Question

Reference sources reveal that a workpiece material has a unit horsepower of 1.6 hp/in3/min. For a turning operation, the cutting velocity is set at 500 ft/min with a corresponding feed of 0.025 in/rev. and a depth of cut of 0.2 in. From the given data, calculate:_______.

Answer 1

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
`$in^3 / min$`

c). Horsepower required = MRR x unit horsepower

= 30 x 1.6

= 48 hp

a). Cutting force,

`$F=(power)/(cuting \ velocity)$`

`$=(48 * 550)/(500 /60)$` (1 hp = 550 ft lbf /sec)

= 3168 lbf

d). Machine HP required

`$=(HP)/(\eta)$`

`$=(48)/(0.9)$`

= 53.33 HP