Mechanics of Cutting

&

,sin(90 ( )) cos( )

sincos( )

u

c c

u

c

ABC ABDtAB

sint talso AB

tt

φ

φ α φ αφ

φ α

∆ ∆

=

= =− − −

=−

: : :

: :

:

c

u

f

s

c

t Chip thicknesst Uncut chip thicknessV Chip Sliding VelocityV Shear VelocityV Cutting Velocity

Shear Angleφ

Fig: Schematic of Geometry of chip formation

Geometry of chip Formation:

φ

90-ф+α = 90-(ф-α)

How to determine φ & rc ? tc should be determined from the chip. tu (= feed) and α are already known.To determine tc with micrometer, is difficult and not so because of

uneven surface. How? (say, f=0.2 mm/rev. An error of even 0.05 mm will cause an error of 25 % in the measurement of tc)

Volume Constancy Condition: Volume of Uncut chip = Volume of cut chip

: /

1

1(1 ) tan

ctan1

uc

c

c

c c

c c

c

c

tr Chip thickness Ratio Coeffinicienttcos cos sin sin

r sinr cot cos r sin

r cos r sin

r osr sin

φ α φ αφ

φ α αα α φ

αφα

=

+=

= += −

∴ = −

φ

90-ф+α= 90-(ф-α)

Substitute the value of tu /tc

from earlier slide and simplify to get:

SHEAR ANGLE AND CHIP THICKNESS RATIO EVALUATION

3Dr. V.K.jain, IIT Kanpur

u u c cL t b L t b=

c c u uL t L t∴ =

, u cc

c u

t Lor rt L

= =

c

u

L = Chip length L = Uncut chip lengthb = Chip width (2-D Cutting)

SHEAR ANGLE AND CHIP THICKNESS RATIO EVALUATION

LENGTH OF THE CHIP MAY BE MANY CENTIMETERS HENCE THE ERROR IN EVALUTION OFrc WILL BE COMPARATIVELY MUCH LOWER.

(rc = Lc / Lu)

4Dr. V.K.jain, IIT Kanpur

Clearance Angle

Work

ToolChip

Ft

Fc

F

N

Fn

Fs

α

α

β

∅

R

79

α

G

EA

B

∅

D

Fα

Δ FAD = (β - α)Δ GAD = φ + (β - α)

FORCE CIRCLE DIAGRAM

Forces in Orthogonal Cutting:• Friction force, F• Force normal to Friction force, N• Cutting Force, FC • Thrust force, Ft • Shear Force, FS •Force Normal to shear force, Fn •Resultant force, R

'

'S N c t

R F N

R F F F F R

→ →

→ → → → →

= +

= + = + =

Force Analysis

A

C

F

D

G

Force Circle Diagram

6

FREE BODY DIAGRAM

Dr. V.K.jain, IIT Kanpur

c st cF F os F inα α= +

c sc tN F os F inα α= −

( )Coefficient of Friction µ

c stan c

t c

c t

F os F inFN F os F sin

α αµ βα α+

= = =−

Friction Angleβ =

tan tan

t c

c t

F FF F

αµα

+=

−1, tan ( )also β µ−=

7Dr. V.K.jain, IIT Kanpur

FORCE ANALySIS

DIVIDE R.H.S. BY Cos α

c sS c tF F os F inφ φ= −

c sN t cF F os F inφ φ= +

,also

c ( )CF R os β α= −

S c ( )F R os φ β α= + −

( )( )

C

S

F cosF cos

β αφ β α

−∴ =

+ −

( ) u uS

t b tShearPlaneArea A bsin sinφ φ

= = ×

8Dr. V.K.jain, IIT Kanpur

Foce Analysis

Δ FAD = (β - α)Δ GAD = φ + (β - α)

