MTT

DEPARTMENT OF MECHANICAL ENGG. NIT RAIPURDEPARTMENT OF MECHANICAL ENGG. NIT RAIPUR

MACHINE TOOL

TECHNOLO

BASU SAHU

NIT RAIPUR BATCH

MACHINE TOOL

TECHNOLOGY

BASU SAHU

NIT RAIPUR BATCH - 2010

DEPARTMENT OF MECHANICAL ENGG. NIT RAIPUR

Page 1

MY HOUNARABLE TEACHERS

DEDICATED TO


Page 2

I am very glad to present this book to NIT Raipur. During

my semester (CSVTU scheme) I felt a lot of difficulties in

this subject due to randomness of topics & unavailability

of simplified concepts, therefore, in this book I have

made a small effort to make concepts in simplified

manner. I hope this book will be very helpful to my

juniors and also CSVTU’s students.

Basu Sahu

Mechanical Engg.

NIT Raipur 2010 batch

PREFACE


Page 3

I gratefully remember the contributions made by

teachers to add some practical aspects in this book & to

clear my hazy concepts regarding design of feed gear

box. I also thank to my junior Vivek Singh Rathor &

Kaushik sahu to give me a positive response during

writing of this book and to urge me.

Basu Sahu

Mechanical Engg.

NIT Raipur 2010 batch

ACKNOWLEDGEMENT


Page 4

Machine Tool Technology

UNIT - I

Cutting Tool – types, requirements, specification & application

Geometry of Single Point Cutting Tool - tool angle, Tool angle

specification system, ASA, ORS and NRS and

inter-relationship.

Mechanics of Metal Cutting

Theories of metal cutting, Chip formation, types of chips, chip

breakers, Orthogonal and Oblique cutting, stress and

strain in the chip, velocity relations, power and energy

requirement in metal cutting.

UNIT - II

Machinability

Concept and evaluation of Machinability, Mechanism of Tool

failure, Tool wear mechanism, Tool life, Tool life

equation, Machinability index, factors affecting machinability.

Thermal Aspects in Machining and Cutting Fluid

Source of heat in metal cutting and its distributions, temp

measurement in metal cutting, function of cutting fluid,

types of cutting fluid.

UNIT – III

Design of Machine Tool Element

Design of Lathe bed, Material and construction feature, various

bed section, analysis of force under headstock, tail


Page 5

stock and saddle, torque analysis of lathe bed, bending of lathe

bed, designing for torsional rigidity, use of

Reinforcing stiffener in lathe bed.

Design of Guide ways, Material and construction features, over

turning diagram, Antifriction guide ways.

UNIT – IV

Design of Speed Gear Box

Drives in Machine Tool, classification, selecting maximum and

minimum cutting speeds, speed loss, kinematic

advantage of Geometric progression, kinematic diagrams,

design of Gear Box of 6,9,12 and 18 speed.

UNIT – V

Design of Feed Gear Box

Elements of feed gear box, classification-Norton drive, draw key

drive, Meander’s drive, Design of feed gear box for longitudinal

and cross feed and for thread cutting.

Machine Tool Installation and Maintenance

Machine Tool installation, Machine Tool Maintenance,

lubrication, reconditioning of machine tool.

Machine Tool Testing

Testing, Geometrical checks, measuring equipment for testing,

acceptance test for Lathe and Radial drilling

machines.


Page 6

Unit (i)

Cutting tool

“ A cutting tool is a piece of metal having a single or a

number of hard (hardened) cutting edges, suitably

shaped.”

Classification of cutting tool

<1> According to number of cutting edges:

(a) Single point tool: e.g. turning, planning, slotting

tool etc.

(b) Double point tool: e.g. drill

(c) Multiple (more than two): e.g. milling cutter,

broaching tool, hobs, gear shaping cutter etc.

<2> According to application:

<a> Form cutting tools—tools for cutting grooves, tapers.

 Thread cutting tools – dies, taps etc.

<c> straight cutting tool – broach, hand saw, power saw etc.


Page 7

Cutting tool material requirements

<1> Toughness: To avoid fracture failure, the tool material

must possess high toughness. “Toughness is the capacity of

material to absorb energy without failing.” It is usually

characterised by a combination of strength and ductility in the

material.

<2> Hot hardness : “ Hot hardness is ability of a material to

retain its hardness at high temp.” It is ability of the cutting tool

to withstand at high temp. without losing its cutting edge. Hot

hardness temp. should be high.

<3> Wear resistance : Hardness is the single most important

property needed to resist abrasive wear during machining .

All cutting tool material must be hard.

<4> Mechanical and thermal shock resistance.

<5> Coefficient of friction between tool and work part should

be low in order to have low tool wear and better surface finish

in work part(s).

<6>Ability to maintain above properties at temperature

occurring during cutting operation.

<7> Ease of fabrication.

<8> It should be in range of favourable cost.


Page 8

Types of Cutting tool materials (specification and there

applications):

� Plain carbon tool steel

� Alloy tool steel

� High speed steel(HSS)

� Sintered or Cemented Carbide

� Ceramic and Oxides

� Cermets

� Diamond

� Cubic Boron Nitride

� UCON

� SIALON

Plain carbon tool steel:

Characteristics:

<a> Typical composition: C=1.2% rest iron

 Hot hardness temperature: 200⁰c

<c>Hardness: Rc = 55—64

<d> Cutting speed(grey CI): 20 m/min


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<e> Toughness : good

<f> Wear resistance : poor

<g> Relative cost: low

<h> Easy to fabrication

� Typical use: Making Form tools.

It has advantage, that it is easy to fabricate and acquire keen

edge, hence useful for form tool which is because of there large

line of contact with the work are operated at low speeds upto

about 12m/min.

The use of plain carbon steel as cutting tool , now-a-days , is

limited mainly on account of loss of hardness beyond 200⁰c.

Low Alloy steels :

Characteristics:

<a> Typical composition:

Si = 0.25%, Cr= 0.25% , Va=0.25% , Mn=1.2%

and rest iron.



<c>Hardness: Rc = 60




<g> Relative cost: slightly higher


� Typical use : Making Form tools

High speed steels(HS

W as its principal

alloying ingredient

(12- 20%)

Ex: T1,T2 etc.

Tungsten type HSS

(T-- Series)


Hardness: Rc = 60—64




<g> Relative cost: slightly higher


Typical use : Making Form tools.

High speed steels(HSS):

W as its principal W is partially or fully

alloying ingredient Replaced by Mo

Ex: T1,T2 etc. Ex.: M1,M2etc.

H.S.S.

Tungsten type HSS

)

Molybdenum type HSS

(M-- Series

Page 10

W is partially or fully

Replaced by Mo

Ex.: M1,M2etc.

Molybdenum type HSS

Series)


Page 11

Alloying element Typical content Functions

<1> Tungsten T Type HSS:12-20 %

M Type HSS:1.5-6%

Increase hot hardness

Improve abrasion resistance

through formation of hard

carbide

<2> Molybdenum T Type HSS: None

M type HSS: 5 -10%

Hot hardness



carbide

<3> Chromium T type HSS:

M type HSS: in both cases

3.75-- 4.5%

Depth harden-ability

during heat treatment



carbide in HSS

corrosion resistance

<4> Vanadium In both cases 1-- 5% Combine with carbon for

wear resistance

Retard grain growth for

better toughness

<5> Cobalt In both cases 0-- 12% Increases Hot hardness

<6> Carbon In both cases 0.75-- 1.5% Principal Hardening

element in steel

Provide available C to Form

Carbide with other alloying

elements for wear

resistance


Page 12

DESIGNATION w Mo Cr V Co C

T1 18% 00% 4% 1% 00% 0.75%

T2 18% 00% 4% 2% 00% 0.85%

M1 1.15% 8.50% 3.75% 1.15% 00% 0.80%

M2 6% 5% 4% 2% 00% 0.85%

M36 6% 5% 4% 2% 8.25% 0.85%

Characteristics:


as shown in above table

and rest iron.


<c>Hardness: Rc = 62—65


<e> Toughness : fair

<f> Wear resistance : fair

<g> Relative cost: high


� Typical use : due to high value of hot hardness temperature

and lower cost as compare to other tool material such as

ceramics, cermets etc. This is widely used for example lathe,

drilling, broaching, milling etc.

� Modes of tool failure

Block diagram of production of cemented carbide tool by P/M

technique

Cemented carbide:

C grade


: due to high value of hot hardness temperature



drilling, broaching, milling etc.

l failure- Flank wear, crater wear.


cemented

carbide

S grade

Page 13

: due to high value of hot hardness temperature



Flank wear, crater wear.


S grade


Page 14

• C grade cemented carbide- The C grade consist of

WC(tungsten carbide) with cobalt as a binder for use in

machining cast & non ferrous metals. In this grade cobalt

concentration is varied from 3-16%. Higher is the cobalt content

greater is to resistance to shock.

� S grade cemented carbide-The S grade consist of WC, TiC

(titanium carbide), TaC with cobalt as a binder for use in

machining steel. (TiC=0-16%, TaC=0-10%).

� TiC reduces the tendency of chips to weld to

tool , increase hot hardness.

� TaC helps to improve resistance to crater wear

& make the structure fine grain.

Note -It should be noted that C grade is not suitable for steel

but this drawback is overcome by

adding TaC in WC in S grade.

Characteristics-

<a> Typical composition: It has been described

above.


Page 15


<c>Hardness: Ra = 87-92


<e> Toughness : poor

<f> Wear resistance : good

<g> Relative cost: very high

� Typical use-turning, drilling, machining, broaching,

� Modes of failure- flank wear, crater wear.

� Limitation-It has strong tendency to form pressure weld at

low cutting speed. C grade cannot

use at low.

Note- It should be noted that all carbides, when

finished, are extremely brittle & weak in their resistance to

impact & shock loading. Due to this, vibrations are very harmful

for carbide tool. The machine tool should be rigid, faster &

more powerful. Light feeds, low speed & chatter are harmful.

Due to the high cost & carbide tool materials & other factors,

cemented carbide are used in the form of inserts, or tips which

are brazed

or clamped to steel shank.


Page 16

� In case of a brazed tip, when it is worn-out, it is

resharpened either the help of special grinding wheel

on a tool & cutting grinder

Drawbacks-

1. Because of the difference in coefficients of expansion of

tip material & tool shank material, the brazing has to be done

very carefully.

2. Since tool is removed (un-mount) from machining

for resharpining process hence involving resetting process.

Mechanically clamped tips are known as “Index-able tips”,

because these have more thane one cutting edge which are

used one by one by indexing the tips. Once all the edge of tips

have been used tool is just thrown away. A rectangular tool bit

(insert, tip) can be used upto 8 times before disposed &

requires no sharpening cost.

Ceramics -

ceramics cutting tools are composed primarily of

fine grained Al2O3 , pressed & sintered at high pressure &

temperature with no blinder into insert form (throw away


Page 17

form). The Al2O3 is usually very pure (99% is typical)

although some manufacturers add other oxides like zirconium

oxide, MgO, NiO, Cr2O3 , TiO & TiC etc to improve the grain

structure, cutting properties & sintering.

Characteristics:


Al2O3=95%, Co=5%

 Hot hardness temperature: 1200̊ °c

<c>Hardness: Ra = 90-95%


<e> Toughness : very poor

<f> Wear resistance : very good

<g> Relative cost: very much high

� Typical use : turning.

� Modes of tool failure- DCL notching, micro chipping, gross

fracture.

� Limitation- low strength, low thermal or mechanical fatigue

strength, not for interrupted cutting


Page 18

Purposes of conversion of tool angles from one system to

another

• To understand the actual tool geometry in any

system of choice or convenience from the geometry of a

tool expressed in any other systems

• To derive the benefits of the various systems of tool

designation as and when required

• Communication of the same tool geometry

between people following different tool designation

systems.

Methods of conversion of tool angles from one system to

another

• Analytical (geometrical) method: simple but tedious

• Graphical method – Master line principle: simple,

quick and popular

• Transformation matrix method: suitable for

complex tool geometry


Page 19

• Vector method: very easy and quick but needs

concept of vectors

Conversion of tool angles by Graphical method – Master

Line principle:

This convenient and popular method of conversion of

tool angles from ASA to ORS and vice-versa is based on

use of Master lines (ML) for the rake surface and the

clearance surfaces.

• Conversion of rake angles

The concept and construction of ML for the tool rake

surface is shown in fig(a).

In Fig., the rake surface, when extended along πX

plane,

meets the tool’s bottom surface (which is parallel to πR)

at point D’ i.e. D in the plan view. Similarly when the

same tool rake surface is extended along πY, it meets the

tool’s bottom surface at point B’ i.e., at B in plan view.

Therefore, the straight line obtained by joining B and D is

nothing but the line of intersection of the rake surface

with the tool’s bottom surface which is also parallel to

πR. Hence, if the rake surface is extended in any


Page 20

direction, its meeting point with the tool’s bottom plane

must be situated on the line of intersection, i.e., BD.

Thus the points C and A (in Fig. 4.1) obtained by

extending the rake surface along πo

and πC

respectively

upto the tool’s bottom surface, will be situated on that

line of intersection, BD.

This line of intersection, BD between the rake surface

and a plane parallel to πR

is called the “Master line of the

rake surface”.

From the diagram in Fig.


Page 21

Where, T = thickness of tool.

The diagram in figure(a) is redrawn in simpler form in

figure(b)for conversion of tool angle.

Conversion of tool rake angle from ASA to ORS:


Page 22

Proof of above equations with the help of above

diagram:

Conversion equations can be combined in a matrix form:

Conversion of rake angles from ORS to ASA


Page 23

These inversion can be derived by inverse of above

transformation matrix:

☺ NOTE: These equations can also be directly derived by

using diagram.

Maximum rake angle (γmax

or γm

)

The magnitude of maximum rake angle (γm

) and the

direction of the maximum slope of the rake surface of

any single point tool can be easily derived from its

geometry specified in both ASA or ORS by using the

diagram of fig(b). The smallest intercept OM normal to

the Master line (Fig. b) represents γmax

or γm

as

OM = cot γm

Single point cutting tools like HSS tools after their

wearing out are often resharpened by grinding their rake

surface and the two flank surfaces.

