DeGarmo's Materials and Processes in Manufacturing
12th Edition
ISBN: 9781118987674
Author: J. T. Black, Ronald A. Kohser
Publisher: WILEY
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Chapter 23, Problem 16RQ
How does a steady rest differ from a follow rest?
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A NC lathe cuts two passes across a cylindrical workpiece under automatic cycle. The operator loads and unloads the machine. The starting diameter of the work is 3.00 in and its length = 10 in. The work cycle consists of the following steps (with element times given in parentheses where applicable):
1 - Operator loads part into machine, starts cycle (1.00 min);
2 - NC lathe positions tool for first pass (0.10 min);
3 - NC lathe turns first pass (time depends on cutting speed);
4 - NC lathe repositions tool for second pass (0.4 min);
5 - NC lathe turns second pass (time depends on cutting speed); and
6 - Operator unloads part and places in tote pan (1.00 min).
In addition, the cutting tool must be periodically changed. This tool change time takes 1.00 min. The feed rate = 0.007 in/rev and the depth of cut for each pass = 0.100 in. The cost of the operator and machine = $39/hr and the tool cost = $2.00/cutting edge. The applicable Taylor tool life equation has parameters: n = 0.26 and…
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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)
Chapter 23 Solutions
DeGarmo's Materials and Processes in Manufacturing
Ch. 23 - How is the tool-work relationship in turning...Ch. 23 - What different kinds of surfaces can be produced...Ch. 23 - How does form turning differ from ordinary...Ch. 23 - What is the basic difference between facing and a...Ch. 23 - Which operations shown in Figure 23.3 do not form...Ch. 23 - Why is it difficult to make heavy cuts if a form...Ch. 23 - Show how equation 23.6 is an approximate equation.Ch. 23 - Why is the spindle of the lathe hollow?Ch. 23 - What function does a lathe carriage have?Ch. 23 - Why is feed specified for a boring operation...
Ch. 23 - Why are depths of cut in boring usually smaller...Ch. 23 - How can work be held and supported in a lathe?Ch. 23 - How is a workpiece that is mounted between centers...Ch. 23 - What will happen to the workpiece when turned, if...Ch. 23 - Why is it not advisable to hold hot-rolled steel...Ch. 23 - How does a steady rest differ from a follow rest?Ch. 23 - What are the advantages and disadvantages of a...Ch. 23 - Why should the distance the cutting tool overhangs...Ch. 23 - Prob. 19RQCh. 23 - How can a tapered part be turned on a lathe?Ch. 23 - Why might it be desirable to use a heavy depth of...Ch. 23 - If the rpm for a facing cut (assuming given work...Ch. 23 - Why is it usually necessary to take relatively...Ch. 23 - How does the corner radius of the tool influence...Ch. 23 - What effect does a BUE have on the diameter of the...Ch. 23 - How does the multiple-spindle screw machine differ...Ch. 23 - Why does boring ensure concentricity between the...Ch. 23 - Why are vertical spindle machines better suited...Ch. 23 - Prob. 29RQCh. 23 - Prob. 30RQCh. 23 - In which figures in this chapter is a dead center...Ch. 23 - Prob. 32RQCh. 23 - In which figures in this chapter showing setups do...Ch. 23 - How many form tools are being utilized in the...Ch. 23 - Prob. 35RQCh. 23 - Select the speed, feed, and depth of cut for...Ch. 23 - Calculate the rpm NS to run the spindle on a lathe...Ch. 23 - The lathe in problem 2 has rpm settings of 20, 30,...Ch. 23 - Calculate the cutting time if the length of cut is...Ch. 23 - Calculate the metal removal rate for machining at...Ch. 23 - Determine the speed, feed, and depth of cut when...Ch. 23 - At a speed of 90 fpm, feed of 0.030 ipr, and depth...Ch. 23 - Calculate the cutting time for a 4-in. length of...Ch. 23 - For a boring operation at V=90,fr=0.030, and...Ch. 23 - A cutting speed of 100 sfpm has been selected for...Ch. 23 - The following data apply for machining a part on a...Ch. 23 - A finish cut for a length of 10 in. on a diameter...Ch. 23 - A workpiece 10 in. in diameter is to be faced down...Ch. 23 - A hole 89 mm in diameter is to be drilled and...
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- Three 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_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_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) L Tm- 1,= Nf N AD, vT" = C %3| AD,L Tm fvarrow_forward
- A cemented carbide tool is used to turn a part with length = 18.0 in and diameter = 3.0 in. The parameters in the Taylor equation are: n = 0.27 and C = 1200. The rate for the operator and machine tool = $33.00/hr, and the tooling cost per cutting edge = $2.00. It takes 3.0 min to load and unload the workpart and 1.50 min to change tools. The feed = 0.013 in/rev. Determine: Cutting speed for maximum production rate, Tool life in min of cutting, and Cycle time and cost per unit of product.arrow_forwardName the milling machine operation in which the special profiles (contours) can be produced in the workpiece.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_forward1. Are there any reference measurements on the drawing? (Yes/No) 2. Are there any cutting planes on this drawing? (Yes/No) 3. Who checked the drawing? Exactly how it is on the print. Thanks!arrow_forwardWhat is the working principle of shaper machine?arrow_forward
- Expert Q&A Done The top surface of a rectangular workpart is machined using a peripheral milling operation. The workpart is 735 mm long by 50 mm wide by 95 mm thick. The milling cutter, which is 60 mm in diameter and has five teeth, overhangs the width of the part equally on both sides. Cutting speed = 80 m/min, chip load = 0.30 mm/tooth, and depth of cut = 7.5 mm. (a) Determine the time required to make one pass across the surface, given that the setup and machine settings provide an approach distance of 5 mm before actual cutting begins and an overtravel distance of 25 mm after 1.2 9. 20 65 73 actual cutting has finished v in seconds. (b) What is the maximum material 3.9 239 removal rate during the cut v in mm3/sec? 0.127 5arrow_forwardWhich if these do not belong the the group Hubbing; roll piercing; heading; coiningarrow_forward2. How does taper turning process differ from turning process?arrow_forward
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