Question

1. Stainless steel rod with a diameter of 15 mm and a length of 15 cm is machined to a diameter of 10 mm. The spindle rotates at a revolution speed of 480 rps. And tool is travelling with an axial speed of 200 mm/min. Calculate the cutting speed, material removal rate, cutting time, power dissipated and cutting force.

Answer 1

Spindle speed(given)N=480 r.p.s

N=480*60

N=28800 r.p.m

a.) cutting speed(V) =3.14*D*N/1000

V= 3.14*15*28800/1000

V=1357.16 m/min.

from the given information,note that the depth of cut(d) and feed(f)is

f=200/28800

f=0.00694 mm/rev.

d=15-10/2

d=2.5 mm.

Davg=15+10/2

Davg=12.5

b.) Material Removal Rate=3.14*Davg*d*f*N

=3.14*(12.5)*(2.5)*(0.00694)*28800

=19622.38 mm3/min. (or) 20*10-6 m3/min.

c.) cutting time (t) =L/f*N

=15/0.00694*28800

= 0.75 min.

The average value of stainless steel from the standard is taken as 4W-S/mm3

   Power required =4*19622.38/60

= 1308 W.

d.) Power Dissipated :
1 watt=60N-m/min

therefore, the power dissipated is =1308*60

= 78480 N-m/min.

e.) cutting force :

torque(t) =power dissipated/2*3.14*N

t  = 78480/2*3.14*28800

t = 0.43 N-m.

cutting force(Fc) =torque*1000/(Davg/2)

Fc =0.43*1000/(12.5/2)

Fc =68.8N.