60 mm diameter steel bar is to be turned down to 56 mm diameter on a standard lathe. the length of the bar to be turned is 250 mm and the tool has a pre-travel of 5 mm. if the machine produces 40 parts per hour find the cutting speed in m/min assuming the feed = 0.2 mm/rev. the depth of cut using an hss tool cannot exceed 1 mm. also calculate the material removal rate.
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If the machine produces 40 parts per hour find the cutting ... to be turned is 250 mm and the tool has a pre-travel of 5 mm. If ...
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Answer:if you get the answer the please
The speed of cutting is .000417 m/min almost, and the rate of material removal is 1011.78 mm^3/min.
Given:
Diameter = 60 mm (steel bar )
Diameter = 56 mm
Bar length = 250 mm
Pre-travel of the tool = 5 mm
Feed = 0.2 mm/rev
Depth cut = 1 mm
Produced parts = 40 per hour
To find:
Cutting speed and
Rate of removal rate.
Solution:
Lets find,
Initial volume = πr^2h
= π (30 mm)^2 (250 mm)
= 706500 mm^3
Final volume = πr^2h
= π(28 mm)^2(250 mm)
= 615440 mm^3
Now, removed material= 706500 -615440
=91060 mm^3
Now ,
Time 1st part = 60 m/h ÷ 40 p/h
= 1.5 m/p
Removed length ,
Length to cut = 250 mm - 5 mm
= 245 mm
Revolutions number = 245 mm / 0.2 mm/rev
= 1225 rev
Time to taken for 1 revolution = 1.5 m/p / 1225 rev
= 0.001224 s/rev
Now ,
We know,
Cutting speed = π × Diameter × Speed ÷ 60
So,
Speed = Cutting speed × 60 ÷ π × Diameter =0.001224 sec/rev × 60 ÷ π × 56 mm = .000417 m/min
Now we can calculate the material removed rate,
Material removal rate = 91060 mm^3 ÷ (1.5 m/part × 60 sec/min)
= 1011.78 mm^3/min
So, the speed of cutting is .000417 m/min almost, and the rate of material removal is 1011.78 mm^3/min.
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