fundamentals of modern manufacturing 4th edition by mikell p groover 2 851

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Part VI Material Removal
Processes

21

THEORY OF METAL
MACHINING

Chapter Contents
21.1 Overview of Machining Technology
21.2 Theory of Chip Formation in Metal Machining
21.2.1 The Orthogonal Cutting Model
21.2.2 Actual Chip Formation
21.3 Force Relationships and the Merchant
Equation
21.3.1 Forces in Metal Cutting
21.3.2 The Merchant Equation
21.4 Power and Energy Relationships in Machining
21.5 Cutting Temperature
21.5.1 Analytical Methods to Compute
Cutting Temperatures
21.5.2 Measurement of Cutting Temperature

The material removal processes are a family of shaping
operations (Figure 1.4) in which excess material is removed
from a starting workpart so that what remains is the desired
final geometry. The ‘‘family tree’’ is shown in Figure 21.1.
The most important branch of the family is conventional
machining, in which a sharp cutting tool is used to mechanically cut the material to achieve the desired geometry.
The three principal machining processes are turning, drilling, and milling. The ‘‘other machining operations’’ in
Figure 21.1 include shaping, planing, broaching, and sawing. This chapter begins our coverage of machining, which
runs through Chapter 24.
Another group of material removal processes is the
abrasive processes, which mechanically remove material by
the action of hard, abrasive particles. This process group,
which includes grinding, is covered in Chapter 25. The
‘‘other abrasive processes’’ in Figure 21.1 include honing,
lapping, and superfinishing. Finally, there are the nontraditional processes, which use various energy forms other
than a sharp cutting tool or abrasive particles to remove
material. The energy forms include mechanical, electrochemical, thermal, and chemical.1 The nontraditional processes are discussed in Chapter 26.
Machining is a manufacturing process in which a
sharp cutting tool is used to cut away material to leave the
1

Some of the mechanical energy forms in the nontraditional processes
involve the use of abrasive particles, and so they overlap with the
abrasive processes in Chapter 25.

483


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Turning and
related operations
Conventional
machining

Drilling and
related operations
Milling
Other machining
operations

Material removal
processes

Abrasive
processes

Grinding
operations
Other abrasive
processes

Mechanical energy
processes

Nontraditional
machining

Electrochemical
machining
Thermal energy
processes

FIGURE 21.1
Classification of material
removal processes.

Chemical
machining

desired part shape. The predominant cutting action in machining involves shear deformation of the work material to form a chip; as the chip is removed, a new surface is
exposed. Machining is most frequently applied to shape metals. The process is illustrated
in the diagram of Figure 21.2.
Machining is one of the most important manufacturing processes. The Industrial
Revolution and the growth of the manufacturing-based economies of the world can be
traced largely to the development of the various machining operations (Historical Note
22.1). Machining is important commercially and technologically for several reasons:

FIGURE 21.2 (a) A cross-sectional view of the machining process. (b) Tool with negative rake angle; compare with
positive rake angle in (a).

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å Variety of work materials. Machining can be applied to a wide variety of work
materials. Virtually all solid metals can be machined. Plastics and plastic composites
can also be cut by machining. Ceramics pose difficulties because of their high
hardness and brittleness; however, most ceramics can be successfully cut by the
abrasive machining processes discussed in Chapter 25.
å Variety of part shapes and geometric features. Machining can be used to create any
regular geometries, such as flat planes, round holes, and cylinders. By introducing
variations in tool shapes and tool paths, irregular geometries can be created, such as
screw threads and T-slots. By combining several machining operations in sequence,
shapes of almost unlimited complexity and variety can be produced.
å Dimensional accuracy. Machining can produce dimensions to very close tolerances.
Some machining processes can achieve tolerances of Æ0.025 mm (Æ0.001 in), much
more accurate than most other processes.
å Good surface finishes. Machining is capable of creating very smooth surface finishes.
Roughness values less than 0.4 microns (16 m-in.) can be achieved in conventional
machining operations. Some abrasive processes can achieve even better finishes.
On the other hand, certain disadvantages are associated with machining and other
material removal processes:

å Wasteful of material. Machining is inherently wasteful of material. The chips
generated in a machining operation are wasted material. Although these chips
can usually be recycled, they represent waste in terms of the unit operation.
å Time consuming. A machining operation generally takes more time to shape a given
part than alternative shaping processes such as casting or forging.
Machining is generally performed after other manufacturing processes such as
casting or bulk deformation (e.g., forging, bar drawing). The other processes create the
general shape of the starting workpart, and machining provides the final geometry,
dimensions, and finish.

21.1

OVERVIEW OF MACHINING TECHNOLOGY
Machining is not just one process; it is a group of processes. The common feature is the
use of a cutting tool to form a chip that is removed from the workpart. To perform the
operation, relative motion is required between the tool and work. This relative motion is
achieved in most machining operations by means of a primary motion, called the cutting
speed, and a secondary motion, called the feed. The shape of the tool and its penetration
into the work surface, combined with these motions, produces the desired geometry of
the resulting work surface.
Types of Machining Operations There are many kinds of machining operations, each
of which is capable of generating a certain part geometry and surface texture. We discuss
these operations in considerable detail in Chapter 22, but for now it is appropriate to
identify and define the three most common types: turning, drilling, and milling, illustrated
in Figure 21.3.
In turning, a cutting tool with a single cutting edge is used to remove material from a
rotating workpiece to generate a cylindrical shape, as in Figure 21.3(a). The speed motion in
turning is provided by the rotating workpart, and the feed motion is achieved by the cutting
tool moving slowly in a direction parallel to the axis of rotation of the workpiece. Drilling is
used to create a round hole. It is accomplished by a rotating tool that typically has two

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Speed motion (tool)

Work

New surface
Speed motion (work)

Cutting tool

Drill
bit

Feed
motion
(tool)

Feed motion

(tool)
Work

(a)

(b)

Speed motion

Rotation
Milling cutter

FIGURE 21.3 The three
most common types of
machining processes:
(a) turning, (b) drilling, and
two forms of milling:
(c) peripheral milling, and
(d) face milling.

New surface
Feed
motion
(work)

Milling cutter

New surface

Feed motion

(work)
Work

Work
(c)

(d)

cutting edges. The tool is fed in a direction parallel to its axis of rotation into the workpart to
form the round hole, as in Figure 21.3(b). In milling, a rotating tool with multiple cutting
edges is fed slowly across the work material to generate a plane or straight surface. The
direction of the feed motion is perpendicular to the tool’s axis of rotation. The speed motion
is provided by the rotating milling cutter. The two basic forms of milling are peripheral
milling and face milling, as in Figure 21.3(c) and (d).
Other conventional machining operations include shaping, planing, broaching, and
sawing (Section 22.6). Also, grinding and similar abrasive operations are often included
within the category of machining. These processes commonly follow the conventional
machining operations and are used to achieve a superior surface finish on the workpart.
The Cutting Tool A cutting tool has one or more sharp cutting edges and is made of a
material that is harder than the work material. The cutting edge serves to separate a chip
from the parent work material, as in Figure 21.2. Connected to the cutting edge are two
surfaces of the tool: the rake face and the flank. The rake face, which directs the flow of the
newly formed chip, is oriented at a certain angle called the rake angle a. It is measured
relative to a plane perpendicular to the work surface. The rake angle can be positive, as in
Figure 21.2(a), or negative as in (b). The flank of the tool provides a clearance between the
tool and the newly generated work surface, thus protecting the surface from abrasion, which
would degrade the finish. This flank surface is oriented at an angle called the relief angle.
Most cutting tools in practice have more complex geometries than those in Figure 21.2.
There are two basic types, examples of which are illustrated in Figure 21.4: (a) single-point
tools and (b) multiple-cutting-edge tools. A single-point tool has one cutting edge and is used

for operations such as turning. In addition to the tool features shown in Figure 21.2, there is
one tool point from which the name of this cutting tool is derived. During machining, the
point of the tool penetrates below the original work surface of the part. The point is usually
rounded to a certain radius, called the nose radius. Multiple-cutting-edge tools have more


