The maximum cutting speed is at the outer diameter D0, and is obtained from the
... The cutting speed with HSS tool is V=80 m/min, the feed is f=0.8 mm/rev, and ...
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 d. The power required if the unit power is estimated to 4 w.s/mm
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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 Df is 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
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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 pass.
SOLUTION: V=80 m/min f=0.8 mm/rev The turning will be done in 2 steps. In first step a length of (50 + 50) = 100 mm will be reduced from Ø100 mm to Ø80 mm and in second step a length of 50 mm will be reduced from Ø80 mm to Ø60 mm.
3. The shaft shown in the figure below is to be machined on a lathe from a Ø 25mm bar. Calculate the machining time if speed V is 60 m/min., turning feed is 0.2mm/rev, drilling feed is 0.08 mm/rev and knurling feed is 0.3 mm/rev.
Turning, drilling, knurling SOLUTION:
Step 2 Turn Ø20 mm from Ø25 mm Tm =
L 45 = = 0.29 min f x N 0.2 x 764
4. A batch of 800 workpieces is to be produced on a turning machine. Each workpiece with length L=120mm and diameter D=10mm is to be machined from a raw material of L=120mm and D=12mm using a cutting speed V=32 m/min and a feedrate f=0.8 mm/rev. How many times we have to resharpen or regrind the cutting tool? In the Taylor’s expression, use constants as n=1.25 and C=175.