page 43

-Compressive strength is high compared to tensile strength, therefore the bits are often brazed to steel shanks, or used as inserts in holders

-These inserts may often have negative rake angles

-Speeds up to 300 fpm are common on mild steels

-Hot hardness properties are very good

-coolants and lubricants can be used to increase tool life, but are not required.

-special alloys are needed to cut steel

•Ceramics

-sintered or cemented ceramic oxides, such as aluminum oxides sintered at 1800°F

-Can be used for turning and facing most metals, except for nimonic alloys and titanium. Mild steels can be cut at speeds up to 1500 fpm.

-These tools are best used in continuous cutting operations

-There is no occurrence of welding, or built up edges

-coolants are not needed to cool the workpiece

-Very high hot hardness properties

-often used as inserts in special holders

•Diamonds

-a very hard material with high resistance to abrasion

-very good for turing and boring, producing very good surface finish

-operations must minimize vibration to prolong diamond life

-also used as diamond dust in a metal matrix for grinding and lapping. For example, this is used to finish tungsten carbide tools

•Cemented Oxides

-produced using powder metallurgy techniques

-suited to high speed finishing

-cutting speeds from 300 to 7500 fpm

-coolants are not required

-high resistance to abrasive wear and cratering

5.8 TOOL LIFE

•Tool life is the time a tool can be reliably be used for cutting before it must be discarded/ repaired.

•Some tools, such as lathe bits are regularly reground after use.

•A tool life equation was developed by Taylor, and is outlined below,

page 44

V × Tn = C

where,

V = cutting velocity in ft./min. T = tool life in minutes

n = a constant based on the tool material C = a constant based on the tool and work

For example, if we are turning a 1” diameter bar, and we have a carbide tool, we want to have the tool last for 1 shift (8 hours) before a change is required. We know that for carbide tools n=0.2, and when the bar was cut with a velocity of 400 ft./min. the tool lasted for 2 hours. What RPM should the lathe be set at?

First find the C value for the equation,

400 × ( 2 × 60) 0.2 = C =

Next, find the new cutting speed required,

V × ( 8 × 60) 0.2 =

Finally, convert cutting velocity to RPM,

RPM =	12---------------× V	=
	π D

• An important relationship to be considered is the relationship between cutting speed and tool life,

page 45

V Tn = C

log ( V Tn) = log C

log V + log Tn = log C

log V + nlog T = log C

log V = – nlog T + log C

This function can be plotted on log scales as a linear function,

log V

intercept (log C)

slope (n)

V1

V2

log T

We can find the slope of the line with a two point interpolation,

log V1 – log V2 n = ----------------------------------

log T2 – log T1

Some examples of values are, (note that this is related to ‘n’)

High Speed Steel Tool	n = 0.10 to 0.125
Carbide Tool	n = 0.125 to 0.25
Ceramic Tool	n > 0.25

• Although the previous equation is fairly accurate, we can use a more complete form of Taylor’s tool life equation to include a wider range of cuts.

page 46

VT	n	d	x	y	= C	where,
			f			d = depth of cut
						f = feed rate
						x, y = calculated constants

5.8.1 The Economics of Metal Cutting

•As with most engineering problems we want to get the highest return, with the minimum investment. In this case we want to minimize costs, while increasing cutting speeds.

•EFFICIENCY will be the key term - it suggests that good quality parts are produced at reasonable cost.

•Cost is a primarily affected by,

-tool life

-power consumed

•The production throughput is primarily affected by,

-accuracy including dimensions and surface finish

-mrr (metal removal rate)

•The factors that can be modified to optimize the process are,

-cutting velocity (biggest effect)

-feed and depth

-work material

-tool material

-tool shape

-cutting fluid

•We previously considered the log-log scale graph of Taylor’s tool life equation, but we may also graph it normally to emphasize the effects.

page 47

cutting velocity

This graph is representative for most reasonable cutting speeds. The velocities at the high and low ranges do not necessarily exhibit the same relationship.

