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Metal Cutting and

Chip FormationUNIT 1 METAL CUTTING AND CHIP

FORMATION

Structure

1.1

Introduction

Objectives

1.2

Material Removal Processes

1.3 Chip Formation

1.3.1 Deformation in Metal Machining

1.3.2 Chip Types

1.3.3 Types of Cutting

1.3.4

Mechanics of Chip Formation

1.3.5 Geometry of Chip Formation (Orthogonal Cutting)

1.4

Force Analysis

1.5

Velocity Relationships1.6

Shear Strain and Shear Strain Rate

1.7

Shear Angle Relationships

1.8

Summary

1.9 Key Words

1.10 Answers to SAQs

1.1 INTRODUCTION

Manufacturing processes can be broadly divided into four categories, viz., primary(casting, forging, moulding, etc), secondary (machining, finishing, etc.), tertiary

(fabricating processes like welding, brazing, riveting, etc.), and fourth level processes

(painting, electroplating, etc.). Secondary manufacturing processes are as important as

any other level processes. These processes involve removal of material in the form of

chips or otherwise, to give the desired shape, size, surface roughness, and tolerance on

the workpiece obtained from the primary manufacturing processes. The machined

components can be used as it is, or one can be assembled (sometimes using fabricating

processes) and if required, given an aesthetic look by electroplating, painting, etc. This

block/unit will discuss the fundamentals of traditional material removal processes (non-

traditional material removal processes are discussed in Block 4). This unit will discuss

basic principles of metal cutting including mechanics of chip formation, velocity and

force analysis, and some of the models proposed to evaluate the shear angle relationships.

Objectives

After studying this unit, you should be able to

• understand classification scheme for various types of material removal processes,

• identify various types of metal cutting processes, types of chip formed,mechanism of chip formation and geometry of chips,

• analyse forces and velocities in cutting process, and

• know various schools of thought regarding shear angle relationships.

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Theory of Metal Cutting 1.2 MATERIAL REMOVAL PROCESSES

Material removal processes can broadly be divided into two categories : traditional and

advanced (non-traditional). Each of these categories can be further sub-divided into bulk

removal processes (or cutting) and finishing processes. The classification of various types

of material removal processes is shown in Figure 1.1(a). This unit will discuss only the

basics of traditional cutting processes.

Traditional Cutting

Traditional cutting processes can be classified as those which produce parts having

surfaces of revolution and those which produce prismatic shapes. Another scheme

for the classification of metal cutting processes is provided in

Figure 1.1(a). This classification is based on the type of motion imparted to the

work and tool. Cutting tools used for material removal are classified in two

categories : single point cutting tool and multi-point cutting tool (cutting tools

having more than one cutting edge). The following section discusses how material

removal takes place by using a single point cutting tool. Similar principles are

applicable to the multiple point cutting tools as well.

Figure 1.1(a) : Classification of Material Removal Processes

The process of metal cutting is effected by providing relative motion between the

workpiece and the hard edge of cutting tool. Such relative motion is produced by a

combination of rotary and translating movements either of the workpiece or of the

cutting tool or both. Depending on the nature of the relative motion, metal cutting

process is called either turning or planning or boring, etc.

For different types of operations, one needs to have different types of machine

tools. For example, lathe for turning, planer for planning, grinder for grinding, etc.

Some of these machines (say, lathe, boring m/c, and drill) generate surfaces of

revolution whereas others (planer, milling m/c, and shaper) make prismatic (or flat

surfaces) parts. With the help of different types of tools, a lathe can perform

various kinds of operations (Figure 1.1(b)).

Conventionally, the translatory displacement of the cutting edge of the tool along

the work surface during a given period of time is called ‘feed’( f ), while the

relative rate of traverse of work surface past the cutting edge is designated as the

‘cutting velocity’ or simply ‘speed’ (V c).

In case of single point turning, V c is the peripheral velocity of the rotating

workpiece in meters per minute. In case of slab milling, it is the peripheral velocity

of the milling cutter in meters/minute.

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Metal Cutting and

Chip Formation

Figure 1.1(b) : Various Operations that can be Performed on a Lathe [Kalpakjian, 1989]

Table 1.1

Operation Motion of Job Motion of Cutting Tool Figure of Operation

Turning on a

lathe

Rotary motion of the

work

Axial movement of the tool

Boring on a

lathe

Work rotation Axial tool movement

Drilling on a

drill machine

Fixed Rotations as well as

translatory feed

Planning Translatory Intermittent Translation

Milling Translatory Rotation

Grinding Rotary/Translatory Rotary

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Theory of Metal Cutting 1.3 CHIP FORMATION

1.3.1 Deformation in Metal Machining

Figure 1.2 shows a schematic diagram of material deformation during cutting, and

subsequently removal of the deformed material from the workpiece by a single point

cutting tool. Because of the relative motion between the tool and the workpiece, material

ahead of the tool face (rake face) is compressed (elastically and then plastically). Further,movement of the tool into the workpiece deforms the work material plastically and

finally separates the deformed material from the workpiece. This separated material

flows on the rake face of the tool called as chip. The chip near the end of the rake face is

lifted away from the tool, and the resultant curvature of the chip is called chip curl.

Figure 1.2 : Schematic Diagram of Chip Deformation

The study of the mechanism of chip formation involves deformation process of the chip

ahead of the cutting tool. Theoretical study of the material deformation in metal cutting is

difficult and therefore experimental techniques have been resorted to for analyzing the

process of deformation in chips. The methods commonly employed for this purpose are :

(i) Use of movie camera for taking pictures of chip.

(ii)

Observing grid deformation during cutting.