The rake face can be easily and correctly ground by using

the values of γm

and the orientation angle, φγ (visualized

in Fig. b) of the Master line.


Page 24

Determination of γm

and φγ from tool geometry specified

in ASA system

In Fig.(b),

γm

and φγ from tool geometry specified in ORS

Similarly from the diagram in Fig.(b), and taking Δ OAC,

one can prove


Page 25

Conversion of clearance angles from ASA system to ORS

and vice versa by Graphical method:

Like rake angles, the conversion of clearance angles also

make use of corresponding Master lines. The Master

lines of the two flank surfaces are nothing but the dotted

lines that appear in the plan view of the tool (Fig.(c)).

The dotted line are the lines of intersection of the flank

surfaces concerned with the tool’s bottom surface which

is parallel to the Reference plane πR. Thus according to

the definition those two lines represent the Master lines

of the flank surfaces.


Page 26

Fig: (d) Master line of principal flank

Fig. (d) shows the geometrical features of the Master line

of the principal flank of a single point cutting tool.

From Fig.(d),

OD = Ttanαx

OB = Ttanαy

OC = Ttanαo

OA = Tcotλ where, T = thickness of the tool shank.


Page 27

The diagram in Fig.(d)is redrawn in simpler form in Fig.(e)

for conversion of clearance angles.

Fig(e): Use of Master line for conversion of clearance

angles

The inclination angle, λ basically represents slope of the

rake surface along the principal cutting edge and hence is

considered as a rake angle. But λ appears in the analysis

of clearance angles also because the principal cutting

edge belong to both the rake surface and the principal

flank.

• Conversion of clearance angles from ASA to ORS

Angles, αo

and λ in ORS = f(αx and α

y in ASA system)


Page 28

Following the same way used for converting the rake

angles taking suitable triangles (in Fig.(b)), the following

expressions can be arrived at using fig(e).

Combining equations in form of matrix:

• Conversion of clearance angles from ORS to ASA

system

αx and α

y (in ASA) = f(α

o and λ in ORS)

Proceeding in the same way using Fig. (e), the following

expressions are derived

The relations (4.14) and (4.15) are also possible to be attained from inversions of Equation 4.13 as indicated in case of rake angles.

Minimum clearance, αmin

or αm

The magnitude and direction of minimum clearance of a

single point tool may be evaluated from the line segment

OM taken normal to the Master line (Fig. (e)) as


Page 29

OM = tanαm

The values of αm

and the orientation angle, φα

(Fig.(e)) of

the principal flank are useful for conveniently grinding

the principal flank surface to sharpen the principal

cutting edge.

Proceeding in the same way and using Fig.(e), the

following expressions could be developed to evaluate the

values of αm

and φα

From tool geometry specified in ASA system

From tool geometry specified in ORS

Similarly the clearance angles and the grinding angles of

the auxiliary flank surface can also be derived and

evaluated.


Page 30

� Interrelationship amongst the cutting angles used in

ASA and ORS

The relations are very simple as follows:

φ (in ORS) = 90o

- φs (in ASA)

and φ1(in ORS) = φ

e (in ASA)

Theory of chip formation:

The portion of the material that has been cut away from the

work material by the cutting tool is called the chip. Chip

formation is a complex phenomenon however, the basic

mechanism by which all type of chips are formed in cutting

metal is shearing process in shearing zone. This zone is some

time also known as primary shear zone. The metal in front of

rake of tool face gets immediately compressed ,first elastically

and the plastically. Plastic deformation can be caused by

yielding, in which case strained layer of material would get

displace over other layer along slip planes which coincide with

the direction of maximum shear stress.


Page 31

� There is always a deformation of metal lying ahead

of cutting edge by process of shear. Here the metal deforms by

shear in narrow zone extending from the cutting edge to the

work surface. This zone is treated as a single plane for the

purpose of mathematical analysis & commonly referred to as

the single plane. The angle which shear plane makes with the

direction of travel of tool is known as the shear plane.

� The degree of plastic flow ahead of cutting tool

determines the type of chip that will be produced. If the

workpiece material is brittle, due to little capacity for

deformation before fracture, the chip will separate along shear

plane to form what is known as a discontinuous or segmented

chip. If material is ductile , due to capacity for plastic flow

before fracture ,will deform along shear plane without rupture.

The plane tends to slip & weld to successive shear plane and

resulting in continuous chip.

� Piispianen presented a model in which undeformed


Page 32

metal is considered as stacks of cards which would slide over

one another as the wedge shaped tool moved under these

cards.

Types of chips:

1. Discontinuous chips

� Regular sharp size chip

� Irregular sharp size chip

2.Continuous chips

� With BUE

� Without BUE

3. Segmented or serrated or shear localised chips

� Discontinuous chips:

<a> Irregular sharp shape and sized:

Condition:


Page 33

• Work material: brittle & hard like grey cast iron.

• Cutting velocity: low

• Cutting feed : high

• Rake angle: positive and small

• Cutting fluid: use of cutting fluid

 Regular sharp shape and sized:

Condition:

• Work material: ductile & hard and work harden-able like

mild steel

• Cutting velocity: low

• Cutting feed : high


Page 34

• Rake angle: negative

• Cutting fluid: absence of cutting fluid

� Continuous chips:

<a> Continuous chips without BUE:

Condition:

• work material: Ductile.

• Cutting velocity: high

• Cutting feed : low

• Rake angle: positive and large

• Cutting fluid: use of cutting fluid (both cooling and

lubricant)


Page 35

 Continuous chips with BUE:

Condition:

• work material: Ductile.

• Cutting velocity: medium to low

• Cutting feed : medium to large

• Rake angle: positive and small

• Cutting fluid: inadequate or absent.

� Serrated chip(inhomogeneous chip)

These chips are semi-continuous in the sense that they posses a

saw teeth appearance that is produced by a cyclical chip

formation of alternating high shear strain followed by low shear

strain. The shear deformation which occurs during the chip


Page 36

formation causes the temperature on the shear plane to rise,

which in turn may decrease the strength of material and cause

further strain if a material is poor conductor. Thus, a large

strain is developed at the point of initial strain. As cutting

process is continued , a new shear plane will developed at same

distance from the first plane and the deformation shifts to this

point. This type of chip is typical of materials in which yield

strength decreases with temperature and which have poor

thermal conductivity such some steels and titanium(Ta) alloys.

Condition:

• work material: Semi-ductile.

• Cutting velocity: low to medium

• Cutting feed : medium to large

• Rake angle: negative

• Cutting fluid: absent


Page 37

Chip breaker

Chips disposal is a problem that is often encountered in

turning and other continuous operations. Long stringy

chips are often generated , especially when turning

ductile materials at high speeds. These chips cause :

� a hazard to machine operator &

� also hazard to the work part finish,

� and they interfere with automatic operation of

turning process.

Chip breaker are frequently used with single point tool to

force the chips to curl more tightly than they would

naturally be inclined to do, thus causation them to

fracture.

Types of chips breaker:

There are two principal form of chip breaker design

commonly used on single point turning tools—


Page 38

<a> In-built type

Obstruction (clamped or attachment ) type

Rake surface


Page 39

Advantages of using chip breaker:

� Safety of operator(s) from the hot , sharp continuous

chip flowing out at high speed.

� Convenience of collection and disposal of chips.

� Chances of damage of finished surface by entangling

or rubbing with the chip eliminated.

� More effective cutting fluid action due to shorter

and varying chip tool contact length.

Disadvantages:


Page 40

� Chances of harmful vibration due to frequent

chip breaking and hitting at the heel or flank of

the tool bit.

� More heat and stress concentration near the

sharp cutting edge and hence chance of its rapid

failure.

� Increase the cost of cutting tool.

� Increase in cutting force due to increase in cutting

edge.Increase in power consumption and that

increase is by 20%.

In- built type chip breaker

In-built type chips breaker

Groove Type

Tilted Vee groove

Circular groove

Step type

parallel step

Angular step

Parallel step with

Nose radius for

Heavy cut


Page 41

In-built breakers are in the form of step or groove at the

rake surface near the cutting edges of the tools. Such

chip breakers are provided either

• after their manufacture—in case of HSS tools like

drills , milling cutters, broaches etc. and brazed type

carbide insert.

• During their manufacture by power metallurgy

process – e.g. throw away type inserts of carbides,

ceramics and cermets.


Page 42


Page 43

Unique characteristics of in-built chip breakers are:

� The outer end of step or groove acts as the heel

that forcibly bend and fracture the running chip

� Simple in configuration , easy manufacturing and

inexpensive

� The geometry of chip breaking feature are fixed

once made (i.e. cannot be controlled)

� Effective only for fixed range of speed and feed for

any given tool work combination.

Analysis of step type built chip breaker:

From geometry of figure:

��


Page 44

� �� = (2� − )

Where

W= width of step

H= height of step

� = radius of curvature of step

Clamped type chip breaker:

Clamped type chip breaker are basically in the principle

of stepped type chip breaker but have the provision of

varying the width of step and/ or the angle of the heel.

Clamped type chip breaker

Fixed geometry type

Variable width(W) only

Variable width (W), height(H) and

shear angle (β)


Page 45


Page 46

Analysis of clamped type chip breaker:

From the geometry:

W=� tan��/2 )

And h =� sin �

Where � = angle of the chip breaking strip

H is taken 1~2 mm greater than h.


Page 47

� Geometry of chip formation

When a wedge shaped tool is pressed against the work piece

chip is produced by deformation of material ahead cutting edge

because shearing action takes place in a the (work piece).

When the tool moves with the velocity V against the work

piece, it shears the metal along shear plane. The depth of cut

t1(which is actually the feed in turning operation) change into

the chip thickness t2. This experiences two velocities


Page 48

component Vc (Velocity of chip relative to tool) and Vs (velocity

of chip relative to work piece along shear plane).

where α is rake angle and

Ø is shear angle

Chip thickness ratio (r): “ the ratio of depth of cut (thickness of

uncut chip) to the chip thickness is called chip thickness ratio.”

Chip thickness ratio, r = ��


Page 49

From ∆��, AB =

��[�� ∅ "�]

� AB = ��$%�(∅ ") ---------(1)

From ∆��&,

AB = ��

�� ∅ --------------(2)

From equation (1) and (2)..


Page 50

� ��

$%��∅ "� �� ∅

�

Now, from velocity diagram :

From ∆�'� and ∆�'(

� BC = Vsin ∅ Vc sin�)� � �∅ � *�#

� �� ∅

$%��∅ "� +,+

Therefore, �� +,

+ - �� ∅$%��∅ "�

�� - �� ∅

$%��∅ "�


Page 51

Now , consider consistency of volume:

Volume of metal removed from work piece

= volume of chip

� t1× /1 × 11 × �1 = 22 × /2 × 12 × �2

where t denotes thickness, b denotes width, l denotes

length , � denotes density

and also /1 = /2 and �1 = �2

� 21 × 11 = 22 × 12

� �� = 3�

3�

Therefore, 2122 = 12

11 = 454 = sin ∅

cos(∅ − *)

Again,

- = �� ∅$%�(∅ ")

� - = �� ∅$%� ∅ $%� "8�� ∅ �� "

� - = 9:� ∅$%� "89:� ∅ �� "

� rcos ∝ +- tan ∅ sin * = tan ∅


Page 52

� tan ∅ = < $%� "� �� "

Analysis of forces in metal cutting☺☺☺☺ (Merchant circle)☺☺☺☺

In figure shown, Fs is shear force and Fn is normal

reaction force acting on shear plane ; and F is friction

force and N is normal reaction farce acting on tool and

chip contact surface. R’ and R are resultant force acting

on shear plane and tool-chip contact surface

respectively.


Page 53

Free body diagram :

� None of four forces componets F,N,Fs and Fn can be

determined directly because their direction in which

they are applied vary with different tool geometries

(phi and alpa)and cutting condition.

� However, R is usually determined, in experimental

work , from measurement of two orthogonal

components—

� Fc(Cutting force)- is in the direction of cutting ,

same direction as cutting speed V.


Page 54

� Ft(Thrust force)- is associated chip thickness

before cut t1. Its direction is perpendicular to

cutting force Fc.

☺ How to measure cutting force Fc and thrust

force Ft?

• Fc and Ft can be measurement by using

suitable dynamometers (with resistance wire strain

gauge) or face transducers (such as piezo electric crystal)

mounted on machine.

• Forces Fc and Ft can also be calculated from

amount of power consumption that occurs during

cutting, often measured by power monitor provided that

the mechanical efficiency of machine tool can

detemined.

Hence F, N , Fs and Fn =function(Fc, Ft, α,Ø,μ)

Where μ=tan � is coefficient of friction between chip and

tool contact surface.

As shown in fugure, resultant of forces Fc and Ft is R

which is same as resultant of forces F and N i.e. F and N

are replaced by forces Fc and Ft.


Page 55

If R is taken as active force then R’ can be assumed as

reactive and vice versa.

Now , let us draw force triangle……


Page 56

These force Triangles can be represent in a circle which is

known as Merchant circle.

☺As we know chord which passes through centre of

circle makes right angle triangle with any point on

periphery of circle. Using this theorem we can draw

merchant circle as shown below…..


Page 57

Hence we can obtained following relationship….

=�� ° =?

�� ° = @,$%�(A ") = @�

��(A ") = @B$%�(A "8∅) =

@C��(A "8∅)

Since F and N are active forcr components as Fc and Ft,

therefore

∑ E52FGH 5IJKILHL2 = ∑ E52FGH 5IJKILHL2

Hence F and N can be expressed in terms of Fc and Ft…

� M = M5 sin * + M2 cos *

� � = M5 cos * − M2 sin *

Since Fs and Ft are reactive force components,

therefore


Page 58

∑ E52FGH 5IJKILHL2 + ∑ -HE52FGH 5IJKILHL2 = 0

� MO = M5 cos ∅ − M2 sin ∅

� ML = M5 sin ∅ + M2 cos ∅ hence

�

Coefficent of friction chip- tool contact surface

P = tan � = @Q = @, �� " 8@� $%� "

@, $%� " @� �� "


Page 59

P @, 9:� "8@�@, @� 9:� "

☺ Note :

Fc is the component which is responsible for

energy consumption since this is along direction of

motion of tool.

