Question

Problem la: A volume of 10 in. long, 5 in. wide and 0.5 in. thick is to be removed by a face milling cutter. The cutter is 3 in diameter and has a six teeth. The maximum depth of cut is 0.3 in. Assume feed -0.1 ipr and cutting speed - 200 fpm. What are the feed speed, spindle rpm and cutting time? Problem 1b: 0.95 The cutting force for turning a stainless-steel shaft using HSS tool can be predicted as F. = 150,000 9.80 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

Sinano Solution 1a) Given Mo Face milling From the data given lingth of work plece width of work piece Wito hold e- loin oto oth = لا 5 ia depth of cut d- 0.3 in thickness t-o sin feed Olipo f cutting speed 200 fom 2001 200X12. 1pm 2400 ipm op 10 Spindle speed Moser D> Cutter diameter TD D= 3 in 2400 of st N * 3 254647 rpm el feed speed (fm) in 2481 - QT - 35 fm - IN 401 0:1 X 254.647 fm - = 25.4647 in/min in/minst

cutting speed (Tm) than Sinic width of workepice higher out dia of cutter we connot mochine single pass to number of passes şa 2 A Time for each in pass (Ti) Ti L fm L=2+CAP tlo CAP - compulsory approch CAP= 을 1.5 2 2% L=1.5+ 10 = 33 13in n number of cut n- t d 0.5 2 0.3 Ti= 13 x 2 25.4647 G90. In 1.02102 min Total cutting time Tm m= Ti & no of paves. 10 210 2 @x 2

m= 2.042 min T) 16.12 In question be find it find in in/min unit of feed speed a to not on given I alredy number of tooth (2) I will also find it in. in/tooth Siocc is given Feed in/ tooth ft = f Z=6 z Norge 0:1 6 0.0166 in/ tooth Summary cial o + spindle speed N=254.647 8pm cro feed speed 25.4647 in/min 07 0.0166 in/ tooth is - 2.042 min cutting time no SDIED