ecomics of metal cutting

1. Economics of metal cutting
Dr. Cho Cho Myint Aye
Mechanical Engineering Department
Yangon Technological University

2. Course objectives
-To know cutting speed, Tool life, cost per
component, time per component for
minimum production cost and min production
time

3. 4.1 Cutting speed
minute
in
life
tool
t
fpm
,
speed
cutting
V
-(4.1)
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
C
Vt
life
tool
and
Speed
4.1.1
n




4. Average values of n
Tool material Work Material n
High Speed Steel Steel 0.125
High Speed Steel Cast Iron 0.140
Cemented Carbide Steel 0.200
Cemented Carbide Cast Iron 0.250
Sintered Oxide Steel 0.500

5. 4.1 Cutting speed
• 4.1.2 Economical Tool life
When cutting speed is low, tools last a long
time and tool cost is low.
At the same time, metal removal is slow, and
the cutting cost and total cost are high.
On the other hand, cutting cost is low, but, tool
cost is high, and makes the total cost are high.

6. 4.1 Cutting speed
• 4.1 Cutting speed
• At some intermediate cutting speed, the total
cost is a minium.
• The total life corresponding to this cutting
speed is the economical tool life.

7. 4.2 Production Costs and production Rates
In practice, a high production rate would
probably mean low production costs.
The production time is defined as the average
time taken to produce one component.
The production cost is defined as the total
average cost of performing the machining
operation on a component using one machine
tool.

8. 4.2 Production Costs and production
Rates
• The total manufacturing costs, apart from the
cost of the material ,involve many items such
as material handling, temporary storage of
semi-finished products and transportation.
• The problems of metal handling will not be
discussed here.
• The main purpose is to analyse those factors
involved in a particular machining operation.

9. 4.3 Choice of Feed
• Feed is the distance moved by the tool relative
to the work piece in the feed direction for
each revolution of the tool or work piece or
each stroke of tool or work piece.

10. 4.4 Choice of Cutting Speed
Two distinct criteria exist for the choice of
cutting speed
(1) Minimum production cost
(2) Minimum production time or Maximum
production rate.

11. 4.4 Choice of Cutting Speed
• 4.4.1 Determination of Cutting speed Vc for
Minimum Production Cost
For cylinderical turning,
Nb= number of components to be produced
Nb can be separated into three items;
(1) The non-productive time , given by Nbt1,
Where t1 is the time taken to load and unload each
component and to return the tool to the
beginning of the cut.

12. 4.4 Choice of Cutting Speed
4.4.1 Determination of Cutting speed Vc for
Minimum Production Cost
(2)The total machining time, given by Nb,tm is
the machining time for each component.
(3) The total time involved in changing worn
tools, given by
Where is the tool changing time
Nt is the number of tool
ct
t t
N
ct
t

13. Continued
)
2
.
4
(
C
component
each
for
C
cost
production
average
The
)
t
N
t
N
t
M(N
costs
operator
and
machine
Total
1
pr
pr
ct
t
m
b
1
b

















t
b
t
ct
b
t
m C
N
N
t
N
N
M
Mt
Mt

14. Continued
.
increases
speed
cutting
as
increase
which
costs
tool
the
are
C
N
N
t
N
N
M
cost
machining
the
is
Mt
cost
productive
non
the
is
Mt
t
b
t
ct
b
t
m
1


15. Continued
n
1
r
r
r
n
r
V
V
t
t
b
given
is
solution
particular
a
for
life
tool
The
V
V
t
t
have
we
relation,
life
tool
s
By taylor'














y

16. Continued
operation
particular
a
for
constant
a
K
eed
cuttingnsp
V
V
K
t
t
t
N
N
m
m
b
t





17. Continued
feed
f
workpiece
of
diameter
the
d
turned
be
length to
the
l
where,
/f
l
πd
K
turning,
al
cylinderic
For
w
w
w
w





