unit 1 metal cutting and chip formation - IGNOU

UNIT 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 Relationships 

1.6 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 (nontraditional 

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. 

Metal Cutting and 

Chip Formation 

5

Theory of Metal Cutting 

6 

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’ (Vc). 

In case of single point turning, Vc 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.

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 

Boring on a 

lathe 

Drilling on a 

drill machine 

Rotary motion of the 

work 

Axial movement of the tool 

Work rotation Axial tool movement 

Fixed Rotations as well as 

translatory feed 

Planning Translatory Intermittent Translation 

Milling Translatory Rotation 

Grinding Rotary/Translatory Rotary 



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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,

Type 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 BUE 

enhances the tool life (or tool cuts longer before regrind). 



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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. 

1.3.3 Types of Cutting 

(a) Tear Type (b) Shear Type 

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

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 oblique 

cutting, 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

operation 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 as 

the 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 of 

orthogonal 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 Vc (velocity of tool with respect to workpiece) to Vf 

(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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(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 tu (equal to feed in turning) is deformed to give chip 

thickness tc which experiences two velocities Vf (chip sliding velocity) and Vs (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 tu, tc 

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 = tu / sin φ 

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

∴ 

t 

t 

u 

c 

sin φ 

= 

cos ( φ − α)

tu 

is called chip thickness ratio or chip thickness coefficient (rc) which can be written 

tc 

as 

1 cosφ 

cosα 

+ sin φsin 

α 

= 

sin φ 

rc 

or, 1 = rc cot φcos 

α + rc 

sin α 

or, 

∴ 

rc 

cos α = ( 1− 

rc 

sin α) 

tan φ 

⎛ r cos 

tan c α ⎞ 

φ=⎜ ⎟ . . . (1.1) 

⎝1− r c sin α⎠ 

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

⎛ t ⎞ 

ratio 

⎜ u r c = 

⎟ , should be known. But to determine tc with a micrometer is somewhat 

⎝ tc 

⎠ 

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 : 

∴ Lctc= Lutu or, 

bLc tc 

= Lutub 

(volume constancy condition) 

t u L 

= 

c 

tc 

Lu 

tu L 

∴ rc 

= = 

t L 

c . . . (1.2) 

c u 

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 (tu / tc) because of the difficulties and inaccuracies involved in 

the measurement of thickness of the deformed chip (tc). 

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, Fs, is the resistance to shear of the metal in 

forming the chip. Fs acts along the shear plane. Force, Fn, 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 Fs 

and Fn 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 approximate 

analysis. 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 Fs and Fn is a right angle, the intersection of these forces lies on 

the 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 Fc and vertical force Ft 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 Fc and Ft 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 laid 

off, and the forces F and N can then be determined. The shear plane angle φ can be 

measured approximately from a photomicrograph or by measuring tc and tu, or length of 

chip and corresponding length of unmachined chip (discussed elsewhere). 

From Figure 1.11, the following vector Eqs. can be written 

ρ ρ ρ 

R' 

= F + N 

ρ ρ ρ ρ ρ ρ 

R = Fs 

+ Fn 

= Fc 

+ Ft 

= R' 

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 Fc, Ft, 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 Fs 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

F = Ft 

cosα + Fc 

sin α 

. . . (1.3) Metal Cutting and 


and, N = Fc 

cos α − Ft 

sin α 

. . . (1.4) 

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

μ = 

where, β is the friction angle. 

or, 

We can also write : 

μ = 

β = 

β = 

From Figure 1.11, we get : 

Also, from Figure 1.11, 

or, 

F 

∴ 

F 

tanβ 

= 

F 

N 

Ft 

+ Fc 

tan α 

Fc 

− Ft 

tan α 

tan 

−1 

( μ) 

F α + α 

= 

t cos Fc 

sin 

Fc 

cosα 

− Ft 

sin α 

. . . (1.5) 

. . . (1.6) 

