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1
Machining processes
BOOKS
1.Manufacturing Technology, Volume 2, P.N. Rao, McGraw
Hill Education (India) Private Limited.
2.Manufacturing Processes for Engineering Materials, S.
Kalpakjain, Addision-Weseley Publishing Company.
3.Fundamentals of Metal Cutting and Machine Tools, B.L.
Juneja, G.S. Sekhon, New Age International Private Limited.
4.Engineering Metrology, Jain, Khanna Publishers
5.CAD/CAM: Computer-Aided Design and Manufacturing by
M.P. Groover & E.W. Zimmers, Jr.
6.CNC Machines by B.S. Pabla & M. Adithan
2
Machine Tools
A power-driven machine that performs a
machining operation, including grinding
• Functions in machining:
– Holds workpart
– Positions tool relative to work
– Provides power at speed, feed, and depth
that have been set
• The term is also applied to machines that
perform metal forming operations
3
Mechanics of machining
4
Important parameters of machining
1. Thickness of uncut layer
2. Thickness of chip
3. Inclination of the chip-tool interface with respect to the face of
the tool (rake angle)
4. The relative velocity of the work piece and the cutting tool
A clearance angle between the work and the flank surface
make the cutting possible.
5
6
Mechanism of chip formation
7
Cutting ratio
8
Shear angle
9
Shear angle
 





sinr1
cosr
tan
cos
sin
r
t
t
rRatio_Cutting
c





10
Shear strain
 
 








cossin
cos
,or
)tan(cot
BD
DCAD
BD
AC
strain_Shear
11
Shear Strain
Shear strain in machining can be computed
from the following equation, based on the
preceding parallel plate model:
 = tan( - ) + cot 
where  = shear strain,  = shear plane angle,
and  = rake angle of cutting tool
Chip velocity, Vc
• Maintaining mass continuity
12
)cos(
sinV
V
rVV
velocity_cuttingV
tVtV
c
f
cf
cwhere
2f1c







13
Velocity diagram
14
 
 
 
  Y
V
cos
cos
Y
V
,,
tY
S
ratestrain
sinV
V
cossin
cos
,or
)tan(cotstrainShear
sin
V
cos
Vs
cos
V
s
s
f





















(90o-Φ+α)
α
Φ
(90o-α)
Vs
V
Vf
Where, ∆Y = thickness of the deformation zone
∆t = time taken
15
Types of chips
• Continuous chip without built-
up edge (BUE)
• Continuous chip with built-up
edge (BUE)
• Serrated chip
• Discontinuous chip
Continuous chip without built-up edge (BUE)
•Ductile material
•Small uncut thickness
•High cutting speed
•Large rake angle
•Suitable cutting fluid
16
Continuous chip with built-up edge (BUE)
•Stronger adhesion between chips and
tool face
•Large uncut thickness
•Low rake angle
17
18
Continuous chip with BUE
Serrated chip
•Appears as saw teeth
•Low conductivity metal
•Strength of the metal decreases
sharply with temperature
•E.g. titanium
19
Discontinuous chip
•Brittle material
•Large uncut thickness
•Low cutting speed
•Small rake angle
20
21
Problem
The following data are available from an orthogonal cutting
operation:
Determine the (a) cutting ratio (b) shear angle (c) shear strain and strain rate on
shear plane(d) chip velocity and (e)shear velocity.
Work material Aluminium Steel
Uncut chip thickness (t1), mm 0.13 0.23
Width of cut, mm 2.5 2.5
Rake angle 5o -5o
Cutting speed, m/s 2 2
Chip thickness (t2), mm 0.23 0.58
Orthogonal cutting
22
23
Orthogonal cutting
• Can be represented by a 2-dimensional figure
• The work move in the plane parallel to the
plane of the paper
• The chip material particles also move in the
plane parallel to the plane of the paper
• No component of velocity in the direction
perpendicular to the plane of the paper
24
Orthogonal Cutting Model
A simplified 2-D model of machining that describes
the mechanics of machining fairly accurately
Figure - Orthogonal cutting: (a) as a three-dimensional process
25
26
Orthogonal cutting
27
28
Oblique cutting
29
30
31
32
Cutting forces
33
• First proposed by Ernst and Merchant
– Trans. ASME, 29, 299,1941.
– Considered idealized case of a single plane
– An approximate is predicted
– Forces on chip from rake face = Forces on work
surface along shear plane
• Forces on tool
– FC -along the direction of cutting velocity v
– FT-normal to the direction of cutting velocity v
34
Forces in orthogonal cutting
35
36
Merchant’s circle diagram
37
38
Forces in Metal Cutting
For orthogonal cutting the forces can be
derived as:
F = Fc sin + Ft cos
N = Fc cos - Ft sin
Fs = Fc cos - Ft sin
Fn = Fc sin + Ft cos
Fc = Fs cos + FN sin
FT = FN cos - Fs sin
39
Forces in Metal Cutting


