Machining -- Surface, Cutting

Edge, and Shaping

• The principle used in all machining processes is to create the surface required by providing suitable relative motions between the cutting tool and the workpiece.

• The cutting edge or edges on the cutting tool remove a layer of work material; the removed material is often called a chip.

• When a shaping tool is reciprocated back and forth along a straight line and a workpiece is incrementally fed beneath the tool in a direction perpendicular to the motion of the tool, a flat surface will be created on the workpiece.

Turning

• A cylindrical surface can be generated by spinning the workpiece and feeding the tool parallel to the axis of workpiece spin in turning.

• The internal or external cylindrical surfaces using the lathe turning with a pointed tool tip can also be categorized as the single point cutting process.

Primary and Secondary Motion

• The primary motion is the main motion provided by a machine tool to cause the relative motion between the tool and workpiece.

• The feed motion, or the secondary motion, is a motion that may be provided to the tool or workpiece by a machine tool which, when added to the primary motion, leads to a repeated or continuous chip removal.

• Usually, the primary motion absorbs most of the total power required to perform a machining operation.

Three Types of Metal Cutting Tools

• Single-point tools

• Multipoint tools

• Abrasive grits

SINGLE POINT

CUTTING PROCESSES

Lathe Turning

Five Types of Lathe Operationstransient surface

primary motionmachined surface

transient surfaceprimary motion

work surface

work surface

work surface

tool

continuous feed motion

continuous feed motion

continuous feed motion

dw dwdm

ap

ap

κr

κr

apκr

(3-a) (3-b)

(3-c)

(3-d)

machined surfaces

dw

ap

κr

(3-e)

transient surface

d wd m

κ r

a pcontinuousfeed motion

primary motion

tool

f

Single Point Cutting Configurations

-- Top view

feed engagement

f

major cuttingedge angle κ

r

tool shank

cross sectionof uncut chip

tool facemajor cuttingedge

f

ac

κr

ac

ap back engagement(depth of cut)

minor cutting edge

minor cuttingedge angle κ’r

toolcorner

undeformedchip thickness

Geometrical Relationships

(1)

where

ac is the un-deformed chip thickness

f is feed

(2)

where

Ac is the cross-sectional area

ap is depth of cut

rc fa κsin =

pc faA =

Machining Time

where

t is the machining time

lw is the length of cylindrical surface

nw is the rotational frequency of the workpiece

Number of revolution of the workpiece is lw/f

w

w

fn

lt =

Cutting speed and MRR

: cutting speed at the tool corner

: maximum cutting speed

wmndπ

wwndπ

Average cutting speed

Material removal rate

)ad(n)dd(n

v pmwmww

av +=+

= ππ

2

( ) ( )[ ]pmwpavcw adnfavAZ +== π

Front view of the tube

dmdw

Specific Cutting Energy, ps

• Energy required for a unit volume of work-material

• ps -- unit of J/m3

• Power for primary motion (the main workpiece spindle for the case of lathe): Pm = Zwps

Approximate Value of ps

cs alog46.030.0plog −=

csalog46.019.0plog −=

cs alog46.004.0plog −=

cs alog46.026.0plog −−=

cs alog46.00.38plog −−=

Surface Roughness: Ra and Rt

dxdx)x(yL

)x(yL

R

L L

a ∫ ∫

−=

0 0

11 : arithmetic average or centerline average roughness

: maximum (peak to valley) height roughness

[ ] [ ])x(ymin)x(ymaxRLxLx

t≤≤≤≤

−=00

f

R tκ

rrκ’

R cotκ

t’rR cot

κ

t

r

rκ’

f

rκ’

R

rκ’Rsin

f/2

R t

(a) (b)

rr

t'cotcot

fR

κκ +=

( )rr

tta

'cotcot

fRfR

fR

κκ +==

=

44222

12

1

R

fRt

8

2

≈

R

fRa

32

2

≈

Example

A 2024-T4 aluminum cylinder of 64 mm diameter is turned on a lathe to reduce its diameter by 5 mm. The length to be machined on the cylinder is 105 mm. The lathe has 2 kW (2.68 HP, Note that 1 HP=745 W) available at the spindle and the HSS cutter has a major cutting edge angle of 70o and a minor cutting edge angle of 6o. Suppose the operation is to be done in one pass while a smooth surface with Ra less than 5 µm is desired, what should be the spindle speed and how much machining time would be needed.

