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,

$F=\frac{power}{cuting \ velocity}$

    $=\frac{48 \times 550}{500 /60}$                (1 hp = 550 ft lbf /sec)

   = 3168 lbf

d). Machine HP required

  $=\frac{HP}{\eta}$

 $=\frac{48}{0.9}$

= 53.33 HP