Machining Fundamentals: An Introduction to the Key Concepts and Processes in Machining Operations

Many components produced by primary manufacturing
processes need machining to get their final shape,
accurate size and good surface finish.
The term machining is used to describe various
processes which involve removal of material from the
workpiece
Definition of Machining (or Metal cutting)
“Machining is an essential process of finishing by which
jobs are produced to the desired dimensions and surface
finish by gradually removing the excess material from the
preformed blank in the form of chips with the help of cutting
tool(s) moved past the work surface(s)”

The ever increasing importance of machining operations is
gaining new dimensions in the present industrial age.
Competition towards the economical manufacture of
machined parts.
Basic objectives of the economical and efficient
manufacturing practice are:
1.Quick Metal Removal or MRR(Material Removal Rate)
2.High class surface finish
3.Economy in tool cost
4.Less power consumption
5.Economy in cost of replacement and sharpening of tools
6.Minimum lead time of machine tools
Importance of machining

What is machine tool?
A machine tool is a machine for shaping or machining metal or other rigid
materials, usually by cutting, boring, grinding, shearing, or other forms of
deformation.
Machine tools employ some sort of tool that does the cutting or shaping.

CLASIFICATION OF CUTTING TOOLS

 For providing cutting action a relative motion between
the tool and work-piece is necessary.
 This relative motion can be provided by:
1. Either keeping the workpiece stationary and moving
the tool
Ex. Shaper, Slotter, Broaching machine etc.
2. By keeping the tool stationary and moving the work
Ex. Planer
3. By moving both in relation to one another.
Ex. Grinding and Milling
How cutting takes place?

Diagrammatic Representation of Material Removal Operations

Cutting Parameters
Cutting speed (V) is the largest of the relative velocities of cutting tool
or workpiece. In turning, it is the speed of the workpiece while in
drilling and milling, it is the speed of the cutting tool.
Cutting speed is the distance traveled by the work surface in unit time
with reference to the cutting edge of the tool.
Cutting speed of a cutting tool can be defined as the rate at which its
cutting edge passes over the surface of the workpiece in unit time.
It is normally expressed in terms of surface speed, referred to as
speed(v) and expressed in meters per minute (m/min)
In turning, it is given by the surface speed of the workpiece
V= rω= r * (2 π N)/60= πDN/1000 m/min D in mm, N in rpm
D= Dia. of w/p
N=rpm of spindle
V=linear velocity
ω = angular velocity = rad/sec

Depth of cut : It is the distance through which the cutting tool is
plunged into the workpiece surface.
Thus it is the distance measured perpendicularly between the
machined surface and the unmachined (uncut) surface or the
previously machined surface of the workpiece.
The depth of cut d is expressed in mm.=(d1-d2)/2 for turning
Feed: The feed is the distance advanced by the tool into or along
the workpiece each time the tool point passes a certain position
in its travel over the surface.
In case of turning, feed is the distance that the tool advances in
one revolution of the workpiece.
Feed f is usually expressed in mm/rev.
Feed in mm/min = Feed in mm/rev x N

Material Removal Rate
MRR  vfd
Roughing(R)
f  0.4 1.25mm / rev
d  2.5  20mm
Finishing(F)
f  0.125  0.4mm / rev
d  0.75  2.0mm
vR  vF
MRR= Volume removed/cutting time=mm3/min or mm3/sec
Volume of material removed per unit time(1 min) or volume of material removed
divided by the machining time
Power required = MRR x Specific cutting energy (Watts/mm3/sec)
m/min * mm/rev * mm
MRR= (Initial weight – Final weight)/Machining time

Single point cutting tool
2 parts of the cutting tool: 1.Shank 2.Flank
3 Faces of shank: 1. Rake face 2.Principle flank face 3. Auxiliary flank face
2 cutting edges: 1. Principle cutting edge 2.Auxiliary cutting edge

• Shank: Main body of tool, it is part of tool which is gripped in tool
holder
• Face: Top surface of tool b/w shank and point of tool. Chips flow
along this surface
• Flank: Portion of tool which faces the work. It is surface adjacent
to & below the cutting edge when tool lies in a horizontal position
• Point: Wedge shaped portion where face & flank of tool meet.
• Base: Bearing surface of tool on which it is held in a tool holder.
• Nose radius: Cutting tip, which carries a sharp cutting point.
Nose provided with radius to enable greater strength, increase
tool life & surface life.
Typical Value : 0.4 mm – 1.6 mm

Tool Terminology
Side relief
angle
Side cutting
edge angle
(SCEA)
Clearance or end
relief angle
Back
Rake
(BR),+
Side Rake
(SR), +
End Cutting
edge angle
(ECEA)
Nose
Radius
Turning
Cutting
edge
Facing
Cutting
edge

shank
Single point cutting tool Terminology
(with primary cutting edge)
primary plan Approach angle or
primary cutting edge angle

Single point cutting tool
(with primary cutting edge at top)

Traditional tool replaced by inserts of carbide or other tool
materials of various shapes and sizes

Front view
Geometry of Positive rake single point cutting tool
Face
(Clearance angle)

Side
view
(side clearance angle)

Rake Angles
Positive Rake
Negative Rake
Zero Rake
For carbide tipped tools-Extra hard
Top face slopes upward away from point
Top face slopes downward away from
point

Geometry of Negative rake single point cutting tool

The most significant terms in the geometry of a cutting tool angles are:
–Rake angle
» Back Rake angle (FV)
» Side Rake angle (SV)
–Relief or clearance angle
» End relief angle (FV)
» Side relief angle (SV)
–Cutting edge angle
» End Cutting edge angle (TV)
» Side Cutting edge angle (TV)
-- Nose Radius (TV)
Nomenclature of Single Point Lathe Tool

Rake Angle:
It is the angle formed between the face of the tool and a plane parallel
to its base
If this inclination is towards the shank, it is known as back rake or top
rake, when it is measured towards the side of the tool, it is called side
rake.
These rake angles guide the chips away from the cutting edge,
thereby reducing the chip pressure on the face and increasing the
keenness of the tool so that less power is required for cutting
An increased rake angle(+ve rake) will reduce the strength of the
cutting edge
Hence tools used for cutting hard metals are given smaller rake
angles where as those used for softer metals contain larger rakes.

Such a rake is usually employed on carbide tipped tools when they are
used for machining extra hard surfaces like hardened steel parts and for
taking intermittent cuts
Nagative rake:
If the face of the tool is so ground that it slopes upwards from the point it
is said to contain negative rake
It obviously reduces the keenness of the tool and increases strength of
the cutting edge
A tool with negative rake will have a larger lip angle, resulting in a
stronger tool
Favourable for tipped tools. Normally varies from 5 to 10 degrees.

Lip Angle(Wedge angle or Angle of keenness)
The Angle between the face and the flank of the tool is known as lip angle
Also called as Angle of keenness of the tool
Strength of the cutting edge is directly effected by this angle
Larger the lip angle , stronger will be the cutting edge and vice versa
Clearance angle remains constant in all the cases, lip angle varies
inversely as the rake angle
Hence, when harder metals are to be machined, a stronger tool is
required, the rake angle is reduced and consequently the lip angle is
increased. This simultaneously calls for reduced cutting speeds, which
is a disadvantage. Therefore, lip angle is kept as low as possible without
making the cutting edge so weak that it becomes unsuitable for cutting

31
Rake angle:
– Ground on a tool to provide a smooth flow of the chip over the
tool so as to move it away from the work piece
Cutting-Tool Terms
 Back Rake angle
Ground on the face of the tool
Influences the angle at which chip leaves the nose of the tool
Generally 8 – 100
Side Rake angle
Ground on the tool face away from the cutting edge
Influences the angle at which the chip leaves the work piece
A lathe tool has 140 side rake.

32
Side Rake
• As Large as possible to allow chips to escape
• Amount determined
– Type and grade of cutting tool
– Type of material being cut
– Feed per revolution
• Angle of keenness
– Formed by side rake and side clearance
Cutting-Tool Terms Contd..

33
Back Rake
• Angle formed between top face of tool and top of tool
shank
– Positive
• Top face slopes downward
away from point
– Negative
• Top face slopes upward
away from point
– Neutral
+ve
-ve

Negative Rake Tools
• Typical tool materials which utilize negative rakes are:
• Carbide
• Diamonds
• Ceramics(At high speeds)
• These materials tend to be much more brittle than HSS but
they hold superior hardness at high temperatures.
• The negative rake angles transfer the cutting forces to the tool
which help to provide added support to the cutting edge.