cos(
F
R s



cos(
cos(F
F s
c
40
Shear Stress
Shear stress acting along the shear plane:
sin
wt
A o
s 
where As = area of the shear plane
Shear stress = shear strength of work material during cutting
s
s
A
F
S 
41
The Merchant Equation



cos(sin
v)cos(wt
vF)(W
c
c
s1
C
Whereas, t1, w, and vc are constants.
For minimum power, differentiating work, W w.r.t Ø, we get:
0sin(sin)cos(cos  
s
,Hence
)2/cos(0)2cos(  
2/2  
Shear angle relations
42
Source Results
Ernst and Merchant 2Ø + λ - α = π / 2 Ø = π/4 + α/2 – λ/2
Merchant’s second solution 2Ø + λ - α = Cm 2Ø =Cm+α/2 – λ/2
Lee and Shaffer Ø + λ - α = π / 4 Ø = π/4+ α - λ
Stabler Ø + λ - α / 2 = π / 4 Ø = π / 4+ α / 2 - λ
43
Specific energy
Total specific energy
• vc = cutting velocity
• w = width of cut
Specific energy for friction
ovf = chip velocity
orc = cutting ratio
o
c
o
c
t
wt
F
vwt
vF
u
c
c

o
ctc
o
c
o
f
f
wt
rFF
wt
Fr
vwt
Fv
u
c
)cossin( 

44
Specific energy for shearing
cvwt
vF
u
o
ss
s 
Thus, the total specific energy
ut = uf + us
# vs = shear velocity
Problem 1
• Assume that in orthogonal cutting the rake angle is 15o and the
coefficient of friction is 0.2. using Merchant’s first solution, determine
the percentage increase in chip thickness when friction is doubled.
Solution: Hint
Ø = π/4 + α/2 – λ/2; r = ? tc = ?
Ø’ = π/4 + α/2 – λ’/2; r = ? t’c = ?
45
46
Problem 2
The following data are available from an orthogonal turning
operation:
Determine the (a) shear angle (b) friction coefficient (c) shear stress and shear strain
on shear plane(d) chip velocity and shear velocity and (e) energies uf, us and ut.
Work material Aluminium Steel
Feed (t1), mm/revolution 0.33 0.13
Width of cut, mm 2.5 2.5
Rake angle 5o -5o
Cutting speed, m/s 2 2
Chip thickness (t2), mm 0.53 0.58
Cutting force, Fc, N 430 890
Thrust force, Ft, N 280 800
47
Heat generation and temperature
profile
• During cutting
– Plastic deformation
• Primary shear zone
• Secondary shear zone
– 99% of energy converted into heat
• Heat taken away
– Chip
» Major portion
– Work
– Tool
– Faster wear
– Failure of tool
48
Heat calculation
• W = total power consumed
= FCvc “vc = cutting velocity
• Wp & Ws are heat generated in primary and
secondary deformation zones
• W = Wp + Ws
• Ws = Fvf = Frc vc “vf = chip velocity
• Wp = W – Frc vc
49
• Temperature at primary shear zone Tp is
Where
• x = fraction of primary heat goes to work
• ρ = density of the work material
• s = specific heat of the work material
• t1, = uncut chip thickness
• w = width of cut
1
p
p
svwt
W)x1(
T



