1
MATERIAL REMOVAL PROCESSES
Theory of Metal Machining
1. Overview2. Theory of Chip Formation
3. Force Relationships4. Power and Energy Relationship
5. Cutting Temperature
2
Introduction• Everyday Experience: Scraping
the ice from your windshield– Edge angle of the ice scraper– Force required depending on the
characteristics of ice• Incentives: Making a ceramic
vase out of clay– Shaping– Removal of excess materials -
‘machining’• Powder Metal or Cast
– Exact dimension– Tolerance & Surface Finish
3
Classification
Material Removal Processes
ConventionalMachining
ChemicalMachining
Thermal Energy Processes
Electrochemical Processes
Other AbrasiveProcesses
Grinding Processes
Other Machining Operations
Milling
Drilling and RelatedOperations
Turning and Related Operations
Abrasive Processes
Nontraditionalmachining
Mechanical Energy Processes
4
Material Removal ProcessesA family of shaping operations through which
undesired excess material is removed from a starting workpart so the remaining part become closer to the desired shape
• Categories:– Machining – material removal by a sharp cutting
tool, e.g., turning, milling, drilling– Abrasive processes – material removal by hard,
abrasive particles, e.g., grinding– Nontraditional processes - various energy forms
other than sharp cutting tool to remove material
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Machining
• A shearing process in which excess materials is removed by cutting tools.– A variety of work materials– ‘Repeatable’ regular geometries– Close tolerance (<0.025μm)– Smooth surface finish (0.4μm)– Waste, Expensive: Cost and Time– Other processes such as casting, forging, and bar
drawing create the general shape– Machining provides the final shape, dimensions,
finish, and special geometric details
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1. Overview
Feed Motion (work)
Work material
Speed motionMilling Cutter
New Surface
Cutting tool
Work
Speed motion(work)
New Surface
Feed Motion (tool)
Speed motion(Tool)
Work
Drill bitFeed Motion (tool)
Feed Motion (work)
Work
New Surface
Speed motion
Milling Cutter
• Types– Turning - Lathe– Drilling – Drill press– Milling – Milling Machine
• Peripheral• Face
• Cutting Tool
Peripheral (End) Milling Face (Slab) Milling
7
Cutting condition• Relative motion between tool and work• Cutting conditions
– Cutting speed, v (m/s) – Surface speed– Feed f (m): the lateral distance traveled by the tool
during one revolution.– Depth of cut d (m)
• Material Removal Rate:– Roughing - removes large amounts of material, at
high feeds and depths, low speeds– Finishing - Achieves final dimensions, tolerances,
and finish, Low feeds and depths, high cutting speeds
dfvMRR =
8
Machine Tools
• A power-driven machine that performs a machining operation– Holds workpart– Positions tool relative to work– Provides power and controls speed, feed, and
depth.– Pumps a Cutting fluid
df
v
9
2. Theory of Chip Formation• Orthogonal Cutting Model
Chip thickness ratio:
( )αφφ−
==cos
sin
s
s
c
o
ll
ttr
Tool
Workls
α
tc
to φ
ShearPlane
Rake angle: αShear angle: φ
ααφ
sin1costanr
r−
=
By rearranging
is always less than 1.0.r
10
Shear Strain in chip
C
( )
( )( ) ( )αφφ
αφφ
αφαφ
φαφγ
−=+
−−
=
+−=+
==
cossincos
sincos
cossin
cottanBD
ADDCBDAC
AB
D
As φ (from 10° to 35°) increase, γ (from 5 to 2) decreases.
