Gearbox Design by Chitale

316 17.1.2 VarIous Motions of Machine Tool System

Todrivethevarious componentsofthe machinetoolsystem,we can have fourmethods:

(i)Mechanical (ii)Electrical

(iii)Hydraulic (iv)Pneumatic.

The choiceofa particular methodof drivewill depend upon many factors such ascost. operating speeds and feeds.power toweightratio.rigidity.reliability.maintenancecosts. intended use. sophistication. and control.

17.1.1 DrIves and Regulatlon of MotIon on Metal-cuttIng MachInes

Metal-cuttingmachines receiveworking motions (speed andfeed) from electricmotors.which usually have constant revolutions per minute. Inorder to fulfil different operations. it isnecessary to find out various numbers ofspindle revolutions as wel!asdifferent values of feed. For these purposes. speed and feed boxes which workby eitherstepped or unstepped principle of regulation are used.

The various elements of a machine toolare assembled together so as to providemaximum rigidity to the system;however.this assembly has revolving.sliding and fixed orstationary components. Generally the drivesof a machine tool arecovered and hidden.but operated bycontrolswhich areaccessibleto the operator. Variouselements of a machinetool are madeintegralorfabricated and assembled together to make a system quitehomogeneousin appearanceand operation.

17.1 INTRODUCTION: MACHINE TOOL SYSTEM

Design of Machine Tool Gear Box

17

II,n¢,nf ... nI/f'-', ...,n¢!' If you multiply thisseries by ¢,you will bave

II¢.IIf. nif •... , nl/l'.n¢!,+1 then this can be written as

Va nn

-=--=k=tan¢ do 1000

which reveals that to getv/np a constant for thesame diameter, we need to change n such that "o,tnt isequal to trd,tlOOO.

lfwehave aGPseries such as

The ideal regulation of speed is stepless drive;but it has ccnain limitations. due to which it cannot beincorporated in all types of machine tools. Some idea regarding this facthas already been given in previous sections. The cost of a gear box increases as the number of speeds increases, and this limitation hampers many users; therefore such machine tools are not commercially prospective.

Inview ofthe above fact, itis customary to design (for asetofspeeds) a gear box which hasa optimum numbers of speeds, with minimum speed loss. Alternative approaches to this problem have posed various solutions, out ofwhich we have topickup the best one.

Ifstepless regulation is notavailable, then the increment of speeds from a minimum level can bearranged inanAP.GPor HPseries; ithas been'proved that tbe GP(Geometric Progression) gives minimum speed loss, ' and the GP series hasother advantages too.

Aswe know that

17.2 FUNDAMENTALS OF MECHANICAL REGULATION

Generally, today the user of machine tools demands beuer quality, improved performance, and higher operating speeds, and this has led to a design system quite complex in nature-consisting of anyone of the drive methods stated above-s-of the machine toolsystem.

The field of machine dynamics, particularly that ofthe load bearing components of asystem such as base, bed, table, saddle, columns andspindlesupport, are important for technological gains, as these components are made up of iron castings, but their design as steel weldments offers functional and economic advantages. The functional advantage is the possibility of using higber speeds and feeds, and the economic advantage is thelow power toweightratio and thus lower cost and ease of handling.

There are two types of motion in a machine tool systems: (i) the main motions, viz. cutting speed and feed, and (ii)the subsidiary motions such as fixing of workpiece, tool seuing, machine control, ere,

Sometimes the primary motion isonlyinone axis asinthe case ofabroaching macbine, butmore often such motions are required intwo or more than two planes. Insuch situations, we prefer tohave an individual drive rather than a common drive. The lineshaft drive is most obsolete. as there is not much control ofspeed in such asystem of speed regulation.

The hydraulic or pneumatic speed regulation devices are used where an infinite number of speeds. within a range of maximum and minimum speed, are required; however, stepless regulation can also be achieved by a DCmotor with a resistance control or mechanical drives using pressure variations, but for a very limited range.

Hydraulic speed regulation is good forstraight line motions,

e

.

g.

broaching, grinding, milling and shaping machines.

V' I-..../!. V'_A V~

-VB

MaxflV

=

constant

=

--"---"-V~ get

LetflVbe the relative loss of the cuuing speed, where

6V= VA -VB VA

The maximum of this value win be with the diameter d.: V' - V'

Max flV= A H

flA

lfwe take this value as constant (asthisisprofitable for the exploitation of machine tools), then wesball Fig.17.1

-d (a)Retation between diameterdandspeed V

_d

(b)Loss ofculling speed with change of diameter

t

II---r<----".t

"

n= constant

f

Inourexample. it isnecessary tohavethespeed corresponding to thepoint A. But wemust work either withthe speed VA'or with the speed VB(seeFig. 17.1). Itisnecessary to work with the speed VBbecause it is near the speed VN

K=tan

I/!

V = 1r11 d A 1000 A

If n isconstant, then this relationship will be a straight line which passes through the origin (see Fig. 17.1).

or

"l; "l;"1; ... ;"4--1; "k;"«I; ...

What is thelaw governing these numbers?

For this purpose, letus examine tberadial diagram. Letustum a sbaft whichhas a diameter d•.According to the theory ofmetal cutting, we know that the cutting speed isgiven asunder.

V = --Ird"" rnImm. 0111 1000

Thus we get another orientation of the speeds, just shifting thefirstspeed to the next higher one and so on, without affecting thestructural changeincase of a geometric series with a common rates 4Jwhicb isnot

possible forAPorHPseries. Preferred numbers can be used in this series, which isnot possible in other cases. At present, weshall consider only rotary motions and other types of motions later on.