− ⎧ + tan α⎫ 

tan 

1 F t F 

. . . (1.7) 

c 

⎨ 

⎬ 

⎩ Fc 

− Ft 

tan α ⎭ 

F s = Fc 

cos φ − Ft 

sin φ 

. . . (1.8) 

F n = Ft 

cos φ + Fc 

sin φ 

F n = Fs 

tan( φ + β − α) 

. . . (1.9) 

F 

c 

s 

c 

s 

= R 

cos(β 

− α) 

F = R cos( 

φ+ 

β − α) 

F 

Shear plane area is equal to : 

c 

cos( 

β −α) 

= 

cos( 

φ + β − α) 

cos( 

β− 

α) 

= Fs 

cos( 

φ + β − α) 

. . . (1.9(a)) 

t b 

A = 

u 

s . . . (1.10) 

sin φ 

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

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

hence, 

t b 

F = 

u 

s τ 

. . . (1.11) 

sin φ 

⎛ t b 

F 

u τ ⎞ cos( β − α) 

c = ⎜ ⎟ 

. . . (1.12) 

⎝ sin φ ⎠ cos( φ + β − α) 

t bτ 

R = 

u 

. . . (1.12(a)) 

sin φcos( 

φ + β − α) 

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From Figure 1.11, 

Ft 

= Rsin( 

β − α) 

t τ 

= ub 

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

sin ( β − α) 

sin φcos 

( φ + β − α) 

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

Ft 

= tan ( β − α) 

. . . (1.14) 

Fc 

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

coefficient of friction can be calculated provided Fc, Ft, α, tu and tc 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., Fs and Fn. Shear stress (τ) can be found as 

F ( cosφ 

− sin φ) 

Mean shear stress ( τ) = 

s F 

= 

c Ft 

sin φ 

As 

btu 

Mean Normal stress (σ) 

where, As = Shear plane area = b tu 

/ sin φ . 

1.5 VELOCITY RELATIONSHIPS 

. . . (1.15) 

Fn sin φ 

= = ( Ft 

cosφ 

+ Fc 

sin φ) 

. . . (1.16) 

As 

btu 

Since the chip is thicker than the uncut chip, the velocity of the chip as it moves along the 

tool 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 Vc 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 Vf (chip flow velocity). The newly cut chip elements 

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

velocity). From the principle of kinematics that the relative velocity of two bodies (here 

tool 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 Δ ABC, we get : 

or, 

V V 

c f Vs 

= = 

. . . (1.17) 

sin ( 90 − ( φ − α)) 

sin φ sin ( 90 − α) 

V V 

c 

f Vs 

= = 

cos( φ − α) 

sin φ cosα 

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

Vc 

sin φ 

= 

cos( 

φ − α) 

V f 

=Vc . rc . . . (1.18) 

whereas the shear velocity Vs is obtained as :

V cosα 

V = 

c 

s 

. . . (1.19) 

cos( 

φ − α) 

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 complicated 

cutting process involving oblique cutting. Due to simplicity and fairly wide applications, 

the continuous chip without BUE has been most extensively studied. There are 

conflicting evidences about the nature of the deformation zone in metal cutting. This has 

led 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 their 

analysis 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 speeds 

most evidences indicate that a thin shear plane is approached. Thin zone model is more 

useful 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 the 

following 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. 

Δs 

γ = = 

Δy 

AB 

CD 

= 

AD 

CD 

+ 

DB 

CD 

= tan ( φ − α) 

+ cot φ 

. . . (1.20) 

sin ( φ − α) 

cosφ 

= + 

cos( 

φ − α) 

sin φ 

sin ( φ − α) 

sin φ + cosφ 

cos( 

φ − α) 

= 

sin φcos( 

φ − α) 



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18 

or, 

cos( 

φ+ 

α −φ) 

cosα 

= 

= 

sin φcos( 

φ − α) 

sin φcos( 

φ − α) 

cosα 

γ = 

. . . (1.21) 

sin φcos( 

φ − α) 

Figure 1.13 : Strain and Strain Rate in Orthogonal Cutting 

Strain may also be expressed in terms of the shear velocity (Vs) and the chip velocity (Vf) 

∴ 

γ= 

V 

Vs 

(from Eq.(1.19)) 

sin φ 

c 

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

⎛ Δs 

⎞ 

⎜ ⎟ 

dγ 

⎝ Δy 

⎠ ⎛ Δs 

⎞ 1 

γ& 

= = = ⎜ ⎟ 

dt Δt 

⎝ Δt 

⎠ Δy 

V V 

γ& s c cosα 

= = 

Δy 

cos( 

φ − α) 

Δy 

Δy is mean thickness of PSDZ . 

