3. The face and flank of involute teeth are generated by a single curve where as in cycloidalgears, double curves (i.e. epi-cycloid and hypo-cycloid) are required for the face and flank
respectively. Thus the involute teeth are easy to manufacture than cycloidal teeth.
In involute system, the basic rack has straight teeth and the same can be cut with simple
tools.
Advantages of Cycloidal gear teeth:
1. Since the cycloidal teeth have wider flanks, therefore the cycloidal gears are stronger than
the involute gears, for the same pitch. Due to this reason, the cycloidal teeth are preferredspecially for cast teeth.
2. In cycloidal gears, the contact takes place between a convex flank and a concave surface,where as in involute gears the convex surfaces are in contact. This condition results in less
wear in cycloidal gears as compared to involute gears. However the difference in wear is
negligible
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3. In cycloidal gears, the interference does not occur at all. Though there are advantages of
cycloidal gears but they are outweighed by the greater simplicity and flexibility of theinvolute gears.
Properties of involute teeth:
1. A normal drawn to an involute at pitch point is a tangent to the base circle.
2. Pressure angle remains constant during the mesh of an involute gears.
3. The involute tooth form of gears is insensitive to the centre distance and depends only onthe dimensions of the base circle.
4. The radius of curvature of an involute is equal to the length of tangent to the base circle.
5. Basic rack for involute tooth profile has straight line form.
6. The common tangent drawn from the pitch point to the base circle of the two involutes isthe line of action and also the path of contact of the involutes.
7. When two involutes gears are in mesh and rotating, they exhibit constant angular velocityratio and is inversely proportional to the size of base circles. (Law of Gearing or conjugate
action)
8. Manufacturing of gears is easy due to single curvature of profile.
The 14Ocomposite system is used for general purpose gears.
It is stronger but has no interchangeability. The tooth profile of this system has cycloidal
curves at the top and bottom and involute curve at the middle portion.
The teeth are produced by formed milling cutters or hobs.
The tooth profile of the 14Ofull depth involute system was developed using gear hobs forspur and helical gears.
System of Gear Teeth
The following four systems of gear teeth are commonly used in practice:
1. 14O Composite system
2. 14 O Full depth involute system
3. 20O Full depth involute system
4. 20O Stub involute system
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The tooth profile of the 20ofull depth involute system may be cut by hobs.
The increase of the pressure angle from 14o to 20o results in a stronger tooth, because thetooth acting as a beam is wider at the base.
The 20o stub involute system has a strong tooth to take heavy loads.
Involutometry
The study of the geometry of the involute profile for gear teeth is called involumetry.
Consider an involute of base circle radius ra and two points B and C on the involute as
shown in figure. Draw normal to the involute from the points B and C. The normal BE and
CF are tangents to the Base circle.
Let
ra= base circle radius of gear
rb= radius of point B on the involute
rc= radius of point C on the involute
and
b= pressure angle for the point B
c= pressure angle for the point C
tb= tooth thickness along the arc at B
tc= tooth thickness along the arc at C
15
ra
Pitch Circle
Addendum Circle
Base Circle
E
F
B
C
Gear
O
A
r
( )2cos
)1(cos
cca
bba
rr
rr
OCF
andOBEFrom
==
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From the properties of the Involute:
Arc AE = Length BE and
Arc AF = Length CF
Similarly:
16
ccbb rr
Therefore
coscos =
( )
=
==
===
functioninvolutecalled
isExpression
Inv
AOEAOBOE
BE
OE
ArcAEAOE
bb
bbb
bbb
b
tan
tan.
tan
tan
ccc
CCc
c
Inv
AOFAOC
OF
BE
OF
ArcAFAOF
===
===
tan.
tan
tan
b
bbb
b
b
r
t
r
tAOBAOD
BpotheAt
2tan
2
int
+=
+=
c
ccc
b
c
r
t
r
tAOCAOD
CpotheAt
2tan
2
int
+=
+=
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Using this equation and knowing tooth thickness at any point on the tooth, it is possible to
calculate the thickness of the tooth at any point
Path of contact:
Considera pinion driving wheel as shown in figure. When the pinion rotates in clockwise,the contact between a pair of involute teeth begins atK(on the near the base circle of pinion
or the outer end of the tooth face on the wheel) and ends atL (outer end of the tooth face on
the pinion or on the flank near the base circle of wheel).
MNis the common normal at the point of contacts and the common tangent to the basecircles. The pointKis the intersection of the addendum circle of wheel and the common
tangent. The pointL is the intersection of the addendum circle of pinion and common
tangent.
The length of path of contact is the length of common normal cut-off by the addendum
circles of the wheel and the pinion. Thus the length of part of contact isKL which is the sum
17
Pitch
Circle
Pinion
Wheel
O2
O1
P
Base Circle
Base Circle
PitchCircle
Addendum
Circles
rra
RA
R
N
K
L
M
Catthicknesstooth
rr
tinvinvt
r
tinv
r
tinv
r
t
r
t
equationsabovetheEquating
c
b
bcbc
c
cc
b
bb
c
ccc
b
bbb
=
+=
+=+
+=+
22
..
2.
2.
2tan
2tan
:
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of the parts of path of contactsKPandPL. Contact lengthKPis called as path of approach
and contact lengthPL is called as path of recess.
ra = O1L = Radius of addendum circle of pinion,
and
R A = O2K= Radius of addendum circle of wheel
r = O1P= Radius of pitch circle of pinion,
and
R = O2P= Radius of pitch circle of wheel.