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FIGURE 21.4 (a) A single-point tool showing rake face, flank, and tool point; and (b) a helical milling cutter, representative
of tools with multiple cutting edges.

than one cutting edge and usually achieve their motion relative to the workpart by rotating.
Drilling and milling use rotating multiple-cutting-edge tools. Figure 21.4(b) shows a helical
milling cutter used in peripheral milling. Although the shape is quite different from a singlepoint tool, many elements of tool geometry are similar. Single-point and multiple-cuttingedge tools and the materials used in them are discussed in more detail in Chapter 23.
Cutting Conditions Relative motion is required between the tool and work to perform
a machining operation. The primary motion is accomplished at a certain cutting speed v.
In addition, the tool must be moved laterally across the work. This is a much slower
motion, called the feed f. The remaining dimension of the cut is the penetration of the
cutting tool below the original work surface, called the depth of cut d. Collectively, speed,
feed, and depth of cut are called the cutting conditions. They form the three dimensions

of the machining process, and for certain operations (e.g., most single-point tool
operations) they can be used to calculate the material removal rate for the process:
RMR ¼ vf d

ð21:1Þ

where RMR ¼ material removal rate, mm3/s (in3/min); v ¼ cutting speed, m/s (ft/min), which
must be converted to mm/s (in/min); f ¼ feed, mm (in); and d ¼ depth of cut, mm (in).
The cutting conditions for a turning operation are depicted in Figure 21.5. Typical
units used for cutting speed are m/s (ft/min). Feed in turning is expressed in mm/rev

FIGURE 21.5 Cutting
speed, feed, and depth of
cut for a turning operation.


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(in/rev), and depth of cut is expressed in mm (in). In other machining operations,
interpretations of the cutting conditions may differ. For example, in a drilling operation,

depth is interpreted as the depth of the drilled hole.
Machining operations usually divide into two categories, distinguished by purpose
and cutting conditions: roughing cuts and finishing cuts. Roughing cuts are used to
remove large amounts of material from the starting workpart as rapidly as possible, in
order to produce a shape close to the desired form, but leaving some material on the piece
for a subsequent finishing operation. Finishing cuts are used to complete the part and
achieve the final dimensions, tolerances, and surface finish. In production machining jobs,
one or more roughing cuts are usually performed on the work, followed by one or two
finishing cuts. Roughing operations are performed at high feeds and depths—feeds of 0.4
to 1.25 mm/rev (0.015–0.050 in/rev) and depths of 2.5 to 20 mm (0.100–0.750 in) are
typical. Finishing operations are carried out at low feeds and depths—feeds of 0.125 to 0.4
mm (0.005–0.015 in/rev) and depths of 0.75 to 2.0 mm (0.030–0.075 in) are typical. Cutting
speeds are lower in roughing than in finishing.
A cutting fluid is often applied to the machining operation to cool and lubricate the
cutting tool (cutting fluids are discussed in Section 23.4). Determining whether a cutting
fluid should be used, and, if so, choosing the proper cutting fluid, is usually included within
the scope of cutting conditions. Given the work material and tooling, the selection of these
conditions is very influential in determining the success of a machining operation.
Machine Tools A machine tool is used to hold the workpart, position the tool relative
to the work, and provide power for the machining process at the speed, feed, and depth
that have been set. By controlling the tool, work, and cutting conditions, machine tools
permit parts to be made with great accuracy and repeatability, to tolerances of 0.025 mm
(0.001 in) and better. The term machine tool applies to any power-driven machine that
performs a machining operation, including grinding. The term is also applied to machines
that perform metal forming and pressworking operations (Chapters 19 and 20).
The traditional machine tools used to perform turning, drilling, and milling are
lathes, drill presses, and milling machines, respectively. Conventional machine tools are
usually tended by a human operator, who loads and unloads the workparts, changes
cutting tools, and sets the cutting conditions. Many modern machine tools are designed to
accomplish their operations with a form of automation called computer numerical

control (Section 38.3).

21.2 THEORY OF CHIP FORMATION IN METAL MACHINING
The geometry of most practical machining operations is somewhat complex. A simplified
model of machining is available that neglects many of the geometric complexities, yet
describes the mechanics of the process quite well. It is called the orthogonal cutting model,
Figure 21.6. Although an actual machining process is three-dimensional, the orthogonal
model has only two dimensions that play active roles in the analysis.

21.2.1 THE ORTHOGONAL CUTTING MODEL
By definition, orthogonal cutting uses a wedge-shaped tool in which the cutting edge is
perpendicular to the direction of cutting speed. As the tool is forced into the material, the
chip is formed by shear deformation along a plane called the shear plane, which is
oriented at an angle f with the surface of the work. Only at the sharp cutting edge of the
tool does failure of the material occur, resulting in separation of the chip from the parent


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FIGURE 21.6 Orthogonal cutting: (a) as a three-dimensional process, and (b) how it reduces to two dimensions in

the side view.

material. Along the shear plane, where the bulk of the mechanical energy is consumed in
machining, the material is plastically deformed.
The tool in orthogonal cutting has only two elements of geometry: (1) rake angle and
(2) clearance angle. As indicated previously, the rake angle a determines the direction that
the chip flows as it is formed from the workpart; and the clearance angle provides a small
clearance between the tool flank and the newly generated work surface.
During cutting, the cutting edge of the tool is positioned a certain distance below
the original work surface. This corresponds to the thickness of the chip prior to chip
formation, to. As the chip is formed along the shear plane, its thickness increases to tc. The
ratio of to to tc is called the chip thickness ratio (or simply the chip ratio) r:
r¼

to
tc

ð21:2Þ

Since the chip thickness after cutting is always greater than the corresponding thickness
before cutting, the chip ratio will always be less than 1.0.
In addition to to, the orthogonal cut has a width dimension w, as shown in Figure 21.6(a),
even though this dimension does not contribute much to the analysis in orthogonal cutting.
The geometry of the orthogonal cutting model allows us to establish an important
relationship between the chip thickness ratio, the rake angle, and the shear plane angle. Let
ls be the length of the shear plane. We can make the substitutions: to ¼ ls sinf, and tc ¼ ls cos
(f À a). Thus,
r¼

ls sin f

sin f
¼
ls cos (f À a) cos (f À a)

This can be rearranged to determine f as follows:
tan f ¼

r cos a
1 À r sin a

ð21:3Þ

The shear strain that occurs along the shear plane can be estimated by examining
Figure 21.7. Part (a) shows shear deformation approximated by a series of parallel plates
sliding against one another to form the chip. Consistent with our definition of shear strain


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FIGURE 21.7 Shear strain during chip formation: (a) chip formation depicted as a series of parallel plates sliding

relative to each other; (b) one of the plates isolated to illustrate the definition of shear strain based on this parallel
plate model; and (c) shear strain triangle used to derive Eq. (21.4).

(Section 3.1.4), each plate experiences the shear strain shown in Figure 21.7(b). Referring to
part (c), this can be expressed as
g¼

AC AD þ DC
¼
BD
BD

which can be reduced to the following definition of shear strain in metal cutting:
g ¼ tan (f À a) þ cot f

Example 21.1
Orthogonal
Cutting

ð21:4Þ

In a machining operation that approximates orthogonal cutting, the cutting tool has a
rake angle ¼ 10 . The chip thickness before the cut to ¼ 0.50 mm and the chip thickness
after the cut tc ¼ 1.125 in. Calculate the shear plane angle and the shear strain in the
operation.
Solution:

The chip thickness ratio can be determined from Eq. (21.2):
r¼

0:50
¼ 0:444
1:125

The shear plane angle is given by Eq. (21.3):
tan f ¼

0:444 cos 10
¼ 0:4738
1 À 0:444 sin 10
f ¼ 25:4


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491

Finally, the shear strain is calculated from Eq. (21.4):
g ¼ tan (25:4 À 10) þ cot 25:4
g ¼ 0:275 þ 2:111 ¼ 2:386

n

21.2.2 ACTUAL CHIP FORMATION
We should note that there are differences between the orthogonal model and an actual
machining process. First, the shear deformation process does not occur along a plane, but
within a zone. If shearing were to take place across a plane of zero thickness, it would imply
that the shearing action must occur instantaneously as it passes through the plane, rather
than over some finite (although brief) time period. For the material to behave in a realistic
way, the shear deformation must occur within a thin shear zone. This more realistic model of
the shear deformation process in machining is illustrated in Figure 21.8. Metal-cutting
experiments have indicated that the thickness of the shear zone is only a few thousandths of
an inch. Since the shear zone is so thin, there is not a great loss of accuracy in most cases by
referring to it as a plane.
Second, in addition to shear deformation that occurs in the shear zone, another
shearing action occurs in the chip after it has been formed. This additional shear is
referred to as secondary shear to distinguish it from primary shear. Secondary shear
results from friction between the chip and the tool as the chip slides along the rake face
of the tool. Its effect increases with increased friction between the tool and chip. The
primary and secondary shear zones can be seen in Figure 21.8.
Third, formation of the chip depends on the type of material being machined and
the cutting conditions of the operation. Four basic types of chip can be distinguished,
illustrated in Figure 21.9:
å Discontinuous chip. When relatively brittle materials (e.g., cast irons) are machined
at low cutting speeds, the chips often form into separate segments (sometimes the
segments are loosely attached). This tends to impart an irregular texture to the
machined surface. High tool–chip friction and large feed and depth of cut promote
the formation of this chip type.
å Continuous chip. When ductile work materials are cut at high speeds and relatively
small feeds and depths, long continuous chips are formed. A good surface finish
typically results when this chip type is formed. A sharp cutting edge on the tool and

Chip

FIGURE 21.8 More
realistic view of chip
formation, showing shear
zone rather than shear
plane. Also shown is the
secondary shear zone
resulting from tool–chip
friction.

Effective

Tool

Primary shear
zone

Secondary shear zone


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Discontinuous chip

Continuous chip

Continuous chip
High shear
strain zone

Tool

Tool

Tool

Low shear
strain zone

Tool

Built-up edge
Irregular surface due
to chip discontinuities
(a)

Good finish typical
(b)

Particle of BUE

on new surface
(c)

(d)

FIGURE 21.9 Four types of chip formation in metal cutting: (a) discontinuous, (b) continuous, (c) continuous with
built-up edge, (d) serrated.

low tool–chip friction encourage the formation of continuous chips. Long, continuous
chips (as in turning) can cause problems with regard to chip disposal and/or tangling
about the tool. To solve these problems, turning tools are often equipped with chip
breakers (Section 23.3.1).
å Continuous chip with built-up edge. When machining ductile materials at low-tomedium cutting speeds, friction between tool and chip tends to cause portions of the
work material to adhere to the rake face of the tool near the cutting edge. This
formation is called a built-up edge (BUE). The formation of a BUE is cyclical; it
forms and grows, then becomes unstable and breaks off. Much of the detached BUE
is carried away with the chip, sometimes taking portions of the tool rake face with it,
which reduces the life of the cutting tool. Portions of the detached BUE that are not
carried off with the chip become imbedded in the newly created work surface,
causing the surface to become rough.
The preceding chip types were first classified by Ernst in the late 1930s [13]. Since
then, the available metals used in machining, cutting tool materials, and cutting speeds
have all increased, and a fourth chip type has been identified:
å Serrated chips (the term shear-localized is also used for this fourth chip type). These
chips are semi-continuous in the sense that they possess a saw-tooth appearance that
is produced by a cyclical chip formation of alternating high shear strain followed by
low shear strain. This fourth type of chip is most closely associated with certain
difficult-to-machine metals such as titanium alloys, nickel-base superalloys, and
austenitic stainless steels when they are machined at higher cutting speeds. However,
the phenomenon is also found with more common work metals (e.g., steels) when

they are cut at high speeds [13].2

21.3 FORCE RELATIONSHIPS AND THE MERCHANT EQUATION
Several forces can be defined relative to the orthogonal cutting model. Based on these
forces, shear stress, coefficient of friction, and certain other relationships can be
defined.
2

A more complete description of the serrated chip type can be found in Trent & Wright [12], pp. 348–367.


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21.3.1 FORCES IN METAL CUTTING
Consider the forces acting on the chip during orthogonal cutting in Figure 21.10(a). The forces
applied against the chip by the tool can be separated into two mutually perpendicular
components: friction force and normal force to friction. The friction force F is the frictional
force resisting the flow of the chip along the rake face of the tool. The normal force to friction N
is perpendicular to the friction force. These two components can be used to define the
coefficient of friction between the tool and the chip:

m¼

F
N

ð21:5Þ

The friction force and its normal force can be added vectorially to form a resultant
force R, which is oriented at an angle b, called the friction angle. The friction angle is
related to the coefficient of friction as
m ¼ tan b

ð21:6Þ

In addition to the tool forces acting on the chip, there are two force components applied
by the workpiece on the chip: shear force and normal force to shear. The shear force Fs is the
force that causes shear deformation to occur in the shear plane, and the normal force to shear
Fn is perpendicular to the shear force. Based on the shear force, we can define the shear stress
that acts along the shear plane between the work and the chip:
t¼

Fs
As

ð21:7Þ

where As ¼ area of the shear plane. This shear plane area can be calculated as
As ¼

to w

sin f

ð21:8Þ

The shear stress in Eq. (21.7) represents the level of stress required to perform the
machining operation. Therefore, this stress is equal to the shear strength of the work
material (t ¼ S) under the conditions at which cutting occurs.
Vector addition of the two force components Fs and Fn yields the resultant force R0 .
In order for the forces acting on the chip to be in balance, this resultant R0 must be equal
in magnitude, opposite in direction, and collinear with the resultant R.

FIGURE 21.10 Forces in metal cutting: (a) forces acting on the chip in orthogonal cutting, and (b) forces acting on
the tool that can be measured.


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FIGURE 21.11 Force diagram showing
geometric relationships between F, N,
Fs, Fn, Fc, and Ft.

None of the four force components F, N, Fs, and Fn can be directly measured in a
machining operation, because the directions in which they are applied vary with different
tool geometries and cutting conditions. However, it is possible for the cutting tool to be
instrumented using a force measuring device called a dynamometer, so that two additional
force components acting against the tool can be directly measured: cutting force and thrust
force. The cutting force Fc is in the direction of cutting, the same direction as the cutting
speed v, and the thrust force Ft is perpendicular to the cutting force and is associated with the
chip thickness before the cut to. The cutting force and thrust force are shown in Figure 21.10
(b) together with their resultant force R00 . The respective directions of these forces are
known, so the force transducers in the dynamometer can be aligned accordingly.
Equations can be derived to relate the four force components that cannot
be measured to the two forces that can be measured. Using the force diagram in
Figure 21.11, the following trigonometric relationships can be derived:
F ¼ F c sin a þ F t cos a

ð21:9Þ

N ¼ F c cos a À F t sin a

ð21:10Þ

F s ¼ F c cos f À F t sin f

ð21:11Þ

F n ¼ F c sin f þ F t cos f

ð21:12Þ

If cutting force and thrust force are known, these four equations can be used to calculate
estimates of shear force, friction force, and normal force to friction. Based on these force
estimates, shear stress and coefficient of friction can be determined.
Note that in the special case of orthogonal cutting when the rake angle a ¼ 0, Eqs. (21.9)
and (21.10) reduce to F ¼ Ft and N ¼ Fc, respectively. Thus, in this special case, friction force
and its normal force could be directly measured by the dynamometer.