Relief or Clearance angle:
– Ground on the end and side faces of a tool to
prevent it from rubbing on the work piece.
– To enable only the cutting edge to touch the work
piece and cut freely without rubbing against the
surface of the job
Side Relief angle:
• Angle ground directly below the cutting edge on
the flank of the tool
End Relief angle:
• Angle ground from the nose of the tool

Cutting edge angle
• Ground on a tool so that it can be mounted in the
correct position for various machining operations.
Side Cutting edge angle
• Allows flank of the tool to approach the work piece first
• Spreads the material over a greater distance on the cutting
edge, thereby thinning out the chip.
• Approximately 150
End Cutting edge angle
• Allows the cutting tool to machine close to the work piece
during turning operations
• Usually 20 – 300

37
Functions:
• Strengthens finishing point of tool
• Improves surface finish on work
• Should be twice amount of feed per revolution
• Too large – chatter;
• Too small – weakens point
• Values: 0.4 mm to 1.6 mm
Nose Radius:
• Rounded tip on the point of the tool

Tool signature
 It is the system of designating the principal angles of a
single point cutting tool.
 The signature is the sequence of numbers listing the
various angles, in degrees, and the size of the nose
radius.
 The two systems widely used are:
1) ASA System
2) ORS System

ASA
 The system most commonly used is American Standards Association (ASA)
 Also called as Machine Reference System as the three planes for describing
the angles based on configuration and axes of machine tool
 The single point tool is designated as:
Side rake angle
End relief angle
Side relief angle,
End cutting Edge
Angle Φe
Side cutting
Edge angle
Φs
Nose radius
Bake rake angle
 The system most commonly us
(ASA), which is:
Bake rake angle, Side rake angle, E
cutting Edge angle, Side cutting Ed
αb - αs - θe - θs - Ce - Cs - r

Example:
A tool may designated in the following sequence:
8-14-6-6-6-15-1
1. Bake rake angle is 8o
2. Side rake angle is 14o
3. End relief angle is 6o
4. Side relief angle is 6o
5. End cutting Edge angle is 6o
6. Side cutting Edge angle is 15o
7. Nose radius is 1 mm
Tool signature (ASA)

When the cutting tool is forced against the work, the metal layer
which is just ahead of tool is compressed.
If the tool is forced further, a condition will be reached, in which
the stress exceeds ultimate shear strength of the given work
material.
This leads shear along the shear plane and cutting off the chip
from the workpiece.
With further movement of the tool, the new layer is compressed
and the cycle is repeated.
The chip formed in the metal cutting operations, undergoes
plastic deformation, it becomes shorter (chip contraction) and
cross-section increases.
Due to contraction, the length of chip is shorter than the length of
the tool travel, along the surface of the work.

Shear zone
Deformation of metal occurs along shear plane. However, in realistic model
the shear deformation occurs within a shear zone(Primary shear zone).
An other shear occurs due to friction between the chip and tool as the chip
slides along the rake face of the tool. This is referred as secondary shear
zone.
Another shear occurs between work and tool interface, which is called as
tertiary shear zone.
If machined at low cutting speed  Shear zone is thick
If machined at high cutting speed Shear zone is thin

Tool
Workpiece
Chip
Heat Generation Zones
(Dependent on sharpness
of tool)
(Dependent on m)
(Dependent on f)
10%
30%
60%

CHIP FORMATION
Tool will cut or shear off the metal, provided
1.Tool is harder than the work metal.
2.Tool is properly shaped so that its edge can be effective in
cutting the metal.
3.The tool is strong enough to resist the cutting pressures.
4.Movement of the tool relative to the material or vice versa,
so as to make cutting action possible.

TYPES OF CHIPS
The chips produced during machining can be broadly classified
as 3 types:
1.Continuous chips
2.Discontinuos chips or Segmental chips
3.Continuous chips with build-up edge

Continuous chips
Continuous chips are formed when machining ductile materials(low
carbon steel, mild steel, copper, aluminium etc) with a cutting tool of
large rake angle and sharp cutting edge.
Chip flows off the tool face in the form of a ribbon
The other favorable conditions which give rise to this type of chips are
High cutting speed
Small feeds and depth of cut
Low friction
Formation of continuous chips are desirable because a smooth
surface will be obtained. They also help in providing higher tool life and
lower power consumption
Long continuous chips can cause problems of chip disposal
These problem can be solved by providing chip breakers(step or
groove in the tool rake face) which allow the chips to be broken into
small pieces so they can be removed easily

Discontinuous Chips
This type of chip is produced when machining brittle material, such as
cast iron and bronze, with a cutting tool having low rake angle.
The following factors favours the formation of discontinuous chips
1.Low to medium cutting speed
2.Large feeds and depth of cut
3.Absence of cutting fluid
Chips are broken into small segments instead of plastic flow of chip
along tool face.
The discontinuous chips may also result if the material is ductile and the
coefficient of friction between chip and tool is very high.
The most of the heat generated is carried by the chip and hence the tool
is heated to a lower temperature. Thus the tool life is longer

Discontinuous Chips Contd..
Stages of formation of Discontinous chips

Continuous chips with BUE
Continuous chips with BUE are formed when machining ductile
metals with a cutting tool of smaller rake angle at lower cutting speed.
The other conditions which give rise to BUE are:
1.Higher values of feed and depth of cut
2.High friction
3.Poor lubrication
4.High cutting pressure and temperature in shear zone
These BUE eventually swept from the tool and remain attached to the
machined surface.
This causes poor surface finish of work surface.
Presence of build up edge increases power consumption.

In machining ductile metals like steels with long chip-tool contact
length(small rake angle), lot of stress and temperature develops in the
secondary deformation zone at the chip-tool interface.
Under such high stress and temperature in between two clean
surfaces of metals, strong bonding may locally take place due to
adhesion similar to welding.
In ductile materials, with lower cutting speeds
small particles of cut chip adheres,
under the action of pressure and
temperature, to the face of the tool.
Continuous chips with BUE

 Such bonding will be encouraged and accelerated if the chip-tool materials
have mutual affinity or solubility. The weldment starts forming as an embryo at
the most favourable location and thus gradually grows
 With the growth of the BUE, the force, F also gradually increases due to
wedging action of the tool tip along with the BUE formed on it.
 Whenever the force, F exceeds the bonding force of the BUE, the BUE is
broken or sheared off and taken away by the flowing chip.
Then again BUE starts forming and growing. This goes on repeatedly.
HOW BUE DEVELOPED?

Effects of BUE
Effects of BUE formation
Formation of BUE causes several harmful effects, such as:
1. It unfavourably changes the rake angle at the tool tip causing
increase in cutting forces and power consumption
2. Repeated formation and dislodgement of the BUE causes fluctuation
in cutting forces and thus induces vibration which is harmful for the
tool, job and the machine tool.
3. Surface finish gets deteriorated
4. May reduce tool life by accelerating tool-wear at its rake surface by
adhesion

Type of chip
Work
material
Cutting
speed
Feed Rake Angle
Continuous Ductile High Small Large
Continuous
with BUE
Ductile Medium High Small
Discontinuous Brittle Low High Small

Why Chip Breakers ?
When carbide tipped tools are used for machining, because of
higher cutting speeds, due to high temperatures, the resulting chip
will be continuous, blue in colour and take the shape of a coil.
Such a chip, if not broken into parts and removed from the
surroundings of the metal cutting area, is likely to adversely effect
the machining results.
1.It may adversely effect the tool life by spoiling the cutting edge,
creating crater and raising temperature
2.Its presence may lead to a poor surface finish on the workpiece
3.If the chip gets curled around the rotating workpiece and/or cutting tool,
it may be hazardous to the machine operator
4.Very large coils offer a lot of difficulty in their removal

To prevent the adverse effects, chip breakers are used. These will break
the produced chips into small pieces. Chip breakers reduces the radius of
curvature of the chip
1.By control of tool geometry: Grinding proper back rake and side rake
according to the feeds and speeds to be used.
2.By obstruction method: By interposing a metallic obstruction in the path
of the coil.
Chip breakers Contd…

1.Groove type: Grinding a groove on the face of the tool, leaving small land
near the tip
2.Step type: Grinding a step on the face of the tool, adjacent to the cutting
edge
3.Secondary rake: Providing a secondary rake on the tool through
grinding, together with a small step
4.Clamp type: Very common with carbide tipped tools. Chip breaker is a
thin and small plate which is either brazed to or held mechanically on the
tool face
2.Step type1.Groove type