(iii)

Examination of frozen chip samples obtained by the use of quick-stop

device.

Experimental study of chip deformation process has revealed that :

(i) During machining of ductile materials, a plastic deformation zone is formed

in front of the cutting edge (Figure 1.2).

(ii) The distinctive zone of separation between the chip and workpiece where

deformation gradually increases towards the cutting edge is called the primary deformation/shear zone. In shear zone extensive deformation

occurs. The width of shear zone is very small.

(iii)

The plastic deformation involved in the formation of chips affects the

hardness of material (strain hardening). Strain hardening increases when a

layer undergoes deformation in the shear zone.

1.3.2 Chip Types

The type of chip obtained from a machining process is characterized by a number of

parameters e.g., the type of tool-work engagement, work material properties and the

cutting conditions.

Ernst has classified the chips obtained in machining processes into three categories :

Type 1 : Discontinuous chip,

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Metal Cutting and

Chip FormationType 2 : Continuous chip, and

Type 3 : Continuous with built-up edge.

When ductile materials at high cutting speed are cut by a single point cutting tool, ribbon

like continuous chip (Figure 1.3(a) and 1.3 (b)) is obtained. The conditions that promote

formation of continuous chips in metal cutting are sharp cutting edge, low feed rate (or

small chip thickness), large rake angle, ductile work material, high cutting speed, and low

friction at chip-tool interface. As shown in Figure 1.2, major deformation takes place in primary shear deformation zone (PSDZ) resulting in the formation of chip. Due to ductile

nature of work material and reasonably high temperature in the PSDZ, the deforming

material flows on the rake face of the tool as continuous mass rather than the one

fractured/ruptured at small distances at the underneath of the chip as in discontinuous

chip.

Continuous chip results in good surface finish, high tool-life, and low power

consumption. But disposal of large coiled chips is a serious problem, for many industries

where tons of chips are produced every week. To get rid of this problem various types of

chip breakers are used which are in the form of step or groove on the rake face of the

tool (Figure 1.4). The chip strikes with this step/groove and gets broken in the form of

small segments. Disposal of such small chips is not a problem .

If the friction between tool and chip while machining ductile materials is high, some part

of the chip gets welded to the rake face of the tool near its cutting edge. The welded

material is extremely hard and its size keeps on increasing with time. Because of the

hardness of the adhered materials onto the cutting edge, it participates in cutting to a

certain extent. That is why it is named as built up edge (Figure 1.5). As the size of the

BUE grows larger, it becomes unstable and it breaks. Some part from the broken BUE is

carried away by the chip as well as on the machined surface (Figure 1.3).

Figure 1.3 : Different Kinds of Chips : (a) Continuous; (b) Photograph of Continuous Chip;

(c) Continuous Chip with Built Up Edge; and (d) Discontinuous Chip

The chip with the adhered parts of the BUE is known as continuous chip with BUE. The

adhered parts of the BUE on the machined surface make the machined surface rough, but

the BUE protects the actual cutting edge of the tool from wear. Thus, cutting with BUEenhances the tool life (or tool cuts longer before regrind).

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Theory of Metal Cutting

Figure 1.4 : Different Types of Chip Breakers

Figure 1.5 : Development of Built Up Edge [Rao, 2000]

Discontinuous or segmented chips are produced while machining brittle materials or

ductile materials at low speeds and high friction conditions. The basic difference between

the mechanism of formation of discontinuous chip and continuous chip is that, instead of

continuous shearing of the material ahead of the cutting tool, rupture occurs

intermittently producing segments of chip (Figures 1.3 and 1.6). These chips are smaller

in length hence easy to dispose off, and give good surface finish on the workpiece.

Discontinuous chips are formed when cutting brittle materials, or cutting ductile

materials at low speed, or cutting with tools of small rake angle.

(a) Tear Type (b) Shear Type

Figure 1.6 : Hypothesized Discontinuous Chip Formation [Rao, 2000]

1.3.3 Types of Cutting

Principally, there are two types of cutting :

(i) Orthogonal cutting, and

(ii)

Oblique cutting.

Orthogonal Cutting

Orthogonal cutting operation is “the simplest type of cutting operation, in which

the cutting edge is straight, parallel to the original plane surface of the workpiece

and perpendicular to the direction of cutting, and in which the length of the cutting

edge is greater than the width of the chip removed (Figures 1.7(a) and (b))”. This

orthogonal cutting is also known as Two Dimensional (2-D) Cutting . A few of the

cutting tools perform orthogonally, such as lathe cut-off tools (Figure 1.7(a)),

straight (not helical) milling cutters, broaches, etc.

In actual machining, majority of the cutting operations (turning, milling, etc.) are

three dimensional (3-D) in nature and are called as oblique cutting. In obliquecutting, the cutting edge of the tool is inclined to the line normal to the cutting

direction, and this angle is known as angle of obliquity. This is also called the

inclination angle, i (Figure 1.7(c)). Oblique cutting can be defined as “the cutting

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Metal Cutting and

Chip Formationoperation in which the cutting edge is straight and parallel to the original surface of

the workpiece, but is not perpendicular to the cutting direction, being inclined to

it”. An angle of interest in this case is the chip flow angle, ηc which is defined asthe angle measured in the plane of cutting face between the chip flow direction and

the normal to the cutting edge (Figure 1.7(c)). Both ‘i’ and ηc are zero in case oforthogonal cutting. The certain practical limitations to orthogonal cutting are

mitigated by three dimensional tooling.