Example 21.2
Shear Stress in
Machining

Suppose in Example 21.1 that cutting force and thrust force are measured during an
orthogonal cutting operation: Fc ¼ 1559 N and Ft ¼ 1271 N. The width of the orthogonal
cutting operation w ¼ 3.0 mm. Based on these data, determine the shear strength of the
work material.
Solution: From Example 21.1, rake angle a ¼ 10 , and shear plane angle f ¼ 25.4 . Shear
force can be computed from Eq. (21.11):
F s ¼ 1559 cos 25:4 À 1271 sin 25:4 ¼ 863 N


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495

The shear plane area is given by Eq. (21.8):
As ¼

(0:5)(3:0)
¼ 3:497 mm2
sin 25:4

Thus the shear stress, which equals the shear strength of the work material, is
t¼S¼

863
¼ 247 N/mm2 ¼ 247 MPa
3:497

n

This example demonstrates that cutting force and thrust force are related to the shear
strength of the work material. The relationships can be established in a more direct way.
Recalling from Eq. (21.7) that the shear force Fs ¼ S As, the force diagram of Figure 21.11
can be used to derive the following equations:
Fc ¼

Sto w cos (b À a)
F s cos (b À a)
¼
sin f cos(f þ b À a) cos(f þ b À a)

ð21:13Þ

Ft ¼

St w sin (b À a)
F s sin (b À a)
¼
sin f cos(f þ b À a) cos (f þ b À a)

ð21:14Þ

and

These equations allow one to estimate cutting force and thrust force in an orthogonal
cutting operation if the shear strength of the work material is known.

21.3.2 THE MERCHANT EQUATION
One of the important relationships in metal cutting was derived by Eugene Merchant
[10]. Its derivation was based on the assumption of orthogonal cutting, but its general
validity extends to three-dimensional machining operations. Merchant started with the
definition of shear stress expressed in the form of the following relationship derived by
combining Eqs. (21.7), (21.8), and (21.11):
t¼

F c cos f À F t sin f
(to w=sin f)

ð21:15Þ

Merchant reasoned that, out of all the possible angles emanating from the cutting
edge of the tool at which shear deformation could occur, there is one angle f that

predominates. This is the angle at which shear stress is just equal to the shear strength of
the work material, and so shear deformation occurs at this angle. For all other possible
shear angles, the shear stress is less than the shear strength, so chip formation cannot
occur at these other angles. In effect, the work material will select a shear plane angle that
minimizes energy. This angle can be determined by taking the derivative of the shear
stress S in Eq. (21.15) with respect to f and setting the derivative to zero. Solving for f, we
get the relationship named after Merchant:
f ¼ 45 þ

a b
À
2 2

ð21:16Þ

Among the assumptions in the Merchant equation is that shear strength of the work
material is a constant, unaffected by strain rate, temperature, and other factors. Because
this assumption is violated in practical machining operations, Eq. (21.16) must be


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considered an approximate relationship rather than an accurate mathematical equation.
Let us nevertheless consider its application in the following example.

Example 21.3
Estimating
Friction Angle

Using the data and results from our previous examples, determine (a) the friction angle
and (b) the coefficient of friction.
Solution: (a) From Example 21.1, a ¼ 10 , and f ¼ 25.4 . Rearranging Eq. (21.16),
the friction angle can be estimated:
b ¼ 2 (45) þ 10 À 2 (25:4) ¼ 49:2
(b) The coefficient of friction is given by Eq. (21.6):
m ¼ tan 49:2 ¼ 1:16

n

Lessons Based on the Merchant Equation The real value of the Merchant equation is
that it defines the general relationship between rake angle, tool–chip friction, and shear
plane angle. The shear plane angle can be increased by (1) increasing the rake angle and
(2) decreasing the friction angle (and coefficient of friction) between the tool and the
chip. Rake angle can be increased by proper tool design, and friction angle can be
reduced by using a lubricant cutting fluid.
The importance of increasing the shear plane angle can be seen in Figure 21.12. If all
other factors remain the same, a higher shear plane angle results in a smaller shear plane
area. Since the shear strength is applied across this area, the shear force required to form
the chip will decrease when the shear plane area is reduced. A greater shear plane angle
results in lower cutting energy, lower power requirements, and lower cutting temperature.

These are good reasons to try to make the shear plane angle as large as possible during
machining.
Approximation of Turning by Orthogonal Cutting The orthogonal model can be used
to approximate turning and certain other single-point machining operations so long as the
feed in these operations is small relative to depth of cut. Thus, most of the cutting will take
place in the direction of the feed, and cutting on the point of the tool will be negligible.
Figure 21.13 indicates the conversion from one cutting situation to the other.

FIGURE 21.12 Effect of shear plane angle f: (a) higher f with a resulting lower shear plane area;
(b) smaller f with a corresponding larger shear plane area. Note that the rake angle is larger in (a), which
tends to increase shear angle according to the Merchant equation.


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Section 21.4/Power and Energy Relationships in Machining

497

FIGURE 21.13
Approximation of turning
by the orthogonal model:
(a) turning; and (b) the
corresponding orthogonal cutting.

TABLE 21.1 Conversion key: turning operation
vs. orthogonal cutting.
Turning Operation
Orthogonal Cutting Model
Feed f ¼
Depth d ¼
Cutting speed v ¼
Cutting force Fc ¼
Feed force Ff ¼

Chip thickness before cut to
Width of cut w
Cutting speed v
Cutting force Fc
Thrust force Ft

The interpretation of cutting conditions is different in the two cases. The chip
thickness before the cut to in orthogonal cutting corresponds to the feed f in turning, and
the width of cut w in orthogonal cutting corresponds to the depth of cut d in turning. In
addition, the thrust force Ft in the orthogonal model corresponds to the feed force Ff in
turning. Cutting speed and cutting force have the same meanings in the two cases.
Table 21.1 summarizes the conversions.

21.4

POWER AND ENERGY RELATIONSHIPS IN MACHINING
A machining operation requires power. The cutting force in a production machining
operation might exceed 1000 N (several hundred pounds), as suggested by Example 21.2.
Typical cutting speeds are several hundred m/min. The product of cutting force and speed

gives the power (energy per unit time) required to perform a machining operation:
Pc ¼ F c v

ð21:17Þ

where Pc ¼ cutting power, N-m/s or W (ft-lb/min); Fc ¼ cutting force, N (lb); and v ¼
cutting speed, m/s (ft/min). In U.S. customary units, power is traditionally expressed as


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horsepower by dividing ft-lb/min by 33,000. Hence,
HPc ¼

F cv
33; 000

ð21:18Þ

where HPc ¼ cutting horsepower, hp. The gross power required to operate the machine

tool is greater than the power delivered to the cutting process because of mechanical losses
in the motor and drive train in the machine. These losses can be accounted for by the
mechanical efficiency of the machine tool:
Pg ¼

Pc
E

or

HPg ¼

HPc
E

ð21:19Þ

where Pg ¼ gross power of the machine tool motor, W; HPg ¼ gross horsepower; and E ¼
mechanical efficiency of the machine tool. Typical values of E for machine tools are
around 90%.
It is often useful to convert power into power per unit volume rate of metal cut. This
is called the unit power, Pu (or unit horsepower, HPu), defined:
Pu ¼

Pc
RMR

or

HPu ¼

HPc
RMR

ð21:20Þ

where RMR ¼ material removal rate, mm3/s (in3/min). The material removal rate can be
calculated as the product of vtow. This is Eq. (21.1) using the conversions from Table 21.1.
Unit power is also known as the specific energy U.
U ¼ Pu ¼

Pc
F cv
Fc
¼
¼
RMR vto w to w

ð21:21Þ

The units for specific energy are typically N-m/mm3 (in-lb/in3). However, the last
expression in Eq. (21.21) suggests that the units might be reduced to N/mm2 (lb/in2).
It is more meaningful to retain the units as N-m/mm3 or J/mm3 (in-lb/in3).