Cutting Models
ORTHOGONAL GEOMETRY OBLIQUE GEOMETRY
Tool
workpiece
Tool
workpiece

Orthogonal and Oblique Cutting
The two basic methods of metal cutting using a single point tool are the
orthogonal (2D) and oblique (3D).
Orthogonal cutting takes place when the cutting edge is straight and
perpendicular to the direction of cutting(90 degree).
Ex: lathe cut-off operation, straight milling etc.
If the cutting edge of the tool is inclined(less than 90 degree) to the line
normal to the cutting direction, the cutting action is known as oblique.
Ex: Turning, Milling
Oblique
Orthogonal

Orthogonal Cutting:
 The cutting edge of the tool remains
normal to the direction of tool feed or
work feed.
 The direction of the chip flow velocity is
normal to the cutting edge of the tool.
 Here only two components of forces are
acting: Cutting Force and Thrust Force. So
the metal cutting may be considered as a
two dimensional cutting.
 Examples are: Parting off operation,
Broaching, Sawing, straight milling
 Shear force acts on smaller area.
Oblique Cutting:
 The cutting edge of the tool remains inclined at an
acute angle to the direction of tool feed or work
feed.
 The direction of the chip flow velocity is at an angle
with the normal to the cutting edge of the tool. The
angle is known as chip flow angle.
 Here three components of forces are acting: Cutting
Force, Radial force and Thrust Force or feed force.
So the metal cutting may be considered as a three
dimensional cutting.
 The cutting edge being oblique, the shear force acts
on a larger area and thus tool life is increased.
 Examples are: lathe turning, drilling etc.,
 Shear force acts on larger area
Feed
Tool
Work
Oblique cutting
Feed
Tool
Work
Orthogonal cutting
Orthogonal and Oblique Cutting

Schematic illustration of a two dimensional cutting process (or) orthogonal cutting
Orthogonal cutting with a well-defined shear plane, also known as the Merchant Model

Orthogonal Cutting Model
(Simple 2D model)
Mechanism: Chips produced by the shearing process along the shear plane
a
t0
f
+
Rake
Angle
Chip
Workpiece
Clearance AngleShear Angle
t c
depth of cut
Chip thickness
Tool
Velocity V
tool

Elements of Metal Cutting
 The outward flow of the metal causes the chip to be thicker after the
separation from the parent metal.
 That is the chip produced is thicker than the depth of cut.

Cutting Ratio(Chip thickness ratio)
Shear angle f may be obtained either from photo-micrographs
(or) assume volume continuity (no chip density change):
Cutting ratio , r = t0
tc
= Lc
L0
i.e. Measure length of chips (easier than thickness)
w
t
L
0
0
0
wc
Lc
ct
Since t0w0L0 = tcwcLc and w0=wc (exp. evidence)
Chip reduction coefficient = 1/r = tc/to
In turning,
Length of uncut chip L0 =πD= π(D1+D2)/2
Where D = Diameter of workpiece
D1=original dia., D2= final dia.
Shaping

tool
Cutting Ratio (or chip thickness ratio)
Contd..
As Sinf =
to
AB
and Cosf-a) =
tc
AB
Chip thickness ratio (r) =
t0
tc
=
sinf
cos(fa)
f
tc
to
fa)
A
B
Chip
Workpiece

Shear Angle f or θ
 )
 )
1 2
1
2
sin , cos
sin sin
cos cos cos sin sin
cos cos sin sin sin
cos cos sin sin
1
sin sin
cos
sin 1
tan
cos
tan
1 sin
t h t h
t h
r
t h
r r
r r
r
r
r
r
  a
 
 a  a  a
 a  a 
 a  a
 
a
a

a

a
  
  
 
 
 
 


How (f - α )?
At A, let the angle between AB and vertical
is x= 90 - f
From triangle APB, angle A = 90 - f + α
So at B, 90-(90 - f + α)= f - α
P
A
B
C
Chip thickness ration r =
2
1
t
t







)-cos(
sin
r
af
f

Shear Plane Length and Angle f
Shear plane length AB =
t0
sinf
Shear plane angle (f) = Tan
-1 rcosa
1-rsina
f
tc
to
fa)
A
B
Chip
tool
Workpiece

Vc
Vf
Analytically,
)90sin(sin))(90sin( afaf 


sfc vvv
afaf cossin)cos(
sfc vvv


)cos(
sinv
v c
f
af
f


rvv cf 







)-cos(
sin
r
af
f
)cos(
cosv
v c
s
af
a


Velocity Relationships Contd..
Vs
Vc
VsVf
α
f
90-α
90
90- fα
Velocity diagram
a b
c
d
Vs= Velocity of shear along shear plane
Vf= Chip flow velocity along the tool face
( frictional velocity)
parallel to face of the tool
Vc=Cutting velocity

Shaping-2D
Turning-Lathe-3D
Fc = Tangential to the rotation of shaft(cutting speed)
Ft = Thrust force or feed force
=opposite to the direction of feed
Fr = Radial force
= will tend to push the tool away from the work,
which may cause chatter
Fc= In the direction of cutting velocity
Cutting Forces
Fc & Ft

Fc=Cutting force, acting in vertical plane and is tangential to the work surface, Also
called tangential force or tangential feed force (67%)
Ft=Ff = Feed force or thrust force or axial feed force, acting in horizontal plane parallel to
the work axis (opposite to the direction of feed) (27%)
Fr= Radial force or radial feed force, also acting in the horizontal plane but along the
radius of the work piece i.e along the axis of the tool, which will push away from the
work(6%)
Fc
Ft=Ff
Turning

Mechanics of chip formation:
Forces acting on chip(2D- orthogonal)
• Friction force F and Normal force to friction N
• Shear force Fs and Normal force to shear Fn
Equilibrium of Chip
It is assumed that the resultant
forces R & R’ are equal and
opposite in magnitude and
direction and are collinear.
For the purpose of analysis, the
chip is regarded as an
independent body held in
mechanical equilibrium by the
action of two equal and opposite
forces
R - the tool exerts upon the chip.
R’ - the work piece exerts upon
the chip
Fs
R’
Fn

Cutting Forces
(2D Orthogonal Cutting)-Shaping
Free Body Diagram
we know:
Tool geometry & type of
Workpiece material
and we wish to know:
F c = Cutting Force
F t = Thrust Force
F = Friction Force
N = Normal Force
Fs = Shear Force
F n = Force Normal to Shear
Tool
Workpiece
Chip
R
R
R’
R’
Fc
Ft
f Fs
Fn
N
F
Angle between R and N is β

Resultant Forces
• Vector addition of F and N = resultant R
• Vector addition of Fs and Fn = resultant R'
• Forces acting on the chip must be in balance:
– R' must be equal in magnitude to R
– R’ must be opposite in direction to R
– R’ must be collinear with R
R’ = - R
Coefficient of Friction
 Coefficient of friction between tool and chip: N
F
m
Friction angle related to coefficient of friction as follows:
m tan

 F, N, Fs, and Fn cannot be directly measured
 Forces acting on the tool that can be measured:
Cutting force Fc and Thrust force Ft
Forces acting on the tool that can be measured
Cutting Force and Thrust Force

Fs = Shear Force, which acts along the shear plane, is the
resistance to shear of the metal in forming the chip.
Fn = Force acting normal to the shear plane, is the backing up
force on the chip provided by the workpiece.
F = Frictional resistance of the tool acting against the motion of
the chip as it moves upward along the tool.
N = Normal to the chip force, is provided by the tool.
NS
/
FFR
FNR




The forces acting on the chip in orthogonal cutting

Merchant’s Circle Diagram
Merchant’s circle diagram is convenient to determine the relation
between the various forces and angles.
In the diagram two force triangles have been combined and R and
R’ together have been replaced by R.
The force R can be resolved into two components Fc and Ft.
Fc and Ft can be determined by force dynamometers.

