Today, February 25th 2.008 Design & Manufacturing

� HW#2 due before the class, #3 out on II the web after the class. � Math Formulae, handoutSpring 2004

� Lab groups fixed, and thank you. � group report!!!

Metal Cutting I � Metal cutting demo � Cutting physics

2.008-spring-2004 S.Kim 1

22.008-spring-2004 S.Kim

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Material removal processes

� Cost: �

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sun �

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A lathe of pre WWII

2.008-spr ng-2004 S.Kim

Expensive $100 -$10,000

Quality: Very h

Flexibility: Any shape under the

Rate: Slow

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Machined Surface

µ inch

1000 ium 500

125

Fine 32 Very fine 16

8

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Surface roughness by machining

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Rough MedAvg.Better than Avg. 63

Extremely fine

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ss VF ⋅:shearingfor Power

cc VF ⋅:input

cVF ⋅:friction viadissipatedPower

VtwVFu

o

sss ⋅⋅

⋅=:shearingfor energy Specific

VtwVFuo

cf ⋅⋅

⋅=:frictionfor energy Specific

VtwVF

VtwVFuu

o

ss

o

cfs ⋅⋅

⋅+

⋅⋅⋅

=+:energyspecificTotal

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Cutting processes

�

� Product quality: surface, tolerance � Productivity: MRR , Tool wear

� Physics of cutting � Mechanics � Force, power

� Tool materials � i

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Cutting Tools

2.008-spr ng-2004 S.Kim

Why do we study cutting physics?

Des gn for manufacturing

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Cutting process modeling

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� What are the forces involved? �

� How do the above relate to power

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Orthogonal cutting in a lathe

Rake angle

To

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Methods: Modeling and Experiments

Key issues How does cutting work?

What affect does material properties have?

requirements, MRR, wear, surface?

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Shear angle

: depth of cut

Shear plane

Assume a hollow shaft

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Cutting tool and workpiece..

Power

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Varying rake angle α:

- α α = 0

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Primary shear zone

Secondary shear zone

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2

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Basic cutting geometry

�

( ) ( )cos

φsin r

ct ot

V cV

− ===

ctcVo ⋅

r <1

V

Vc

tcto

Continuity

to

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Velocity diagram in cutting zone

( ) ( ) ( )φsin cV

αcos sV

cos V

== −

( ) ( )cos

φsin r

ct ot

V cV

− ===

( ) ( )cos

φVsin cV

− =

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We will use the orthogonal model

Shear angle

α φ

t V = ⋅

Cutting ratio:

: chip thickness : depth of cut

2.008-spr ng 2004 S.Kim

α φ

α φ

α φ

Law of sines

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Forces and power

� FBD at the tool-workpiece contact � What are the forces involved

� Thrust force, Ft � Cutting force, Fc � Resultant force, R � Friction force, F � Normal Force, N � Shear Force, Fs, Fn

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ian

Fc

α

φ

R

Ft

Fc

Fn

Fs

F

N α

β

Ft β−α

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E. Merchant’s cutting diagram

Source: Kalpajk

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( ) ( ) α )βRcos(φφsintFφcoscFsF =⋅−⋅=

( ) ( )φcostFφsincFnF ⋅+⋅=

( )βF ⋅=

( )βN ⋅= ( )βtanµ =

β =

FBD of Forces

( ) ( )αtantFcF

αtancFtF

N F

µ ⋅+

==

( )α-βsinRFt ⋅=

( )α-βcosRFc ⋅=

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� What does this mean: �

� Merchantto minimize cutting force or max. shear stress

�

22 45 βαφ = o

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− +

sin R cos R

Angle Friction

2 µ 0.5 :Typcially < <

⋅ −

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Analysis of shear strain

Low shear angle = large shear strain ’s assumption: Shear angle adjusts

Can derive: − +

3

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Shear angle

( ) α ­βcosRFc

( ) ( ) α)βRcos(φφsintFφcoscFsF =⋅=

α)βcos(φ/α)cos(βFscF −=

As.σssF =

α)βcos(φ/α)cos(βφsin A

σcF s −=

0dFc/dφ =

2245 βαφ = o

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� ion increases �

�

� ion via shear �

�

2 β

2 α

45φ o

−+=

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⋅ =

− + ⋅ −

− +

strength) shear plane. shear of (area

− +

− +

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Things to think about

As rake angle decreases or frictShear angle decreases Chip becomes thicker Thicker chip = more energy dissipatMore shear = more heat generation Temperature increase!!!

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Power

ss VF ⋅

VF c ⋅

c ⋅

V VF u

o

ss s ⋅ ⋅

⋅ =

Vu

o

c f ⋅ ⋅

⋅ =

V VF

Vuu

o

ss

o

c fs ⋅ ⋅

⋅ +

⋅ ⋅ ⋅

=+

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imate)

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: shearing for Power

: input Power

V F : friction via dissipated Power

t w : shearing for energy Specific

t w V F : friction for energy Specific

t w t w V F : energy specific Total

MRR

=> shearing + friction

Experimantal data 2.008-spr ng-2004 S.Kim Kalpak an

Specific energy (rough est

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Example � i�

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2.0

1.01.25

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Cutting zone pictures BUE

serrated di

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Consider the turning with a rod, from 1.25 nch diameter to 1.0. to; depth of cut, 0.005 inch f; feed rate, 0.025 inch/rev

; spindle speed, 1000 rpm uf;spec c energy for fr ct on us;specific energy for shear, 0.35 hp min/in1 hp = 550 ft. bf/s

How many passes? Tools speed? Time to make this part? V c max? Max power needed? Initial cutting force?

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continuous secondary shear

scontinuous

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