mechanics of cutting - iit kanpurhome.iitk.ac.in/~vkjain/l6-ta-202 mechanics of cutting.pdf ·...
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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