50
• Ts = temperature rise at secondary shear zone
• To = initial temperature of the work piece
• Total temperature, T is given by
T = To + Ts + Tp
The maximum temperature is along the rake
face of the tool.
51
Temperature profile when cutting
with a single cutting tool
52
53
54
55
56
57
58
59
60
61
62
63
64
CNC Shaper Machine
65
66
67
68
69
70
71
72
73
74
75
76
77
Classification of Cutting Tool
1. Single-Point Tools
– One cutting edge
– Turning uses single point tools
– Point is usually rounded to form a nose radius
2. Multiple Cutting Edge Tools
– Also called multipoint cutting tools
– More than one cutting edge
– Motion relative to work usually achieved by
rotating
– Drilling and milling use rotating multiple cutting
edge tools.
78
Single Point Cutting Tools
79
80
81
82
Nomenclature of single point cutting tool
83
84
85
86
Signature of right hand single point
cutting tool
( American Standard System)
10-20-7-6-8-5-0.8
Multipoint cutting tools
87
88
89
90
91
Broaching tools
92
Broaching tools
93
BORING TOOL TYPES AND NOMENCLATURE
Problem
The following data are available from a turning operation:
* s stands for (second digit of serial number)
Determine the (a) shear angle (b) friction coefficient (c) shear stress and shear strain
on shear plane(d) chip velocity and shear velocity and (e) energies uf, us and ut.
94
Work material Aluminium Steel
Tool signature 0,s*,6,7,10,0,9 0,-s* ,6,7,15,0,9
Feed, mm/revolution 0.15 0.15
Depth of cut, mm 2.5 2.5
Cutting speed, m/s 2 2
Chip thickness, mm 0.23 0.58
Cutting force, Fc, N 430 890
Thrust force, Ft, N 280 800
Oblique cutting
95
96
97
98
99
100
101
102
103
104
Orthogonal cutting
1- Cutting tool travel in the
direction perpendicular to the
cutting edge.
2-The cutting edge clear either
end of work piece.
3- Chip flows in the direction
perpendicular to the cutting
edge.
4-Two mutually perpendicular
cutting forces act on the work
piece.
Oblique cutting
1-cutting edge travels,
making an angle with the
normal of cutting edge.
2-The cutting edge may or
may not clear either end of
work piece.
3- chip flows, making an
angle with normal of cutting
edge.
4-Three mutually
perpendicular forces are
involved.
105
Ff
Fr
Ft
Z
X
Y
Fc
Orthogonal plane
R
Turning tool
Cylindrical job
Rake face
Forces acting during oblique turning

106
Cutting forces acting during oblique
turning operation
• Feed force, Ff - along z-axis
• Radial force, Fr - along x-axis, radial direction
• Cutting force, Fc- along y-axis perpendicular to
x-z plane
• Merchant circle diagram is drawn for the forces
acting in orthogonal plane, i.e. the plane containing
cutting force Fc and Ft ( resultant of Ff and Fr)
107
Normal rake angle (αn)
Where αb and αs = back and side rake angles
Ψ =side cutting edge angle
Shear angle (φn)
nc
nc
n
sinr1
cosr
tan



108
Z
X
Y
Turning tool
Cylindrical job
Rake face
Forces acting during oblique turning
Depth
ofcut=
d
Feed = f
Chip
ψ
ψ
109
Depthofcut=d
Feed = f
ψ
110
cos
d
wcut.of.Width 
cosftthickness.chip.Uncut 1 
cc1 fdvvwtmrrrate.removal.Metal 
Where,
Cutting velocity = vc = πDN
D = Average diameter of the
work
N = RPM
ψ = side cutting edge angle
d, f = depth of cut, feed/
revolution
111
volume.unit/consumed.Energy..........
)U(energy.Specific c

fd
F
mrr
vF
U
ccc
c 
ccvFW
Wnconsumptio.Power


112
Machining time (tc)
(for a single cut)
fN
t
L
c 
Where,
L = Length of the work along the axis
F = feed/ revolution
N = rpm of the work
113
Problem
The following data are available from a turning operation:
* s stands for (second digit of serial number)
Determine the (a) shear angle (b) friction coefficient (c) shear stress and shear strain on
shear plane(d) chip velocity and shear velocity and (e) energies uf, us and ut.
Work material Aluminium Steel
Tool signature 9,s*,6,7,10,15,9 mm 9,s*,6,7,10,15,9 mm
Depth of cut, mm 2.5 2.5
Cutting speed, m/s 2 2
Chip thickness, mm 0.23 0.58
Cutting force, Fc, N 430 890
Thrust force, Ft, N 280 800
Thanks
114

