EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 8220100793431
Author: KALPAKJIAN
Publisher: PEARSON
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Textbook Question
Chapter 23, Problem 24QLP
Describe the difficulties that may be encountered in clamping a workpiece made of a soft metal in a three-jaw chuck.
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Identify and explain the method of milling process in which the stresses on the teeth is maximum at the end of the cut with neat diagram.
The end of a large tubular workpart is to be faced on a NC vertical boring mill.
The part has an outside diameter of 38.0 in and an inside diameter of 24.0
in. If the facing operation is performed at a rotational speed of 40.0 rev/min,
feed of 0.015 in/rev, and depth of cut of 0.180 in, determine (a) the cutting
time to complete the facing operation and the cutting speeds and metal
removal rates at the beginning and end of the cut.
In the lathe machine, the diameter of workpart 35mm is turned at 100 rpm,
depth of cut 0.125 mm, length of continuous chip for one revolution of
workpart = 60mm, length undeformed chip= 100.53mm, rake angle = 35°,
coefficient of friction= 1.5 and resultant cutting force=215.5 N.
Calculate:
1- Horizontal cutting force of the tool on the workpart and vertical cutting force
required to hold the tool against the work.
2- The forces components applied against the chip by the tool.
3- Velocity of chip relative to the tool and workpart.
4- The percentage of total energy dissipated due to friction along the shear plane,
and the friction at the tool-chip interface.
Chapter 23 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 23 - Describe the types of machining operations that...Ch. 23 - What is turning? What kind of chips are produced...Ch. 23 - What is the thrust force in turning? What is the...Ch. 23 - What are the components of a lathe?Ch. 23 - (a) What is a tracer lathe? (b) What is an...Ch. 23 - Describe the operations that can be performed on a...Ch. 23 - Why were power chucks developed?Ch. 23 - Explain why operations such as boring on a lathe...Ch. 23 - Why are turret lathes typically equipped with more...Ch. 23 - Describe the differences between boring a...
Ch. 23 - How is drill life determined?Ch. 23 - What is the difference between a conventional...Ch. 23 - Why are reaming operations performed?Ch. 23 - Explain the functions of the saddle on a lathe.Ch. 23 - Describe the relative advantages of (a)...Ch. 23 - Explain how external threads are cut on a lathe.Ch. 23 - Prob. 17RQCh. 23 - Explain the reasoning behind the various design...Ch. 23 - Note that both the terms tool strength and...Ch. 23 - (a) List and explain the factors that contribute...Ch. 23 - Explain why the sequence of drilling, boring, and...Ch. 23 - Why would machining operations be necessary even...Ch. 23 - A highly oxidized and uneven round bar is being...Ch. 23 - Describe the difficulties that may be encountered...Ch. 23 - (a) Does the force or torque in drilling change as...Ch. 23 - Explain the similarities and differences in the...Ch. 23 - Describe the advantages and applications of having...Ch. 23 - Assume that you are asked to perform a boring...Ch. 23 - Explain the reasons for the major trend that has...Ch. 23 - Describe your observations concerning the contents...Ch. 23 - The footnote to Table 23.12 states that as the...Ch. 23 - In modern manufacturing, which types of metal...Ch. 23 - Sketch the tooling marks you would expect if a...Ch. 23 - What concerns would you have in turning a powder...Ch. 23 - The operational severity for reaming is much lower...Ch. 23 - Review Fig. 23.6, and comment on the factors...Ch. 23 - Explain how gun drills remain centered during...Ch. 23 - Comment on the magnitude of the wedge angle on the...Ch. 23 - If inserts are used in a drill bit (see Fig....Ch. 23 - Refer to Fig. 23.11b, and in addition to the tools...Ch. 23 - Calculate the same quantities as in Example 23.1...Ch. 23 - Estimate the machining time required to rough turn...Ch. 23 - A high-strength cast-iron bar 8 in. in diameter is...Ch. 23 - A 0.30-in.-diameter drill is used on a drill press...Ch. 23 - In Example 23.4, assume that the workpiece...Ch. 23 - For the data in Problem 23.45, calculate the power...Ch. 23 - A 6-in.-diameter aluminum cylinder 10 in. in...Ch. 23 - A lathe is set up to machine a taper on a bar...Ch. 23 - Assuming that the coefficient of friction is 0.25,...Ch. 23 - A 3-in.-diameter, gray cast iron cylindrical part...Ch. 23 - Would you consider the machining processes...Ch. 23 - Would it be difficult to use the machining...Ch. 23 - If a bolt breaks in a hole, it typically is...Ch. 23 - An important trend in machining operations is the...Ch. 23 - Review Fig. 23.8d, and explain if it would be...Ch. 23 - Boring bars can be designed with internal damping...Ch. 23 - A large bolt is to be produced from extruded...Ch. 23 - Make a comprehensive table of the process...
