Degarmo's Materials And Processes In Manufacturing
13th Edition
ISBN: 9781119492825
Author: Black, J. Temple, Kohser, Ronald A., Author.
Publisher: Wiley,
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Textbook Question
Chapter 23, Problem 10RQ
Why is feed specified for a boring operation typically less than that specified for turning if the MRR equations are the same?
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In face milling, assume that the Diameter of the cutter is 200 mm, the width of the block is 60 mm, and the length of the block is 344 mm, if the depth of cut is 3 mm, = 0.6 mm/min, and N = 100 rpm. The cutter has 12 inserts, and the workpiece material is bronze. Calculate the estimated power required, the cutting force, and the torque.
Find the machining time required to turn a mild steel rod from 65mm to 58 mm over a length of 100 mm by using a carbide insert. If the approach length and over run length is taken as 5 mm, Cutting speed as 20 m/min and feed is =0.2 mm/rev, and the depth of cut is 0.5mm
A job has to be machined in shaping and the process parameters are given below:
Length of the job=120 mm
Speed of the motor=640 rpm
Cutting speed=248 m/min
Tool allowance before cutting =31.5 mm
Tool allowance after cutting=10 mm
Determine the cutting to return stroke ratio for the above operation and draw the arrangement of machining with tool head and allowances.
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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- Figure and table given below shows the design details of a taper turning operation with a cutting speed of 22 m/ minute, feed 0.2 mm/ revolution; Depth of cut of 0.5 mm. (c) Time for n passes to complete taper turning in minutes .arrow_forwardAssuming that the coefficient of friction is 0.25, calculate the maximum depth of cut forturning a hard aluminium alloy on a 15-kW lathe (with a mechanical efficieny of 80%) at awidth of cut of 6 mm, rake angle of 0˚, and a cutting speed of 90 m/min, shear strength of 150 MPa.arrow_forwardPlease give the correct solution A cylinder of 25 mm diameter and 100 mm length is turned with a tool, for which the relation VT0.25 = 55 is applicable. The cutting velocity is 22 m/min. For a tool feed of 0.046 mm/rev, the number of tool regrinds required to produce 425 cylinders is?arrow_forward
- to perform a rough cut single point facing operation in a lathe of 5 mm diameter cylinder workpiece made of bronce . Feed rate is 0.5 mm/rev , depth of cut of 4.5mm, and Specific Energy of 3 W-s/mm^3. What is the Material Removal Rate in mm^3/s?arrow_forwardIn an orthogonal cutting operation an 8 mm deep groove is to be turned on a 50 mm diameter steel bar. Spindle speed is 300 rpm and a feed rate of 0.25 mm/rev is given to the tool. Produced chips have a width of 2 mm. Calculate the material removal rate at the beginning and at the end of the cut. Can u help me please?arrow_forwardOrthogonal turning of a mild steel tube with a tool of rake angle 10° is carried out at a feed of 0.14 mm/rev. If the thickness of the chip produced is 0.28 mm, Calculate the values of shear angle and shear strain.arrow_forward
- 8 A job has to be machined in shaping and the process parameters are given below: a. Length of the job=120 mm b. Speed of the motor=640 rpm c. Cutting speed=264 m/min d. Tool allowance before cutting =30 mm e. Tool allowance after cutting=15mm Determine the cutting to return stroke ratio for the above operation and draw the arrangement of machining with tool head and allowances.arrow_forwardIn orthogonal turning of a bar 100 mm diameter with a feed of 0.25 mm/rev., depth of cut = 4 mm, cutting velocity = 90 m/min. it is observed that the main cutting force is perpendicular to friction force acting at the chip tool interface and cutting force is 1500 N. find (i) rake angle (ii) normal force.arrow_forwardAssume that, in orthogonal cutting, the rake angle is 20 and the friction angle is 35 at thechip-tool interface. Determine the percentage change in chip thickness when the frictionangle is 50. Note that Merchant’s equation is more preferable.arrow_forward
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