1.7 SHEAR ANGLE RELATIONSHIPS 

. . . (1.22) 

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 (Fc) 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 

the planes, however, the shear stress will be greater than on any other plane, and as Fc is 

further 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 Fc 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( 

β − α) 

Fc 

= Fs 

cos 

τ tub 

= 

cos( 

β − α) 

{ φ + ( β − α) 

} sin φ cos( 

φ + β − α) 

× 

(Eq.1.12) 

Here, except φ all other parameters can be taken as constant during machining (assuming 

that no strain hardening takes place). It would give the condition for the minimum energy 

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

Therefore, 

d Fc 

d φ 

= τ 

t 

u 

⎡cosφ 

cos( 

φ+ 

β − α) 

−sin 

φ( 

φ+ 

β − α) 

⎤ 

b cos( 

β − α) 

⎢ 

= 0 

2 2 

⎥ 

⎢⎣ 

sin φcos 

( φ+ 

β − α) 

⎥⎦ 

cosφ cos ( φ +β−α) −sin φsin ( φ+β−α ) = 0 

or, cos (2 φ +β−α ) = 0 

or, (2 φ+β−α ) = 

π 

2 

π 1 

Hence, φ = − ( β − α) 

. . . (1.23) 

4 2 

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 and 

the 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 

1 

φ = − ( β− 

α) 

2 2 

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). 

π 

φ = + ( α − β) 

. . . (1.25) 

4 

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 for 



19


20 

the shear angle (φ) which included an additional angle θ, which depends on the size of 

the built up edge, 

π 

φ = + ( α − β) 

+ θ 

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 : 

π 

φ = + ( α −β ) + η ′ 

4 

While deriving the above relation, they assumed that the shear plane is not a plane of 

maximum 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, 

(iv) temperature effect, 

(v) diffusion, 

(vi) oxidation, and 

(vii) local heating, etc. 

Example 1.1 

Solution 

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’. 

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

Here, Δs is in the direction of force, Δy is in the direction ⊥ to the force. 

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

Δ s 

shear strain is defined as and hence in cutting (Figure 1.13), 

Δy 

Δs 

AB AD DB 

γ= = = + = tan ( φ−α ) + cot φ 

Δy 

CD CD CD 

We want the condition when γ should be minimum. Hence, differentiate γ with 

respect to φ and equate the derivative equal to zero. 

dγ 

d 

∴ = 

dφ 

dφ 

sec 

2 

∴sec 

{ tan ( φ − α) 

+ cot φ} 

= 0 

( φ− 

α) 

+ ( −cosec 

2 

( φ − α) 

= cos ec 

2 2 

or, ∴sin 

φ = cos ( φ− 

α) 

. 

Take the under root to both sides, 

∴ ± 

sin φ = ± cos( 

φ − α) 

2 

2 

φ) 

= 0 

φ 

or, sin φ = cos( 

φ − α) 

. . . (A) 

= cos φ cosα 

+ sin φ sin α 

∴1 = cosα 

cot φ + sin α 

. . . (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 

If, γ = 1, 

then 

t sin φ 

γ = 

u 

c = 

tc 


sin φ 

1 = 

cos( 

φ − α) 

∴ sin φ= cos ( φ−α) 

. . . (C) 

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. 



21


22 

Example 1.2 

Solution 

In orthogonal turning operation with +10° back rake angle tool, the following 

observations were made: cutting speed =160 m/min, width of cut = 2.5 mm, 

Fc = 180 kgf, Ft = 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. 

