Tool Life Numerical Problems

Tool Life Numerical
Problems
Problem-1. Determine percentage change in cutting speed required to give
50% reduction in tool life (i.e., to reduce tool life to 1/5 of its previous
value). Take n = 0.2.
Solution: Given:
Tool life in Ist case = 1
Tool life in IInd case = 1/5
Exponent n = 0.2
To find:
Percentage change in cutting speed.
Formula used:
Taylor’s tool life equation
VTn = C
where, V= Cutting speed (m/min)
T= Tool life (min)
C = Machining constant
Sol: Using the Taylor’s tool life equation
Result:
The percentage increase in cutting speed is 38% by doing half tool life.
Problem:2. Compare the tool life of two cutting tools (HSS and carbide) at
a speed of 30 m/min. The tool life is 130 min. The tool life equation for
HSS tool is given by VT1/7 = C1 and for carbide VT1/5= C2 at a cutting
speed of 24 m/min.
Sol: Given:
• Tool life T= 128 min
• VT1/7 = C1 (for HSS) at 24 m/min
• VT1/5 = C2 (for carbide) at 24 m/min
To find:
• (i) Tool life for HSS = T1 at V = 30 m/min
• (ii) Tool life for carbide = T2 at V=30m/min
• (iii) T1/T2.
Formula used:
• Taylor’s tool life equation VTn = C.
Procedure:
(i) Tool life for HSS = T 1
(ii) Tool life for carbide = T2
Taylor’s equation for carbide,

(iii) Comparing tool life for HSS and carbide
• T1/T2 = 27/42.5 = 0.63 Ans.
Results:
(i) Tool life for HSS at 30m/min is 27 minutes.
(ii) Tool life for carbide at 30m/min is 42.5 minutes.
(iii) T1/T2 =0.63.
Problem 3:
Find the tool life equation if a tool life of 80 min is obtained at a cutting
speed of 30 m/min and 8 min at 60 m/min.
Solution: Given:
• Tool life T1 = 80 min, cutting speed V1 = 30 m/min
• Tool life T2 = 8 min, cutting speed V2 = 60 m/min.
To find:
• Tool life equation
Formula used:
• Taylor’s tool life equation
• VTn = C
Procedure:
• Applying V1T1n = V2T2n
• 30(80)n = 60 (8)n
• => n = 0.3
Hence, Tool life equation becomes VT0.3 = C.
Result:
• Tool life equation VT0.3 = C.
Problem 4:
For machining medium carbon steel, the conditions are:
Cost of operating machine = Rs 0.3 per min
Total cost of tool change = Rs 8.
The cutting speed = 40 m/min.
Tool life = 50min
Exponent index n = 0.2
Find optimum tool life and cutting speed at optimum tool life i.e., optimum
cutting speed.
Solution: Given:
• Cost of operating machine = C1 = Rs 0.3 per min.
• Total cost of tool change = C2 = Rs 8.
• Cutting speed = V = 40 m/min
• Tool life = T = 50 min
• Exponent index = n = 0.2
To find:
(i) Optimum tool life (TopT)
(ii) Optimum cutting speed (VopT)
Formula Used:
Procedures:
(i) Finding optimum tool life TOPT
(ii) Finding optimum cutting speed (VOPT)
Results:
(i) Optimum Tool life, TOPT = 106 min.
(ii) Optimum cutting speed, VOPT = 34.41 m/min
1. A tool has a life of 9 min when cutting at 250 m/min. Calculate the cutting speed for
the same tool to have a tool life of 160 min. Take Taylor tool life parameter n=0.22.
(Ans: 132.73 m/min)
2.Calculate the percentage change in cutting speed required to give 60 percent reduction
in tool life. Take n=0.22
(Ans: Increase in cutting speed by 22.3%)
3. Tool life is expressed by the relation VT^0.125= C1 for H.S.S. tool and VT^0.2= C2
for tungsten carbide. Assuming that at a speed of 24 m/min, the tool life for both the
tools is 170 minutes; compare the life of tools at a speed of 30 m/min.
(Ans: Tool life: HSS=28.52
min, Tungsten carbide=55.7 min)
4. A mild steel workpiece is being machined by two different tools A and B under
identical machining conditions. The tool life equations for these tools are: For tool A:
VT^ 0.32 = 42.5 and For tool B: VT^ 0.45 = 88.6Where V and T are in m/sec and sec
respectively. Determine the cutting speed above which tool B will give better tool life.
(Ans: 7.0 m/sec)

5.Tool life of 9 min was obtained during straight turning of a 24 mm diameter steel bar
at300 rpm with HSS tool. When the same bar was turned at 250 rpm, the tool life
increased to 48.5 min. What will be the tool life at a speed of 280 rpm?
(Ans: 17.06 min)
6. For a specified roughing cut on a lathe, the following information is known: Taylor
tool life exponent = 0.30 and Taylor equation constant = 900, cutting speed of 120
m/min, feed rate of 0.2 mm/rev, work piece diameter of 50, and cut length of 800 mm.
Calculate the percentage of the tool life that remains after the roughing cut is completed.
7. The outside diameter of a cylinder made of titanium alloy is to be turned. The initial
diameter is 500 mm and turning length is 1000 mm. The turning is to be carried out
using a cemented carbide tool having
Taylor’s tool life parameters n=0.23 and C=400.
Cutting conditions are: feed = 0.4 mm/rev and depth of cut = 3.0 mm. Compute the
cutting speed that will allow the tool life to be just equal to the machining time of this
part.
(Ans: 201.5 m/min)

Estimation of Machining Time
Necessity Of Estimation Or Determination Of Machining
Time Requirement For Particular Operations.
The major aim and objectives in machining industries are:-
• reduction of total manufacturing time, T
• increase in MRR, i.e., productivity
• reduction in machining cost without sacrificing product quality
• increase in profit or profit rate, i.e., profitability.
All those objectives are commonly and substantially governed by the total
machining time per piece, Tp, where again,
• Ti and TCT could have been spectacularly reduced by development and
application of modern mechanisation or automation.
• The tool life, TL has been substantially enhanced by remarkable
developments in the cutting tool materials.
Therefore, the actual cutting or machining time Tc remains to be controlled

as far as possible for achieving the objectives and meeting the growing
demands.
Hence, it becomes extremely necessary to determine the actual machining
time, Tc required to produce a job mainly for,
• assessment of productivity
• evaluation of machining cost
• measurement of labour cost component
• assessment of relative performance or capability of any
machine tool, cutting tool, cutting fluid or any special or new
techniques in terms of saving in machining time.
The machining time, Tc required for a particular operation can be determined
• roughly by calculation i.e., estimation
• precisely, if required, by measurement.
Major Factors That Govern Machining Time
The factors that govern machining time will be understood from a simple case
of machining. A steel rod has to be reduced in diameter from D1 to D2 over a
length L by straight turning in a centre lathe as indicated in Fig.
where, t = depth of cut in one pass, mm.
But practically the value of t and hence np is decided by the machining
allowance kept or left in the preformed blanks. Usually, for saving time
and material, very less machining allowance is left, if not almost
eliminated by near – net – shape principle.
Hence, number of passes used is generally one or maximum two : one for
roughing and one for finishing.
However, combining equation, we get
Material Removal Rate :

Tool life is most affected by machine.......
A. Cutting speed
B. Tool geometry
C. Feed and depth
D. Microstructure of material being cut
E. Not using coolant and lubricant
Answer: Cutting speed
Tool life is said to be over if.......

A. poor surface finish is obtained
B. Sudden increase in power and cutting force with chattering take place
C. Overheating and funning due to friction start
D. All of the above
E. None of the above
Answer: All of the above

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Tool Life Numerical

Taylor’s equation for carbide,

(Ans: 132.73 m/min)

(Ans: Increase in cutting speed by 22.3%)

(Ans: 7.0 m/sec)

(Ans: 17.06 min)

(Ans: 201.5 m/min)

Therefore, the actual cutting or machining time Tc remains to be controlled

Material Removal Rate :

Tool life is said to be over if.......

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