Topic 1 Material Removal Rate and Tool Wear When slot millin

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

Solution

Given:

Work material : Carbon steel

Tool material : Carbide

From table:-

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

As, Q= a_e*a_p*V_f

Here a_p is depth of cut where as a_e will be width of cut.

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

Converting ft and inches into metre

a_p= 0.2 inches = 0.00508 m

Let , Traditional surface speed, V_f1= 350 fpm = 106.68 m/min

High speed machining surface speed, V_f2= 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 = (a_e*a_p*V_f1 - a_e*a_p*V_f1) = [a_e x 0.00508x(457.2 - 106.68)}]

= 1.78 a_e m³/min

suppose if   a_e = 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 / V_f1 )^{(1 / n)} - ( C / V_f2 )^{(1 / n)}

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