316

17.1.2 VarIous Motions of Machine Tool SystemTo drive the various components of the machine tool system, we can have four methods:

(i) Mechanical (ii) Electrical(iii) Hydraulic (iv) Pneumatic.The choice of a particular methodof drive will depend uponmany factors such as cost. operating speeds

and feeds. power to weight ratio. rigidity. reliability. maintenance costs. intended use. sophistication. andcontrol.

17.1.1 DrIves and Regulatlon of MotIon on Metal-cuttIng MachInes

Metal-cutting machines receive working motions (speed and feed) from electric motors. which usually haveconstant revolutions per minute. In order to fulfil different operations. it is necessary to find out variousnumbers of spindle revolutions as wel! as different values of feed. For these purposes. speed and feed boxeswhich work by either stepped or unstepped principle of regulation are used.

The various elements of a machine tool are assembled together so as to provide maximum rigidity to thesystem; however. this assembly has revolving. sliding and fixed or stationary components. Generally thedrivesof a machine tool are covered and hidden. but operated by controls which are accessible to the operator.Variouselements of a machine tool are made integral or fabricated and assembled together to make a systemquite homogeneous in appearance and operation.

17.1 INTRODUCTION: MACHINE TOOL SYSTEM

Design of Machine Tool Gear Box

17CHAPTER

The Mechangers

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

II.IIf. nif ... , nl/l'. n!,+1

then this can be written as

Va nn-=--=k=tando 1000

which reveals that to get v/np a constant for the same diameter, we need to change n such that "o,tnt is equalto trd,tlOOO.

lfwe have a GP series such as

The ideal regulation of speed is stepless drive; but it has ccnain limitations. due to which it cannot be incorporatedin all types of machine tools. Some idea regarding this fact has already been given in previous sections. Thecost 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 of the above fact, it is customary to design (for a set of speeds) a gear box which has a optimumnumbers of speeds, with minimum speed loss. Alternative approaches to this problem have posed varioussolutions, out of which we have to pickup the best one.

Ifstepless regulation is not available, then the increment of speeds from a minimum level can be arrangedin anAP. GP or HP series; it has been 'proved that tbe GP (Geometric Progression) gives minimum speed loss, 'and the GP series has other advantages too.

As we know that

17.2 FUNDAMENTALS OF MECHANICAL REGULATION

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

The field of machine dynamics, particularly that of the load bearing components of a system such asbase, bed, table, saddle, columns and spindle support, are important for technological gains, as these componentsare 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 isthe low power to weight ratio 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 andfeed, and (ii) the subsidiary motions such as fixing of workpiece, tool seuing, machine control, ere,

Sometimes the primary motion is only in one axis as in the case of a broaching macbine, but more oftensuch motions are required in two or more than two planes. Insuch situations, we prefer to have an individualdrive rather than a common drive. The line shaft drive is most obsolete. as there is not much control of speedin such a system 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 beachieved by a DC motor with a resistance control or mechanical drives using pressure variations, but for avery limited range.

Hydraulic speed regulation is good for straight line motions, e.g. broaching, grinding, milling and shapingmachines.

318 TEXTBOOKOf PRODUCllON ENGINEERING

The Mechangers

V'I-..../!.V'A

V~-VBMax flV = constant = --"---"-V~

get

Let flV be the relative loss of the cuuing speed, where

6V= VA -VBVA

The maximum of this value win be with the diameter d.:V' - V'Max flV= A H

flA

lfwe take this value as constant (as this is profitable for the exploitation of machine tools), then wesball

Fig. 17.1

-d(a) Retation between

diameter d and speed V

_d(b) Loss of culling

speed with change of diameter

tII------r

Ray diagram or speed spectrum for geometric series.

It is the law of the geometrical progression with the common 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,2n. = n,l/fnx= n,'-'

Let us call the ratio nm."lnmin as the range of regulation denoted by R.n n~=R=.....:L''min nl

nx= n,'-'R = '-'= x-rR

orsoor

no

vt At.V

~B~HY~~~--~--~~--~~-

Let nk_' = n._, =.!..n. n._'9 "

So, if max ~Vis constant, then the segmentA'B' must be constant too, for all of the values of rpm of themachine tool as shown in Fig. 17.1. Then the full radial diagram will be as shown in Fig. 17.2. This radialdiagram is useful for determining the number of revolutions if d and V are known.

