TA202A - Manufacturing Processes II

Mechanics of Machining

Lecture 4

Niraj Sinha

Department of Mechanical Engineering

IIT Kanpur

nsinha@iitk.ac.in

Machining Processes

2

Example part to be made on a mill-turn center

Sequence of operations

Process Planning For A Component

3

Most machining operations are geometrically complex and 3D

https://www.youtube.com/watch?v=JoVQAn7Suto

Machining Processes

Mechanics of Cutting

4

Simple 2D orthogonal cutting can help explain the general mechanics of metal removal

https://www.youtube.com/watch?v=mRuSYQ5Npek&t=21s

Tensile Testing

5

@ ACMS, IITK

Milling

MachiningParameters

Work Material Cutting Tool Machining Conditions

Machine Tool

Cutting Process

Product

Measurements

Determinations

Metal type Purity BCC, FCC, HCP Predeformation(work hardening prior to machining)

Machining Conditions

Cutting parameters • Depth of cut • Speed • Feed

Environment • Lubricant

(cutting fluid) • Oxygen • Temperature

Workholder• Fixtures • Jigs • Chucks • Collets

Cutting speed (v) – primary motion

The speed at which the work moves with respect to the tool

Feed (f) – secondary motion

Depth of cut (d) – penetration of tool below original work surface

Machining Conditions in Turning

Feed Rate - fr

f = feed per rev

Depth of Cut - d

Machining Time - Tm

L = length of cut

Material Removal Rate - MRRoDπ

vN

Spindle Speed - N

v = cutting speed

Do = outer diameter

fNf r

2fo DD

d

r

mf

LT

dfvM R R

Selecting Depth of Cut

Depth of cut is often predetermined by workpiece

geometry and operation sequence

– In roughing, depth is made as large as possible to maximize

material removal rate, subject to limitations of

horsepower, machine tool and setup rigidity, and strength

of cutting tool

– In finishing, depth is set to achieve final part dimensions

Determining Feed

• In general: feed first, speed second

• Determining feed rate depends on:

– Tooling – harder tool materials require lower feeds

– Roughing or finishing - Roughing means high feeds,

finishing means low feeds

– Constraints on feed in roughing - Limits imposed by cutting

forces, setup rigidity, and sometimes horsepower

– Surface finish requirements in finishing – select feed to

produce desired finish

Optimizing Cutting Speed

• Select speed to achieve a balance between high

metal removal rate and suitably long tool life

• Mathematical formulas are available to determine

optimal speed

• Two alternative objectives in these formulas:

1. Maximum production rate

2. Minimum unit cost

Maximum Production Rate

• Maximizing production rate = minimizing cutting time per unit

• In turning, total production cycle time for one part consists of:

1. Part handling time per part = Th

2. Machining time per part = Tm

3. Tool change time per part = Tt/np , where np = number of pieces cut in one tool life

Total time per unit product for operation:

Tc = Th + Tm + Tt/np

Cycle time Tc is a function of cutting speed

Cycle Time vs. Cutting Speed

Minimizing Cost per Unit

In turning, total production cycle cost for one part consists of:

1. Cost of part handling time = CoTh , where Co = cost rate for operator and machine

2. Cost of machining time = CoTm

3. Cost of tool change time = CoTt/np

4. Tooling cost = Ct/np , where Ct = cost per cutting edge

Total cost per unit product for operation:

Cc = CoTh + CoTm + CoTt/np + Ct/np

Again, unit cost is a function of cutting speed, just as Tc is a function of v

Unit Cost vs. Cutting Speed

Cutting Fluids

Any liquid or gas applied directly to machining operation to

improve cutting performance

• Two main problems addressed by cutting fluids:

1. Heat generation at shear zone and friction zone

2. Friction at the tool-chip and tool-work interfaces

• Other functions and benefits:

– Wash away chips (e.g., grinding and milling)

– Reduce temperature of workpart for easier handling

– Improve dimensional stability of workpart

Cutting Fluid Functions

Cutting fluids can be classified according to function:

