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 24, Problem 1P
You have selected a feed per tooth and a cutting speed for a face milling process, using Table 244. Reasonable values for feed and speed are 0.010 in. per tooth and 200 sfpm. The cutter is 8 in. in diameter, as shown in Figure 24.10. Compute the input values for the machine tool.
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A two-spindle drill cuts two holes at the same time, one 1/2 inch and one 3/4 inch. The workpiece is 1.0 inches thick. Both drills have point angles of 118 degrees and the cutting speed for the material is 300 ft/min. The rotational speed of each drill can be set individually but the feed rate for both holes must be set to the same value because they move together into the material. The feed rate is set so that the total metal removal rate of both drills combined does not exceed 1.50 in3/min. Determine (a) maximum feed rate (in/min) that can be used, (b) individual feeds (in/rev) for each hole, and (c) cutting time for the operation.
Chapter 24 Solutions
DeGarmo's Materials and Processes in Manufacturing
Ch. 24 - Suppose that you wanted to machine cast iron with...Ch. 24 - How is the feed per tooth related to the feed rate...Ch. 24 - Why must the number of teeth on the cutter be...Ch. 24 - Why is the question of up or down milling more...Ch. 24 - For producing flat surfaces in mass-production...Ch. 24 - Milling has a higher metal removal rate than...Ch. 24 - Which type of milling (up or down) is being done...Ch. 24 - Why does down milling dull the cutter more rapidly...Ch. 24 - What parameters do you need to specify in order to...Ch. 24 - In Figure 24.2b, the tool material is carbide....
Ch. 24 - What is the advantage of a helical-tooth cutter...Ch. 24 - What would the cutting force diagram for Fc look...Ch. 24 - Could the stub arbor-mounted face mill shown in...Ch. 24 - In a typical solid arbor milling cutter shown in...Ch. 24 - Make some sketches to show how you would you set...Ch. 24 - Make some sketches to show how you would set up a...Ch. 24 - Explain how controlled movements of the work m...Ch. 24 - Prob. 18RQCh. 24 - What is the purpose of the hole-circle plate on a...Ch. 24 - You have selected a feed per tooth and a cutting...Ch. 24 - How much time will be required for a milling...Ch. 24 - If the depth of cut is 0.35 in., what is the metal...Ch. 24 - Prob. 4PCh. 24 - Calculate the spindle rpm and table feed (ipm) for...Ch. 24 - A gray cast iron surface 6 in. wide and 18 in....Ch. 24 - Prob. 7PCh. 24 - In Figure 24.12, the feed is 0.006 in. per tooth....Ch. 24 - Suppose you want to do the job described in...Ch. 24 - The Bridgeport vertical-spindle milling machine is...Ch. 24 - The KC Machine Works, which does job shop...
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- solvearrow_forward1. In Figure 1, the feed is 0.006 in. per tooth. The cutter is rotating at an rpm that will produce the desired surface cutting speed of 125 fpm. The cutter diameter is 3 in. The depth of cut is 0.15 in and width of the cut is 2 in. The block is 5 in. wide and 12in long. Determine a. The feed, in inches per minute. b. Approach distance for this type of milling c. Cutting time to make one pass across the surface d. If an additional approach distance of 0.5 in is provided at the beginning of pass and an over travel distance is provided at the end of pass equal to the cutter radius plus 0.5 in, what is the duration of the feed motion? Speesi neten Cite Citter praiten t staof cit Solld hellcal plain milling cuttar Wout Helical teeth TUp w Figure: 01arrow_forwardA shaft groove with a length of 250 mm and a depth of 12 mm is cut by a 20 mm diameter slot cutter. What is the processing time if the cutting depth is to be 4 mm. The feed is 25 mm/min. Over run is 1/2 the cutter diameter.arrow_forward
- Solve first Question Only.arrow_forwardPlease answer all partsarrow_forwardA surface of 43 mm x 246 mm will be machined by using a shaper and 7 mm deep stock will be removed from the surface. The shaper is operating at 24 strokes/minute and feed is 0.7 mm/stroke. If the maximum stroke length is 827 mm and allowable depth of cut is 0.4 mm for both of rough and finish cuts, what is the total machining time, in minutes ?arrow_forward
- Turning Operation In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in 4.8 min. The piece is 400 mm long and 150 mm in diameter. Using a feed = 0.43 mm/rev and a depth of cut = 4.0 mm, what cutting speed must be used to meet this machining time requirement? Hint: Re-arrange the Tm equation to find v in m/min Tm= T.DO.L v.f 1.522arrow_forwardThe face milling cutter of D = 200 mm used to cut a block which is 200 mm H, 1=480 mm, if f- 0.7 mm/rev is used and the overrun of the cutter is 12 mm and cutting speed 320 mm/min. Calculate the machining time. Cuttinarrow_forwardIt takes only 12 seconds to mill 0.5" x 5.5" keyway by means of a 40-tooth cutter at a feed rate of 0.249 mm/tooth. Compute the required speed of the cutter in rpm. 70 O65 O 60 75 Aarrow_forward
- operation, the shear IS cIose to, In turning operation, spindle speed is set to provide a cutting speed of 2.2 m/: The feed and depth of cut are 0.32 mm and 2.1 mm, respectively. The tool rake angle is 15°. After the cut, the deformed chip thickness is measured to be 0.43 mm. Using the Orthogonal model as an approximation of turning operation, the shear strain is close to, In turning operation, spindle speed is set to provide a cutting speed of 2.4 m/arrow_forwardA high-speed heavy duty plain milling cutter 4 inches in diameter with 12 teeth is to be used. Each tooth s expected to remove 0.006 inch. What is the feed per revolution? 2. The feed per revolution is 0.0072 inch/rev and the machine is set for 120 rev/min. Find the feed in inch per minutearrow_forwardWhat is the feed speed (mm/min) of an end milling cutter if it mills a distance of 300 mm in ten minutes? What is the feed per tooth if it is rotating at 800 rpm?arrow_forward
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