Ft is not responsible for cutting operation.

Variation of Fc and Ft with respect to cutting

velocityV, feed f and depth of cut d.

Work done and power required:

Let W = Total work done in cutting operation

=M5 × 4

& W1 =work done in shear

=MO × 4O


Page 60

& W2 = work done against fricton

=M × 45

∴ � = �1 + �2

� W=Fc× 4

� W = @B $%�(A ")$%�(∅8A ") × 4 ……………(1)


Page 61

As 1O × / = ��×S�� ∅ = TU

�� ∅

From equation(1)………….

� W= VB×TB $%�(A ")$%�(∅8A ") × 4

� W= VB×TU×$%�(A ")��(∅)×$%�(∅8A ") × 4

� W(Ø) = WXQYZTQZ��(∅)×$%�(∅8A ")

Condition for min. power required Wmin�

� [\[∅ = 0

� cos(∅) cos(∅ + � − *) − sin(∅) sin(∅ + � − *)=0

� 2∅ + � − * = ]/2 This relationship is known as

merchant relation.


Page 62

☺ What does merchant relation indicates?

• � ↑ FJK1FHO ∅ ↓ JEFL2EFLFL` * 5ILO2EL2

• * ↑ FJK1FHO ∅ ↑ JEFL2EFL` � 5ILO2EL2

Since shear angle Ø increases then the tool force and

power requirement decrease.

Wmin =2aO × �I × 4 × tan ∅

And Power consumed (H.P.) = W(in kgf m/min)/4500

☺ NOTE: It has been experimentally observed that the

nature of variation of cutting force Fc and chip thickness

ratio r with cutting speed V and rake angle α tallies with

that indicated in figure.


Page 63

Specific cutting energy:

E= \bU3cde = \

(S×��)+

� E = @,×+

S×��×+

� E = @,

S×�� (fLF2 �/OgfF-H JJ, hIf1H/5f/H JJ)

“ Specific cutting energy based on unit it is some

times also called as specific cutting pressure; Specific

cuttin force / area”.

☺ Condition for minimun energy consumption is given

by various equations by different scientists as follows:

� Ernst & Merchant

2∅ + � − * = ]/ 2

� Merchant’s second solution

2∅ + � − * = 'J

� Lee & Shaffer

∅ + � − * = ]/4

� Stabler

2∅ + 2� − * = ]/2


Page 64

Shear strain:

The chips are considered to be consisting of series of

plate like elements of thickness ∆j and displacement

through distance ∆O relative to each other.


Page 65

Shear strain can be defined as:

k ∆B∆l

� k TQ8QmXQ

� k TQXQ � Qm

XQ

� nℎHE- n2-EFL k = cot(∅) + tan(∅ − *)

Condition for minimum k:

� [p[∅ = 0

� −(csc ∅)� + (sec(∅ − *))� = 0

� 2∅ − * = )�

� kJFL = 2 cot(∅)

Shear stress:

shear stress can be defined as

aO = MO�O

� aO = @,,UB∅ @�BrC∅TU/BrC∅ where �I = / × 21

Strain energy per unit volume:

Strain energy per unit volume = aO × k


Page 66

Unit (ii)

MACHINABILITY

“Machinability of a metal is a term which is using in

comparative analysis of ease of machining of that metal

to another one on the basis of important parameters

which are

• Machining force and power consumption

• Surface finish and

• tool life”

Evaluation of machinablity of a metal:

<1> Machinability forces and power consumption :

Machining requiring larger forces results in large power

consumption and similarly machining requiring smaller

forces results in small power requirement and higher

Machinability.

Larger machining force� Smaller Machinability

Smaller machining force�Higher Machinability

<2> Surface finish: In some situations, the major concern

can be over the quality of finish and , depending on the


Page 67

severity of this problem, the machinablity may be low or

high.

Higher surface finish and low severity � Higher

machinablity

Lower surface finish and High severity� Lower

machinibility

<3> Tool life: The length of the period for which a tool

can be used is defined as the tool life. This criterion is

also linked up with the productivity and economics and

can be very good index for an overall judgement of a

machining operation.

More Tool life� Higher Machinability

Lower Tool Life� Lower Machinability

There are other minor factors which are also used in

evaluating machinability of a metal:

<4> Material removal rate:

High Material Removal Rate(MRR)� Higher machinablity

Low Material Removal Rate(MRR)� Lower machinablity

<5> Heat generated during cutting:


Page 68

Higher heat generation during cutting � Lower

machinability

Lower heat generation during cutting � Higher

machinability

<6> Tool life between grinds:

Higher tool life between grinds � Higher machinability

Lower tool life between grinds � Lower machinability

<7> Shape and size of chips:

Chips with BUE formation� Lower machinability

Chips without BUE formation � Higher machinability

� Factors affecting machinablity:

<1> Machine variables:

• Power

• Torque

• Accuracy of machine and

• Rigidity of the machine and holding device.

Machine variable indirectly affecting the machinability.


Page 69

The machine should be rigid and have sufficient power to

withstand the induced cutting forces and to minimise

deflection. If not so then, both tool life and surface finish

are affected , and to limit the cutting forces , the speed ,

feed and depth of cut have to be limited.

<2> Tool life:

• Tool material

• Tool geometry

• Nature of engagement of tool with work continuous

or intermittent.

The cutting tool has to optimised to obtain a reasonable

value of tool life and remove maximum material. Proper

tool geometry is essential for efficient machining and it is

chosen depending on the work material and machining

conditions. Surface finish and other parameters have

greatly influenced by the tool geometry. Rake angle and

nose radius in large improvement of surface finish and

other have little influence. Tool rigidity affects tool life ,

surface finish and dimensional accuracy.


Page 70

<3> Cutting condition:

• Cutting speed

• Cutting depth

• Feed rate

• Cutting fluid

Cutting speed has the greatly influence on tool life . the

surface finish, normally , is improved by increase in the

cutting speed , due to continuous reduction of the built

up edge. Dimension of cut and cutting fluid also influence

on machinability.

<4> Work material variables:

• Hardness

• Microstructure

• Tensile strength

• Chemical composition

• Methods of production of work material.


Page 71

MACHINABILITY INDEX

“Machinablity index is ratio (Expressed in %)of cutting

speed of metal investigated for 20 min. tool life to

cutting speed of standard steel for 20 min. tool life .”

Let machinability index = I

Cutting speed of metal investigated

for 20 min. tool life = Vi

Cutting speed of standard steel

for 20 min. tool life = Vs

I(%)=(Vi / Vs ) × 100

Standard steel:

Carbon – 0.13% (max)

Manganese – 0.06 to 1.1 %

Sulphur -- 0.08 to 0.03%

Rest is iron.


Page 72

Standard steel has 100% machinability index.

Material Machinability index(%)

Low carbon steel 55 – 60

Aluminium alloy 390 – 1500

Magnesium alloy 500 – 2000

Carbon steel C – 40 125

C – 45 100

C – 55 80

C – 85 50


Page 73

Cutting fluid

“Cutting fluid can be defined as any liquid and gaseous

substance which is applied to a tool during cutting

operation to facillate removal of chip.”

Function of cutting fluid:

<1> Act as Cooling agent : It cools the tool and work

piece. The heat produced is carried away by the fluid by


Page 74

supplying adequate quantity of cutting fluid. This makes

possible more accurate production and measurement.

<2> Acts as lubricant : It lubricates the cutting tool and

thus reduces the coefficient of friction between the chip

and tool. This increases tool life.

<3> Acts as Antiwelding agent : The cutting fluid

prevent the intimate contact between the surface of chip

and tool face.

<4> Acts as anticorrosion agent: It prevent corrosion of

work and tool and also machine.

<5> Acts as washing agent : It washes away the chips

from the tool.

<6> Breaking of chips: It causes the chips to break up

into small pieces.


Page 75

� Properties of cutting fluid

� It should have a high specific heat , high

conductivity and high film coefficient.

� It should possess good lubricating properties to

reduce frictional forces and to decrease the

power consumption.

� It should be odourless.

� It should non corrosive to work and machine.

� It should be non – toxic to operating personnel.

� It should have low viscosity to permit free flow

of the liquid.

� It should be stable in use and storage.

� It should permit clear view of work which is

specially desirable in precision work.

� It should be safe particularly with regards to fire

land accident hazards.


� Types of cutting fluids:

Water based cutting fluid:

� Soft soap + water

� Soluble oil(1 to 5%)+ water

fluid(emulsion)

The ratio of soluble oil to water depends upon the

application of cutting fluid and ranges from about

1:5 to 1:50.

Turning , Drilling & Reaming

Milling : 1:10

Grinding : 1:50

Water based cutting fluid


ypes of cutting fluids:

Water based cutting fluid:

Soft soap + water � water based cutting fluid

Soluble oil(1 to 5%)+ water� water based cutting

fluid(emulsion)



Drilling & Reaming: 1:25

Milling : 1:10

Grinding : 1:50

Cutting fluid

Water based cutting fluid Oil based cutting fluid

Mineral oils

Straight fatty oils

Page 76

water based cutting fluid

water based cutting



Oil based cutting fluid

Mineral oils

Straight fatty oils

Blended oil

Sulphurised oil

Chlorinated oil


Page 77

� Specific properties

Cooling properties : excellent

Lubricating properties : poor

Antiwelding properties : poor

Oil based cutting fluids:

There are mainly two types of oil :

Fixed oil(Fatty oil) and Mineral oil.

<1>Straight fatty oils(Fixed oil):

Fixed oil consists of animal , fish & vegetable oils. Chiefly

used fixed oils are Lard oil , Sperm or whale oil & olive ,

cotton seed & Linseed oil. Turpentine oil distilled from

vegetable oils is also used. Fixed oils have greater oiliness

than mineral oils(i.e. fixed oil have excellent Lubricating

properties) but are not stable and rapidly lose their

lubricating properties by forming gummy deposits and

decomposes when heated . Neither are they satisfactory

coolants as they have a high viscosity.


Page 78


Cooling properties : fair

Lubricating properties : excellent

Antiwelding properties : fair

Application: Suitable for light duty.

Lard oil is mainly used during thread

cutting with taps and dies.

<2> Mineral oils:

Mineral oils are primarily composed of hydrocarbons of

different structures and molecular weights derived from

petroleum oil.


Cooling properties : fairly good

Lubricating properties : good

Antiwelding properties : fair

Application: Suitable for light duty

machining operation such as turret

and capstan lathes and single spindle

automatics where free cutting brasses

and steels are being machined.


Page 79

<3> Compounded or blended oils:

Mineral oil + Fatty oil � Blended oil

Since fatty oils have excellent lubricating properties but

unstable nature it can not be used in heavy duty to make

assure using in heavy duty this is blend with mineral oils.

The film strength of fatty oils is retained ever when

diluted with 75% mineral oil. As a result they are much

cheaper and more fluidity than straight fatty oils.


Cooling properties : good


Antiwelding properties : good

Application : suitable for heavy duty

machining operations such as

threading on capstan and turret lathes,

thread milling and medium capacity

automatic lathe.

☺ NOTE: To improve Antiwelding properties of oil based

cutting fluids sulphur and chlorine are added and these

are known as “Extreme pressure” (E.P.)additive. These


Page 80

additive form solid films of iron sulphide and iron

chloride between tool face and chips. These films are

easily sheared & have high melting points. Thus, they

help in preventing the chips from welding to nose of tool

where the pressure is maximum. Hence Antiwelding

properties prevents the formation of built up edge on

cutting tool .( Sulphur is most commonly used as EP

additive then chlorine)

<4> Sulphurised oils:

� Mineral oil + about 5% of sulphur�

Sulphurised cutting oil


Cooling properties : fair

Lubricating properties : Excellent

Antiwelding properties : Excellent

Application: Heavy duty machining

operation such as lathe works, gear

cutting and thread grinding.


Page 81

<5>Chlorinated oils

� Mineral oil + about 3% chlorine � Chlorinated oils


Cooling properties : Fairly good



Application: suitable in heavy duty.

<5>Extreme pressure (EP) cutting oil

� Mineral oil + Fixed oil + sulphur(sulphur may be in

free form, combined or both ; upto 5%) +

chlorine(upto 3%) � E.P. cutting oil


Cooling properties : good

Lubricating properties : Excellent


Application: heaviest duty machining

operation


Page 82

☺ NOTE: How cutting fluids are applied to machines?

The cutting fluid may be applied to cutting tool in

following ways:

• By hand , using a brush.

• By means of drip tank attached to the machine

body.

• By means of pump.

Selection of cutting fluid on the basis of following

factors:

� Cutting speed

� Feed rate

� Depth of cut

� Cutting tool material

� Workpiece material

� Velocity of cutting fluid

� Expected cutting tool life

� Cost of cutting tool life

� The life of cutting fluid and loss of cutting fluid

during operation.


Page 83

Sources of heat in metal cutting:

According to the first law of thermodynamics “ when

work is transformed into heat , the quantity of heat

produced is equivalent to work.”

This heat will be generated when conversion of

mechanical energy takes place . The main source of heat

cutting are shown in figure….

<1>Primary shear Zone: here maximum heat is

dissipated because of the main plastic deformation of

metal. A large portion of this heat is goes to the chip.

Contribution of this zone in total heat generation is

about 80 – 85% .

<2>Secondary shear Zone: here heat is generated mainly

due to friction between rack surface of tool and chips

and partially due to secondary deformation of built up

edge. From this source also, the chip takes away a major

portion of heat. Contribution of this zone in total heat

generation is about 15 – 20% .


Page 84

<3>Work tool interface: here heat is generated due to

job rubs against the flank surface of tool. But with sharp

tool , contribution of this sources to the heating

phenomenon is insignificant. Contribution of this zone in

total heat generation is about less than 5% .

☺The direction of maximum heat flows these zones to

chip or work piece is indicated in figure.

Distribution of heat in chip , tool and work versus

cutting speed

It is found that distribution of heat in Chips, Workpiece

and Tool is in ratio of 80:10:10, when cutting speed with

carbide cutter at speed above 30 metre/min.


