Question

The main cutting force acting on a tool during the turning (orthogonal cutting) operation of a metal is 400 N. The turning was performed using 2 mm depth of cut and 0.1 mm/rev feed rate. The specific cutting pressure (in N/mm²) is

1. 1000
2. 2000
3. 3000
4. 4000

Answer 1

Correct Answer - Option 2 : 2000

Concept:

For orthogonal turning, t = f

Because t = f sinλ and λ = 90         

\({\rm{P}} = \frac{{{{\rm{F}}_{\rm{c}}} \times {\rm{V}}}}{{{\rm{fdV}}}}\)

Where, P = cutting pressure, F_c = cutting force, V = velocity of tool, f = feed rate, d = depth of cut,

λ = principle cutting edge angle

Given: F_c = 400 N, d= 2 mm, f = 0.1 mm/rev

Calculation:

\({\rm{P}} = {\rm{}}\frac{{400}}{{0.1 \times 2}} = 2000{\rm{}}\,\,{\rm{N}}/{\rm{m}}{{\rm{m}}^2}\)