φ
α
φ
α
φ−α
A B
CD
( )( ) βαβαβα
βαβαβαsinsincoscoscossincoscossinsin
m=±±=±Using
11
Velocity
( ) φααφ
φαπαφπ
sincoscos
sin2
sin2
sin
cs
cs
VVV
VVV
==−
=⎟⎠⎞
⎜⎝⎛ −
=⎟⎠⎞
⎜⎝⎛ +−
Tool
Work
α
φV Vs
Vc
φ90-α
90-φ+αVs
Vc
V
φ-αα
( ) yV
yVs
Δ−=
Δ=
αφαγ
coscos
&
where Δy is the finite thickness of the shear plane, typically 0.03mm.Shear Strain rate is around 103-105sec-1
12
Actual Chip Formation
Tool
Work
Effective φ
Primary Shear Zone
(a) Discontinuous chip•Brittle materials at low cutting speed•High tool-chip friction and large feed and depth
(b) Continuous chip•Ductile materials with high speeds and small feed and depth of cut
(c) Continuous chip with built-up edge•Ductile material at low to medium speeds
(d) Serrated chip•Difficult-to-machine metals at high cutting speeds
Secondary Shear Zone
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The ‘Real’ Cutting Force
R
Ft
Fn
Ft
Frσ
Cutting Forces are measured with Dynamometer.
•Area: A=bh where b=chip width
h=chip thickness•Temperature (500-1000oC)• Pressure (1000-3000 MPa)
Shear
Normal
Sticking Zone Sliding Zone
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3. Force Relationships
Tool
Work
Tool
Work
α
φF
N
R
R’Fn
Fs
Ft
Fc
R”
β
Fc: Cutting ForceFt: Thrust Force
F: Friction ForceN: Normal Force
Fs: Shear ForceFn: Normal Force
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Force Diagram
α
φ
β
Ft
Fc
Fn
Fs
F
N α
φφφφαααα
cossinsincossincoscossin
tcn
tcs
tc
tc
FFFFFFFFNFFF
+=−=−=+=
( )
( )( )
( )( )( )αβφφ
αβαβφφ
αβαβφφ
ταβφφ
−+−
=
−+−
=
−+=
−+=
cossinsincossincos
cossin
cossin
st
sc
os
s
FF
FF
wt
FR
β−α
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Cutting Force• Cutting Force:
• Thrust Force:
( )( ) [ ]
( )( ) bhKbhF
mmNKbhKbhF
tst
ccsc
=⎥⎦
⎤⎢⎣
⎡−+
−=
=⎥⎦
⎤⎢⎣
⎡−+
−=
αβφφαβτ
αβφφαβτ
cossincossin
/cossin
cos 2
h f
db
Top View
( )
( ) φφαβφφ
τφφαβφ
cossinsinsin
sincoscos
tcn
stcs
FFRF
bhFFRF
+=−+=
=−=−+= ( )( )αβ
αβ−=−=
sincos
RFRF
t
c
Tool
Work
α
tc
toφ
Ft
Fc R
Kc and Kt must be calibrated through machining experiments.
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Chatter Analysis• Mechanical vibration
0=++ kxxcxm &&&k c
F kx xc&
F )(tx
( )φω
+==++
wtXtxtFkxxcxm o
sin)(Assumesin&&&
( )
( ) ( ) tjjo
tjj
tj
tjjo
eeFtFeXecjmkXetx
eeFkxxcxm
ωαωφ
φω
ωα
ωω ==+−
=
=+++
2
)(Assume
&&&
Or using complex harmonic functions
( )( ) ( )
n
o
rrr
rrkFXw
ωω
αζφ
ζ
=
+−
−=
+−==Φ
−
where1
2tan
21
11
21
2221 r
r
φ
( )wΦ
-90
-180
0
Magnitude ratio:
Phase:
Free Vibration:
Forced Vibration:
1
2
kmc 2=ζ
and mk
n =ω
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The Merchant Equation• Shear stress:
• Shear Plane Area:
• Shear stress:
• Merchant’s Assumption: Shear plane angle will form to minimize energy
• After differentiating τ w.r.t φ, Merchant’s Equation:
s
s
AF
=τ
φsinwtA o
s =
φφφτ
sinsincos
wtFF
o
tc −=
2245 βαφ −+=
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Implication of Merchant’s Eq.
• An increase in rake angle causes the shear plane angle to increase.
• A decrease in friction angle cause the shear plane angle to increase.