Page 85

Maximum temperature along the rake surface:

The total power consumption (i.e. total rate of heat generation)

during machining is � M5 × 4. If the rates of the heat

generation in the primary and the secondary deformation zones

are W1and W2, respectively, then

W = W1 + W2.

Again,

�2 M × 45 M × -4

Where Vc is the chip velocity along chip tool interface ,V is

cutting velocity of tool and F is frictional force acting on chip

tool interface. We can write

�1 M5 × 4 � M × -4

Thus, when enough information is available, the rate of heat

generation in the primary deformation zone (i.e. the shear plane)

and the secondary deformation zone (i.e. the rake face) can be

found out.

When a material particle moves across the primary deformation

zone, the temperature rise is given by


Page 86

θp = (1-Λ)W1 /ρcVt1b

where

Λ = fraction of primary heat which goes to work piece,

ρ = density of material,

c = specific heat of the material,

t1 = uncut thickness,

b= width of cut,

Since the computation of Λ needs an elaborate analysis, we will

give here only the result which agrees well with the experimental

results. It has been found that Λ is a function of the shear angle Φ

and a non-dimensional quantity, namely,

Θ = ρcVt1/k

k being the thermal conductivity of the material. For wide range

of work materials and machining conditions,


Page 87

Λ = 0.15 ln{27.5/(Θ tanΦ)}.

Hence, if Φ is known or determined, Λ & θp can be estimated

with the help of above equations

The maximum temperature rise θs when the material particle

passes through the secondary deformation zone along the rake

face of the tool can

θs ≈ 1.13s[�Θt2/l� × �W2/ρcVbt1�]

where l is the length of contact between the tool and chip. The

corresponding average temperature rise is obtained from the

equation

(θs)ave = W2/ρcVbt1

It has been found that

l/t2 = [1+tan(Φ-α)]

hence θs =1.13z {�|$}~9��[�89:�� "�]


Page 88

The final temperature is given as

θ = θ0+θp+θs

where θ0 is the initial temperature of the work piece. This

maximum temperature is along the rake face of the tool.

☺NOTE: Experimental Determination of chip Tool

interface Temp.:

Several methods have been used for measuring

temp. at the chip tool interface

<1> By calorimetric method used by A.O.Schimdt

<2> By decolorizing agent used by Bickel

<3> Tool work thermocouple technique

<4> By infra-ray detection

<5> By probing the distribution of hardness &

investigating the structural transformation used by

Loladze. ect.


Page 89

Tool Wear And Tool life

Tool wear causes the tool to lose its original shape so

that in time tool ceases to cut efficiently or even fails

completely. After a certain degree of wear, the tool has

to be resharpened for further use. The following basic

causes which can operates singly or in various

combinations, produce tool wear:

(a)Attrition Wear:

At relatively low cutting speeds, the flow of the material

past the cutting edge is irregular and less streamlined.

Sometimes built-up edge may be formed and contact

with the tool may not be continuous. Under these

conditions, fragments of the tool are torn intermittently

from the tool surface. This phenomenon is called

attrition. This type of wear progresses slowly in the case

of continuous cutting but with interrupted cutting or

where vibrations are severe due to lack of rigidity of the

machine tool or uneven work surfaces, it leads to rapid

destruction of cutting edge. As the cutting speed is

increased, the flow of metal becomes uniform and

attrition disappears. When carbides are used at low


Page 90

cutting speeds where built up edge is likely to form, the

wear is largely controlled.

(b) Diffusion wear:

Diffusion wear occurs because of the diffusion of metal

and carbon atoms from the tool surface into the work

material and the chips. Wear by diffusion due to the high

temperature and pressure developed at the contact

surfaces in metal cutting and rapid flow of chip and the

work surfaces past the tool. The rate of diffusion wear

depends upon metallurgical relationship between the

tool and the work material. It is one of the major causes

of wear and is of special significance in the case of

carbide tools.


Page 91

(C) Abrasive wear:

The abrasive action of the work on the tool is basically

due to two principal effects:

• The inherently hard constituents present in the

microstructure of the material being cut.

• The strain hardening induced in chip and work by

the cutting process, and the formation of built up

edge.

Abrasive wear is caused when these hard particles are

swept over the tool surface. The action is much more

pronounced on the tool flank because of rigid backing

provided by the work. However, it is still not clear

whether this type of wear is of much significance except

when work material contains greater concentrations of


Page 92

hard particles such as pocket of sand on the surface of

castings, etc. In the case of carbides, a fine grain size and

lower percentages of cobalt help in reducing the abrasive

wear.

(d)Electrochemical Wear: This type of wear may occur

when ions are passed between the tool and the work

piece causing an oxidation of the tool surface and a

consequent breakdown of the tool material in the region

of the chip tool interface. Not much data is available

regarding this type of wear in the metal cutting in order

to evaluate its relative importance.

(e) Chemical Wear: This type of wear due to interaction

between the tool and work material. While machining

some types of plastics with carbide, rapid wear on the


Page 93

face occurs owing to chemical action. This may be

accelerated in certain cutting fluid environments, where

the fluid is active with respect to the tool.

(f) Plastic deformation: When high compressive stresses

act on the tool rake face, the tool may be downwards;

this deformation takes place primarily in the nose area of

the insert and reduces the relief angle. This is a

deformation rather than a wear process, but it

accelerates other wear processes which reduces the tool

life. Deformation leads to sudden failure of the tool by

fracture or localised heating. The occurrence of plastic

deformation is in itself an indication of the over stressing

of the tool material. The severity should be construed as

a limit of the cutting conditions which should never be

exceeded.


Page 94

(g)Thermal Cracking: Owing to thermal cyclical stresses

at the cutting edges, short cracks called comb cracks are

often observed in the tool, running at right angles to the

cutting edge. These occur mainly while machining with

carbide tools and are caused by the alternating

expansion and contraction of surface layers of cutting

tool because of interruption in cutting. If these comb

cracks are deep enough, they separate the worn part of

cutting edge into several sectors. Cracks may also

appears in the transverse direction within these sectors

located in several layers one over other. This may be due

to cyclical compressive stresses which occurs in

interrupted cutting. The comb crack and transverse

cracks lead o chipping of cutting edges and premature

failure of the tool. Tougher grades of carbide with less


Page 95

sensitively to thermal fatigue and uniform rate of cooling

should be adopted for avoiding these thermal cracks.

� Geometry of tool wear:

The progressively wear of the cutting tools can take

two forms:

(i) Flank Wear: Wear on the tool flank

characterised by the formation of wear land as a

result of the newly cut surface rubbing against

the tool flank; and

(ii) Crater Wear: Tool wear on the rack face

characterised by the formation of a Crater or


Page 96

Depression , as a result of chip flowing over the

tool rack face.

� Flank wear: As a rule, the extent of flank wear is

considered a dependable criterion for judging the life of

the cutting edge. In the case of carbide tools, through

proper alloying of tungsten carbide with titanium and

tantalum carbides, sufficiently resistance to crater is

obtained so the most of tools do not fail by cratering or

deformation, before a reasonable amount of flank wear

is obtained on the flank of the tool. The flank wear can

be more easily observed and measured than other type

of wear and relatively easy to predict when a given

amount of wear will be reached once the wear rate has

been established. The development of flank wear initially

assume a high rate followed by more or less linear

increase and finally rises rapidly when the amount of

wear crosses beyond the critical value.

A typical case of flank wear development is shown

figure. It cab be seen from this figure that the graph can

be divided into three definite regions A,B and C.


Page 97

� Region A: In region A , the wear grows rapidly within

short period of time because during the initial contact of

sharp cutting edge with the work piece, the peaks of

micro unevenness at cutting edges are rapidly broken

away.

� Region B: In the region B, the wear progresses at a

uniform rate.

� Region C: In region C, the wear rate is rapid and may

lead to catastrophic failure of the tool. In general, it has

been found that the most economical wear land at which


Page 98

to remove the tool and resharpen is just before the start

of rapidly increasing portion of curve.

In this case of high speed steels, the linear rate of

wear section of curve is quite flat and final failure occurs

suddenly. This phenomenon occurs at low speed

normally used in cutting with high speed tools and is the

result of protection of the cutting edge by the formation

of the built-up edge.

Since there is very little measurable increase in flank

wear, until catastrophic failure occurs, this of failure is

usually used as a criterion for the end of tool life in such

tool materials. Using catastrophic failure as the end point

of the tool life eliminates the need for frequent flank

wear measurement during tool testing.


Page 99

The wear land on the flank will not be generally uniform

along the entire cutting edges length. Depending upon

the machining conditions of them are generally

observed:

1. Excessive wear at nose end of flank is brought about

by deformation of the tool material which reduces the

relief in area, thus increasing the rate of wear. This can

also be brought about if the crater on the rake face


Page 100

breaks through the nose area.

2. Irregularities in the wear along the whole cutting

edge length due to minute chipping or attrition of cutting

edge.

3. Excessive wear at line of depth of cut. This can be

either due to the work hardened surface caused by the

previous cut or heat treat scales or by other abrasive


Page 101

material on the work piece.

� Crater wear: As the cutting speed in increased, the

tendency of the cutting tool fail by cratering is increased.

The tool chip interface temperature increases with

cutting speed, and at these higher temp. the rate of

material removed from the tool increases. The potion of

crater wear indicates that the wear in this region is

primarily due to the diffusion or chemical reaction

between the tool and chip material. At low cutting

speeds a crater may be formed owing to the action of the

chip flowing over the tool rake face. In the narrow region

close to the tool cutting edge is protected from the

action of the chip by the presence of a stable built up


Page 102

edge. This cavity of crater has its origin not along the

cutting edge, but at some distance away from it and

within the chip contact

area. It is to be noted that the maximum tool chip

interface from the cutting edge, and in this region the

crater in initiated. As the crater wear progress with time,

it becomes wider, larger and deeper , and approaches

the edges of tool. If the crater wear allowed to proceed

too far, the cutting edge becomes too weak and breaks

down suddenly. The depth of crater and the distance of


Page 103

the centre line of the crater from cutting edge are

measured for the quantitative assessment of crater wear.

� Tool life: Tool life may defined as “ the effectively

cutting time between resharpenings.”

When the wear reaches a certain value of the tool is

not capable of further cutting unless it resharpened.

Tool life is most important criterion for assessing the

performance of the tool material, machinablity of work

material and for determining cutting condition. Flank

wear is generally considered as the decisive factor of

the tool life. However at higher cutting speeds and

metal removal rates, the tool failure may also be

caused by cratering. Both type of wear are strongly

influenced by the cutting speed (V).

Taylor’s tool life equations: The relationship between

cutting speed(V in m/min.) and tool life (T in min.) can

be determined from plots of wear rates at various

cutting speeds. The tool life values for different cutting

speeds are read from the graph for any stipulated


Page 104

value of wear.

The values of tool life and cutting speeds shows a

straight line relationship when plotted on log-log

graph.


Page 105

It has been shown by Taylor that the relationship

between the tool life and the cutting speed can be

represented by the equation:

4�C '

Where C and n are constant depending on the tool

and work material, tool geometry , and cutting

condition(Except speed).


Page 106

Tool materials Typical value of n for 4�C = '

Low alloy steels 0.14 – 0.16

HSS 0.25

Cemented carbide 0.30

Ceramics 0.40

Coated carbides TiC or TiN 0.35

Al2O3 Coated carbide 0.40

☺ NOTE: when machining through the production of

continuous chips without a built-up edge, the Taylor

equation can be written as:

4�C21�� = '′

Where V = cutting speed(m/min.)

T = tool life (min.)

t1 = depth of cut (mm.)

w = width of work piece (mm.)

p, q, n and C’ are constants criterion


Page 107

Conversion of Turning operation to Orthogonal cutting

model:

Turning operation Orthogonal cutting model

Feed f (mm/rev) Thickness of uncut chip t1 (mm)

Depth d (mm) Width of cut b (mm)

Cutting velocity V

(m/min)

Cutting speed V (m/min)

Cutting force

Fc (kgf)

Cutting force Fc (kgf)

Feed force Ff (kgf) Feed force Ff (kgf)

Problem(1):

During turning a mild steel component with an

orthogonal tool a feed of 0.75 mm/rev is used at 50

R.P.M. if the chip thickness is 1.5 mm, determine the chip

thickness ratio. Also find the length of chip removed in

one minute, if the work diameter is 50 mm before the cut

is taken. Assume a continuous chip.

Solution- Given:


Page 108

Feed in turning operation (f) = 0.75 mm/rev

Hence t1= thickness of uncut chip =0.75 mm

t2= thickness of chip =1.5 mm

D = Diameter of work piece = 50 mm.

and N = R.P.M. = 50

Now, r = Chip thickness ratio = ��

� r = �.��.� = 0.5

and l1 =Length of chip before cutting = ]DN

� l1 = π × 50 × 50 = 7850 mm

now, l2 = length of chip removed per minute can be

determined from

r = ��

� 0.5 = ��

��

� l2 = 3925 mm (Ans)


Page 109

Problem(2):

In an orthogonal turning operation :

Cutting speed = 80 m/ min

Cutting force = 20 kgf

Feed force = 8kgf

Back rake angle = 15° Feed = 0.2 mm/ rev.

Chip thickness = 0.4 mm

Determine the following :

(a) Shear angle.

(b) Work done in shear.

(c) Shear strain.

Solution- given :

Cutting speed ( V ) = 80 m/ min

Cutting force (Fc) = 20 kgf

Feed force or thrust force (Ft) = 8kgf


Page 110

Back rake angle (α ) = 15° Feed (f) = 0.2 mm/ rev.

hence thickness of uncut chip (t1) = 0.2 mm

Chip thickness (t2) = 0.4 mm

Now , r = chip thickness ratio = �� =

�.��.� = 0.5

Let φ = shear angle

tan ∅ - cos *1 � sin *

� tan ∅ �.�×$%� ��°� �.�×�� °

� tan ∅ = 0.554753

� ∅ = 29°

And let Fs be shear force

then Fs = M5 cos ∅ − M2 sin ∅

� Fs = 20 cos 29° − 8 sin 29°

� Fs= 13.61 kgf

Now let Vs be velocity of chip with respect to

work piece


Page 111

4O 4 cos *cos�∅ � *�

� 4O �� $%� ��°$%�(��° ��°)

� 4O = 79.64J/JFL

Hence Work done in shear W1 = MO × 4O

� W1 =13.61× 79.64

� W1 = 1083�`�. J/JFL

Shear strain = cot ∅ + tan(∅ − *)

Shear strain = cot 29° + tan(29° − 15°) = 2.05 (ans.)

