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A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feed and depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energy of the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power. The main or tangential cutting force (in N) is ________
(Important - Enter only the numerical value in the answer)
Correct answer is '1000'. Can you explain this answer?
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To calculate the specific cutting energy, we need to know the cutting power and the material removal rate.
The cutting power can be calculated using the formula:
Power (P) = Cutting force (Fc) x Cutting velocity (Vc)
The cutting force can be calculated using the formula:
Cutting force (Fc) = Specific cutting energy (U) x Material removal rate (Q)
The material removal rate can be calculated using the formula:
Material removal rate (Q) = Feed (f) x Depth of cut (d) x Number of cutting edges (n)
Given data:
Cutting velocity (Vc) = 120 m/min
Feed (f) = 0.25 mm/rev
Depth of cut (d) = 4 mm
Diameter of the bar (D) = 100 mm
First, we need to calculate the material removal rate (Q).
Number of cutting edges (n) = 1 (for orthogonal turning)
Q = f x d x n
Q = 0.25 mm/rev x 4 mm x 1
Q = 1 mm³/rev
Next, we can calculate the cutting force (Fc).
Fc = U x Q
Fc = U x 1 mm³/rev
Finally, we can calculate the cutting power (P).
P = Fc x Vc
P = (U x 1 mm³/rev) x 120 m/min
However, we are missing the value for the specific cutting energy (U) in this problem statement, so we cannot calculate the cutting power or the specific cutting energy without that information.
The cutting power can be calculated using the formula:
Power (P) = Cutting force (Fc) x Cutting velocity (Vc)
The cutting force can be calculated using the formula:
Cutting force (Fc) = Specific cutting energy (U) x Material removal rate (Q)
The material removal rate can be calculated using the formula:
Material removal rate (Q) = Feed (f) x Depth of cut (d) x Number of cutting edges (n)
Given data:
Cutting velocity (Vc) = 120 m/min
Feed (f) = 0.25 mm/rev
Depth of cut (d) = 4 mm
Diameter of the bar (D) = 100 mm
First, we need to calculate the material removal rate (Q).
Number of cutting edges (n) = 1 (for orthogonal turning)
Q = f x d x n
Q = 0.25 mm/rev x 4 mm x 1
Q = 1 mm³/rev
Next, we can calculate the cutting force (Fc).
Fc = U x Q
Fc = U x 1 mm³/rev
Finally, we can calculate the cutting power (P).
P = Fc x Vc
P = (U x 1 mm³/rev) x 120 m/min
However, we are missing the value for the specific cutting energy (U) in this problem statement, so we cannot calculate the cutting power or the specific cutting energy without that information.
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A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feedand depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energyof the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.The main or tangential cutting force (in N) is ________(Important - Enter only the numerical value in the answer)Correct answer is '1000'. Can you explain this answer?
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A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feedand depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energyof the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.The main or tangential cutting force (in N) is ________(Important - Enter only the numerical value in the answer)Correct answer is '1000'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feedand depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energyof the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.The main or tangential cutting force (in N) is ________(Important - Enter only the numerical value in the answer)Correct answer is '1000'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feedand depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energyof the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.The main or tangential cutting force (in N) is ________(Important - Enter only the numerical value in the answer)Correct answer is '1000'. Can you explain this answer?.
A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feedand depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energyof the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.The main or tangential cutting force (in N) is ________(Important - Enter only the numerical value in the answer)Correct answer is '1000'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feedand depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energyof the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.The main or tangential cutting force (in N) is ________(Important - Enter only the numerical value in the answer)Correct answer is '1000'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feedand depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energyof the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.The main or tangential cutting force (in N) is ________(Important - Enter only the numerical value in the answer)Correct answer is '1000'. Can you explain this answer?.
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