A peripheral milling operation is performed on the top surface of a rectangular workpart which is 400 mm long and 60 mm wide. The milling cutter, which is 80 mm in a diameter and has five teeth, overchanges the width of the part on both sides.
Cutting speed = 70 m/min, chip load = 0.25 mm/tooth, and depth of cut = 5.0 mm.

Determine:
a) The actual machining time to make one pass across the surface.
b) The maximum material removal rate during the cut.

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

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hope this is what you're talking about

answer:

200k

explanation: