EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 9780100793439
Author: KALPAKJIAN
Publisher: YUZU
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Textbook Question
Chapter 21, Problem 40QLP
Explain whether it is desirable to have a high or low (a) n value and (b) C value in the Taylor tool-life equation.
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In a production turning operation, the workpart is 60 mm in diameter and 500 mm long. A feed of 0.75 mm/rev is used in the operation. If cutting speed=9 m/s, the tool must be changed every 4 workparts; But if cutting speed=5 m/s, the tool can be used to produce 50 pieces between tool changes. Determine the Taylor tool life equation for this job. (use the equations given below for solution)
In a production turning operation, the workpart is 60 mm in diameter and 500 mm
long. A feed of 0.75 mm/rev is used in the operation. If cutting speed-9 m/s, the tool
must be changed every 4 workparts; But if cutting speed=5 m/s, the tool can be used
to produce 50 pieces between tool changes. Determine the Taylor tool life equation for
this job. (use the equations given below for solution)
L
Tm- 1,= Nf
N
AD, vT" = C
%3|
AD,L
Tm
fv
(a) Taylor's equation is used to predict the life of a tool. Explain the main parameters
that effect a tool life with the help of a graph.
Chapter 21 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 21 - Explain why continuous chips are not necessarily...Ch. 21 - Name the factors that contribute to the formation...Ch. 21 - What is the cutting ratio? Is it always less than...Ch. 21 - Explain the difference between positive and...Ch. 21 - Explain how a dull tool can lead to negative rake...Ch. 21 - Comment on the role and importance relief angle.Ch. 21 - Explain the difference between discontinuous chips...Ch. 21 - Why should we be interested in the magnitude of...Ch. 21 - What are the differences between orthogonal and...Ch. 21 - What is a BUE? Why does it form?
Ch. 21 - Is there any advantage to having a built-up edge...Ch. 21 - What is the function of chip breakers? How do they...Ch. 21 - Identify the forces involved in a cutting...Ch. 21 - Explain the characteristics of different types of...Ch. 21 - List the factors that contribute to poor surface...Ch. 21 - Explain what is meant by the term machinability...Ch. 21 - What is shaving in machining? When would it be...Ch. 21 - List reasons that machining operations may be...Ch. 21 - Are the locations of maximum temperature and...Ch. 21 - Is material ductility important for machinability?...Ch. 21 - Explain why studying the types of chips produced...Ch. 21 - Prob. 22QLPCh. 21 - Tool life can be almost infinite at low cutting...Ch. 21 - Explain the consequences of allowing temperatures...Ch. 21 - The cutting force increases with the depth of cut...Ch. 21 - Why is it not always advisable to increase the...Ch. 21 - What are the consequences if a cutting tool chips?Ch. 21 - What are the effects of performing a cutting...Ch. 21 - Prob. 29QLPCh. 21 - Prob. 30QLPCh. 21 - Prob. 31QLPCh. 21 - Prob. 32QLPCh. 21 - Comment on your observations regarding Figs. 21.1...Ch. 21 - Prob. 34QLPCh. 21 - Comment on your observations regarding the...Ch. 21 - Why does the temperature in cutting depend on the...Ch. 21 - You will note that the values of a and b in Eq....Ch. 21 - Prob. 38QLPCh. 21 - Prob. 39QLPCh. 21 - Explain whether it is desirable to have a high or...Ch. 21 - The Taylor tool-life equation is directly...Ch. 21 - Prob. 42QLPCh. 21 - Why are tool temperatures low at low cutting...Ch. 21 - Can high-speed machining be performed without the...Ch. 21 - Prob. 45QLPCh. 21 - Prob. 46QLPCh. 21 - State whether or not the following statements are...Ch. 21 - Let n = 0.5 and C = 400 in the Taylor equation for...Ch. 21 - Assume that, in orthogonal cutting, the rake angle...Ch. 21 - Prob. 50QTPCh. 21 - Prob. 51QTPCh. 21 - Using trigonometric relationships, derive an...Ch. 21 - An orthogonal cutting operation is being carried...Ch. 21 - Prob. 54QTPCh. 21 - Prob. 55QTPCh. 21 - Prob. 56QTPCh. 21 - Show that, for the same shear angle, there are two...Ch. 21 - With appropriate diagrams, show how the use of a...Ch. 21 - In a cutting operation using a 5 rake angle, the...Ch. 21 - For a turning operation using a ceramic cutting...Ch. 21 - In Example 21.3, if the cutting speed V is...Ch. 21 - Using Eq. (21.30), select an appropriate feed for...Ch. 21 - With a carbide tool, the temperature in a cutting...Ch. 21 - The following flank wear data were collected in a...Ch. 21 - The following data are available from orthogonal...Ch. 21 - Prob. 66QTPCh. 21 - Design an experimental setup whereby orthogonal...Ch. 21 - Describe your thoughts on whether chips produced...Ch. 21 - Recall that cutting tools can be designed so that...Ch. 21 - Recall that the chip-formation mechanism also can...Ch. 21 - Prob. 73SDPCh. 21 - Describe your thoughts regarding the recycling of...Ch. 21 - List products that can be directly produced from...Ch. 21 - Obtain a wood planer and some wood specimens. Show...Ch. 21 - It has been noted that the chips from certain...Ch. 21 - As we have seen, chips carry away the majority of...