Raydiagram or speed spectrum forgeometric series. Itis the law ofthe geometrical progression with thecommon ratio

.p

.

nk=nk_, ¢ where ¢ is a constant Therefore we have the number of speeds of a virtual gearbox

as

follows:

n,

n2=n,.¢

n3 =n2· ¢= n,¢2 n. =n,l/f nx

=

n,¢'-'

Let uscall theratio nm."lnmin asthe range of regulation denoted by R.

n n ~=R=...:L ''min nl nx=n,¢'-' R

=

¢'

-

'

¢=

x

-rR

or so or no

v

t

A t.V

~ B

~HY

~~

~

--

~--

~

~--~~-Let nk_'

=

n._,

=.!..

n.

n

.

_'9

"

So,ifmax~Visconstant, thenthesegmentA'B' mustbe constant too,forall of the values of rpmofthe machine tool as shown in Fig. 17.1.Then the fullradial diagram will be

as

shown in Fig. 17.2.This radial diagram isuseful for determining the number ofrevolutions ifd and V areknown.

Then

= 1_ 1rd~ nk-l1000=1_ n.- 1

1rd~

n

.

1000 nk

Sothe ratio

n

.

-

1 mustbe constant for V to be constant.

E,

3 -= -~ 10 (I7.3) (17.4) (17.2) ¢E,= 10

1/>

=

E

<.ho

Letus take logarithms of expressions.17.1 and 17.2.thenwe have E, logI/>

=

log 2

=

0.3 E,log

1/>

=

log 10= 1.0 Dividing Eq.(17.3)by Eq.(17.4).we get

The condition of the second factor maybeexpressed mathematically as follows: If the serieshasmember

II

;

.

thenitshould have anumbersuchthat

ny' =lOnx' 1J' =mt'l 1J' y '1" x 1011'="'£'11'_y 'Y' x or (17.1) or

What requirements should the standardvaluesofthe constant of geometrical progressionsatisfy? This isan important point of debate. Thedenominator of geometricalprogression should be chosenby taking into considerationthe followingfactors:

(a)The possibility of applying multi-speedasychronousor synchronous motors (e.g. with the number of revolutionsas 3000, 1500, 750,etc.);

(b)The decimalsystemof a series.

The condition as expressed in the first factor may be expressed mathematicallyas follows: Ifthe seriesof speeds has a member

"

.<),

then itmust have a member"y'such that

"

,

=

211,1

11 =_y 11_xl. ",_I"£, where_I"",£,=2

",E,

nx3='f' '''x3

l

'=

2

¢=

£#2

17.3 DEVELOPMENT OFSERIES OF NUMBERS

Evenifthe values"m", and"mi. arethesame, thenumber of speeds will be moreiftheconstant

.p

is smaller.When choosing the denominator of the geometrical progression, we take into consideration the following two factors:

1.Desirability of having ahigher number of speeds, asthen itisnecessary tohave smallervaluesof

.p;

2.Aspiration to have a compact structure of the speedbox,asthen it isnecessary to have highervalues

of

.p

.

This gives us a lesser number ofspeeds.

The values of

.p

arestandardized,andthe most common ones are 1.06, 1.12,1.25, 1.41, 1.58. 1.87. and 2.

system. or !!A. log" k= 1+ --log¢' k=I+logR log¢'

Let us assume that k=2£1·3£2, asthis isone ofthe requirements of design ofa speed regulation system.

The value ofk could be any number, as EJ and E2arewhole numbers starting from zero.

Thus, K= 2, 3. 4,6.8,9, 12,16, 18,24,27, 36,OUIofwhich the mostwidely used numbers are 3, 4, 6,

8,9, 12, 16,11,24 and 36. Therefore, these are the stepsof astepped regulation or mechanical regulation or or ¢,=k-Jf,i;

V;;

I

n

.

log ¢= -I og-k-I nJ Wehave

Therefore, according to common understanding, if E2=40, 20, 10 or5, EI would be 12,6,3 or 1.5. Hence, thestandard values of ¢ from Eq.(17.2)would be

¢4(J= ~ =J¥2=1.06

tf>w

=~ =

ifi

= 1.12

¢to= JlflO=

Vi

=1.26

¢S

=~ = J,rz =1.58

However, the series of twonumbers for the values of ¢was found to be insufficient; therefore. this series

is tobe supplemented with the following values of the series of number 2:

¢=..J2 =1.41 ¢

=

t/iO

=

1.78

Theadvantages of standardisation are the following:

(i)The decimal system of series is enough to erect aseries from 10to 100.All other numbers ofthis

series may be obtained by multiplication or division of 10, 100,etc., and it is convenient for calculations.

(ii)Ifwe putdowneverysecond member oftheseries with¢40=1.06.then we geta series with ¢20= 1.12; if we repeat this after every third member. the series with ¢IOwill be obtained.

(iii)The revolutions per minute of a synchronous motor can be fitted into this series very well. As

mentioned earlier. R=Range of regulation = n... /nm;n. and ifthe number of speeds in the gear box ofa machine tool is denoted by k,then

n)=e). el'o "6 =

e

3.C

s

1t

o

1( -=el T2 T .2.=e2 T, Ts -=e) . T6 " _-T_=e4, T8 T9 -=es 1(0 Lei us assume Hence

Similarly. wecanobtain

and "I =el.e4

"

o

"5 =e2.eS"O Therefore "4 =el .3es"o ''1=/l,¢=¢=e2 ·e4/l0 _ ~ "l nl e) ·e4"o el e2=e,.if>

Solution 10 !hissystem of transmission of speeds at 6steps,werequire 10gears asshown isFig.17.3.We

canproceedforanalyticalinvestigationofthekinematic chainofgearsasfollows:

Seriesof number of revolutions:

To 10

•

Fig. 17.3 Design ofgear box-kine maliclayout diagram.

n, ...ns III II 1r+'_2.