Problem(3):

Calculate the power consumed during cutting of a low

carbon steel bar 40mm. diameter if cutting force is 150

kgf at 200 rpm.

Solution :

Given:

Diameter of bar (D) = 40 mm


rpm (N) = 200


Page 112

let velocity of cutting tool = V

then V = ] � � = ] × � �� × 200 = 25.12 m/min

work done by tool (W) = M5 × 4

� W = 150 × 25.12 �`�. J/JFL

� W = 150 × 25.12 × �.�� . J/OH5

� W = 616.068 hIf1H/OH5 = 0.616068 kw

∵ 1�� = 1.341 (

� Power required = 0.616068 × 1.341 = 0.826 HP.

Problem(3)

In orthogonal cutting of a material the feed force is

80 kgf and cutting force is 150 kgf.

Calculate the following :

(a) Compression and shear force on shear plane.

(b) Coefficient of friction of the chip on the tool face.

Take chip thickness ratio as 0.3 and rake angle as

8∘.


Page 113

Solution-

Given: Feed force(Ft) = 80 kgf


Chip thickness ratio(r) = 0.3

Rake angle (α) = 8∘

(a)

Fs = shear force on shear plane

= M5 cos � − Ft sin �

Fn = Compressive force on shear plane.

=M5 sin ∅ + M2 cos ∅

Let- φ= shear angle .

tan � = < $%� �

� < �� = U. $%� �

� �. �� = 0.3

Φ = 17∘

Fn = M5 sin ∅ + M2 cos ∅

� ML = 150 sin 17° + 80 cos 17°

� ML =120.36kgf (Ans.)

Fs = M5 cos � − Ft sin �


Page 114

� Fs = 150 cos 17 � 80sin 17

Fs = 120.06 kgf (Ans.)

(b)

coefficient of friction of chip on tool face=

P M5 tan ∝ �M2M5 � M2 tan *

P 150 tan 8° � 80150 � 80 tan 8 °

� P 0.7285 (Ans.)

Problem(4): During a metal cutting test under orthogonal

conditions it was found that the cutting force is 110 kgf

and Feed force is 100 kgf when cutting speed at 165

m/min. the rake angle of tool is 10∘. If the depth of cut

was 3.50 mm. and feed is 5 cuts/ mm. and chip thickness

is 0.4 mm.

Determine followings:


Page 115

(1) Chip thickness ratio(r)

(2) Shear angle (φ)

(3) Velocity of chip along shear plane(Vs)

(4) Velocity of chip along chip tool interface(Vc)

(5) Force acting on chip along shear plane (Fs)

(6) Force acting on chip along normal to shear plane

(Fn)

(7) Force acting on chip along tool chip interface(F)

(8) Force acting on chip along normal to chip

interface(N)

(9) Power Required to tool(W) and specific energy

consumption

(10) Work done by tool in shearing action (W1) and

specific shearing energy.

(11) Work done by tool against friction force(W2) and

specific frictional energy.

(12) Maximum metal removal rate (MMR max.)

(13) H.P./ 5J /OH5 .

(14) Shear stress

(15) Shear strain

(16) Strain energy per unit volume.

(17) Normal Pressure on chip


Page 116

(18) If overall efficiency of setup is 80% then find power

required for motor .

Solution :

Given :


Feed force or thrust force (Ft) = 100 kgf

Cutting speed (V) = 165 m/min.

Rake angle (α) = 10∘

Depth of cut (d) = width of chip (b) = 3.50 mm

Feed (f) = 1/5 mm/rev. = 0.2 mm/rev

hence (t1) = 0.2 mm

and chip thickness (t2) = 0.4mm.

now,

(1) Chip thickness ratio (r) = �� =

�.��.� = 0.5

(2) Let shear angle be φ then,


Page 117

� tan ∅ = < $%� ¡� < �� ¡

� tan ∅ = �.� $%� ��°� �.� �� °

� ∅ = 28.33°

(3) 4O = $%� ∅$%�(∅ ") 4

� 4O = $%� ��. °$%�(��. ° ��°) × 165J/JFL.

� Vs = 153 m/min.

(4) 45 = 4 × - = 165 × 0.5 J/JFL. � 45 = 82.5 J/JFL.

(5) MO = M5 cos � − Ft sin �

� MO = 110 cos 28.33° − 100 sin 28.33°

� MO = 49.37 �`�

(6) ML = M5 sin ∅ + M2 cos ∅

� ML = 110 sin 28.33° + 100 cos 28.33°

� ML = 140.22 �`�

(7) M = M5 sin * + M2 cos *

� M = 110 sin 10° + 100 cos 10°


Page 118

� M = 117.58 �`�

(8) � = M5 cos * − Ft sin *

� � = 110 cos 10° − 100 sin 10°

� � = 90.96 �`�

(9) Power required to tool (W) = M5 × 4

= 110 × 165 �`�. J/JFL

= 110 × 165 × 9.8160 × 1000 �� = 2.968 ��

( = 2.968 × 1.341 = 3.98

and Specific energy consumption =

W/ volume of chip per sec

and volume of chip per sec = 21 × / × 4

= �.�� × .�

�� × �� J /OH5

= 1.925 × 10 �J /OH5

= 1.9255J /OH5

Hence specific energy consumption

= �.��.��×��¢£ = 1.542&h/J


Page 119

(10) Work done by tool in shearing action (W1)

MO × 4O

� �1 = 49.37 × 153 × �.��×��

� �1 = 1.235��

And specific shearing energy

= W1/ volume of chip per sec

= �.� ��.��×��¢£ ��/J /OH5 = 0.642 &h/J

(11) Work done against frictional force(W2)

= M × 45

� �2 = 117.58 × 82.5 × �.��×��

� �2 = 1.586��

And specific frictional energy

= W2/ volume of chip per sec

= �.��.��×��¢£ ��/J /OH5 = 0.8239 &h/J


Page 120

(12) Maximum material removal rate(MMRmax.)

21 × / × 4

�.�� × .�

�� × �� J /OH5

1.925 × 10 �J /OH5

= 1.9255J /OH5

(13) H.P./ 5J /OH5 = 3.98/1.925 = 2.07

(14) Shear stress aO = MO/�O

And �O = TU�� ∅ = S×��

�� ∅

= 3.5 × 0.2sin 28.33° = 1.475JJ�

= 1.475 × 10 �J�

Hence aO = ��. �×�.��.��×��¢£ �/J� = 328.334 × 10��/J�

(15) Shear strain

k = tan(∅ − *) + cot ∅

� k = tan(28.33° − 10°) + cot 28.33° = 2.186


Page 121

(16) Strain energy per unit volume = aO × k

328.334 × 10� × 2.186 �/J�

= 717.795 × 10��/J�

(17) Since efficiency of setup is 80% hence

Power required for motor = 3.98/0.80 H.P.

Power required for motor = 4.975H.P.

Problem(5):

A tool life of 80 min. is obtained at a speed of 30 m/min

and 8 min. at 60 m/min.

Determine the following:

(1) Tool life equation

(2) Cutting speed for 10 min. tool life

(3) Tool life for 40 m/min cutting speed

(4) Change in cutting speed required to give 50%

reduction in tool life.

(5) If length of bar (L) is 50 mm , diameter of bar 20

mm. and feed is 0.2mm/rev and cutting speed

40m/min. determine no. of components produced

in 8 hour shift.


Page 122

Solution :

T1 = 80 min. and V1 = 30 m/min.

T2 = 8 min. and V2 = 60 m/min.

(1) 41 × �1C 42 × �2C '

� L 1I` �+�+�� /1I` �Z�

Z�� =0.3

And ' 30 × 80�. 111.7

� Tool life equation 4 × ��. 111.7

(2) � 10JFL. � 4 ��.�

��¤.¥ 55.98 J/JFL

(3) V = 40m/min

� � = ��.�� /� .

= 30.67 min.

(4) Given

T = 0.5 T1

� 4 × ��. = 41 × �1�. = '

� 4 = +��.�¤.¥ = 1.2341

� Change in cutting speed required

= (V-V1)/V1 x 100%

� Change in cutting speed required = 23%

(5)


Page 123

Let N = spindle speed

D = dia. of bar = 20mm = 0.02m.

]�� 40J/JFL

� ∴ � 2000 -KJ

And feed f = 0.2 mm/rev

Length of piece L = 50mm

Cutting time per piece = ¦§×Q

= ��×��.�×�� OH5 = 7.5OH5

No. of components produced in 8 hours shift

=�× ��

�.� = 3840

Problems from GATE

Problem (1)

Maximum shear strain in orthogonal turning with a

cutting tool of zero rake angle is

(a) 0 (b) 1 (c) 2 (d) 0.5


Page 124

Solution:

Shear strain k tan�∅ � *� � cot ∅

And condition for maximum shear strain 2∅ � * )�

Here * 0 then ∅ )�

∴ k tan )� + cot )

� = 2 ans(c)

Problem (2)

In machining experiment tool life was found with cutting

speed in following manners :

Cutting speed(m/min) Tool life (min)

60 81

90 36

Q(1) The exponent (n) & constant (C) of Taylor’s tool

equation are:

(a) n = 0.5 and C = 540 (b) n = 1.0 and C = 4860

(c) n = -1 and C = 0.74 (d) n = -0.5 and C = 1.115


Page 125

Q(2) What is the percentages increase in tool life when

cutting speed is halved—

(a)200 (b) 100 (c) 300 (d) 400

Solution :

(1) 41 × �1C 42 × �2C '

� 60 × 81C 90 × 36C

� L 3U¨�©¤£¤�

3U¨�ª«¥£� 0.5

And 60 × 81�.� '

� C = 540

Hence 4�C 540

Ans (a)

(2)

41 × �1�.� 42 × �2�.� 540

And V2 = ½ V1

∴ �+�+�� Z�

Z��.�


Page 126

� 2 �Z�Z��.�

� Z�Z� 4

� Z� Z�

Z� 3 ans (c)

Problem (3) A batch of 10 cutting tools could produce

500 components while working at 50rpm with a tool feed

0.25 mm/rev & depth of cut 1mm.

A similar batch of 10 tools of same specification could

produce 122 components while a feed of 0.25 mm/rev at

1mm depth of cut.

How many components can be produce with one cutting

tool at 60 rpm—

(a) 29 (b) 31 (c) 37 (d) 42

Solution:

Let K = tool life coefficient per component per tool

Hence �1 �� ¬ 50¬ and �2 ��

�� ¬ 12.2¬


Page 127

41 50 × 0.25 12.5 and 42 80 × 0.25 20

Now, 41 × �1C 42 × �2C '

� 12.5 × �50¬�C 20 × �12.2¬�C

� L 0.333

∴ 12.5 × �50¬��. �60 × 0.25� × ��¬��.

� N = 29 ans(a)

Problem (4)

In a machining operation , doubling the cutting speed

reduces the tool life to 1/8th

original value. Exponent in

Taylor’s tool life will be—

(a) 1/8 (b) 1/2 (c) 1/4 (d) 1/3

Solution:

� 4�C '

� 4 × �C 24 × �Z��C '

� n = 1/3 ans(d)


Page 128

problem (5)

In a typical metal cutting operation using a cutting tool of

positive rake angle α = 10° . It was observed that shear

angle was 20°. The friction angle is

(a) 45° (b) 30° (c) 60 ° (d) 40°

Solution :

Given α = 10° and ∅ 20°

2∅ � � � * 90°

� � 60 ° ans(c)

Problem(6)In an orthogonal machining operation

Uncut chip thickness 0.5 mm.

Cutting speed 20m/min

Rake angle 15°

Width of cut 5mm

Chip thickness 0.7mm

Thrust force 200N

Cutting force 1200N

Assume Merchant’s theory.


Page 129

Q(1) The value of shear angle and shear strain are:

(a) 30.3° & 1.98 (b) 30.3° & 4.23

(c) 40.2° & 2.97 (d) 40.2° & 1.65

Solution:

- �� .�

�.�

� r = ��

�

And tan ∅ < $%� "� < �� "

Putting values α = 15° and r �� in above equation then

∅ 40.2°

And shear strain k tan�∅ � *� � cot ∅

Hence k tan�40.2° � 15°� � cot 40.2°

k 1.65 ans (d)

Q(2) coefficient of friction at tool chip interface is—

(a) 0.46 (b) 0.23 (c) 0.85 (d) 0.96

solution:


Page 130

P M5 tan * � M2M5 � M2 tan *

� P �� 9:� ��°8�� 9:� ��°

� P 0.46 ans (a)

Q(3) The fraction of total energy dissipated due to

friction at tool chip interface is—

(a) 30% (b) 42% (c) 55% (d) 70%

Solution:

Power input to tool

� M5 × 4 1200 × �� E22 400 �E22

Power dissipated due to friction

�2 = M × 45

and M = M5 sin * + M2 cos * EL 45 = -4

putting values; we have

W2 = 119.94 watt

∴ \�\ = ��.��

�� = 0.30 ans(a)


Page 131

Design of lathe bed:

Steps<1> Material selection bed for : The composition of

material & the subsequent heat treatment decide the

strength and other requirements of the machine tool

beds.

Beds are generally made up of ordinary cast iron ,

though nodular & alloy cast iron have also been used for

the purpose.

During solidification of lathe bed casting(single piece)

some distortion taking place due to cooling stress set up.

To avoid this , a very common practice of natural

seasoning, called ageing is prevalent. For this reason , the

bed casting are rough machined and then left in open for

considerable time , usually a couple of years and then

machined to required size for final assembly.

Step <2> Selection of type of cross section for lathe bed:

A lathe bed is subjected to different types of stresses

during the operation, such as tensile , compressive,

torsional and bending stress etc. which cause distortion

of lathe bed structure so design of cross section should

be adequate to provide static & dynamic stiffness to


Page 132

counteract the adverse effect of above stresses.