Example 21.4
Power
Relationships in
Machining

Continuing with our previous examples, let us determine cutting power and specific

energy in the machining operation if the cutting speed ¼ 100 m/min. Summarizing the
data and results from previous examples, to ¼ 0.50 mm, w ¼ 3.0 mm, Fc ¼ 1557 N.
Solution:

From Eq. (21.18), power in the operation is

Pc ¼ (1557 N)(100 m/min) ¼ 155; 700 N À m/min ¼ 155; 700 J/min ¼ 2595 J/s ¼ 2595 W
Specific energy is calculated from Eq. (21.21):
U¼

155; 700
155; 700
¼
¼ 1:038 N-m/min3
100(103 )(3:0)(0:5) 150; 000

n

Unit power and specific energy provide a useful measure of how much power (or
energy) is required to remove a unit volume of metal during machining. Using this
measure, different work materials can be compared in terms of their power and energy
requirements. Table 21.2 presents a listing of unit horsepower and specific energy values
for selected work materials.
The values in Table 21.2 are based on two assumptions: (1) the cutting tool is sharp,
and (2) the chip thickness before the cut to ¼ 0.25 mm (0.010 in). If these assumptions are
not met, some adjustments must be made. For worn tools, the power required to perform
the cut is greater, and this is reflected in higher specific energy and unit horsepower values.
As an approximate guide, the values in the table should be multiplied by a factor between
1.00 and 1.25 depending on the degree of dullness of the tool. For sharp tools, the factor is

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Section 21.4/Power and Energy Relationships in Machining

499

TABLE 21.2 Values of unit horsepower and specific energy for selected work
materials using sharp cutting tools and chip thickness before the cut to = 0.25 mm
(0.010 in).
Specific Energy U or
Unit Power Pu
Unit Horsepower
Brinell
HPu hp/(in3/min)
Material
Hardness
N-m/mm3
in-lb/in3
Carbon steel

Alloy steels

Cast irons
Stainless steel
Aluminum

Aluminum alloys
Brass
Bronze
Magnesium alloys

150–200
201–250
251–300
200–250
251–300
301–350
351–400
125–175
175–250
150–250
50–100
100–150
100–150
100–150
50–100

1.6
2.2
2.8
2.2
2.8
3.6
4.4
1.1
1.6

2.8
0.7
0.8
2.2
2.2
0.4

240,000
320,000
400,000
320,000
400,000
520,000
640,000
160,000
240,000
400,000
100,000
120,000
320,000
320,000
60,000

0.6
0.8
1.0
0.8
1.0
1.3
1.6

0.4
0.6
1.0
0.25
0.3
0.8
0.8
0.15

Data compiled from [6], [8], [11], and other sources.

1.00. For tools in a finishing operation that are nearly worn out, the factor is around 1.10,
and for tools in a roughing operation that are nearly worn out, the factor is 1.25.
Chip thickness before the cut to also affects the specific energy and unit horsepower
values. As to is reduced, unit power requirements increase. This relationship is referred to as
the size effect. For example, grinding, in which the chips are extremely small by comparison to
most other machining operations, requires very high specific energy values. The U and HPu
values in Table 21.2 can still be used to estimate horsepower and energy for situations in which
to is not equal to 0.25 mm (0.010 in) by applying a correction factor to account for any
difference in chip thickness before the cut. Figure 21.14 provides values of this correction
Chip thickness before cut to (in.)
0.005 0.010 0.015 0.020 0.025 0.030

0.040

0.050

1.6
1.4

FIGURE 21.14 Correction
factor for unit horsepower
and specific energy when
values of chip thickness
before the cut to are
different from 0.25 mm
(0.010 in).

Correction factor

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1.2
1.0
0.8
0.6
0.4
0.2
0.125 0.25 0.38 0.50 0.63 0.75 0.88 0.1
Chip thickness before cut to (mm)

1.25


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Chapter 21/Theory of Metal Machining

factor as a function of to. The unit horsepower and specific energy values in Table 21.2 should
be multiplied by the appropriate correction factor when to is different from 0.25 mm (0.010 in).
In addition to tool sharpness and size effect, other factors also influence the values of
specific energy and unit horsepower for a given operation. These other factors include rake
angle, cutting speed, and cutting fluid. As rake angle or cutting speed are increased, or when
cutting fluid is added, the U and HPu values are reduced slightly. For our purposes in the
end-of-chapter exercises, the effects of these additional factors can be ignored.

21.5 CUTTING TEMPERATURE
Of the total energy consumed in machining, nearly all of it ($98%) is converted into heat.
This heat can cause temperatures to be very high at the tool–chip interface—over 600 C
(1100 F) is not unusual. The remaining energy ($2%) is retained as elastic energy in the chip.
Cutting temperatures are important because high temperatures (1) reduce tool life,
(2) produce hot chips that pose safety hazards to the machine operator, and (3) can cause
inaccuracies in workpart dimensions due to thermal expansion of the work material. In this
section, we discuss the methods of calculating and measuring temperatures in machining
operations.

21.5.1 ANALYTICAL METHODS TO COMPUTE CUTTING TEMPERATURES
There are several analytical methods to calculate estimates of cutting temperature.
References [3], [5], [9], and [15] present some of these approaches. We describe the
method by Cook [5], which was derived using experimental data for a variety of work
materials to establish parameter values for the resulting equation. The equation can be
used to predict the increase in temperature at the tool–chip interface during machining:

DT ¼

0:4U vto 0:333
rC K

ð21:22Þ

where DT ¼ mean temperature rise at the tool–chip interface, C (F ); U ¼ specific energy
in the operation, N-m/mm3 or J/mm3 (in-lb/in3); v ¼ cutting speed, m/s (in/sec); to ¼ chip
thickness before the cut, m (in); rC ¼ volumetric specific heat of the work material, J/mm3C (in-lb/in3-F); K ¼ thermal diffusivity of the work material, m2/s (in2/sec).

Example 21.5
Cutting
Temperature

For the specific energy obtained in Example 21.4, calculate the increase in temperature
above ambient temperature of 20 C. Use the given data from the previous examples in this
chapter: v ¼ 100 m/min, to ¼ 0.50 mm. In addition, the volumetric specific heat for the work
material ¼ 3.0 (10À3) J/mm3-C, and thermal diffusivity ¼ 50 (10À6) m2/s (or 50 mm2/s).
Solution: Cutting speed must be converted to mm/s: v ¼ (100 m/min)(103 mm/m)/(60 s/
min) ¼ 1667 mm/s. Eq. (21.22) can now be used to compute the mean temperature rise:


0:4(1:038)  1667(0:5) 0:333
C
¼ (138:4)(2:552) ¼ 353 C
DT ¼
50
3:0(103 )

n

21.5.2 MEASUREMENT OF CUTTING TEMPERATURE
Experimental methods have been developed to measure temperatures in machining.
The most frequently used measuring technique is the tool–chip thermocouple. This
thermocouple consists of the tool and the chip as the two dissimilar metals forming the


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References

501

RC-130B Titanium (T = 479v 0.182)
1600

FIGURE 21.15
Experimentally measured
cutting temperatures
plotted against speed
for three work materials,
indicating general
agreement with
Eq. (21.23). (Based on
data in [9].)3

Cutting temperature, °F

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1200
18-8 Stainless steel (T = 135v 0.361)
800
B1113 Free machining steel (T = 86.2v 0.348)
400

200

400

800

600

1000

Cutting speed (ft/min)

thermocouple junction. By properly connecting electrical leads to the tool and workpart (which is connected to the chip), the voltage generated at the tool–chip interface
during cutting can be monitored using a recording potentiometer or other appropriate
data-collection device. The voltage output of the tool–chip thermocouple (measured in
mV) can be converted into the corresponding temperature value by means of calibration equations for the particular tool–work combination.
The tool–chip thermocouple has been utilized by researchers to investigate the
relationship between temperature and cutting conditions such as speed and feed. Trigger
[14] determined the speed–temperature relationship to be of the following general form:

T ¼ K vm

ð21:23Þ

where T ¼ measured tool–chip interface temperature and v ¼ cutting speed. The
parameters K and m depend on cutting conditions (other than v) and work material.
Figure 21.15 plots temperature versus cutting speed for several work materials, with
equations of the form of Eq. (21.23) determined for each material. A similar relationship
exists between cutting temperature and feed; however, the effect of feed on temperature
is not as strong as cutting speed. These empirical results tend to support the general
validity of the Cook equation: Eq. (21.22).

REFERENCES
[1] ASM Handbook, Vol. 16, Machining. ASM International, Materials Park, Ohio, 1989.
[2] Black, J, and Kohser, R. DeGarmo’s Materials and
Processes in Manufacturing, 10th ed. John Wiley &
Sons, Inc., Hoboken, New Jersey, 2008.

