CHAPTER (3)

Metal Cutting and Cutting Tools

Dr. Ahmed Abou El-Wafa

3.1 Clearance Angle:

The clearance angle may be defined as the angle between the flank face of the tool and a tangent to the work surface originating at the cutting edge.

FLeading

edge

(a) Heel fouls work & prevents leading edge from entering.

P

Clearance necessary to allow cutting to take place.

(b) Heel cleared a way allowing leading edge to enter the work.

F

P

Clearance angle: 6-8º

2

P

FPrimary Clearance

angle

Secondary Clearance

angle

(b) Internal cylinder

Effect of workpiece shape on the clearance angle.

(a) External cylinder

P

F

Clearance angle : 5-7º

An excessive clearance angle will not increase efficiency and will merely weaken the tool.

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3.2 Rake Angle:

The rake angle is the angle between the top face of the tool and the normal to the work surface at the cutting edge.

Tool angles showing the rake angle.

- Although increasing the rake angle improves the cutting action, it tends to weaken the tool. Therefore, the choice of a suitable rake angle becomes a compromise between adequate strength of tool and good cutting action.

- Generally, for work materials that give a continuous chip, the greater the work-material strength, the smaller the rake angle.

Chip

Rake angle

positive

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3.3 Chip Formation

3.3.1 Types of chips

Brittle materials: e.g. C I

Ductile materials: e.g. st.

With BUE-Tool-chip friction causes portions of chip to adhere to rake face- BUE forms, then breaks off, cyclically

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

Serrated chip

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3.3.2 Chip breakers

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3.4 Single Point Cutting Tools

Features of a single point cutting tool.

Section B-B at 90º to cutting edge

3.4.1 The straight-edge cutting tool

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3.4.2 The parting-off tool

Such a tool must be set on or very slightly above the center, but never below center, or the work will climb on top of the tool just before it is parted off. The tool is naturally rather weak and can easily be broken if this happens.

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3.4.3 Tool setting

Effect of tool setting.

(a) Tool above center. Tool tends to rub.

(b) Tool below center. Work tends to climb over tool.

(c) Tool on center but inclined upward at front. Tool rubs.

(d) Tool on center but inclined downward at front. Work tends to drag tool in.

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3.5 Multi-Point Cutting Tools

3.5.1 Reamers

The clearance angle of reamer is very small. A reamer must obviously not be ground on its clearance faces or its size will be destroyed, and it is ground on its rake faces, i.e., along its flutes.

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3.5.2 Taps

A tap.

Example 3.1:

Calculation of drill size required for producing M48x3 thread;

Depth of thread = 0.54*pitch = 0.54*3 = 1.62 mm.

90% depth of thread = 1.62*0.9 = 1.46 mm.

Tapping size = 48-(2*1.46) = 45.08 mm diameter.

- The next convenient drill size above this should be used.

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3.6 Copying and Generating Processes

- Copying process: in this process the surface is dependent for its shape on the shape of the cutter, (as an example form tools).

- Generating process: in this process the surface is independent on the shape of the tool, but depends on the relative motions of work and tool.

W.P

Tool

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3.7 Cutting Fluids

3.7.1 Types of cutting fluids

(1)WaterA poor lubricant, has little wetting action, it also encourages rusting and is suitable only as a coolant during tool grinding.

(2) Soluble OilsWith water, they form an intimate mixture, or emulsion, by adding emulsifying agents. While the oil prevents rusting, it is suitable for light cutting operations, i.e. low metal removal rates.

(3) Mineral OilsThey are suitable for heavier cutting operations, i.e. high metal removal rates, because of their much better lubricating properties. Sulfur compounds are added to prevent the chip welding to the rake face and forming BUE. They should not be used on Cu and its alloys since they have corrosive effect on them, but are suitable for steels.

(4) Vegetable OilsThey are little used since they are liable to decompose and smell, and may become a health hazard.

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3.7.2 Functions or uses of coolants or cutting fluids

The important functions of cutting fluids are given as follows;

(i) Cutting fluid washes away the chips and hence keeps the cutting region free.

(ii) It helps in keeping freshly machined surface bright by giving a protective coating against atmospheric, oxygen and thus protects the finished surface from corrosion.

(iii) It decreases wear and tear of cutting tool and hence increases tool life.

(iv) It improves machinability and reduces power requirements.

(v) It prevents expansion of workpieces.

(vi) It cools the tool and workpiece and removes the generated heat from the cutting zone.

(vii) It decreases adhesion between chip and tool, provides lower friction and wear, and a smaller built-up edge.

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3.8 The Mechanics of Metal Cutting

3.8.1 Introduction

15

w

wctc

to

Terminology in orthogonal cutting.