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Mechanism of Metal cutting

  • 2. BOOKS 1.Manufacturing Technology, Volume 2, P.N. Rao, McGraw Hill Education (India) Private Limited. 2.Manufacturing Processes for Engineering Materials, S. Kalpakjain, Addision-Weseley Publishing Company. 3.Fundamentals of Metal Cutting and Machine Tools, B.L. Juneja, G.S. Sekhon, New Age International Private Limited. 4.Engineering Metrology, Jain, Khanna Publishers 5.CAD/CAM: Computer-Aided Design and Manufacturing by M.P. Groover & E.W. Zimmers, Jr. 6.CNC Machines by B.S. Pabla & M. Adithan 2
  • 3. Machine Tools A power-driven machine that performs a machining operation, including grinding • Functions in machining: – Holds workpart – Positions tool relative to work – Provides power at speed, feed, and depth that have been set • The term is also applied to machines that perform metal forming operations 3
  • 5. Important parameters of machining 1. Thickness of uncut layer 2. Thickness of chip 3. Inclination of the chip-tool interface with respect to the face of the tool (rake angle) 4. The relative velocity of the work piece and the cutting tool A clearance angle between the work and the flank surface make the cutting possible. 5
  • 10. 10 Shear strain             cossin cos ,or )tan(cot BD DCAD BD AC strain_Shear
  • 11. 11 Shear Strain Shear strain in machining can be computed from the following equation, based on the preceding parallel plate model:  = tan( - ) + cot  where  = shear strain,  = shear plane angle, and  = rake angle of cutting tool
  • 12. Chip velocity, Vc • Maintaining mass continuity 12 )cos( sinV V rVV velocity_cuttingV tVtV c f cf cwhere 2f1c       
  • 13. 13
  • 14. Velocity diagram 14         Y V cos cos Y V ,, tY S ratestrain sinV V cossin cos ,or )tan(cotstrainShear sin V cos Vs cos V s s f                      (90o-Φ+α) α Φ (90o-α) Vs V Vf Where, ∆Y = thickness of the deformation zone ∆t = time taken
  • 15. 15 Types of chips • Continuous chip without built- up edge (BUE) • Continuous chip with built-up edge (BUE) • Serrated chip • Discontinuous chip
  • 16. Continuous chip without built-up edge (BUE) •Ductile material •Small uncut thickness •High cutting speed •Large rake angle •Suitable cutting fluid 16
  • 17. Continuous chip with built-up edge (BUE) •Stronger adhesion between chips and tool face •Large uncut thickness •Low rake angle 17
  • 19. Serrated chip •Appears as saw teeth •Low conductivity metal •Strength of the metal decreases sharply with temperature •E.g. titanium 19
  • 20. Discontinuous chip •Brittle material •Large uncut thickness •Low cutting speed •Small rake angle 20
  • 21. 21 Problem The following data are available from an orthogonal cutting operation: Determine the (a) cutting ratio (b) shear angle (c) shear strain and strain rate on shear plane(d) chip velocity and (e)shear velocity. Work material Aluminium Steel Uncut chip thickness (t1), mm 0.13 0.23 Width of cut, mm 2.5 2.5 Rake angle 5o -5o Cutting speed, m/s 2 2 Chip thickness (t2), mm 0.23 0.58
  • 23. 23 Orthogonal cutting • Can be represented by a 2-dimensional figure • The work move in the plane parallel to the plane of the paper • The chip material particles also move in the plane parallel to the plane of the paper • No component of velocity in the direction perpendicular to the plane of the paper
  • 24. 24 Orthogonal Cutting Model A simplified 2-D model of machining that describes the mechanics of machining fairly accurately Figure - Orthogonal cutting: (a) as a three-dimensional process
  • 25. 25
  • 27. 27
  • 29. 29
  • 30. 30
  • 31. 31