tool life

•There are two basic conditions to trade off,

-Low cost - exemplified by low speeds, low mrr, longer tool life

-High production rates - exemplified by high speeds, short tool life, high mrr

*** There are many factors in addition to these, but these are the most commonly considered

	cost per	total	tool
			cost
	unit	cost
	volume
	cutting
	cost
			overhead
	minimum		cutting speed
	cost

• A simplified treatment of the problem is given below for optimizing cost,

page 48

First lets look at costs for a cutting tool over the life of a tool,

where,

and,

c3 where,

and,

Ct = c1 + c2 + c3

Ct = cost per cutting edge

c1 = the cost to change a tool

c2 = the cost to grind a tool per edge c3 = the cost of the tool per edge

c1	= t1 ×	Rc
c2	= t2 ×	Rs
		-----
		N1

=		CT
	--------------------------------
	N1	× ( N2 + 1)

t1 = tool change time

t2 = tool grind time in minutes

Rc = cutting labour + overhead cost Rs = grinding labor + overhead cost CT = cost of the original tool

N1 = the number of cutting edges to grind N2 = the maximum number of regrinds

Cc = Rc × T

where,

Cc = cutting operation cost over life of tool, per edge T = tool life

page 49

Next, lets consider the effects of metal removal rate,

QT = V × T × f × c

where,

QT = metal removal rate per edge V = cutting velocity

f = tool feed rate

c = depth of width of the cut

consider the life of the tool,

V × Tn	= C (Taylors toolifeequation)
V =	C
	-----
	Tn

Now combine tool life (2) with the mrr (1),

QT = V × T ×	f ×	C	× T ×	f ×	C × f × c
		c = -----			c = --------------------
		Tn			Tn – 1

(1)

(2)

At this point we have determined functions for cost as a function of tool life, as well as the metal removal rates. We can now proceed to find cost per unit of material removed.

Cu =	Cc + Ct	=	Tn – 1		( Rc × T + Ct)
	-----------------		--------------------
	QT		C × f ×	c

Using some basic calculus, we can find the minimum cost with respect to tool life.

dCu

1

×

n ×

T

n – 1

+ Ct

× ( n – 1) ×

T

n – 2

--------- =

-------------------- ( Rc

) = 0

dT

C × f × c

Rc × n × T = –Ct × ( n – 1)

T =

–Ct × ( n – 1)

=

Ct

1 – n

-------------------------------

-----

-----------

Rc × n

Rc

n

• We can also look at optimizing production rates,

page 50

There are two major factors here when trying to increase the mrr. We can have a supply of tools by the machine, and as the tools require replacement, the only down-time involved is the replacement of the tool.

This gives us an average rate of production,

Rp =	QT
	-------------
	T + t1

where,

Rp = average rate of production

recall from before that,

Cfc

Q = -----------

T Tn – 1

now substituting in gives,

	Cfc
	-----------
	Tn – 1	= Cfc( T	n	+ t1)	–1
Rp = -----------------
	T + t1

We can now optimize the production rate,

dRp

n

+ t1)

–2

+ ( nT

n – 1

+ t1) (

T

n

–1

0

-------- = Cfc[ – ( T

+ t1) ] =

dT

( Tn + t ) –2

= ( nTn – 1 + t ) ( Tn

+ t ) –1

1

1

1

1 = ( nTn – 1 + t ) ( Tn + t )

1

1

1 = nT2n – 1 + nt

1

Tn – 1

+ t

1

Tn

+ t2

1

log ( 1) =

log ( nT2n – 1)

+ log ( nt

1

Tn – 1) + log ( t

1

Tn)

+ log ( t2)

1

0 = log ( n) + ( 2n – 1) log ( T) + log ( nt1) + ( n – 1) log ( T) + log ( t1) + nlog ( T) + log ( t21)0 = 2log ( n) + ( 4n – 2) log ( T) + 4log ( t1)

log ( T)	log ( n) + 2log ( t1)
	= ------------------------------------------
	1 – 2n