Radius of the base circle of pinion = O1M = O1P cos= r cos
and
radius of the base circle of wheel = O2N = O2P cos = R cos
From right angle triangle O2KN
Path of approach:KP
Similarly from right angle triangle O1ML
Path of recess:PL
Length of path of contact =KL
18
( ) ( )
( ) 222
2
2
2
2
cosRR
NOKOKN
A =
=
sinsin2 RPOPN ==
( ) sincos222 RRR
PNKNKP
A=
=
( ) ( )
( ) 222
2
1
2
1
cosrr
MOLOML
a =
=
sinsin1 rPOMP ==
( ) sincos222 rrr
MPMLPL
a=
=
( ) ( ) ( ) sincoscos 222222 rRrrRR
PLKPKL
aA ++=
+=
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Arc of contact: Arc of contact is the path traced by a point on the pitch circle from thebeginning to the end of engagement of a given pair of teeth. In Figure, the arc of contact is
EPForGPH.
Considering the arc of contact GPH.
The arc GPis known as arc of approach and the arcPHis called arc of recess. The anglessubtended by these arcs at O1 are called angle of approach and angle of recess respectively.
Length of arc of approach = arc GP
Length of arc of recess = arcPH
Length of arc contact = arc GPH = arc GP + arc PH
Contact Ratio (or Number of Pairs of Teeth in Contact)
The contact ratio or the number of pairs of teeth in contact is defined as the ratio of the
length of the arc of contact to the circular pitch.
Mathematically,
Where: and m = Module.
19
M
L
K
N
R
RA
ra
r
Addendum
Circles
Pitch
Circle
Base Circle
P
O1
O2
Pinion
Pitch
Circle
H
FE
G
Gear
Profile
Wheel
coscos
KPapproachofpathofLenght==
coscos
PLrecessofpathofLenght==
coscoscoscos
contactofpathofLengthKLPLKP==+=
CP
contactofarctheofLengthratioContat =
mpitchCircularPC ==
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Number of Pairs of Teeth in Contact
Continuous motion transfer requires two pairs of teeth in contact at the ends of the path
of contact, though there is only one pair in contact in the middle of the path, as in Figure.
The average number of teeth in contact is an important parameter - if it is too low due to theuse of inappropriate profile shifts or to an excessive centre distance.The manufacturing
inaccuracies may lead to loss of kinematic continuity - that is to impact, vibration and noise.
The average number of teeth in contact is also a guide to load sharing between teeth; it is
termed the contact ratio
Length of path of contact for Rack and Pinion:
20
R
r
RACKc
T
a
bh
Pc
PITCH LINE
Base Circle
c
RACK
PINION
PITCH LINE
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Let
r = Pitch circle radius of the pinion = O1P
= Pressure angle
ra. = Addendu m radius of the pinion
a = Addendum of rack
EF= Length of path of contact
EF= Path of approach EP + Path of recess PF
From triangle O1NF:
Exercise problems refer presentation slides
Interference in Involute Gears
21
coscos
sinsin
:
)3(
)2(sin
)1(sin
11
1
1
rPONO
rPONP
NPOtriangleFrom
NPNFPFrecessofPath
aEPapproachofPath
EP
a
EP
AP
==
==
==
==
==
( ) ( )
( )
( )
sincossin
sincos
)3(
cos
2
1222
2
1222
2
12222
12
1
2
1
rrra
PFEPEFcontactoflengthofPath
rrrPFracessofPath
equationtheinvaluesNFandNPngSubstituti
rrNOFONF
a
a
a
+=
+==
==
==
PitchCircle
Pinion
WheelO2
O1
P
Base Circle
Base Circle
PitchCircle
AddendumCircles
r
ra
RA
R
N
K
L
M
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Figure shows a pinion and a gear in mesh with their center as O 1andO2 respectively.MNisthe common tangent to the basic circles andKL is the path of contact between the two
mating teeth.
Consider, the radius of the addendum circle of pinion is increased to O1N, the point ofcontactL will moves fromL toN. If this radius is further increased, the point of contactLwill be inside of base circle of wheel and not on the involute profile of the pinion.
The tooth tip of the pinion will then
undercut the tooth on the wheel at the
root and damages part of the involuteprofile. This effect is known as
interference, and occurs when the teeth
are being cut and weakens the tooth at itsroot.
In general, the phenomenon, when the tip
of tooth undercuts the root on its mating
gear is known as interference.
Similarly, if the radius of the addendum circles of the wheel increases beyond O 2M, then thetip of tooth on wheel will cause interference with the tooth on pinion. The points M and N
are called interference points.
Interference may be avoided if the path of the contact does not extend beyond interferencepoints. The limiting value of the radius of the addendum circle of the pinion is O1N and of
the wheel is O2M.
The interference may only be prevented, if the point of contact between the two teeth is
always on the involute profiles and if the addendum circles of the two mating gears cut thecommon tangent to the base circles at the points of tangency.
When interference is just prevented, the maximum length of path of contact is MN.
Methods to avoid Interference
1. Height of the teeth may be reduced.
22
Wheel
Undercut Pinion
sinrMPapproachofpathMaximum ==sinRPNrecessofpathMaximum ==
( ) sinRrPNMPMN
MNcontactofpathoflengthMaximum
+=+==
( )( )
tan
cos
sinRr
RrcontactofarcoflengthMaximum +=
+=
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2. Under cut of the radial flank of the pinion.