[3] Boothroyd, G., and Knight, W. A. Fundamentals of
Metal Machining and Machine Tools, 3rd ed. CRC
Taylor and Francis, Boca Raton, Florida, 2006.
[4] Chao, B. T., and Trigger, K. J.‘‘Temperature Distribution at the Tool-Chip Interface in Metal

3
The units reported in the Loewen and Shaw ASME paper [9] were  F for cutting temperature and ft/min
for cutting speed. We have retained those units in the plots and equations of our figure.


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[5]

[6]

[7]

[8]
[9]

15:44:5

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Chapter 21/Theory of Metal Machining
Cutting,’’ ASME Transactions, Vol. 77, October
1955, pp. 1107– 1121.
Cook, N.‘‘Tool Wear and Tool Life,’’ ASME Transactions, Journal of Engineering for Industry,
Vol. 95, November 1973, pp. 931–938.
Drozda, T. J., and Wick, C. (eds.). Tool and Manufacturing Engineers Handbook, 4th ed., Vol. I,
Machining. Society of Manufacturing Engineers,
Dearborn, Michigan, 1983.
Kalpakjian, S., and Schmid, R. Manufacturing Processes for Engineering Materials, 4th ed. Prentice
Hall/Pearson, Upper Saddle River, New Jersey, 2003.
Lindberg, R. A. Processes and Materials of Manufacture, 4th ed. Allyn and Bacon, Inc., Boston, 1990.
Loewen, E. G., and Shaw, M. C.‘‘On the Analysis of
Cutting Tool Temperatures,’’ ASME Transactions,

Vol. 76, No. 2, February 1954, pp. 217–225.

[10] Merchant, M. E., ‘‘Mechanics of the Metal Cutting
Process: II. Plasticity Conditions in Orthogonal Cutting,’’ Journal of Applied Physics, Vol. 16, June 1945
pp. 318–324.
[11] Schey, J. A. Introduction to Manufacturing Processes, 3rd ed. McGraw-Hill Book Company, New
York, 1999.
[12] Shaw, M. C. Metal Cutting Principles, 2nd ed. Oxford University Press, Oxford, UK, 2005.
[13] Trent, E. M., and Wright, P. K. Metal Cutting, 4th ed.
Butterworth Heinemann, Boston, 2000.
[14] Trigger, K. J.‘‘Progress Report No. 2 on Tool–Chip
Interface Temperatures,’’ ASME Transactions,
Vol. 71, No. 2, February 1949, pp. 163–174.
[15] Trigger, K. J., and Chao, B. T.‘‘An Analytical Evaluation of Metal Cutting Temperatures,’’ ASME
Transactions, Vol. 73, No. 1, January 1951, pp. 57–68.

REVIEW QUESTIONS
21.1. What are the three basic categories of material
removal processes?
21.2. What distinguishes machining from other manufacturing processes?
21.3. Identify some of the reasons why machining is
commercially and technologically important.
21.4. Name the three most common machining
processes.
21.5. What are the two basic categories of cutting tools in
machining? Give two examples of machining operations that use each of the tooling types.
21.6. What are the parameters of a machining operation
that are included within the scope of cutting
conditions?
21.7. Explain the difference between roughing and finishing operations in machining.

21.8. What is a machine tool?
21.9. What is an orthogonal cutting operation?

21.10. Why is the orthogonal cutting model useful in the
analysis of metal machining?
21.11. Name and briefly describe the four types of chips
that occur in metal cutting.
21.12. Identify the four forces that act upon the chip in the
orthogonal metal cutting model but cannot be
measured directly in an operation.
21.13. Identify the two forces that can be measured in the
orthogonal metal cutting model.
21.14. What is the relationship between the coefficient of
friction and the friction angle in the orthogonal
cutting model?
21.15. Describeinwords whatthe Merchant equationtells us.
21.16. How is the power required in a cutting operation
related to the cutting force?
21.17. What is the specific energy in metal machining?
21.18. What does the term size effect mean in metal cutting?
21.19. What is a tool–chip thermocouple?

MULTIPLE CHOICE QUIZ
There are 17 correct answers in the following multiple choice questions (some questions have multiple answers that are
correct). To attain a perfect score on the quiz, all correct answers must be given. Each correct answer is worth 1 point. Each
omitted answer or wrong answer reduces the score by 1 point, and each additional answer beyond the correct number of
answers reduces the score by 1 point. Percentage score on the quiz is based on the total number of correct answers.
21.1. Which of the following manufacturing processes
are classified as material removal processes (two
correct answers): (a) casting, (b) drawing, (c) extrusion, (d) forging, (e) grinding, (f) machining,

(g) molding, (h) pressworking, and (i) spinning?

21.2. A lathe is used to perform which one of the
following manufacturing operations: (a) broaching,
(b) drilling, (c) lapping, (d) milling, or (e) turning?
21.3. With which one of the following geometric forms is
the drilling operation most closely associated:


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Problems

21.4.

21.5.

21.6.

21.7.

21.8.

(a) external cylinder, (b) flat plane, (c) round hole,

(d) screw threads, or (e) sphere?
If the cutting conditions in a turning operation are
cutting speed ¼ 300 ft/min, feed ¼ 0.010 in/rev, and
depth of cut ¼ 0.100 in, which one of the following
is the material removal rate: (a) 0.025 in3/min,
(b) 0.3 in3/min, (c) 3.0 in3/min, or (d) 3.6 in3/min?
A roughing operation generally involves which one
of the following combinations of cutting conditions: (a) high v, f, and d; (b) high v, low f and d;
(c) low v, high f and d; or (d) low v, f, and d, where v ¼
cutting speed, f ¼ feed, and d ¼ depth?
Which of the following are characteristics of the
orthogonal cutting model (three best answers):
(a) a circular cutting edge is used, (b) a multiplecutting-edge tool is used, (c) a single-point tool is
used, (d) only two dimensions play an active role in
the analysis, (e) the cutting edge is parallel to the
direction of cutting speed, (f) the cutting edge is
perpendicular to the direction of cutting speed, and
(g) the two elements of tool geometry are rake and
relief angle?
The chip thickness ratio is which one of the following:
(a) tc/to, (b) to/tc, (c) f/d, or (d) to/w, where tc ¼ chip
thickness after the cut, to ¼ chip thickness before
the cut, f ¼ feed, d ¼ depth, and w ¼ width of cut?
Which one of the four types of chip would be
expected in a turning operation conducted at low

21.9.

21.10.

21.11.

21.12.

21.13.

503

cutting speed on a brittle work material: (a) continuous, (b) continuous with built-up edge,
(c) discontinuous, or (d) serrated?
According to the Merchant equation, an increase
in rake angle would have which of the following
results, all other factors remaining the same (two
best answers): (a) decrease in friction angle,
(b) decrease in power requirements, (c) decrease
in shear plane angle, (d) increase in cutting temperature, and (e) increase in shear plane angle?
In using the orthogonal cutting model to approximate a turning operation, the chip thickness before
the cut to corresponds to which one of the following
cutting conditions in turning: (a) depth of cut d,
(b) feed f, or (c) speed v?
Which one of the following metals would usually
have the lowest unit horsepower in a machining
operation: (a) aluminum, (b) brass, (c) cast iron, or
(d) steel?
For which one of the following values of chip thickness before the cut to would you expect the specific
energy in machining to be the greatest:(a) 0.010 in,
(b) 0.025 in, (c) 0.12 mm, or (d) 0.50 mm?
Which of the following cutting conditions has the
strongest effect on cutting temperature: (a) feed or
(b) speed?