Figure 1.7 : (a); (b) Orthogonal Cutting System; and (c) Oblique Cutting System

Generally for the mathematical analysis of the mechanics of metal cutting, orthogonal

cutting is considered because it is simpler than the oblique cutting. The results so

obtained can be used for oblique cutting operations.

1.3.4 Mechanics of Chip Formation

Plastic deformation is the main factor that governs formation of chips. Initially,

researchers (Merchant and others) proposed that deformation of the material takes place

along a plane (called shear plane) just ahead of the cutting tool and runs up to free

surface of the workpiece (Figure 1.8(a)). Once the deforming material crosses the shear plane, it slides along the rake face of the tool due to the velocity of cutting tool (relative

motion between the tool and workpiece). This hypothesis of a shear plane is useful from

the analysis of metal cutting point of view but has theoretical drawbacks. Here, the

transition from the un-deformed to the deformed material takes place along a shear plane

by changing cutting velocity from V c (velocity of tool with respect to workpiece) to V f (chip velocity relative to the tool). For this change to take place, the acceleration across

the plane (plane thickness equal to zero) has to be infinite. This also applies to the

stress-gradient across the shear plane. Due to the above anomaly, researchers (Oxley and

others) experimentally studied the deformation zone by freezing the cutting process with

the help of a quick stop device. When they studied the deformed zone under a

microscope, they found that the deformation takes place within a finite zone (thin or thick

depending upon various governing parameters). This is called as primary shear

deformation zone (PSDZ) (Figures 1.8 (b) and (c)). They also found that under certain

machining conditions, deformation also takes place at the tool-chip interface

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Theory of Metal Cutting (Figures 1.8 (b) and (c)). This deformation is known as secondary shear deformation

zone (SSDZ).

Figure 1.8 : (a) Shear Plane; (b) Primary and Secondary Shear Deformation Zone in Chip Formation;

and (c) Frozen Chip Obtained by a Quick Stop Device

1.3.5 Geometry of Chip Formation (Orthogonal Cutting)

Figure 1.9 shows a simple geometry of chip formation in case of continuous chip

(type 2). The uncut chip thickness t u (equal to feed in turning) is deformed to give chip

thickness t c which experiences two velocities V f (chip sliding velocity) and V s (shear

velocity) along the tool face and shear plane, respectively. From this geometry, it is

possible to calculate the shear angle (φ) in terms of measurable or known quantities t u , t c and α.

Figure 1.9 : Geometry of Continuous Chip Formation

From right angle triangles, ABC and ABD (BD is perpendicular to AD drawn from B),

AB = t u / sin φ

Also, AB = t c / sin (90 − (φ −α)) = t c / cos (φ −α)

∴ α)(cos

sin

−φφ=

c

u

t

t

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Metal Cutting and

Chip Formationc

u

t

t is called chip thickness ratio or chip thickness coefficient (r c) which can be written

as

φφ+φ

=sin

αsinsinαcoscos1

cr

or, α+αφ= sincoscot r cc1 r

or, φα−=α tan)sin1(cos cc r r

∴ cos

tan1 sin

c

c

r

r

αφ =

− α

. . . (1.1)

To determine the shear plane angle (φ) for a given cutting condition the chip thickness

ratio

=

c

uc

t

t r , should be known. But to determine t c with a micrometer is somewhat

inaccurate. Hence, an indirect approach to this problem is to assume that the density of

metal during the cutting process does not change. Hence, the volume of uncut chip is

equal to the volume of metal removed (or deformed chip). Since the width of chip (b) is

equal to the width of metal being cut (in orthogonal cutting), therefore :

bt Lt bL uucc = (volume constancy condition)

∴ c c u u L t L t =

or,u

c

c

u

Lt

Lt =

∴ ucc u

t Lr

t L= = c . . . (1.2)

where, Lc is length of chip, and Lu is corresponding length of material removed from the

workpiece (or uncut chip length). Lc can be easily measured, and it ( Lc / Lu) will give

more accurate results than (t u / t c) because of the difficulties and inaccuracies involved in

the measurement of thickness of the deformed chip (t c).

1.4 FORCE ANALYSIS

Let us analyse the forces acting on the chip in orthogonal cutting. These are shown in

Figure 1.10 (a) and are as follows : Force, F s, is the resistance to shear of the metal in

forming the chip. F s acts along the shear plane. Force, F n, is normal to the shear plane

and is a backup force on the chip provided by the workpiece. Force N acting on the chip

is normal to the cutting face of the tool and is provided by the tool. Force F is frictional

resistance offered by the tool to the chip flow. The latter force acts downwards against

the motion of the chip as it slides upwards along the tool face.

Figure 1.10 (b) shows the free body diagram of the forces acting on the chip. Forces F s

and F n are represented by the resultant R, and F and N are replaced by the resultant R'.

This means that only two combined forces are acting on the chip, i.e., R and R'. There are

external couples on the chip which curl it, and they may be negated in this approximateanalysis. If equilibrium is to exist when a body is acted upon by two forces, they must be

equal in magnitude, and be collinear. Hence, R and R' are equal in magnitude, opposite in

direction and collinear (Figure 1.10).

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Figure 1.10 : (a) Force Components Acting on a Chip; and (b) Free Body Diagram of a Chip

Figure 1.11 shows a composite diagram in which the two force triangles of

Figure 1.10, have been superimposed by placing the two equal forces R and R' together.

Since the angle between F s and F n is a right angle, the intersection of these forces lies onthe circle with diameter R as shown. Also, F and N may be replaced by R to form the

circle diagram (Figure 1.11).