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- How do you select a machine tool for a given application? Please list the four characteristics for the selection. Then explain your answer with an application (example).arrow_forwardQ1:- Put a mark of True or false for below:- (a):- The milling cutting tool has a single cutting edge? (b):- At the break-even point, the total revenue equals the total cost? (c):- The right-hand rule is used to define machine axes? (d):- Design error is a simple mistake? (e):- The quality of products is poor in mass production? (f):- Allowances must be left before turning? (g):- Up-ward milling is better than down-ward milling?arrow_forwardA HSS tool is used to turn a steel workpart that is 300 mm long and 80 mm in diameter. The parameters in the Taylor equation are: n=0.13 and C= 75 (m/min) for a feed of 0.4 mm/rev. The operator and machine tool rate = $30.00/hr, and the tooling cost per cutting edge = $4.00. It takes 2.0 min to load and unload the workpart and 3.50 min to change tools. Determine: a. Tutting speed for maximum production rate, b. Tool life in min of cutting, and c. Cycle time and cost per unit of product.arrow_forward
- A HSS tool is used to turn a steel workpart that is 300 mm long and 80 mm in diameter. The parameters in the Taylor equation are: n = 0.13 and C = 75 (m/min) for a feed of 0.4 mm/rev. The operator and machine tool rate = $30.00/hr, and the tooling cost per cutting edge = $4.00. It takes 2.0 min to load and unload the workpart and 3.50 min to change tools. Determine: Tutting speed for maximum production rate, Tool life in min of cutting, and Cycle time and cost per unit of product. determine cutting speed for minimum cost.arrow_forwardIn a production turning operation, the workpart is 60 mm in diameter and 500 mm long. A feed of 0.75 mm/rev is used in the operation. If cutting speed=9 m/s, the tool must be changed every 4 workparts; But if cutting speed=5 m/s, the tool can be used to produce 50 pieces between tool changes. Determine the Taylor tool life equation for this job. (use the equations given below for solution)arrow_forwardThree tool materials are to be compared for the same finish turning operation on a batch of 100 steel parts: high speed steel, cemented carbide, and ceramic. For the high speed steel tool, the 170 Taylor equation parameters are: n = 0.125 and C = 70. The price of the HSS tool is $15.00 and it is estimated that it can be ground and reground 15 times at a cost of $1.50. Tool change time = 3 min. Both carbide and ceramic tools are in insert form and can be held in the same mechanical toolholder. The Taylor equation parameters for the cemented carbide are: n = 0.25 and C = 500; and for the ceramic: n = 0.6 and C = 3,000. The cost per insert for the carbide = $6.00 and for the ceramic = $8.00. Number of cutting edges per insert in both cases = 6. Tool change time = 1.0 min for both tools. Time to change parts = 2.0 min. Feed = 0.25 mm/rev, and depth = 3.0 mm. The cost of machine time = $30/hr. The part dimensions are: diameter = 56.0 mm and length = 290 mm. Setup time for the batch is 2.0…arrow_forward
- A turning operation is performed with HSS tooling on mild steel, with Taylor tool life parameters n = 0.12, C = 60 m/min. Work part length = 450 mm and diameter = 80 mm. Feed = 0.20 mm/rev. Handling time per piece = 4.0 min, and tool change time = 1.5 min. Cost of machine and operator = $27/hr, and tooling cost = $2 per cutting edge. Find the a. cutting speed for maximum production rate and b. cutting speed for minimum cost Equations used n *-=c(")* Vmax = C 1-n Tt Vmin = C =c(₁" n 1 n Co n CoTt + Ct narrow_forwardIn an orthogonal cutting test, the cutting force and thrust force were observed to be 1000N and 500 N respectively. If the rake angle of tool is zero, What is the coefficient of friction in chip-tool interface ?arrow_forward2 1.4 1 The Taylor tool-life equation is directly applicable to flank wear. Explain whether or not it can be used to model tool life if other forms of wear are dominant.arrow_forward
- 8. During the turning of a 20mm-diameter steel bar at a spindle speed of 400 rpm, a tool life of 20 minute is obtained. When the same bar is turned at 200 rpm, the tool life becomes 60 minute. Assume that Taylor"s tool life equation is valid. When the bar is turned at 300 rpm, the tool life (in minute) is approximately. (A) 25 (B) 32 (C) 40 (D) 50arrow_forwardFind the machining time, in seconds, and the rate of material removing in mmA3;sec for a turning operation having the following information: 1- Wp diameter is 80mm, 2- the length is 0.12m, 3- the cutting speed is 80m/min, 4- feed i50.5 mm/rev and 5- the depth of cut is 0.002m.arrow_forwardA high-speed steel tool is used to turn a steel workpart that is 295 mm long and 75 mm in diameter. The parameters in the Taylor equation are: n = 0.15 and C = 80 (m/min) for a feed of 0.5 mm/rev. The operator and machine tool rate = $29.00/hr, and the tooling cost per cutting edge = $3.90. It takes 2.1 min to load and unload the workpart and 3.40 min to change tools. Determine (a) cutting speed for maximum production rate, (b) tool life in min of cutting, and (c) cycle time and cost per unit of product (d) cutting speed for minimum cost and tool life in min of cuttingarrow_forward
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