PROBLEMS
Chip Formation and Forces in Machining
21.1. In an orthogonal cutting operation, the tool has a
rake angle ¼ 15 . The chip thickness before the cut ¼
0.30 mm and the cut yields a deformed chip thickness ¼ 0.65 mm. Calculate (a) the shear plane angle
and (b) the shear strain for the operation.
21.2. In Problem 21.1, suppose the rake angle were
changed to 0 . Assuming that the friction angle
remains the same, determine (a) the shear plane
angle, (b) the chip thickness, and (c) the shear
strain for the operation.
21.3. In an orthogonal cutting operation, the 0.25-in
wide tool has a rake angle of 5 . The lathe is set
so the chip thickness before the cut is 0.010 in.
After the cut, the deformed chip thickness is measured to be 0.027 in. Calculate (a) the shear plane
angle and (b) the shear strain for the operation.
21.4. In a turning operation, spindle speed is set to provide
a cutting speed of 1.8 m/s. The feed and depth of cut
of cut are 0.30 mm and 2.6 mm, respectively. The tool
rake angle is 8 . After the cut, the deformed chip

thickness is measured to be 0.49 mm. Determine (a)
shear plane angle, (b) shear strain, and (c) material
removal rate. Use the orthogonal cutting model as
an approximation of the turning process.
21.5. The cutting force and thrust force in an orthogonal
cutting operation are 1470 N and 1589 N, respectively. The rake angle ¼ 5 , the width of the cut ¼
5.0 mm, the chip thickness before the cut ¼ 0.6, and
the chip thickness ratio ¼ 0.38. Determine (a) the

shear strength of the work material and (b) the
coefficient of friction in the operation.
21.6. The cutting force and thrust force have been
measured in an orthogonal cutting operation
to be 300 lb and 291 lb, respectively. The rake
angle ¼ 10 , width of cut ¼ 0.200 in, chip thickness
before the cut ¼ 0.015, and chip thickness ratio ¼
0.4. Determine (a) the shear strength of the work
material and (b) the coefficient of friction in the
operation.


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21.7. An orthogonal cutting operation is performed
using a rake angle of 15 , chip thickness before
the cut ¼ 0.012 in and width of cut ¼ 0.100 in. The
chip thickness ratio is measured after the cut to be
0.55. Determine (a) the chip thickness after the cut,
(b) shear angle, (c) friction angle, (d) coefficient of

friction, and (e) shear strain.
21.8. The orthogonal cutting operation described in previous Problem 21.7 involves a work material whose
shear strength is 40,000 lb/in2. Based on your answers
to the previous problem, compute(a) the shear force,
(b) cutting force, (c) thrust force, and (d) friction
force.
21.9. In an orthogonal cutting operation, the rake angle ¼
À5 , chip thickness before the cut ¼ 0.2 mm and
width of cut ¼ 4.0 mm. The chip ratio ¼ 0.4. Determine (a) the chip thickness after the cut, (b) shear
angle, (c) friction angle, (d) coefficient of friction,
and (e) shear strain.
21.10. The shear strength of a certain work material ¼
50,000 lb/in2. An orthogonal cutting operation is
performed using a tool with a rake angle ¼ 20 at
the following cutting conditions: cutting speed ¼
100 ft/min, chip thickness before the cut ¼ 0.015 in,
and width of cut ¼ 0.150 in. The resulting chip
thickness ratio ¼ 0.50. Determine (a) the shear
plane angle, (b) shear force, (c) cutting force and
thrust force, and (d) friction force.
21.11. Consider the data in Problem 21.10 except that
rake angle is a variable, and its effect on the forces
in parts (b), (c), and (d) is to be evaluated.
(a) Using a spreadsheet calculator, compute the
values of shear force, cutting force, thrust force, and
friction force as a function of rake angle over a
range of rake angles between the high value of 20
in Problem 21.10 and a low value of À10 . Use
intervals of 5 between these limits. The chip thickness ratio decreases as rake angle is reduced and
can be approximated by the following relationship:

r ¼ 0.38 þ 0.006a, where r ¼ chip thickness and a ¼

21.12.

21.13.

21.14.

21.15.

21.16.
21.17.
21.18.

rake angle. (b) What observations can be made
from the computed results?
Solve previous Problem 21.10 except that the rake
angle has been changed to À5 and the resulting
chip thickness ratio ¼ 0.35.
A carbon steel bar with 7.64 in diameter has a
tensile strength of 65,000 lb/in2 and a shear strength
of 45,000 lb/in2. The diameter is reduced using a
turning operation at a cutting speed of 400 ft/min.
The feed is 0.011 in/rev and the depth of cut is
0.120 in. The rake angle on the tool in the direction
of chip flow is 13 . The cutting conditions result in a
chip ratio of 0.52. Using the orthogonal model as an
approximation of turning, determine (a) the shear
plane angle, (b) shear force, (c) cutting force and
feed force, and (d) coefficient of friction between

the tool and chip.
Low carbon steel having a tensile strength of
300 MPa and a shear strength of 220 MPa is cut
in a turning operation with a cutting speed of 3.0 m/s.
The feed is 0.20 mm/rev and the depth of cut is
3.0 mm. The rake angle of the tool is 5 in the
direction of chip flow. The resulting chip ratio is
0.45. Using the orthogonal model as an approximation of turning, determine (a) the shear plane angle,
(b) shear force, (c) cutting force and feed force.
A turning operation is made with a rake angle of
10 , a feed of 0.010 in/rev and a depth of cut ¼ 0.100
in. The shear strength of the work material is
known to be 50,000 lb/in2, and the chip thickness
ratio is measured after the cut to be 0.40. Determine the cutting force and the feed force. Use the
orthogonal cutting model as an approximation of
the turning process.
Show how Eq. (21.3) is derived from the definition
of chip ratio, Eq. (21.2), and Figure 21.5(b).
Show how Eq. (21.4) is derived from Figure 21.6.
Derive the force equations for F, N, Fs, and Fn
(Eqs. (21.9) through (21.12) in the text) using the
force diagram of Figure 21.11.

Power and Energy in Machining
21.19. In a turning operation on stainless steel with hardness ¼ 200 HB, the cutting speed ¼ 200 m/min,
feed ¼ 0.25 mm/rev, and depth of cut ¼ 7.5 mm.
How much power will the lathe draw in performing
this operation if its mechanical efficiency ¼ 90%.
Use Table 21.2 to obtain the appropriate specific
energy value.

21.20. In Problem 21.18, compute the lathe power requirements if feed ¼ 0.50 mm/rev.
21.21. In a turning operation on aluminum, cutting
speed ¼ 900 ft/min, feed ¼ 0.020 in/rev, and depth
of cut ¼ 0.250 in. What horsepower is required of

the drive motor, if the lathe has a mechanical
efficiency ¼ 87%? Use Table 21.2 to obtain the
appropriate unit horsepower value.
21.22. In a turning operation on plain carbon steel whose
Brinell hardness ¼ 275 HB, the cutting speed is
set at 200 m/min and depth of cut ¼ 6.0 mm. The
lathe motor is rated at 25 kW, and its mechanical
efficiency ¼ 90%. Using the appropriate specific
energy value from Table 21.2, determine the maximum feed that can be set for this operation. Use of
a spreadsheet calculator is recommended for the
iterative calculations required in this problem.


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Problems
21.23. A turning operation is to be performed on a 20 hp
lathe that has an 87% efficiency rating. The roughing cut is made on alloy steel whose hardness is in
the range 325 to 335 HB. The cutting speed is 375 ft/

min, feed is 0.030 in/rev, and depth of cut is 0.150 in.
Based on these values, can the job be performed on
the 20 hp lathe? Use Table 21.2 to obtain the
appropriate unit horsepower value.
21.24. Suppose the cutting speed in Problems 21.7 and
21.8 is 200 ft/min. From your answers to those
problems, find (a) the horsepower consumed in
the operation, (b) metal removal rate in in3/min,
(c) unit horsepower (hp-min/in3), and (d) the specific energy (in-lb/in3).
21.25. For Problem 21.12, the lathe has a mechanical
efficiency ¼ 0.83. Determine (a) the horsepower
consumed by the turning operation; (b) horsepower
that must be generated by the lathe; (c) unit horsepower and specific energy for the work material in
this operation.
21.26. In a turning operation on low carbon steel (175
BHN), cutting speed ¼ 400 ft/min, feed ¼ 0.010 in/
rev, and depth of cut ¼ 0.075 in. The lathe has a
mechanical efficiency ¼ 0.85. Based on the unit
horsepower values in Table 21.2, determine (a) the
horsepower consumed by the turning operation
and (b) the horsepower that must be generated
by the lathe.
21.27. Solve Problem 21.25 except that the feed ¼ 0.0075 in/
rev and the work material is stainless steel (Brinell
hardness ¼ 240 HB).
21.28. A turning operation is carried out on aluminum (100
BHN). Cutting speed ¼ 5.6 m/s, feed ¼ 0.25 mm/
rev, and depth of cut ¼ 2.0 mm. The lathe has a
mechanical efficiency ¼ 0.85. Based on the specific
energy values in Table 21.2, determine (a) the cutting power and (b) gross power in the turning

operation, in Watts.