It observed that box section is more suitable as it is

strong in bending & torsion ,& it can be easily produced.

To avoid massive section in castings carefully designed

systems of ribbing are used to offer the maximum

resistance to bending & torsional stresses.

<3>Step ribs in lathe beds:

Ribs are used to provide stiffness against bending &

torsional stresses. There are mainly two types of ribs are

used—

• Diagonal ribs

• Parallel ribs


Page 133

� Characteristics of Diagonal ribs:

� Relative torsional stiffness wrt. Bed without rib

=2.48

� Relative weight wrt. bed without rib= 1.38

L

t

d

t1


Page 134

� Relative torsional stiffness per unit weight

=2.48/1.38= 1.80

Diagonal ribs are quite commonly used in lathe beds.

Diagonal ribbing provides greater torsional stiffness .

� Characteristics of Parallel ribs:

� Relative torsional stiffness wrt. Bed without rib

=1.34

� Relative weight wrt. bed without rib= 1

� Relative torsional stiffness per unit weight


Page 135

=1.34/1= 1.34

Use of Stiffeners (Ribs) in lathe bed:

In lathe bed mainly parallel & diagonal ribs or stiffeners

are used & mainly depends upon—

<a> thickness of main members of bed beam (t)

 ratio of width & length of bed (b/l=ψ)

<c> depth to length ratio (d/l=β)

<d> number of ribs or stiffeners, Z

<e> thickness of rib(t1) & t/t1=ŋ

t

L

d

t1


Page 136

Step <4> Design of slide-ways:

The specific functional feature of slide ways dictate the

following additional design considerations:

<a> It should be possible to provide effective lubrication

without much difficulty.

 There must be provision for compensations of

possible wear, &

<c> The material of the slide ways should have high wear

resistance.

� Material: grey cast iron or steel.

� Shape: Flat & V shaped slide ways are generally used

in lathe bed.


Page 137


Page 138

Step<5> An important point t be born in mind is that an

accurate location & proper levelling of bed , during

installation & afterwards, plays an important role. Even

very strong bed are observed to have been distorted if

they are placed on unlevelled foundation. This twisting of

bed effects the accuracy of work very serious. The bed

small therefore, be tested for level both lengthwise as

well as crosswise.


Page 139

Force analysis in lathe bed:

In shown figure:

� Px = Axial force acting on work material

Py= Transverse component (thrust force)

Pz = Tangential component

� Force PxH, PyH, and PzH are acting on live centre

� Force PxT, PyT, and PzT are acting on dead centre


Page 140

� Forces PyH1 and PzH1 are acting on bearing

housing1

� Forces Px , PyH2 and PzH2 are acting on bearing

housing2

� Forces PzH1B and PyH1B are acting on bolt no.1 in

head stock side

� Forces PzH2B and PyH2B are acting on bolt no. 2 in

head stock side

� Forces PyT1B and PZT1B are acting on bolt in tail

stock side

� Forces PyT2B and PZT2B are acting on heel.

Force analysis on Work Piece(Only considering force)


Page 141

Shifting of forces from peripheral point to centre point

of work piece(WP)

Considering Forces and Moments on work piece due to

sifting of tool forces from peripheral point to centre:


Page 142

Mx =Moment due to Pz (& Moment Vector is in X

direction) = Pz × �/2

and Mz = Moment due to Px ((& Moment Vector is in Z

direction) = Px . �/2

Appling equilibrium condition:

∑ &j 0 � PzH. L � Pz . ± � � . ²/2 0

�

∑ M³ 0

� PzH�Pz �PzT �� 0

PzH = Pz�´¦� � \

�

FBD of work piece


Page 143

�

Similarly , ∑ &³ = 0 � PyH× L − Py × ± − &³ = 0

�

∑ Mj = 0 � PYH−Py +PYT = 0

� PyT = Py�¦ ´¦ � − (± � ¶

�·�

�

And ∑ M± = 0

� PxH − Px − PxT = 0 and PxT = K is tightening force

�

Hence Forces acting on the headstock centre:

PzH = Pz�´¦� − \

�

PyH = Py�´¦� + (± � ¶

�·�

PxH = Px + PxT= Px + K

PzT = Pz�¦ ´¦ � − \

�

PyT = Py�¦ ´¦ � − (± � ¶

�·�


Page 144

PyH = Py�´¦� � (± � ¶

�·�

PxH Px + PxT= Px + K

Forces acting on tailstock centre:

PzT = Pz�¦ ´¦ � � \

�

PyT = Py�¦ ´¦ � � (± � ¶

�·�

PxT K

FORCE ANALYSIS ON BEARING HOUSING AND BOLTING:

Free body diagram of head stock spindle and bearings

housing and bolts are shown in figure.


Page 145

Now, using equilibrium equations: (applying in head

stock spindle)

∑ &j 0

� PZH1× m �PZH×(m+m’) = 0

� PZH1 ¹º»�¼8¼?�¼

Now, ∑ M³ 0

� PZH2�PZH� PZH1= 0

� PZH2 ¹º»×¼?¼

Now, ∑ &³ 0

� PYH1× m �PYH×(m+m’) = 0

� PYH1 ¹½»�¼8¼?�¼

Now, ∑ Mj 0

� PYH2�PYH� PYH1= 0

� PYH2 ¹½»×¼?¼

Hence forces acting on bearing housing are:


Page 146

PZH1 ¹º»�¼8¼?�¼

PZH2 ¹º»×¼?¼

PYH1 ¹½»�¼8¼?�¼

PYH2 ¹½»×¼?¼

CALCULATION OF FORCES ON BOLTS:

Fig : PxH is shifted to lathe bed and moment My is considered

due to shifting of force PxH


Page 147

Hence now equilibrium equations:

∑ M³ 0

� PZH1�PZHB1�Fy = 0

� PZHB1 ��PZH1�Fy )

� PZHB1 ��PZH1� ¹¾».»¼ )

and

� PZHB2�PZH2�Fy = 0

� PZHB2 �PZH2�Fy )

� PZHB2 �PZH2� ¹¾».»¼ )

Fig: My is converted into Fy by applying moment


Page 148

And ∑ Mj 0

� PyHB1�PyH1 = 0

� PyHB1 − PyH1

And

� PyHB2−PyH2 = 0

� PyHB2= PyH2

Hence forces at supports(bolting ):

PZHB1= −(PZH1+ ¹¾»×»¼ )

PZHB2= (PZH2+ ¹¾»×»¼ )

PyHB1= − PyH1

PyHB2= PyH2

Forces on tail stock support (bolting) and Heel:


Page 149

Applying equilibrium equations:

∑ &j 0

� PZT. �n � n′� �PZTB1.(n) + PXT. �H� = 0

� PZTB1 = PÀÁ��8�?� ) + PXT �»

�)

∑ M³ 0

� PZT�PZTB1� PZTB2 = 0

� PZTB2 = PÀÁ��?� ) + PXT �»

�)


Page 150

∑ &³ 0

� PYT× �n � n′� �PyTB1×(n) = 0

� PYTB1 = PÂÁ��8�?� )

And ∑ Mj 0

� PYT�PYTB1� PYTB2 = 0

� PYTB2 = PÂÁ��?� )

Hence forces acting on rear bottom end (Heel)

PZTB1 = PÀÁ��8�?� ) + PXT �»

�)

PZTB2 = PÀÁ��?� ) + PXT �»

�)

PYTB1 = PÂÁ��8�?� )

PYTB2 = PÂÁ��?� )


Page 151

FORCES ACTING ON LATHE BED:

The lathe bed receives forces through:

• The headstock and tailstock

• The saddle on which the cutting tool is mounted

� Forces through the headstock and tailstock


Page 152

Since Px, Py and Pz are acting on cutting tool, these

forces can be considered as active forces and forces

acting on lathe bed can considered as reactive forces.

∑ &±�-HE52FGH� ∑ &±�E52FGH� (applying on point C)

� VA× B= PÀ�Ã8¶� ) + PY �H�

� VA = PÀ�Ã8¶�Ã ) + Py �»

Ã)

∑ &±�-HE52FGH� ∑ &±�E52FGH� (applying on point A)

� Vc× B= �PÀ�Ã ¶� ) + PY �H�

� Vc = �PÀ�Ã ¶�Ã ) + Py �»

Ã)


Page 153

∑ &³�-HE52FGH� ∑ &³�E52FGH� (applying on point of

intersection of AC and BD line)

� VB = VD (assume)

� VB× L = PÄ�H)

� VB = VD = Px �»·)

∑ &j�-HE52FGH� ∑ &j�E52FGH� (applying on point of

intersection of AC and BD line)

� HB = HD (assume)

� HB× L = PÄ�D/2)

� HB = HD = Px � ¶�·)

∑ Mj�-HE52FGH� ∑ Mj�E52FGH�

� Hc = Py


Page 154

Mt1 = PÀ�¶�) + PYH �H�

Mt2 = PÀ�¶�) + PY �H�

Mt3 = PYT �H�


Page 155

Now , ∑ &± Mt1 – Mt2 + Mt3 = 0

Bending analysis on lathe bed:

Various simplified assumption are made:

<1> m = m’ = a

<2> n = n’ = b

<3> PzH = PzT = Pz/2

<4> PyH = PyT = Py/2

<5> Distance between headstock centre and tailstock

centre is L and loading is done at L/2.


Page 156


Page 157

Unit(iv)

Machine tool spindle speed:

To machine work of any diameter D at cutting speed V,

the spindle speed is given by

� �� +)m rpm.

Where D is in mm and V is in m/min.

� � ∝ 4/�

The spindle speed can be obtained by two types of

drives:

(a) Stepped Drive

(b) Step-less Drive

� Stepped Drive: “ In stepped drive the output

spindle speeds are obtainable in finite number

through single or multistage with there

corresponding torque.”

NxT = constant

� N1T1=N2T2 = N3T3


Page 158

� Stepless Drive: “ In stepless drive the output spindle

speed are obtainable in infinite numbers within an

finite range of speed with their corresponding

Torque.”


Page 159

� In a stepped drive following factors should be

decided first –

<1> Maximum spindle speed(Nmax): Depends upon

Vmax, Dmin

<2> Minimum Spindle speed (Nmin): Depends upon

Vmin, Dmax.

<3> Number of spindle speed steps(n)

<4> The no. of subdivisions of steps.

<5> The number of stages in which the steps are to be

obtained.

☺ The output spindle speeds form a series which may be

any one of following types—

(a) Arithmetic Progression(A.P.)

(b) Geometrical Progression(G.P.)

(c) Logarithmic Progression(L.P.)


Page 160

� Maximum possible speed loss:

Let us consider the various speed in some progression

in several steps be N1,N2,N3......Nn

� � �� +,)m rpm.

Where Vc is given and D is given.

Let us assume that corresponding to certain diameter

(D), the required rpm(N)for accurate cutting is not

available i.e. N is unknown for given Vc. In this case loss

in speed is given as follows:

� Loss of cutting speed = Vc – Vp-1


Page 161

� & % loss in cutting speed = }$ }Æ �

+,

=Ç¶È Ç¶ÈÆ �

)mQ

=È ÈÆ �

Q

1 � Q� �Q

∴ Maximum possible speed loss in between two steps

Np-1 & Np is thus given by:

(1− Q� �Q� ) × 100%

<a > In case of G.P.


Page 162

�2�1 �3

�2 �4�3 �5

�4 … �K�K � 1 . . . �L

�L � 1 ∅

where

∅ common ratio in G.P.

�2 � �1�1 �3 � �2

�2 ⋯ �K�K � 1 . . . �L

�L � 1 �1 � 1

∅�

i.e. Maximum possible speed loss in between any

two available range of speeds is constant & given as-

(1� �∅� × 100%

Hence in G.P.

<1> %age of max. Loss for any available range of speeds

is constant.

<2> %age max. Loss is not a function of dia.(D) for given

Vc.

<3> No crowding of speed even at higher speed.


Page 163

 In case of A.P.

N2= N1+∆ N3= N2+∆ N4= N3+∆

Np= Np-1+∆ ................................ Nn= Nn-1+∆

Where ∆ = common difference in A.P. series

Q� Q�Q� ∆

Q� Q Q�Q� = ∆

Q Q� Q� �

Q� = ∆Q� QC QC �

QC = ∆QC

As N2 < N3 < N4...............<Np.......<Np-1

�2 − �1�1 > �3 − �2

�2 > ⋯ > �K�K − 1 >. . . > �L

�L − 1


Page 164

Hence

<1> % of maximum loss for any two available range of

speed is not constant & is larger at lower speed.

<2> % of maximum loss is a function of dia.(D) for given

Vc.

<3> Considerable crowding of speeds at higher speeds.

� Speed spectrum:

Maximum loss in speed between two speed Np-1 & Np

is given as:

Max. Loss in speed 1 � Q� �Q�

&for given dia. Dp

� Dp = ��×+dÍ´

)Q� = ��×+drC)Q� �

� Q� �

Q� = +drC+dÍ´

<a> In case of G.P.

Let us assume Vmax is constant for all dia. range

� Q� �

Q = �∅ = 5ILO2EL2


Page 165

� Q� �

Q� = +drC+dÍ´ = 5ILO2EL2

� As Np increases or decreases ;

Since Vmax is constant ,therefore, Vmin is always

constant

 In case of A.P.

Let us assume Vmax is constant for all dia. range

� Q� �

Q� = +drC+dÍ´

� & Np= Np-1+∆

� ∴ +drC+dÍ´ = 1 − ∆

Q�

� As speed increases i.e. Np increases i.e.

(1 − ∆Q�) increases �

+drC+dÍ´ increases

Since Vmax is constant ,therefore, Vmin increases

As Np↑ � Vmin ↑

& as Np = ∞ � Vmin = Vmax (only theoretically )

Hence considerable crowding of speed Vmax and Vmin at

higher speeds and theoretically coincide when Np= ∞ as

shown in figure.


Page 166

Selection of maximum & minimum speeds and feeds

Let Dmax = maximum diameter of work piece

Dmin = minimum diameter of work piece

Vmax = maximum cutting speed.

Vmin = minimum cutting speed.