Solution:

Using a safety factor of 1.5 on the desirable Ra, Eq. (10) gives:

From Fig. 4, the specific cutting energy for aluminum alloy at an

undeformed chip thickness of ac = fsinκr = 0.13sin(70o) = 0.12 mm is

1.1 GJ/m3. Using a safety factor of 1.5 on the lathe power against

possible motor noise and vibration, Eqs. (5) and (6) give:

From Eq. (3), the required machining time is

( ) mmff

oo 13.0

6cot70cot45.1

105 6

≤⇒+

≥× −

( ) ( )3 3

9 32000 5 10 5 101.1 10 0.13 10 π 0.064 19.3 Hz 1158 rpm

1.5 2 2w w

n n− −

− × ×≥ × × − ⇒ ≤ =

( )

3

3

105 1042 s

0.13 10 19.3t

−

−

×≥ =

×

( ) ( )[ ]pmwpavcw adnfavAZ +== π

Facing

Machining time

Maximum cutting speed vmax:

Maximum metal removal rate Zw,max:

w

wm

fn

dt

2=

wwdnv π=max

wwpw dnfaz π=max,

Example -- FacingThe following figure shows the facing of a hollow cylinder with

inner diameter (di) of 20 mm and outer diameter (do) of 50 mm.

This operation is performed on a computer numerical controlled

(CNC) machine that reads the instantaneous tool position. Based on the tool position, the tool infeed per revolution is varied

during the cut to maintain a constant material removal rate of

549 mm3/sec throughout the process. As ap of 1.5 mm is used,

what is the total machining time (in s)? If a constant infeed

process is exercised with a maximum material removal rate of 549 mm3/s at the beginning of cut, what is the total machining time required?

a p

dd o

i

Solution

For constant Zw:

Computer Controlled FacingTypical Facing

infeed

MRRinfeed

MRR

machining

time

machining

time

( ) ( ) ( )s 5.4

549

20504

5.1

4removed material of 2222

=−

=−

==

ππ

w

io

p

w z

dda

z

volumet

If a constant “f” is maintained with 3

,max 549 mm / swz =

,max

,max

2 2

o i o iw p w o p o m p o

m w

d d d dz fa n d a d t a d

t zπ π π

− −= = ⇒ =

( ) ( )( )

50 20 π 1.5 50 6.44 s

2 549mt

−⇒ = =

Other Single Point Processes

• Shaping

• Boring

• Planing

In turning a mild steel tube with 2.5 mm thick wall in an orthogonal cutting configuration (0 deg lead angle) with 0.25 mm/rev feed, 0 deg rake angle and 200 m/min cutting speed, the average chip thickness was 0.75 mm and the cutting and thrust force was 600 Nand 900 N, respectively. Calculate:

• (i) From the cutting geometry, find the cutting ratio, γc, and shear angle, φ.

• (ii) From the Merchant force circle, find the force components normal and parallel to the shear plane (Fn and Fs), normal and parallel to the rake face (N and F), friction angle, β, and coefficient of friction, µ.

• (iii) Estimate the shear stress, τ, across the shear plane.• (iv) Use the Merchant’s equation to estimate the shear angle and

compare to the measured shear angle.• (v) From the velocity triangle, find the chip velocity and velocity

across shear plane.• (vi) Estimate the specific cutting energy, specific shear energy and

specific friction energy.

Concluding Remarks

• Single point machining is the most basic machining process

• It is an important first step to understand the machining processes

• Most of machining simulation is still based on single point cutting

• Multipoint machining is next!