3. Centre distance may be increased. It leads to increase in pressure angle.
4. By tooth correction, the pressure angle, centre distance and base circles remain unchanged,but tooth thickness of gear will be greater than the pinion tooth thickness.
Minimum number of teeth on the pinion avoid Interference
The pinion turns clockwise and drives the gear as shown in Figure.
Points M and N are called interference points. i.e., if the contact takes place beyond M and
N, interference will occur.
The limiting value of addendum circle radius of pinion is O1N and the limiting value of
addendum circle radius of gear is O2M. Considering the critical addendum circle radius ofgear, the limiting number of teeth on gear can be calculated.
Let
= pressure angle
R = pitch circle radius of gear = mT
r= pitch circle radius of pinion = mt
T & t= number of teeth on gear & pinion
m = module
aw = Addendum constant of gear (or) wheel
23
Pitch
Circle
Pinion
WheelO2
O1
P
Base Circle
Base Circle
Pitch
Circle
Max.
AddendumCircles
r
ra
RA
R
N
K
L
M
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ap = Addendum constant of pinion
aw. m = Addendum of gear ap. m = Addendum of pinion
G = Gear ratio = T/t
From triangle O1NP, Applying cosine rule
Limiting radius of the pinion addendum circle:
Addendum of the pinion = O1N - O1P
Addendum of the pinion = O1N - O1P
The equation gives minimum number of teeth required on the pinion to avoid interference.
If the number of teeth on pinion and gear is same: G=1
1. 14O Composite system = 12
24
( )
( )
sinsin
sin21sin2sin
1
sin2sin
90cossin2sin
cos2
2
222
2
222
2222
222
11
22
1
2
1
RPOPN
r
R
r
Rr
r
R
r
Rr
RrRr
RrRr
PNOPNPONPPONO
==
++=
++=
++=
++=
+=
2
1
22
1
2
1 sin212
sin21
++=
++=
t
T
t
Tmt
r
R
r
RrNO
++=
++=
1sin212
2sin21
2
2
1
2
2
1
2
t
T
t
Tmt
mt
t
T
t
Tmtmap
( )( )
++
=
++=
1sin21
2
1sin212
2
12
2
1
2
GG
at
t
T
t
Tta
p
p
( )
+
=1sin31
2
2
12
pa
t
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2. 14 O Full depth involute system = 32
3. 20O Full depth involute system = 184. 20O Stub involute system = 14
Minimum number of teeth on the wheel avoid Interference
From triangle O2MP, applying cosine rule and simplifying, The limiting radius of wheel addendumcircle:
Addendum of the pinion = O2 M- O2P
25
PitchCircle
Pinion
WheelO2
O1
P
Base Circle
Base Circle
PitchCircle
Max.AddendumCircles
r
ra
RA
R
N
K
L
M
2
1
2
2
1
2
2
sin212
sin21
++=
++=
T
t
T
tmT
R
r
R
rRMO
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The equation gives minimum number of teeth required on the wheel to avoid interference.
Minimum number of teeth on the pinion for involute rack to avoid Interference
The rack is part of toothed wheel ofinfinite diameter. The base circle
diameter and profile of the involute
teeth are straight lines.
Let
26
++= 1sin21
2
2
1
2
T
t
T
tmTmaw
++= 1sin21
2
2
1
2T
t
T
tTaw
++
=
1sin211
1
2
2
1
2GG
aT W
PITCH LINE
Pc
Th
a
bRACK
c
PITCH LINE
PINION
RACK
c
M
L
HP
K
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t= Minimum number of teeth on the pinion
r= Pitch circle radius of the pinion = mt
= Pressure angle
AR.m = Addendum of rack
The straight profiles of the rack are tangential to the pinion profiles at the point of contactand perpendicular to the tangentPM. Point L is the limit of interference.
Addendum of the rack:
Backlash:
The gap between the non-drive face of the pinion tooth and the adjacent wheel tooth is
known as backlash.
If the rotational sense of the pinion were to reverse, then a period of unrestrained pinionmotion would take place until the backlash gap closed and contact with the wheel tooth re-established impulsively.
Backlash is the error in motion that occurs when gears change direction. The term "backlash"
can also be used to refer to the size of the gap, not just the phenomenon it causes; thus, onecould speak of a pair of gears as having, for example, "0.1 mm of backlash."
A pair of gears could be designed to have zero backlash, but this would presuppose
perfection in manufacturing, uniform thermal expansion characteristics throughout thesystem, and no lubricant.
Therefore, gear pairs are designed to have some backlash. It is usually provided by reducingthe tooth thickness of each gear by half the desired gap distance.
In the case of a large gear and a small pinion, however, the backlash is usually taken entirely
off the gear and the pinion is given full sized teeth.
27
( )
2
2
2
2
sin
2:ceinterferen
sin2
sin
sin
sinsin
sin
R
R
AtavoidTo
mt
r
OP
OP
PLLHmA
=
=
=
=
===
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Backlash can also be provided by moving the gears farther apart. For situations, such as
instrumentation and control, where precision is important, backlash can be minimisedthrough one of several techniques.
