A 15.0 in by 2.0 in work part is machined in a face milling operation using a 2.5 in diameter cutter with a single carbide insert. The machine is set for a feed of 0.010 in/tooth and a depth of 0.20 in. If a cutting speed of 400 ft/min is used, the tool lasts for 3 pieces. If a cutting speed of 200 ft/min it used, the tool lasts for 12 parts. Determine the Taylor tool life equation. (20 points)

N1 =v/πD = 400(12)/2.5π = 611 rev/min fr = Nfnt = 611(0.010)(1) = 6.11 in/min A=D/2 = 2.50 / 2 = 1.25 Tm = (L+2A)/fr = (15 + 2(1.25))/6.11 = 2.863 min T1 = 3Tm = 3(2.863) = 8.589 min when v1 = 400 ft/min N2 = 200(12)/2.5π = 306 rev/min fr = Nfnt = 306(0.010)(1) = 3.06 in/min Tm = (15 + 2(1.25))/3.06 = 5.727 min T2 = 12Tm = 12(5.727) = 68.724 min when v2 = 200 ft/min n = ln (v1/v2)/ln(T2/T1) = ln (400/200)/ln (68.724/8.589) = 0.333 C = vTn = 400(8.589 )0.333 = 819

Amakamerem Amakamerem · Answer 2 · 2020-04-30T02:36:15+02:00

Answer:

C = 787.2

Explanation:

The Taylor tool life is referred to as the duration of the actual cutting time after which the tool can no longer be used.

To Quantify the end of a tool life we equate it to a limit on the maximum acceptable flank wear.

For the tool life equation, With the slope, n and intercept, c, Taylor derived the simple equation as

VTn = C where.

n is called, Taylor's tool life exponent. The values of both 'n' and 'c' depend mainly upon the tool-work materials and the cutting environment (cutting flake application)

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