Force circle to determine various forces acting in the cutting zone.
Fn

The rake angle (α) can be measured from the tool, and forces F
and N can then be determined.
The shear angle (f) can be obtained from it’s relation with chip
reduction coefficient. Now Fs & Fn can also be determined.
∅
Work
Tool
Chip
Clearance Angle
Ft
Fc
F
N
Fn
Fs
α
α
β
(β - α)
R

Clearance Angle
Procedure to construct a merchants circle diagram
Work
Tool
Chip
Ft
Fc
F
N
Fn
Fs
α
α
β
∅
R

Procedure to construct a merchants circle diagram
• Set up x-y axis labeled with forces, and the origin in the
centre of the page. The cutting force (Fc) is drawn
horizontally, and the tangential force (Ft) is drawn
vertically. (Draw in the resultant (R) of Fc and Ft.
• Locate the centre of R, and draw a circle that encloses
vector R. If done correctly, the heads and tails of all 3
vectors will lie on this circle.
• Draw in the cutting tool in the upper right hand quadrant,
taking care to draw the correct rake angle (α) from the
vertical axis.
• Extend the line that is the cutting face of the tool (at the
same rake angle) through the circle. This now gives the
friction vector (F).
• A line can now be drawn from the head of the friction
vector, to the head of the resultant vector (R). This gives
the normal vector (N). Also add a friction angle (β)
between vectors R and N. Therefore, mathematically, R =
Fc + Ft = F + N.
• Draw a feed thickness line(t1) parallel to the horizontal
axis. Next draw a chip thickness line(t2) parallel to the tool
cutting face.
• Draw a vector from the origin (tool point) towards the
intersection of the two chip lines, stopping at the circle.
The result will be a shear force vector (Fs). Also measure
the shear force angle between Fs and Fc.
• Finally add the shear force normal (Fn) from the head of
Fs to the head of R.
• Use a scale and protractor to measure off all distances
(forces) and angles.
CHIP
WORK
TOOL
α
α
f
β
F
N
FC
Ft
Fn
Fs
RFn
Known: t1 or t2, α , Fc, Ft

Merchant’s Circle Diagram
Work
Tool
Chip
Clearance
Angle
Ft
Fc
F
N
Fn
Fs
α
α
β
∅
(β - α)
R

Shear Force System –Fs & Fn
Relationship of various forces acting on the chip with the horizontal and
vertical cutting forces from Merchant circle diagram
∅
Work
Tool
Chip
Clearance
Angle
Ft
Fc
F
N
Fn
Fs
α
α
β
(β - α)
R
E
O
G

From Triangle OEG, angle E= β ( Angle between R & N)
Angle at G= 900 , So angle at O = 90 – β (Angle between R & F)
From Triangle ECO,
Angle at O = 90 – (90- β) – α = β – α
Angle at E = 90 – (β – α) = 90 - β+ α
From Triangle EAO,
Angle at E = 90 - (φ + β –α )
So angle CED = (90- β+ α) – (90-(φ + β –α)) = φ
ff
ff
cossin
sincos
tCN
N
tCS
S
FFF
DEBCDEADAEF
FFF
CDOBABOBOAF




)tan( af  SN FF
Also:
Shear Force System –Fs & Fn

Draw a perpendicular from C to OG, meeting at K
Draw a perpendicular from E to CK, meeting at L
Consider triangle COK, angle at O =(90-β) +(β-α)
= 90- α
Angle between R & N is β
So, in triangle GOE, angle O = 90- β
We already know in triangle ECO, angle O= β - α
There fore, In triangle KCO, angle C = 90-(90- α) = α
In triangle CEL, angle at C= 90 - α (As angle between Fc and Ft is 90)
So, angle at E = 90 -(90 - α) = α
F= OK + KG = OK + EL = Fc (Sinα) + Ft (Cos α)
N = CK - CL = Fc(Cos α) – Ft (Sin α)
µ = N
F
= Tanβ =
a
a
FtTanFc
FtFcTan


From Triangle ECO, Tan(β-α) = Fc
Ft
Frictional Force System: F & N
angleFrictionWhere
N
F
tan
frictionoftcoefficienThe



m

Relationship of various forces acting on the chip with the horizontal and
vertical cutting force, from Merchant circle diagram
∅
Work
Tool
Chip
Clearance
Angle
Ft
Fc
F
N
Fn
Fs
α
α
β
(β - α)
R
c
t
F
F
Tan  )( a
c
t
s
S
tCN
tCS
tC
tC
F
F
Tan
RFAlso
F
FFF
FFF
FFN
FFF







)(
)cos(,
)tan(
cossin
sincos
sincos
cossin
a
af
af
ff
ff
aa
aa
)cos( af  RFs

Shear Area (As) = OA × AC
= (to / SinØ ) × w
= w to / SinØ
w
C
Where w= width of chip
SinØ = to/OA
More the shear angle, more the denominator, less the shear area, more the shear stress

Stresses
On the Shear plane:
On the tool rake face:
 = Normal Force / Area =
N
tc w
(often assume tc = contact length)
 = Shear Force / Area =
F
tc w
Note: s = y = yield strength of the material in shear
Normal Stress = s = Normal Force / Area =
FnSinf
t o w
Shear Stress = s = Shear Force / Area = Fs Sinf
t0 w

0
0
0
sin)cos()sec(
sin
)cos()sec(
sin
,
)cos()sec(
)sec(
)cos(
wt
F
wt
F
wt
Awhere
A
F
FF
FR
RF
c
s
c
s
s
s
s
s
cs
c
s
fafa

f
afa

f

afa
a
af









It is assumed that f adjusts itself to give minimum work.
For a given set of cutting condition, to, w and α are all constants.
It is also assumed that β is independent of f.
)( a 
Theory of Ernst and Merchant (1944)-Shear Stress
∅
Work
Tool
Chip
Clearance
Angle
Ft
Fc
F
N
Fn
Fs
α
α
β
(β - α)
R
We can express Fs and R as

Earnst-Merchant
(Principle of Minimum
Energy Consumption)
Lee & Shaffer
(slip line field theory)
C af2
Where C=Machining constant
(Over estimate)
- 2nd solution
As R= Fs / Cos(φ +β –α) we know Fs=Ƭ x As and Substituting in Fc

s
n
s
s
A
F
Mean
A
F
Mean


s
s
stressnormal
stressshear


faf
a

sin)cos(
cos

Shear Strain
where  = shear strain,
φ = shear angle, and
a = rake angle of cutting tool
= fSinV
V
c
s

The Power consumed or work done per sec in cutting:
CCC vFP 
Power required in Metal cutting
If Fc is in Kg, Vc is in m/min
Power = (Fc x Vc)/4500 HP
= (Fc x Vc)/(4500x1.36) KW
Fc= Cutting force in N
Vc=Cutting speed in m/sec
Watts
Power = (Fc x Vc)/1000 KW

Specific cutting Energy
To get better picture of the efficiency of the metal-cutting operation, it
is necessary to have a new parameter which does not depend upon
the cutting process parameters.
The specific cutting energy is such a parameter which can be
obtained by dividing the total work done with the material removal
rate(MRR)
The material removal rate(MRR) = Vc w t0
Specific cutting Energy, ut, is defined as the total energy per unit
volume of material removed, or ratio of work done to MRR
00 wt
F
vwt
vF
u C
c
cC
t 

Approximate specific-energy requirements in cutting operations.
MATERIAL SPECIFIC ENERGY*
W-s/mm3
hp-min/in3
Aluminum alloys
Cast irons
Copper alloys
High-temperature alloys
Magnesium alloys
Nickel alloys
Refractory alloys
Stainless steels
Steels
Titanium alloys
0.4-1.1
1.6-5.5
1.4-3.3
3.3-8.5
0.4-0.6
4.9-6.8
3.8-9.6
3.0-5.2
2.7-9.3
3.0-4.1
0.15-0.4
0.6-2.0
0.5-1.2
1.2-3.1
0.15-0.2
1.8-2.5
1.1-3.5
1.1-1.9
1.0-3.4
1.1-1.5
* At drive motor, corrected for 80% efficiency; multiply the energy
by 1.25 for dull tools.
Source “ Manufacturing Processes for Engineering Materials”, 4th edition,
Kalpakjian, Schmid, Prentice Hall 2003

tc
= Lc
L0
Chip thickness ratio (r) =
t0
tc
=
sinf
cos(fa)
-1 rcosa
1-rsina
)cos(
sinv
v c
f
af
f


)cos(
cosv
v c
s
af
a


rvv cf 
Vs= Velocity of shear along shear plane
Vf= Chip flow velocity along the tool face
( frictional velocity)
parallel to face of the tool
Vc=Cutting velocity