Let
r= standard pitch circle radius of pinion
R = standard pitch circle radius of wheel
c = standard centre distance = r +R
r= operating pitch circle radius of pinion
R= operating pitch circle radius of wheel
c= operating centre distance = r + R
= Standard pressure angle
= operating pressure angle
h = tooth thickness of pinion on standard pitch circle=p/2
h= tooth thickness of pinion on operating pitch circle
Let
28
M'
N'
RR'
r'r
Base Circle
Base Circle
P
O1
O2
M
N
R
RA
rar
P
O1
O2
Wheel
Pinion
Standard
(cutting)
Pitch Circle
Standard
(cutting)Pitch Circle
c
Standard
(cutting)
Pitch Circle
Standard
(cutting)Pitch Circle
c
c' OperatingPitch Circle
'
'
Figure a
Pinion
Figure b
Wheel
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H = tooth thickness of gear on standard pitch circle
H1 = tooth thickness of gear on operating pitch circle
p = standard circular pitch = 2 r/ t = 2R/T
p = operating circular pitch = 2 r1/t = 2R1/T
C= change in centre distance
B = Backlash
t = number of teeth on pinion
T = number of teeth on gear.
Involute gears have the invaluable ability of providing conjugate action when the gears'
centre distance is varied either deliberately or involuntarily due to manufacturing and/ormounting errors.
On the operating pitch circle:
Substituting h and H in the equation (1):
29
===
=
=
==
1'cos
cos
'cos
cos'
'cos
cos'
cos'cos'
'''
ccccccNow
cc
cc
c
c
R
R
r
r
)1(''' BHhp
BacklashthicknesstoothofsumpitchOperating
++=+=
+=
+=
R
hinvinvRH
r
hinvinvrh
tryinvolutomeBy
2'..'2'
2'..'2'
:
( ) ( )
Binvcinvc
c
c
c
chp
BRrinvRrinvR
R
r
rhp
BR
hinvinvR
r
hinvinvrp
++
+=
++++
+=
+
++
+=
'.'2.'2''
'
'''.2''.2''
'
2'..'2
2'..'2'
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There is an infinite number of possible centre distances for a given pair of profile shifted
gears, however we consider only the particular case known as the extended centre distance.
Non Standard Gears:
The important reason for using non standard gears are to eliminate undercutting, to prevent
interference and to maintain a reasonable contact ratio.
The two main non- standard gear systems:
(1) Long and short Addendum system and
(2) Extended centre distance system.
Long and Short Addendum System:
The addendum of the wheel and the addendum of the pinion are generally made of equal
lengths.
Here the profile/rack cutter is advanced to a certain increment towards the gear blank andthe same quantity of increment will be withdrawn from the pinion blank.
Therefore an increased addendum for the pinion and a decreased addendum for the gear is
obtained. The amount of increase in the addendum of the pinion should be exactly equal tothe addendum of the wheel is reduced.
The effect is to move the contact region from the pinion centre towards the gear centre, thusreducing approach length and increasing the recess length. In this method there is no change
in pressure angle and the centre distance remains standard.
Extended centre distance system:
30
[ ]
[ ]
[ ]
.'.'2'
'2
.'.'2'2
22'2
.'.'2'
2
invinvcc
crr
tB
invinvccc
tr
trB
invinvcc
chpB
+
=
+=
+=
[ ]
[ ]
.'.'2
.'.'2'
'2
invinvcBBacklash
invinvcr
rrr
tB
==
+
=
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Reduction in interference with constant contact ratio can be obtained by increasing the centre
distance. The effect of changing the centre distance is simply in increasing the pressureangle.
In this method when the pinion is being cut, the profile cutter is withdrawn a certain amount
from the centre of the pinion so the addendum line of the cutter passes through the
interference point of pinion. The result is increase in tooth thickness and decrease in toothspace.
Now If the pinion is meshed with the gear, it will be found that the centre distance has been
increased because of the decreased tooth space. Increased centre distance will have twoundesirable effects.
NOTE: Please refer presentation slides also for more figure, photos and exercise
problems
References:
1. Theory of Machines and Mechanisms by Joseph Edward Shigley and JohnJoseph Uicker,Jr.McGraw-Hill International Editions.
2. Kinematics and Dynamics of Machines by George H.Martin. McGraw-Hill
Publications.
3. Mechanisms and Dynamics of Machinery by Hamilton H. Mabie and Fred W.
Ocvirk.John Wiley and Sons.
4. Theory of Machines by V.P.Singh.Dhanpat Rai and Co.
5. The Theory of Machines through solved problems by J.S.Rao. New ageinternational publishers.
6. A text book of Theory of Machines by Dr.R.K.Bansal. Laxmi Publications (P)
Ltd.
7. Internet: Many Web based e notes
31
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Chapter 5: Gears Trains
A gear train is two or more gear working together by meshing their teeth and turning each other in a
system to generate power and speed. It reduces speed and increases torque. To create large gear ratio,
gears are connected together to form gear trains. They often consist of multiple gears in the train.
The most common of the gear train is the gear pair connecting parallel shafts. The teeth of this type
can be spur, helical or herringbone. The angular velocity is simply the reverse of the tooth ratio.
Any combination of gear wheels employed to transmit motion from one
shaft to the other is called a gear train. The meshing of two gears may beidealized as two smooth discs with their edges touching and no slip
between them. This ideal diameter is called the Pitch Circle Diameter
(PCD) of the gear.
Simple Gear Trains
The typical spur gears as shown in diagram. The direction of rotation is reversed from one gear to
another. It has no affect on the gear ratio. The teeth on the gears must all be the same size so if gear Aadvances one tooth, so does B and C.
The velocity v of any point on the circle must be the same for all the gears, otherwise they would beslipping.