– Coolants - designed to reduce effects of heat in machining

– Lubricants - designed to reduce tool-chip and tool-work friction

• Water used as base in coolant-type cutting fluids

• Most effective at high cutting speeds where heat generation and high

temperatures are problems

• Most effective on tool materials that are most susceptible to

temperature failures (e.g., HSS)

• Usually oil-based fluids

• Most effective at lower cutting speeds

• Also reduces temperature in the operation

Coolants

Lubricants

Work Material Cutting Tool Machining Conditions

Machine Tool

Cutting Process

Product

Measurements

Determinations

Cutting Tool Classification

1. Single-Point Tools– One dominant cutting edge

– Point is usually rounded to form a nose radius

– Turning uses single point tools

2. Multiple Cutting Edge Tools– More than one cutting edge

– Motion relative to work achieved by rotating

– Drilling and milling use rotating multiple cutting edge tools

Figure: (a) Seven elements of single-point tool geometry; and (b) the tool

signature convention that defines the seven elements

Single-Point Tool Geometry

Tool MaterialsTool failure modes identify the important properties that a tool material should possess:

– Toughness - to avoid fracture failure

– Hot hardness - ability to retain hardness at high temperatures

– Wear resistance - hardness is the most important property to resist abrasive wear

High speed steel (HSS) Cemented carbides Cermets Coated carbides Ceramics Synthetic diamonds Cubic boron nitride

Failure of Cutting Tools and Tool Wear

• Fracture failure

– Cutting force becomes excessive and/or dynamic, leading to brittle fracture

• Temperature failure

– Cutting temperature is too high for the tool material

• Gradual wear

– Gradual wearing of the cutting tool

Preferred Mode of Tool Failure: Gradual

Wear

• Fracture and temperature failures are premature failures

• Gradual wear is preferred because it leads to the longest possible use of the tool

• Gradual wear occurs at two locations on a tool:

– Crater wear – occurs on top rake face

– Flank wear – occurs on flank (side of tool)

Figure - Diagram of worn cutting tool, showing the principal

locations and types of wear that occur

Figure -

(a) Crater wear, and

(b) flank wear on a cemented carbide tool, as seen through a toolmaker's microscope

(Source: Manufacturing Technology Laboratory, Lehigh University, photo by J. C. Keefe)

Figure - Tool wear as a function of cutting time

Flank wear (FW) is used here as the measure of tool wear

Crater wear follows a similar growth curve

Figure - Effect of cutting speed on tool flank wear (FW) for three cutting speeds, using a tool life criterion of 0.50 mm flank wear

Figure - Natural log-log plot of cutting speed vs tool life

Taylor Tool Life Equation

This relationship is credited to F. W. Taylor (~1900)

CvT n

where v = cutting speed; T = tool life; and n and C are parameters that

depend on feed, depth of cut, work material, tooling material, and the

tool life criterion used

• n is the slope of the plot

• C is the intercept on the speed axis

Variables Affecting Tool Life

Cutting conditions.

Tool geometry.

Tool material.

Work material.

Cutting fluid.

Vibration behavior of the machine-tool work system.

Built-up edge.

Work Material Cutting Tool Machining Conditions

Machine Tool

Cutting Process

Product

Measurements

Determinations

Cutting Process: Chip Formation

• More realistic view of chip formation, showing shear zone rather than shear plane

• Also shown is the secondary shear zone resulting from tool-chip friction

Four Basic Types of Chip in Machining

1. Discontinuous chip

2. Continuous chip

3. Continuous chip with Built-up Edge (BUE)

4. Serrated chip

• Brittle work materials

• Low cutting speeds

• Large feed and depth of cut

• High tool-chip friction

Optics and Lasers in Engineering, Volume 49, Issue 2, February 2011, Pages 240–247

Discontinuous Chip

• Ductile work materials

• High cutting speeds

• Small feeds and depths

• Sharp cutting edge

• Low tool-chip friction

Journal of Materials Processing Technology, Volume 121, Issues 2–3, 28 February 2002, Pages 363–372