Nmax ��×+dÍ´)mdrC & Nmin = ��×+drC

)mdÍ´

� Rn = range ratio of spindle speed = QdÍ´QdrC

= +dÍ´+drC × mdÍ´

mdrC

� ÎL = ÎG × Î

Where ÎG = +dÍ´+drC = ratio of cutting speed

And Î = mdÍ´mdrC = ratio of work piece diameter.

� Maximum cutting speed for which machine

Tool is designed depends upon several factors such as—

manufacturing process to be employed, the material of

work piece , cutting tool geometry and tool life and also

surface finish of work piece.


Page 167

S.No. Machine Tool Rn

01 Centre lathe 40 – 60

02 Boring machine 40 – 60

03 Milling machine 30 – 50

04 Drilling machine 15 – 30

05 Shaping and planning 10

06 Grinding machine 1 – 10

For machine tool ratio , range ratio , Rn , where the work

pieces made of alloy of steels to light alloys are machined

could have a high value.

Ex(1) Vmin for cemented carbide on alloy steels

= 100 m/min

Vmax for cemented carbide on light alloy steel

= 2000 m/min

∴ ÎL +dÍ´+drC = ��

�� = 20

Ex(2) Vmin for HSS on alloy steels = 20m/min

Vmax for HSS on light alloy steel = 500m/min


Page 168

∴ ÎL +dÍ´+drC = ��

�� = 25

Basic rules for layout of Gear boxes having sliding

clusters:

The rules applied while designing gears boxes with sliding

clusters are as follows:

<1> Transmission ratio(r) in the gear box is limited by

�� ≤ - ≤ 2

<2> For stable operation the range ratio (Rn) of any stage

should not be greater than 8

ÎL ≤ 8

<3> One set of gear must be completely disengaged

before the other set begins to come into mesh.

<4> The sum of teeth of mating gears in given stage must

be the same for same module in clustered set.

<5> The axial gap between adjacent gear box must be

equal to at least two gear width.


Page 169

<6> The minimum difference between the no. of teeth of

adjacent gears must be 4.

∆Ð ≥ 4

<7> The minimum no. of teeth on smallest gear be

greater than or equal to 17

Ð ≥ 17

<8> Least no. of shafts , gears & Levers should be used.

<9> Gear box should be minimum possible size. Both

radial as well as axial dimensions should be as small as

possible.


Page 170


Page 171

� Speed structure diagram

“This diagram represent the speed at output as well as

intermediate shaft of gear box is developed from

kinematic arrangement of the drive.”

• Shaft are shown in vertical equidistant &

parallel lines.

• Speeds are plotted vertical on a logarithmic

scale with log ∅ as unit.

• Transmission engaged at definite speed of

driving & driven shaft shown on the diagram by rays

connecting the points on the shaft lines representing

these speeds.

• Obviously for transmission ratio of 1 the ray is

horizontal. It is inclined up for transmission ratio greater

than 1 & inclined down for transmission ratio less than 1.


Page 172

� Speed structure diagrams indicates the distributive

connection between input and output points & are

two types--

(a) Wide(Open) diagram: path do not cross each

other

(b) Narrow(Crossed) diagram: Path cross each other.


Page 173

� Selection of best ray diagram of a Gear box:

The design of a Gear box should be optimum wrt. the

size, no. of gears, shafts etc. and the cost. The size(cost)

is controlled and kept minimum by the following three

rules—

(a)Ray restriction rule:

If Zmin = 20 and Zmax = 120, then (∅´)max = 120/20 =6

This means that the horizontal projection of any ray

should be equal to or less than 6.

(b)Stage restriction:

Consider a 3 speed stage which derives three speeds at

the output from one input speed as shown in fig.

It is seen that (i)min =N1/N0=1/4 and (i)max =N3/N0=2

Therefore, the stage range Rn

Rn = (i)max/(i)min = N3/N1=∅´=8


Page 174

Hence ∅´ ≤ 8 where x is intervening space between the

ray diagram between the min. and max. ray.

(c)Shaft optimisation:

Mt∝ �Q

& Ó� × [/�)[Ô / � = VB

§B

Where aO is shear strength and �O is factor of safety and

d is diameter of shaft and Mt is torsional bending applied

on shaft.

∝ &2 ∝ 1/�

� [�[� = Q�

Q� = ∅�/

Shaft diameter also consider a geometrical progression

having progression ratio ∅�/

☺ NOTE: Optimisation of Gear box takes several steps

and thinking about each and every subdivisions in

structure and also a management in space for shafts and

gears. Hence whatever we are going to solve problem for

gear box such as speed 18,12,9 and 6 basically we are


Page 175

here more interested in only design and calculation of

teeth not in optimisation of gear box. Therefore, design

of each student may have their own design of feed gear

box.

Design procedure of Feed gear box:

Step (1)

Find Nmax and Nmin.

Step(2)

Find common ratio ∅ using ∅ �QdÍ´ÓdrC��/�C ��

and standardised as per ISA standard as follows:

R 20 series:

R20/n means ∅ 10C/��

e.g. R 20/4 series ∅ 10�/�� 1.6

∅ 1.12 ∅ 1.25 ∅ 1.4 ∅ 1.6 ∅ 2

R 20 R 20/2 R20/3 R20/4 R20/6


Page 176

Machine Tool ∅

<1> general purpose

machine tools

(a) Large 1.25

(b) Medium 1.4

(c) Small 1.6

<2> heavy duty machine

tool & automates

1.12

Step(3)

Find speeds in between Nmax and Nmin using G.P

Step(4) Draw Ray diagram using an

arrangement(subdivision) keeping in mind stage

restriction.(Here We are not talking about optimised Ray

diagram but you can optimise ray diagram on the basis of

no. of gear required and it should also noted that Rn and

shaft size constraint optimisation of ray diagram. )

Step(5) Calculate teeth(Z) of gears.(Method for

calculation See in problem)

F �2�1 Ð1

Ð1′


Page 177

Min. no teeth should be 17 and max. no. of teeth should

180 for spur gears.

Step(6) Draw kinematic arrangement of feed gear box.

Design of 18 speed gear box:

Problem: specification of feed gear box are as follows:

(1) Max. dia. to be turned = 300mm

(2) Min. dia. to be turned = 20 mm

(3) Vmax = 60m/min

(4) Vmin = 18.8m/min

Solution:

Step (1):

� �JE±. +dÍ´)mdrC ��

)×�.�� 1000-KJ

And �JFL. +drC)mdÍ´ ��.�

)×�. �� 20-KJ Step(2):

∅ = Õ�JE±&JFLÖ

�/(C �)


Page 178

� ∅ �� /��

= 1.2587

� Standard value of progression ratio

∅ 1.25 OH-FHO Î20/2

Step (3):

N1= Nmin, N2= N1× ∅, N3= N2× ∅ and so on.

Hence output speeds are as follows:

20,25,32,40,50,65,80,100,125,160,200, 250, 315,

400, 500, 625, 800, 1000.

Step(4):

For 18 speed drive there are various arrangement

(sub division)such as

Possible subdivision No. of gears required

× × × × Ø (× + × + Ø) × 2 =16

× × Ø × × (× + × + Ø) × 2 =16

Ø × × × × (× + × + Ø) × 2 =16

Ù × × (× + 6) × 2 =18

× × Ù (× + 6) × 2 =18

The sub-divisions 3× 6 , 6× 3 may be rejected on

the basis of no. of gear required than other

combination that is in no. 18.


Page 179

(Here basically we are not concern about

optimisation of ray diagram rather drawing of an

arrangement.)

Take subdivision 2× 3 × 3

Step(5) calculation of no. of teeth.

� For 1st

part drive(shaft I/P- I.S.II)

i1 = ��

�� ×�

��×� ��

∴ Z1 = 28 , Z1’ = 105

Hence for direct drive, Z1= 28, Z1’ = 105

Now , for same module,

(Z1+Z1’)/2 direct = (z1/2+Zi+Z1/2)reverse


Page 180

Where , Z1= no. of teeth on the driver gear

Z1’= no. of teeth on the driven gear

Zi= no.of teeth o the idler gear.

Therefore, for the reverse drive, the following no. of

teeth satisfy the relationship—

Z1 = 20, Zi = 19, and Z2 = 75

� For 2nd

part drive(shaft I.S.II- I.S.III)

i2 = �

�.��Ú 0.328 � ...............1+3=4

i3= �.��Ô� = 2.44 = ��

� ...............12+5=19

LCM of 4 and 19 is k = 76and also

SZmin = E× � = 76 ; taking E= 1

∴ Z2 = 19 , Z2’ = 57

Z3 = 48 , Z3’ = 20

� For 3rd

part drive(shaft I.S.III- I.S.IV)

i4 = �

�.��£ = 0.262 = �� ...............1+4=5

i5= ��.��¥ = �

� ...............1+2 = 3

i6= ��.��¤ = �

� ...............1+1 = 2

LCM of 5,3and 2 is k = 30 and

SZmin= E× � = 120 ; taking E= 4


Page 181

∴ Z4 = �×��

� 24 , Z4’ = �×��

� 96

Z5 = �×��

40 , Z5’ = �×��

80

Z6 = �×��

� 60 , Z6’ = �×��

� 60

� For 4th

part drive(shaft I.S.III- o/p)

i7 = �

�.��Û �� ...............8+5=13

i8= ��.��« = �

� ...............4+5=9

i9= ��.��¤ = �

� .............. 1+1=2

LCM of 2,9 and 13 is k = 234 and


∴ Z9 = �×� �

� = 117 , Z9’ = �×� �

� = 117

Since the value of teeth are too large(although this

value is acceptable) therefore we have to correct

value of i7:

i7= ��.��Û = �

........................2+3=5

and LCM of 2,9 & 5 is K = 90 and also

SZmin = E× � = 90; taking E= 1

∴ Z7 =�×��

� = 36 , Z7’ = ×��

� = 54

Z8 = �×��

� = 40 , Z8’ = �×��

� = 50

Z9 =�×��

� = 45, Z9’ = �×��

� = 45


Page 182

Step (6): kinematic arrangement of 18 speed gear box

with reverse drive:


Page 183


Problem(2): Design of layout of gear box for a milling

machine to provide 12 output speeds ranging from 160

rpm to 2000 rpm. Input speed 1440 rpm. Choose

standard speed ratio & construct the structural diagram

& kinematic arrangement.

Solution:

Given:

Input speed = 1440 rpm

No. of output speeds required (n)= 12

Nmax = 2000 rpm

Nmin = 160 rpm Step (1)

Step(2): ∅ �QdÍ´ÓdrC��/�C ��

� ∅ �� /��

= 1.2581


∅ 1.25 OH-FHO Î20/2


Page 184

Step (3):



160, 200, 250, 315, 400, 500, 625, 800, 1000, 1250,

1600, 2000

Step(4):

For 12 speed drive there are various arrangement

(sub division)such as

Possible subdivision No. of gears required

Ø × Ø × × �2 � 2 � 3� × 2 14

× × Ø × Ø �× � 2 � 2� × 2 14

Ø × × × Ø �2 � × � 2� × 2 14

Ü × × (4 + ×) × 2 =14

× × Ü (3 + 4) × 2 =14

Ø × Ù (2 + 6) × 2 =16

Ù × Ø (6 + 2) × 2 =16

The sub-divisions 2 × 6 , 6× 2 may be rejected on

the basis of no. of gear required than other

combination that is in no. 16

4 × 3 , 3 × 4 are also dropped due to restriction of

min shaft size.


Page 185

(Here basically we are not concern about

optimisation of ray diagram rather drawing of an

arrangement.)

Take subdivision 2× 2 × 3

Step(5): Calculation of no. of teeth:

� For 1st

part drive(shaft I/P- I.S.I)

i1 = ��

�� .4340 ��

∴ Z1 = 22 , Z1’ = 50


Page 186

� For 2nd

part drive(shaft I.S.I- I.S.II)

i2 = �

�.��¥ .512 �� ...............20+39=59

i3= �.��¥� = 1.953 = �

�� ...............20+39=59

LCM of 59 and 59 is k = 59 and also


∴ Z2 = 20 , Z2’ = 39

Z3 = 39 , Z3’ = 20

� For 3rd

part drive(shaft I.S.II- I.S.III)

i4 = �

�.��« = �� ...............4+5=9

i5= �.��Û� = ��

�� ...............25+16=41

LCM of 9 and 41 is k = 369 and


∴ Z4 = �× ��

� = 164 , Z4’ = �× ��

� = 205

Since the value of teeth are too large(value are not

acceptable) therefore we have to correct value of

i5:

i5= �.��Û� = �

� ........................8+5=13

and LCM of 9 & 13 is K = 117 and also


∴ Z4 = �×��

� = 52 , Z4’ = �×��

� = 65


Page 187

Z5 = �×��

� 72 , Z5’ = �×��

� 45

� For 4th

part drive(shaft I.S.III- o/p)

i6= ��.��Û = �

� ...............8+5=13

i7= ��.��« = �

� ...............4+5=9

i8 = �

�.��¤ = �� ...............1+1=2

LCM of 2,9 and 13 is k = 234 and


∴ Z8 = �×� �

� = 117 , Z8’ = �×� �

� = 117

Since the value of teeth are too large(although this

value is acceptable) therefore we have to correct

value of i6:

i6= ��.��Û = �

........................2+3=5

and LCM of 2,9 & 5 is K = 90 and also

SZmin = E× � = 90; taking E= 1

∴ Z6 = �×��

� = 36 , Z6’ = ×��

� = 54

Z7 = �×��

� = 40 , Z7’ = �×��

� = 50

Z8= �×��

� = 45 , Z8’ = �×��

� = 45


Page 188

Step (6) Kinematic arrangement of 12 speed

feed gear box:


Problem(3):Design of layout of gear box for a large

general purpose m/c tool to provide 9 output speeds

ranging from 120 rpm to 720 rpm. Input speed 600 rpm.


Page 189

Choose standard speed ratio & construct the structural

diagram & kinematic arrangement.