18. Continued
(A)
C
Mt
M
n
1
n
V
V
cost
production
minimum
for
V
speed
cutting
the
find
To
n
t
ct
tr
r
c
c
















19. Continued
(B)
t
t
n
1
n
V
V
time
production
minimum
for
V
speed
cutting
the
find
To
n
ct
r
r
p
p





















20. Continued
• 4.4.3 Determination of Tool tc for minimum
production cost i.e. Economical Tool life
• By Taylor’s tool-life relationship,
(C)
M
C
t
n
n
1
t
V
V
t
t
cost
production
minimum
for
life
tool
V
V
t
t
t
ct
c
n
1
r
r
c
n
1
r
r























 
















21. 4.4.4 Determination of tool life tp for minimum
production time
ct
p
n
1
p
r
r
p
t
n
n
1
t
V
V
t
t






 











22. 4.5 Estimation of factors needed to determine the
optimum cutting conditions
4.5.1. The operator and machine rate (M)
t
t
o
o M
100
overheads
m/c
Percent
M
W
100
overheads
operator
Percent
W
M 















Wo = the operator’s rate of pay(Wo)
The overheads associated with his or her employment (i.e. worker’s
benefit provided by the company ;costs of providing the working
facilities; costs of the administrators necessary to employ the
worker etc.)

23. Continued
• The depreciation of the machine tool Mt
period
on
Amortizati
year
per
hours
work
of
Number
tool
machine
of
cost
Initial
Mt



24. • 4.5.2.The tool cost(Ct) i.e. cost per sharp
cutting edge
life
tool
during
edges
cutting
of
no.
Available
possible)
regrinds
no.of
grind
per
(Cost
C
tool
of
cost
C T
t




25. • (a) For regrindable tools with single cutting
edge such as high speed and carbide-tipped
brazed tools,
1)
(N
life
tool
during
edges
cutting
of
no.
Available
1
N
possible
regrinds
of
no.
N
grind
per
Cost
C
C
tool
of
Cost
C
edge
sharp
with
supplied
tools
assuming
,
1
N
)
N
(C
C
C
g
g
g
g
T
T
g
g
g
T
t










26. • 4.5.2.The tool cost(Ct) i.e. cost per sharp
cutting edge
life
tool
during
edges
cutting
of
number
N
grind
per
Cost
C
tool
of
Cost
C
edge
sharp
with
initially
supplied
not
tool
assuming
,
C
N
C
C
g
g
T
g
g
T
t






27. • (b)For regrindable tools with multiple cutting
edge (Ne) such as carbide-tipped clamped
tools.
e
g
g
g
T
t
N
1)
(N
)
N
(C
C
C





For regrindable tools with multiple cutting edge (Ne)

28. • (c)For disposable tools with multiple cutting
edge such as throw away carbide tools.
ls
insert too
disposable
N
of
use
a
for
last
can
ch
holder whi
tool
of
cost
the
is
C
where
,
N
N
C
N
C
C
I
H
I
e
H
e
T
t




29. • 4.5.3.The tool changing time ct
t
This is the total time required to put a new
tool on the job and resume machining. It
consists of time required;
(a)To remove the tool from the machine
(b)To place a new tool in the machine
(c)To set the new tool in its correct position,
and
(d)To restart the machining process,

30. insert
per
used
edges
cutting
of
number
Average
insert
replace
to
time
1)
insert
per
used
edges
cutting
of
no.
(average
index
to
Time
tct





31. Example 4.1
• In a machining operation, the tool material is cemented
carbide and the workpiece material is cast iron, and the
basic information is as followed.
• Price of insert plus tool-holder depreciation 1.80 $
• Total cutting edges in life of insert 6
• Machine operator’s rate 4.00$/hr
• Machine overhead rate 8.00$/hr
• Tool-changing time 1 min
• Determine the tool life for minimum production cost and
maximum production rate for disposable tool.