N
F
m
m tan
c
t
s
S
tCN
tCS
tC
tC
F
F
Tan
RFAlso
F
FFF
FFF
FFN
FFF







)(
)cos(,
)tan(
cossin
sincos
sincos
cossin
a
af
af
ff
ff
aa
aa
CCC vFP  The material removal rate(MRR) = Vc (w t0)
If Fc is in Kg, Vc is in m/min
Power = (Fc x Vc)/4500 HP
= (Fc x Vc)/(4500x1.36) KW
Fc= Cutting force in N
Vc=Cutting speed in m/sec
Power(Pc) = (Fc x Vc)/1000 KW

Problem1.
In orthogonal turning of a 50mm dia. mild steel bar on a lathe, the following data
were obtained:
Rake angle= 15 deg, cutting speed=100 m/min, feed= 0.2 mm/rev
Cutting force=180 kg, feed force=60kg.
Calculate the shear plane angle, coefficient of friction, cutting power, chip flow
velocity and shear force, if the chip thickness =0.3 mm
r = t1/t2 = 0.2/0.3 =0.667
Shear plane angle = 370 55’
Coefficient of friction = 0.66
Cutting power = (cutting force in Kg x cutting speed in m/min)/4500 HP = 4 HP
Chip flow velocity = cutting velocity x r = 66.7 m/min
Shear force =105.2 Kg
-1 rcosa
1-rsina
In turning, feed is the distance that the tool advances in one revolution of the workpiece,
So, assume feed=t0
tc
= Lc
L0

Problem2. A bar of 75mm diameter is reduced to 73 mm by a cutting tool while cutting
orthogonally. If the mean length of the chip is 73.9 mm, find the cutting ratio. If the
rake angle is 15 degrees, what is the shear angle?
Length of uncut chip L1 = π(d1+d2)/2 = π(75+73)/2 = 232.4779 mm
Cutting ratio r = t1/t2 = Lc/Lo= 73.9/232.4779 = 0.3179
Shear angle φ = Tan-1
a
a
sin1
cos
r
r

= 190
tc
= Lc
L0

Problem 3:
In orthogonal cutting test with a tool of rake angle 10 degrees, the following
observations were made:
Chip thickness ratio= 0.3
Horizontal component of the cutting force=1290 N
Vertical component of the cutting force = 1650 N
From the Merchants theory, calculate the various components of the cutting
forces and the coefficient of friction at the chip-tool interface
r= 0.3 a =10
a
a
f
sin1
cos
r
r
Tan


φ = 17.310
Fc=1290 N Ft=1650 N
F=1848.94
N=983.88
µ = F/N =1.8792
β = 620
R=
22
12901650 
Fs= 740.63 N
Fn=1959.1N
c
t
s
S
tCN
tCS
tC
tC
F
F
Tan
RFAlso
F
FFF
FFF
FFN
FFF







)(
)cos(,
)tan(
cossin
sincos
sincos
cossin
a
af
af
ff
ff
aa
aa

Problem 4: In orthogonal turning of a mild steel bar of 60mm diameter on a lathe a
feed of 0.8 mm was used. A continuous chip of 1.4 mm thickness was removed at a
rotational speed of 80 rpm of work. Calculate the chip thickness ratio and chip
reduction ratio and total length of the chip removed in one minute
t1= uncut chip thickness
=feed rate in mm/rev = 0.8 mm
t2 = cut chip thickness = 1.4 mm
K= 1/r =1.75
r =t1/t2=0.57 -> r is always less than 1
Also r = L2/L1
L2 = r x L1 =0.57 x πD -- (length in one rpm)
With 80 rpm, L2 = 0.57 x π x 60 x 80 =8595 mm

Problem5: Mild steel is being machined with a cutting tool of rake angle100.The width
of the cut is 3mm and uncut thickness is 0.2 mm. The coefficient of friction between
the tool and chip is 0.5 and shear stress of the work material is 400 N/mm2.
Determine a)Shear angle b)shear force c) cutting and thrust force
Sol: From Merchant’s relationship
Shear angle ᶲ = 35.25
Shear force Fs =456.41 N
)cos( af  RFs
R=699.5 N
Fc= 666.94 N
Ft= 210.9 N

Problem 6. A seemless tube 32 mm outer diameter is turned on the lathe. The cutting
velocity of tool relative to workpiece is 10m/min. Rake angle 35 degrees, depth of cut
0.125 mm, length of chip 60 mm. Horizontal cutting force of the tool on workpiece is
200 N. Vertical cutting force required to hold the tool against work is 80 N. Calculate
1. Coefficient of friction
2. Chip thickness ratio
3. Shear plane angle
4. Velocity of chip relative to tool
5. Velocity of chip relative to workpiece
Vc=10 m/min a =350 L2 = 60mm
Assuming no expansion along width, r =t1/t2 = L2/L1 = 60/ πD = 60/ π x 32 = 0.597
Coefficient of friction =µ = a
a
tan
tan
FtFc
FcFt


= 1.5282
r= 0.597 Shear angle ᶲ =36.64
Vf = r x Vc =5.97 m/minVs = )cos(
cos
af
a

Vc
= 8.195 m/min
a
a
f
sin1
cos
r
r
Tan



Due to plastic deformation, heat is generated
Tool failure is observed by:
1. Extremely poor surface finish on w/p
2. Higher consumption of power
3. Work dimensions not being produced as specified
4. Over heating of cutting tool
5. Appearance of a burnishing band on the work surface
Cutting tool may fail due to one or more of the following reasons
1.Thermal cracking and softening
2.Mechanical chipping
3.Gradual wear i) Crater wear ii)Flank wear
1.Thermal cracking and softening: Due to lot of heat generated during metal
cutting, the tool tip and the area closer to the cutting edge becomes very hot.
Every tool material has a certain limit to which it can withstand the elevated
temperature without losing its hardness.
If that limit is crossed, the tool material starts deforming plastically at the tip
and adjacent to the cutting edge under the action of the cutting pressure and
the high temperature. Thus, the tool loses its cutting ability and is said to have
failed due to softening.
Tool Wear & Tool Life

Main factors: High cutting speed, High feed rate, excessive depth of cut.
Smaller nose radius and choice of a wrong tool material
Tool Material Temp. range(oC)
Carbon tool steels 200-250
High Speed Steels(HSS) 560-600
Cemented Carbides 800-1000
Because of fluctuations in temperatures, tool material is subjected to local
expansion and contraction. This gives rise to the setting up of temperature stresses
or thermal stresses due to which cracks are developed in the material. These are
known as thermal cracks which proceed from the cutting edge and extend inwards,
as shown in fig.
1.Thermal cracking and softening Contd..

2.Mechanical chipping:
Mechanical chipping of the nose and/or the cutting edge of the tool are
commonly observed causes of tool failure.
Reasons:1.Too high cutting pressure 2.Mechanical Impact 3.Excessive wear
4.Too high vibrations and chatter 5.weak tip and cutting edge
More pronounced in carbide tipped and
diamond tools due to high brittleness of toll
material
3.Gradual wear
After some time of use, tool loses some weight or mass, implying that it has lost
some material from it which is due to wear.
i) Crater wear: Occurs at tool face, at a small distance ‘a’ from its cutting
edge. Occurs in ductile materials like steel and steel alloys, in which
continuous chip is produced.
 Forms a crater or depression at the tool chip interface.

Due to pressure of the hot chip sliding up the face of the tool. The metal from
the tool face is supposed to be transferred to the sliding chip by means of the
diffusion process.
Size: breadth ‘b’ and depth ‘d’
Continued growth of crater will result in weak cutting edge and finally lead to
tool failure
Crater wear Contd..

ii) Flank wear: At flank below the cutting edge.
Due to abrasion between the tool flank and w/p and excessive heat generated as
a result of the same.
More pronounced while machining brittle material, because the cut chips of such
materials provide a lot of abrasive material readily.
The entire area subjected to flank wear is known as wear land(WL). This type of
wear mainly occurs on the tool nose and front side relief faces as shown.
Magnitude of this wear depends upon the relative hardness of the w/p and tool
materials at the time of cutting operation
Effects:
w/p loses its dimensional accuracy
Energy consumption increases
Poor surface finish

Tool life
Tool life represents the useful life of the tool, expressed generally
in time units from the start of a cut to some end point defined by
failure criterion.
A common method of forecasting tool wear is to use Taylor’s
equation; his study on tool life was done in 1907.
Frederick W. Taylor
1856-1915
Taylor thought that there is an optimum cutting speed for
best productivity. This is reasoned from the fact that at low
cutting speeds, tools have higher life but productivity is low,
and at higher speeds the reverse is true.
This inspired him to check up the relationship of tool life
and cutting speed. Based on the experimental work he
proposed the formula for tool life.
Def: The time interval for which the tool works satisfactorily between two
successive grindings(sharpening)