Continuous Chip

• Ductile materials

• Low-to-medium cutting speeds

• Tool-chip friction causes portions of chip to adhere to rake face

• BUE forms, then breaks off, cyclically

Springerimages.com

Continuous with BUE

• Semi-continuous - saw-tooth appearance

• Cyclical chip forms with alternating high shear strain then low shear strain

• Associated with difficult-to-machine metals at high cutting speeds

Springerimages.com

Serrated Chip

Roughing vs. Finishing Cuts

For increasing the production, several roughing cuts

are usually taken on a part, followed by one or two

finishing cuts

– Roughing - removes large amounts of material from

starting workpart

• Some material remains for finish cutting

• High feeds and depths, low speeds

– Finishing - completes part geometry

• Final dimensions, tolerances, and finish

• Low feeds and depths, high cutting speeds

Orthogonal Cutting Geometry

73

Assumptions:1. Cutting edge is perfectly sharp2. Uncut chip thickness is constant

and << than width3. Width of tool > width of

workpiece4. Continuous chip with no built up-

edge5. Uniform cutting along edge6. 2D plane strain deformation7. No side spreading of material8. Uniform stress distribution on

shear plane9. Forces primarily in directions of

velocity and uncut chip thickness

Orthogonal Cutting Geometry

74

Primary shear zone:Material ahead of tool is sheared to form a chip

Secondary shear zone:Sheared material (chip) partially deforms and moves along the rake face

Tertiary zone:Flank of tool rubs the newly machined surface

Simple 2D orthogonal cutting can help explain the general mechanics of metal removal

Tool

Workpiece

Chip

Primary Shearing Zone

75

𝐹𝑡𝑐

𝐹𝑓𝑐

𝐹𝑛

𝐹𝑠

𝐹𝑣

∅𝑐

𝐹𝑠 = 𝐹𝑡𝑐 cos𝜙𝑐 − 𝐹𝑓𝑐 sin𝜙𝑐 ;

𝐹𝑛 = 𝐹𝑡𝑐 sin 𝜙𝑐 + 𝐹𝑓𝑐 cos𝜙𝑐 ;𝐹𝑠𝐹𝑛

=cos𝜙𝑐 −sin𝜙𝑐

sin 𝜙𝑐 cos𝜙𝑐

𝐹𝑡𝑐𝐹𝑓𝑐

Force components in primary shear zone

𝐹𝑢

𝛼𝑟 𝛼𝑟- rake angle; 𝜙𝑐 - shear angle𝐹𝑡𝑐 - tangential force/cutting force;𝐹𝑓𝑐 - feed force/thrust force

𝐹𝑠 - force acting along the shear plane𝐹𝑛 - normal force acting on the shear plane𝐹𝑣 - force acting on the rake face𝐹𝑢 - frictional force

• Forces acting on the tool that can be measured: Cutting force Ftc and Thrust force Ffc

• Other forces cannot be directly measured

Chip ratios – basic characteristics

76

sin𝜙𝑐 =ℎ

𝑙𝜙

ℎ

ℎ𝑐

∅𝑐

𝛼𝑟

∅𝑐 − 𝛼𝑟

𝑙𝜙

cos 𝜙𝑐 − 𝛼𝑟 =ℎ𝑐𝑙𝜙

ℎ = 𝑙𝜙sin𝜙𝑐 ℎ𝑐 = 𝑙𝜙cos 𝜙𝑐 − 𝛼𝑟

𝑟𝑐 =ℎ

ℎ𝑐=

sin𝜙𝑐

cos 𝜙𝑐 − 𝛼𝑟

&

&

Chip thickness ratio

From mass (volume flow rate) conservation, chip thickness ratio ≅ chip length ratio

𝑟𝑐 =ℎ

ℎ𝑐=𝑙𝑐𝑙= 𝑟𝑙 No side spreading assumption, 𝑟𝑤 = 1

𝜙𝑐 = tan−1𝑟𝑐 cos 𝛼𝑟

1 − 𝑟𝑐sin 𝛼𝑟

Velocity Relations

80

𝑉𝑠 = 𝑉cos 𝛼𝑟

cos 𝜙𝑐 − 𝛼𝑟

𝑉𝑠

𝑉

𝑉𝑐

𝑉𝑠𝑉𝑐

𝑉

𝛼𝑟

𝜙𝑐

𝜙𝑐 − 𝛼𝑟

𝛼𝑟

𝜙𝑐

𝜙𝑐 − 𝛼𝑟Chip velocity, 𝑉𝑐 (acknowledging conservation of volume-flow rate):