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- Your manager has asked you to submit a report based on lead screws and screw jacks on plant premises. You need to include following calculations for the report. 1. The lead screw of a lathe has Acme threads of 50 mm outside diameter and 8 mm pitch. The screw must exert an axial pressure of 2500 N in order to drive the tool carriage. The thrust is carried on a collar 110 mm outside diameter and 55 mm inside diameter and the lead screw rotates at 30 r.p.m. Determine (a) the power required to drive the screw; and (b) the efficiency of the lead screw. Assume a coefficient of friction of 0.15 for the screw and 0.12 for the collar. 2. The lead screw of a lathe has Acme threads of 60 mm outside diameter and 8 mm pitch. It supplies drive to a tool carriage which needs an axial force of 2150 N. A collar bearing with inner and outer radius as 30 mm and 60 mm respectively is provided. The coefficient of friction for the screw threads is 0.12 and for the collar it is 0.10. Find the torque…arrow_forwardIt is required to standardise 11 speeds from 72 to 720 rpm for a Machine tool.Specify the speeds.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_forward
- Please answer all parts within 30 mins to get a thumbs up! Please show neat and clean work . I will upvote for sure .arrow_forwardQ1. Calculate the time required to machine a workpiece 170 mm long, 60 mm diameter to 165 mm long 50 mm diameter. The workpiece rotates at 440 rpm, feed is 0.3 mm/rev and maximum depth of cut is 2 mm. Assume total approach and overtravel distance as 5 mm for turning operation. Answer Q2. A gray cast iron surface 280 wide and 540mm long may be machined either on a vertical milling machine, using a 100mm - diameter face mill having eight inserted HSS teeth, or on a horizontal milling machine using an HSS slab mill with eight teeth on a 200-mm. diameter. Which machine has the faster cutting time? The values of feed per tooth and cutting speed for both processes are 0.4mm/tooth and 80m/min, respectively. The depth of cut = 3.0 mm and assume A and O equal to 5. Answerarrow_forward2. Through hole of 12 mm diameter is to be drilled in a steel plate of 20 mm thickness, drilled spindle speed is 20 m/min, feed 0.25 mm/rev and drill point angle is 120°, the amount of material removed at the first 10 sec after full engagement of drill will be mm/sec. (a) 2300, (b) 2400, (c) 2500, (d) 2600, (e) 2700arrow_forward
- 1. A 7.5 -mm-diameter drill is used on a drill press operating at 300 rpm. If the feed is 0.125 mm/rev, what is the MRR? What is the MRR if the drill diameter is doubled? 2. Assume that the work piece material is high-strength aluminum alloy and the spindle is running at N = 500 rpm. Estimate the torque required for this operation.arrow_forwardPlease answer all partsarrow_forward3. Through hole of 12 mm diameter is to be drilled in a steel plate of 20 mm thickness, drilled spindle speed is 20 m/min, feed 0.25 mm/rev and drill point angle is 120°, the amount of material removed at the first 10 sec after full engagement of drill will be .. (a) 2300, (b) 2400, (c) 2500, (d) 2600, (e) 2700 mm /sec.arrow_forward
- The outside diameter of a cylinder made of steel is to be turned. The starting diameter is 120 mm and the length is 1400 mm. The feed is 0.3 mm/rev and the depth of cut is 2.5mm. The cut will be made with a cemented carbide cutting tool whose Taylor tool life parameters are: n= 0.33 and C=500. Units for the Taylor equation are min for tool life and m/min for cutting speed. Compute the cutting speed that will allow the tool life to be just equal to the cutting time required to complete this turning operation.arrow_forwardA turning operation is performed on C1008 steel (a ductile steel) using a tool with a nose radius = 1.3 mm. Cutting speed = 61 m/min and feed = 0.27 mm/rev. Compute an estimate of the surface roughness in this operation. (Hint: the ratio of actual to ideal roughness can be read on the figure below) Equations used R₁ = f² 32NR Ra = rai Ri Actual Theoretical Ratio= 2.4 2.2 2.0 1.8 1.4 1.2 1.0 0 Ductile metals Cast irons Free machining alloys 100 I 30.5 200 Cutting speed-ft/min I 61 300 91.5 Cutting speed - m/min T 400 I 122arrow_forwardOrthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The following conditions are used: cutting velocity is 180 m/min, feed is 0.20 mm/rev, depth of cut is 3 mm, chip thickness 0.5. The orthogonal rake angle is 7°. Apply ratio Merchant's theory for analysis.arrow_forward
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