32
(Idler gear)
GEAR 'C'GEAR 'B'GEAR 'A'
v
v
CBA
.
module
module
meshwould notrwise theygears othe
alle same formust be th
and
t
D=m =
in rpmN = speedmeter,circle diaD = Pitch
r,on the geaof teetht = number
r=D
cle. v =on the cirvelocityv = linear
.r velocity= angula
= m tDand= m tD;= m tD
t
D=
t
D=
t
Dm =
CCBBAA
C
C
B
B
A
A
2
CCBBAA
CCBBAA
CCBBAA
CCBBAA
C
C
B
B
A
A
tNtNtN
revoftermsinor
ttt
tmtmtm
DDD
DDDv
==
====
==
===
min/
222
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Application:a) to connect gears where a large center distance is required
b) to obtain desired direction of motion of the driven gear ( CW or CCW)
c) to obtain high speed ratio
Torque & Efficiency
The power transmitted by a torque T N-m applied to a shaft rotating at N rev/min is given by:
In an ideal gear box, the input and output powers are the same so;
It follows that if the speed is reduced, the torque is increased and vice versa. In a real gear box, poweris lost through friction and the power output is smaller than the power input. The efficiency is defined
as:
Because the torque in and out is different, a gear box has to be clamped in order to stop the case or
body rotating. A holding torque T3 must be applied to the body through the clamps.
The total torque must add up to zero.T1 + T2 + T3 = 0
If we use a convention that anti-clockwise is positive and clockwise is negative we can determine the
holding torque. The direction of rotation of the output shaft depends on the design of the gear box.
Compound Gear train
33
60
2 TNP
=
GRNN
TTTNTN
TNTNP
===
==
2
1
1
22211
2211
60
2
60
2
11
22
11
22
602
602
TN
TN
TN
TN
InPower
outPower=
==
GEAR 'A'
GEAR 'B'
GEAR 'C'
GEAR 'D'
CompoundGears
A
C
BD
Output
Input
Compound gears are simply a chain of simple gear
trains with the input of the second being the output of the
first. A chain of two pairs is shown below. Gear B isthe output of the first pair and gear C is the input of thesecond pair. Gears B and C are locked to the same shaft
and revolve at the same speed.
For large velocities ratios, compound gear trainarrangement is preferred.
The velocity of each tooth on A and B are the same so:
A tA = B tB -as they are simple gears.
Likewise for C and D, C tC = D tD.
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Reverted Gear train
The driver and driven axes lies on the same line. These are used in speed reducers, clocks and
machine tools.
IfR and T=Pitch circle radius & number of teeth of the gear
RA + RB = RC + RD and tA + tB = tC + tD
34
C
D
A
B
DB
CA
C
DD
A
BBCA
C
DD
CA
BB
A
C
D
D
C
A
B
B
A
t
t
t
t
t
t
t
t
T
tand
t
t
ttand
tt
=
=
==
==
( )( )
GRt
t
t
t
OutN
InN
aswritten
bemayratiogearThe
NSince
GRt
t
t
t
shaftsametheonareCandBgearSince
C
D
A
B
C
D
A
B
D
A
CB
==
=
==
=
:
2
CA
DB
D
A
tt
tt
N
NGR
==
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Epicyclic gear train:
Epicyclic means one gear revolving upon andaround another. The design involves planet andsun gears as one orbits the other like a planet
around the sun. Here is a picture of a typical gear
box.
This design can produce large gear ratios in a
small space and are used on a wide range of
applications from marine gearboxes to electricscrewdrivers.
Basic Theory
Observe point p and you will see that gearB also revolves once on its own axis. Any object orbiting
around a center must rotate once. Now consider thatB is free to rotate on its shaft and meshes with C.
Suppose the arm is held stationary and gear Cis rotated once. B spins about its own center and the
35
Arm 'A'
B
C
Planet wheel
Sun wheel
Arm
B
C
The diagram shows a gear B on the end of an arm. Gear
B meshes with gear C and revolves around it when thearm is rotated. B is called the planet gear and C the sun.
First consider what happens when the planet gear orbits
the sun gear.
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number of revolutions it makes is the ratioB
C
t
t. B will rotate by this number for every complete
revolution ofC.
Now consider that C is unable to rotate and the armA is revolved once. GearB will revolveB
C
t
t+1
because of the orbit. It is this extra rotation that causes confusion. One way to get round this is toimagine that the whole system is revolved once. Then identify the gear that is fixed and revolve it
back one revolution. Work out the revolutions of the other gears and add them up. The following
tabular method makes it easy.
Suppose gearCis fixed and the arm A makes one revolution. Determine how many revolutions the
planet gearB makes.
Step 1 is to revolve everything once about the center.Step 2 identify that Cshould be fixed and rotate it backwards one revolution keeping the arm fixed as
it should only do one revolution in total. Work out the revolutions ofB.
Step 3 is simply add them up and we find the total revs ofCis zero and for the arm is 1.
Step Action A B C
1 Revolve all once 1 1 1
2Revolve Cby 1 revolution,
keeping the arm fixed0
B
C
t
t+ -1
3 Add 1B
C
t
t+1 0
The number of revolutions made byB is
+
B
C
t
t1 Note that if C revolves -1, then the direction ofB
is opposite soB
C
t
t+ .
Example: A simple epicyclic gear has a fixed sun gear with 100 teeth and a planet gear with 50teeth. If the arm is revolved once, how many times does the planet gear revolve?