32. Solution
• For Tool material –cemented carbide and the
workpiece material-cast iron , n=0.25 (from
table)
1min
t
$
20
.
0
60
00
.
8
00
.
4
60
hour
per
overhead
hour
per
Labor
M
$
0.30
6
1.80
C
insert
of
life
in
edges
cutting
tool
on
depreciati
holder
tool
insert
of
price
C
ct
t
t











33. For the minimum production cost , by(C)
Ans
7.5min
0.25
3
0.2
0.3
1
0.25
0.25
1
t
M
C
t
n
n
1
t
c
t
ct
c
















 














 


34. For the maximum production rate , by(D);
Ans 3min
1
3
t
n
n
1
t ct
p 








 


35. Example 4.2
• A large batch of steel shafts are to be rough-turned to
3 in diameter for 11.8 in of their length of a feed of
0.00985 ipr. A brazed-type carbide tool is to be used,
and taylor’s tool life speed relationship is given by
where Vis in fpm and t is in min. The initial cost of the
machine was $ 10800 and is to be amortized over 5
years. The operator’s wage is $5.40/hr, and the
operator and machine overheads are 100 percent.
Tool-changing and resetting time on the machine is 5
min. and the cost of regrinding the tool is $2.00.
800

n
Vt

36. • The initial cost of a tool is $ 6.00 and on the
average ,it can be ground 10 times altogether.
Finally, the non-productive time for each
component is 2 min. The machine is to be
used on 8 hour shift per day, 5 days per week
and 50 weeks per years . Find the cost and
production time per component
corresponding to the two criteria of minimum
production cost and minimum production
time.

37. Solution
dw = 3 in ,lw= 11.8 in, feed = 0.00985 ipr. where Vis in fpm
and t is in min.
The initial cost of the machine = $ 10800
Amortized period = 5 years
The operator’s wage = $5.40/hr,
the operator and machine overheads = 100 percent
Tool-changing and resetting time on the machine = 5 min
The cost of regrinding the tool = $2.00

38. Solution
The initial cost of a tool = $ 6.00
Ng =10 times
the non-productive time for each component
,t1= 2 min.
The machine is to be used on 8 hour shift per day,
5 days per week and 50 weeks per years .
To find the cost and production time per
component corresponding to the two criteria of
minimum production cost and minimum
production time

39. Solution
8)
(P
t
t
N
N
.3)
4
(
t
t
t
t
t
t
N
N
t
t
r
t
,
r
cost t
minimum
for
time
Production
)
2
.
4
(
C
N
N
t
N
N
M
Mt
Mt
r
C
nent
cost/compo
Tool
8)
(p
fV
l
πd
V
K
t
,
t
cost,
production
minimum
for
time
Machining
m
b
t
ct
c
mc
mc
1
ct
b
t
mc
1
c
p
c
p
t
b
t
ct
b
t
m
1
c
p
c
w
w
c
mc
mc






































40. Solution
8)
(P
t
t
N
N
)
3
.
4
(
t
t
t
t
t
t
N
N
t
t
r
t
,
r
cost t
minimum
for
time
Production
)
2
.
4
(
C
N
N
t
N
N
M
Mt
Mt
r
C
nent
cost/compo
Tool
8)
(P
fV
l
πd
V
K
t
,
t
time,
production
minimum
for
Conditions
m
b
t
ct
c
mp
mp
1
ct
b
t
mp
1
p
p
c
p
t
b
t
ct
b
t
m
1
p
p
p
w
w
p
mp
mp














































41. Solution
ct
c
mp
mp
1
ct
b
t
mp
1
p
p
c
p
t
t
t
t
t
t
N
N
t
t
r
t
,
r
cost t
minimum
for
time
Production







42. Solution
/min
0.216
$
M
0.018
100/100
0.018
5.4/60
100/100
5.4/60
M
M
0
percent/10
M
W
0
percent/10
W
M
M,
rate
operator
and
Machine
0.018/min
$
5
60
50)
5
(8
10800
M
period
Amortized
year
per
hours
working
of
No.
tool
machine
of
cost
Initial
M
tool
machine
the
of
on
depreciati
the
M
t
t
o
o
t
t
t























43. min
2
.
51
)
216
.
0
/
6
.
2
5
(
25
.
0
25
.
0
1
M
C
t
n
n
1
t
(C)
equation
using
by
,
cost
production
minimum
for
lifet
tool
Optimum
min
5
t
time
resetting
and
changing
Tool
2.6
$
2
10
6
C
N
C
C
C
ngedge),
(i.e.cutti
tool
sharp
a
providing
of
Cost
t
ct
c
c
ct
g
g
T
t
t




















