|Nov18-07, 02:50 AM||#1|
manufacturing (machining time for turning)
Following is a problem that I have made some progress on but am now stuck. I can see that I probably need to use the orthogonal model to approximate turning but I cannot see how do it, there seems to be too many unknowns. I am confused by how n and C relate to the model or how to use the information given in part (c) of the question. Finally, the deflection... I don't know what that corresponds to, is it the chip size?
Thanks for any help on this.
You are using a 1” micrometer.
Katrin 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 centres , and the overall efficiency of the lathe is 75%.
The log is made from medium carbon 4345 alloy steel. The steel manufacturer, some basic experimentation, and established knowledge of the product and its manufacture have provided the following info:
(a) 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.
(b) Two test cuts were run, one at V = 60sfpm
Where T = 100 min and another at V = 85 sfpm where T = 10 min
(c) 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 )
(d) 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 forging, 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
What I have already worked out:
(1) in the taylor tool life equation, n = 0.151, C = 120.3
(2) Brinell Hardness of 300 – 400 corresponds to Specific Energy (U) 3.6 – 4.4 Nm/mm3 – I used an average value of 4 Nm/mm3 in subsequent calcualtions
(3) depth of cut = 2 in = 50.8 mm (i'm not in the states)
(4) Pg =
Pc = (50)(0.75) = 37.5
(5) Pc = URmr
Rmr = 37.5/4 = 9.375 mm3/s
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