(i) Chip thickness ratio, rc = 

t u 0. 

20 

= = 0.74 

tc 

0. 

27 

rc = 0.74 

(ii) Shear angle, φ = tan -1 ⎡ rc 

cosα 

⎤ 

⎢ ⎥ 

⎣1− 

rc 

sin α ⎦ 

= tan -1 ⎡ 0. 

74 cos 10 ⎤ 

⎢ 

⎥ 

⎣1− 

0. 

74sin 

10⎦ 

φ = 39.94 o 

(iii) Friction angle, β = tan -1 

⎛ F ⎞ 

-1 

μ = tan 

⎜ ⎟ 

⎝ N ⎠ 

(iv) R = 

= 

F 

N 

F + α 

= 

t Fc 

tan 

Fc 

− Ft 

tan α 

= 

Fc 

cos( β − α) 

50 + 180 tan10 

180 − 50 tan10 

= 0.477 

β = tan -1 (0.477) 

β = 25.52 o 

180 

cos( 25. 

52 −10) 

R = 186.81 kg 

(v) FS = R cos (φ + β − α) 

= 186.8 cos (39.9 + 25.5 −10) 

FS = 106.07 kg 

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

= tan 29.9° + cot 39.9° 

= 0.575 +1.196

Example 1.3 

Solution 

γ = 1.771 

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 15s. 

Arrow in the figure shows the tool movement. 

To find, 

Revolution/second (Ns) = 600/60 = 10 

(i) V15 = cutting speed at the end of 15 seconds of facing operation. 

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

πDΝ 

t 

(i) Vt = 

s 

1000 

where, Ns t = 15 ×10=150 rev.(# of revolutions made by the the work 

at the end of 15s) 

Vt = cutting speed at time 't' 

D = d + 2f Ns t (Figure above) 

D = Diameter of the workpiece at which the tool tip will be after the 

time of machining =15s. In one revolution of the workpiece, the 

diameter at which the tool will be cutting, will increase by 2f. (or in 

one revolution the diameter to which the tool tip reaches is increased 

by 2f). 

where, f → feed rate 

∴ D 15 = 15 + (2 × 0.20 × 10 × 15) 

V 

∴ 15 

= 75 mm 

= 

π × 75 × 10 

1000 

V15 = 2.36 m/s 

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



23


24 

Example 1.4 

Solution 

V 15 = total area machined × depth of cut 

π 2 2 3 

= (D − d ) × 1 mm 

4 

π 2 2 

= (75 −15 ) × 1 

4 

V 15 = 4241 mm 3 

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

was 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 help 

of a dynamometer, feed force was measured as 460 N. Workpiece was rotating at 

450 revolution per minute. Determine chip velocity, shear strain, shear strain rate 

and mean width of PSDZ. 

We know from Eq.(1.12) that 

Vc 

V V 

s 

f 

= 

= 

sin( 90 + α − φ) 

sin( 90 − α) 

sin φ 

∴ Vf = Vc 

Vs = Vc 

sin φ 

sin ( 90 + α − φ) 

sin ( 90 − α) 

sin ( 90 + α − φ) 

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

r 

tan φ = c cosα 

1− 

rc 

sin α 

0. 

4cos20 

= 

1− 

0. 

4sin 

20 

= 0.436 

φ = tan -1 0.436 

. . . (A) 

. . . (B) 

φ = 23.54 o . . . (C) 

Vc = 

πDN π× 

100 × 450 

= 

1000 1000 

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

We also know, 

Vc =141.37 m/min . . . (D) 

141. 

37 × sin 23. 

54 

Vf = 

sin ( 90 + 20 − 23. 

54) 

Vf = 56.56 m/min 

141. 

37 × sin( 90 − 20) 

Vs = 

sin ( 90 + 20 − 23. 