I"

_flo r-

L

T,

2i

r- -

r-r-

r-r--

X X X- X

Xf--T.

T.

$

'r

I-

-

-2

•

....

_...

'-

T

Example 17.1 Design a gear box having 6speeds.i.e., 3x2and3 shafts.

Flg.17.4 Symmetric ray diagrams. Shaft nos.

1st

II III II III

Main Gearing Gearing Main

group group group group n. n. n. n. n, n, n, n, n,

_n.

<I> n, n,

(a)Crossed type speedlayoul (b)Open type speedlayout

nx

=

n1tP'C-t

logII,=log111+(x- I)log4>

Therefore. from theabove formulaprepresenting astraight line equation. itis possibleto draw line

diagramswhichare calledray diagrams.These aresbown in Fig. 17.4.

Let us draw the ray diagram of the gear box having6speeds,i.e.6=3x 2.

~

=

¢

=

es es=e.¢

nl e4

In this case,the double block of thegears isthe main group, and the trebleblock ofgears is the first gearinggroup.

The aboveanalysisiscalled the methodof investigatingkinematic chains.

In cases where the number of sets ismore than 2, such analysisbecomes difficultbecause ofbulky

calculations.

Example 17.2 Showthe graphicalmethodofinvestigationofspeed regulationbydrawingraydiagrams.

Solution In general weknow that

!!1.=¢3 = el·eS·~ =eS

111 el·e.·~ e.

es=e•.

if

The link betweenthe numberof revolutionsand the denominator 4> is called the main set.

The first gearing set is that which bas the power of denominatorequal to thenumber of independent changes of speed in themain set.

11

T

7

nl=--~ =el·e4~

whereRisthe range of regulation.

Sometimes a ray diagram, besidesproviding thisinformation, showsthevalues of thetransmission

ratios, and thenit is drawnasymmetrically.These are showninFigs. 17.5(a)and(b).

Tbe step betweenthemotor andthefirst shaft withnospeed iscalled thecompensatory step; it provides thepossibility of framing a standardseries of rpm's.

Aoobvious question arises: Which arrangement or diagram is better?The one inFig. 17.5(a) or 17.5(b)?

The first arrangement,shown inFig. 17.5(a),is better, becausethe sizeof thegearboxsystem in this case is smaller,andfasterspeeds are available;thusless torqueistransmitted.Hence, smallershaftdiameters and modules of gears arepreferable.

Example 17.3 Plot a ray diagram for a 12-speed gear box, which is a structure of 12 speeds, i.e., 12=2 x 3 x 2.

Solution Referto Fig.17.6.

If Po is the numberofindependentshaftingsof tbemain group; P, isthe number of independentshaftings ofthe 1st gearing group; P2is tbe numberof independent shaftingsof the2nd gearing group;

then the powervalue is givenas follows: for themaingrouP-Po;

for thefirstgearing group-p,: forthe secondgearinggroup-po.P,:

forthethird gearinggroup-po'p,pz.

The followinglimitationsare accepted in machinetool designforseparatetransmissionratiosofgear

drives:

In the kinematic chainofgeardrive

Flg.17.5 Asymmetricray diagrams.

2nd method (b) 1st method (a) n, "'" 9

<,

c _{1:1 1:1}

•

~:2

1£-..._

•

\"'"

1:1 4 1:4\ 1~ 3

\

1:1 2 ~

,

M M M M M M "'" 9c

<,

1:1 1:1 no_~2

-,

~

""""

""""""

1:~""

""

n,. III II Motor III II Motor

The ray diagram represents the method of gear arrangement, and also provides the values of motor

speed and itspower for various kinematic pairs as well as arrangements ofgroups. The ray diagram does not

Overlapping speeds are obtained by reducing the power by oneor more degrees in one of the gearing groups.

Letus take up a mechanism of 12=2·3·2speeds. the main group of whichisthe travelblock of gears.

17.4 RAY DIAGRAM FOR OVERLAPPING SPEEDS

Let us consider the limitations of value

.p

by takingupanexample of 4-shaft mechanism inthe kinematic

chain of motions.

From the

structural diagram

as

givenearlier (12 = 2·3·2)it is obvious that

e = e .t/> main group

'. = ') .¢ Is: gearing group

and therefore '7=e6.¢ 2nd gearing group

Allvalues oft/>are suitable for the main group.

For the first gearing group,R

=

e/

"

3

=

1/1';

the maximum value of¢is asfollows:

for the main motion,¢=~ =1.67;

for the cbain of feed motion,¢= ~ = 1.94;

for the second gearing group, R=

.

-/

e

)

=

rf/';

and therefore the maximum value of ¢ isasfollows:

forthechain offeed motion,¢= ~ = 1.41;

for the chain of feed motion,t/>=~ =1.93.

Itappears thatdisplacement of gearing group will change nothing. Hence, the value oft/>is limited and

depends on the number of shans and on the number of stepsof revolution are obtained here.

1

- $e$2.8

2

4> ;'

l

Fig.17.6 Raydiagram for 12-speedgearbox.

In the chain of feed motion

Main 1stgearing 2ndgearing group group group

Po p,

Let us decrease thepower of the first gearing group by 1 unit.

Thus, theoverlapping oftwo stage speeds is obtained, and we have ¢= ~ =1.41

¢= ~ =1.56

Fig. 17.8 Raydiagramforoverlappingspeedsin 'cross'structure.

