Topic #1: Material Removal Rate and Tool Wear When slot milling carbon steel wit

Question

Topic #1: Material Removal Rate and Tool Wear When slot milling carbon steel with a carbide cutting tool the traditional surface speed is normally around 350 fpm, while the surface speed in high speed machining is around 1500 fpm. The tool is .75 inches in diameter and the depth of cut (ap) is 2 inches. What would be the difference between the two speeds for Material Removal Rate (MRR), also known as Q (see lecture slides), and tool life? (tool life equation values for C and n can be found in lecture slides) lecture slides), and to lool life equation values for C and n can be found in lecture slides) C. Tool Life (min)- 1/n

Explanation / Answer

Given:

Work material : Carbon steel

Tool material : Carbide

From table:-

=> ( n =0.25 & C = 500 m/min)

As, Q= ae*ap*Vf

Here ap is depth of cut where as ae will be width of cut.

As ae is not given therefore taking it as a variable.

Converting ft and inches into metre

ap= 0.2 inches = 0.00508 m

Let , Traditional surface speed, Vf1= 350 fpm = 106.68 m/min

High speed machining surface speed, Vf2= 1500 fpm = 457.2 m/min

and Q1 and  Q2 be the MRR for Traditional surface speed and High speed machining surface speed respectively

Therefore, Difference in MRR = Q1 - Q2 = (ae*ap*Vf1 - ae*ap*Vf1) = [ae x 0.00508x(457.2 - 106.68)}]

= 1.78 ae m3/min

suppose if   ae = 1 inch = 0.0254 m

Then answer will be , Difference in MRR = Q1 - Q2 = 1.78x0.0254 m3/min = 0.045228  m3/min = 1.597 ft3/min

PART -II

Let T1 and T2 be the Tool life for Traditional surface speed and High speed machining surface speed respectively.

Difference in Tool life = T1 - T2 = ( C / Vf1 )(1 / n) - ( C / Vf2 )(1 / n)

= ( 500 / 106.68 )(1 / 0.25) - ( 500 / 457.2 )(1 / 0.25) = 481.119 min

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.