Figure 1.11 : Force Circle Diagram

The horizontal cutting force F c and vertical force F t can be measured in a machining

operation by the use of a force dynamometer. The electric strain gauge type of transducer

is used in the dynamometer. After F c and F t are determined, they can be laid off as in

Figure 1.11 and their resultant is the diameter of the circle. The rake angle α can be laidoff, and the forces F and N can then be determined. The shear plane angle φ can bemeasured approximately from a photomicrograph or by measuring t c and t u, or length of

chip and corresponding length of unmachined chip (discussed elsewhere).From Figure 1.11, the following vector Eqs. can be written

' R F F F F R

N F ' R

t cn s

ρρρρρρ=+=+=

+=

Merchant represented various forces in a force circle diagram in which tool and reaction

forces have been assumed to be acting as concentrated at the tool point instead of their

actual points of application along the tool face and the shear plane. The circle has the

diameter equal to R (or R' ) passing through tool point.

After F c, F t , α and φ are known, all the component forces on the chip may be determined

from the geometry. For instance, the average stress on the shear plane can be determined

by using force F s and the area of the shear plane. Another useful quantity is the

coefficient of friction (µ ) between the tool and chip. Using force circle diagram, it can be

shown that

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Metal Cutting and

Chip Formationα+α= sincos ct F F F . . . (1.3)

and, α−α= sincos t c F F N . . . (1.4)

Then, the coefficient of friction (µ ) is calculated as

α−αα+α

==β=µsincos

sincostan

t c

ct

F F

F F

N

F . . . (1.5)

where, β is the friction angle.

or,α−

α+=µ

tan

tan

t c

ct

F F

F F . . . (1.6)

We can also write :

α−

α+=β

µ=β

−

−

tan

tantan

)(tan

1

1

t c

ct

F F

F F . . . (1.7)

From Figure 1.11, we get :

. . . (1.8) φ−φ= sincos t c s F F F

φ+φ= sincos ct n F F F

. . . (1.9))tan( α−β+φ= sn F F

Also, from Figure 1.11,

α)β(cos

)αβ(cos

α)β(cos

α)(βcos

−+φ−

=∴

−+φ=

−=

s

c

s

c

F

F

R F

R F

or,)αβ(cos

)αβ(cos

−+φ−

= sc F F . . . (1.9(a))

Shear plane area is equal to :

φ=

sin

bt A u s . . . (1.10)

If τ be the shear strength of the work material, then,

τφ

=sin

bt F u s . . . (1.11)

Substituting in Eq.(1.9 (a)), we get

bt F uc

)cos(

)cos(

sin α−β+φ

α−β

φ

τ= . . . (1.12)

hence, )cos(sin α−β+φφ

τbt u= R . . . (1.12(a))

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Theory of Metal Cutting From Figure 1.11,

)(sin)(cossin

)sin(

α−βα−β+φφ

τ=

α−β=

bt

R F

u

t

(From Eq. 1.12(a)) . . . (1.13)

From Eqs. (1.12) and (1.13), we can write :

)(tan α−β= F

F

c

t . . . (1.14)

From the above analysis, unknown forces in the force circle diagram and the value of

coefficient of friction can be calculated provided F c , F t , α , t u and t c are known measured.

During machining operations, chips are formed as a result of plastic deformation. Hence,

chips experience stresses and strains. At shear plane, two normal forces simultaneously

act, i.e., F s and F n. Shear stress (τ) can be found as

Mean shear stress φ

φ−φ

==τ sin)sincos(

)( u

t c

s

s

bt

F F

A

F

. . . (1.15)

Mean Normal stress (σ)u

ct s

n

bt F F

A

F φφ+φ==

sin)sincos( . . . (1.16)

where, A s = Shear plane area = φ / t u sinb .

1.5 VELOCITY RELATIONSHIPS

Since the chip is thicker than the uncut chip, the velocity of the chip as it moves along thetool face must be less than the cutting speed (assuming volume constancy during cutting,and width of cut before machining and after machining remains same). Different

velocities during cutting can be estimated as follows :

Assume that the cutting velocity of the tool relative to the workpiece is V c which is

known before hand. The chip slides along the cutting (rake) face of the tool with a

velocity relative to the tool equal to V f (chip flow velocity). The newly cut chip elements

move relative to the workpiece along the shear plane with a velocity equal to V s (shear

velocity). From the principle of kinematics that the relative velocity of two bodies (heretool and chip) is equal to the vector difference between their velocities relative to the

reference body (the workpiece). Employing this principle, Figure 1.12 has been drawn.

Using sine rule, from ∆ ABC, we get :

)90(sinsin))(90(sin α−=

φ=

α−φ−

V V

V s f c . . . (1.17)

or,α

=φ

=α−φ cossin)cos(

s f c V V V

The chip flow velocity along the tool rake face is given by

)(cos

sin

α−φ

φ= c f

V V =V c . r c . . . (1.18)

whereas the shear velocity V s is obtained as :

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Metal Cutting and

Chip Formation)(cos

cos

α−φ

α= c s

V V . . . (1.19)

Figure 1.12 : Velocity Relationship

1.6 SHEAR STRAIN AND SHEAR STRAIN RATE

Since most practical cutting processes are geometrically complex, let us first study the

orthogonal machining and extend the theory of orthogonal cutting to more complicatedcutting process involving oblique cutting. Due to simplicity and fairly wide applications,

the continuous chip without BUE has been most extensively studied. There areconflicting evidences about the nature of the deformation zone in metal cutting. This hasled to two basic schools of thought in the approach to the analysis.