505

21.29. Solve Problem 21.27 but with the following changes:
cutting speed ¼ 1.3 m/s, feed ¼ 0.75 mm/rev, and
depth ¼ 4.0 mm. Note that although the power used
in this operation is only about 10% greater than in
the previous problem, the metal removal rate is
about 40% greater.
21.30. A turning operation is performed on an engine
lathe using a tool with zero rake angle in the
direction of chip flow. The work material is an
alloy steel with hardness ¼ 325 Brinell hardness.
The feed is 0.015 in/rev, depth of cut is 0.125 in and
cutting speed is 300 ft/min. After the cut, the chip
thickness ratio is measured to be 0.45. (a) Using the
appropriate value of specific energy from Table
21.2, compute the horsepower at the drive motor, if
the lathe has an efficiency ¼ 85%. (b) Based on
horsepower, compute your best estimate of the
cutting force for this turning operation. Use the
orthogonal cutting model as an approximation of
the turning process.
21.31. A lathe performs a turning operation on a workpiece of 6.0 in diameter. The shear strength of the
work is 40,000 lb/in2 and the tensile strength is
60,000 lb/in2. The rake angle of the tool is 6 . The
cutting speed ¼ 700 ft/min, feed ¼ 0.015 in/rev, and
depth ¼ 0.090 in. The chip thickness after the cut is
0.025 in. Determine (a) the horsepower required in

the operation, (b) unit horsepower for this material
under these conditions, and (c) unit horsepower as
it would be listed in Table 21.2 for a to of 0.010 in.
Use the orthogonal cutting model as an approximation of the turning process.
21.32. In a turning operation on an aluminum alloy workpiece, the feed ¼ 0.020 in/rev, and depth of cut ¼
0.250 in. The motor horsepower of the lathe is 20 hp
and it has a mechanical efficiency ¼ 92%. The unit
horsepower value ¼ 0.25 hp/(in3/min) for this aluminum grade. What is the maximum cutting speed
that can be used on this job?

Cutting Temperature
21.33. Orthogonal cutting is performed on a metal whose
mass specific heat ¼ 1.0 J/g-C, density ¼ 2.9 g/cm3,
and thermal diffusivity ¼ 0.8 cm2/s. The cutting
speed is 4.5 m/s, uncut chip thickness is 0.25 mm,
and width of cut is 2.2 mm. The cutting force is
measured at 1170 N. Using Cook’s equation, determine the cutting temperature if the ambient temperature ¼ 22 C.
21.34. Consider a turning operation performed on steel
whose hardness ¼ 225 HB at a speed ¼ 3.0 m/s,
feed ¼ 0.25 mm, and depth ¼ 4.0 mm. Using values
of thermal properties found in the tables and
definitions of Section 4.1 and the appropriate

specific energy value from Table 21.2, compute
an estimate of cutting temperature using the
Cook equation. Assume ambient temperature ¼
20 C.
21.35. An orthogonal cutting operation is performed on a
certain metal whose volumetric specific heat ¼ 110
in-lb/in3-F, and thermal diffusivity ¼ 0.140 in2/sec.

The cutting speed ¼ 350 ft/min, chip thickness before the cut ¼ 0.008 in, and width of cut ¼ 0.100 in.
The cutting force is measured at 200 lb. Using
Cook’s equation, determine the cutting temperature if the ambient temperature ¼ 70 F.


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Chapter 21/Theory of Metal Machining

21.36. It is desired to estimate the cutting temperature for
a certain alloy steel whose hardness ¼ 240 Brinell.
Use the appropriate value of specific energy from
Table 21.2 and compute the cutting temperature by
means of the Cook equation for a turning operation in which the cutting speed is 500 ft/min, feed is
0.005 in/rev, and depth of cut is 0.070 in. The work
material has a volumetric specific heat of 210 in lb/
in3-F and a thermal diffusivity of 0.16 in2/sec.
Assume ambient temperature ¼ 88 F.
21.37. An orthogonal machining operation removes
metal at 1.8 in3/min. The cutting force in the
process ¼ 300 lb. The work material has a thermal
diffusivity ¼ 0.18 in2/sec and a volumetric specific

heat ¼ 124 in-lb/in3-F. If the feed f ¼ to ¼ 0.010 in
and width of cut ¼ 0.100 in, use the Cook formula
to compute the cutting temperature in the operation given that ambient temperature ¼ 70 F.

21.38. A turning operation uses a cutting speed ¼ 200 m/
min, feed ¼ 0.25 mm/rev, and depth of cut ¼ 4.00 mm.
The thermal diffusivity of the work material ¼ 20 mm2/
s and the volumetric specific heat ¼ 3.5 (10À3) J/mm3C. If the temperature increase above ambient temperature (20 F) is measured by a tool–chip thermocouple
to be 700 C, determine the specific energy for the
work material in this operation.
21.39. During a turning operation, a tool–chip thermocouple was used to measure cutting temperature.
The following temperature data were collected
during the cuts at three different cutting speeds
(feed and depth were held constant): (1) v ¼ 100 m/
min, T ¼ 505 C, (2) v ¼ 130 m/min, T ¼ 552 C,
(3) v ¼ 160 m/min, T ¼ 592 C. Determine an
equation for temperature as a function of cutting
speed that is in the form of the Trigger equation,
Eq. (21.23).


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22

MACHINING
OPERATIONS AND
MACHINE TOOLS

Chapter Contents
22.1 Machining and Part Geometry
22.2 Turning and Related Operations
22.2.1 Cutting Conditions in Turning
22.2.2 Operations Related to Turning
22.2.3 The Engine Lathe
22.2.4 Other Lathes and Turning Machines
22.2.5 Boring Machines
22.3 Drilling and Related Operations
22.3.1 Cutting Conditions in Drilling
22.3.2 Operations Related to Drilling
22.3.3 Drill Presses
22.4 Milling
22.4.1 Types of Milling Operations
22.4.2 Cutting Conditions in Milling
22.4.3 Milling Machines
22.5 Machining Centers and Turning Centers
22.6 Other Machining Operations
22.6.1 Shaping and Planing
22.6.2 Broaching
22.6.3 Sawing
22.7 Machining Operations for Special Geometries
22.7.1 Screw Threads
22.7.2 Gears
22.8 High-Speed Machining

Machining is the most versatile and accurate of all manufacturing processes in its capability to produce a diversity
of part geometries and geometric features. Casting can also
produce a variety of shapes, but it lacks the precision and
accuracy of machining. In this chapter, we describe the
important machining operations and the machine tools
used to perform them. Historical Note 22.1 provides a brief
narrative of the development of machine tool technology.

22.1 MACHINING AND PART
GEOMETRY
To introduce our topic in this chapter, let us provide an
overview of the creation of part geometry by machining.
Machined parts can be classified as rotational or nonrotational (Figure 22.1). A rotational workpart has a cylindrical or
disk-like shape. The characteristic operation that produces
this geometry is one in which a cutting tool removes material
from a rotating workpart. Examples include turning and
boring. Drilling is closely related except that an internal
cylindrical shape is created and the tool rotates (rather
than the work) in most drilling operations. A nonrotational
(also called prismatic) workpart is block-like or plate-like, as
in Figure 22.1(b). This geometry is achieved by linear motions
of the workpart, combined with either rotating or linear tool
motions. Operations in this category include milling, shaping,
planing, and sawing.
Each machining operation produces a characteristic
geometry due to two factors: (1) the relative motions between the tool and the workpart and (2) the shape of the
cutting tool. We classify these operations by which part
shape is created as generating and forming. In generating,
the geometry of the workpart is determined by the feed

trajectory of the cutting tool. The path followed by the tool
during its feed motion is imparted to the work surface in
order to create shape. Examples of generating the work
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