OVerlapping

~----

+-~

~

~

~~

~~

~

----

+-+-

--+-

~

~~~

~

--

-4

---+--

~~

~

~

--

-4

--

---+~~~~

t::$

~

~3

:

~~

~+-~

~~~

*-

~~

~~~~n4 2ndgearing Maingroup groupmustbe 1, ¢S 1st gearing groupmustbe ¢.z

2ndgearing group 1st gearing group Maingroup

Fig. 17.7 'Open'structure raydiagram.

The ray diagram foroverlapping speeds in 'cross structure' isshown inFig. 17.8.

"

,

n,o

"

"

r<

/'

/

f<

<,

/

<,

/

t<'--v

-,

k:::""'"

-,

_/

<

K

-.

•

'2 "-

K/

1 1

/

T

"

IV n'2 III II

(a) (b)

Flg.17.10 Ray diagrams forkinematiclayoutof Fig.17.9.

-,

\\

\\\

\:0<:

0<

~ ~

/~

///

//

/

/

//

///

/"

//

,/

X/

<X

XX

<~

X'\

<,

_I

_{X'\. \}

1

\\\

\\

\

I

V

III II V

I

V

III II

The disadvantage incase ofthearrangement shown inFig. 17.IO(a)isthat a return step isnot available. This arrangement isquite often used. The disadvantage incase of the kinematic arrangement in Fig. 17.IO(b) isthatvisual control is difficult, but here a stage of return isobserved, which isan advantage.

Fig. 17.9 Kinematiclayoutforversatile geartrain.

--I

V

---

+L++~~~

-r- -X X 11-

'-+--+-

+-

-+'x

t--- X1---- -'-- '--- r- -X X X X

--

x~====

hxl=

=1

t- - X

--

v

III _ ~ __

'--A barrel of gears is incorporated in the kinematic chain.Itis freelysuspended and gives astage of returnspeeds. Let 12

=

3·2·2 be the kinematic cbain with a stage of return. The structural diagram for this mechanism can be drawn up by anyone of the methods shown in Figures 17.9 and 17.10 below. The ovedappings are also oblained here.

17.5 RAY DIAGRAMS FOR RETURNSTEP OFSPEED

Regulation with overlapping speeds should be applied incases when (i) It is necessary to increase the value of !p, and

(ii)It is necessary to avoid bigh speeds.

There arethreemethods:

(i)Method ofleastcommonmultiple;

(ti)Method of difference; (iii) Constructivemethod.

17.7 DETERMINATION OF NUMBER OF TEETH ON GEARS OF STEPPED CONTROL MECHANISMS Fig.17.11 (a) 'Cross'structurelayout L---_'_--~n, 2ndgearing Main group group (b)'Semi-open'structure layout n, 2ndgearing group Main group

I-~M

-

---:

".j

n" f---"I<E=--~n, II

n"

n,.

n.

"s

,.-

{

,.,

group _"s "s n,

n

,

.

/

II

III

IIII

r=:::-

X/I

X

X

I<:::::::"

_X'\\

'\.'\.'\.\

\\'\

'\.'\

\

tstgearing{ group II III n" n"

While choosinga multispeedinput, it isnecessary to considerthe following:

(a) Economicaspect

(b) Smallersize and simpler control.

Inthiscase,the value of¢ is limitedto 2.That is ¢m =2

Themotorshouldbeconsidered eitherthe maingroup oras one ofthe groups.

Iftreated as the maingroup,411in the 15tgearingis41P,in the 2nd gearing groupis41N, ans so on.

Inthe general case,it is

qI

.

It means that ¢=

lfi.

or

qI

=2.

Hence 41

=

1.06

=

J~ ¢

=

1.12

=

~

¢

=

1.26

=

ifi

¢

=

1.41

= J2

¢

=

2

The values¢

=

1.58and41

=

1.78arenotsuitable.

Themotor shouldbeassumed as an elementary mechanism,andthe structuraldiagram shouldbedrawn

for a 12-specdkinematic chain,suchthat12

=

2·2·3,andas showninFig. 17.11,which shows the'cross'and

'semi-open' structure layouts.

T4_---TOeJ 1

+

<!:l

c

ez

="d

T

J=

T

oe

z

1

+

ez

a

e) = -b Let

By analogy. we can also find

T, (1 )

To

=

T,+-

=T

,

1+

-el el To _I =e

T

Z

'

Thus, in the same way we can find

T.

Flg.17.12 Designing layoutwith least common multiple. T.

A

eland

e

z

are thus known tous.

If T

o

is equal to a certain number. we can

find T

o

= T

I

+

T

z.

Now,

e

l

T

z=

T

,;

therefore

T

o

=

e,

T

z

+

r

,

=

T

z

(

e

,

+

1).

'

r

3

T

o

=T

I

+T

2

=T

J

+T

.

e

l

=

2i..

ez

=

T

J

T

z

T4

17.7.1 Method of Least Common Multiple

Let us assume that all gears have thesame module. From Fig. 17.12,we have

2A

=

m(TI+ T

z)

=

m(

T

J

+

T

J

=

mTs

where

T

o

isthetotal number of teeth,

a 1j=--·ToK a+b 1 ez=- tn,=2mm tn2=3mm 2.5 b=2 c=2 d=5 a+b=3 c+d=7 I e= -I 2 a= 1 Example 17.5 Let LetustakeK= 6.ThenT, =42, T2=84,TJ= 36,andT.=90. Ifthemodulesof a pair of gearsaredifferent,then we have

2Ao=m(T,+T2)

=

m(TJ+T.)

Ifit isnecessaryto calculate by one module, saym ..then

13

+T4=5..(1j +T

z

)=5..To tnz "'2 and LCMTo=2 {a_c+d+b₌

=

₇

3

}

b=2 I e=-_{, 2} 2 e2=

5'

c=2 d=5 then a= I Example 17.4 Let a a 1j=To ·K=Tox--·K

b

(

I+ ~) (a+

b

)

whereToistheleast commonmultiple of thevaluesabandcd:

Kisthe setfor obtaining a minimumnumberof teeth ofthe smallestgear.

I b

T

z

=To--·K=Tox--·K

I+~ a+b

b

and To=T, +T. =T2+Ts

e=

E.

e=

T

2

'

T

,

2

75

1j+t.T e2= T4- t.T Also Since

Fig. 17.13 Kinematiclayout with gear details.

_l_

2.5m

T

T2m

•

T2 To

;5..

-

r2--

'-

,_

rr

-X X X T._

T.

'- T• where Tmin~5 teeth T2=T,+

sr

Is

=

T4+"'T

This method bas the disadvantage over thefirst method that itdoes not takeintoaccount thesituation inwhich it isnecessary that 17.7.2 Method of Difference Let ustake K

=

5.Then 52 T4=--21·K=lOK 73 22 T3=--·21·K =4K 73 2 T2 =-21·K=14K 3 c

m

,

13

=----·ToK c+dlllz d

m

,

T4=----·ToK c+dlllz

where To

=

LCM ofvalues a+b,c+dandm/"'z'

Inthisexample, the LCM is3, 7, and 213To