Many workers, such as Piispanen, Merchant, Kobayashi and Thomson have favoured the

thin plane model while Palmer and Oxley, and Okushima and Hitomi have based theiranalysis on thick deformation region (Figure1.8). Experimental evidences indicate that

the thick zone model may describe the cutting process at low speeds, but at high speedsmost evidences indicate that a thin shear plane is approached. Thin zone model is moreuseful in practical cutting and its analysis is simpler hence it has received more attention.

Thin Zone Model

Merchant developed an analysis based on the thin shear plane model. He made thefollowing assumptions :

• The tool tip is sharp and no rubbing or ploughing occurs between the tool

and the workpiece.

• The deformation is two dimensional, i.e., no side spread.

• The stress on the shear plane is uniformly distributed.

• The resultant force R on the chip applied at the shear plane is equal, opposite

and collinear to the force R' applied to the chip at the tool-chip interface.

Strain and strain rate are determined as follows :

To derive an expression for shear strain, the deformation can be idealized as a process of block slip (or preferred slip planes), as shown in (Figure 1.13). Shear

strain (γ) is defined as the deformation per unit length.

CD

DB

CD

AD

CD

AB

y

s+==

∆∆

=γ

φ+α−φ= cot)(tan . . . (1.20)

φφ

+α−φα−φ

sin

cos

)(cos

)(sin=

)(cossin)(coscossin)(sin

α−φφ α−φφ+φα−φ=

)(cossin

cos

)(cossin

)(cos

α−φφα

=α−φφ

φ−α+φ=

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Theory of Metal Cuttingor,

)(cossin

cos

α−φφα

=γ . . . (1.21)

Figure 1.13 : Strain and Strain Rate in Orthogonal Cutting

Strain may also be expressed in terms of the shear velocity (V s) and the chip velocity (V f )

∴ sin

γ =φ

s

c

V

V (from Eq.(1.19))

Therefore, shear strain rate (γ ) in cutting is given by

yt

s

t

y

s

dt

d

∆

∆∆

=∆

∆∆

=γ

=γ1&

γ

y

V

y

V c s

∆α−φ

α==

)(cos

cos

∆

. . . (1.22)

∆ y is mean thickness of PSDZ .

1.7 SHEAR ANGLE RELATIONSHIPS

A number of attempts have been made to study the mechanics of cutting process. In

designing a metal cutting operation, it would be helpful to predict the position of the

shear plane (angle φ ). Attempts have been made to derive a fundamental relationship of

the shear plane angle φ in terms of rake angle (α ) and friction angle ( β ). Several theories

have been proposed to establish a relationship between φ , α and β . Some of the theories

have been discussed below.

Ernst-Merchant derived a relationship using the minimum energy criterion, that is, the

shear plane is located where the least energy is required for shear. The derivation of

Ernst-Merchant equation is based on the following assumptions :

(i) cutting is orthogonal,

(ii)

the shear strength of the metal along the shear plane is independent of the

magnitude of compressive (normal) stress acting on that plane,

(iii) the chip is continuous type with no built up edge, and

(iv) the energy of separation of chip elements is neglected and the minimum

energy criterion establishes the plane on which shearing deformation occurs.

As the cutting progresses in the beginning, the cutting force ( F c) increases gradually, the

shear stress on various planes ahead of the tool also increases. However, the shear stress

will not be same on all the planes ahead of the tool because the shearing components of

the forces on the planes are not the same, nor is the extent of areas the same. On one of

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Metal Cutting and

Chip Formationthe planes, however, the shear stress will be greater than on any other plane, and as F c isfurther increased, the shear stress will reach the yield strength in shear of the material

being cut and plastic deformation will occur along that plane, thus forming the chip. The

cutting force required to cause shear deformation along that plane will then be the lowest

cutting force.

Once the shear deformation begins along one plane, the cutting force

cannot exceed that minimum value.

To determine shear-plane angle φ, express the cutting force F c in terms of φ, differentiate

it with respect to φ, equate the derivative to zero, and solve it for the angle φ as follows :

{ } )(cos)(cos

sin)(cos

)(cos

α−β+φα−β

×φ

τ=

α−β+φα−β

= bt

F F u

sc (Eq.1.12)

Here, except φ all other parameters can be taken as constant during machining (assumingthat no strain hardening takes place). It would give the condition for the minimum energy

if the derivative of F c with respect to φ is equated to zero.

0)(cossin

)(sin)(coscos)(cos

22 =

α−β+φφ

α−β+φφ−α−β+φφα−βτ=

φbt

d

F d u

c

Therefore, cos cos ( ) sin sin ( ) 0φ φ + β − α − φ φ + β − α =

or, cos (2 ) 0φ + β − α =

or, (2 )2

πφ + β − α =

Hence, )(2

1

4α−β−

π=φ . . . (1.23)

where, φ, β and α are shear angle, friction angle and rake angle, respectively.

Eq. (1.23) indicates that the shear angle φ is a unique function of the tool rake angle andthe angle of friction in metal cutting.

Merchant further introduced a modification to this theory and assumed that the shear

strength of a polycrystalline metal is affected by temperature, rate of shear, shear strain

(plastic) and the stress acting normal to the shear plane. While it is known that the normal

compression stress on a plane does not affect the shear strength of a single crystal

however, the shear strength of polycrystalline material is affected. The modified Eq. is

)(2

1

2α−β−=φ

C . . . (1.24)

where, ‘C ’ depends on the slope of the shear strength vs. compressive stress curve for the

given material. 'C ' is also known as machining constant.

In 1949, another approach to the analytical solution of the shear plane angle was made by

Lee and Shaffer. They assumed that the material being cut behaves as an ideal plastic

which does not strain harden. It was assumed that the shear plane coincides with the

direction of the maximum shear stress (Figure 1.14). Based on these assumptions, they

applied slip line field theory and derived the relationship given by Eq. (1.25).

