337: Materials & Manufacturing
Processes
Lecture 6:
Machining Operations
and Machinability
Chapter 22 and 24
This Time
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2
Parameters
Material Removal Rate
Power Requirements
Surface Finish
Machinability
Turning
Single point cutting tool removes material from
a rotating workpiece to form a cylindrical
shape
3
Turning

A single point cutting tool removes material from a
rotating workpiece to generate a rotationally symmetric
shape
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Machine tool is called a lathe
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Types of cuts:
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4
Facing
Contour turning
Chamfering
Cutoff
Threading
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Workholding methods:
 Holding the work between
centers
 Chuck
 Collet
 Face plate
Turning Parameters Illustrated
5
Primary Machining Parameters

Cutting Speed – (v)
 Primary motion
 Peripheral speed

ft/min
mm/rev
mm/tooth
in/rev
in/tooth
Feed – (f)
 Secondary motion
 Turning:
 Milling:

m/s
Depth of Cut – (d)
 Penetration of tool below original work surface
 Single parameter
mm

Resulting in Material Removal Rate – (MRR)
MRR = v f d
mm3/s
in3/min
where v = cutting speed; f = feed; d = depth of cut
6
in
Machining Calculations: Turning
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Spindle Speed - N
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Feed Rate - fr
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7
Do = outer diameter
Df = final diameter
Machining Time - Tm
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f = feed per rev
Depth of Cut - d
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v = cutting speed
Do = outer diameter
L = length of cut
Mat’l Removal Rate - MRR
v
N
π Do
fr  N f
Do  Df
d
2
L
Tm 
fr
MRR  v f d
(rpm)
(mm/min -or- in/min)
(mm -or- in)
(min)
(mm3/min -or- in3/min)
Example
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8
In a production turning operation, the foreman
has decided that a single pass must be
completed on a cylindrical workpiece in 5.0
min. The piece is 400 mm long and 150 mm in
diameter. Using a feed = 0.30 mm/rev and a
depth of cut = 4.0 mm, what cutting speed
must be used to meet this machining time
requirement?
Example: Solution
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Tm = L/fr = L/Nf =
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v = pDoL/fTm
pDoL/vf
= p(0.4)(0.15)/(0.30)(10-3)(5.0)
= 0.1257(103) m/min
= 125.7 m/min
9
Power and Energy Relationships

Power requirements to perform machining can
be computed from:
Pc = Fc v
N-m/s (W)
ft-lb/min
where:
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10
Pc = cutting power;
Fc = cutting force; and
v = cutting speed
Customary U.S. units for power are Horsepower (= 33000 ft-lb/min)
Power and Energy Relationships

The Gross machine power (Pg) available is:
Pc = Pg• E
where E = mechanical efficiency of machine tool
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Typical E for machine tools =  80 - 90%
Note: Textbook relationship is same -
Pc
Pg 
E
11
HPc
HPg 
E
Unit Power in Machining
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
Useful to convert power into power per unit volume
rate of metal cut
Called the unit power, Pu or unit horsepower, HPu
Pc
Pu 
MRR
HPc
HPu 
or
MRR
where MRR = material removal rate
 Tool sharpness is taken into account multiply by 1.00 – 1.25
 Feed is taken into account by multiplying by factor in Figure
21.14
12
Specific Energy in Machining

Unit power(Pu) is also known as the specific energy (U),
or the power required to cut a unit volume of material:
Pc
Fc
U  Pu 

MRR t ow
where
t0 = un-deformed chip thickness;
w = width of the chip; and
Fc = cutting force
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13
Units for specific energy are typically N-m/mm3 (J/mm3)
or in-lb/in3
Table 21-2 (p. 497) in the text approximates specific
energy for several materials based on est. hardness
Example
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14
In a turning operation on
stainless steel with hardness =
200 HB, the cutting speed =
200 m/min, feed = 0.25 mm/rev,
and depth of cut = 7.5 mm. How
much power will the lathe draw
in performing this operation if its
mechanical efficiency = 90%.
From Table 21.2, U = 2.8 Nm/mm3 = 2.8 J/mm3
Pc
Pu 
MRR
Example: Solution
MRR = vfd
= (200 m/min)(103 mm/m)(0.25 mm)(7.5 mm)
= 375,000 mm3/min = 6250 mm3/s