53) 

Vs = 133.11 m/min

γ = tan (φ − α) + cot φ 

= tan (23.54 − 20) + cot (23.54) 

∴ γ = 2.357 

γ • 

Vc cosα 

= 

. . . (E) 

ds. 

cot ( φ − α) 

Here, we do not know the value of ds. Using Lee and Shaffer’s theory, ds can be 

derived as [Jain and Pandey, 1980] 

Therefore, from Eq. (E) 

1 f sin ( 90 + α − φ) 

ds = 

2 2 sin φ sin ( 45 − α + φ) 

= 

1 0. 

275 sin ( 90 + 20 − 25. 

54) 

. 

2 2 sin ( 25. 

54) 

sin ( 45 − 20 + 25. 

54) 

ds ≈ 0.324 mm 

• 141 . 37× 

1000 cos20 

γ = 

× 

0. 

324 cos( 

23. 

53 − 20) 

• 

γ = 6830 s -1 

Note that the shear strain rate in metal cutting is very high as compared to the one 

obtained in classical deformation test. 

Example 1.5 

Solution 

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). 

→ 

Specific cutting pressure = c F 

btu 

From Eq. (1.12), 

t τ 

Fc = u b cos( β − α) 

. 

sin φ cos( φ + β − α) 

It is given that, 

. . . (A) 

. . . (B) 

(β − α) + 2 φ = π/2 . . . (C) 

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

tu bτ 

cos( 

π/ 

2 − 2φ) 

Fc = . 

sin φ cos( 

φ + π/ 

2 − 2φ) 

t τ 

= u b sin 2φ 

. 

sin φ sin φ 

t τ 

Sp. cutting press = u b sin 2φ 

. . 

sin φ sin φ 

2sin 

φcosφ 

= τ 

sin φsin 

φ 

1 

btu 

Sp. cutting press = 2 τ cot φ proved 

Example 1.6 



25


26 

Solution 

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 = 7 ο , spindle speed = 500 RPM. 

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

velocity. 

t 

Chip thickness ratio = 

u l 

= 

c 62. 

50 

= . . . (A) 

tc 

lu 

lu 

We know, undeformed chip length 

lu = π D Nr 

= π × 40 × 1 

= 120.66 mm 

From (A), rc = 62.5/120.66 = 0.479 

From Eq. (1.1), 

Substitute the values, 

rc = 0.479 

φ = tan −1 ⎡ rc 

cosα 

⎤ 

⎢ ⎥ 

⎣1− 

rc 

sin α ⎦ 

φ = tan −1 ⎡0. 

494⎤ 

⎢ ≈ 

0. 

939 

⎥ 

⎣ ⎦ 

β = 

φ = 27.7 ο 

Ft 

+ Fc 

tan α 

Fc 

− Ft 

tan α 

Substitute the values in the above equation 

πDN 

Cutting velocity, Vc = 

1000 

= 

0. 

526 

(Eq. 1.7) 

85 + 220 tan 7 

β = ≈ 0. 

53 

ο 

220 − 85tan 

7 

β = 28.12 ο 

π × 40 × 500 

1000 

Vc = 62.83 m/min 

Vf = Vc 

sin φ 


0. 

465 

= 62.83 × 

0. 

935 

Vf = 31.22 m/min. 

o

SAQ 1 

Vs = Vc × 

= 

62. 

83 

sin ( 90 − α) 

sin ( 90 + α − φ) 

Vs = 66.66 m/min. 

sin ( 90 − 7) 

× 

sin ( 90 + 7 − 27. 

7) 

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 effect 

on surface roughness. 

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

1.8 SUMMARY 

(a) >1, (b)


28 

1.9 KEY WORDS 

Chip : It is the material which is separated from the 

workpiece when the tool moves into the 

workpiece. 

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 cutting 

edge is straight, parallel to the original plane 

surface of the workpiece and perpendicular to the 

direction of cutting. 

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

cutting edge at the tool is inclined to the line 

normal 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 the 

orthogonal 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.

(v) Determine the condition when 

φ = tan − 1 (rc) 

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

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

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

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

(vii) Find the ratio of Fc / Ft 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 and 

shear 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 = 15 o , β = 30 o , 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 10 o . Calculate the 

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

relationship used. 



