Kinematics of Machines Compiled T.v.govindaraju

7/30/2019 Kinematics of Machines Compiled T.v.govindaraju

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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

1

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.

3


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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

4

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

9

Figure 5-2 Two gearing tooth profiles


10/49

( )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

11

Figure 4.3 Involute curve


12/49

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

==


16/49

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

+=

+=


17/49

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

:


18/49

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 ++=

+=


19/49

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 ==


20/49

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


21/49

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


22/49

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 +=

+=


23/49

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


24/49

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


25/49

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


26/49

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


27/49

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

=

=

=

=

===


28/49

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


29/49

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'


30/49

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

==

+

=


31/49

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


32/49

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


33/49

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.


34/49

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

==


35/49

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.


36/49

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


37/49

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


38/49

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


39/49

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


40/49

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


41/49

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

41


42/49

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

+=

42


43/49

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


44/49

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


45/49

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


46/49

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


47/49

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


48/49

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


49/49

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 ==

Date post:	14-Apr-2018
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Kinematics of Machines Compiled T.v.govindaraju

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