Pc = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s
= 17,500 W = 17.5 kW
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Accounting for mechanical efficiency, Pg
= 17.5/0.90 = 19.44 kW
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15
What if feed changes?
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Facing
Tool is fed
radially inward
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Contour Turning
Instead of feeding the
tool parallel to the axis
of rotation, tool follows
a contour that is other
than straight, to create
a contoured form
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Chamfering
Cutting edge cuts
an angle on the
corner of the
cylinder, forming a
"chamfer"
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Cutoff
Tool is fed radially
into rotating work
at some location to
cut off end of part
20
Threading
Pointed form tool is
fed linearly across
surface of rotating
workpart parallel to
axis of rotation at a
large feed rate to
create threads
21
Engine Lathe
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Boring
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Difference between boring and turning:
 Boring is performed on the inside diameter of an
existing hole
 Turning is performed on the outside diameter of an
existing cylinder
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In effect, boring is an internal turning operation
Boring machines
 Horizontal or vertical - refers to the orientation of the
axis of rotation of machine spindle
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Drilling
Used to create a round hole, usually by means
of a rotating tool (drill bit) that has two cutting
edges
24
Through Holes vs. Blind Holes
Through-holes - drill exits the opposite side of work
Blind-holes – drill does not exit work on opposite side
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Two hole types: (a) through-hole, and (b) blind hole
Machining Calculations: Drilling
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Spindle Speed - N
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Feed Rate - fr
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v = cutting speed
D = tool diameter
f = feed per rev
Machining Time - Tm
 Through Hole :
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t = thickness
 = tip angle
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(rpm)
fr  N f
(mm/min -or- in/min)
t  D tan90 
Tm 
fr
 Blind Hole :

v
N
πD
d = depth
Mat’l Removal Rate - MRR
1
2
Tm 

2

(min)
d
fr
π D 2 fr (mm3/min -or- in3/min)
MRR 
4
Milling
Rotating multiple-cutting-edge tool is moved
slowly relative to work to generate plane or
straight surface
 Two forms: peripheral milling and face milling
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Milling

Machining operation in which work is fed past
a rotating tool with multiple cutting edges
 Axis of tool rotation is perpendicular to feed
direction
 Creates a planar surface; other geometries
possible either by cutter path or shape
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Milling Parameters Illustrated
Two forms of milling:
(a) peripheral milling, and (b) face milling
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Slab Milling
The basic form
of peripheral
milling in which
the cutter width
extends beyond
the workpiece
on both sides
(tool axis parallel
to machined
surface)
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Conventional Face Milling
Cutter overhangs work
on both sides
(tool axis perpendicular to
machined surface)
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Machining Calculations: Milling
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Spindle Speed - N
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Feed Rate - fr
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v = cutting speed
D = cutter diameter
f = feed per tooth
nt = number of teeth
Machining Time - Tm
 Slab Milling:
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L = length of cut
d = depth of cut
(rpm)
v
N
πD
fr  N nt f
L  d D  d 
Tm 
fr
LD
Tm 
w = width of cut
fr
2nd form is multi-pass
 Face Milling:
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(mm/min -or- in/min)
Mat’l Removal Rate - MRR
-or-
MRR  w d fr
(min)
L  2 w D  w 
Tm 
fr
(mm3/min -or- in3/min)
Example
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33
A face milling operation is used to machine 5
mm from the top surface of a rectangular piece
of aluminum 400 mm long by 100 mm wide.
The cutter has four teeth (cemented carbide
inserts) and is 150 mm in diameter. Cutting
conditions are: v = 3 m/s, f = 0.27 mm/tooth,
and d = 5.0 mm. Determine the time to make
one pass across the surface.
Example: Solution
Tm 
fr  N nt f
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LD
fr
N
v
πD
Example: Solution
N = (3000 mm/s)/150p = 6.37 rev/s
fr = 6.37(4)(.27) = 6.88 mm/s
N
v
πD
fr  N nt f
LD
Tm 
fr
Tm = (400 + 150)/6.88 = 80 s = 1.33 min.
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You should have learned
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36
Parameters
Material Removal Rate
Power Requirements
Surface Finish
Machinability
Assignment
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HW 2 (Due Tuesday):
 CH 21,22 and 24 Problems
 In Assignments folder
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Next Time
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Casting
Chapter 10
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