=

21.

I 1j=-21·K=7K 3 a

12

=--·ToK a+b

The ray diagram.gearboxlayout and structuraldiagram for Example 17.7are showninFigures 17.14. 17.15and17.16respectively.

Solution liz = 130 "4=280 "6=585 liS=1240 "9=1800

,

{l8OO

.

=

tp

= 9-

V

90

= ~20 = 1.455

Example 17.7 Designa gearboxof amachine tool(turret)having 9 spindle speedsrangingfrom90to 1800rpm.The gearbox should be a compactone.Also

(a) Represent thespeeds graphically.

(b) Drawthestructural diagram.

(c) Show thelayoutof the gear box.

(d) Find Outthe numbers ofteeth on various gears.

"

,

=

90

"3= 190

"

s

= 400

n.,

= 850

The shortcomings oftheprevious two methods are that the diameters ofshafts,keyways, etc.arenot taken

into account. The constructive method consists of determiningteethnumbers,whilethe above factors arealso taken into consideration.

17.7.3 Constructive Method

T2=T, + 6T=35 1j 30·5 T4=-=--=75

e,

2 5'~(1

+

.!..)

1j= 5 2 =3·10=30 I 2 5-4 2 5 Therefore, 6T= 5 I e2= -2 e2-e,

Inthe samewaywe can determineT2• T3•etc.

The kinematiclayoutwith geardetailsisshown in Fig. 17.13.The corresponding raydiagram is shown

in Fig. 17.17. Example 17.6 or 1j

+

6T

e,

(1j

+

6T)

°

z

= 7j _6T 1j- 0,·6T

e

,

T,ez-e,ez 6T= e,T,+0, ·6T 0,.6T(1+"2) 1j Therefore, since

~:

-

---{

"

:~

80/38 57/5

Flg.17.16 Structuraldiagram for Example 17.7.

III II

90

Fig.17.15 Gear boxlayoutfor Example 17.7.

190 r

_{!2

1 ~ 2 3

...

'C::::: _{~ ~}

'"

fe

90

rs

r-

~57 '-

_r;;-

30

'a

'"

'9

'"

..

..

30

~ 'i2

I-r-

_f;-j

4s

r,o

--~ 57 '-90

Input Intermediate Output

Flg.17.14 Ray diagram for Example 17.7. 130

190

1240

Example 17.8 (a) A manufacturing concern takes upthe demand of supplying turret lathes to its customers

having 9speeds powered by a 8 kW motor. The speed range isfrom 90to 1500 rpm. Design a suitable gear

box giving alldetails.

(b)After a few years, the customers demand that the working speed range maybe increased to2500 rpm, as lower speeds are rarely used. Suggest aworkable alteration in the gears so as tomeet this demand

without changing the structure of the gear box. Solution (a) Gear box forturret lathe:

Data -l.Spindle speeds: 9 2. Range: 90 to 1500 3.Capacity: 8 kW 4. Preferred Number: 1.06, 1.12, 1.26, 1.41, 1.58, 1.78,2. Since

Since the centre distance is toremain constant,

T,+TIO= T)+Til =TI2=120 Ts

=

I(the speed ratio)

7i1

Tg=Til =57 T9 I T9=30

7i2

=

3.1

7i

2

=90 or Sowetake

n

o

=

190

Now choose narrow diagram for compactness. For speeds n.,nsand n6,i.e.,286, 402 and 585

1i

=3.0 T.

Let ~=~

.. ~=~

Since the centre distance between the shafts carrying these two gears is toremain constant,

TI+T. =T2+ Ts =T)+T6 = 120 But T2 =2.1 Ts Ts=38 Similarly T)=67

Now 2nd stage bas alltbe ratios

as

3.1.

!J....

=3.1

1io

n

o

=200 or

no

=280

Let the motor rpm be 1400. Using belt and cone pulley arrangement from motor 10 the spindle and taking the reduction to be 7 or 5,i.e.

Now. as thecentre distanceremainsconstant, T)+T. =T, +T2=60 T;

T4

=1.41 60 T.=-=24.9=~ 2.41 Tl=~ To

2.

=T)+T.=T,+T2=6O

T

6

The gear arrangement is shown io Fig. 17.18.

~

---

r-~~~

135

L- -L

~OO

Flg.17.17 Ray diagramfor Fig. 17.13 and Example 17.8(a).

Let the minimum number of teeth on thegear be 20.Then

2i.

=

I.

99

T2

=@1

T2 1500 1062 755 535 380 "0 =270 270 191

The nearest preferred number in the list of 1.41,Outof the given listed numbers.

The variousspeeds in descending ordersball be:

n

,

=1500 n;=755

ns

= 380 ", =191 "2=1062 ". =535 "6=270 n8=135

n9

=

96

According to the considerations of slip andstrength etc.,the speed ratio allowable ar the pulleys are

5 to 7.

So the primaryspeed is between

1440n

=

206

and

1440/5

=

288

.

Assuming motor rating to be

444 V

at

1440

rpm and

Hence we take 270

as

the primary speed.For economy and compactoess,narrow ray diagram will be

preferred.

(b) Now,

if

after a few years the company wants to rearrange the speed to have maximum speed of

2500 rpm approximately, the raydiagram shifts towards thehigh value ofspeeds.

Then, using the same preferred number

t500 x 1.41=2118 (i)

2118 x 1.41 = 2980 (ii)

There is no restriction on minimum speeds as theyare rarely used.

Let us assume that the minimum number of gear teeth is20.

1j =3.95

72

T2

=

1lQ]

T,=IW T,+T2=99 =T)+T. =Tl+T6 35125 20156 8E~0~50120~ 30135 38/38

Flg.17.19 Structuraldiagram forExample 17.8(a).

The structural diagram is shown inFig. t7.19.

!i

=2.8

1jo

T,=~

Therefore

Flg.17.18 Geararrangementfor Example17.8(a).

Now forspeeds 1500, 1062.755

37

--

-

-

-

-

-

-

-

-

-

-

--

n

20 38

r-I-

--1

--- J

'~I.

, 'I

T9

=1

li2

66/33 38/28 ~7~0~5M0~ 73/28 20/50

Fig. 17.21 StructuraldiagramforExample17.8(b). T1

+

T,o=Ta+T'1=

t;

+

T'2=76 Ta

=

2.8 Til

=

20 Ts

=

~

11,

!J...

=

2.8 TIO

=~

T1

=

56

110

Tg

=

1 Ta

=

Til

=1lID

11

1

~ =2.8 T,z=20 Ts=56

11

2

The structural diagram isshown in Fig. 17.21. Porsecond stage

Fig. 17.20 Raydiagramfor Example 17.B(b).

r---.----.,,

2980

f----+-=~.,j

2

1

1

8

f-

--

+---r.~H'500

~--

~

