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

5

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

6

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

13

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

14

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

15

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

16

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.

17

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

18

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 −+=

19

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 −+=

20

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

21

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 =

22

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

23

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

24

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

25

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 =