  • 33. 33 • First proposed by Ernst and Merchant – Trans. ASME, 29, 299,1941. – Considered idealized case of a single plane – An approximate is predicted – Forces on chip from rake face = Forces on work surface along shear plane • Forces on tool – FC -along the direction of cutting velocity v – FT-normal to the direction of cutting velocity v
  • 35. 35
  • 37. 37
  • 38. 38 Forces in Metal Cutting For orthogonal cutting the forces can be derived as: F = Fc sin + Ft cos N = Fc cos - Ft sin Fs = Fc cos - Ft sin Fn = Fc sin + Ft cos Fc = Fs cos + FN sin FT = FN cos - Fs sin
  • 39. 39 Forces in Metal Cutting   cos( F R s    cos( cos(F F s c
  • 40. 40 Shear Stress Shear stress acting along the shear plane: sin wt A o s  where As = area of the shear plane Shear stress = shear strength of work material during cutting s s A F S 
  • 41. 41 The Merchant Equation    cos(sin v)cos(wt vF)(W c c s1 C Whereas, t1, w, and vc are constants. For minimum power, differentiating work, W w.r.t Ø, we get: 0sin(sin)cos(cos   s ,Hence )2/cos(0)2cos(   2/2  
  • 42. Shear angle relations 42 Source Results Ernst and Merchant 2Ø + λ - α = π / 2 Ø = π/4 + α/2 – λ/2 Merchant’s second solution 2Ø + λ - α = Cm 2Ø =Cm+α/2 – λ/2 Lee and Shaffer Ø + λ - α = π / 4 Ø = π/4+ α - λ Stabler Ø + λ - α / 2 = π / 4 Ø = π / 4+ α / 2 - λ
  • 43. 43 Specific energy Total specific energy • vc = cutting velocity • w = width of cut Specific energy for friction ovf = chip velocity orc = cutting ratio o c o c t wt F vwt vF u c c  o ctc o c o f f wt rFF wt Fr vwt Fv u c )cossin(  
  • 44. 44 Specific energy for shearing cvwt vF u o ss s  Thus, the total specific energy ut = uf + us # vs = shear velocity
  • 45. Problem 1 • Assume that in orthogonal cutting the rake angle is 15o and the coefficient of friction is 0.2. using Merchant’s first solution, determine the percentage increase in chip thickness when friction is doubled. Solution: Hint Ø = π/4 + α/2 – λ/2; r = ? tc = ? Ø’ = π/4 + α/2 – λ’/2; r = ? t’c = ? 45
  • 46. 46 Problem 2 The following data are available from an orthogonal turning operation: Determine the (a) shear angle (b) friction coefficient (c) shear stress and shear strain on shear plane(d) chip velocity and shear velocity and (e) energies uf, us and ut. Work material Aluminium Steel Feed (t1), mm/revolution 0.33 0.13 Width of cut, mm 2.5 2.5 Rake angle 5o -5o Cutting speed, m/s 2 2 Chip thickness (t2), mm 0.53 0.58 Cutting force, Fc, N 430 890 Thrust force, Ft, N 280 800
  • 47. 47 Heat generation and temperature profile • During cutting – Plastic deformation • Primary shear zone • Secondary shear zone – 99% of energy converted into heat • Heat taken away – Chip » Major portion – Work – Tool – Faster wear – Failure of tool
  • 48. 48 Heat calculation • W = total power consumed = FCvc “vc = cutting velocity • Wp & Ws are heat generated in primary and secondary deformation zones • W = Wp + Ws • Ws = Fvf = Frc vc “vf = chip velocity • Wp = W – Frc vc
  • 49. 49 • Temperature at primary shear zone Tp is Where • x = fraction of primary heat goes to work • ρ = density of the work material • s = specific heat of the work material • t1, = uncut chip thickness • w = width of cut 1 p p svwt W)x1( T   
  • 50. 50 • Ts = temperature rise at secondary shear zone • To = initial temperature of the work piece • Total temperature, T is given by T = To + Ts + Tp The maximum temperature is along the rake face of the tool.
  • 51. 51 Temperature profile when cutting with a single cutting tool
  • 52. 52
  • 53. 53
  • 54. 54
  • 55. 55
  • 56. 56
  • 57. 57
  • 58. 58
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  • 68. 68
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  • 70. 70