29


30 

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

Kalpakjian, S. (1989), Manufacturing Engineering and Technology, Addison-Wesley 

Publishing Co., New York. 

Pandey, P. C. and Sing, C. K. (1998), Production Engineering Science, Standard 

Publishers 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.

THEORY OF METAL CUTTING 

This block comprises three units. In this block, we are going to discuss about the 

fundamentals of traditional material removal processes. Non-conventional material 

removal processes are discussed in the last block. 

In Unit 1, traditional material removal processes have been classified. In this unit, basics 

of chip formation has been discussed in details. This unit also discusses force analysis, 

velocity relationships, and shear strain. 

Unit 2 gives the thermal aspects in metal cutting. During metal cutting large amount of 

heat is produced which leads to high temperature resulting into high tool wear rate. In 

this unit, measurement of temperature of tool, chip and workpiece has been studied in 

detail. 

Finally, Unit 3 deals with tool wear, tool life, machinability and tool material. In this 

unit, different types of wear and criterion to evaluate tool life are discussed. 

Machinability and methods to evaluate machinability have also been described. 



31


32 

MACHINING TECHNOLOGY 

Manufacturing Technology course provides learners with the sound knowledge of 

conventional as well as non-conventional (advanced) manufacturing techniques. In 

Manufacturing Technology, you have already studied the principles of metal casting, 

metal forming, principles of metal cutting and welding technology. In Machining 

Technology, you are going to be introduced with other manufacturing processes. The 

main aim while manufacturing a component is to achieve the desired component with the 

required quality in the most economical manner. This objective can be achieved only 

when the manufacturer has acquired sufficient knowledge of manufacturing concepts like 

principles of metal cutting, machining operations, abrasive machining processes and nonconventional 

material removal processes. This course on Machining Technology 

comprises four blocks. 

Block 1 (Theory of Metal Cutting) comprises 3 units. In this block, fundamentals of 

traditional material removal processes have been introduced. In Unit 1, concepts like 

material removal processes, chip formation and force analysis have been discussed. 

Unit 2 elaborates the thermal aspects of metal cutting processes. It gives a complete 

understanding of thermal phenomenon in machining. It deals with theoretical and 

experimental methods to evaluate temperature distribution in the tool, work and 

tool-work interface. In Unit 3, tool wear, tool life, machinability and tool materials have 

been discussed. 

The emphasis in Block 2 is on abrasive machining processes. Surface integrity in general, 

and surface finish in particular of a finished component is very important. Abrasive 

finishing and its working principles are well explained in Unit 4. This unit also deals with 

the specifications of grinding wheels and its selection for the given workpiece 

requirements. Various kinds of grinding operations (face grinding, surface grinding, 

cut-off grinding and others) and the mechanics of grinding are elaborated in Unit 5. Apart 

from grinding there are various other finishing processes like lapping, honing, and superfinishing. 

The characteristics and working principles of these processes are dealt with in 

Unit 6. 

Block 3 has three units which discusses various surface finishing processes. Unit 7 deals 

with the surface characteristics and product performance. In this unit, surface finish 

parameters, their importance and different types of surfaces produced by machining have 

been discussed. Unit 8 describes the conventional surface finishing processes like honing, 

lapping, deburring etc. in detail. Last unit, i.e. Unit 9, discusses the 

micro-machining processes like ion-beam machining, electron beam machining etc. in 

detail. 

Block 4 deals with non-conventional (or advanced) material removal processes. It 

comprises four units. In the first unit of this block, these processes have been introduced 

and classified in three major categories. Unit 11 of this block deals with mechanical 

advanced machining processes, e.g. abrasive jet machining (AJM), ultrasonic machining 

(USM) and abrasive flow machining (AFM). Unit 12 introduces thermal processes like 

laser beam machining (LBM), electric discharge machining (EDM) and allied processes. 

In the last unit of this block, i.e. Unit 13, electrochemical machining (ECM) method has 

been elaborated.

unit 1 metal cutting and chip formation - IGNOU

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