Question

A slab-milling operation is being carried out on a block on a 200 mm-long, 150 mm-wide annealed mild steel block at a feed f= 0.20 mm/tooth and depth of cut d= 2.5 mm. The cutter is D= 45 mm in diameter, has 20 straight teeth, rotates at N=110 rpm and by the definition, is wider than the block to be machined. Calculate ; I.material-removal, II.estimate the power and torque required for this operation, III.calculate the cutting time

Answer 1

Solution - Given (Llength of block = 200mm width of block &W) = 150mm (f) feed = 0.20 mm/tooth depth of cut (d) = 2.5mm Diametel of cutter (1) = 45 mm No. of teeth (2) = 20 Spindle speed (N) = 110 rpm. Ke = 1950 MPa for mild steel with f = 0.2mm/tooth. MRR = ? fm = = Nxzx f. 110x20x0.20 = 440 mm/min MRR2 wx dx fm = 150x2.5 x 440 (mm/min) Material removal = les x 103 mm3/min rate cutting time a? = 165 cm3/min r. culting time (Tm) = LIA fm A A length of approach Jacord) $2.5(45.215) = 10.30mm Tm- (200 + 10.30) mm = 0.4779 min mm/min Time for culting = 0.4779 min - 28.678 seconds

I Power & Torque Required = ? net Power (Pc) - wxdx fm xke (kW) [Ke=1950 60 x 100 Mia / Pc = (150 m) x (2.5 mm2x1440 mm/min ) x(950 Nimeni) 60 x 100 5,362,5 W Pc = 5.3625 RW Torque = lex 30x103 - Их М (om) As (TV = P.) 5362.5 x 60 465.76 Nm T = Pcx60 QTN 2x XX 110 Torque required - 465.76 Nm.