Can you help me estimate the minimum machining time for this forging problem?

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The discussion centers on estimating the minimum machining time for a large forging that needs to be turned down from a 10” diameter to 6”. The user seeks help, offering $50 for individual assistance, emphasizing the urgency for their grade. Key details include the material specifications, tool life equations, and the importance of maintaining deflection below .005” during machining. The conversation suggests using test cut data to derive the tool life constants and optimal cutting speed. The user also requests the removal of redundant posts in engineering forums to adhere to guidelines.
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Ok i need help on this for my grade, very crucial.
I have 100 posts on another forum, not much but I hope it will give me a little credibility.
I am willing to pay 50$ via paypal if you are willing to talk with me individually, but only if you know what you are doing and how to solve this problem. Again, i need this desperately for my grade.

Book is Degarmo, Black, Kosher " Materials and Processes in manufacturing " pg 582,9th ed.

Typed problem out..
Estimating machining time for turning

You are using a 1” micrometer.
Katin would like to know the minimum time required to machine a large forging. The 8 ft long forging is to be turned down from an original diameter of 10” to a final diameter of 6”. The forgin has a BHN of 300-400. The turning is to be performed on a heavy duty lathe, with a 50 hp motor and continuously variable speed drive on the spindle. The work will be held down between centers , and the overall efficiency of the late is 75%.

The log is made from medium carbon 4345 alloy steel. The steel manufacturer, some basic experimentation, nad established knowledge of the product and its manufacture have provided the following info:

a tool life equation developed for the most suitable type of tool material at a feed of .02ipr and a rake angle of alpha = 10 degrees. The equation VT^n = C generall fits the data, with V= cutting speed and T = time in minutes to tool failure.


Two test cuts were run, one at V = 60sfpm
Where T = 100 min and another at V = 85 sfpm where T = 10 min


The dynamic shear strength of the material is on the order of 125,000 psi. Jay decides to make two test cuts at the standard feed of .02 ipr. He assumes that the chip thickness ratio varies almost linearly between the speeds of 20 and 80 fpm, the values being .4 at the speed of 20 fpm and .6 at 80 fpm. The chip thickness values were determined by micrometer measurements to determine the value of Rc ( Rc is chip ratio )

The log will be used as a roller in a newspaper press and must be precisely machined. If the log deflects during the cutting more than .005” the roll will end up barrel shaped.


How should I proceed to estimate the minimum time required to machine this forgin, assuming that one finishing pass will be needed when the log has been reduced to 6” in diameter? The deflection due to cutting forces must be kept below .005” at the mid log location.

Assume Fc * .5 = Ff and Ft * .5 = Fr and that Fr causes the deflection
 
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We'll see what we can do. Meanwhile, please delete redundant posts in the engineering forums. The guidelines stipulate one post. Thank you for your cooperation.


Meanwhile - from this part:
a tool life equation developed for the most suitable type of tool material at a feed of .02ipr and a rake angle of alpha = 10 degrees. The equation VT^n = C generall fits the data, with V= cutting speed and T = time in minutes to tool failure.

Two test cuts were run, one at V = 60sfpm
Where T = 100 min and another at V = 85 sfpm where T = 10 min
One should be able to determine n and C - one has two equations and two unknowns. Ostensibly, this is relevant information.
 
For the first part, you can start by plotting on a log scale the tool life vs. speed at a given allowable amount of tool wear. That will give you a slope which will be the n exponent in the Taylor tool life equation. You can then calculate the C for each trial and use the average of those to calculate your "optimal" cutting speed.

I'm going to have to look up some things for the next parts.
 
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