16

Merchant circle

Fn

R

17

3.8.2 Measurement of tool forces

Tangential force Ft

Radial force Fr

Turning dynamometer18

Calibration Curve

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3.8.3 Force relationship

Where: r = chip thickness ratio (or cutting ratio); to = thickness of the chip prior to

chip formation; and tc = chip thickness

after separation• Chip thickness after cut always greater

than before, so chip ratio always less than 1.0

c

o

tt

r

sincos

tanr

r

1

Where: = rake angle

bt

tr

c

o 1.21cos

sin

20

sinwt

A os

- Coefficient of friction between tool and chip: where β is: the friction angle

- Shear stress acting along the shear plane:

where S is: the shear strength

where As is: the shear plane area

– to is: cut depth

– w is: cutting edge width,

– Φ is: shear plane angle

NF tan

s

s

AF

S τ =

3.2122

45

- Of all the possible angles at which shear deformation can occur, the work material will select a shear plane angle that minimizes energy, given by;

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• F, N, Fs, and Fn cannot be directly measured

• The only forces that can be measured are the forces acting on the tool:– Cutting force Fc

– Thrust force Ft

- Then;

F = Fc sin + Ft cos

N = Fc cos ‑ Ft sin

Fs = Fc cos ‑ Ft sin

Fn = Fc sin + Ft cos- From Merchant circle:

Fc = Fs cos(β - α) / [cos( + β - α)]

Ft = Fs sin(β - α) / [cos( + β - α)]

Shear Strain: = tan( - ) + cot -22

Power calculations

- Units for specific energy are typically N.m/mm3 or J/mm3 (in.lb/in3)

1 kW = 1.34 hp

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wvt

vF

R

PPU

o

c

MR

cu

MRR

- Power is force times speed:Pc = Fc v (hp or kW)

Where: v is the cutting speed

- The cutting horsepower ishpc = Fc v / 33,000 (hp)

- Due to efficiency losses (η about 90%), the gross horsepower (hpg) required is; hpg = hpc / η

- The unit horsepower is; hpu = hpc / MRR (hp.min/mm3) Unit power is also known as the specific energy U

Example 3.2

In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine; (a) the machining shear strain, (b) shear stress, and (c) cutting horsepower.

SOLUTION

(a) (Shear strain):

r = 0.02/0.045 = 0.444Shear plane angle: tan = r cos / [1 – r sin ]

tan = 0.444 cos 10 /[1 – 0.444 sin 10] = 25.4°

Shear strain: = tan( - ) + cot = tan(25.4 - 10) + cot 25.4 = 2.386

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SOLUTION (cont.)

(b) (Shear stress):

Shear force: Fs = Fc cos - Ft sin Fs = 350 cos 25.4 - 285 sin 25.4 = 194 lb

Shear plane area:As = to w / sin As = (0.02) (0.125)/sin 25.4= 0.00583 in2

The shear stress is τ = 194/0.00583 = 33,276 lb/in2

(c) (Cutting horsepower):

Cutting hp: hpc = Fc v / 33,000

hpc = (350) (200) / 33,000 = 2.12 hp

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Tool wear (a) Crater wear and (b) Flank wear.

3.9 Tool Failure and Tool Life

- Tool wear

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I: Initial wear

II: Steady state wear

III: Rapid wear

The criteria recommended by ISO3685:1993 to define the effective tool life for cemented carbides tools, high-speed steels (HSS) and ceramics are:Cemented carbides:1. VBB = 0.3 mm, or2. VBB,max = 0.6 mm, if the flank is irregularly worn, or;HSS and ceramics:1. Catastrophic failure, or;2. VBB = 0.3 mm, if the flank is regularly in region B; or3. VBB,max = 0.6mm, if the flank is irregularly in region B.

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- Tool life

Taylor’s equation:

Another form of Taylor’s equation:

n

1

2

2

1

T

T

v

v

Where:

V: Cutting speed (m/min.)

TL: Tool life (min.)

n, C : Constants depend on machining conditions, e.g., feed, depth of cut, coolant, etc.

CTv n

L

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The outside diameter of a cylinder made of titanium alloy is to be turned. The starting diameter = 500 mm and the length = 1000 mm. Cutting conditions are feed = 0.4 mm/rev and depth of cut = 3.0 mm. The cut will be made with a cemented carbide cutting tool whose Taylor tool life parameter n = 0.23 and C = 400 (m/min). Compute the cutting speed that will allow the tool life to be just equal the cutting time for this part.

Example 3.3

SOLUTION

v TLn = C

(1)

Cutting time (Tm) =

n

1

L v

CT

Nf

Lo

, where: N = (1000 * v) / D

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v1000f

DLT o

m

(2)

For Tool life (TL) = Cutting time (Tm) i.e., Equating (1) & (2):

vf1000

DL

v

C on

1

n

1on

11

Cf1000

DLv

V = 202.18 m/min.

f1000

DL

v

Cv o

n

1

n

1

23.0

123.0

11

4004.01000

5001000v

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Web site:

http://www.staff.zu.edu.eg/awafa/

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