~~~

~

1

00

2

1---,,+**

-

--17

55

~..;;;<r*

-":...

o--i535 h~O::_+"':>"':~~380 ~~--+-~~~270

f-

--

+

--

~1

9

1

The raydiagram is shown inFig. 17.20.

T)

=

iTIl

T4=~ Ts

=i§§]

Fig.17.22 Beltand pulley drive.

.

_

--

FB}--

-,

,

,

,

._--

t}P

---17.9.1 Belt and Cone Pulley Drive

The belt and cone pulley drive (see Fig. 17.22)is supposed to be the simplest and oldest mechanism for transmitting power from one shaft to another. It has certain inherent disadvantages such as large size. small range ofspeeds. and that the torque transmitted bythe driven shaft is proportional tothe speed atwhich it is rotating. However. ithas the advantage of being simple in design and cheap. Thus it is still found useful for various purposes.

If the kinematic chain between theinputand output shafts isof mechanical type, we willget stepped regulation; however. steplessregulation can also be obtained. but it isnot so commonly used. The methods of stepped regulation are as follows:

(a) Belt and cone pulley drive;

(b) Belt and puUey with back gear drive; (c) Gear box drives.

For

steples

s

regulation there arc various methods. asmentioned previously and as willbediscussed in Chapter 18.

17.9 MECHANICAL REGULATION OF DRIVES

The theoretical calculations of the speeds and the number of teeth have been duly provided in the preceding section; however, the actual speeds available onthe machine toolgenerally vary by±¢n in such a way that

!fin

is equal to zero. Not only this. it is limited to±10% (0-1). Ifdeviation exceeds this limit, re-design should be done.

After all speeds, the number of teeth are determined, the number of gears and layout of gear box. location of input. output shaft, etc. are taken up. The design calculation of a gear wheel is taken up after the forces acting on the spindle of a machine tool have been estimated. Researchers have given empirical relationships for estimating these forces, and these have been presented in previous chapters.

The gear tooth must be strong enough to resist bending due to pitch line pressure, and also hard enough to resist thesurface wear that takes place during rotations. In view of this, suitable gear material should be chosen.

Ag. 17.27 V-belteo pulley.

Fig. 17.26 Transmission from

speed box to splndle-Method 2.

Fig. 17.25 Transmission from speedboxtospindle-Method 1. A V-belted pulley is shown inFig. 17.27. Speed box

,

,

, ,

, ,

, ,

, ,

,

,

T

,

~ , Flat beH

Wedge· belt drives are usually employed in the first variation and flat-belt drives in the second one. Ag.17.24 Transmission from gear box to spindle.

Ag. 17.23 Transmission from motorto gear box.

Speed box

Speedbox

There aremany types of

belts

as follows:

(i) Flat belt;

(ii)Wedge belt;

(iii)Round belt(for low power).

If a speed box is used there, the belt drive may be situated either in the place where the motion is

transmitted to the speedbox orintheplace of transmitting the motion from the speed box to thespindle (see Figures 17.23and 17.24).Two methods oftransmission fromspeed boxto spindle areshown in Figures 17.25 and 17.26.

Fig.17.29 Useofinternal gearofsmalldiameter.

(iii) Unloading of the.first shaft of thespeed box.

T,'tOgears

in

m

e

sn

(ii)Thespindle does notparticipateinuansmluing theforce(seeFig.17,29):and Fig. 17.28 Spocialpulleyorgearforpowertransmission.