  • 71. 71
  • 72. 72
  • 73. 73
  • 74. 74
  • 75. 75
  • 76. 76
  • 77. 77
  • 78. Classification of Cutting Tool 1. Single-Point Tools – One cutting edge – Turning uses single point tools – Point is usually rounded to form a nose radius 2. Multiple Cutting Edge Tools – Also called multipoint cutting tools – More than one cutting edge – Motion relative to work usually achieved by rotating – Drilling and milling use rotating multiple cutting edge tools. 78
  • 80. 80
  • 81. 81
  • 82. 82
  • 83. Nomenclature of single point cutting tool 83
  • 84. 84
  • 85. 85
  • 86. 86 Signature of right hand single point cutting tool ( American Standard System) 10-20-7-6-8-5-0.8
  • 88. 88
  • 89. 89
  • 90. 90
  • 93. 93 BORING TOOL TYPES AND NOMENCLATURE
  • 94. Problem The following data are available from a turning operation: * s stands for (second digit of serial number) Determine the (a) shear angle (b) friction coefficient (c) shear stress and shear strain on shear plane(d) chip velocity and shear velocity and (e) energies uf, us and ut. 94 Work material Aluminium Steel Tool signature 0,s*,6,7,10,0,9 0,-s* ,6,7,15,0,9 Feed, mm/revolution 0.15 0.15 Depth of cut, mm 2.5 2.5 Cutting speed, m/s 2 2 Chip thickness, mm 0.23 0.58 Cutting force, Fc, N 430 890 Thrust force, Ft, N 280 800
  • 96. 96
  • 97. 97
  • 98. 98
  • 99. 99
  • 100. 100
  • 101. 101
  • 102. 102
  • 103. 103
  • 104. 104 Orthogonal cutting 1- Cutting tool travel in the direction perpendicular to the cutting edge. 2-The cutting edge clear either end of work piece. 3- Chip flows in the direction perpendicular to the cutting edge. 4-Two mutually perpendicular cutting forces act on the work piece. Oblique cutting 1-cutting edge travels, making an angle with the normal of cutting edge. 2-The cutting edge may or may not clear either end of work piece. 3- chip flows, making an angle with normal of cutting edge. 4-Three mutually perpendicular forces are involved.
  • 105. 105 Ff Fr Ft Z X Y Fc Orthogonal plane R Turning tool Cylindrical job Rake face Forces acting during oblique turning 
  • 106. 106 Cutting forces acting during oblique turning operation • Feed force, Ff - along z-axis • Radial force, Fr - along x-axis, radial direction • Cutting force, Fc- along y-axis perpendicular to x-z plane • Merchant circle diagram is drawn for the forces acting in orthogonal plane, i.e. the plane containing cutting force Fc and Ft ( resultant of Ff and Fr)
  • 107. 107 Normal rake angle (αn) Where αb and αs = back and side rake angles Ψ =side cutting edge angle Shear angle (φn) nc nc n sinr1 cosr tan   
  • 108. 108 Z X Y Turning tool Cylindrical job Rake face Forces acting during oblique turning Depth ofcut= d Feed = f Chip ψ ψ
  • 110. 110 cos d wcut.of.Width  cosftthickness.chip.Uncut 1  cc1 fdvvwtmrrrate.removal.Metal  Where, Cutting velocity = vc = πDN D = Average diameter of the work N = RPM ψ = side cutting edge angle d, f = depth of cut, feed/ revolution
  • 112. 112 Machining time (tc) (for a single cut) fN t L c  Where, L = Length of the work along the axis F = feed/ revolution N = rpm of the work
  • 113. 113 Problem The following data are available from a turning operation: * s stands for (second digit of serial number) Determine the (a) shear angle (b) friction coefficient (c) shear stress and shear strain on shear plane(d) chip velocity and shear velocity and (e) energies uf, us and ut. Work material Aluminium Steel Tool signature 9,s*,6,7,10,15,9 mm 9,s*,6,7,10,15,9 mm Depth of cut, mm 2.5 2.5 Cutting speed, m/s 2 2 Chip thickness, mm 0.23 0.58 Cutting force, Fc, N 430 890 Thrust force, Ft, N 280 800