Let be the strength of work materialτ

uS S

t bF Asin

τ τφ

= =

C( )F

( )ut b cos

sin cosτ β αφ φ β α

−= + −

,and 1( )

ut bRsin cosτφ φ β α

= × + −

( ) s ( )( )

ut

t b sinF R insin cosτ β αβ αφ φ β α

−= − = ×

+ −

tan( )t

c

FF

β α= −

9Dr. V.K.jain, IIT Kanpur

Foce Analysis

( ) Schip

S

FMean Shear Stress tA

= ( )On Chip

( c s )sin

c t

u

F os F inb t

φ φ φ−=

( ) Nchip

S

FMean Normal StressA

σ = ( )On Chip

( c s )sin =

t c

u

F os F inb t

φ φ φ+

10Dr. V.K.jain, IIT Kanpur

Foce Analysis

VELOCITY ANALYSIS

: cV Cutting velocity of tool relative to workpiece

: fV Chip flow velocity

: sV Shear velocity

Using sine Rule:

sin(90 ( )) sin sin(90 )fc sVV V

φ α φ α= =

− − −

cos( ) sin cosfc sVV V

φ α φ α= =

−

sin cos( )

cf c c

Vand V V rφφ α

= = ⋅−

cos coscos( ) cos( )

c ss

c

V VVV

α αφ α φ α

= ⇒ =− − 11

Shear Strain & Strain Rate

Two approaches of analysis:

Thin Plane Model:- Merchant, PiisPanen, Kobayashi & Thomson

Thick Deformation Region:- Palmer, (At very low speeds) Oxley, kushina, Hitoni

Thin Zone Model: Merchant

ASSUMPTIONS:-

• Tool tip is sharp, No Rubbing, No Ploughing

• 2-D deformation.

• Stress on shear plane is uniformly distributed.

• Resultant force R on chip applied at shear plane is equal, opposite and

collinear to force R’ applied to the chip at tool-chip interface. 12

Expression for Shear StrainThe deformation can be idealized as a process of block slip (or preferred slip planes)

( ) deformationShearStrainLength

γ =

s AB AD DBy CD CD CD

γ ∆= = = +∆

tan( ) cotφ α φ= − +

sin( )sin cos cos( ) ,sin cos( )

φ α φ ϕ φ αφ φ α

− + −−

cossin cos( )

αγφ φ α

∴ =−

13Dr. V.K.jain, IIT Kanpur

In terms of shear velocity (V s ) and chip velocity (Vf), it can be written as

cos sincesin cos( )

s s

c c

V VV V

αγφ φ α

∴ = = −

( )Shear strain rate γ•

1syd s

dt dt y yγγ

•

∆ ∆ ∆ = = = ∆ ∆

s c coscos( )

V Vy y

αφ α

= =∆ − ∆

, : where y Mean thickness of PSDZ∆14Dr. V.K.jain, IIT Kanpur

Expression for Shear Strain rate

Shear angle relationship

• Helpful to predict position of shear plane (angle φ)

• Relationship between-Shear Plane Angle (φ)Rake Angle (α)Friction Angle(β)

Several TheoriesEarnst-Merchant(Minimum Energy Criterion):Shear plane is located where least energy is required for shear.

Assumptions:-• Orthogonal Cutting.• Shear strength of Metal along shear plane is not affected by

Normal stress.• Continuous chip without BUE.• Neglect energy of chip separation.

15Dr. V.K.jain, IIT Kanpur

Assuming No Strain hardening:

Condition for minimum energy, 0cdFdφ

=

u 2 2

cos cos( ) sin sin( ) cos ( )sin cos ( )

cdF t bd

φ φ β α φ φ β ατ β αφ φ φ β α

+ − − + −= − + − 0=

cos cos( ) sin sin( ) 0φ φ β α φ φ β α∴ + − − + − =

cos(2 ) 0φ β α+ − =

22πφ β α+ − =

1 ( )4 2πφ β α∴ = − −

16Dr. V.K.jain, IIT Kanpur

Shear angle relationship

17

THANK YOU

Dr. V.K.jain, IIT Kanpur