)(4

β−α+π

=φ . . . (1.25)

As a modification, later on Lee and Shaffer considered the effect of a small built up edge

or nose, and its effect on the stress field referred to above and arrived at an expression forthe shear angle (φ ) which included an additional angle θ , which depends on the size of

the built up edge,

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20

Theory of Metal Cuttingθ+β−α+

π=φ )(

4

In 1952, Shaw, Cook and Finnie extended the Lee and Schaffer theory by further

analytical and experimental investigations, and arrived at the following relationship :

( )4

π′φ = + α − β + η

While deriving the above relation, they assumed that the shear plane is not a plane ofmaximum shear. Here, η′ is established by the analytical method and it is not constant. η′ is the angle between the shear plane and the direction of the maximum shear stress. To

determine the value and sign of the η′, it is necessary to draw the Mohr’s circle diagram.

Figure 1.14 : Shear Plane Model of Lee and Shaffer

Based on the experimental study of the mechanics of chip formation and the flow of

grains in the material during cutting, Palmer and Oxley observed that the deformation

does not take place along a plane, rather it takes place in a narrow wedge shaped zone.

But for analytical simplicity, it was considered as a parallel sided shear zone

(Figure 1.15).

Figure 1.15 : Shear Zone Model by Oxley

A further contribution towards the solution of this problem was made by R Hill in 1954,

who analyzed the state of stress at the shear zone, using a new principle “On the limits set by plastic yielding to the intensity of singularities of stress”. But in 1959, Eggleston,

Herzog and Thomsen tried to show by their test results that none of the three Eqs. (by

Ernst and Merchant, Lee and Shaffer, and Hill) was correct which implies that metal in

the shear zone under the existing conditions of stress, high rates of strain and elevated

temperature does not behave as ideal plastic solid. Since no single criterion is applicable

to the shear angle relationship in metal cutting, and since a satisfactory theory has not

been advanced at present to explain the experimental observations adequately, the

challenge exists for a closer solution to the problem of angle relationship. This problem is

so tedious because the complexity is created by the simultaneous presence of so many

variables at a time, for example :

(i)

plastic deformation,(ii)

work hardening,

(iii)

external and internal friction,

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21

Metal Cutting and

Chip Formation(iv)

temperature effect,

(v) diffusion,

(vi) oxidation, and

(vii) local heating, etc.

Example 1.1

Show that in case of ideal orthogonal cutting operation the shear strain undergone by the chip during its removal from the workpiece would be minimum if the chip

thickness ratio is ‘1’.

Solution

In Figure 1.13 the shear strain in general and shear strain in cutting are shown.

Here, ∆ s is in the direction of force, ∆ y is in the direction ⊥ to the force.

Shear strain in another term of interest is associated with the cutting process. The

shear strain is defined as y

s

∆∆

and hence in cutting (Figure 1.13),

tan ( ) cot s AB AD DB y CD CD CD

∆γ = = = + = φ − α + φ∆

We want the condition when γ should be minimum. Hence, differentiate γ withrespect to φ and equate the derivative equal to zero.

{ } 0cot)(tan =φ+α−φφ

=φγ

d

d

d

d ∴

sec 0)eccos()( 22 =φ−+α−φ

∴ φ=α−φ 22 eccos)(sec

or, .)(cossin 22 α−φ=φ∴

Take the under root to both sides,

)(cossin α−φ±=φ±∴

or, )(cossin α−φ=φ . . . (A)

αφ+αφ= sinsincoscos

α+φα= sincotcos1∴ . . . (B)

Question is that at the condition (A) whether the chip thickness ratio is 1 or not.

We know that chip thickness ratio is given by

)cos(

sin

α−φφ

==γc

uc

t

t

If, γ = 1,

then)(cos

sin

α−φ

φ=1

∴ . . . (C) sin cos ( )φ = φ − α

By comparing Eqs. (A) and (C), we find that both are the same. Hence, it is proved

that shear strain will be minimum only when the chip thickness ratio is unity.

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22

Theory of Metal Cutting Example 1.2

In orthogonal turning operation with +10° back rake angle tool, the followingobservations were made: cutting speed =160 m/min, width of cut = 2.5 mm,

F c = 180 kgf, F t = 50 kgf, deformed chip thickness = 0.27 mm, tool chip contact

length = 0.63 mm and feed rate = 0.20 mm/rev.

Determine the following : chip thickness ratio, shear angle, friction angle, resultant

force, shear force and shear strain.Solution

(i) Chip thickness ratio, r c =c

u

t

t =

27.0

20.0= 0.74

r c = 0.74

(ii) Shear angle, φ = tan-1

α−

α

sin1

cos

c

c

r

r

= tan-1

− 10sin74.0110cos74.0

= 39.94o

(iii) Friction angle, β = tan-1µ = tan-1

N

F

α−α+

=tan

tan

t c

ct

F F

F F

N

F

=10tan50180

10tan18050

−

+

= 0.477

β = tan-1 (0.477)

= 25.52o

(iv) R =)cos( α−β

c F

=)1052.25cos(

180

−

R = 186.81 kg

(v) F S = R cos (φ + β − α)

= 186.8 cos (39.9 + 25.5 −10)

F S = 106.07 kg

(vi) Shear strain = tan (φ − α) + cot φ

= tan 29.9° + cot 39.9°

= 0.575 +1.196

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23

Metal Cutting and

Chip Formation

γ

= 1.771

Example 1.3

A cylindrical bar has a blind hole of 15 mm diameter. Its face is being turned

(facing operation) from inner diameter to the outer periphery (Figure given below)at a speed of 600 RPM, feed = 0.20 mm/rev., and depth of cut =1.0 mm. Calculate

the cutting speed (m/s) and total volume removed at the end of 15 s.