44. fpm
299
5.12
1
800
t
t
V
V
,
V
speed
cutting
economical
the
Thus,
ly.
respective
1min,
and
C
as
taken
be
always
can
andt
V
C,
Vt
ip
relationsh
speed
life
tool
s
taylor'
In the
0.25
n
c
r
r
c
c
r
r
n


















45. 407fpm
15
1
800
t
t
V
edV
Cuttingspe
min
15
5
3
t
n
n
-
1
tp
,
t
time
production
minimum
for
life
tool
Also,
0.25
n
p
r
r
p
ct
p






























46.    
23
.
0
$
)
6
.
2
5
216
.
0
(
12
.
5
/
15
.
3
C
Mt
t
t
C
Mt
N
N
cost
Total
0.68
$
3.15
0.216
Mt
cost
Machining
0.432
$
2
0.216
Mt
cost
productive
-
Non
8)
-
(P
-
-
-
min
3.15
12
299
0.00985
11.8
3
π
fV
l
πd
V
K
t
,
t
cost,
production
minimum
for
time
Machinincg
t
ct
c
mc
t
ct
b
t
mc
1
c
w
w
c
mc
mc

























47. 8)
(P
t
t
N
N
min
5
.
5
5
2
.
51
/
15
.
3
15
.
3
2
r
t
.3)
4
(
t
t
t
t
t
t
N
N
t
t
r
t
,
r
cost t
minimum
for
time
Production
34
.
1
$
23
.
0
68
.
0
432
.
0
r
C
)
2
.
4
(
C
N
N
t
N
N
M
Mt
Mt
r
C
nent
cost/compo
Tool
,
t
cost,
production
minimum
for
time
Machining
m
b
t
c
p
ct
c
mc
mc
1
ct
b
t
mc
1
c
p
c
p
c
p
t
b
t
ct
b
t
m
1
c
p
mc






































48.    
570
.
0
$
)
6
.
2
5
216
.
0
(
15
/
32
.
2
C
Mt
t
t
C
Mt
N
N
cost
Total
0.501
$
32
.
2
0.216
Mt
cost
Machining
0.432
$
2
0.216
Mt
cost
productive
-
Non
8)
-
(P
-
-
-
min
32
.
2
12
407
0.00985
11.8
3
π
fV
l
πd
V
K
t
,
t
time,
production
minimum
for
Conditions
t
ct
c
mc
t
ct
b
t
mc
1
p
w
w
c
mp
mp

























49. 8)
(P
t
t
N
N
min
1
.
5
5
15
/
32
.
2
32
.
2
2
r
t
.3)
4
(
t
t
t
t
t
t
N
N
t
t
r
t
,
r
cost t
minimum
for
time
Production
5
.
1
$
570
.
0
501
.
0
432
.
0
r
C
)
2
.
4
(
C
N
N
t
N
N
M
Mt
Mt
r
C
nent
cost/compo
Tool
,
t
cost,
production
minimum
for
time
Machining
m
b
t
p
p
ct
c
mp
mp
1
ct
b
t
mp
1
p
p
p
p
c
p
t
b
t
ct
b
t
m
1
c
p
mc






































50. • If the company receives $1.75 for each
component, then
• For minimum cost condition,
• Profit rate=(Priced received-cost)/production
time
=(1.75-1.34)/5.5=$ 0.0745 $/min
Profit/year= profit rate x time
= 0.0745x 8 x 5 x 50 x 60 = $ 8940

51. • If the company receives $1.75 for each
component, then
• For minimum production time condition,
• Profit rate=(Priced received-cost)/production
time
=(1.75-1.5)/5.1=$ 0.049 $/min
Profit/year= profit rate x time
= 0.049x 8 x 5 x 50 x 60 = $ 5882

52. Comparison
Min.Prod.Cost Min.Prod.Time Remark
Cutting Speed 299 fpm 407 fpm
Tool life 51.2 min 15 min
Cost/component $ 1.34 $ 1.5 11.9%prod cost
increase
Time/component 5.5 min 5.1 min 7.3 % prod.time
reduction
Profit/year $ 8940 $ 5822 34.9 % profit
reduction