Solution: Given:



Nmax = 720 rpm

Nmin = 120 rpm Step (1)


� ∅ ��/��

= 1.251


∅ 1.25 OH-FHO Î20/2

Step (3):



120, 150, 188, 235, 290, 370, 460, 580, 720


Page 190

Step(4): subdivision: 3× 3


� For 1st


i1 = �� .483~ ��

��

∴ Z1 = 24 , Z1’ = 50

� For 2nd


i2 = �

�.��¥ .512 �� ...............1+2=3

i3= �� = 1 = �� ...............1+1=2


Page 191

i4= �.��¥� = 1.953 = �

� ...............2+1=3



∴ Ð2 = �×��

= 18 , Z2’ = �×��

= 36

Z3 = �×��

� = 27 =Z3’

Z4 = �×��

= 36, Z4’ = �×��

= 18

� For 3rd

part drive(shaft I.S.II-o/p)

i5 = �

�.��« = �� ...............4+5=9

i6= �� ............................1+1=2

i7 = �.��

� = �� ................5+4=9



∴ Z5 = �×��

� = 40 , Z5’ = �×��

� = 50

Z6= �×��

� = 45 , Z6’ = �×��

� = 45

Z7 = �×��

� = 50 , Z7’ = �×��

� = 40

Step (6) Kinematic arrangement of 9 speed feed gear

box:


Page 192


Page 193


Problem(4): Design gear box to provide 6 output speeds

ranging from 150 and 800rpm. Where input speed is 600

rpm.

Solution:

Given:



Nmax = 800 rpm

Nmin = 150rpm Step (1)


� ∅ ��/��

= 1.398


∅ 1.4 OH-FHO Î20/3


Page 194

Step (3):



150, 210, 290,410, 570,800

Step(4): subdivision: 2× 3


� For 1st



Page 195

i1 = �� .683~ ��

�

∴ Z1 = 21 , Z1’ = 30

� For 2nd


i2 = �

�.�Û .510 �� ...............1+2=3

i3= �.�� = �

� ...............7+5=12


SZmin. = E× � = 60; taking E= 5

∴ Ð2 = �×��

= 20 , Z2’ = �×��

= 40

Z3 = �×��

�� = 35 , Z3’=�×��

�� = 25

� For 3rd

part drive(shaft I.S.II-o/p)

i4= �

�.� = �� ...............7+5=12

i5= �� ............................1+1=2

i6 = �.�� = �

� ................7+5=12



∴ Z5 = �×��

�� = 25 , Z5’ = �×��

�� = 35


Page 196

Z6= �×��

� 30 , Z6’ = �×��

� 30

Z7 = �×��

�� 35 , Z7’ = �×��

�� 25

Step (6) Kinematic arrangement of 6 speed feed gear

box:


Page 197

Unit (V)

Norton gear box:

The motion from driving shaft(input shaft) is transmitted

to driven shaft(output shaft) tumbler gear. The gears on

input shaft(1), all gears are fixed (keyed)& hence the of

transmission equal to number of gears. To engage the

required transmission b/w driving shaft (1)and driven

shaft (3) ,the tumbler gear is mounted on pin(2)in the

arm is disengaged by swiveling the arm about its axis and

moved along driven shaft(3). When the tumbler gear

comes against the required gear of gear cone, the arm is


Page 198

swiveled back to mesh the tumbler gear and it is then

locked


Page 199

Swivelling of tumbler gear about axis

to engage and disengage with input

shaft gears

Change of position of tumbler

or driven gear by axial

movement to mesh with

required fixed gears arm


Page 200

Transmission ratio (TR) are:

F1= Þ�Þ� . Þ�

ÞU Þ�ÞU F2=

Þ�Þ� . Þ�

ÞU Þ�ÞU F3=

Þ Þ� . Þ�

ÞU Þ ÞU

F4= Þ�Þ� . Þ�

ÞU Þ�ÞU F5 =

Þ�Þ� . Þ�

ÞU Þ�ÞU

Advantages :

� compact design ; 10-12 transmission may be

obtained in one group

� Small dimensions or small no. of gears as (n+2)gears

are required to obtain n feed rate value.

� Simple control as all transmission are engaged by

single lever.

TR i1 i2 i3 i4 i5


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Disadvantages:

� Insufficient rigidity and accuracy of meshing.

� Poor lubrication

� Possibilities of dirt penetrating into the gear box

through slots in the housing.

Feed gear box with Meander’s mechanism:

It consists of identical double cluster gears mounted on

driving shaft(1) and intermediate shaft (2). Only the first

gear block on the driving shaft is rigidly mounted , the

rest of the gear blocks on shafts (1)&(2) are mounted

freely. Transmission to driven shaft (3) takes place

through a tumbler gear Zt which is mounted on a pin in

arm. The arm can be rotate about shaft (3) and can slide

along its axis(spline). The feed rates obtained at output

shaft from a geometrical progression.


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Transmission ratio can be determined as:

Fig (a)


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F1 ß�ß� . ß�

ß� . ß� ßU 5 �ß�

ß��

{let 5 ß�ßU}

F2 ß�ß� . ß�

ß� . ß� ßU ß�

ßU 5 �ß�ß��

�

Fig (b)

Fig (c)


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F3 ß�ß� . ß�

ß� . ß� ßU 5 �ß�

ß��

F4 ß�ß� . ß�

ß� . ß�ßU 5 �ß�

ß��

Fig (d)

Fig(e)


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F5 ß�ß� . ß�

ß� . ß�ß� . ß�

ß� . ß� ßU 5 �ß�

ß��

F6 ß�ß� . ß�

ß� . ß�ß� . ß�

ß� . ß� ßU 5 �ß�

ß��

Fig(f)


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F7 ß�ß� × ß�

ß� × ß�ß� × ß�

ß� × ß�ß� × ß�

ß� × ß�ßU = 5 �ß�

ß��

F8 = ß�ß� × ß�

ß� × ß�ß� × ß�

ß� × ß�ß� × ß�

ß� × ß� ßU = 5 �ß�

ß��

Advantages:

� Compactness in design.

� Simple control as all transmission are engaged or

disengaged by single lever.

� Large transmission can be obtained because they are

in geometrical progression(G.P.).

Disadvantages:

� Insufficient accuracy and rigidity.


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� Poor lubrication.

� Large no. of dimensions are required for n

transmission 2n gears are required.

Feed gear box with gear cone and sliding key:

This design consists of continuously meshing gear pairs.

The gears on driving shaft(1) are all rigidly fixed where as

those on driven shaft (2) are mounted freely. A sliding,

spring loaded key travel in the keyways of driven

shaft(2). The transmission from shaft (1) to shaft(2) can

be achieved through any of gear pairs by shifting key

with the help of pulling rod from one gear to another

gear. Thus at a time only one gear of shaft(2) transmits

torque while other rotates freely.


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Advantages:

� Compact design; enables 8—10 transmission in a

single group.

� Simple control as all transmissions are engaged by

single lever, &

� Helical gear may be used for smooth operation.

Disadvantages:

� Inability to transmit large torque.

� Poor rigidity of driven shaft due to long key ways.

� The key may get stuck due to cocking

� Poor lubrication of gears on shafts as they are thin.


Page 209

Machine Tool Installation:

While installation a machine tool the following procedure may

be adopted:

� Check the machine tool including all accessories for any

damage which might have been caused to any part during

transit.

� Prepare the foundation as per drawing. Foundation should

be checked for

• Level at top surface

• Position of foundation bolts

• Strength

� Lift machine tool and place it on the foundation after

inserting foundation bolts in legs of machine.

� Check level of bed of machine. Levelling is done with the

help of levelling pads, wedges and shoes.

� Tighten nuts on foundation bolts.

Function of foundation:

� Foundation transmits static load(weight) and dynamic

load(vibration and shock) of machine through bed or legs

of machine to soil.


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Types of foundation:

<a> Ordinary machine shop foundation:

Types of anchor bolts:


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 Sand base foundation:

☺In this type of foundation ,machine bed should not be heavy

so much.

<c> Vibration isolation foundation:


Page 212

Preventive Maintenance:

“ The aim of preventive maintenance is to reduce wear &

tear and to prevent interruption in production.”

A proper preventive maintenance schedule should be followed

such as under:

1. Daily checks: The following duties should be performed

by the operator.

a. To clean the machine

b. To check lubricating oil level & oil flow in sight

glasses.

c. To check coolant level.

d. To keep the maintenance department informed of

even the minor defect noted in the performance

of the machine.

2. Weekly checks

The following check should be carried out by the

maintenance department.

a. To check all lubrication level

b. To check coolant

c. To check all filters


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d. To check hydraulic & pneumatic lines.

3. Monthly checks

a. To check spindle drive belts for wear

b. To check hydraulic pump & hydraulic oil

c. To check movement of all axis under manual dial

in control

4. Six monthly check

a. To check machine alignment

b. To replace oils & filters

NOTE:

Actually preventive maintenance schedule varies for each

machine tool and those are decided by maintenance engineer.

But the basic aim is to reduce wear and tear and increase life

cycle of machine and components. Some times we wait for

failure of certain components that is known as Breakdown

maintenance.


Page 214

Reconditioning:

“Reconditioning is Rebuilding of machine after certain life

span of machine to meet same productivity and is

determined by frequent corrective maintenance.”

� A rough estimate of the cost is prepared & generally it is

undesirable to reconditioning the machine tool if the cost of

parts to be replaced & repaired exceeds 50% of prevailing cost

of the new equipment.

� Another very important aspect governing the decision to

estimate the time required to put the machine back in normal

operation considering the availability of reconditioning facilities

& spare required for replacement. This will give the figures of

production loss due to direct or inter related effect on

manufacturing process. Sometimes the replacement proves to

be more economical than to recondition the machine.


Page 215

Replacement:

All wearing parts in the machine or the parts subjected to

fatigue are replaced normally before failure in a preventive

maintenance system.

NOTE:

A schedule is prepared depending upon life of the part.

Reconditioning becomes undesirable when cost of replaced &

repaired of parts exceeds 50% of prevailing cost of new

equipment. Secondly, if time of reconditioning of machine tool

is much more & is affected on production cost very much then

there is direct replacement is advisable instead of

reconditioning. Third important factor for replacement of

machine tool is obsolete. If Obsolesce machine tool does not

compete the present requirement of production in terms of

productivity, cost of production, accuracy in terms of

productivity, cost of production, accuracy required then we

have to replace the machine tool.


Page 216

Testing of Machine Tool

☺The name of Dr.G.Schlesinger has for many years been linked

with subject of machine tool test and alignments. Many year

ago he established certain acceptance tests and tolerances in

Germany for machine tools which were to be supplied to Russia

under contact.

☺When machine tool testing is carried out?

� After manufacturing, repairs or overhaul each machine tool

must be tested as per alignment test chart in order to check

whether it meets the requirement of specification or not.

� Acceptance test

[ Alignment test + performance test] = Acceptance test

i.e. collectively Alignment test and Performance test is known

as Acceptance test.

� Alignment test (also known as Geometrical test):

Geometrical test cover the grade of manufacturing

accuracy of machine tool i.e. the accuracy with which

machine tool has been assembled. They check on the

relationship of various elements of machine tool when it

is idle and unloaded. Such test includes :

� Flatness of base plate and table

� Square-ness of table


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� Parallelism, flatness, straightness of guiding & bearing

surface, movement of various component of machine tool

in various direction

� True running of spindle

� Perpendicularity of spindle sleeve with table

etc.

� Performance test (also known as Practical test):

Practical test are used to check the working accuracy of

machine tool by checking accuracy of finished

components. This test is performed after one hour of trial

run so that spindle get sufficiently warmed up.

� Testing equipments:

The various tools & equipment used for carrying out the

acceptance tests are as follows:

(a)Dial gauge: Dial gauge is widely used in alignment test. Dial

gauge should have clearly readable graduation on sufficient

large scale. Its graduation need not be finer than 0.01 mm &

initial plunger should vary between 40 – 100 gm.

Application in metrology:

(b) Spirit Level: Both horizontal & frame type spirit levels are

used. It should have a sensitivity varying from 0.03 to 0.05 mm

/m for each division. The bearing surface of spirit levels be as


Page 218

long as possible & for medium sized machine tool testing, the

length of spirit level bearing surface should not be less than

200mm.

Application in metrology: straightness of bed,

(c) Straight edges and squares: These are made of Cast Iron or

Steel should be well rolled & seasoned. A square must have a

wider bearing surface. The standard square have a tolerance of

± 0.01 mm & precision square

± 0.005 mm.

Application in metrology: straightness

(d) Test mandrel: test mandrels are made to a length may vary

from 100 to 300 mm. They are accurately turned and ground.

The mandrel must be as light as possible otherwise deflection

may occur due to its weight.

Application in metrology:

(e)Autocollimator: autocollimator in conjunctions with block,

deflector and optical square is very sensitive instrument for

checking deflection of long beds in horizontal , vertical or

inclined planes.

Application in metrology: straightness,


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(f)Waviness meter: waviness meter with 50:1 magnification is

useful in reading & examining the surface waviness.

Basic concept of mechanics

(1) Moment & Axis of Moment:

Moment is shown by arc having a particular direction indicated

by a single arrow whereas the Axis of Moment is shown by a

line segment having a particular direction indicated by double

arrow.

If you have direction of rotation of Moment the you can find

Axis of Moment using right hand thumb rule i.e.


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Curl your fingers of right hand in the

direction of Moment , your thumb will show Axis of Moment.

(2) Axis of Moment and Force:

If you know the force and direction of displacement i.e. the

displacement of force to the point upon which you want to find

moment of force then you can find Axis of Moment by using

right hand rule as shown in figure.

& × M


Page 221

(3) Shifting of force:

Whenever a force which is acting on a body is sifted to

another point within that body then a Moment will be always

come into role along with that Force.


Page 222

(4) Equilibrium Equations:

According to Newton’s III rd Law action force(Active Force) is

always equal and opposite of reaction force(Reactive force).

á �52FGH MI-5H � á ÎHE52FGH MI-5H 0

But it should be noted that if active components are replaced

by another forces in such a way that their resultant force is

remains same then

Equivalent FBD


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á �52FGH MI-5H = á �H� �52FGH MI-5H

Similarly we can write for reactive components also; basic

thing is that resultant of reactive forces must not change.


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