Then

= 1 _ 1rd~ nk-l 1000 = 1_ n. - 11rd~ n. 1000 nk

So the ratio n. - 1 must be constant for V to be constant.n.

The Mechangers

E, 3-=-~ 10

(I7.3)(17.4)

(17.2)

E, = 10

1/>= E = log 2 = 0.3E,log 1/>= log 10= 1.0

Dividing Eq. (17.3) by Eq. (17.4). we get

The condition of the second factor may be expressed mathematically as follows:If the series has member II;. then it should have a number such that

ny' = lOnx'

1J ' = mt'l 1J 'y '1" x

1011'="''11'y 'Y' x

or (17.1)

or

What requirements should the standard values of the constant of geometrical progression satisfy? This is animportant point of debate. The denominator of geometrical progression should be chosen by taking intoconsideration the following factors:

(a) The possibility of applying multi-speed asychronous or synchronous motors (e.g. with the numberof revolutions as 3000, 1500,750, etc.);

(b) The decimal system of a series.The condition as expressed in the first factor may be expressed mathematically as follows:If the series of speeds has a member ".

system.

or

!!A.log"

k= 1+--log'

k=I+logRlog'

Let us assume that k = 2132, as this is one of the requirements of design of a speed regulation system.The value of k could be any number, as EJ and E2 are whole numbers starting from zero.

Thus, K = 2, 3. 4, 6. 8,9, 12, 16, 18, 24,27, 36, OUI of which the most widely used numbers are 3, 4, 6,8,9, 12, 16, 11,24 and 36. Therefore, these are the steps of a stepped regulation or mechanical regulation

or

or,=k-Jf,i;V;;I n.log = -Iog-

k-I nJ

We have

Therefore, according to common understanding, if E2 = 40, 20, 10 or 5, EI would be 12,6,3 or 1.5.Hence, the standard values of from Eq. (17.2) would be

4(J= ~ = J2 = 1.06

tf>w = ~ = ifi = 1.12to = JlflO = Vi = 1.26S = ~ = J,rz = 1.58

However, the series of two numbers for the values of was found to be insufficient; therefore. this seriesis to be supplemented with the following values of the series of number 2:

=..J2 = 1.41

= t/iO = 1.78The advantages of standardisation are the following:(i) The decimal system of series is enough to erect a series from 10 to 100.All other numbers of this

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

(ii) Ifwe put down every second member of the series with 40= 1.06. then we get a 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. Asmentioned earlier. R = Range of regulation = n... /nm;n. and if the number of speeds in the gear boxof a machine tool is denoted by k, then

, ~ ir ". . '"

The Mechangers

n) = e) . el'o"6 = e3 . Cs1to

1(-=elT2T.2.= e2T,

Ts-=e). T6

" T,-=e4T8

T9-=es1(0

Lei us assume

HenceSimilarly. we can obtain

and

"I = el . e4"o"5 = e2 . eS"O

Therefore

"4 = el .3 es"o''1 = /l, == e2 e4/l0 _ ~"l nl e) e4"o ele2=e,. if>

Solution 10!his system of transmission of speeds at 6 steps, we require 10gears as shown is Fig. 17.3.Wecan proceed for analytical investigation of the kinematic chain of gears as follows:

Series of number of revolutions:

To 10Fig. 17.3 Design of gear box-kine malic layout diagram.

n, ... nsIII

II

1 r+' _2.

I" flor- L T, 2ir- - r-r- r-r-- X X X- X Xf--

T. T. 'r I- -$ -2 .......'-T

Example 17.1 Design a gear box having 6 speeds. i.e., 3 x 2 and 3 shafts.

T Ts T

DESIGN Of MACtlNE TOOL ClEAR BOX 321 ~; .' :: ,j"~

The Mechangers

Flg.17.4 Symmetric ray diagrams.

Shaft nos.1st

II III II IIIMain Gearing Gearing Maingroup group group group

n. n.

n. n.

n, n,

n, n,

n, n.

n, n,(a) Crossed type speed layoul (b) Open type speed layout

nx = n1tP'C-tlog II, = log 111+ (x - I) log 4>

Therefore. from the above formula prepresenting a straight line equation. it is possible to draw linediagrams which are called ray diagrams. These are sbown in Fig. 17.4.