Solution

Arrow in the figure shows the tool movement.

Revolution/second ( N s) = 600/60 = 10

To find,

(i)

V 15 = cutting speed at the end of 15 seconds of facing operation.

(ii) 15V = volume of material removed at the end of 15 seconds.

(i) V t =1000

t D Ν sπ

where, N s t = 15×10=150 rev.(# of revolutions made by the the workat the end of 15s)

V t = cutting speed at time 't '

D = d + 2f N s t (Figure above)

D = Diameter of the workpiece at which the tool tip will be after thetime of machining =15s. In one revolution of the workpiece, thediameter at which the tool will be cutting, will increase by 2 f . (or inone revolution the diameter to which the tool tip reaches is increased

by 2 f ).

where, f → feed rate

∴ 15 15 (2 0.20 10 15) D = + × × ×

= 75 mm

∴ 1575 10

1000

V π × ×=

V 15 = 2.36 m/s

(ii) Volume of the material removed in 15s.

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24

Theory of Metal Cutting15V = total area machined × depth of cut

=4

π ( D

2 − d 2) × 1 mm3

=4

π(75

2 −152) × 1

15V = 4241 mm3

Example 1.4

During orthogonal turning of a pipe of 100 mm diameter, the rake angle of the toolwas 20

o. The ratio of the cutting force to feed force was 3.0.The feed rate, depth of

cut and chip thickness ratio were 0.275, 0.687 and 0.4 respectively. With the helpof a dynamometer, feed force was measured as 460 N. Workpiece was rotating at450 revolution per minute. Determine chip velocity, shear strain, shear strain rate

and mean width of PSDZ.

Solution

We know from Eq.(1.12) that

φ=

α−=

φ−α+ sin)90sin()90sin( f sc

V V V

∴ V f = V c)90(sin

sin

φ−α+φ

. . . (A)

V s = V c)90(sin

)90(sin

φ−α+α−

. . . (B)

But, we do not know the values of φ and V c. They can be evaluated as follows :

tan φ =α−

αsin1

cos

c

c

r

r

=20sin4.01

20cos4.0

−

= 0.436

φ = tan-10.436

= 23.54o . . . (C)

V c = 1000

450100

1000

××π

=

π DN

V c =141.37 m/min . . . (D)

Substitute the values of φ and V c in Eqs.(A) and (B).

V f =)54.232090(sin

54.23sin37.141

−+×

V f = 56.56 m/min

V s =)53.232090(sin

)2090sin(37.

−+−×141

V s = 133.11 m/min

We also know,

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25

Metal Cutting and

Chip Formation γ = tan (φ − α) + cot φ

= tan (23.54 − 20) + cot (23.54)

∴ γ = 2.357

)(cot.

cos

α−φα

=γ•

ds

V c . . . (E)

Here, we do not know the value of ds. Using Lee and Shaffer’s theory, ds can bederived as [Jain and Pandey, 1980]

ds =)45(sin

)90(sin

sin22

1

φ+α−φ−α+

φ f

=)54.252045(sin

)54.252090(sin

)54.25(sin

275.0.

22

1

+−−+

ds ≈ 0.324 mm

Therefore, from Eq. (E)

•

γ =324.0

100037.141 × ×)2053.23(cos

20cos

−

•

γ = 6830 s-1

Note that the shear strain rate in metal cutting is very high as compared to the oneobtained in classical deformation test.

Example 1.5

Prove that the specific cutting pressure in an ideal orthogonal cutting is given by

τ cot φ, provided 2 φ + β − α = π/2 holds good (τ shear stress).→

Solution

Specific cutting pressure = c

u

F

bt . . . (A)

From Eq. (1.12),

F c =φτ

sin

bt u .)cos(

)cos(

α−β+φα−β

. . . (B)

It is given that,

(β − α) + 2 φ = π/2 . . . (C)

Substitute the value of (β − α) from (C) in (B),

F c =)22/(cos

)22/(cos.

sin φ−π+φφ−π

φ

τbt u

=φτ

sin

bt u .φφ

sin

2sin

Sp. cutting press =φ

τ

sin

bt u .φφ

sin

2sin.

ubt

1

= τφφφφ

sinsin

cossin2

Sp. cutting press = 2 τ cot φ proved

Example 1.6

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26

Theory of Metal Cutting Following data were recorded during orthogonal machining :

Bar diameter = 40 mm, depth of cut = 0.125 mm, length of chip obtained= 62.5 mm/rev, horizontal cutting force = 220 kgf, vertical cutting force = 85 kgf,

α = 7ο, spindle speed = 500 RPM.

Find out friction angle, chip thickness ratio, shear angle, chip velocity and shearvelocity.

Solution

Chip thickness ratio =c

u

t

t =

uu

c

l l

l 50.62= . . . (A)

We know, undeformed chip length

l u = π D N r

= π × 40 × 1

= 120.66 mm

From (A), r c = 62.5/120.66 = 0.479

r c = 0.479

From Eq. (1.1),

φ = tan−1

α−

α

sin1

cos

c

c

r

r

Substitute the values,

φ = tan−1 526.0939.0

494.0≈

= 27.7ο

β =α−

α+

tan

tan

t c

ct

F F

F F (Eq. 1.7)

Substitute the values in the above equation

β = 53.07tan85220

7tan22085 o≈

−

+ο

= 28.12ο

Cutting velocity, V c =1000 DN π

=1000

50040 ××π

V c = 62.83 m/min

V f = V c)cos(

sin

α−φφ

= 62.83 × 935.0

465.0

V f = 31.22 m/min.