• The analysis from orthogonal cutting can be used in a typical turning if the feed is small relative to depth of cut.
Effect of shear plane angle φ : (a) higher φ with a resulting lower shear plane area; (a) smaller φ with a resultinglarger shear plane area.
ToolTool
2245 βαφ −+=
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Turning vs. OrthogonalFeed f Uncut Chip thickness to
Depth d Width of cut w
Cutting speed v Cutting speed v
Cutting force Fc Cutting force Fc
Feed force Ff Thrust force Ft
Tool
Work
α
tc
toφ
FtFc
d
f
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4. Power & Energy Relation• Power (energy per unit time)
• Horse power
• Unit Power
• Specific energy
000,33VF
HP
VFP
cc
cc
=
=
wtF
wvtvF
MRRP
PUo
c
o
ccu ====
000,33Phpc =
Ehp
hpEP
P cg
cg == or
with mechanical efficiency E=90%
P in ft-lb/min
MRRP
P cu =
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Size Effect & Energy Distribution
1 2 3
100
50
0P
ropo
rtion
of E
nerg
y
Tool
Work
Chip
Cutting Speed (m/s)
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.125 0.25 0.5 1.0Chip thickness before cut (mm)
Cor
rect
ion
Fact
or
0.01 (in)0.04
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Specific Energy for various work materials (to=0.25mm)
Materials BrinellHardness
Specific Energy (U)N·m/mm3 In-lb/in3 Hp/in3/min
Carbon Steel 150-200200-250251-300
1.62.22.8
240,000320,000400,000
0.60.81.0
Alloy Steels 200-250251-300301-350351-400
2.22.83.64.2
320,000400,000520,000640,000
0.81.01.31.6
Cast iron 125-175175-250
1.11.6
160,000240,000
0.40.6
Stainless steel 150-200 2.8 400,000 1.0
Aluminum 50-100 0.7 100,000 0.25
Aluminum Alloys 100-150 0.8 120,000 0.3
Magnesium Alloys 50-100 0.4 60,000 0.15
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Problem 21.30A lathe performs a turning operation on a work piece of 6in diameter. The shear strength of the
work=40,000lb/in2. The rake angle of the tool =10o. The machine settings are: rotational speed=500rev/min, feed=0.0075in/rev. and depth=0.075in. The chip thickness after the cut is 0.015in. Determine: (a) the horsepower required (b) the unit horsepower for this material, (c) the unit horsepower with the correction factor (1 for to=0.01in.) Use the orthogonal model.
( )( )
hp2000,33
)785(6.83000,33
minft/785rev/ft)12/6min(/rev500
lb6.83)104.433.28cos(
)103.43cos(5.47cos
coslb5.47)00119.0(000,40
in00119.03.28sin075.00075.0
sin
4.433.282
10452;22
45
3.2810sin5.01
10cos5.0arctan;sin1
costan
5.0015.00075.0
2
===
===
=−+
−=
−+−
=
===
=⋅
==
=⎟⎠⎞
⎜⎝⎛ −+=−+=
=⎟⎠⎞
⎜⎝⎛
−=
−=
===
vFHP
rv
FF
SAF
wtA
rr
tt
r
c
sc
ss
os
o
πωαβφαβ
φ
ββαφ
φα
αφ
o
o
(a) To get HP, Fc and v are needed
(b) HPu
/min)hp/(in375.03.5
2minin3.5)075.0)(0075.0(785
3
3
===
===
MRRHPHP
vfdMRR
u
(c) HPu with the correction factor Fig. 21.14
/min)hp/(in326.015.1375.0 3===
fHP
HP uu
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5. Cutting Temperature• Cook’s dimensional analysis
• Experimental Measurement– Tool-chip thermocouple– Trigger’s results
• RC-130B Ti (T=479v0.162)• 18-8 Stainless steel (T=135v0.361)• B113 Free machining steel (T=86.2v0.348)
333.0
4.0 ⎟⎠⎞
⎜⎝⎛=
Kvt
CUT o
ρ
mKvT =