The 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.
Answers
The 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.
Answer:
The material removal rate is 49 mm^3/min.
Explanation:
To find the cutting speed, we need to use the formula:
Cutting speed (V) = πDN/1000
where D is the diameter of the workpiece in mm, N is the spindle speed in RPM, and π is a constant value of approximately 3.14.
Given,
Diameter of the bar before turning = 60 mm
Diameter of the bar after turning = 56 mm
Length of the bar = 250 mm
Pre-travel = 5 mm
Feed rate = 0.2 mm/rev
Depth of cut = 1 mm
Number of parts produced per hour = 40
First, let's calculate the spindle speed (N) using the following formula:
N = (cutting speed x 1000) / πD
where π is a constant value of approximately 3.14.
Diameter of the bar before turning = 60 mm
Diameter of the bar after turning = 56 mm
So, the average diameter = (60+56)/2 = 58 mm
Cutting speed (V) = (π x 58 x 40) / 1000 = 7.28 m/min
Now, let's calculate the spindle speed (N) using the above formula:
N = (cutting speed x 1000) / πD = (7.28 x 1000) / (π x 58) = 398.43 RPM (approx.)
Next, let's calculate the material removal rate (MRR) using the formula:
MRR = feed rate x depth of cut x length of cut x number of cutting edges
where number of cutting edges = 1 (for HSS tool)
Length of cut = (total length of the bar) - (pre-travel) = 250 - 5 = 245 mm
MRR = 0.2 x 1 x 245 x 1 = 49 mm^3/min
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