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27

Metal Cutting and

Chip FormationV s = V c × )90(sin

)90(sin

φ−α+α−

)7.27790(sin

)790(sin83.62

−+−

×=

V s = 66.66 m/min.

SAQ 1

Write the most appropriate option from the given ones

(i) In actual practice, chip thickness ratio is

(a) > 1, (b) < 1, (c) = 1.

(ii) In oblique cutting, the number of forces that act on the tool are

(a) one, (b) two, (c) three, (d) none of these.

(iii) Which of the following is the chip removal process?

(a) rolling, (b) extruding, (c) die casting, (d) broaching, (e) none of these.(iv) Time taken to drill a hole through a 2.5 cm thick plate at 3000 RPM at a

feed rate 0.025 mm/rev. will be

(a) 20 s, (b) 10 s, (c) 40 s, (d) 50 s.

(v) Shear plane angle is the angle between

(a) shear plane and the cutting velocity vector, (b) shear plane and tool face,

(c) shear plane and horizontal plane, (d) rake face and vertical plane.

(vi) In orthogonal cutting, the cutting edge should be

(a) straight, (b) parallel to the original plane surface of the workpiece,

(c) normal to the direction of cutting, (d) all of these, (e) none of these.(vii) Continuous chip with BUE

(a) yields good surface finish, (b) yields poor surface finish, (c) has no effecton surface roughness.

(viii) The ratio of cutting velocity to chip velocity is usually

(a) >1, (b)

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Theory of Metal Cutting 1.9 KEY WORDS

Chip : It is the material which is separated from theworkpiece when the tool moves into theworkpiece.

Primary Shear Deformation : Finite zone (thin or thick depending upon

Zone (PSDZ) various governing parameters) within which

deformation takes place.

Secondary Shear Deformation : Deformation which takes place at tool-chip

Zone (SSDZ) interface.

Orthogonal Cutting : Two dimensional (2-D) cutting in which cuttingedge is straight, parallel to the original plane

surface of the workpiece and perpendicular to thedirection of cutting.

Oblique Cutting : Cutting operations are 3-D in nature. In this type

cutting edge at the tool is inclined to the linenormal to the cutting direction.

1.10 ANSWERS TO SAQs

SAQ 1

(i) (b)

(ii) (c)

(iii) (d)

(iv) (a)

(v) (a)

(vi) (d)

(vii) (a)

(viii) (b)

EXERCISES

Q 1. (i) Derive a relationship to calculate shear angle in terms of measurable/known parameters.

(ii) Draw force circle diagram proposed by Merchant for orthogonal cutting

conditions showing different forces acting on tool, chip, and work system.From the diagram, derive the expression for

(a) shearing force on the shear plane,

(b) friction force on the tool face in terms of cutting force, thrust force,rake angle, and shear angle.

(iii) Define orthogonal cutting. Draw Merchant's force circle diagram for theorthogonal cutting.

(iv) Using the Figure in Q.1 (ii) (a), derive the expression for friction force.

What are the factors which affect the formation of different types of chip

obtained in cutting.

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Metal Cutting and

Chip Formation(v) Determine the condition when

φ = tan− 1 (r c)

where, φ is the shear angle and r c is the chip thickness ratio.

(Hint : in the original Eq., Substitute tan φ = r c) (Ans. α = 0)

(vi) Determine the condition for which chip flow velocity is equal to the cutting

velocity, assuming α = 0. (Ans. φ = 45o)

(vii) Find the ratio of F c / F t for an imaginary case of machining if α = β = π/4.

Q 2. Mild steel rod is being turned at the speed of 27.3 m/min. Feed rate used is

0.25 mm/rev, and deformed chip thickness is equal to 0.30 mm. Rake angle andshear angle of the tool are 20

o and 30

o, respectively. Calculate the shear flow

velocity.

Q 3. For orthogonal cutting of a M.S. rod, the following data are obtained : width of

cut = 0.125'', feed = 0.007'' per rev., α = 15o, β = 30o, and machining constant,C = 70

o.The dynamic shear strength of the work material = 80000 lb/in

2.

Calculate Fc and Ft .

Q 4. During orthogonal cutting of a tube at 100 m/min, the tangential force(in the direction of cutting velocity) measured by the 3-D dynamometer is

200 kgf, and the axial force is 100 kgf. Assume the rake angle as 10o. Calculate the

work in shearing the metal if the shear angle = 30o. Also, derive the velocity

relationship used.

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BIBLIOGRAPHY

Armarego, E. J. A. and Brown, R. H. (1969), The Machining of Metals, Prentice Hall,

Englewood Cliffs, NJ.

Jain, V. K. and Pandey, P. C. (1980), An Analytical Approach to the Determination of Mean Width of Primary Shear Deformation Zone (PSDZ) in Orthogonal Machining ,

Proc. 4th

International Conference on Production Engineering, Tokyo, pp 434-438.

Kalpakjian, S. (1989), Manufacturing Engineering and Technology, Addison-WesleyPublishing Co., New York.

Pandey, P. C. and Sing, C. K. (1998), Production Engineering Science, StandardPublishers Distributors, Delhi.

Rao, P. N. (2000), Manufacturing Technology : Metal Cutting and Machine Tools, Tata

McGraw-Hill Publishing Co. Ltd., New Delhi.

Shaw, M. C. (1984), Metal Cutting Principles, Oxford, Clarendon Press.

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