Chapter 6
ECONOMICS OF
METAL
CUTTING
OPERATIONS
Prof. Dr. S. Engin KILIÇ
MFGE 307 THEORY OF MANUFACTURING TECHNOLOGY II
Content
Economics of Metal Cutting
Operations
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
Optimization Criteria
Choice of Cutting Conditions
Unit Production Time
Unit Production Cost
Photos from internet sites.
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Economics of Machining
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Production cost and production rate are vitally important for a
manufacturer.
Increasing production rate means producing more from the
available resources.
Decreasing production cost means less expenditure for the
same volume of production.
If the conditions are so selected to maximize the production
rate and minimize the production cost, his/her profit can be
maximized.
It is however not possible to find a common set of
manufacturing conditions to satisfy this.
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Economics of Machining (2)
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A common practice is to treat each case separately
and find the corresponding conditions, then to make a
compromise in between.
Production of a component involves several machining
operations using a variety of machine tools.
Only one operation to be performed on one machine
will be considered and the conditions leading to min.
production cost and max. production rate (min. prod.
time) will be evaluated.
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Assumptions
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One operation-one machine, hence manual times
between operations and transportations between
machines are ignored.
Components are ready at the side of the machine
before operation and to be stacked at the side of the
machine after operation.
Appropriate tool and cutting fluid have already been
chosen.
Depth of cut has been selected at maximum value
(limited by the total stock to be removed, chatter
vibrations, cutting force, etc.)
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Optimization Criteria

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Unit production time: average time taken to produce one
component.
Unit production cost: total average cost of producing one
component:

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As cutting speed and feed rate are increased, both tend to decrease
due to an increase in metal removal rate.
At the same time, they may tend to increase due to increased
frequency of tool changes (increasing tool wear rate).
Optimum conditions do exist where the measure of the selected
criterion is minimum.
Profit rate: it is very difficult to attribute revenue to individual
operations and profit rate cannot be expressed as an explicit
function of cutting conditions.
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Choice of Cutting Conditions
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Effects of depth of cut, feed rate and cutting speed on metal
removal rate are the same.
Cutting speed has the highest adverse effect on tool life
amongst the cutting conditions.
Common practice is to select depth of cut at its maximum value,
then choose feed rate as high as possible considering the
limitations such as the available feeds on the machine, surface
finish requirement, force level which the cutting edge can
withstand, etc.
Cutting speed will be selected as the one to optimize the
selected criterion.
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Unit Production Time
Unit production time = {time for machining} + {time for tooling}
tpr = tf
tf = tl + t c
+
ttp
ttp = nt tch
where
tl = loading and unloading time/piece
tc= actual machining time /piece
tch= tool changing time (time/edge)
nt = fraction of tool life to cut one piece (edge/piece)
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ME 303 - Section 05c
8
Unit Production Time
tpr = tl + tc + nt tch
For turning operation:
V
tc 
w
 volume
V  volume of material to be removed 
 piece
 volume
w  chip removal rate
 time

tc  machining time






 time 
 piece 


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Actual Machining Time
V
di
l


d
4
2
0

 d i2 .l
w  d . f .v
d0
d0  di
d  depth of cut 
2
f  feed
v  cutting speed
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Actual Machining Time (2)


d 0  di d 0  di   l
d 0  di   l
tc  4
 2
d 0  di   f  v
f v
2
 d 0  di
B
if
B
 l  tc 
2
f
v
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Actual Machining Time (3)
t
Nt 
tc
 N t  no. of pieces to be machined

during life of one cutting edge

 pieces  
 time 
 edge   t  tool life
 edge 





 time 
t c  machining time
 piece 



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Unit Production Time
Taylor’s Equation:
1
n
C 
vt  C  t     t r
v 
n
1 n 
nt 
1
Nt
tc


t
B
v
 v
tr C
1
n
 nt 
B v
C
1
n
n
tr
Unit production time as a function of the cutting speed :
t pr
B

