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 26, Problem 21RQ
To determine
The three types of offsets.
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Give 1 example of methods of section
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60.00
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A workpiece turning at 20 rad/sec has a diameter of 4 inches. Find the peripheral speed of the workpiece.
Chapter 26 Solutions
DeGarmo's Materials and Processes in Manufacturing
Ch. 26 - What human attribute is replaced by a CNC machine?Ch. 26 - Give an everyday example of a household device or...Ch. 26 - Explain how a toaster could be made a closed-loop...Ch. 26 - Prob. 4RQCh. 26 - Prob. 5RQCh. 26 - Prob. 6RQCh. 26 - What kind of operations on a milling machine...Ch. 26 - How is a machining center different from a milling...Ch. 26 - Prob. 9RQCh. 26 - Many manufacturers have purchased large machining...
Ch. 26 - Why was it necessary for machine tool builders to...Ch. 26 - Prob. 12RQCh. 26 - Prob. 13RQCh. 26 - Some of the functions performed by the operator in...Ch. 26 - Prob. 15RQCh. 26 - Prob. 16RQCh. 26 - Prob. 17RQCh. 26 - Prob. 18RQCh. 26 - Prob. 19RQCh. 26 - Prob. 20RQCh. 26 - Prob. 21RQCh. 26 - What is an encoder?Ch. 26 - What does a post processor do?Ch. 26 - What is a peck-drilling subroutine for a CNC...Ch. 26 - What is pocket milling, and what kind of milling...Ch. 26 - How are touch probes used in CNC machines to...Ch. 26 - What are G words used for in CNC?Ch. 26 - What major structural changes are introduced into...Ch. 26 - What process steps are eliminated by the UHSMC in...Ch. 26 - Draw a large circle and imagine you are (the tool)...Ch. 26 - Prob. 1P
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- My Home Duty 2: Found in a cylindrical turning process of a 24mm round shaft, the life of the turning stylus is 9 minutes when the rotary speed was 300 rpm and using several HSS. When turning the same shaft at a rotational speed of 250 minutes, the life of the pen increased to 48.5 minutes. Find the life of the pen when the rotational speed becomes rpm 280 5:09arrow_forwardApplying the technique of tool radius compensation write a program to cut the external profile of the part shown below using the following data: PART TOOL S (rpm) Depth (mm) F (mm/min) 400 130X130X10 mm T1 08 Flat End Mill 2000 10 9 130 P15 P16 R4 P4 P3 P13 P14 R40 P2 PL R30 8 130 20 R4! P12 P11 P7 P8 PO (0, -45, 2) 40 80 Fig.: A Clamping Jaw P5 P6 P10 35 P9 08arrow_forwardkindly answer itarrow_forward
- Consider a milling process, where a pocket with width (X) of 400mm, Length (Y) 300mm, and depth (z) of 15 mm. The cutting tool is a 12 mm diameter flat end mill with 4 cutting edges. For some reason, the tool can be radially immersed in the material at most 50% of the tool diameter. The spindle speed is selected as 2000 rpm, where the max allowable axial cutting depth is 2.5 mm. a) Calculate the number of total cutting steps, indicating which feed direction would you prefer x or y. It may be preferred to have feed rate can be along X direction Total # of cutting steps: If feed direction is along X, the total number of cutting steps is 294 b) Calculate the minimum achievable cycle time if the feed rate is 0.15 mm/rev/tooth. Approx cycle time: 98 minutesarrow_forward4. For the drawing below answer the following: A. The name of the device? B. Its function? C. The parts name? D. How does it work? E. What are the operation parameters? Explain with the help of sketches? F. How is the machine speed changed? 3.arrow_forwardoperation, 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_forward
- On the attached drawing, is the allowance between "shaft C" and "hole D" positive or negative?arrow_forwardSub-parts d) and e) pleasearrow_forwardWrite the NC part program for the part. Holes are threaded and requires two tools; T1 is a 0.5" HSS drill bit and T2 is a 0.5" tapping bit. The rest of the features are cut using a 0.75" HSS milling tool, T3. Feed rate and spindle speed are 10 ipm and 2000 rpm. The material is low- carbon steel. Part thickness is 0.5”. Use dotted lines two show the path of tools movements. Set up proper operation conditions as discussed in the class today.arrow_forward
- The following data were obtained from a time study performed on a horizontal milling machine: Average manual effort time per cycle: 4.62 minutes. Average cutting time (electric feed): 3.74 minutes. Average performance rating: 115%. Backlash due to machine (power supply): 10%. Slack due to fatigue: 15%. What is the standard time for the operation?arrow_forwardDefine the term layout and briefly explain the purpose for layout.arrow_forward51. Draw a feature control frame for a position tolerance. The tolerance zone is to be .024" diameter regardless of feature size. The tolerance is related to primary datum feature A, secondary datum feature C, and tertiary datum feature F. (Note that the datum feature references are not A, B, C.) 52. Complete a feature control frame for a position tolerance. The tolerance zone is to be .031" diameter when the feature is at maximum material condition. The tolerance is related to primary datum feature D, secondary datum feature C, and tertiary datum feature G. (Note that the datum feature references are not A, B, C.)arrow_forward
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