Answer:
N = 278.5 rpm
F = 348.15 mm/min
machine time = 10.34 seconds
material removal rate = 33420000 mm³/min
Explanation:
given data
length L = 400 mm
width W = 60 mm
diameter D = 80 mm
no of cutting n = 5
velocity V = 70 m/min = 70000 mm/min
chip load P = 0.25 mm/tooth
depth = 5 mm
solution
we know velocity that is express as
velocity V = D N ........1
N =
N = 278.5 rpm
and
now we get feed rate in milling operation is
F n P N
F = 5 × 0.25 × 278.5
F = 348.15 mm/min
and
now we get actual machine time to make 1 pass across surface of worl is
machine time =
machine time =
machine time = 0.1723 min = 10.34 seconds
and
max material removal rate during these cut is
material removal rate = A × d × N
material removal rate = 400 × 60 × 5 × 278.5
material removal rate = 33420000 mm³/min