Three common ways of expressing tool life
1. Time period in minutes between two successive grindings
2. In terms of no. of components machined(C1) between two successive
grindings. This method is commonly used when the tool operates
continuously, as in case of automatic machines.
3. In terms of volume of material removed between two successive grindings.
This mode of expression is commonly used when tool is primarily used for
heavy stock removal.
Volume of material removed per unit time is a practical one and can be easily
applied as
Vol. of metal removed per minute = π D N (d f) mm3/min
If ‘T’ be the time in minutes to tool life
total vol. of metal removed to tool failure = π D N (d f) (T) mm3
Where D= dia. of w/p in mm
d= depth of cut in mm f= feed rate in mm
N= No. of revolutions of work per minute(rpm)
VOLUME per unit time = Velocity x area
NfaC1
b = C

Factors affecting tool life:
1.Cutting speed
2.Feed & Depth of cut
3.Tool geometry
4.Tool material
5.Work material
6.Natute of cutting
7.Regidity of machine tool and work
8.Use of cutting fluids
Effect of cutting speed
-Maximum effect on tool life is of cutting speed
-Tool life varies inversely as the cutting speed, i.e higher the cutting speed,
smaller the tool life
Taylor’s Empirical Equation:VT n =C
Where, T = tool lifetime; usually in minutes
V = cutting velocity, m/min
C = constant; (numerically equal to cutting speed that gives the tool life of one minute)
n = constant; which depends on finish, tool & work material, called tool life index, <1
V1T1
n = V2 T2
n

n = An exponent, whose value depends on the tool, called tool life index or
exponent. Upto a certain extent, its value is also influenced by some other variables
like tool material, cutting conditions etc.
C= A constant, called machining constant, or Taylors Constant
Which is numerically equal to the cutting speed in meters per minute that would give
a tool life of one minute
Each combination of workpiece, tool material and cutting condition has its own n
and C values, both of which are determined experimentally

Sl No. Work materials Values of ‘C’ for different Tool Materials
HSS Carbide Ceramic
1 Carbon steel 40 – 100 200 – 160 2500
2 Cast Iron 30 – 60 100 – 150 9000
3 Stainless Steel 20 – 35 120 – 200
4 Titanium 10 – 20 100 – 150
5 Tungsten 120 - 160 400 - 600
Tool life (C and n values)
For HSS n= 0.08-0.2
For cemented
carbides n= 0.2 - 0.6
For ceramics
n= 0.5 to 0.8

Two curves having same
slope(n).
 B makes higher intercept on
the axis.
‘C’ for B is higher than for A

By increasing the feed and depth of cut, tool life will decrease, because it
increases the cutting forces.
 A General relationship between V, T, d & f is given by the empirical formula:
Where
V= cutting speed in m/min
T= Tool life in min
f = feed rate in mm/min
d = depth of cut in mm
For a given tool life(T), the relationship among other variables is given by the
empirical formula
V= cutting sped for given tool life
C=Constant
Tool life equation does not take the effective parameters such as feed(f) and depth
of cut(d).
The modified Taylor’s tool life equation is V Tn fx dy=C or
 For steel, x= 0.15 to 0.2 and y= 0.2 to 0.4
ba
df
C
V 
min/
257
8.036.019.0
m
dfT
V 
yxn
dfT
C
V 

V1=23.5 m/min T1=10 min
V2=19.5 m/min T2=52.5 min
V3=22 m/min T3=?
VTn =C
V1T1
n = V2 T2
n
n=
2
1
log
1
2
log
T
T
V
V
=0.11 C= 30.27
V3T3
n =C
Therefore T3 = 18.2 min
Problem1: During turning of a 25 mm diameter steel bar on a lathe at 23.5 m/min with
an HSS tool, a tool life of 10 min was observed. When the same bar was turned at
19.5 m/min, the tool life increased to 52.5 min. What will be the tool life at a speed of
22m/min

2. Following data is available on cutting speed and tool life
V=150 m/min T=60 min
V=200 m/min T=23 min
Determine the Taylors constant and tool life exponent
VTn =C
150(60)n =C
200(23)n=C
150(60)n = 200(23)n
(60/23)n = 200/150 = (2.6087)n
n log(2.6087) = log 1.33
n= 0.3
C=512.31

3. Spindle speed(N) Feed(f) Components
250 0.1 311
250 0.125 249
300 0.125 144
This is the record of number of identical components produced at different
speeds and feeds. Estimate the no. of components that can be produced at a
spindle speed of 350 rpm and feed 0.15 mm/rev
NfaC1
b = C
250(0.1)a(311)b = C
log 250 + a log 0.1 + b log 311 = log C
a= 0.3318
b = 0.333
C = 787.46
350(0.15)0.3318 C1
0.333 = 787.46
C1 = 75.59
So, 75 components will be produced

Problem 4. A carbide tool with MS workpiece was found to give tool life of 2 hrs,
while cutting at 0.5 m/min. Compute the tool life if the same tool is used at speed of
25% higher than previous one. Also determine the value of cutting speed if the tool
is required to have tool life of 3 hrs. Assume Taylor’s exponent ‘n’ to be 0.27
C=1.82116
Tool life with 25% higher speed = 52.51 min
If tool life is 3 hrs, cutting speed = 0.44815 m/min

5. The following data were recorded while turning a workpiece on a lathe: Cutting
speed=25 m/min, feed =0.3 mm/rev, depth of cut=2.0 mm, tool life=100 min. The
following tool life equation is given for this operation: VT0.12 f0.7 d 0.3 =C
If the cutting speed, feed and depth of cut are all increased by 25% each, and also
collectively, what will be their effect on the tool life and draw conclusions.
VT0.12 f0.7 d 0.3 =C
C= 25 x 1000.12 x 0.30.7 x 20.3 = 23
 If V increased by 25%, V= 25x1.25=31.25 m/min
T0.12 = 3.07.0
dVf
C
T= 15.27 min
 If feed increases by 25 %, new feed=0.3 x 1.25 = 0.375 mm/rev, Then T=27.13 min
 If depth of cut increases by 25%, new d=2x1.25=2.5 mm, then T=56.29 min
 With cumulative effect of all three increments, T=2.42 min

Cutting Fluids
Cutting fluids are used for decreasing power requirement and increasing heat
dissipation.
Cutting fluids generally used are
1.Neat oils(mineral) + Extreme pressure additives(EP additives)
2.Water based fluids (soluble oils or emulsions(immiscible))
Oil to water ratio between 1:5 to 1:50 for machining and
upto 1:80 for grinding
High forces (345 MPa) & Temperatures (900 0C) can be generated at work tool interface
Chlorinated paraffins, phosphorous compounds, sulphurized additives, polymeric
esters are typical Extreme Pressure additives, which are added into neat oils
New materials, changes in equipment design has led to faster machining speeds and
higher cutting temperatures to help improve productivity
Extreme pressure (EP) additives prevent seizure conditions caused by direct metal-
to-metal contact between the parts under high loads.
Base oils:mineral(neat), synthetic, vegetable and animal

Short and medium chain chlorinated paraffins are considered carcinogenic, toxic to
marine life and will not be authorized to be used in products that will affect health and
environment
Use of chlorinated paraffins will cause detrimental effect on the environment
because, use and disposal of chlorinated cutting fluids have low solubility in water and
environment persistence.

Properties of Cutting fluids
1. High thermal conductivity
2. Low viscosity
3. It should not react with machining components
4. Easily available
5. It should not fume
6. It should not foam
7. It should not give bad odour
For many years metal removing industry has used neat oils that contain chlorine.
Ongoing changes in machine tool design and global health and environmental related
concerns are limiting their use and leading to new opportunities for non-chlorinated
products.
Neat cutting oils are fluids usually based on mineral oils and used for cutting without
further dilution i.e. as supplied by the manufacturer. They are generally blends of
mineral oils and other additives.