  t l    t ch v
v

1 n
n
1
B

 1
tr
Cn
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Unit Production Cost
Unit production cost = {cost of machining} + {cost of tooling}
Cpr
Cf
=
( R0
=
+
Rm )
cost of machining
per unit time
(cost of time)
Cf
(
tl
+
+
Ctp [cost/piece]
tc )
time for
machining
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Unit Production Cost (2)
R0 = Operator’s wage rate + overheads
[money paid/time]
Rm= Machine’s cost (depreciation) rate + overheads
Ctp =
[cost/time]
nt [ tch ( R0 + Rm ) + Ct ]
tool changing
cost
Ct = cost of each sharp cutting edge
tch = tool changing time
tool
cost
[cost/edge]
[time/edge]
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Cost Parameters
cost of tool
Ct 
 cost of grinding cutting edge
no. of cutting edges usable
R 0  operator' s wage rate  labor overhead rate as % of op.' s wage rate
Rm 
cost of machine
 machine overhead rate as % of mach. rate
amortizati on time
Cost of time: R0 + Rm , usually expressed in cost/hour
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Unit Production Cost
C pr  R 0  R m t l  t c   n t t ch R 0  R m   C t 
But from the earlier analysis:
1 n 
B
tc 
v
and
nt 
B v
C
1
n
n
tr
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Unit Production Cost (2)
Substituting these in the main equation, unit production cost can be
expressed as a function of the cutting speed:
C pr
B

 R0  Rm  t l    t ch R0  Rm   C t v
v

1 n
n
1 B
  1
tr
Cn
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Optimum Conditions


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In order to get the optimum conditions (i.e. the
optimum cutting speed), one needs to optimize the
objective function according to the selected criterion.
It is necessary to find the point where the slope of
the objective function is zero, (i.e. the partial
derivative of the objective function w.r.t. the cutting
speed) is zero.
Solving for the cutting speed to satisfy that condition
will give the optimum cutting speed for the selected
criterion.
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Minimization of Unit Production Cost
C pr
v
0
Solving for the cutting speed gives the cutting speed for
minimum cost, vmc and the corresponding tool life tmc:
 n

( Ro  Rm )tr
vmc  C 

1  n [tch ( Ro  Rm )  Ct ] 
n
 C
tmc  tr 
 vmc
1/ n



Ct 
1 n 
tmc 
tch 

n 
Ro  Rm 
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Maximization of Production Rate
t pr
v
0
Solving for the cutting speed gives the cutting speed for maximum
production rate, vmp and the corresponding tool life tmp:
 n tr 
vmp  C 
 
1  n tch 
n
 C
t mp  t r 
 vmp

1/ n




1 n
t mp 
tch
n
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unit prod. cost
production rate
high
efficiency
range
vmc
Production rate
Unit production cost
Machining at High Efficiency
vmp Speed
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Problem 1
A coated carbide cutting tool has 60 min tool life when
cutting mild steel at 1.2 m/s. If Taylor’s tool-life
exponent is 0.25.
a) find Taylor’s tool-life constant
b) find the tool life if the cutting speed is doubled.
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Problem 2
A mill roll that is 2 m long and 50 cm in diameter is to be rough turned
to a diameter of 49.2 cm. Work material is alloy steel with a hardness of
300 BHN. The labour and the machine costs including the overheads are
$20.00 per hour for each. The loading and unloading time is 10 min.
There are two alternatives for the tool material to be used:
i) an HSS tool which costs $2.00 and can be ground 10 times with a cost
of $1.20 per grind. Removing and resetting the tool requires 4 min.
ii) a triangular throwaway carbide tool having 3 cutting edges costs
$6.00. Indexing the tool takes 2 min.
a) Which type of tool should be used?
b) What would be the unit production cost if cost minimization criterion
is applied when WC tool is used? Assume that the recommended
cutting speeds correspond to a tool life of 90 min and Taylor’s tool
life exponent is 0.25 for WC tools.
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Problem 2
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Problem 3
A cylindrical component of 60mm x 120mm is to be rough
turned orthogonally from a carbon steel bar stock of 66mm
using a WC throw-away type insert tool. The following data
apply for the operation:
• Time taken to load and unload work parts = 60 s
• Time taken to index a cutting edge = 20 s
• Cutting speed = 2.5 m/s
• Specific cutting energy = 1.3 GN/m2 (const. for all chip
thickness values)
• Motor power = 5 kW
• Machine efficiency = 80%
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Problem 3 (Cont’d)
a) Find the feed rate in mm/rev for maximum power utilization.
b) If the cutting force is not allowed to exceed 1300 N, find the
allowable feed rate.
c) For the feed rate found in (b), it has been experimentally found
that the tool life becomes 2400 s when the cutting speed is 4 m/s
and the tool life exponent is 0.25. Using this information and the
feed rate found in (b), calculate the optimum cutting speed to
maximize the production rate.
d) Check if the available power is enough to perform the operation at
the conditions set in (c). If not, adjust the cutting speed to make
the operation feasible.
e) Find the total production time/part for the conditions found in (d).
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