AMEM201 LEC9 PROBLEMS FIT（AMEM201 LEC9问题健康）.pdf

Turning formulas 1. A cylindrical stainless steel rod with length L=150 mm, diameter D0 = 12 mm is being reduced in diameter to Df =11 mm by turning on a lathe. The spindle rotates at N = 400 rpm, and the tool is travelling at an axial speed of υ=200 mm/min Calculate: a. The cutting speed V (maximum and minimum) b. The material removal rate MRR c. The cutting time t 3 d. The power required if the unit power is estimated to 4 w.s/mm SOLUTION: a. The maximum cutting speed is at the outer diameter D0, and is obtained from the expression V = π D0 N Thus, Vmax = (π) (12) (400) = 15072 mm/min The cutting speed at the inner diameter D is f Vmin = (π) (11) (400) = 13816 mm/min b. From the information given, the depth of cut is d = (12 – 11) / 2 = 0,5 mm and the feed is f = υ/ Ν f = 200 / 400 = 0,5 mm/rev thus the material removal rate is calculated as 3 3 MRR = (π) (Davg) (d) (f) (N) = (π) (11,5) (0,5) (0,5) (400) = 3611 mm /min = 60,2 mm /s c. The cutting time is t = l / (f. N) = (150) / (0,5) (400) = 0,75 min d. The power required is Power = (4) (60,2) = 240,8 W 2. The part shown below will be turned in two machining steps. In the first step a length of (50 + 50) = 100 mm will be reduced from Ø100 mm to Ø80 mm and in the second step a length of 50 mm will be reduced from Ø80 mm to Ø60 mm. Calculate the required total machining time T with the following cutting conditions: Cutting speed V=80 m/min, Feed is f=0.8 mm/rev, Depth of cut = 3 mm per pa