Engineering 470
Product/Process Design
Assignment #2
Cut # |
RPM |
Cut Depth (in.) |
Cut Feed (in.) |
Tool Life (min.) |
1 |
400 |
0.8 |
0.01 |
40 |
2 |
500 |
0.8 |
0.01 |
28 |
3 |
400 |
1.0 |
0.01 |
36 |
4 |
400 |
0.8 |
0.02 |
32 |
5 |
450 |
0.9 |
0.015 |
???????? |
Solution:

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Using Mathcad 7.0 the given data was substitutes into the Tool Life Equation and the equations were then solved simultaneously:
Initial guesses for the exponentials were:
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The simultaneous equations are written as:

Finding values for the initial guesses and then solving for the Tool Life at T5 gives:

Solving for the Log(T)=D we get,
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This denotes the Tool Life with the given parameters at stage 5
Solution
For the suggested feed and speed, the table for high speed drills in the notes gives for ¼ in. hole to be:
Feed – 0.004 in./rev.
Speed – 3055 ft/min.
For the time the operation will take the equation used is:
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Where T is drill time, L is cut length, F is feed rate and N is revolutions.
Using MathCad to solve gives:


This is the time for the operation
For the MRR the equation used is:
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Mathcad solves this giving:
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This is the MRR
Cut # |
RPM |
Tool Life (min.) |
1 |
500 |
30 |
2 |
500 |
25 |
Using Mathcad the following equation was solved:

Choosing some arbitrary initial guesses and solving gives:

The matrix shows the values for n & C
Looking at Fig 2.1 discuss the processes and the machinery by which the specimen can be made and describe how the process you selected can be changed for economical production as the number of specimens to be made increases.
It is necessary to machine on a lathe, which gives the ability to face, turn and taper the part. Holes can be drilled using a drill and also can be tapped at the same station. To produce in large quantities castings could produce blanks, which could then be drilled and tapped on stand-alone workstations that are programmed to drill and tap according to certain parameters describing the part.