Solution:
Step Action A B C
1 Revolve all once 1 1 1
2 Revolve Cby 1 revolution,keeping the arm fixed
050
100+ -1
3 Add 1 3 0
Gear B makes 3 revolutions for every one of the arm.
The design so far considered has no identifiable input and output. We need a design that puts an input
and output shaft on the same axis. This can be done several ways.
36
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Problem 1: In an ecicyclic gear train shown in figure, the arm A is fixed to the shaft S. The wheel Bhaving 100 teeth rotates freely on the shaft S. The wheel F having 150 teeth driven separately. If the
arm rotates at 200 rpm and wheel F at 100 rpm in the same direction; find (a) number of teeth on the
gear C and (b) speed of wheel B.
Solution:
TB=100; T F=150; N A=200rpm; NF=100rpm:
CgearsonteethofNumberT
T
TTT
rrr
gearsallforsameisuletheSince
C
C
CBF
CBF
=+=
+=+=
25
2100150
2
2
:cirlcepitchthetoalproportionisgearson theteethofnumberthe
:mod
The gear B and gear F rotates in the opposite directions:
37
Arm A
C
SB100B
F150
C 200 rpm
100 rpm
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350200
200100
150
100
)exp(
=
=
=
=
=
=
E
B
AB
AF
F
B
AB
AF
ArmF
ArmL
F
B
NN
NN
NN
T
T
traingearepicyclicforressiongeneralNN
NN
NN
NNTValso
T
TvalueTrain
The Gear B rotates at 350 rpm in the same direction of gears F and Arm A.
Problem 2: In a compound epicyclic gear train as shown in the figure, has gears A and an annulargears D & E free to rotate on the axis P. B and C is a compound gear rotate about axis Q. Gear A
rotates at 90 rpm CCW and gear D rotates at 450 rpm CW. Find the speed and direction of rotation of
arm F and gear E. Gears A,B and C are having 18, 45 and 21 teeth respectively. All gears having
same module and pitch.
Solution:
TA=18 ; T B=45; T C=21;NA = -90rpm; ND=450rpm:
DgearonteethT
TTTT
rrrr
and
D
CBAD
CBAD
84214518
:cirlcepitchthetoalproportionisgearson theteethofnumberthe
:gearsallforsamearepitchmoduletheSince
=++=++=++=
Gears A and D rotates in the opposite directions:
38
Q
P
C Arm F
D
A
E
B
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Annular 'A'
Spider 'L'
Sun Wheel 'S'
Planet Wheel 'P'
CWrpmArmofSpeedN
N
N
NN
NN
T
T
T
T
NN
NN
NN
NNTValso
T
T
T
TvalueTrain
F
F
F
FA
FD
D
C
B
A
FA
FD
ArmF
ArmL
D
C
B
A
==
=
=
=
=
=
9.400
90
450
8445
2118
Now consider gears A, B and E:
EgearonteethofNumberT
T
TTT
rrr
E
E
BAE
BAE
=
+=
+=
+=
108
45218
2
2
Gears A and E rotates in the opposite directions:
CWrpmEgearofSpeedN
N
NN
NN
T
T
NN
NNTValso
T
TvalueTrain
E
E
FA
FE
E
A
FA
FE
E
A
==
=
=
=
=
72.482
9.40090
9.400
108
18
Problem 3: In an epicyclic gear of sun and planet type shown in figure 3, the pitch circle diameter of
the annular wheelA is to be nearly 216mm and module 4mm. When the annular ring is stationary, the
spider that carries three planet wheelsPof equal size to make one revolution for everyfive revolutionof the driving spindle carrying the sun wheel.
Determine the number of teeth for all the wheels and the exact pitch circle diameter of the annular
wheel. If an input torque of 20 N-m is applied to the spindle carrying the sun wheel, determine the
fixed torque on the annular wheel.
39
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Solution: Module being the same for all the meshing gears:
TA = TS + 2TP
teethm
AofPCDTA 54
4
216===
OperationSpider
arm LSun Wheel S
TS
Planet wheel P
TP
Annular wheel A
TA = 54
Arm L is fixed &
Sun wheel Sisgiven +1 revolution
0 +1P
S
T
T
A
S
A
P
P
S
T
T
T
T
T
T
=
Multiply by m
(S rotates throughm revolution)
0 m mT
T
P
S mT
T
A
S
Add n revolutions
to all elementsn m+n m
T
Tn
P
S mT
Tn
A
S
IfL rotates +1 revolution: n = 1 (1)
The sun wheel S to rotate +5 revolutions correspondingly:
n + m = 5 (2)From (1) and (2) m = 4
WhenA is fixed:
teethT
TTmT
Tn
S
SA
A
S
5.134
54
40
==
==
But fractional teeth are not possible; therefore TS should be either 13 or 14 and TAcorrespondingly 52 and 56.
Trial 1: Let TA = 52 and TS = 13
teethTT
T SAP 5.194
1352
2=
=
= - This is impracticable
Trial 2: Let TA = 56 and TS = 14
teethTT
T SAP 214
1456
2=
=
= - This is practicable
40
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Figure 4
C Arm
B
D
A
H
G
E
F
TA = 56, TS = 14 and TP= 21
PCD ofA = 56 4 = 224 mm
Also
Torque onL L = Torque on S S
Torque onL L = mN= 1001
520
Fixing torque on A = (TL TS) = 100 20 = 80 N-m
]
Problem 4: The gear train shown in figure 4is used in an indexing mechanism of a milling
machine. The drive is from gear wheels A and
B to the bevel gear wheel D through the geartrain. The following table gives the number of
teeth on each gear.
