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9 days ago

7

The Machine Shop has received an order to turn three alloy steel cylinders. Starting diameter = 250 mm and length = 625 mm. Feed

= 0.30 mm/rev, and depth of cut = 2.5 mm. A coated carbide cutting tool will be used whose Taylor tool life parameters are n = 0.25 and C = 700, where the units are min for tool life and m/min for cutting speed. Compute the cutting speed that will allow the tool life to be just equal to the cutting time for the three parts.

1 answer:

alex41 [274]9 days ago

5 0

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

....................1

where D refers to diameter, L refers to length and f refers to feed while V represents speed $\frac{\pi DL}{1000*f*V}$

Thus, we derive

Tc = $\frac{\pi (250)*625}{1000*0.3*V}$

Tc = $\frac{1636.25}{V}$

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 × $\frac{1636.25}{V}$

According to the Taylor tool formula, cutting speed is expressed as

$VT^{0.25} = 700$

× V × 8.37 = 700

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

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Answer:

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Explanation:

Initially, we create a small list of alphabets from A to J (Line 1).

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