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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Chapter 40, Problem 6P
To determine
To determine the lot size to justify the use of jig.
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ASSIGNMENT-2
Tool Box
Objective: To perform cutting and forming processes on a sheet metal to produce a tool box
with SIZE of 445mmL x 281mmW x 220mmH, with handle holes diameter 20mm.
Note : assume any missing dimensions
a) List the tools used
b) Write the steps to preapare
c) Sketch each steps
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…
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
5
Chapter 40 Solutions
Degarmo's Materials And Processes In Manufacturing
Ch. 40 - What are the two primary functions of a...Ch. 40 - Prob. 2RQCh. 40 - An early treatise defined a jig as a device that...Ch. 40 - Does an ordinary vise qualify as a fixture? Why or...Ch. 40 - What basic criteria should be considered in...Ch. 40 - In any part drawing, what are the critical...Ch. 40 - What difficulties can result from not keeping...Ch. 40 - Prob. 8RQCh. 40 - Which of the basic design principles relating to...Ch. 40 - What are the two reasons for not having drill...
Ch. 40 - Why does the use of down milling often make it...Ch. 40 - Prob. 12RQCh. 40 - A large assembly jig for an airplane-wing...Ch. 40 - Explain why the use of a given fixture may not be...Ch. 40 - Prob. 15RQCh. 40 - In the clamps shown in Figure 40.20, what is the...Ch. 40 - What are other common types of clamp?Ch. 40 - What is the purpose of dimensioning the...Ch. 40 - Prob. 19RQCh. 40 - In Figure 49.9. why arent there three points put...Ch. 40 - Which set of locators in Figure 49.9 establishes...Ch. 40 - What are the four stages of the SMED methodology...Ch. 40 - Prob. 23RQCh. 40 - How could the intermediate jig concept be used in...Ch. 40 - Develop a sequence of operations to produce the...Ch. 40 - Suppose in the sample problem at the end of this...Ch. 40 - Notice that the holes in the part in Figure 49.9...Ch. 40 - Prob. 6PCh. 40 - A part has two holes located for drilling by...
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- 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_forwardCheck the picturearrow_forwardThe Conner Company uses a two-step process for making wooden boards with holes in them. Step 1 is drilling and Step 2 is sanding. Drilling is done using an electronic drill that takes 8 minutes to set up and ½ minute to drill each hole. Step 2 involves a sanding machine with a 20 minute setup time that processes each board (regardless of how many holes are in it) in 10 minutes. If all orders are for a single batch of 10 boards, and each board has 14 holes what is the capacity of this 2-step process? (Assume that both machines only need to be set up once.)arrow_forward
- Task 4 Here students need to calculate (a) The cutting speed (b) Turning time (c) Amount of material removed. For a specimen as shown below, if spindle rotates at N = 360 rpm, number of pass = 03, feed rate= 0.2mm/rev, if density of material=7.85× 10-6 Kg/mm³. Appendix Volume for Cylinder = rr?h Volume of frustum= nh (a?+b²+ab)/3 radii 'a' & b' height(Length) 'h'. b-a Taper angle : = tan20 h TDN Cutting speed:V : meters/min 1000 Where: V = Cutting speed in Meter/minute. T= 3.14 D = Diameter of work piece in mm N = rpm Cutting Time Length of the job to be turned NO.of cuts Time to turn mins. Feed/Rev rpm Amount of Material Removed = Original Mass – Machined Mass Original Size of Component 380 mm Machined Component 80 mm 100 mm 120 mm 6 20mm + 25mm + 40mm + 25mm 030mmarrow_forwardA production process at Kenneth Day Manufacturing is shown in the figure below. The drilling operation occurs separately from, and simultaneously with, the sawing and sanding, which are independent sequential operations. A product needs to go through only one of the three assembly operations (the operations are in parallel). 2.1 units/hr Sawing 7 units/hr Sanding Drilling 6 units/hr Welding 2.5 units/hr a) Sawing is the bottleneck. b) The bottleneck time is 28.57 minutes per unit (round your response to two decimal places). c) The throughput time of the overall system is minutes (round your response to two decimal places). Assembly 0.8 units/hr Assembly 0.8 units/hr Assembly 0.8 units/hrarrow_forward3)arrow_forward
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