Spindle Pulley Ofgeaf

x

x

unloading barrel

The second variation is applied if the S))C(.'<1box (situnted below) and thespindle tire separate units. In Ihiscase the machine tool is more steady. and hence thepossibility of vibrations is reduced. This scheme is used formachinesof higher precision,Itnecessitatestheapplicationof Oatbells (asitisdifficulttoplace

wedgebeltshere),

There arc three rncrhods of regulating thestressof aben in such drives. They are:

(i)Bymovingamotor (inadvanceof swivel): [ii}Belt-tightening pulleys;

(iii)Special construction or aregulated pulley (one.of the1\\'Oin Pig. 17,28). Merethan3Sl015

a

re

usuallynorprovided.

The bending of shafts can becliuumucd bymaking the belt transmit thetorquealone. This isdone iirst of allforspindles byan)' Oneof the.following three mctbods:

(i) Unloading of thebarre) (Sc,'C Fig. 17.28);

Fig.17.33 Geardrive wilh Slappedpulley.

•

•

• •

"

.

"

.

"

.

"

.

"

.

_Q_x

'

'"

----v-It

___n__

~

.

"

"

.

"

.

..

.

..

.

•

•

f\typical example ofsuch amechanism is shown in Fig. 17.33.wlth thestructural diagram inFig. 17.:l4 and

theray diagraminFig. 17.35.

17.9.2 Belt Pulley Drive with Back Gear

To improve the transmission system, one more shafl is added as shown in Pig. 17.29. and by this the oscilkuions ofone chain are absorbedbythe other,

Fig. 17.32 Indined pulley

beltIransmission.

Fig.17.31 Horizontalpulley bell Iransmission.

Fig. 17.30 Verticalpulley

belltransmisslon.

-

_-

₉ Endless belt drives

o

I

9

I

o

+

Inde~i;igningbelt and especially chain transmissions. great attention should be paid lor their location.

Generally. theyCanbe tocared intwo ways:

(i) Vertlcaltocatlon(see Fig. 17.30).

(ii) Horizontal location (see fig. 17.31).

The horizontal tocaucn

or

transmlssicn is better as the gravity force .f;changes the character

or

belt

stress. andthe character of cscilia lions arc mere favourable man inthe vertical locations. Itisclearly revealed

inchain transmissionswith largedistance betweencentres. OUIhorizontally located beltsrequire larger

dimensions orthe chainor belt: therefore angular locationofbelts ispreferableasshown in Fig. 17.32.

Advantages

(i)Smoothness ofspindle rotation

(H)Absence ofvibration

This type of system has the following advantagesin cornparison to plain belt and pulley drive mechanisms:

t

.~

am

gtOtsl) 1$1gCtlJioggtOVJ) Fig.17.35 Raydfagramfor gea( Clrive\\lith stopped plJlQy.

II n, n, ~

-c:

_~ t

<--

r----r---.

'--v-' III IV III II TI dId

'

do

T, Fig. 17.34 Structuraldiagramforgeardrive\vith steppedpulley. II IV Countershaft

I-

__

I

:

:

+H+

A

E)

T, II~=111)

-~ T4

The gear wheels 7'1and Z"arc either made of one 8C'.trblank or arc pcrruancruly connected witheach Other.There are 2. :\ aml sctdcm 4 gear wheels. Thesearc keyed together to formaSliding gear.

There arcrunny IY1)CS01'mechanical drives wlth gears as fo110\'1s:

(i)Gear box withsliding gears (ii) Norton gc.ar box with idler gear (ii) Meander's gear mechanism

(i'l)Gearboxwith clutches

(v) Gearboxwlrhdrivekey

(vi}Reversing gear box.

Gelfr box with sliding gctJrs

f\typical example ofthis type ofmechanism is shown in f'i~17..36.The ~earblock shlft in splines On the

shari with roundedcogsas shown in here.A keyedblockof gearsmeshalternmetvwith different gears

mourned onthe other shaft. 17.9.3 GearBox Dtlves Hs =T lTdllto kW I 60.102.1000 H{I II V =---2....£.nlpm , 1000 Jrd2110 V~=---nlpln

-

10

0

0

nd 11{j VI=--'--n11)Jll 1000 1 HSI =TV.- k\V 60.102 1 Hs..=

tv, ---

kVl

-

-

(>0

.

1

02

1 fls) =TV,-- k\V 60.102 (iii)Safetyinuse {iv}Compactness and greater number of speeds. Disadvantages (i)Cannot beauromarizcrt.

(ii)Distribution 01'power isdirectly propcnicnal tothe diameter of the stepped pulley.

IfTis the tension ~n('!H isthepower on the spindle. then

(a) KinomatiC layOvl of Norton gc)ArOOx, 2, /

,

/ _.__ Throw-on pinion /

x

X X X

,

.

-

-.i

.

,-

)

.

.

,

II T, T,T

,

T, r, eT, T, Advantages These are:

(i) Muchhighervaluesor power canbe rmnsmiued:

[ii]Suchdrives give constant powerexceptforfrictionallosses;

(iii) Highefficiency;

(iv) Providealargerangeofspeeds: ('I)Can beautomated.

Disadvantages

(i)Ditficuhy inopemnon;

(ii)Itispossible10shirlontywhenthemachineisstopped.

Inorder to111Ukcthe..shirtings easier. cogs are made with roundings. as shown inFig. 17.36. Norton gear box mechanism

ThismechanismisshowninFig.17.37(;) and(b). andisgenerallyusedforlowspeeds.i.e..forfeedboxes.

ThevariousspeedsIh~1canbeobtainedarcshowninthesame tlgurc.