How many revolutions does D makes for one
revolution of A under the following
situations:a. If A andB are having the same speed and same directionb. IfA andB are having the same speed and opposite direction
c. If A is making 72 rpm andB is at rest
d. If A is making 72 rpm andB 36 rpm in the same direction
Solution:
Gear D is external to the epicyclic train and thus C and D constitute an ordinary train.
OperationArm
C (60)E (28) F (24) A (72) B (72) G (28) H (24)
Arm or C is fixed& wheelA is given
+1 revolution
0 -124
28= +1 -1 +1
6
7
24
28 =
Multiply by m
(A rotates throughm revolution)
0 -m m6
7 +m -m +m m
6
7
Add n revolutions
to all elementsn n - m mn
6
7 n + m n - m n + m mn
6
7+
Gear A B C D E F
Number of
teeth72 72 60 30 28 24
Diametral
pitch in mm08 08 12 12 08 08
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P2
A2
A1
P1
P
S1
S2
Q
Figure 5
(i) For one revolution ofA: n + m = 1 (1)ForA andB for same speed and direction: n + m = n m (2)
From (1) and (2): n = 1 and m = 0
IfCor arm makes one revolution, then revolution made byD is given by:
CD
D
C
C
D
NN
T
T
N
N
2
230
60
=
===
(ii) A andB same speed, opposite direction: (n + m) = - (n m) (3)
n = 0; m = 1
When Cis fixed andA makes one revolution,D does not make any revolution.
(iii) A is making 72 rpm: (n + m) = 72
B at rest (n m) = 0 n = m = 36 rpm
Cmakes 36 rpm and D makes rpm7230
6036 =
(iv) A is making 72 rpm andB making 36 rpm
(n + m) = 72 rpm and (n m)= 36 rpm
(n + (n m)) = 72; n = 54
D makes rpm10830
6054 =
Problem 5: Figure 5 shows a compoundepicyclic gear train, gears S1 and S2 being
rigidly attached to the shaft Q. If the shaft
P rotates at 1000 rpm clockwise, while
the annular A2 is driven in counterclockwise direction at 500 rpm,
determine the speed and direction ofrotation of shaft Q. The number of teeth
in the wheels are S1 = 24; S2 = 40; A1 =
100; A2 = 120.
Solution: Consider the gear trainPA1S1:
Operation ArmP A1
(100) S1 (24) Operation ArmP
A1
(100)
S1 (24)
Arm P is fixed &
wheelA1 is given+1 revolution
0 +1
6
25
24
100 1
1
=
+P
P
ORArm P is fixed
& wheelA1 is
given -1revolution
0 -1
1
1
1
1
S
A
P
A
+=
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Multiply by m(A1 rotates through
m revolution)
0 +m m6
25 0 -1
6
25
24
100=
Add n revolutions
to all elementsn n+ m mn
6
25
Add +1
revolutions to
all elements
+1 06
31
6
25=+
IfA1 is fixed: n+ m; gives n = - m
1
631
625
1
31
6
31
61
SP
S
P
NN
nn
n
N
N
=
==+
=
Now consider whole gear train:
Operation A1
(100)A2
(120)S1 (24),S2 (40)
and Q Arm P
A1 is fixed &
wheelA2 is given
+1 revolution
0 +1
3
40
1202
2
=
+P
P
31
18
31
63
=
Multiply by m
(A1 rotates through
m revolution)
0 +m m3m
31
18
Add n revolutions
to all elements
n n+ m mn 3 mn
31
18
WhenPmakes 1000 rpm: mn31
18 = 1000 (1)
andA2 makes 500 rpm: n+ m = -500 (2)
from (1) and (2): 100031
18500 = mm
( ) ( )
rpmnand
rpmm
m
449500949
949
4931500100031
===
=+
NQ = n 3 m = 449 (3 -949) = 3296 rpm
Problem 6. An internalwheel B with 80 teeth is
keyed to a shaft F. A fixed
internal wheel C with 82teeth is concentric with B. A
Compound gears D-E
meshed with the two
43
F
DE
C
A
BArm C
B
E
D
A
B80
C82
D28NA=800rpm
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internal wheels. D has 28
teeth and meshes withinternal gear C while E
meshes with B. The
compound wheels revolvefreely on pin which projects
from a arm keyed to a shaftA co-axial with F. if thewheels have the same pitch
and the shaft A makes 800
rpm, what is the speed of the
shaft F? Sketch thearrangement.
Data: tB = 80; tC = 82; D = 28; NA = 800 rpm
Solution: The pitch circle radius is proportional to the number of teeth:
Since the wheel C is fixed and the arm (shaft) A makes 800 rpm,
44
Egearonteethofnumber
t
t
tttt
rrrr
E
E
EBDC
EBDC
=
=
=
=
26
802882
m+nnAdd n
revolutions to all
elements
+m0
Multiply by m
(B rotates
through m
revolution)
+10Arm is fixed & Bis given ONE
revolution (CW)
D (28)E(26)
C (82)Compound Gear wheel
B (80)ArmOperation
m+nnAdd n
revolutions to all
elements
+m0
Multiply by m
(B rotates
through m
revolution)
+10Arm is fixed & Bis given ONE
revolution (CW)
D (28)E(26)
C (82)Compound Gear wheel
B (80)ArmOperation
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Problem 7: The fig shows an Epicyclic gear train. Wheel E is fixed and wheels C and D are
integrally cast and mounted on the same pin. If arm A makes one revolution per sec (Counterclockwise) determine the speed and direction of rotation of the wheels B and F.