2.Straight oils: are used in machining operations in an undiluted form. They are
composed of a base mineral or petroleum oil and often contains lubricants such
as fats, vegetable oils and esters as well as extreme pressure additives such as
Chlorine, Sulphur and Phosphorus.
Straight oils provide the best lubrication and the poorest cooling characteristics
among cutting fluids
1.Water
2.Straight oils (neat oil + EP)
3.Soluble oils & Emulsions (water based cutting fluids, Mineral oil + Emulsifier)
4.Synthetic fluids (no petroleum or mineral oils)
5.Semisynthetic fluids(Synthetic & soluble oils)
 Suitable only for low cutting speed. These are of three types:
1.Mineral oil: Kerosene, low viscosity petroleum fraction.
2.Fatty oil: Lard oil (Animal oil like pig)
3.Combination of mineral and fatty oil
1.Water: Water provided good cooling effect but is not a good lubricant.
Water is hardly used as cutting fluid because of its corrosiveness.
Addition of sulphur or chlorine compounds(with EP) reduces chances of chip welding
on the tool rake face. Also to improve corrosion protection.

3.Soluble Oils or water miscible fluids : These fluids form an emulsion when
mixed with water. The concentrate consists of a base mineral oil and emulsifiers
which help to produce a stable emulsion.
The emulsifier breaks the oil into minute particles and disperses them
throughout water. These cutting fluids have excellent lubricating properties. It has
milky appearance.
They are used in a diluted form (usual concentration = 3 to 10%) and provide
good lubrication and heat transfer performance.
They are widely used in industry and are the least expensive among all cutting
fluids.
These are also called water based cutting fluids.
These comprises of mineral oil or fat mixtures and emulsifiers added to water.

4.Synthetic Fluids contain no petroleum or mineral oil base and instead are
formulated from inorganic and organic compounds along with additives for
corrosion inhibition.
They are generally used in a diluted form (usual concentration = 3 to 10%).
Synthetic fluids often provide the best cooling performance among all cutting
fluids.
Synthetic lubricants can be manufactured using
chemically modified petroleum components rather than
whole crude oil, but can also be synthesized from other
raw materials.
it provides superior mechanical and chemical properties
to those found in traditional mineral oils

5.Semi-synthetic fluids are essentially combination of synthetic and soluble
oil fluids and have characteristics common to both types.
The cost and heat transfer performance of semi-synthetic fluids lie between
those of synthetic and soluble oil fluids.

HSS: Need to be cooled as they lose their hardness at high temp (above 6000C)
Carbide tipped tools:Do not require any cutting fluid as they retain hardness at high
temp.
Cutting tool material
Workpiece material
Hard and brittle materials such as cast iron(1.8-4.5% carbon) and brass are
usually machined dry.
Soft materials such as mild steel & wrought iron need cutting fluids with good
cooling capacity and antiwelding properties
Material Cutting fluids
Aluminium Kerosene, soluble oil
Brass Dry machining(paraffin oil may be used)
Cast Iron Dry machining(compressed air may be used to blow away the chips)
Malleable CI Dry machining or soluble oil
Wrought Iron Soluble oils, sulphurised oil, Lard oil
Steel Soluble oils, Sulphurised oil, Mineral oil

Cast Iron: Cutting fluids are not used because, it produces graphite flakes, this
mixes with the cutting fluid and affects the machining area. So either no cutting
fluid is used or compressed air is used
Steel: For low cutting speeds, neat oil and EP additives are used. For medium
cutting speeds, water emulsions in the ratio of 1:10. For high cutting speeds water
emulsions in the ratio 1:100
Aluminium: Very soft material. At high speed nothing is required. But at low
speed, there will be a tendency of build up edge formation, so neat oils(Kerosene)
with EP additives are used
Magnesium: Reacts with water at high temperatures and burns. So only neat oils
are used
Brass and Bronze: EP additives present in the cutting fluid reacts with the
material and produces dull surface. So only neat oils like kerosene is used

Dry machining requires either coated tools or ceramic/cubic boron nitride
cutting tool materials to withstand the intense heat generated by the process.
The coatings with a low friction coefficient and low thermal conductivity work
best at isolating a tool from heat and TiAlN-based coatings are recommended
for dry machining of cast ferrous materials, including cast irons.
http://www.afsinc.org/multimedia/contentMC.cfm?ItemNumber=11371
With the continued development of advanced tool coatings, high-speed dry
machining of cast iron has become possible.
Dry Machining
Heat dissipation without coolant requires high-performance tool coatings,
heat-resistant tool materials and high pressure through spindle air.
For high-speed dry machining of cast iron, the tools must have:
high hardness at high operating temperatures to resist abrasive wear;
high structural strength to resist cutting forces at high chip loads and high
operating temperatures;
high fracture toughness, resistance to thermal shocks and chemical stability
with respect to the workpiece.

Methods of Application of cutting fluids:
Hand applications: use of brush or oil-can to high pressure applications
For high production: 1.Flood method 2.Jet Method 3. Mist methods
3.Mist :cutting fluid is atomised
by a jet of air and the mist is
directed at the cutting zone
1.Flood
2.Jet

4.2 Cutting Fluids
• Enhances the machining quality while
reducing the cost of machining.
• A large variety of cutting fluids based on
organic and inorganic materials have been
developed.

4.2 Cutting Fluids
• The mist and vapor generated is harmful
for the operator.
• Direct exposure of cutting fluids has been
responsible for a number of skin cancer
cases.
• Recycled or disposed of in a manner that
is not harmful to the environment.

Concerns Associated with
Metal Working Fluid(MDF)
Use
BOD-Biochemical oxygen demand FOG- Fats Oils & Grease

Cutting Fluid
• Coolant consumption is estimated higher than
100 million gallons per year in U.S (c 1996).
• U.S. consumed 2 billion gallons in 2000.
• The cost of purchasing and disposing cutting
fluid is about 48 billion dollars a year
• Why such large costs?
–Purchase: $5-$16/gallon concentrate
–Maintenance:$0.20-$1.20/gallon
–Disposal: $0.25-$2/gallon 142

Machining, 15
Oil pressure pump, 25
Coolant, 33
Cooler, mist collector,
etc, 16
Centrifuge, 11
Energy breakdown % for Machining
(Courtesy Toyota Motor Corp)
Dahmus, J., & Gutowski, T. (2004, November). An environmental analysis of machining.
In ASME International Mechanical Engineering Congress and RD&D Exposition, Anaheim,
California, USA.
143

Cutting
Fluid
• Cutting fluids cost 7-17% of the
manufacturing cost in German automotive
industry compared to tool costs that are
quoted as being 2 – 4%.
0
5
10
15
Cutting Fluid cost Tool cost
% of Machining cost
144

Cutting Fluid
• Use dry cutting
• Minimum Quantity Lubrication (MQL)
–Drop lets (without compressed air)
–Mist with compressed air
• Vegetable based cutting fluids
• Use nano cutting fluids
• Use nano coated tools in drycutting
145

Machinability
It is not possible to evaluate machinability in terms of precise numerical
values, hence expressed as a relative quantity

The criteria for determining the machinability are:

Machinability Index
Machinability of a material is a relative quantity and machinability of
different materials are compared in terms of their machinability indexes.
SAE 1212 (AISI 1212)-c% 0.12 is taken as standard material for testing machinability.

Increasing the cutting speed, feed and depth of cut, the tool life shortens
and therefore tooling cost increase and therefore, and so the total
production cost also increases.

1.How to find optimum cutting speed for minimum cost
Total cost and other cost components like non productive cost (labour costs &
overheads), tool regrinding, tool changing cost and machining cost are calculated
and plotted for a batch of components as shown in fig.
It is observed that tooling cost increases and machining cost decreases with
increase in cutting speed
#Effect of variations in cutting speed on various cost factors
Cutting speed Vs Total cost

1. Non-productive cost(Idle cost): It is not effected by variations in cutting speed and
therefore remains unchanged. (Direct labour cost + Overheads)
2. Machining cost: It reduces with increase in cutting speed
3.Tool changing cost: It increases with increase in cutting speed
4. Tool regrinding cost: This also increases with increase in cutting speed
5.Total cost: It reduces with increase in cutting speed until the latter attains its
optimum value(V0) and beyond that it increases with increase in cutting speed
At ‘P’ :
Minimum cost of production. Cutting speed
corresponding to this point gives the optimum
cutting speed(V0) for economical production
and tool life corresponding to this optimum
speed will be the most economical tool life.
Similarly, the production cost per piece (Km)
is the minimum cost per piece.