Fig.17.36 Mechanismvlilh gears.

T,

n1

=no

-T, T,

n,

=

n,,-

_T,

'

-0-0-1I---.:t

[}J.-t

i:....---

n1 or

na

T'

'T'

T

,

_ eI) 212)

,.

._-

--J d

z

Z2 24 · To l) =-T,

Ifthethrow-out SIOI)is pushedinand if thebackgearis stopped. the spindleof themachinetools gels

only these3rpms.when thesropis pulled outandthegearboxisengaged.thespindlegets 3additional rpms.

(/, Jl~=

n

.~

- (12 · T.

1

,=

-r,

· T. J~

=--

'/~

The advantageofIhis mechanismis cornpncrness,whilethe shonccming is low rigfdltywhich leadsto failure totransmit large power. It is generally used for various feed boxes.

The driving shaft getseither clockwise- Ofcounter-clockwise reversing, depending ontheposition

o

r

the double cone-clutch (on theright Or Ontheleft side).

The steppedpulley gets one of thethreerpm's.

r,

Fig. 17.38 Structuralrepresentationforspeeds on secondshaft. · To

'

1=

_T

-

_, · To ~=T e · To ~=_T,

-The gear wheel is fixedon the shaft bymeans of the sliding key(see I";·ig.1.7.38). ItnKI)'beclutched

(through theidler) with <myone.of the gC'.drwheels which arcFixed on theshaft

T,

(b)Feedboxwithtumbler(Norton's)gear.

Fig. 17,37 Nortongear box mechanism,

**11

('r,)' II~= T~ "0

"

,

=

(~J'

"0

Fig. 17.39 Moonder's gearmechanism.

T, T, T, T, T, T, X

x

-

r--

_n"

f-11-

-

r--

-

l-f-

-

f--

-T, T, T, T,

'i;

T, T, T,

-:;-

-;=-III

]

[

W

2 4 8 3 5 7

I

_I

r

_I

r

I

e

II III

Thisback-geariscalled asingleback-gear. There.are double andeventreble back-gears.but theyare

seldomused.TIleyalso work bythepreceding principle.

Meander's gearmechanism

Itissimilarto the rumblergearboxrncchanlsm,but thisrncrhndofspeedregulationis slightlybcucrthanthe previousone inhaving slightly higher speedsandbeingquite rigid incomparison to rumbler gear box mechanism. The arrangement isshown in

r

i

~.

17.39.

Gear boxIvithdrive key

A typical arr'.tngelllent isshown inFig. 17.42. In thisdesign. thegears to bemarked arc placed inakey which

slides inor out asshewn in the Pig. 17.42.

T, 7j

"r =-/I()~ltl =-"<.1

Tl T~

This type is good forcompactness and automatic speed regulation. but it is quite expensive and the

speed loss isquite high,

Elementary mechanisms wuh clutches: Refer 10Fig. 17.41. Instead ofgear chuchcs, il ispossible 10usc

friction clutches which facilitatc shifting without stopping amachine tool. The mechanical efficiency inthis case is10\\'<.'rthan in thecase of IIICehnnisI11Swithchain of gears. because theidle pair ofthe gear consumes

a pal'!of thepower \\,llhout giving nnyuseful work. These mechanisms are applied inboth speed and feed

boxes.

Fig. 17.40 Mull~la.(flsc Iriotiondutch.

'"

=(

~

)

~

"0

n

,

=(

~

r'

1I

0

11uses gecmerricat progression with the denominator TI17·l•

°fhi:advantages. thefield of usage and theshortcomings arc the same as inNorton's mechanism.

Mecbanisms with the chain of gears areused invarious feed boxes. speed boxes and inother machine

-toolmechanisms.

Gear box driveIvith clutches

'lllis arraugcruentis used(sec Fig, 17.40) where speed is regulated by the usc of mechanical or rrtcuon

clutches. Furthermore, the use

or

more than one clutch can increase therange ufspeed regulation to amuch

higher value. Aclutch isengaged Ordisengaged bymoving the key. 346 TEXTBOOK OF PRODUCTION ENClINEER.lNG

Thesemechanisms oreusedonly invarious feedboxes.

Theadvantagesandshcnccmings arcthesameasinNorton'sandMeander'srncchanlsms. Speedreversing mechanism

Sometimes thi:direction ofspeedis to be reversedwithout wastingmuchtime: for thispurpose.an arrangement isused asshowninFigures 17.43and17.44. T, II~

=

II~I -- 1~ Fig. 17.42 Gearbox vJithdrivekey. Fig. 17.41 cone-type Iricliondulch.

I. Discuss the various motionsofmachinetoolsystc-ms.

2. wfuu arethefundamentalprinciples

o

r

mechanical regulation'! 3, Distinguish between ;Jmy diagram andaspeed diagram.

4. \Vhy arethespeedsarrangedin GP't

5. Describe theprinciple of steppedspeedregulation as al)plied 10machine tools,

6. Discuss themeihodsof stepped regulation. 7. Discuss the advantages and limhations of gear boxdrive. 8. Write snon notes On (i) Norton gearboxmccbantsrn. (ii) Meander's gear meehanism. Fig. 17.43 Reversing mechanism. (c)Usingtumblergears

'c

'-E {b)Usinghelicalgesl$ l-II-

1-

"

I-'B

r

III

II

I....:

-.t

-

Ft-j:

:._",

~ ,- X

1--

---1

X

1-

'

A

e

o

Jewciutcn

e

(a)Us,;ngspurgears

c

'"

e

II X

1-

-

---1

X A ~

'Y

e

!!....

~

V .y

X

"

X

I

,

A (0)

Fig.17.44 Reversing mechanism with gears,

REVIEW QUESTIONS