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

~ = = es es= e.nl e4

In this case, the double block of the gears is the main group, and the treble block of gears is the firstgearing group.

The above analysis is called themethod of investigating kinematic chains.Incases where the number of sets is more than 2, such analysis becomes difficult because of bulky

calculations.Example 17.2 Show the graphical method of investigation of speed regulation by drawing ray diagrams.Solution In general we know that

!!1.=3 = eleS~ = eS111 ele.~ e.es = e. if

The link between the number of revolutions and the denominator 4> is called the main set.The first gearing set is that which bas the power of denominator equal to the number of independent

changes of speed in the main set.

11 T7nl=--~ =ele4~T2 T8

The Mechangers

where R is the range of regulation.

Sometimes a ray diagram, besides providing this information, shows the values of the transmissionratios, and then it is drawn asymmetrically.These are shown in Figs. 17.5(a)and (b).

Tbe step between themotor and the first shaft with nospeed is called the compensatory step; it providesthe possibility of framing a standard series of rpm's.

Ao obvious question arises:Which arrangementor diagram is better?The one inFig. 17.5(a) or 17.5(b)?The first arrangement, shown in Fig. 17.5(a), is better, because the size of the gearbox system in this

case is smaller, and faster speeds are available; thus less torque is transmitted. Hence, smaller shaft diametersand modules of gears are preferable.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 Refer to Fig. 17.6.

IfPo is the number of independent shaftings of tbe main group;P, is the number of independent shaftings of the 1st gearing group;P2 is tbe numberof independent shaftings of the 2nd gearing group;

then the power value is given as follows:for the main grouP-Po;for the first gearing group-p,:for the second gearing group-po.P,:for the third gearing group-po'p,pz.The following limitations are accepted in machine tool design for separate transmission ratios of gear

drives:In the kinematic chain of gear drive

Flg.17.5 Asymmetricray diagrams.

2nd method(b)

1st method(a)

n,

"'" 9c

Overlapping speeds are obtained by reducing the power by one or more degrees inone of the gearing groups.Let us take up a mechanism of 12= 232speeds. the main group of which is the travel block of gears.

17.4 RAY DIAGRAM FOR OVERLAPPING SPEEDS

Let us consider the limitationsof value .pby taking up anexample of 4-shaft mechanism in the kinematicchain of motions.

From the structural diagram as given earlier (12 = 232) it is obvious thate = e . t/> main group'. = ') . Is: gearing group

and therefore '7 = e6 . 2nd gearing groupAll values of t/> are suitable for the main group.For the first gearing group, R = e/"3 = 1/1'; the maximum value of is as follows:for the main motion, = ~ = 1.67;

for the cbain of feed motion, = ~ = 1.94;for the second gearing group, R = .-/e)= rf/';

and therefore themaximum value of is as follows:for the chain of feed motion, = ~ = 1.41;

for the chain of feed motion, t/> = ~ = 1.93.It appears that displacement of gearing group will change nothing. Hence, the value of t/> is limited and

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

1- $e$2.82

4> ;' lFig. 17.6 Ray diagram for 12-speed gearbox.

In the chain of feed motion

Main 1stgearing 2ndgearinggroup group groupPo p,

K OF PRODUCTION ~

The Mechangers

Let us decrease the power of the first gearing group by 1 unit.Thus, the overlapping of two stage speeds is obtained, and we have

= ~ = 1.41

= ~ = 1.56

Fig. 17.8 Ray diagram for overlapping speeds in 'cross' structure.OVerlapping

~----+-~~~~~~~~----+-+---+-~~~~

~---4-----+--~~~~---4-----+~~~~t::$~~3:~~~+-~~~~*-~~~~~~n4

2nd gearingMaingroup groupmustbe1, S

1stgearinggroupmustbe

.z

2nd gearing group 1st gearing group Main group

Fig. 17.7 'Open' structure ray diagram.

The ray diagram for overlapping speeds in 'cross structure' is shown in Fig. 17.8.

",

n,o""

r