The question is incomplete. We have to calculate :
a). the cutting force
b). volumetric metal removal rate, MRR
c). the horsepower required at the cut
d). if the power efficiency of the machine tool is 90%, determine the motor horsepower
Solution :
Given :
Cutting velocity (v) = 500 ft/min
= 500 x 12 in/min
= 6000 in/min
Feed , f = 0.025 in/rev
Depth of cut, d = 0.2 in
b). Volumetric material removal rate, MRR = v.f.d
= 6000 x 0.025 x 0.2
= 30 [tex]$in^3 / min$[/tex]
c). Horsepower required = MRR x unit horsepower
= 30 x 1.6
= 48 hp
a). Cutting force,
[tex]$F=\frac{power}{cuting \ velocity}$[/tex]
[tex]$=\frac{48 \times 550}{500 /60}$[/tex] (1 hp = 550 ft lbf /sec)
= 3168 lbf
d). Machine HP required
[tex]$=\frac{HP}{\eta}$[/tex]
[tex]$=\frac{48}{0.9}$[/tex]
= 53.33 HP
