Question

0.95 Problem 1b: The cutting force for turning a stainless-steel shaft using HSS tool can be predicted as Fe = 150,000 £9.50 ap The recommended speed is 200 fpm and the feed is 0.03 ipr. The diameter of the workpiece is 3 inches and the final diameter is 2.8 in. The part is 4 in. long. A 10 hp motor is used to drive the machine, which has an efficiency of 0.85. What is the maximum depth of cut, based on the power consumption constraint? How long will it take to finish the workpiece?

Answer 1

PSSPPU solution speed N 200 fit/min fed f = 0.03 in/rev diameter of wooke piele Do = 3inchs. finial diameter df = 2.8in length = 4 in motor power p = tohp. machine efficiency = 0.85% 3 Pmachine Pmotor Pmache=0.85810 8.SH.P = = 0.80 Fc= 0.95 150,000 ooxf (99) Pmache FEX Vc ПX 3х у Ve TON = 60 60 8.5 = FCX 200 60 Fcx 1.016 (m/s) 8.5x746

Fc= 6241.14 N Fc=1463.06409 abf Vc = TON 60 f = 0.000762m/rev 1403.06409 = 150,000 X 150,000 X (6:03) Xcap 0.80 0.95 ap = 0.140 inch Ap= depth of unt = 0.140 inch TON 200 60 N = 4.24413 revl see 254.64 rev/min 4 = 31.41 Sec the t = L EN 0.03x254.64 = 31.41 sechat

no of cult Dog- Do ad = no of with 3-2.8 2X0.140 no of of cutt = 0.8 ~ 1 To teal tine no of ult x time (with t - 1x 31,41 sec t = 31.41 Sec