Solution:
Data: tB = 20; tC = 35; tD = 15; tE = 20; tF = 30 NA = 1rps-(CCW)
45
rpmm
m
nm
rpmn
42.761
080041
14
13
40
041
14
13
40
800
=
=+
=+
=
rpmFshaftofSpeedBgearofSpeed
rpmnmBgearofSpeed
58.38
58.3880042.761
===+=+=
Arm
B20
C35
D15
E20 F30
m+nnAdd n
revolutions to
all elements
+m0
Multiply by m
(B rotates
through m
revolution)
+10
Arm is fixed &
B is given
ONE
revolution
(CW)
C (35)D (15)
F (30)E (20)
Compound Gear
wheelB (20)ArmOperation
m+nnAdd n
revolutions to
all elements
+m0
Multiply by m
(B rotates
through m
revolution)
+10
Arm is fixed &
B is given
ONE
revolution
(CW)
C (35)D (15)
F (30)E (20)
Compound Gear
wheelB (20)ArmOperation
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Since the wheel E is fixed and the arm A makes 1 rps-CCW
Problem7: In the gear train shown, the wheel C is fixed, the gear B, is keyed to the input shaft andthe gear F is keyed to the output shaft.
The arm A, carrying the compound wheels D and E turns freely on the out put shaft. If the input
speed is 1000 rpm (ccw) when seen from the right, determine the speed of the output shaft. The
number of teeth on each gear is indicated in the figures. Find the output torque to keep the wheel Cfixed if the input power is 7.5 kW.
Solution:
Data :
tB= 20; tC = 80; tD = 60; tE= 30; tF= 32; NB = 1000 rpm (ccw) (input speed); P= 7.5 kW
46
429.07
301
3
7
03
71
===
=+=
mm
nmandrpsn
)(667.1429.09
141
9
14
)(571.01429.0
CCWmnFgearofSpeed
CCWrpsnmBgearofSpeed
===
==+=
D60
C80
B20F32
E30
Input
Shaft
Output
Shaft A
m+nn
Add n
revolutions toall elements
m0
Multiply by m
(B rotates
through m
revolution)
+10Arm is fixed &B is given +1
revolution
E (30)D (60)
F (32)C (80)
Compound Gear
wheelB (20)
InputArmOperation
m+nn
Add n
revolutions toall elements
m0
Multiply by m
(B rotates
through m
revolution)
+10Arm is fixed &B is given +1
revolution
E (30)D (60)
F (32)C (80)
Compound Gear
wheelB (20)
InputArmOperation
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Input shaft speed = 1000 rpm (ccw)
i.e., gear B rotates 1000 rpm
47
2008001000
80025.1
1000
025.01000
04
1;
1000
=+=
==
=
=
=+
n
m
mm
mnfixedisCGear
nm
)(50
5016
5800200
16
5
CWrpmFshaftoutputtheofSpeed
mnFofSpeed
+=
=+=
=
NmT
T
TNPpowerInput
B
BB
59.7110002
607500
60
1000210005.7
60
2
=
=
=
==
B
NmTT
NTNT
NfixedisCSince
NTNTNT
equationenergytheFrom
F
F
FFB
C
CCFFB
8.1431050100059.71
0
0:
0
;
+==+
=+
=
=++
B
B
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Figure 6
A2
A1
S2
P2P1
S1
The Torque required to hold the wheel C = 1360.21 Nm in the same direction of wheel
Problem 8: Find the velocity ratio of two co-
axial shafts of the epicyclic gear train as shownin figure 6. S1 is the driver. The number of teeth
on the gears are S1 = 40,A1 = 120, S2 = 30,A2 =
100 and the sun wheel S2 is fixed. Determine also
the magnitude and direction of the torquerequired to fix S2, if a torque of 300 N-m is
applied in a clockwise direction to S1
Solution: Consider first the gear train S1,A1 andA2 for which A2 is the arm, in order to find the
speed ratio ofS1 toA2, whenA1 is fixed.
(a) Consider gear train S1,A1 andA2:
OperationA2
(100)
A1(120)
S1(40)
A2 is fixed &wheelA1 is given
+1 revolution
0 +140
120=
Multiply by m
(A1 rotates through
m revolution)
0 +m m3
Add n revolutions
to all elementsn n+ m mn 3
A1 is fixed: nm =
48
NmT
T
TTT
equationtorquetheFrom
C
C
CF
21.1360
08.143159.71
0
:
=
=++
=++B
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2AT
2ST2A
1S
1ST
21
2
1
4
43
AS
A
S
NN
n
nn
N
N
=
=+
=
(b) Consider complete gear train:
Operation A1(120) A2(100) S1(40) S2(30)
A1 is fixed & wheel S2 is given+1 revolution
01100
30=
5
64
10
3= +1
Multiply by m(A1 rotates through m revolution)
0 m10
3= m5
6 +m
Add n revolutions to all elements n mn10
3 mn
5
6 n+ m
S2 is fixed m = - n
13
22
13
10
5
11
10
356
2
1 ==+
+=
nn
nn
N
N
A
S
Input torque on S1 = TS1 = 300 N-m, in the direction of rotation.
Resisting torque onA2;
rotationofdirectiojntoopposite
mNTA
== 7.50713
22300
2
Referring to the figure:
)(7.2073007.5072 CWmNTS ==