𝑉𝑐𝑉=𝑙𝑐/∆𝑡

𝑙/∆𝑡= 𝑟𝑙 ⇒ 𝑉𝑐 = 𝑟𝑙𝑉

𝑟𝑐 = 𝑟𝑙 =sin𝜙𝑐

cos 𝜙𝑐 − 𝛼𝑟

𝑉𝑐 =sin𝜙𝑐

cos 𝜙𝑐 − 𝛼𝑟𝑉

Shear velocity, 𝑉𝑠 is vector sum of 𝑉𝑐 and 𝑉

Shear Strain and Material Movement

81

Undeformed chip section 𝐴0𝐵𝑜𝐴1𝐵1moves with velocity 𝑉

When one element traverses the shear plane in time ∆𝑡:

point 𝐴1 moves to point 𝐴2, point 𝐴0 moves to point 𝐴1

point 𝐵1 moves to point 𝐵2, point 𝐵0 moves to point 𝐵1

Undeformed chip section 𝐴0𝐵𝑜𝐴1𝐵1hence becomes deformed chip with section 𝐴1𝐵1𝐴2𝐵2

Hence chip is shifted from expected position of 𝐵2′ 𝐴2

′ to 𝐵2𝐴2 because of shearing

Shear Strain and Shear Strain Rate

82

∆𝑥

𝑦

𝛾

tan 𝛾 =∆𝑥

𝑦𝛾 =

∆𝑥

𝑦𝛾 =

∆𝑠

∆𝑑=𝐴2𝐴2

′

𝐴1𝐶=𝐴2

′𝐶

𝐴1𝐶+𝐶𝐴2𝐴1𝐶

= cot𝜙𝑐 + tan 𝜙𝑐 − 𝛼𝑟Smallangles

Shear strain rate

ሶ𝛾 =𝛾

∆𝑡

Assuming shear zone increment is ∆𝑠 and the thickness of the shear deformation zone is ∆𝑑

𝛾 =∆𝑠

∆𝑑𝑉𝑠 =

∆𝑠

∆𝑡ሶ𝛾 =

𝑉𝑠∆𝑑

= 𝑉cos𝛼𝑟

∆𝑑 cos 𝜙𝑐 − 𝛼𝑟

∆𝑑 → 0 → thin shear plane → very large strain rates

Primary Shearing Zone

83

Endres, Adv. Mcing processes

84

Assuming uniform shear stress distribution on the shear plane and that all shear plane material is plastically shearing

𝜏𝑠 =𝐹𝑠𝐴𝑠

𝐴𝑠 = 𝑏ℎ

sin𝜙𝑐

𝐹𝑠 = 𝐹𝑡𝑐 cos𝜙𝑐 − 𝐹𝑓𝑐 sin𝜙𝑐

𝐴𝑠 - shear plane area𝑏 – width of cut (depth of cut in turning)ℎ - uncut chip thickness 𝜙𝑐 - shear angle𝐹𝑠 - force acting along the shear plane𝐹𝑛 - normal force acting on the shear plane

Normal stress on the shear plane

𝜎𝑠 =𝐹𝑛𝐴𝑠

𝐹𝑛 = 𝐹𝑡𝑐 sin𝜙𝑐 + 𝐹𝑓𝑐 cos𝜙𝑐

Primary Shearing Zone

Orthogonal Cutting Geometry

85

Primary shear zone:Material ahead of tool is sheared to form a chip

Secondary shear zone:Sheared material (chip) partially deforms and moves along the rake face

Tertiary zone:Flank of tool rubs the newly machined surface

Secondary Shear Zone

86

𝐹𝑣 = 𝐹𝑡𝑐 cos𝛼𝑟 − 𝐹𝑓𝑐 sin 𝛼𝑟 ;

𝐹𝑢 = 𝐹𝑡𝑐 sin 𝛼𝑟 + 𝐹𝑓𝑐 cos 𝛼𝑟 ;𝐹𝑣𝐹𝑢

=cos𝛼𝑟 −sin 𝛼𝑟sin 𝛼𝑟 cos 𝛼𝑟

𝐹𝑡𝑐𝐹𝑓𝑐

Force components in secondary shear zone

𝜇𝑎 = tan𝛽𝑎 =𝐹𝑢𝐹𝑣

Average friction coeff on rake face:

Force Circle Diagram

87

Tool

𝐹𝑡𝑐

𝐹𝑓𝑐

𝐹𝑛

𝐹𝑠

𝐹𝑣

∅𝑐

𝐹𝑢

𝛼𝑟∅𝑐𝐹𝑐

𝛽𝑎 − 𝛼𝑟

𝛼𝑟

𝛽𝑎 − 𝛼𝑟 𝛽𝑎

88

Tool

𝐹𝑡𝑐

𝐹𝑓𝑐

𝐹𝑛

𝐹𝑠

𝐹𝑣

∅𝑐

𝐹𝑢

𝛼𝑟∅𝑐

𝛽𝑎 − 𝛼𝑟

𝛼𝑟

𝛽𝑎 − 𝛼𝑟 𝛽𝑎

𝐹𝑠 = 𝐹𝑡𝑐 cos𝜙𝑐 − 𝐹𝑓𝑐 sin𝜙𝑐 ;

𝐹𝑛 = 𝐹𝑡𝑐 sin𝜙𝑐 + 𝐹𝑓𝑐 cos𝜙𝑐 ;

Force components in primary shear zone

Force components in primary shear zoneusing the force circle diagram

𝐹𝑠 = 𝐹𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟 ;𝐹𝑛 = 𝐹𝑐 sin 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟 ;

𝐹𝑐

Force components in secondary shear zone using the force circle diagram

𝐹𝑣 =? ; 𝐹𝑢 =?

Force Circle Diagram

Total Power Consumed in Cutting

89

𝑃𝑡𝑐 = 𝑃𝑠 + 𝑃𝑢

𝑃𝑡𝑐 = 𝐹𝑠𝑉𝑠 + 𝐹𝑢𝑉𝑐

𝑉𝑠 = 𝑉cos 𝛼𝑟

cos 𝜙𝑐 − 𝛼𝑟

Total power consumed in cutting is sum of energy spent in shear and friction zones

𝐹𝑢 = 𝐹𝑡𝑐 sin 𝛼𝑟 + 𝐹𝑓𝑐 cos 𝛼𝑟𝐹𝑠 = 𝐹𝑡𝑐 cos𝜙𝑐 − 𝐹𝑓𝑐 sin 𝜙𝑐

𝜙𝑐 - shear angle?𝐹𝑠 - force acting along the shear plane?𝐹𝑢 - frictional force?

𝑉𝑐 = 𝑉sin𝜙𝑐

cos 𝜙𝑐 − 𝛼𝑟

Force Prediction for Power Consumption

90

𝑃𝑡𝑐 = 𝐹𝑠𝑉𝑠 + 𝐹𝑢𝑉𝑐

𝐹𝑠 = 𝐹𝑡𝑐 cos𝜙𝑐 − 𝐹𝑓𝑐 sin𝜙𝑐 𝐹𝑠 = 𝐹𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟

𝐹𝑠 = 𝜏𝑠𝐴𝑠 = 𝜏𝑠 𝑏ℎ

sin 𝜙𝑐

Total power consumed in cutting is sum of energy spent in shear and friction zones

Primary chip generation mechanism is shearing

𝑃𝑠 = 𝐹𝑠𝑉𝑠

𝑉𝑠 = 𝑉cos 𝛼𝑟

cos 𝜙𝑐 − 𝛼𝑟

Shear force can also be expressed as

Resultant cutting force can be now be expressed in terms of shear stress, friction and shear angles, width of cut, and feed rate as follows:

𝐹𝑐 =𝐹𝑠

cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟= 𝜏𝑠 𝑏ℎ

1

sin𝜙𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟Know everything but for 𝜙𝑐

91

Resultant cutting force can be now be expressed in terms of shear stress, friction and shear angles, width of cut, and feed rate as follows:

𝐹𝑐 =𝐹𝑠

cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟= 𝜏𝑠 𝑏ℎ

1

sin𝜙𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟

Nature always takes the path of least resistance, so during cutting 𝜙𝑐 takes a value such that least amount of energy is consumed, i.e. since 𝜏𝑠, 𝑏, ℎ, and 𝛽𝑎 and 𝛼𝑟 are given and do not

change, power consumed becomes:

𝐹𝑡𝑐 = 𝐹𝑐 cos 𝛽𝑎 − 𝛼𝑟 = 𝜏𝑠 𝑏ℎcos 𝛽𝑎 − 𝛼𝑟

sin𝜙𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟

𝑃𝑡𝑐 = 𝑉𝐹𝑡𝑐

Recalling the FCD

Power consumed during cutting:

𝑃𝑡𝑐 𝜙𝑐 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

sin 𝜙𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟

Shear Angle – from Merchant’s Energy Principle

Shear Angle – from Merchant’s Energy Principle

92

Nature always takes the path of least resistance, so during cutting , 𝜙𝑐 takes a value such that least amount of energy is consumed, i.e. since 𝜏𝑠, 𝑏, ℎ, and 𝛽𝑎 and 𝛼𝑟 are given and do not

change, power consumed becomes:

𝑃𝑡𝑐 𝜙𝑐 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

sin𝜙𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟

𝑃𝑡𝑐 𝜙𝑐 will be a minimum when the denominator is a maximum, hence differentiate denominator w.r.t 𝜙𝑐 and equate it to zero:

cos𝜙𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟 − sin𝜙𝑐 sin 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟 = 0

cos 2𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟 = 0

2𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟 =𝜋

2 𝜙𝑐 =𝜋

4−𝛽𝑎 − 𝛼𝑟

2

Force Prediction

93

𝐹𝑡𝑐 = 𝐹𝑐 cos 𝛽𝑎 − 𝛼𝑟 = 𝜏𝑠 𝑏ℎcos 𝛽𝑎 − 𝛼𝑟

sin𝜙𝑐 cos 𝜙𝑐 + 𝛽𝑎 − 𝛼𝑟𝑃𝑡𝑐 = 𝑉𝐹𝑡𝑐

𝜙𝑐 =𝜋

4−𝛽𝑎 − 𝛼𝑟

2

Shear angle using Merchant’s minimum energy principle

Force prediction Power consumed

• Shear angle prediction with these and other such models are not very accurate

• However, they provide important relationships between tool geometry and shear

angle – which is important for tool design

• To increase shear plane angle ₋ Increase the rake angle ₋ Reduce the friction angle (or reduce the coefficient of friction)

• Higher shear plane angle means smaller shear plane which means lower shear force, cutting forces, power, and temperature

94

• Contact mechanics between flank face and finished surface depends on

tool wear, preparation of cutting edge and friction characteristics of tool

and workpiece

• Assume total friction force on flank face is 𝐹𝑓𝑓

• And force normal to flank face in 𝐹𝑓𝑛

• Assume pressure (𝜎𝑓) on the flank face to be uniform (a gross

oversimplification)

𝐹𝑓𝑓

𝐹𝑓𝑛

𝛾𝑓

Tertiary Deformation Zone

Tertiary Deformation Zone

95

𝐹𝑡𝑒 = 𝐹𝑓𝑛 sin 𝛾𝑓 + 𝐹𝑓𝑓 cos 𝛾𝑓 ;

𝐹𝑓𝑒 = 𝐹𝑓𝑛 cos 𝛾𝑓 + 𝐹𝑓𝑓 sin 𝛾𝑓;

𝐹𝑓𝑛 = 𝜎𝑓𝑉𝐵𝑏

𝑉𝐵

𝛾𝑓

𝜇𝑓 = 𝐹𝑓𝑓/𝐹𝑓𝑛

Normal force on the flank face

Because the tool rubs on the finished surface, there is friction

Resolving contact forces into tangentialand feed directions

Any measured forces will include forces due to shearing and tothe tertiary deformation process, i.e. rubbing/ploughing at theflank of the cutting edge

𝐹𝑡 = 𝐹𝑡𝑐 + 𝐹𝑡𝑒;𝐹𝑓 = 𝐹𝑓𝑐 + 𝐹𝑓𝑒;