Various costs
2. Cutting or Machining cost per piece(R3) =
K1 x machining time/piece
3. Tool changing cost per piece (R4)=
K 1 x Tool failures per w/p x Tool changing time for each failure(Tc)
1. Idle cost or Non-productive cost per piece (R2)= K1 x Ti
4. Tool regrinding cost per piece(R5) =
cost of tool per grind (K2 ) x Total no. of tool failures per w/p (Tx)
K1 = Idle cost per piece per min, in Rs/min (Labor cost + Overheads)
Ti = Idle time per piece in min
K2=cost of tool per regrind in Rs.
Material cost(R1), does not depend on cutting conditions and remains constant

Let K1 = Idle cost per piece(non productive cost) in Rs/min= Direct labour cost + Overheads
K2=cost of tool per regrind in Rs.
Ti= Idle time per piece, in minutes
Tc=Tool changing time, in minutes
V= Cutting speed, in m/min=
f= Feed rate, in mm/rev
D= Diameter of w/p in mm
L= Length of machining done in mm
1. Idle cost per piece(R2)= idle cost(Rs/min) x Idle time per piece(min) = K1 x Ti
2. Cutting or Machining cost per piece (R3)= K1 x machining time/piece
Machining time /piece = Length of machining
Feed rate in mm/rev x rpm
=
1000
DN
Tx = Total number of tool failures per workpiece
fV
DL
1000

=
Tr= Tool regrinding cost per piece=K2 x no. of tool failures per workpiece
Machining time = Length/ (feed in mm/min)

Therefore, cutting cost or machining cost/piece = K1 x
fV
DL
1000

3. Tool changing cost/piece (R4)=
Idle cost(K1) x No.tool failures per w/p (Tx) x Tool changing time for each failure(Tc)
No. of tool failures per w/p(Tx ) = Machining time per piece/ Tool life(T)
Machining time/piece
= K1 x
n
n
V
C
T 1
1
From VTn =C
So, Tool changing cost/piece (R4) = K1 x
4.Tool regrinding cost per piece Tr (R5)
= cost of tool per grind(K2) x No. of tool failures per w/p(Tx)
Tx =
n
n
fC
DLV
1
1
1
1000


n
n
fC
DLV
1
1
1
1000


x Tc
Tx =

Tr
Now,
total cost per piece(R) is given by
The sum of all the four cost components + material cost(R1)
R= R1+ Idle cost/piece + Machining cost/piece + Tool changing cost/piece + Tool regrinding cost/piece
K1 x Ti + + + Rupees
For minimum cost :
n
n
fC
DLV
1
1
1
1000


= K2 x
R =
0


V
R
R1 + +
01
1
1
1 2
1
1
2
2
1
1
1
2
1

































n
n
n
n
c
V
n
fC
DLK
V
n
fC
DLTK
f
DLVK 

On simplification
n
c
pt
K
KTK
n
C
V











 








1
21
0
1
1Cutting speed
Tool life (Tmc ) = 




 







1
21
1
1
K
KTK
n
c

2.Optimizing cutting parameters for maximum production(mp)
Sometimes, the rate of production is more important than the cost per piece.
A large rate of production may result in a better return, and therefore it is also
useful to investigate the conditions leading to the highest possible rate of
production
The maximum production rate can be achieved if the total time required per
piece is reduced to a minimum. For this, we shall keep the feed at highest
possible value and search for the optimum velocity
Total time required per piece(T)=
setting and idle time/piece(Ti )+ machining time/piece(Tm )+ tool changing time/piece
For maximum production : 0


V
T
Cutting speed Vmp= n
cT
n
C












1
1
Tool life Tmp = cT
n






1
1
Tc=Tool changing time, in minutes
fV
DL
1000

c
m
i T
T
T
fV
DL
TT 

n
C
V
T
1








Relationships among Cutting speed & Production rate
The relationship between cutting speed and production rate (no. of pieces
produced per unit time) is shown below.
Cutting speed Vs Total time
Various time curves and maximum production curve
No. of products per unit time= 1/T

When the cutting speed is low, the production rate (pieces produced per unit time)
is also low because of lesser amount of metal removal per unit of time.
But, as the cutting speed increases, the production rate also increases and
continuous to increase upto a point(Pm) where the rate of production is maximum.
The corresponding cutting speed(Vmp) represents the optimum value of cutting
speed at which the rate of production will be highest.
Any further increase in the cutting speed will lead to a quicker wearing of tool,
more frequent changing of tool, more down time and therefore, a reduced rate of
production
From graph, the lowest point on the total time curve gives the minimum time
taken for production of each piece. The corresponding cutting speed(Vmp)
represents the optimum value of cutting speed at which the total time taken in
production of the component will be minimum.

Conclusions
1.Non-productive time: It remains constant and is not effected by variations in cutting
speed
2.Machining time: It reduces with increase in cutting speed
3.Tool changing time: It increases with increase in cutting speed
4.Tool regrinding time: It also increases with increase in cutting speed
5.Total time: It reduces till the cutting speed reaches the value Vmp and beyond that
it increases with increase in cutting speed
6.Production rate: It increases with increase in cutting speed till the latter attains the
optimum value(Vmp). Beyond this, it decreases with increase in cutting speed
Our interest is producing the components at maximum rate and at minimum cost.
If we replot the minimum cost and maximum production curves, as shown:

Cutting speed V0, at which the total production cost is minimum, is not the same
as that at which the production rate is maximum. The former is lesser than latter.
The area lying in between these two values of cutting speeds is known as High
Efficiency Range(Hi-E Range) and the cutting speeds lying in this range are either
economical or more productive.
 For efficient and economical production of a workpiece, the cutting speed should
always be selected within this range only.
Minimum cost
and
Maximum production curves

An ideal tool material is the one which will remove the largest volume of
work material at all speeds
Higher the hot hardness and toughness in the tool material, the longer the tool life

Low alloy steels
• High carbon tool steel is the oldest cutting tool materials, having C
content ranging from 0.7 – 1.5% carbon.
• Shaped easily in the annealed condition and subsequently hardened by
quenching and tempering. • Hv ~ 700 after quenching and tempering.
For low cutting speed due to a drop in hardness above 150oC.

Inherent unreliability of ceramic tooling limits its use to specialist cutting
operation.

CBN -A new generation cutting tool material
Polycrystalline cubic boron nitride, CBN, is a material with excellent hot hardness
that can be used at very high cutting speeds. It also exhibits good toughness and
thermal shock resistance.
Modern CBN grades are ceramic composites with a CBN content of 40-65%. The
ceramic binder adds wear resistance to the CBN, which is otherwise prone to
chemical wear. Another group of grades are the high content CBN grades, with 85%
to almost 100% CBN. These grades may have a metallic binder to improve their
toughness.
CBN is brazed onto a cemented carbide carrier to form an insert.
CBN grades are largely used for finish turning of hardened steels, with a
hardness over 45 HRc. Above 55 HRc, CBN is the only cutting tool which can
replace traditionally used grinding methods. Softer steels, below 45 HRc, contain a
higher amount of ferrite, which has a negative effect on the wear resistance of
CBN.
CBN can also be used for high speed roughing of grey cast irons in both turning
and milling operations.

Cermet
A cermet is a cemented carbide with titanium based hard particles. The name
cermet combines the words ceramic and metal. Originally, cermets were composites
of TiC and nickel.
Modern cermets are nickel-free and have a designed structure of titanium
carbonitride Ti(C,N) core particles, a second hard phase of (Ti,Nb,W)(C,N) and a W-
rich cobalt binder.
In comparison to cemented carbide, cermet has improved wear resistance and
reduced smearing tendencies. On the other hand, it also has lower compressive
strength and inferior thermal shock resistance.
Applications
Cermet grades are used in smearing applications(stains with greasy substance)
where built-up edge is a problem. Its self-sharpening wear pattern keeps cutting
forces low even after long periods in cut. In finishing operations, this enables a long
tool life and close tolerances, and results in shiny surfaces.
Typical applications are finishing in stainless steels, nodular cast irons, low
carbon steels and ferritic steels. Cermets can also be applied for trouble shooting
in all ferrous materials.
http://www.sandvik.coromant.com/

Machining Fundamentals: An Introduction to the Key Concepts and Processes in Machining Operations

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Machining Fundamentals: An Introduction to the Key Concepts and Processes in Machining Operations

Similar to Machining Fundamentals: An Introduction to the Key Concepts and Processes in Machining Operations (20)

More from Kunduru Srinivasulu Reddy

More from Kunduru Srinivasulu Reddy (6)

Recently uploaded

Recently uploaded (20)

Machining Fundamentals: An Introduction to the Key Concepts and Processes in Machining Operations