All cutting force expressions presented up until now were only for shearing, but in reality, edge forces also exist, hence edge forces must be subtracted from measured tangential and feed

forces before applying laws of orthogonal cutting mechanics

Orthogonal and Oblique Cutting Geometry

96

Cutting velocity is inclined at an acute angle 𝑖 to the cutting edge

Cutting velocity is perpendicular to cutting edge

Altintas, Mfg. Automation

Prediction of cutting forces in oblique cutting

97

𝐹𝑡 = 𝐾𝑡𝑐𝑏ℎ + 𝐾𝑡𝑒𝑏;𝐹𝑓 = 𝐾𝑓𝑐𝑏ℎ + 𝐾𝑓𝑒𝑏;

𝐹𝑟 = 𝐾𝑟𝑐𝑏ℎ + 𝐾𝑟𝑒𝑏;

𝐾𝑡𝑐 =𝜏𝑠 cos 𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

𝐾𝑓𝑐 =𝜏𝑠 sin 𝜃𝑛

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

𝐾𝑟𝑐 =𝜏𝑠𝑏ℎ tan 𝜃𝑖 − cos 𝜃𝑛 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin 𝜙𝑛

𝐹𝑟𝑐 =𝜏𝑠𝑏ℎ tan 𝜃𝑖 − cos 𝜃𝑛 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

𝐹𝑡𝑐 =𝜏𝑠𝑏ℎ cos 𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛𝐹𝑓𝑐 =

𝜏𝑠𝑏ℎ sin 𝜃𝑛cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

Force in direction of cutting speed

Expressing force components as a 𝑓 𝜏𝑠, 𝜙𝑛, 𝜙𝑖 , 𝜃𝑖, 𝜃𝑛 :

Force in direction of thrust

Force in direction of normal

Rewriting forces in the convenient form of:

Cutting Temperature

• Approximately 98% of the energy in machining is

converted into heat

• This can cause temperatures to be very high at the

tool-chip

• The remaining energy (about 2%) is retained as

elastic energy in the chip

Cutting Temperatures are Important

High cutting temperatures

1. reduce tool life

2. produce hot chips that pose safety hazards to the

machine operator

3. can cause inaccuracies in part dimensions due to

thermal expansion of work material

Cutting Temperature

Analytical method derived by Nathan Cook from dimensional analysis using experimental data for various work materials

where T = temperature rise at tool-chip interface; U = specific energy;v = cutting speed; to = chip thickness before cut; C = volumetricspecific heat of work material; K = thermal diffusivity of work material

333040

..

K

vt

C

UT o

• Experimental methods can be used to measure

temperatures in machining

– Most frequently used technique is the tool-chip

thermocouple

• Using this method, Ken Trigger determined the

speed-temperature relationship to be of the form:

T = K vm

where T = measured tool-chip interface temperature,

and v = cutting speed

Cutting Temperature

Example

Consider a turning operation performed on steel whose hardness = 225 HB at a speed = 3.0 m/s, feed = 0.25 mm, and depth = 4.0 mm. Using values of thermal properties and the appropriate specific energy value from tables, compute an estimate of cutting temperature using the Cook equation. Assume ambient temperature = 20C.

Solution: From Table, U = 2.2 N-m/mm3 = 2.2 J/mm3

From Table, = 7.87 g/cm3 = 7.87(10-3) g/mm3

From Table, C = 0.11 Cal/g-C. From note “a” at the bottom of the table, 1 cal = 4.186 J.

From Table, thermal conductivity k = 0.046 J/s-mm-C

Thus, C = 0.11(4.186) = 0.460 J/ g-C

C = (7.87 g/cm3)(0.46 J/g-C) = 3.62(10-3) J/mm3-C

Also, thermal diffusivity K = k/C

K = 0.046 J/s-mm-C /[(7.87 x 10-3 g/mm3)(0.46 J/g-C)] = 12.7 mm2/s

Using Cook’s equation, to = f = 0.25 mm

T = (0.4(2.2)/3.62(10-3))[3(103)(0.25)/12.7]0.333 = 0.2428(103)(59.06)0.333

= 242.8(3.89) = 944.4 C

Final temperature, taking ambient temperature in account T = 20 + 944 = 964C